ratio
trigonometry
algebra
polynomial
symmetry
absolute value
Hi

Suppose $\alpha$ and $\beta$ are in $(0,\frac{\pi}{2})$ and are rational multiples of $\pi$. Is it possible that $\frac{\cos{\alpha}}{\cos{\beta}}$ is a rational number? (The $(0,\frac{\pi}{2})$ restriction is only to avoid the obvious examples—where the ratio is $1$ or $1/2$ or $2$ in magnitude.)

What I already know: If $\cos{\theta}$ is rational and $\theta$ is a rational multiple of $\pi$, then $\left vert|\cos{\theta}\right vert|$ is either 0,1, or $\frac{1}{2}$. This is because $2\cos{n\theta}$ is a monic polynomial  $P_n$ in $2\cos{\theta}$ with integer coeffecients, and so if $\theta$ is $\frac{p}{q}\pi$, then $2\cos{\theta}$ is a root of $P_{2q} = 2$, so it must be an integer. Of course, if those polynomials were irreducible, it would be easy to prove that the above ratio cannot be rational, but I don't know if they are irreducible or not.

For each $n$, there is a monic polynomial $p_n$ of degree $n$ with integer coefficients such that $p_n(2\cos\theta)=2\cos n\theta$. Using this, we can show that if $a$ is twice the cosine of a rational multiple of $\pi$, it is an algebraic integer between $-2$ and $2$ such that all of its conjugates are real numbers between $-2$ and $2$.

If $a$ and $b=r\frac mn a$ are both algebraic integers, then $\frac1n a$ is an algebraic integer, and its conjugates are $\frac1n$ times the conjugates of $a$. Therefore, if $\frac a2$ and $\frac b2$ are both cosines of rational angles, $\frac1n a$ is an algebraic integer between $-\frac2n$ and $\frac2n$, with all of its conjugates in that same interval. This is only possible if $n=1$ or $a=\pm2$. Similarly, either $m=1$ or $b=\pm2$, and the examples you have already noted are the only ones.

[quote="jmerry"]For each $n$, there is a monic polynomial $p_n$ of degree $n$ with integer coefficients such that $p_n(2\cos\theta)=2\cos n\theta$. [/quote] Agreed.

[quote]Using this, we can show that if $a$ is the cosine of a rational multiple of $\pi$, it is an algebraic integer between $-2$ and $2$ such that all of its conjugates are real numbers between $-2$ and $2$.[/quote] Did you mean $a$ is [i]twice[/i] the cosine of ...? It is an algebraic integer because it satisfies a monic polynomial, but why are its conjugates in that interval, and why are they even real? Am I missing something obvious?

[quote]If $a$ and $b=r\frac mn a$ are both algebraic integers, then $\frac1n a$ is an algebraic integer, and its conjugates are $\frac1n$ times the conjugates of $a$. Therefore, if $a$ and $b$ are both cosines of rational angles, $\frac1n a$ is an algebraic integer between $-\frac2n$ and $\frac2n$, with all of its conjugates in that same interval. [/quote] By $r\frac mn$, do you mean "$r+\frac mn$, where $\gcd(m,n)=1$ and $m\leq n$?" In any case, I understand the statement about $\frac 1n a$ and its conjugates.
[quote]This is only possible if $n=1$ or $a=\pm2$. Similarly, either $m=1$ or $b=\pm2$, and the examples you have already noted are the only ones.[/quote] I don't see why, sorry.

The conjugates of $a$ lie in the same intervall since they are the other zeros of the desired polynomial: $cos(nx)$ is periodically $\mod \frac{2\pi}{n}$, so we know all zeros, namely $2 \cos(\theta), 2\cos(\theta+\frac{2\pi}{n}), 2\cos(\theta+\frac{4\pi}{n}), ..., 2\cos(\theta+\frac{2\pi (n-1)}{n})$.

For the last part: the norm of $\frac{a}{n}$ is the product over it's conjugates and it's an integer. But $n>2$ would give that this norm has absolute value $<1$, so $a=0$.

Great, I think I understood that. Let me just write down the complete proof, and kindly tell me if something is wrong:

If $a$ is two times the cosine of a rational multiple of $\pi$, then $a$ in an algebraic integer, all of whose conjugates are real numbers in $(-2,2)$ (assume that $a$ is neither $-2$ nor $2$ nor $0$, because those cases are easy).
 If $b$ is also twice the cosine of a rational multiple of $\pi$ and $b=\frac mn a$ with $\gcd(m,n)=1$, then as $\frac 1n a$ is a $\mathbb Z$-linear combination of $a$ and $b$, it is also an algebraic integer, and has all conjugates in $(\frac{-2}{n},\frac{2}{n})$ (and nonzero). But the product of all these conjugates is an integer (the constant term of the polynomial for $\frac 1n a$, upto sign), so if $n\geq2$, then each of them is of magnitude less than $1$, so their product cannot be an integer. Hence, $n = 1$, i.e, $b$ is an integer multiple of $a$. But then, by symmetry, $a$ must also be an integer multiple of $b$, so $a$ and $b$ are equal (upto sign).

[quote="shreevatsa"] Did you mean a is twice the cosine of ...?[/quote]
Fixed. There was another related mistake farther down.


MATHCOUNTS
analytic geometry
geometry
area of a triangle
Heron\u0027s formula
absolute value

There are multiple ways to do this problem and I would like to see a number of different solutions on this one!

What is the area of the triangle with coordinates (-1, -1), (3, 0), and (1, -9)?

it is a nice problem, and i urge you to generalize to any coordinates, and maybe even to an n-sided polygon. (you can just present a method/algorithm, not a full-fledged formula).

[hide]I think the determinant way is the easiest for this one. Finding the sides for heron's is a bother and could get messy, while for Pick's you have to draw out the diagram and count all the boundary, interior points, a method easy to mess up on when you are making a rough drawing with your hand.[/hide]

1. High school students should really sit on the sidelines and let middler school students try to attack these MATHCOUNTS problems [i]before[/i] chiming in -- even with spoilers.  There is plenty of time to comment later and it's best for MATHCOUNTS students to be forced to wade blindly through the possibilities themselves first.

2. Fortunately nobody has given an answer yet, so we can see how the MATHCOUNTS students can apply these methods.

3. There are still yet unmentioned methods for solving this problem. :)

Hmm...as an middle schooler, the only method I understood was the Heron's formula, and even that I doubt all middle schoolers learn.

Heron's formula: [hide] :sqrt: s(s-a)(s-b)(s-c) whereas s=(a+b+c)/2[/hide]

You could also draw an auxiliary line such that the line goes through both the vertex (any one of them) and form a right angle with the side opposite the vertex you chose, but that'll be even messier...

You could even go and move the triangle so the triangle's only in the first quadrant and put one of the vertex in the origin, then apply some formula to it, but sadly I do not know of such formula...

This is easy with Heron's, but I'm pretty sure your supposed to do it otherwise...

Hint: [hide] Note that the sides are 2 :rt: 17, :rt:17, and :rt5::rt:17[/hide]

The area can be computed with determinants using the coordinates with a generalized method for finding the area.

Solution: (-1,-1)
             (3,0)
             (1,-9)
             (-1,-1)     (list the first coordinate again)
multiply the 1st coordinate on the top with its alternate pair one row down, seperate the sides you multiply by and add them independently. Subtract your sum on the right by the sum on your left, find the absolute value of that and divide by two.

Now assuming that the coordinates can be replaced with variables such as (X1,Y1)
(X2,Y2)
(X3,Y3)
(X1,Y1)

our formula would be: (x1y2+x2y3+x3y1)-(x2y1+x3y2+x1y3)/2, and finally finish by finding its absolute value as the answer.
Correct me if I am wrong but my answer was [b]12[/b][/i]

[hide="Solution using Shoestring"]
Placing these points in clockwise order:
-1,-1
3,0
1,-9
-1,-1
Now multiplying diagonally,
We get a sum of 6 and - 28 in each.
The absolute value of each is 22.
Now we divide by 2.
So we get that the area is 11.
[/hide]


function
number theory proposed
number theory

For a natural number $ n$ we denote $ 1\cdot11\cdot111\cdot\ldots\cdot\underbrace{1\ldots1}_{n}$ with $ \left[n\right]!$. For example $ \left[4\right]!\equal{}1\cdot11\cdot111\cdot1111$. Prove that for every natural $ m$ and $ n$ $ \left[m\right]!\cdot\left[n\right]!$ divides $ \left[m\plus{}n\right]!$.

Let $ x_k\equal{}\frac{10^k\minus{}1}{9}$, then $ A\equal{}\frac{[m\plus{}n]!}{[m]![n]!}\equal{}\frac{x_{m\plus{}1}...x_{m\plus{}n}}{x_1...x_n}.$
Obviosly $ x_k|x_l$ if and only if $ k|l$.  Therefore $ A\in Z$.


geometry
perimeter
inequalities
search
triangle inequality

[quote="da ban man"]How many distinct, nondegenerate triangles are there with integer side lengths such that its perimeter is 30.[/quote]

[hide]Brute-force!
No, but something like it should work.
Suppose the sides are $ a,b,c$ with $ a\le b\le c$.
If one of the side lengths is $ 1$, then by the Triangle Inequality,
$ 1+b>c$
$ c-b<1$,
clearly impossible since they are integers and their sum is $ 29$, which is odd.

If one of the side lengths is $ 2$, then
$ c-b<2$, and we get that
$ (b,c) = (14,14)$
is the only possible solution.

Experimenting more, we get
$ (a,b,c)=(2,14,14)$
$ (a,b,c)=(3,13,14)$
$ (a,b,c)=(4,13,13)$
$ (a,b,c)=(4,12,14)$
$ (a,b,c)=(5,12,13)$
$ (a,b,c)=(5,11,14)$
$ (a,b,c)=(6,12,12)$
$ (a,b,c)=(6,11,13)$
$ (a,b,c)=(6,10,14)$
$ (a,b,c)=(7,11,12)$
$ (a,b,c)=(7,10,13)$
$ (a,b,c)=(7,9,14)$
$ (a,b,c)=(8,11,11)$
$ (a,b,c)=(8,10,12)$
$ (a,b,c)=(8,9,13)$
$ (a,b,c)=(8,8,14)$
$ (a,b,c)=(9,10,11)$
$ (a,b,c)=(9,9,12)$
$ (a,b,c)=(10,10,10)$

$ 1+1+2+2+3+3+4+2+1 = \boxed{19}$[/hide]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1358695267&t=157412[/url]
is almost the exact same problem, and it's posted by you...


N is an odd integer such that $-59<N<59$. How many integers are there which have the form $\frac{3N}{7}+1$?

[hide]N has to be divisible by 7 for it to be an integer. We want the number of odd N that are divisible by 7, between -59 and 59. There are 8 such numbers.[/hide]

[hide]
N must be a multiple of 7, 8 odd negative numbers work, 8 odd positive ones, and 0 works. However, 4 negatives, 4 positives, and 0 aren't odd so your answer is $\boxed {8}$[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

What is the smallest integer $n$ s.t. $G$ is a non abelian group of order $n$ and $\mathrm{Aut}(G)$ is abelian?

Hint:
[hide]$n\leqslant 128$ [/hide]

Any other hints? I tried the smaller orders but I didn't found such a group.

The non-abelian groups with order less than 20 are:
Order 6: $S_{3}$ ($Aut(S_{3})\cong S_{3}$)
Order 8: $D_{4}$ ($Aut(D_{4})\cong D_{4}$) and $Q_{8}$ ($Aut(Q_{8})\cong S_{4}$)
Order 10: $D_{5}$ ($Inn(D_{5})\cong D_{5}$)
Order 12: $A_{4}$ ($Aut(A_{4})\cong S_{4}$), the nontrivial semidirect product $\mathbb{Z}_{3}\ltimes\mathbb{Z}_{4}$ and $D_{6}$ (both inner automorphism groups are isomorphic to $S_{3}$)
Order 14: $D_{7}$ ($Inn(D_{7})\cong D_{7}$
Order 16 is lengthy but if I didn't made an error, all nonabelian groups with this order have nonabelian automorphism groups
Order 18: $D_{9}$ ($Inn(D_{9})\cong D_{9}$), $S_{3}\times\mathbb{Z}_{3}$ (the automorphism group contains a subgroup isomorphic to $S_{3}$) and the nontrivial semidirect product $(\mathbb{Z}_{3}\times\mathbb{Z}_{3})\ltimes\mathbb{Z}_{2}$ (there is only one that is not isomorphic to $S_{3}\times\mathbb{Z}_{3}$ and its center is trivial, so the inner automorphism group is isomorphic to the group itself and hence nonabelian)

How did you determine the automorphism groups?

Hi.

$Aut(S_{n})~=S_{n}$ for all $n\neq 2,6$ and $Aut(A_{n})~=S_{n}$ for n=4,5 and all $n>6$ are a (well?) known theorems in group theory. I proved $Aut(Q_{8})~=S_{4}$ and $Aut(D_{4})~=D_{4}$ some time ago.
For the other groups I used the theorem $Inn(G)~=G/Z(G)$ and calculated the center.

The earliest known example is due to G. A. Miller, ``A non-abelian group whose of isomorphisms is abelian'' in Messenger Math. $\pmb{ 43}$, 124--125, 1913.

His example is a group of order 64. The group has four generators $a,b,c,d$ and relations \[a^{8}=b^{2}=c^{2}=d^{2}=[a,c]=[b,c]=[c,d]=[d,a]=1, bab=a^{5},bdb=cd.\] The autormphism group is an elementary abelian group of order $2^{7}.$


I don't know if this is the right place to place a thread on vectors, in case not you are of course welcome to move it to the right subforum - would probably have done it anyway :)

First of, i have to say i don't know how to make a vector sign, but until i find out how to do it i will just write $a>$ which here is a vector. How would you solve this problem?

$|a>|=3$, $|a>+b>|=5$ and $a\cdot b=2$

Find $|b>|$ and the angle between $a>$ and $b>$.

$|\vec{a}+\vec{b}|=5$

square both side you get $|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}\cdot\vec{b}|=25$

So $|\vec{b}|=\pm 2\sqrt{3}$

And for the angle , just use

$\cos\theta =\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$

OK, i see how you get from $|\vec a+\vec b|=5$ to $|\vec a|^2+|\vec b|^2+2|\vec a\cdot \vec b|=25$, but i don't see how you get from here to $|\vec b|=\pm2\sqrt3$. BTW: What does $|\vec a \cdot \vec b|$ actually mean? I know that $\vec a\cdot \vec b=|\vec a|\cdot |\vec b|\cdot \cos \theta$, but i am not sure that is the same as $|\vec a \cdot \vec b|$, is it?

the question give us $|\vec{a}|=3$ and $\vec{a}\cdot\vec{b}=2$ .So just substitute inside and get the answer .

well i guess it shud be $2\vec{a}\cdot\vec{b}$ rather than a putting a modulus :P . just some typo

I don't see how you get $|\vec b|=\pm2\sqrt3$, but nvm with that, because i get $|\vec b|=\pm\sqrt{12}$ which obviously is the same.

So the end result is:

$\theta=\pm78.90$

Thx for your help ;)


Find all the triplets $ (s,t,x), s,t,x \in N, t>s$  such that:
$ (1) ((t^{2}\minus{}s^{2})y)^{2}\plus{}64sty$ is a perfect square and $ 2|t\minus{}s$
$ (2) ((t^{2}\minus{}s^{2})y)^{2}\plus{}8sty$ is a perfect square

([u]not[/u] necessarily both)

I think there are no solutions. Can anyone prove it?

There exist solutions.
In particular, $ t\equal{}2$, $ s\equal{}1$, $ x\equal{}1$ imply
\[ ((t^2\minus{}s^2)x^2)^2\plus{}8stx \equal{} 5^2.\]

You're right, my claim is too strong. Here's a (hopefully) correct one-

There are no triplets $ (s,t,x)$ of natural numbers, $ t > s$, such that:
$ (1) ((t^{2} \minus{} s^{2})x)^{2} \plus{} 4stx$ is a perfect square and $ 2|t \minus{} s$.
$ (2) ((t^{2} \minus{} s^{2})x)^{2} \plus{} 8stx$ is a perfect square and $ 2|t \minus{} s$.

These problems are equivalent to a problem from a recent math olympiad, so they're suppose to be true. But I might have done something wrong.

[quote="bambaman"]There are no triplets $ (s,t,x)$ of natural numbers, $ t > s$, such that:
$ (1) ((t^{2} \minus{} s^{2})x)^{2} \plus{} 4stx$ is a perfect square and $ 2|t \minus{} s$.[/quote]
\[ ((18^2\minus{}16^2)\cdot 1)^2 \plus{} 4\cdot 16\cdot 18\cdot 1 \equal{} 76^2\]


A set of five positive integers has a median of 3 and a mean of 11. What is the maximum possible value of the set's largest element?

We know the numbers are like this:

_ _ 3 _ x

and they all add up to 55. To make the lasrgest number as large as possible, we try to make the other numbers as small as possible. For the first two blanks, we fill them with 1's. The 4th blank may not be lower than 3, so the smallest possible value of that would be 3. Thus we get:

1+1+3+3+x=55
So the largest value of x is 47.

Maximize largest element by making the rest smaller.  1,1,3,3,47, max is 47.


Say ABCD is a quadrilateral so that a circle can be inscribed in this which touches the quadrilateral at E, F, G and H on AB, BC, CD, DA.

Prove that, DA, BC and GE are concurrent.
Prove that, AC, BD, EG, FH are concurrent.

Hmm... i think the first one can't be true. because if we take a triangle ABC and its one excircle (opposite to A say), then we can make another side of quadrilateral arbitrarily so that A, BC and circles tangent point and the other point can't be collinear.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=77663 for part #2 and more. For part #2, see also http://www.mathlinks.ro/Forum/viewtopic.php?t=19549 .

  darij


For every positive a, b, c and d, show that

(a+b)(b+c)(c+d)(a+d) is at least equal to 16abcd.

First of all, this isn't number theory (mods: this could go to Intermediate, probably).

Second of all,

[hide="Hint"]Multiplying AM-GM for all four pairs on the left hand side will give right hand side[/hide]


Okay, this is our team :

1. Beka Ergemlidze
2. Lasha lakirbaia
3. Nika Machavariani
4. Lasha Peradze
5. Tsotne Tabidze
6. Levan Varamashvili

~~~~~~~~~~~~~~

Good luck, boys!!!


The transitions will happen at:
$0, \pm 1, \pm 2, \pm 3, \dots$
$0, \pm 1, \pm\sqrt{2}, \pm\sqrt{3}, \dots$
$0, \pm 1, \pm\sqrt[3]{2}, \pm\sqrt[3]{3}, \dots$.

Looking at the values of $\ [x]$, $[x^2]$, $[x^3]$, it's easy to see that there won't be any solution with $x<-1$ or $x>\sqrt{3}$.

The hard part is to order the intervals. I must confess I used a calculator  :blush:
$0<1<\sqrt[3]{2}<\sqrt{2}<\sqrt[3]{3}<\sqrt[3]{4}<\sqrt[3]{5}<\sqrt{3}<\dots$.

Looking at the values at every interval, the only solutions are:
$x \in [-1,0) \cup [0,1) \cup [\sqrt[3]{2},\sqrt{2}) \cup [\sqrt[3]{3},\sqrt[3]{4})$.

[quote="lordWings"]The hard part is to order the intervals. I must confess I used a calculator  :blush:
$0<1<\sqrt[3]{2}<\sqrt{2}<\sqrt[3]{3}<\sqrt[3]{4}<\sqrt[3]{5}<\sqrt{3}<\dots$..[/quote]

I don't understand are you joking  ;)  ?

To order that you just need to rise all the terms at the power 6 (all the value are positiv so you can do that) and it becomes easy that $0<1<2^2=4<2^3=8<3^2=9<4^2=16<5^2=25<3^3=27$


There are 2 boxes with $ n$ pencils each one.

Randomly we remove a pencil of these boxes, and is not import of what box we remove the pencil.

In a inesperd moment, a box of them this empty one...

Calculate the probability that in the other box (nonempty) they are $ k$ pencils.

[quote="jorgeston"]In a inesperd moment, a box of them this empty one... [/quote]

Er, what?

 :?:

EDIT: Googling "inesperd", I found that your question has been answered elsewhere: http://www.mathhelpforum.com/math-help/advanced-probability-statistics/33980-probability-problem.html


How do you find the equation of a sequence 7, 23, 55, 109, 191,307, 463, ...?

[hide="hint"]
Look at those 3rd differences...

[/hide]

i have read one article regarding this,  i just don't what next. yup, its a third degree polynomial. i have already got the equation but i don't know what to do next.

I tried to pattern it to what I had read and the equation is

$ f(x) \equal{} 7_xC_0 \plus{} 16_xC_1 \plus{} 16_xC_2 \plus{} 16_xC_4$

What's next? How do I simplify this into a 3rd degree polyonmial.

That's a fourth-degree polynomial (did you mean $ _x C_3$ for the last one?)  Now, $ {x \choose 0} \equal{} 1, {x \choose 1} \equal{} x, {x \choose 2} \equal{} \frac {x(x \minus{} 1)}{2}, {x \choose 3} \equal{} \frac {x(x \minus{} 1)(x \minus{} 2)}{6}$.  Remember the definition of a combination!

an equation of the sequence is a third degree polynomial (cubic).

$ S_n \equal{} n^3 \plus{} 2n^2 \plus{} 3n \plus{} 1$


trigonometry
algebra
polynomial
geometry
trig identities
Law of Sines
geometry proposed

$ ABC$ a triangle such that $ AB\equal{}AC$,the bissector of $ \angle B$ intersect $ (AC)$ at $ D$,we suppose that:$ BC\equal{}BD\plus{}AD$.
Find $ \angle A.$

Here is what I have so far:
Let $ \angle DBC \equal{} x$, AD=n, BD=m. By Law of Sines on $ \triangle BDC$,
$ DC \equal{} m/2cosx$.
So by LOS on $ \triangle ABC$, we have
$ sin4x/(m \plus{} n) \equal{} sin2x/(n \plus{} (m/2cosx))$
Also, by LOS on $ \triangle ABD$,
$ sin4x/m \equal{} sinx/n$, hence $ n \equal{} msinx/sin4x$.
Substituting, we have
$ sin 4x/(m \plus{} msinx/sin4x) \equal{} sin2x/(msinx/sin4x \plus{} m/2cosx)$.
M cancels, this simplifies to
$ 1 \plus{} 2cos2x \equal{} 2cosx \plus{} 1/(2cos2x)$.
If $ cosx \equal{} p$,
$ 16p^4 \minus{} 8p^3 \minus{} 12p^2 \plus{} 4p \plus{} 1 \equal{} 0$.
My calculator says this polynomial is 8(2p-1)(p-cos140)(p-cos100)(p-cos20).
So x=20, and $ \angle A \equal{} 100$, I can't prove it without a calculator.

Here is a better method:
Extend BD to E so DE=AD and form quadrilateral AEBC. If $ \angle ABE \equal{} x$, by angle chase, we can find all angles in the figure in terms of x.
Virgil Nicula showed me a property that in any quadrilateral (let's use this one as an example), $ sinABEsinCAEsinBECsinACB \equal{} sinCBEsinBACsinAEBsinECA$.
So we have:
$ sinx \cdot sin(3x/2) \cdot sin(90 \minus{} x/2) \cdot sin2x \equal{} sinx \cdot sin4x \cdot sin(3x/2) \cdot sin(90 \minus{} 5x/2)$
$ cos(x/2) \cdot sin2x \equal{} sin4x \cdot cos(5x/2)$
$ cos(x/2) \equal{} 2cos2x \cdot cos(5x/2)$
$ cos(x/2) \equal{} 2cos2x \cdot (cos2x \cdot cos(x/2) \minus{} sin2x \cdot sin(x/2))$
$ (2cos^2(2x) \minus{} 1) \cdot cos(x/2) \equal{} 2sin2xcos2x \cdot sin(x/2)$
$ cos 4x \cdot cos(x/2) \equal{} sin4x \cdot sin(x/2)$
$ tan4x \cdot tan(x/2) \equal{} 1$
$ 4x \plus{} x/2 \equal{} 90m$ (m is odd)
$ x \equal{} 20m$
So x is 20, $ \angle A$=100
x can't be higher than that (like 60) or it wouldn't make a triangle.

$ a \equal{} BC, b \equal{} CA \equal{} AB.$
$ BC \equal{} BD \plus{} AD \Longleftrightarrow$ $ a \equal{} \frac {\sqrt {ab(a \plus{} 2b)a}}{a \plus{} b} \plus{} \frac {b^2}{a \plus{} b}.$
Choosing $ b \equal{} 1,$ this becomes $ a(a \plus{} 1) \equal{} a \sqrt {a \plus{} 2} \plus{} 1.$ Squaring to get rid of $ \sqrt {\ \ },$ $ a^4 \plus{} a^3 \minus{} 3a^2 \minus{} 2a \plus{} 1 \equal{} 0.$
This has obvious root $ a \equal{} \minus{} 1,$ not acceptable. (The only rational roots can be $ \pm 1,$ this is the thing to look for.) Factoring it out by long division, $ a^3 \minus{} 3a \plus{} 1 \equal{} 0.$ We need $ \cos B \equal{} \frac {a}{2b} \equal{} \frac {a}{2}$ $ \Longrightarrow$ $ \cos 3B \equal{} 4 \cos^3 B \minus{} 3 \cos B \equal{} \minus{} \frac {_1}{^2},$ $ 3B \equal{} k \cdot 120^\circ,$ where $ k \mod 3 \neq 0.$ $ B \equal{} k \cdot 40^\circ < 90^\circ$ $ \Longrightarrow$ $ B \equal{} 40^\circ$ or $ 80^\circ$ for $ k \equal{} 1, 2.$ Suppose $ B \equal{} 80^\circ > 60^\circ,$ then $ BD \plus{} AD > AB > BC,$ not acceptable. (In fact, then $ AD \minus{} BD \equal{} BC.$) The only possible result is $ B \equal{} 40^\circ,$ $ A \equal{} 180^\circ \minus{} 2B \equal{} 100^\circ.$


linear algebra
matrix
real analysis
real analysis unsolved

Solve the determinant and then the differential equation.

[quote="eureka-123"]Find equation of all such possible curves which pass through (1,0) and are given by the determinant eqn::[/quote]
the second matrix is not square. what do you mean by determinant????? :wink:  :wink:


:) Where would u most likely Stop, Drop, and Roll?

Any discussion?

I like stop drop and roll when the surface is not hard and when the surface does not stink. If there was a pile of dollar bills on the ground (I mean MOUNTAINS) I would.

i put other.  u see me everyday and i'm pretty sure you know by now you know what i'd do with stop, drop and roll.  if you don't, just ask me at school.


Below is a piece of news I just read. Maybe some of you already know this but I found it interested to share it to others:

[quote][b]Did Jesus Ask Judas to Betray Him?[/b]

[i]Updated 3:48 PM ET April 6, 2006[/i]

It is a mystery 2,000 years in the making, buried in the desert and fueled by centuries of debate and doubt, theft and deceit. The question: Was there ever a Gospel according to Judas? And if there was, what did it reveal?

The mystery began to unravel almost 30 years ago, according to a new National Geographic Channel documentary.

[b]Watch the full story on ABC's "Primetime" Thursday at 10 p.m. ET and "Nightline" at 11:35 p.m. ET and get more information from National Geographic by clicking here. Watch the full documentary "The Gospel of Judas" at 8 p.m. ET Sunday, April 9, on the National Geographic Channel.[/b]

A farmer looking for treasure in a cave in Egypt instead found a decaying leather-bound book, a codex.

Because the text was written in ancient Coptic, the farmer did not know what it was, but he figured he could sell it. He did, to an antiquities dealer, but still no one knew its secret text -- and no one ever would if it continued to disintegrate before it could be translated.

Five years after it was found, the first glimpse of the book's meaning was gleaned when scholar Stephen Emmel was asked to look at the document.

"We were told we were not allowed to make any photographs, we were not allowed to make any notes," Emmel told "Primetime." "I leafed through and by chance spotted a dialogue between Jesus and Judas and his disciples and in particular the name Judas came up again and again. Judas said ... the Lord said, blah, blah, blah."

Could it be a Gospel according to Judas? Scholars had long believed such a Gospel existed but that it had been banned by the early church, called blasphemous and ordered destroyed. Had a copy somehow survived, telling the story of the most reviled man in history?

[b]A Great Betrayal[/b]
"[Judas is] the one who handed over his friend," said Marvin Meyer, co-chair of the religious studies department at Chapman University. "He's the one who brought about the crucifixion, and he's the one who is damned for all time."

The Bible says Judas betrayed Jesus for 30 pieces of silver. The Bible also tells how the other disciples let Jesus down: Peter denied him three times but is nevertheless honored with the basilica in Rome. It is Judas alone who is not forgiven, condemned to the seventh level of hell in Dante's "Inferno," eaten head-first by Lucifer.

"Often they think of him as somebody who was greedy, avaricious, who was more interested in making money than in being faithful to his master," said Bart Ehrman, chair of the religious studies department at the University of North Carolina.

And Judas over the centuries also became a symbol of anti-Semitism.

"Traditionally in Christian circles, Judas in fact has been associated with Jews," Ehrman said. "Of being traitors, avaricious, who in fact, betray Jesus, who are Christ-killers. And this portrayal of Judas of course also leads then to horrendous acts of anti-Semitism through the centuries."

But what if there were more to the story? The first task was to see if the document was genuine. For 16 years, the Codex sat crumbling in the most unlikely of places -- a safe-deposit vault in a Citibank on Long Island, N.Y., until the year 2000, when it was purchased by a former antiquities dealer, Frieda Nussberger-Tchacos.

"I think the circumstances of this manuscript coming to me were predestined," she said. "Judas was asking me to do something for him."

[b]Searching for Answers[/b]
Finally, with the funding of the National Geographic Society and two foundations, a dream team of scientists and scholars was assembled to determine if the document was really an ancient text.

Verification was an enormous challenge. The 13 pages of papyrus, with writing on the back and front, were in a thousand pieces. The box with what might be the lost Gospel was shipped to Swiss restorer Florence Darbre.

"Your heart is beating a little bit faster the first time you take it in your hand ever," she said, "in your hand, yes, because you can see it but you don't know how he is, how brittle he is, how delicate it is."

Painstakingly, Darbre and her partner fit the tiny pieces together. But just because it looked authentic, it still had to be proved whether it was written 1,800 years ago. For that, Emmel, the scholar, and Rodolphe Kasser, another expert, were brought in.

"I've looked at hundreds of papyri, Coptic papyri, in my career, and this is absolutely typical of ancient Coptic manuscripts," Emmel said. "I'm completely convinced. The number of people who could create such a text is very small," he said, perhaps just 25. Kasser estimated there were just four or five people who could create a convincing imitation of an ancient Coptic document.

"We're two of them," Emmel said. "Now, who would create such a thing? Someone would have to know Coptic better than we do. And there isn't anybody who knows Coptic better than we do. I'm sorry."

But the final test of authenticity meant taking a radical step -- destroying tiny pieces of the document to carbon date it.

"I will have to burn the Gospel of Judas in order to be able to date it," said Timothy Jull, a carbon-dating expert at the University of Arizona's physics center.

Fifteen hours later, he had a final answer. The text was real. The Gospel according to Judas, written between the third and fourth century, was believed to be a copy of a much older document written in Greek in the second century.

"Radio carbon dating of the papyrus from the Gospel of Judas confirms that it's from the third to the fourth century A.D.," Jull said, "and this supports the authenticity of the Gospel of Judas."

[b]Did Jesus Ask to Be Betrayed?[/b]
"You can't fake a codex like this," said Elaine Pagels, a professor at Princeton University and one of the world's foremost experts on ancient religious texts, especially the so-called Gnostic or secret Gospels, which include the Gospel of Judas, all written in the first and second centuries and banned by the early church.

"It's crumbling; it's a particular kind of papyrus; it's a particular kind of script," Pagels said. "It would be absolutely not worth anyone's while, and far too difficult, to try to fake this kind of text. This is a genuine ancient text."

So what is in the Gospel of Judas? It is a dialogue that claims to be a conversation between Jesus and Judas in which Jesus asks Judas to betray him.

"Judas has the terrible task of taking it upon himself to turn him over to the authorities for this reason," Pagels said. "Now, the Gospel of Judas also has Judas say to Jesus in fear and terror that he has a dream that the other disciples will hate him and will stone him to death, will attack him.

"And Jesus says, 'Yes, in fact, they will think that you are a terrible person because of what you did. This is part of the burden that you bear. But they will be wrong about that.' So it is an extraordinary transformation of the ordinary understanding of Judas Iscariot."

Pagels said the text shows that Christ, in fact, asked Judas to betray him for an undisclosed reason. "The Gospel of Judas does suggest that the betrayal of Jesus is not a reprehensible act, not the act of a traitor, you know, the worst villain in the history of the world, but that it's a secret mystery between him and Jesus," she said.

Herbert Krosney, author of "The Lost Gospel: The Quest for the Gospel of Judas Iscariot," one of two new books and a National Geographic Channel documentary about the Gospel of Judas, said the finding is significant.

"Judas is not the betrayer," Krosney said. "Judas is, rather, the favored disciple of Jesus. He is the one whose star shines in the heavens and in the skies, and Judas, therefore, becomes unique. He is Jesus' best friend rather than his betrayer."

[b]A New Perspective[/b]
Today the Gospel of Judas got its first public outing at a news conference, and it is on display at the National Geographic Society in Washington, D.C. It will eventually return to Egypt to be housed in Cairo's Coptic Museum. It is also available online, in Coptic and English, and is the cover story of the new National Geographic magazine.

But while the document is a real one, is what it claims also true? Did the New Testament Gospels of Matthew, Mark, Luke and John get it wrong? Did Jesus askJudas to betray him?

"I don't think that we have in this Gospel what I would call historical proof," Pagels said. "We also don't have that in the other Gospels."

She said there may never be an answer. "I don't see how we would, although you see we could always find next week or in 50 years other evidence that we don't have now."

In the end, science may have its answers, but questions of spirit and soul cannot be analyzed like a piece of papyrus. If Judas did not betray Jesus and was part of a grand plan, does that change anything for Christian theology?

The discovery does not alter the belief of evangelical scholar Ben Witherington that Judas did indeed betray Jesus. "Well, it would mean among other things that Jesus had some kind of death wish, for a start," Witherington said. "And it would raise some questions about his character.

"I would think there are some questions of integrity that would be raised about that, for him to sort of script it in such a way that he's using his disciples to go and set this up would suggest a sort of level of hands-on intention to it, where he's not just submitting to the will of God in his life," Witherington said. "He's actually got his hands on the wheel, and he's driving the wheel of history in a particular direction. And some would find that troubling."

The scholars interviewed by "Primetime" agreed that the real importance of the Gospel of Judas is the window it provides into what some early Christians were thinking. But they acknowledged that some in the organized church will not like the discovery of this Gospel.

"Absolutely, they won't," Pagels said. "Some will be very offended, and they'll say this proves that all of those texts are rubbish, because it's an utterly preposterous idea that Judas could have been involved in a secret mystery with Jesus."

But that would miss the point, she added. "The Christian message is a message about faith and hope and, you know, the relationship between God and human beings. It's not a matter of historical fact."

Pagels said she hoped the find will have a big impact.

"I would hope that people appreciate the excitement of this discovery and recognize that it's all right to ask the kinds of questions that sometimes they're afraid to ask, and say, 'What else didn't we know about the early Christian movement? Could, for example, Judas be forgiven?'" she said.

"And when people start asking that question, they'll realize that it doesn't destroy faith, it actually can strengthen it. But it's a different kind of faith; it's informed by what we understand about our past." [/quote]

This article misses a major point when it talks about how hard this gospel would be to fake.

What is really meant is that it would be really hard to fake it today.  If only 20 people could fake it today, how many could fake it 100, 300, 800 years ago?  It is illogical to assume that such a document is a copy of a prior document.  We have the history of copying documents for the church documents.  We also have many copies showing up across the world for most of the church documents that are included in the Bible.  This doesn't apply to this paper.  It is also illogical to assume that if it is a fake it could only have been faked by those living today.

The dating, if correct, says that it was created in the second or third century.  There would be many people who would have a reason to come up with something discrediting the Christian story, or redeeming the name of Judas.  (Remember the descendent of Dr. Mudd (who treated Mr. Booth, Abe Lincon's killer) who spent his entire life trying to get a legal clearing of Dr. Mudd?  There may have been many who could have made such a document as a fake, joke, or fiction; which lay undiscovered until recently, where the context of the document not available.

Just because something is old doesn't make it necessarly true.  Those creating it didn't say, "Now let's put this here where it will be found thousands of years later.

I don't know, but let's not jump to illogical conclusions in looking at on document.

on a comical side note : where there lots of people called judas before that?

It's the same thing with Adolf : the name practically vanished the last half century :oops:

i thought judas was a pretty common name back then

It was common.  Read on if you want a full answer, rather than just the conclusion.

Thanks to computer science (or a concordance) one can easily determine that the name did not appear once in the old testament, but had the following appearances in the new testament.

One of Jesus' half brothers (different father) was named Judas (Mark 6:3).

What were the names of the twelve apostles?

I suspect not one in ten thousand Christians can name them all, even though they are listed in the Gospels.  See Luke 6:13 (written by Luke a physician who accompanied Paul, neither of who were among the original 12 Apostles, but together wrote far over half of the new testament), when the 12 are first called Apostles by Jesus.

Simon (Peter)
Andrew
James
John (the Gospel writer and youngest Apostle)
Philip
Bartholomew
Matthew (the Gospel writer and former tax collector)
Thomas (the doubter)
James (not Jesus' half brother)
Simon (the Zealot - this was a sect of religion at the time, not a character statement)
Judas (son of James, yet another James)
Judas Iscariot (the Betrayer)

The list of the twelve original Apostles in Matthew 10:3 is the same except Judas (son of James) is replaced by Thaddeus.

The Gospel of John (14:22) reports a discussion between "Judas (not Judas Iscariot)" and Jesus, after Jesus rose from the dead.

Acts 1:13 reports Judas (son of James) as one of the original Apostles present when Matthias was selected by lot from those with Jesus from the beginning.  (This is not surprising, as Acts is written by Luke, who wrote the Gospel with his name.)

In Acts 5:37, a Pharisee named Gamaliel argued in the Sanhedrin (Jewish high court) for the lives of the 12 disciples, after Jesus' death.  Part of his argument was that others' movements died out after their leader died, and this Jesus sect should be treated the same way.  One of the prior revolts was lead by a person designated as "Judas the Galilean."  (Gamaliel was one of Paul's teachers.)

In Acts 9:11, after Paul is struck blind on the road to Damascus, he is taken to the house of a man called Judas on Straight Street, where Jesus tells a Christian named Ananias to go find and bless Paul, who has been beating, killing and imprisoning Christians and who has a legal warrant to collect Christians from Damascus and take them back to Jerusalem for trial.  (Ananias was initially reluctant to go and find Paul.)

In Acts 15:22, the church in Jerusalem, led by Peter, sent two of its members to go with Paul back to Antioch to tell them that God did not require that the Jewish law applied to Christians.  One of these is "Judas (called Barsabbas)."  The accompanying letter designated him only as "Judas."

Maybe too much Bible for some, but this exhaustive description will allow all the gentle readers determine for themselves if they are convinced that the name Judas was common at the time and in the area where Jesus lived.


Given are $\triangle ABC$ prove that
$sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}\le \frac{3}{2}+\frac{r}{2R}(4cos\frac{A}{2}.cos\frac{B}{2}.cos\frac{C}{2}-(cos\frac{A}{2}+cos\frac{B}{2}+cos\frac{C}{2}))$ :lol:

Are you sure about $\frac{r}{2R}\geq \frac{1}{3}$ ??

It's well known that $OI^{2}= R^{2}-2Rr \geq 0$

so $R\geq 2r$. Therefore, $\frac{r}{2R}\geq \frac{1}{3}$ is wrong.

ok you are right Chang Woo-JIn, because my second ineq is true, too and it is stronger than the first, but maybe it is easier than the first!  :)  :)

[quote="gemath"]Maybe it is
$sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}\le \frac{3}{2}+\frac{1}{3}(4cos\frac{A}{2}.cos\frac{B}{2}.cos\frac{C}{2}-(cos\frac{A}{2}+cos\frac{B}{2}+cos\frac{C}{2}))$
Are anybody interested in?  :)[/quote]
I'm very sorry it is flase :blush:,therefore I'm not sure the first is true but the second must be $sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}\le \frac{3}{2}+\frac{1}{6}(4cos\frac{A}{2}.cos\frac{B}{2}.cos\frac{C}{2}-(cos\frac{A}{2}+cos\frac{B}{2}+cos\frac{C}{2}))$
I also want to find the inequality stronger than the basic inequality I made some mistake therefore hope anybody are interested in it excuse for me  :blush:


limit
integration
calculus
calculus computations

That limit is equal to $ \lim_{k\to\infty}\sum_{i\equal{}0}^{k\minus{}1}\frac{1}{k}\cdot \sqrt{1\plus{}\frac{i}{k}}\equal{}\int_{0}^1\sqrt{1\plus{}x}\,\text{dx}$.


We know that since Al Gore dropped out. (slander), there are alot of democrats running. Lets see who you like the most. EVEN REPUBLICANS. Currently the order of the poll (dean, gephardt, ....) is the rank they are in.
I personally like Kerry, a decorated Vietnam veteran.

Hmm, if only we could vote for more than 1 ...  like, all of them.  Honestly, in terms of fundamental beliefs, I'd take any of the 9.  And, actually, just about anybody.  Heck, I could name Republicans who I'd feel glad to have as president instead of W.

By the way, did any of you hear the results from D.C.?  They held an unofficial primary, and all the main candidates except Dean asked not to be included, so it was only Dean (42%), Sharpton (35%), Braun and Kucinich.

wasn't that DC primary mostly for protesting the citizens' civil rights because their votes don't count in the real thing etc.

Yes.  They do get one or two electoral votes, though.  (That is, they help elect the president, even though they aren't represented by a voting member of congress.)

I'd say anyone but John Kerry, I've seen two of his ads and they were peeuuu! He clearly said he only "approved" of this ad and he didn't write it himself and he filled it with scenarios in which poor or the sad are shown and not specific political acts he said he'll do.

Still I'd rather have him then our current president...then again everyone knows he'll still win so *sigh*

[quote="JBL"]Lieberman?!  Lieberman has the most votes so far?  That's like voting for Zell Miller for Democrat, or Jim Jeffords for Republican ...[/quote]
Ah, but it's not like the 2004 Democratic Candidate even matters. I think we can all admit that.

All I know is I don’t like Sharpton. Liebermans ok I guess. Id like never to see a Clinton in office again even if it was Hillary, who I like, I just don’t want to see them in office ever again. I know they are good for the U.S. but when it comes down to it they are awful with foreign affairs (that’s just my belief.) I don’t know if Id like to see Bush get reelected or not because I have a list of his pros and cons; one of the pros is his secretary of state. I like Powell he is a smart man, one of the smartest secretary’s we’ve ever had. I like him a lot I just wish he ‘d have chosen a different party to run with. I like some of the things about democrats but then again I like some of the things about republicans.

Do yall know who has more power than the president?

Please try making your "sentences" easier to read; I had to skip your whole post because it was giving me a headache.

Ok its fixed. I can use proper grammer when I want to. I tire of it real fast because I use it so much at school. I guess thats what you would call lazy. I have the TAKS test as proof of being grammatically correct. For whoever doesnt know what that is, it is a state mandatory test that we have to take in Texas its horrible. Thats all the teachers teach about. Oh you have to do this and you have to that in order to pass this test. You know what, Ive never once used one of their stratagies and I make academic recognition on it every time they are like the easiest tests in the world. I made a perfect score on my writing and grammer exam last year and in 4th grade (those are the only two years you take the writing exam.) I just am lazy I guess. Oh well all I have to worry about this year is passing the American history exam thats like gonna be hard for me because all I have memorized is my 50 states and their capitals, the first presidential cabinent, and the preamble to the Declaration of Independence. I dont even have my presidents memorized; how many of them were there again?

Clinton was the 42nd, I believe.  Depending on whether you count Cleavland once or twice :).

news update
i hear it is a four guy race now that kerry and edwards have catched up (the two others are gehpart, den) . LETS GO FOR KERRY.
dean has lost a few points. oh and never hear about clak and lieberman in iowa its because they aren't capiangin there.

I personally would like to see Bush reelected.

aalindsey-there have been 44 presidents counting Bush in office now.  I know you should be able to name them because I can.  Washington, Adams, Jefferson,Madison, and Monroe in order just to name a few.

[quote="JBL"]That's interesting, given that 50% of the country is Democratic and as of now, at most 55% of Americans on polls would like to see him re-elected.  That's not too good for a sitting president.[/quote]
Ah who cares, I bet on Bush.

Hilary is smart by waiting unlike Gore :lol:

ugh I can now clearly see Bush is making U.S. education even worse than it is now, and I thought it can't get any worse...memorization can get you no where but they have to make you memorize or else you'll realize the horrible truth about the U.S. and it's dirty dirty past...

I'd say Bush is the strongest though I'll beat him soon enough...:twisted:

HUH?!?

BHorse, you're making us Republicans look, well, not good.

Huh?

We republicans?  I am not either!

I am independent! besides,  I can't vote yet!

I am not making republicans look bad.

Clinton? Sure he's a pressy boy but Bush is just downright dirty when it comes to recruiting, not only did he avoid getting recruited during Vietnam or Korea (I think it's Vietnam but I'm not sure; I read it in Time like an year ago so maybe it's online somewhere) using his dad's political power he also made it [b]look[/b] like he actually was in that war by going to Texas aviation school or something and he took photos of him in a plane which never went to the war field at all. Besides he cares more about his own life than any other so don't be surprised if he decides to sacrifice you guys for money or oil or whatever he want.

tare Bush never ran away from those wars I know that at least he may not have participated but then again I don't know if there was major recruting for those war.  I just know he didn't run away my parents were army.

BHorseMath, there was no draft during Desert Storm...plus, wasn't he governor of Arkansas at that time? It would have been completely unnecessary for a man his age to have been worried about being drafted. On Vietnam, he was a Rhodes Scholar, which would have made him exempt (as long as he was a college student), so he never draft-dodged.

Um, it was in [b]Time magazine[/b] and I know your parents are in the Texan army but I'm still going to take Time's word for it since I know Time pretty well but I don't know your parents at all :lol:

Tare, all writers have their own bias, so you can't always take what Time says by face value. Although it's my favorite news magazine too.

First of all:  Clinton was far too old to be in Desert Storm.  You're thinking of Vietnam.  Neither Clinton nor Bush, nor, for that matter, most politicians, went to Vietnam.  Mostly, this is due to the political influence of their families.  (Kerry is a notable exception.)  I wasn't exactly able to figure out in a quick search of the internet how Clinton avoided the draft.  Partly it was due to a student deferment while he was at college, but beyond that it's a little unclear.  Roughly, I think he was able to put it off until Nixon suspended the draft and thus was never drafted.  I do know that Bush's family pulled political strings to get him a spot in the Texas National Guard that he ended up not showing up for for 15 months or so.

[quote="Tare"]I understand that but it wasn't in the editorial section it was like the second biggest topic there and I'm sure that it is a fact rather than an opinion sort of a thing.[/quote]

hmm...don't think you bother checking the "setting the record straight" where TIME corrects its errors, which there are quite a few of...

BBOO YAH I WAS THE ONLY ONE WHO VOTED FOR KERRY AND I WAS THE ONE WHO CREATED THE POLL. BOY AM I PHYSIC. Kerry wins the Big Tuesday and Edwards drops out. Kerry wins all except for one: Vermont.

Hehe,  but he got beat nearly 2-to-1 by a man no longer running in the race.  (On the other hand, he did get nearly 75% of the vote in MA.)

Anyhow, Kerry 2004 -- yeehaw!

(By the way, I think you meant "psychic," not "physic."

On the news the analyst people were saying that Vermont was not anti-Kerry; they were just pro-Dean.b

I am voting for either Poppin' Fresh or the Jolly Green Giant.

Would it be any worse to have a fictional president than a real one?

FYI -- Poppin' Fresh has mad math skills and would likely cause an international economic boom.

For a moment there I thought you were churchill :lol:

I have no idea who they are, but I'm still rooting for Kelly...I'm choosing the red pill :lol:


inequalities unsolved
inequalities
algebra

Let $f(a,b,c)=\frac{1}{\sqrt{a}}+\frac{3}{\sqrt{b}}+\frac{8}{\sqrt{3c+2a}}-\frac{16\sqrt{2}}{\sqrt{3(a+b+c)}}$.
It is easy to show that $f(a,b,0) \ge 0$, $f(a,b,c)\ge 0$ when $c\to \infty$.
$\frac{\partial f}{\partial c}= 0 \iff a+3c=3b$.
So it is enough to show that $f(3(b-c),b,c) \ge 0$.
Let $b-c=x, b=y$.
$f(3(b-c),b,c)\ge 0 \iff \frac{1}{\sqrt{x}}+\frac{3\sqrt{3}}{\sqrt{y}}\ge \frac{8}{\sqrt{x+y}}$, it is obvious by jensen.


limit
inequalities
calculus
search
quadratics
calculus computations

Hello there,

I am trying to prove the following limit using the delta-epsilon definition. I have worked backwards to find a suitable epsilon value but am stuck on how to write the actual proof using the minimum of this value and my assumed delta value.

In class, my instructor simply ends the proof with a new statement under (10) below saying that if $ \delta \text{=min}\left\{ \text{1, }\frac{57\varepsilon }{20} \right\}$, then QED. However, I want to go beyond this and actually arrive at $ |f(x) - L | = \abs{\frac{x + 3}{x^2 + x + 4} - \frac{2}{3}} < \varepsilon$.

Thank you very much.

---

\begin{align}
  & \text{Prove the following statements directly using the formal }\varepsilon \text{,}\delta \text{ definition}\text{.} \\
 & \text{ii) }\underset{x\to 1}{\mathop{\lim }}\,\frac{x+3}{{{x}^{2}}+x+4}=\frac{2}{3} \\
 & ---- \\
 & \text{My work is below}\text{. I am working backwards from the epsilon inequality} \\
 & \text{to find a suitable delta}\text{.} \\
 & \text{As question states: }\text{ }0<\left| x-1 \right|<\delta \Rightarrow \left| \frac{x+3}{{{x}^{2}}+x+4}-\frac{2}{3} \right|<\varepsilon  \\
 & \left| \frac{3(x+3)-2({{x}^{2}}+x+4)}{3({{x}^{2}}+x+4)} \right|<\varepsilon \Rightarrow \left| \frac{(2x+1)(x-1)}{3({{x}^{2}}+x+4)} \right|<\varepsilon  \\
 & \text{Let }\left| x-1 \right|<\frac{1}{2}. \\
 & \text{Then }\frac{1}{2}<x<\frac{3}{2}. \\
 & \therefore \left| \frac{(2\left( \frac{3}{2} \right)+1)(x-1)}{3({{\left( {}^{1}\!\!\diagup\!\!{}_{2}\; \right)}^{2}}+\frac{1}{2}+4)} \right|=\left| \frac{5(x-1)}{{}^{57}\!\!\diagup\!\!{}_{4}\;} \right|<\varepsilon \Rightarrow \left| x-1 \right|<\frac{57\varepsilon }{20} \\
 & \text{Proof: } \\
 & \text{Let }\delta \text{=min}\left\{ \text{1, }\frac{57\varepsilon }{20} \right\} \\
 & \Rightarrow \left| x-1 \right|<\frac{1}{2}\text{ and }\left| x-1 \right|<\frac{57\varepsilon }{20} \\
\end{align}

Oh, this is a typical request! Some authors, knowing how computations will go, actually do this (but it is utterly unnecessary, just aesthetic).

Start with $ \varepsilon' \equal{} \frac {20} {57} \varepsilon < 1$ and go through exactly the same steps. You will notice that at the end, you obtain exactly $ \varepsilon$  :)

But as I said, completely unnecessary; any expression in $ \varepsilon$ that $ \to 0$ with $ \epsilon \to 0$ is allright.

My own pedagogical opinion is that I'm not happy with even posing this problem in the first place. There's something distorted and artificial about trying to apply the raw $ \epsilon$-$ \delta$ formalism to such a specific problem. Maybe, just once, for the sake of illustration, you do this for $ \lim_{x\to 2}x^2\equal{}4.$ But nothing messier than that.

Very early in your study of the definition of limit, you prove the main arithmetical theorem about limits: the limit of a sum is the sum of the limits; the limit of a product is the product of the limits; and the limit of a quotient is the quotient of the limits provided the denominator doesn't tend to zero (and all provided both original limits exist). Once you have that theorem, you [b]use[/b] it - you would certainly use it here.

As an analogue from a Calculus I course: would you ask that $ \frac{d}{dx}\,\frac{x\plus{}3}{x^2\plus{}x\plus{}4}$ be computed as $ \lim_{h\to0}\frac1h\left[\frac{x\plus{}h\plus{}3}{(x\plus{}h)^2\plus{}(x\plus{}h)\plus{}4}\minus{}\frac{x\plus{}3}{x^2\plus{}x\plus{}4}\right]\,?$

Thank you for your responses.

I am not sure how to go through the algebra with this problem because I can't seem to bound $ ({x}^{2} \plus{} x \plus{} 4)$ from $ |{x \minus{} 1}| < \frac{1}{2}$. I am writing to write my proof like the one at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1603977789&t=204887, where the bound on $ |x \plus{} 3|$ isn't too hard to derive from assuming that $ |x \minus{} 3| \equal{} 1$
Could I in fact bound that above quadratic expression?

This is indeed a more challenging problem than usual; my guess is that my instructors want to have us face more involved expressions with the formal definition of a limit.


geometry
parallelogram
rotation
complex numbers
geometry proposed

I want a solution with complex numbers to the following problem. Please help .

Let a triangle $ABC$ with $\angle{A}>45$ and $\angle{B}>45$.
In the interior of the triangle we construct a right-angled and isosceles triangle $ABD$ with $\widehat{ADB}=90$ .
Now in the exterior of the triangle we construct the right-angled and isosceles triangles $BCE$ and $ACZ$ with $\widehat{BEC}=90$ and $\widehat{AZC}=90$ respectively .Prove that
$DECZ$ is a paralellogram .

Assume $A, B, C$ are positively oriented. Let $M$ be the midpoint of $AB$. Then $MD$ is the rotated image of $MB$ for $90^\circ$ CCW - expressing that with complex numbers, we get $d-m=i(b-m)\iff d={a+b\over 2}+i{b-a\over 2}$, since $m={a+b\over 2}$

Similarly, $e={b+c\over 2}+i{b-c\over 2}, z={a+c\over 2}+i{c-a\over 2}$

Now $e-d+z={b+c-a-b+a+c\over 2}+i{b-c-b+a+c-a\over 2}=c$, which yields $e-d=c-z$. The conclusion follows.


superior algebra
superior algebra solved

a tricky problem (one of those with short enounces :D) let R be a ring such that xn = x. Prove that for each x,y we have xyn-1=yn-1x.

[quote="Valentin Vornicu"]let R be a ring such that xn = x. Prove that for each x,y we have xyn-1=yn-1x.[/quote]

For n = 2

x+y=(x+y)= x+xy+yx+y => xy=-yx
a+1= (a+1) = a+2a+1 => 2a=0 => -a=a

xy=yx

[quote="Valentin Vornicu"] let R be a ring such that xn = x. Prove that for each x,y we have xyn-1=yn-1x.[/quote]

Remind me the  theorem:Let R a ring with a unit element such that for any x\inR, there exist a positive integer nx such that x  nx=x, then R is commutative

Biography of Jacobson
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Jacobson.html

[quote="Diogene"]Warning ! It seems, in the V.V. problem the ring hasnt  unit element for the multiplicative operation ... Am I right ?[/quote]

Unit or not unit element that's the question, V.V will tell us

one does not need an unit in the ring to prove the property.

Well, clearly n>=2 ==> x^(2n-2) = x^(n-1)
Also we have that x^2 = 0 ==> x^n = 0 ==> x = 0

Let z = x^(n-1) and y be in the ring

(zyz-yz)^2 = (zyz-zy)^2 = 0 ==> yz = zy ==> x^(n-1) y = y x^(n-1)


Would taking this class only look frivolous on a college application? Would it look the same if I took a class like this somewhere else? This is one of the subjects I want to study over the summer, and I wouldn't mind getting credit for it at my high school.

You are talking about [url=http://epgy.stanford.edu/courses/english/EG20/]EG20[/url], right? That's a good course. It wouldn't look silly on a high school transcript, and the knowledge is useful.

[quote="tokenadult"]You are talking about [url=http://epgy.stanford.edu/courses/english/EG20/]EG20[/url], right?[/quote]
Yep, that's it. Thanks.


Hello, I have found an approximation to $\sum_{k=1}^n\frac{1}{k}$.  Would you say that this is something good.  If it is, does it worth being reviewed by some people?  I'm not sure about such things.  By the way, I also have an algebraic proof for it, including some calculus. Woudl you please assist me?  Thanks in advance.

Masoud Zargar

How good¿
Something near to best possible was given in http://www.mathlinks.ro/Forum/viewtopic.php?t=46075 ...

well, I do not know much calculus; I have never studied it in depth and neither taken a course, but my formula gives it to an accuracy of approximately $\pm 0.06$, and as $n\rightarrow\infty$, the accuracy increases.  The formula I have found is obvious that it can be improved.  I'll post it here later.

[url=http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html]Euler-Maclaurin[/url] might be helpful.
$\sum_{i=1}^{n}\frac{1}{i}\sim \ln n+\gamma+\frac{1}{2n}-\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{1}{n^{2k}}$. Please notice that this is not an equality. What it really means is $\lim_{n \to \infty}n^{2m}\left[\sum_{i=1}^{n}\frac{1}{i}-\left(\ln n+\gamma+\frac{1}{2n}-\sum_{k=1}^{m-1}\frac{B_{2k}}{2k}\frac{1}{n^{2k}}\right)\right]=-\frac{B_{2m}}{2m}$


Suppose your college does not offer a particular math course that interests you.  Are there some good sources for correspondence or online courses in advanced math?

As an example, where might one find such a course in Game Theory? Or Analysis of Time Series?

Thank you.

Just because a course isn't in the catalog doesn't mean that you can't take it---many college faculty have areas of expertise not reflected in any course offered in their departments.  So, I have two suggestions.  (1) Ask around in the department to see if anyone is able and willing to offer an independent study in the topics you want to learn.  (2) Check out MIT's Open Courseware and see if the topics you want are covered there.

you may also want to consider some sort of study abroad (one example is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68721]BSM[/url])

Hi,

Thank you for the responses to my original question, but none of them is quite what I was hoping for.

I am looking for a place that gives correspondence or online courses in advanced math.

I did check the MIT site, and that is close, but not really an active course with an instructor who will answer questions.

Any more ideas?

Thank you.

There might be a very small number of colleges that offer math classes beyond calculus/linear algebra on-line.  The only program that I know of is Stanford's EPGY [url]http://www-epgy.stanford.edu/courses/math/[/url].  They offer several courses that are common to many college-level math programs.  Nothing real esoteric though - partial differential equations and complex analysis are the most advanced courses.  High-school graduates (older than 18) might not be able to take the EPGY university-level courses - I'm not that familiar with the program's admission requirements.

There's probably not much demand for on-line advanced math courses among students already attending colleges.  There's not much demand for the traditional-delivery of these courses on most campuses either.  I think it would be tough for most colleges to find a critical mass of quality students to take these kinds of classes on-line.


Identify the units on the variable x

x=1/2 at^2 (at^2 is atfter the whole 1/2 part, not on bottom or top)

a is measured in meters/sec^2 and t is measured in seconds

[quote="shobber"]$x=\frac12 a t^2$
Newton's second law with initial velocity $v_0=0$
The unit $=(m)(s)^2\cdot (s)^2=m$[/quote]
it's $\frac{m}{s^2}*s^2$ :lol:


Suppose $ f$ is absolutely continuous on $ (a,b)$ and that $ f'(x) \equal{} 0$ almost everywhere (in the sense of Lebesgue). Then $ f$ is constant.
one proof is in Royden's book.
but
One approach is to define a signed measure on $ (a,b)$ by
$ \mu(c,d) \equal{} f(d) \minus{} f(c)$ , and apply the Radon-Nikodym theorem.
but how is this done?
can anyone write something in more detalis?
Thanks.

Try page 8 in [url=http://www.emis.de/journals/DM/v81/art9.pdf]this[/url] link.  Is this what you're hoping for?

The proof I'm familiar with uses Vitali coverings.


logarithms
calculus
derivative
calculus computations

Find an expression for $\frac{dy}{dx}$ if:
$y^{2}=x^{x}$

Taking natural logarithm of both sides and use the differentiation of implicit function.

Implicit differentiation: \[y^{2}=x^{x}\Longrightarrow 2y\frac{dy}{dx}= \frac{d}{dx}\exp(x\log x)\\ \Longrightarrow \frac{dy}{dx}= 1/2 x^{ x/2}\left( \ln \left( x \right)+1 \right)\]

[hide="My Solution"]
Take logs of both sides:
$2lny=xlnx$

$\frac{2}{y}\frac{dy}{dx}=1+lnx$

$\frac{dy}{dx}=\frac{y(1+lnx)}{2}$
[/hide]

Ottem, your answer is correct.
coffeym, $\ln y^{2}=\ln |y|^{2}=2\ln |y|.$


Let $n \geq 3$ be a positive integer. Prove that the sum of the cubes of all natural numbers, coprime and less than $n$, is divisible by $n$.

[hide="Solution"] We wish to show that the sum of the cubes of the members of $U_{n}$ is zero.

This is rather simple.  The additive inverse of the cube of a member of $U_{n}$ is the cube of its additive inverse (edit: in other words, $(a+b) | (a^{3}+b^{3})$), so we add each number to its additive inverse (invariant under cubing) for a sum of zero.

For $n \ge 3$ we must have $2 | \varphi(n)$ and so such a pairing always exists.  (Or to be more explicit:  For $n \ge 3$ we have $\frac{n}{2}\not \in U_{n}$, so every member of $U_{n}$ has an additive inverse different from itself.) [/hide]


abstract algebra
superior algebra
superior algebra unsolved

[quote="JustSee12"]This is from the Dummit and Foote section on conjugacy and the class equation:
Let $ G$ be a finite group and choose a system of representatives $ g_1, \cdots, g_r$ for the conjugacy classes of $ G$. Show that if the $ g_i$ pairwise commute then $ G$ is abelian.[/quote]

If $ G$ is non-abelian, then $ |G|>r$ and, without loss of generality, we may assume that $ Z(G)\equal{}\{g_1, \cdots, g_s \},$ for some $ 1 \leq s < r.$ Since the $ g_i$ pairwise commute, we have $ [G: C(g_j)] \leq \frac{|G|}{r},$

for all $ j.$ But then $ |G|\equal{}s\plus{}\sum_{j\equal{}s\plus{}1}^r[G: C(g_j)] \leq s\plus{}\frac{r\minus{}s}{r}|G|,$ which gives us the contradiction $ |G| \leq r.$


vector
geometry
circumcircle
analytic geometry

[quote="PenguinIntegral"][quote="pkerichang"][hide]
Let $z_{1}, z_{2}, z_{3}$ be three vectors in complex plane that start at the origin.  Connect the three endpoints we get a triangle with circumcenter at the origin.  Let the end point of $z_{1}$ be $A$, $z_{2}$ be $B$, and $z_{3}$ be $C$, then by simple analytic geometry (and simple algebra arrangement) the given equation is the same as:

$\min|k(z_{2}-z_{3})+z_{3}-z_{1}|=\frac{bc}{2R}$

(for notation sake, I change the $a$ in the original equation to $k$, because $a$ will stand for a side of my triangle).

Now look at the expression in the min.  We see that $z_{3}-z_{1}$ is a vector parallel to and the same length as the side $b$, and $z_{2}-z_{3}$ is parallel and the same length as the side $a$.  using tip to tail vector addition, we see that the minimum expression becomes the minimum length from $A$ to $BC$, which is the length of the altitude from $A$ to $BC$ by simple geometry.

So, we wish to prove that $h_{A}= \frac{bc}{2R}$.  Recall that $\frac{ah_{A}}{2}= \triangle$, and also $\triangle = \frac{abc}{4R}$, equate and solve we see that the equation indeed is true.

If $z_{1}$ equals to either $z_{2}$ or $z_{3}$, plug in $k=1$ we get both sides to be zero, so the equation holds all the time.
[/hide][/quote]
Congratulations on writing something about vectors that I understand. Your proof was very clear and intuitive.[/quote]

Thanks.  The first time I saw this one I feel "There must be some geometry in this or else I'm not going to bother with all the nasty algebra stuff", and it comes out nice :)


Using Am-Gm:
$b\geq a\geq c\geq d$
(add $d^{3}, a^{4}, b^{5}$ to each side of each equation repectively then powermean)
we now need to prove $d\geq b$ to maintain equality
but from first eq. $3d^{3}\geq 3c^{3}, c\geq d \implies c=d \implies 2d^{3}\geq 2a^{3}\implies d=a \implies d=b=a=c$

Does anyone look at these problems at all? I think my solution might be wrong but right idea.

What do you think, gollywog?


Daca in triunghiul $\ ABC$ avem $\ m(\angle ACB)=120$, demonstrati ca centrul cercului inscris triunghiului $\ ABC$ este situat pe [i]Dreapta lui Euler[/i] $\triangle ABC$.

Problema e greşită, deci e un pic cam greu de rezolvat. Vezi urmatoarele doua linkuri pentru varianta corecta: [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=277944&highlight=incenter+line+euler#277944]Link 1[/url] şi [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=445562&highlight=incenter+line+euler#445562]Link 2[/url].

perfect_radio, eu asa am vazut-o scrisa intr-o cartea a domnului Cristinel Mortici.
[hide="post scriptum"]Am inteles ce ziceau oamenii aia...in linkurile date de tine.Cat de cat.. :) [/hide]


function
LaTeX
number theory unsolved
number theory

Find the minimum of natural number $ n$ such that there exist a function which is a nonconstant function $ f: Z\mapsto R_{+}\cap\{0\}$ for $ x,y\in{Z}$,we have:
 a) $ f(xy)=f(x)f(y)$
 b)$ 2f(x^{2}+y^{2})-f(x)-f(y)\in\{0,1,2,...,n\}$
And for this above value $ n$, find all which function $ f(x).$

Edited by N.T.TUAN
[code]Find the minimum of natural number $n$ such that there exist a function which is a nonconstant function $f: Z\mapsto R_{+}\cap\{0\} for $x,y\in{Z}$,we have:
 a) $f(xy)=f(x)f(y)$
 b)$2f(x^{2}+y^{2})-f(x)-f(y)\in\{0,1,2,...,n\}$
And for this above value $n$, find all which function $f(x).$[/code]

[quote="Nbach"]Find the minimum of natural number $ n$ such that there exist a function which is a nonconstant function $ f: Z\mapsto R_{+}\cap{0}$ for $ x,y\in{Z}$,we have:

a) $ f(xy)=f(x)f(y)$
b) $ 2f(x^{2}+y^{2})-f(x)-f(y)\in{0,1,2,...,n}$

And for this above value $ n$,find all which function $ f(x)$.[/quote]

Tip:  Regular text looks horrible in $ \text{\LaTeX}$.

[quote="Nbach"]Find the minimum of natural number $ n$ such that there exist a function which is a nonconstant function $ f: Z\mapsto R_{+}\cap\{0\}$ for $ x,y\in{Z}$ [...][/quote]
$ \mathbb{R}_{+}\cap \{0\}\subseteq \{0\}$, so what!? :huh: And what's then $ \mathbb{R}_{+}$? I would guess the set of all positive real numbers, but I feel somewhere there's something absolutely wrong...

suppose that $ f: \ Z\to B$ :D

$ f$ nonconstant gives $ \exists a: \ f(a)\neq f(0)$
so $ f(0)f(a)=f(0)=f(0)f(0)$ gives $ f(0)=0$
and $ y=0$ gives $ 2f(x^{2})-f(x)-f(0)=f(x)(2f(x)-1)\in\{0,1,2,...,n\}$
then $ \forall x\in Z: \ f(x)\in H=\{\frac{1\pm \sqrt{8a+1}}{4}|\ a\in \{0,1,2,...,n\}\}$
we take $ n_{min}=a$
then for some $ x\in Z$: $ f(x)=\frac{1\pm \sqrt{8a+1}}{4}$ and we have $ (f(x))^{2}=f(x^{2})\in H$ so we must have $ \frac{(1\pm \sqrt{8a+1})^{2}}{16}=\frac{1\pm \sqrt{8b+1}}{4}$ for some $ b\in \{0,1,2,...,a\}$
 :arrow: $ 8a-2\pm 2\sqrt{8a+1}=\pm \sqrt{8b+1}$

it's nothing, but it's can gives an idea for some one  :wink:

Trust me, I'm still wondering how to interpret the problem. Maybe it should be $ \mathbb{R}_{+}\!\cup \{0\}$, and not $ \mathbb{R}_{+}\!\cap \{0\}$, I don't know... :maybe: Anyway, let me say your $ B$ is fantastic! :D

OK,I'm so sorry.It's the first time I use LaTeX...
It's    $ R_{+}\cup{0}$.

[quote="Nbach"]Find the minimum of natural number $ n$ such that there exist a function which is a nonconstant function $ f: \mathbb{Z}\mapsto \mathbb{R}_{+}\cup\{0\}$ for $ x,y \in \mathbb{Z}$, we have:
 a) $ f(xy)=f(x)f(y)$
 b) $ 2f(x^{2}+y^{2})-f(x)-f(y)\in\{0,1,2,...,n\}$[/quote]
Well, I'm proving the minimum is attained for $ n = 1$. Indeed, since $ f$ is multiplicative, we need $ f(0) = f(0)^{2}$ and $ f(1) = f(1)^{2}$. Hence $ f(0), f(1) \in \{0, 1\}$. Suppose $ f(0) = 1$. Then $ 1 = f(x \cdot 0) = f(x) \cdot f(0) = f(x)$, for any $ x \in \mathbb{Z}$, i.e. $ f$ is constant. So necessarily $ f(0) = 0$. In a similar way, admit $ f(1) = 0$. Then $ f(x) = f(1 \cdot x) = f(1) \cdot f(x) = 0$, for any $ x \in \mathbb{Z}$, viz $ f$ is constant again. So a fortiori $ f(1) = 1$. Then $ 2 f(1^{2}+0^{2})-f(1)-f(0) = 1$, yielding $ n \ge 1$. Now let
\[ g(x) = \left\{ \begin{array}{ll}0 & \mbox{ if }x = 0 \\ 1 & \mbox{ if }x \in \mathbb{Z}, x \ne 0 \end{array}\right.. \]
Then clearly $ g$ is a multiplicative function $ \mathbb{Z}\to \mathbb{R}_{+}\! \cup \{0\}$ and $ 0 \le 2g(x^{2}+y^{2})-g(x)-g(y) \le 1$, for any $ x, y \in \mathbb{Z}$. Thus $ n = 1$ and the claim is proved. :wink: So we are left with sieving for any other function $ f: \mathbb{Z}\mapsto \mathbb{R}_{+}\!\cup\{0\}$, apart from $ g$, such that $ f(xy) = f(x) f(y)$ and $ 0 \le 2f(x^{2}+y^{2})-f(x)-f(y) \le 1$, whenever $ x, y \in \mathbb{Z}$. First, note that $ f$ is uniquely determined by its values on the set $ \mathbb{P}= \{p \in \mathbb{N}: p \mbox{ is prime}\}$, because of its multiplicativity (here $ \mathbb{N}= \{1, 2, \ldots\}$).

Second, by way of contradiction, take for granted there is $ x_{0}\in \mathbb{Z}$ such that $ f(x_{0}) \not \in \{0, 1\}$. Since $ f(x_{0}^{n}) = f(x_{0})^{n}$, for any $ n \in \mathbb{N}= \{1, 2, \ldots\}$, then $ 0 \le f(x_{0})^{2n}-f(x_{0})^{n}= f(x_{0}^{2n}+0^{2})-f(x_{0}^{n})-f(0) \le 1$, i.e. $ 0 \le f(x_{0})^{n}\cdot (2f(x_{0})^{n}-1)) \le 1$. Anyway this is absurd, as either $ f(x_{0})^{n}\to 0^{+}$ or $ f(x_{0})^{n}\to+\infty$, as $ n \to \infty$ (recall that $ f(\mathbb{Z}) \subseteq \mathbb{R}_{+}\!\cup \{0\}$ and hence $ f(x_{0}) > 0$). Thus in fact $ f(\mathbb{Z}) = \{0, 1\}$. Now, $ 1 = f(1) = f(-1)^{2}$. Therefore $ f(-1) = 1$ and $ f(x) = f(-1) \cdot f(-x) = f(-x)$, for any $ x \in \mathbb{Z}$, i.e. $ f$ has to be an even function.

At this point, admit $ f(2) = 0$. Then $ 2f(1^{2}+1^{2})-f(1)-f(1) = 2f(2)-2 =-2$. But actually from (b): $ 0 \le 2f(1^{2}+1^{2})-f(1)-f(1) \le 1$. Hence necessarily $ f(2) = 1$. Now consider that
\[ \forall x, y \in \mathbb{Z}: \begin{array}{ll}\left\{ \begin{array}{l}2f(x^{2}+y^{2})-f(x)-f(y) \ge 0 \\ f(x^{2}+y^{2}) = 0 \end{array}\right.\quad\Longrightarrow\quad f(x) = f(y) = 0 & \quad\qquad (1) \\ & \\ \left\{\begin{array}{l}2f(x^{2}+y^{2})-f(x)-f(y) \le 1 \\ f(x) = f(y) = 0 \end{array}\right.\quad\Longrightarrow\quad f(x^{2}+y^{2}) = 0 & \quad\qquad (2) \end{array}\]
and set $ \mathcal{S}= \{n \in 2\mathbb{N}+1: \exists^{no}a, b \in \mathbb{N}_{\;\!0}\mbox{ s.t. }\!\gcd(a,b) = 1\;\wedge\;n = a^{2}+b^{2}\}$ and $ \mathcal{D}= \{n \in \mathcal{S}: f(n) = 0\}$ (here as usual, $ \mathbb{N}_{\;\!0}= \mathbb{N}\cup \{0\}$). Note $ 1 \not\in \mathcal{D}$, though $ 1 \in \mathcal{S}$. By way of contradiction, suppose $ \mathcal{D}\ne \emptyset$ and let $ n_{0}= \min(\mathcal{D})$ (recall that $ \mathbb{N}$ is well-ordered). Then there must exist $ a, b \in \mathbb{N}$ such that $ n_{0}= a^{2}+b^{2}$ and $ \gcd(a,b) = 1$. Thus $ f(a) = f(b) = 0$ from (1). Moreover, as $ n_{0}$ is odd, we lose no generality assuming $ a = 2 \alpha$, for some $ \alpha \in \mathbb{N}$.

Then $ f(\alpha) = f(2) f(\alpha) = f(a) = 0$ and $ f(\alpha^{2}+b^{2}) = 0$ from (2). Letting $ \alpha^{2}+b^{2}= 2^{k}u$, where $ u \in 2\mathbb{N}+1$ and $ k \in \mathbb{N}_{0}$, it follows $ f(u) = 0$. Besides, $ u$ is representable by the form $ c^{2}+d^{\;\!2}$, for suitable $ c, d \in \mathbb{N}$. In fact, $ u \mid (\alpha^{2}+b^{2})$ and $ \gcd(\alpha, b) = 1$. Hence there exists no prime $ p \in \mathbb{N}$ such that $ p \equiv 3 \bmod 4$ and $ p \mid u$, which is enough to grant that $ u$ and any of its divisors $ w \in \mathbb{N}$ can be expressed as a sum of squares of positive integers. In particular, this is the case when
\[ w = \left(\frac{c}{\gcd(c,d)}\right)^{\!2}+\left(\frac{d}{\gcd(c,d)}\right)^{\!2}, \]
whether $ c, d \in \mathbb{N}$ and $ u = c^{2}+d^{\;\!2}$. But actually we got an absurd, since then $ w \in \mathcal{D}$ and $ w < n_{0}= \min(\mathcal{D})$ in the same time. So necessarily $ f(n) = 1$, whenever $ n \in \mathcal{S}$. In particular, $ f(p) = 1$, if $ p \in \mathbb{N}$ is a prime $ \equiv 1 \bmod 4$.

Indeed, we're near to the end, don't give up! :rotfl: Well, note there can't exist two distinct primes $ p, q \in \mathbb{N}$ such that $ p \equiv q \equiv 3 \bmod 4$ and $ f(p) = f(q) = 0$. Otherwise, it would be $ f(p^{2}+q^{2}) = 0$ on the one hand [according to (1)] and $ f(p^{2}+q^{2}) = 1$ on the other [since apparently $ (p^{2}+q^{2})$ has a divisor $ w \in \mathcal{S}$ - guess what? :wink: ], absurd! So either $ f \equiv g$ or there is one only prime $ q \in \mathbb{N}$, with $ q \equiv 3 \bmod 4$, such that $ f(q) = 0$.

Indeed, we want to prove the last posit is consistent. In fact, it's enough to prove it doesn't clash [btw, is "clash" the right verb to use in this context? :maybe: Thanks!] with b). So fix $ x, y \in \mathbb{Z}$ and let $ h(x,y) = 2f(x^{2}+y^{2})-f(x)-f(y)$ for notational convenience. If $ q \mid (x^{2}+y^{2})$, then necessarily $ q \mid \gcd(x,y)$ (it's a well-known fact!) and trivially $ t(x,y) = 0$. Hence suppose $ q \nmid (x^{2}+y^{2})$. Then again $ t(x,y) = 0$, whether $ q \nmid xy$, and $ t(x,y) = 1$, if $ q \mid x$ or $ q \mid y$ (all is quite easy to check, so I'm leaving out details). Thus everything is right and we're done. Nice problem! :)

Is it possible to know who's the author of the problem? Thanks! :)

OK.Gabriel Dospinescu is author of this above problem.

O.K., thank you so much. :) Btw, note I've just edited my first post to fix a minor detail.


The set A contains 10 distinct integers. The smallest of these integers is -6 and the largest is 5.  The sum of the integers in set A is 1. How many possible set As are there?

well, the range for the remaining 8 is -5,-4,-3,-2,-1,0,1,2,3,4 or 10 numbers

so, 10*9*8*7*6*5*4*3 /me is too lazy to do calculations.

Since they're distinct integers and two of the spots are already taken, we have 8 integers left to choose from between $ \minus{}5$ and $ 4$ inclusive.  There are $ 10$ integers there, and assuming that permutations of a set are the same as the set, then there are obviously $ \binom{10}{8}\equal{}45$ possible sets.


trigonometry
logarithms
function
inequalities
real analysis
real analysis unsolved

Prove that
\[ \sum_{i \equal{} 1}^{n}\frac {Cos(ix)}{i} \geq{ \minus{} 1}
\]

The sum can be computed exactly:

$ \sum_{k \equal{} 1}^{\infty}\frac {\cos kx}{k} \equal{} \minus{} \text{Re}(\log|1 \minus{} e^{ix}|) \equal{} \minus{} \ln\left|2\sin\frac {x}2\right|.$

The minimum occurs when $ x \equal{} \pi$ and the minimum value is $ \minus{} \ln2,$ which is greater than $ \minus{} 1.$

This computation comes from a combination of the power series identity $ \sum_{k \equal{} 1}^{\infty}\frac {z^k}{k} \equal{} \minus{} \log(1 \minus{} z)$ for $ |z| < 1$ together with Abel's Theorem for extending the function to the boundary of the disk.

Erm, the question was about partial sums, not only about the full series ;)

Oops, sorry, misread that. Looking at it numerically, I certainly believe it is true, but I don't have the proof yet.

One simple way to approach the problem is to notice that $ \frac 34\plus{}\cos x\plus{}\frac{\cos2x}2\plus{}\dots\plus{}\frac{\cos nx}n$ has almost convex Fourier coefficients, so if we move everything down by $ \frac 1n(\cos x\plus{}\cos 2x\plus{}\dots\plus{}\cos nx)$, we'll get at least $ \minus{}\frac 34$ for the resulting sum. Now if we could bound the sum of pure cosines by $ \minus{}\frac n4$ from below, we would be done. Unfortunately, it is not the case and we get only $ \minus{}7/4$ this way if we use the trivial estimate $ \minus{}n$. On the other hand, we are fine with this approach if $ \pi>|x|>10/n$, say, so the problem really occurs only at a small neighborhood of the origin, where it is easy to deal with: the first $ cn$ terms give you something of order $ \log n$ and the remaining terms sum to a  constant at worst, so for large $ n$ we get what we need. All estimates are effective, so now one needs just to play with a computer a bit to finish the problem, but I hate this way. There should be much neater approach.

This inequality was proved by Young in 1913.  I can give the exact reference if you want me to.

You can read about this and similer inequalities in Mitrinivic, Pekaric, Fink, Classical and new inequalities in analysis, Kluwer, 1993 on page 615.  The original reference is W.H.Young, On a certain series of Fourier, Proc.London Math.Soc. 2(11), 1913, 357-366.


probability
number theory
relatively prime

Ten points in the plane are given, with no three collinear.  Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely.  The probability that some three of the segments form a triangle whose vertices are among the ten given points is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[hide="Solution"]
There are $\binom{10}{2}=45$ segments.

There are $\binom{45}{4}$ ways to choose four segments.

There are $\binom{10}{3}$ ways to choose three points for a triangle.  After we choose this triangle we have $45-3=42$ segments left to choose from.

Thus our answer is $\frac{42\binom{10}{3}}{\binom{45}{4}}=\frac{16}{473}.$  Our answer is thus $16+473=\boxed{489}.$[/hide]

At a quick glance, you are overcounting (obviously).  I think this is occuring because it doesn't seem that you are taking into account the overcount produced since you can start with AB, BC, or CA and still form triangle ABC.

So, divide by $3!$? It's still off, though... :(

[quote="bob123"][quote="chess64"][hide]In order for a triangle to be formed, two of the segments must have endpoints on the same point, let it be $A$. Also, the two other endpoints, $B$ and $C$, must be connected. The fourth segment can be any other one.

The probability that a second segment $BC$ shares the same endpoint as the first one, $AB$, is $8/44=2/11$, since there are $\binom{10}{2}=45$ total segments. The probability that a third segment has endpoints $B$ and $C$ is $1/43$. We don't care about the fourth segment; there are $42$ possibilities for it. So our answer is $42(2/11)(1/43)=84/473$.[/hide]

What's wrong with this solution?[/quote]

Though there are 42 possibilities for the last part, the probability of getting a last part is 1 ($42/42$ since it doesn't matter). Also the $8/44$ should $16/44=4/11$ since it it can stem from either A or B. So your answer should be $4/11\cdot1/43=4/473$, which is $477$ and still $4$ times off of what joml got... Hmmm...?[/quote]

Yeah, that's what I got before but then I changed it because it was wrong...but it's still wrong :stink:

Not so sure...

[hide]
Case 1: We form the triangle with the first 3 segments we choose.
The probability for this was already calculated by bob123: $\frac{4}{473}$

Case 2: We form the triangle only after the $4th$ segment is chosen.

Subcase 1: First segment can be any segment, second segment we choose shares an endpoint with the first, third segment shares no endpoint with the other two, fourth segment completes triangle.

Probability = $1\times\frac{16}{44}\times\frac{42}{43}\times\frac{1}{42}=\frac{4}{473}$

Subcase 2: First segment can be any segment, second segment does not share an endpoint with the first, third segment shares an endpoint with at least one of the two previous segments, fourth segment completes the triangle.

This is one of the parts i'm not sure about. We have another two subcases to deal with: the third segment shares an endpoint with exactly one of the previous two segments, or with both. In the case of the former (of which there are four possible segments), we then have 2 choices for the fourth segment whereas we only have 1 choice for the latter.

Probability for the first case: $\frac{44-16}{44}\times\frac{4}{43}\times\frac{2}{42}=\frac{4}{11\times43\times3}$

Probabilitiy for the second case: $\frac{44-16}{44}\times\frac{24}{43}\times\frac{1}{42}=\frac{4}{11\times43}$

But when I add all these probabilities together, I get a different answer...[/hide]

[quote="kops723"][quote="joml88"][quote="kops723"]atm, i cant see the problem with chess's solution, but joml, i believe you are over counting as well.
you took into account the fact that order doesnt matter for the first 3 segments, but what about the last one?
i believe, you have to divide your answer by 4, so you end up with [hide="this"]$\frac{4}{473}$ so $m + n = \boxed{477}$[/hide] am i wrong?[/quote]

You are wrong :)  Look at my solution again.  I am not taking order into consideration when I chose the first three segments.  I choose the whole triangle as a first step.  I don't break the triangle down into three separate components.  So my answer is valid...kalva agrees http://www.kalva.demon.co.uk/aime/aime99.html[/quote]

i think you misunderstood my argument, but if you have a reference youre probably right, but i understand that you didnt take into account how you chose the segments of the triangle, but how come the way you did doesnt imply that you [i]have[/i] to choose the 3 segments of the triangle first and only [i]then[/i] can you choose the 4th segment, instead of being able to choose the 4th segment as any of the 4 picks[/quote]

huh?  I think it does imply that.  There are 42 choices for the last segment...does not that imply that the other 3 sides were already decided ;)

Can someone provide a solution that calculates the probability directly, choosing one segment at a time?

[quote="kops723"]no, you still misunderstand me. let me give an example:
by your method, clearly we can pick the 4 segments like this(i'll call the points p1, p2... p10):
p2p4, then p7p4, then p2p7, then p1p8

but it doesnt seem to me that your allowing for the fact that that is the same as:
p2p4, p1p8, p7p4, lastly p2p7... in which you are placing the abitrary segment pick in between two other picks[/quote]

Please reread my solution carefully.

We want to find how many ways we can choose 4 segments such that 3 of them make a triangle.  What I am doing is first choosing the triangle.  Then I choose the fourth segment.  By choosing the whole triangle I am in no way, shape, or form considering the order that the segments are chosen.  Neither am I considering whether the triangle or extra line segment was chosen first (I would have to multiply my answer by 2 if that were the case).

Let's first pick the three points of the triangle. There are $\binom{10}{3}$ ways to do this. Then, out of the $\frac{10 \cdot 9}{2} = 45$ segments there are $42$ of them left, so we multiply by $42$. The total amount of ways is $\binom{45}{4}$. Dividing, we get a result of $\frac{16}{473} \implies \boxed{489}$.


function
real analysis
real analysis theorems

Does anyone know why harmonic functions are called "harmonic"? There should be some reason for this name but I couldn't think of any when asked this question two days ago...   :oops:

[url=http://www.mathpages.com/home/kmath214/kmath214.htm]This webpage[/url] explains the name as follows: solving Laplace's equation by separating variables, one obtains the equation of harmonic motion. (The latter is probably named after harmonics in music.)

Why is the harmonic series called harmonic? I don't see how it connects either to harmonic functions or to harmonic oscillator.

Well, the harmonic series most certainly has to do with the music concept of harmonics. The overtones produced by a vibrating string (or any sound wave) are given by $1,\frac{1}{2}, \frac{1}{3}, ...$ of the fundamental wavelength. Thus producing the harmonic series (both mathematically and musically).


search
combinatorics proposed
combinatorics

Prove that an equilateral triangle could not decomposed to more than on equilateral triangles with different size.

it was discussed on forum do the search to find omid

Dear sam-n, I searched but I didn't find anything. (Maybe it because of my bad search). Still I suggest try this challenging problem.


HI!

For the titlepage of my work I need one word to be very large.
One letter should be about 4cm high. I tried to get that done by using \fontsize but
it's not getting taller than aprox. 1.5 cm.

Waht can I do?

Thanks
Bye

\usepackage{scalefnt}

...

\begingroup \scalefont{1000} word\endgroup will make the word as big as it can be. (I think) I've had the same problem...


calculus
integration
trigonometry
calculus computations

Prove that $\sum_{n=1}^{\infty}\int_{\frac{1}{n+1}}^{\frac{1}{n}}{\left|\frac{1}{x}\sin \frac{\pi}{x}\right| dx}$ diverge for $x>0.$


what state is going to win Nationals this year at National Mathcounts?

north carolina?

how are you assuming that california will not win? or florida? or new york?

on the bright side a non-massachusetts resident spelled massachusetts right.

but we all know it's too early to predict these things well, although deep deep down we know massachusetts will obviously win.

Indiana isn't that good this year. People like me might make Nats

uhh...

I think Indiana isn't that bad. Next year I think we have a legitimate shot at winning it all. This year it looks like I'm the only eighth grader [not a guarantee for people that will go OMG you assumed you'd make nats and bug me about it -_-], with an outside shot of jeffery shen and a more likely but still not >50% shot of joseph botros making it.

I think our top 3 is very solid in terms of winnability, just our fourth might be our achilles heel. I'm very interested in finding out who he/she will be, or if one of the "solid 3" will make a slipup at state, which is possible. I'll probably screw up  :rotfl:. More likely though is me getting 2nd -- I have a knack of always getting 1 worse than my goals -- mathcounts 2 years in a row [9th, 5th when i was aiming for 8th, 4th], history day[runner up, aiming for 2nd place who is the other person that makes nationals], and science fair [honorable mention aka 4th, goal was 3rd] -_-.

and indianamath, don't turn into ubemaya 2.0  :P

well, i just said i MIGHT make it. at least i have a better shot than jeffrey.

By "ubemaya 2.0" i meant "don't put yourself down too much"... ubemaya is king of putting himself down. You said "even people like me".

[quote="perfect628"]MA has four people who were at nats last year, all eighth graders now. We have a decent shot.[/quote]

Uh, I thought I had a roommate from Massachusetts that was an 8th grader...I am not so sure that I was ignorant of the fact that he was a 7th grader the whole time.

Yes, but a kid who made it to nats last year from HI now lives in MA

owned

seriously what is it with this north carolina crap? BFG what are you basing your predictions on?

[quote="BFG"][quote="perfect628"]MA has four people who were at nats last year, all eighth graders now. We have a decent shot.[/quote]

You do not know if they can all get into the team or not.

It is too early to have this poll. You are able to do a good prediction ONLY after each team has been formed.

My rough prediction:

CA, WA, MI, and NC have the best shots.[/quote]

Well, I know that, but we'll probably have at least 2-3.

[quote="BFG"][quote="13375P34K43V312"]

BFG what are you basing your predictions on?

[/quote]

Based on your signature. I am truly a member of the Arnav appriciation club. You are the fake one. ARML.[/quote]
Just because they are good at high school math doesn't mean they can do anything at all in MC.
Look at NH (though that's mostly because of exeter I suppose)

im all with the whole

its too early to know.  i mean, we dont even know who will be on nats teams for sure.

[quote="LawOfSigns"]I think it will be California because so many middle schoolers are there.  I too got 5th in district mathcounts last year (4th to 5th by countdown) but I'm going to Florida state this year...any idea what 4th place cutoff is?  (The thread for this topic only shows top 10 cutoff)[/quote]

well, last year it was around 35.  not too high.  but this year you get to play vs bpms and ragnarok and myself.

no offense, but I think this topic is pointless.

States aren't even over.  :D

[quote="bpms"][quote="Walk Around The River"][quote="perfect628"]States aren't even started.[/quote]

i thought states started in the beginning of march...so like a week ago?[/quote]
So, a week from now?[/quote]

what do you mean?  wasnt today like march 12th...?

EDIT:  okay, apparently its still february.  i dont really pay attention to that stuff.

lmfao......march 12th :rotfl:  :rotfl:  :rotfl:

Dude, why is Illinois up there? Take it down, we suck. I only voted for it because I WISH we'll win...

[quote="Fanatic"]
More likely though is me getting 2nd -- I have a knack of always getting 1 worse than my goals -- mathcounts 2 years in a row [9th, 5th when i was aiming for 8th, 4th], history day[runner up, aiming for 2nd place who is the other person that makes nationals], and science fair [honorable mention aka 4th, goal was 3rd] -_-.

[/quote]

Aim for 0th.
 :lol:

Please don't spam :)
Thank you!

[hide]
Someone's gonna say that I'm one to talk :D[/hide]
-Jorian

Really?
I always get my goal times about $\frac{1}{2}$ rankwise.
50-22
10-3
7-4
57-27
1-1
1-2 (the only time I place lower than my goal)

[quote="espark52"]:o  u didn't put west virginia on ur poll list!! i feel insulted...[/quote]
Lets look at who VA has.
They're only returnee got $\frac{2}{3}$ of my score at nats.
Inyoung is a beast, but she can't carry the team alone.

Maybe I'm missing someone, but I doubt it.

FL should be on the poll, as well as NJ.
MD and NC aren't going to win, I don't think.
IL doesn't have any returnees, I doubt they're going to win.
CA should definitely be on the poll, they have Albert Gu and Alexander Lin contending for the team. They're actually the favorites since they have no returnees this year.

YOU DIDN'T PUT NY ON THE LIST! I FEEL INSULTED!

neither did they with Connecticut, but that's not the point

and by the way, they did, it's under the "other" section

has a team not from the states ever done particulary well?

-jorian

This thread has degenerated to a venue for SPAM and dissing your competitors.

Thread locked.


calculus
integration
calculus computations

Evaluate $ \int_{0}^{\infty }\frac {dx}{ ( x \plus{} \sqrt {1 \plus{} x^2 } ) ^ n }$ .

 please help me

Note that we need $ n>1$ in order for the integral to converge, as the integrand looks like a constant times $ x^{\minus{}n}$ for large $ x.$

Make the substitution $ x\equal{}\sinh t,$ from which $ dx\equal{}\cosh t\,dt.$

$ \int_0^{\infty}\frac1{(x\plus{}\sqrt{x^2\plus{}1})^n}\,dx\equal{}\int_0^{\infty}\frac{\cosh t}{(\sinh t\plus{}\cosh t)^n}\,dt\equal{}\int_0^{\infty}\frac{\frac12(e^t\plus{}e^{\minus{}t})}{e^{nt}}\,dt$

$ \equal{}\frac12\int_0^{\infty}e^{\minus{}(n\minus{}1)t}\plus{}e^{\minus{}(n\plus{}1)t}\,dt\equal{}\frac12\left(\frac1{n\minus{}1}\plus{}\frac1{n\plus{}1}\right)\equal{}\frac{n}{n^2\minus{}1}.$

We don't even need for $ n$ to be an integer for this to make sense - just that $ n>1.$

Put $ t \equal{} x \plus{} \sqrt {x^2 \plus{} 1}$ and solving for $ x$ we get $ x \equal{} \frac {1}{2}\left( t \minus{} \frac {1}{t} \right)$ and then $ \frac {1}{2}\int_{1}^{\infty }{\left( t^{ \minus{} n} \plus{} t^{ \minus{} (n \plus{} 2)} \right)\,dt} \equal{} \frac {n}{n^{2} \minus{} 1}.$

Essentially the same solution.


last year i was given nice problem but i failed to solve now i remember it from dusty part of of my memory :)
suppose that we have $f:\mathbb{Z}^2\rightarrow \mathbb{Z}^2$.such that $\frac{|f(x)-f(y)|}{|x-y|}$ is bounded prove that there exists $r$ such that each circle of radius $r$ in the plane contain at least one point of $f(\mathbb{Z}^2)$.

sam-n,first congratulation for the iran team in imo $4$th of the world.
Now $f:\mathbb{Z}^2 \longrightarrow \mathbb{Z}^2$
So what is the meaning of $f(x)-f(y)$,
shouldnet it be $f(x,y)-f(y,x)$ :?:

P.S well done to all of the iran contestantes
Mr.Hatami(By idea Omid Hatami) can you say who did the full mark? :D

yes that's the best result since 1999 and about your question $x$ and $y$ are points in $\mathbb{Z}^2$.

How do you divide two elements of $ \mathbb{Z}^2 $?

Sam_n , If by $f(x)-f(y)$ you means the distances between the $f(x)$ and $f(y)$ then i thing it is wrong  :?

For every points in the plane like $A=(x,y)$ let the $f(A)=B=(0,0)$ and we have that $\frac{f(x)-f(y)}{x-y}$is bounded (it is actually zero ) so if there exist a
$r$ then get away from the $(0,0)$ more than $r$ so the circle wont have any point of $f(\mathbb{Z}^2)$.

Am i saying something stupid? :?

He means bounded both from above and away from zero. In other words, there are two positive constants $c_1,c_2>0$ s.t. $c_1<\frac{|f(x)-f(y)|}{|x-y|}<c_2$ (where $|\cdot|$ is the usual distance in the plane).

oops :blush:  yes i'm really mindless :(  thanx grobber ,i meant the euclidean distance between points.

are you sam nariman the student of dr hesabi school

Hmm, i am not sam to answer this but at least i can ALGEBRA
No , he is not a student , he isa man studing in the university(guess which one :sharif :) ) in math faculty.
Am i right sam ? :lol:


Hmmm....My index is a 230.

AMC 10B 150.0,   AIME 8

But...my name is not on the list!!

(By the way, I took is as an international....in South Korea -_-;;;)

Well, I DO know that international papers will be scored and evaluated later than other

papers, but it still worries me....It has been 4 days since the preliminary list, and my

name isn't on the list yet Hmm.....Any ideas as to what will happen to me??

I know that they wrote "Some late papers are coming in", but it is still not that cheerful to

see my name being left out  -_-;;;Maybe I should send an e-mail on the 14th....

err are you Korean? the USAMO is for residents of the U.S. or for U.S. citizens living abroad only, because it's the qualifying test for the U.S. national team. Countries have their respective national teams, thus a different string of contests for qualification.

meh I'm Korean, and took the AMC in Japan. You were kinda lucky to be able to have taken the AIME at all.. my materials didn't even arrive.

Nice scores, btw :wink:

Dont fret.

I know you might be excited about making the USAMO, but there is no need to worry. I've been in the same situation for three years, and I've always been able to take the test.

[quote] err are you Korean? the USAMO is for residents of the U.S. or for U.S. citizens living abroad only, because it's the qualifying test for the U.S. national team. Countries have their respective national teams, thus a different string of contests for qualification.

meh I'm Korean, and took the AMC in Japan. You were kinda lucky to be able to have taken the AIME at all.. my materials didn't even arrive.

Nice scores, btw  [/quote]

-----I am an American -_-;;  Able to be qualified for the USAMO;
I'm just TAKING it in Korea...hehe...[/quote]

Please see this topic:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198964[/url]


Show that there is no continious function f:R-->R satisfying f[x-f(x)]=x/2, for real x.

Here it is:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=absurd&t=29819[/url]


geometry
geometric transformation
reflection
rectangle
Pythagorean Theorem
AMC

A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=611[/img]

$ \textbf{(A) } 20\pi \qquad \textbf{(B) } 25\pi \qquad \textbf{(C) } 30\pi \qquad \textbf{(D) } 40\pi \qquad \textbf{(E) } 50\pi$

[hide]Each side of the square has length $\sqrt{40}=2\sqrt{10}$.  Draw a radius from the midpoint of the side along the diameter (the center of the circle) to one of the points where the square touches the circle.  By Pythag. Theorem we find the square of the radius is 50, so the circle has area 50*pi, so the semicircle has area 25*pi. B.[/hide]

[hide]
If the area of the square is $40$, then the length of a side is $\sqrt{40}$. We draw an auxiliary line segment from the center of the semicircle to a vertex of the square, forming a right triangle. The center of the semicircle bisects the side of the square on the diameter, so one leg of this triangle has length $\frac{\sqrt{40}}{2}=\sqrt{10}$. The other leg, obviously, has length $\sqrt{40}$. By the Pythagorean Theorem, the length of the hypotenuse is $\sqrt{10+40}=\sqrt{50}$ so the area of the semicircle is $\frac{50 \pi}{2}=25 \pi$.[/hide]

I just reflect the picture in the diameter. Then I have a rectangle with sides $a$ and $2a$ its diagonal is a diameter of a circle. By Pitagorean Theorem:
$d^2=a^2+(2a)^2$. From now the answer is obvious

Wait why was this topic revisited after one month?

I have no idea. But it was the Problem of the Day on March 17, even though it was posted on February 17. :?