[quote="Interval"]Question: The rules of a certain game allow a team to score either 3 points
or 8 points.  A team's final score will be any combination of these points.
For example,  14=3+3+8 and 27=8+8+8+3  Which numbers cannot be a team's
final score?[/quote]
[hide]First by using Chicken Mgnuggets formula the highest you can't get is 13. So keep ckecking below 13 to see that the numbers that can't work are 1,2,4,5,7,10,and 13.[/hide]

M[b]c[/b]Nuggets. McNuggets.

Search it on google if you're not familiar with it.

Chicken McNugget Theorem:

The largest integer that can not be expressed as a sum of x's and y's is $xy-x-y$, if x and y are relatively prime.

Whoah, so if we were to use that theorem on a contest it would be o.k. to call it the Chicken McNugget theorem? Is there some "proper" way of saying it? (Kind of like the 'star trek' theorem and similar idiosyncratic theorem names).

It's called the Chicken McNugget Theorem because its actual statement is something like

"If there are infinitely many boxes of $n$ chicken McNuggets and infinitely many boxes of $m$ chicken McNuggets, where $n$ and $m$ are relatively prime, the largest finite amount of chicken McNuggets that cannot possibly be attaied is $mn-m-n$."

Which is why it's called the chicken McNugget Theorem.[

[quote="bpms"][quote="mysmartmouth"]It's kind of like an AoPS name, with a little more background (kinda like Simon's Favorite Factoring Trick).  If you use it on a proof, I wouldn't recommend calling it the McNugget theorem.  But I don't think it really has a real name otherwise.[/quote]
No it's not just the aops name. It's on wolfram I think.[/quote]

according to people from my school last year, the first contest problem that ever asked that problem was asking about chicken mcnuggets, so thats where it came from.

then again, they might have lied to me   :|


And just as you think you understand it all in this chapter you encounter another problem.

Ok, i have this fraction which i first of rewrite, from which i must find the numbers that can not be used (can't remember the english word for it, i hope you get it anyway once you see the fraction):

$\frac{1}{x}+\frac{5}{x-2}=0$

$\frac{6x-2}{x(x-2)}=0$

If i want to find the numbers that i can't used for this fraction i have to find the numbers for x that makes the denominator equal 0 as you can't divide with 0. But on the other hand you used the denominator when you want to find the limits of the fraction, where you will see what numbers makes either the nominator and the denominator equal 0.

$x=0$

and

$x-2=0$

$x=2$

So these numbers are both used as the limits for the denominator and the numbers which can't be used due to the law that you can't divide by 0.

I believe i have misunderstood something about the law of dividing with 0, so can anyone guide me, plz?

Might have been the wrong word. 'Where does the function intersect with 0', that might have been a better word. But that was not really the problem, i can find the intersections. But you know if you have this fraction:

$\frac{2}{x^2-1}$

Then x can't be +1 as it would make the denominator 0. But unless there is something i've misunderstood both +1 and -1 will intersect with 0. So we have kind of a problem with +1 one since you can not use +1 but on the other hand +1 does intersect 0.

Hope you get me now.

x=1 does not intersect with zero. It makes the denominator 0, with causes you to divide by zero, which is not allowed.

So in this equation:

$\frac{6x-2}{x(x-2)}=0$

... you would say that the only intersections is:

$6x-2=0$

$6x=2$

$x=\frac{1}{3}$

Is this right?


The Asdf family wants to go from city A to city B. they can go via 2 routes, one which is 40 miles longer than the other. They shorter route requires a $ \$ 2.50$ toll. Gas currently costs $ \$2.50$ per gallon.(i really hope we can get there soon). Their car runs on 40 miles per gallon(i also hope that we get there pretty soon). If the cost of going on either route is the same, what is the length of the shorter route. express your answer in miles.

test \$

Solvable?

The shorter route has $ \$2.50$ more in tolls; the longer route has $ \$2.50$ more in gas

40mpg is probable, $ \$$2.50 gas is not

There is some typo in the problem or something.

Blah guess and check FTW!
First number: 1337

LOng route: 2.5(1337+40)/40=86.0625

Short route: 2.5(1337)/40+2.5=83.5625+2.5=86.0625

Thus the shorter route is 1337 miles.

Blah guess and check FTW!
Second number: 42

LOng route: 2.5(42+40)/40=5.125

Short route: 2.5(42)/40+2.5=2.625+2.5=5.125

Thus the shorter route is 42 miles.

I can use your argument too.

Plug in $ x$ as your amount of miles...and apparently it all works out.....

Me?  No...

Sorry for posting if you're talking to abacadaea.


induction
combinatorics unsolved
combinatorics

In a certain language, EH and HH are the only two-letter words, and longer words are formed by replacing all of the Hs in a word with words.  For example, we can replace the H in EH by HH to get EHH.  Then, can replace the first and second Hs in EHH by EH and EHH, respectively, to get EEHEHH.

Prove that the number of $n$-letter words is
\[ \frac{2^n + 2(-1)^n}{3}. \]

[EDIT: This was a short version of the solutin below, but it was too short to be understandable.]

Could you post your solution?  I don't really understand.

I'll just translate author's solution, in which A=E and B=H.
Each finite sequence of A, B we'll call a  [b]string[/b]. In the folowing $\dots$ stands for a suitible (may be void) string and $***$ stands for string consisting of the same lettes, $*x*$ stands for string consisting of $x$ the same letters.
Let|s prove that arbitrary string is a word (of considered language), IFF it fulfils following condition:
P: string ending with B either begins with A or begins (or is even made of) even number of B's.
Neccesarity of P is quite obvious: it's fulfiled by both AB and BB of lenght  2 and is fulfiled by any word constructed according to the task, if P is fulfiled by words, in which the B's are replaced.
Now by induction to n we'll prove that each string of lenght n fulfils P is a word. It's obvious for n=1 and n=2. If n>2, the sring of lenght n fulfiling P is in one of the forms
$AA\dots B$,$AB\dots B$,$B*2k-2*BA\dots B$, $B*2k*B$, where $2\leq 2k \leq n-2$. Our task is to prove that these four can be constructed according to the task.
The word $AA\dots B$ can be constructed as $A(A\dots B)$ from the word AB. Word $AB\dots B$ can be constructed as $A(B\dots B)$ from the word AB or $(AB)(\dots B)$ from the word BB, which depends on the parity of number of B's after A. The word $B***BA\dots B$ can be constructed as
$(B***B)(A\dots B)$ from the word BB, the word $B*2k*B$ as $(B***B)(BB)$ from BB. So the prove by induction is complete.
Now using the condition P we'll prove that the number $p_n$ of words of lenght n realy fulfils the formula
$p_n=\frac{2^n+2(-1)^n} 3$,
which is obviously true   for $n=1$ and $n=2$.  Thus the formula will be proved by inductin, if we prove, that for each n the equality $p_{n+2}=2^n+p_n$, which is obviously fulfiled by numbers
$\frac 1 3 \left(2^n+2(-1)^n\right)$. The equality $p_{n+2}=2^n+p_n$ follows imidiately from the fact, that according to condition P a word of lenght n+2 is either of the form $A\dots B$, where $\dots$ is arbitrary string of lenght n (there are $2^n$ such strings), or of the form $BB\dots$, where $\dots$ is arbitrary word of lenght n.

My longest AoPS post ever is complete, I hope you can understand it, although my English is horrible.


In triangle ABC, BAC = 120. Let the angle bisectors of angles A;B and C meet the opposite sides in D;E and F respectively.
Prove that the circle on diameter EF passes through D.

According to http://www.mathlinks.ro/Forum/viewtopic.php?t=1336 , the condition < BAC = 120° implies < EDF = 90°. Thus, the circle with diameter EF passes through the point D.

  Darij


advanced fields
advanced fields unsolved

Let $G, H$ be two connected graphs, such that $L(G)=L(H)$, where $L(G)$ indicates the line graph of $G$.Up to isomorphism, prove or disprove that $G=H$.

One can find a counterexample for the question, for instance let $G$ be a triangle and $H$ be a star. But can anyone prove it when $G, H$ have exactly one cycle?


complex numbers
geometry solved
geometry

ABCDE is a convex pentagon. A' is chosen so that B is the midpoint of AA', B' is chosen so that C is the midpoint of BB' and so on. Given A', B', C', D', E', how do we construct ABCDE using ruler and compasses?

We could simply write down the equations corresponding to the complex numbers $a,b,c,d,e$ and $a',b',c',d',e'$. These are of the form $2b-a=a'$ and so on. We solve this linear system with five equations and get $a,b,c,d,e$ in terms of $a',b',c',d',e'$, and after that it's easy. This has nothing to do with the fact that the polygon is a regular pentagon, of course :).


Is there a formula for $\cos\left(\frac{\theta}{3}\right)$ in terms of $\cos(\theta)$?
or
more generally $\cos\left(\frac{\theta}{n}\right)$ for $n \in \mathbb{N}$.

There is a formula for $\cos n\theta$ in terms of $\cos \theta$, where "in terms of" means "as a polynomial of degree $n$."  So the "formula" for $\cos \frac\theta n$ is equivalent to solving some degree-n polynomial.


[quote="10000th User"]:lol: This question appeared on the 1st MIMC[/quote]
Sorry, I didn't realize that.  Thanks for the links chess64.


Please, help me!
Let $ \frac {\pi} {6} \le x_1 , x_2 , x_3 , x_4 \le \frac {\pi} {4}$. prove that:
\[ (cotx_1 \plus{} cotx_2 \plus{} cotx_3 \plus{} cotx_4)(\frac {1} {cotx_1} \plus{} \frac {1} {cotx_2} \plus{} \frac {1} {cotx_3} \plus{} \frac {1} {cotx_4}) \le \frac {4(\sqrt {3} \plus{} 1)^2} {\sqrt {3}}\]

Let $ y_i\equal{}\cot x_i$. So $ 1\le y_i\le\sqrt{3}$
To maximize $ f(x)\equal{}(x\plus{}A)(\frac1x\plus{}B)$ with positive $ A$ and $ B$ for $ x\in[1,\sqrt{3}]$, we can easily get $ x\equal{}1$ or $ x\equal{}\sqrt{3}$.
Hence, we have
$ \sum y_i\sum\frac1{y_i}\le \max_{m\plus{}n\equal{}4}(m\plus{}n\sqrt{3})(m\plus{}\frac n{\sqrt{3}})$.
It is not hard to check the RHS is maximized when $ m\equal{}n\equal{}2$.

[quote="xxp2000"]Let $ y_i \equal{} \cot x_i$. So $ 1\le y_i\le\sqrt {3}$
To maximize $ f(x) \equal{} (x \plus{} A)(\frac1x \plus{} B)$ with positive $ A$ and $ B$ for $ x\in[1,\sqrt {3}]$, we can easily get $ x \equal{} 1$ or $ x \equal{} \sqrt {3}$.
Hence, we have
$ \sum y_i\sum\frac1{y_i}\le \max_{m \plus{} n \equal{} 4}(m \plus{} n\sqrt {3})(m \plus{} \frac n{\sqrt {3}})$.
It is not hard to check the RHS is maximized when $ m \equal{} n \equal{} 2$.[/quote]
Thank you very much!


calculus
derivative
function
trigonometry
calculus computations

Hi , I have the following math problem I'm studying :

f(x) = (cos4x) / (1 - sin4x)

I need to find the derivative of that function .

So far , I have this ;

f'(x) = f(x)g'(x) - g(x)f'(x) / (g(x))²
f'(x) = (cos4x)(0 - cos4x)(4) - (1 - sin4x)(-sin4x)(4) / (1-sin4x)²
f'(x) = (cos4x)(-4cos4x) + (1-sin4x)(4sin4x) / (1-sin4x)²

is this correct ?
can i simplify it further ?

thank you for your help .

$f(x) = \frac{\cos4x} {1 - \sin4x}$

You are can kind of double using your functions.

$\frac{df(x)}{dx} = \frac{(-4\sin 4x)(1-\sin 4x) - (-4 \cos 4x)(\cos 4x)}{(1-\sin4x)^2} = \frac{-4 \sin 4x + 4}{(1-\sin 4x)^2} = \frac{4}{1-\sin 4x}$

Thus, that is correct and yes you can simplify it further.

thank you for your help .
i appreciate it .

[i]We make a mockery of things ![/i]

[b]The first method.[/b] $g(t)=\frac{\cos t}{1-\sin t}\Longrightarrow g(t)\cdot (1-\sin t)=\cos t\Longrightarrow$

$g'(t)\cdot (1-sin t)-g(t)\cdot \cos t =-\sin t\Longrightarrow$

$g'(t)\cdot (1-\sin t)-\frac{\cos^2t}{1-\sin t}+\sin t=0\Longrightarrow$

$g'(t)\cdot (1-\sin t)-\frac{(1-\sin^2 t)}{1-\sin t}+\sin t=0\Longrightarrow$

$g'(t)\cdot (1-\sin t) -(1+\sin t)+\sin t=0\Longrightarrow\boxed {\ g'(t)=\frac{1}{1-\sin t}\ }$

[u]Remark.[/u] $g(t)=\frac{1+\sin t}{\cos t}\Longrightarrow$$g'(t)=\frac{\cos^2t+\sin t(1+\sin t)}{\cos^2t}=\frac{1+\sin t}{1-\sin^2t}=\frac{1}{1-sin t}.$

Therefore, $f(x)=g(4x)\Longrightarrow f'(x)=g'(4x)\cdot (4x)'\Longrightarrow \boxed {\ f'(x)=\frac{4}{1-\sin 4x}\ }$

[b]The second method.[/b] $f(x)=\frac{\cos^2 2x-\sin^2 2x}{(\cos 2x-\sin 2x)^2}=\frac{\cos 2x+\sin 2x}{\cos 2x-\sin 2x}=\frac{1+\tan 2x}{1-\tan 2x}=$

$\tan \left(\frac{\pi}{4}+2x\right)\Longrightarrow f'(x)=\frac{2}{\cos^2 \left( \frac{\pi}{4}+2x\right)}=\frac{4}{1+\cos \left(\frac{\pi}{2}+4x\right)}=\frac{4}{1-\sin 4x}.$


Evaluate:
\[\dbinom{n}{0}+\dbinom{n}{6}+\dbinom{n}{12}+\ldots+\dbinom{n}{6k}+\ldots \]

(EDIT: ignore this edit. :D )

I don't know what your hint means, but [hide]roots of unity filter[/hide] should work.

[quote="cosinator"]how do you get that, altheman?[/quote]
[hide="The way I do it:"]
You can show, for the six solutions to $x^{6}=1$, that $\sum x^{n}$ for positive integers $n$ is equal to zero except when $6 | n$. (how?) Then, just see that $1, w^{1}, \ w^{2}, \ \ldots, \ w^{5}$ are a rearrangement of those six roots.[/hide]

You may want to refer to the description in the following topic:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=678662#678662[/url]

Especially the post by paladin8


Express in scientific notation the number that is one-tenth as large as $ 2 \times 10^6$.


geometry
vector
analytic geometry
integration
calculus

The shape of an insulator of charge $q$ is defined as the area between $y =12x^{2}$ and $y=12x^{3}$.  Find the field(as a vector) for any given point $(x,y)$ on the coordinate plane.

For example the electric field at any point $(x,y)$ of a point charge of charge $q$ is $\frac{kqx}{(x^{2}+y^{2})^{3/2}}i+\frac{kqy}{(x^{2}+y^{2})^{3/2}}j$

Just wondering: is there an elegant solution to the problem?  :o The only solution I see is a brute-force onslaught...

what about using superposion for all of the points?

$\int_{0}^{\infty}\int_{12x^{2}}^{12x^{3}}\frac{kqx}{(x^{2}+y^{2})^{3/2}}i+\frac{kqy}{(x^{2}+y^{2})^{3/2}}j dy dx+\int_{-\infty}^{0}\int_{12x^{3}}^{12x^{2}}\frac{kqx}{(x^{2}+y^{2})^{3/2}}i+\frac{kqy}{(x^{2}+y^{2})^{3/2}}j dy dx$

I'm not sure if thats what you meant by brute force approach, but i dont see any other way to doing it

Yes, that's the brute force I meant. Are these integrals analytically doable?

the area is infinite and the charge is finite so the charge per area is zero and our electric field is also zero.

if we changed q to charge per area then our electric field at point p would look like this: (my above equation is dumb)

$kq\int_{0}^{\infty}\int_{12x^{2}}^{12x^{3}}[\frac{1}{[(x-p_{x})^{2}+(y-p_{y})^{2}]^{3/2}}(x-p_{x})i+\frac{1}{[(x-p_{x})^{2}+(y-p_{y})^{2}]^{3/2}}(y-p_{y})j]dydx+kq\int_{-\infty}^{0}\int_{12x^{3}}^{12x^{2}}[\frac{1}{[(x-p_{x})^{2}+(y-p_{y})^{2}]^{3/2}}(x-p_{x})i+\frac{1}{[(x-p_{x})^{2}+(y-p_{y})^{2}]^{3/2}}(y-p_{y})j]dydx-k\frac k{q}{(p_{x}+p_{y})^{3/2}}(p_{x}i+p_{y}j)$

now are these integral solvable, that is a very good question


I have the 1984 nationals sprint all cool in $\LaTeX$ but there not in that traditional MATHCOUNTS format. (you know, boxes around the edge, blanks on the side)
Is there a way to do it?
Help is appreciated

Yes, and it doesn't involve Latex (I don't think).  I have a word template, but it's old and might need a BIT of adjusting!  Feel free to delete all credits at the end!


geometry
combinatorics theorems
combinatorics

Let P be a lattice polygon. Assume there are I(P) lattice points in the interior of P, and B(P) lattice points on its boundary. Let A(P) denote the area of A. Then
A(P) = I(P) + B(P)/2 - 1

If anyone wants to see a proof, it can be found on [url=http://www.cut-the-knot.org/ctk/Pick_proof.shtml]Cut-the-knot[/url], [url=http://en.wikipedia.org/wiki/Pick_theorem]Wikipedia[/url] etc.

[hide="P. S."]Why does this happen:

[url=http://en.wikipedia.org/wiki/Pick's_theorem]Wikipedia[/url] etc. (I know it is becuse of the ', but how can it be corrected?)[/hide]

Hei, this is old. But it is very useful.


geometry
circumcircle
angle bisector
geometry unsolved

Let $ABC$ be a triangle inscribed in circle $R$ .Also let the angle bisector line of $A,B,C$ intersect the circumference of circle $R$ at$A',B,'C'$ respectively . Let $AA' \cap BC=N$, $C'A'\cap BB'=P$. Also denote the orthocenter as $I$ . Now , let $O$ be the circumcenter of triangle $IPC'$ , and $OP\cap BC=M$. If $BM = MN$ , and $\angle BAC = 2\angle ABC$ , find all the angle $\angle A ,\angle B,\angle C$ in the triangle $ABC$

Well , i post this question before in Pre-Olympiad but still un-solved . So I will post it here for others to solve .Thanks .


John and Jim are playing a game.  At the beginning of the game, there is a pile of 76 pennies.  Whoever takes the last penny loses and the other player wins.  On each turn they take at least one penny but no more than seven pennies.  If John goes first, how many pennies should he take on his first turn to ensure that he will win.

First, John should take 3, leaving the pile at 73. Every time after, when Jim picks $ x$ pennies, John should take $ 8\minus{}x$ pennies, reducing the pile by 8's until Jim is left with a pile of 1 and loses.

bruce, use hides

[hide]

He should take 3 pennies

it would leave it at 73

Each time, John would take 8-Jim. So it would go down in 8s.

to 65, 57, 49, 41, 33, 25, 17, 9, then Jim is left with 1 and he loses.[/hide]

[hide]

He should take 3 since then there would be 73 left and then he make up for the numder of pennies that Jim did not take to make it 8. So at the end when it is Jim's turn he takes the last one.

[/hide]

[hide]He should take 3 because if he takes 3, there would be 73 which is 1 more than a multiple of 8. Each turn, he takes 8-Jim.[/hide]

[quote="the future"]So John will be guaranteed of a win no matter what the starting pile is unless its  1 mod 8 right?[/quote]

well we can have 0 mod 8-then he can take 7 to create 1 mod 8 to win
1 mod 8-he cant win beucase he cant take anything from 1-7 to make it 1 mod 8
2 mod 8-take 1
3mod 8-take 2...

so forth take n-1 for n mod 8 where n<8.

So yea for 1 mod 8 you cant win so there is a 7/8 chance total for John to win on any pile.

[quote="Ihatepie"]No, unless it is 0 mod 8. If it is any other number, Bruce will win every time.[/quote]
uhh no..because something like 72, John can take 7 to being with to make it 65.

No matter what from there, John will be second and will just match Jim's taking by taking a total of 8 every time (so he will take 8-Jim), so at the end Jimwill take the final 1 since hes first now...after 65-57-49...-1


A circle K just fits (tangent to exactly three sides) inside a 6x4 rectangle. What is the radius of the largest circle which can fit in the rectangle but is completely outside K?

[hide="Solution"]
Draw in the circle. I'll assume the 6 unit side is horizontal.

Connect centers of the circles, and draw vertical lines from the centres of the circles to the sides. Draw a horizontal line to view a right triangle.

$ (4-r)^{2}+(2-r)^{2}= (2+r)^{2}\implies r = 4(2-\sqrt{3})$
[/hide]


let O is the center of circumcircle of ABC triangle.M is a point on AB arc.K & N are 2 point on AB such that $ MK \bot OA and MN \bot OB$.
(MK,AC)=L and (MN,BC)=P.
what is <MLP ?  :)

[quote="stvs_f"]
what is <MLP ?  :)[/quote]

What do you mean with this ? the measure ?


trigonometry
inequalities
geometric inequality
inequalities unsolved

Let $X,Y,Z$ points on sides $BC,CA,AB$ of a triangle.

Prove that

$\frac{(AX)^2}{(BX)(CX)}+\frac{(BY)^2}{(CY)(AY)}+\frac{(CZ)^2}{(AZ)(BZ)} \geq 9$

This one is easy.

[b]Lemma 1.[/b] We have

$\frac{AX^2}{BX\cdot CX} \geq \frac{\sin B \cdot \sin C}{\sin^2 \frac{A}{2}}$.

[i]Proof.[/i] By the Sine Law in triangles BAX and CAX, we have

$\frac{AX^2}{BX\cdot CX}=\frac{AX}{BX}\cdot\frac{AX}{CX} = \frac{\sin\measuredangle ABX}{\sin\measuredangle BAX}\cdot \frac{\sin\measuredangle ACX}{\sin\measuredangle CAX} = \frac{\sin\measuredangle ABX \cdot \sin\measuredangle ACX}{\sin\measuredangle BAX \cdot \sin\measuredangle CAX} = \frac{\sin B \cdot \sin C}{\sin\measuredangle BAX \cdot \sin\measuredangle CAX}$.

Let $t = \measuredangle BAX - \measuredangle CAX$. Then, since $A = \measuredangle BAX + \measuredangle CAX$, we have $\measuredangle BAX = \frac{A+t}{2}$ and $\measuredangle CAX = \frac{A-t}{2}$. Denoting $p=\frac{A}{2}$ and $q=\frac{t}{2}$, this becomes $\measuredangle BAX = p+q$ and $\measuredangle CAX = p-q$. Thus,

$\sin\measuredangle BAX \cdot \sin\measuredangle CAX = \sin\left(p+q\right) \cdot \sin\left(p-q\right)$
$= \left(\sin p \cos q + \cos p \sin q \right) \cdot \left(\sin p \cos q - \cos p \sin q \right)$
$= \left(\sin p \cos q\right)^2-\left(\cos p \sin q\right)^2 \leq \left(\sin p \cos q\right)^2 = \sin^2 p \cos^2 q$.

Since $\cos^2 q \leq 1$ (as $-1\leq \cos q \leq 1$), we thus have

$\sin\measuredangle BAX \cdot \sin\measuredangle CAX \leq \sin^2 p = \sin^2 \frac{A}{2}$,

and thus

$\frac{AX^2}{BX\cdot CX}=\frac{\sin B \cdot \sin C}{\sin\measuredangle BAX \cdot \sin\measuredangle CAX} \geq \frac{\sin B \cdot \sin C}{\sin^2 \frac{A}{2}}$.

This proves Lemma 1.

By Lemma 1,

$\frac{AX^2}{BX\cdot CX} \geq \frac{\sin B \cdot \sin C}{\sin^2 \frac{A}{2}}$.

Similarly,

$\frac{BY^2}{CY\cdot AY} \geq \frac{\sin C \cdot \sin A}{\sin^2 \frac{B}{2}}$;
$\frac{CZ^2}{AZ\cdot BZ} \geq \frac{\sin A \cdot \sin B}{\sin^2 \frac{C}{2}}$.

Thus, in order to prove

$\frac{AX^2}{BX\cdot CX} + \frac{BY^2}{CY\cdot AY} + \frac{CZ^2}{AZ\cdot BZ} \geq 9$,

it is enough to show

$\frac{\sin B \cdot \sin C}{\sin^2 \frac{A}{2}} + \frac{\sin C \cdot \sin A}{\sin^2 \frac{B}{2}} + \frac{\sin A \cdot \sin B}{\sin^2 \frac{C}{2}} \geq 9$.

Now, if $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC, and x = s - a, y = s - b, z = s - c, then a = y + z, b = z + x and c = x + y and s = x + y + z. Moreover,

$\frac{\sin B \cdot \sin C}{\sin^2 \frac{A}{2}} = \frac{\sin B}{\sin A} \cdot \frac{\sin C}{\sin A} \cdot \frac{\sin^2 A}{\sin^2 \frac{A}{2}} = \frac{b}{a} \cdot \frac{c}{a} \cdot \frac{\left(2\sin\frac{A}{2} \cos\frac{A}{2}\right)^2}{\sin^2 \frac{A}{2}}$
$= \frac{bc}{a^2} \cdot 4\cos^2 \frac{A}{2} = \frac{bc}{a^2} \cdot 4\frac{s\left(s-a\right)}{bc} = \frac{4s\left(s-a\right)}{a^2} = \frac{4\left(x+y+z\right)x}{\left(y+z\right)^2}$.

Similarly,

$\frac{\sin C \cdot \sin A}{\sin^2 \frac{B}{2}} = \frac{4\left(x+y+z\right)y}{\left(z+x\right)^2}$;
$\frac{\sin A \cdot \sin B}{\sin^2 \frac{C}{2}} = \frac{4\left(x+y+z\right)z}{\left(x+y\right)^2}$.

Hence, it remains to prove

$\frac{4\left(x+y+z\right)x}{\left(y+z\right)^2} + \frac{4\left(x+y+z\right)y}{\left(z+x\right)^2} + \frac{4\left(x+y+z\right)z}{\left(x+y\right)^2} \geq 9$,

or, equivalently,

$\frac{x}{\left(y+z\right)^2} + \frac{y}{\left(z+x\right)^2} + \frac{z}{\left(x+y\right)^2} \geq \frac{9}{4\left(x+y+z\right)}$.

This was shown on http://www.mathlinks.ro/Forum/viewtopic.php?t=5084 ; equivalently, it is problem 7 in "Old and New Inequalities". By the way, I found this inequality by proving a geometric inequality equivalent to the one you posted, Manlio: it's problem 7 at http://www.mathlinks.ro/Forum/viewtopic.php?t=5178 .

  Darij

Hope I understand your problem right.
By Stewart therem we have:

AX2/BX*CX =  b2/BX*a +c2/CX*a - 1

So we need to prove:
∑ (b2/BX*a +c2/CX*a) ≥ 12.

Using  AM-GM inequality for 6 numbers it is enough to prove:

(a2b2c2/BX*CX*BY*CY*AZ*BZ)^1/6 ≥ 2

or ((a2/BX*CX)*(b2/AY*CY)*( c2/AZ*BZ)) ≥ 64.

But BX*CX ≤ a2/4 by AM-GM
so a2/BX*CX ≥ 4.
Putting this one we get the desired inequality. q.e.d.

Thank you very much. Now I will study them.


superior algebra
superior algebra unsolved

let G be nonempty subset with associative binary operation such that corresponding to each g in G ,there exists unique g` such that gg`g=g.
show G is a group.

Well I've seen this problem before. So I hide the hint (one can even call it the whole solution :wink: ) to let other guys give other solutions, if possible. I myself don't know any other approach.
[hide="Hint"][b]i)[/b]If $e$ is an idempotent in $G$ then $e^{*}=e$.
[b]ii)[/b]If $a^{*}x=a^{*}\; , x, a \in G$, then $x=aa^{*}$.
[b]iii)[/b]For all $a\in G, \; a^{*}aa^{*}=a^{*}$ and $a^{**}=a$.
[b]iv)[/b]For all $a,b\in G,\; x=(ba^{*})^{*}$ is a solution of $xb=a$.[/hide]

Here is another approach.

All formulas are meant to be valid for all occuring variables ranging over $G$.

Assumptions:
$E(g)$. $gg'g = g$. (existence)
$U(g,h)$. $ghg = g \rightarrow g' = h$. (uniqueness)

Intermediate steps:
Lemma 1. $(xx')' = xx'$. (uses $U(xx',xx')$)
Lemma 2. $(xy)' xx' = (xy)'$. (uses $U(xy,(xy)'xx')$)
Lemma 3. $xy (xy)' = xx'$. (uses $U(xy(xy)',xx')$, $L1(xy)$)
Lemma 4. $y xx' = (x(yx)')' = y$. (uses $U(x(yx)',yxx')$, $U(x(yx)',y)$, and L2, L3 with $x,y$ swapped)
Lemma 5. $xx' = yy'$. (uses $U(xx',yy')$, L4, L1)
Assumption E is used for $g=x$ and $g=xy$.

Conclusion:
$e : = xx'$ is a constant with $g = ge$ (by L4) and $gg' = e$ (by definition) for all $g \in G$.

LiMa


geometry
3D geometry
sphere
tetrahedron
vector
analytic geometry
geometry unsolved

$S$ is a sphere center $O. G$ and $G'$ are two perpendicular great circles on $S$. Take $A, B, C$ on $G$ and $D$ on $G'$ such that the altitudes of the tetrahedron $ABCD$ intersect at a point. Find the locus of the intersection.

Let $O$ be the origin and the $z$-axis is perpendicular to $G$.   Notice that
\[\vec {A}\cdot \vec {A}=\vec {B}\cdot \vec {B}=\vec {C}\cdot \vec {C}=\vec {D}\cdot \vec {D}\]
and that the orthocenter $H$ of $\triangle ABC$ is
\[\vec{A}+\vec{B}+\vec{C}\]
It is well-know that if $ABCD$ has an orthocenter then the foot of $D$ to $\triangle ABC$ coincides with the orthocenter of $\triangle ABC$.  Thus,
\[\vec{D}=\vec{A}+\vec{B}+\vec{C}+\vec{Z}\]
where $\vec{Z}$ is a vector that is parallel to the $z$-axis.  We also have that the orthocenter, $K$ of $ABCD$ is
\[\vec{A}+\vec{B}+\vec{C}+\vec{Z'}\]
wher $\vec{Z'}$ is another vector that is parallel to the $z$-axis.  By definintion of the orthocenter, $\vec{AK}$ is perpendicular to $\triangle BCD$ so $\vec{AK}$ is perpendicular to $\vec{BD}$.  Thus,
\begin{align*}
0&=\vec{AK}\cdot \vec{BD}\\
&=(\vec K-\vec A)\cdot (\vec D-\vec B)\\
&=(\vec B+\vec C+\vec {Z'})\cdot (\vec A+\vec C+\vec Z)\\
&=(\vec B+\vec C)\cdot (\vec A+\vec C)+\vec {Z'}\cdot \vec Z\\
&=\frac{1}{2}(|\vec{A}+\vec{B}+\vec{C}|^2-|\vec{A}|^2)+|\vec {Z'}||\vec Z|\\
|\vec {Z'}|&=\frac{\frac{1}{2}(|\vec Z|^2)}{\vec Z}\\
&=\frac{1}{2}{|\vec Z|}\end{align*}
Thus, the locus of $K$ is a subset the curve formed by reducing the $z$-coordinate of each possible $D$ in half.  Since we can construct for each $D$, a triangle $ABC$ such that the foot of $D$ to $\triangle ABC$ is the orthocenter of $ABC$, and that is can be proven that $ABCD$ does have an orthocenter, the locus of $K$ is this curve.

The locus is a concentric sphere with the same center O and radius R/4 where R is the radius of the great circles.


function calculus trigonometry algebra domain

Is there anyway to determine if an equation is an equation of a function without graphing it and applying the vertical line test, or guessing numbers to check?

I am just curious. If anyone knows any method to determine this, please post.

First of all, you need to make sure that the denominator (if there is one) is never zero. In other words, make sure there aren't undefined points. Also, make sure that there aren't more than 1 y value for an x-value.

And... you can always use some basic calculus if things aren't that obvious.  :)

like $x^2+y^2=1$ is not a function because for $x=0$ there are 2 values for $y$, $-1$ and $1$
but $yx=1$ is a function because if you choose $x$ there is at most 1 values for $y$, if there is one it is $y=\frac1{x}$
is that what you're talking about?

[quote="Altheman"]you isolate y, and if there is a way that there could be 2 y values for 1 x value, then it is not a function (like if you have a plus or minus)[/quote]
If you isolate y and there is a plus/minus in the expression of x this means that the function isn't injective, but it can still be a function.

[quote="jli"]is

$f(x)=$
$\sin{\frac{1}{x}}$ for $x\ne 0$
$0$ for $x=0$

a funtion?[/quote]
yes, it is.

[quote="4everwise"]But it isn't a continuous function.  :)[/quote]
Yes it is. It's not continuous over all real numbers, but it's continuous in all points in its domain, and so it is a continuous function. Generally all products/sums/quotients/compositions of continuous functions are continuous ($f(x) = 1$ is continuous $g(x) = x$ is continuous and $\sin$ is continuous, and so $h(x) = \sin 1/x$ is continuous) .

[quote="4everwise"]But it isn't a continuous function.  :)[/quote]
And you can't plot it clearly.


[quote="tiredepartment"]In the coordinate plane, the endpoints of a chord of a circle have coordinates $(-1-2\sqrt{3},4)$ and $(-1+2\sqrt{3},4)$.  If the degree measure of the minor arc determined by this chord is $120^{o}$, determine the corodinates of all possible locations for the center of this circle.[/quote]
How would you approach this problem?  Help anyone?

Thanks.

We see that the distance between the points given is $2\sqrt{3} - (-2\sqrt{3}) = 4\sqrt{3}$.  So a $120-30-30$ triangle forms, with sides $4,4,$ and $4\sqrt{3}$ (this can easily be confirmed by separating the triangle into two $30-60-90$ triangles).  Thus, the altitude, or the distance from the center of the circle to the midpoint of the chord, is $2$.

Two possible coordinates, then, would be $\boxed{(-1,6) \ \text{and} \ (-1,2)}$.


geometry
rectangle
circumcircle
perpendicular bisector
geometry proposed

In a rectangle $ABCD$ be a rectangle and $BC = 2AB$, let $E$ be the midpoint of $BC$ and $P$ an arbitrary inner point of $AD$. Let $F$ and $G$ be the feet of perpendiculars drawn correspondingly from $A$ to $BP$ and from $D$ to $CP$. Prove that the points $E,F,P,G$ are concyclic.

From the right angle triangle $\triangle PAB$ with the altitude AF, we get $BF \cdot BP = BA^2 = BE^2$, which means that the line BC is a tangent to the circumcircle (O) of the triangle $\triangle EFP$ at the vertex E, i.e., this circumcircle is centerd on the midline of the rectangle ABCD through the midpoint E of BC. Likewise, from the right angle triangle $\triangle PDC$ with the altitude DG, we get $CG \cdot CP = CD^2 = CE^2$, which again means that the line BC is a tangent to the circumcircle (O') of the triangle $\triangle EGP$ at the vertex E, i.e., this circumcircle is also centerd on the midline of the rectangle ABCD through the midpoint E of BC. But since all circles passing through the points E, P are centered on the perpendicular bisector of the segment EP, only one such circle exists, which is also centered on the rectangle midline through E. Thus the circumcircles $(O) \equiv (O')$ of the triangles $\triangle EFP, \triangle EGP$ are identical. In other words, the quadrilateral EFPG is cyclic.

[i]Yetti, [/i]I think that you are not angry cross me ! I continue the your solution differently: from the relations $BE^2=BF\cdot BP,\ CE^2=CG\cdot CP$ results $\triangle BEF\sim \triangle BPE,\ \triangle CEG\sim \triangle CPE\Longrightarrow$
$\widehat {BEF}\equiv \widehat {BPE},\ \widehat {CEG}\equiv \widehat {CPE}\Longrightarrow m(\widehat {FEG})+m(\widehat {FPG})=180^{\circ}.$


[quote="mihail911"]i don't understand this proof...iv been slaving over it for several days but im not good enough at number theory (this is a proof after a number theory chapter) to find the solution.

[b]Denote by $ p.k$  the kth prime number. show that $ p1p2...pn+1$ cannot be the perfect sqare of an integer[/b]

please give me assistance on how to approach this problem, not the answer!

thanks.. :)[/quote]

uh relatively prime to all the primes.. (uh..yes like paladin (below) pointed how, this has nothing to do with the problem..although its true)...I guess what you do is look at mods..because perfect squares are always some speical case mod a number..(look at the post below paladins)

[quote="paladin8"]Too bad that doesn't say anything about the problem at hand.[/quote]

yea my bad, but looking at that its what i generally think of..I guess you could try out differnt mods

like squares are always 0 mod 3, 1 mod 3..
or 0 mod 4, 1 mod 4..which i think works becuase its always going to be even (the $ p1p2p3..pn$ part) but 4 will never divide it...so $ p1p2p3p4..pn$cant be 0 mod 4 or 3 mod 4..so add 1 it cant be 1 mod 4 or 4 mod 4 which  is 0 mod 4, so it cant be a perfect square...

Answer

Lets assume that $ p_{1}p_{2}p_{3}\ldots p_{k}+1$ is a perfect square

$ \Rightarrow p_{1}p_{2}p_{3}\ldots p_{k}+1 = m^{2}, m \in Z$

$ \Rightarrow p_{1}p_{2}p_{3}\ldots p_{k}= m^{2}-1$

$ \Rightarrow p_{1}p_{2}p_{3}\ldots p_{k}= (m-1)(m+1)$

Hence $ p_{1}p_{2}p_{3}\ldots p_{k}$ can be expressed as the product of two integers which differ by two.

Lets consider all possible ways in which $ p_{1}p_{2}p_{3}\ldots p_{k}$ can be split into two integers which multiply to give $ p_{1}p_{2}p_{3}\ldots p_{k}$.

One of these integers must contain the prime 2 making this integer even, and the other integer must either be 1, or consist entirely of odd primes, making this integer odd. But an odd and an even never differ by two. Contradiction

Hence $ p_{1}p_{2}p_{3}\ldots p_{k}+1$ is never a perfect square.

sludgethrower, your solution is correct (and a lot better than this)..but like I said theres a lot easier solution...you could simplify it by..

p1=2,otherwise pk is not even, so p1p2..pn must be 2 mod 4.  Thus p1p2p3..pn+1 is 3 mod 4.

No perfect square can be 3 mod4.

because if the p is 0 mod 4, p^2 is 0 mod 4
if p is 1 mod 4 p^2 is 1 mod 4
if p is 2 mod 4, p^2 is 0 mod 4
if p is 3 mod 4, p^2 is 1 mod 4.

[quote="mihail911"]by what you're saying ordinaryboy, does that mean that any odd integer multiplied by two will result in a number that is 2 mod 4?[/quote]

yea..2 times any odd integer because any odd integer is 1 mod 4 or 3 mod 4. so 2=2 mod 4. 2 mod 4 * 1 mod 4 is 2 mod 4.  2 mod 4*3 mod 4 is 6 mod 4 which is 2 mod 4.. but you must only have one 2.


arithmetic sequence
number theory unsolved
number theory

Consider the set of all rationals numbers x satisfying x^2 +1 = y^2, where y is also a rational number. These numbers are closely related to Pythagorean triples.

For example, 3/4, 5/12, 8/15 would all be such numbers. So would 4/3, 12/5 and 15/8.

I need to find elements of the set that form an algebraic sequence; hopefully having 5 terms or more, ideally 7 or more.

Sorry; typo. I meant an algebraic sequence.

P.S. Fixed the original post.

Same question.  What's an algebraic sequence?

I'm am idiot.

Let me try again:

Consider the set of all rationals numbers x satisfying x^2 +1 = y^2, where y is also a rational number. These numbers are closely related to Pythagorean triples.

For example, 3/4, 5/12, 8/15 would all be such numbers. So would 4/3, 12/5 and 15/8.

I need to find elements of the set that form an [b]arithmetic sequence[/b]; hopefully having 5 terms or more, ideally 7 or more.

They both start with "a"...


Alicia's average score on her five tests is 88 points. The score range for each test is 0 points to 100 points, inclusive. What is the lowest possible score that Alicia could have earned on one of the five tests? Please explain how you solved it.

Since her average is 88 on five tests, she earned a total of $ 5(88)\equal{}440$ points.  If she got 100 on four of the tests, then she still has 40 points left over, so 40 is the lowest score she could have gotten.


[hide]Okay so draw a line from E to C. Then from E draw perpendiculars to each side. The one that connects to side DE call that w. The one that connects to AB call that one x. The one that connects to CB call y and the one that connects to CD call that one z. So angle EWD will equal 90 and and angle DEW will be equal to 70. Angle WEA will equal 10 and so will EAX. Angles EAW and AEX will equal 80. Angles ZEC and ECY will equal 35. angles ECZ and CEY will equal 55. After you look at this you see that rectangles EXBY and ZDWE are congruent because they have congruent sides. So X equals 20.[/hide]

I am no speak english.
shape designate,
 please


Six cylinders are stacked in a pyramid with three cylinders on the bottom, two in the middle and one on top as shown in the diagram.  If the radius of each cylinder is 6cm, what is the height of this stack?  Give your answer in whatever form makes you happiest :)

[hide]I got $12\sqrt{3}+12$. You are given the radii of the circle, so you can go and a triangle with these measurements. If you use two bottom radii for a leg of a right triangle(=12), and use one radius, a diameter, and another radius as the hypotenuse (=24), you get a right triangle which you should identify as a 30:60:90 triangle, with the missing side being the $\sqrt{3}$ side. You then have to add in the two radii that you ommited, (=12)giving you the answer of $12\sqrt{3} + 12$[/hide]

[hide="my answer"] Form an equilateral triangle with the side length of 24. The height of the triangle is 12 sqrt 3. So the height of this stack is just 12 sqrt 3 +12.[/hide]

[hide]18sqrt(3)  Height of one can is 6sqrt(3) because we make a right triangle that's 30 60 90 where 30 side is 6 and the 90 side is 12.  We then know that the height, the side that's missing is 6sqrt(3).  THree cans stacked like this is 18sqrt(3).
I had a hard time with this one because I was overthinking it.[/hide]

[hide]12 :sqrt: 3 + 12. The equilateral triangle with sides 24 in the middle, formed by connecting the c3enters of the outer circles, has a height 12 :sqrt: 3, plus 6 at the top and bottom, for my answer.[/hide]

[hide]make a 30-60-90 triangle with the medium long side going down the center

the little side is 12 so the medium side is $12\sqrt{3}$

add 12 to that and get $12+12\sqrt{3}$[/hide]


Solve the equation
$ 2(x^2\minus{}3x\plus{}2)\equal{}3 \sqrt{x^3\plus{}8}$, where $ x\in R$

[quote=RIP]$ 4(x^2\minus{}3x\plus{}2)^2\minus{}9(x^3\plus{}8)\equal{}(x^2\minus{}6x\minus{}4)(4x^2\minus{}9x\plus{}14)$ hence $ x\equal{}3\pm\sqrt{13}$.[/quote]

Nice sol! I did the same thing.


limit
real analysis
real analysis unsolved

How to prove that
\[ \lim_{n\to \infty}\prod_{k = 1}^{n}k^2 \sum_{p = 1}^{n}\dfrac{1/p}{\prod\limits_{\begin{subarray}{l} k = 1 \\
k\neq p \end{subarray}}^{n}(k^2 - p^2)} = \dfrac{1}{2}
\]

$ a_p\equal{}\frac{1/p}{\prod\limits_{k\equal{}1 \atop k\neq p}^n (k^2\minus{}p^2)}$ can be written as $ \frac{2\, (\minus{}1)^{p\minus{}1} \, p}{(n\plus{}p)!\, (n\minus{}p)!}$ . See this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5494]link[/url].

Furthermore $ \sum_{p\equal{}1}^n a_p\equal{}\frac{1}{n!^2}\, \frac{2n}{4n\minus{}2} \, \Rightarrow \, n!^2\, \sum_{p\equal{}1}^n a_p\longrightarrow \frac12$ .


modular arithmetic
quadratics
number theory unsolved
number theory

Let F(n) = 2^(2^n) +1.

Let p | F(n), show that p = 1 mod 2^(n+1) and find the order of 2 modulo p.

The order of $2\pmod p$ is clearly $2^{n+1}$, since $p|2^{2^{n+1}}-1$, so the order must be a power of $2$, but $p\not |\ 2^{2^n}-1$.

Actually, this is also the proof that $2^{n+1}|p-1$: the order of $2\pmod p$ is $2^{n+1}$, and this order divides $p-1$, so that's that :).

Yeah, I wanted to write that, but I forgot :).

If $n\ge 2$, we have $8|p-1$, so $2$ is a quadratic residue $\pmod p$. Let $t\in\mathbb Z,\ t^2=2\pmod p$. We have $t^{2^{n+1}}\ne 1\pmod p,t^{2^{n+2}}=1\pmod p$, so the order of $t\pmod p$ is precisely $2^{n+2}$, which must then divide $p-1$.


linear algebra
matrix
linear algebra unsolved

Is there a matrix $A$ such that $A^n=J(0,n)$ where J is the jordan matrix.

Nope. The Jordan block with 0 on the diagonal is nilpotent, thus so is A and thus $ A^n=0$, contradiction.

Can you tell me where i can read about Jordan form of matrices? (It looks very usefull :D )

Matrix Analysis-Roger Horn and Charles Johnson.This book is the best.

Cezar, please post the problem in the right section next time, ok? :)


modular arithmetic
algebra
polynomial
number theory
relatively prime
prime numbers

I have a few questions about two proofs of Wilson's.

[hide="first proof"]if we have p!+p, we can factor it into p((p-1)!+1), rite?
(p-1)! +1 must have a unique factorization rite?
we know 2,3,4,.....p-1 are not factors since (p-1)! +1 is 1mod(those numbers)
therefore, (p-1)! +1 must be divisible by p.[/hide]second proof: http://www.artofproblemsolving.com/Wiki/index.php/Wilson%27s_Theorem

1. In the first proof, why does the fact that we know $ 2,3,4,.....p\minus{}1$ are not factors since $ (p\minus{}1)! \plus{}1$ is 1mod(those numbers)
imply that $ (p\minus{}1)! \plus{}1$ must be divisible by $ p$?

2. In the second proof, why does $ \gcd( (p \minus{} 1)!, p) > 1$ imply  $ (p\minus{}1)! \neq \minus{}1 \pmod{p}$? and why do we care about that second statement? The equals should probably be congruence but isn't that the opposite of what we want?

3. In the last step of the proof with inverses:[quote]
"Now, if we omit 1 and , then the others can be grouped into pairs whose product is congruent to one,
$ 2\cdot3\cdot4\cdots(p\minus{}2) \equiv 1\pmod{p}$"[/quote]
a. Do we omit 1 because multiplying the left side by 1*1 is useless and do we omit p-1 because it is its own inverse?

b. How do we know that there isn't some number within [2,p-2] that is its own inverse? Wouldn't that mess things up?

1. I have absolutely no idea what the first proof is doing. Looks like it makes a bunch of observations and then says QED.

2. Do you know [url=http://www.artofproblemsolving.com/Wiki/index.php/Bezout%27s_Lemma]Bezout's Lemma[/url]? (if you don't, you should!) Note that $ (p\minus{}1)! \equiv \minus{}1 \pmod{p}$ implies that $ (p\minus{}1)! \plus{} kp \equal{} \minus{}1$ for some integer $ k$. Then $ (\minus{}1)(p\minus{}1)! \plus{} (\minus{}k)p \equal{} 1$, and by Bezout's Theorem $ (p,(p\minus{}1)!) \equal{} 1$. What is stated in the proof is the contrapositive of that.

This part of the proof is proving the converse of Wilson's Theorem, that it doesn't hold for any composite $ p$. That's probably why it looks like the opposite.

3a. We exclude $ 1$ and $ p\minus{}1$ because those are always the only residues whose inverses are themselves.

3b. A residue $ x$ is its own inverse mod a prime $ p$ if and only if $ x^2 \equiv 1 \pmod{p}$. $ 1,\minus{}1$ obviously satisfy this. If there were another, then $ x^2 \minus{} 1$ would have three factors, which is silly. (That polynomial factorizations work in the integers mod $ p$ is a consequence of the existence of inverses)

[quote="MellowMelon"]2. Note that $ (p \minus{} 1)! \equiv \minus{} 1 \pmod{p}$ implies that $ (p \minus{} 1)! \plus{} kp \equal{} \minus{} 1$ for some integer $ k$. Then $ ( \minus{} 1)(p \minus{} 1)! \plus{} ( \minus{} k)p \equal{} 1$, and by Bezout's Theorem $ (p,(p \minus{} 1)!) \equal{} 1$. What is stated in the proof is the contrapositive of that.
[/quote]

1. I don't get how we are applying bezout's theorem. What are we taking a linear combination of and why does that imply that  $ (p,(p \minus{} 1)!) \equal{} 1$?

2. I thought that given relatively prime numbers you can get a linear combination of them equal to 1. Didn't you use the converse of that? Why is that the contrapositive?

3. If anyone understands the first proof or thinks yous knows what he was getting at, please respond.

Bezout's theorem states that given integers $ a, b$ there exist integers $ x, y$ such that $ ax \plus{} by \equal{} 1$ [i]if and only if[/i] $ \gcd(a, b) \equal{} 1$.  We are applying the "only if" direction to $ a \equal{} p, b \equal{} (p \minus{} 1)!$, which is the converse if the "if" direction; this works because Bezout's theorem is an "if and only if" statement.

What is the source of the first proof?  As far as I can tell, it is grossly incomplete.  There is simply not enough work there; the observations as they are written apply to $ k(p \minus{} 1)! \plus{} 1$ for any $ k$.


Let ABC be an acute-angled triangle. Prove that:
$ cos^{2}A\plus{}cos^{2}B\plus{}cos^{2}C\geq4(cos^{2}Acos^{2}B\plus{}cos^{2}Bcos^{2}C\plus{}cos^{2}Ccos^{2}A)$

[b]Here is mine solution[/b]:
$ \sum cos^{2}A\ge 4(\sum cos^{2}Acos^{2}B)$
$ \Rightarrow \sum \frac {cosA}{cosBcosC}\ge 4(\frac {cosAcosB}{cosC})$

Put $ x \equal{} \frac {cosBcosC}{cosA}$, $ y \equal{} \frac {cosAcosC}{cosB}$, $ z \equal{} \frac {cosAcosB}{cosC}$.
The inequality becomes:
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\ge 4(x \plus{} y \plus{} z)$, $ (*)$.
where $ x,y,z > 0$ and $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$.
[i]Proof[/i]
Put $ x \equal{} \frac {a}{b \plus{} c}$, $ y \equal{} \frac {b}{c \plus{} a}$, $ z \equal{} \frac {c}{a \plus{} b}$.

$ (*)\Rightarrow \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c}\ge 4(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b})$.
Indeed, by [b]AM-GM[/b], we get:
$ \sum \frac {4a}{b \plus{} c}\le \sum(\frac {a}{b} \plus{} \frac {a}{c}) \equal{} \sum \frac {a}{b \plus{} c}$. $ \blacksquare$
Our proof is done! :lol:


[quote="Rust"]posted before.[/quote]i can't  find  it ,  can u  send  link at that.
thanks !!! :D

If x is natural, then
\[ \sum_{k\equal{}1}^{x^3\minus{}1}[\sqrt[3] k]\equal{}\sum_{k\equal{}1}^{x\minus{}1}k(3k^2\plus{}3k\plus{}1)\equal{}\frac{x^2(x\minus{}1)(3x\plus{}1)}{4}.\]

let $ f(x) \equal{} \sum_{i \equal{} 1}^{x^3 \minus{} 1}{\sqrt[3]{i}}$
and assume that $ x\in \mathbb{N}$, so
$ f(x) \equal{} (2^3 \minus{} 1)*1 \plus{} (3^3 \minus{} 2^3)*2 \plus{} \ldots$
I think the solution's $ x \equal{} 5$

$ \bullet$ $ \displaystyle \left[ \sqrt [3]{1}\right] \plus{} \displaystyle \left[ \sqrt [3]{2}\right] \plus{} ... \plus{} \displaystyle \left[ \sqrt [3]{x^3 \minus{} 1}\right] \equal{} 400 \Longleftrightarrow \sum_{k \equal{} 1}^{x \minus{} 1}k\left[(k \plus{} 1)^3 \minus{} k^3\right] \equal{} 400$

$ \bullet$ Let $ S_k \equal{} k[(k \plus{} 1)^3 \minus{} k^3] \equal{} 3k^2 \plus{} 3k \plus{} 1$. We have $ \displaystyle \left[ \sqrt [3]{1}\right] \plus{} \displaystyle \left[ \sqrt [3]{2}\right] \plus{} ... \plus{} \displaystyle \left[ \sqrt [3]{x^3 \minus{} 1}\right] \equal{} 400 \Longleftrightarrow \sum_{k \equal{} 1}^{x \minus{} 1}S_k \equal{} 400$

$ \bullet$ Because $ S_k > 0, k\in N^*$, $ S_1 \equal{} 7, S_2 \equal{} 38, S_3 \equal{} 111, S_4 \equal{} 244$ and $ S_1 \plus{} S_2 \plus{} S_3 \plus{} S_4 \equal{} 7 \plus{} 38 \plus{} 111 \plus{} 244 \equal{} 400$

So, the equation have a root is $ \boxed{x \equal{} 5}$

[quote="borislav_mirchev"]It's bulgarian math olympiad III round problem I don't know that year exactly between 1972-1976...[/quote]

Yes. You see at this link: http://www.mathlinks.ro/viewtopic.php?search_id=1567451195&t=201119


geometry
parallelogram
geometry unsolved

$ ABP$ is an isosceles triangle with $ AB\equal{}AP$ and $ \angle PAB$ acute. $ PC$ is the line through $ P$ perpendicular to $ BP$, and $ C$ is a point on this line on the same side of $ BP$ as $ A$. (You may assume that $ C$ is not on the line$ AB$.) $ D$ completes the parallelogram $ ABCD$. $ PC$ meets $ DA$ at $ M$.

Prove that $ M$ is the midpoint of $ DA$.

Construct the parallelogram $ PBAB'$, see that $ \triangle APB'$ is isosceles, hence $ PC$ passes through the midpoint of $ AB'$ and, $ CPB'D$ being parallelogram, this giving $ \triangle PBC \cong \triangle B'AD$, it follows that $ PC$ passes through the midpoint of $ AD$ as well.

Best regards,
sunken rock

Let $ N$ be the midpoint of $ BP$. Then we have $ AN \perp BP$, so  $ AN \parallel MP$. Thus, $ AN$ must pass through the midpoint of $ BC$. Thus, as $ AN \parallel CM$, $ M$ must be the midpoint of $ AD$.


LaTeX
trigonometry
calculus
derivative
limit
AoPS Books
calculus computations

In AoPS vol. II, the text talks about evaluating a trignometric limit like (sorry for lack of LaTeX i don't know how to represent limits) lim            sinx/x
  x>infinity

and other trig limits using the squeeze methods. can't you just take the derivative and then evaluate?

If you mean L'Hopital's Rule, you need the limit to be of indeterminate form. Although the denominator goes to infinity as $x\to\infty$, you have $-1\leq\sin x\leq 1$, as $\lim_{x\to\infty}$, so it isn't an indeterminate form.

i am going to assume that you mean $\lim_{x\to 0}\frac{\sin(x)}{x}=1$
sure you could use l'hopitals rule, but AoPS books assume that the readers do not know calculus. Even if you do know calculus it is still good to know how to use the squeeze theorem.


algebra
polynomial
number theory
greatest common divisor
superior algebra
superior algebra solved

[quote="liyi"]$F$ is a field. $\mbox{ch} F = p$. $f(x)$ is irreducible in $F[x]$. $K$ is an extension of $F$. Prove that $f(x)$ has a multiple root if and only if $f(x)$ is a polynomial of $x^p$.[/quote]

The "if" part of the statement is obvious.
Consider $f$ with a multiple root. Then $f$ and $f'$ are not
coprime - I mean their $gcd$ is different from $1$. But
then  $gcd(f,f') |f$ which is absurd because $f$ is irreducible
in $F$. Thus $f'=0$. BUT $f \neq 0$. Now suppose than one
monic of $f$ is not of the form $p \cdot k$, with $k\in \mathbb{N}$.
then we get a monic in $f'$ with a non-zero coefficient ! This is absurd.
So all the monic of $f$ are of the form $c_{p\cdot k} \cdot x^{p\cdot k}$.
This gives the "only if" part.

Fabrice

[quote="fmathurin"]Consider $f$ with a multiple root. Then $f$ and $f'$ are not
coprime - I mean their $gcd$ is different from $1$. But
then  $gcd(f,f') |f$ which is absurd because $f$ is irreducible
in $F$. Thus $f'=0$.[/quote]

$f$ has a multiple root in $K$ $\Rightarrow$ $\deg\gcd(f,f')\geq 1$ in $K$. How to prove that in $F$?

But when youy asked the question you assumed that $f\in K[x]$ had a root in an extension of $K$! :? That means that the root is algebraic. This is the question I was answering: given $f,g\in K[x]$ which share a root in some extension of $K$, they have a common divisor in $K[x]$. What exactly is unclear here?

Eh.. the original problem say that $K$ is an extension of $F$ (it doesn't say it's an algebraic extension).

[quote="liyi"][quote="grobber"]In general, if $f,g\in K[x]$ have a common root $\alpha$ in some field extension of $K$, then they're both divisible by the minimal polynomial of $\alpha$ over $K$, and that's where that result you asked about comes from.[/quote]
Hmm. If $\alpha$ is a transcendental number? There is no minimal polynomial for it on $K$  :?[/quote]

Correct-me if I am wrong, but if $f(\alpha) = 0$, there are very few
chances for $\alpha$ to be transcendental over $K$ ......
You should give us the precise statement,
I do believe my solution is correct with the original problem.

Fabrice

[quote="jmerry"]Forget $K$- we aren't using it anyway except to provide a space for the roots to exist. Define $E$ instead as a splitting field for $f$ over $F$. This splitting field is an algebraic extension and has finite degree.[/quote]
And then?


function
algebra
domain
inequalities
triangle inequality
real analysis
real analysis unsolved

I am having a gigantic brain cramp.  I learned epsilon-delta continuity proofs a couple of years ago, but I am having trouble proving that f(x) = x^2 is a continuous function using the e-d method   .  Help please...  :blush:

Have you tried googling? :)
http://www.uclic.ucl.ac.uk/topology/def_continuity.html

Using the definition from that page. Let $\epsilon >0$ . First we require $|x - y| < 1$ which means that $x-1 < y < x+1$. Then $|x + y| < |2x+1|$. So if we now also require $|x-y| < \frac{\epsilon}{|2x + 1|}$ we get that $|x^2 - y^2| = |(x-y)||(x+y)| < \epsilon$.

edit: I saw $y$ as the variable, unlike Kent.

Let's try to prove continuity at the point $y.$

$f(x)-f(y)=x^2-y^2=(x-y)(x+y).$

We're going to control the size of $(x-y)$ but what can we do about $(x+y)?$

Start by assuming that $|x-y|<1.$ (The $1$ is arbitrary; we need to pick something.)

Then $|x+y|\le|x-y+2y|\le|x-y|+|2y|<2|y|+1.$

Now, for a given $\epsilon>0,$ choose $\delta = \min\left(\frac{\epsilon}{2|y|+1},1\right).$

If $|x-y|<\delta,$ then

$|x^2-y^2|=|x-y|\cdot|x+y|<\frac{\epsilon}{2|y|+1}\cdot(2|y|+1)=\epsilon$

and we are done.

Note that the choice of $\delta$ depended on $y$ as well as on $\epsilon,$ so we have not proved uniform continuity, and in fact, if the domain is all of $\mathbb{R},$ this function is not uniformly continuous.

Ah, essentially simultaneously posted nearly identical solutions. Nearly identical because how else are you going to do it?

I don't think my solution is right though, because the inequality I used doesn't work for negative $x$. I just assumed it would work without thinking about it. It is better to use the triangle inequality like you did.


The product of two consecutive negative integers is 2550. What is the sum of the two integers?

The two numbers are 50 and 51, so the sum is 101

Except that it says NEGATIVE integers.

This makes the integers $ \minus{}51$ and $ \minus{}50$, so the sum is $ \minus{}101$


inequalities
function
calculus
calculus computations

In the new school book for class 8 , I read:

' the inequality 0x<10 is an [color=red]identity[/color] '! Do you use in your country this name(term) (identity) for such inequalities ? Untill now I used to write : the ineq. holds for every x but not ''it is an identity''
 Please see your books, discuss the question and tell me your opinion.

  Thank you !!!

 babis

We don't use this terms. You are right:inequality holds for all x from $R$.
In university, In Algebra there exist some fields when this inequality is wrong. :wink:

[quote="stergiu"]Thanks !

 But have you ever read that this ineq. is called '' identity''? Because in Greece some people want to make us all feel crazy ! I'd like to hear the opinion from other countries too !
 Babis[/quote]
No, I never read it.

The word "identity" is sometimes used if we have a relation with a variable involved, but the validity of the relation does not depend on the value that we assign to the variable.  I think I've most often seen it occurring in sentences which end, "... so the function $f(x)$ is identically 0," or the like.


a and b are positive intergers.  gcd(a,b)=1. Prove that phi\(a^n+b^n) divided by n

   i cannot use latex. Who can help me!

have a look [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]here[/url]

I have post this problem . Strong problem :
$ 2n|\varphi(a^n\plus{}b^n)$
Where $ a,b\in N$ and $ ab>1$


Let positive real numbers $ a,b,c$ satisfying $ a\plus{}b\plus{}c\equal{}1$. Prove that
$ \frac {a^2b^2}{c^3(a^2 \minus{} ab \plus{} b^2)} \plus{} \frac {b^2c^2}{a^3(b^2 \minus{} bc \plus{} c^2)} \plus{} \frac {c^2a^2}{b^3(c^2 \minus{} ca \plus{} a^2)}\ge \frac {3}{ab \plus{} bc \plus{} ca}$

[quote="thaithuan_GC"]Let positive real numbers $ a,b,c$ satisfying $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ \frac {a^2b^2}{c^3(a^2 \minus{} ab \plus{} b^2)} \plus{} \frac {b^2c^2}{a^3(b^2 \minus{} bc \plus{} c^2)} \plus{} \frac {c^2a^2}{b^3(c^2 \minus{} ca \plus{} a^2)}\ge \frac {3}{ab \plus{} bc \plus{} ca}$[/quote]
Let $ ab\equal{}z,$ $ ac\equal{}y$ and $ bc\equal{}x.$ Hence, your inequality is equivalent to
$ \sum_{cyc}\frac{z^3}{x^2\minus{}xy\plus{}y^2}\geq\frac{3(xy\plus{}xz\plus{}yz)}{x\plus{}y\plus{}z},$ which is obvious.

[quote="carlebee"][quote="arqady"][quote="thaithuan_GC"]Let positive real numbers $ a,b,c$ satisfying $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ \frac {a^2b^2}{c^3(a^2 \minus{} ab \plus{} b^2)} \plus{} \frac {b^2c^2}{a^3(b^2 \minus{} bc \plus{} c^2)} \plus{} \frac {c^2a^2}{b^3(c^2 \minus{} ca \plus{} a^2)}\ge \frac {3}{ab \plus{} bc \plus{} ca}$[/quote]
Let $ ab \equal{} z,$ $ ac \equal{} y$ and $ bc \equal{} x.$ Hence, your inequality is equivalent to
$ \sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2}\geq\frac {3(xy \plus{} xz \plus{} yz)}{x \plus{} y \plus{} z},$ which is obvious.[/quote]

$ \sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2}$ :arrow:  how did this come  :?:  ı found $ \sum_{cyc}\frac {z^2}{(x^2 \minus{} xy \plus{} y^2)c}$ I'm feeling that i couldnt see an easy thing[/quote]
bring the xy+yz+zx term on left hand denominator, then everything will fall in place.
or else multiply your c in denominator by (a+b+c) and rewrite c^2=xy/z.simplify.

[quote="carlebee"][quote="arqady"][quote="thaithuan_GC"]Let positive real numbers $ a,b,c$ satisfying $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ \frac {a^2b^2}{c^3(a^2 \minus{} ab \plus{} b^2)} \plus{} \frac {b^2c^2}{a^3(b^2 \minus{} bc \plus{} c^2)} \plus{} \frac {c^2a^2}{b^3(c^2 \minus{} ca \plus{} a^2)}\ge \frac {3}{ab \plus{} bc \plus{} ca}$[/quote]
Let $ ab \equal{} z,$ $ ac \equal{} y$ and $ bc \equal{} x.$ Hence, your inequality is equivalent to
$ \sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2}\geq\frac {3(xy \plus{} xz \plus{} yz)}{x \plus{} y \plus{} z},$ which is obvious.[/quote]

$ \sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2}$ :arrow:  how did this come  :?:  ı found $ \sum_{cyc}\frac {z^2}{(x^2 \minus{} xy \plus{} y^2)c}$ I'm feeling that i couldnt see an easy thing[/quote]
$ a\equal{}\sqrt{\frac{yz}{x}},$  $ b\equal{}\sqrt{\frac{xz}{y}}$ and $ c\equal{}\sqrt{\frac{xy}{z}}.$
Id est, $ \sum_{cyc}\frac {a^2b^2}{c^3(a^2 \minus{} ab \plus{} b^2)}\ge \frac {3}{ab \plus{} bc \plus{} ca}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {a^5b^5}{a^2 \minus{} ab \plus{} b^2}\ge \frac {3a^3b^3c^3(a\plus{}b\plus{}c)}{ab \plus{} bc \plus{} ca}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{z^5}{\frac{yz}{x}\minus{}z\plus{}\frac{xz}{y}}\geq\frac{3\left(\sqrt{xyz}\right)^3\frac{xy\plus{}xz\plus{}yz}{\sqrt{xyz}}}{x\plus{}y\plus{}z}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2}\geq\frac {3(xy \plus{} xz \plus{} yz)}{x \plus{} y \plus{} z}.$ :wink:

Ok maybe this question is very dumb but why $ \sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2}\geq\frac {3(xy \plus{} xz \plus{} yz)}{x \plus{} y \plus{} z}$ is obvious?

Look here, you are discussing the same question in two different topics

http://www.mathlinks.ro/Forum/viewtopic.php?p=1335989#1335989

Daniel

[quote="Flakky"] why $ \sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2}\geq\frac {3(xy \plus{} xz \plus{} yz)}{x \plus{} y \plus{} z}$ is obvious?[/quote]
For example:
$ \sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2} \equal{} \sum_{cyc}\frac {z^4}{x^2z \plus{} y^2z \minus{} xyz}\geq\frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\sum(x^2y \plus{} x^2z \minus{} xyz)}.$
But $ \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\sum(x^2y \plus{} x^2z \minus{} xyz)}\geq\frac {3(xy \plus{} xz \plus{} yz)}{x \plus{} y \plus{} z}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^5 \plus{} x^4y \plus{} x^4z \minus{} x^3y^2 \minus{} x^3z^2 \minus{} 6x^3yz \plus{} 5x^2y^2z)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2x^5 \plus{} 2x^4y \plus{} 2x^4z \minus{} 2x^3y^2 \minus{} 2x^3z^2 \minus{} 12x^3yz \plus{} 10x^2y^2z)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^5 \minus{} x^4y \minus{} xy^4 \plus{} y^5) \plus{} 3\sum_{cyc}(x^4y \minus{} x^3y^2 \minus{} x^2y^3 \plus{} xy^4) \plus{}$
$ \plus{} \sum_{cyc}(z^3x^2 \minus{} 2z^3xy \plus{} z^3y^2) \minus{} 5\sum_{cyc}(x^3yz \minus{} 2x^2y^2z \plus{} y^3xz)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x \minus{} y)^2(z^3 \minus{} 5xyz \plus{} x^3 \plus{} 4x^2y \plus{} 4xy^2 \plus{} y^3)\geq0$ and $ z^3 \minus{} 5xyz \plus{} x^3 \plus{} 4x^2y \plus{} 4xy^2 \plus{} y^3\geq z^3 \plus{} 5\sqrt {x^3y^3} \plus{} 5\sqrt {x^3y^3} \minus{} 5xyz\geq$
$ \geq(3\sqrt [3]{25} \minus{} 5)xyz\geq0.$
But the following reasoning is better.
$ \sum_{cyc}\frac {z^3}{x^2 \minus{} xy \plus{} y^2}\geq x \plus{} y \plus{} z\geq\frac {3(xy \plus{} xz \plus{} yz)}{x \plus{} y \plus{} z}.$ :wink:


geometry
trigonometry
perpendicular bisector
angle bisector

AB and AC are tangent to circle O at B and C, respectively, and CE is perpendicular to the diameter BD. Show that $ BE\cdot BO\equal{}AB\cdot CE$.

(CPIG)

Here's another one of my long, algebraically tedious proofs.  I'm assuming that point E is on the diameter BD.

[hide]
Let $ x$ be the length of the radius of circle O.  Obviously, the length of BO is $ x$.  Additionally, let $ \theta$ be the measure of $ \angle{BAC}$.  Because quadrilateral ABOC is cyclic, with two right angles at B and C, $ \text{m}\angle{BOC}=180-\theta$.  Because $ \angle{BOC}$ and $ \angle{COE}$ are supplementary, $ \text{m}\angle{COE}=\theta$.  Considering $ \triangle{OCE}$, with OC a radius of circle $ O$, we can assign OE a length of $ x\cos{\theta}$ using trigonometry definitions and multiplying by $ x$.  Thus BE will have a length of $ x+x\cos{\theta}$.

Likewise considering $ \triangle{OCE}$, we can assign CE a length of $ x\sin{\theta}$.  Finally, considering $ \triangle{ABO}$, we can assign AB a length of $ x\cot{(\theta /2)}$.

All that's left is algebra and trigonometry.

\begin{align*}
(BE)(BO) &= (AB)(CE)\\
(x+x\cos{\theta})(x) &= (x\cot{(\theta /2)})(x\sin{\theta})\\
x^2(1+\cos{\theta}) &= x^2\frac{\cos{(\theta /2)}}{\sin{(\theta /2)}}(2\sin{(\theta /2)}\cos{(\theta /2)})\\
&= x^2(2\cos^2{(\theta /2)})\\
&= x^2(2\cos^2{(\theta /2)}-1+1)\\
x^2(1+\cos{\theta}) &= x^2(\cos{\theta}+1)
\end{align*}
[/hide]

Nice Philip xDDD

[hide="Another Way"]
(Please draw the figure, or this may go over your head...)

Note that $ \angle ABO\equal{}90\equal{}\angle CBO$, by definition. Now note that $ AO$ is the perpendicular bisector of $ BC$, since $ AB\equal{}AC$ and $ BO\equal{}CO$ (which implies that ABOC is a kite) and hence also the angle bisector of $ BOC$. Now, let $ \angle AOB\equal{} x$. Since $ \angle ABC$ intercepts arc $ BC$ and since $ AO$ is the angle bisector of $ \angle BOC$, $ \angle ABC\equal{}\frac{1}{2}\angle BOC\equal{}x\Rightarrow \angle CBE\equal{}(90\minus{}x)\Rightarrow \angle BCE\equal{}x$. Since $ \angle ABO\equal{}90\equal{}\angle CBO$, and since $ \angle BCE\equal{}x\equal{}AOB$, $ ABC\sim BEC$. Thus, $ \frac{BO}{AB}\equal{}\frac{CE}{BE}$ and we are done.

[/hide]


without any doubts StarCraft is much, much better :D

I play both. I play warcraft more now because I got it for my birthday. I played a little bit of starcraft yesterday and I sucked. I like Warcraft better now.

Starcraft is awesome! Warcraft 3 is pretty good too. But remember, no other real-time strategy beats Starcraft.

Starcraft is a classic. No debating that one.

[quote="miraculouspostmaster"]I enjoy Warcraft 3 more simply because of the whole RPG-style hero play and the storyline.
However, Starcraft is one of my favorites.[/quote]

I agree with miraculouspostmaster. Warcraft is a bit like starcraft, age of empires and a bit of diablo II combined. The age of empires comes from the fact that you have to upgrade your main building. The starcraft comes from the all "aliens and other races" kinda thing. Then we have the diablo II. I say its kinda like diablo II because of your hero. The hero levels up and chooses abilities.

i like starcraft but i've never played Warcraft III

starcraft is great...but i heard warcraft is pretty good too...

Starcraft is the best!!!!!!!!!! Blizzard made the three races perfectly balanced. Warcraft III has better graphic but the different races aren't well-developed.

Ooh ooh I like witchcraft!  Oh wait, that wasn't one of the options.  Well from what I've seen Warcraft looks a lot cooler, although I own neither game and have only played each once or twice.

im pretty good at starcraft but i suck at warcraft plus i cant play cause i cant get the patch to work.
most people like starcraft more like me but warcraft 3 graphics are better, you cant argue that.  what i dont get is that people play warcraft 3 more when they like starcraft more.

starcraft owns warcraft :D  I haven't played in a while though, I'll play some of you in an online game just tell me when :D BRING IT MWHAHAHAHAHAH lol i'm not that good :)

elvenchamp777

unfortunately, last time i played SC i got owned :-(

ill bring it im really good with zerg :)

If anyone has a warcraft/starcraft account please post your battle.net names here.

Warcraft: 4rcanefox

starcraft but i suck so badly that im below "noob"

[quote="Jhyon8149"]Starcraft is the better game! I mean, it's an old game but you still have millions of people playing it. I remember when I visited Korea last year and on TV, they had people playing Starcraft. Kind of extreme but the point is, people are still crazy over Starcraft. Not as much with Warcraft ALTHOUGH it's a great game as well. Man, Blizzard is amazing.....[/quote]The amazing thing about Starcraft (that makes it more playable and more fun than WarCraft) is that all 3 races are equal in battle, although very different in behaviour. It was one of the first RTS games to present 3 completely different races that battle each other. All other previous attempts (like Command and Conquer, Warcraft II, etc.) were really lacking this (very) important feature.

Warcraft >j00 :P .

Battle.Net name = Hallucinations

Um...I played Lemmings for about half an hour yesterday 'cause I was bored. :D  I've never heard of Starcraft but I have a friend who is overly obsessed with Warcraft. A couple of them, actually. More like ten. Me? I've played neither.


geometry circumcircle inradius trigonometry LaTeX vector logarithms

Let $O$ be the circumcenter, $G$ the centroid, $R$ the circumradius, $r$ the inradius and $S$ the area of a triangle $ABC$. Prove that

[b]1)[/b] $OG^2=R^2-\frac{1}{9}\left(a^2+b^2+c^2\right)$
[b]2)[/b] $\sin x\cdot\sin y\cdot\sin z\leq\sin^3\frac{1}{3}\left(x+y+z\right)$
[b]3)[/b] $4RS=abc$
[b]4)[/b] $\cos A+\cos B+\cos C=1+\frac{r}{R}$

[color=red][Moderator edit: Corrected grammar and LaTeX mistakes.][/color]

I don't think this should be here...

1) follows from $3OG = OH$ and $OH^2 = 9R^2 - \left(a^2 + b^2 + c^2\right)$. The latter formula can be proven by vectors or trigonometry...

2) follows from the concavity of $f(x) = \ln{\sin{x}}$.

3) follows from $S = \frac{ab \sin{\gamma}}{2}$ and $c = 2R \sin{\gamma}$.

4) follows from $S = pr = \frac{abc}{4R}$ (where $p$ is the semiperimeter) and from Heron's formula $S = \sqrt{p(p - a)(p - b)(p - c)}$. Maybe there is a shorter way.

I proved number 4 with the Law of Cosine and the post it right in this forum. Maybe darij can find it.

for the number $4$ :

http://www.mathlinks.ro/Forum/topic-45622.html

here's my solution on 4 (it's from scratch):

AO = OB = R, so AOB is isosceles
<AOB = 2<BCA, so <OAB = (180-2<BCA)/2 = 90-<BCA
the same goes for the other two angles
summing the areas by the sine rule:
S = (a R sin(90-<CAB))/2 + (b R sin(90-<ABC))/2 + (c R sin(90-<CBA)/2
<=> 2S = R(a cos A + b cos B + c cos C)
<=> a cos A + b cos B + c cos C = 2S/R

Using the areas of the three triangles formed by linking the incenter with the vertices gives:
S = ar/2 + br/2 + cr/2
<=> 2S = r(a+b+c)

So
a cos A + b cos B + c cos C = r/R (a+b+c)

Summing a = b cos C + c cos B gives:
a+b+c = (b+c) cos A + (a+c) cos B + (a+b) cos C
<=> a+b+c = (a+b+c)(cos A + cos B + cos C) - (a cos A + b cos B + c cos C)
<=> a+b+c = (a+b+c)(cos A + cos B + cos C) - r/R (a+b+c)
<=> 1 = cos A + cos B + cos C - r/R

QED


[i]Find all natural numbers n such that $n^{n+1}+(n+1)^n$ is divisible by 5.[/i]

I think there no solutions but cant express that its too complicated ... for me

Davron

We can find solutions only modulo $20$ :arrow:
Answer is $n=20k+6$

Pavel please explain the proof, show your work but well done...

Davron

It looks really interesting, how can we look mod 20? Can somebody explain please?

I can explain. Well, there isn't any mistery in looking $\mod{20}$. You look $\mod{5}$ for obvious reason. You look $\mod{4}$ because of the exponents: they are perodic $\mod{4}$ (I mean that $a^{k} \equiv a^{k+4} \mod{5}$). Because of that it's simple case bash: you fix $n\equiv 1 \mod{5}$ and then you check what is $n \mod{4}$ and you see that $n \equiv 2 \mod{4}$ so $n \equiv 6 \mod{20}$. Then you fix $n \equiv 2 \mod{5}$ etc etc... I hope it's clear.


Hallo!
Ich wollte fragen, ob jemand vom Känguruwettbewerb mit nach Sachsen kommt!
Es wäre doch schön, wenn man vorher ungefähr wüsste wer da so hinkommt!
Ich selber habe mich jetzt einfach mal angemeldet...
(Ich weiß, ein schrecklicher Wettbewerb, aber für ein paar weniger Stunden Unterricht kann man sich zu vielem breitschlagen lassen!)

Ich komme auch!  :)
Mal schauen was daraus wird, bisher war ich immer in den Auslandsseminaren...  :)

Kann wohl doch nicht kommen (blödes Bremenseminar...)  :(  :(  :(

Jaja, jetzt fällt es aus  :(  Und ich soll übermorgen nach Berlin (wie nett, mich heute zu fragen ;) )

Ja, denn es haben nur 3 zugesagt... Und das sind zu wenig... Nun ja schade, denn ich bezweifel, dass ich übermorgen für 5 Tage nach Berlin kann!


Here is a great problem that I had in an oral examination: prove that we cannot prove that the equation $x^n+y^n=z^n$ has no trivial solution using congruences, that is working modulo different primes. Isn't it great?

Schur's theorem states that for all $n$ and sufficientely large prime $p$, the equation $x^m+y^m = z^m \mod [p]$ has a non trivial solution (that is $x,y,z$ are all coprime with $p$).

This is a direct consequence of the first (historically speaking) result in Ramsey theory  :)

Pierre.

Yup, that's the idea. Of course, I imagined you will answer in 4 minutes, Pierre. ;)  You should have let the others a little bit. :D

Well, Ramsey theory is just an hint...I didn't give full solution  :)

[Moderator edit: See http://www.mathlinks.ro/viewtopic.php?t=40985 .]

Pierre.


inequalities
real analysis
real analysis theorems

Let A be the set of real numbers x in [0,1] for which there are infinitely many pairs
(p,q) of positive integers s.t. |x-p/q|<=1/q^3.Show that m(A)=0

Just compute the darn thing: for all $n$, ${m(A)\le\sum_{q=n}^{\infty}\sum_{p=1}^q}\frac2{q^3}= \sum_{q=n}^{\infty}\frac2{q^2}<\frac2{n-1}$.
That goes to zero, and we are done. The first sum is an overestimate for the measure of the set on which at least one such inequality is true for $q\ge n$.


Where are the threads re: the AHSIMC Gold Team and the AHSIMC updates??

Were they deleted?  And if so, why?

If not, please point me to where they were moved within this message board.

I haven't hunted for the threads myself, but you can always do a search on your own name (to see all threads you have participated in) or search for keywords.

Yeah, I did that...they didn't turn up, and in fact the search said "7 threads" that AHSIMC was mentioned in, but only turned up 5...which probably means that the two got deleted.

Its not a huge deal, I just didn't check the board yesterday during the discussion of the Gold Team registering and was wondering why that thread, as well as my update thread, got deleted...

We've had a couple of junior-hacker problems...maybe one of them deleted it?  This forum isn't auto-pruned, which is odd.

Those threads have reappeared.  Just a temporary database problem that has now been fixed.

ok, great.  sorry for the hassle, and thanks for fixing it!


Let ABC a triangle and I its incenter.Denote D the tangency point with BC and $A_1$ the intersection of AD with the circumcircle.Prove that if $\displaystyle \frac{BA_1}{CA_1}=\frac{a}{a+c-b}$ then b=c.
(Own)

we have $\frac{BA_1}{CA_1} = \frac{sin(BCA_1)}{sin(CBA_1)} =$

$\frac{sin(DAB)}{sin(DAC)} = \frac{b(p-b)}{c(p-c)}$.

so, we have $b(p-b)(a+c-b) = ac(p-c)$

this is the same that $(b-c)((a-b)^2 - c(a+b)) = 0$.

see that $c > |a-b|$ and $a+b > |a-b|$, so $(a-b)^2 - c(a+b)$ isn't zero.

this implies $b-c = 0$  :)


Prove that for complex inner-product space V, every normal operator on V has a square root.

A matrix is normal if and only if it is unitarily diagonalizable. You can take square roots of diagonal matrices.


In a 10-mile race, Janet covered the first 2 miles at a constant rate. She then sped up and rode her bike the last 8 miles at a rate that was 0.5 miles per minute (WOW) faster. Janet's overall time would have been 2 min. faster had she ridden her bike the whole race at the faster pace. What was Janet's average speed (in miles per minute) for the whole race?
Solutions please. I know the answer.

[hide]
we'll talk in terms of time.
$d=rt \Longrightarrow t=\frac{d}{r}$
define $r$ the initial speed and $r+\frac{1}{2}$ the faster speed
$\frac{2}{r}+\frac{8}{r+\frac{1}{2}}=2+\frac{10}{r+\frac{1}{2}}$
Solve this and you get $r=\frac{1}{2}$.
So the total time is $t=\frac{2}{\frac{1}{2}}+\frac{8}{\frac{1}{2}+\frac{1}{2}}=12$ minutes
Speed = $\frac{10\text{ miles}}{12\text{ minutes}}= \frac{5}{6}\frac{\text{miles}}{\text{minute}}$
[/hide]


[color=darkblue]Not all teams have taken Olympiad 1 yet.[/color]
[i][b]Please don't give anything away!
Hold back your comments.[/b][/i]
[color=darkblue]We can only maintain this forum if everyone cooperates.[/color]

Thanks.

Will there be a time before mid-December when it will be safe to discuss overall impressions of the difficulty of this contest compared to prior contests, or when partial results will be available to PICOs?

Thanks!

[quote="mathmom"]Will there be a time before mid-December when it will be safe to discuss overall impressions of the difficulty of this contest compared to prior contests, or when partial results will be available to PICOs?[/quote]

December 13 seems safe.


Let A and B are 2 unitary matrices. Prove that: |det(A+B)|<=2^n

(Sorry, it is 2^n, not 2)

It's false. The identity is unitary, and the determinant of $ 2I$ is $ 2^n$ where $ n$ is the size.


inequalities
symmetry
inequalities theorems

Let $a,b,c$ be positive numbers such that $ab+bc+ca=\frac{1}{3}.$

Prove that $\frac{a}{a^2-bc+1}+\frac{b}{b^2-ca+1}+\frac{c}{c^2-ab+1}\geq \frac{1}{a+b+c}.$

Hint: We need Cauchy Schwarz in Engel form and some identity for this. Very nice inequality.

[quote="kunny"]I think so. Sorry for that do you know Bergstrom Inequality?[/quote]
Bergstrom? Well, it's a strange name.
[hide="Can you show it?"]I tried to find in PlanetMath,... but no result.[/hide]

Please take a look.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17200]www.mathlinks.ro/Forum/viewtopic.php?t=17200[/url]

Go to page 4, and vist post #72 onward.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cauchy&t=7835]a very old post of mine[/url]

Yes, I obtained the [b]equality[/b] at the last step. I will show you my solution.

Maybe my solution is same as yours.

[b]Let[/b] $a,b,c$ [b]be positive numbers such that [/b]$ab+bc+ca=\frac{1}{3}.$

[b]Prove that [/b]$\frac{a}{a^2-bc+1}+\frac{b}{b^2-ca+1}+\frac{c}{c^2-ab+1}\geq \frac{1}{a+b+c}.$

Proof:

Use the inequality $\frac{p^2}{x}+\frac{q^2}{y}+\frac{r^2}{z}\geq \frac{(p+q+r)^2}{x+y+z}\ (p,q,r,x.y.z>0),$ since $a,b,c>0$ and $ab+bc+ca=\frac{1}{3},$

and use the equality $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca),$ we have

$\frac{a}{a^2-bc+1}+\frac{b}{b^2-ca+1}+\frac{c}{c^2-ab+1}$

$\geq \frac{(a+b+c)^2}{a^3+b^3+c^3-3abc+a+b+c}$

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

$=\frac{a+b+c}{a^2+b^2+c^2-ab-bc-ca+1}$

$=\frac{a+b+c}{a^2+b^2+c^2+\frac{2}{3}}.$

Here $(a+b+c)^2-\left(a^2+b^2+c^2+\frac{2}{3}\right)$

$=2(ab+bc+ca)-\frac{2}{3}=0\Longleftrightarrow \frac{a+b+c}{a^2+b^2+c^2+\frac{2}{3}}\geq \frac{1}{a+b+c}.$ which completes the proof.

kunny

Here's my solution at this inequality, Kunny. :)

Wao, Splendid! stancioiu sorin. :lol:

The symmetry of equation gives us the insight.

Why is it so splendid? All three proofs are exactly the same.

I don't think they are [b]EXACTLY[/b] the same! :)

The calculations differ, the idea is exactly the same.

Please learn to accept it :D Your beloved Bergström inequality IS Cauchy-Schwarz ;)

Let's finish it! Arne and stancioiu sorin.

[b]Cauchy-Schwarz[/b]-[b]Buniakovsky[/b]-[b]Bergstrom[/b]

Mathematics is a culture. :lol:

kunny


function
real analysis
real analysis unsolved

Could somebody help me with this problem?

Suppose that $ f_n$: [a,b] --> R and $ f_n$ converges uniformly to f. Which of the following discontinuity properties of the function $ f_n$ carries over to the limit function? (Prove or disprove)
1) No discontinuities
2) At most 10 discontinuities
3) At least 10 discontinuities
4) Finitely many discontinuities.

At least 10 discontinuities. I'll do it with at least 3, but you should see how to modify it.

Define $ f_n\equal{}\begin{cases} \frac{1}{n} \ \text{for} \ \minus{}1\leq x\leq 0\\
 \frac{1}{n^2} \ \text{for} \ 0<x\leq 1\\
 \frac{1}{n^3} \ \text{for} \ 1< x\leq 2\\ \end{cases}$

Then $ f_n$ converges uniformly to $ f\equal{}0$ on $ [\minus{}1, 2]$, but there are three discontinuities in any $ n>1$.

Actually this works for finitely many in the obvious way as well.

An other one :  $ f_n(x)\equal{} \frac{1}{n}  \ if \    x\in Q (rational) \  ;\  f_n(x)\equal{} 0 \ if\  x\notin Q $  $ f_n$ converges uniformly to  $ f\equal{}0$ , but $ f_n$ is  discontinuous $ \forall x$
 :cool:

[quote="Diogene"]An other one :  $ f_n(x) \equal{} \frac {1}{n} \ if \ x\in Q (rational) \ ;\ f_n(x) \equal{} 0 \ if\ x\notin Q$  $ f_n$ converges uniformly to  $ f \equal{} 0$ , but $ f_n$ is  discontinuous $ \forall x$
 :cool:[/quote]
Thanks. This Dirichlet function works for 4 cases right?

[quote="HilbertThm90"]At least 10 discontinuities. I'll do it with at least 3, but you should see how to modify it.

Define $ f_n \equal{} \begin{cases} \frac {1}{n} \ \text{for} \ \minus{} 1\leq x\leq 0 \\
\frac {1}{n^2} \ \text{for} \ 0 < x\leq 1 \\
\frac {1}{n^3} \ \text{for} \ 1 < x\leq 2 \\
\end{cases}$

Then $ f_n$ converges uniformly to $ f \equal{} 0$ on $ [ \minus{} 1, 2]$, but there are three discontinuities in any $ n > 1$.

Actually this works for finitely many in the obvious way as well.[/quote]
Thank you very much for your help.


geometry
calculus
integration
function
calculus computations

the base of an object is the first quadrant region enclosed by the x-axis, the line y=x-3 and the line x=5.
if the cross section of the volume, perpendicular to the x-axis are squares, write an integral expression for the volume?

Haven't done this in a while so I hope this is right:
We are using square cross-sections so we must integrate each infinitesimal area over the region.  It is not hard to see that our region of integration is $ x\equal{}3$ to $ x\equal{}5$.  The area of a square is of course $ s^{2}$ and in this case, the side is determined by the function $ f(x)\equal{}x\minus{}3$ (the boundary you gave me).  Then
$ V\equal{}\int_{3}^{5}f^{2}(x)\ dx\equal{}\int_{3}^{5}(x\minus{}3)^{2}\ dx$
which is the integral expression we desire.


Are there scores for the individuals, teams, guys, etc somewhere?

I think Danny meant more detailed results for specific teams/people. Mr. C has ours, and to anyone else with the same question, they should have been emailed to your coach.


How would you find the equation of the tangent plane to the parametric surface x=-3r*cos(t), y=5r*sin(t), z=r at the point (-3*sqrt(2),5*sqrt(2),2) when r=2, and t=pi/4. Thanks for your help.


Solve for x:
(x^2 - 5x + 5)^x^2-9x+20 =1

a) 4 and 5 b) 1 and 4 c) 1,4 and 5 d) 2 and 3 e) 1,2,3,4 and 5

answer is e

Do you mean

$ (x^2 \minus{} 5x \plus{} 5)^{x^2 \minus{} 9x \plus{} 20} \equal{} 1$

?

$ 1^{(\text{any number)}} \equal{} 1$
AND
$ \text{(any number)}^0 \equal{} 1$
So we only need $ x^{2} \minus{} 5x \plus{} 5 \equal{} 1$ or $ x^{2} \minus{} 9x \plus{} 20 \equal{} 0$

And from there its just factoring

And don't forget $ \minus{}1$ to an even power.

Uh, plug in 3 and 4; if they both work out to 1, that eliminates every choice but E...

IF x^2-5x+5 =1
this give me x=1,4

or x^2-9x+20=0
and this give me x=4,5

therefore the solution is c ={1,4,5}
thank you

You forgot $ (\minus{} 1)^n$, for any even integer $ n$. Letting $ x^2 \minus{} 5x \plus{} 5 \equal{} \minus{} 1$, $ x \equal{} 2$ or $ 3$. Plugging these into $ x^2 \minus{} 9x \plus{} 20$, we have even integers, so the set is $ \boxed{{1,2,3,4,5}}$.


search
linear algebra
matrix
vector
geometry
videos
number theory

Any cheap Dover books that are sort of an advanced introduction to linear algebra?  Preferably the theorem lemma, lemma lemma theorem type, I'd like to have some knowledge of linear algebra, but don't really care about solving problems or anything.

http://www.ericweisstein.com/encyclopedias/books/LinearAlgebra.html

If you do a text search for Dover, there are lots of them on that page. As well as many other inexpensive linear algebra books.

[quote]I really don't know anything about it, but I think if you know the basics of matrices then you know really introductory linear algebra.[/quote]

Not based on the linear algebra classes you take in college.

A nice linear algebra book is [i]Linear Algebra Done Right[/i] by Sheldon Axler. It is published by Springer-Verlag, so it isn't that expensive. (I think I paid about $30 for it.) It's one of the textbooks (and I think the primary one) for the upper division linear algebra sequence here at UCSB.

Another amazing book, if you can get it, is [i]Finite-Dimensional Linear Analysis[/i] by Glazman and Ljubic (possibly with some variant on the spelling of the latter). It has no proofs; instead, it has only definitions and (if I recall correctly) 2405 problems! You really have to be motivated if you are going to learn from this book, but I think it is worth it. (I'm still near the beginning, so I have yet to reap the rewards this book offers.)

[quote="ComplexZeta"]A nice linear algebra book is [i]Linear Algebra Done Right[/i] by Sheldon Axler. It is published by Springer-Verlag, so it isn't that expensive. (I think I paid about $30 for it.) It's one of the textbooks (and I think the primary one) for the upper division linear algebra sequence here at UCSB.[/quote]

I'm looking into this one.  Would it work as an introduction?  It sounds perfect, I'm just afraid I don't have the prereqs to understand it.

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0387982582/qid=1089254251/learninfreed]Linear Algebra Done Right[/url] by Sheldon Axler (which I have checked out from the U of MN Mathematics Library at the moment) is a very interesting book. Inside, the author says that the book is designed for a SECOND course in linear algebra at the undergraduate level. Follow the link I provided to read the Amazon.com reviews: it is apparent that many undergraduates find this book too hard, perhaps because they haven't developed enough "mathematical maturity" before taking a course using that book. I like the book, but I don't understand it very well.

Yeah...but I know a little linear algebra...I dunno if it would work or not to do that book.  If all they're asking for is mathematical maturity I'm ok with that.

And, because you're dying to know why, I need the algebra book because a professor at University of Illinois at Chicago has said that he'll do an independent study in number theory with me next year if I learn some algebra.  So here I am, rushing to cram in as much algebra as possible by September.

[quote="ComplexZeta"]A nice linear algebra book is [i]Linear Algebra Done Right[/i] by Sheldon Axler. It is published by Springer-Verlag, so it isn't that expensive. (I think I paid about $30 for it.) It's one of the textbooks (and I think the primary one) for the upper division linear algebra sequence here at UCSB.[/quote]

After reading the customer reviews on Amazon, this sounds like the book I would get if I were ever to study linear algebra theoretically. (Which likely won't be the case, since I'll be studying engineering.) I get queasy looking at matrices. (Case in point: it took forever for me to master cross-product.) On the other hand, I absorbed the content at Mathcamp's Linear Algebra class quite naturally, partially because Mira emphasised that matrix is but a convenient notation for vectors, as far as linear algebra is concerned.

Speaking of which, I have yet to dig into your group theory notes... Hmm...

[quote="Magnara"]And, because you're dying to know why, I need the algebra book because a professor at University of Illinois at Chicago has said that he'll do an independent study in number theory with me next year if I learn some algebra.  So here I am, rushing to cram in as much algebra as possible by September.[/quote]

If you aim to study number theory soon, I'm not sure that linear algebra is at all what the professor had in mind. (But once you ask him, you'll know what he had in mind.) [racking brain:] There has to be some other kind of book, I think, that would be more to the point for preparing for what you desire to do. I'll have to think for a while and see if I can remember what it would be. Meanwhile, you can always ask the professor for advice.

There is

"college algebra," further development of high school algebra (much the same syllabus as "precalculus"), which is all that EPGY requires as a prerequisite for taking the EPGY number theory course,

"linear algebra," the algebra of vector spaces, which is chock full of implications for higher mathematics but not a subject that I see pointed to often as a prerequisite for studying number theory,

and

"abstract algebra," a generalization of algebras that every math major eventually has to learn.

PERHAPS you need only learn the first. But check with the professor to be sure.

Magnara, I think Axler's book is the book for you. I agree with tokenadult's comment that by algebra, the professor probably means abstract algebra rather than linear algebra. You can read the [url=http://www.albanyconsort.com/algebra/220Anotes.pdf]group theory notes that Osiris was referring to[/url], but they aren't easy. (I can't understand it, and I wrote it. :))

Yeah, it seemed unlikely to me that linear algebra would be useful for number theory.  But he kept implying it.  The book he actually recommended was S. Lang's "Algebra", which I requested from the UIC library and should arrive soon.  But, every review on Amazon basically said "This is a great book if you know a lot of algebra going into it."  So, we'll see.

Lang's [i]Algebra[/i] is not about Linear Algebra. (He did write a book on linear algebra also, but that's a different one.) It's not easy reading at all.

[quote="Magnara"]The book he actually recommended was S. Lang's "Algebra", which I requested from the UIC library and should arrive soon.[/quote]

Okay, what ComplexZeta said. Lang's book Algebra is about "abstract algebra," NOT about linear algebra. It is a highly advanced book, so in rare colleges it is assigned to undergraduate classes, and in most places it is used by graduate students.

After racking my brain, I think the book you might find most helpful is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0387982892/learninfreed]Numbers and Geometry[/url], by John Stillwell, also published by Springer-Verlag. Numbers and Geometry is one of my all-around favorite books about math.

Yeah, [i]Numbers and Geometry[/i] is a good book.

There is a dover book with an introduction to abstract algebra that I'm reading through now slowly so that I have to do less work next year.

...finding website...

Okay,
[url=http://search.store.yahoo.com/cgi-bin/nsearch?follow-pro=1&vwcatalog=doverpublications&catalog=doverpublications&query=Abstract%20Algebra]here[/url] is a list of their books on "abstract algebra."  Actually, you might want to do a further search there just for "algebra," because there are some with only that in the title, I believe.

The one I have is called "Elements of Abstract Algebra," which I like and which has a reasonable number of problems as well as proofs of important theorems and all the definitions and such that you should need.

While not a book this video series is well done:  Its is the Linear Algebra I at MIT.

http://ocw.mit.edu/OcwWeb/Mathematics/18-06Linear-AlgebraFall2002/VideoLectures/index.htm

Rob Sperry


Let be $ p,q$ the prime divisors of $ 2009$ . Find all the numbers prime with $ 2009$ , from the interval $ (p^2,q^2)$ .

I'm not sure if this is what you mean but:

the prime divisors are 7 and 41, so the numbers you are asking for are all the numbers between 49 and 41^2=1681 that aren't mulitples of 7 or 41


$ (n\plus{}1)(n\minus{}2)x_{n\plus{}1}\equal{} n(n^{2}\minus{}n\minus{}1)x_{n}\minus{}(n\minus{}1)^{3}x_{n\minus{}1}$ for all n>2
$ x_{2}\equal{} 1,x_{3}\equal{} 1$
Find all  n such that $ a_{n}$ is an integer.

Call $ y_{n}\equal{} nx_{n}$ we have: $ y_{2}\equal{} 2,y_{3}\equal{} 3$ and:
$ (n\minus{}2)y_{n\plus{}1}\equal{} (n^{2}\minus{}n\minus{}1)y_{n}\minus{}(n\minus{}1)^{2}y_{n\minus{}1}$
$ \Leftrightarrow (n\minus{}2)(y_{n\plus{}1}\minus{}y_{n})\equal{} (n\minus{}1)^{2}(y_{n}\minus{}y_{n\minus{}1})$
Call $ z_{n}\equal{} y_{n}\minus{}y_{n\minus{}1}$ we have $ z_{3}\equal{} 1$ and:
$ z_{n\plus{}1}\equal{}\frac{(n\minus{}1)^{2}}{n\minus{}2}z_{n}$
Then $ z_{n}\equal{}\frac{(n\minus{}2)^{2}}{n\minus{}3}.\frac{(n\minus{}3)^{2}}{n\minus{}4}...\frac{2^{2}}{1}z_{3}\equal{} (n\minus{}2)(n\minus{}2)! \equal{} (n\minus{}1)!\minus{}(n\minus{}2)!$
Then $ y_{n}\equal{} (n\minus{}1)!\plus{}1\Leftrightarrow x_{n}\equal{}\frac{(n\minus{}1)!\plus{}1}{n}$
Then $ n$ is a prime number because if $ n\equal{}p.q,1<p<n\Rightarrow (n\minus{}1)!\minus{}1\vdots p\Rightarrow 1\vdots p$, contradiction!


if we have a non-rectangular hyperbola that has its center at the origin and its 2 axes paralell to the x and y axes

we then rotate that hyperbola 45 degrees (direction doesnt matter). the new assymptotes of this hyperbola are slanted since it's non-rectangular
is there a way to express the equation of this new hyperbola given its new assymptotes(lets just say the new assympotes are y=ax and y=bx)?

$ (y\minus{}ax)(y\minus{}bx)\equal{}k$ for non zero constant $ k$.

woah, thanks!
but why does it work so?

edit: nvm, i sort of get it.

In case you need more clarification, it's because there's no way for $ y \equal{} ax$ or $ y \equal{} bx$ and the equation to be satisfied (plus the graph does tend to those asymptotes).


geometry
calculus
number theory
linear algebra

Show that if $a\neq c$, $a^x=c^q$, and $c^y=a^z$, then $xy=qz$.

By the way does anybody know where I can get really old problems from the AHSME (like 1951 for example) and their answers. :?

[hide]$c=a^{\frac x q}$

$c=a^{\frac z y}$

$\frac x q=\frac z y$

$xy=qz$[/hide]

[hide]Take the yth power of $a^x=c^q$, so $a^{xy}=c^{qy}$.
Take the qth power of $c^y=a^z$, so $c^{qy}=a^{qz}$.
Thus $a^{xy}=a^{qz} \implies xy=qz$.[/hide]

I didn't post the first ones because I couldn't solve them. But yeah, I didn't get the solutions manual. :(

take logs on both sides ,

$xlog{ a} = qlog{ c}$
and
$ylog{ c} = zlog{ a}$

Substitute value of $log{ a}$ from second eqn in first eqn

Therefore
$\frac{xylog{ c}} z = qlog{ c}$

=> $xy = qz$

any flaw please let me know.

[quote="leonariso"][hide]
$a^x=c^q$
$c^y=a^z$
Multiplying we get
$(ac)^{xy}=(ac)^{qz}$
$xy=qz$
[/hide][/quote]
I think u are wrong

[quote="mathgeniuse^ln(x)"][quote="lotrgreengrapes7926"]Another AoPS Volume 1, Chapter 1. :D I'm guessing you didn't get the Solution Manual.  :P[/quote]

How do you recognize this so easily?  This is sort of random, but do you know what books will be coming soon from AoPS.[/quote]
Intro to Number theory in 2007
Intermediate Counting and Prob.
Into to Alg.
That's all I know of right now.

I think Interm Alg is coming out before Intro Alg, but I could be wrong.

[quote="ragnarok23"]
Intro to Number theory in 2007
Intermediate Counting and Prob.
Into to Alg.
That's all I know of right now.[/quote]

Intro to Number Theory is coming out at the end of this month (27th). Intermediate Counting and Prob.,
Into to Alg., and Intermediate Alg. are coming out in early 2007 :D

[quote="mathgeniuse^ln(x)"]
How do you recognize this so easily? [/quote]
I was just doing these problems last week.  :D

I pmed Mr. Rusczyk a couple of weeks ago, he gave me the following info:

June 2006: Intro to NT
End of 06: Interm C&P
Beginning of 07: Intro/Interm Algebra
Spring/summer 07 (i think): Interm Geometry
End of 07/Beginning of 08: Interm NT
Sometime in 09: Linear Algebra, Calculus

He said all of these books have been started.

Apparently the Olympiad series is not going to come out for another few years.

Woah! There is going to be a Calculus book?! :o


limit
integration
calculus
calculus computations

I need some help in computing in determining the fallowing sum :
$\sum_{n=1}^{\infty} \ k*k! =  ?$

[quote="Buffalo"]$\sum_{k=1}^{\infty}k*k!=\sum_{k=1}^{\infty}\left(\left(k+1\right)!-k!\right)=\sum_{k=1}^{\infty}\left(k+1\right)!-\sum_{k=1}^{\infty}k!=-1$[/quote]

And if I want to find the formula when :

$\sum_{k=1}^{n} \ k*k! =  ?$

(depending on n, something like (n+1)!-1 )

Uh, buffalo, did you stop to wonder how a sum of positive terms could turn out negative? One cannot ignore convergence quite that blatantly.

Yes, that is a useful manipulation, and reveals this as a telescoping sum.

$\sum_{k=1}^nk\cdot k!=(n+1)!-1.$

$\sum_{k=1}^{\infty}k\cdot k!=\lim_{n\to\infty}\sum_{k=1}^nk\cdot k!= \lim_{n\to\infty}(n+1)!-1=\infty.$

[quote="Kent Merryfield"]Uh, buffalo, did you stop to wonder how a sum of positive terms could turn out negative? One cannot ignore convergence quite that blatantly.[/quote]
There is a unique linear mapping $S: Sequence(\mathbb{C})\rightarrow \mathbb{C}$ s.t. $S(a\oplus T)=a+S(T)$, here $(a\oplus T)_n=If[n==0,\ a,\ T_{n-1}]$, we call $S(T)$ the (regularized)sum/value of T, If $\sum_{n=0}^{\infty}T_n$ exists, we have $S(T)=\sum_{n=0}^{\infty}T_n$, we don't want to invent an unnecessary symbol, thus even when $\sum_{n=0}^{\infty}T_n$ does not exist, we still write $\sum_{n=0}^{\infty}T_n$, which represents $S(T)$.
You can get some interesting results like $\sum_{n=0}^{\infty}2^n=-1$, $\sum_{n=0}^{\infty}(-1)^n n!=\int_{0}^{\infty}\frac{e^{-x}}{1+x}dx$

[quote="Buffalo"] There is a unique linear mapping $S: Sequence\left(\mathbb{C}\right)\rightarrow \mathbb{C}$ s.t. $S\left(a\oplus T\right)=a+S\left(T\right)$ [/quote]
Hmmm... I doubt this statement. What is $1+1+1+\dots$ then? (the problem is that this sequence $T$ is the same as $1\oplus T$, so we must have $S(T)=1+S(T)$, which is impossible). I agree that there are plenty of methods that allow to assign some finite values to some divergent sums but  I have never heard of a universal summation method like the one postulated by Buffalo.


For complex nos. z and w, prove that $|z|^2w - |w|^2z = z-w$, if and only if $z=w$ or $z\bar{w} = 1.$

[hide="general method to use"]

Let $z = a + bi$ and let $w = f + gi$.  Then you need to show each of the following:

1)if $z=w$ the given equation holds (just expand the equation using the substitutions above)
2)if $z\bar{w}=1$ the given equation holds (again, expand and substitute known information)

3)assume that the equation holds, then show that either $z=w$ must hold or $z\bar{w}$ must hold, again by expanding and substituting values for variables.

Once you have shown those things you'll have completed the proof.  good luck.

[/hide]

I think the third one is sufficient enough i.e., to prove either one of the condns. should hold for the given statement to be true.

But this is turning out to be difficult in this problem, especially by the means suggested.

That does not prove that if the conditions are true, then the equation must hold. All 3 parts are necessary.

Those are trivial to verify, though. Just note that $|z|^2 = z\overline{z}$. For the third part, try assuming that neither of the required things are true, but the given equation is, and look for a contradiction.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89469
Don't double-post!

[quote="Immanuel Bonfils"]$\,\,\,(x^2+y^2+1)a\,-\,(a^2+b^2+1)x\,=\,0$

$\,\,\,(x^2+y^2+1)b\,-\,(a^2+b^2+1)y\,=\,0$

that can be looked at as an homogeneous system of equations in $\,\,x^2+y^2+1\,\,\,\textrm{and}\,\,\,a^2+b^2+1,\,\,\,$

that can't be null (trivial solution), so must it be the determinant $\,\,\,ay-bx\,=0.\,\,$[/quote]
This is totally invalid -- the "coefficients" and "variables" aren't independent!

Calculuslover's solution is a really bad way to go about this question.  Xev's method is the way to go, and the result is pretty immediate from there.

Sorry but, in this case, when the "variables" can't be null, the assertion still stands.
      Let's put it in other words .

We can write the system like that

$\,\,\,(x^2+y^2+1)a\,=\,(a^2+b^2+1)x\,\,$

$\,\,\,(x^2+y^2+1)b\,=\,(a^2+b^2+1)y\,\,$

and divide member  by member to obtain

$\,\,\frac{a}{b}\,=\,\frac{x}{y}.\,\,$

$\begin{gathered}
  |z|^2 w - |w|^2 z = z - w\mathop {}\limits^{}  \Leftrightarrow \mathop {}\limits^{} z\bar zw - w\bar wz = z - w\mathop {}\limits^{}  \Leftrightarrow \mathop {}\limits^{}  \hfill \\
   \hfill \\
  (1)\mathop {}\limits^{} \mathop {}\limits^{} \mathop {}\limits^{} z(\bar zw - 1) = w(z\bar w - 1)\mathop {}\limits^{}  \Leftrightarrow \mathop {}\limits^{} zw(\bar z - \bar w) = z - w\mathop {}\limits^{} \mathop {}\limits^{} \mathop {}\limits^{} (2) \hfill \\
\end{gathered} $

If  z=w or   $z\bar w = 1$ ,  then  $\mathop {}\limits^{} \mathop {}\limits^{} $ (1) ,(2)  is true.

 Conversely:

$(1),(2)\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} \left. \begin{gathered}
  |z||\bar zw - 1| = |w||z\bar w - 1| \hfill \\
  {\text{  }}|zw||z - w| = |z - w| \hfill \\
\end{gathered}  \right\}\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} z = w\mathop {}\limits^{} or\mathop {}\limits^{} z\bar w = 1$

[color=darkred][b][u]PP[/u].[/b] Let $z$ , $w$ be two complex numbers. Prove that $|z|^2w - |w|^2z = z-w\ \iff\ z=w\ \vee\ \bar{w}z = 1$ .[/color]

[b][u]Proof[/u].[/b] Observe that if $wz=0$ , then the equivalence is truly. Thus can suppose that $wz\ne 0$ .

$\blacktriangleright\ \boxed{\Longleftarrow}$ Prove easily that $z\in\{w,\frac {1}{\overline w}\}\implies |z|^2w - |w|^2z = z-w$ .

$\blacktriangleright\ \boxed{\Longrightarrow}$ Suppose $z\ne w$ . Observe that $|z|^2w - |w|^2z = z-w\ \iff\ wz(\overline z-\overline w)=z-w\ \stackrel{(z\ne w)}{\implies}\ |wz|=1$ ,

 i.e. ${\boxed{w\overline z=\frac 1{\overline wz}}\ (*)}$ . Hence $|z|^2w - |w|^2z = z-w\ \iff\ z(w\overline z-1)=w(\overline wz-1)\ \stackrel{(*)}{\implies}$

${z(\frac 1{\overline wz}}-1)=w(\overline wz-1)\iff\overline wz=1\ \vee\ w\overline w+1=0\iff \overline wz=1$ .

[quote="Virgil Nicula"]Observe that if $wz=0$ , then the equivalence is truly. Thus can suppose that $wz\ne 0$ .[/quote]
This part is redundant as $wz=0\implies w=0\vee z=0$ and either one implies the other is also zero, thus covered under $z=w$.