conics
parabola
algebra
polynomial
ellipse
analytic geometry

i was very struggle when i am doing this question, don't know what to do with this:

Given any three points, (a,b), (c,d), and (e,f), find the equation of a parabola passing through all three.

help please!!!!!!!!!!!!1

I think basically it's going to be a system of equations.

The general form of a parabola is y=ax^2+bx+c so if they gave you the three points (-1,0),(1,0), and (0,1) you could plug them into the general formula and solve for a,b, and c.

Ex:
a+b+c=0
a-b+c=0
c=1

I know it's a simple case, but this is how I do it.

[quote="Kent Merryfield"]It's not calculus. This is an algebra question, and it should be moved somewhere else. I'll try this form, although the moderators there might have different ideas.[/quote]

I think you can use Lagrange's interpolation formula.

Yup, which is a formula for an (n-1)-degree polynomial through n points.

$P(x) = \frac{(x-x_2)(x-x_3)\cdots (x-x_n)}{(x_1-x_2)(x_1-x_3)\cdots (x_1-x_n)}y_1
\\+
\frac{(x-x_1)(x-x_3)\cdots (x-x_n)}{(x_2-x_1)(x_2-x_3)\cdots (x_2-x_n)}y_2
\\+
\cdots
\\+
\frac{(x-x_1)(x-x_2)\cdots (x-x_{n-1})}{(x_n-x_1)(x_n-x_2)\cdots (x_n-x_{n-1})}y_n$

Written in summation form:
\[P(x) = \sum_{j=1}^n\left(y_j\prod_{k=1\\k\ne j}^n \frac{x-x_k}{x_j-x_k}\right)\]

Where, of course, the n points are \[(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)\].

Holy cow.... Can you use it in an example?

Hehe, yeah, it looks horrible, but once you realise whats happening its actually quite easy to remember and use.

Lets see.. lets find the equation of a parabola that goes through (1,2), (2,6) and (3,4).
We take the first x value, 1.
On the top of the fraction we have $(x-2)(x-3)$, ie (x-xvalue) for each of the other points. On the bottom we have the same thing but 1 in place of x, ie $(1-2)(1-3) = 2$. Then we multiply by the y value to get $(x-2)(x-3)$ as our first term.

Repeating that for the second part: $\frac{(x-1)(x-3)}{(2-1)(2-3)}\times 6 = -6(x-1)(x-3)$.
Repeating that for the third part: $\frac{(x-1)(x-2)}{(3-1)(3-2)}\times 4 = 2(x-1)(x-2)$.

Adding those all up, our parabola is $(x-2)(x-3)-6(x-1)(x-3)+2(x-1)(x-2) = -3x^2+13x-8$.

Sorry to point out the obvious, but did any one realize that:
 [quote][size=150]There are  infinite  number of parabolas which can pass through a given set of three points?[/size][/quote]

Don't we all remember  that old nursery rhyme:  To uniquely define .. a curve ...
For a line you need two points, for a circle three points, for a parabola  four, for an ellipse five ... and so on.:)

[quote="TripleM"]Yeah, well most people don't care about non-rectangular (is that the word?) parabolas :)[/quote]

Huh?? :?  No, for most people  $y = x^2 $ is as much a parabola as $x=y^2$ And the  truth is parabola's don't care in  which direction you choose your x or y  axis. :)

The polynomial  fit is nice,  but it is  a different problem.  To find  a parabola,    given a set of points,  you need 4 points (and not 3) to uniquely determine it.

No, parabolas don't have to have their directrix parallel to one of the x or y axes (e.g, parabolas with an xy term or in polar coordinates)

And for Gyan, there is only an infinite number of parabolas if the three points are not collinear :D

[quote="paladin8"][quote="Gyan"]To find  a parabola,    given a set of points,  you need 4 points (and not 3) to uniquely determine it.[/quote]

No... you can determine a parabola with 3 points. Just plug them in. (x1, y1) (x2, y2) (x3, y3) for the parabola [tex] ax^2+bx+c [/tex]

[tex] ax_1^2 +bx_1+c = y_1 \\
ax_2^2 +bx_2+c = y_2 \\
ax_3^2 +bx_3+c = y_3 [/tex]

you have 3 equations with 3 variables, thus one solution.[/quote]

As  $blah^3$ pointed out ,  let us see you try these three points:  (0,0), (1,1),(1,-1) ... by the way all lie on a parabola  $y2=x$  :D

[quote]And for Gyan, there is only an infinite number of parabolas if the three points are not collinear Mr. Green[/quote]

You mean three [b]distinct [/b] points, right ? You are, as always, correct!.


calculus
geometry
3D geometry
sphere
function
trigonometry
calculus computations

Ok, this problem was givin to us by our Calculus prof. and I was curious if anyone here knew how to solve it, If I can solve the problem using cal. I will be given 10 bonus points on the next test. Here is the problem.

Suppose you were a genie and you made the earth a perfect sphere with a circumference of exactly 25,000 miles. In this perfect scenario there are no mountains or vallys or even waves, the earth is a perfect sphere. So, you, the genie go down to the store and purchase some sting that is long enough to go around the earth at the equator , but when you begiin to lay the string around the earth you notice that you have exactly 3 extra feet of string, so the problem is this: what is the distance in inches that the string hovers above the earths surface?

I know this problem can be solved using geometry but that isn't worth any points on the test, we just covered differentals so I am assuming that we need to use to solve this problem, if anyone is willing to help it will be greatly appreciated! Thanks in advance!

It's a famous old puzzle. It's also not really calculus, but I'll let the post stay here in the "computations and tutorials" section.

Two hints:

1. There's only one formula that matters, and you learned it several years ago in geometry.

2. The fact that the circumference of the earth is 25,000 miles and not 10 times that or a tenth of that is entirely irrelevant.

Circumference $C$ is a function of radius $R$: $C(R) = 2 \pi R$. Differentiating: $\text dC = 2\pi\ \text dR$. Then $\text dR = \frac{\text dC}{2 \pi} = \frac{3\ \text{ft}}{2 \pi} = \frac{36\ \text{in}}{2 \pi} \doteq 5.73\ \text{in}$.

I just wanted to thank yall for helping me through this problem, Like I said I was able to solve this problem with basic geometry but not with calcululs. Thanks again to all who conributed!

[quote="fedja"]There is a much more interesting version of this problem in SPRA: under the same assumptions, after you put the rope around the Earth, you pinch it at one point and pull this point as far away from the surface as possible. How far will that be? (this one really requires calculus; the answer should be given with precision 1% or less)[/quote]
Assume that the Earth circumference is $C = 25,000\ \text{mi} =  25,000\ \text{mi} \cdot 1760\ \text{yd/mi} \cdot 3\ \text{ft/yd} = 132 \times 10^6\ \text{ft}$ and that the rope length is $L = C + \triangle L = C + 3\ \text{ft}$. Let $O$ be the Earth center, $P$ the highest point of the pulled up rope, $Q$ the intersection of the line $OP$ with the Earth surface, $S, T$ the tangency points of straight segments of the rope with the Earth surface, $R = OS = OT$ the Earth radius and $\phi = \measuredangle SOP = \measuredangle TOP$.

$L = 2(\pi - \phi)R + PS + PT =2(\pi - \phi)R + 2 \cdot PT = 2R(\pi - \phi + \tan \phi) =$

$= \frac{C}{\pi} (\pi - \phi + \tan \phi) = C + \frac{C}{\pi} (\tan \phi - \phi) = C + \triangle L$

$\tan \phi - \phi - \frac{\pi \cdot \triangle L}{C} = 0$

This equation for the angle $\phi$ has to be solved numerically, for example, by Newton's method:

$\phi \doteq 0.34281518^o$

The height $\triangle h = PQ$ of the point $P$ above the Earth surface is then

$\triangle h = \sqrt{R^2 + R^2 \tan^2{\phi}} - R = R \left(\sqrt{1 + \tan^2{\phi}} - 1 \right) = \frac{C}{2 \pi} \left( \frac{1}{\cos \phi} - 1 \right)$

$\triangle h \doteq \frac{132 \times 10^6}{2 \pi} \left( \frac{1}{0.9999821004} - 1 \right) \doteq 376.050\ \text{ft}$

Since $\triangle h \ll R, \triangle L \ll C = 2 \pi R$, it is possible to approximate $\triangle h, \triangle L$ by Taylor's expansion with high accuracy. Let $h = OP$ be the distance of the point $P$ from the Earth center.

$L(\phi) = C + \frac{C}{\pi} (\tan \phi - \phi)$

$h(\phi) = \frac{R}{\cos \phi} = \frac{C}{2 \pi \cos \phi}$

$\triangle L = L(\phi) - L(0) = L'(0) \cdot \phi + \frac 1 2 \cdot L''(0) \cdot \phi^2 + \frac 1 6 \cdot L'''(0) \cdot \phi^3 + ... =$

$= \frac{C}{\pi} \left( \frac{1}{\cos^2{0}} - 1 \right) \cdot \phi + \frac{C}{2 \pi} \cdot \frac{2 \sin 0}{\cos^3{0}} \cdot \phi^2 + \frac{C}{6 \pi} \cdot \frac{2(\cos^2{0} + 3 \sin^2{0})}{\cos^4{0}} \cdot \phi^3 + ... \doteq \frac{C}{3 \pi} \cdot \phi^3$

$\triangle h = h(\phi) - h(0) = h'(0) \cdot \phi + \frac 1 2 \cdot h''(0) \cdot \phi^2 + \frac 1 6 \cdot h'''(0) \cdot \phi^3 + ... =$

$= \frac{C}{2 \pi} \cdot \frac{\sin 0}{\cos^2{0}} + \frac{C}{4 \pi} \cdot \frac{\cos^2{0} - 2 \sin^2{0}}{\cos^3{0}} \cdot \phi^2 - \frac{C}{12 \pi} \cdot \frac{6 \sin^3{0}}{\cos^4{0}} \cdot \phi^3 + ... \doteq \frac{C}{4\pi} \cdot \phi^2$

Eliminating the angle $\phi$, we get

$\left( \frac{3 \pi\ \triangle L}{C} \right)^2 = \left( \frac{4 \pi\ \triangle h}{C} \right)^3$

$\triangle h = \left(\frac{9C}{64 \pi} \triangle L^2 \right)^{\frac 1 3} = \left(\frac{9 \cdot 132 \times 10^6}{64 \pi} \cdot 3^2 \right)^{\frac 1 3} \doteq 376.048\ \text{ft}$

The difference from the result obtained previously is only about $-0.0005\ \%$.


function
topology
real analysis
real analysis solved

Let A be the ring (C([0,1]),+,*), where C([0,1]) is the set of all continuous functions f:[0,1]->R. Let I be an ideal of A. Show that there is an x_0 s.t. f(x_0)=0 for all f in I.

What's the matter, if I = A !!! Have you some details about I ?

Terribly sorry about that! I =/= A (forgot to mention this).

Hi Grobber,

Let Z(f)={z / 0 =< z =< 1 , f(z)=0} , S(f) = [0,1] - Z(f)

Suppose f in I , such that Z(f) = {empty}  ===> g = 1/f is a well defined function in A  .
Then : g*f=1 is in I ===> h*1 = h is in I, for all h in A ===> I=A (contradiction) .
So for any f in I , Z(f) is not empty.

With the usual topology restricted to [0,1], we have
f continuous ===> Z(f) is a closed set ===> S(f) is an open set

Suppose now, for any u in [0,1], there is Fu in I such that F(u)<>0 .
Manifestly u is in S(Fu) so Union(u in [0,1]){S(Fu)} = [0,1] .
But [0,1] is compact so it is covered by a finite family : [0,1] = S(F1) U...U S(Fn).
Because of ideal I we have : G = (F1)^2 + ... + (Fn)^2 is in I  ===> Z(G) is not empty
So there is v such that G(v)=0 ===> F1(v)=..=Fn(v)=0
v is in [0,1] ===> there is a Fk s.t. v is in S(Fk)  ===> Fk(v) <>0 (contradiction)

Finally there is a x0 in [0,1] such that f(x0)=0 for all f in I

 Hey Grobber, have a proof without the choice axiom ? ..

I don't think so... I did pretty much the same thing, only I didn't pay attention to using the choice axiom (I think I used it without realizing it). :D


For every  $ i \in N$, $ a_{i}(n),$ is a positive series, bounded by 1 from above, who converges to 0.
Does the series $ b_{m}\equal{}\sum_{i\equal{}1}^{\infty} \frac{a_{i}(m)}{2^i}$ converge to 0?

Yes, it does.  Given an $ \epsilon>0$ choose an integer $ N$ such that $ 2^{-N}<\frac{\epsilon}{2}$ and an $ M$ such that $ \sum_{i=1}^{N}{\frac{a_i(m)}{2^i}}<\frac{\epsilon}{2}$ for each $ m>M$ (guaranteed to exist since $ a_i(M) \to 0$ as $ M\to \infty$ and $ N$ is finite).  We then have
\[ b_m < \frac{\epsilon}{2}+\sum_{i=N+1}^\infty{\frac{a_i(m)}{2^i}<\epsilon \ \ \ \forall m>M}\]

[quote="JoeBlow"]Yes, it does.  Given an $ \epsilon > 0$ choose an integer $ N$ such that $ 2^{ - N} < \frac {\epsilon}{2}$ and an $ M$ such that $ \sum_{i = 1}^{N}{\frac {a_i(m)}{2^i}} < \frac {\epsilon}{2}$ for each $ m > M$ (guaranteed to exist since $ a_i(M) \to 0$ as $ M\to \infty$ and $ N$ is finite).  We then have
\[ b_m < \frac {\epsilon}{2} + \sum_{i = N + 1}^\infty{\frac {a_i(m)}{2^i} < \epsilon \ \ \ \forall m > M
}\]
[/quote]

Cool... I was hoping for a proof but waiting for a counterexample.  :blush:


[quote="pda"]Let f(x) = [x] (the greatest integer less than or equal to x) and g(x) be the cube root of x. If f(g(1)) + f(g(2)) + f(g(3)) + ... + f(g(n)) = 2n, what is n?[/quote]

EDIT: Darn, Altheman beat me to it. But here's a more elementary solution.
[hide="Click for solution"]Yay, let's look for a pattern.

Basically, you want:
$ \lfloor \sqrt[3]{1}\rfloor+\lfloor \sqrt[3]{2}\rfloor+\lfloor \sqrt[3]{3}\rfloor+\cdots+\lfloor \sqrt[3]{n}\rfloor = 2n$
[hide="Click for messy work"]$ n = 1 \quad 1 =^{\textrm{?}}2$
$ n = 2 \quad 2 =^{\textrm{?}}4$
$ n = 3 \quad 3 =^{\textrm{?}}6$
$ \vdots \textrm{(Note that the differences are getting larger, so we move on)}$
$ n = 7 \quad 7 =^{\textrm{?}}14$
$ n = 8 \quad 9 =^{\textrm{?}}16$
$ n = 9 \quad 11 =^{\textrm{?}}18$
$ n = 10 \quad 13 =^{\textrm{?}}20$
$ n = 11 \quad 15 =^{\textrm{?}}22$
$ \vdots \textrm{(Note that the differences are staying constant, so we should move on)}$
$ n = 26 \quad 45 =^{\textrm{?}}52$
$ n = 27 \quad 48 =^{\textrm{?}}54$
$ n = 28 \quad 51 =^{\textrm{?}}56$
$ n = 29 \quad 54 =^{\textrm{?}}58$
$ n = 30 \quad 57 =^{\textrm{?}}60$
$ n = 31 \quad 60 =^{\textrm{?}}62$
$ n = 32 \quad 63 =^{\textrm{?}}64$
$ n = 33 \quad 66 =^{\textrm{?}}66$[/hide]
Aha!, we get a solution at $ \boxed{n=33}$[/hide]


function
integration
calculus
real analysis
real analysis unsolved

I'm jumping from a plane. The air resistance and my weight are simplified to a constant k=0,15, gravity is g=9,81 and I have the following
$ \frac{dv_1}{dt}\plus{}0,15v_1(t)\equal{}9,81 ,v_1(0)\equal{}0$
$ v_1(t)\equal{}\frac{327}{5}\minus{}\frac{327}{5}e^{\frac{\minus{}3t}{20}}$

Now I need to find a function for the height above ground level so I do this:
$ s_1(t)\equal{}\int v_1(t) dt$
$ s_1(t)\equal{}\frac{327t}{5}\plus{}436e^{\frac{\minus{}3t}{20}}\plus{}C$

I have that $ s_1(0)\equal{}3500$ and $ \frac{ds}{dt}\equal{}\minus{}v_1$. Because the direction downwards is positive I get the proper function for the height:
$ s_1(t)\equal{}\minus{}\frac{327t}{5}\minus{}436e^{\frac{\minus{}3t}{20}}\plus{}3936$

Ok, I hope everything so far is correct.
Next problem is to calculate the speed and time at the moment I release my parachute, I do that after 1000m of falling, that is at 2500m above ground.
$ s_1(t)\equal{}2500\minus{}>t\equal{}21.7$  the speed is $ v_1(21.7)\equal{}62.8768$

So now my problem really gets interesting; I have opened my chute and I got a new resistance constant k= 1.5. I now get new functions for my velocity and speed.

$ \frac{dv_2}{dt}\plus{}1,5v_2(t)\equal{}9,81 ,v_1(?)\equal{}?$ What will be the correct differential equation here? What are the condition for this one? I tried to have that $ v_2(21.7)\equal{}62.8768$, but it got ugly. Any ideas?
(I need to calculate the speed of landing, the momemt I hit the ground, I know how to do this, but it all comes down to get that second eq. right.thanks)

[quote]I tried to have that $ v_2(21.7) \equal{} 62.8768$, but it got ugly.[/quote]
Ugly or not, this is the right thing to do. You'll want to use a computer to finish the problem.

Some of my calculus students solved a similar problem for their term project, except that the time of parachute release was variable (and it did not open instantly), and the objective was to find the maximal time of free fall that allows for safe landing. And then to find what happens if the maximal safe time is exceeded just a little bit. :(


Twelve people are seated around a circular table. In how many ways can six pairs of people engage in handshakes so that no arms cross?

(Nobody is allowed to shake hands with more than one person at once.)

[hide]
It's just the $6th$ Catalan number: $132$. There seem to be plenty of descriptions of these numbers which do a better job than I can, so I won't type out an explanation.
[/hide]

:huh: How is that the answer? Can you show how you got it?

[hide]
Clearly, there must be an even number of people around a table in order for each person to shake hands with someone else. So let $t(n)$ be a table with $n$ pairs of people.

We can quickly write out the cases for smaller tables, and see that $t(1)=1$, $t(2)=2$.

To find the number of cases for larger tables, we can use recursion.

For a table with $6$ people, label them in order $A, B, C, D, E, F$. We can't have $A$ shake hands with a person an even number of seats away, or else the table would be divided into two sections with an odd number of people in them. So $A$ can only shake hands with $B, D, F$.

If $A$ shakes hands with $B$ or $F$, the case breaks down into the $4$ person case; if $A$ shakes hands with $D$, it breaks down into two $2$ two person cases.

The total number of possible arrangements is equal to \[2t(2)+t(1)t(1)\] So $t(3)= 2\times 2+1 = 5$.

We can solve for $t(4)$ and $t(5)$ in the same way; $t(4)=14$, $t(5) = 42$.

For $12$ people \[t(6) = t(5)+t(1)t(4)+t(2)t(3)+t(3)t(2)+t(4)t(1)+t(5)\] Plugging in the corresponding values, we get $t(6)=132$.
[/hide]


um....heres something that MIGHT help. $ n^{2}\equal{}(n\minus{}1)\plus{}2n\minus{}1$. or, in english, the nth square is equal to the (n-1) square plus the nth odd number. $ 1: 1 2: 1\plus{}3\equal{}4 3: 4\plus{}5\equal{}9 4: 9\plus{}7\equal{}16 5: 16\plus{}9\equal{}25...$ and so on.

$ 5^{2}\neq (5\minus{}1)\plus{}2(5)\minus{}1$
$ 25\neq 4\plus{}10\minus{}1$
$ 25\neq 13$
$ \mathbb{Q.E.D.}$

Although reading your example makes it a little clearer. But then again, 43 and 165 aren't perfect squares. Your post confuses me, Abacadea.

I think he meant $ (x\minus{}1)^{2}\plus{}2n\minus{}1$, but I don't get the last line.

I think abacadaea meant
1:  1
2:  1+3=4
3: 4+5=9
4: 9+7=16
5:  16+9=25

That decoding was like a Mathcounts problem in itself :)

I would suggest just doing it.  Really.  The numbers will become your familiar friends in time :)

i agree with torajirou. just memorize them, this is one of the few areas in competition math where the easiest way to do something is through brute-memorization.
for example, although abacadae did make a valid expression that relates the squares (after erdog's correction) it still requires you to know what the previous square before that is before that one. that doesn't really help when you have a REALLY BIG SQUARE.

well, my point was, the sum of the first n odd numbers is $ n^{2}$. i dont think that got to you guys. so just add the nth odd the the n-1 square but doing it t0rajir0u's way works too.

Well, for the squares what I think is just get a calculator, punch in numbers and familiarize yourself with the numbers.

 For example, we all know that 289 is a square of 17 and 361 is a square of 19.

I am not saying and discouraging new to not use the formula.

Try out the way that you can get adequainted fast and makes it easy for you.

I don't think you have to memorize cubes up to 30, but if it's required I would find kind of similarities between them to memorize better.

If you want n^2 and n ends in the number 5, then mutliply n(n+1) and add 25 at the end.

15^2=(1*2)*100+25

25^2=(2*3)*100+25

this is probably familiar anyway...

[quote="moogra"][quote="13375P34K43V312"][quote="mathirons"]Does anyone know how to memorize squares up to thirty-five and cubes up to thirty? I need to do it quickly for a mathcount worksheet![/quote]

What do you mean how? You just kind of...do it, I guess. Don't just sit there look at the list, and try to recite it, if you use the squares and cubes enough, they will eventually become part of you. I never spent time trying to just cram everything in all at once, I just did Mathcounts for 3 years and I know my squares and cubes decently well.[/quote]

that's how I used to do it. I would just keep saying it to myself and eventually I remember it. It also works for pi.[/quote]
Or if you're feeling especially nerdy you could set it to music and sing along  :)

wow, i never thought of that archimedes :D

[quote="archimedes1"][/quote][quote="moogra"][quote="13375P34K43V312"][quote="mathirons"]Does anyone know how to memorize squares up to thirty-five and cubes up to thirty? I need to do it quickly for a mathcount worksheet![/quote]

What do you mean how? You just kind of...do it, I guess. Don't just sit there look at the list, and try to recite it, if you use the squares and cubes enough, they will eventually become part of you. I never spent time trying to just cram everything in all at once, I just did Mathcounts for 3 years and I know my squares and cubes decently well.[/quote]

that's how I used to do it. I would just keep saying it to myself and eventually I remember it. It also works for pi.[/quote][quote="archimedes1"]
Or if you're feeling especially nerdy you could set it to music and sing along  :)[/quote]

actually there is a pi song  :rotfl:
I saw it on google once

Strangly, songs don't help me remember much.

[quote="abacadaea"]theres like 5. heres my favorite:

http://www.vvc.edu/ph/TonerS/mathpi.html[/quote]

i luved that !!!!
it's a spoof of ms. american pie, right? for a second, i thought it was a weird al yankovic song. :D


calculus
inequalities solved
inequalities

Let a,b,c > 0 and 21ab+2bc+8ca<=12
Find the min value of  P(a,b,c)=1/a +2/b+3/c

Namdung posted a variation of this problem, I solved that and Namdung gave also a solution, then someone else (I'm sorry, I don't remember who) posted exactly this problem. So, the problem is already in the solved problems section.

Well, technically there was no solution to this problem itself, although I suppose harazi's and Namdung's solutions can be adapted to solve this problem: [url=http://www.mathlinks.ro/viewtopic.php?t=1535]minima without calculus[/url]


That's uncyclopedia for you.  Funny as always.

[url=http://uncyclopedia.org/wiki/IQ:Sample_test]IQ Test[/url]

Fixed your code.

Your syntax for the URL was a bit messed up before but now it's fixed.

What's syntax? Is that sin + tax? So governemnt make you pay when you are bad? Like a fine! But fine's aren't fine...

No, that's syntax, exactly as I wrote it.  It refers to, in this case, $ \text{\LaTeX}$ code or URL tags.

Where was the word syntax derived from? :huh:

Isn't "syn" the Greek root for "math" or something?

And tax means language! So we have to pay taxes because we don't know English? :maybe:

I think it's exclamation, like the Proclamation of 1980.

Oops. I must have mixed it up with the Vietnam War. Both involved people, right? :maybe:

I believe the Proclamation of 1763 involved colonists. Do colonists count as people?

What are you people talking about?

http://en.wikipedia.org/wiki/Syntax

Well, colonists are from England, right? England stole Hong Kong, so they don't count as people. :mad:

I believe colonists come from France...

England stole Hong Kong?

Yeah. I believe they stole it in 1849 and returned it in 1999.

Wow, thats a long time. The Japanese must have been really mad if England kept Hong Kong for that long....

No, Oceania is a region, Australia being in Oceania.  That is,
$ \mbox{Australia}\subseteq\mbox{Oceania}$ but $ \mbox{Oceania}\nsubseteq\mbox{Australia}$


analytic geometry
geometry
similar triangles

[quote="AndrewTom"]Two poles XY and ZW of heights a and b respectively above the ground are vertical. A wire joins X, the base of XY, to W, the top of ZW, and another wire joins Z, the base of ZW to Y, the top of XY. The wires cross at P. How high is P above the ground?[/quote]

[hide] $ \frac {ab}{a\plus{}b}$ [/hide]

Use similar triangles to try and derive this.

Answer:

ab/(a + b)

Let the coordinates of the following be:
 X = (0,0), Y = (0,a), Z= (c,0), and W = (0,b)

y = (b/c)x for line XW

y = (-a/c)x + a for line YZ

Equating them and solving for x produces x = ac/(a + b).

Substituting this back into y = (b/c)x gives
y = ab/(a + b).  Or, the height of P above the ground
is ab/(a + b).

*** Note:  Because of the nature of this problem, it was
expected that the coefficients on a and b would be the
same in the answer as they were represented in the problem
in an even-handed type way with regard to each other.

You can avoid coordinates by trying to show that $ 1/a\plus{}1/b\equal{}1/x$, where $ x$ is our desired height.
Hint: Use similar triangles/homothety by abusing the fact that the poles are parallel.


pigeonhole principle
geometry
parallelogram
algorithm
combinatorics solved
combinatorics

1)
Does the set {1;2;...;3000} contain a subset A consisting of 2000 numbers
such that  x belongs to A implies 2x does not belong to A

Austrian-Polish mah competition 1987
---------------------------------------------------------------------------
2)
For n>1, let 2n chess pieces be placed at the centers of 2n squares
of an nxn chessboard.Show that there are four pieces among them
that formed the vertices of a parallelogram.
If 2n is replaced by 2n-1, is the statement still true in general ?
---------------------------------------------------------------------------
3)
The set {1;2;....;49} is partitioned into three subsets.
Show that at least one of the subsets contains three different numbers
a,b,c such that a+b=c

I think I have a proof for the second one.

Let's assume that the pieces are placed on k columns, c1, c2, .., ck. Let ai be the number of pieces on column ci. The distance between 2 pieces on the same column is the difference of position. For column ci we consider the distances between the lowest piece and all the others. All these distances are (for column ci) different since they are taken from the same piece. We have ai-1 distances on column ci. In total we have   \sum from i=1 to k of (ai-1) distances = 2n-k >= 2n-n=n. But on a column there are only n-1 distinct distances (1 through n-1) so by applying pigeonhole we get at least 2 equal distances (2 pairs of pieces the same distance apart); but these 2 pairs can't be on the same column, since on a column all the distances we considered are distinct, so we have a paralellogram (2 opposite sides are formed by the 2 pairs: they are parallel since they are placed on columns and we proved that they are equal, meaning that the distances are equal for the 2 pairs of pieces)

For the second part the answer is no. We just put n pieces on a main diagonal and the other n-1 all on a line (we only have n-1 spaces left since one is occupied by a piece from the first group which was placed on the diagonal). Think this is correct?

And I have solution for the first one

Consider 1001 nonoverlapping pairs of form (a,2a)
1500,3000
1499,2998
...
751,1502

375,750
...
188,376

93,186
...
47,94

23,46
...
12,24

5,10
4,8
3,6

1,2

There are 998 numbers from {1,...,3000} which is not in pairs  ==>  for any 2000 numbers no less then 1002 lay in pairs  ==>  two numbers in the same pair

3) There is a similar one with 2 subsets of {1,2,...,9} But the solution I looked up doesn't have pigeonhole principle. It just considers wich numbers cannot be at the same subset and gets contradiction.

Hmm...

I proved, using pigeon-hole principle and an adequat algoritm, that for each integer n \geq 2, if the set {1,2,...., [n*e*(n!)] + 1 } is divided into n disjoint parts then at least one part contains three distinct numbers x,y,z such that x+y = z,
where [.] the integer part, and e is the solution of Ln(x) = 1.

Unfortunately, the proof 5 pages so I will not give it here....

Reference :
P.Bornsztein, 'On an extension of a theorem of Schur' , Acta Arithmetica, 101.4 (2002), p. 395-399.

Pierre.

That means that there is no nice elegant simple elementary proof? :(

Let us assume for the sake of contradiction that the problem is false. Assume WLOG that at least [49/3]+1 = 17 elements of {1,2,..,49} are in A. Let these numbers be a_1<a_2<..<a_17. Clearly a_2-a_1,a_3-a_1,...,a_17-a_1 cannot be in A since else we would be done. So there are at least 8 numbers in the set B (WLOG). If they are b_1=a_i1-a_1 , b_2=a_i2-a_1 , .. , b_8=a_i8-a_1, then it is clear that b_2-b_1,b_3-b_1,..,b_8-b_1 cannot be in B, and neither in A (why?). So they must be a member of C. Let c_1=b_2-b_1,c_2=b_3-b_1,..,c_7=b_8-b_1. The numbers c_2-c_1,c_3-c_1,..,c_7-c_1 cannot be in C, in B, or in A (why?) and the problem is over.

Here is a much more difficult problem which I couldn't solve: {1,2,..,3n} is partitioned into three sets A,B,C such that |A|=|B|=|C|=n. Prove that there are x in A,y in B and z in C such that one is the sum of the other two. I have seen a solution somewhere but it's somewhat artificial and maybe someone can post a nicer solution in here!

Hi Pierre, I'm interested in this problem. Can you post your solution?
Thanks

[quote="pbornsztein"]About the problem of Andrei, there is a more general result, which I think has been given as an exercise in the 1987 Schweizer contest :

Let {1,2,...,n} be 3-colored, such that each color is used at least n/4 times. Then, the equation x+y = z has a tricolor solution.

If you are interested, I may give a (long) solution.

Pierre.[/quote]


integration
calculus
calculus computations

$ \int \frac{dx}{\sqrt{x} (9\plus{}\sqrt[3]{x})}$

Using the substitution $ x\equal{}t^6$ we have to compute:

$ \int \frac{6t^5}{t^3(9\plus{}t^2)}dt\equal{}6 \int \frac{t^2}{9\plus{}t^2}\equal{}6 \int \left( 1\minus{}\frac{9}{9\plus{}t^2}\right)dt \equal{}6(t\minus{}3\arctan \frac{t}{3}) \plus{}\mathcal{C}$.

Therefore we get that the initial integral is $ 6(\sqrt[6]{x}\minus{}\arctan \frac{\sqrt[6]{x}}{3})\plus{}\mathcal{C}$


abstract algebra
group theory
superior algebra
superior algebra unsolved

[quote="sastra81"]to determine only  homomorphism  with kernel and image from Z50 to S3(THE PERMUTATION)
THANKS
SASTRA[/quote]
I don't understand you. What is Z50? (There are exactly $5$ groups of order $50$)
Is it the cyclic group $C_{50}=\mathbb{Z}/50\mathbb{Z}$ of order $50$?
And what do you mean by "homomorphism with kernel and image"?
Every homomorphism has a kernel and image.

Anyway there are exactly $4$ homomorphisms $C_{50}\rightarrow S_{3}$ with images $1,\langle(1,2)\rangle,\langle(1,3)\rangle,\langle(2,3)\rangle$.
($S_{3}$ has no other subgroups of order $|50$ and $C_{50}$ has only one subgroup of index $2$)


[quote="Captain Sugar"]hello. i am in 8th grade and plan to be an aerospace engineer.  what should i take in college to help me get there? i am also trying to get to OSSM, oklahoma school of science and math, for my junior and senior years. they specialize in science and math and offer some college-level courses.[/quote]

Whatever college you go to will have that course very well mapped-out for you.  You shouldn't have to worry about that now.  In the math area, you'll need linear algebra and multivariable calculus, and you'll want to learn you way around a computer.

Actually, i am an aerospace major and linear algebra is not required. However, i think some aspects of it might be covered in the upper division courses, because i know that we cover Laplace transforms in ARO 322 (Aerospace Feedback and Control Systems), yet we do not take the Laplace Transforms and Fourier Series course.

And as for Captain Sugar, take as many math and science courses as you can in high school, namely physics. Also take computer programming and/or a graphic design class if your high school offers them.


search
number theory proposed
number theory

Let $ f(n) = 4^n + 499^n$ (n=1,2,3,... ). For each $ k\in \mathbb N$, let $ k$ integers $ s_1,s_2,...,s_k > k$ . For each $ s_i$ (i=1,2,...,k), let 1 prime divisor $ P(s_i)$ of $ f(2^{s_i})$. Prove that for positive integer $ t\le k$, we have that $ (2^t)|\sum^k_{i=1}P(s_i) \iff \(2^t)|k$

Really I don't under stand this problem . Can you explain more clearly ?

Edited:
That is a particular case of the following problem:

http://www.mathlinks.ro/viewtopic.php?search_id=1327755543&t=217286


linear algebra
matrix
induction
vector
linear algebra unsolved

:oops:
Let $A$ in $M_n(R)$ whose trace is $0$.
a)Show that there exits an inverse matrix $P$ in $M_n(R)$ such that the matrix $P.A.P^-1$ has its diagonal as $(0,0,...,0)$ ($n$ times).
b)Show that for any matrix A satisfied $trace(A)=0$: $A=CD-DC$ with some matrix $C,D$ in $M_n(R)$.
--------------------------------------------------------------------------------------------
-I don't know why can show b) is true?Because in my idea, $W=ker(tr)$ with $W$=$<${$CD-DC$ $/C,D$ $in$ $M_n(R)$}$>$.Or $W$={$CD-DC/C,D$ $in$  $M_n(R)$}??? :mad:  :(

for b/, you just have to prove that if a matrix $M$ has all its coefficients equal to $0$ on the diagonal, then $M=AB-BA$. Try to choose for example $A=diag(1,2,..,n)$ and you will easily conclude.

a).Let prove the fact by induction on $n$. If $A \neq \lambda I$, then we can find linearly independent pair of vectors $e_1,e_2=Ae_1.$ It gives us a zero in the $(1,1)$'th position of the matrix $A$.
Now suppose we've already constructed vectors $e_1,e_2,...,e_k$, s.t. $a_{11}=0,...,a_{kk}=0$. Suppose we managed to take a vector $e_{k+1} \in V_k=<e_1,e_2,...,e_k>$ such that $e_{k+2}$ [b]not[/b] in $<V_k,e_{k+1}>$. In this case we get one mor null in position $(k+1,k+1)$. It it is impossible to choose such  $e_{k+1}$ we must have that $\forall x$ not in $V_k$, $Ax \in <V_k,x>$. If for all such $x$ $Ax \in V_k$ then we can take any base complementing $e_1,...,e_k$. If $Ax=\lambda_x x+...$, where $...$-is some vector from $V_k$. If $\lambda_x=l$ is the same for all $x$ then $trA=(n-k)\lambda=0 \rightarrow \lambda=0$ and we are done. If for some vectors $u,v$ we get $Au=\lambda u$, $Av=\alpha v$, $\lambda \neq \alpha$, then we can take $e_{k+1}=u+v$,$e_{k+2}=Ae_{k+1}=A(u+v)=\lambda u+ \alpha v$ and it is readilt that $e_{k+1}$ and $e_{k+2}$ are linearly independent. So, in each of the cases we add one more zero to the main diagonal which prove our problem.

Nice,thx eugene! :)
[quote]alekk wrote:
for b/, you just have to prove that if a matrix  has all its coefficients equal to  on the diagonal, then . Try to choose for example  and you will easily conclude [/quote]
 :oops:  with matrix has all its coefficients equal to $0$ on the diagonal,it 's ok.But in the general case,I haven't still done it.(sorry!).Can u write it clearly?Thanks...
I 'm doubting about this conclusion...

$A$ an arbitrary matrix with zero trace. $M=P^-1AP$ has null coefficients on the diagonal: $M=CD-DC$.
Hence $A=..$

i know that b) is true...(seen in in cool book) but could someone please post a full solution ? i don't seem to be able to nail it

[quote="dreammath"]$M$=$CD-DC$--->$CD-DC$=$P^-1AP$--->$PCDP^-1 - PDCP^-1$=$A$---->...?[/quote]

You must put something and then take it.

See here: http://www.cs.berkeley.edu/~wkahan/MathH110/trace0.pdf .


geometry
rectangle
IMO Shortlist
Extremal combinatorics
combinatorial geometry
combinatorics

In the plane we consider rectangles whose sides are parallel to the coordinate axes and have positive length. Such a rectangle will be called a [i]box[/i]. Two boxes [i]intersect[/i] if they have a common point in their interior or on their boundary. Find the largest $ n$ for which there exist $ n$ boxes $ B_1$, $ \ldots$, $ B_n$ such that $ B_i$ and $ B_j$ intersect if and only if $ i\not\equiv j\pm 1\pmod n$.

[i]Proposed by Gerhard Woeginger, Netherlands[/i]

Assume $ n\ge7$ and there exist $ n$ boxes $ B_1,\ldots,B_n$ such that $ B_i$ and $ B_j$ intersect if and only if $ i\not\equiv j\pm1\pmod{n}$.

$ B_1$ and $ B_2$ does not intersect, but both intersect $ B_{n\minus{}3}$ and $ B_{n\minus{}2}$. So $ B_{n\minus{}2}$ and $ B_{n\minus{}3}$ are [i]bridges[/i] which connect $ B_1$ and $ B_2$. But $ B_n$ intersects $ B_{n\minus{}2}$, $ B_{n\minus{}3}$ and $ B_2$. Therefore the three borders of $ B_n$ must coincide with sides of $ B_2$, $ B_{n\minus{}3}$ and $ B_{n\minus{}2}$. This makes an empty space bordered by $ B_1,B_{n\minus{}3},B_{n\minus{}2},B_n$ (see attachment C1a.gif). Note that $ B_3$ intersects $ B_1,B_{n\minus{}3},B_{n\minus{}2},B_n$. Therefore the space is exactly $ B_3$. But now $ B_3$ cannot intersect any other boxes, which is impossible since $ B_3$ is supposed to intersect $ n\minus{}2\ge5$ boxes. We get a contradiction.

Therefore $ n\le6$. The following figure (C1b.gif) shows that the answer is indeed 6.

I have just realized a mistake in the above reasoning. :blush:  The empty space is not [i]exactly[/i] $ B_3$, but only a part of it. But this hole can be fixed easily. Consider $ B_{n\minus{}1}$. It must intersect $ B_1,B_2,B_3,B_{n\minus{}3}$ but not $ B_{n\minus{}2},B_n$. Clearly, we cannot put $ B_{n\minus{}1}$ anywhere.

Please tell me if there are still some mistakes.

[quote="Johan Gunardi"]I have just realized a mistake in the above reasoning. :blush: [/quote]

I think you just fixed the mistake nicely, and I think your solution is now right.  However, I think that your reasoning may be simplified and made more clear if you just transform it
[hide] into 1-D, by considering the projections of the boxes onto the coordinate axis, so that the condition of the problem transforms into projection $ X_i$ on the x-axis intersecting all projections $ X_j$ for $ |i \minus{} j|\neq1$ modulus $ n$, idem. for projections $ Y_i$ on the y-axis, and either $ X_i\cap X_{i \plus{} 1}$ or $ Y_i\cap Y_{i \plus{} 1}$ must be empty for each $ i$.  You can then find at most $ 3$ empty intersections for the projections on each axis, leading to a maximum of $ 6$ boxes...[/hide]

[quote="Ivob"]I don't understand one thing in daniel's solution. Why can we have at most three empty intersections on each axis?[/quote]

Here goes the complete proof of this result:

Let $ [a_1,b_1], [a_2,b_2],\dots,[a_n,b_n]$ be a set of at least $ 4$ intervals (the problem is trivial for $ n\leq3$) in the real axis, such that $ [a_i,b_i]\cap[a_j,b_j]\neq\emptyset$ if $ |i \minus{} j|\neq1$ (assume cyclic notation, or modulo $ n$ notation, such that $ i \equal{} n \plus{} i$).  Now, call $ A \equal{} \max\{a_i\}$, and $ B \equal{} \min\{b_i\}$, such that wlog $ a_2 \equal{} A$ (otherwise we rotate the coefficients cyclically).  Then,...
[hide]
... if $ B\geq A$, we are done, since $ [A,B]\subset[a_i,b_i]$ for each $ i \equal{} 1,2,\dots,n$, and no pairwise intersection is empty.  Otherwise, $ A > B$, and since $ [a_i,b_i]\cap[a_2,b_2]\neq\emptyset$ for all $ i \equal{} 4,5,\dots,n$, we must have $ b_i\geq A$ for all $ i \equal{} 4,5,\dots,n$.  Moreover, wlog $ B \equal{} b_1$, since we may invert the order in which we number the intervals starting from $ i \equal{} 2$ without altering the problem.  But $ [a_1,b_1]\cap[a_j,b_j]$ for all $ j \equal{} 3,4,5,\dots,n \minus{} 1$, and $ a_i\leq B$ for all $ i \equal{} 3,4,\dots,n \minus{} 1$.  We conclude that $ [A,B]\subset[a_i,b_j]$ for all $ i \equal{} 4,5,\dots,n \minus{} 1$, and the intersecion of these sets with any other set is nonempty.  The only intersections that remain, and could potentially be empty, are $ [a_n,b_n]\cap[a_1,b_1]$, $ [a_1,b_1]\cap[a_2,b_2]$ and $ [a_2,b_2]\cap[a_3,b_3]$, just three of them, qed.
[/hide]

Daniel's solution was very nice, and I suppose even my silly solution has its merits.

Straightforward (my) approach: As above, we provide the example for $ n \equal{} 6$ and claim that $ n\ge 7$ does not work. Note that if we prove that it's impossible to arrange $ n\ge 7$ boxes so each borders exactly $ n \minus{} 3$ of the others, we are done; so suppose on the contrary that this is possible. Every box borders exactly a fixed number greater than four others; implying that if we consider a given box, for each box bordering it, there must exist another box bordering it which does not border the first  - this is easy to show through contradiction. But this holds for any box, so consider one pair of the boxes which do not border each other, $ A$ and $ B$. $ A$ must also border at least three other boxes, so consider the boxes which $ A$ does not border which border these; $ C$, $ D$, and $ E$. Likewise, $ B$ must also border at least three other boxes, so consider the boxes which $ B$ does not border which border these; $ F$, $ G$, and $ H$.

But there must only be two distinct boxes among $ B,C,D,E$ and similarly for $ A,F,G,H$. This now degenerates into some unfortunate casework on which of these are the same, all of which lead to a contradiction - i.e. if they're both two and two then let $ B \equal{} C,D \equal{} E,A \equal{} F,G \equal{} H$; then $ B,A$ don't border, $ D,G$, don't border. But then eachof the extremities - $ D$, $ G$ - would have to be not bordering at least three of the boxes or they would be on both sides of $ A,B$ respectively at the same time, so we have a contradiction. We can argue similarly for the other cases. Hopefully there's a slick way to bypass this which I can't find.

EDIT: Actually I think the extremities argument works for all cases except when $ B\equal{}C\equal{}D\equal{}E$ or $ A\equal{}F\equal{}G\equal{}H$ (WLOG the first), in which case we can argue using the four boxes which border both $ B$ ($ \equal{}C\equal{}D\equal{}E$) and $ A$ and note that they must all border, because the ones not between any other ones would be not bordering more than two boxes otherwise. Yet this only accounts for six boxes, and any additional boxes would not border at least one of the edge ones, so we have a contradiction there too.

[quote="fwolth"]Daniel's solution was very nice, and I suppose even my silly solution has its merits.
[/quote]

The solution is not silly, and unless I am mistaken it has the merit of being right!  Good job fwolth!   :)

[quote="Johan Gunardi"]I have just realized a mistake in the above reasoning. :blush:  The empty space is not [i]exactly[/i] $ B_3$, but only a part of it. But this hole can be fixed easily. Consider $ B_{n \minus{} 1}$. It must intersect $ B_1,B_2,B_3,B_{n \minus{} 3}$ but not $ B_{n \minus{} 2},B_n$. Clearly, we cannot put $ B_{n \minus{} 1}$ anywhere.

Please tell me if there are still some mistakes.[/quote]

Hi, in your original post, you wrote that B3 must intersect $ B_1, B_{n\minus{}3}, B_{n\minus{}2}, B_n$.  For n=7, $ B_{n\minus{}3}\equal{}B_4$ which B3 should not intersect?  Am I misunderstanding something?

it was a typo, and i think it is not important because i have revised my solution.

Hello everyone.

I've attached an image for n=7.

I'm pretty sure there's something wrong with this construction but I can't pinpoint exactly where.

Perhaps someone can help me spot it.

Thanks!

[quote="leader"]Assume $n\ge 7$
Consider the projections of $B_{1},B_{2},...,B_{n}$ on the $x,y$ axis. Call those segments $X_{1},X_{2},...X_{n}$ $Y_{1},Y_{2},...,Y_{n}$ respectively. Now let 2 of those segments make a [i]good[/i] pair if they are disjoint. Now since $B_{i}$ and $B_{i+1}$ are disjoint for each $i(mod n)$ we have that for each $i$ one of the pairs $(X_{i},X_{i+1}),(Y_{i},Y_{i+1})$ are good. Now WLOG we assume that there is more or the same good pairs in $X_{1},X_{2},...,X_{n}$ than there is in $Y_{1},Y_{2},...,Y_{n}$. This gives us at least $n/2$ good pairs in $X_{1},....,X_{n}$. Now choose some good pair $X_{i},X_{i+1}$(since all $X_{i},X_{k}$ where $k-i\not = 1 (mod n)$ are not good pairs) and let WLOG $X_{i}$ be left of $X_{i+1}$. Than all points which are not $X_{i-1},X_{i},X_{i+1},X_{i+2}$ contain the right end of $X_{i}$ and left end of $X_{i+1}$ so they are all connected to each other and don't form any good pairs. The only possible good pairs are from points $X_{i-2},X_{i-1},...,X_{i+3}$(Rename those to segments $a,b,c,d,e,f$ in that order.
If $b,c$ and $d,e$ are connected that are atmost $3$ good pairs.
If $b,c$ and $d,e$ are not connected then $b$ containt left end of $d$ (and so is connected to $a$) and $e$ contains the right end of $c$(and so is connected to $f$) which gives again at most $3$ good pairs.
Now WLOG $b,c$ are connected and $d,e$ are not then $b$ contains the left end of $d$ and so is connected to $a$ giving at most $3$ good pairs.
We have now proved that there are at most $3$ good pairs so $n/2\le 3$ or $n\le 6$ giving a contradiction.
It is not hard to find examples for $n=1,2,3,4,5,6$[/quote]

Rectangles 4 and 5 touch each other. However, just move the right side of rectangle 4 a bit to the left so it lies in rectangle 7 but not 5, and you have a valid solution for $n=7$.

Not quite. 7 also intersects 1, and that's not so easy to fix. There is no construction for $n = 7$.

Oh, I just read modulo $n$; I only thought $i \neq j \pm 1$.

[hide=Solution?]
I claim that $n = 6$. Suppose $n=7$ or greater. Then we must have intersections between $(1,3),(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,6),(4,7),(5,7)$.

Consider triples of intersections; clearly for $B_i$ to intersect $B_j$, $B_j$ to intersect $B_k$, and $B_k$ to intersect $B_i$, they must share a common point as they are rectangles parallel to the coordinate axes. Therefore we have triples $(1,3,5),(1,3,6),(1,4,6),(2,4,6),(2,4,7),(2,5,7),(3,5,7)$.

Consider the intersection of rectangles $B_3$ and $B_5$. Clearly $B_1$ must contain points from this intersection; furthermore, $B_1$ cannot contain all the points from this intersection, as $B_7$ must also contain some points from this intersection. Assume without loss of generality that $B_3$ and $B_5$ intersect "vertically", and also assume that $B_1$ is to the left of $B_7$.

Now, as $B_4$ must also intersect $B_1$ and $B_7$, and cannot intersect either $B_3$ or $B_5$, it must be above or below both of them. Without loss of generality, assume that it is below. In this case, it follows that $B_5$ is lower than $B_3$, as $B_2$ must intersect each of $B_5,B_4,B_7$.

Now consider $B_6$. Firstly suppose that its upper side is below $B_5$'s lower side. If so, it cannot possible intersect $B_3$, contradiction. Now assume $B_6$'s lower side is above $B_5$'s upper side; in this case, it cannot intersect $B_4$, contradiction. Finally, assume that one of its sides lies in between the $y$-coordinates of $B_5$'s upper and lower sides; in this case, it must either lie to the left or to the right of $B_5$. If it is to the left of $B_5$, it cannot intersect $B_2$, and if it is to the right, it cannot intersect $B_1$, so we are done.

The construction for $n = 6$ is fairly easy, so I will omit it because I do not wish to upload an image.[/hide]
Can someone please check this? My solution appears to be solid enough, but I believe that it may have a hole or two somewhere...

edit: this is almost certainly wrong rip

The largest possible  $n$ is $6$. Construction is shown below.

Consider the two rectangles $B_2$, $B_{n}$. If they are both entirely on one side of $B_1$, then we see that it would be impossible to put rectangles $B_3$ and $B_{n-1}$ such that both of them intersect $1$ and one of $B_2$ or $B_n$.

Then for $B_2$ there exists a side of $B_1$ for which $B_2$ is on the opposite side of that side from $B_1$, and similarly another corresponding side of $B_1$ for $B_n$. Then consider the corner formed by these two sides.

If there exists two rectangles $B_i$ and $B_{i+1}$ where $4 \le i \le n-3$, then they must intersect $B_1$, $B_2$,$B_n$ which means that they must intersect at that corner of $B_1$. But this is impossible since $B_i$ doesn't intersect $B_{i+1}$.

We see that such two rectangles exist if $n \ge 7$.

We claim that the largest such $n$ is $6$.
The construction for $n = 6$ is easy enough, so I'll omit it.

Call two non-intersect rectangles RLE to each other if one lies completely to the right of the other, and TDE to each other if one lies completely above the other. For convenience, we'll omit 'to each other' from now on.
Note than any two non-intersecting cannot be both not RLE and not TDE.
Note that if $A, B$ are RLE, and $C, D$ intersect both $A, B$, and if $C, D$ are non-intersecting, then $C, D$ have to be $TDE$, and cannot be $RLE$. Analogous result holds for $A, B$ that are TDE.

Let $n \ge 7$. WLOG let rectangles $1, 2$ be TDE. Then as $4, 5, 6$ intersect both $1, 2$, the pairs $4, 5$ and $5, 6$ are RLE. As $1, n$ intersect both $4, 5$, the rectangles $1, n$ cannot be RLE. As $2, 3$ intersect both $5, 6$, the rectangles $2, 3$ are TDE. Then as $6, 7$ intersect both $2, 3$, the rectangles $6, 7$ have to be RLE.

Case : $3, 4$ are RLE -
As $6, 7$ intersect both $3, 4$, the rectangles $6, 7$ cannot be $RLE$. Contradiction.

Case : $3, 4$ are TDE -
As $1, n$ intersect both $3, 4$, the rectangles $1, n$ have to be $RLE$. Contradiction.

Consider a graph where the nodes are boxes, and an edge joins any two boxes that intersect. Note that if this graph is not planar, then it is clearly impossible to draw boxes in the plane with the desired property.

For $n \ge 8$, it is trivial to find a $K_{3, 3}$ graph within the graph, so therefore it is not possible for $n \ge 8$ (because $K_{3, 3}$ isn't planar).

For $n = 7$, if you combine $B_2$ with $B_6$ and combine $B_4$ with $B_7$, then the graph becomes $K_5$. Hence this graph is not planar, so it is also not possible for $n = 7$.

For $n = 6$ see someone else's construction.

Why does planar -> not possible to draw the bloxes
[img]https://media.discordapp.net/attachments/986049049834713139/996219174672146533/ISL2008C1.png[/img]
We now prove that $n\ge 7$ is not possible. Project each box $B_i$ to segments $X_i,Y_i$ on the $x$ and $y$ axes respectively. Note that $B_i,B_j$ intersect if and only if $X_i,X_j$ and $Y_i,Y_j$ both intersect.

First, we claim that for each $i$ taken mod $n$ either $X_i,X_{i+1}$ intersect or $Y_i,Y_{i+1}$ does. Assume otherwise, then $X_{i+3},X_{i+4}$ intersects both $X_{i}$ and $X_{i+1}$ so $X_{i+3}$ and $X_{i+4}$ must both cover the space between $X_{i},X_{i+1}.$ As a result, $X_{i+3},X_{i+4}$ intersect and similarly so do $Y_{i+3},Y_{i+4}$, a contradiction.

Suppose WLOG $Y_1$ and $Y_2$ intersect. Then, $X_4,X_5,X_6$ both cover the gap between $X_1$ and $X_2$ and so $Y_4,Y_5$ and $Y_5,Y_6$ are disjoint while $Y_4,Y_6$ intersect. $Y_3$ intersects $Y_1$, $Y_5$, $Y_6.$ Suppose WLOG that $Y_5$ is below $Y_4$ and $Y_6$.

If $Y_6$ has a part that lies below $Y_4$ then take $B_7$ which must intersect $B_2,B_3,B_4,B_5$ but be disjoint from $B_6.$ Obviously then $B_7$ cannot intersect $B_1$ and there would be no place to put $B_3.$ Induce a similar argument for the other case, and we're done.

For $n=6$ see constructions already listed.

For $n\ge 7$ note that $B_1$ and $B_3$ intersect. Consider the intersection $R$ of $B_1$ and $B_3$. Clearly there exists a side of $R$ which is intersected by both $B_5$ and $B_6$, as otherwise, there would be no place to put $B_2$. Suppose that $B_3$ intersects $B_6$ at a point farther from the corner of $B_6$ contained in $R$ compared to where $B_1$ intersects $B_6$ (here, we take the intersection which lies on the side perpendicular to the side which both $B_5$ and $B_6$ intersect $R$ at). Then note that $B_4$ is blocked from $B_1$ by $B_3$ (otherwise, it wouldn't intersect $B_7$), so we are done by symmetry. $\blacksquare$

[hide=Note]This is really handwavy, especially the last part. It's due to the fact that $B_1$ and $B_6$ "block" $B_7$ and the fact that $B_3$ and $B_5$ block $B_4$.

I don't find this solution here, please see if it is wrong or not.Denote the boxes by their numbers

Say $n \geq 7$.Now consider $n-2,2$ clearly they intersect each other now $n$ intersects both so should be on the same side.Now $n-3$ intersect $n,2$ so $n-3$ should be on the opposite side of $2$ w.r.t $n-2$.Now $n-1$ intersects $2,n-3$ but not $n$ and $1$ intersects $n-2,n-3,n-1$.So $1$ and $n-1$ should be on the opposite side of $2,n-2,n-3$ w.r.t $n$.Now $3$ intersects $n-2,n-1,1,n$.Now $3$ must be on the opposite side of $n-2$ w.r.t $2$.Now $n-1$ if intersects $3$ the $n-2$ height must be in a lesser extend than a lower most portion of $n-1$.Which means height of $n-2$ is less than $2$ ,but that is not possible since height of $n-2,n-3$ need to be more than $2$ so that $1$ could intersect both of them without intersecting $2$ and thus we have a contradiction.

I will try to attach some documents to make this feasible to understand.


Suppose that I have an unlimited supply of identical math books, history books, and physics books.  All are the same size, and I have room on a shelf for 8 books.  In how many ways can I arrange eight books on the shelf if no two books of the same type can be adjacent?

$ 3 \times 2^7$

_ _ _ _ _ _ _ _

The first book can be anything, so 3, but for the next book the choices should be limited to anything but the first one, so 2.

[hide]We can choose any book for the first position on the shelf. This is $3$ possibilities. After placing this book, we can choose either of the other two types of books to place directly after it. Similarly, there will be two choices for the third book once the second has been chosen. There will be two choices for each book after the first. So, the total number of ways to arrange the books on the shelf will be: $$3\times2\times2\times2\times2\times2\times2\times2=3\cdot2^7=\boxed{384}.$$[/hide] :-D  :-D

[quote=StellarG]What about repeating some of the ways?[/quote]

What do you mean?


http://mathstuff.rootforge.org
http://www.maths.mq.edu.au/~wchen/lndpnfolder/lndpn.html
http://www.justpasha.org/math/links/books/online.html
http://www.combinatorics.org/Software/index.html
http://www.princeton.edu/~ploh/olympiad.shtml
http://de.geocities.com/darij_grinberg/
http://mathdb.org/mainpage/e_main.htm
http://www.math.ust.hk/excalibur/
http://www.madras.fife.sch.uk/maths/enrichment/index.html

opoios exei kai aytos link me mathimatiko periexomeno as ta bazei
edo

Sti selida mou http://www.math.uoc.gr/~ags/links.htm exw arketes endiaferouses istoselides me periexomeno kiriws gia mathimatikous diagwnismous. Enjoy...

Alexandros

http://mathworld.wolfram.com
http://www.cut-the-knot.org
http://www.mathsnet.com
http://kr.cs.ait.ac.th/~radok/math/mat/intro.htm

Επίσης, [url=http://archives.math.utk.edu/topics/]εδώ[/url] και [url=http://www.angelfire.com/ab3/GIANNNO/edu/subject/math.htm]εδώ (ελληνικό)[/url] θα βρείτε πολλά χρήσιμα links

Ενα τοπικ που πραγματικα πρεπει να υπάρχει...
Και μαλιστα να ειναι και σε περιοπτη θεση... :wink:

Προσθετω μερικα λινκσ :

http://www.gallup.unm.edu/~smarandache/  (πολυ ωραιο... :) )
http://math.berkeley.edu/~desouza/pb.pdf.html
http://www.mathpropress.com/competitions.html
http://www.mymathforum.com/

[hide]Δε βλεπω να στελνετε λινκσ...[/hide]

Να προσθέσω και εγώ ένα εξαιρετικό για όσους ξέρουν ρωσικά

http://www.mccme.ru

Λανς

[quote="DuLac"]Να προσθέσω και εγώ ένα εξαιρετικό [b]για όσους ξέρουν ρωσικά[/b]

http://www.mccme.ru

Λανς[/quote]

Εχει και English version! :D

Η πλειοψηφία των βιβλίων που μπορεί κανείς να κατεβάσει είναι στα Ρωσικά!

Λανς

Εχεις δίκιο.... :(

http://www.math4u.de/
http://www.nzamt.org.nz/olympiad/competitions.htm
http://www.math.toronto.edu/oz/turgor/archives.php
http://www.imo.org.yu/

Στην υγειά μας :lol: !

Άλλο ένα και από εμένα:

http://www.planetmath.org

Εξαιρετικό site για γεωμετρία είναι το agutie.homestead.com/files/Geoproblem_B.htm

http://lostmathematicians.blogspot.com/

Σε τελείως διαφορετικό κλίμα, αποτελεί ένα ευχάριστο διάλειμμα.


algebra
polynomial
function
parameterization
calculus
inequalities
analytic geometry

Suppose p(x,r) is a polynomial, where r is a parameter. For each r, you could evaluate the roots of p. As r changes, the roots of p changes continuous. What is the theorem that proves this?

Example:

$ p(x,r) \equal{} 4r^2x^3 \minus{} x^2 \plus{} 6x \plus{} r \plus{} 9$

Rouche's theorem. Surround the roots small circles on which the polynomial does not vanish. If the perturbation is small enough, the number of complex zeroes inside each circle (counted with their multiplicities) will be preserved. This is exactly what the correct continuity statement here would say.

What does it mean by "vanish"?

I am imagining this:

$ f(z) \equal{} z^3 \plus{} 5z^2 \minus{} 9 \plus{} r$
$ g(z) \equal{} z^3 \plus{} 5z^2 \minus{} 9 \plus{} r \plus{} \Delta r$

$ f(z)$ is a complex function that can be infinitely differentiated. The roots of $ f(z)$ lies on a complex plane. Suppose $ x$ is a root of $ f$, and $ y$ is a root of $ g$. Suppose a draw a small circle around $ x$ on the complex plane, then if the following is true:

$ |f(z) \minus{} g(z)| < |f(z)| \plus{} |g(z)|$, for all $ z$ on the circle..... (Eq.1)

then $ g(z)$ has a root in the neighborhood of $ f(z)$.

But what shows that I could always find a $ \Delta r$ small enough such that Eq.1 is true? As $ \Delta r$ goes to 0, the LHS goes to 0, and the RHS goes to $ 2|f(z)|$?

"doesn't vanish on $ E$" is just a commonly used mathematical slang expression for "is not $ 0$ at every point of $ E$".

(Eq. 1) should really be $ |f(z) \minus{} g(z)| < |f(z)|$.

Otherwise, yes, the point is that, for this particular case, $ |\Delta r|$ can be taken less than the minimum of $ |f(z)|$ on the circle.

What is the relation between a polynomial, its root, and a contour that circles the root? What theorem is that?

----
[b]Q2[/b]

Is the following description correct?

Given $ f(z) \equal{} z^5 \plus{} 3z^3 \plus{} 7$, at a root, $ z^5 \equal{} \minus{}3z^3 \minus{} 7$, and $ |z^5|\equal{}|3z^3\plus{}7|$. This relation is a necessary condition. Suppose $ f(z)$ has a root $ x$ that lies outside the circle of radius 2 at the origin, then:

$ |x^5|\equal{}|x|^5 \geq 2^5 \equal{} 32$
$ |3x^3\plus{}7| \leq 3|x|^3 \plus{} 7 \leq 31$

This shows that for any $ x$ with magnitude greater than 2, the relation is impossible to hold, therefore it cannot be a root. Therefore, all roots of $ f(z)$ must be inside the radius 2 circle.

I don't think this description uses Rouche's theorem, but is the logic correct?

----
[b]Q3[/b]
This is the problem I had before I asked the question:

$ p(z) \equal{} \frac{1}{z}((z\plus{}b)^n \minus{} b^n) \plus{} rb^{n\minus{}1}$

The parameters $ b$ and $ r$ are related: $ b \equal{} rx_0$. So $ r$ is everywhere. I group $ p(z)$ into two polynomials:
$ p(z) \equal{} p_1(z) \plus{} p_2(z)$
$ p_1(z) \equal{} \frac{1}{z}((z\plus{}b)^n \minus{} b^n)$
$ p_2(z) \equal{} rb^{n\minus{}1}$

The roots of $ p_1(z)$ are on the open-left-half-plane. And if $ r$ goes to 0, the effect of $ p_2(z)$ would go to 0, so the roots of $ p(z)$ would approach the roots of $ p_1(z)$, which are all on the OLHP. Therefore, I am tempted to say that there is maximum value $ r_{max}$ for $ r$, such that whenever $ 0<r<r_{max}$, the roots of $ p(z)$ would stay on the OLHP. While I could observe that this is true, I don't know how to prove it because $ p_1(z)$ has $ b$, which is related to $ r$. So as $ r$ goes to zero, the roots of $ p_1(z)$ go to zero also. How do I show that for a small but non-zero $ r$, the roots of $ p(z)$ would stay on the left before some of them cross the imaginary axis? How do I show that $ r$ doesn't drag the root to the right at the onset?

Interesting.  Consider $ p(x,r) \equal{} x^3 \minus{} x \minus{} r$ .  When $ r \equal{} 6$ the unique real root is  $ 2$.  When $ r \equal{} \minus{} 6$ the unique real root is $ \minus{} 2$.  But there is no continuous $ z(r)$ with $ z(6) \equal{} 2$, $ z( \minus{} 6) \equal{} \minus{} 2$, and $ p(z(r),r) \equal{} 0$ for $ \minus{} 6 \le r \le 6$.

Can we fix this if we allow the use of [i]complex[/i] roots?  (As in the arguments already given...)  Or not?

Another problem can occur when the leading coefficient has value $ 0$ for some value of $ r$.  To fix this you may have to allow $ \infty$ to count as the root.  Say $ p(x,r) \equal{} rx \minus{} 1$ and let $ r$ move from $ \minus{} 1$ to $ 1$ through the interval $ [ \minus{} 1,1]$.  Then the root moves from $ \minus{} 1$ to $ 1$ by way of $ \infty$.

Oops, I was indeed including complex roots. I suppose this should be in the Complex analysis part. I just though that the continuity has something to do with calculus, and this heading said calculus.

[color=green][Yes, perhaps it should be in complex analysis - moderator.][/color]

[quote="Twinbee"]What is the relation between a polynomial, its root, and a contour that circles the root? What theorem is that?[/quote]
This is not a "theorem", just a definition of what continuity means in this case. The precise statement is the following.

Let $ p(x,r)$ be a polynomial in $ x$ whose coefficients are continuous functions of $ r$. Let $ z_1,\dots,z_k$ be its complex roots and $ m_1,\dots,m_k$ be the corresponding multiplicities. Then, for any small $ \varepsilon>0$, there exists $ \delta>0$ such that the circles centered at $ z_k$ of radius $ \varepsilon$ contain exactly $ m_k$ zeroes of $ p(x,s)$ (counted with multiplicity) when $ |s\minus{}r|<\delta$.

[quote]
Is the following description correct?
[/quote]
No. You use the inequality $ |x|\ge 2$ in the first estimate and the inequality $ |x|\le 2$ in the second. So, what is proved is that there are no zeroes with $ |x|\equal{}2$, not that there are no zeroes with $ |x|\ge 2$ as claimed.

[quote]
$ p(z) \equal{} \frac {1}{z}((z \plus{} b)^n \minus{} b^n) \plus{} rb^{n \minus{} 1}$
The parameters $ b$ and $ r$ are related: $ b \equal{} rx_0$. So $ r$ is everywhere.
[/quote]
Scaling is your friend. If you take $ z\equal{}bw$, you'll get $ \frac{(w\plus{}1)^n\minus{}1}{w}\plus{}r$ to analyze, which is much simpler.

[b]Re: The relation between polynomial, its roots, and a contour that encircles it.[/b]

I was asking this question because in the read I read Rouche's Theorem, it was like this:

Given $ f(z)$ and a disturbance $ g(z)$, let $ x$ be a root of $ f(z)$, and let $ K$ be a contour around $ x$ on the complext plane. Then, if $ |f(z)| > |f(z)\minus{}g(z)|$ for all $ z$ on $ K$, $ h(z): \equal{}f(z)\plus{}g(z)$ has the same number of roots as the number of $ f(z)$ roots encircled by $ K$.

In this description, I don't see anything about evaluating anything such that $ f(z)\equal{}0$ or $ h(z)\equal{}0$. So I was wondering what relation there was. I assume that the relation is in the proof of Rouche's Theorem. But I was also assuming that there was a simple answer for example:

Imagine the 1D case and a polynomial $ f(x)$. I claim that if, $ f(\minus{}4)<0$ and $ f(4)>0$, then $ f(x)$ has a root in $ (\minus{}4,4)$. This is obvious if I know beforehand that $ f(x)$ is continuous. I was wondering whether there is something similarly obvious about the Rouche's Theorem.

It might be as obvious as "If you ate a ham sandwich, then you ate meat." I ask, "Why?", and you would say, "Because ham is meat." But what is the relation in Rouche? How does Rouche's theorem know something about the location of the roots without asserting that the roots are continuous?

[b]Re: The problem in terms of $ \epsilon$ and $ \delta$[/b]

The question was why are the roots of a polynomial parameterized by a variable are continuous. This question can be rephrased in terms of $ \epsilon$ and $ \delta$:

Claim:
Let $ f(z)$ be a polynomial with roots at $ \{x_i\}$. Let $ g(z,r)$ be a polynomial such that $ g(z,0)\equal{}0$. Let $ h(z) \equal{} f(z) \plus{} g(z,r)$. Then, for any $ \epsilon>0$, there exists $ \delta>0$ such that $ h(z)$ has a root in $ ball(x_i,\epsilon)$ whenever $ |r|<\delta$.

--------
[b]Re: Q2[/b]
Another attempt:

Assume that $ f(z)\equal{}z^5\plus{}3z^3\plus{}7$ has a root that lies on or outside the center circle of radius 2. Assume that such a root has magnitude $ |x| \equal{} 2 \plus{} h$, $ h>0$. Rearrached:

$ |z^{5}|\equal{}|3z^{3}\plus{}7|$

LHS:
$ |x^5| \equal{} (2\plus{}h)^5$. When $ h\equal{}0$, LHS = 32.

RHS:
$ |3x^3\plus{}7| \leq 3|x|^3 \plus{} 7 \equal{} 3(2\plus{}h)^3 \plus{} 7$. When $ h\equal{}0$, RHS = 31.

Since the two numbers do not match, $ f(z)$ cannot have a root $ x$ with magnitude 2. If $ f(z)$ has a root outside the circle, the difference between the LHS and RHS must shrink.

$ \frac{d}{dh}LHS \equal{} 5(2\plus{}h)^4$
$ \frac{d}{dh}RHS \equal{} 3(2\plus{}h)^2 < 5(2\plus{}h)^4$

Since the RHS started smaller and always grow slower than the LHS, there is no h where the LHS=RHS. Therefore, $ f(Z)$ has no root outside the circle.

--------
[b]Re: Q3[/b]

I tried scaling but to me that has the same conceptual problem. Say I scaled it and get:
$ q(w) \equal{} \frac{(w\plus{}1)^{n}\minus{}1}{w}\plus{}r \equal{}: q_1(w) \plus{} r$, $ z \equal{} bw$, $ b\equal{}rx_0$
If I just look at $ q(w)$, I would be tempted to say that, "since as r goes to 0, the roots of $ q(w)$ goes to the roots of $ q_1(w)$, and the roots of  $ q_1(w)$ are on the OLHP, therefore, the roots of $ q(w)$ starts on the left and reaches the right later as r increases."

But, on the other hand, I also know that as r goes to 0, the roots of $ q_1(w)$ goes to the origin. An analogy:

Imagine you are trying to hit a nail into a wooden board with a hammer only once. I say that, "if the force is not strong enough, the nail does not go into the wood. Therefore, there is a minimum force $ f_{min}$ that is required to hit the nail in." So far so good. Now imagine that everytime you increase your force, the size of the nail changes accordingly. If you increase the force, the nail becomes bigger and more blunt. If you decrease the force, the nail becomes smaller and thus sharper. Since the nail changes itself, how do you know that the nail would not go in regardless what force you use?

[quote="Twinbee"]
Given $ f(z)$ and a disturbance $ g(z)$, let $ x$ be a root of $ f(z)$, and let $ K$ be a contour around $ x$ on the complext plane. Then, if $ |f(z)| > |f(z) \minus{} g(z)|$ for all $ z$ on $ K$, $ h(z): \equal{} f(z) \plus{} g(z)$ has the same number of roots as the number of $ f(z)$ roots encircled by $ K$.
[/quote]
Where have you read this? The correct condition should be $ |f(z)| > |g(z)|$.
[quote]
But what is the relation in Rouche? How does Rouche's theorem know something about the location of the roots without asserting that the roots are continuous?
[/quote]
You mean "how is Rouche's theorem proved?" ? It is a much longer story. If you want, we can go over it but I'll need to know your background in complex analysis first.

[quote]
Let $ f(z)$ be a polynomial with roots at $ \{x_i\}$. Let $ g(z,r)$ be a polynomial such that $ g(z,0) \equal{} 0$. Let $ h(z) \equal{} f(z) \plus{} g(z,r)$. Then, for any $ \epsilon > 0$, there exists $ \delta > 0$ such that $ h(z)$ has a root in $ ball(x_i,\epsilon)$ whenever $ |r| < \delta$.
[/quote]
This is just a less precise form of what I wrote above with slightly changed notation, isn't it?

[quote]
Another attempt:
[/quote]
Now it is fine (though you can simplify this argument quite a bit).
[quote]
But, on the other hand, I also know that as r goes to 0, the roots of $ q_1(w)$ goes to the origin.
[/quote]
 :what?: They do not depend on $ r$ at all!
[quote]
Imagine you are trying to hit a nail into a wooden board with a hammer only once...[/quote]
Great analogy! This is exactly what scaling is about. $ w$ is the relative length of the part of the nail driven in, not the absolute one. It doesn't change when the resistance and the force change proportionally. So, if I know that my force is just enough to drive half of the nail in, I do not need to know the exact length to say that the cap of the nail will be still above the board.

[b]Re: Q3[/b]

The way I understand it now is that suppose I find the roots of $ q(w)$, to get back to the roots of $ p(z)$, I would multiply those roots by $ b$. Since $ b>0$, if a root of $ q(w)$ is on the left, the corresponding root of $ p(z)$ is still on the left. That's why I ate ham if I ate a ham sandwich.

[b]Re: Rouche[/b]

I got the equation from your #4 when you told me to remove one of the $ |g(z)|$. I don't have a background in complex analysis, that was why I didn't even know that the question was related to complex analysis. I thought it was just calculus.

Suppose I have polynomial $ f(z)$ and $ x$ is a root. Since $ f(z)$ is continuous, if I evaluate at $ x\plus{}\Delta x$, I would get some $ f(x\plus{}\Delta x)$ that is close to $ f(x)\equal{}0$. In the local region around $ x$, $ f(x\plus{}\Delta x)$ all gets evaluated into some very small number.

Can Rouche's Theorem be applied in this situation:

$ f(x) \equal{} \minus{}1$. $ f(x)$ has no root.
$ g(x)$ is a smooth polynomial approximation of a horizontal line with a bump at the center: $ g(0)\equal{}2$, $ g(k)\equal{}0.1$. $ k\in K$, and $ K$ is the unit circle.
$ h(x) \equal{} f(x) \plus{} g(x) \equal{} x^2\minus{}1$. h(x) has two roots around 0.

$ |f(k)| \equal{} 1$.
$ |g(k)| \equal{} 0.1$, therefore, $ |f(k)| > |g(k)|$. So $ h(x)$ would have the same number of roots as $ f(x)$ have inside $ K$, which is none. ?!

[quote="Twinbee"]
$ f(x) \equal{} \minus{} 1$. $ f(x)$ has no root.
$ g(x)$ is a smooth polynomial approximation of a horizontal line with a bump at the center: $ g(0) \equal{} 2$, $ g(k) \equal{} 0.1$, $ k\in K$, and $ K$ is the unit circle.
[/quote]
I'd like to see this polynomial approximation first ;)  (note that it should be a polynomial in the complex variable $ z\equal{}x\plus{}iy$, not just a polynomial in two real variables $ x$ and $ y$ and the identity $ g(k)\equal{}1$ should hold on the entire circle, not just at $ \pm 1$).

Isn't it still be true that a smooth function can be approximated by a polynomial? So from this alone we could say that it exists, and it is probably some kind of Taylor series expansion of something.

Is it also true that $ g(z)$ need not contain any parameter $ r$ such that if $ r$ goes to 0, $ |g(z)|$ goes to 0? (So this is different from the problem I was originally asking about, since my perturbation function would be $ g(z,r)$, but it seems that Rouche's Theorem should be applicable to $ g(z)$, as long as $ |g(k)| < |f(k)|$. )

[quote="Twinbee"]Isn't it still be true that a smooth function can be approximated by a polynomial? [/quote]
True on the interval $ [\minus{}1,1]$, but the Weierstrass theorem says absolutely nothing about the behaviour of the approximating polynomial on the circle.
[quote]
Is it also true that $ g(z)$ need not contain any parameter $ r$ such that if $ r$ goes to 0, $ |g(z)|$ goes to 0? ... it seems that Rouche's Theorem should be applicable to $ g(z)$, as long as $ |g(k)| < |f(k)|$. )
[/quote]
It is correct.

Then is it true that a complex polynomial can never have a local maximum or minimum (a dimple)? If it has a local maximum on the real direction, it would have a slope in the imaginary direction (like the side of a cone) or a local minimum on the imaginary direction (like a saddle)?

Yes, this is exactly what happens. It is called "the maximum principle".

For the maximum principle, suppose I want K to encircle the entire RHS of the complex plane. When I evaluate the $ g(k)$, is it ok to only use the points along the imaginary axis instead of also including the conceptual semi-circle that encircles the RHS?

You should be careful here. If you know that your function doesn't grow exponentially fast, then yes. Otherwise, $ e^{az}$ is a counterexample for every positive real $ a$. Now the key words are "the Phragmen-Lindelof theorem".

It makes so much sense. /joy

How do you pronounce "Phragmén-Lindelöf" ?

[quote="Twinbee"]
How do you pronounce "Phragmén-Lindelöf" ?
[/quote]
Erm... How am I supposed to type the phonetic transcriptions? :? Anyway, since one of them is Swedish and the other one Finnish (if my memory doesn't betray me here), we'd better wait until some scandinavian forum member will enlighten us: I know how the names would sound in Russian and English but after I heard once how a Chinese person pronounced "Пафнутий Львович Чебышёв", I prefer to reduce my saying foreign names aloud to the absolute minimum :P.
[quote="t0rajir0u"]
The idea is that $ p(x, r) : \mathbb{C}^2 \to \mathbb{C}$ is (Lipschitz, hence) uniformly continuous, so one can prove that the graph of $ y \equal{} p(x, r)$ changes very little if $ r$ changes very little,
[/quote]
Yes, but I'd like to see how exactly you pass from little change in the graph to little change in the roots.

What is the meaning of "graph"? What is the difference between a "graph" and a "map" or a "relation"?

The way I read it is that:

We know that $ p(x,r)$, that was not part of the argument. The roots of a polynomial are nothing more than locations where $ p(x,r) \equal{} 0$. Since $ p$ is continuous, if $ r$ changes continuously, the points where $ p(x,r)\equal{}0$ will also change continuously.

[quote="t0rajir0u"]Rouche's theorem is unnecessary.  The idea is that $ p(x, r) : \mathbb{C}^2 \to \mathbb{C}$ is uniformly continuous, so one can prove that the graph of $ y \equal{} p(x, r)$ changes very little if $ r$ changes very little, which is well in accord with our intuition.

(I mention this mostly to emphasize that this isn't fundamentally a complex-analytic problem; everything carries over to $ \mathbb{R}$ as long as $ \Delta r$ is small enough so that a "phase transition" in which roots change multiplicity doesn't occur.)[/quote]
Continuity isn't really enough; a spot that's too flat can cause trouble. We want the inverse function theorem, which tells us that the roots are a differentiable function of the coefficients, except where a derivative isn't invertible.
Of course, that derivative fails to be invertible precisely at the phase transitions. We absolutely need to be complex there, and Rouche's theorem is the cleanest way to handle them. Since it also works nicely everywhere else, the cleanest proof just uses Rouche and leaves out the other arguments.

Point taken.

[quote="Twinbee"]What is the meaning of "graph"? What is the difference between a "graph" and a "map" or a "relation"?[/quote]
A graph is a topological object; here what I mean is the object that, for fixed $ r$, consists of the points $ (x, y) \in\mathbb{C}^2$ such that $ y \equal{} p(x, r)$.  One can define a topology on the space of all such objects, for example via the supremum norm, and talk about continuity of the function taking $ r$ to the corresponding graph $ G(r)$.  At least this is the way I would like to think about the problem conceptually.  But I accept that if I tried to make this argument rigorous it might be unnecessarily tedious.

How do you pronounce Rouche? Is it one syllable or two?

It's two, and that should really be accented as Rouché.


Two hundred students participated in a mathematical contest. They had six problems to solve. It is known that each problem was correctly solved by at least 120 participants. Prove that there must be 2 participants such that each problem was solved by at least one of the two students.

My failed attempt with fubini's principle:
[hide]
Let $A$ be the set of all pairs of students and $B$ be the set of problems. $|A|= \binom{200}{2}$ and $|B|=6$. We proceed by contradiction. Suppose that every pair $a \in A$ solved at most $5$ problems (by "solve," we mean that at least one guy in the pair got it). Let $S$ be a subset of $A \time B$ s.t. $S(a_{i}, *) = \{(a_{i},b) \in S\}$ and $S(*, b_{i})$ defined analogously. Then
\[|S| = \sum |S(*, b_{i})| = \sum|S(a_{i}, *)| \]
(Fubini's principle). Well $\sum S(a_{i}, *)$ counts the number of problems solved by team $a_{i}$, which is less than 6 by assumption, so $|S| \leq 5*\binom{200}{2}$.

Meanwhile, $\sum S(*, b_{i})$ counts the number of teams that solve each problem. Of the $\binom{200}{2}$ teams, only the ones picked from the $\leq 80$ who didn't solve the problem, so $S(*, b_{i}) \geq \binom{200}{2}-\binom{80}{2}$ and $|S| \geq 6\binom{200}{2}-6 \binom{80}{2}$. Well that's all good, but there's no contradiction! I should have gotten something like $12931414 \leq |S| \leq 3$[/hide]

I know there are easier solutions, but this problem appeared in a chapter called "fubini's principle" so I thought I should try it.

[hide="One of the Easier Solutions"]
Picture a 200 by 6 table, with rows representing the 200 students and columns representing the 6 problems.  Put an x in a box if the corresponding problem was solved by the corresponding student.

There must be at least $6*120=720$ x's in the table, so by Pigeonhole there must be 4 in one row.  The other two columns have a total of 240 x's in them, so by Pigeonhole there are two x's in the same row.  These two rows represent two students who together solved all the problems.
[/hide]

[quote="Ravi B"]I didn't study your argument, but didn't you reach a contradiction?  Your upper bound is smaller than your lower bound, right?[/quote]

Ah, right. I thought $\binom{200}{2}$ was $10,000$ :wink:

I see.  By the way, what book contained this problem?

Counting strategies, by Zuming Feng and Titu Andreescu.

Oh, that's the "Path to Combinatorics" book, right?  Sounds like a good book.  Andreescu and friends have been writing an amazing number of books.

[quote="WatsonLadd"][quote="E^(pi*i)=-1"][hide="One of the Easier Solutions"]
Picture a 200 by 6 table, with rows representing the 200 students and columns representing the 6 problems.  Put an x in a box if the corresponding problem was solved by the corresponding student.

There must be at least $6*120=720$ x's in the table, so by Pigeonhole there must be 4 in one row.  The other two columns have a total of 240 x's in them, so by Pigeonhole there are two x's in some other row.  These two rows represent two students who together solved all the problems.
[/hide][/quote]
Fixed a typo for you.[/quote]

Um, that wasn't a typo.  Your statement is true, but so is mine, and they both solve the problem.


Suppose five days before the day after tomorrow was Friday.  What day was yersterday?

[quote="jimli"]3 days before today was friday.  today was monday.[/quote]
it asked for yesterday not today

[quote="gr8sk8r"]i got something different.

[hide="my answer"]5 days after friday=wednesday
wednesday was the day after tomorrow, so that's 2 days ahead. so then, it is monday. therefore yesterday was Sunday.[/hide][/quote]
My solution exact kinda not really


Well, it's a pretty good competition as far as I'm concerned...congratulations to Kevin Yang for winning the state championship (for the third consecutive time) :)  :)
Did any of you go to the "Masters" competition earlier today? Or earlier this year?

Yeah, I agree. Then again, most of the proctors are volunteers, so you can't really blame them. I didn't go this year, but I got 15th or 16th in the fifth grade division last year.

[quote="sjaelee"]this contest always has sucky procters, I went to masters[/quote]
What always frustrates me are the ones that read the college bowl problems wrong, then say that they didn't. :|

They made me get a score of 68 but I was supposed to get an 85!! :(

Good for you and thanks for reviving a 5 year old thread.


If we can show $ \left( \frac{\log_3 (a\plus{}1)}{\log_3 a}\right)^2>\frac{\log_3 (a\plus{}2)}{\log_3 a}$ for $ a>1$, then we will have
\[ (2A/N)^2>\frac{\log_3 5}{\log_3 3}\cdot \frac{\log_3 7}{\log_3 5}\cdot \cdots \cdot \frac{\log_3 81}{\log_3 79}\equal{}4\]
and so $ \boxed{A>N}$.

The assertion at the beginning is equivalent to $ \log_a (a\plus{}1)>\log_{a\plus{}1} (a\plus{}2)$. For if $ x\equal{}\log_a (a\plus{}1)$, then $ x>1$ and
\[ a^x\equal{}a\plus{}1\Rightarrow (a\plus{}1)^x\equal{}(a\plus{}1)\left( \frac{a\plus{}1}{a}\right)^x>\frac{(a\plus{}1)^2}{a}>a\plus{}2\]
so that $ x>\log_{a\plus{}1} (a\plus{}2)$.

[i]Very nice solution Scorpius119
    Congratulation :lol:
 [/i]
[i] but we can give a nice solution to the ineq $ log_{n}(n \plus{} 1) > log_{n \plus{} 1}(n \plus{} 2)$, $ n \in N$ , $ n \geq 2$
    Just AM - GM[/i] :wink:
  [i]We have  $ \sqrt { (log_{n \plus{} 1 }(n) .log_{n \plus{} 1}(n \plus{} 2)) }\leq \frac { log_{n \plus{} 1 }(n) \plus{} log_{n \plus{} 1}(n \plus{} 2)}{2}$
   = $ \frac {log_{n \plus{} 1}(n(n \plus{} 2))}{2} \equal{} \frac {log_{n \plus{} 1}((n \plus{} 1)^2 \minus{} 1)}{2} < \frac {log_{n \plus{} 1}((n \plus{} 1)^2)}{2}$
   = $ \frac {2}{2} \equal{} 1$
  Deduce
$ (log_{n \plus{} 1 }(n) . log_{n \plus{} 1 }(n \plus{} 2)) < 1$

  =>$ log_{n \plus{} 1 }(n \plus{} 2) < \frac {1}{log_{n \plus{} 1 }(n)} \equal{} log_{n}(n \plus{} 1)$[/i]


geometry
parallelogram
geometric transformation
reflection
invariant

There is a parallelogram $ ABCD$ with $ AC\equal{}3$ and  $ AD\equal{}\sqrt{2}$ such that its area is $ 3$.
Now you can choose one of the points A,B,C or D. Let's say A. Then you can reflect it on one of the other points and you obtain A' and the new quadrilateral A'BCD. (There is no A anymore!!!)
This is repeated very often.
Prove or disprove that the area of the quadrilaterals is always a multiple of three 3.

[quote="gradient"]
This is repeated very often.
[/quote]

What do you mean? that you reflect $ A$ to get $ A'$, then $ A'$ to get $ A''$ ...

like this???

 :)

Not necessarily. You have the new points A', B, C and D. From these points you choose again one point (you can take A', but you needn't) and reflect it on one of the other three points!
I hope, it's clear now.  :)

And note: To take A at first is just an example, you can also take B,C or D.

It is very easy to show the validity for using two vertexes of the quadrilateral being face to face with each other. Then the area is an invariant and thus remains $ A\equal{}3\equal{}3 \cdot 1$.
But I can't prove it generally (I believe, the theorem is true).  :(


$  \begin{array}{c} x + \frac {1}{y} = y + \frac {1}{z} = z + \frac {1}{x} \\
xyz = 1\Rightarrow z = \frac {1}{xy} \end{array} \}\Rightarrow  x + \frac {1}{y} = y + xy = \frac {1}{xy} + \frac {1}{x}|.xy\Rightarrow \\
\Rightarrow x^2y + x = xy^2 + x^2y^2 = 1 + y\Rightarrow x(xy + 1) = xy^2(1 + y) = 1 + y\Rightarrow \\
\Rightarrow xy^2 = 1 \ (1) \ \ \mbox{ or: } 1 + y = 0.\ (2)$

[b]1).[/b]
$ \mbox{If } xy^2 = 1 = xyz\Rightarrow y = z\ \ (3) \mbox{ and } x + \frac {1}{y} = y + \frac {1}{z}
\Rightarrow x + \frac {1}{z} = y + \frac {1}{z}\Rightarrow x = y. \ \ (4)$
$ (3);(4)\Rightarrow x = y = z \mbox{ and } xyz = 1\Rightarrow x = y = z = 1.$

[b]2).[/b]
$ \mbox{If } 1+y=0\Rightarrow y=-1 \mbox{ and } x+\frac{1}{y}=y+\frac{1}{z}\Rightarrow x-1=\frac{1}{z}-1\Rightarrow \\
\Rightarrow x=\frac{1}{z}\Rightarrow xz=1=xyz\Rightarrow y=1. (contradiction!)$

$ \mbox{Then, unique solution is: } x=y=z=1.$


ratio
quadratics
geometry
analytic geometry
vector
Gauss

Let $ABCD$ be a convex quadrilateral with $AD\not\parallel BC$. Define the points $E=AD \cap BC$ and $I = AC\cap BD$. Prove that the triangles $EDC$ and $IAB$ have the same centroid if and only if $AB \parallel CD$ and $IC^{2}= IA \cdot AC$.

[i]Virgil Nicula[/i]

nice problem.. complex numbers is good to this?

Won't it be too simple for grade 9 student? I solved it immediately tonight...

If the points $E, A, D$ and $E, C, B$ follow on the lines $DA, BC$ in this order, the triangles $\triangle EDC, \triangle IAB$ do not have a common point and the proposition is trivial. Assume that the points $E, D, A$ and $E, B, C$ follow on the lines $DA, BC$ in this order. Assume that the lines $AB, CD$ are not parallel - they intersect at a point $F \equiv AB \cap CD$. WLOG, assume that the points $A, B, F$ and $D, C, F$ follow on the lines $AB, CD$ in this order. Let $M, N$ be the midpoints of the segments $AB, CD$ and $K \equiv EM \cap AB$. In a central projection of the line $AB$ to the line $CD$, the cross-ratio of 4 points is preserved:

$\frac{\frac{ND}{NC}}{\frac{FD}{FC}} = \frac{\frac{KA}{KB}}{\frac{FA}{FB}}$

Let $B'$ be the intersection the line $AB$ of a line through the point $C$ parallel to the line $DA$. Obviously, $FB > FB'$. Since $\frac{ND}{NC} = 1$,

$\frac{KA}{KB} = \frac{\frac{FA}{FB}}{\frac{FD}{FC}} = \frac{FA}{FD} \cdot \frac{FC}{FB} < \frac{FA}{FD} \cdot \frac{FC}{FB'} = 1$

Consequently, $KA < MA, KB > MB$ and the points $A, K, M, B$ follow on the segment $AB$ in this order. Let $P, Q$ be the intersections of the line $EI$ with the lines $AB, CD$, respectively. Project the point $F$ to infinity. The projected lines $A'B', C'D'$ become parallel and the projected points $P', Q'$ become the midpoints of the projected segments $A'B', C'D'$. Again, the cross ratios of 4 points are preserved:

$\frac{\frac{PA}{PB}}{\frac{MA}{MB}} = \frac{\frac{P'A'}{P'B'}}{\frac{M'A'}{M'B'}}$, $\frac{\frac{FA}{FB}}{\frac{MA}{MB}} = \frac{\frac{F'A'}{F'B'}}{\frac{M'A'}{M'B'}}$

Since $\frac{MA}{MB} = \frac{P'A'}{P'B'} = 1$ and $\frac{F'A'}{F'B'} \rightarrow 1$,

$\frac{PA}{PB} = \frac{M'B'}{M'A'} = \frac{FA}{FB} > 1$

Consequently, $PA > MA, PB < MB$ and the points $A, M, P, B$ follow on the segment $AB$ in this order. Similarly, it can be shown that the points $C, N, Q, D$ follow on the segment $CD$ in this order. Thus the segments $PQ, KN$ do not have a common point and consequently, the median segments $EN, IM$ of the triangles $\triangle ECD, \triangle IAB$ do not have a common point either. Hence, these 2 triangles cannot have a common centroid.

Assume that the lines $AB \parallel CD$ are parallel and let $M, N$ be the intersections of the line $EI$ with the lines $AB, CD$. From similarity of the triangles $\triangle EBA \sim \triangle ECD$, we have

$\frac{DA}{DE} = \frac{BC}{BE}$

By Ceva's theorem for concurrent cevians $EM, AC, BD$ of the triangle $\triangle ECD$

$\frac{ND}{NC} \cdot \frac{BC}{BE} \cdot\frac{DE}{DA} = 1$

and consequently $ND = NC$, i,e., $N$ is the midpoint of the segment $CD$. From similarity of the triangles $\triangle EBA \sim \triangle ECD$, $M$ is also the midpoint of the segment $AB$. Thus the median lines $EN \equiv IM$ of of the triangles $\triangle EBA, \triangle ECD$ are coincident.

Let a line through the point $D$ and parallel to the line $EN$ intersect the lines $AC, BC$ at points $K, L$. Since $ND = NC$, from similarity of the triangles $\triangle ICN \sim \triangle KCD$ and $\triangle ECN \sim \triangle LCD$ we deduce that $IK = IC$ and $EL = EC$. The cross ratio of 4 points is preserved in a central projection of the line $AC$ to the line $BC$ from the point $D$. Thus

$\frac{\frac{AC}{AI}}{\frac{KC}{KI}} = \frac{\frac{EC}{EB}}{\frac{LC}{LB}}$

$\frac 1 2 \frac{AC}{AI} = \frac 1 2 \frac{LB}{EB}$

$\frac{AI + IC}{AI} = \frac{LE + EB}{EB} = \frac{EC + EB}{EB}$

$1 + \frac{IC}{AI} = 1 + \frac{EC}{EB}$

$\frac{IC}{AI} = \frac{EC}{EB}$

From similarity of the triangles $\triangle EBM \sim \triangle ECN$ and $\triangle IAM \sim \triangle ICN$, it follows that

$\frac{IN}{IM} = \frac{IC}{AI} = \frac{EC}{EB} = \frac{EN}{EM}$

$\frac{IM}{IN} = \frac{EN - MN}{EN} = 1 - \frac{MN}{EN}$

Let $G, H$ be the centroids of the triangles $\triangle IAB, \triangle ECD$. Since the centroids cuts the medians $IM, EN$ in the ratio $IG : GM = 2 : 1$, $EH : HN = 2 : 1$. For coincidence of the centroids $G, H$, it is necessary and sufficient that

$HN + GM = MN = IN + IM$

$\frac{EN}{3} + \frac{IM}{3} = IN + IM$

Substituting $EN = \frac{MN}{1 - \frac{IM}{IN}} = IN \cdot \frac{1 + \frac{IM}{IN}}{1 - \frac{IM}{IN}}$, we obtain a quadratic equation for the ratio $x = \frac{IM}{IN}$

$\frac 1 3 \cdot \frac{1 + x}{1 - x} + \frac{x}{3} = 1 + x$

$1 + x + x(1 - x) = 3(1 - x^2)$

$x^2 + x - 1 = 0$

$x = \frac{-1 \pm \sqrt{5}}{2}$

The negative root is not acceptable and $x = \frac{IM}{IN} = \frac 1 g$, where $g = \frac{1 + \sqrt 5}{2}$ is the golden ratio. From similarity of the triangles $\triangle IAM \sim \triangle ICN$,

$\frac{IC}{AI} = \frac{IN}{IM} = \frac 1 x = g$

$\frac{AC}{IC} = \frac{AI + IC}{IC} = \frac 1 g + 1 = g$

and as a result,

$\frac{IC}{AI} = g = \frac{AC}{IC}$

$IC^2 = AI \cdot AC$

[quote="Myth"]Ops  :huh:[/quote]

What do you mean, Myth :?: Did I make errors, misunderstand the problem, or is the proof just too long ?

yetti

[quote="perfect_radio"][quote="Myth"]An easy task for mass point geometry.[/quote]

may i ask what is mass point geometry ?[/quote]

i have found on the net that mass point geometry is another name for barycentric coordinates

i'd like to know how do you use the coordinates to solve it  :?

Are there somewhere on the net some articles or books (but only if i can download them, not to buy) where i can find things like mass geometry, or complex numbers geometry (i know only the general stuf about this two)...? Chavez you said that complex numbers are good for this problem if you have solution via complex number can you post it
Thanx

m@re

Thanks, Yeti!. May be tediously a vectorial solution or the Yeti's solution, however the conclusion of this problem is interestingly and it has a great use in physics in the equilibrium problems. Sir Chavez, afterwards I will present my [i]vectorial[/i] solution, abviously ! See and http://www.mathlinks.ro/Forum/viewtopic.php?t=47616

Let $I$ be the origin of the vectorial system, i.e. for any $X$ we have the vectors $X=\overrightarrow{IX},\ I=\overrightarrow 0$. Thus,

$\ (\exists ) m,n\in R^{+}- \{ 0,1\},\ IC=m\cdot IA,\ ID=n\cdot IB\Longleftrightarrow C=-mA,D=-nB$.

Therefore, $E\in AD\cap BC\Longleftrightarrow (\exists )x,y\in R,\ x,y\not\in \{ 0,1\}$ such that

$E=(1-x)B+xC=(1-y)A+yD\Longleftrightarrow E=-xmA+(1-x)B=$ $(1-y)A-ynB\Longrightarrow$

$x=\frac{1+n}{1-mn},\ y=\frac{1+m}{1-mn},\ E=\frac{m(1+n)}{mn-1}\cdot A+\frac{n(1+m)}{mn-1}\cdot B.$

The triangles $ABI$ and $CDE$ have the same centroid $\Longleftrightarrow E+C+D=A+B\Longleftrightarrow $

$\frac{m(1+n)}{mn-1}\cdot A+\frac{n(1+m)}{mn-1}\cdot B-mA-nB=A+B\Longleftrightarrow$

$mn^2=2n+1\ \wedge \ nm^2=2m+1\Longleftrightarrow m=n\ \wedge \ n^3=2n+1\Longleftrightarrow m=n\ \wedge \ n^2=n+1$

$\Longleftrightarrow AB\parallel CD\ \wedge\ \left( \frac{IC}{IA}\right) ^2=\frac{IC}{IA} +1\Longleftrightarrow AB\parallel CD\ \wedge \ IC^2=IA\cdot AC.$

you write $\overrightarrow{AB}$

[code]\overrightarrow{AB}[/code]

i don't have time right now to look over your solution (seems nice and correct at first sight), but i'll do it tomorrow

Thank you, perfect_radio !. With you I am learning better LaTeX.

The solutions are great! Are there any more interesting ones? (I mean, looking at what Myth said...)

Which is the official solution to this problem ?The vectorial one ?There exist solution with plain geometry ?Thanks for all

The official solution is Virgil Nicula's one. This is actually his problem... as you can well see.

[yeah, you're right. typo]

Is there a synthetic one shorter than Yetti?

[size=125][hide="Lemma 1"][/size]
[quote]Let $ABCD$ be a trapezoid $(AB \parallel CD)$.
Let the points $E\in AD \cap BC$ and $I \in AC \cap BD$.
$M,N$ are the midpoints of $AB,CD$ respectively.
It is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=52603]known[/url] that the points $E,M,I,N$ are collinear.

Prove that the division $(E,M,I,N)$ is harmonic (that is, $\frac{EM}{EN}= \frac{IM}{IN}$)

[color=brown]P.S. It can be applied also to a parallelogram $ABCD$[/color][/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=106811

[/hide]
[size=125][hide="Lemma 2"][/size]
[quote]Let $(E,M,I,N)$ be an harmonic division with ratios $\frac{EM}{EN}= \frac{IM}{IN}= \lambda$.

$Z$ is a point $\in (MI)$ which divides the segments $MI,NE$ in the same ratio $\frac{ZM}{ZI}= \frac{ZN}{ZE}= \mu$

Then $\boxed{\mu = \frac{1}{2}(\frac{1}{\lambda}-\lambda)}$[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=106817

[/hide]

[size=125][hide="Proof"]
[hide="(<=)"][/size]
Suppose that $AB \parallel CD$ and $IC^{2}=IA \cdot AC$. We will show that the triangles $EDC,IAB$ have the same centroid

from the [b]Lemma 1[/b] we get that $\frac{EM}{EN}= \frac{IM}{IN}= \lambda$, for some $\lambda>0$

From the parallel lines $AB,CD$ we get that $\frac{IA}{IC}= \frac{IM}{IN}$

So, $IA = \lambda \cdot IC$

$\boxed{IC^{2}=IA \cdot AC}\Rightarrow$

$\lambda^{2}\cdot IC^{2}= \lambda \cdot IC(1+\lambda) IC \Rightarrow$

$\lambda^{2}= \lambda(1+\lambda) \Rightarrow \boxed{\lambda = \frac{\sqrt{5}-1}{2}}$, because $\lambda>0$

As [b]yetti[/b] noticed, $\lambda=\frac{1}{\phi}$, where $\phi$ is the golden ratio.

From the [b]Lemma 2[/b] we get that the point $Z$ wich divides the segment $MI$ in the ratio $\mu=\frac{1-\lambda^{2}}{2\lambda}$, divides the segment $NE$ in the same ratio

$\lambda=\frac{1}{\phi}\Rightarrow$
$\mu=\frac{1}{2}(\frac{1}{\lambda}-\lambda) = \frac{1}{2}(\phi-\frac{1}{\phi})=\frac{1}{2}$.

So $\frac{ZM}{ZI}= \frac{ZN}{ZE}= \frac{1}{2}$, which means that the point $Z$ is the centroid of $\triangle IAB$ and $\triangle ECD$.

[/hide]
[size=125][hide="(=>)"][/size]

We suppose that the triangles $EDC$ and $IAB$ have the same centroid.
We'll prove that $AB\parallel CD$ and $IC^{2}=IA \cdot AC$

Let $G$ be te common centroid.

From the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=81254]Gauss line theorem[/url] we get that the points $M,N$ and the midpoint of $EI$ are collinear.

But $\frac{GM}{GI}= \frac{GN}{GE}\Rightarrow MN \parallel EI$

The midpoint of $EI$ belongs both to the lines $MN,EI$, so the only possible result is that the points $M,N,E,I$ are collinear.

So the median $EN$ of the triangle $ECD$ is also median for the triangle $EAB$.
And $A\in (ED, B\in (EC$
This concludes that $AB \parallel CD$. So $ABCD$ is a trapezoid.

According to the [b]Lemma 1[/b] we have $\frac{EM}{EN}= \frac{IM}{IN}= \lambda$, for aome $\lambda>0$
According to the [b]Lemma 2[/b], there is a unique point $Z$ in $(MI)$ which divides the segments $MI,NE$ at the same ratio, and the ratio is $\mu = \frac{1-\lambda^{2}}{2\lambda}$

$G$ is the one with this property, and the value of $\mu$ is $\frac{1}{2}$

$\frac{1}{2}(\frac{1}{\lambda}-\lambda) = \frac{1}{2}\Rightarrow$

$\frac{1}{\lambda}-\lambda) = 1 \Rightarrow$

$\lambda^{2}+\lambda-1 = 0 \Rightarrow \lambda = \frac{-1+\sqrt{5}}{2}$, because $\lambda$ is positive

So $\frac{IA}{IC}= \frac{-1+\sqrt{5}}{2}$, which yields $IC^{2}=IA \cdot AC$

[/hide][/hide]

[quote="Pontios"]From the [size=134][b]Gauss line theorem[/b][/size] we get that the points $M,N$ and the midpoint of $EI$ are collinear. But $\frac{GM}{GI}= \frac{GN}{GE}\Rightarrow MN \parallel EI$.
The midpoint of $EI$ belongs both to the lines $MN,EI$, so the only possible result is that [b]the points $M,N,E,I$ are collinear.[/b][/quote]

[color=darkblue][i]Congratulations, Pontios[/i], for your nice solution ! It is and the background (spring) of the my idea for set up this problem.
Now, Siluan and Indybar have a nice and short synthetical solution. [i]In the their name, thanks, Pontios ![/i][/color]

[quote="Virgil Nicula"][color=darkblue][i]Congratulations, Pontios[/i], for your nice solution ![/color][/quote]
[color=blue][b]Congratulations for your nice problem!!![/b][/color]  :)

here I've got a solution with barycentric coordinates  :D

Let $A=(1: 0: 0)$, $B=(0: 1: 0)$, $C=(0: 0: 1)$, $D=(x: y: z)$. So the line $BD=[-z: 0: x]$  and the line $AC=[0: 1: 0]$. This follows that $I=(x: 0: z)$. Similary we have that the line $BC=[1: 0: 0]$ and $AD=[0:-z: y]$. Thus $E=(0: y: z)$.

As we have got the points let us do it.

let $G_{1}$ and $G_{2}$ be the barycenters of $\triangle ABI$ and $\triangle ECD$ so $3G_{1}=(1: 0: 0)+(0: 1: 0)+\frac{(x: 0: z)}{x+z}$ and $3G_{2}=\frac{(0: y: z)}{y+z}+(0: 0: 1)+\frac{(x: y: z)}{x+y+z}$ then if $G_{1}=G_{2}$

Let's asume that $x+y+z=1$, it means that $D=(x: y: z)$ is in normalized coordinates.

$\frac{(2x+z: x+z: z)}{(x+z)}=\frac{(xy+xz: y^{2}+yz+y: z^{2}+2z+yz+y)}{(y+z)}$

Thus $2x+z=x(x+z)$, $y+z=y^{2}+yz+y$ and $z(y+z)=(z^{2}+2z+yz+y)(x+z)$.

so the solutions are $x=\frac{(1+\sqrt[]{5})}{2}$, $y=-\frac{(1+\sqrt[]{5})}{2}$, $z=1$. This follows $\overrightarrow{CD}=(x: y: z-1)=(\frac{(1+\sqrt[]{5})}{2}:-\frac{(1+\sqrt[]{5})}{2}: 0)=-\frac{(1+\sqrt[]{5})}{2}(-1: 1: 0)=\frac{(1+\sqrt[]{5})}{2}\overrightarrow{AB}$. That's $AB//CD$

Let $\lambda$ be $\frac{(1+\sqrt[]{5})}{2}$, so $x=\lambda$, $y=-\lambda$ and $z=1$

Now, we have got that $I=(\lambda: 0: 1)$ and $E=(0:-\lambda: 1)$.

$IC^{2}=S_{A}(\lambda)^{2}+S_{B}(0)^{2}+S_{C}(0)^{2}=\lambda^{2}S_{A}$

Now $IA^{2}=S_{A}(\lambda-1)^{2}+S_{B}(0)^{2}+S_{C}(1)^{2}$ and $AC^{2}=S_{A}+S_{C}$ so $IA^{2}.AC^{2}=((\lambda-1)^{2}S_{A}+S_{C})(S_{A}+S_{C})$ and with some betrayals we get $IC^{4}=IA^{2}AC^{2}$ and we get $IC^{2}=IA.AC$.

$\bigstar$ Capitán Mandarina

What do u think?  :icecream:

$\bigstar$ Capitán Mandarina

[quote=Virgil Nicula]This problem had officially [b]four[/b] solutions: syntetic (with Ceva, Menelaus,...), vectorial, with complex numbers and analytical. The student makes choice of the best suitable method. In my opinion, the best is the vectorial method !.[/quote]

Can you post here those official?