I was wondering if someone can help me on this question (sorry, but this is my first post and I am not sure how to embed the images in the message):

[i][color=red]mod edit to post problem[/color][/i]

Given $a,b,c$ solve the following system of equations for $x,y,z:$

\begin{eqnarray} x^2-yz &=& a^2\\ y^2-zx &=& b^2\\ z^2-xy &=& c^2\end{eqnarray}

Anyone?  I am in desperate need of help on this.

Thanks...

You're not given actual numerical values for a, b, and c?

Try to subtract equations from each other... (1) - (2), (2) - (3) and (1) - (3).

Then factor lots of stuff.

I don't really understand in terms of what you're trying to solve for x, y, and z. No matter what, you're going to get some kind of algebraic expression, so what are we trying to get?

You have to solve for x,y, and z in terms of a,b, and c.  And I have gotten about 20 different equations by adding, subtracting, mutliplying, and factoring the three equations but I can never solve for x,y, or z interms of a,b and c. boxedexe, unfortunately I don't know what year it is from :(

Maybe this should be moved to a higher forum?

Please post the problem in the Message Body of the post instead of attaching a document.

Sorry to keep pushing this but I really need some help on this question (p.s. thnks to the mod who wrote out the question).

Thanks...

You can turn it into a much nicer set of similtanious equations if you subtract 2 from 1, 3 from 2, and 1 from 3. Then factor. Kinda like what arne said.

In fact, now that i read it again, its exactly what Arne said!

Ya, but you can play around with the equations to get two more like that.  It is however, very ugly.

[quote="seamusoboyle"][quote="Singular"]It's 1997 COMC last question.

Solution:  Add equations (1), (3) together, and subtract twice equation (2).  Result is:

$(x + z)^2 - (x+z)y + 2b^2 - a^2 - c^2 - 2y^2 = 0$,

a quadratic in $x+z$.    It's ugly from here, because the solutions (x,y,z) are ugly.  In the COMC, we had $a^2 + c^2 = 2b^2$.[/quote]

But does this not give you $x+z$ in terms of $y$? Which isnt really any good....[/quote]

here is an example of how it might be done

y in terms of x+z, then sub in for y, so we have 2 vars 2 eqns, use one equation to get x in terms of z, now we have 1 var 1 eqn....

---

however it is too ugly for any contest, both the method of soln and the answer. that is why we need 2b^2 = a^2 + c^2 for the contest to make it simple.


geometry
3D geometry
combinatorics proposed
combinatorics

Prove that for every number n natural there exist a finite sequence with n terms and the elements 1,2 or 3 such that no digit or group of digits doesnt repeat consecutively.(for example 2323)

This has strong connections with the Thue-Morse sequence:

Start with $0$ and continue by making the replacements $0\rightarrow 01,\ 1\rightarrow 10$. It's well-known that his sequence is cube-free, i.e. it has no sequences of the type $AAA$, where $A$ is a word.

I thought long and hard about obtaining a square-free sequence with $3$ symbols, $0,1,2$ (instead of $1,2,3$ :)). by modifying the Thue sequence. I thought I had found one, but didn't manage to show that it was Ok. However, I found an example on Mathworld:

[url]http://mathworld.wolfram.com/Thue-MorseSequence.html[/url]

Check the last paragraph of the article and you'll find a solution to this beautiful problem. I'm still trying to show that my own sequence is square-free, but I'm not even sure that's true..:(.

Here's how I obtained my sequence: I took the Thue sequence and replaced every digit in a position of the form $1+2^{2n-1}+k2^{2n}$ with $2$ instead of whatever was there at the beginning ($0$ or $1$). Actually, I think tyhe solution presented on Mathworld works on pretty much the same principle.

I don't have proofs for any of the important things: the fact that the Thue sequence is cube-free and the fact that the other sequence is square-free.

Didn't ne1 else find this interesting? Maybe you've got some other nice results..

I read on Mathworld that not only is the Thue sequence cube-free, but it contains no subsequences of the type $AAa$, where $a$ is the first character of the word $A$. After giving some thought (during a long and boring train ride :)), I realized that this isn't that hard to prove. However, I think it would be pointless to post a proof, as I don't think it helps with the initial problem xxxxtt posted.

The whole thing comes down to this:

In the Thue sequence consider the sequence of words of $2$ letters, $01,\ 11,\ 10,\ ..$ etc. Make the substitutions $01\rightarrow 0,\ 10\rightarrow 1,\ 00\rightarrow 2,\ 11\rightarrow 2$. We need to show that the sequence obtained after these substitutions is square-free, i.e. it contains no substrings of th type $AA$, where $A$ is a word.

P.S.

This sequence is almost the same as the one I defined. The only difference is that in this one w replace with $2$ the letters in the positions of the form $2^{2n-1}+k2^{2n}$ instead of $1+2^{2n-1}+k2^{2n}$.

Sorry to come back to this, when nobody seems to be interested in it :)..

I saw another solution (belonging to Thue): start with $0$ and then continue by the mappings $0\rightarrow 12,\ 1\rightarrow 102,\ 2\rightarrow 0\ (*)$. Quite amazingly, afterwriting it down I saw that the same sequence can be obtained as follows: assume we start $0$, and the next term of the sequence of sequences is $12$. Then we get from step $n$ to step $n+1$ as follows: we change the last letter of the $n$'th sequence (there's only one choice for a change) and then concatenate the sequence we obtain and the "negative", i.e. the one obtained by leaving the $0$'s unchanged and interchenging $1$'s and $2$'s. In particular, we see that all the sequences obtained by the mappings $(*)$ have the consecutive powers of $2$ as lengths.

I don't have a solution of the fact that the sequences obtained by these completely different methods are identical, and I don't have a proof of the fact that the sequence is a solution to xxxxtt's problem, but maybe my observation helps.


not sure, but i'll give it a try

[hide]
x^(a^2 - b^2) = z
lnz = (a^2 - b^2)lnx

from
alnx = ln2
-blnx = ln3
we get
a^2lnx = aln2
-b^2lnx = bln3

add the 2 equations:
(a^2 - b^2)lnx = ln(2^a) + ln(3^b)
lnz = ln(2^a * 3^b)
z = 2^a * 3^b

then..?
[/hide]

Hmm...

Maybe you're thinking of it in a too difficult perspective?

[hide="hint"]
x^(a^2 - b^2) = x^(a^2) * x^(-b^2)
[/hide]

[quote="SnowStorm"][hide]
$x^{a^2} = x^a * x^a
[/hide][/quote]

9 out of 10 exponent rules agree, that step is not correct.  The 10th one is named "The exponent rule everyone wishes were true but never, ever, ever is, unless you happen to be extraordinarily lucky."

I don't think there is any numerical value for x^(a^2-b^2)
Only an algebraic one, because
x^(a^2-b^2) = x^((a+b)(a-b))=(x^(a-b))^(a+b)
which equals to 6^(a+b) = 6^(logx(2) - log x(3)) = 6^(logx(2/3))
now if u look at this expression, only x remains.
since definite numerical values are only found with both x and a or both x and b
in an equation or expression, the final algebraic expression remains a variable
or a function of x.

[quote="Treething"]
[hide="hint"]
x^(a^2 - b^2) = x^(a^2) * x^(-b^2)
[/hide][/quote]

and then?

IssacChao and Xirtamax have both found correct solutions.  (If anyone is wondering, the answer is not a numerical constant -- different choices for x yield different results.)  However, their results are quite different looking.  Can anyone prove that they are equal?

Issac -- Treething's hint can be used to find your solution slightly more quickly: [hide]$x^{a^2 - b^2} = x^{a^2}\cdot x^{-b^2} = \left(x^a\right)^a\cdot \left(x^{-b}\right)^b$ and from there it should be clear how to get to the answer you got.[/hide]

Heys guys, what up? To end this, i got the answer as a constant.... sorry i dont have latex....

[hide]Firstly x^a=2    =>  (a)(logx)=log2 and x^b=1/3  => (b)(logx)=-log(3)
                          =>  a=(log2)/(logx)   and    b=-(log3)/(logx)

now x^(a^2-b^2) = x^(a.a)/x^(b.b)

=(x^a)(x^a)(x^a).......(x^a) (a times)/(x^b)(x^b)(x^b)......(x^b) (b times)

=2.2.2......2 (a times)/(1/3)(1/3)(1/3).......(1/3) (b times)

=6a/b

=6(log2/logx)/-(log3/logx)

=-(6)(log2)/(log3) and you can play around with this and get a nice single log but ill let you do that by yourself! Goodnight![/hide]

Theres a couple of mistakes in here..

Firstly, [hide="Mistake 1"]$x^{-b}=\frac{1}{3}$, not $x^b$.[/hide]
Secondly, [hide="Mistake 2"]\[\frac{\overbrace{2\cdot 2\cdot 2\cdots 2}^{a\text{ times}}}{\underbrace{\frac{1}{3}\cdot\frac{1}{3}\cdots\frac{1}{3}}_{b\text{ times}}} = 2^a 3^b\]
I don't know where you got $\frac{6a}{b}$ from.
[/hide]

As its been said before, there is no single answer here.

By the way, you don't need to 'have' latex to use latex on the site.

[quote="JBL"]I refer you to the 5th comment on this thread which comments on the use of the exponent rule $a^{(b^c)} = (a^b)^c$.[/quote]

That is wrong! :D

[quote="ssupermath"]x^a=2
x^(-b)=3 then find x^(a^2-b^2)=?[/quote]
[quote="JBL"]IssacChao and Xirtamax have both found correct solutions. (If anyone is wondering, the answer is not a numerical constant -- different choices for x yield different results.) However, their results are quite different looking. Can anyone prove that they are equal?[/quote]
[hide=""Finding a and b and answering JBL s question""]$a=\log_x2$, $b=-\log_x3$.
$a^2=(\log_x2)(\log_x2)=\log_x2^{\log_x2}$ and $b^2=(-\log_x3)(-\log_x3)=\log_x3^{\log_x3}$.
$x^{a^2-b^2}=x^{(\log_x2^{\log_x2})-(\log_x3^{\log_x3})}=x^{(\log_x2^{\log_x2})}\cdot x^{(\log_x3^{-\log_x3})}=2^{\log_x2}\cdot3^{-\log_x3}=2^a\cdot3^b=6^{\log_62^a+\log_63^b}=6^{\log_x\frac23}$
Challenge: Prove that $\log_62^a+\log_63^b=\log_x\frac23$[/hide]

[quote="SnowStorm"]$\displaystyle x^{a^2} = x^a * x^a = 2 * 2 = 4$
$\displaystyle x^{-b^2} = x^{-b} * x^{-b} = 3 * 3 = 9$[/quote]
Nice try but no, these 2 steps are not true in general.

[quote="seamusoboyle"]Heys guys, what up? To end this, i got the answer as a constant.... sorry i dont have latex....

[hide]Firstly x^a=2 => (a)(logx)=log2 and x^b=1/3 => (b)(logx)=-log(3)
=> a=(log2)/(logx) and b=-(log3)/(logx)

now x^(a^2-b^2) = x^(a.a)/x^(b.b)

=(x^a)(x^a)(x^a).......(x^a) (a times)/(x^b)(x^b)(x^b)......(x^b) (b times)

=2.2.2......2 (a times)/(1/3)(1/3)(1/3).......(1/3) (b times)

=6a/b

=6(log2/logx)/-(log3/logx)

=-(6)(log2)/(log3) and you can play around with this and get a nice single log but ill let you do that by yourself! Goodnight! [/hide][/quote]
This is not true in general either.
[hide=""Why""]Because $a$ and $b$ are more likely non-integer real numbers and therefore does not hold for most values of $x$. How can you count '$a$ times' or '$b$ times' if $a$ or $b$ are not integers?[/hide]

[quote="TripleM"][hide=""Mistake 1""]$\displaystyle x^{-b}=\frac{1}{3}$, not $x^b$.[/hide][/quote]
[hide=""Mistakes and mistakes equal doubly mistakes""]$\displaystyle x^{-b}\neq\frac{1}{3}$.[/hide]

[quote="kunny"][quote="JBL"]I refer you to the 5th comment on this thread which comments on the use of the exponent rule $\displaystyle a^{(b^c)} = (a^b)^c$[/quote]That is wrong! :D[/quote]
You are the one wrong. Please read carefully what JBL said. He is [b]referring[/b] you to read the 5th comment.


inequalities
induction
inequalities proposed

there is a nice trick for 2)
prove with induction that $ \prod_{k\equal{}2}^n \left(1\plus{}\frac{1}{k^2}\right)\le 2\minus{}\frac{2}{n}$


calculus
derivative
analytic geometry
graphing lines
slope
symmetry
function

Hello everyone.

I am looking over my text book for an up coming exam I have and I need help on this problem.

This is as the problem reads:
Show that the sum of the x- and y- intercepts of any tangent line to the curve sqrt(x) + sqrt(y) = sqrt(c) is equal to c.

I have no idea how to attack this problem.  I tried numerous times at solving it, but i can never get c in the end.
Can someone help!  Thanks!

i have some question regarded to the notation in ur first step .

$ \frac{dx}{2\sqrt{x}}+\frac{dy}{2\sqrt{y}}=0$

can u write like this ?
i know it is rite , but i don't understand the way in which u wrote in ur equation.

i am used to differentiating   both side with  respect to x ,  and the notation would look like ,

2x +  d( sqrt{y})/ dx = 0 ,

and then get to ur second step .

can u help me understand ur notation ?
can we differentiate both sides with respect to both dx and dy at the same time?
i am a little confused , can u help me out ?

( if there is any problem with my english , plz correct and inform me , thanks)

Call it adopting an 18th century mind-set. For the first hundred years or so of calculus, there were no functions and no derivatives - there were only variables and differentials. And implicit differentiation wasn't some oddity tucked away in one little section of a textbook: at that time, all differentiation was implicit differentiation.

We have here a typical implicit differentiation setup: the relationship between $x$ as $y$ is given by an equation $F(x,y)=C.$

In the notation you're comfortable with, we differentiate this with respect to $x$:

$\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}=0$

Solving this for $\frac{dy}{dx}$ we get

$\frac{dy}{dx}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$

What I did was to start out writing

$\frac{\partial F}{\partial x}\cdot dx+\frac{\partial F}{\partial y}\cdot dy = 0$

Now, we could solve that for $\frac{dy}{dx}$ and get the exact same expression as above; but since our interest was actually the tangent line, we can simply proceed to replace $dx$ by $x-x_0$ and $dy$ by $y-y_0$:

$\frac{\partial F}{\partial x}(x_0,y_0)(x-x_0)+ \frac{\partial F}{\partial y}(x_0,y_0) (y-y_0)=0.$

Try thinking that way - you'll worry about it not being rigorous, but it's really about as well-justified as any other part of ordinary calculus.

[quote="Kent Merryfield"]$\sqrt{x}+\sqrt{y}=\sqrt{c}.$

Find the differentials:

$\frac{dx}{2\sqrt{x}}+\frac{dy}{2\sqrt{y}}=0,$ which can be manipulated to

$\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}.$

Now, take an arbitrary but particular point $(x_0,y_0)$ that is on the curve.

The tangent line at that point is

$y-y_0=-\frac{\sqrt{y_0}}{\sqrt{x_0}}(x-x_0).$

Now divide both sides by $\sqrt{y_0}$ and manipulate it around to the following symmetric form. (With a clearer intuitive understanding of differentials and slopes, you might arrive at this form more directly):

$\frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}}= \frac{x_0}{\sqrt{x_0}}+\frac{y_0}{\sqrt{y_0}}= \sqrt{x_0}+\sqrt{y_0}=\sqrt{c}.$

Now for the intercepts:

If $x=0$, then $y=\sqrt{c}\sqrt{y_0}$

If $y=0$, then $x=\sqrt{c}\sqrt{x_0}$

The sum of the intercepts is

$\sqrt{c}\sqrt{y_0}+\sqrt{c}\sqrt{x_0}= \sqrt{c}\left(\sqrt{x_0}+\sqrt{y_0}\right)= \sqrt{c}\cdot\sqrt{c}=c.$

Two after the fact hints:

1. Students often go wrong on tangent line problems by failing to fix the "arbitrary but particular point." A symptom of this would be an alleged equation for the tangent line that involves only $x$ and $y$ and not $x_0$ and $y_0$ (or similar arbitrary labels) and which, if examined closely, is not actually the equation of a line.

$\frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}}=\sqrt{c}$ [b]is[/b] the equation of a line.

2. The original equation was symmetric in its treatment of $x$ and $y.$ While you don't have to maintain this symmetry throughout your solution, it is often a very good idea to try to keep your equations as symmetric as you can. This guides your algebraic manipulations along paths with fewer errors and more intermediate checks.[/quote]

Taking parametric equations for the curves like these, helps even better

Here you can try $x=c\cos^2t,   y=c\sin^2t$
and  $(c\cos^2t,   c\sin^2t)$ as any point on the curve.


Let $ a,b,c,\alpha$ be reals such that $ \alpha \not= c-b$ and $ c\not =0$. Assume that sequences $ (u_{n}),(v_{n})$ are defined by $ u_{1}=a,u_{n+1}=\frac{u_{n}^{2}+bu_{n}}{c},v_{n}=\sum_{k=1}^{n}\frac{u_{k}}{u_{k+1}+b-c}(\forall n\geq 1)$. If $ \lim_{n\to\infty}u_{n}=\alpha$ then find $ \lim_{n\to\infty}v_{n}.$

And what is its answer? Are you thinking?  :maybe:


I cannot think of an example of a metric on N the set of integers that makes it into a connected topological space. Would finding some sort of path between the integers help to prove the topological space is path-connected and therefore connected?

A metric on a countable set $ S$ with more than one element cannot give rise to a connected topology.

Choose an arbitrary point $ x$ in $ S$. For each positive real number $ r$, the sets $ \{y: d(x,y)<r\}$ and $ \{y: d(x,y)>r\}$ are open. There is some element $ Y\in S$ with $ d(x,Y)\equal{}R>0$. In the interval $ (0,R)$, there are uncountably many numbers. Since $ S$ is countable, the set $ D(x)\equal{}\{d: d(x,y)\equal{}d\text{ for some y}\}$ is countable, and it does not contain all of $ (0,R)$. Choose $ r\in (0,R)$ not in $ D(x)$; we then have that the two sets $ \{y: d(x,y)<r\}$ and $ \{y: d(x,y)>r\}$ are disjoint open sets which cover the whole space, and $ S$ is not connected.

Now, if you drop the "metric" requirement, there are connected topologies. My favorite is the cofinite topology: the open sets are the empty set and the sets with finite complement. It's connected because any two nonempty open sets intersect.


calculus integration trigonometry function limit algebra domain

Calculate

$\int_{0}^{+\infty}\prod_{n\geq 1}\left(\frac{1}{2}\left(\cos(3x/4^n)+\cos(x/4^n)\right)\right)dx$

We use several trig identities to simplify the product first. First, we have that $\cos 3u=4\cos^3 u-3\cos u$, so $\frac12(\cos 3u+\cos u)=2\cos^3 u-\cos u=\cos u\cos 2u$.

The infinite product then becomes $\prod_{n=1}^{\infty}\cos\left(\frac{x}{2^n}\right)$
We have $2^m\sin\left(\frac{x}{2^m}\right)\prod_{n=1}^m\cos\left(\frac{x}{2^n}\right)=\sin x$, so
 $\prod_{n=1}^m\cos\left(\frac{x}{2^n}\right)=\frac{\sin x}{2^m\sin\frac{x}{2^m}}.$

Taking the limit as $m\to \infty$, $\prod_{n=1}^{\infty}\cos\left(\frac{x}{2^n}\right)=\frac{\sin x}{x}$

Now we just need $\int_0^{\infty}\frac{\sin x}{x}$, which I don't recall at the moment

I think it's $\frac {\pi}{2}$ via a contour integral.

Now that I remember, here's a quick way to do it. Consider $\int \frac{e^{iz}}{z} dz$ on a contour in the upper half-plane consisting of a semicircle of radius $R$ centered at 0, a circle of radius $\epsilon$ centered at zero, and two large chunks of the $x$-axis. Orient the curve so that the $x$-axis portion is pointed in the positive direction.

The integral on the outer semicircle clearly goes to zero. The integral on the inner circle tends to $-\pi\cdot i$ since $\frac{dz}{z}=i\cdot d\theta$. The whole integral tends to zero because the function is analytic on the upper half-plane except at 0. Therefore, using only the imaginary part, $\int_{-\infty}^{\infty}\frac{\sin x}{x}=\pi$. This function is even, so $\int_{0}^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}.$

jmerry used complex analysis for that last calculation. I'll use the occasion as an excuse to put up my favorite real analysis approach.

The first step is to integrate by parts, using $1-\cos x$ as the primitive of $\sin x:$

$\int_0^{\infty}\frac{\sin x}x\,dx=\int_0^\infty \frac{1-\cos x}{x^2}\,dx.$

This has the side benefit of proving that the improper integral converges, since this last integral is both positive and absolutely convergent.

Now write $\frac1{x^2}=\int_0^\infty te^{-tx}dt$ and substitute that into the integral, getting:

$\int_0^\infty \int_0^\infty (1-\cos x)te^{-tx}dt\,dx.$

To an analyst of my training, that double integral can mean only one thing: we [b]will[/b] interchange the order of integration. In this case, that's justified by Tonelli's Theorem since the integrand is nonnegative.

[Grubby calculus details omitted. Consult a table of Laplace transforms if necessary.]

$=\int_0^\infty\frac{dt}{1+t^2} = \frac{\pi}2.$

For $\int_{0}^{+\infty}\frac{\sin x}{x}dx$

Consider the Laplace transform

$f(t)=\int_{0}^{+\infty}\frac{e^{-tx}\sin x}{x}dx$
 for $t\geq 0$

derivate $f$, expression of $f'$ without integral, and finally find $f$

Moubinool's argument does require the continuous analogue of Abel's theorem. In particular, given that $\int_0^{\infty}\frac{\sin x}xdx$ converges, we must show that $\lim_{t\to0^+}f(t)$ equals that number.

That theorem is true, so the approach does work. I'm just saying we should note that it's needed.

This function and its relatives (I tend to use $\frac{\sin\pi x}{\pi x}$.) is one of the two most important functions of Fourier analysis, competing with the Gaussian for top honors.

Moubinool's solution can be proved in elementary calculus, which is used in most calculus and mathematical analysis books.

Of course, Merryfield's solution needs less computation... However... Tonelli's theorem is used and it is beyond the domain of mathematical analysis.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Please help me to solve the following problems as specific as possible :
1. Let A is the additive group of dyadic rationals  $m2^{n}, m, n \in Z$, and X = <x> is an infinite cyclic group acting on A via multiplication by 2. And lat G is the semidirect product of X and A. Prove that G is finitrly generated, but G is not polycylcic.
2. Let G is polycyclic, prove that G is not supersolvable.
3. G is supersolvable, prove that G is not nilpotent.

I found that in 3, $S_{3}$ is supersolvable, but it's not nilpotent, because $Z(S_{3})=1$.
And in 2, I've solved such as: Consider $A_{4}$ with its composition : $1\lhd <(13)(24)>\lhd V_{4}\lhd A_{4}$, with $V_{4}$ is Klein group. We have :
$A_{4}/V_{4}= C_{3}, V_{4}/<(13)(24)\geq C_{2}, <(13)(24)>/1 = C_{2}$, so $A_{4}$ is polycyclic group.
Assume that $A_{4}$ is supersolvable, because $A_{4}$ is not abelian, its normal groups are 1, $V_{4}$ and $A_{4}$, so its normal cyclic series is $1 \lhd V_{4}\lhd A_{4}$, but $V_{4}/1=V_{4}$ is not cylic, contrary with hypothesis,so $A_{4}$ is not supersolvable.  (Is it right?)
But I can solve problem 1, please help me!


geometry perimeter rhombus trigonometry parameterization algorithm search

The perimeter of a rhombus is 24cm and its area is 18sq.cm. What are the measures of its angles?

Nice work, the angles are 75 and 15.

But this problem was intended for middle schoolers, but I thought it deserved a place in teh high schools.
Can you do it without trig?

[quote="towersfreak2006"]Label one diagonal as $2a$ and the other diagonal as $2b$. Then the area of the rhombus is $2ab$ and the perimeter is $4\sqrt{a^2+b^2}$, so we have:\[4\sqrt{a^2+b^2}=24\]\[2ab=18\]Then $a^2+b^2=36$ and $ab=9$, so $\frac{a}{b}+\frac{b}{a}=4$. Note that $\frac{a}{b}$ is the tangent of half of one of the rhombus's angles while $\frac{b}{a}$ is the tangent of half the other, so let $\tan\left(\frac{\theta}{2}\right)=\frac{a}{b}=x$.\[x+\frac{1}{x}=4\]\[x=\frac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt{3}\] Which value of $x$ do we choose? If we note that $2+\sqrt{3}=\frac{1}{2-\sqrt{3}}$, then we realize, it doesn't really matter, since whichever value of $x$ we choose, the tangent of half the other angle will be the other value of $x$. (Gee, that sounded really confusing.)
Thus, $\theta_1=2\tan^{-1}\left(2+\sqrt{3}\right)$ and $\theta_2=2\tan^{-1}\left(2-\sqrt{3}\right)$.[/quote]

couldnt understand what that was?
Simple sol:
each side is $6cm$ so $b*h=area of rhombus$ so $h=3cm$ so using some trigonometry $sinx=3/6$ which makes $x=30$ so you get the answer of 1 angle so you can get other angles which are $150,30,150$...
Did I do a mistake?

[quote="towersfreak2006"]Label one diagonal as $2a$ and the other diagonal as $2b$. Then the area of the rhombus is $2ab$ and the perimeter is $4\sqrt{a^2+b^2}$, so we have:\[4\sqrt{a^2+b^2}=24\]\[2ab=18\]Then $a^2+b^2=36$ and $ab=9$, so $\frac{a}{b}+\frac{b}{a}=4$. Note that $\frac{a}{b}$ is the tangent of half of one of the rhombus's angles while $\frac{b}{a}$ is the tangent of half the other, so let $\tan\left(\frac{\theta}{2}\right)=\frac{a}{b}=x$.\[x+\frac{1}{x}=4\]\[x=\frac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt{3}\] Which value of $x$ do we choose? If we note that $2+\sqrt{3}=\frac{1}{2-\sqrt{3}}$, then we realize, it doesn't really matter, since whichever value of $x$ we choose, the tangent of half the other angle will be the other value of $x$. (Gee, that sounded really confusing.)
Thus, $\theta_1=2\tan^{-1}\left(2+\sqrt{3}\right)$ and $\theta_2=2\tan^{-1}\left(2-\sqrt{3}\right)$.[/quote]
You can find angle
$a^2+b^2=36$, $2ab=18$, combine $(a+b)^2=a^2+2ab+b^2=36+18=54$, $(a-b)^2=a^2-2ab+b^2=36-18=18$
$a+b=3\sqrt{2}\sqrt{3}$, $a-b=3\sqrt{2}$
Solve $a=3\sqrt{2}(\sqrt{3}+1)$, $a=3\sqrt{2}(\sqrt{3}-1)$
Law of cosine to find $\cos \theta$
$\cos \theta=\frac{6^2+6^2-(2b)^2}{2\cdot 6^2}=\frac{\sqrt{3}}{2}$ and you get $\theta=30$

[quote="riddler"]couldnt understand what that was?
Simple sol:
each side is $6cm$ so $b*h=area of rhombus$ so $h=3cm$ so using some trigonometry $sinx=3/6$ which makes $x=30$ so you get the answer of 1 angle so you can get other angles which are $150,30,150$...
Did I do a mistake?[/quote]

Yes this is best simple solution. Very good

What about no trig  :(

But I do understand the solutions,
10000th user (congrats on this!), for a long time I was looking for part of your solution.

Oh, the sides of a rhombus are all the same length! That certainly makes things easier.  :blush:

The area on a side of one diagonal (half of the rhombus) is $\frac{18}{2}=9$. From perimeter $24$, all sides are $6$. Then the area of the triangle is $\frac{1}{2}(6*6*\sin{\theta})=9$, so $\sin{\theta}=\frac{1}{2}$ and $\theta=30^\circ$ or $\theta=150^\circ$.[list][/list]

*yawn* I guess the non trig solution will never come. I guess I will take the trig solutions  :)

EDIT: I found the non trig solution, sort of. Based on the other solutions.

[quote="236factorial"][quote="10000th User"][quote="236factorial"]*yawn* I guess the non trig solution will never come. I guess I will take the trig solutions  :)[/quote]
*yawn*? You mocking me?  :mad:
NO non trig solution!  :mad:[/quote]

Yes there is. Notice that in the right triangle, 6 is twice of 3, which is a 30-60-90 triangle. That was Algebra I.
I'm just tired and bored. I'm probably going to sleep.
*going offline*[/quote]
my cousin just explained to me what happened here. i just asked him to log off so i can post. basically there is NO non trig. WHY?
you already assumed the angle is 30-60-90 and not only that, you can only make 30-60-90 if you drop a perpendicular from an opposite vertex. and yet, you ARE assuming that angle is 30. you can't do that, you are trying to find the angle (the unknown). want me to prove you wrong? say the area is 19. you no longer have that 30 angle, yet the sides are still length 6 with perimeter 24, and please, respect other users. there is a reason why they are replied because they see you are wrong and want to correct you, not to watch you lie in ignorance. if you still got a problem with it, PM me

edit: to confirm all of you and to my cousin as well, YES there is a nontrig solution. like i said before, drop a perpendicular from an opposite vertex. treat the side of rhombus as base of rhombus, then the height found by riddler is 3, so we have a right triangle with leg 3 and hypotenuse 6. this is clearly a 30-60-90 triangle and since the angle opposite of shortest leg is 30, one angle of rhombus is 30.

towersfreak2006's solution is nice because it determines 2 angles at once using sine and proves that adjacent angles are supplementary.
to 236factorial: even though you got it, i still got a point if the area wasn't  18.

The difference in numerical parameters of a problem can make all the difference in the difficulty of the solution.  The following problems range from elementary to middle school to high school.

Without using a calculator, solve to 8 decimal places:

a.  $\sqrt{4}$

b.  $\sqrt{10.24}$

c.  $\sqrt{\pi}$

Attached is a diagram which illustrates the middle school, non-trigonometric solution, previously found by contributors to this thread.

a and b are obvious, but what about c?
 Oh yeah, divide and average or something like that.

There is actually a rather interesting square root algorithm somewhere on the internet that I should have learned at some point or another, but haven't. It's not brought up very often, as far as I know. Look it up though!

Yes, I know it, but you can search it up as it is very hard to explain. It's something like long division.

There are lots of ways to find square roots, including your little calculator.

[quote="rcv"]Without using a calculator, solve to 8 decimal places:

a.  $\sqrt{4}$

b.  $\sqrt{10.24}$

c.  $\sqrt{\pi}$

[/quote]
a. 2
b. 3.2
c. Very hard! $\pi=3.14159265$. I use old method square root:
[code]
3.14159265             (2xquotient+digit)xdigit        Digit
1                                    _                   1
2 14                                2_                   7
1 89
  2515                             34_                   7
  2429
    8692                          354_                   2
    7084
    160865                       3544_                   4
    141776
     19089
[/code]
I only find $\sqrt{\pi}=1.7724$ but I dont know more digits of $\pi$  :blush: [color=red]rcv[/color], my answer is OK?

Try using $\pi=3.14159265358979$, you might be able to get enough digits using that.

[quote="shobber"]what is [b]rhombuses[/b]? I can't find it in my dictionary...[/quote]

The correct word should be "rhombi" plural of "rhombus" It's a quadrilateral with 4 congruent sides. (A square is always a rhombus, but not all rhombi are squares).

Guys!  I intended this problem as an illustration that the numbers chosen for a problem can make a big difference in difficulty.  I didn't really intend it as a challenge.

However, those interested can find a nice description of the classic (long-division style) square root algorithm at [url]http://www.mathpath.org/Algor/algor.square.root.htm[/url].  The site includes a link to a Web page that will demonstrate the algorithm.

[quote="shobber"]what is rhombuses?[/quote]
The proper question to ask is "What are rhombuses?"

If your dictionary is excessively abridged, please look [url=http://www.m-w.com/cgi-bin/dictionary?book=Dictionary&va=rhombus&x=0&y=0]here[/url]


function
trigonometry
algebra proposed
algebra

Find all continuous functions $f: \mathbb{R}\to\mathbb{R}$ such that \[f(f(f(x)))+f(x)=2x\;\;\forall x\in\mathbb{R}.\]

Very similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=128763]this problem[/url]. (with only a few more difficulties)

[b](*)[/b]  $f(f(f(x)))+f(x)=2x$

[b]1)[/b] $f$ is strictly increasing with $f(0)=0$:
$\forall x,y\in\mathbb{R}: f(x)=f(y)\Rightarrow x={f(f(f(x)))+f(x)\over 2}={f(f(f(y)))+f(y)\over 2}=y$
So $f$ is injective and by the intermediate value theorem strictly monotonic.
If $f$ were strictly decreasing, then
$\forall x,y\in\mathbb{R}: x<y\Rightarrow f(x)>f(y)\Rightarrow f(f(x))<f(f(y))\\\hspace*{1.2in}\Rightarrow f(f(f(x)))>f(f(f(y)))\\\hspace*{1.2in}\Rightarrow x={f(f(f(x)))+f(x)\over 2}>{f(f(f(y)))+f(y)\over 2}=y$
Contradiction!
So $f$ is strictly increasing and
$f(0)>0\Rightarrow f(f(0))>f(0)>0\Rightarrow f(f(f(0)))+f(0)>2f(0)>0$
$f(0)<0\Rightarrow f(f(0))<f(0)<0\Rightarrow f(f(f(0)))+f(0)<2f(0)<0$
both contradicting (*).
This gives $f(0)=0$.

Denote $f^{n}(x)=\underbrace{f(f(\cdots f(x)\cdots))}_{n\text{ times}}$ for all $x\in\mathbb{R},\ n\in\mathbb{N}_{0}$. (here $f^{0}(x)=x$)

Then
[b]2)[/b] $xf^{n}(x)>0$ for all $x\in\mathbb{R}^{*}$ and $n\in\mathbb{N}_{0}$:
1) gives $x>0\Rightarrow f(x)>0\Rightarrow f(f(x))>0\Rightarrow \cdots$
and  $x<0\Rightarrow f(x)<0\Rightarrow f(f(x))<0\Rightarrow \cdots$.
This gives the result.

Now (*) gives $xf^{n+3}(x)+xf^{n+1}(x)-2xf^{n}(x)=0$ for all $x\in\mathbb{R},n\in\mathbb{N}_{0}$.
Also $x^{3}+x-2=(x-1)(x^{2}+x+2)$ has roots $1$ and ${-1\pm\sqrt{7}i\over 2}$. Set $\zeta={-1+\sqrt{7}i\over 2}$.
Then for each $x\in\mathbb{R}$ there exist $a,b,c\in\mathbb{C}$ with
[b](**)[/b]  $xf^{n}(x)=a+b\zeta^{n}+c\overline{\zeta}^{n}$ for all $n\in\mathbb{N}_{0}$.
Then $xf^{n}(x)=\overline{xf^{n}(x)}=\overline{a}+\overline{b}\;\overline{\zeta}^{n}+\overline{c}\zeta^{n}$ for all $n\in\mathbb{N}_{0}$
and since $\{1\}_{n\in\mathbb{N}_{0}}$, $\{\zeta^{n}\}_{n\in\mathbb{N}_{0}}$ and $\{\overline{\zeta}^{n}\}_{n\in\mathbb{N}_{0}}$ are linear independent over $\mathbb{C}$, we get $a=\overline{a}\in\mathbb{R}$ and $c=\overline{b}$.

Now $|\zeta|=\sqrt{2}$ and $\zeta=\sqrt{2}e^{2\pi\lambda i}=\sqrt{2}(\cos 2\pi\lambda+i\sin 2\pi\lambda)$ for some $\lambda\in(0,1)$ and $\cos 2\pi\lambda=-{\sqrt{2}\over 4}$.
If $b=0$, then $c=\overline{b}=0$ and (**) gives $xf(x)=a=x^{2}$ and $f(x)=x$ using 1).
Assume otherwise for some $x\in\mathbb{R}^{*}$.
Then $d=|b|\in\mathbb{R}_{+}$ and $b=de^{2\pi\mu i}$ for some $\mu\in[0,1)$, and
$xf^{n}(x)=a+b\lambda^{n}+\overline{b}\;\overline{\lambda}^{n}=a+d\cdot 2^{n/2}(e^{2\pi(n\lambda+\mu)i}+e^{-2\pi(n\lambda+\mu)i}) \\\hspace*{1.7in}=a+2d\cdot 2^{{n/2}}\cos 2\pi(n\lambda+\mu)$
and using 2)
[b](***)[/b]  $0<{xf^{n}(x)\over 2^{{n/2}}}={a\over 2^{{n/2}}}+2d\cos 2\pi(n\lambda+\mu)$  for all $n\in\mathbb{N}_{0}$.

[b]3)[/b] For all $\alpha,\beta\in\mathbb{R}$ with $\alpha<\beta$ there exist $n,m\in\mathbb{N}$ with $\alpha<n\lambda-m<\beta$:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137062]Here[/url] I already determined all $r\in\mathbb{Q}$ with $\cos 2\pi r$ algebraic of degree $\leq 2$. So $\cos 2\pi\lambda=-{\sqrt{2}\over 4}$ gives $\lambda\not\in\mathbb{Q}$.
Assume $\tau: =\inf \{n\lambda-m: \ n,m\in\mathbb{N},\ n\lambda-m>0\}>0$. ($\tau$ exists, as $\lambda>0$)
Then $(k-1)\tau<1\leq k\tau$ for some $k\in\mathbb{N}$, and then $\tau<{\tau+1\over k}$
and $\tau\leq n\lambda-m<{\tau+1\over k}$ for some $n,m\in\mathbb{N}$ by definition of $\tau$.
This gives $0\leq k\tau-1\leq kn\lambda-km-1<\tau$ with $kn,km+1\in\mathbb{N}$.
By definition of $\tau$ we must have $kn\lambda-km-1=0$ and $\lambda={km+1\over kn}\in\mathbb{Q}$. Contradiction!
So $\inf \{n\lambda-m: \ n,m\in\mathbb{N},\ n\lambda-m>0\}=0$ and $0<n\lambda-m<\beta-\alpha$ for some $n,m\in\mathbb{N}$.
Also $\alpha+p>0$ for some $p\in\mathbb{N}$, and then $(q-1)(n\lambda-m)\leq \alpha+p<q(n\lambda-m)$ for some $q\in\mathbb{N}$.
This gives $\alpha<qn\lambda-qm-p\leq \alpha+(n\lambda-m)<\beta$ with $qn,qm+p\in\mathbb{N}$.

And now for all $k\in\mathbb{N}$ there exist $n_{k},m_{k}\in\mathbb{N}$ with ${1\over 2}-\mu<n_{k}\lambda-m_{k}<{1\over 2}-\mu+{1\over k}$,
and then obviously $n_{k}\mathop{\longrightarrow}_{k\to \infty}\infty$ and $n_{k}\lambda-m_{k}\mathop{\longrightarrow}_{k\to \infty}{1\over 2}-\mu$.
And therefore (***) gives
$0<{xf^{n_{k}}(x)\over 2^{n_{k}/2}}={a\over 2^{n_{k}/2}}+2d\cos 2\pi(n_{k}\lambda+\mu)\\\hspace*{0.97in}={a\over 2^{n_{k}/2}}+2d\cos 2\pi(n_{k}\lambda-m_{k}+\mu)\mathop{\longrightarrow}_{k\to \infty}2d\cos\pi=-2d<0$
Contradiction!

So $f(x)=x,\ x\in\mathbb{R}$ is the only continuous solution of (*).

olorin,
If $f$ is increasing on $\mathbb{R}$ then $f(x)=x\forall x\in\mathbb{R}$. In fact,

If $f(x_{0})>x_{0}$ then $f(f(x_{0}))>f(x_{0})>x_{0}$ and $f(f(f(x_{0})))>...>x_{0}$ therefore $f(f(f(x_{0})))+f(x_{0})>x_{0}+x_{0}=2x_{0}$. Contradiction!
If $f(x_{0})<x_{0}$, similary.

Am I wrong? :P Do you agree? :D

[quote="N.T.TUAN"]Am I wrong? :P Do you agree? :D[/quote]
F*** :mad: This happens to me all the time.
Thanks  :lol:


i must find the "proof of maxwell equations in covariant form" and "proof of maxwell's two homogeneous equations"...

please help me quickly...

its very important for me...

thanks...


The integers $ 1,2,3...n^2$ are placed in an n*n board. Prove that there exist 2 adjecent integers whose difference is atleast n.

I thought i had a pigeonhole solution, but that appearently didnt work.

Hint: How many pairs of adjacent squares are there? [hide="Answer to hint"]$ 2n(n \minus{} 1)$[/hide]
Next, what is the average (positive) difference of each pair of adjacent squares?

I'm not sure if that hint will help... but it's a start.

I have managed to find a board where the average positive difference is less than < n - 1.


geometry
trigonometry
incenter
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

Does anyone have any ideas on how to prove Hero's formula or Area=(1/2)ab*sin(c)?  This is a very interesting question and I've started toying around with it...I'll work on it more over teh weekend but anyone have any ideas??? Sigh...I was never very good with trig proofs :oops:

I vaguely remember seeing the proof of Hero's formula before but I'm not entirely sure how. On the other hand, the proof of A=1/2 ab sin C is relatively easy: just draw in one altitude of the triangle and use A=1/2 bh.

I'll have another think about Hero's formula.

[quote="akatookey"]Does anyone have any ideas on how to prove Hero's formula or Area=(1/2)ab*sin(c)? [/quote]

The latter is pretty simple.  Like TripleM said, you want to drop an altitude h to a side BC.  We know that sin(C) = h/b => h = bsin(C).  We also know that Area = ah/2.  Stick in bsin(C) for h and that gives you exactly what you want.

Hero(n)'s formula is quite a bit messier to derive.  You want to start from absin(C)/2.  Then replace sin(C) by sqrt(1 - cos(C)^2).  Next use the law of cosines to write cos(C) in terms of the three sides of the triangle.  Once you have that, there's a bunch of slightly annoying factoring and maybe a tiny bit of cleverness .  Eventually you get Heron's formula.  If that wasn't clear enough, let me know and I'll go through the whole thing when I have some time.

Extra challenge: can you get to Hero(n) without all that nasty trig?

From what i know (and check the outstanding planetmath web site at  http://planetmath.org/?op=getobj&from=objects&name=HeronsFormula) this is not the Heron formula.

Heron's original proof relied on constructing the incenter and then showing that the area of a triangle is the semiperimeter times the inradius. The proof is quite involved and difficult to come up with. It is also extremely clever. It can be found in William Dunham's [i] Journey Through Genius.[/i]

Hero's formula can be found my manipulating the area sine formula, if I remember correctly.

Let me try it and get back to you in the morning.  :P

x biz kil l, your sig is inaccurate. In Hoc Signo Vinces means "by this sign you shall [b]conquer[/b], not prosper. I love nitpicking  :wink:

-Isaac

I did this one in 6th grade, before I knew about trig or Hero(n)'s formula (I didn't convert it to s(s-a)..., I did that later).

Start with triangle ABC. Label the sides a, b, and c. Call the altitude dropped to c d, and call the line segment from the point of intersection of d and c to corner B x. Then we have two right triangles, with equations:

x^2 + d^2 = a^2
d^2 + (c-x)^2 = b^2

Subtracting the second from the first, we get -2cx + c^2 = b^2 - a^2. Solving for x, we get x = (c^2+a^2-b^2)/(2c). Plugging that into the first equation, we get ((c^2+a^2-b^2)/(2c))^2 + d^2 = a^2. Moving everything to the right and expanding, we get
d^2 = a^2 - (c^4 + b^4 + a^4 +2 (ac)^2 - 2(ab)^2 - 2(bc)^2)/(4c^2)
d^2 = (-c^4 - b^4 - a^4 + 2(ac)^2 + 2(ab)^2 + 2(bc)^2)/(4c^2)
Instead of playing around with this, go to area = cd/2. Then area^2 = ((cd)^2)/4, and we can plug in d:

Area^2 = (-c^4 - b^4 - a^4 + 2(ac)^2 + 2(ab)^2 + 2(bc)^2)/(16)
Area = sqrt(-c^4 - b^4 - a^4 + 2(ac)^2 + 2(ab)^2 + 2(bc)^2)/(16)

This was as far as I got in 6th grade. It can be factored into
sqrt(((a+b+c)/2)((a+b-c)/2)((a-b+c)/2)((-a+b+c)/2))
Then defining the semiperimeter gives Hero(n)'s formula.

I've also read about the incircle method, but I prefer to use my own proof.


limit
trigonometry
calculus
calculus computations

[color=red][b]without[/b][/color] using L Hospital, show that  :D

$\lim_{n\to 0}\left( \frac{n^{6000}-\sin^{6000}{n}}{n^{2}\cdot\sin^{6000}{n}}\right) \;=\;\boxed{1000}$

Rewriting it as (by dividing by $n^{6000}$)
\[\frac{1-\left(\frac{\sin{n}}{n}\right)^{6000}}{n^{2}\left(\frac{\sin{n}}{n}\right)^{6000}}\]
EDIT: See post below


algebra
polynomial
inequalities
function
algebra unsolved

Let $f$ be unitary polynomial. If $x_1<...<x_n \in \mathbb{Z}$. Prove that there is a $k$ so that:

$|f(x_k)|\geq \frac{n!}{2^n}$

This has been discussed before, I'm sure, but I cannot find. The idea is to use Lagrange's interpolation formula and to equal the leading coefficient in the two expressions. This, combined with modulus inequality will lead to contradiction. Please, try it like this and if it's still unclear I will post formulae.

Thanks Harazi  :) I think I have it so I'm posting it to make sure it's alright:

from Lagrange interpolation formula (and comparing leading coefficients) we have:

$\sum _{k=1}^n  \frac {f(x_k)}{\prod _{j \not =k}(x_k-x_j)}=1$

suppose that for all $k$ we have $|f(x_k)| < \frac{n!}{2^n}$, then (taking modulus function as well) we have:

$\frac{n!}{2^n} \sum _{k=1}^n \frac{1}{|\prod _{j \not =k}(x_k-x_j)|} > 1$

which is contradiction because:

$\sum _{k=1}^n \frac{1}{|\prod _{j \not =k}(x_k-x_j)|} < \frac{1}{(n-2)!}$ (very weak bound but I think it's enough here)

I don't understand your last inequality.

Sorry, the last inequality is completely wrong  :blush: I think this is correct (and attainable when $x_i = i$)

$\sum _{k=1}^n \frac{1}{|\prod _{j \not =k}(x_k-x_j)|} \leq \sum _{k=1}^n \frac{1}{(n-k)!(k-1)!}$

Though now I'm not so sure if I get cotradiction


A path zig-zags from (1,0) to (0,0) along line segments (P_n)(P_(n+1)), where P_0 is (1,0) and P_n is (2^−n, (−2)^−n), for n > 0. Find the length of the path.


This code produces an error message:
[code]
[asy]
size(200);
pair A=(0,0);
dot(A);
label("A",A,E);
[/asy]
[/code]
Why? :?
[asy]size(200); pair A=(0,0); dot(A); label("A",A,E);[/asy]

When running the code on my local Asymptote environment I don't get a dot, but a circle (with no label). If I add "import graph;" command to it, it creates the picture that it's supposed to however it also gives 2 warnings: "x scaling in picture is unbounded, y scaling in picture is unbounded". I have no idea how to correct it (or even what's wrong).

Actually, it is much worse than just a problem with this particular code: I cannot produce [b]any[/b] code with label command that would work on the site :( Can you? (the old ones do not count: when I try to quote them verbatim, they come out OK, but if I change just one line in them, I get an error message)

Can you give me an example of a code with label that works on your computer but doesn't work on the site?

Here goes: it is your own example to which I added one command (dot(A); )
[asy]size(100); pair A,B,C,O,I; A=origin; B=2*right; C=1.5*dir(70); O=circumcenter(A,B,C); I=incenter(A,B,C); draw(A--B--C--cycle); dot(O); dot(I); dot(A); draw(circumcircle(A,B,C)); draw(incircle(A,B,C)); label("I",I,W); label("O",O,S);[/asy]

Errrr, it was a small bug that took me over 2 hours to find. Now it's fixed :)


Has anyone solved this with java?
I'm not sure if it can be done with there compiler.  You can't use arraylist so you cannot have any kind of dynamic list of points (unless you make some kind of extensive linked list but that prolly wouldnt run in time).  So inclusion/exclusion is pretty much out.  Using a 2 dimensional array is also out because of memory and processing it one row at a time is also out because of time.

the point is to store the rectangles instead of pixels

I know, but I'm saying that when you store the rectangles you find an intersection point and then create new rectangles that result from the intersection.  It's difficult to store these .  I think I figured it out though.


abstract algebra
superior algebra
superior algebra unsolved

Show if every element of group G is equal to its own inverse, then G is Abelian

:cool: Trivial : $a+b+a+b=0 \Longrightarrow a+b+a=b \Longrightarrow b+a=a+b $  :cool:


linear algebra
matrix
linear algebra unsolved

Show that it's impossible to fill a $2n \times 2n$ matrix with numbers $\{1,2,3,\ldots, 2n\}$ such that every number in the set has appeared at most once in every row, every colum and every diagonal that makes an angle of 45 degrees with the rows. To be more precise it's impossible to have all of the following at the same time:

\[a_{ij}\in \{1,2,3, \ldots, 2n\}\]
\[i \neq j \rightarrow a_{ik}\neq a_{jk}\]
\[i \neq j \rightarrow a_{ki}\neq a_{kj}\]
\[i+j=k+l \rightarrow a_{ij}\neq a_{kl}\]

[quote="eternalbliss"]It seems like an overkill to me. Is there a sudoku (any nxn) in which every row, column and the two main diagonals have distinct permutations of ${1,2,...,n}$?

If there is, I'd like to see an example of it.[/quote]

I don't know the answer to your question. However the problem I posted seems to be optimal i.e. it is possible that all conditions are satisfied except that only one of these 'quasi'-diagonals (I don't have a better name for them) fail to contain distinct numbers.


trigonometry
function
logarithms
algebra
polynomial
complex analysis
complex numbers

proof:  $ cosz \equal{} \prod\limits_{n\equal{}0}^{\infty}[1\minus{}(\frac{2z}{(2n\plus{}1)\pi})^2]$. help me. thanks

A path:
Step 1: Show that the product $ f(z)$ converges absolutely for all $ z$, and uniformly on compact sets. Conclude that the limit is an entire function.
Step 2: Show that $ f$ has simple zeros in exactly the same places as $ \cos z$, that it's even, and that $ f(0)\equal{}1$.
Step 3: Show that $ f(z\plus{}\pi)\equal{}\minus{}f(z)$
[hide="Hint for step 3"]Write $ 1\minus{}\left(\frac{z}{(n\plus{}\tfrac12)\pi}\right)^2\equal{}\frac1{(n\plus{}\tfrac12)^2}\left(\minus{}z\plus{}\left(n\plus{}\frac12\right)\pi\right)\cdot\left(z\plus{}\left(n\plus{}\frac12\right)\pi\right)$, and shuffle these factors between terms. Be careful here that the terms you move around balance in the limit.[/hide]

From the above, $ f(z)\equal{}g(z)\cos z$, where $ g$ is an even entire function with no zeros, $ g(0)\equal{}1$, and $ g(z\plus{}\pi)\equal{}g(z)$ for all $ z$. We can even easily see that $ g$ is real-valued on $ \mathbb{R}$. Is that enough to ensure that $ g(z)\equal{}1$ everywhere? No. Functions like $ e^{\sin^2(z)}$ are still possible. That leads to...

Step 4: Show that $ f$ doesn't grow too fast: $ |f(z)|\le e^{\alpha|z|^2}$ for some constant $ \alpha$.

Step 5: $ \ln g(z)$ is an entire function, and its real part is never greater than $ \alpha|z|^2$. It must therefore be a polynomial of degree no more than $ 2$. Since it's also periodic, it must be constant and we're done.

Now, how many of these pieces can you understand and do yourself?


Solve equation:
$ \sum^n_{i\equal{}1}\frac{1}{sin2^ix}\equal{}0(with n\in Z^{\plus{}}$

Hm... I´ve already seen the solution for very similar to this one.

Use the identity $ sinx \equal{} cotg(\frac{x}{2}) \minus{} cotgx$. Actually, I don´t know how someone could have guessed it. Does somebody know?


calculus derivative function integration Gauss vector topology

Let $ u \in C^{2}(\Omega) \cap C^{1}(\overline\Omega)$ be a harmonic function on a bounded open connected subset $ \Omega$ of $ \mathbf R^{n}$ ($ n\ge2$). Show that there exist at least two points in $ \partial\Omega$ at which $ \frac{\partial u}{\partial\nu}$ vanishes.

I think that the idea is pretty clear. For example, if the bounded open connected subset is a circle, because $ \int_{\partial \Omega}\nabla u . \vec n ds = 0 = \int_{0}^{2 \pi}\nabla u(r,\theta ) . \vec n(r, \theta) ds$ the conclusion follows quite easily. It suffices to prove that if $ f$ is a $ 1$-periodic continuous function such that $ \int_{[0,1]}f(x) dx = 0$, then $ f$ has at least $ 2$ zeroes in $ [0,1]$, wich is obvious. Anyone for a real proof in the general case ?

The integral of the normal derivative of a harmonic function over the boundary is always zero (just apply Gauss-Ostrogradsky, or the Divergence Theorem, or whatever you want to call it to the vector field $ \nabla u$). If that normal derivative were to vanish in at most one point, it would have the same sign everywhere, with only one zero. Clearly then, it's impossible for the integral to be zero.

Why if $ \partial u/\partial\nu$ vanishes at exactly one point in $ \partial\Omega$, then $ \partial u/\partial\nu$ does not change sign in $ \partial\Omega$?

The boundary of $ \Omega$ is a closed $ n-1$-manifold (you didn't mention smoothness, but I'm assuming as much, since you need it in order to define $ \frac{\partial u}{\partial\nu}$). Now try to prove that if the boundary of a non-empty open subset $ U$ of a closed $ m$-manifold $ M$ ($ m\ge 1$) consists of precisely one point $ p$, then $ U=M\setminus p$. Then apply this to $ M=$the boundary of $ \Omega$ and $ U=$the set where the normal derivative is positive or negative (one of these sets is certainly non-empty).


topology
linear algebra
matrix
linear algebra unsolved

Let $A$ be a matrix. Proof without using topology arguments that $A+tI_{n}$ is diagonalisable for infinitely many values of $t$.

That's obviously wrong: just consider a nilpotent nonzero matrix $A$.

i don't get why it's wrong , can you explain more?

Have you ever seen a respectful matrix all of whose eigenvalues are equal
 and which is diagonalisable, but not a homotethy? I haven't seen something like that...

maskman,
To make it more concrete from harazi's comment take $n=2$ and $A=\begin{pmatrix}0 & 1\\ 0 & 0 \end{pmatrix}$


iF THE ROOTS OF THE EQUATION-     x^3+x+1=0  are a,b,c,
then find the value of 1/(a^3)+1/(b^3)+1/(c^3)

[hide]When we use the given equation, we see that $\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$
equals
\[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\]
Which is equal to
\[\frac{bc+b+c+1+ac+a+c+1+ab+a+b+1}{(a+1)(b+1)(c+1)}\]
And Vieta's formulas tell us that $a+b+c=0$, $abc=1$ and $ab+ac+bc=-1$
So the answer is $2$[/hide]

Jan, that's wrong.
[hide="The correct answer is"]$-4$[/hide]

[quote="Jan"]When we use the given equation, we see that $\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$
equals
\[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\]
[/quote]
Isn't it $-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)$?

[quote="Jan"]And Vieta's formulas tell us that $a+b+c=0$, $abc=1$ and $ab+ac+bc=-1$
So the answer is $2$[/quote]
I think it's $a+b+c=0$, $abc=-1$ and $ab+ac+bc=1$, which give us $-2$ for the answer... Did I do something wrong? :|

Chess64, can you post your solution?

[b]EDIT[/b]- Now I see my mistake. I did my arithmetic wrong :oops:
Yes, the answer is $4$.

It follows pretty much by expansion and then everything you said...

$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} = -\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)$

and put a common denominator to get

$-\frac{ab+bc+ca+2(a+b+c)+3}{abc+ab+bc+ca+a+b+c+1} = -\frac{1+2(0)+3}{-1+1+0+1} = -4$.

Oops, yeah, you're right. Never mind :blush:

Please look very carefully to the question. The given equation is $-x^{3}+x+1=0$, with a $-$ in front, although it's not very clear.
So if take that in account, I think my solution is pretty correct...
If that '-' wouldn't be there, the solution would indeed be -4

Vieta's formula is: $s_{i}=(-1)^{i}.\frac{a_{n-i}}{a_{n}}$
With $s_{1}=r_{1}+r_{2}+r_{3}+...$, $s_{2}=r_{1}r_{2}+r_{1}r_{3}+...$
($r_{n}$ the roots of the polynomial)

So in the equation $-x^{3}+x+1=0$ we get
$a+b+c=s_{1}=0$
$ab+ac+bc=s_{2}=-1$
$abc=s_{3}=1$

So look carefully at the question, I still think the answer is $2$

Ah, I interpreted that '-' to be a dash, not a negative sign. An ambiguous question indeed, though it is rare to leave a space between a negative sign and what it modifies.

The question does not have a -ve sign.it is a dash.thats why i left the space


Find the smallest positive integer which is divided by 89 and consists of 6, 7.

example) 66767 consists of 6, 7 but is not divided by 89, so 66767 is not our consideration.


I only received answer from one team: Irrationally Outraged Without ARML which is from Iowa. I received a Relay, a Team, and two individuals. So, because only one team turned in (there were 4 teams registered though; Florida had two teams The Balla Squad and The Triumvirate of Asianness; one GA team, and obviously, Iowa team), Iowa team won the first Reality Mock ARML. Good job and thanks for participating!

By same reason, I'm not going to write up "official" solution and etc. You guys (both people who participated and people who just read about it) can look at the problems and post your solutions. I might come and give hints but I can assure that before real ARML, you should be able to find solutions to all the problems in this thread.

Err, yes.  Sorry about that :blush: .  I've been a little busy...

My bad dude, we did everything except the team round... that seemed to take too much time coordination.

[quote="quantum leap"]So do we just post any anwers we got on this thread?[/quote]

Yeah and I'll confirm it for you.

Woohoo! Go Iowa!

Just like what's going to really happen at ARML, right?  :lol:


Well i really like this problem:
Let $1 \le k\le n$ be two natural numbers.
and $a_1,...,a_k$ be numbers that

$\sum_{i=1}^k a_i=\sum_{i=1}^k a_i^2=...=\sum_{i=1}^k a_i^k=n$
then prove:
$\prod_{i=1}^k (x+a_i)=\sum_{i=0}^k \binom{n}{k} x^{k-i}$

are [tex]a_1,...,a_k[/tex] integers , reals ?

Though the way to tackle the problem is quite obvious (consider the polynomial with $-a_1,...,-a_k$ as roots) I think something is wrong with the right side of equality we're about to prove:

$\sum_{i=0}^k \binom{n}{k} x^{k-i}$

I think it'd rather be something like $\sum_{i=0}^k \binom{k}{i} x^{k-i}$ (just a guess, I'm not sure about it)

[b]lomos_lupin[/b] can you check it, please?

The problem is straigtforward with Newton formulae.

Megus ,I have checked it .The original form is correct.
you have to prove :
$\prod(x+a_i)=x^k+\binom{n}{1}x^{k-1}+\binom{n}{2}x^{k-2}+...+\binom{n}{k}x^{k-k=0}=\sum_{i=1}^k \binom{n}{i} x^{k-1}$
And $a_1,...,a_n \in \mathbb{R}$.

To Harazi=you are good, man.

I was thinking that [b]megus[/b] is somehow right and then i reached this :
That megus suggested is what problem is asking for cause :
we can prove $k=n$ like this :
[hide="proof"]
$\frac{\displaystyle (\sum_{i=1}^k a_i)=n}{k}\le \sqrt{\frac{\displaystyle (\sum_{i=1}^k a_i^2)=n}{k}}$ So we have $\frac{n^2}{k^2} \le \frac{n}{k}$ which means $n\le k$ but we have
$k \le n$ So $k=n$.[/hide]


advanced fields
advanced fields theorems

I am confused with the definition of $ H_0(X,A)$ (where $ A \subset X$).   The definition of $ H_n(X,A)$, for $ n \geq 1$, is

\[ H_n(X,A)\equal{}\frac{ker \{\partial_n: \Delta_n(X,A) \rightarrow \Delta_{n\minus{}1}(X,A)\}}{Im \{ \partial_{n\plus{}1}: \Delta_{n\plus{}1}(X,A) \rightarrow \Delta_n(X,A)\}}\]

However, when $ n\equal{}0$, $ \Delta_{\minus{}1}(X,A)$ doesn't even make sense, so what is the definition for $ H_0(X,A)$.   Moreover, what is the definition for $ H_{n}(X,A)$, when $ n \leq 0$.

Thank you.

All the chain groups past dimension 0 are taken to be 0


From the set of numbers {1,2,3,...20} one by one two numbers,

a) we are subtracting from first second. What probability that the difference is larger is from 2?
b) we are subtracting from second first. What probability that the difference is larger or equal is 4?

Are you sampling with replacement or without replacement? In other words, is your pair of numbers allowed to be $ (8,8)?$


quadratics
algebra
IMO Shortlist
algebra unsolved

Given $a,b\in{R}$  and the equation $x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least one real root, determine the minimum value of $a^{2}+b^{2}$.

I get $4/5$.
My solution is long and not very elegant. I hope there aren't any errors.
[hide]Let's define $P(x)=x^{4}+ax^{3}+bx^{2}+ax+1$.

$P(0)=1$, and for $x\neq 0$, $P(1/x)=P(x)/x^{4}$, so if $z$ is a root of $P$, $1/z$ is a root too.

Therefore, $P(x)=(x-z_{1})(x-1/z_{1})(x-z_{2})(x-1/z_{2})=(x^{2}-t_{1}+1)(x^{2}-t_{2}x+1)$,
where $t_{1}=z_{1}+1/z_{1}$ and $t_{2}=z_{2}+1/z_{2}$.

Identifying coefficients, $a=-(t_{1}+t_{2})$, and $b=2+t_{1}t_{2}$.
It follows that $t_{1}$ and $t_{2}$ are the solutions of the quadratic equation $t^{2}+at+(b-2)=0$, so we can say that
$t_{1}=\frac{-a+\sqrt{a^{2}-4b+8}}{2}$, and $t_{2}=\frac{-a-\sqrt{a^{2}-4b+8}}{2}$.

From $t_{1}=z_{1}+1/z_{1}$, or the equivalent quadratic equation ${z_{1}}^{2}-t_{1}z_{1}+1=0$, we can get back the values of
$z_{1}=\frac{t_{1}+\sqrt{{t_{1}}^{2}-4}}{2}$, and $\frac{1}{z_{1}}=\frac{t_{1}-\sqrt{{t_{1}}^{2}-4}}{2}$.
The same can be done with $t_{2}$ and $z_{2}$.

Now, if we want one of the roots to be real, we must have  at least one real $t_{i}$ with ${t_{i}}^{2}-4\geq 0$, that is $|t_{i}|\geq 2$.
That means $a^{2}-4b+8\geq 0$ and
$\frac{-a+\sqrt{a^{2}-4b+8}}{2}\geq 2$, equivalent to $2a+b+2\leq 0$, or
$\frac{-a-\sqrt{a^{2}-4b+8}}{2}\leq-2$, equivalent to $-2a+b+2\leq 0$.

The minimum $a^{2}+b^{2}$ (geometrically, the square of the lowest distance to the origin) will be reached at the boundary of each of the symmetrical lines $\pm 2a+b+2=0$.
Taking for instance $2a+b+2=0$, the perpendicular that goes through the origin is $a-2b=0$, and the intersection is at $a=-4/5$, $b=-2/5$ (we can check that $a^{2}-4b+8\geq 0$), so the minimum $a^{2}+b^{2}$ is $4/5$.[/hide]

My solution is simlar to lordWings's:
$P(x)=0$
$\leftrightarrow(x+\frac1x)^{2}+a(x+\frac{1}{x})+b-2$
so $g(y)=y^{2}+ay+b-2$ should have a root $\leq-2$or$\geq 2$

[quote="silouan"]Ithink the problem is :
If the equation $x^{4}+ax^{3}+2x^{2}+bx+1=0$  has at least one real root then  $a^{2}+b^{2}\geq 8$[/quote]

No... the original problem is just what I've posted.

[quote="lordWings"]

 $|t_{i}|\geq 2$.
That means $a^{2}-4b+8\geq 0$ and
$\frac{-a+\sqrt{a^{2}-4b+8}}{2}\geq 2$, equivalent to $2a+b+2\leq 0$, or
$\frac{-a-\sqrt{a^{2}-4b+8}}{2}\leq-2$, equivalent to $-2a+b+2\leq 0$.

[/quote]

$|t_{i}|\geq 2$ doesn't necessarily  mean that $\frac{-a+\sqrt{a^{2}-4b+8}}{2}\geq 2$ or $\frac{-a-\sqrt{a^{2}-4b+8}}{2}\leq-2$. For example, if $\frac{-a-\sqrt{a^{2}-4b+8}}{2}\geq 2$, it follows that $\frac{-a+\sqrt{a^{2}-4b+8}}{2}\geq 2$. And if $\frac{-a+\sqrt{a^{2}-4b+8}}{2}\leq-2$, then $\frac{-a-\sqrt{a^{2}-4b+8}}{2}\leq-2$.

Further more, I wonder if $\frac{-a+\sqrt{a^{2}-4b+8}}{2}\geq 2$ is equivalent to $2a+b+2\leq 0$. In the simplification, you might get $\sqrt{a^{2}-4b+8}\geq a+4$, but this doesn't mean that $a^{2}-4b+8\geq (a+4)^{2}$ unless $a+4\geq0$, which hasn't been checked yet.

Answer to your first comment: I'll mark the important part that maybe you missed:
"If we want at least one of the roots to be real, we must have [b]at least one[/b] real $t_{i}$ with $|t_{i}|\geq 2$".
With that in mind, I wrote the less strict conditions so that there's at least one $|t_{i}|\geq 2$.
Your two extra conditions would force the two roots to be $|t_{i}|\geq 2$, which is more strict than needed.

Answer to your second comment: You're right, I should have checked more carefully all the cases before squaring.
-If $a\geq 0$, the less strict condition is $a^{2}-4b+8\geq 0$ and $\frac{-a-\sqrt{a^{2}-4b+8}}{2}\leq-2$.
-If $a\leq 0$, the less strict condition is $a^{2}-4b+8\geq 0$ and $\frac{-a+\sqrt{a^{2}-4b+8}}{2}\geq 2$.
The combined condition can be written $a^{2}-4b+8\geq 0$ and $\sqrt{a^{2}-4b+8}\geq 4-|a|$.
-If $|a|\leq 4$, the condition is $\sqrt{a^{2}-4b+8}\geq 4-|a|\geq 0$, equivalent to $-2|a|+b+2\leq 0$, and we get the min. distance to the origin at $|a|=4/5$, $b=-2/5$, where $a^{2}+b^{2}=4/5$.
-If $|a|>4$, we'd have $a^{2}+b^{2}>16$, which is always worse.

orl already posted this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58582]problem[/url], with the source
IMO ShortList 1973, Sweden 2, IMO 1973, Day 1, Problem 3.
There are nice solutions there, using inequalities.


geometry
analytic geometry
college
ratio

Prove Varignon's Theorem using mass points:
If the midpoints of consecutive sides of a quadrilateral are connected, the resulting quadrilateral is a parallelogram.

Taken from Tom Rike's speech on Mass Points and Barycentric Coordinates (Berkeley Math Circle).

In addition, when exactly can one use mass points on non-degenerate, non-triangular polygons?

I had put this in the Intermediate Topic forum, but nobody was able to answer my question.  And so, I pose it here.

just to tell you,my mass point threads have barely any replys and i checked old threads on mass points, they usually dont even have a reply

so im going to assume aops people dont like mass points

and sorry i cant answer you original question

Label the vertices of the quadrilateral $ A$,$ B$,$ C$, and $ D$. Let $ E$ be the midpoint of $ AB$, $ F$ the midpoint of $ BC$, $ G$ the midpoint of $ CD$, and $ H$ the midpoint of $ DA$.

Assign the mass of 1 to each of the vertices $ A$,$ B$,$ C$, and $ D$. The points $ E$,$ F$,$ G$, and $ H$ will then all have mass 2, and the center of mass of the quadrilateral $ ABCD$ will be located at $ 1A\plus{}1B\plus{}1C\plus{}1D \equal{} 4M$ for some point $ M$. We now recalculuate this center of mass in two ways.

$ 1A\plus{}1B\plus{}1C\plus{}1D \equal{} (1A\plus{}1B)\plus{}(1C\plus{}1D) \equal{} 2E \plus{} 2G \equal{} 4M$ and $ 1A\plus{}1B\plus{}1C\plus{}1D \equal{} (1B\plus{}1C)\plus{}(1A\plus{}1D) \equal{} 2F \plus{} 2H \equal{} 4M$

So $ 2E \plus{} 2G \equal{} 4M$ and $ 2F \plus{} 2H \equal{} 4M$ and, by the definition of mass point addition, the point $ M$ is located on both of the line segments $ EG$ and $ FH$; i.e. at their intersection. Furthermore, the ratio $ EM: MG \equal{} 2: 2$ and $ FM: MH \equal{} 2: 2$. From here it is not hard to prove that $ EFGH$ is a parallelogram.

Here's the way I look at it.  Following gauss202's notation, we know that
\begin{align*}
  E &= \frac{A + B}{2} \\
  F &= \frac{B + C}{2} \\
  G &= \frac{C + D}{2} \\
  H &= \frac{D + A}{2}.
\end{align*}
Therefore
\[ E + G = \frac{A + B}{2} + \frac{C + D}{2} = \frac{A + B + C + D}{2}
           = \frac{B + C}{2} + \frac{A + D}{2} = F + H.
\]
But it's well known that $ E + G = F + H$ implies that $ EFGH$ is a parallelogram.  For example, we can rewrite the equation as $ E - F = H - G$, which means the sides $ \overline{EF}$ and $ \overline{GH}$ are parallel and equal.


analytic geometry
conics
parabola
algebra
polynomial
Vieta
function

On the plane, a Cartesian coordinate system is chosen. Given points $ A_1,A_2,A_3,A_4$ on the parabola $ y \equal{} x^2$, and points $ B_1,B_2,B_3,B_4$ on the parabola $ y \equal{} 2009x^2$. Points $ A_1,A_2,A_3,A_4$ are concyclic, and points $ A_i$ and $ B_i$ have equal abscissas for each $ i \equal{} 1,2,3,4$.
Prove that points $ B_1,B_2,B_3,B_4$ are also concyclic.

Suppose that $ (a,ka^2),(b,kb^2),(c,kc^2),(d,kd^2)$ are some concyclic points on the parabola $ y \equal{} kx^2$, where $ k\neq 0$. We will prove that
\[ a \plus{} b \plus{} c \plus{} d \equal{} 0
\]
Let $ \omega(O,R)$ be the circle circumscribed around the chosen points and assume that $ O \equal{} (m,n)$.

So $ (a \minus{} m)^2 \plus{} (ka^2 \minus{} n)^2 \equal{} R^2$, and the same equation holds for $ b,c$ and $ d$.

Therefore, the polynomial $ P(x) \equal{} k^2x^4 \minus{} 2knx^2 \plus{} x^2 \minus{} 2xm \plus{} m^2 \plus{} n^2 \minus{} R^2$ has 4 real roots, namely $ a,b,c,d$.
Since the coefficient in front of $ x^3$ equals $ 0$, by Vieta's formula we get:
\[ a \plus{} b \plus{} c \plus{} d \equal{} 0
\]
Now apply this result for cases $ k \equal{} 1$ and $ k \equal{} 2009$ and we get the desired result.

P.S: Actually this is mine solution, try to figure out the official one. It is a bit harder though.

yeah my solution is similar, I reach the same equation but after that I take e new point and get it's coordinates as functions of the coordinates of the center of the first circle in such a way, that the new equation for the Bi-s and the new point (if it's K then KBi^2=const) is const - it's enoug the coeficients in frotn of x^2 and x to be equal in the 2 equations ;)

Let $ N_i(x_i,ax_i^2)$. I proved that whether $ N_1,N_2,N_3,N_4$ is concyclic or not does not depend on $ a$. This is easy. :D

[quote="Umut Varolgunes"]Erken, I think your solution contains a bug. Yes it's true that you have proved; if $ (a,ka^2), (b,kb^2), (c,kc^2), (d,kd^2)$ are concyclic, $ a \plus{} b \plus{} c \plus{} d \equal{} 0$. But you also must prove that if $ a \plus{} b \plus{} c \plus{} d \equal{} 0$, $ (a,ka^2), (b,kb^2), (c,kc^2), (d,kd^2)$ are concyclic; for proving that the points on $ 2009x^2$ parabola are concyclic. There is a little bit more analysis.[/quote]
hmm...I personally disagree that this might be considered as a bug. To be honest, I put it down for an obvious fact, that doesn't require any further explanations, though it might be a bit awkward.

- If $ a,b,c,d$ are [i]different[/i] coordinates of the points on $ y\equal{}x^2$, then we should prove that the circumcircle of a triangle with vertices at 3 points from $ \{a,b,c,d\}$ intersects the graph of $ y\equal{}2009x^2$ at precisely $ 4$ points - if that is proven, then the rest easily follows from the made observations - sorry for lite notations and explanations.

Assume the contrary, i.e the circumcircle is tangent to the graph. Clearly, the graph of both equations can be imagined as two [i]branches[/i], and two basic cases are possible, two of the points lie on each of the branches. and 1 point on one brach and 3 points on another. Suppose that $ a,b$ and $ c,d$ are on different branches. So circumcircle of $ \{a,b,c\}$ is tangent to the graph, and so is the circumcircle of $ \{a,b,d\}$. It yields that $ c\equal{}d$, contradiction.
Now if $ a$ and $ b,c,d$ are on different branches then the circumcircle of $ a,b,c$ is tangent to the graph at $ a$ and so is $ a,c,d$,so all these points are concyclic.

I actually think, this is exactly what a bug is. A minor mistake, which can be fixed easily; but if it isn't fixed, the solution can't be counted as completed. You may have already done it on your own, but there was nothing written in your first post about it. Now everything is more clear, thanks for paying attention.


conics
parabola
geometry unsolved
geometry

1.Line FV2.Put on this line the poin F2 simmetric of F wrt V2.Line perpendicular to FV2 at F2 is the diretrix
2.The same way find the other diretrix
3.The intersections points of the parabolas will be on the angle bissector of the diretrixes.
4.The points will be the centers of the circles tangents to the diretrix and go to focus F


How do you find the upper and lower bounds of the SOLUTIONS of a polynomial? and how is this method proved. Well, my text says to use synthetic division on the polynomial and if all the coefficients of the quotient and the remainder are positive then no root is greater than c, which is the number being tested. This is supposed to be the upper bound. The lower bound is simply a tested number which produces a quotient and remainder of alternating signs. But my problem is proving this. Descartes' rule of signs is obviously what we start with but i still can't find the answer.

[quote="Geeimatree"]my text says to use synthetic division on the polynomial and if all the coefficients of the quotient and the remainder are positive then no root is greater than c, which is the number being tested. [/quote]

You seem to be talking about http://www.math.csusb.edu/math110/src/poly/root_bound.html .  I hope it helps.


geometry
3D geometry
prism
calculus
integration

Lucifa has a badly misshapen brick (it's still a rectangular prism, though) which has sides of integral dimensions. If eight edges have the same length and the volume is 166, what is the surface area?

the answer was premeditated

[hide]um...if the prism was 1by1by166 it would have 8 sides of 1 and 4 sides of 166 therefore fitting the requirements...if I am right then...nice brick! (I can whack people with it)

edit: sorry forgot surface area $166*4+2=666$[/hide]

[hide]$x^2*y = 166$

Find $2*x^2 + 4xy$

Since the only factors of 1, 2, 83, and 166, x must be 1 and y 166. $2*x^2 + 4xy = 2 + 4*664$ or $\boxed{664}$[/hide]

[hide]Dimensions are 1x1x166, so the surface area is 2x1+4x166=666.[/hide]

I took 1 and 166 as the factors of 166 and it worked for the volume. After that its just 2(1^2)+2(166)+2(166)=666 :diablo:

Does anyone else think that Daesun should write the contests?

[quote="Treething"][quote="fireemblem13"]who's Daesun?
what contests? :huh:[/quote]

Daesun = Treething = that guy who used to be moderator once upon a time, then got banned, then got reinstated as a member, then got banned again, and now he's back.[/quote]

Don't forget, he won nats!  :D

[quote="Treething"][quote="fireemblem13"]who's Daesun?
what contests? :huh:[/quote]
Daesun = Treething = that guy who used to be moderator once upon a time, then got banned, then got reinstated as a member, then got banned again, and now he's back.[/quote]
How did you get banned the first time? Just wondering.

[quote="Remorseless Thinking"][quote="Treething"][quote="fireemblem13"]who's Daesun?
what contests? :huh:[/quote]

Daesun = Treething = that guy who used to be moderator once upon a time, then got banned, then got reinstated as a member, then got banned again, and now he's back.[/quote]

Um, yeah? How'd you get banned twice?[/quote]

same way people die twice

mkay... how did treething get banned the first time??
I know about the picture of MCrawford but I can only imagine what else daesun did...

What picture? Why did treething get banned?:what?:
Am I the only one who is completely in the dark?  :huh:

[hide]I'm going to make one of these to start whacking people with it.

It is a 1*1*166, so the surface area is 666, which does equal 37*18.

Nice questions that you create Treething.[/hide]


geometry
trapezoid
quadratics
Pythagorean Theorem
algebra

Source: MMPC Part 1, Practice

I found this problem very interesting:

[b]An isosceles trapezoid of area $A$ is circumscribed about a circle. If the radius of the circle is $r$, then find the lengths of the parallel sides in terms of $A$ and $r$.[/b]

Whoops.  :blush:

I meant isosceles trapezoid. I fixed it.  Sorry. :(

[hide="I tried"]Let $b_1,b_2$ be the lengths of the shorter base and longer base, respectively. $(b_1<b_2)$

The height of the trapezoid is $2r$. Area of trapezoid yields $A=\dfrac12(2r)(b_1+b_2)$.

$b_1+b_2=\dfrac{A}{r}$

Use the fact that a vertex is equidistant from the point of tangencies and that it is isosceles. Pythagorean theorem yields

$\left(\dfrac{b_2-b_1}2\right)^2+(2r)^2=\left(\dfrac{b_1+b_2}2\right)^2$

$b_1b_2=(2r)^2$

Solve for x: $x^2-\dfrac{A}{r}x+(2r)^2=0$

$b_1=\dfrac{A}{2r}-\dfrac{\sqrt{A^2-16r^4}}{2r}$

$b_2=\dfrac{A}{2r}+\dfrac{\sqrt{A^2-16r^4}}{2r}$[/hide]

Where did you get [code]\left(\dfrac{b_2-b_1}2\right)^2+(2r)^2=\left(\dfrac{b_1+b_2}2\right)^2 [/code]?

Edit: I don't know how to do the code thing.....

[quote="10000th User"]Please draw a diagram, don't just read my answer! If you draw the diagram, Computer Nuke and riddler, both of your questions will be answered.[/quote]
Ive tried that but I cant get how you got the height :(  please help me!
you can have different height trapezoids in the circle of different height why $2r$? :?
Is it the circle is IN the trapezoid?

The height is $2r$ and the sum of bases is $\dfrac{A}{r}$, now knowing tangents from a point are equal (by calling the tangents $x$ and $y$), therefore the two bases are $2x$ and $2y$, you know that $2x+2y=\dfrac{A}{r}$, therefore the side is $x+y$ which is $\boxed{\dfrac{A}{2r}}$.

That should be right, but correct me if I'm wrong :)

we are not looking for the sum, but the individual sides...

[quote="chess64"][quote="riddler"][quote="10000th User"]
The height of the trapezoid is $2r$.
[/quote]
how did you get that??
 :huh:[/quote]

??? isn't that obvious? :huuh: i didn't even draw a diagram but i can visualize it. the radius is $r$, so the diameter is $2r$...and remember that the circle is obviously inscribed in the trapezoid....[/quote]
Sorry I thought the trapezoid was inscribed in a circle. :D

[quote="Iversonfan2005"][quote="frt"]Iversonfan2005, your idea was interesting, but you are supposed to find the lengths of two parallel sides ($2x$ and $2y$) ;)[/quote]

sorry about that, I was looking for the sides, not the bases.

Well, after algabraic manipulation using right triangles, I found:

$x+y=\dfrac{A}{2r}$ and $xy=r^2$, and I'm trying to put those together....[/quote]
If you have $x+y=\dfrac{A}{2r}$ and $xy=r^2$, we can use the fact that $x$ and $y$ are the roots of the quadratic equation $w^2-\left(\dfrac{A}{2r}\right)w+r^2=0$. Mutiplying the roots by 2, you get the lengths (since the lengths are $2x$ and $2y$).

The last quadratic equation is basically the same as 10000th User's equation (except that we have to multiply the answers by 2).


H is inside triangle ABC, angle BHC+A=180 degrees and angle AHC+B=180 degrees.

show that H is the orthocenter.

[hide="solution"]
Let $ A'$ be the intersection of $ AH$ with the circumcircle of the triangle.
Then $ \widehat{BHC}\equal{}\widehat{BA'C}$
$ \widehat{AHC}\plus{}\widehat{B}\equal{}180,\ \widehat{AHC}\plus{}\widehat{CHA'}\equal{}180\Rightarrow\widehat{CHA'}\equal{}\widehat{B}$
$ \widehat{ABC}\equal{}\widehat{AA'C}\Rightarrow\widehat{CHA'}\equal{}\widehat{HA'C}\Rightarrow\widehat{BHA'}\equal{}\widehat{BA'H}$
Then triangles $ BHA'$ and $ CHA'$ are isosceles$ \Rightarrow HA'\perp BC$
In the same way $ BH\perp AC$
[/hide]

sorry could you explain a bit more? i havn't really properly learned circle geometry yet.

I do not follow, either.

Why is line 2 true?

In line 3, shouldn't it be A'HC in the first figure named?

Please explain more.

He first asserted that $ \angle{BHC}=\angle{BA'C}$. We let $ \angle{BAC}= x$, so $ \angle{BHC}= 180-x$, and since $ ABA'C$ is cyclic, $ BA'C = 180-\angle{ABC}= 180-x$, so $ \angle{BHC}=\angle{BA'C}$.

The next line is quite straightforward, the first equality is given, the second is becuase $ H%Error. "inAA" is a bad command.
'$, then $ a = b,a = c\Rightarrow b = c$.

The first equality of the third line is because $ \angle{ABC}$ and $ \angle{AA'C}$ subtend arc $ AC$. But from the third equality of the second line, $ \angle{B}=\angle{ABC}=\angle{CHA'}$, and also $ \angle{ABC}=\angle{AA'C}=\angle{HA'C}$ from the first equality of the third line, so therefore $ \angle{CHA'}=\angle{HA'C}$. By symmetry, we can get the other equality, since there is nothing special about $ C$ and we can apply the same to $ B$.

Therefore, from the equal angles, $ BH = BA'$ and $ CH = CA'$, and $ BHCA'$ is a kite, so $ HA'\perp BC$, and the same can be done by drawing $ BH$ to hit the circumcircle at $ B'$, and we get $ BH\perp AC$, and we're done.

Very nice solution red_dog!

uh...(sorry i am posting question too hard for myself)

why is arc BHC=arc BA'C?.  BC isn't the diameter,...?  i'm 100% sure i'm missing something obvious.

oh sorry i was reading redog's post. i think he typoed.

yeah thanks for clearing things up for an idiot(me).


quadratics
number theory unsolved
number theory

How to solve $ x^2 \minus{} 2y^2 \plus{} x \minus{} 4y \equal{} 2$ in integers?

$ x^2 \minus{} 2y^2 \plus{} x \minus{} 4y \equal{} 2\Longleftrightarrow (2x\plus{}1)^2\minus{}8(y\plus{}1)^2\equal{}1$ follows Pell equation.


algebra
polynomial
linear algebra
matrix
topology
linear algebra unsolved

Let $A=(a_{ij})_{i,j=1,\ldots ,n}\in M_{n}(\mathbb{R})$. Prove the following statements:

     a) \[ tr(C^{-1}AC)=tr(A)\forall C\in M_{n}(\mathbb{R}),  \]
          invertible matrix.

     b) for any $A\in L(\mathbb{R^n},\mathbb{R^n})$ there is an unique real number $tr(A)\in\mathbb{R}$
         such that \[ tr(A)=tr(A_{\mathcal{B}}),  \] for any basis $\mathcal{B}$ included in $\mathbb{R^n}$.

for a) since the characteristic polynomial of $A$ and $C^{-1}AC$ is the same and $trA$= the sum of the eigenvalues the conclusion follows.

Or, for a VII-th degree student, just use that $tr(AB)=tr(BA)$. Of course, b) is obvious from a), which is itself obvious. ;)

[quote="harazi"]$tr(AB)=tr(BA)$[/quote]

Thanx harazi, now I know why there are not matrices A,B such that AB-BA=I (I is identity matrix) :)

Isn't there in fact a theorem saying that with $A$ and $B$ square

$AB$ and $BA$ have exactly the same characteristic polynomial
I dislike that theorem, because or professor avoided the singular case by continuity, which does not work for all fields

Fredbel6, never mind that it does not work for all fields. charpoly(AB,t)-charpoly(BA,t) is some algebraic expression of elements of matrices A and B. If it vanishes on an open set of matrices and all real, say, t, then all its coefficients (which are a priori integers) are equal to zero. The last fact does not depend on a choice of field, so it proves a theorem in all fields.

[quote]The last fact does not depend on a choice of field, so it proves a theorem in all fields. [/quote]

That is usually when I get paranoid.  The argument doesn't seem to be water proof :huh: .  The same things happens when they are trying to make a determinant like criterion for singularity of quadrics in projective spaces of even dimension , over a field with characteristic two (and every element square)


If e+ \sqrt 2 is rational, e+ \sqrt 2 = a/b

(e-a/b)^2 = 2

e2  + \alpha e + \beta = 0 with \alpha,\beta rational different than 0

but (1;e;e2) is independant over Q

So e +  \sqrt 2 is irrational

i know how to prove \sqrt 2 is irrational, but it's quite hard to prove that e is a transcendent number, moubi :D:D:D

I have a nice proof of  (1;e;e) is independant over the field Q,
wich use that e is irational, and Taylor-Lagrange formulae for
f(x) = a.ex + b.e-x

i will post later.
Now I am going to an exhibition of the great Norvegian mathematician Niels Henrik Abel.

Chers,

moubinool

Suppose there exist x,y,z rational not all 0 with  x+ye+ze=0

by multipling by a integer we have a,b,c positive integers with

ae+be^(-1) = c we can also suppose a>0

f(x)  = a.ex + b.e-x  with Taylor-Lagrange

formulae there exist wn in ]0;1[

c = f(1) =a.e+b.e-1 =

a.\sum_'{k=0 to n }1/k! + b\sum_{k=0 to n}(-1)^k/k! +

(a.ewn +(-1)n+1 .b.ewn)/((n+1)!)

by multipling by n! the last equality  the number

An =  (a.ewn +(-1)n+1 .b.ewn)/(n+1) is an

integer

if b >0 the sequence A_(2n+1) of integer tends to 0 impossible
if b<0 consider the sequence A_(2n) to get the same contradiction.

As Valentin said the transcendent of e is not obvious
http://mathworld.wolfram.com/e.html


A finite number of circular discs cover an area $ T$ on the plane together. Prove that it is possible to select a few discs that do not overlap and cover an area of at least $ T/9$ together.

I don't understand the question very well; is it "Prove that it is possible to select a few discs that do not [overlap and cover an area of at least $ T/9$ ]together." or "Prove that it is possible to select a few discs that do not [overlap] and cover an area of at least $ T/9$ together." ?

Whenever you can think of multiple interpretations of a problem and you're not sure which one is right, check to see if one of them doesn't make sense.  Your first interpretation is clearly false, since you could just take one disc (hopefully I am understanding your interpretation correctly).

So I believe the problem is asking you to prove that there exists some subset of the discs that satisfies the following two properties:

1.   The discs in the subset do not overlap.
2.   The discs in the subset cover a total area of $ T/9$.

And if they all overlap witch each other, what then ?

"A few discs" just means one or more disc.  So even if all the discs overlap, you can just take one disc.

More precisely:

For a set $ S$ of disks in the plane, denote by $ |S|$ the area of the union of the disks in the set $ S$.

Let $ D$ be a such set of disks in the plane.  Show that there exists some subset $ E$ of the disks in $ D$ such that $ |E| \le \frac{|D|}{9}$.

[hide]lets try the greedy algorithm.

take the largest circle, if it is greater than 1/9 the total area then stop. If not then lets show that there must be another circle that doesn't intersect it.  Any circle that intersects it must be centered at most 2r away from its center and have all its area within 3r, so the total area is less than 9 times the first circle, so there must be some circle that doesnt intersect it.  Now lets add the largest leftover cicle that doesn't intersect out original. Now if the are is greater then 1/9 the total then stop, if not then find the next largest circle that doen't intersect them, which must exist by a similar analysis.[/hide]


find polynomial of several integer variables with integer coefficient whose range consists of certain negative numbers and the numbers $ceiling(\frac{(2+\sqrt{3})^n}{2}$(n=1,2,...)

So, can anyone solve this problem?

[quote="sam-n"]Find polynomial of several integer variables with integer coefficients whose range consists of certain negative numbers and the numbers $\left \lceil\frac{(2+\sqrt{3})^n}{2} \right \rceil$ ($n=1,2,\ldots$)[/quote]

[quote="mathmanman"]So, can anyone solve this problem?

[quote="sam-n"]Find polynomial of several integer variables with integer coefficients whose range consists of certain negative numbers and the numbers $\left \lceil\frac{(2+\sqrt{3})^{n}}{2}\right \rceil$ ($n=1,2,\ldots$)[/quote][/quote]

$P(x,y) = x (1-2(3+x^{2}+y^{2}-4xy)^{2})-(100000+4x^{2})(3+x^{2}+y^{2}-4xy)^{2}$


function
topology
real analysis
real analysis unsolved

Can the set of extreme values of a continous function $ f: [a,b]\to \mathbb{R}$ be more then countable?

Yes.

Among solutions are the functions, which are constant on some subinterval $ [c,d]\subset[a,b]$. All the values on $ (c,d)$ are extremes.

By extreme values i mean the actual values of the function, that is, if $ x_0$ is the local minima(maxima), $ y_0\equal{}f(x_0)$ is the extreme value. Otherwise i wouldn't ask  :lol:

Yes. Define

$ w(x) \equal{} \begin{cases} x \plus{} \sqrt{x \minus{} x^2} & 0 < x < 1 \\ 0 & x \leq 0 \\ 1 & x \geq 1 \end{cases}$

and put

$ f(x) \equal{} \sum_{n\equal{}0}^{\infty} \sum_{k\equal{}0}^{3^n \minus{} 1} \frac{1}{4^n} w \left( 3^{n\plus{}1} x \minus{} (3k\plus{}1) \right).$

Weierstrass M-test assures that $ f$ is a continuous function. Also, if $ C$ is a Cantor set, then $ f$ is strictly increasing on $ C$ and every point in $ C$ is a local minimum of $ f$. So every element in $ f(C)$ is an extreme value of $ f$. Since $ C$ is uncountable. so is $ f(C)$.

Thanks sos440! Actually, i was sure i could prove it is at most countable, but my argument was wrong.
Did you construct this example yourself or is it from the book? If so, can you tell the sourse?

I constructed it by myself, based on the observation that there are continuous functions whose extrem points consist a totally disconnected perfect set.

Ok, I'll try to work out some details myself. Well, for the set of extreme points there is something easier: for a closed set $ A$ define $ f(x): \equal{}\rho (x,A)$, then $ A$ is exactly the set of local minima of $ f$. In particular, Cantor set fits in here.


calculus
integration
function
limit
real analysis
real analysis unsolved

Let $f: [0,1]\to\mathbb{R}$ be a continuous function such that $f$ is derivable in $0$ and $f(0)=0$.
 Prove that \[ \lim_{n\to\infty}\int_{0}^{1}nf(x^{n})dx=\int_{0}^{1}\frac{f(x)}{x}dx.  \]

a change of variable $u=x^n$ and dominated convergence give immediately the result.

Yes, alekk, I solved it in the same way, but it would be better if post full solutions from time to time. ;)

[quote="infy"]let $y=x^n\implies\ dx=\frac{dyy^\frac{1}{n}}{ny}$ after variables substitution obtain the result[/quote]

Is very nice ;)


inequalities
geometry
function
real analysis
real analysis unsolved

Say $ f$ is continuos on $ [a,b]$ and increasing in all points from $ [a,b] / S$, where $ S$ is a countable set. Prove $ f$ is increasing on $ [a,b]$.

Hint: Put $ T \equal{} [a, b]/S$. Then $ T$ is dense in $ [a, b]$. Also, for all $ x, y \in T$, $ x <y \Rightarrow f(x) \leq f(y)$. Combine two facts with continuity gives the desired result.

Actually when I say $ f$ is increasing at point $ x$, I mean:
$ \exists \delta > 0 : f(a) < f(x)$ for  $ x \minus{} \delta < a < x$ and $ f(a) > f(x)$ for $ x < a < x \plus{} \delta$
Basicly I want to derive the global property of $ f$ (increases in $ [a,b]$) from the local properties (increases in all points from $ [a,b] / S$)
So I don't think it is that simple.

In the end I found the solution myself:

Assume $ f$ is not strictly increasing then
$ \exists c,d$ such that $ c < d$ and $ f(c) \geq f(d)$
Later I'll need strict inequality $ (f(c) > f(d))$ so assume it is not possible to find such $ c$ and $ d$.
Then $ f(c) \equal{} f(d)$. But $ f$ is continuous so it attains both its minimum and maximum on $ [c,d]$ and these values must both equal $ f(c)$, otherwise we could find $ c' < d'$ such that $ f(c') > f(d')$. So $ f$ is constant on $ [c,d]$ hence it is not increasing at any point from $ [c,d]$ which is contradiction.

Now we may assume $ f(c) > f(d)$. Take some $ q \in (f(d),f(c))$ and $ q_1 : \equal{} \frac {f(c) \plus{} q}{2}$ (imagine these as horizontal lines on graph of $ f$).
By continuity: $ \exists x_1 \in (c,d)$ such that $ f(x_1) \equal{} q_1$.
Now inductively $ q_n : \equal{} \frac {q_{n \minus{} 1} \plus{} q}{2}$ and again we find $ x_n \in (x_{n \minus{} 1},d)$ such that $ f(x_n) \equal{} q_n$.

The sequence $ {x_n}$ is increasing and bounded hence converges. Say $ x_n \rightarrow x$. Again by continuity $ f(x) \equal{} q$ (because $ q_n \rightarrow q$). Now notice that $ x$ is one of the points, where $ f$ is not increasing (no $ \delta$-area can exist). But we can repeat this process for any $ q$ from $ (f(d),f(c))$ and find new points, where $ f$ is not increasing. It is easy to realize that all such points are distinct (remember $ f(x) \equal{} q$). So we can find uncountably many points, where $ f$ is not increasing. Contradition.

I'd appreciate any ideas how generalize this claim e.g. would it suffice if the set on which f is increasing was dense?

[quote="Kenny_cz"]
I'd appreciate any ideas how generalize this claim e.g. would it suffice if the set on which f is increasing was dense?[/quote]

Well, it begins to be a bit of a monologue, but I found solution to this as well, if anyone is interested.  :wink:

The answer is negative. Just take f(x) = -C(x) + x/100, where C(x) is the Cantor function (derived from the Cantor set). This function is increasing almost everywhere, but in the end it decreases.


pigeonhole principle
combinatorics unsolved
combinatorics

Suppose you have 16 integers (positive, not necessarily distinct) that sum to 30. Show that there is always some subset of these integers that sums to n, for n=1,2,.....,29.

I know it may seem odd, but can anyone think of an inductive approach?

It is clear that our sequence consists of at least two $1s$ and at most fifteen $1s$.
We prove it by induction.The case $n=1$ is obvious.Suppose there exists exactly $k$ ones in our sequence, and let $\alpha$ denote the least element in our sequence which is not one.
We will show that $\alpha \le k+1$, hence we can replace some($=\alpha$) ones in our collection by $\alpha$ to obtain a $1$ and as a result the next collection is established.
Proof: $\alpha (16-k)\le 30-k$ so $\alpha \le \frac{14}{16-k}+1 \le k+1$ .
(What can we say when $\alpha$ is contained in the collection? )


induction
geometry
trapezoid
cyclic quadrilateral
geometry solved

Hi to all!! I want to introduce another skills .. dividing .....

_______________

Show that an arbitrary inscribed quadrangle can be divided into [tex]n   (\succeq {4})[/tex] inscribed quadrangles.
_______________

Isn't it nice??

Have a nice day all !   :D  :D

i guess it's sufficient to prove that one can draw the four perpendiculars to the four sides from an interior point, and get four inscriptible quadrilaterals again (they have opposite angles which sum is pi)... of course, any line parallel to one side (which intersects two opposite sides) divides the quadrilater into tho quadrilaterals, one of which is surely incriptible... some induction here, and that's all right :)
isn't it?

For n>4 it's [url=http://www.kalva.demon.co.uk/imo/isoln/isoln722.html]IMO 1972, problem 2[/url].

Let ABCD be the given quadrilateral, so that <DAB is its smallest inner angle. Let's construct an isosceles trapezoid ABEF, with E, F inside ABCD, and EF//AB (and EF sufficiently near AB). Then, construct another isosceles trapezoid AFGD with FG//AD, and finally the isosceles trapezoid DGHC, with GH//DC. It's obvious that also BEHC is an isosceles trapezoid. Also EFGH is cyclic, since its sides are parallel to the sides of ABCD.
So, we have solved the problem for n=5. Then, we might divide any isosceles trapezoid with a line parallel to its base, so getting one more cyclic quadrilateral, and solving the problem for n>5.
For n=4, we may choose a point P inside ABCD and join it to its sides with 4 segments, in such a way that the so formed quadrilaterals are all cyclic. It's trivial to show that there's always a P that allows this construction.

[quote="ma_go"]
any line parallel to one side (which intersects two opposite sides) divides the quadrilater into tho quadrilaterals, one of which is surely incriptible
[/quote]
Yes, but what about the other quadrilateral?


I won't put the problem up until I get a confirmation. Can we discuss?

You should ask Dr. Titu Andreescu. I'm not sure if AwesomeMath will use "Test A" again in Regular or Late Registration again :?:

I'm pretty sure tests won't be reused...

Scroll down; they've all already been posted in this forum.


Consider the intervals $ A\equal{}[\minus{}1,2[,B\equal{}]\minus{}\infty,1[,C\equal{}]0,\plus{}\infty[.$ The conjunction $ C\minus{}(A\cap B)$ is equal:

a) {}
b) [-1,0]
c) [-1,1[
d) [1, + $ \infty$]
e) ]1,+8[

Answer is D.

[hide]The intersection of $ A$ and $ B$ is the set of elements that are in both $ A$ and $ B$. Call this set $ S$; $ S$ contains all numbers between $ \minus{}1$ and $ 1$; $ [\minus{}1, 1]$. Note that set $ C$ comprises $ [0, \infty]$, so the numbers in $ C$ not in $ S$ are the numbers $ [1, \infty]$[/hide]

What part do you want me to clarify? I'm not sure if there's really any other way of solving it, you just find the intersection of sets $ A$ and $ B$ and remove the intersection of that set with $ C$ to yield the answer.


conics
parabola
analytic geometry
geometry
graphing lines
slope
linear algebra

Let A(-1,1) and B(3,9) be two points on the parabola  $y=x^2$.Take another point $M(m,m^2)$ on the parabola lying between A and B.Let H be the point on the line segment joining A to B that has the same x-coordinate as  M.Show that if the length of MH is k units,then triangle AMB has area 2k square units.Does this relationship still hold if M is not between A and B?

[hide="Proof"]I think this proof holds for all points M, not just between A and B.

Then line connecting A(-1,1) and B(3,9) has slope (9-1)/(3+1)=2, so the equation is y=2x+3.  The distance k between  point M(x, x^2) and the point on that line with the same x-coordinate, (x, 2x+3), is abs(2x+3-x^2).

By using the thingy where you find the area of a triangle in a coordinate plane with vertices (a, b), (c, d), and (e, f) by finding the determinant of the matrix,

  a     b    1
  c     d    1
  e     f     1

with the triangle formed by A(-1, 1), B(3, 9), and M(x, x^2), you use the determinant of,

 -1     1      1
  3     9      1
  x     x^2  1

which equals
=abs(1/2)[(-1*9*1+1*1*x+1*3*x^2)-(x*9*1+x^2*1*-1+1*3*1)]
=abs(1/2)[(3x^2+x-9)-(-x^2+9x+3)]
=abs(1/2)[4x^2-8x-12]
[b]=abs(2x^2-4x-6)[/b]

And since k=abs(2x+3-x^2)=abs(-x^2+2x+3)=abs(x^2-2x-3) and [b]2k=abs(2x^2-4x-6)[/b], the area of the triangle ABM=2k[/hide]

Generalization:  take any three points, $P(x, x^2), Q(y, y^2)$ and $R(z, z^2)$.  Let the vertical distance between $P$ and line $QR$ be $d$ and let the area of triangle $PQR$ be $A$.  Find a nice relationship between the numbers $d, A, y,$ and $z$.

$d$ is the vertical distance between the point and the line, not the perpendicular distance.  (Also, you're missing a square sign.)

Because the perpendicular "base" of the triangle is 4, shouldn't this follow intuitively?


Let $a,b,c$ be positive integers for which $ac=b^2+b+1$.Prove that the equation $ax^2-(2b+1)xy+cy^2=1$ has integer solution :D  :D It isn't hard but i love it and I want another solution .

I have a solution:Show that [tex]\delta[/tex] of 2 equation of variable [tex]x[/tex] and [tex]y[/tex]  is perfect square(since the given condition)!

Well if I am not wrong then the $\delta^2 = 4c-3x^2$ and $\delta^2 = 4a-3y^2$. From here how can you conclude that those are square number?