The non-negative integers are divided into three groups as follows: \[ A=\{0,3,6,8,9,\ldots\},\qquad B=\{1,4,7,11,14,\ldots\},\qquad C=\{2,5,10,13,\ldots\} \] Explain.

[hide]A: numbers that can be written with only curved lines
B: numbers that can be written with only straight lines
C: numbers that have to be written with both curved and straight lines [/hide]

I can't believe I got this.

EDIT: Ew. Fixed poor phrasing on my part.

[quote="catcurio"][quote="lightrhee"][quote="towersfreak2006"]The problem does not say so.[/quote]
Does this mean that sky9073's answer is wrong? :?[/quote]towersfreak2006 didn't indicate whether sky9073's answer was right or wrong...[/quote]
Well.... If 12 is not in C, sky9073's answer is wrong. However, that answer seems to be the only thing that works so far. :D

Yeah, I can't think of anything else that works either. But the problem says:

[quote="towersfreak2006"]The non-negative integers are divided into three groups[/quote]So doesn't that mean that 12, being a non-negative integer, has to be in one of the 3 groups?  :?

The problem doesn't appear to state that the numbers are listed in increasing order, but it seems implied.... :ninja:

Edit:  I think this works :)

Construct the sets $P = \{0, 3, 6, 8, 9\}, \qquad Q = \{1, 4, 7 \}, \qquad R = \{2, 5 \}$ according to the curved-straight line drawing rules.  Then construct $A, B, C$ as follows:

$A$ is the set of all numbers that can be written using only members of $P$ OR using at least one member of $Q$ and one member of $R$.

$B$ is the set of all numbers that can be written using only members of $Q$ OR using at least one member of $P$ and one member of $R$ (disallowing the first digit to be zero).

$C$ is the set of all numbers that can be written using only members of $R$ OR using at least one member of $P$ and one member of $Q$ (again disallowing the first digit to be zero).

Edit #2:  I just realized this ignores numbers that are written with members from all three sets, so I will put all those numbers in $C$ since it makes the most sense  :P

sky9073's answer is correct.  It is the solution that is intended.

This problem reminds me of another similar question, but with a different idea.

The English alphabets are divided into two sets as follows:
A={A,E,F,H,I,L,M,N,O,R,S,U,X,Y}
B={B,C,D,G,J,K,P,Q,T,V,W}
Which set would Z fall into?

There might be a few mistakes. Correct me if you know this problem.


Recently I saw a geometry problem that was solved using the angle bisector theorem a weird way... Usually, in a triangle ABC, the angle bisector of A splits BC into BE and EC respectively. Then AB/BE=AC/EC, but the solution was like this: AB/AC=BE/EC=1.... Why? How did they get AB/AC and BE/EC through the angle bisector theorem?

Multiply both sides of the equation by $ BE/AC$ :wink:


function
IMO Shortlist
number theory
algebra unsolved
algebra

[quote="lasha"]Find all functions f:N--N such, that for any m,n from N,
                                  (m^2+n)is divisible by f(m)^2+f(n).
Not too difficult :blush:[/quote]

IMO Shortlist 2004(?). But this is the wrong place for posting problems!

 darij

[quote="lasha"]darij grinberg wrote:IMO Shortlist 2004(?). But this is the wrong place for posting problems!
Why do you think so?[/quote]
You posted it into the topic for results of the IMO 2007!

A correct place would be the Number Theory subsection. But you don't need to post this problem anymore, since it has already been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=27188 .

  darij


calculus
integration
trigonometry
function
calculus computations

How do I determine convergence/divergence for that kind of integral?

I tried Diricihlet and it wouldnt work here.

And the comparsion methods are only used on positive functions...

What to do?

$\int_{2}^\infty \frac{\sin{x}}{x+2\cos{x}}dx$

Thanks.

Extremely unrigorous, but the infinite series sum that corresponds to this is, very roughly, asymptotic to an alternating harmonic series, so it should converge.

Is there a way to make this rigorous?

It does seem that you should be able to ignore the $2 \cos x$ term, since it is negligible as $x$ becomes large and should not affect convergence.  I think the integral is evaluatable without that term... ?

[quote="mlok"]You know that $\int_{2}^{\infty}\frac{\sin x}{x}dx$ converges by Dirichlet. And the difference \[\int_{2}^{\infty}\left(\frac{\sin x}{x+2\cos x}-\frac{\sin x}{x}\right)dx\] converges even absolutely, by comparison.[/quote]

I dont understand a few things....

What gives us a "hint" that we should use that $\frac{\sin x}{x}$ ?

Also why did you subtract exactly ? and what are we comparing the new expression to?

Thanks :)

[quote="RealTime"]What gives us a "hint" that we should use that $\frac{\sin x}{x}$ ? [/quote]
Here's the hint you want: what mlok did was to implement the idea:

[quote="t0rajir0u"]It does seem that you should be able to ignore the $2 \cos x$ term, since it is negligible as $x$ becomes large and should not affect convergence. I think the integral is evaluatable without that term... ?[/quote]
Now, for your second question:

[quote="RealTime"]Also why did you subtract exactly ? and what are we comparing the new expression to? [/quote]
We're trying to deal with $u$ - a limit, or an integral, or an estimate. We sense that $u$ is "near" $v,$ which we understand. So we write $u=v+(u-v),$ and hope that we will be able to show that $u-v$ is "small" in some useful way.

mlok didn't display the details; had he done so, they might have looked something like this:

$\frac{\sin x}{x+2\cos x}-\frac{\sin x}x=\frac{x\sin x-(x+2\cos x)}{x(x+2\cos x)}= \frac{-2\sin x\cos x}{x(x+2\cos x}$

Note that we have $x>2.$ Thus we can say that $x+2\cos x\ge x-2.$ Crudely estimating that neither sine nor cosine ever exceeds one, we get

$\left|\frac{-2\sin x\cos x}{x(x+2\cos x}\right| \le \frac2{x(x-2)}$

That estimate has problems near $2$ but the original integral has no such problems, so we don't care about that. If $x\ge 4,$ then $x-2>\frac{x}{2}$ and we can simplify our estimate to

$\left|\frac{-2\sin x\cos x}{x(x+2\cos x}\right| \le \frac4{x^{2}}$

Now we have an absolutely convergent integral. We estimate that

$\left|\int_{2}^{\infty}\frac{-2\sin x\cos x}{x(x+2\cos x}\,dx\right|\le \int_{2}^{4}\text{ a continuous function}+\int_{4}^{\infty}\frac4{x^{2}}\,dx$

And that converges to a finite number.

[quote="mlok"]You know that $\int_{2}^{\infty}\frac{\sin x}{x}dx$ converges by Dirichlet.[/quote]

Can anyone tell me what Dirichlet states? Thanks.

If $f$ has a bounded antiderivative and $g(x)$ decreases to zero as $x\to\infty$, then $\int_{a}^{\infty}f(x)g(x)\,dx$ converges. Rationale: integration by parts gives \[\int_{a}^{b}f(x)g(x)\,dx=Fg\bigg|_{a}^{b}-\int_{a}^{b}F(x)g'(x)dx\] where $F'=f$. The integral on the right is controlled by $(g(a)-g(b))\max|F|$.

A belated thought: the proof of the Dirichlet's test shows that you don't really need $g$ to be decreasing. All you need to know about $g$ is (1) $g(x)\to 0$ as $x\to\infty$; and (2) $\int_{a}^{\infty}|g'(x)|dx$ converges. The function $g(x)=\frac{1}{x+2\cos x}$ satisfies both (1) and (2). This gives an alternative solution of the problem.


The Eureka Hovercraft Corporation wanted to hold hovercraft races as an advertising stunt. The hovercraft supports itself by blowing air downward, and has a big fixed propeller on the top deck for forward propulsion. Unfortunately, it has no steering equipment, so that the pilots found that making high speed turns was very difficult. The company decided to overcome this problem by designing a bowl shaped track in which the hovercraft , one up to speed, would coast along in a circular path with no need to steer. they hired an engineer to design and build the track, and when he finished, he hastily left the country. When the company held their first race, they found to their dismay that the craft took exactly the same time $T$ to circle the rack no matter what is speed. Find the equation for the cross section of the bowl in terms of $T$.

Let $\alpha$ be the angle of a tangent to the bowl cross section f(r) with the horizontal at a radius r. The centripetal force component on the hovercraft in the direction of this tangent is $F_P = mg \sin \alpha$ downward, while the centrifugal force component in this direction is $F_F = m \omega^2 r \cos \alpha$ upward, where m is the hovercraft mass, g the gravitational acceleration and $\omega = \frac{2\pi}{T}$ the hovercraft angular velocity. Since the forces $F_P = F_F$ balance,

$mg \sin \alpha = m \omega^2 r \cos \alpha$

$\frac{\text df(r)}{\text dr} = \tan \alpha = \frac{\omega^2 r}{g}$

and since $\omega = \frac{2\pi}{T}$ is constant,

$f(r) = \frac{\omega^2 r^2}{2g} = \frac{2 \pi^2 r^2}{gT^2}$

Great solution, yetti!!! Nice to see you back at Physics forum after a Looooong time. How about looking around and solving some of my theoretical problems that NOBODY seems to be able to solve!@!!!


Euler
geometry
incenter
geometric transformation

That point is called Jerabek center, it lies on nine point circles
[url=http://forumgeom.fau.edu/FG2004volume4/FG200401.pdf]here[/url] is that result.  :)

But I don't like a solution by coordinate  :D My solution use nondot angle.  Anyone can post that solution to here  :D

[quote="zaizai-hoang"]Let acute triangle $ABC$ with altitudes are $AA',BB',CC'$. Prove that:
Euler lines of triangles $AB'C', CA'B', BC'A'$ are concurrent at a point  :D[/quote]

The center of a triangle wich have the same propiety are

Fermat (concur at the centroid of the triangle), Circuncenter and Incenter (concur at the Schiffer point... look at darij website)

An elementary proof of this problem, is as follows: (In my drawing AB = 11.5, AC = 13.0, BC = 14.5).

We denote as K1, K2, K3, the midpoints of the segments AH, BH, CH, where H is the orthocenter of the given triangle ABC. These points are circumcenters of the triangles AB'C', BA'C', CA'B', respectively.

Let N be the circumecenter of the triangle K1K2K3 (nine point circle of ABC). If D, E, are the intersection points of the sidelines K2K3, K1K2, of KiK2K3 respectively, from the segment line HN (Euler line of ABC and also of K1K2K3), then as well known (and easy to prove), the intersection point P of the segment lines A'D, C'E, lies on the circle (N).

(The points A', B', C', are the reflexions of H, with respect to the lines K2K3, K1K3, K1K2 respectively and so, the  lines A'D, C'E, as the reflections of the line HN passes through H, intersect reach other at a point so be it P, lies on (N), circumcircle of K1K2K3).

Let C'', Nc be the reflections of K3, N respectively, with respect to the sideline K1K2 of the triangle K1K2K3. The point C', is the orthocenter of the triangle C''K1K2 and so, the segment line NcC', is it;s Euler line (because of Nc is the circumecenter of C''K1K2).

We denote as H3 the orthocenter of the triangle CA'B' and as V the intersection point of the circle (N), from the line H3K3 (Euler line of the triangle CA'B'). We will prove that the point V, is the antidiametric point (antipode) of P. It is enough to prove that <NVK3 = <PC'K3.

From the isosceles triangle NK3V (NK3 = NV), we have that <NVK3 = <NK3V, (1).

From HC' // NNc, => <PC'K3 = <NNcC', (2).

Because of the triangles K1K2K3, C''K1K2, are congruent and the triangles K1K2K3, CA'B' are similar (easy to prove), we conclude that the triangles C''K1K2, CA'B', are also similar.

Hence we have that <NNcC' = <NK3V, (3). (Because of their Euler lines NcC', H3K3, form congruent angles with their homologous midperpendiculars NNc, NK3, respectively).

From (1), (2), (3), => <NVK3 = <PC'K3, (4).

From (4) we conclude that the lines NV, PV, are coincided and hence, the point V is the antidiametric point of P.

By the same way we can prove that the Euler lines of the triangles AB'C', BA'C', pass through the point V and the proof is completed.

Kostas Vittas.

PS. This problem is the first result which V. Thebault’s theorem states. The second result of this theorem is that the one of the segments A'V, B'V, C'V, is the sum of the other two (in above configuration, B'V = A'V + C'V).

The nice solution vittasko.There is a general for this problem (it is not of me)
Given are $\triangle ABC,I$ inside it, and $A'B'C'$ is cevian triangle of $I, d_{a}$ is the line through midpoin of $IA$ and centroid of triangle $AB'C'$, similar $d_{b},d_{c}$ provethat $d_{a},d_{b},d_{c}$ are concurrent at $J$ more $J$ lie in the nine point conic base on $I$
There is a coordinate for this solution but it isn't nice, do you have a synthetic proof similar that problem ?

Sorry gemath, I have not any solution. I have iust now learned from you, about this interesting generarization of the problem.

Kostas Vittas.


hey MNrule is there really a way to solve this problem without using softwares..??
if it is so ,then plz tell me coz it is pointless to waste time on such a question which cant be solved :lol:

This series is called harmonic progression
and there is no way to directly find the sum using any formula,it is a diverging series...


What is the value of $ 3\cdot 3 \plus{} 3(3 \plus{} 3) \minus{} 3^3$?


There is a bridge in a dark forest. You are part of a group of people that wants to cross the bridge in the shortest time possible, and it is the middle of the night. However, the bridge can only hold a certain weight, and you only have one flashlight, meaning that you have to bring people back and forth. If you want to bring multiple people over the bridge, the group can only go as fast as the slowest person.

The input file, "bridge.in", contains the following input:

Line [b]1[/b] contains X, the maximum weight the bridge can carry. You may assume that X <= 1000.
Line [b]2[/b] contains P (P <= 40), the number of people that need to cross.
Lines [b]3[/b] through [b]P+2[/b] contain two numbers each, representing one person - the first number being W, the weight of the person, and the second number is M, the number of minutes it takes for the person to cross the bridge. You may assume that W <= 200, and M <= 200.

The output file, "bridge.out", should contain one number, which is the minimum number of minutes total that it takes for all the people to cross. (You may assume that this number will be less than one million.) If it is not possible for everybody to cross, output 0.

Example:

[code]
200
4
100 1
100 2
100 5
100 10
[/code]

The output is 17.

[code]
200
4
150 3
100 4
100 6
100 10
[/code]

The output is 0.

How do you do this problem...?

This would be equivalent to solving s variation of the knapsack problem, which is NP-hard. (Which, of course, takes forever to solve)

Finding an algorithm to allow the most people to cross in the shortest amount of time would rely on a weighted heuristic based on weight/speed.

Obviously, if the heaviest person + lightest person have total weight exceeding bridge load, there is no solution.

Also, it is obvious that when sending a person back to transfer the flashlight, you would only send one person, and that person should be fast.

The problem gets complicated here; I don't see an easy way out, and any intuitive code would not give you an optimal solution.


Ring Theory
superior algebra
superior algebra solved

Describe all maximal ideals of Z[x].
Show (3, x^2+x+1) is not  a prime ideal.

Thanks in advance.

[quote="mathproblem"]Describe all maximal ideals of Z[x].
[/quote]
There can only be one maximal ideal in a ring. In this case, the ideal $(2,x)$ is a maximal ideal in $\mathbb{Z}[x]$ because its quotient ring is the field $\mathbb{Z}/2\mathbb{Z}$. This follows when we apply the following theorem:

"Let $R$ be a commutative ring, then is the ideal $M$ maximal if and only if the quotient ring $R/M$ is a field"

[quote="kouboy"]"There can only be one maximal ideal in a ring"

@yannick
i'm afraid its not true. the answer lies in the last line of what you wrote. take for instance $\mathbb{Z}$ the ring of integers. then $\mathbb{Z_{P}}$ is maximal for every prime p[/quote]

Ohoh. My mistake. I only saw the theory of maximal ideals a few hours ago. The word 'maximal' gave me the impression that there was only one.

Thanks for remarking

[quote="yannick"]
Ohoh. My mistake. I only saw the theory of maximal ideals a few hours ago. The word 'maximal' gave me the impression that there was only one.

Thanks for remarking[/quote]
FYI:

There is a slight semantic difference between the use of the words maximum and maximal.

A is  maximal means there is nothing larger than A (there can be other objects that are not comparable (by inclusion) to A that are also maximal).

What you have descibed above is a maximum (an object A that is larger than everything else). In terms of ordering ideals (or sets) this implies there is only one.

[quote="mathproblem"]Describe all maximal ideals of Z[x].[/quote]
Done before on the forum.

[quote="mathproblem"]Show (3, x^2+x+1) is not  a prime ideal.[/quote]

Let $X$ denote the class of $x$ in the factor ring.
Then $X-1$ is a non-trivial nilpotent so the factor ring is not a domain...


Find all polynomials with real coefficients $ P(x);Q(X)$ and $ R(X)$such that
$ \sqrt{P(x)}\minus{}\sqrt{Q(x)}\equal{}R(x)$ for all $ x$

There are many.
Take, for example, $ P(x) \equal{} 16x^2 Q(x) \equal{} x^2 R(x) \equal{} 3x$
I guess you mean P(x) and Q(x) are non-square polynomials.

[quote="friendlist"]Find all polynomials with real coefficients $ P(x);Q(X)$ and $ R(X)$such that
$ \sqrt {P(x)} \minus{} \sqrt {Q(x)} \equal{} R(x)$ for all $ x$[/quote]
Are you a Vietnamese,it is a problem in the new MYM,the result is a genaral Polynomials


find the maximum of : $ \sin(x_1)\sin(x_2).......\sin(x_n)$

such that :  $ \tan(x_1)\tan(x_2).......\tan(x_n)\equal{}1$

very easy  :)

[hide="Hidden by mod"]$ tan(x_1)*...tan(x_n) \equal{} 1$
$ < \equal{} >$
$ sin(x_1)*sin(x_2)*....*sin(x_n) \equal{} cos(x_1)*..cos(x_n)$ (1)
so we want the max of the sin(x_1)*sin(x_2)*...sin(x_n) such that the (1) stands.
the (1) stand only when sin(x_1)=cos(x_1) etc
and that is only when
$ sin(x_1) \equal{} \frac {\sqrt {2}}{2} \equal{} sin(x_2) \equal{} ... \equal{} sin(x_n)$
or
$ sin(x_1) \equal{} \minus{} \frac {\sqrt {2}}{2} \equal{} sin(x_2) \equal{} ... \equal{} sin(x_n)$
so :
$ \minus{} (\frac {\sqrt {2}}{2})^n < \equal{} sin(x_1)*sin(x_2)*...sin(x_n) < \equal{} (\frac {\sqrt {2}}{2})^n$[/hide]

i think that you're wrong  :wink:

for $ n\equal{}3$    try  $ x_1\equal{}\frac{\pi}{3}$  and  $ x_2\equal{}\frac{\pi}{6}$  and  $ x_3\equal{}\frac{\pi}{4}$

we have  $ \sin(\frac{\pi}{3})\sin(\frac{\pi}{6})\sin(\frac{\pi}{4})\equal{}\cos(\frac{\pi}{3})\cos(\frac{\pi}{6})\cos(\frac{\pi}{4})$

[hide="Slick-ish"]
the product of the sines=product of cosines.
Use QM-GM on $ \sin(x_{1})...\sin(x_{n})$ and $ \cos(x_{1})...\cos(x_{n})$ to get that
$ \sqrt{\frac{1}{2}} \geq \sqrt[n]{\sin(x_{1})...\sin(x_{n})}$
So the maximum is $ 2^{\minus{}.5n}$[/hide]


function
algebra
domain
modular arithmetic
geometry
geometric transformation
trigonometry

How can I show:

If $f$ is a $C^1$ function on the circle and $a_{N}(n)=\frac{1}{N}\sum^{N}_{k=1}{f(e^{2\pi i\frac{k}{N}})e^{-2\pi i\frac{kn}{N}}}$ then $\left|a_{N}(n)\right|\leq\frac{C}{\left|n\right|^2}$ for some constant $C, 0<\left|n\right|\leq\frac{N}{2}$

??????

Well I haven't started doing these problems yet, but it looks just like what we proved before for $C^1$ functions on the unit circle in the continuous case, so you might just want to go back to that proof and see if you can rework it in these terms. But I suck at algebra so maybe this is totally wrong...   :?

I'll take a crack at it. And after I'm done, I'll move it out of the complex analysis section, since there will be no complex analysis used.

I also assume you meant $a_N(n)$ rather than $a_n(n).$ I'll change that notation anyway.

Suppose $f$ is a $C^1$ function periodic of period $2\pi.$ I'd rather write $f(x)$ that $f(e^{ix}).$ Let $M=\sup_x|f'(x)|.$

Fix a positive integer (a sampling rate) $N$ and let $u(k)=f\left(2\pi\frac kN\right).$

Note, by the Mean Value Theorem, that $|u(k+1)-u(k)|\le\frac{2\pi M}N.$

With that last estimate, the function $f$ will disappear from the scene - we're done with it, and with all references to any continuous domains.

For notational simplicity, let $\omega=e^{2\pi i/N}.$ Define

$\widehat{u}(n)=\sum_{k=0}^{N-1}u(k)\omega^{-kn}.$

Now let $v(k)=u(k+1)-u(k).$ (Note: such computations on the domain as $k+1$ should always be interpreted $\pmod{N}.)$ Translation of a function results in pointwise multiplication of the (discrete) Fourier transform by an exponential. We can show the following:

$\widehat{v}(n)=(\omega^n-1)\widehat{u}(n).$

But we also have that

$|\widehat{v}(n)|\le \frac1N\sum_{k=0}^{N-1}|v(k)| \le\frac1N\sum_{k=0}^{N-1}\frac{2\pi M}N=\frac{2\pi M}N$

Hence,

$|\widehat{u}(n)|\le\frac{2\pi M}{N|\omega^n-1|}$

Now, $|e^{i\theta}-1|=\text{chord}(\theta)=2\sin\frac{\theta}2$ so

$|\omega^n-1|=|e^{2\pi in/N}-1|=2\sin\left(\frac{\pi n}N\right)$ and

$|\widehat{u}(n)|\le\frac{\pi M}{N\sin\left(\frac{\pi n}N\right)}$

Hmm - that didn't get there. That estimate looks like $\frac1{|n|},$ not $\frac1{|n|^2}.$ Come to think of it, in the continuous case, having $f\in C^1$ would get us Fourier coefficients decay at a $o(n^{-1})$ rate, but merely $C^1$ would not be enough to get us $O(n^{-2}).$

That's true. For some reason I was thinking we had proved a continuous function has Fourier coefficients that are $O(n^{-1})$ but actually we just proved they are $o(1)$, which gives us the very estimate Professor Merryfield offered.

To be honest I can't see anywhere in his proof that a better estimate might be made. I guess we can ask Sinai...

check out the book... page 236... looks like sinai made another one of his infamous typos.

the problem asks to show $\left|a_{N}(n)\right|\leq\frac{C}{\left|n\right|}$ ..... not $\left|a_{N}(n)\right|\leq\frac{C}{\left|n\right|^2}$

these things have cost me about a week of my life so far


Hehe, it's the [hide="give-away"]Subway Series![/hide]  Very classy :)

Awww... no subways within a couple hundred miles of where I live.

[quote="alan"]what is the next term?

4 14 34 42 59 __

Hint: [hide]may not be fair to those who don't live in a certain city :D[/hide][/quote]

Within a mile radius of my home, there are three metro stations, nevertheless I am still clueless about the series (Probably it has to be with Boston's metro system :? ).

A good guess, but Boston doesn't number it's subway stations by street numbers.  Viewing Alan's profile might help, as might the website [hide]www.mta.info[/hide].


function
logarithms
calculus
derivative
calculus computations

given the function $ f(x)\equal{}x^3\minus{}\cfrac{x^2}{\ln{x}}\minus{}x\plus{}1$, find the values of $ x$ where $ f(x)$ has a horizontal tangent.

I solved for the derivative: $ f'(x)\equal{}3x^2\minus{}\cfrac{2x\ln{x}\minus{}x}{(\ln{x})^2}\minus{}1$

but I have no idea how to start solving for $ x$ after setting that to $ 0$..

There is a unique such value, near $ \frac {3}{10}$, but I don't think you can express it in terms of elementary functions.  (That is, "giving up on solving that equation exactly" is my chosen course of action.  Mathematica's, too.)

ahh.. and I'd spent a good portion of an hour digging through my text/trying to solve the equation.. lol

Thanks for the help, JBL :)


Find all pairs $ (x,y)$ of real numbers which satisfy $ x^3\minus{}y^3\equal{}4(x\minus{}y)$ and $ x^3\plus{}y^3\equal{}2(x\plus{}y)$.

[b]Case:[/b]  $ x \equal{} y$.  Then $ 2x^3 \equal{} 4x \implies x \equal{} 0, \pm \sqrt {2}$.

[b]Case:[/b]  $ x \equal{} \minus{} y$.  Then $ 2x^3 \equal{} 8x \implies x \equal{} 0 \pm 2$.

[b]Case:[/b]  $ |x| \neq |y|$.  Then $ x^2 \plus{} xy \plus{} y^2 \equal{} 4, x^2 \minus{} xy \plus{} y^2 \equal{} 2$.  Letting $ u \equal{} x \plus{} y, v \equal{} xy$ we have $ u^2 \minus{} v \equal{} 4, u^2 \minus{} 3v \equal{} 2 \implies v \equal{} 1, u \equal{} \pm \sqrt{5}$.  Then $ x, y$ are roots of $ t^2 \minus{} (\pm \sqrt{5})t \plus{} 1$.

arent $ x,y$ roots of

$ t^2 \minus{} \sqrt{5} \plus{} 1$

 :(

Yes, sorry; but both signs are necessary.  If $ (x, y)$ is a solution then so is $ (\minus{}x, \minus{}y)$.


In rectangle $ADEH$, points $B$ and $C$ trisect $\overline{AD}$, and points $G$ and $F$ trisect $\overline{HE}$. In addition, $AH=AC=2$. What is the area of quadrilateral $WXYZ$?

$\text{(A) }\frac{1}{2}\qquad \text{(B) }\frac{\sqrt{2}}2\qquad \text{(C) }\frac{\sqrt{3}}2 \qquad \text{(D) }\frac{2\sqrt{2}}3 \qquad \text{(E) }\frac{2\sqrt{3}}3$

I dunno.  :maybe: I just copied it from [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2006]here[/url]

There is a figure that i not functioning.

The problem says see the figure.


After getting answer in less than 1 minute :D , I'll give really good hints.

First, find the one thing that is not defined in the problem using the area formula for trapezoid.

Second, using it, find the side of AB and CD.

Third, find the perimeter.

Fourth, the answer would be in form of:

$a + b \sqrt {c}$

That is big hint.

Why don't people answer? Too easy?

[hide]Answer 24+8 :sqrt: 2

My attempt to an explanation

Trapezoid area is fount ba adding the to parallel sides and dividing by 2 and then multiplying by height.

So 8+16/2=12

Area is given, so 12h=48, h=4.

Since AB=CD, then the trapezoid can be seperated into a rectangle and two 45-45-90 triangles, the rectangle is 8x4, and the triangles make a square of area 16, therefore having side lenths of 4 and hypotenuse (spelling correction anyone?)  of  :sqrt: 32. By adding, we get 8+16+2  :sqrt: 32 = 24+8 :sqrt: 2[/hide]

As we can see, I'm probably not the go-to guy for proofs  :blush: , either case, feel free to point out any errors, or suggest ways this can be simplified.

[hide]The perimeter of the trapezoid will be 24 + 8 :sqrt: 2.

Proof: The formula of the area of a trapezoid is the length of the two parallel sides multiplied by the height and divided in half.

We know that the top and bottom sides are 8 and 16 respectively. So:

H= height

(8+16) x H/ 2= 48
            24 x H= 96
                    H =4
Now we can see the height of the trapezoid is 4. Using the height, we can divide the trapezoid into two right angled triangles and a square. The triangles sides will be 4 and 4. So the length of the hypotenuse will be:

x2 = 42 + 4 2
x 2= 32
                        x = :sqrt: 32

So the four sides are, top--> 8, bottom --> 16, sides  :sqrt: 32 each.

16+ 8 + 2 :sqrt: 32
= 24 + 2  :sqrt: 32
= 24 + 2 :sqrt: 16x2
= 24 + 8 :sqrt: 2

Thus, the perimeter is 24 + 8 :sqrt: 2

                    [/hide]

[hide][quote]
A trapezoid has BC || AD and AB = CD. BC = 8, AD = 16, and the area of the trapezoid is 48. What is the perimeter of the trapezoid?
[/quote] A=48, then h*(8+16)/2=48. So h=4. AB= :sqrt: h 2 +16=4* :sqrt: 2
PErimeter is equal to 8* :sqrt: 2+24[/hide]


[quote="graciebarn256"]If I will be e-mailing my solutions in, do I need to put my name/round and problem #s on each PDF page?[/quote]

Yes.


Let $ a,$ $ b$ and $ c$ are nonnegative numbers,  no two of which are zeros. Prove that:
$ \frac{a}{b^{2}\plus{}c^{2}}\plus{}\frac{b}{a^{2}\plus{}c^{2}}\plus{}\frac{c}{a^{2}\plus{}b^{2}}\ge\frac{3(a\plus{}b\plus{}c)}{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}ac\plus{}bc}.$

By Cauchy Schwarz
\[ \frac{a}{b^{2}\plus{}c^{2}}\plus{}\frac{b}{a^{2}\plus{}c^{2}}\plus{}\frac{c}{a^{2}\plus{}b^{2}}\geq\frac{(a\plus{}b\plus{}c)^{2}}{\sum{ab(a\plus{}b)}}.\]

Hence it remains to prove
\[ (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca)\geq 3\sum_{cyc}{ab(a\plus{}b)},\]
which is Schur.


prove $ tan \frac {3\pi }{11} \plus{} 4sin \frac {2 \pi}{11} \equal{} \sqrt {11}$

This problem has been posted many times.

[url=http://www.mathlinks.ro/viewtopic.php?t=204958]Value of formula[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=14937]Old Trigonometry Identity[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=12775]A Few Trig Exercises[/url]

and
http://www.mathpropress.com/cmj/pages/page224.html

[b]paper related to this problem[/b]
[url=http://math.tulane.edu/~vhm/papers_html/sqrt.pdf]An elementary trigonometric equation[/url]
[url=http://www.maa.org/pubs/cmj.html]College Mathematical Journal[/url] 39, 2008, 394-398.
We present some solutions of the equation A tan t + B sin t = C with A, B and C^2 rational numbers.
Papers by [url=http://math.tulane.edu/~vhm/]Victor Hugo Moll[/url]


In your signature, are you allowed to put things such as "Check out my blog!" if you have a blog, on say, Blogger or something?

Seeing that it happens in just about every signature there is, I guess it is definitely safe to conclude that you are allowed to advertise your blog given the strict condition that the link leads to something still within the AoPSilic realm.

I actually believe that you are not allowed to put any foriegn urls in your signature.

[quote]This is a block of text that can be added to posts you make. There is a 255 character limit on 3 line(s), font size 10 to 12. In your text, no more than 75 characters without space too. No images. No more than 1 image(s). No more than 2 url(s). [b]Only internal links allowed.[/b] Only 3 smilies are allowed. No code tags. No hide tags. No quote tags. No list tags. No colors. No centered LaTeX tags. No youtube tags.[/quote]

Yea, DeeKay was right. Only AoPS links. But apparently, the number one poster broke the rules i guess.

Actually, I'd take that as defining the limit of the rules. t0rajir0u formerly linked to an AoPS blog, and now mentions the url of his new blog without making it a link.

So you can, as long as the site is math-ish?
I'm kinda confused.

My understanding is that it is still OK to put external links in the web site section of your profile and to put something like "Check my web site! (see my profile for the link)" in the signature [b]without[/b] including a clickable external URL into the signature itself. In principle, you can then create a web page that refers to several other web pages that ... and invite the reader to explore the whole tree this way. As long as this possibility is not abused, I do not see anything wrong with it and so far the imposed rules do not formally prohibit this option.

On the other hand, this is just a logical conclusion from what is written in plain English in the signature rules. There may be some unwritten regulations as well (+common sense rules, of course). It would be nice if some admin would clarify that (the search for "signature rules" brings up a few old threads but they do not seem to go beyond what I said above).

You do not obtain it; you set it up yourself. Go to your profile, click on "signature control panel", and fill out the text box with whatever you want to be your signature. Then click "preview" to see what it'll look like. Then, if you are happy with the result, click "submit".

For more FAQ, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280499]here[/url].

To answer your signature question, see the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1513442#1513442]first post[/url] on the same thread.

[quote="myyellowducky82"][quote="fedja"]My understanding is that it is still OK to put external links in the web site section of your profile and to put something like "Check my web site! (see my profile for the link)" in the signature [b]without[/b] including a clickable external URL into the signature itself. In principle, you can then create a web page that refers to several other web pages that ... and invite the reader to explore the whole tree this way. As long as this possibility is not abused, I do not see anything wrong with it and so far the imposed rules do not formally prohibit this option.
[/quote]
So you can't put an actual link into the signature, but you can mention it, correct?[/quote]
Yes, as long as it doesn't lead to a website that's disturbing or inappropriate in nature.

Just rely upon your common sense and taste but don't be surprised if admins decide to rely upon theirs in the case you show a clear lack of one of those :P


vector
function
calculus
integration
calculus computations

Show that $\vec F=\nabla f$ for some function $f$. Use the Fundamental theorem of Calculus for line integrals.

A way
[hide]

The field clearly has the potential $U=(x^{2},y^{2},z^{2})$
Thus we get the integral to be
$U(\bar{a})-U(\bar{0})=a\cdot a$

[/hide]


In above solution, we use not only parity ingeniously but also coloring art in solving mathematical problem. Coloring idea is flashed so naturally that we don't notice it. It is also very close with us and almost present everywhere in our life such as chessboard or simply the floor which we are standing... This method doesn't require much knowledge and calculation skills, it mainly require giveing a specific model and also the flexility, creativity in thinking. However, it is just difficulty and advantage of combinatorial in general and coloring in particular. So using coloring to solve problems may be supposed to be a good method in combinatorial geometry. Because of these reasons, we hope that this document can offer readers a not only overview but also detailed corner about using coloring as the art of problem solving. Now let us present the following kinds of problem

   This is part of the beginning of my mathematical subject. Maybe there are some mistake and some sentences do not sound natural.
Could someone please spend time correcting it for me?
Thanks a lot!

I'm sorry but there are a lot of mistakes.  It would take me a very long time to try to completely understand what you're trying to say and correct them, but in the first few sentences alone:

In the above solution, we use not only parity in a clever manner, but also coloring art in mathematical problem solving. The idea of coloring is so natural that we don't even notice it.  It is so natural that it is present almost everywhere in our lives, such as chessboards or even the floor on which we stand.

These edits may not be good because I'm having trouble understanding what you're trying to say.

Thank you, Thank you very much!! :)
I think that your sentenses are what I want to say  :lol:
Thank you again.

Thank you for spending time reading it.
I have corrected it a little and there must be some mistake but I hope you could understand it.
     In the above solution, we use not only parity in a clever manner, but also coloring art in mathematical problem solving. The idea of coloring is so natural that we don't even notice it, although it is present almost everywhere in our lives, such as chessboards or even the floor on which we stand… Moreover, this method doesn’t require us wide knowledge about mathematical theorems, lemmas as well as good calculation-skill with many complicated expressions. But, it demands the creative thinking, which can’t be modeled mechanically. However, that is the difficulty and advantage of combinatorics in general and coloring in particular. So using coloring to solve problems may be supposed to be a good method in combinatorial geometry. Because of these reasons, we hope that this document can offer readers a not only overview but also detailed view about using coloring as an art of problem solving. Now let us present the following kinds of problem:

Thanks!

[quote="JBL"]In the above solution, we not only use parity in an ingenious way but also [b]coloring art in solving mathematical problem[/b].  The idea of coloring occurred so naturally that we didn't even notice it.  It is [b]also very close with us[/b] and present almost everywhere in our lives, such as in the pattern of the chessboard or simply the floor on which we are standing... This method doesn't require much knowledge or calculation skills; it mainly requires [b]giving a specific model[/b] and having flexibility and creativity in thinking. However, it is [b]just difficulty and advantage of combinatorial in general and coloring in particular[/b]. So using coloring to solve problems may be supposed to be a good method in combinatorial geometry. For these reasons, we hope that this document can offer readers not only an overview but also a detailed [b]corner[/b] about using coloring [b]as the art of problem solving[/b]. Now let us present the [b]following kinds of problem[/b]

I've made a number of corrections (not necessarily the same ones that JRav made).  I bolded phrases, sentences and words for which I can't figure out precisely what you intended to say -- some of them look like literal translations of idiomatic expressions, for example, and in such a case it can be impossible to guess the original meaning.  Some changes were purely stylistic, i.e., what you had before was grammatically correct but I changed it to make it clearer or to sound better.[/quote]

I think that:
_"the art of problem solving" maybe refer to the good or excellent methods to solve problem

_"following kinds of problem" mean the different types of problem such as Coloring chessboard, Coloring graph,...
I can't replace the word "kinds"  :(

[quote="Kyoshiro"]Moreover, this method doesn’t require us wide knowledge about mathematical theorems, lemmas as well as good calculation-skill with many complicated expressions. But, it demands the creative thinking, which can’t be modeled mechanically. However, that is the difficulty and advantage of combinatorics in general and coloring in particular. So using coloring to solve problems may be supposed to be a good method in combinatorial geometry. Because of these reasons, we hope that this document can offer readers a not only overview but also detailed view about using coloring as an art of problem solving. Now let us present the following kinds of problem:[/quote]
Here are some more changes that I've made.  Make what you want of them.
Moreover, this method doesn't require too much knowledge of mathematical theorems or lemmas, nor does it require exceptional calculation skills with complicated expressions.  However, it does demand creative thinking, which can't be accurately modeled physically.  (I can't really understand your next sentence).  Therefore, using coloring may be useful for solving problems in combinatorial geometry.  For these reasons, we hope this document will offer readers not only an overview, but also a detailed look at using coloring as a problem solving technique...

[quote="Kyoshiro"]In the above solution, we use not only parity in a clever manner, but also coloring art in mathematical problem solving. The idea of coloring is so natural that we don't even notice it, although it is present almost everywhere in our lives, such as chessboards or even the floor on which we stand… Moreover, this method doesn’t require us wide knowledge about mathematical theorems, lemmas as well as good calculation-skill with many complicated expressions. But, it demands the creative thinking, which can’t be modeled mechanically. However, that is the difficulty and advantage of combinatorics in general and coloring in particular. So using coloring to solve problems may be supposed to be a good method in combinatorial geometry. Because of these reasons, we hope that this document can offer readers a not only overview but also detailed view about using coloring as an art of problem solving. Now let us present the following kinds of problem:[/quote]

Here's my rewrite of what I've quoted above.  I made some changes again ("a not only overview" -> "not only an overview") that are grammatically essential.  The bolded part still sounds very awkward in English, and I'm not sure precisely how to reword it.

In the above solution, we not only use parity in a clever manner, but also the technique of [b]coloring in mathematical problem solving[/b]. The idea of coloring is so natural that we don't even notice it, although it is present almost everywhere in our lives, such as chessboards or even the floor on which we stand… Moreover, this method doesn’t require us to have wide knowledge about mathematical theorems and lemmas or skill at calculating many complicated expressions. But it demands creative thinking, which can’t be modeled mechanically.  That is both the difficulty and advantage of combinatorics in general and coloring arguments in particular. So using coloring to solve problems may be supposed to be a good method in combinatorial geometry.  For these reasons, we hope that this document can offer readers not only an overview but also a detailed look at using coloring as an art of problem solving. Now let us present the some examples of problems in which coloring can be used:

Thank you so much JBR and JRav!
It was really kind of you to help me!
Sorry to bother you with my paragraph, which is full of mistake!:blush:

[quote="JRav"][quote="Kyoshiro"]Thank you so much JBR and JRav! [/quote]
I don't think JBL will be too happy you mis-spelled his name. :wink:[/quote]
Oops :blush: :D.
Sorry JBL.
Here is my paragraph:

     In the above solution, we not only use parity in a clever manner, but also the technique of coloring in mathematical problem solving. The idea of coloring comes to us so naturally that we don't even notice it, although it is present almost everywhere in our lives, such as chessboards or even the floor on which we stand…

     Moreover, this method doesn’t require us to have a wide knowledge of mathematical theorems and lemmas or skill at calculating many complicated expressions. It mainly requires giving a specific model and being flexible and creative when thinking which can’t be modeled mechanically. This is both the difficulty and the advantage of combinatorics in general and coloring arguments in particular. So using coloring to solve problems may be supposed to be a good method in combinatorial geometry. For these reasons, we hope that this document can offer readers not only an overview but also a detailed look at using coloring as an art of problem solving. Now let us present some examples of problems in which coloring can be used:


How many of the first 100 positive integers are neither perfect squares nor perfect cubes?

In other words, we want to find $ 100\minus{}(\text{\# of perfect squares}\plus{}\text{\# of perfect cubes}\minus{}\text{\# of perfect sixth powers})$

$ \equal{}100\minus{}(10\plus{}4\minus{}2)\equal{}\boxed{88}$.

Ah okey dokey. Although it wouldnt matter cuz its a square.


Let $n$ blue lines, no two of which are parallel and no three concurrent, be drawn on a plane. An intersection of two blue lines is called a blue point. Through any two blue points that have not already been joined by a blue line, a red line is drawn. An intersection of two red lines is called a red point, and an intersection of red line and a blue line is called a purple point. What is the maximum possible number of purple points?

I think that my proof works out:

Consider a red line, it may instersect a blue line on a blue point or a purple point. By the construction of that red line, it intersects at least four blue lines on a blue point, so there are at most $n-4$ purple points for each red line.

It's easy to see that any four blue lines determines exactly three red lines.
So we have $\frac{n(n-1)(n-2)(n-3)}{8}$ red lines, and then the answer is  $\frac{n(n-1)(n-2)(n-3)(n-4)}{8}$ purple points.

PS: Please tell me if i made any mistake.


geometry
trapezoid
rectangle
AMC
AMC 10
Pythagorean Theorem
algebra

In trapezoid ABCD with bases AB and CD, we have AB=52, BC=12, CD=39, and DA=5. The area of ABCD is (highlight if you want to know the choices)

[hide](A) 182,  (B) 195,  (C) 210,  (D) 234,  (E) 260[/hide]

Beautiful problem.  Here's my answer (highlight):

[hide](C) 210[/hide]

And a solution  :mrgreen: :

[hide]Question: In trapezoid ABCD with bases AB and CD, we have AB=52, BC=12, CD=39, and DA=5. Area of ABCD=?

Call the height of trap ABCD x.  Draw a perpendicular down from C and D to AB; each of these is x units long.  Also, these two lines form right triangles with the two "legs" of the trapezoid.  So, we have split trapezoid ABCD into three parts: two triangles and a rectangle.

Triangle #1 has hypotenuse 5 and height x.  By the pythagorean theorem, then, its base is  :sqrt: 25 - x:^2:

Triangle #2 has hypotenuse 12 and height x.  Similarly, its base is  :sqrt: 144 - x:^2:

The rectangle has length 39 and height x.

AB, with length 52, can be split into three parts: the base of triangle 1, the base of triangle 2, and the base of the rectangle.  Hence, base 1 + base 2 + rect length = 52.

Therefore,

:sqrt:(25 - x:^2:) + 39 + :sqrt:(144 - x:^2:)  = 52

:sqrt:(144 - x:^2:)  = 13 + :sqrt:(25 - x:^2:)

144 - x:^2: = (13 + :sqrt:(25 - x:^2:)):^2:

144 - x:^2: = 169 - 26 * :sqrt:(25 - x:^2:) + 25 - x:^2:

144 = 194 - 26 * :sqrt:(25 - x:^2:)

26 * :sqrt:(25 - x:^2:) = 50

:sqrt:(25 - x:^2:) = 50/26

25 - x:^2: = 625/169

x:^2: = 25 - 625/169 = 3600/169

x = 60/13

Area ABCD = 1/2 * h * (b1 + b2)

= 1/2 * 60/13 * (52+39)

= 210 (C) [/hide]

[hide]Just want to point out a slight error in mathfanatic's work -

(25 - x^2)^1/2 + 39 + (144 - x^2)^1/2 = 52 does not give us (144 - x^2)^1/2 = 13 + (25 - x^2)^1/2, however, you corrected yourself in the 4th line of your working anyway, so it doesn't really matter.[/hide]

Wow.  You read the whole solution and found the error in there... I'm impressed.  I'll correct that. (I should change the + to a -, right?)

I guess I could claim that I put that mistake on purpose as a test (like my math teacher does - "Just checking to make sure you're paying attention" - his exact words every time), but I'm still in awe.  Bravo!  :P

i only read the answer, the solution would require thinking on my part

*shrug* I always read solutions, just to see if there's a different way of doing the problem...

i do too, but this was an exception, i was felling tired at the time

that was the last question on an amc10 to those who may fell a tidbit discouraged

Haha, when I did the problem at first, I labeled my trapezoid, A,B,D and C. Then threw me off completely.

ah, so that's syntax error's name.  This was a good problem.  Mathfanatic's solution was very assiduous!  I arrived at the solution using about 5 lines, but that's because I did some of the reasoning in my head.

-interesting_move

Even though I know what "assiduous" means, I'm not sure whether it's a good thing or a bad thing.  Am I being insulted or complimented on how long my answers are?

But then again, when I write out as many lines as I did that time, I'm bound to make a stupid little type.  So AM I assiduous?

I don't know of anyway to do this in 5 lines.

It can be done in not very many lines if instead of considering each of the two right triangles separately, you completely remove the rectangle and squish the triangles together. Then you get a 5-12-13 triangle left over. (One sign that you're doing a problem right is that you get nice results after you do something you think is clever.) The length of the altitude to the hypoteneuse is then 5*12/13, so that's the height of the rectangle as well. Then you can find the areas of the triangle and the rectangle separately and add them. This probably won't be five lines, and I'm leaving out details, but it's shorter.

First of all, assiduous normally has a positive connotation so it's a compliment for you, mathfanatic.

Also, I think you could do this problem in like 5 lines if you drew a diagram (but of course, this would depend on your definition of a line).  Since the longer base is 52 and the shorter 39, their difference (which is also the sum of the 2 legs of the 2 right triangles that are part of the longer base) is 13.  From here, you can do what ComplexZeta said, or you could set up a system of equations to solve for the height:

x^2+h^2=5^2; (13-x)^2+h^2=12^2
x^2+h^2=25; 169-26x+x^2+h=144
Subtract the two equations to get 169-26x=119.
x=25/13 and thus (25/13)^2+h^2=25 and h=sqrt(3600/169)=60/13.
So the area is 1/2(60/13)(39+52)=210.

-interesting_move[/i]

Then again, you could always just draw a diagram and do all your work on scratch paper and then answer with just one line: x = 210!

 :wink:  mathfanatic


could someone please explain to me the dervitation of them. thanks in advance

Explain what¿
More accurate questions please.

oh ok, basically i'm just asking for an introduction to them and their uses

See at ''Filaseta M. Irreducible polynomials theory (Math788F, 1998)''. It has got a chapter on
 cyclotomic polynomials . Good luck!  :wink:

thanks alot for the reference however do you know of any online links?

Ok, I sent to you the link of that book. Good luck!  :)


geometry
trapezoid
perpendicular bisector

Let $ ABCDE$ be a regular convex pentagon whose center is $ O$. The line $ AO$ intersect $ BE$ at $ M$ and $ DC$ at $ N$. Prove that the points $ A,M,O$ and $ N$ form a harmonic division.

Let $ P$ and $ Q$ be the midpoints of $ AE$ and $ BC$, respectively.

Lemma 1: $ P$, $ M$, and $ Q$ are collinear.

Proof: Note that in $ \triangle EAB$, $ PM//AB$. In $ \triangle BEC$, $ MQ//EC$. But since $ EC//AB$, $ PM//MQ$, so $ P$, $ M$ and $ Q$ are collinear.

Lemma 2: $ P$, $ O$, and $ C$ are collinear.

Proof: Let $ CO\cap AE\equal{}P'$, but notice $ CE\equal{}CA$, and $ AO\equal{}OE$, so $ CO$ is the perpendicular bisector of segment $ AE$. Hence $ P'$ is the midpoint of $ AE$, so $ P\equiv P'$, and we are done.

It follows by SAS congruency that $ \triangle PNC\cong \triangle PQC$ so $ \angle OPN\equal{}\angle CPN\equal{}\angle CPQ\equal{}\angle OPM$. So

$ \angle MPO\equal{}\angle NPO$ (1)

Additionally from the proof of lemma 2, $ OP\perp AE$, so

$ \angle OPA\equal{}90^\circ$ (2)

Lines (1) and (2) imply that $ A$, $ M$, $ O$, and $ N$ form a harmonic range by a common lemma.

Let $ O\equal{}0$ and $ A\equal{}1$, then $ B\equal{}z$, $ C\equal{}z^2$,... where $ z$ is the first fifth root of unity. Easily, we have that

$ 2M\equal{}z\plus{}z^{\minus{}1}$, and $ 2N\equal{}z^2\plus{}z^{\minus{}2}$.

We want to show that $ \frac{ \frac{AM}{MO} }{\frac{AN}{NO}}\equal{}1$.

$ AM*NO\equal{}MO*AN$

Then note that $ AM^2\equal{}AM*\overline{AM}$...the condition is equivalent to:

$ (z\plus{}z^{\minus{}1}\minus{}2)(z\plus{}z^{\minus{}1}\minus{}2)(z^2\plus{}z^{\minus{}2})(z^2\plus{}z^{\minus{}2})$
$ \equal{}(z\plus{}z^{\minus{}1})(z\plus{}z^{\minus{}1})(z^2\plus{}z^{\minus{}2}\minus{}2)(z^2\plus{}z^{\minus{}2}\minus{}2)$

or $ (z\minus{}1)^4*(z^2\plus{}z\plus{}1)(z^4\plus{}z^3\plus{}z^2\plus{}z\plus{}1)\equal{}0$ which is true.

Very nice Altheman. There are another solution using trig.

We must prove that $ \dfrac{MA}{MO}\equal{}\dfrac{NA}{NO}$.

Let $ \ell$ be the side of the pentagon and $ r$ the radius of the circle circunscribed. We know that

$ \ell\equal{}\dfrac{r}{2}\cdot\sqrt{10\minus{}2\sqrt 5}$.

From $ \Delta AEP$:

$ AE^2\equal{}MA\cdot AP\Longrightarrow \boxed{MA\equal{}\dfrac{r}{4}\cdot (5\minus{}\sqrt 5)}$

From $ \Delta OND$:

$ NO\equal{}r\cdot \cos36^{\circ}\Longrightarrow \boxed{NO\equal{}\dfrac{r}{4}\cdot (\sqrt 5 \plus{}1)}$

$ MO\equal{}r\minus{}MA \Longrightarrow \boxed{MO\equal{}\dfrac{r}{4}\cdot (\sqrt 5 \minus{}1)}$

And

$ NA\equal{}MA\plus{}MO\plus{}NO\Longrightarrow \boxed{NA\equal{}\dfrac{r}{4}\cdot (5\plus{}\sqrt 5)}$

Plugging in $ \dfrac{MA}{MO}\equal{}\dfrac{NA}{NO}$ we are donne.

Here is the figure for the first Altheman's solution.

Let $ P$ be the midpoint of $ AB$, $ S$ - the midpoint of $ AC$. Clearly $ P,S,N$ are collinear (the lie on the midline of trapezoid $ ABCD$).
As $ OP \perp AB$, $ OS \perp AC$ and $ BM \perp AN$ $ AS,OP,BM$ are concurrent (3 altitudes in AOB).
So by Ceva we have    $ \frac {AP \cdot BS \cdot OM}{PB \cdot SO \cdot MA} \equal{}1$

and by Menelaus    $ \frac {AP \cdot BS \cdot ON}{PB \cdot SO \cdot NA} \equal{}1$
or $ \frac {MO}{MA} \equal{} \frac {NO}{NA}$.
The last 3 lines are a proof of a well-known lemma.


linear algebra
matrix
linear algebra unsolved

Prove or disprove
   If $ X$ and $ X\minus{}AA^*$ are positive definite Hermitian, so is
   $ X\minus{}A^*A$.

[hide]
If $ X,A$ are matrices of order $ 2$, that statisfies $ X\equal{}{a}_{ij}$ where $ {a}_{11}\equal{}3$,$ {a}_{12}\equal{}{a}_{21}\equal{}0$ and $ {a}_{22}\equal{}1$, $ A\equal{}{b}_{ij}$ where $ {b}_{11}\equal{}0$,$ {a}_{12}\equal{}\sqrt {2}$,$ {a}_{21}\equal{}0$ and $ {a}_{22}\equal{}0$, then $ X\minus{}AA*\equal{}I$ but $ X\minus{}A*A\equal{}{c}_{ij}$ where
$ {c}_{11}\equal{}3$,$ {c}_{12}\equal{}{c}_{21}\equal{}0$ and $ {c}_{22}\equal{}\minus{}1$, we can prove that $ X$ is a positive definite Hermitian matrix and so is $ X\minus{}AA*$, but $ X\minus{}A*A$ is not.
We can generalize it for $ nxn$ matrices.
[/hide]
Sorry if I am wrong because I have just studied about positive definite matrix.


function
calculus
derivative
real analysis
real analysis unsolved

Let $ \Lambda_\alpha([0,1])$ be the space of Hölder continuous functions of exponent $ \alpha$ on $ [0,1]$; that is, of functions $ f$ such that $ \sup_{x\neq y\in[0,1]} \frac{|f(x)\minus{}f(y)|}{|x\minus{}y|^\alpha}$ is finite.
Let $ \lambda_\alpha([0,1])$ be the set of all $ f\in\Lambda_\alpha([0,1])$ such that $ \frac{|f(x)\minus{}f(y)|}{|x\minus{}y|^\alpha}\rightarrow 0$ as $ x\rightarrow y$ for all $ y\in[0,1]$.

Show that:
(a) If $ \alpha<1$, then $ \lambda_\alpha([0,1])$ is an infinite-dimensional closed subspace of $ \Lambda_\alpha([0,1])$.
(b) If $ \alpha\equal{}1$, then $ \lambda_\alpha([0,1])$ contains only constant functions.

(Clearly, $ \lambda_\alpha([0,1])$ is a closed subspace of $ \Lambda_\alpha([0,1])$.)

For $ \alpha<1,$ note that $ C^1$ - that is, the set of continuously differentiable functions - is contained in $ \lambda_{\alpha},$ and $ C^1$ is infinite dimensional.

For $ \alpha\equal{}1,$ does being in $ \lambda_{1}$ require that a function have a derivative and that its derivative is zero?


search
inequalities proposed
inequalities

Prove that:
$ x\plus{}y\plus{}z\leq xyz\plus{}2$
Where $ x,y,z$ are some reals satisfying: $ x^2\plus{}y^2\plus{}z^2\equal{}2$

Already posted [url=http://www.mathlinks.ro/viewtopic.php?t=163928]here[/url]
And solved by me :lol:

It's OK,but next time try to search yourself,OK?


geometry
geometric transformation
reflection
geometry unsolved

Let $ ABC$ be a triangle and $ I$ be its incenter. Let $ \omega_1$, $ \omega_2$, $ \omega_3$ be the incircles of triangles $ BCI$, $ ACI$, $ ABI$, respectively. Let $ \omega_3$ touches $ AB$ at $ T$. Prove that the another inner common tangent of $ \omega_1$ and $ \omega_2$ passes through $ T$.

It is equivalent to prove that the reflection of $ T$ in the line connecting the centers of $ \omega_1$ and $ \omega_2$ lies on $ CI$. I don't know if it helps. It is hard to deal with common tangents...

See this post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3680[/url]


algebra
polynomial
coefficients
number theory
degree
IMO
IMO 1974

[quote="orl"]Let $P(x)$ be a polynomial with integer coefficients. We denote $\deg(P)$ its degree which is $\geq 1.$ Let $n(P)$ be the number of all the integers $k$ for which we have $(P(k))^{2}=1.$ Prove that $n(P)- \deg(P) \leq 2.$[/quote]

This problem had been proposed on AMM , i think!  :) . Here's a simple way: We need a lemma:
[b]Lemma[/b]: Let $P(x)$ be a polynomial with integer coefficients which is not constant. Then if $P(x)$ obtains $1$ (or $-1$)  as its values for at least four times then $P(x)\neq -1$ ( or $P(x)\neq 1$) for all $x$.
[b]Proof[/b]. Assume that $P(a)=P(b)=P(c)=P(d)=1$ for $a,b,c,d$ distince. Then if there's $u$ which $P(u)=-1$ then $2=P(a)-P(u)=...$ so $P(x)-P(u)-2=(x-a)(x-b)(x-c)(x-d)Q(x)$ where $Q(x)$ is a polynomial with the integer coefficients! So
$-2=(u-a)(u-b)(u-c)(u-d)Q(u)$ which is impossible cause $-2$ can not presents as product of more than three distince numbers! This proved the lemma!

Back to our problem: For convinet put $m=n(P)$ and $n=\deg P$. Firstly if $n\leq 2$ then $m-n\leq2$. Assume $n\geq3$. If equation $P(x)=1$ with more than three integer points (ie.. at least $4$) then equation $(P(x))^2=1$ implies $P(x)=1$ so $m\leq n$, ie... $m-n\leq 0\leq2$. The same case for equation $P(x)=-1$. So $m\leq 6$. If $n\geq4$ then $m-n\leq 6-n\leq 2$. Now assume that $n=3$. In this case if $m\leq 5$ then $m-n\leq 2$.

So let us show that $m<6$. In fact if $m=6$ then $P(x)-1=0$ has three integers distince roots, and the same for $P(x)+1=0$. So
$P(x)-1=k_1(x-a_1)(x-a_2)(x-a_3)$ and $P(x)+1=k_2(x-b_1)(x-b_2)(x-b_3)$ where $a_i$ distince and $b_j$ distince and all with $k_1,k_2$ are integers! Then
$k_2(x-b_1)(x-b_2)(x-b_3)-k_1(x-a_1)(x-a_2)(x-a_3)=2$ for all $x$. So $k_1=k_2=k$.
Finally, we have
$2=k(a_i-b_1)(a_i-b_2)(a_i-b_3)$ for $i=1,2,3$ and because that $\pm1$ can not presents as products of three distince numbers so $k=\pm1$, we may assume $k=1$. Because $2=(-2)\cdot 1\cdot -1$ so
$\{-2,1,-1\}=\{a_i-b_1,a_i-b_2,a_i-b_3\}$
This means $3a_i-(b_1+b_2+b_3)=-2+1-1=-2$. So we must have $3a_1=3a_2=3a_3$ which follows $a_1=a_2=a_3$, which contracts!. So $m\leq 5$ and we're done

[hide=Sketch of my solution]Consider two polynomials $F(x)$ and $G(x)$ with same coefficients but different constant term $c_1$ and $c_2$, with $c_1=c_2+2$. Let $r_1$ and $r_2$ be the two integers root of $F$ and $G$ respectively. Then, $r_1-r_2\mid 2$ or. Now, consider the smallest integer $k$ satisfying $P(k)=1$ or $P(k)=-1$. Both of these equations have no more than $\deg (P)$ roots. Now, with the above note, if $P(x)=1$ has $t$ roots then $P(x)=-1$ has at most $t+2$ root. And hence, we get QED.[/hide]

a very silly question..... :P  :blush:
why would p(x)^2=1 be at least 4 times??
I mean...it should have maximum 2 roots(1 and -1)
 :maybe:  :help:


Why are our post counts going down??...Tare had like 1865 last night and now its a little more than 1700...almost everyone's post count has gone down...

This may have to do with some culling in the Etc. forum.  If posts are manually culled, then post counts go down.

I agree. Some (well a lot of) threads were deleted manually yesterday or around that time. The post counts doesn't decrease when the post's auto-pruned, but it does go down if the moderator/administrator deletes it themselves.

It would be a lie to say this was completely unexpected after all [b]those[/b] spamming, but oh well.


who here is waiting for it to come out? YEAH! :D Snake rules!

I was about to start a topic like this. When is it coming out?

MARCH NINTH. CAN'T WAIT! GETTING IT AT MIDNIGHT. RESERVED IT IN DECEMBER. YAY!!!!

Hm wait does the Wii even have online capabilities (that is, you can play other people, not if you can surf Google  :roll: )?

Yep. SSBB will utilize the Wii's full WiFi capabilities. You can play friends(whom you have in your friend list) with pre-programmed messages and a name, or others with just the game.

Who was your favorite character in Mélee?

i like kirby b/c he can swallow enemies.. :D

Wow I just realized I have all of Spring Break to finish SSBB. No sleep for me. Before, I was just going to not do any homework and fail all my classes. Hopefully we wont be travelling anywhere

Jigglypuff obviously pwns, and so does Kirby. I also really liked Roy,[hide="SPOILER"]which is why I'm mad Marth is in Brawl and not him.[/hide]Anyway, CF is also pretty good.

I always pwn with link or samus or jigglypuff.

Strange, nobody's said anything about Fox (the awesomest of all), i'm not going to get brawl until next week cause my mom says it's a distraction before State (MC).

All right Ive been Brawling for two days now and have unlocked 10/14 characters. Go me. Are we allowed to post Brawl Friend Codes on here? Should we?

I, personally don't see anything wrong with it, since you need both parties to enter friend codes in order to see status/play together.

Mine's 4854-6150-2058.

I think we should make a separate topic for SSBB friend codes.  It'll probably attract more attention.

http://www.onemorelevel.com/game/super_smash_flash

That is [i]the[/i] best game.

Pit>Marth>Eveyone else but Yoshi>Yoshi, in a nutshell.

Pit is pretty good, from what I've seen.

Maybe we should make a "AoPS SSBB FC Archive" or something ocne we get more FCs. That would be much easier to find FCs.

The characters that I'm sure have a bright future coming are:

Marth
Space Animals (in case you don't know, that means Fox, Falco, Wolf)
Dedede

Well, dedede probably won't be high tier, but definitely a high middle tier alongside Meta Knight. This is pretty amazing, considering that this is pretty much the first time in smash history that a slow and bulky character wasn't completely crap. Same for Ike.

I have no idea but that thing can do a whole lot of damage. Getting chomped by it is like a KO.

Im hooked on Subspace Emissary
Im also hooked onto SSBB's soundtrack

[quote="andersonw"]I went to my friend's house yesterday to play brawl, and it was really awesome. Now I wish I had a Wii. :lol:

Also: Who names a giant robot Porky???[/quote]

It's a fat guy controlling the Robot. He's from the Earthbound/Mother series.

I just beat Subspace Emissary yesterday.
Toon Link seems like a good character.


geometry
trigonometry
projective geometry

[b]Έστω κύκλος (Ο,R), ένα σταθερό σημείο Α του κύκλου και ε η εφαπτομένη του κύκλου στο Α.Θεωρούμε πάνω στην ε δυό μεταβλητά σημεία Μ, Ν που είναι τέτοια ώστε ΑΜ*ΑΝ=ct.Έστω Ρ το σημείο επαφής του κύκλου και της εφαπτομένης από το Μ, και Σ το σημείο επαφής του κύκλου και της εφαπτομένης από το Ν.Να αποδείξετε ότι η ΡΣ διέρχεται από ένα σταθερό σημείο.[/b]

Μάλιστα ..... Το σχόλιό  μου είναι ότι η άσκηση είναι φοβερή και (σχεδόν ακατόρθωτη με απλή γωμετρία...) .Όμως Νίκο όπως ξέρεις υπάρχουν και άλλα κόλπα.....   :D
Μακάρι να τα βρω όμως :rotfl:

Φυσικά και υπάρχουν κι άλλα κόλπα.....Όμως θα αρκεστώ στο να πω ότι με αυτή την άσκηση μπορώ να δείξω σε όλους τη δύναμη της ευκλείδειας γεωμετρίας!!Βέβαι επειδή δεν τη βρήκα με τη λύση και την έλυσα μόνος μου μπορεί να είναι και πιο απλή απ' όσο φαντάζομαι....
Καλή Επιτυχία σε όλους!

Υ.Γ.Τη χαρακτήρισα warm up διότι έπονται πολύ πιο δύσκολες.....

Την έλυσα με αναλυτική όμως με δύο σελίδες υπολογισμούς..... :o
Very ungly .Το μόνο θετικό στην υπόθεση είναι ότι βρήκα ότι το σταθερό σημείο είναι το σημείο τομής της ΟΑ με την ΡΣ .Από εκεί περνάνε όλες οι άλλες ευθείες ΡΣ  καθώς τα Ρ, Σ κινούνται ανάλογα με τα Μ,Ν .Επίσης έλυσα με απλή γεωμετρία και αν θέλετε θα βάλλω τη λύση μου την απλή και τετριμμένη περίπτωση όπου Για δύο θέσεις των Μ,Ν έστω Μ',Ν' ισχύει ότι Ν'Α=ΑΜ' και ΝΑ=ΜΑ  :|
Από εκεί και πέρα θα το προσπάθησω και για τις άλλες περιπτώσεις και θα ανακοινώσω προόδους (αν έχω από δω και πέρα)....  :lol:

Εντάξει έστω ότι δε βαρέθηκες να κάνεις δύο σελίδες πράξεις....Πουθενά δεν έκανες λάθος????Ένα + αντί για - ή κάτι τέτοιο βρε αδερφέ????

Περιμένω <<απλή>>- sorry εννοώ  κομψή λύση

Τα αποτελέσματα που βρήκα σωστά δεν είναι ? Τώρα πρέπει επιτέλους να λύσω μια άσκησή σου με απλή ευκλείδια γεωμετρία .

ΟΤΑΝ ΣΑΣ ΛΕΕΙ Ο ΡΑΠΑΝΟΣ ΑΠΛΗ ΦΕΡΤΕ ΤΟΝ ANDRESCU  ΤΟΝ ANDRICA KAI ΤΟΝ PELIKAN MAZI ΝΑ ΤΗΝ ΛΥΣΟΥΝ ΧΑΙΡΕΤΩ ΜΕΤΑ ΠΟΛΛΗΣ ΦΙΛΙΑΣ ΤΟΝ ΑΓΑΠΗΤΟ ΜΟΥ ΝΙΚΟ ΡΑΠΑΝΟ ΕΚ ΠΑΤΡΑΣ

[u]Προς Σιλουανό[/u]
Σιλουνέ σωστό είναι το αποτέλεσμα που βρήκες!Συγχαρητήρια!!Βέβαια με τόσα λίγα στοιχεία ήταν πασιφανές ότι πρόκειται γι' αυτό το σημείο.....Όλη η ιστορία γίνεται για να το αποδείξεις με <<στοιχειώδη>> τρόπο!

[u]Προς τον αγαπητό Αμβρόσιο[/u]
Ρε τι είναι η γεωμετρία???Γυναίκα δεν είναι??

[b][i][u]ΑΜΑ ΤΗΝ ΠΑΙΔΕΨΕΙΣ ΛΙΓΟ, ΘΑ ΒΓΕΙ!![/u][/i][/b]

[b][i][u]ΝΙΚΟΣ[/u][/i][/b]

ΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩ ΜΑΓΕΙΑ ΜΑΓΕΙΑ ΜΑΓΕΙΑ ΥΠΟΚΛΙΝΟΜΑΙ ΣΤΟ ΤΑΛΕΝΤΟ ΣΟΥ ΑΡΧΟΝΤΑ
ΧΑΙΡΕΤΩ

Κι εγώ λέω τι μου θυμίζει... http://www.mathlinks.ro/Forum/viewtopic.php?t=44854

[b][color=blue]Περίπτωση κατά την οποία τα Μ,Ν βρίσκονται εκατέρωθεν του Α.[/color][/b]
Έστω Ε το αντιδιαμετρικό του Α.

Τότε θα είναι $\Sigma E \parallel NO,PE\parallel MO$.
Επομένως οι γωνίες $NOM,PE\Sigma$ είναι ίσες.

Επίσης $\widehat{P\Sigma E} = 90-\widehat{P\Sigma A} = 90-\widehat{MOA} = \widehat{OMA} = \widehat{OMN}$

Άρα τα τρίγωνα $PE\Sigma,NOM$ είναι όμοια

Για απλοποίηση των συμβολισμών θέτουμε $\widehat{MOA} = \phi, \widehat{NOA}=\theta$
Τότε $\tan\phi\tan\theta = \frac{AM\cdot AN}{R^2}=ct$

Γράφω τις ίσες γωνίες των όμοιων τριγώνων:

$\widehat{PE\Sigma} = \widehat{NOM}$

$\widehat{E\Sigma P} = \widehat{OMN} = 90^o-\phi$

$\widehat{\Sigma PE} = \widehat{MNO} = 90^o-\theta$

Τα σημεία Ρ,Σ είναι προφανώς σε διαφορετικό μέρος της ΟΑ.
Έστω λοιπόν Β το σημείο τομής του ευθυγράμμου τμήματος ΡΣ με την ευθεία ΟΑ.

$\triangle BE\Sigma: \frac{\sin\widehat{E\Sigma B}}{EB} =\frac{\sin\widehat{BE\Sigma}}{B\Sigma} \Rightarrow \frac{\sin(90^o-\phi)}{EB} = \frac{sin\theta}{B\Sigma}\Rightarrow \boxed{\frac{\cos\phi}{EB} = \frac{\sin\theta}{B\Sigma}}$
αφού $\widehat{BE\Sigma} = \widehat{AE\Sigma} = \widehat{NOA} = \theta$

.. και ομοίως

$\triangle BEP: \frac{\sin\widehat{BPE}}{EB} =\frac{\sin\widehat{PEB}}{BP} \Rightarrow \frac{\sin(90^o-\theta)}{EB} = \frac{\sin\phi}{BP} \Rightarrow \boxed{\frac{cos\theta}{EB} = \frac{\sin\phi}{BP}}$

Πολλαπλασιάζοντας παίρνουμε:

$\frac{\cos\phi \cos\theta}{EB^2} = \frac{sin\phi \sin\theta}{BP \cdot B\Sigma}\Rightarrow BP \cdot B\Sigma = EB^2\tan\phi \tan\theta$

Από τη δύναμη του σημείου Β παίρνουμε $BP \cdot B\Sigma = EB \cdot BA \Rightarrow BA = EB\tan\phi \tan\theta$

Άρα $\frac{BA}{EB} = \tan\phi \tan\theta = \sigma\tau a \theta\epsilon\rho o$

Για την περίπτωση στην οποία τα Μ,Ν βρίσκονται στο ίδιο μέρος του Α, η ΡΣ τέμνει την ΟΑ στο Κ, το οποίο είναι το αντίστροφο του Β.
Ένας τρόπος θα ήταν να δείξουμε ότι
[color=blue]
Έστω κύκλος κ, ένα σταθερό σημείο Α του κ και ε η εφαπτομένη του κύκλου στο Α
Σημεία Μ,Ν επί της ευθείας ε και Ν' το συμμετρικό του Ν ως προς το Α.
ΜΡ,ΝΣ,Ν'Σ' τα εφαπτόμενα τμήματα στον κύκλο
Τότε τα σημεία τομής της ευθείας ΟΑ με τη ΡΣ και τη ΡΣ' είναι αντίστροφα
[/color]

(στη γενική περίπτωση, χωρίς να μας ενδιαφέρει το γινόμενο ΑΜ.ΑΝ)

Δεν το έχω αποδείξει, οπότε χρησιμοποίησα τον ίδιο τρόπο με πριν.
Στο σχήμα έχω πάρει το συμμετρικό του Ν (το ονομάζω και πάλι Ν, όχι Ν').
Επίσης, φαίνεται η θέση του Β όπως ήταν στο αρχικό σχήμα.

Ας υποθέσουμε ότι ΑΝ>ΑΜ.

Η διαφορά είναι ότι τα Ρ,Σ τώρα βρίσκονται στο ίδιο μέρος της ΟΑ.

Είναι και πάλι $\widehat{P\Sigma E} = \widehat{OMN}$ αφού από τον κύκλο έχουμε $\widehat{P \Sigma E} = 180^o-PAE$ και $\widehat{PAE} = \widehat{AMO}$

Επίσης, η γωνία ΕΣΡ είναι $\widehat{E\Sigma P} = 180 - \widehat{AMO} = 180-(90-\phi) = 90+\phi$

Η ΡΣ τέμνει την ΟΑ ανν $P\Sigma \not \parallel OA$
Είναι παράλληλες αν ισχύει $\sin(2\theta)=\sin(2\phi) \Leftrightarrow 2\phi+2\theta=180^o \Leftrightarrow \phi+\theta=90^o$

Αυτό βέβαια σημαίνει ότι $\tan \phi \tan \theta = 1$
επομένως η περίπτωση $AM\cdot AN=R^2$ δεν έχει λύση όταν τα Μ,Ν είναι στο ίδιο μέρος του Α.

Στο νόμο των ημιτόνων στα τρίγωνα ΚΕΣ,ΒΕΡ δεν αλλάζει τίποτα αφού (πχ στο ΚΕΣ)
$E\Sigma K = 180-E\Sigma P = 90-\phi$, και
$KE\Sigma = 180-AE\Sigma = \theta$

Τελικά, παίρνουμε και πάλι

$KP \cdot K\Sigma = KE^2\tan\phi \tan\theta \Rightarrow$

$KE \cdot KA = KE^2\tan\phi \tan\theta \Rightarrow$

$KA = KE\tan\phi \tan\theta \Rightarrow$

$\frac{KA}{KE} = \tan\phi \tan\theta$

με μόνη διαφορά ότι το Κ είναι στο εξωτερικό του τμήματος AE

[color=brown][b][u]Παρατήρηση[/u][/b]
Θέτοντας $w$ το σταθερό γινόμενο $\tan\phi\tan\theta = \frac{AM\cdot AN}{R^2}$

μπορούμε να γράψουμε $BO = R \frac{w-1}{w+1}$ και $KO = R \frac{w+1}{w-1}$

απ' όπου βλέπουμε ότι $BO \cdot KO=R^2$, άρα τα Β,Κ είναι αντίστροφα![/color]

Λουκάς

[b][color=blue]Και μία τριγωνομετρική λύση[/color][/b]

Έστω o κύκλος $(O,1)$, το σημείο $A(1,0)$ και σημεία $M(1,m),N(1,n)$ έτσι ώστε $mn = w = \sigma\tau{a}\theta\epsilon\rho{o}$

Υπάρχουν γωνίες $\phi,\theta \in (-\frac{\pi}{2},\frac{\pi}{2})$ τέτοιες ώστε $\tan{\phi} = m, \tan{\theta}=n$

Παρατηρούμε ότι $\cos\phi, \cos\theta >0$.

Επίσης, αν $\phi = \theta$, τότε τα Ρ,Σ ταυτίζονται. Επομένως υποθέτουμε $\phi \neq \theta$

Από τα $M,N$ φέρουμε τα εφαπτόμενα τμήματα $MP,N\Sigma$ αντίστοιχα.
Ως γνωστόν θα είναι $P(\cos{2\phi},\sin{2\phi})$ και $\Sigma(\cos{2\theta},\sin{2\theta})$
Έχουμε $2\phi,2\theta \in (-\pi,\pi)$

Θα δείξουμε ότι [u]αν[/u] η ευθεία $P\Sigma$ τέμνει τον άξονα $x'x$, τότε τον τέμνει σε σταθερό σημείο και θα βρούμε αυτό το σημείο συναρτήσει του $w$.

[b][hide="Ας δούμε πρώτα πότε η ΡΣ δεν τέμνει τον άξονα x"][/b]
Θα πρέπει τα Ρ,Σ να έχουν την ίδια τεταγμένη, δηλαδή $\sin 2\phi = \sin 2\theta \Leftrightarrow \boxed{\phi=\theta}$ ή $\boxed{2\phi+2\theta = \pm\pi}$
Η πρώτη σχέση έχει απορριφθεί από την αρχή.

Η δεύτερη δίνει $\phi+\theta = \pm\frac{\pi}{2}$, άρα $\tan\phi \tan\theta=1$.

Έχουμε λοιπόν $P\Sigma \parallel x'x \Leftrightarrow w=1$
Στην περίπτωση αυτή μάλιστα, η ευθεία ΡΣ δε θα είναι σταθερή (έχει εξίσωση $y = \sin 2\phi$),
άρα [u]δεν περνά από σταθερό σημείο[/u].[/hide]

Στα παρακάτω θα θεωρούμε ότι $w\neq 1$
Έστω ότι η ευθεία $P\Sigma$ τέμνει τον άξονα $x'x$ στο σημείο $T(x,0)$.

Από την ορίζουσα των σημείων Ρ,Σ,Τ παίρνουμε:

$\left | \begin{array}{ccc} \cos 2\phi & \sin 2\phi & 1 \\ \cos 2\theta & \sin 2\theta & 1 \\ x & 0 & 1 \\ \end{array} \right |=0 \Leftrightarrow$

$\sin 2\phi \cos 2\theta - \sin 2\theta\cos 2\phi = x(\sin 2\phi - \sin 2\theta) \Rightarrow$

$\sin(2\phi-2\theta) = x\cdot 2 \sin(\phi-\theta)\cos(\phi+\theta) \Rightarrow$

$2\sin(\phi-\theta)\cos(\phi-\theta) = x\cdot 2 \sin(\phi-\theta)\cos(\phi+\theta) \Rightarrow$

$\cos(\phi-\theta) = x\cdot \cos(\phi+\theta) \Rightarrow$

$\cos\phi\cos\theta+\sin\phi\sin\theta = x(\cos\phi\cos\theta-\sin\phi\sin\theta) \Rightarrow$

$1+\tan\phi\tan\theta = x(1-\tan\phi\tan\theta) \Rightarrow$

$1+w = x(1-w) \Rightarrow$

$\boxed{x = \frac{1+w}{1-w}} \ \ (2)$

Η τελευταία διαίρεση μπορέι να γίνει, αφού υποτέθηκε ότι $w\neq 1$

Πιο πάνω διαιρέσαμε με το $\sin(\phi-\theta)$.
Αυτό μπορεί να γίνει αφού το ημίτονο αυτό είναι διάφορο του μηδέν
Πράγματι έχουμε:

$\phi,\theta \in (-\frac{\pi}{2},\frac{\pi}{2})\Rightarrow (\phi-\theta) \in (-\pi,\pi)$

και αφού $\phi \neq \theta \Rightarrow \sin(\phi-\theta)\neq 0$

ΣΥΓΧΑΡΗΤΗΡΙΑ ΣΤΟΝ ΦΙΛΟ ΜΑΣ ΤΟΝ ΠΟΝΤΙΟ ΓΙΑ ΤΙΣ 2 ΚΟΜΨΕΣ ΜΑ ΛΙΓΟ ΠΕΡΙΠΛΟΚΕΣ ΛΥΣΕΙΣ
ΠΑΡΑΘΕΤΩ ΕΠΙΓΡΑΜΜΑΤΙΚΑ ΜΙΑ ΑΛΛΗ ΠΙΟ ΑΠΛΗ ΠΡΟΣΠΑΘΗΣΤΕ ΝΑΔΕΙΞΕΤΕ ΟΤΙ  ΡΟ'*Ο'Σ/Ο'Α'*Ο'Α=ΣΤΑΘ=ΑΜ*ΑΝ/R^2 ME OMOIOTHTA ΤΡΙΓΩΝΩΝ ΟΠΟΥ Ο' Η ΤΟΜΗ ΡΣ ΚΑΙ ΑΑ' ΟΠΟΥ Α' ΤΟ ΑΝΤΙΔΙΑΜΕΤΡΙΚΌ ΤΟΥ Α ΟΚ? ΤΟΤΕ ΘΑ ΙΣΧΥΕΙ ΤΟ ΖΗΤΟΥΜΕΝΟ ΛΟΓΩ ΣΤΑΘΕΡΗΣ ΔΥΝΑΜΗΣ
ΧΑΙΡΕΤΑΙ
ΠΣ ΤΟ ΕΛΕΞΑΜΕ ΜΕ ΤΟΝ ΡΑΠΑΝΟ ΓΙΑ ΟΣΟΥΣ ΑΜΦΙΣΒΗΤΟΥΝ ΤΗΝ ΟΡΘΟΤΗΤΑ ΤΗΣ ΛΥΣΗΣ

Στην ουσία η λύση σου είναι σχεδόν ίδια με του Λουκά  ;)

Παιδιά, πάρα πολύ ωραίες οι λύσεις αν και διάβαζα δύο-δύο τις γραμμές διότι δεν με  βολεύει το διάβασμα από την οθόνη.
Βέβαια, η άσκηση βγαίνει σε μισή γραμμή χρησιμοποιώντας το Θεώρημα Caley!!!

Νίκ είσαι σίγουρος ότι το θεώρημα λέγεται Caley  : ? ???
Έψαξα και δεν το βρήκα .Αν λέγεται Caley μπορείς να μας πεις τι υποστηρίζει αυτό το θεώρημα

Ε ΑΙ ΣΤΟΝ .............. ΚΟΡΑΚΑ ΜΕ ΤΙΣ ΛΥΣΕΙΣ ΜΕ ΤΟ ΤΑΔΕ Η ΤΟ ΑΛΛΟ ΘΕΩΡΗΜΑ ΕΙΝΑΙ ΕΚΝΕΥΡΙΣΤΙΚΟ ΟΠΩΣ ΚΑΙ Ο ΑΝΤΡΕΣΚΟΥ ΜΕ ΤΟ FIVE COIN THEOREM  ΜΑΣ ΕΧΕΤΕ ΠΡΗΞΕΙ ΚΑΛΚΟ ΘΑ ΕΙΝΑΙ ΝΑ ΜΑΣ ΠΕΙΤΕ ΟΣΟΙ ΞΕΡΕΤΕ ΓΙΑΤΙ ΕΓΩ ΔΕΝ... ΠΟΥ ΝΑ ΤΑ ΒΡΟΥΜΕ ΑΥΤΑ
ΧΑΡΕΤΑΙ

Χαιρετώ το Σιλουανό και τον επιθετικό Αμβρόσιο....
Σιλουανέ το θεώρημα μπορεί να λέγεται και Casey....(υπάρχουν και οι δύο εκδοχές)
[b][u]ΘΕΩΡΗΜΑ CALEY[/u]
Tο απόλυτο της διαφοράς των δυνάμεων ενός τυχαίου σημείου του επιπέδου από δύο δοσμένους κύκλους, είναι ίσο με το διπλάσιο της διακέντρου επί την απόσταση του σημείου από το ριζικό άξονα.
[/b]
Ρε Αμβρόσιε αυτό είναι γνωστό θεώρημα.Μπορεί να μη χρησιμοποιείται συχνά, αλλά παρ' όλα αυτά είναι γνωστό.
Π.χ. το Θεώρημα Desargues δεν το έχεις χρησιμοποιήσει ποτέ αλλά το ξέρεις....

[b][u]Νικος[/u][/b]

ΧΑΙΡΕΤΩ ΚΙ ΕΓΩ ΤΟΝ ΕΚΦΡΑΣΤΙΚΟ ΜΟΥ ΦΙΛΟ ΝΙΚΟ ΚΙ ΛΕΩ
ΤΟ ΘΕΩΡΗΜΑ C....(ΤΟ ΛΕΩ ΕΤΣΙ ΕΠΕΙΔΗ ΔΕΝ ΕΙΝΑΙ ΕΠΙΒΕΒΑΙΩΜΕΝΟ ΤΟ ΟΝΟΜΑ) ΜΟΥ ΦΑΙΝΕΤΑ ΟΤΙ ΤΟ ΕΧΩ ΞΑΝΑΣΥΝΑΝΤΗΣΕΙ ΟΧΙ ΩΣ ΘΕΩΡΗΜΑ ΜΕ ΟΝΟΜΑ ΑΛΛΑ ΕΤΣΙ ΤΩΡΑ ΓΙΑ ΤΟ 2Ω ΘΕΩΡΗΜΑ ΠΟΥ ΠΡΟΑΝΕΦΕΡΕΣ ΔΕΝ ΤΟ ΓΝΩΡΙΖΩ ΚΑΝΕ ΤΟΝ ΚΟΠΟ ΝΑ ΤΟ ΑΝΑΦΕΡΕΙΣ ΜΗΠΩΣ ΚΑΙ ΤΟ ΓΝΩΡΙΖΩ ΑΛΛΑ ΠΑΛΙ ΑΠΟΦΕΥΓΕΙΣ (ΓΙΑΤ??????) :huh:  :huh:  :mad: ΝΑ ΑΠΑΝΤΗΣΕΙΣ [u][b]ΠΟΥ[/b][/u] ΤΑ ΒΡΙΣΚΕΙΣ ΑΥΤΑ
ΧΑΙΡΕΤΩ ΜΕΤΑ ΠΟΛΛΗΣ ΦΙΛΙΚΟΤΗΤΑΣ

Mπορείς να ρωτήσεις του Παπαμιχάληδες και θα σου πουν τα διπλά από αυτά που θα σου λεγα εγώ.....
Το Desargues είναι πολύπλοκο στη διατύπωση....με σχήμα μπορώ να στο δείξω αλλά με λόγια.....
Μια φορά το έχω χρησιμοποιήσει μόνο αλλά η άσκηση ξευτιλίστηκε...
Αυτό το θεώρημα για να αποδειχθεί θέλει 3 Μενέλαους και γι' αυτό είναι πολύ ισχυρό!!!

Υ.Γ. Θα σε πάρω τηλ το βράδυ για να μου δώσεις το FAX.

Bye

Αμβρόσιε
  δες στο συνημμένο αρχείο το θεώρημα Desargues .

Λοιπόν όσον αφορά το πρώτο παντού το θεώρημα λέγεται  casey και αν έλεγες casey θα το καταλάβαινα γιατί είναι πολύ γνωστό . Επίσης Νίκο (Ράπανε) μάλλον δε θυμάσαι  με ποιον μιλούσες τώρα τελευταία για το θεώρημα Desarques .Νομίζω ότι μιλούσες για αυτό με το Γιώργο(Δέσκο) .Και μιας και έχεις ασχοληθεί (και για όσους έχουν ασχοληθεί με προβολική γεωμετρία ) εκεί το θεώρημα αποδεικνύεται πολύ εύκολα νομίζω . :)

Ρε έχεις δίκιο για το όνομα-έτσι πρέπει να είναι το σωστό αλλά μου έχει κολλήσει το Caley τι να κάνουμε???
Θα το διορθώσω.....
Όσο για το Desargues μιλούσα με το Δέσκο πριν τα προφορικά..Για την ακρίβεια μου εξηγούσαν (Δέσκος και Κουφογιάννης) το πως θα διατυπώσω το θεώρημα σε ενδεχόμενη ερώτηση στα προφορικά!!Όπως είπα και πριν μπορώ να το δείξω αλλά βαριέμαι να το λέω....
Παιδιά άλλη μία άσκηση θα βάλω και μετά (μετά από τις 3 Μαίου) θα γράψω κι εγώ καμιά λύση για να μη φωνάζετε...

Nick

ΕΥΧΑΡΙΘΣΤΩ ΠΑΡΑ ΠΟΛΥ ΟΛΑ ΤΑ ΠΑΙΔΙΑ ΓΙΑ ΤΙΣ ΔΙΑΦΩΤΙΣΤΙΚΕΣ ΠΛΗΡΟΦΟΡΙΕΣ ΘΑ ΗΘΕΛΑ ΝΑ ΕΥΧΗΘΩ ΣΕ ΟΛΟΥΣ ΚΑΛΟ ΠΑΣΧΑ ΚΑΙ ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΣΕ ΟΣΟΥΣ  ΠΑΝΕ ΚΥΠΡΟ-ΚΙΖΙΝΑΟΥ
ΧΑΙΡΕΤΑΙ

hi my dears...

as i saw in the diagrams, it seems that the problem is good.

could any body please trans late it for me?

i want u to share ure nice problems here with me ;)

[quote="silouan"]I can do it .
Let a circle (Ο,R),and a constant point Α of the circle and $e$ be the tangent of the circle at the point $A$.Let two varietes points on $e$ named  $M,N$ such that $AM\cdot AN=ct$
Let $P$ the point of tangecy of the circle with the tangent from $M$ and $S$ the point of tangency of the circle with the tangent from $N$ .Prove that $PS$ passes through a fixed point .

Please use inversion to solve it  ;)[/quote]

Thanks a lot my dear Silouan…

I thought and found a solution by Inversion. its really a good problem.

Ok……………………………………………………………………………………………………………………….

Lt me invert the shape round point $A$, and with radius, $AM.AN$.

Points $M,P,S,N,O$, goes to points $M',P',S',N',O'$, respectively.

Circle $APS$, is tangent to line $X$, then, after inversion, line $P'S'$ is

parallel to line $X$.

The lines $MP,NS$ are tangent to circle $APS$, at points $P,S$,

respectively, then , after inversion,

Circles $AM'P',AN'S'$, are tangent to line $P'S'$, at $P',S'$ respectively.

We know that, quadrilaterals $AMPO,ANSO$, are cyclic , then after

inversion, points $M',P',O'$ are collinear.

And points $N',S',O'$, are collinear, too.

And the conclusion is that, $O'$ is the intersection of $M'P',N'S'$.

as an inversion theorem said, $O'$, is the respect of $A$, to line $P'S'$.

Now in the new problem, we should prove that: $AP'S'$, goes through a

fixed point.

Ok……………………………………………………………………………………………………………………….

Suppose $AO'$ intersects $P'S'$, at $K$. it is very clear that triangles

$O'KP',O'AM'$ , are similar to each other.

And $O'KS',O'AN'$ are also similar to each other.

And we have:

$\frac{KP'}{AM'}=\frac{KS'}{AN'}=\frac{1}{2}$

we know that: $AM'.AN'=constant$

if we divide the equality at $4$, we have:

$\frac{AM'}{2}.\frac{AN'}{2}=\frac{constant}{2}= constatnt$

but:  $\frac{AM'}{2}=KP'$, and $\frac{AN'}{2}=KS'$.

And the conclusion is: $KP'.KS'=constant$.

Suppose circle $AP'S'$ intersect $AO'$, at $Z$, with power of the point,

we find that, $KA.KO'=KP'.KS'=constant$.

And because of $A$ is fixed, point $Z$, must be fixed, too.

And it is proved that by moving $S',P'$, circle $AP'S'$, goes through a

fixed point($Z$).

I really enjoyed it.

It has another proof by harmonic and pole and polar.

And I don’t know whether u discussed it in this topic or not.

If not, I can send that solution too.

How ever u made me glad, thanks a lot again , my dear friends.

[quote="silouan"]Very nice Ashegh .Your solution with inversion is really nice .
In The previous topics the man who solved the exercise solved in three ways . The one was trigonometrical and the other was with analitic geometry .The third solution was by the proposer Nick Rapanos and it was one-line solution .Just ysing Casey's theorem .  ;)[/quote]

really good ideas. ;)

then i will post my proof by pole and polar and harmonic, later. :D


A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $\$4.95$ for the father and $\$0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $\$9.45$, which of the following could be the age of the youngest child?

$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$

[hide]B. 2. Because the father is charged 4.95 you subtract that from 9.45=4.50. You divide 4.50 by .45 which is 10. The sum of the three children's age is 10. The younger child's age must be even because otherwise the twin's birthday wouldnt be whole numbers. You only have 2,4 left. You subtract 4 from 10 you get 6. and divide by 2. you get 3. This is impossible because the younger child would be older than the twins. All you have left is 2 which should work.[/hide]

[hide]
First, take off the father, 9.45-4.95=4.50.
4.50/45=10. Thus, the sum of the children's ages is 10.
Since there are twins, the younger child's age has to be even, leaving 2 and 4 left.
4 doesn't work, since the twins would be 3 each, but the twins are older.
So, the answer is B.[/hide]

[quote="Silverfalcon"]A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $\$4.95$ for the father and $\$0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $\$9.45$, which of the following could be the age of the youngest child?

A. 1
B. 2
C. 3
D. 4
E. 5[/quote]

[hide="Solution"]
9.45 - 4.95 = 4.50, so the sums of the ages must be 10.

Now, the younger child's age must be even (see other posts for reason).

The younger child can be 0, 2, 4, 6, 8, or 10.

If the younger child is 4 or older, then the younger child is older than both of the twins, which doesn't work out.

The younger child can be either 0 or 2, however, 0 is not an answer choice.

[b]B[/b]
[/hide]


Let $f:\mathbb{R} \to \mathbb{R}$ be a Lebesgue-measurable function. Suppose that $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Prove that $f$ is a continuous function. Conclude that $f$ has the following form $f(x)=cx$, where $c \in \mathbb{R}$.

A proof is sketched in post number $2$ from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32514]this topic[/url].

I'm looking for a more detailed solution for this problem.


modular arithmetic
Gauss
quadratics
search
algebra
functional equation
number theory proposed

Find all primes $ p$ such that the equation
$ x^{2}\equiv 10\bmod p$

[hide="what i have"]
$ p\equiv 1,4 \bmod 5$ or $ p\equiv 1\bmod 4$[/hide]
however, the book says that these are not correct. can anyone point me in the correct direction.

$ (5/p)=1$ when $ p\equiv \pm 1 \pmod{5}$
$ (5/p)=-1$ when $ p\equiv \pm 2 \pmod{5}$

$ (2/p)=1$ when $ p\equiv \pm 1 \pmod{8}$
$ (2/p)=-1$ when $ p\equiv \pm 3 \pmod{8}$

So, we have that
$ (10/p)=1\cdot 1=1$ when $ p\equiv \pm 1 \pmod{5}$ and $ p\equiv \pm 1 \pmod{8}$
$ (10/p)=(-1)\cdot(-1)=1$ when $ p\equiv \pm 2 \pmod{5}$ and $ p\equiv \pm 3 \pmod{8}$

[quote="maokid7"]Find all primes $ p$ such that the equation $ x^{2}\equiv 10\bmod p$ [omissis][/quote]
I guess you mean "find all primes $ p$ such that the equation $ x^{2}\equiv 10 \bmod p$ has a solution in $ \mathbb{Z}/p\mathbb{Z}$", don't you? :wink: If I'm right, then say $ p$ is [i]good[/i] when fulfilling the condition in hand. Obviously $ 2$ and $ 5$ are [i]good[/i]. So admit $ \gcd(10,p) = 1$. Then for sure $ p \equiv r \bmod 10$, for some $ r \in \{\pm 1, \pm 3\}$. Hence
\[ \left(\frac{10}{p}\right) = \left(\frac{2}{p}\right) \cdot \left(\frac{5}{p}\right) = (-1)^{(p^{2}-1)/8}\cdot \left(\frac{|r|}{5}\right), \]
according to Gauss' lemma (on the quadratic character of $ 2$) and Gauss' (quadratic) reciprocity law. Thus $ p$ is [i]good[/i] iff $ p \equiv \pm 1 \bmod 8$, whenever $ p \equiv \pm 1 \bmod 10$; or $ p \equiv \pm 3 \bmod 8$, whenever $ p \equiv \pm 3 \bmod 10$. So finally iff it's $ \equiv \pm 3 \bmod 40$ or $ \equiv \pm 13 \bmod 40$ or $ \equiv \pm 1 \bmod 40$ or $ \equiv \pm 31 \bmod 40$. Quite boring?! :|

[quote="baz"]I agree, my solutions were equivalent to saying $ (10/p)=1$ iff $ p\equiv \pm 1,\pm 3, \pm 9, \pm 13 \pmod{40}$.[/quote]
Yes, I know! But I began writing down my solution before yours appearing on the forum, there's nothing else. :wink: You know, I'm so slow when writing in English... :oops:

:rotfl: well funny but maybe not  :rotfl:  that funny...but hey did you understand my other post?

Arrrgh!!! :mad: :rotfl: Well, I think I will try to understand your results tomorrow, in the late morning. Now I'm going to bed, that awful functional equation ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=896896#896896]here[/url]) has almost killed me! :lol:

thanks but how did you manage to find the solutions to
$ \left(\frac{2}{p}\right)=\pm 1$?

since i was able to solve $ \left(\frac{5}{p}\right)=\pm 1$ but not the other. Really im wondering why you chose $ \bmod 8$ and not $ \bmod 4$

So, you are wondering how to calculate $ (2/p)$.  The only proof I know is a corollary of the Gauss Lemma and shows that $ (2/p)=(-1)^{(p^{2}-1)/8}$ which equals what I claimed.  I do not know of an easier proof.  Maybe someone else has a self contained proof.  Also, you might be able to find this if you search online.  :)
P.S.  It is Legendre symbol not Lagrange.

Oh wow how did i manage to forget about that. I just read about the Gauss lemma and recreated the proof for it.  :rotfl:
thanks.


trigonometry
geometry
circumcircle
similar triangles
angle bisector
geometry unsolved

pls help

the convex hexagon $ ABCDEF$ is inscribed in a circle .so that $ AB.CD.EF\equal{}BC.DE.FA$
prove that the diagonals $ AD,BE,CF$ are concurrent
ishfaq

This problem is the inverse (and easy to prove), of the well known theorem which states that in every cyclic hexagon $ ABCDEF,$ whose the main diagonals $ AD,$ $ BE,$ $ CF$ are concurrent, then the result that $ (AB)\cdot (CD)\cdot (EF) \equal{} (BC)\cdot (DE)\cdot (FA),$ it holds.

Easy to prove by the similarity of the triangles with bases the opposite side-segments of the given hexagon and as their common vertex, the concurrency point of its diagonals.

We often use this result as an argument for the solution of other problems and probably it has already been posted in this forum before.

Unfortunately I don’t have in mind any reference about the solution of the problem you presented and if it wouldn’t be answered, I will post here next days, the well known its simple proof.

Kostas Vittas.

or you can use ceva's theorem for $ \triangle ACE$.

can you kindly describe the process of solving it using ceva????i cant understand it how to apply

according to the $ \sin$ form of ceva's theorem in $ \triangle ACE$ the concurrency of $ AD,EB,CF$ is equivalent to:

$ \frac {\sin \angle CAD}{\sin \angle DAE}\cdot\frac {\sin \angle AEB}{\sin \angle BEC}\cdot\frac {\sin \angle ECF}{\sin \angle FCA} \equal{} 1$ (*)

but note that $ A,B,C,D,E,F$ lie on a circle so we have:

$ \frac {\sin \angle CAD}{\sin \angle DAE} \equal{} \frac {CD}{DE}$

and similary:

$ \frac {\sin \angle AEB}{\sin \angle BEC} \equal{} \frac {AB}{BC}$

$ \frac {\sin \angle FCE}{\sin \angle FCA} \equal{} \frac {EF}{FA}$

hence (*) holds iff:

$ \frac {CD}{DE}.\frac {AB}{BC}.\frac {EF}{FA} \equal{} 1\iff CD.AB.EF \equal{} DE.BC.FA$

which is true by the hypothesis,so $ AD,BE,CF$ are concurrent iff $ CD.AB.EF\equal{}DE.BC.FA$...

[quote="vittasko"][b][size=100][color=DarkBlue]In every cyclic hexagon $ ABCDEF,$ its main diagonals $ AD,$ $ BE,$ $ CF$ are concurrent at one point, if and only if the result that $ (AB)\cdot (CD)\cdot (EF) \equal{} (BC)\cdot (DE)\cdot (FA),$ it holds[/color][/size][/b].[/quote]
[b][size=100]PROOF[/size][/b] (well known).

$ \bullet$  From the similar triangles $ \bigtriangleup PAB,$ $ \bigtriangleup PED,$ $ \Longrightarrow$ $ \frac {AB}{DE} \equal{} \frac {PA}{PE}$ $ ,(1)$

From the similar triangles $ \bigtriangleup PCD,$ $ \bigtriangleup PFA,$ $ \Longrightarrow$ $ \frac {CD}{FA} \equal{} \frac {PC}{PA}$ $ ,(2)$

From the similar triangles $ \bigtriangleup PEF,$ $ \bigtriangleup PCB,$ $ \Longrightarrow$ $ \frac {EF}{BC} \equal{} \frac {PE}{PC}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ (AB)\cdot (CD)\cdot (EF) \equal{} (BC)\cdot (DE)\cdot (FA)$ $ ,(4)$

$ \bullet$  When $ (4)$ is true, we denote as $ P$ the intersection point of $ AD,$ $ BE$ and let $ C'$ be, the intersection point of the circumcircle $ (O),$ from the line segment $ FP.$

So, it is enough to prove that $ C'\equiv C.$

Because of $ P\equiv AD\cap BE\cap C'F$ $ \Longrightarrow$ $ (AB)\cdot (C'D)\cdot (EF) \equal{} (BC')\cdot (DE)\cdot (FA)$ $ ,(5)$

From $ (4),$ $ (5)$ $ \Longrightarrow$ $ \frac {CD}{C'D} \equal{} \frac {BC}{BC'}$ $ \Longrightarrow$ $ \frac {BC}{CD} \equal{} \frac {BC'}{C'D}$ $ ,(6)$

Let $ N$ be, the intersection point of the sidesegment $ BD$ of the triangle $ \bigtriangleup CBD,$ from the angle bisector of $ \angle C.$

So, we have that $ \frac {BC}{CD} \equal{} \frac {BN}{ND}$ $ ,(7)$

Similarly, if we denote as $ N'$, the intersection point of the sidesegment $ BD$ of the triangle $ \bigtriangleup C'BD,$ from the angle bisector of $ \angle C',$

we have also $ \frac {BC'}{C'D} \equal{} \frac {BN'}{N'D}$ $ ,(8)$

From $ (6),$ $ (7),$ $ (8)$ $ \Longrightarrow$ $ \frac {BN}{ND} \equal{} \frac {BN'}{N'D}$ $ \Longrightarrow$ $ N'\equiv N$ $ ,(9)$

From $ (9)$ and because of the angle bisectors of the angles $ \angle C$ and $ \angle C',$ intersect the circumcircle $ (O),$ at the same point, so be it $ M,$ we conclude that $ MN\equiv MN'$ $ \Longrightarrow$ $ MC\equiv MC'$ $ \Longrightarrow$ $ C'\equiv C$ and the proof is completed.

Kostas Vittas.

Thank [b]vittasko[/b] . Your post is very nice.


linear algebra
matrix
inequalities
linear algebra solved

If $A,B,C \in M_2(R)$, and $AB=BA,BC=CB,AC=CA$ such that $0 \ge det(AB+BC+CA)$.prove that:
$det[(A^2-BC)^2+(B^2-AC)^2+(C^2-BA)^2] \ge 0$

Silviu, do you have solution for this one?  ;)

Now I don't have a solution for this one,do you have one?

No, but I've seen it in the proposed section area...  ;)

Let's work in $M_2(C)$ the matrix commute,
they are trigonalisable in the same basis with two eigenvalues conjugate

$A$ is similar to a triangular superior matrix with $diag(a,a*)$ where $a*$ is the conjugate of $a$
The same thing for $B,C$

So

$M=(A^2-BC)^2+(B^2-AC)^2+(C^2-BA)^2$ in the common basis M is similar to a

triangular superior matrix with diagonal $diag(w,w*)$ where

$w=(a^2-bc)^2+(b^2-ac)^2+(c^2-ba)^2$

$detM=|w|^2\geq 0$

But I am sure there is a solution  without trigonalisation using some algebraic identity
if you know post it

but i think that you have not use the inequality verified by A,B and C (the det is negatif )!! :|


let k, a, b, c be non-negative real numbers. prove that (a/(b+c))^k+(b/(c+a))^k+(c/(a+b))^k>=min{2,3/2^k}

It's special case of IMO Longlist 1989.

[url]http://www.mathlinks.ro/viewtopic.php?t=180491[/url]

If $ k \ge 1$, it is very easy(b/c Power mean inequality)

but I don't know solution when $ k \le 1$ ;;


[tex]\textrm{Prove that for } 0 \leq a,b,c \leq 1, \frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} +[/tex] [tex] (1-a)(1-b)(1-c) \leq 1$[/tex]

Query: why is [tex]\textrm{Prove that for } 0 \leq a,b,c \leq 1, \frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} + (1-a)(1-b)(1-c) \leq 1[/tex] unparseable, even though the only difference is a split after the + sign?

Not a formal proof, but [hide]The left hand side is a sum of linear and convex functions when taken in each variable, so the maximum should occur at the endpoints of the intervals of the variables. So we just try 0,1 for each of a,b,c. for these 8 options, we find 1 to be the maximum.[/hide]

[quote="ComplexZeta"]You have something funny going on with the $ signs. You shouldn't use them.[/quote]

Removing them doesn't change anything.


Should I bring my own compass/protractor/calculator/etc., or will they be readily available?

You bring your own.  Though I doubt you'll ever need a calculator.

I was wondering if they would confiscate a compass at an airport...who knows? It is possible if you have a metal one.

Well, you pack it.  In your luggage.  Don't be stupid like me and put it in your carry-on and have the flight people detain you and tell you "Now, little girl.  You don't look like a terrorist, but these sharp pointy things are weapons.  You can either go back out, mail it to yourself, or we will take it and destroy it."

Just as a fun fact...I brought an expensive compass to MathPath last year and gave it to Ubemaya to hold after a class because I didn't have a pocket nor a pencil case, then he put it in his backpack and I only realized that he never gave it back after camp ended. Then I didn't remember to tell him until like three months after camp and he said he couldn't find it.

[quote="Piggypi314"]Ubemaya is awesome.  And that automatically allows him to lose any expensive compasses he wants to.

deathbypapercutz (3:02:58 PM):	teapot
Ubemaya (3:02:59 PM):	TEAPOT
Ubemaya (3:03:00 PM):	TEAPOT
Ubemaya (3:03:00 PM):	TEAPOT
Ubemaya (3:03:01 PM):	TEAPOT
Ubemaya (3:03:06 PM):	I'M A LITTLE TEAPOT SHORT AND STOUT
Ubemaya (3:03:09 PM):	HERE IS MY HANDLE HERE IS MY SPOUT
Ubemaya (3:03:14 PM):	WHEN I GET ALL STEAMED UP HEAR MY SHOUT
Ubemaya (3:03:17 PM):	TIP ME OVER AND POUR ME OUT
deathbypapercutz (3:03:26 PM):	wow
Ubemaya (3:03:22 PM):	I'M A CLEVER TEAPOT YES IT'S TRUE
deathbypapercutz (3:03:28 PM):	do you do this every time
Ubemaya (3:03:25 PM):	HERE'S AN EXAMPLE OF WHAT I CAN DO
deathbypapercutz (3:03:30 PM):	someone says teapot
Ubemaya (3:03:32 PM):	I CAN CHANGE MY HANDLE TO MY SPOUT
Ubemaya (3:03:37 PM):	TIP ME OVER AND POUR ME OUT YAY!

And.  Just to keep this about compasses.  I think mine ran away from home.  From disuse and/or misuse.[/quote]

um abuse


[quote="chess64"]Using the area of a regular pentagon, prove that $4\sin \frac{2\pi}{5} + \tan \frac{2\pi}{5} = 5\cot \frac{\pi}{5}$.[/quote][hide]Let the regular pentagon $ABCDE$ have center $O$ and sides length $2$. Drop a perpendicular from $O$ to $AB$, with intersection $H$. It's easy to calculate $m\angle HOB$; it is $\frac{\pi}{5}$. Since $HB$ is half of $AB=2$, then the area of $\bigtriangleup HOB$ is half the area of $\bigtriangleup HOA$. Since $OH=\cot\left(\frac{\pi}{5}\right)$, then the area of $\bigtriangleup BOA$ is simply $OH=\cot\left(\frac{\pi}{5}\right)$ via one half times the product of base and height. Finally, the area of the pentagon is $5\cot \frac{\pi}{5}$.

Now, triangulate the pentagon by drawing segments $DA$ and $DB$ to form $\bigtriangleup DEA$, $\bigtriangleup DAB$, and $\bigtriangleup DBC$. Again, it is easy to calculate $m\angle DAB=\frac{2\pi}{5}$, so then $DH$ has length $\tan \frac{2\pi}{5}$. Thus, the area of $\bigtriangleup DAB$ is $\tan \frac{2\pi}{5}$ through \[ A=\frac{1}{2}(bh). \] Drop a perpendicular from $E$ to $DA$ to intersect at $P$. Then $m\angle EAP=\frac{\pi}{5}$. Hence, \[ AP=2\cos\frac{\pi}{5} \] \[ EP=2\sin\frac{\pi}{5}. \] The area of $\bigtriangleup PEA$ is then \[ \frac{1}{2}\left(2\cos\frac{\pi}{5}\right)\left(2\sin\frac{\pi}{5}\right)=\sin \frac{2\pi}{5}. \] Thus, the area of $\bigtriangleup DEA$ is $2\sin \frac{2\pi}{5}$, as is the area of $\bigtriangleup DBC$. Then the area of the pentagon is $4\sin \frac{2\pi}{5} + \tan \frac{2\pi}{5}$.

The pentagon never changed, so its area surely could not have. Thus, \[ 4\sin \frac{2\pi}{5} + \tan \frac{2\pi}{5} = 5\cot \frac{\pi}{5}. \][/hide]


geometry
circumcircle
trigonometry
angle bisector
geometry solved

[quote="April"]Let be given an acute triangle $ABC$. Let $X$ be the point of intersection of tangents to circumcircle of triangle $ABC$ at $B$ and $C;\,M$ be the midpoint of $BC$.
Prove that: $\frac{AM}{AX}=\cos\widehat{BAC}$[/quote]

See e. g. my proof of http://www.mathlinks.ro/Forum/viewtopic.php?t=26242 .

  Darij

Let D be the midpoint of the arc $ BC$. It is well known that $ AX$ is the $ A$ simmedian, hence $ AM$ and $ AX$ are symmetrical about $ AD$, so $ \frac {AM}{AX} \equal{} \frac {MD}{DX}$, but $ CD$ is the angle bisector of $ \angle BCX$, with $ \frac {CM}{CX} \equal{} \frac {MD}{DX}$, but $ \cos \widehat {MCX} \equal{} \cos \widehat {BAC} \equal{} \frac {CM}{CX}$.
Further, $ \frac {CM}{CX} \equal{} \frac {OM}{OC}$.

Best regards,
sunken rock


vector
trigonometry
geometry
parallelogram
parameterization
algebra
system of equations

The cross product of two 3D vectors is the vector orthogonal to both of the operands. So if $a \times b = c$, then $c$ is orthogonal both to $a$ and $b$.

The direction of the cross product can either be determined mathematically or with the "right hand rule" which I will not explain here.

The magnitude of the cross product is equal to the product of the magnitudes of the two vectors times the sine of the angle between them. So if two vectors have no angle between them (they are parallel), the cross product is 0, which makes sense as the vector orthogonal to both is not unique. If the two vectors are orthogonal, the magnitude of the cross product is maximized.

Hope I addressed the parts you were confused about.

If you're looking for some kind of a physical interpretation, the magnitude of the cross product of two vectors $\vec{v}, \vec{u}$ is the area of the parallelogram given by $0, \vec{v}, \vec{u}, \vec{v}+\vec{u}$.  The determinant formula follows from this interpretation.

[quote="mathnerd314"][quote="bthings_2000"]I'm trying to find a really good explanation of what a cross product of 3D vectors really is and, especially, how the formula for it (using determinants) is derived. Mathworld and such have failed to provide really good explanations. Can anyone help me out?[/quote]

What do you get when you cross an elephant with a banana?

$\parallel Elephant\parallel \cdot \parallel Banana\parallel \cdot \sin \theta$ in a direction mutually perpendicular to the two, as determined by the right-hand rule.

I'm dead serious, that's how I remember it.[/quote]

I don't know why I found that so funny, but I just laughed for like a minute

[quote]I don't know why I found that so funny, but I just laughed for like a minute[/quote]

Very old classic joke also goes as
"What do you get when you cross a mosquito with a mountain climber?

[hide]That's not valid,  you  don't have cross product of a vector and a scaler  [/hide]
Another  common physical application, is force on a moving charge which is  =  charge (Velocity X magnetic field)

Thanks for the help so far! More questions:

1) Is the vector mutually perpendicular to the two vectors used to generate the cross product unique?

2) Two planes (usually) meet in a line. But when you set the equation of two planes equal, you get the equation of another plane. How does the new plane relate to the two used to generate it?

Example:
$2x+y-z=5$
$3x-7y-2z=3$

Therefore,
$2x+y-z-2=3x-7y-2z\Rightarrow-x+8y+z=2$

What does the plane $-x+8y+z=2$ have to do with the two used to generate it?

Well... given two lines, there's only one line in 3d space that can be perpendicular to both of them. The direction of this line depends on the direction of the vectors.

Oh, and don't take my (bad attempt at) a joke too seriously. :D

hehe, don't worry about it; I have a shirt that's basically the same thing: It's "What do you get when you cross a *picture of cat* with a *picture of lightbulb*?

|*picture of cat*||*picture of lightbulb*|$\sin{\theta}$

I love the shirt and I've known what cross product is and how to use it for a while, I'm just trying to understand it better.

[quote="bthings_2000"]Thanks for the help so far! More questions:

1) Is the vector mutually perpendicular to the two vectors used to generate the cross product unique?[/quote]
If the vectors aren't parallel, then there are two directions that the result can go in (one negative the other). The one that the cross product uses is determined by the right-hand rule or by mathematically computing it.

[quote="bthings_2000"]

2) Two planes (usually) meet in a line. But when you set the equation of two planes equal, you get the equation of another plane. How does the new plane relate to the two used to generate it?

Example:
$2x+y-z=5$
$3x-7y-2z=3$

Therefore,
$2x+y-z-2=3x-7y-2z\Rightarrow-x+8y+z=2$

What does the plane $-x+8y+z=2$ have to do with the two used to generate it?[/quote]
Not much. You completely left out the fact you were setting both of those planes equal to 0, so that plane is actually equal to something else also. That method done correctly would get as your first step $2x+y-z-5 = 3x-7y-2z-3 = 0$. As you can see, that's not too useful. The correct way to solve it is to make one of the three variables a parameter and then find the other 2 variables in terms of it. Here's a way to solve it with a bunch of basic techniques for system of equations problems.

Let $z = t$
$2x+y-t=5$
$y = 5+t-2x$
$3x-7(5+t-2x)-2t=3$
$3x+14x=38+7t$
$x = \frac{38+7t}{17}$
$y = \frac{9+3t}{17}$
$z = t$
Those last three lines parametrically define the line of intersection if you allow $t$ to run across the real line. If I made a math error in that, forgive me. Just trying to show the general technique.


If $a^2 + b^2 + c^2 = 1$, what is the maximum value of $(a - b)^2 + (b - c)^2 + (c - a)^2$?

Can anyone do this?  Somehow this is supposed to be nice and easy, but save Lagrange multipliers, I can't do this.  I can't get very far with it either.  I can find the minimum....  I need work on inequalities solving.

since nothing is said about the sign of $a,b,c$ ...

It is $3 - (a+b+c)^2$ so max is $3$ reached when $a+b+c=0$ (for instance when $a=-b=1/\sqrt{2},c=0$)

[quote="chess64"]Well if you expand it, you get $2-2(ab+ac+bc)$.... :D

AM-GM yields $\frac{a^2+b^2+c^2}{3} \ge \sqrt[3]{a^2b^2c^2}$, or $abc \le \frac{1}{\sqrt{27}}$...

I have $\approx 0$ experience with inequalities. :( The only inequalities I know are trivial and AM-GM :D[/quote]

Like tµtµ said, you can't simply use AM-GM when there is nothing restricted about $a,b,c$. Remember, you can only use AM-GM iff the requirements are given S.T. $a,b,c$ are positive reals.

Little bit more explanation on how tµtµ got $3-(a+b+c)^2$.. Recall that $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac)=1+2(ab+bc+ac)$ which implies $(a+b+c)^2-1 = 2(ab+bc+ac)$. So, expanding the original equation gives $2-2(ab+bc+ac) = 2-(ab+bc+ac)^2+1 = 3-(ab+bc+ac)^2$.

Hope this helps..  :)