vector
algebra
polynomial
superior algebra
superior algebra unsolved

Let $ \alpha \equal{} \sqrt{2} \plus{} \sqrt{3}$.  The four vectors $ 1, \alpha, \alpha^2, \alpha^3$ are linearly independent (because the minimal polynomial of $ \sqrt{2} \plus{} \sqrt{3}$ is of degree four), so they form a basis of $ \mathbb{Q}[ \sqrt{2} ][ \sqrt{3} ]$ (which is a vector space of dimension four over $ \mathbb{Q}$).

[quote="t0rajir0u"]Let $ \alpha \equal{} \sqrt {2} \plus{} \sqrt {3}$.  The four vectors $ 1, \alpha, \alpha^2, \alpha^3$ are linearly independent (because the minimal polynomial of $ \sqrt {2} \plus{} \sqrt {3}$ is of degree four), so they form a basis of $ \mathbb{Q}[ \sqrt {2} ][ \sqrt {3} ]$ (which is a vector space of dimension four over $ \mathbb{Q}$).[/quote]

But how do you prove that  $ (\mathbb{Q}( \sqrt {2} ))( \sqrt {3} )$ is a vector space of dimension 4 without using the fact that it is equal to $ \mathbb{Q}(\sqrt {2},\sqrt {3})$ ? $ \mathbb{Q}(\sqrt {2})$ over $ \mathbb{Q}$ is a vector space of dimension 2, so if we can show that $ (\mathbb{Q}(\sqrt {2} ))( \sqrt {3} )$ over $ \mathbb{Q}(\sqrt {2} )$ is a vector space of dimension 2 that would complete the proof, but how do you do that (without assuming that $ (\mathbb{Q}(\sqrt {2} ))(\sqrt {3}) \equal{} \mathbb{Q}(\sqrt {2},\sqrt {3})$)?

assuming absolutely trivial  facts should be ok.

Umm--maybe you could explain why it is trivial?

One way to define $ \mathbb{Q}[ \alpha, \beta, ... ]$ is as the intersection of every field containing $ \mathbb{Q}$ and $ \alpha, \beta, ...$.  In other words, it is the "smallest" field containing $ \mathbb{Q}, \alpha, \beta ...$.  It should be clear that $ (\mathbb{Q}[ \sqrt{2} ])([ \sqrt{3} ])$ is the "smallest" field containing $ \mathbb{Q}, \sqrt{2}, \sqrt{3}$, which is $ \mathbb{Q}[ \sqrt{2}, \sqrt{3} ]$.

In other words, both fields are precisely the set $ \{ a \plus{} b \sqrt{2} \plus{} c \sqrt{3} \plus{} d \sqrt{6}, a, b, c, d \in \mathbb{Q} \}$.

In other other words:  $ \sqrt{3} \not \in \mathbb{Q}[ \sqrt{2} ]$, so $ ( \mathbb{Q}[ \sqrt{2} ])( [ \sqrt{3} ] ) : \mathbb{Q}[ \sqrt{2} ]$ is at least $ 2$ and is also at most $ 2$.

[quote="t0rajir0u"]One way to define $ \mathbb{Q}[ \alpha, \beta, ... ]$ is as the intersection of every field containing $ \mathbb{Q}$ and $ \alpha, \beta, ...$.  In other words, it is the "smallest" field containing $ \mathbb{Q}, \alpha, \beta ...$.  It should be clear that $ (\mathbb{Q}[ \sqrt {2} ])([ \sqrt {3} ])$ is the "smallest" field containing $ \mathbb{Q}, \sqrt {2}, \sqrt {3}$, which is $ \mathbb{Q}[ \sqrt {2}, \sqrt {3} ]$.

In other words, both fields are precisely the set $ \{ a \plus{} b \sqrt {2} \plus{} c \sqrt {3} \plus{} d \sqrt {6}, a, b, c, d \in \mathbb{Q} \}$.

In other other words:  $ \sqrt {3} \not \in \mathbb{Q}[ \sqrt {2} ]$, so $ ( \mathbb{Q}[ \sqrt {2} ])( [ \sqrt {3} ] ) : \mathbb{Q}[ \sqrt {2} ]$ is at least $ 2$ and is also at most $ 2$.[/quote]

Sorry--I just learned what it meant to be algebraic last week.

I agree with the first paragraph. I think you're making assumptions in the second and third that I have not seen proved before.
For example, it seems intuitive, but how do you prove that $ \sqrt {3} \not \in \mathbb{Q}( \sqrt {2} )$? Let me try.

OTC, assume $ \sqrt {3} \in \mathbb{Q}(\sqrt {2} )$. Then $ \sqrt {3} \equal{} \sum_{i\equal{}0}^n q_{i} 2^{i/2}$. Somehow that should imply that $ \sqrt{3}$ is rational (and I know how to prove it is not), but how?

[quote="FibonacciFan"]how do you prove that $ \sqrt {3} \not \in \mathbb{Q}( \sqrt {2} )$?[/quote]

First show that $ \mathbb{Q}[ \sqrt {2} ]$ is the set $ \{ a \plus{} b \sqrt {2}, a, b \in \mathbb{Q} \}$ (since $ (\sqrt {2})^2 \equal{} 2 \in \mathbb{Q}$).  Suppose by contradiction that $ \sqrt {3} \in \mathbb{Q}[ \sqrt {2} ]$.  Then

$ \sqrt {3} \equal{} a \plus{} b \sqrt {2} \implies$
$ 3 \equal{} (a^2 \plus{} 2b^2) \plus{} 2ab \sqrt {2}$.

This implies that $ ab \equal{} 0$, from which we have either $ \sqrt {3} \equal{} a$ or $ \sqrt {3} \equal{} b \sqrt {2}$, both of which are absurd.

Edit:  This whole discussion can be generalized significantly.  In particular, $ \mathbb{Q}[ \sqrt{2}, \sqrt{3}, \sqrt{5}, ... \sqrt{p_n} ] : \mathbb{Q} \equal{} 2^n$, where $ p_n$ is the sequence of primes.

[quote="FibonacciFan"]Why must it be true that $ \mathbb{Q}(\sqrt {2} \plus{} \sqrt {3}) \equal{} (\mathbb{Q}(\sqrt {2}))(\sqrt {3})$?

That the field on the left must be contained inside the field on the right is obvious because $ \sqrt {2}$ and $ \sqrt {3}$ are both elements of the field on the right. I don't see why the inclusion the other way around is necessarily true.[/quote]

Well, if you want to show the other inclusion you have to show that $ \sqrt {2}$, $ \sqrt {3} \in \mathbb{Q}(\sqrt {2} \plus{} \sqrt {3})$, and a rather roundabout way of doing that is as follows.

$ {({2}^{1/2}\plus{}{3}^{1/2})}^{2} \equal{} 5 \plus{} 2*{6}^{1/2} \Rightarrow {6}^{1/2}\in \mathbb{Q}(\sqrt {2} \plus{} \sqrt {3})$
$ {6}^{1/2}*({2}^{1/2}\plus{}{3}^{1/2}) \minus{} 2*({2}^{1/2}\plus{}{3}^{1/2}) \equal{} {2}^{1/2} \Rightarrow {2}^{1/2}\in \mathbb{Q}(\sqrt {2} \plus{} \sqrt {3})$
$ {2}^{1/2}\plus{}{3}^{1/2} \minus{} {2}^{1/2} \Rightarrow {3}^{1/2}\in \mathbb{Q}(\sqrt {2} \plus{} \sqrt {3})$

Actually, now that I think about it, it'd be better to use $ (2 \minus{} 3)/({2}^{1/2}\plus{}{3}^{1/2}) \equal{} {2}^{1/2}\minus{}{3}^{1/2}$
Hope that helps.


Segment AB is parallel to segment CD inside circle R.

The radii are: RC, RD, RA and RB.

If minor arc AC = 79 degrees and RC = segment CD, find the measure of angle ARB.

Congruent chords have congruent arcs and parallel chords have congruent arcs.

This is my work: d = degrees

79d + 79d + 2x = 360d

158d + 2x = 360d

2x = 360d - 158d

2x = 202d

x = 101d

Minor arcs = the degree of the central angle, right?

So, my answer is angle ARB = 101 degrees.

However, the book's answer is 142 degrees.

How can I find 142 degrees?  Is this right?  If so, what am I doing wrong?

Thanks

I loaded a attachment.

$\angle CRB = \angle DRA = 79^{\circ}$

And $%Error. "biguptriangle" is a bad command.
ABR$ is equilateral.

So ARB is $142$ after subtraction.

After subtracting what?

What minus what = 142 degrees?


ARML
geometry
3D geometry
modular arithmetic
algebra
binomial theorem
factorization

Well, this forum is dead.  Maybe some ARML practice problems will help.  They're really easy, though.

1) What is the units' digit of the sum of the squares of the first 2000 odd positive integers?

2) What are the last two digits of $11^{2007}$?

3) Let $N$ be the sum of all four digit numbers that begin with $1$ that have the property that the sum of any two consecutive digits is a multiple of $3$.  What is the remainder when $N$ is divided by $4$?

4) Suppose $N=x2007y$ is a six digit positive integer with two unknown nonzero digits $x$ and $y$.  If $N$ is a multiple of $9$ and $11$, determine $x$ and $y$.  Give your answer as an ordered pair $(x,y)$.

5) How many distinct prime factors does $3^{12}-2^{12}$ have?

6) What is the smallest prime factor of $3^{9}-2^{9}$?

#1- What's the units' digit of the sum of the first 2000 odd numbers.
nth odd number=2n-1.So, sum of them will equal to  S(summation-from 1 to 2000){(2n-1)^2}=S. {4n^2-4n+1}`=S.4n^2-S.4n+1.
S.4n^2=4n(n+1)(2n+1)/6..................(1)
By putting n=2000,(1)=4*1000*667*4001.(which will have 0 as last digit).
S.(4n+1)=4n(n+1)+n=4*2000*2001+2000.again a 0 as last digit.so the answer will equal to 0.

#2-What are the last two digits of 11^2007.
11^2007=(10+1)^2007.
Using binomial theorem-                                                      10^(2007)+2007*10^2006+..+2007C2005*10^2+2007C2006*10+1
All of which except the last 2 are divisible by100.And, therfore has more than two 0's in the end.200C2006*10=2007*10=20070.
so, the last two digits are 71.

#4-x2007y is a six digit integer and is divisible by 9&11.Find x,y.
-x+2+0+0+7+y=9(k+2).where k is an integer.
or,x+y=9k.Now, x+y can't be greater than 18 cause a digit can be atmost 9.
So, x+y=0 or 9 or 18..................(1)
And x-2+0-0+7-y=11k
or x+5-y=11k.x-y can atmost be 9 and therefore the L.H.S. of the equation can be atmost 14.
so,x+5-y=0 or 11.
or, y=x+5 or x-6.
substituting this values of y in (1) we can find the values of x & y.

6) What is the smallest prime factor of $ 3^9 \minus{} 2^9$?

[hide]We can begin factoring this as a difference of cubes:

$ (3^3)^3 \minus{} (2^3)^3 \equal{} (3^3 \minus{} 2^3)(3^6\plus{}6^3\plus{}2^6)$

The smallest prime cannot be 2 because the difference between an odd number and an even number must be odd. Also, the prime cannot be 3 because we have a multiple of three minus something which is not a multiple of 3.

We have a factor of 19 from $ 3^3\minus{}2^3\equal{}19$. And $ 3^6\plus{}6^3\plus{}2^6 \equal{} 1009$ is prime. (For example, 1009 can't be divisible by 7: $ (3^6\plus{}6^6\plus{}2^6)\equal{}(3^3 \plus{} 2^3)^2 \minus{} 6^3 \equal{} 35^2 \minus{} 216$ and 216 is not divisible by 7. We know 11 does not divide 1009 by divisibility rules, and quick checks show that 13 and 17 don't either.)

So 19 is the answer.[/hide]

#3 (as far as i got)

[hide]
Ummm the second digit can be $ 3a\minus{}1$, the third $ 3b\plus{}1$ and then fourth $ 3c\minus{}1$
only the last 2 digits matter so u just need that there are 3 possible values for the second digit so for each possible combinations of the last 2 digits, you have 3 of them
and thats the point where i gave up =/

[/hide]

[hide="Solution to #3"]
The hundreds digit can be any of {2,5,8}
The tens digit can be any of {1,4,7}
The ones digit can be {2,5,8}

This gives a total of 27 possible numbers.  Their arithmetic mean is 545 (taking the means of the digits), so their sum is
\[ 27 \cdot 545 \equiv 27\cdot1 \equiv \boxed{3} \pmod{4}\]
[/hide]

1: good way:
We know sum of 1st n odds is n^2. So, sum of 1st 2000 is 2000^2, and the last digit of 2000 is 0, so the answer is 0^2=0.


Hi
Can any one say me how I can gain :Gazeta Matematica Archive .
Is This Journal has download site such as :Crux Mathematicoroum.?
                                                            Thanks.

no, you must buy it, i think it costs ~100, for more details try here, if you understand romanian  :maybe: [url]http://www.gazetamatematica.ro[/url]


In how many ways can  5 students be selected from a group of 6 students?

This is just $ \binom{6}{5}\equal{}\frac{6!}{5!1!}\equal{}\boxed{6}$ ways.

the  we have to place  6 objects  in 5 places ........  so ans will be 6

You can select a group of $5$ students from $6$ students by selecting which of the $6$ students won't be in the group.  There are $6$ students that can be selected to not be in the group, so $\boxed{6}$ groups of $5$ students can be selected.

[quote=Hermia][hide=Solution]Each of the students could be the one not in the group, so there are a total of 6 groups.[/hide][/quote]

To elaborate on that, [hide]choosing 5 people out of 6 is the same as choosing the one person who isn't chosen, so just 6c1 so 6[/hide]

pls don't post uncontributive solutions @ppl posting after me

[hide= Precise Solution]

There are $6$ students, $5$ need to be selected. $1$ student will not be selected. Since there is $6$ students, the answer is $6$ as that is is the amount of ways to choose the student that is not selected.[/hide]


function
superior algebra
superior algebra solved

Does there exist a nonnegative and periodical function defined on the reals such that the arithmetic mean of the numbers f(1), f(2),...,f(n) tends to zero when n approaches infinity, but not all natural numbers are zeros of f? I believe the answer is yes, but i couldn't find an example of such function.

An example:
f(n)=1/n, f has period  p=(\sqrt 2)/2.
For the rest of [0, \infty), define f so that it is periodic:
For all x in [0,p), if x=n-k*p (because p is irrational, this writing is unique if it exists) , for some k and n in N, take f(x)=f(n). If else take f(x) arbitrarily. Now extend f to [0,\infty) so that it is periodic.

Nice solution. Now the problem  is completely solved.


analytic geometry
graphing lines
slope
Pythagorean Theorem
geometry

Two questions:

(1) Show that the points A = (2, 5), B = (6, 1) and C = (8, -1) lie on a circle with center (-1, 2).  What is the radius if this circle?

(2) Show that the points A = (-2, 0), B = (-4, 4) and C = (8, 5) are the vertices of a right triangle in two ways:

(a) By using the converse of the Pythagorean Theorem.

(b) By using slopes of the lines joining the vertices.

There is no need to make so many successive topics
if you have 3 questions all at once, put them in one topic

[hide]
I don't think that those points even lie on a circle with that center

let (-1,2) be O

$OA = \sqrt{18}$
$OB = \sqrt{50}$
$OC = \sqrt{90}$

EDIT:
A, B,C are collinear
so there is no circle that passes through those 3 points
[/hide]

[hide="first one"]A, B, and C are collinear and different, so no such circle exists.[/hide]

[hide="2nd one hints"]The Pythagorean theorem is $a^{2}+b^{2}=c^{2}$.  You can find $a$ and $b$ by finding the difference in $x$ and $y$ in two points.  Then, you can get the distance between them.

Simply find the slopes between each pair of points and see if 2 happen to be negative reciprocals of each other.    If so, you have a right triangle[/hide]

@sharkman:  please show your work (of what you have done) if you are needing help on these problems, or do not post them.  Also, please group very similar problems together.

@chickendude:  problems should only be grouped together if they are related.


trigonometry
calculus
calculus computations

solve the folowing equation:
[img]http://i36.tinypic.com/2073hub.jpg[/img]
thanks

hello, at first the following formulas are very useful:
$ \tan(2x)\equal{}\frac{2\sin(x)\cos(x)}{\cos(x)^2\minus{}\sin(x)^2}$
$ \sin(5x)\equal{}5\cos(x)^4\sin(x)\minus{}10\cos(x)^2\sin(x)^3\plus{}\sin(x)^5$.
Sonnhard.


The number of triples $(a,b,c)$ of positive integers which satisfy the simultaneous equations \begin{align*} ab+bc &= 44,\\ ac+bc &= 23, \end{align*} is

$\textbf{(A) }0\qquad
\textbf{(B) }1\qquad
\textbf{(C) }2\qquad
\textbf{(D) }3\qquad
\textbf{(E) }4$

[quote="236factorial"][hide]We can subtract and factor to get $a(b-c)=21$. $a=1,3,7,21$ and there are four solutions. [/hide][/quote]

is that all you really need to do? i got to that part... then i got skeptical about it and thought it was wrong...  :?

[hide="Answer"]$ab-ac=a(b-c)=21$. Therefore, we have $a={1,3,7,21}$, so there are four possible solutions.

$\boxed{E}$[/hide]

[hide]$ab+bc=44$ and $ac+bc=23$, by subtracting we get $a(b-c)=21$, now the factors of $21$ can be $21 \cdot 1, 3 \cdot 7, 1 \cdot 21, 3 \cdot 7$, so there can be up to $4$ values of $a$, thus $4$ triples.[/hide]

You subract both equations from one another and factor, thus you get $a(b-c)=21\implies d(21)=4$.  Therefore, there are 4 triples...$\boxed{E}$.

Masoud Zargar

[quote="YooSam"]Guys, I don't get it.

From equation (2): $ac+bc=23$ $\implies c(a+b)=23$

But $23$ is a prime, $a,b,c$ are positive integers, so $c$ must be a factor of $23$, which gives us two choices $1, 23$.

So I think the number of triples $(a,b,c)$ is $2$. What am I thinking wrong here?[/quote]

yeah, i'd like an explanation of this too please... this is what confused me

Ignore my solution, I'm not even certain about it :blush:

Everyone else did similar things... tells me how much they copy  :P  jk

[hide]goin from YooSam's idea I'll argue what I think...
from the second equation, you can factor it so that $c\left(a+b\right)=23$. Since $23$ is prime, you can assume that $c$ is either $1$ or $23$ and the same goes for $a+b$. For the ease of calculation, let's say that $c=1$ so, $a+b=23$ which implies that $b=23-a$. Then you can substitute this into the first equation to get a quadratic equation: $a\left(23-a\right)+23-a\implies -a^2+22a+23=44$. Simplifying gives $a^2-22a+21=0$ and solving gives $a=1$ or $a=21$. So the only triples you can get are $\left(1,22,1\right)$ and $\left(21,2,1\right)$. So I think the answer is $\boxed{C}$[/hide]

think about this, it is 21 on the AMC, there is some difficulty to it, you guys state 4 values for $a$, but forget that you need positive integers for the other solutions, and you have to check for those because knowing the value for a defines b and c because you have 2 different equations with the same variables

[hide]

looking at the second thing, we have that $c(a+b)=23$
factoring, we see that $c=1,23$, and $a+b=23,1$
using this to solve, we get the triples

$(21,2,1),(1,22,1),(-1,2,23),(-21,22,23)$

and we eliminate the last 2 because they are not all positive integers

so the answer is $\boxed{C=2 \, \text{possibilities}}$

[/hide]

Here is how I did it, I dunno whether it is correct or not
[hide]
$b(a+c) = 44$
$c(a+b) = 23$
So $c = 1$ and $a+b = 23$ or $c = 23$ and $a+b = 1$.
Taking these one at a time and using the factors of 44:
$1, 4$
$4, 11$
$2, 22$

$b(a+1) = 44$
And the only values that work are when $b = 2$ and $a = 21$ or when $b = 22$ and $a = 1$

In the other case, when $a+b = 1$ either $a$ or $b$ must be zero, so this possibility cannot satisfy the positive integer requirement and the answer is $\mbox{C}$[/hide]

Edit: Seems correct as it is what the people above me have done.

C is the correct answer.

This is an advice to ALL of you who didn't get this one right. Even though it seems to be obvious for $a$ to exist in four ways, it's better if you find the "triple" because that's what the problem's asking. If you tried $a = 3,7$, you should immediately see that this is impossible and the answer is NOT 4.

I saw people posting wrong answers earlier but I waited to see when the right answer would come along.  :)

There was a nagging suspicion that I should have checked the values, but I didn't listen to it. :oops:


$ a,b,c\geq 0$,prove that
\[ \frac{2a(b\plus{}c)}{a^2\plus{}bc}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}ca}\plus{}\frac{c(a\plus{}b)}{c^2\plus{}ab}\geq 2\]

I was inspired from http://www.mathlinks.ro/viewtopic.php?t=320828
By the way,happy new year. :)

[quote="Tourish"]$ a,b,c\geq 0$,prove that
\[ \frac {2a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\geq 2\]
I was inspired from http://www.mathlinks.ro/viewtopic.php?t=320828
By the way,happy new year. :)[/quote]
Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are nonnegative real numbers, no two of which are zero, then
$ \frac{a(b\plus{}c)}{a^2\plus{}bc}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}ca}\plus{}\frac{c(a\plus{}b)}{c^2\plus{}ab} \ge 2.$

By the way, the following similar inequality holds, too:

If $ a,$ $ b,$ $ c$ are nonnegative real numbers, no two of which are zero, then
$ \sqrt{\frac{a(b\plus{}c)}{a^2\plus{}bc}}\plus{}\sqrt{\frac{b(c\plus{}a)}{b^2\plus{}ca}}\plus{}\sqrt{\frac{c(a\plus{}b)}{c^2\plus{}ab}} \ge 2.$

[quote="can_hang2007"][quote="Tourish"]$ a,b,c\geq 0$,prove that
\[ \frac {2a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\geq 2\]
I was inspired from http://www.mathlinks.ro/viewtopic.php?t=320828
By the way,happy new year. :)[/quote]
Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are nonnegative real numbers, no two of which are zero, then
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab} \ge 2.$

By the way, the following similar inequality holds, too:

If $ a,$ $ b,$ $ c$ are nonnegative real numbers, no two of which are zero, then
$ \sqrt {\frac {a(b \plus{} c)}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {b(c \plus{} a)}{b^2 \plus{} ca}} \plus{} \sqrt {\frac {c(a \plus{} b)}{c^2 \plus{} ab}} \ge 2.$[/quote]

For the first one,it's equalent to
\[ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\plus{}\sum{a^2(b\plus{}c)}\plus{}2a^2b^2c^2\geq 0\] :roll:
the second....I am trying now.. :blush:

[quote="can_hang2007"]
If $ a,$ $ b,$ $ c$ are nonnegative real numbers, no two of which are zero, then
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab} \ge 2.$
[/quote]
I found that the following also ture:
\[ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab} \ge 2\plus{}\frac{8a^2b^2c^2}{(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}\]

[quote="Tourish"][quote="can_hang2007"]
If $ a,$ $ b,$ $ c$ are nonnegative real numbers, no two of which are zero, then
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab} \ge 2.$
[/quote]
I found that the following also ture:
\[ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab} \ge 2 \plus{} \frac {8a^2b^2c^2}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}\]
[/quote]

Yes, i really like it. I have found the following identity:

\[ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab} \minus{} 2 \minus{} \frac {8a^2b^2c^2}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}\equal{}\frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\plus{}abc(a(b\minus{}c)^2\plus{}b(c\minus{}a)^2\plus{}c(a\minus{}b)^2)}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}\]

[quote="Tourish"][quote="can_hang2007"]
If $ a,$ $ b,$ $ c$ are nonnegative real numbers, no two of which are zero, then
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab} \ge 2.$
[/quote]
I found that the following also ture:
\[ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab} \ge 2 \plus{} \frac {8a^2b^2c^2}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}\]
[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-18-unknown-author.html

The following inequality is true, too:
$ x \plus{} y \plus{} z \ge 2 \plus{} xyz,$
where
$ x \equal{} \frac {a(b \plus{} c)}{a^2 \plus{} bc},\quad y \equal{} \frac {b(c \plus{} a)}{b^2 \plus{} ca},\quad z\equal{}\frac {c(a \plus{} b)}{c^2 \plus{} ab}.$

Nice,Can!
\[ \frac{a(b\plus{}c)}{a^{2}\plus{}bc}\plus{} \frac{b(c\plus{}a)}{b^{2}\plus{}ca}\plus{}\frac{c(a\plus{}b)}{c^{2}\plus{}ab}\geq 2\plus{}\frac{abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}{(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}\]
Let $ x\equal{}\frac{a}{b},y\equal{}\frac{b}{c},z\equal{}\frac{c}{a}$,then $ xyz\equal{}1$
\[ \Longleftrightarrow \frac{1\plus{}xz}{x\plus{}z}\plus{}\frac{1\plus{}zy}{z\plus{}y}\plus{}\frac{1\plus{}yx}{y\plus{}z}\geq 2\plus{}\frac{(x\plus{}1)(y\plus{}1)(z\plus{}1)}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\]
\[ \Longleftrightarrow \sum{(1\plus{}xz)(z\plus{}y)(z\plus{}x)}\geq 2(x\plus{}y)(y\plus{}z)(z\plus{}x)\plus{}(1\plus{}x)(1\plus{}y)(1\plus{}z)\]
\[ \Longleftrightarrow (x\plus{}y\plus{}z)^2\plus{}(xy\plus{}yz\plus{}zx)^2\geq 2(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)\]
which is obvious..... :lol:

[quote="Tourish"]Nice,Can!
\[ \frac {a(b \plus{} c)}{a^{2} \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^{2} \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^{2} \plus{} ab}\geq 2 \plus{} \frac {abc(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}\]
Let $ x \equal{} \frac {a}{b},y \equal{} \frac {b}{c},z \equal{} \frac {c}{a}$,then $ xyz \equal{} 1$
\[ \Longleftrightarrow \frac {1 \plus{} xz}{x \plus{} z} \plus{} \frac {1 \plus{} zy}{z \plus{} y} \plus{} \frac {1 \plus{} yx}{y \plus{} z}\geq 2 \plus{} \frac {(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\]

\[ \Longleftrightarrow \sum{(1 \plus{} xz)(z \plus{} y)(z \plus{} x)}\geq 2(x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} (1 \plus{} x)(1 \plus{} y)(1 \plus{} z)\]

\[ \Longleftrightarrow (x \plus{} y \plus{} z)^2 \plus{} (xy \plus{} yz \plus{} zx)^2\geq 2(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)\]
which is obvious..... :lol:[/quote]

$ \frac {a(b \plus{} c)}{a^{2} \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^{2} \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^{2} \plus{} ab} \minus{} 2 \minus{} \frac {abc(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}$
$ \equal{} \frac {(b \minus{} c)^2(a \minus{} c)^2(a \minus{} b)^2}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}.$

[quote=Tourish][quote="can_hang2007"]
By the way, the following similar inequality holds, too:
If $ a,$ $ b,$ $ c$ are nonnegative real numbers, no two of which are zero, then
$ \sqrt {\frac {a(b \plus{} c)}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {b(c \plus{} a)}{b^2 \plus{} ca}} \plus{} \sqrt {\frac {c(a \plus{} b)}{c^2 \plus{} ab}} \ge 2.$[/quote]
$ \sqrt {\frac {a(b \plus{} c)}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {b(c \plus{} a)}{b^2 \plus{} ca}} \plus{} \sqrt {\frac {c(a \plus{} b)}{c^2 \plus{} ab}} \ge 2\plus{}\frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
holds for all $ a,b,c\geq 0$.It can be proved by AM-GM easily. :lol:[/quote]

[img]http://s6.sinaimg.cn/middle/006ptkjAzy77FRdBEdna5&690[/img]


Let $ x,y,z$ be sides of triangle and   $ {{{{{}}}}}\frac{1}{1\plus{}x}\plus{}\frac{1}{1\plus{}y}\plus{}\frac{1}{1\plus{}z} \geq 1$
Prove that  \[ x\plus{}y\plus{}z \geq \sum_{cyc}\frac{(x\plus{}y\minus{}z)(x\minus{}y\plus{}z)}{2}\]

appolo, it is a known inequality.

Positive real numbers $ a,b,c$ such that
\[ \frac1{1\plus{}a\plus{}b}\plus{}\frac1{1\plus{}b\plus{}c}\plus{}\frac1{1\plus{}c\plus{}a}\geq1.\]
Prove that $ a\plus{}b\plus{}c\geq ab\plus{}bc\plus{}ca$.

[quote="hitek"]appolo, it is a known inequality.

Positive real numbers $ a,b,c$ such that
\[ \frac1{1 \plus{} a \plus{} b} \plus{} \frac1{1 \plus{} b \plus{} c} \plus{} \frac1{1 \plus{} c \plus{} a}\geq1.
\]
Prove that $ a \plus{} b \plus{} c\geq ab \plus{} bc \plus{} ca$.[/quote]
Thanks  :lol:


trigonometry
algebra
polynomial
algebra open

for  what  values  of  a  is  $sin(a)$  rational?

$\sin \alpha$ is rational at the following values of $\alpha$:

$\sin (0 + k\pi) = 0, k = 0, 1, 2, 3...$

$\sin (\displaystyle\frac{\pi}{2} + 2k\pi) = 1, k = 0, 1, 2, 3...$

$\sin (\displaystyle\frac{3\pi}{2} + 2k\pi) = -1, k = 0, 1, 2, 3...$

$\sin (\displaystyle\frac{\pi}{6} + 2k\pi) = \sin(\displaystyle\frac{5\pi}{6} + 2k\pi) = \displaystyle\frac{1}{2}, k = 0, 1, 2, 3...$

$\sin (\displaystyle\frac{7\pi}{6} + 2k\pi) = \sin(\displaystyle\frac{11\pi}{6} + 2k\pi) = \displaystyle\frac{-1}{2}, k = 0, 1, 2, 3...$

For values of $\alpha$ that are rational multiples of $\pi$, these are the only rational values of $\sin \alpha$, since other rational multiples of these values yield irrational results by half-angle formulas, etc.  A more detailed discussion of why this is so can be found  [url=http://www.mathpages.com/home/kmath460.htm]here[/url].

For arbitrary values of $\alpha$,  $\sin \alpha$ can take on any real value $-1 \le \sin \alpha \le 1$, including all rational values in that range.  So:

$\sin \alpha\text{ is rational iff }\arcsin \displaystyle\frac{m}{n} = \alpha\text{ for some }m \le n \in \mathbb{Z}$

You should probably clarify what values of $\alpha$ you consider meaningful.

The theorem of Lindemann-Weierstrass gives that $sin(a)$ is transcendental for all algebraic numbers $a \neq 0$. Additionally, adding rational multiples of $\pi$ to such an $a$ will not change transcendence.

Although... the link I provided only proves that $\sin \alpha$ can't be rational for integer values of $\alpha$ in degrees except for multiples of 30 degrees, which, even when combined with Lindemann-Weierstrass, doesn't answer the question of whether $sin (\displaystyle\frac{m\pi}{n})$ can be rational for values other than the ones I gave.  It seems like it still wouldn't be, but I would like to see a proof.

That $sin\left( \frac{m \pi}{n} \right)$ is just rational for the given ones is not so easy I think:

1.: Assume that $m,n$ are coprime. Looking at the algebraic degree of the primitive $n$-th roots of unity will do: because of $sin(x)^2+cos(x^2)=1$, their degree has to be $\leq 2$, so $\phi (n) \leq 2$, giving the result.

2.: Note that $sin\left( \frac{m \pi}{n} \right) = \frac{e^{\frac{m \pi i}{n}}-e^{-\frac{m \pi i}{n}}}{2i}$. Now $e^{\frac{m \pi i}{n}}$ and $e^{-\frac{m \pi i}{n}}$ are $(2n)$-th roots of unity, thus they and so also their difference is an algebraic integer. Now also $-i$ is an algebraic integer, so also $\frac{e^{\frac{m \pi i}{n}}-e^{-\frac{m \pi i}{n}}}{i}$ is one. But all rational algebraic integers are integers itself, so $2sin\left( \frac{m \pi}{n} \right)$ is an integer, giving the result.

I'm still looking for an elementary solution.

Yes, that is the solution I know as well, although it presupposes a bit too much abstract algebra.

A solution presented by Ravi B on this forum uses the fact that $2\cos(nt)$ is a monic polynomial in $2\cos t$ - from there it's trivial to conclude via the rational root theorem.

Value ??
[b]$cos\frac{\pi}{7}=?$[/b]
help me!!!

[quote="mszew"]Is this what you need?

In[4]:=
N[Cos[Pi/7],20]

Out[4]=
0.90096886790241912624[/quote]
No
It isn't  that I Want

----------
exact

Check http://en.wikipedia.org/wiki/Heptagon

and use the fact that:
$ Cos(2\theta)=2Cos^{2}(\theta)-1$


$\sum_{k \equiv 1 (\mod 3)}\binom{n}{k}$ and if $k \equiv 2 (\mod 3)$

Let $\omega=\mathrm{e}^{\frac{2\pi i}{3}}$. It is easy to see that for any integer $l$ the value of expression $\frac{1+\omega^{l}+\omega^{2l}}{3}$ is $1$ when $n\mid l$ and $0$ when $n\nmid l$. Thus
\begin{eqnarray*}\sum_{k\equiv j\pmod{3}}\binom{n}{k}&=& \sum_{n\mid k-j}\binom{n}{k}\\ &=& \frac13\cdot\sum_{n\mid k-j}\binom{n}{k}\cdot\left(1+\omega^{k-j}+\omega^{2(k-j)}\right)\\ &=& \frac13\cdot\sum_{k}\binom{n}{k}\cdot\left(1+\omega^{k-j}+\omega^{2(k-j)}\right) \\ &=& \frac13\cdot\left(\sum_{k}\binom{n}{k}\right)+\frac13\cdot\left(\omega^{-j}\cdot \sum_{k}\binom{n}{k}\omega^{k}\right)+\frac13\cdot\left(\omega^{-2j}\cdot \sum_{k}\binom{n}{k}\omega^{2k}\right)\\ &=& \frac13\cdot\left( 2^{n}+\omega^{-j}\cdot \left(1+\omega\right)^{n}+\omega^{-2j}\cdot \left(1+\omega^{2}\right)^{n}\right)\\ &=& \frac13\cdot\left( 2^{n}+\omega^{-j}\cdot \left(-\omega^{2}\right)^{n}+\omega^{-2j}\cdot \left(-\omega\right)^{n}\right)\\ &=&\frac13\cdot\left( 2^{n}+(-1)^{n}\cdot\left(\omega^{2n-j}+\omega^{n-2j}\right)\right)\\ &=& \frac13\cdot\left( 2^{n}+(-1)^{n}\cdot\left(\omega^{-(j+n)}+\omega^{j+n}\right)\right)\\ &=& \frac13\cdot\left( 2^{n}+(-1)^{n}\cdot 2\mathrm{Re}\left(\omega^{j+n}\right)\right)\\ &=& \frac13\cdot\left( 2^{n}+(-1)^{n}\cdot 2\cos\left(\frac{2\pi}{3}\cdot (j+n)\right)\right). \end{eqnarray*}
The last expressions allows us to evaluate the desired sum if we know residues of $n$ and $j$ modulo $3$:
$\left\{\begin{array}{lll}\sum_{k\equiv j\pmod{3}}\binom{n}{k}=& \frac13 \cdot\left(2^{n}+2\cdot (-1)^{n}\right) & \text{ for }3\mid n+j\\ \sum_{k\equiv j\pmod{3}}\binom{n}{k}= & \frac13 \cdot\left(2^{n}-(-1)^{n}\right) & \text{ for }3\nmid n+j.\end{array}\right.$

[quote="john lai"]hid the solution[/quote]
Why?
PS. The idea is taken from [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=119683[/url].


calculus
integration
logarithms
trigonometry
complex numbers
real analysis
real analysis unsolved

Show that
\[ \int_0^{2\pi } {\ln \left| {1 \minus{} e^{it} } \right|dt}  \equal{} 0.
\]
 :P

kunny, the original problem had a logarithm.

For $ |z| < 1,\ \minus{} \ln|1 \minus{} z| \equal{} \text{Re}\left(\sum_{n \equal{} 1}^{\infty}\frac {z^n}n\right).$

We can argue (a summation-by-parts argument) that the series converges for $ |z| \equal{} 1,z\ne1,$ and Abel's Theorem identifies the sum. Hence:

$ \ln|1 \minus{} e^{it}| \equal{} \minus{} \text{Re} \left(\sum_{n \equal{} 1}^{\infty}\frac {e^{int}}n\right)$

$ \equal{} \minus{} \sum_{n \equal{} 1}^{\infty}\frac {\cos nt}{n}$

Integrate that from $ 0$ to $ 2\pi$ and each term integrates to zero; hence the sum integrates to zero.

[We still need an argument to justify the exchange of integral and limit; we can rely on the fact that we're talking about the Fourier series of an $ L^2$ function. Or we can write explicit formulas for the partial sums and show that we can quote the Dominated Convergence Theorem.]

Writing $ z \equal{} e^{it}$, this becomes $ \int_{|z| \equal{} 1}\frac {\text{Re}(\ln(1 \minus{} z))}{iz}\,dz \equal{} \text{Re}\left(\int_{|z| \equal{} 1}\frac {\ln(1 \minus{} z)}{iz}\,dz\right)$

Dropping the real part, $ \frac1{i}\int_{|z| \equal{} 1}\frac {\ln(1 \minus{} z)}{z}\,dz \equal{} 0$ since $ \frac {\ln(1 \minus{} z)}{z}$ is analytic inside the unit disk, and the logarithmic branch point on the boundary isn't a big enough effect to change the integral. (We can imagine a sequence of paths that cut that point out of the circle- the integrals are all zero, and they tend to the integral on the circle.)


search
inequalities unsolved
inequalities

Let $(a_{n})_{n=1}^{\infty}$ a sequence of positive real numbers and $\alpha_{n}=\frac{a_{1}+...+a_{n}}{n}$.Prove that for every natural number N we have:$\displaystyle\sum_{n=1}^{N}\alpha_{n}^{2}\leq 4\sum_{n=1}^{N}a_{n}^{2}$

guys,please try this inequlity.It's really hard for me :)

tank you very much blah blahblah.Can someone give another proof or try to prove hardy for p=2
 :)

[quote="rem"]
Let's prove that:
$(\sum_{i=1}^{k}a^{2}_{i}\sqrt{i}) (\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}})\ge \frac{1}{4}(\frac{\sum_{i=1}^{k}a_{i}}{k}$

why? why does the sigma disappear?


$ P(A|B)$ is the conditional probability $ A$ GIVEN $ B$ and is calculated $ \frac {P( A \text{ and } B)}{B}$
So $ P(A|B) \equal{} \frac {0.6 \cdot 0.4}{0.4}$

This assumes $ A$ and $ B$ are independent.
So:
$ P(A or B)$ DOES equal $ 1$ (you can sum them).

[quote="thekiterunner"]I has just hoping if anyone could verify that the following was correct.

If P(A)=0.6, P(B)=0.4 find P(A|B), P(B|A), P(A or B), P(A and B)

I got the following answers:

P(A|B)=0
P(B|A)=0
P(A or B)=1
P(A and B)=0

Can anyone tell me if these answers are correct as they look a little odd[/quote]IF you are told $ A$ and $ B$ are mutually exclusive (like, for example, $ A\equal{}$rain today and $ B\equal{}$no rain today), then it is not odd at all.
P(A|B)=probability that it will rain today if it does not rain today=$ 0$
P(A or B)=probability that it will either rain or not rain today=$ 1$
P(A and B)=probability that today it will rain and not rain at all=$ 0$


geometry
geometric transformation
homothety
geometry theorems

Could somebody give some general strategies for attacking problems of collinearity concyclicity and concurrency. I'm really stuck trying to solve such problems.
      Ashwath

i my self solve most of colinearity and concurrenty problems by using pole and polar and harmonic.

but i thinks the strategies arent out of these notes:

1)menelaus theorem

2)ceva theorem

3)pole and polar

4)harmonic

5)and also recently by using inversive geometry

thanks!  but haven't you also forgot to mention homothecy?
   Ashwath
P.s. I find Inversion a little difficult to comprehend., but nevertheless I'm working on it.  :)

Homothety is one of my personal favourites, together with spiral similarity and the construction of pedal triangles, pedal quadrilaterals, etcetera.

Ashwath, please don't leave such a big "blank" gap in your posts.


For what value of $ \theta$ does  $ cos \theta \equal{} \theta$ ?

Using a graphing calculator, I found the answer is $ \sim .73908513$.
I do not think it possible to algebraically manipulate the equation to solve for $ \theta$. Instead, if you wanted to solve it by hand, I suppose you could use Newton's method to get an approximation.

$ \cos 0\equal{}1$ and $ \cos \frac\pi 2\equal{}0$, so we'd expect the answer to lie something near halfway between these.

$ \theta\equal{}1\minus{}\frac{\theta^2}{2}\plus{}\frac{\theta^4}{4}\plus{}\cdots\approx1\minus{}\frac{\theta^2}{2}$.

This is easy enough to solve and gets you the answer $ \theta\equal{}\sqrt{3}\minus{}1\equal{}0.732051$, which is pretty close...

If you introduced a quartic term you'd get even closer but it would be rather hard to solve.

[quote="gokussj3"][quote="mismatchtea"]Using a graphing calculator, I found the answer is $ \sim .73908513$.
I do not think it possible to algebraically manipulate the equation to solve for $ \theta$. Instead, if you wanted to solve it by hand, I suppose you could use [b]Newton's method to get an approximation.[/b][/quote]

What is that? :maybe:[/quote]

[url=http://www.google.com/search?q=Newton's+method]http://www.google.com/search?q=Newton's+method[/url]


Meewhee to the rescue!!
First, you click the profile button on the toolbar above. (see attachment #1)
Then, you click Personal on the side toolbar (see attachment #2)
Then, you should see at the bottom a choice of... well I don't know what to call it. (see attachment #3)
Then click save just below that.
Hope it helped!

Another profile question: In the signature control panel, it says: No images. Then it says No more than 1 image. What's up with that? Are we allowed to post an image in our signiture?

This has been mentioned before. I don't think images are allowed (try it). Other than that, ignore it.

I tried, and images aren't allowed. Or maybe it was just because the image I tried was too big...

I've tried images of practically all sizes; maybe images were allowed before...

(see attachment)
Note: there was only 1 image I attempted to use as an example, why does it say 2?  :huh:

You had 2 images in your signature, the maximum number allowed is 0 (aka "none").


algebra
function
domain
geometry
3D geometry
sphere
topology

Ahlfors defines Riemann surfaces using sheaves, in which case a Riemann surface is defined as a space of germs of some global analytic function. In other words, he defines defines them such that a Riemann surface appears as the natural domain of a possibly multi-valued function $ f(z)$.

For example, for $ \log(z)$, you get the covering space of the punctured plane as its Riemann surface.

My question is this: You can define Riemann surfaces abstractly using atlases, i.e. as a one-dimensional complex manifold. Can all Riemann surfaces (in this abstract sense) be realized as the domain of a multi-valued function, as in Ahlfors above?

An obvious counterexample is the Riemann sphere $ S$ which is a one-dimensional complex manifold but cannot be realized as the Riemann surface in the sense of sheaves over $ \mathbb C$ (just because the projection to $ \mathbb C$ would be a non-constant holomorphic function on $ S$).

I didn't do much thinking on this so I may very well be wrong, but I think any compact Riemann surface cannot be realized in this way.


inequalities
geometry
circumcircle
inradius
algebra
polynomial
inequalities proposed

let $ABC$ be a triangle with sides $a,b,c$ and circumradius $R$ and inradius $r$
then prove :
\[\frac{R}{2r}\ge \left(\frac{64a^{2}b^{2}c^{2}}{(4a^{2}-(b-c)^{2})(4b^{2}-(c-a)^{2})(4c^{2}-(a-b)^{2})}\right)^{2}\]
i will post my soln later.

no one interested in such a nice inequality.

It's a good inequality I think. It makes me remember
$R\geqslant 2r$
and yours is obviously stronger.

But I'm not very good at geometrical inequality :roll:

ok here is my solution :
we know that $R/2r=\frac{abc}{(c+a-b)(b+a-c)(b+c-a)}$
now employ the same substitution \[a=y+z\ ;b=z+x;\ c=x+y\]
then $(4a^{2}-(b-c)^{2})=4(y^{2}+z^{2}+2yz)-(y-z)^{2}=(3y^{2}+3z^{2}+10yz)$
so it becomes to prove :
\[(3y^{2}+3z^{2}+10zy)^{2}(3x^{2}+3y^{2}+10xy)^{2}(3x^{2}+3z^{2}+10zx)^{2}\ge64^{2}.2^{3}(x+z)^{3}(z+y)^{3}(x+y)^{3}zxy\]
now we will prove that
\[(3x^{2}+3y^{2}+10xy)^{2}\ge 32\sqrt{xy}.(x+y)^{3}\]
because of homogeinity assume xy=1;
then it becomes to prove
\[(3x^{2}+3/x^{2}+10)^{2}\ge 32(x+1/x)^{3}\]
put p=x+1/x;then $p\ge2$
so it becomes to prove
\[(3p^{2}+4)^{2}\ge 32p^{3}\iff 9p^{4}-32p^{3}+24p^{2}+16\ge0\]
$p-2$ seems to be a double root of above polynomial and we find the factorization\[f(p)=(p-2)^{2}.(9p^{2}+4p+4)\ge0\] is obvios .now multiplying similar two inequality we get the result.


geometry
rectangle
circumcircle
ratio
symmetry
geometry proposed

[quote="vittasko"]Let $BCEF$ be, a rectangle erected on the side segment $BC,$ of a given triangle $\bigtriangleup ABC,$ outwardly to it. Let also $B',$ $C',$ $D'$ be, the intersection points of the circumcircle  $(O)$ of $\bigtriangleup ABC,$ from the segment lines $AE,$ $AF,$ $AD$ respectively, where $AD,$ is the altitude of $\bigtriangleup ABC.$

Prove that the segment lines $BB',$ $CC'$ and the diameter of $(O),$ through the point $D',$ are concurrent at one point.

Kostas Vittas.[/quote]
Let's look at the triangel $BD'C$.We need to prove that $CC',BB'$ and $D'O$ are concurrent.From the Cheva's theorem we can say that it is sufisant to prove that
\[\frac{sin\angle{BCC'}}{sin\angle{C'CD'}}\frac{sin\angle{D'BB'}}{sin\angle{B'BC}}\frac{sin\angle{CD'O}}{sin\angle{OD'B}}=1\]
but we have that
\[\frac{sin{\angle{BCC'}}}{sin{\angle{C'CD'}}}=\frac{sin{\angle{BAC'}}}{sin{\angle{D'AC'}}}=\frac{sin{\angle{BAF}}}{sin{\angle{BFA}}}=\frac{BF}{AB}\]
similary we can see that
\[\frac{sin{\angle{D'BB'}}}{sin{\angle{B'BC'}}}=\frac{AC}{CE}\]
and you can esiely prove that
\[\frac{sin{\angle{CD'O}}}{sin{\angle{OD'B}}}=\frac{AB}{AC}\]
and therefor
\[\frac{sin\angle{BCC'}}{sin\angle{C'CD'}}\frac{sin\angle{D'BB'}}{sin\angle{B'BC}}\frac{sin\angle{CD'O}}{sin\angle{OD'B}}=\frac{BF}{AB}\frac{AC}{CE}\frac{AB}{AC}=1\]

Tigran Sloyan  :wink:

Thank you dear Tiks, for your interest and nice solution. Although it is a trigonometric way, however I like it, because of it is a concise solution and I will preserve it in the future, with your name. So, I would like to know it, if it is possible.

I will sent you soon the synthetic proof I have in mind, before it’s posting in the forum (after two months), as a least indication of honour.

Best regards, Kostas Vittas.

[quote="vittasko"]Thank you dear Tiks, for your interest and nice solution. Although it is a trigonometric way, however I like it, because of it is a concise solution and I will preserve in the future, with your name. So, I would like to know it, if it is possible.

I will sent you soon the synthetic proof I have in mind, before it’s posting in the forum (after two months), as a least indication of honour.

Best regards, Kostas Vittas.[/quote]

You'r welcome dear [b]vittasko[/b].Actually,I have alread preserved your name in my mind as a new great geometre in the Mathlinks  :wink:  .
Vittas I may don't get your question correctly,but if you were asking my name,my name is [b]Tigran[/b](and my family name or last name is Sloyan :) ).
By the way I like your countery very much!I was there in the summer of 2004(as a member of the Armenian team for the IMO 2004) and spent there 10 unforgetable days :wink: .

Also thanks for a nice problem :wink: .

Tigran

Thank you Tiks for your compliments about me and my country. Hellas (Greece), is truly a beautiful country and I love it very much, as the same everybody loves his country.
About me, I think that I owe to you and all, an explanation. I am not a mathematician (I am an architect) and so, my geometry knowledge is fewer than you and more of mathlinkers.
I try to solve the problems (also the difficult of them), based on simple ideas, comprehensible to the mean mind. I like geometry and I am learning every day, new thinks from this forum. I will be happy if sometimes there is something interest, in what I wrote and I am looking forward to your remarks and advices.

Kostas Vittas.

Dear all my friends.

I would like to present the synthetic proof I have in mind about this problem and sorry for the late answer.

We denote as $K,$ $L,$ the intersection points of the circumcircle $(O)$ of $ABC,$ from the segment lines $BF,$ $CE,$ respectively. It is clear that these points are the antidiametric points (= antipodes ) of $C,$ $B$ respectively, and so, the quadrilateral $BCLK,$ is a rectangle.

Because of $\angle LAC = \angle LBC,$ $\angle CAE = \angle CBB'$ and $\angle EAD' = \angle B'BD',$ we conclude that the pencils $A.LCED',$ $B.LCB'D',$ have equal

double ratios and so we have that $(A.LCED') = (B.LCB'D')$ $,(1)$

Similarly we have that $(A.KBFD') = (C.KBC'D')$ $,(2)$

But $(A.LCED') = (L,C,E)$ $,(3)$ $($ because of $LE\parallel AD'$ $)$ and similarly $(A.KBFD') = (K,B,F)$ $,(4)$

Because of $KL\parallel BC\parallel EF,$ $\Longrightarrow$ $(L,C,E) = (K,B,F)$ $,(5)$

From $(1),$ $(2),$ $(3),$ $(4),$ $(5),$ $\Longrightarrow$ $(B.LCB'D') = (C.KBC'D')$ $,(6)$

From $(6)$ and because of the pencils $B.LCB'D',$ $C.KBC'D',$ have the sideline $BC$ of $ABC,$ as their common ray, we conclude that the points

$O\equiv BL\cap CK,$ $M\equiv BB'\cap CC',$ $D'\equiv BD'\cap CD',$ are collinear and the proof is completed.

$\bullet$ This proof is dedicated to [b]Loukas Sidiropoulos[/b].

Kostas Vlittas.

[color=darkblue][b]Very nice the Vitasko's problem and also very nice the Tiks' and Vitasko's proofs.[/b][/color]

[color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ABC$ be a triangle inscribed in the circle $w$. Denote the point $D\in w$ for which $AD\perp BC$. For two mobile points $X\in AB$, $Y\in AC$ so that $XY\parallel BC$ define the intersections $M\in BC\cap DX$, $N\in BC\cap DY$ and the second intersections $E$, $F$ of the lines $AM$, $AN$ respectively with the circumcircle $w$ of the given triangle. Ascertain the geometrical locus of the intersection $L\in BF\cap CE$.

[u]Notation.[/u] I denoted $XY$- line.[/color]

Dear all, I apologize because I forgot to post the relative figure.

Kindly please a moterator, to incorporate this figure in my previous message.

Thanks a lot, Kostas Vittas.

[quote="vittasko"]Let $BCEF$ be a quadrilateral for which $BF\perp BC$ and $BF\parallel CE$. It is erected on the side segment $BC$ of a given triangle $ABC$ outwardly to it. Let $B',$ $C'$ be the second intersections of the circumcircle  $w(O)$ of the triangle $ABC$ with the lines $AE$, $AF$ respectively. Denote the point $D\in w$ for which $AD\perp BC$. Prove that the lines $BB'$, $CC'$, $DO$ are concurrently if and only if $BF=CE$ ([b][u]Kostas Vittas[/u][/b]).[/quote]
[color=darkred][b][u]Lemma[/u][/b] (prove easily). Let $ABCDEF$ be a cyclical convex hexagon. Then

$AD\cap BE\cap CF\ne\emptyset\Longleftrightarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$.[/color]

[color=darkblue][b][u]Proof.[/u][/b] Denote the second intersection $D'$ of the ray $(DO$ with the circle $w$.

Observe that $\{\begin{array}{c}D'B=AC\ ,\ D'C=AB\\\\ \triangle ABF\sim \triangle DC'B\Longrightarrow\frac{C'D}{C'B}=\frac{AB}{BF}\\\\ \triangle ACE\sim\triangle DB'C\Longrightarrow\frac{B'C}{B'D}=\frac{CE}{AC}\end{array}$.

Apply the above lemma to the hexagon $BC'DB'CD'$ :

the lines $BB'\cap CC'\cap DO\ne\emptyset\Longleftrightarrow$ $BB'\cap CC'\cap DD'\ne\emptyset$ $\Longleftrightarrow$

$BC'\cdot DB'\cdot CD'=C'D\cdot B'C\cdot D'B$ $\Longleftrightarrow$ $\frac{C'D}{C'B}\cdot\frac{B'C}{B'D}\cdot\frac{D'B}{D'C}=1$ $\Longleftrightarrow$

$\frac{AB}{BF}\cdot\frac{CE}{AC}\cdot\frac{AC}{AB}=1$ $\Longleftrightarrow$ $BF=CE$.

[b]Remark.[/b] This proof is not trigonometrically and is identically with the [u]Tiks' proof[/u].[/color]

The main reason this concurrency takes place is the fact of symmetry
 of the rectangle wrt diameter orthogonal to the side BC as well as the
 fact  of  the point A' ( symmetric of A in this diameter) being the antipode
 of D'.

  Thank you.

   M.T.


function
real analysis
real analysis unsolved

Let $f$ be the function defined by $f(x) = x^{2006}$ if $x$ is rational and $f(x) = 0$ if $x$ is irrational.

a) Show that $f$ is continuous in exactly one point.
b) Is $f$ also differentiable in this point?
c) If so, is it twice differentiable in that point?

It's hard for me to believe they asked you this question in class exactly as it is, because it doesn't make sense. In order to define the notion of "twice differentiable" in a point, you need differentiability (and hence continuity) in a neighborhood of that point, but you just told us that the function is continuous in exactly one point, so.. :)

The third question is a bit silly indeed :)

The first and second question can be shown(but not rigorous) with a simple graph(sketch). This function is both continuous and defferentiable at $x_0 = 0$.
An other way is to work with sequences which tend to 0.

[u]babis[/u]

But it's easy to give a rigorous argument, right? :)


algebra
polynomial
quadratics
function
quadratic formula
superior algebra
superior algebra unsolved

Find an isomorphism from $ \mathbb{F}_{5} [X]/(X^2\plus{}X\plus{}1)$ to  $ \mathbb{F}_{5}(\sqrt 2)$
I don't know if there such isomorphism exists.  We have to send $ X$ to an element $ a\plus{}b\sqrt 2$ such that
$ (a\plus{}b\sqrt 2)^2\plus{}a\plus{}b\sqrt 2\plus{}1 \equal{}0$ mod $ 5$.
if I expand everything I find:
$ a^2\plus{}2b^2\plus{}a\plus{}1\plus{}(2ab\plus{}b)\sqrt2$.
solving $ (2ab\plus{}b)\sqrt2 \equal{}0$ gives either $ b\equal{}0$ mod $ 5$ or $ a\equal{}2$ mod $ 5$.
In both cases I can't find $ b$.
What am I doing wrong?
Thanks for helping!

oh I was working with $ x^2 \plus{} x \plus{} 2$ :S Sorry. Thanks!!

is there an elegant methode to find such number?.
For example, if we want to find an isomorphism
 from: $ \mathbb{F}_7(X)$/$ (X^2\plus{}2X\plus{}2)$ to $ \mathbb{F}_7(X)$/$ (X^2\plus{}X\plus{}3)$. (the polynomials are irred. )
One way is to follow the same methode as above. But is there a more clever way?

One way is to see how they might be related by "solving" for $ x$ using the quadratic formula (as long as you're not in a field containing $ \mathbb{F}_2$).

For the first one, $ x^2 \plus{} 2x \plus{} 2 \equal{} 0\Rightarrow x \equal{} \minus{} 1\pm \sqrt { \minus{} 1}$. For the second,
\[ x^2 \plus{} x \plus{} 3 \equal{} 0\Rightarrow x \equal{} \frac { \minus{} 1\pm \sqrt { \minus{} 11}}{2}
\]
The second one can be "simplified" to
\[ \frac { \minus{} 1\pm 2\sqrt { \minus{} 1}}{2} \equal{} \minus{} \frac {1}{2}\pm \sqrt { \minus{} 1} \equal{} \minus{} 4\pm \sqrt { \minus{} 1}
\]
Although this isn't exactly rigorous, it says that the second $ x$ should be obtained from the first by subtracting 3. Indeed, the function $ \phi$ mapping $ P(x)\plus{}(x^2\plus{}2x\plus{}2)$ to $ P(x \minus{} 3)\plus{}(x^2\plus{}x\plus{}3)$ is an isomorphism from $ \mathbb{F}_7[x]/(x^2 \plus{} 2x \plus{} 2)$ onto $ \mathbb{F}_7[x]/(x^2 \plus{} x \plus{} 3)$.


logarithms
inequalities proposed
inequalities

[quote="Zamfirmihai"][log(a) from b]^3 / (c+d) +....and so on( 4 therms) >= 1/max(a^2, b^2,c^2,d^2)

a,b,c,d are from [1/2;1)[/quote]

i forgot to say:D composed by me


Prove that there is no even-degree poly $ P$ such that all the primes are values of $ P$ for integer arguments.

The series $ \sum_{k \equal{} \minus{} \infty, P(k) \neq 0}^{\infty} \frac {1}{P(k)}$ converges (edit: and only contains a finite number of negative terms).


Mary has five marbles with different colors for each of them.  Find the color of marble B.

Colors are blue, black, red, green, and orange.  Marble A, C, and Es colors are not black.  Marble A, C, and Bs colors are not blue.  Marble A, D, and Cs colors are not orange.  Marble A is green. Marble A, D, and C are not black.

By the way, I made this problem by myself =) It's not gonna be much challenging, though.

nice problem... but do you need all of the information?

[hide]A,C,E are not black, so B or D must be black

then the last piece of info, A,D,C are not black implies that B has to be black...

[/hide][/hide]

Oh, I hadn't seen it..
Then new addtional problem for that.

Find the rest of marbles' colors.

Well, uhm...

[hide]D is blue, C is red, A is green , E is orange[/hide]

Nice job.  Can you show your work if you don't mind?

Sorry about that, so here we go:

[hide]What I did was make a chart. The columns were labeled Blue, Black, Red, Green, Orange.  I put the letters B and D under Black; B, D, under Blue; B, E, under Orange, A under Green, and then I erased D under black and just kept B. Since B is Black, from the above you can conclude E is orange (simply setting up the table will let you see this). Since E is orange, D must be Blue.  Since we have found the 4 colors, C is red. It all comes down to making the chart.[/hide]

[hide]a- green

b- black

c- red

d- blue

e- orange[/hide]

i used to love these chart problems. especially those with a weird chart that has 2 charts connected. too bad i don't do them anymore

Nice work, everyone..
I didn't think to make chart but here's easy tip to think the answer.

Colors are blue, black, red, green, and orange. Marble A, C, and Es colors are not black. Marble A, C, and Bs colors are not blue. Marble A, D, and Cs colors are not orange. Marble A is green. Marble A, D, and C are not black.

Rearrange like this:
ACE aren't black
ACB aren't blue
ACD aren't orange
ACD aren't black
A is green.

Then C isn't green, black, blue, orange, which means it's red.
Then D isn't green, red, orange, black, which means it's blue.
Then B isn't red, blue, green, blue, which means black.
Then E is orange.

I found it was nice to make problems like this.  It's pretty fun, you know...


Prove that if the rectangular solid has volume of 8 and its edges are in geometric progression, its surface area is numerically equal to the sum of its edges.

Don't look in the spoiler unless you have to.

[hide]Edges are $a,ar,ar^2$

Volume is $a^3r^3=8\Rightarrow ar=2$

Surface area is $2(a^2r+a^2r^3+a^2r^2)$

Sum of edges is $4(a+ar+ar^2)$

Using the reduction $ar=2$ a few times, we get:

$\sum edges=4(a+2r+2)$
$SA=2(2a+4r+4)=4(a+2r+2)$[/hide]

Beautiful solution, blahblahblah.

This was actually an AHSME question #25 to be exact but I wanted to see what the proof looked like. :)

How do you get a 'proof' question on the AHSME?

Technically, it was not a proof problem. The question gave the volume and its surface area value, and asked you to define the sum of the edges. And on the answer key, there was note that if the volume is 8 and the edges are in geometric progression, the surface area = sum of the edges. I saw that and wondered if this was possible to prove.

So, basically, there was no proof question on AHSME.

Ah.  In any event, provided that you notice $ab=2$ and have the spatial skills to figure out what the sum of the edges and the sum of the surface areas are, it's pretty easy.

Yeah, I thought that was neat too.

Old AHSME sure included many identities. I remember seeing one Theorem that I didn't hear before but in my "Geometry Revisited" book by Coexter on AHMSE. I can't remember the problem nor the theorem because I really didn't cover much of "Geometry Revisited" but anyway, AHSME used to include just about every theorem in the Geometry. :D


quadratics
number theory
advanced fields
advanced fields theorems

can some one help me to find a book or a document about quadratic field ? because it's very important for me to know this concept.

thank u ! :D

the book is called: Number Fields by Marcus of CSU San Louis Obispo. It is out of print but i heard many number theorists used this book to start their number theory studies. if you like to solve diophantine equations using advanced algebra go get this book. another one by R.Mollin is also pretty good for beginning learners.


limit
real analysis
real analysis unsolved

hi,
consider $ \varphi : \mathbb{R} \rightarrow \mathbb{R}$, in 0 differentiable with $ \varphi (0) \equal{} 0$. calculate
\[ \lim_{x \rightarrow 0}{\frac{1}{x}(\varphi (x) \plus{} \varphi (\frac{x}{2}) \plus{} ... \plus{} \varphi (\frac{x}{n}))},\]
$ n \in \mathbb{N}^{*}$.


Prove that any polynoamial of the form $f(x)=(x-a_{0})^{2}...(x-a_{n})^{2}+1$where $a_{0}...a_{n}$ are integers cannot be factorized into twon ono trivial polynomails each with integer coeefficients


1) Triangle $ABC$ has integer-length sides, and $AC=2007$. The internal bisector of $\angle BAC$ meets $BC$ at $D$. Given that $AB=CD$, determine $AB$ and $BC$.

2) Let $ABC$ be an acute-angled triangle with $AB>AC$ and $BAC=60^{\circ}$. Denote the circumcentre by $O$ and the orthocentre by $H$ and let $OH$ meet $AB$ at $P$ and $AC$ at $Q$. Prove that $PO=HQ$.

2) Because $\angle BOC=\angle BHC=120^\circ$, $BCHO$ is cyclic. By angle chasing, $\triangle BOC,\triangle BPH$ and $\triangle HQC$ are isosceles with base angle of $30^\circ$. For those triangles, base length = $\sqrt{3}$ of length of the other sides. Applying Ptolemy to $BCHO$,

$BC\cdot HO+BO\cdot CH=BH\cdot CO$
$(\sqrt{3}R)(HP-PO)+(R)(\sqrt{3}HQ)=(\sqrt{3}HP)(R)$ where $R=OB$
$HQ=PO$

[hide="For the record"][quote="DPopov"]Oops...[/quote]
[quote="DPopov"]What was I thinking.[/quote][/hide] With CA = 2007, AB = 2007, BC = 4014, we cannot make a triangle.

$c = AB = CD = \frac{ab}{b+c},$ where a = BC, b = CA, c = AB are positive integers, $b = 2007 = 223 \cdot 3^{2}.$ Then $c^{2}+bc = ab,$ $c^{2}+223 \cdot 3^{2}\cdot c = 223 \cdot 3^{2}\cdot a,$ $c = 223 \cdot 3n,$ $n > 0$ is a positive integer. Substituting it in, $223^{2}\cdot 3^{2}\cdot n^{2}+223^{2}\cdot 3^{3}\cdot n = 223 \cdot 3^{2}\cdot a,$ $a = 223 (n^{2}+3n).$ As a result, $(a, b, c) = 223 (n^{2}+3n,\ 3^{2},\ 3n).$ $a+b > c$ always. $b+c > a \Longrightarrow 3^{2}+3n > n^{2}+3n,$ $n < 3.$ $c+a > b \Longrightarrow n^{2}+6n > 9.$ $P(n) = n^{2}+6n-9 = 0$ has roots $n_{12}= 3 (-1 \pm \sqrt{2}),$ hence $n < n_{1}= 3(-1-\sqrt{2})$ or $n > n_{2}= 3(-1+\sqrt{2}).$ Combining with $n > 0$ integer, $n \ge 2.$ Combining with $n < 3$ obtained previously, $n = 2,$ $a = 223 \cdot (n^{2}+3n) = 2230,$ $c = 223 \cdot 3n = 1338.$


Suppose that  x  and  y  are two real numbers such that
$ x+2^{x}=y+2^{y}$

 Prove that
$ x+\sin x= y+\sin y$

[hide]
Let $ f: \mathbf{R}\to\mathbf{R},f(x)=x+2^{x}$.
$ f$ is strictly ascending, so $ f$ is injective. Then $ f(x)=f(y)\Rightarrow x=y$
[/hide]


A circle passes through the vertex of a rectangle $ABCD$ and touches its sides $AB$ and $AD$ at $M$ and $N$ respectively. If the distance from $C$ to the line segment $MN$ is equal to $5$ units, find the area of rectangle $ABCD$.

[quote="Rushil"]A circle passes through [b]the vertex[/b] of a rectangle $ABCD$[/quote]

You have probably forgotten to say which vertex, but obviously it must be the vertex C.

We will show a more general result:

[color=blue][b]Theorem 1.[/b] Let C, M, N be three points on a circle k, and O the center of this circle k. The tangents to the circle k at the points M and N meet at a point A, and we denote by B, D and H the orthogonal projections of the point C on the lines AM, AN and MN. Then, $CB\cdot CD=CH^2$.[/color]

[i]Proof of Theorem 1.[/i] We will use directed angles modulo 180°.

Since < CHM = 90° and < CBM = 90°, the points H and B lie on the circle with diameter CM. Thus, < CBH = < CMH. Since the line AN is the tangent to the circle k at the point N, by the tangent-chordal angle theorem, < (CN; AD) = < CMN. Thus, < CBH = < CMH = < CMN = < (CN; AD) = < CND. Finally, since < CHN = 90° and < CDN = 90°, the points H and D lie on the circle with diameter CN, so that < CND = < CHD. Thus, < CBH = < CHD. Similarly, < CDH = < CHB.

Now, from < CBH = < CHD and < CHB = < CDH, it follows that the triangles CBH and CHD are directly similar; thus, CB : CH = CH : CD, so that $CB\cdot CD=CH^2$. This proves Theorem 1.

Now, this clearly applies to your problem: Your points B and D are indeed the orthogonal projections of the point C on the lines AM and AN, since the angles < ABC and < ADC are 90° (as the quadrilateral ABCD is a rectangle), and we have CH = 5, since CH is the distance from the point C to the line segment MN; thus, Theorem 1 yields $CB\cdot CD=CH^2=5^2=25$; but the area of the rectangle ABCD clearly equals to $CB\cdot CD$, and thus this area equals 25.

  darij


ratio
inequalities
algebra
system of equations

Minimise ((a+3b)(b+4c)(c+2a))/(abc) for real numbers 0<=a<=b<=c.

By the way, this may help you understand about making sure your bound is attainable, fierytycoon.

The solution below involves the intuivitive steps I took when I first looked at the problem, and making that intuition into a rigorous proof w/o weighted AM-GM.

[hide]From the setup of the problem, we easily deduce that only the ratio of a:b:c matter.

Thus, we set b=1 WLOG, and attempt to solve from here.

Our New equation is ((a+3)(1+4c)(c+2a))/ac.

Lemma 1: If this expression increases in value for any increase dC of a valid value of c (c>=1)

Proof: when a=1, then the expression will also increase for a similar change with any a.

Let us consider what would happen to (a+3)(1+4c)(c+2a) as  c increases by dC with a fixed a in

the valid range (0<a<=1).    (a+3)(1+4c)(c+2a) will increase by a factor of ((4c+1+dC)/(4c+1))((

2a+c+dC)/(2a+c)) >= ((4c+1+dC)/(4c+1))((2+c+dC)/(2+c)) (since 0<a<=1).

Thus if ((a+3)(1+4c)(c+2a))/ac = ((4)(1+4c)(c+2))/c increases as some c increases by some dC, ((a+3)(1+4c)(c+2a))/ac will increase as some c increases by some dC FOR ALL A.  End lemma.

By taking a=1, it is easy to see that the resulting expression, (4)(1+4c)(c+2)/c is minimum on the valid range of c when c=1.

Thus, we know that ((a+3)(1+4c)(c+2a))/ac is minimum when b=c=1.  This reduces the problem to finding the minimum of ((a+3)(5)(1+2a))/a, which can be easily found to be a=1 on the valid range of a.

So we have a=b=c => min is 60.

[/hide]

Yea, its pretty messy.  One of these days I'm gonna have to practice doing inequalities for the USAMO.

[hide]quite messy, though using the fact that the inequality was homogenous is a good one (therefore you can arbitrarily set some value to be 1, such as the sum or product of the three numbers)

The weighted AM-GM also has an intuitive application. Looking at the problem statement, and trying a few values for a, b, and c tells us that the minimum is probably when a=b=c and the value would then be 60. Therefore, its easy to see we can't use normal AM-GM in the top, since that would require a = 2c and all those things for the equality condition to exist (however, this system is not linearly independent so we aren't going to solve it anytime soon). Therefore, to exploit the a=b=c, we use weighted AM-GM, whose equality condition IS a=b=c, and it works wonders for us.

As it turns out, the a<=b<=c condition, which appeared at first as a hindrance, simplifies the problem greatly, since you can later toy with exponents to obtain nice symmetry.

Two things I like to do on inequalities:

1. Maintain (or obtain) symmetry.

2. For nonstrict inequalities, whenever you use an inequality like AM-GM, target the equality case (in other words, make sure the equality case is ACHIEVABLE). It simplifies things quite a bit.

[/hide]

Nukular wrote:[hide]
use weighted AM-GM,
and you get

Quantity in question >= 60 a55/60b57/60c68/60/(abc)>=60abc/(abc) = 60.

Which holds when 0<a=b=c

[/hide]

In this proof, he just expanded out the left side (which yields [hide]25abc + 2a2b + 6ab2 + 8a2c + 4ac2 + 3b2 + 12bc2[/hide]). He then treated [hide]25abc[/hide] as [hide](abc + abc + ... + abc)[/hide], and so on. Applying AM-GM to these terms yields a RS of [hide]a55/60b57/60c68/60[/hide].

I don't see how he can claim that the bound is attainable, though. And, I am a little bit shaky about he just got [hide]60abc/abc[/hide] in the next step. Aren't you supposed to argue that AM-GM has an obtainable equality when all the terms in the expression on the left side are equal, i.e. [hide]abc = a2b = ab2 = a2c = ac2 = b2c = bc2, from which you get a = b = c != 0[/hide]? Then RS = [hide]60a55/60a57/60a68/60 / a3 = 60.[/hide]

Osiris wrote:[hide]
In this proof, he just expanded out the left side (which yields 25abc + 2a2b + 6ab2 + 8a2c + 4ac2 + 3b2 + 12bc2). He then treated 25abc as (abc + abc + ... + abc), and so on. Applying AM-GM to these terms yields a RS of a55/60b57/60c68/60.

I don't see how he can claim that the bound is attainable, though. And, I am a little bit shaky about he just got 60abc/abc in the next step. Aren't you supposed to argue that AM-GM has an obtainable equality when all the terms in the expression on the left side are equal, i.e. abc = a2b = ab2 = a2c = ac2 = b2c = bc2, from which you get a = b = c != 0? Then RS = 60a55[/hide]

[hide]

Firstly i didnt expand the LHS:

((a+3b)(b+4c)(c+2a))/(abc)

so a + 3b >= 4a^(1/4)*b^(3/4) = x

     b + 4c >= 5b^(1/5)*c^(4/5) = y

     c + 2a >= 3c^(1/3)*a^(2/3) = z

and multiplying them together: xyz = 60a^(55/60)b^(57/60)c^(68/60) >= 60a^(55/60)b^(57/60)a^(5/60)b^(3/60)c = 60abc

so the LHS >= xyz / (abc) >= 60abc/(abc) = 60.

So we know the LHS >= 60. Now we show it is attainable, by setting a=b=c=1.

This gives us 4*5*3/1 = 60. So its attainable and a lower bound, and therefore must be the minimum value.

[/hide][hide][/hide]

Wait a minute, I still don't get it:

[hide]So we have xyz = 60a^(55/60)b^(57/60)c^(68/60). But how do we know that we must make the inequality weaker by stating that 60a^(55/60)b^(57/60)c^(68/60)  >= 60abc so that equality can be achieved in ((a+3b)(b+4c)(c+2a))/(abc) >= 60 (other than noting that that yields the nicest nubmers without ugly fractional exponents)? That is, why do we know not to go ahead and say something random like 60a^(55/60)b^(57/60)c^(68/60)  >= 60a^(55/60)b^(57/60)c^(67/60) instead?[/hide]

You don't make them weaker. You started by applying AM-GM to each of the terms (a + 3b), (b + 4c) and (c + 2a), to find their minimums. Then set the minimums to x, y and z. Then we must have that LS :ge: 60 xyz / abc. However, since x = a1/4b3/4, y = b1/5c4/5, z = c1/3a2/3, you find that xyz = a55/60b57/60c68/60.

Take a look at the condition under which the minimum can be achieved, then you will see that a = b = c. Then xyz = abc (amazing!) So LS :ge: 60 abc / abc = 60. Equality is achieved when a = b = c.

[quote]Take a look at the condition under which the minimum can be achieved, then you will see that a = b = c. Then xyz = abc[/quote]

Well, you have taken AM-GM thrice, so equality for each of those three times occurs when a = 3b, b = 4c and c = 2a. But solving this system of equations, we find (a,b,c) = (0,0,0), which does not work since the LHS is divided by abc. So how can we see that equality occurs when a = b = c?

take this example (we didn't use standard AM-GM)

here is what we REALLY did to a+3b (called weighted AM-GM.. u will see why)

a + 3b = a + b + b + b

Apply AM-GM to the RHS:

(a+3b)/4 >= a1/4b3/4

so a+3b >= 4a1/4b3/4

with equality condition a=b=b=b or simply a=b.

the other eq. conditions are a=c and b=c making the net condition
a=b=c which is doable.

Oh, whoops. Sorry, my bad, forgot that we were using weighted AM-GM and so the equality case is different. Thanks.


If [tex]a, b, c, d > 0[/tex], prove that

[tex]
\displaystyle \sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \geq \left(\frac{abc+abd+acd+bcd}{4}\right)^{1/3}
[/tex]

We proceed as follows:

a^3+b^3+c^3  :ge: 3abc
a^3+b^3+d^3  :ge: 3abd
....
b^3+c^3+d^3  :ge: 3bcd

(all by AMGM)

Summing, we have:

a^3+b^3+c^3+d^3  :ge: abc+abd+acd+bcd

But

((a^3+b^3+c^3+d^3)/4)^(1/3)  :le: ((a^2+b^2+c^2+d^2)/4)^(1/2) because P(1/2)  :ge: P(1/3).

....

Wow, second time in a week that I've messed up a Power Mean question.

[quote]((a^3+b^3+c^3+d^3)/4)^(1/3)<=((a^2+b^2+c^2+d^2)/4)^(1/2)[/quote]

Im pretty sure that inequality is flipped

[quote="blahblahblah"]((a^3+b^3+c^3+d^3)/4)^(1/3)  :le: ((a^2+b^2+c^2+d^2)/4)^(1/2) because P(1/2)  :ge: P(1/3).[/quote]

Nope.  The expressions you wrote there are P(2) and P(3), respectively, not P(1/2) and P(1/3).  (For those of you who don't know what we're talking about, this is the Power Mean Inequality, a generalization of AM-GM.)  So your final inequality should be reversed, so your proof doesn't hold.

*Edit* Ach, Zabelman beat me to it :)

What do P(1/2) and P(1/3) mean? (By context P(1/n) seems to mean the nth root of a^n + b^n + ..., but that strikes me as somewhat odd)

zabelman is correct, the inequality should be:

((a^3+b^3+c^3+d^3)/4)^(1/3) >= ((a^2+b^2+c^2+d^2)/4)^(1/2)

by the Power Mean Inequality. So your solution does not work. :(

edit: a lot of people beat me to it, lol ;)

....I would like to see a solution to this.

We'll be using the inequality [tex]xy\le\frac{(x+y)^2}{4}[/tex] a couple times. (Do you see why it's true?)

We get

[tex]\displaystyle abc+bcd+cda+dab[/tex]

[tex]\displaystyle  = ab(c+d) + (a+b)cd[/tex]

[tex]\displaystyle \le \frac{(a+b)^2}{4}(c+d)+(a+b)\frac{(c+d)^2}{4}[/tex]

[tex]\displaystyle  = \frac{(a+b+c+d)(a+b)(c+d)}{4}[/tex]

[tex]\displaystyle  \le \frac{(a+b+c+d)(a+b+c+d)^2}{16}[/tex]

[tex]\displaystyle  = \frac{(a+b+c+d)^3}{16}[/tex]

Therefore,

[tex]\displaystyle \frac{abc+bcd+cda+dab}{4}\le\frac{(a+b+c+d)^3}{64}[/tex]

[tex]\displaystyle \left(\frac{abc+bcd+cda+dab}{4}\right)^{1/3}\le\frac{a+b+c+d}{4}[/tex]

Then by AM-QM, we have

[tex]\displaystyle \frac{a+b+c+d}{4}\le\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}[/tex]

Therefore,

[tex]\displaystyle \frac{abc+bcd+cda+dab}{4}\le\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}[/tex]

as desired. [tex]\displaystyle \blacksquare[/tex]

[quote="zabelman"] (Proof)

[/quote]

Very nice, and simple.

zabelman gave a very nice and clean proof!

Actually, AM >= RHS is a special case of the Theorem of Symmetric Mean.

I am pointing out some references here for people who want to study more about the Symmetric Mean.

Kiran Kedlaya's Inequality notes (Maclaurin's Theorem on p.9). Using his notation in the notes, it's [tex]d_1\ge d_3^{1/3}[/tex].

The Theorem of Symmetric Mean also appears as problem 286 on p.68 of The USSR Olympiad Problem Book.

Write it out the theorem for four numbers, it is:

[tex]\displaystyle\frac{a+b+c+d}{4}\geq \left(\frac{ab+ac +ad+bc +bd+cd}{6}\right)^{1/2}[/tex][tex]\displaystyle\geq \left(\frac{abc+bcd+cda+dab}{4}\right)^{1/3}
\geq\left({abcd}\right)^{1/4}
[/tex]

[quote="Fierytycoon"]Zach, I'm just curious: about how long did it take you to discover that solution?[/quote]

After I realized the clever way to write abc+bcd+cda+dab as ab(c+d)+(a+b)cd (which was after about 15 mins), everything just fell together. Of course I did just complete the Inequalities course here on AoPS, so that might have something to do with it ;)


Gauss
geometry
3D geometry
number theory
divisibility tests

In general, a sum and difference of two irrational numbers is irrational.  This is not the case when the two "cancel", like $ 1\plus{}\sqrt{2}\plus{}(1\minus{}\sqrt{2})\equal{}2$, but that is not the case in any of your problems.

I'll show the proof for $ \sqrt{2}$:
- Suppose the opponent,($ \sqrt{2}$ is rational) so that $ \sqrt{2}\equal{}\frac{p}{q}$ , where $ gcd(p,q)\equal{}1$.
We get:
$ 2q^2\equal{}p^2$
From here it's obvious that $ p\equal{}2p_1$ where $ p_1 \in N$
substituting $ p\equal{}2p_1$ into $ 2q^2\equal{}p^2$ we get
$ q^2\equal{}2p_1^2$
From here it's obvious that $ q\equal{}2q_1$ where $ q_1 \in N$
Therefore both $ p,q$ are multiple of $ 2$ which is contradition of $ gcd(p,q)\equal{}1$.
So we're done.
PS: The other cases are similar.I think you can easily prove them  :D

Thanks. $ \sqrt{2}$ and $ \sqrt{3}$ are the easy and well-known parts. What about the others?

We can prove that $ \sqrt {3}$ is irrational in much the same way as we prove that $ \sqrt {2}$ is irrational, but using divisibility by $ 3$ rather than by $ 2$.

To prove that $ \sqrt {6}$ is irrational in the same way, we can use divisibility by $ 2$ or by $ 3$.

This then enables us to prove that $ \sqrt {2} \plus{} \sqrt {3}$ is irrational as follows:

Suppose that $ \sqrt {2} \plus{} \sqrt {3}$ were rational and equals the rational number $ q$, say.

Then, squaring gives $ 2 \plus{} 3 \plus{} 2\sqrt {6} \equal{} q^{2}$, so that $ \sqrt {6} \equal{} \frac {q^{2} \minus{} 5}{2}$ which means that $ \sqrt {6}$ is rational, a contradiction. Therefore $ \sqrt {2} \plus{} \sqrt {3}$ is irrational.

I'm guessing that, from Gauss' theorem on rational roots, since the equations $ x^{3} \minus{}2\equal{}0$ and $ x^{3} \minus{} 4\equal{}0$ have no integer solutions, $ 2^{\frac{1}{3}}$ and $ 2^{\frac{2}{3}}$ are irrational so we can do something similar to the previous part and cube this (having assumed it's rational) to get $ 6\plus{}6(2^{\frac{1}{3}} \plus{} 2^{\frac{2}{3}}) \equal{} q^{2}$ leading to a contradiction as before.

And the last part?

- Not every time multiplying two irrational numbers the result is irrational.Example:
$ 2^{\frac {1}{2}}*2^{\frac {2}{3}}$,which both are irational numbers when we multiply them we get rational number as in this case:
$ 2^{\frac {1}{3}}*2^{\frac {2}{3}} \equal{} 2$

PS:Anyway I will work on others,maybe i get a result,because i didn't solve them before.

Thanks obel1x.

I was editing my previous post while you were typing.

Now I'm fairly sure that the same trick will work for the last part. Any takers?

[quote="Obel1x"]$ 2^{\frac {1}{2}}*2^{\frac {2}{3}} \equal{} 2$[/quote]  You might want to reconsider this particular claim.  Also, \cdot instead of * looks much nicer in LaTeX.

JRav, nice job of restating the question as the answer  :roll:

Has anyone had a go at the last part yet?

Proving that $ 2^{\frac {1}{3}} + 2^{\frac {2}{3}}$ is very simple.

Assume that $ x$ is rational and equal to $ 2^{\frac {1}{3}} + 2^{\frac {2}{3}}$.

So,

${ x = 2^{\frac {1}{3}} + 2^{\frac {2}{3}} = 2^{\frac {1}{3}}(1 + 2^{\frac {2}{3}})}$

$ x^3 = 2(1 + 2^{\frac {1}{3}})^3 = 2(1 + 3 \cdot 1 \cdot 2^{\frac {1}{3}} + 3 \cdot 2^{\frac {2}{3}} + 2)$.

${ x^3 = 2(3 + 3(2^{\frac {1}{3}} + 2^{\frac {2}{3}}})) = 6 + 6(2^{\frac {1}{3}} + 2^{\frac {2}{3}})$.

$ x^3 = 6 + 6x$

$ x^3 - 6x - 6 = 0$

The left side is a polynomial. Let's call it $ W(x)$.

Each rational solution of this equality is also an integer solution since the coefficient standing near the highest power of $ x$ is equal to $ 1$.

Each integer solution must be a divisor of free coefficient.

It easy to see that none of the numbers $ W(1),W( - 1),W(2),W( - 2),W(3),W( - 3),W(6),W( - 6)$ is equal to $ 0$,

which means that the equation doesn't have an integer solution and $ x$ must be irrational.

[b]Excersice[/b]

Use this method to show that both $ (\sqrt{3} - \sqrt{2})^{\frac{1}{100}}$ and $ (\sqrt{3} - \sqrt{2})^{\frac{1}{100}}$ are irrational.[/b]

Thanks sage!. I've already done $ 2^{\frac{1}{3}} \plus{} 3^{\frac{2}{3}}$. I'm hoping to see a proof of the irrationality of $ (\sqrt{3} \plus{} \sqrt{2})^{\frac{1}{100}} \plus{} (\sqrt{3} \minus{} \sqrt{2})^{\frac{1}{100}}$.

It seems clear that if $ (\sqrt{3} \plus{} \sqrt{2})^{\frac{1}{100}}$ were rational, then $ \sqrt{3} \plus{} \sqrt{2}$ would be too but we know that $ \sqrt{3} \plus{} \sqrt{2}$ is irrational.

The same holds for $ (\sqrt{3} \minus{} \sqrt{2})^{\frac{1}{100}}$.

But what about their sum: $ (\sqrt{3} \plus{} \sqrt{2})^{\frac{1}{100}} \plus{} (\sqrt{3} \minus{} \sqrt{2})^{\frac{1}{100}}$?

Hint: Prove that if $ x \plus{} \frac{1}{x}$ is rational, then $ x^{100} \plus{} \frac{1}{x^{100}}$ is rational.

Thanks Ravi B. I have $ x \plus{} \frac {1}{x}$ rational implies $ (x \plus{} \frac {1}{x})^2 \equal{} x^{2} \plus{} \frac {1}{x^{2}} \plus{} 2$ is rational.

Similarly for  $ (x \plus{} \frac {1}{x})^{3}$ and so on.

For $ n$ odd, ($ x \plus{} \frac {1}{x})^{n} \equal{} x^{n} \plus{} \frac {1}{x^{n}} \plus{} \sum_{1}^{\frac{n \minus{} 1}{2}} \binom{n}{r}(x^{n \minus{} 2r} \plus{} \frac {1}{x^{n \minus{} 2r}})$ and similarly for $ n$ even, except that this has a term $ \binom{n}{\frac {n}{2}}$ and so we have the result by mathematical induction.

I hope this is OK. Is it?

Lets assume that $ x$ is rational and $ x \equal{} (\sqrt{2} \plus{} \sqrt{3})^{\frac{1}{100}}$

So we have,

$ x \equal{} (\sqrt{2} \plus{} \sqrt{3})^{\frac{1}{100}}$

$ x^{200} \equal{} (\sqrt{2} \plus{} \sqrt{3})^{2} \equal{} 5 \plus{} 2\sqrt{6}$

$ x^{200} \minus{} 5 \equal{} 2\sqrt{6}$

$ (x^{200}\minus{}5)^2 \equal{} 24$

$ x^{400} \minus{} 10x^{200} \plus{} 1 \equal{} 0$

Each rational solution of this equation must be an integer solution and a divisor of free coefficient.

But we see that after substituting $ x \equal{} \minus{}1$ or $ x \equal{} 1$ we don't get $ 0$ so the equation has no integer solution and $ x$ must be irrational.

You've shown that $ (\sqrt{3} \plus{} \sqrt{2})^{1/100}$ is irrational, but how does that show that the sum $ (\sqrt{3} \plus{} \sqrt{2})^{1/100} \plus{} (\sqrt{3} \minus{} \sqrt{2})^{1/100}$ is irrational?


If the line $y = mx + 1$ intersects the ellipse $x^2 + 4y^2 = 1$ exactly once, then the value of $m^2$ is
\[ \text{(A)}\ \frac{1}{2}\ \ \ \ \text{(B)}\ \frac{2}{3}\ \ \ \ \text{(C)}\ \frac{3}{4}\ \ \ \ \text{(D)}\ \frac{4}{5}\ \ \ \ \text{(E)}\ \frac{5}{6} \]

Ok here goes:

[hide]So plugging in the value of y into the equation of the ellipse you get x^2 +4(mx+1)^2=1.  Expanding we get x^2 + 4m^2x^2+8mx+3=0.  Then factoring the first two terms we get (1+4m^2)x^2+8mx+3=0.  Now since these two graphs intersect only once.  We see the discriminant of this equation must be 0.  So 64m^2-4(1+4m^2)(3)=0.  Solving this equation for m^2 we see m^2 is equal to 3/4.[/hide]

[hide="hint"]
put $y=mx+1$ into the equation, since the two curves has only one intersection, so $\Delta=0$, then solve for $m$[/hide]


Loan repayment help
hi,
im stuck with this problem please help,

Anne borrows 3200. Loan is paid 7 months later by a lump sum that includes 168 in interest. What was the simple rate of interest on the loan?

Im not sure which formulate to use? its not the simple interest formula right?i dont think it is

thanks

When I saw "lump sum", I first thought of compond interest...

Am I right?

No, it isn't compound interest. The problem stipulates that it is simple interest. The formula for simple interest is $ I \equal{} Prt$, where $ I$ is simple interest, $ P$ is principal (i.e. the initial amount), $ r$ is the rate of simple interest (usually per year), and $ t$ is time (usually in years).

Hmmm...at the end of the line giving the problem, it says, "What was the simple rate of interest on the loan?" So what should we use?

In case you didn't notice the sarcasm there, it's simple interest.

$ I\equal{}Prt$

$ 168\equal{}(3200)r(7)$

$ .0075\equal{}r$

So the rate is 0.75% per month.


algebra
difference of squares
special factorizations

So like if $k = 3k'$, where $k'$ is odd,

Then $2^{k}-1 = 2^{3k'}-1 = (2^{k'}-1)(2^{2k'}+2^{k'}+1)$?

I mean, this doesn't differ much from finding a counter example.
It's just a form in which there are infinite counter examples.

A counterexample is sufficient proof, but any composite $k$ will produce a composite value of $2^{k}-1$ because if $d | k$ then $2^{d}-1 | 2^{k}-1$.

Edit:  The fun part is the counterexamples where $k$ is prime.  The smallest one is $2^{11}-1 = 23 \cdot 89$.

[quote="1=2"]If k is even, it's diference of squares. If k is odd, it could be a multiple of 3, and then difference of cubes. So there are infinite possibilities that it isn't prime.

What if k is prime? How do you know that there are a finite or infinite solutions to make it composite?[/quote]

Primes of the form $2^{k}-1$ are Mersenne Primes, and whether or not there are a finite number of Mersenne primes is an open question. So I doubt there's an easy proof that there are a finite number of composite numbers of the form $2^{p}-1$ (as this would imply that there are an infinite number of Mersenne Primes).

[quote="Mday"]I want to prove that not all numbers of the form $2^{k}-1$, where $k$ is odd, are primes. How can I do that?

I know I can show that if $2^{k}-1$ is a prime, then $k$ must be a prime but I dont want to use that fact since I know there's a more simple solution). I also dont just want a counter example (which can be found pretty easy).[/quote]
simply provide a counterexample. you are proving "not all", not "not any". if $k=1$ then $2^{k}-1=1$ which is not prime. if $k=9$ then $2^{k}-1=511$ which is not prime. etc

[quote="rjt"][quote="I Am Me"]Also, for a more simple general (but not all-inclusive) counterexample, notice that if $k$ is even, then $2^{k}-1$ will factor as a difference of squares into $(2^\frac{k}{2}+1)(2^\frac{k}{2}-1)$, and, since $k$ is even and thus divisible by $2$, both terms will be integers.[/quote]

Yes, although the question states that $k$ is odd. If $k$ could be even, we could also use Fermat's Little Theorem.[/quote]

 :oops_sign:

Oops. Helps if I actually read what I'm answering.


function
trigonometry
complex analysis
complex analysis unsolved

Prove that the function $f(z) = \cos{\sqrt{z}}$ is of order $\frac{1}{2}$.

Beware: The following is nonsense. I forgot that $\sqrt{z}$ is not a meromorphic function... The argument works for $\cos(g(z))$ for meromorphic $g$ though, so I'll keep it there. Maybe I'll think of a way to save it.

Using the standard notation of Nevanlinna-theory:
i) $T\left(r,\frac{\alpha f+\beta}{\gamma f+\delta}\right) = T(r,f)+O(1)$
ii) $T(r,f^{n}) = nT(r,f)$

Here $f$ is a meromorhpic function,$n$ is a natural number, $\frac{\alpha z+\beta}{\gamma z+\delta}$ is a Möbius transform. So taking integer powers and Möbius transforms does not affect the order of a meromorphic function.

Now our particular $f$ is the composition of $e^{i\sqrt{z}}$ with $\frac{1}{2}\left(w+\frac{1}{w}\right) = \frac{1+(w-1)^{2}/(w+1)^{2}}{1-(w-1)^{2}/(w+1)^{2}}$. So the order of $f$ is the same as that of $e^{i\sqrt{z}}$ which is trivially $1/2$.

Note that by Hadamard's theorem it follows that the genus of $\cos \sqrt{z}$ is 0. This makes it very easy to establish the product formula for $\cos \sqrt{z}$


inequalities
search
inequalities unsolved

[quote="libra_gold"]$a,b,c>0$, try to prove,
$\frac{a^{3}}{2a^{2}-ab+2b^{2}}+\frac{b^{3}}{2b^{2}-bc+2c^{2}}+\frac{c^{3}}{2c^{2}-ca+2a^{2}}\geq \frac{a+b+c}{3}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66271

Thanks a lot. But i wonder how to search post. (I've tried to search, but failed to find this link)

I have a same problem. Search does not work some months. :(


How to prove Ramanujan`s fromula:
$ 1 \plus{} \frac {1}{1\cdot 3} \plus{} \frac {1}{1\cdot 3\cdot 5} \plus{} \frac {1}{1\cdot 3\cdot 5\cdot 7} \plus{} \frac {1}{1\cdot 3\cdot 5\cdot 7\cdot 9} \plus{} ... \plus{} \frac {1}{1 \plus{} \frac {1}{1 \plus{} \frac {2}{1 \plus{} \frac {3}{1 \plus{} \frac {4}{1 \plus{} \frac {5}{1 \plus{} ...}}}}}} \equal{} \sqrt {\frac {\pi\cdot e}{2}}$
:oops:

[quote="JRav"]He's Ramanujan and we just take it fore granted that he's right? :lol:[/quote]

I can`t find the solution of this formula  :(
Maybe someone help me... I want to see the proof.

[quote="Yuriy Solovyov"]
I can`t find the solution of this formula  :(
Maybe someone help me... I want to see the proof.[/quote]

phew  !... it took me almost 3 days to finally show it.  :maybe:

it requires some out of the box thinking, so i will give you a clue...

essentially you need to show (which is not too hard i think), by some   clever means that

$ \sum_{n\equal{}0}^{\infty}\dfrac{x^n}{1\cdot 3\cdot 5\cdot 7\cdot 9\cdot \; \ldots\; \cdot (2n\plus{}1)}$

$ \;\equal{}\;e^{x/2}\cdot \sqrt{\;\dfrac{\pi}{2\,x} \;}\;\minus{}\;\dfrac {1}{x \plus{} \dfrac {1}{1 \plus{} \dfrac {2}{x \plus{} \dfrac {3}{1 \plus{} \dfrac {4}{x \plus{} \dfrac {5}{1 \plus{} \dfrac {6}{x \plus{} \dfrac {7}{1 \plus{}\dfrac {8}{x \plus{} \dfrac {9}{1 \plus{} \dfrac {10}{x \plus{} \dfrac {11}{1 \plus{}  \ldots}}}}}}}}}}}}$

then put $ x\equal{}1$ to get what you want.

:o   :D i'll try to do something with this sum... :spider:


a,b>0

(a+b) 2 /2 +(a+b)/4  \geq a \sqrt b + b \sqrt a

This is equivalent to (a^2+b^2)/2+(a+b)/4+ab>=a*b^1/2+b*a^1/2.
To prove this we add the following 4 ineqs obtained by AM-GM:

a/8+ab/2>=(a*b^1/2)/2

b/8+ab/2>=(b*a^1/2)/2

a^2/2+b/8>=(a*b^1/2)/2

b^2/2+a/8>=(b*a^1/2)/2 and we're done.


[quote="neelnal"]Box A contains black balls and box B contains white balls. Take a certain number of balls from A and place them in B. Then take the same number of balls from B and place them in A. Is the number of white balls in A then greater, equal to, or less than the number of black balls in B?

Again, rather easy problem from country olympiad. Have fun![/quote]

Objection. You don't know whether the black and white balls are mixed homogenously or not, and since they're discrete quantities, you can't do that.

[hide="However..."]If it were water and [insert other liquid here], and they were mixed evenly, it would work. They would be the same.[/hide]

Lol. Good point. Well, then looks like Brazil Olympiad 1980 had a glitch!

[quote="neelnal"]Box A contains black balls and box B contains white balls. Take a certain number of balls from A and place them in B. Then take the same number of balls from B and place them in A. Is the number of white balls in A then greater, equal to, or less than the number of black balls in B?

Again, rather easy problem from country olympiad. Have fun![/quote]
[quote="seamusoboyle"]Er, i dont see any problem with this.

Notice in the liquid problem, it dosnt matter if the mixtures are homogenious.[/quote]
The problem says that after putting the balls from A into B, then you take out that many balls from B into A, so the number of white and black balls picked out from B aren't necessarily the same but in the liquid problem they are...unless that wasn't your point? :huh:

There are $a$ black balls in box $A$ and $b$ white boxes in box $B$. In the first step we transfer $n$ black balls from $A$ to $B$, and in the second step we transfer $k$ black and $n-k$ white balls back from $B$ to $A$. Therefore we finally have $n-k$ white balls in $A$ and $n-k$ black balls in $B$.


-Rolling a Ball Bearing-
A ball bearing is placed on an incline plane and begins to roll. The angle of elevation of the plane is θ.
The distance (in meters) the ball bearing rolls in t seconds is

s(t) = 4.9(sin θ)t^2

a. Determine the speed of the ball bearing.

b. What value of θ will produce the maximum speed at a particular time.

Hello
[hide="Kindly note"]
Ur word rolling is very much disturbing me.As its contain a lot more.
When u say a body is rolling u should have mention whether it is slipping or not if we see it with rotational prospective.Even then I considering to be simplest possible case.Ur equation is telling a simple translational motion. If we draw a FBD.It would be better if u can explain r u dealing with rotational motion or with translational motion or with both.Sorry for not helping u for ur question.  [/hide]
Thank u.

1.
$ S(t) \equal{} 4.9 \sin(\theta) \cdot t^2$
$ S'(t) \equal{} 9.8 \sin(\theta) \cdot t$

2.
$ a\equal{}g \sin{\theta}$
Clearly the largest acceleration will produce the greatest velocity
$ \Rightarrow \theta \equal{} 90^o$


Given that $ x^2 \plus{} y^2 < 3x \plus{} 4y \minus{} 5,$ show that $ \frac{y}{x}$ lies between $ \frac{1}{2}$ and $ \frac{11}{2}$.

Thanks Russian_man. I didn't think to start with $ y\equal{}kx$.


Two independent particles perform simple harmonic motion with the same amplitude and frequency along a common axis .

They pass one another

(1) when travelling in oppposite directions ,and
(2) as they pass their displacement is half their amplitude

Find the phase difference between them.

I hope someone answers this post. FIVE of my previous posts are still unanswered-so disappointing!

just model one particle's position as $Asin(\omega t)$ and the other's as $Asin(\omega t + \phi)$

Then, because the point is question is $\frac{A}{2}$, $\omega t=\frac{\pi}{6}$

So you figure, okay, the second particle's phase makes it at the same point...
So what angle plus $\frac{\pi}{6}$ would do that?

[hide]
I believe it is $\frac{2\pi}{3}$
[/hide]


function
calculus
derivative
algebra
polynomial
limit
real analysis

Here is a very difficult problem. Let f be an infinitely diferentiable function defined on an open interval such that all its derivatives in any point are positive. Prove that we can develop the function in Taylor series around each point of the interval.

Do you mean that the function is analytic around the interval?

First of all , a trivial equality : $\displaystyle f(x) = T_n(x,x0) + L_n(x,x_0)$ where $T_n$ is the Taylor polynomial and $L_n$ is the Lagrange reste .

If $x>x_0$ then  $T_{n+1} > T_n$ and $L_n >0 $ because  all the derivative are positive .  So $T_n$ is convergent because it is bounded by $f(x)$ .

So the elementary term of the sequence/serie $T_n$ converges to 0 :

$\displaystyle \forall_{ x>x_0}  \lim_{n\to\infty}\frac{(x-x_0)^nf^{(n)}_{(x_0)}}{n!} = 0 \  , \  $ so in general  $\displaystyle \  (1)  \  \forall_{ x,x_0}  \lim_{n\to\infty}\frac{(x-x_0)^nf^{(n)}_{(x_0)}}{n!} = 0 $

But $\ :  \ \displaystyle |L_n(x,x0)| = \frac{|(x-x0)|^{n+1}f^{(n+1)}_{(u(x,x0))}}{n!} \ , \ $ where $|u-x_0| \leq |x-x_0|$

Manifestly $0 \leq  f^{(n+1)}_{(u(x,x0))} \leq  f^{(n+1)}_{(max(x,x0))} $ and from $ (1) $ we have $\displaystyle \forall_{x,x0}\lim_{n \to \infty}L_n(x,x_0) = 0$

 :cool: And now , could can conclude ? Do you follow me ? :cool:

:?  Damned, I made a little mistake, did you see it  :?   The correct Lagrange rest is :

$ L_n(x,x0) = \frac{(x-x0)^{n+1}f^{(n+1)}_{(u(x,x0))}}{(n+1)!}  $

 :cool:

Dear Diogene, I know I am too idiot to understand such rafinated arguments, but could you please try to explain it a little bit more clearly? Use more words, please. Thanks a lot!

What arguments are unclear for you, harazi?

Ah, sorry, I really am idiot, I missinterpreted a formula. Sorry, I understood finally.


calculus
derivative
trigonometry
complex numbers

Using the not so legitimate sum,
$ \sum_{n \equal{} 0}^{\plus{}{\infty}} \exp {in{\theta}} \equal{} \frac{1}{2} \plus{} \frac{1}{2} i\cot {\frac{{\theta}}{2}}$

I was able to obtain (by differentiation) valid expressions for the analytic continuation of the $ {\zeta}$ series for negative integers(and 0)

Putting $ {\theta} \equal{} {\pi} \plus{} {\theta}$
in the real part of the above equation I have,
$ \sum_{n \equal{} 1}^{\plus{}{\infty}} (\minus{}1)^{n\minus{}1} \cos {n{\theta}} \equal{} \frac{1}{2}$

I proceeded to integrate this equation 0 to $ {\theta}$ 2k times in the hopes of finding the general formula for $ {\zeta}(2k)$

However I am stuck at,
$ \frac{(\minus{}1)^{k}{\theta}^{2k}}{2{(2k)}!} \equal{} \sum_{n \equal{} 1}^{\plus{}{\infty}} (\minus{}1)^{n\minus{}1} n^{\minus{}2k} \cos {n{\theta}}  \plus{} \sum_{n \equal{} 1}^{\plus{}{\infty}} (\minus{}1)^{n\minus{}1} n^{\minus{}2k} \sum_{r \equal{} 0}^{k\minus{}1} (\minus{}1)^{r\minus{}1} \frac{n^{2r}}{{(2r)}!} {\theta}^{2r}$

No idea what to make of that troublesome finite series or how to introduce the Bernoulli numbers to the mix!

Any assistance would be greatly appreciated! Thanks!

PS- people who'll feed me the convergence issues shouldn't even bother replying!

Just for closure, I recovered a recursive identity from the above which is equivalent to the 'closed form' with Bernoulli numbers. $ \theta \equal{} \pi$ collapses it to the result! :)


Să se determine toate fracţiile de forma $\frac{a}{b}$ $(a,b\in\mathbb{N^{*}})$, cu proprietatea că, adunând şi la numărător şi la numitor acelaşi număr raţional $x>0$, obţinem o putere cu exponent natural a fracţiei date.

[color=darkblue]Definim : "[u][b]numerele reale $x,y$ au acelasi semn[/b][/u]" daca si numai daca

"$xy>0$ [u][b]sau[/b][/u] $x=y=0$" si vom nota $x\ .as.\ y$. Altfel spus,

numerele reale $x,y$ au aceeasi pozitie fata de $0$, adica $\underline{\mathrm{sgn}\ (x)=\mathrm{sgn}\ (y)}$.

[b][u]Demonstratie.[/u][/b] Presupunem ca exista $x>0$ si $n\in N$ astfel incat

$\frac{a+x}{b+x}=(\frac{a}{b})^{n}\ \ (*)$. Daca $n\in \{0,1\}$, atunci rezulta $a=b$. Fie $n\ge 2$.

[u]Metoda 1.[/u] Relatia $(*)$ este echivalenta cu relatia

$M\equiv ab(b^{n-1}-a^{n-1})=x(a^{n}-b^{n})\equiv N$, $M=N$, $M\ .as.\ (b-a)$ si

$N\ .as.\ (a-b)$, adica $(a-b)^{2}\le 0$ ceea ce inseamna $a=b$.

[u]Metoda 2.[/u] Notam $U=\frac{a+x}{b+x}-\frac{a}{b}$ si $V=(\frac{a}{b})^{n}-\frac{a}{b}$. Se observa ca $U=V$,

$U\ .as.\ (b-a)$ si $V\ .as.\ (a-b)$. In concluzie, $(a-b)^{2}\le 0$, adica $a=b$.

[b][u]Exemple competitive[/u][/b] de utilizare a relatiei de echivalenta $.as.$ ([u]clasa a X - a[/u]) :

$1.\blacktriangleright \max \{a^{x},a^{-x}\}\ge a^{y}$, unde $y=\max\{x,\frac{1}{x}\}$ si $0<a\ne 1$.

$2.\blacktriangleright$ Pentru $a>1$, $b>1$ si $a\ne b$ sa se rezolve urmatoarele ecuatii logaritmice :

$2.1.\ \ \log_{a}(a+x)=\log_{b}(b+x)$.

[u]Aplicatie.[/u] $\log_{x^{2}+3x+4}|x+2|=\log_{x^{2}+5x+9}|x+3|\ .$

$2.2\ \ (a+x)^{\log_{a}b}-(b+x)^{\log_{b}a}=b-a$ ([b]Dinu Serbanescu[/b]).

$3.\blacktriangleright$ Sa se rezolve sistemul de ecuatii : $\{\begin{array}{c}\log_{3}x+\log_{2}y=2\\\\ 3^{x}-2^{y}=23\end{array}$

[b][u]Observatii.[/u][/b] Presupunem ca $0<a,b\ne 1$. Atunci :

$\{\begin{array}{c}x,y\in \mathrm R\Longrightarrow \{\begin{array}{c}(a^{b}-a)(b^{a}-b)>0\\\\a^{x}-b^{x}\ .as.\ (a-b)(x-1)\\\\(a^{x}-b^{x})\ .as.\ x(a-b)\end{array}\\\\ x,y>0\Longrightarrow \{\begin{array}{c}(\log_{a}x-\log_{a}y)\ .as.\ (a-1)(x-y)\\\\ \log_{a}b\ .as.\ (a-1)(b-1)\\\\ \log_{a}x-\log_{b}x\ .as.\ (a-1)(b-1)(b-a)(x-1)\end{array}\end{array}$[/color]

La nivelul clasei a $VII$ -a , era suficient sa precizezi ca daca

$ab\left(b^{n-1}-a^{n-1}\right)=x\left(a^{n}-b^{n}\right)$ si $a \not= b$ , atunci observam ca $E1 = b^{n-1}-a^{n-1}$ si $E2 = a^{n}-b^{n}$ au semne diferite $\Rightarrow$  $ab$ si $x$ au semne diferite . Ceea ce este fals ...  $\Rightarrow$ afirmatia facuta este falsa ... deci $a = b$ , iar daca n= 1 .. se deduce tot  $a = b$ ??

[quote="turcas_c"]La nivelul clasei a $VII$ -a , era suficient sa precizezi ca daca ... [/quote]

[color=darkblue]De acord, insa am considerat ca aceasta problema este un pretext pentru a introduce un instrument de atac pentru semnul unei expresii algebrice care inlatura discutia pentru fiecare caz in parte.

Iti ofer un exemplu (daca esti in clasa a X - a) : sa se rezolve ecuatia $\boxed{|x|^{2x^{2}-3x+1}\le 1}$.
Se observa ca $\{-1,0,1\}\subset S$, unde $S$ este multimea solutiilor ecuatiei date. Fie $0<|x|\ne 1$. Notam $E=|x|^{2x^{2}-3x+1}-|x|^{0}$. Astfel, $E(x)\ .as.\ (|x|-1)\left[(2x^{2}-3x+1)-0\right]\ .as.$ $\ (|x|-1)(|x|+1)(2x^{2}-3x+1)=$ $(x-1)^{2}(x+1)(2x-1)$. In concluzie, $E(x)\le 0$ $\Longleftrightarrow$ $(x+1)(2x-1)\le 0$ $\Longrightarrow$ $S=\left[-1,\frac{1}{2}\right]\cup\{1\}\ .$ Sper ca acum ai inteles si ai remarcat ... bunele mele intentii.

[b]P.S.[/b]Te rog, transmite omagiile mele Domnului profesor si editor de buna carte matematica, Mircea Lascu din Zalau (Editura GIL - Gandeste Inteligent si Liber)[/color]

Va multumesc mult domnule profesor pentru rabdarea si pasiunea cu care ne impartasiti unele exemple . Sunt clasa a IX-a , dar am inteles ideea si va sunt recunoscator . :) Daca o sa am ocazia o sa-i transmit domnului prof. Mircea Lascu  urarile dumneavoastra de bine .


induction
function
integration
Putnam
college contests

I am stuck with this one, and no solutions are on the net ( on the site I checked ). Need help thanks.

Just to put the problem here:

Let $r,s$ and $t$ be integers with $0 \leq r$, $0 \leq s$ and $r+s\leq t$. Prove that
\[ \frac{\binom s0}{\binom tr}+ \frac{\binom s1}{\binom{t}{r+1}} + \cdots+ \frac{\binom ss}{\binom{t}{r+s}} = \frac{t+1}{(t+1-s)\binom{t-s}{r}}. \]

This is from one of the prof at Mcgill:

Go by induction on s, for fixed t and all r at once. The induction hypothesis is that, for all r such that r+s is less than or equal to t, h(r,s) is the expression on the right. s=0 is trivial.
[h(r,s) is defined to be the
horrible sum on the left.]
Using the sum formula for ``s+1 choose r'' (it's ``s choose r''+``s choose r-1''), show that
h(r,s+1)=h(r,s)+h(r+1,s). From there and the IH, it's easy.

I like this problem.  The 1985-2000 book has the solution mentioned above and 1 or 2 others.  However it doesn't have the following straight forward solution, probably because people are not expected to know the Beta Function: \[(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx\&=\cfrac{1}{\binom{n}{k}}\] So

\begin{eqnarray*}\cfrac{\binom{s}{0}}{\binom{t}{r}}+\cdots+\cfrac{\binom{s}{s}}{\binom{t}{r+s}}&=& (t+1)\int_{0}^{1}\sum_{k=0}^{s}\binom{s}{k}x^{r+k}(1-x)^{t-r-k}dx\\ &=&(t+1)\int_{0}^{1}x^{r}(1-x)^{t-r-s}\sum_{k=0}^{s}\binom{s}{k}x^{k}(1-x)^{s-k}dx\\ \ &=&(t+1)\int_{0}^{1}x^{r}(1-x)^{t-r-s}dx\\ \ &=&\cfrac{t+1}{(t+1-s)\binom{t-s}{r}}\\ \ \end{eqnarray*}


You may now discuss Round 4 problems and solutions.

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number theory
relatively prime
number theory open

[u][b]The author of this posting is : Krishna Kothoor[/b][/u]
____________________________________________________________________

Euler's theorem: If N is any natural number and $\phi\left(N\right)$ is the number of intergers in the sequence 1,2,3,....N-1 which are relatively prime to N then $a^{\phi\left(N\right)}-1$ is divisible by N where a is any integer which is relatively prime to N.

My conjecture:
For any prime number p there exists no k less than $\phi\left(p\right)$ such that $2^k-1$ is divisible by p.

I considered several examples of powers of 2 but only upto a limit because after that calculations become pretty laborius, but i am very confident about this conjecture being true.
Does any one agree with this?
Any comments or validations are welcomed.

um 7 = 2^3 - 1? unless im mistaking your question

i do belive zscool is right.  r u sure u stated the idea correctly?

Any $k$ less than $\varphi (p)$ means any $k<p-1$. Obivously that statement is not true, as the order of 2 modulo $p$ is not always $p-1$.

Let's try thinking about this if $p$ is not a prime number (then $\varphi(p)<p-1$ ... altough I doubt it there too.

:(   seems that doesn't work either.

$n=21$

$2^6 \equiv_{21} 1 \Rightarrow n|2^6-1$

hehe... it's true for even numbers  :D

the "conjecture" says ord(2) = \varphi (p) which I don't believe is true in general.
It seems like it for some examples just beause 2 is small.
In fact, I seem to remember having encountered that claim as a joke, because 2 seems to be the first example everybody tries when trying to find a generator for the unit group.