[quote="math92"]in New York a taxi departs the airport every 4.5 minutes.  How many taxis depart the airport between 1:15 and 4:15?[/quote]

[hide]40[/hide]

[hide]There are 180 minutes between 1:15 and 4:15.
180÷4.5=40 taxi departures[/hide]

Aha, but don't forget that the answers may vary too for the following similar problem:

How many leap years are there in an interval of 5 years?

In that case there could be one, two (2000-2005) or even zero (centuries aren't leap years unless they're divisible by 400, so 1898-1902).

math92, please specify when a given taxi leaves and we will answer more precisely.

sorry about the problem but that is how it is written in my book.  Say the first taxi leaves at 1:15. :)

[hide]1:15 - 4:15 = 3 hrs, or 180 minutes.

If a taxi leaves every four and a half minutes, then it'll be:

180/4.5, or 40 taxis.[/hide]

The answer could be any number 40 higher than a multiple of 160 (40, 200, 360, etc.) Specify the AM and PM's and day.

Ah. Right!!!!! Yes, I think he means all on the same day, in a time span of 3 hours, or else it would be indeterminate.

Okay then, 3 hours is 180 minutes, divide by 4.5 to obtain 40 taxis.
I hope  :)

It can be 41 if a taxi starts at 1h15 and we count both this taxi and the one at 4h15.

1:15 pm and 4:15 pm on the same day is what the problem means.  The answser in my book says
[hide]
40 taxis[/hide]

I think the problem still needs to be more clear. Does the taxi leaving at 1:15 or 4:15 count? That can make a difference from 39-41.

[quote="math92"]in New York a taxi departs the airport every 4.5 minutes.  How many taxis depart the airport between 1:15 and 4:15?[/quote]

Please specify AMs and PMs.


Express each product as a sum containing ONLY sines and cosines.

(1) cos(4x) cos(2x)

(2) sin(x/2) cos(5x/2)

No, the other one changes from sums to products. This one is reversed.

EDIT: Never mind, the exact same problems *have* been posted... Oops.

They're essentially the same formulas, just written with different variables.


can you calculate $ log_22003$ without a calculator? if so, how?

if $ log_2 2003 \equal{} x$, $ 2^x\equal{}2003.$

I don't it's possible figuring the value of x out without the calculator.

i kinda know what logs are. i just need a possible way to solve it w/o a calculator

Depends.  What do you mean by "calculate"?

There are ways to compute the numerical value of $ \log_2 2003$ to an arbitrary degree of precision that mimic the technique your calculator uses.  (Have you ever wondered what that technique is?  That's the more interesting question.)  Of course, it'll take a lot of time.

Well the best way I can think of is using differentials which yields
$ \log_2{2003}\approx 11\minus{}\sum^{\infty}_{n\equal{}1} \frac{1}{n}\cdot\frac{45^n}{2048^n}$
That seems to be about $ .01$ greater than what we need though.

hmmm


Θεωρούμε στο επίπεδο $ \frac {n(n \plus{} 1)}{2}$ σημεία έτσι ώστε να σχηματίζουν ισόπλευρο τρίγωνο στην πλευρά του οποίου υπάρχουν $ n$ σημεία. Να βρείτε τον ελάχιστο αριθμό σημείων που πρέπει να μετακινήσουμε, ώστε να σχηματίζεται πάλι το ίδιο ισόπλευρο τρίγωνο.

''Την άσκηση αυτή την ανακάλυψα σε ένα παλιό βιβλίο του Κοντογιάννη αλλά έχει λίγο πολύπλοκη λύση και έτσι ελπίζω μήπως κάποιος απο σας με βοηθήσει. :maybe:

[quote="filosofimenos"]Μπορεις να γινεις λιγο πιο σαφης. Απο να σημεια που αναφερεις υπαρχουν καποια που βρισκονται στα ακρα των πλευρων του? Γιατι αν ειναι ετσι αλλαζουν τα πραγματα.[/quote]

Κοίτα η άσκηση λέει πως τα σημεία είναι τοποθετημένα έτσι ώστε να σχηματίζουν ισόπλευρο τρίγωνο...

 Άρα, 3 από τα σημεία λογικά θα αποτελούν τις κορυφές του τριγώνου.

Μου φαινεται πολυ απλο αρα μαλλον εχω καταλαβει κατι λαθος. Αν εχω καταλαβει ομως σωστα τοτε
ν+ν+ν-1-2=ν(ν+1)/2 αρα ν=2 ή ν=3
Στην πρωτη περιπτωση προφανως πρεπει να μετακινηθει 1 σημειο ενω στην δευτερη 3.


trigonometry
Euler
real analysis
real analysis unsolved

Compute $ \sum_{n \equal{} 0}^{\infty}(\sum_{k \equal{} n}^{ \plus{} \infty}\frac {\sin k}{k})$

hello, mathematica gives
$ \sum _{n\equal{}0}^{\infty } \frac{i\left(B_{e^{\minus{}i}}(n,0)\minus{}B_{e^i}(n,0)\right)}{2}$
Sonnhard.

Euler would do:
\[ \equal{} lim_{\lambda \to 1^\minus{}} \sum_{n \geq 1} \sum_{k \equal{} n}^{\infty} \frac {\sin(k) \lambda^k}{k} \equal{} lim_{\lambda \to 1^ \minus{} } \sum_{p \geq 1} \sin(p) \lambda^p \equal{} \Re(\frac {\exp(i)}{1 \minus{} \exp(i)})\]
:D


[quote="ML2008"]Let a,b,c>0: $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{b} \plus{} \sqrt [3]{{abc}} \ge \frac {{10}}{{9({a^2} \plus{} {b^2} \plus{} {c^2})}}$[/quote]
It's easy by Am-Gm  :lol:
$ LHS \ge \frac{1}{\sqrt[3]{abc}}\plus{}\sqrt[3]{abc} \ge \frac{10}{3} \ge RHS$


calculus
integration
linear algebra
matrix

Post a popular culture item that has a math term in it.

examples:
Metroid "Prime"
Joy "Division"
Bizarre Love "Triangle"

An "integral" part of...
Two paths "converge"...

[quote="AIME15"]Actually, that DOES refer to exponents in math.[/quote]
Sometimes it is used by a math moron to refer to all sorts of rapid growth.

I'm sorry, but I don't understand what we do?

In "POPULAR CULTURE" :D

TRANSFORMERS!

What about "The Matrix"?

I am sorry, but how does a metroid have a prime in it?

It doesn't, it's just a phrase from popular culture that has a math word in it.  Doesn't mean it has anything to do with math, it just has the term.


function
calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

Find the interval on which the function f(x)=4x^3+2x-1
is concave up?

It's concave up when $f''(x)>0$. The form of $f''$ (which I leave for you to compute) makes the rest completely transparent.

Thanks for your help. But I have no idea where to go with this. They took up our calculus books at school. I remember making some sort of chart and having critical values and testing the first second and third derivative to see when the function is concave up/concave down. However, i really don't remember what exactly what we did. Could you possibly point me to some material that deals with this so i can figure it out.
Thanks

is it :
$f''(x)= 24x$ so the function is concave down for all negative x values or $(-\infty,0)$ and concave up for all x values $(0, \infty)$

You don't need the chart - you need visualization and common sense. The first derivative is slope - $f$ is increasing in any interval in which $f'(x)>0$ and $f$ is decreasing in any interval in which $f'(x)<0.$ The second derivative controls concavity, which is curvature - upward curvature like a smile, downward curvature like a frown. A place where $f'(x)=0$ is a place where the graph has a horizontal tangent. That might be a maximum, a minimum, or neither - try to see where it fits into the scheme.

For this particular function, $f'(x)>0$ everywhere, so it is always increasing. You can have a increasing or decreasing, concave up or concave down, in any combination. This is all visual - draw it!

If you were standing in my office asking this, I'd make you go to the whiteboard and start drawing sketches of the graph; I suspect mlok would do the same.

I see now! I was used to doing so much harder ones earlier in the year that i didnt even realize that this one was so simple! Thanks!!


A rectangular piece of paper $ ADEF$ is folded so that corner $ D$ meets the opposite edge
$ EF$ at $ D'$ forming a crease $ BC$ where $ B$ lies on edge $ AD$ and $ C$ lies on edge $ DE$. If
$ AD \equal{} 25 cm$, $ DE \equal{} 16 cm$ and $ AB \equal{} 5 cm$, ﬁnd $ BC^ 2$.

We have
$ 20^2\plus{}DC^2\equal{}BC^2$
$ 16^2\plus{}CE^2\equal{}DC^2$
$ DE\equal{}CE\plus{}DC$

Equating the area of the rectangle gives
$ 25DE \equal{} 7DE \plus{} 20DC \plus{} 8CE
\implies 18DE \equal{} 18(CE\plus{}DC) \equal{} 20DC \plus{} 8CE
\implies 10 CE \equal{} 2 DC
\implies CE \equal{} DC/5$

So we have
$ 16^2 \plus{} DC^2/25 \equal{} DC^2
\implies DC^2 \equal{} 25*16^2/24$
$ BC^2 \equal{} 20^2 \plus{} 25*16^2/24$
Simplify.

mathwizarddude, i believe your wrong.

[u]Here is my solution[/u]

Here is my diagram

http://img72.imageshack.us/my.php?image=paperdiagramks0.jpg

We construct the diagram as the question dictates, as we do, we note that

$ BD \equal{} BD' \equal{} 20$ and

$ DC \equal{} D'C$

this is because its the same side just moved

We also note that

$ BD \equal{} 20$ as $ AD \equal{} 25$ and $ AB \equal{} 5$

We now construct a line perpendicular to $ AD$ at point $ B$.

Let the point on $ FE$ made by the perpend. of $ AD$ at $ B$ be $ G$ (i didnt put this on my diagram)

We have a right angle triangle with sides of $ 16$ and hypotenuse $ 20$

We find $ GD'$ using Pythagoras theorem to be $ 12$.

Now,

$ FG \plus{} FD' \plus{} D'E \equal{} AD \equal{} 25$

$ FG \equal{} AB \equal{} 5$ and
$ GD' \equal{} 12$

$ \therefore D'E \equal{} 8$

Now,

Let $ DC \equal{} D'C \equal{} n$

$ \therefore DE \equal{} 16 \minus{} DC \equal{} 16 \minus{} n$

Now we have the equation

$ 8^2 \plus{} (16 \minus{} n)^2 \equal{} n^2$

solving gives,

$ 8^2 \plus{} 16^2 \plus{} n^2 \minus{} 32n \equal{} n^2$   subtracting $ n^2$  and rearranging gives

$ 32n \equal{} 8^2 \plus{} 16^2$

$ 32n \equal{} 320$

$ \therefore n \equal{} 10$

Now we have another equation for which we can get $ BC^2$

$ BC^2 \equal{} 20^2 \plus{} 10^2$

$ \therefore BC^2 \equal{}\boxed{500}$

Oops, sorry, my diagram shows $ D'E \equal{} 16$.

Edit:
In that case we have, by equating area,
$ 25*16 \equal{} 7*16 \plus{} 20DC \plus{} 1/2 D'E(16\minus{}DC)$
where $ D'E \equal{} 20 \minus{} \sqrt{20^2\minus{}16^2} \equal{} 20 \minus{} \sqrt{144} \equal{} 8$
Solve for $ DC$ and we have $ BC^2 \equal{} 20^2 \plus{} DC^2$.


find integer numbers $x;y;z$ satisfies $x^3+y^3+z^3=3xyz+1$

Just a tip: Your equation $x^3+y^3+z^3=3xyz+1$ is equivalent to $x^3+y^3+z^3-3xyz=1$. But since $x^3+y^3+z^3-3xyz=\frac12\left(x+y+z\right)\left(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right)$, this transforms into $\frac12\left(x+y+z\right)\left(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right)=1$. And noting that the number $\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2$ is always $\geq 2$ unless all three numbers x, y, z are equal, and equals 2 only in the case when two of the numbers x, y and z are equal and the third one differs from them by 1, you can easily solve this equation.

The answer: The solutions of your equation are

x = 1, y = 0, z = 0;
x = 0, y = 1, z = 0;
x = 0, y = 0, z = 1.

  darij


IMO Shortlist
number theory unsolved
number theory

$A$ is a non-empty set of positive integers. There are positive integers $b_{1}, b_{2}, ... , b_{n}, c_{1}, c_{2}, ... , c_{n}$such that each set $b_{i}A+c_{i}=$ {$b_{i}a+c_{i}: a \in A$} $\subseteq A$, and the $n$ sets $b_{i}A+c_{i}$ are pairwise disjoint. Prove that $\frac{1}{b_{1}}+\frac{1}{b_{2}}+...+\frac{1}{b_{n}}\le 1$

You [i]know[/i] the source, so, is it that hard to :
-[color=green]click[/color] on the "Resources" link : http://www.mathlinks.ro/Forum/resources.php
-[color=green]click[/color] on the "IMO Shortlist" link : http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17
-[color=green]click[/color] on the year, here 2002 : http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2002
-and eventually, [color=green]click[/color] on your problem (I'll let this part to you..  :roll: )
[hide=":?:"] Grunz! [/hide]

i've have post this problem because there don't give solutions in those links !!!so i 'm waiting for a solution . :(

There is |no reason|  to repost anything!
[hide=" :roll: "]Grunz![/hide]


Solve the following equation:
\[ \sqrt3\cos^3 x \minus{} 3\cos^2 x \minus{} 3\sqrt3\cos x\plus{}1\equal{}0\]

[quote="md45"]put cos(X) = t;

transfer last 2 terms to RHS
factorize
so, t=0 or root(3)t - 3t -3root(3)+1 = 0;
solve the quadratic for t;
put back cos(x)=t ;

voila[/quote]
If $ t \equal{} 0$ then $ 1 \equal{} 0.$ Voici! :wink:

[quote="md45"]how ?

t=0 means cos(x)=0

i.e x = (2n+1) pie/2[/quote]
If $ \cos x\equal{}0$ then $ \sqrt3\cos^3 x \minus{} 3\cos^2 x \minus{} 3\sqrt3\cos x\plus{}1\equal{}1.$ :wink:

sorry
i made a mistake in factorization
i ll try again :blush:

i think the question is the expansion

( a + b )^3 = a^2 + b^2 + 3ab^2 + 3ba^2

but I can' t find a and b

do you have the answer of the Q?

[quote="md45"]i think the question is the expansion

( a + b )^3 = a^2 + b^2 + 3ab^2 + 3ba^2

but I can' t find a and b

[/quote]
No! Here is another idea. :P

The equation has only nice roots. :wink:

let $ t\equal{}\cos x$
and solve it by Cardano Method

OK. Here is my solution:
Let $ \cos x \equal{} \tan\alpha.$ :wink:

Let's do it.
$ \sqrt3\sin^3 \alpha \minus{} 3\sin^2 \alpha \cos \alpha \minus{} 3\sqrt3\sin \alpha \cos^2 \alpha\plus{}\cos^3 \alpha\equal{}0$
And now? :huh:

[size=117][color=darkred][b]Very nice, Arkady ![/b][/color][/size]

[quote="arqady"]Solve the equation $ \sqrt3\cos^3 x \minus{} 3\cos^2 x \minus{} 3\sqrt3\cos x \plus{} 1 \equal{} 0$ .[/quote]
[b][u]Proof.[/u][/b] We"ll use the substitution $ \boxed {\ \cos x \equal{} \cot y\ }$ . Thus, $ \sqrt3\cos^3 x \minus{} 3\cos^2 x \minus{} 3\sqrt3\cos x \plus{} 1 \equal{} 0$ $ \Longleftrightarrow$ $ \sqrt 3\cos x \equal{} \frac {3\cos^2 \minus{} 1}{\cos^2x \minus{} 3}$ $ \Longleftrightarrow$

$ \sqrt 3\cot y \equal{} \frac {3\cot^2y \minus{} 1}{\cot ^2y \minus{} 3}$ $ \Longleftrightarrow$ $ \sqrt 3\cot y \equal{} \frac {3 \minus{} \tan^2y}{1 \minus{} 3\tan^2y}$ $ \Longleftrightarrow$ $ \sqrt 3 \equal{} \frac {3\tan y \minus{} \tan^3y}{1 \minus{} 3\tan ^2y}$ $ \Longleftrightarrow$ $ \sqrt 3 \equal{} \tan 3y$ because $ \tan 3y \equal{} \tan (y \plus{} 2y) \equal{}$

$ \frac {\tan y \plus{} \tan 2y}{1 \minus{} \tan y\tan 2y} \equal{} \frac {\tan y \plus{} \frac {2\tan y}{1 \minus{} \tan^2y}}{1 \minus{} \tan y\cdot\frac {2\tan t}{1 \minus{} \tan^2y}} \equal{} \frac {3\tan y \minus{} \tan^3y}{1 \minus{} 3\tan ^2y}$ . Therefore, $ 3y\in \frac {\pi}{3} \plus{} \pi\cdot Z$ $ \Longleftrightarrow$ $ y\in \frac {\pi}{9} \plus{} \frac {\pi}{3}\cdot Z$ $ \Longleftrightarrow$ $ \cos x\in \cot\left(\frac {\pi}{9} \plus{} \frac {\pi}{3}\cdot Z\right)$ $ \Longleftrightarrow$

 $ x\in \pm\arccos\left[\cot\left(\frac {\pi}{9} \plus{} \frac {\pi}{3}\cdot Z\right)\right] \plus{} 2\pi\cdot Z$ , i.e. $ \boxed {\ x\in \left\{ 2s\pi\pm\arccos\left[\cot\frac {(3k \plus{} 1)\pi}{9}\right]\ \boxed{\|}\ \{k,s\}\subset Z\ \right\}\ }$ .

[size=117][color=darkred][b]Know someone how I can do this boxed symbol $ \|$ or $ |$ more high in interior ? Thank you.[/b][/color][/size]

[code]$\left|\ \frac{a+b}{c+d}\right|$.[/code]

You would get $ \left| \frac{a\plus{}b}{c\plus{}d}\right|$.

[quote="kunny"][code]$\left|\ \frac{a+b}{c+d}\right|$.[/code]

You would get $ \left| \frac {a \plus{} b}{c \plus{} d}\right|$.[/quote]
And this applies for ()'s also...
[code]$(\frac{1}{2})$[/code]
gives $ (\frac{1}{2})$, while
[code]$\left(\frac{1}{2}\right)$[/code]
gives $ \left(\frac{1}{2}\right)$.

Thank you ! But I knew these. But between the big acolades there is my symbol (little).

 I wish in $ \boxed {\ x\in \left\{ 2s\pi\pm\arccos\left[\cot\frac {(3k \plus{} 1)\pi}{9}\right]\ \boxed{\|}\ \{k,s\}\subset Z\ \right\}\ }$

same length for the boxed $ \|$ as at symbol $ \{$ or as at this $ \}$   !

You could try some hack... for instance this seems to work:

$ \boxed {\ x\in \left.\left\{ 2s\pi\pm\arccos\left[\cot\frac {(3k \plus{} 1)\pi}{9}\right]\ \right|\ \{k,s\}\subset Z\ \right\}\ }$

(the idea is that you use an invisible bracket such that TeX automatically chooses the right height.)

Thank you very much, [b]Bertram[/b] !  I understood the trick : [b]....\left.\left\{ ...[/b]


Find the distance from the point (1,1,1) to the plane containing (1,0,0), (0,2,0), and (0,0,4)

You have a pyramid  and you can find the altitude from there.

This is a standard vector exercise.  Find the normal vector $ \textbf{N}$ of the plane with cross products and find a vector $ \textbf{V}$ from the point to the plane.  The distance from the point to the plane is then the projection of $ \textbf{N}$ onto $ \textbf{V}$, which is given by $ \frac{ | \textbf{V} \cdot \textbf{N} |}{ | \textbf{N} | }$.

sorry can someone explain it easier? btw how do u find like the equation of the plane given the points?

Let $ \mathbf{x} \equal{} \left<x, y, z\right>$.  One way to write the equation of a plane is as

$ \mathbf{N} \cdot (\mathbf{x} \minus{} \mathbf{A}) \equal{} 0$;

in other words, the plane passes through $ \mathbf{A}$ and consists of all of the vectors from $ \mathbf{A}$ that are perpendicular to a vector $ \mathbf{N}$ (called the normal vector of the plane).

It's easy to find a point on the plane; it remains to find $ \mathbf{N}$.  We are given three points $ \mathbf{A}, \mathbf{B}, \mathbf{C}$ on this plane; we can then choose, for example, $ \mathbf{N} \equal{} (\mathbf{A} \minus{} \mathbf{B}) \times (\mathbf{B} \minus{} \mathbf{C})$.  This is a vector perpendicular to the plane as desired.


[quote="Inspired By Nature"]Natural numbers a and b are relatively prime.Prove that the greatest common divisor of $(a+b)$ and $(a^{2}+b^{2})$ is either 1 or 2. :P[/quote]
first of all you must say that a#b
((a+b),(a^2+b^2))=d
dla+b & dl(a+b)^2 that shows that:
dl2ab
now if a,b#2 we all know d#ax & d#by
that shows that d=<2 as we know ifa,b#2 we have odd answer ,so the answer is d=2
[size=150]now if a or b=2 [/size]
for example a=2
we have dl2ab &we all know that d#ax & d#by that shows that dl2 & d is odd so d=1.

$ a$ and $ b$ are said to be relatively prime or coprime if and only if $ 1\equal{}\gcd(a,b)$.

It was made in 2006. A lot happens in 2 years. Actually, that's not a logical explanation. It's probably just like that Socc dude who had 1113482435 posts according to the Memberlist a while ago.

Speaking of Socs, my essay is now 7 pages long and has 2500 words. /me is epic failure

[url]dictionary.com[/url] is your friend (not literally)

I have a..... :roll: friend :roll: ????? :wow:


12 points are located in space so that 7 of them lie on a single plane (no 3 are collinear) and the remaining five are in general position, meaning that none of them lies on the planes determined by any other 3 points. How many distinct planes pass through at least 3 of the 12 points?

[hide="I got"]
300.
12*(5*4)+(5*4*3).
Don't know if that's right.
[/hide][/hide]

If they are all not co-planar, then the number of planes is $ {12\choose 3}$.  We must now subtract $ {7\choose 3}$ to account for the number that are coplanar.

$ 220\minus{}35\plus{}1\equal{}186$.


How many integers are greater than $ \sqrt{27}$ and less than $ \sqrt{72}$?

hello, from the inequality $ \sqrt{27}<x<\sqrt{72}$ we get $ x\equal{}6$ or $ x\equal{}7$ or $ x\equal{}8$ as the solutions.
Sonnhard.


Given

$x^3 - 3x = y$

$y^3 - 3y = z$

$z^3 - 3z = x$

Find all sets of solutions $[x, y, z]$

[hide]
Just an approach:

$x^3-3x=y$
$x(x^2-3)=y$

For nontrivial solutions of x,y,z,
$x^2-3=\frac{y}{x}$

Apply cyclically, and likewise,
$y^2-3=\frac{z}{y}$ and $z^2-3=\frac{x}{z}$

$\frac{y}{x}\cdot\frac{z}{y}\cdot\frac{x}{z}=1 \Rightarrow (x^2-3)(y^2-3)(z^2-3)=1$

Solutions include (0,0,0), (2,2,2), (-2,-2,-2), but I'm not sure of any more or how to prove it.
[/hide]

I'm pretty sure theres no others. If x>2, then y>x, z>y, and x>z which is contradiction. Also if 0<x<2, y<x, z<y, and x<z, contradiction again. Same goes with negative numbers.

Given: $a = 2 \cos x$

Take $a^3 - 3a = b$ and we have

$b = 8 \cos^3 x - 6 \cos x = 2(4\cos^3 x - 3 \cos x) = 2 \cos 3x$

Try $\left|x\right| > 2$ and see that the equation can't work, so $\left|x\right| \leq 2$ so we can make a substitution.

So, set $x = 2 \cos \theta$
$\Rightarrow y = 2\cos 3\theta \Rightarrow z = 2\cos 9\theta \Rightarrow x = 2\cos 27\theta$

So, solve $2\cos \theta = 2\cos 27\theta$ and plug in for the set of values $(x,y,z) = (2\cos \theta, 2\cos 3\theta, 2\cos 9\theta)$

We get $27\theta \equiv \pm \theta \bmod{2\pi}$

$\Rightarrow \theta = \frac{n\pi}{13}, \frac{m\pi}{14}$
for any $m,n$ integers.

Very nice solution!   :coolspeak:   :thumbup:

So, we have two sets of solutions in the complex plane

One is the set of x-coordinates of the points of a regular 26-gon.
The other set is the set of x-coordinates of a regular 28-gon.

Both the polygons are inscribed in the circle $(O,2)$ and $(2,0)$ is their common vertice


complex analysis
complex analysis unsolved

Let $ Tz\equal{}\frac{az\plus{}b}{cz\plus{}d}, ad\minus{}bc\not\equal{}0$  Find necessary and sufficient conditions that $ T(D)\equal{}D$ where $ D$ is the unit circle $ \{z: |z|\equal{}1\}$

any hint will be appreciated. Thanks

Mobius transformations are determined by the images of three points.  Circles are determined by three points.

thanks for your reply but your answer is too general  :(

It's a hint to a procedure, not an answer. Pick three convenient points on the circle, and look at where they go.


counting
derangement
function
algebra
polynomial
combinatorics unsolved
combinatorics

$N$ couples meet for a party and they exchange gifts.
Each of the $2N$ writes their name on a piece of paper and puts it into a hat. They take turns drawing names and give their gift to the person whose name they drew.
It is not allowed for anyone to draw either their own name or the name of their partner.
How many different permissible drawings are there?

Let $f(n, k)$ be the number of ways that $n$ people can exchange presents such that no one gives a present to herself and none of the people in the $k$ couples ($n \geq 2k$) gives a present to his or her significant other.  Then $f(n, 0) = D_{n}$, the $n$th derrangement number, while $f(2n, n)$ is the value we're trying to calculate.

For any $k \geq 1$, $n \geq 2k$, consider the situations counted by $f(n, k)$.  Pick two people, $x$ and $y$, who are coupled with each other.  $f(n, k-1)$ counts those ways to give gifts where we ignore the restriction on $x$ and $y$; $f(n-2, k-1)$ counts those ways to give gifts in which $x$ and $y$ exchange gifts; $f(n-1, k-1)$ counts those ways in which $x$ gives to $y$, but $x$ receives from someone else (and so $y$ gives to someone else), and it also counts the same situation with $x$ and $y$ reversed.  None of these later three cases has any overlap, so

$f(n, k) = f(n, k-1)-2f(n-1, k-1)-f(n-2, k-1)$.

Since we know $f(n, 0)$, this gives us a complete recursive definition of $f$, but I don't know if it simplifies to a nice closed form.  Where did you get this problem from?

[url=http://www.research.att.com/~njas/sequences/A000316]Here[/url] is the link to the sequence in the OEIS, which has a generating function for the sequence.

It also seems like this problem has a natural generalization to polygamous relationships: choose a partition of $[n] = \{1, 2, \ldots, n\}$.  How many permutations of $[n]$ are there such that no point is sent to a member of its own partition class?

And here is the (recursive) solution to the generalization:

Let $f(n_{1}, n_{2}, \ldots, n_{m})$, $n_{i}\geq 0$, be the number of permutations of the set $\{a_{1,1}, a_{1,2}, \ldots, a_{1,n_{1}}, a_{2, 1}, \ldots, a_{2, n_{2}}, \ldots, a_{m, n_{m}}\}$ such that no $a_{i, j}$ is sent to $a_{i, k}$ for any $j, k$.

Then $f$ is defined recursively as follows:

1) If any 0s appear as arguments of $f$, remove them; the value is unchanged.

2) If only 1s appear as arguments of $f$, $f(1, 1, \ldots, 1) = D_{m}$, the number of derrangements of $m$ letters.

3) Otherwise, $f$ has one argument which is at least 2.  Then $f(n, \ldots) = f(n-1, 1, \ldots)-(n-1)^{2}f(n-2, \ldots)-2(n-1)f(n-1, \ldots)$.

The argument is essentially the same: we remove a single point from one of our classes and count all resulting permutations.  Then we subtract those in which our special point both maps onto and is mapped onto by other members of his class (the $(n-1)^{2}f(n-2, \ldots)$ term, which really is $(n-1)f(n-2, \ldots)$ where there is a transposition and $(n-1)(n-2)f(n-2, \ldots)$ where three distinct elements are involved) and those in which either our element maps onto another element of his class or is mapped onto by another element of his class, but not both ($(n-1) f(n-1, \ldots)$ each).

The values the original question asked us to find were $f(2, 2, \ldots, 2)$.

It seems that the only way to solve even the general case you sugest is to use [url=http://mathworld.wolfram.com/RookPolynomial.html]Rook Polynomials[/url].


x, y, z positive
$x+y+z = 3$
Prove that
\[\sqrt x+\sqrt y+\sqrt z \ge xy+yz+zx.\]

We have:
\[\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx \] \[\iff 2\left(\sqrt x+\sqrt y+\sqrt z\right)\ge 9-\left(x^{2}+y^{2}+z^{2}\right) \] \[x^{2}+y^{2}+z^{2}+2\left(\sqrt x+\sqrt y+\sqrt z\right)\ge 9 \]
Which is trivial because:
\[x^{2}+\sqrt x+\sqrt x\ge 3x \] \[y^{2}+\sqrt y+\sqrt y\ge 3y \] \[z^{2}+\sqrt z+\sqrt z\ge 3z \]

[quote="djuro"]x, y, z positive
$x+y+z = 3$
Prove that
\[\sqrt x+\sqrt y+\sqrt z \ge xy+yz+zx. \]
[/quote]

Prove that:
$\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\ge xy+yz+zx$ ?

[quote="NguyenDungTN"]
Prove that:
$\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\ge xy+yz+zx$ ?[/quote]

$\sqrt[3]{x}\geq \sqrt{x}$  etc. (for 0<=x<=1) so your ineq. follows from what was proven here.

[quote="kar"][quote="NguyenDungTN"]
Prove that:
$ \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\ge xy+yz+zx$ ?[/quote]

$ \sqrt[3]{x}\geq \sqrt{x}$  etc. (for 0<=x<=1) so your ineq. follows from what was proven here.[/quote]

And what if $ x\geq 1$? :P

[quote="NguyenDungTN"]
Prove that:
$ \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\ge xy+yz+zx$ ?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34428
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35839
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81983

:oops:  Sorry, my mistake. I though it's x+y+z = 1 and not 3.

[b]Djuro[/b] The origin of this problem is not Serbia as you wrote
It appeared once before in one Russian competition and Serbia just copied it.

Evo ti link [url]http://www.kalva.demon.co.uk/russian/rsoln/rsol0216.html[/url]


Let us choose arbitrarily $ n$ vertices of a regular $ 2n$-gon and colour them red. The remaining vertices are coloured blue. We arrange all red-red distances into a nondecreasing sequence and do the same with blue-blue distances. Prove that the two sequences hence obtained are identical.

[hide]Let the $ 2n$-gon be $ A_1A_2...A_{2n}$.  Consider all line segments $ A_iA_{i\plus{}k}$ for a fixed $ k$ and a variable $ i$.  These $ 2n$ line segments have equal length.  By virtue of the problem statement, half of these have $ A_i$ colored red, and the other half have $ A_i$ colored blue.  Suppose that of the $ n$ ones which have $ A_i$ colored red, $ t$ of them have $ A_{i\plus{}k}$ colored blue.  Then, there are $ n\minus{}t$ of them which have $ A_i$ and $ A_{i\plus{}k}$ colored red.  Also, we know that there are exactly $ t$ blue vertices $ A_{i\plus{}k}$ which have $ A_i$ colored red.  Thus, there are exactly $ n\minus{}t$ blue vertices $ A_{i\plus{}k}$ with $ A_i$ colored blue.  Now, we have shown that the number of $ A_iA_{i\plus{}k}$ with $ A_i$ and $ A_{i\plus{}k}$ both red is the same as the number of $ A_iA_{i\plus{}k}$ with both $ A_i$ and $ A_{i\plus{}k}$ colored blue.  Now it is obvious that the two nondecreasing sequences are equal due to the fact that the number of pairs of monochromatic vertices with given length $ l$ is constant no matter whether the color is red or blue. [/hide]


Let a, b real number [tex]a \ne b[/tex]. Show that
[tex]P(x) = (a - b)x^n + (a^2 - b^2)x^{n - 1} + ... (a^n - b^n)x + (a^{n + 1} - b^{n + 1} )[/tex] has at most one real root.


inequalities
logarithms
search
inequalities unsolved

I think Inequalities Solved Problems is a not proper section for this unsolved problem ;)
I will move it.

Since $b_1b_2...b_k\geq a_1a_2...a_k$ for every $1\leq k\leq n$
Using A.M.-G.M we have $\sum_{i=1}^k \frac{b_i}{a_i} \geq k$ for every $1\leq k\leq n$
$\displaystyle \sum_{i=1}^n b_i =\sum_{i=1}^{n-1} (a_i-a_{i+1})( \frac{b_1}{a_1}+...+\frac{b_i}{a_i}) + a_n( \frac{b_1}{a_1}+...+\frac{b_n}{a_n}) = A$
Given that   $a_{1}\geq a_{2}\geq ...\geq a_{n}>0$ we have $a_i-a_{i+1}\geq 0 $ and $ a_n\geq 0$
Therefore $\displaystyle A\geq \sum_{i=1}^{n-1} (a_i-a_{i+1})(k) + a_n(n)= \sum_{i=1}^n a_i$
Q.E.D.

may i know why we can assume them to be equal?

[quote="Myth"]Because we can always reduce $b_n$ in order to obtain the equality $b_1...b_n=a_1...a_n$. Note that sum $b_1+...+b_n$ will decrease ;)[/quote]

thx!
Last night i tried the same approach but i constructed a new $a_{n+1}$ and $b_{n+1}$ and then i got stuck... now everything's clear, thx!

Here is an other solution :
put $b_{i}=\lambda _{i}a_{i}$
then we have to prove that :$\left( \lambda _{1}-1\right) a_{1}+\left( \lambda _{2}-1\right) a_{2}+...+\left( \lambda _{n}-1\right) a_{n}\geq 0$ (1)
and we have :$\lambda _{1}\lambda _{2}...\lambda _{k}\geq 1$ for every $1\leq k\leq n$
put $S_{k}=\left( \lambda _{1}-1\right) +\left( \lambda _{2}-1\right) +...+\left( \lambda _{k}-1\right) $ , for $1\leq k\leq n$
we have
$\left( \lambda _{1}-1\right) a_{1}+\left( \lambda _{2}-1\right) a_{2}+...+\left( \lambda _{n}-1\right) a_{n}=S_{1}\left( a_{1}-a_{2}\right) +...+S_{n-1}\left( a_{n-1}-a_{n}\right) +S_{n}a_{n}$
to prove (1) we have to prove that : $S_{k}\geq 0$ for every $1\leq k\leq n$
we have
$S_{k}=\lambda _{1}+\lambda _{2}+...+\lambda _{k}-k\geq 0\Leftrightarrow \frac{\lambda _{1}+\lambda _{2}+...+\lambda _{k}}{k}\geq 1$
and since $\frac{\lambda _{1}+\lambda _{2}+...+\lambda _{k}}{k}\geq \root{k}\of{\lambda _{1}\lambda _{2}...\lambda _{k}}\geq 1$ ( AM $\geq $GM)
we are done

Oh yeeh I have forgotten

What is caramata's inequality ?  :)

But it is the same as [S]unn[Y]'s solution...  :?

Search forum for Karamata or Caramata. Or go directly to the Inequalities Theorems section.

[quote="indybar"][quote="Myth"]Very nice!

I just want to say that it is a Caramata's inequality for $f(x)=e^x$. Indeed, let $x_i=\ln b_i$ and $y_i=\ln a_i$. Then $x$ majorizes $y$ (we may WLOG assume $x_1+...+x_n=y_1+...+y_n$). Applying Caramata we obtain the result.[/quote]

How does this solution use  $b_{1}b_{2}...b_{k}\geq a_{1}a_{2}...a_{k}$?[/quote]

used to prove the majorization


Let $ a_1,a_2,....,a_n$ be the positive integer($ a_i<1000$ where 1≤i≤n).Knowing that $ lcm(a_i,a_j)>1000$ for all 1≤i,j≤n(i is not equal to j).Prove that :
$ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\frac{1}{a_3}\plus{}...\plus{}\frac{1}{a_n}<2$

clearly the number of multiples of $ a_i$ among $ 1,2,\dots, 1000$ is $ \left[\frac {1000}{a_i}\right]$, furthermore, neither of $ (a_i,a_j)$ have a common multiple among $ 1,2,\dots, 1000$.

So $ \sum_{i \equal{} 1}^{n}\left[\frac {1000}{a_i}\right]\leq 1000$.

Clearly, $ \left[\frac {1000}{a_i}\right] > \frac {500}{a_i}$ - this is because $ [x] > x \minus{} 1\geq\frac {x}{2}$, for all $ x\geq 2$ and if $ 1 < x < 2$ then $ 1 \equal{} [x] > \frac {x}{2}$ so

$ \sum_{i \equal{} 1}^{n}\frac {500}{a_i} < 1000$, or

$ \sum_{i \equal{} 1}^{n}\frac {1}{a_i} < 2$


Prove that $ \lfloor (1 \plus{} \sqrt {3})^{2n}\rfloor$ is odd for all positive integers $ n$, and that $ \lfloor (1 \plus{} \sqrt {3})^{2n\minus{}1}\rfloor$ is divisible by $ 2^n$ for all positive integers $ n$.

[hide="Part 1"]$ (1\plus{}\sqrt3)^2\equal{}4\plus{}2\sqrt3$, so $ \lfloor (1 \plus{} \sqrt {3})^{2n}\rfloor \equal{} \lfloor (4\plus{}2\sqrt3)^n\rfloor$
Now consider $ (4\plus{}2\sqrt3)^n \plus{}(4\minus{}2\sqrt3)^n$. This is equivalent to
$ \left( 4^n \plus{} \binom{n}{1} 4^{n\minus{}1} (2\sqrt3)^1 \plus{} \cdots \plus{} \binom{n}{n\minus{}1}4^1(2\sqrt3)^{n\minus{}1} \plus{} (2\sqrt3)^n \right)\plus{}\left( 4^n \minus{} \binom{n}{1} 4^{n\minus{}1} (2\sqrt3)^1 \plus{} \cdots \plus{} \binom{n}{n\minus{}1}4^1(2\sqrt3)^{n\minus{}1} \minus{} (2\sqrt3)^n \right)$
All of the terms with a square root cancel out, and the other terms are doubled, giving an even integer. However, we must still subtract $ (4\minus{}2\sqrt3)^n$ from this number. $ 4\minus{}2\sqrt3<1$, so $ (4\minus{}2\sqrt3)^n<1$ for $ n>0$. Therefore, $ (4\plus{}2\sqrt3)^n$ will be just below an even integer, and $ \lfloor (4\plus{}2\sqrt3)^n\rfloor$ will be odd.[/hide]

[hide="hint for a slick solution"]Hmmm... Consider the recursive sequence defined by F_0=F_1=2 and F_n=2F_(n-1)+2F_(n-2).  What can be said about the number of times 2 divides F_n based of of the recursion? What is the explicit form for F_n?[/hide]

[hide]Let $ t_k \equal{} (1 \plus{} \sqrt {3})^{k} \plus{} (1 \minus{} \sqrt {3})^{k}$. Using characteristic equation, we find the following recursive relation
$ t_k \equal{} 2t_{k \minus{} 1} \plus{} 2t_{k \minus{} 2}$
with $ t_0 \equal{} 2, t_1 \equal{} 2$.
Hence, $ t_k$ are always positive integers. Since $ \minus{} 1 < 1 \minus{} \sqrt {3} < 0$, $ \lfloor (1 \plus{} \sqrt {3})^{2n} \rfloor \equal{} \lfloor t_{2n} \minus{} (1 \minus{} \sqrt {3})^{2n} \rfloor \equal{} t_{2n} \minus{} 1$. Similarly $ \lfloor (1 \plus{} \sqrt {3})^{2n \minus{} 1} \rfloor \equal{} t_{2n \minus{} 1}$.

[b]Part 1:[/b] Because for $ n\geq 1$, $ t_{2n} \equal{} 2t_{2n \minus{} 1} \plus{} 2t_{2n \minus{} 2}$ is even, therefore $ \lfloor (1 \plus{} \sqrt {3})^{2n} \rfloor \equal{} t_{2n} \minus{} 1$ is odd. $ \blacksquare$

[b]Part 2:[/b] We'll proceed by indcution.
[u]Claim:[/u] $ 2^n|t_{2n},t_{2n \minus{} 1}$ for all positive integers $ n$
[u]Base:[/u] $ 2|t_1, t_2$.
[u]Inductive hypothesis:[/u] suppose the claim holds for $ n$, namely $ 2^n|t_{2n},t_{2n \minus{} 1}$. Then $ t_{2n \plus{} 1} \equal{} 2(t_{2n} \plus{} t_{2n \minus{} 1})$ is divisible by $ 2^{n \plus{} 1}$, and so is $ t_{2n \plus{} 2} \equal{} 2(t_{2n \plus{} 1} \plus{} t_{2n})$.

Therefore, $ \lfloor (1 \plus{} \sqrt {3})^{2n \minus{} 1} \rfloor \equal{} t_{2n \minus{} 1}$ is divisble by $ 2^n$. $ \blacksquare$[/hide]


[quote="dakyru"]this was posted in the problem solving marathon thread awhile back.[/quote]

  hopefully, no one remembers it.....

Anyways, I'm gonna try to solve this.

[hide="Solution"]
The parity of the numbers are the only things that matter, so the numbers can be replaced with 1, 0, 1, 0, etc. in the place of 1, 2, 3, 4, etc. up to 1994.

1 - 0 = 1
0 - 1 = 1
1 - 1 = 0
0 - 0 = 0

Subtracting two numbers will do one of the following things:

Replace two 1s with a 0
(An odd number minus an odd number is an even number.)

Replace two 0s with a single 0
(An even number minus an even number is also an even number.)

Take a 1 and a 0 and delete the 0, leaving only a 1
(An odd number minus an even number, or vice versa, leaves an odd number.)

There are 997 1s and 997 0s.

You can replace 996 1s with 498 0s. That leaves one 1 and a crapload of 0s. You can to replace all of the 0s with a single 0, leaving a 1 and a 0. You can delete the 0, leaving a single 1. The number 1 in mod 2 means an odd number, which pretty much closes up the problem.
[/hide]

You havent rigorously (is that the correct term?) solved the problem.  You've simply solved for one condition, you need to show the same result applies to the other conditions as well...

[hide="solution"]invariance principle:
the sum $S$of all numbers is $\frac{1994*1995}{2}$ which is odd.
now what remains invariant: the sum (modulo 2), because
you take the numbers $a,b$ with $a>b$ now $S \rightarrow S+(a-b)-a-b=S-2*b$
so $S$ remains odd until the end.
[/hide]

[quote="G-UNIT"]Heres a problem I really enjoyed from one of my Australian gr.8 math books.

  The numbers 1, 2, 3, . . ., 1994 are written on a blackboard.  Any two of these numbers are chosen, they are both erased, and their difference is written up instead.  This process is continued until only one number remains on the board.  Prove that the final number remaining is odd.

  In the rare event no one solves this, I'll post a few hints tommorow or sometime . . .[/quote]

I've seen this before.

[hide]
Invariance!

Let's look at the sum of the numbers. There are $1994/2=997$ even numbers and $n$ odd numbers, where $2n-1=1993 \implies n=997$. Thus the sum is $\text { even } + \text { odd } = \text { odd. }$ When we take 2 numbers and replace them by their difference, the sum doesn't change. If we have even - odd, we get odd, and even+odd=odd also, so it's still the same. We also have odd-odd=odd+odd (mod 2) and even+even=even-even (mod 2) so we can't change the parity of the sum by changing their difference; therefore the parity of the last number must be equal to the parity of the sum, so it must $\textbf { odd.}$
[/hide]


Does this mean that a function can take in numbers from Z and returns numbers from R?

f : Z -> R

(I think that's how the notation goes)

[quote]Can you think of another way of describing f: N -> R?[/quote]

Huh? What do you mean?

I think what he means is, what is another mathematical object that is essentially the same as a function f: N -> R.


In how many ways can 4 distinct beads be placed on a bracelet that has a knot in it?  (Beads cannot slide past the knot.)

Normally, if the bead can slide past the knot, there would be $ \frac{4!}{4\cdot2}\equal{}3$ different ways. This time, instead of dividing by 4, you divide by 3 because the knot stops the beads, so there are 4 ways.

I think the answer is 12. That's $ \frac{4!}{2}$.
4! for the possible arrangements of the beads, and 2 because flips are indistinguishable.


1. Susan, Bob, and Albert are painting a house.

Susan can paint it in 6 hours by herself, Bob 8 hours by himself, and Albert 12 hours by himself.

Susan and Albert work together for 1 hour and decide to take a break.

How long will it take Bob to finish the job by himself?

2. There are three couples at a circular table. What is the probability that no person sits next to his or her spouse?

1: In 1 hour, Susan paints $\frac {1}{6}$th of the house and Albert paints $\frac {1}{12}$th. So if they work together for an hour, they paint $\frac {1}{6} + \frac {1}{12} = \frac {1}{4}$th of the house in that hour. So $\frac {3}{4}$ of the house has not been painted. Bob paints $\frac {1}{8}$th of the house in one hour, so it takes him $\frac {\frac {3}{4}}{\frac {1}{8}} = \frac {3}{4} \times 8 = 6$ hours to finish painting the house by himself.

2.
there must be 3 person do not sit next to 1 person
so the probobility is $\displaystyle\frac{3}{6-1}=\frac{3}{5}$

Ah Hopitac, but what about the other 2 couples and their spouses (Using your line of reasoning - ).

Chess, what if one couple sits together and the other two are seperate?

Hint: Use PIE.

[hide="Solution"]If all the couples sit together the seating arrangement looks like this: ...AABBCC... So the probability is 2*2*2*2/5!=2/15

Next we find the chance that two couples sit together. That means the seating arrangement has to look something like this: ...AABCCB... Once we're given which couple is separate, there are only four ways that the other couples can sit. So the probablility that only couple X is seperate is 4/5!=1/30. (Divided by 5! because we're looking at it from the point of view of one member of couple X).

If only one couple sits together: ...BCAABC... The chance that only couple Y is intact is 2*2*2*2/120=16/120=2/15

So if Z is the probability that we're looking for: 1-Z=4/30+1/30+4/30=3/10, so Z=7/10.[/hide]

[quote="chess64"][quote="white_horse_king88"]2. There are three couples at a circular table. What is the probability that no person sits next to his or her spouse?[/quote]

[hide]We can just find the P that each couple sits together. The first can choose a seat in $6$ ways, the second can choose a seat in $2$ ways, etc.

$1-\frac{(6 \cdot 2)(4 \cdot 2)(2 \cdot 1)}{6!}=\frac{11}{15}$

EDIT: aah, i have a feeling it's wrong... :([/hide][/quote]

I think you're forgeting that the $\frac{11}{15}$ probability includes the chance that some couples are sitting next to each other and some may not be.  My reasoning:

There are 6 ways to choose a seat for the first person.  There are 3 ways to choose a seat for his spouse, but there are two cases.  One is if the spouse sits directly across from him, and the other case is if she sits one seat away.  If she sits directly across from him, there are 4 ways to seat the person from another couple.  Then there are 2 ways to seat his spouse, and 2 ways left to arrange the remaining couple.  Looking at the second case, if the spouse sits one seat away, there are 2 instances of this.  There are still 4 choices for the first person from the second couple, but only one choice for his spouse.  Two arrangements are left for the last couple.  So:

$P = \frac{6(4 \cdot 2 \cdot 2 + 2(4 \cdot 1 \cdot 2))}{6!} = \frac{4}{15}$

Coincidentally, it seems the chance for no couples sitting next to each other is the same as the probability for all couples sitting next to each other.

Hmm, I don't know what's wrong then.  There's not 3 ways to seat the spouse because 2 of the ways leave the remaining couple sitting next to each other.  If you go around and place the person in each of the other 3 seats, you'll find there's only 1 corresponding seat for the spouse that still leaves arrangements for the last couple to remain separate.


What is the coefficient of restitution? I do not understand the explanation in [hide="this"]:
Some balls bounce higher than others. In most ball sports the bounce height is specified by the rules of the game for a given drop height onto a given surface. In tennis for example, an approved ball dropped from a height of 100 in (2.54 m) onto a slab of concrete must bounce to a height between 53 and 58 inches when tested at a temperature of 21 C. If it bounces higher or lower than that, players will probably complain that there is something wrong with the ball.

100 inches is just out of reach of most people. If you want to test a ball yourself, there are two perfectly acceptable alternatives (although not officially approved). One is to drop the ball from a smaller height, say 80 or 90 inches.  Another is to throw the ball up to about 100 inches and let it fall from whatever height it reaches. In either case, the best way of measuring the drop and bounce heights is to film the bounce with a video camera and then measure the heights from the film. If D  is the drop height and B is the bounce height then the ratio B/D must be between 53/100 = 0.53 and 58/100 = 0.58 for a tennis ball.
In practice, B/D for any given ball varies only slightly with the drop height D. If B/D = 0.55 say for a ball dropped from 100 in, then B/D will be about 0.56 when dropped from a height of 80 in, or 0.54 when dropped from a height of 120 in.

The coefficient of restitution (COR) is defined for tests like these as (rebound speed)/(incident speed). Since fall and bounce heights are proportional to the speed squared, the formula is

      		      COR = Square Root of (B/D)

If B/D = 0.56 then COR = 0.75. The COR drops slightly as the incident speed increases[/hide]
I found it [url=http://www.physics.usyd.edu.au/~cross/BouncingBalls.ppt]here[/url]
What does the COR have to do with and what is it's use?

The coefficient of restitution is a number that determines how high the ball will bounce back up if you drop it. It's like it's bounciness. The closer the coefficient is to 1, the higher it bounces.


Solve for $ x: x\plus{}\frac{1}{x} \le \minus{}2$

[hide="Solution"]
I believe it holds for $ x \le \minus{}2$
But $ x\plus{}\frac{1}{x}\equal{} \frac{x^2 \plus{} 1}{x} \le \minus{}2$
$ x^2 \plus{} 1 \le \minus{}2x$
$ x^2 \plus{} 2x \plus{} 1\equal{} (x\plus{}1)^2 \le 0$

Thus x can only equal -1. But $ \minus{}3\plus{}\frac{1}{\minus{}3} \le \minus{}2$
So I am now confused on what I did that was illegal.[/hide]

You multiplied by $ x$ without considering its sign.  If $ x$ is negative, then you have to flip the sign of the inequality, which gives you the right answer ($ x < 0$).

Would the answer not be x is less than or equal to -1?

[hide]By adding 2 to both sides then squaring to get (x + 1/x + 2)^2
(so that it shouldn't matter whether x is negative, I could be wrong)

you get (x^4 + 4x^3 + 6x^2 + 4x + 1)  less than or equal to 0

factorises to (x+1)^4 less than or equal to 0[/hide]

[quote="PLANETX1970"]By adding 2 to both sides then squaring[/quote]

Squaring an inequality where one or both sides are negative is invalid, and the inequality $ (x \plus{} 1)^4 \le 0$ has only the solution $ x \equal{} \minus{} 1$.

[hide="A correct solution"] When $ x > 0$, both $ x$ and $ \frac {1}{x}$ will be positive, so there are no solutions.

When $ x < 0$, both $ x$ and $ \frac {1}{x}$ are negative.  Multiplying by $ x$, we obtain

$ x^2 \plus{} 1 \ge \minus{} 2x \Leftrightarrow$
$ x^2 \plus{} 2x \plus{} 1 \ge 0 \Leftrightarrow$
$ (x \plus{} 1)^2 \ge 0$

which is true for all $ \boxed{ x < 0 }$. [/hide]
Intuition:  If $ x$ satisfies the inequality, then so does $ \frac{1}{x}$ because $ x \plus{} \frac{1}{x} \equal{} \frac{1}{x} \plus{} x$.  For example, $ x \equal{} \minus{}2$ satisfies the inequality, and therefore so does $ x \equal{} \minus{} \frac{1}{2}$.

the right answer is x<0, you must consider all values of x when you solve the inequation.  :)

How can be $ (x\plus{}1)^2<0$     It can only be $ (x\plus{}1)^2\equal{}0$ for $ x\equal{}\minus{}1$
$ x\equal{}\minus{}1$ is the answer of the inquality $ (x\plus{}1)^2\equal{}0$ and not for $ (x\plus{}1)^2(<\equal{})0$   :lol:


quadratics
function
real analysis
real analysis theorems

Why is there no maximum principle for a hyperbolic PDE? An Explanation or a pointer will be appreciated.

There are several ways to look at this, depending on how "maximum principle" is understood.

1) for hyperbolic PDE we also prescribe velocity at $ t \equal{} 0$. If the velocity is nonzero, then $ \sup_x u(x,t)$ may increase with $ t$ for a while, even if $ u(\cdot, 0)\equiv 0$.

2) If the initial velocity is $ 0$ and you have only one spacial dimension, then $ \sup_x u(x,t)\le \sup_x u(x,0)$, which can be interpreted as maximum principle. This is because
$ u(x,t) \equal{} \frac12 h(x \minus{} ct) \plus{} \frac12 h(x \plus{} ct)$ where $ h(x) \equal{} u(x,0)$ - d'Alembert's formula.

3) With $ n > 1$ spacial dimensions one also has such a formula for solutions of the wave equation (see, e.g. the book "PDE" by Evans). But you will find that they don't give you the same conclusion as in $ n \equal{} 1$. For an intuitive explanation, start with a circular wave supported near the set $ \{|x| \equal{} 1\}$. When the wave reaches the origin, it will spike up considerably, due to focusing. (You can try this at home, just don't use sulfuric acid or things like that...)

If you're talking about maximum principles, it makes sense to put both dimensions on the same footing; instead of space and time, we have two spatial dimensions.

Now, we have a linear differential equation $ Q\left(\frac{d}{dx},\frac{d}{dy},\cdots\right)\equal{}0$, where $ Q$ is a nonzero (symmetric) quadratic form. If $ Q$ is positive definite (or negative definite), we call the operator/equation elliptic. If $ Q$ is semidefinite (singular, but the rest of the eigenvalues have the same sign), we call it parabolic. If $ Q$ has eigenvalues of both signs, we call it hyperbolic.

So, where do maximum principles come from? If $ f$ has a local minimum at $ a$, its Hessian $ Hf(a)$ is positive semidefinite, and $ Q\left(\frac{d}{dx},\frac{d}{dy},\cdots\right)\equal{}\text{tr}(Q\cdot Hf(a))$. Now, if $ A$ is positive definite and $ B$ is positive semidefinite, $ \text{tr}(AB)\equal{}\text{tr}(\sqrt{A}\cdot B\cdot\sqrt{A})\equal{}\text{tr}(\sqrt{A}\cdot\sqrt{B}\cdot\sqrt{B}^T\cdot\sqrt{A}^T)\equal{}\text{tr}(CC^T)>0$
So, if we have an elliptic operator evaluated at an extremum, we can't get zero unless everything's flat. With hyperbolic operators, there's a dimension that goes against the grain, where we can do something different, and the above argument goes nowhere. The maximum/minimum principle is unusual, a special property of positive definite forms.

In practice, it's easy to find functions with local extrema that satisfy any given hyperbolic equation. For the simplest case $ \left(\frac{\partial^2}{\partial x^2}\minus{}\frac{\partial^2}{\partial y^2}\right)f\equal{}0$, $ x^2\plus{}y^2$ is a solution with a minimum at zero. Just take a linear combination of squares with different signs of $ Q$.

Thank you all. I will read Protter and Weinberger's book on the topic as someone suggested that it is a good book on the subject.

[quote="TZF"]The non-dispersive wave equation is an example of a hyperbolic PDE.[/quote]

What is this supposed to mean?!


algebra
polynomial
geometry
function
advanced fields
advanced fields unsolved

Hello.
The set $ (x,x) \in \mathbb{R}$ \ $ {(1,1)}$ is not an affine variety. Can anyone prove this?

[quote="Nukular"]But it is. Take (x,x) and map it to (x-1, 1/(x-1)); this gives an isomorphism between itself and V(xy-1) in R^2. (The inverse morphism is given by (x,y) -> (x+1,x+1)).[/quote]
Are you SUre?
which polynomial does it belong to?
the book of David Cox : An introduction  to Computational  Algebraic Geometry and COmmutative algebra , page 13, problem 8 says it is not!

OK, looking at Cox's book, it seems to be asking to show that your set S is not V(f_1, ..., f_n) for some polynomials f_1 thru f_n. I think this is different from saying it is not an affine variety; but this is more a question of semantics.

However, yes, if f(x,x) = 0 for all x different from 1, then certainly f(1,1) = 0, for then g(x) = f(x,x) has infinitely many roots, and so must be identically zero. Therefore S is not V(f_1, ..., f_n) for any polynomials.

However, this same proof would show that R \ {0} is not affine, even though it is isomorphic to an affine variety as a space with functions.

[quote="Nukular"]But it is. Take (x,x) and map it to (x-1, 1/(x-1)); this gives an isomorphism between itself and V(xy-1) in R^2.[/quote]
This is a projective morphism, not an affine one.  Affine isomorphism is a stronger condition than projective isomorphism, which is in turn a stronger condition than isomorphism of function fields (birational equivalence).

How is it not an affine isomorphism? Regular functions pull back to regular functions, and its continuous. (Sorry for being a pest!)

More directly, is it not isomorphic as a sheaf of rings to R \ {0}, via (x,x) -> (x-1)? That certainly is continuous and regular functions pull back.

It looks like we're just arguing semantics here, but I understand an affine isomorphism to be a polynomial map, which that clearly isn't.  In any case, I agree with your interpretation of the intent of this question.

Ah OK. It seems like the vocabulary + notation aren't well agreed upon for some reason; it's like every introductory text has it's own terminology.


algebra
polynomial
Vieta
quadratics
algebra unsolved

If A , B $ \in \mathbb{R} \sim \mathbb{Q}$ . s.t A+B $\in \mathbb{Q} $ and AB $\in \mathbb{Q} $,

Can We claim that A , B are of the form $ x \pm \sqrt {y} \ where\ x,y \in \mathbb{Q} $

Well, by Vieta they are real roots of a rational quadratic equation. So it's clearly true.

Yes. Aren't A and B the roots of a quadratic equation with rational coefficients?

CAn we prove this absolutely analtically. Are you sure that quadratic eqn proof is watertight

It's really obvious that it's 'watertight' (check it yourself, please!) ...
But what do you understand when speaking of an 'absolutly analytical prove'¿ (this is the forum for Algebra...  ;)  , but iff you find an even simplier prove then tell us)


Factor:
\[ 2x^2 \minus{}7xy \plus{}3y^2 \plus{} 9x \minus{}7y \plus{}4\]

I'm really bewildered and don't know how to approach these types of problems.

Hints and insights would be appreciated!

[hide="Hint"]$ 2x^2 \minus{}7xy \plus{}3y^2 \plus{} 9x \minus{}7y \plus{}4 \equal{} 2x^2 \plus{} (\minus{}7y\plus{}9)x \plus{} (3y^2\minus{}7y\plus{}4)$[/hide]

I suppose you could try this. Let
\[ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4\ \equal{} \ (ax \plus{} by \plus{} c)\left(\frac2ax \plus{} \frac3by \plus{} \frac4c\right)\]
Then compare the coefficients of $ x,$ $ y$ and $ xy$ and solve for $ a,$ $ b$ and $ c.$

[hide]I got $ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4 \equal{} (2x \minus{} y \plus{} 1)(x \minus{} 3y \plus{} 4).$[/hide]

[hide="Hint"]In general, when asked to factor something, you can always guess what the factors will look like.

In particular for this problem, $ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4 \equal{} (ax \plus{} by \plus{} c)(dx \plus{} ey \plus{} f)$.
[hide="Hint"]Now you can expand and solve the resulting system of equations, or choose convenient values for $ x$ and $ y$, since you know this must be satisfied for all $ x,y$.

[hide="Solution"]Well, let $ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4 \equal{} (2x \plus{} by \plus{} c)(x \plus{} ey \plus{} f)$.

When $ y \equal{} 0$, this is equivalent to

$ 2x^2 \plus{} 9x \plus{} 4 \equal{} (2x \plus{} 1)(x \plus{} 4) \equal{} (2x \plus{} c)(x \plus{} e)$.

So now this is $ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4 \equal{} (2x \plus{} by \plus{} 1)(x \plus{} dy \plus{} 4)$.

Let $ x \equal{} 0$ to get

$ 3y^2 \minus{} 7y \plus{} 4 \equal{} (3y \minus{} 4)(y \minus{} 1) \equal{} ( \minus{} y \plus{} 1)( \minus{} 3y \plus{} 4) \equal{} (by \plus{} 1)(dy \plus{} 4)$, so we're done:

$ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4 \equal{} \boxed{(2x \minus{} y \plus{} 1)(x \minus{} 3y \plus{} 4)}$.[/hide][/hide][/hide]

[b]Edit:[/b] Thanks. :)

math154,

in your first hint you have
[quote]
In general, when asked to factor something, you can always guess what the factors will look like.

In particular for this problem, $ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4 \equal{} (ax \plus{} by \plus{} c)(d \plus{} ey \plus{} f)$.
[/quote]

But it would be $ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4 \equal{} (ax \plus{} by \plus{} c)(dx \plus{} ey \plus{} f)$.

Also, you should be consistent with the variables.  What you
started calling "e" and "f" in the first hint, you labeled as "d" and "f,"
respectively, when you referred to them in the third hint.

[quote="Kouichi Nakagawa"][hide="Hint"]$ 2x^2 \minus{} 7xy \plus{} 3y^2 \plus{} 9x \minus{} 7y \plus{} 4 \equal{} 2x^2 \plus{} ( \minus{} 7y \plus{} 9)x \plus{} (3y^2 \minus{} 7y \plus{} 4)$[/hide][/quote]
[hide="Second Hint"]$ 3y^2 \minus{} 7y \plus{} 4 \equal{} (3y\minus{}4)(y\minus{}1)$[/hide]

We can use cross-multiply method.

[quote="Arrange your tan"]

Also, you should be consistent with the variables.  What you
started calling "e" and "f" in the first hint, you labeled as "d" and "f,"
respectively, when you referred to them in the third hint.[/quote]

Firstly, you should at least say the correct thing if you're going to go through and nitpick every post; he labeled them as $ d$ and $ e$ respectively, not $ d$ and $ f$.

Secondly, that's because he didn't need to find the coefficient of $ x$ in the factorization, so he just shifted the variables over. I am fairly sure that anyone who understands what math154's first hint is suggesting would not be confused by the variable shifting.

Also, people make typoes. You probably should not point out everything, unless it's extremely confusing.

Sorry for the relatively off topic post, but I was feeling too lazy to pm. Rate me as you wish.


inequalities
LaTeX
inequalities proposed

Prove that if $ a_i>\frac{1}{n^2} $ and $ a_1+...+a_n=1 $ then we have the inequality $ \frac{n^2\cdot a_1-1}{1-a_1}\cdot\frac{n^2\cdot a_2-1}{1-a_2}\cdot...\cdot\frac{n^2\cdot a_n-1}{1-a_n}\leq n^n $

this one isnt that hard (sorry i dunt noe latex)
let b(k) = a(k) - 1/n^2.
b(1) + b(2) ... + b(n) = (n-1)/n.
we want to show that:
product of n^2*b(k) / (n^2-1)/n^2 - b(k) from k = 1 to n <= n^n.
after some algebra, it comes down to showing
product of (k/b(k) - 1) form k = 1 to n >= n^n.
this follows from the fact that (k/a1 - 1) (k/a2 - 1) ... (k/a(n) - 1) >= t^n where
t=k/(1/n)*(a1+a2...) - 1

Sincerely, I did not understand anything you have written there.

Anyway, your idea leads to a very nice solution:
  The inequality becomes $ [b_2+...+b_n+\frac{b_1+...+b_n}{n}]\cdot...\cdot[b_1+...+b_{n-1}+\frac{b_1+...+b_n}{n}]\geq n^nb_1\cdot...\cdot b_n $. This is quite clear. Take $ b_1\cdot...\cdot b_n=1 $ ( I renounced to the condition $ b_1+...+b_n=\frac{n-1}{n} $ ) and apply AM-GM twice for any right paranthesys!

here is my solution :
so
((na2-1).....(nan-1))^(1/(n-1))<=(n(1-a1)-(n-1))/(n-1)=(n/n-1))(1-a1)-1 let's note n/(n-1)=x
do the left of the ineq is <=   Product(x-(1/(1-ai)))<=(a-(sum(1/(1-ai)))/n)^n
which is <= (a-n/(n-1))^n  ("cause sum (1/(1-ai))>=n/sum(1-ai)")
so
a-n/(n-1)=n

i hope ut's clear cause really it's very hard to write it


Let $P(x),Q(x),R(x)$ be polynomials with rational coefficients. $P(x)$ has an odd degree and is irreducible. $P(x)$ divides $Q(x)^{2}+Q(x)R(x)+R(x)^{2}$. Show that $P(x)^{2}$ divides $Q(x)^{2}+Q(x)R(x)+R(x)^{2}$.

It's long ago that I've seen a more striking example that someone didn't even try to look if this was already posted before!

And no, I will not give any link, and I hope noone else does.

$P(x)$ has a real root $\lambda\in\mathbb{R}$, as $\deg P$ odd.
Then
(*) $Q(\lambda)^{2}+Q(\lambda)R(\lambda)+R(\lambda)^{2}=0$
and if $R(\lambda)\not =0$, then $\left({Q(\lambda)\over R(\lambda)}\right)^{2}+{Q(\lambda)\over R(\lambda)}+1=0$ and ${Q(\lambda)\over R(\lambda)}={-1\pm\sqrt{3}i\over 2}\not\in\mathbb{R}$. Contradiction!
So $R(\lambda)=0$ and (*) gives $Q(\lambda)=0$.
Then, as $P,Q,R\in\mathbb{Q}[x]$ with $P(x)$ irreducible, we get
$P(x)|Q(x)$ and $P(x)|R(x)$ and $P(x)^{2}|Q(x)^{2}+Q(x)R(x)+R(x)^{2}$.

[quote="olorin"]
Then, as $P,Q,R\in\mathbb{Q}[x]$ with $P(x)$ irreducible, we get
$P(x)|Q(x)$ [/quote]

How does this come? What about the complex roots of $P(x)$?

Concrete

http://www.mathlinks.ro/Forum/viewtopic.php?t=138609
 or
http://www.mathlinks.ro/Forum/viewtopic.php?p=317877#p317877


This may be a very simple problem, but I still need a bit of help and more explaining.

The world's largest ball of string is about 2 m in radius. To the nearest order of magnitude, what is the total length $ L$ of the string in the ball?

[hide="Calculations"] Let us assume the ball is spherical with radius $ R \equal{} 2 m$ . The string in the ball is not closely packed (there are uncountable gaps between adjacent sections of string). To all for these gaps, let us somewhat overestimate the cross-sectional area of the string by assuming the cross section is square, with an edge length $ d \equal{} 4 mm$. Then, with a cross-sectional area of $ d^2$ and a length $ L$, the string occupies a total volume of
V= (cross-sectional area)(length) = $ d^2L$.
This is approximately equal to the volume of the ball, given by $ \frac{4}{3}$ $ \pi$$ R^3$, which is about $ 4R^3$ because $ \pi$ is about $ 3$.
Thus, we have
$ d^2L \equal{} 4R^3$,

or $ L \equal{} \frac{4R^3}{d^2} \equal{} \frac{4(2m)^3}{(4 \times 10^{\minus{}3}m)^2}$

     $ \equal{} 2 \times 10^6 m \approx 10^6 m \equal{} 10^3 km.$     (Answer)

(Note that you do not need a calculator for such a simplified calculation) To the nearest order of magnitude, the ball contains about 1000 km of string...

[/hide]

[b]Questions[/b] :
1. What exactly is a cross-sectional area? (My dad has explained it and I've looked it up, but I am still not [b]really[/b] sure as to what it is)

2. I am also not sure how they get that the string occupies  (cross-sectional area) (length) = $ d^2L$
    Did they get that because the string goes around the cross-sectional area and is not really inside it?  and what is the length L in it?

confused... :noo:

Kind of like this, i guess : [url]http://www1.istockphoto.com/file_thumbview_approve/1837318/2/istockphoto_1837318_ball_of_string.jpg[/url]

Ah.. i just realized what the cross-sectional area is... (i think)  :)


Hello the question I have is as follows:

A user for a secure internet service has a 20 digit key phrase. The key phrase is known by the user only and each digit in the key phrase has a value 0-9. When the user attempts to log onto the secure service he/she is asked for 3 random digits from the keyphrase, for example digit 1, digit 2 and digit 3. After they have entered the 3 numbers correctly they are logged into the secure service.

Now there is a keylogger, which observes the data being input, it logs the digit numbers and their values. It will do this each time the same user attempts the enter the secure site.

So how many observations would the keylogger have to make before it has a 50% sucess rate of logging into the secure service?

I'm assuming this problem is more complicated than I initially thought. I haven't done probability in ages, so I'm sort of stuck. Can anyone help or give me some pointers.

Thanks.

[hide="hint"]$ 50\%\equal{}\frac{number\ of\ 3\minus{}digits\ combinations\ he\ has}{total\ amount\ of\ possible\ 3\minus{}digits\ combinations\ out\ of\ 20\minus{}digits}$ [/hide]


there are 64 , ( 1*1 ) squares in a chess board
you put a coin in the first square
you 2 coins in the second square
4coins in the third
8coins in the fourth
( going in Binary sequence )

how many coins will there be on the board if you count all of them?

[hide]The number of coins on the squares is $2^0, 2^1, ... 2^{63}.$  The total number of grains $S$ is given by:

$S = 2^0 + 2^1 + ... + 2^{63}.$

Multiplying both sides by 2 and then subtracting the former equation from the new one (so that a large number of terms cancels out), we get an expression for the final sum:

$2S = 2^1 + 2^2 + ... + 2^{64}.$
$2S - S = (2^1 + 2^2 + ... + 2^{64}) - (2^0 + 2^1 + ... + 2^{63}).$
[b]$S = 2^{64} - 1$.[/b][/hide]

[hide="my half answer"]
the equation would look like this:
$x=2^0+2^1+2^2+2^3+2^4+ \cdots +2^63$
I don't know how to solve this though[/hide]

well we know how it works, but we are looking for a neat way of solving this :P

LuCy4EvA, Barnacle, and math92 are correct
shinwoo what do you mean a nice way of solving it? 3 solvers already posted simple solutions, i wonder what can be simpler than that...  :roll:  ;)

well done guys

and lucy, have you seen this problem before? it;s really rare for people to find such patterns easily and solve it using it

so nice job :10:

( I actually only needed genral concept or patterns ^.^ )

ive never seen the exact problem, but i guess i got freaked out at the beginning about how to do the problem, then figured that there's probably a short cut, then remembered something i learned about the powers of 2.... guess that's how i got it...

[quote="shinwoo"]well done guys
...
( I actually only needed genral concept or patterns ^.^ )[/quote]
Is it really that hard to find that [hide]$a+ar+ar^2+...+ar^n=\frac{a(r^{n+1}-1)}{r-1}$ not its not[/hide]

[b][color=red]<Note from moderator RCV:  Yes!  It is hard!  This forum is for *beginning* contest-level mathematics.  Your formula would be suitable for the MATHCOUNTS Forum.>[/color][/b]


Do all gold winners of USAMTS get invitations to AIME tests?

I heard qualifying AIME through USAMTS is a "worse way" than through AMC.  Why is that?

I guess this question may be up many times so if you can provide a link to another post, that would be fine. (If you can answer, that would be better)

Thanks

[quote="tekno10m"]Go here

http://www.artofproblemsolving.com/USAMTS/AoPS_U_FAQ.php#aime

and

http://www.artofproblemsolving.com/USAMTS/AoPS_U_AbPrizes.php[/quote]

Thanks, but I actually read those before.

The second page says those who did "well" up to 3rd round would be invited.  How well is well?

Also, I heard
if I qualify through USAMTS, the score is equivalent to 100 on AMC when it is calculated for USAMO.  Can this be verified?

[quote="JSRosen3"]If you qualify for AIME through USAMTS, you get the equivalent of 100 points on the AMC. Therefore, it's much more beneficial to qualify through the AMC contests.[/quote]

That is what I heard.  So I wanna take AMC but my school doen't sponsor that test.  It really makes me feel depressed. (My previous high school used to)

Ypu could try to get the school to order the test, or take it at another school.

[quote="Ultimatemathgeek"]

The second page says those who did "well" up to 3rd round would be invited.  How well is well?
[/quote]

I believe in past years, the cutoff was 68/75 through the first 3 rounds.  The cutoff hasn't been set for Year 16 of the USAMTS, but it likely will be around there.


i just wanted to know because i am kinda into sports.

I like playing basketball, but I'm not very good at it.

good choice furious but for 99.9% of the people on this site like math as a sport but i have decided that math doesnt count because its almost given

soccer and table tennis. I can't stand watching sports.

Tennis.

It's the only sport I can actually watch on TV without going crazy.

[quote="mwahahahaha"]hmm well there's a lot that are quite fun
personally...i like full-contact tackle basketball, full-contact team tackle baseball, full-contact tackle football, full-contact tackle ultimate frisbee, full-contact tackle soccer...

but in the world of internationally recognized sports...
tennis, table tennis, and prolly rugby cuz that's full-contact tackle  :D[/quote] i like playing table tennis  but i dont like watching tennis it gets boring the ball goes back and forth. i think mwahahahaha just like hurting people who plays full contact baseball anyway

I like table tennis (whenever I get a chance, I don't have a table), tennis, and basketball.  Actually I think I'm pretty good at basketball considering I'm only like 5'3''.

Hehe, I like full contact Football and Soccer (full contact and non).

tennis is so boring to watch that it gets annoying especially with the funny sounds

If table tennis is a sport, it's my favorite. No table though :(

For playing, basketball.  For watching, hockey.

yeah i just play table tennis at taiwanese school

i love tennis, soccer, motorsport, and cycling.

Mine is swimming. I could watch and swim all day. :D

I play tennis, but I don't really like watching it. All it is is a ball going back and forth.


Census

A census-taker knocks on a door, and asks the woman inside how many children she has and how old they are.

“I have three daughters, their ages are whole numbers, and the product of the ages is 36,” says the mother.

“That’s not enough information,” responds the census-taker.

“I’d tell you the sum of their ages, but you’d still be stumped.”

“I wish you’d tell me something more.”

“Okay, my oldest daughter Annie likes dogs.”

What are the ages of the three daughters?

[hide="solution"]
Ages are 2, 2, 9
The triples of ages that multiply to 36 are (1,1,36), (1,2,18), (1,3,12), (1,4,9), (1,6,6), (2,2,9), (2,3,6), and (3,3,4).  The sums of these triples are 38, 21, 16, 14, 13, 13, 11, and 10.  Since the sum of the ages is not enough information to determine the ages, the sum must be 13.  Their ages are either (1,6,6) or (2,2,9), but the last bit of information shows that there must be an oldest daughter, so the only possibility is (2,2,9).
[/hide]
This problem also appears in the beginning of ACoPS.

Why the sum must be $ 13$ because its come in the middle?

The sum has to be 13 because that's the only sum that appears more than once.  If the sum was anything else, then the woman could've just told the census taker the sum and he could've figured out their ages from that.


The letters of the alphabet are each assigned a random integer value, and $ H\equal{}10$. The value of a word comes from the sum of its letters' values. If MATH is 35 points, TEAM is 42 points and MEET is 38 points, what is the value of A?

First, we can reverse the digits of TEAM to MEAT, and subtract MEET from it. We have A-E=4. Then we plug in 10 for H in Math, and subtract MAT=25 from MEET=38. We have 2E-A=13. We add the two equations, so E=17. Therefore, A=21


function
integration
calculus
calculus computations

1. Let $f(x)=\int^{x}_{1}\sqrt[3]{1+t^{2}}dt$

Find $\frac{d}{dx}\left[f^{-1}(x)\right]\mid _{x=0}$.

2. Given points A(2,1) and B(5,4), find the point P in the interval [2,5] on the x-axis that maximizes angle APB.
This problem looks easy, but the method I used turned out to be extremely ugly.

[quote="236factorial"]1. Let $f(x)=\int^{x}_{1}\sqrt[3]{1+t^{2}}dt$

Find $\frac{d}{dx}\left[f^{-1}(x)\right]\mid_{x=0}$.

2. Given points A(2,1) and B(5,4), find the point P in the interval [2,5] on the x-axis that maximizes angle APB.
This problem looks easy, but the method I used turned out to be extremely ugly.[/quote]

1.
Denote $g(x)=f^{-1}(x)$.
$f(g(x))=x$, thus $f'(g(x))\times{g'(x)}=1$.
And $f'(x)=\sqrt[3]{1+x^{2}}$, $g(0)=1$, so $f'(g(0))=\sqrt[3]{2}$.
Therefore $g'(0)=\frac{1}{f'(g(0))}=\frac{1}{\sqrt[3]{2}}$.

[quote="MysticTerminator"]2. Look for a tangent circle or something; no calculus...[/quote]

tangent circle? I finally did figure it out with calculus, although there was more algebra involved than calculus....  :maybe:


Given $ a,k\in\mathbb{N}$, prove that the expression \[ \ \frac{a^{4k}\minus{}1}{a\plus{}1} \] can [u]never[/u] be a perfect square.

If $ k\equal{}2^mr$, were $ r$ - odd, then \[ \frac{a^{4k}\minus{}1}{a\plus{}1}\equal{}(a\minus{}1)\frac{a^{2r}\minus{}1}{a^2\minus{}1}\prod_{i\equal{}1}^{m\plus{}1}(a^{r2^i}\plus{}1).\]
If it is square $ a^{2r}\plus{}1$ must be square - contradition.

Can you please explain your solution a bit more. Why does $ a^{2r}\plus{}1$ need to be a perfect square?  :huh:

Thanks!

If $ a^{2r} \plus{} 1 \equal{} x^2$, then $ (a^r \minus{} x)(a^r \plus{} x) \equal{} 1$- contradition.
If $ x^2\equal{}\prod q_i$, were $ (q_i,q_j)\equal{}1$, then $ q_i\equal{}y_i^2$.


Gauss
function
quadratics
logarithms
algebra
polynomial
number theory

Hi,

I have three related problems, which I don't know how to approach.  They involve classical Gauss sums, but in my number theory class, we skipped over the section that dealt with these.  I would really appreciate any help and guidance.

1.  Let $ g(\chi)$ be the classical Gauss sum $ \sum_{x\equal{}1}^{p\minus{}1} \chi(x) \zeta^x$, $ \chi$ the Legendre symbol ($ \chi(x)\equal{}0$ if $ p \mid x$, $ \chi(x)\equal{}\minus{}1$ if $ x$ is not a quadratic residue mod $ p$, and $ \chi(x)\equal{}1$ if $ x$ is a quadratic residue mod $ p$), and $ \zeta\equal{}e^{2 \pi i/p}$, for $ p$ a prime.  Define $ P\equal{}\prod(1\minus{}\zeta^n) \prod(1\minus{}\zeta^r)^{\minus{}1}$ where $ n,r$ run over respectively the nonsquares and squares modulo $ p$.  Show that $ P\equal{}\text{exp}(g(\chi)L(1,\chi))$ where $ L(1, \chi)\equal{}\sum_{n\equal{}1}^{\infty} \chi(n)n^{\minus{}1}$ is the Dirichlet $ L$-function associated to $ \chi$.

2.  Using the previous exercise, compute $ L(1, \chi)$ where $ \chi$ is the nontrivial quadratic character modulo $ 5$.

3.  The notation being as in the previous exercises, show that $ P \neq 1$ (and thus $ L(1, \chi) \neq 0$) as follows.  Choose $ C$ a nonsquare modulo $ p$.  Prove that $ P\equal{}1$ implies $ 1\plus{}x\plus{} \cdots \plus{}x^{p\minus{}1} \mid \prod \left( \frac{1\minus{}x^{Cr}}{1\minus{}x^r} \right)\minus{}1$.  Obtain a contradiction by specializing $ x$.

I really don't even know where to begin with these, so please help!

Try asking one question at a time and not getting impatient when your question isn't answered within a few hours.

Well, I thought it appropriate to post all three since they were part of one problem.  As for becoming impatient, how could I not be when it seems that everyone on this site purports to have such vast knowledge in mathematics, that it would take mere minutes to answer my question?  On the other hand, I spent hours on these and got nowhere.

[quote="musicmantribalx"] it seems that everyone on this site purports to have such vast knowledge in mathematics, that it would take mere minutes to answer my question?[/quote]
Sorry, who exactly purports this?

[quote="musicmantribalx"]Well, I thought it appropriate to post all three since they were part of one problem.[/quote]
Try to put yourself in the shoes of the person who you hope to help you.  Reading long questions takes time, which not everybody has at the particular moment you want the question answered.  I went to read in bed before you posted your question, and you were complaining by the time I woke up.

1) I'll spend some more time later on this one if nobody posts a solution.  It looks like expanding out $ \ln P$ as a power series in $ \zeta$ is the way to go, though there are a couple of complications.

2) is a simple computation once you assume the result of 1), since $ g(\chi)$ and $ P$ are easy to compute explicitly. (depending on what kind of answer you're looking for, you may need to solve explicitly for $ \zeta$, which you can do by substituting $ \tau \equal{} \zeta \plus{} \zeta^{ \minus{} 1}$ in $ \zeta^2 \plus{} \zeta \plus{} 1 \plus{} \zeta^{ \minus{} 1} \plus{} \zeta^{ \minus{} 2} \equal{} 0$.)

3) Notice that multiplication by $ C$ sends squares to non-squares and vice-versa, so that $ \prod_r \left( \frac {1 \minus{} x^{Cr}}{1 \minus{} x^r} \right) \minus{} 1$ is $ 0$ when $ x \equal{} \zeta$. (I am simply rewriting the statement $ P \equal{} 1$ here)  Since it is clearly a polynomial, this means that it must be divisible by the minimal polynomial for $ \zeta$, which is $ 1 \plus{} x \plus{} \cdots \plus{} x^{p \minus{} 1}$.

Finally, by setting $ x \equal{} 1$, we find that $ p \mid C^{\frac {p \minus{} 1}{2}} \minus{} 1$, contradicting the fact that $ C$ is not a square.

First, I believe that you mixed up the two products in your $ P$, but that has no effect on the proof of 3) and only a little effect on the answer to 2).  Also, you probably want to specify that $ p$ is an odd prime to avoid some slight problems. :)

So, we will show that $ P \equal{} \exp( \minus{} g(\chi)L(1,\chi))$.  I'm going to play a little fast and loose with infinite sums, but it's not much work to make the following rigorous.  Let $ \sum_s$ denote the sum over squares and $ \sum_r$ denote the sum over non-squares, while $ \sum_{n'}$ denotes the sum over positive integers $ n$ not divisible by $ p$, and $ \sum_{n''}$ denotes the sum over positive integers $ n$ divisible by $ p$.

$ \ln P \equal{} \sum_s {\ln(1 \minus{} \zeta^s)} \minus{} \sum_r {\ln(1 \minus{} \zeta^r)} \equal{} \sum_r \sum_{n \equal{} 0}^{\infty} \frac {\zeta^{rn}}{n} \minus{} \sum_s \sum_{n \equal{} 0}^{\infty} \frac {\zeta^{sn}}{n}$

$ \equal{} \sum_{n \equal{} 0}^{\infty} (\sum_r \frac {\zeta^{rn}}{n} \minus{} \sum_s \frac {\zeta^{rn}}{n}) \equal{} \sum_{n \equal{} 0}^{\infty} (\sum_{m \equal{} 1}^{p \minus{} 1} \minus{} \zeta^{mn}\frac {\chi(m) \minus{} 1}{2} \minus{} \sum_{m \equal{} 1}^{p \minus{} 1} \zeta^{mn}\frac {\chi(m) \plus{} 1}{2})$

$ \equal{} \minus{} \sum_{n \equal{} 0}^{\infty} \sum_{m \equal{} 1}^{p \minus{} 1} \zeta^{mn}\chi(m) \equal{} \minus{} \sum_{n'} \sum_{m \equal{} 1}^{p \minus{} 1} \frac {\zeta^{mn}\chi(mn)\chi(n)}{n} \minus{} \sum_{n''} \sum_{m \equal{} 1}^{p \minus{} 1} \frac {\chi(m)}{n}$

$ \equal{} \minus{} \sum_{n'} \frac {\chi(n)}{n}\sum_{m \equal{} 1}^{p \minus{} 1} \zeta^{m}\chi(m) \minus{} \sum_{n''} \frac {1}{n} \cdot 0 \equal{} \minus{} (\sum_{n} \frac {\chi(n)}{n})(\sum_{m \equal{} 1}^{p \minus{} 1} \zeta^{m}\chi(m))$

$ \equal{} \minus{} g(\chi)L(1,\chi)$

2) $ \frac {1}{\sqrt {5}} \ln(\frac {3 \plus{} \sqrt {5}}{2})$

I dropped the $ \frac {1}{n}$ term in a couple of the sums, and mixed up an $ s$ with an $ r$, but it should be clear anyways.  The second and third lines should read:

$ \equal{} \sum_{n \equal{} 0}^{\infty} (\sum_r \frac {\zeta^{rn}}{n} \minus{} \sum_s \frac {\zeta^{sn}}{n}) \equal{} \sum_{n \equal{} 0}^{\infty} \frac {1}{n} (\sum_{m \equal{} 1}^{p \minus{} 1} \minus{} \zeta^{mn}\frac {\chi(m) \minus{} 1}{2} \minus{} \sum_{m \equal{} 1}^{p \minus{} 1} \zeta^{mn}\frac {\chi(m) \plus{} 1}{2})$

$ \equal{} \minus{} \sum_{n \equal{} 0}^{\infty} \frac {1}{n} \sum_{m \equal{} 1}^{p \minus{} 1} \zeta^{mn}\chi(m) \equal{} \minus{} \sum_{n'} \sum_{m \equal{} 1}^{p \minus{} 1} \frac {\zeta^{mn}\chi(mn)\chi(n)}{n} \minus{} \sum_{n''} \sum_{m \equal{} 1}^{p \minus{} 1} \frac {\chi(m)}{n}$


Its not so much a question of visualization as much as knowing what to count.
[hide="hint"]
The easiest way to count all the space diagonals is subtract face diagonals (and edges) from all $ \binom{26}{2}$ segments. The face diagonals are somewhat easier to count as its given how many of each shape there are as faces.
[/hide]

Hope that helps :)

[quote="AwesomeToad"]A convex polyhedron $ P$ has $ 26$ vertices, $ 60$ edges, and $ 36$ faces, $ 24$ of which are triangular, $ 12$ are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $ P$ have?

I really have trouble visualizing this.[/quote]

I read somewhere in this advanced geometry book on how to visualize such things. Well, what it said was you think about views from the top, side and bottom and then sort of "push" them on to the 2D plane. It works so-so.

I don't have time to get it now. I just need some help on the specific problem currently.

[quote="AwesomeToad"]A convex polyhedron $ P$ has $ 26$ vertices, $ 60$ edges, and $ 36$ faces, $ 24$ of which are triangular, $ 12$ are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $ P$ have?

I really have trouble visualizing this.[/quote]

visualizing it is harder than solving the problem.

Why making the easier(est??) AIME question more difficult?

The answer is

# of lines between any 2 vertices-# of edges-#of diagonal of the faces

$ \equal{}\dbinom{26}{2}\minus{}60\minus{}12(2)\equal{}\boxed{241}$

Wow I missed that.

 :mad: Thanks for help. So this was a counting problem??


Hi!

I wonder if anyone knows how to prove that there are not fibonacci numbers which are odd perfect numbers.

Recall that it is not yet known if odd perfect numbers exist. But, can we assure that the fibonacci sequence does not contain odd perfect numbers?

but it's proved that if an odd perfect number exist
then it's of the forme $ p^{4a\plus{}1}q^2Q^{2}$
where:
$ p\ prime\equiv 1[4]$
$ q\ prime$
$ Q\in\mathbb{N}$

[quote="aviateurpilot"]but it's proved that if an odd perfect number exist
then it's of the forme $ p^{4a \plus{} 1}q^2Q^{2}$
where:
$ p\ prime\equiv 1[4]$
$ q\ prime$
$ Q\in\mathbb{N}$[/quote]
Strange form. For all $ m>1$ we get $ m\equal{}qQ$. Therefore we can write $ p^{4a\plus{}1}m^2$, were $ p\equal{}1\mod 4$ prime, m>1.

aviateurpilot, that's true. But can we find such a number in the fibonacci sequence?

[quote="Rust"][quote="aviateurpilot"]but it's proved that if an odd perfect number exist
then it's of the forme $ p^{4a \plus{} 1}q^4Q^{2}$
where:
$ p\ prime\equiv 1[4]$
$ q\ prime$
$ Q\in\mathbb{N}$[/quote]
Strange form. For all $ m > 1$ we get $ m \equal{} qQ$. Therefore we can write $ p^{4a \plus{} 1}m^2$, were $ p \equal{} 1\mod 4$ prime, m>1.[/quote]
yes rust, but i wanted say $ p^{4a \plus{} 1}q^4Q^{2}$
and if the forme $ p^{4a\plus{}1}m^2$ if $ m\equal{}\prod_{i\equal{}1}^{h}p_i^{b_i}$ where $ p_i$ primes; wa have $ (b_1,b_2,....,b_h)\neq (1,1,....1)$


What does $\prod$ mean?  I've seen it called the product symbol, so is it like $\sum$, except the product of the terms of the series rather than the sum?

Yes.  For example, you can write $n! = \prod_{k=1}^{n} k$.

I have no idea.  Try http://mathworld.wolfram.com/Coproduct.html

Okay, normally I'm pretty good at understanding mathworld, but I didn't get that one at all...


I'm trying to start doing the IMO problems from 1960 and work myself up to the present. I was wondering if the problems steadily got harder so that I'd be "walking up the staircase" for problem difficulty; I see that the 1960s problems are fairly easy. But I was wondering if it suddenly got harder in a particular year. Could anybody answer this?

I would say that most of the process of 'getting harder' took place during the 1980s - though I wouldn't specify an exact year. This aside, problems seem to slowly get harder over time throughout, and medal boundaries are usually lower now than they were at some times in the past.
I think also that problems have been ordered more accurately according to difficulty since around this period - though this could be down to changing fashions and increased levels of training. For example, 83/6 can now be killed off very easily with the Ravi Substitution and Muirhead - techniques known to many IMOers nowadays, but probably very few at the time - whilst problem 4 of the same IMO is, in my opinion, quite hard, and does not fall so easily to such tricks.

Also note that the #1 and #4 of recent IMOs are still considered very easy.


Μια και οι περισσότεροι ανάμεσά σας αγαπούν τη γεωμετρία να μία που δεν είναι και τόσο εύκολη :wink: . Η λύση που έχω υπ' όψιν μου δεν είναι και τόσο κομψή (ως προς τα θεωρήματα που έχω χρησιμοποιήσει είναι τεχνοποιημένη), και είναι και μακροσκελής (σε διαγωνισμό δεν υπήρχε ελπίδα να προλάβω :( ) και επειδή δε γνωρίζω καλά αρμονικά, πολικές και τέτοια μήπως σας έρθει καμία λύση που να τα χρησιμοποιεί και να είναι σύντομη :) .

Σε τρίγωνο $ ABC$ γράφουμε κύκλους με κέντρα τα μέσα $ O1$,$ O2$ και $ O3$ των $ BC$, $ CA$, $ AB$ αντίστοιχα που διέρχονται από το ορθόκεντρο $ H$. Αν ο $ (O1)$ τέμνει τη $ BC$ στα $ R2$, $ R3$, ο $ (O2)$ την $ CA$ στα $ S1$, $ S3$ και ο $ (O3)$ την $ AB$ στα $ E1$, $ E2$, να αποδείξετε ότι τα $ R2$, $ R3$, $ S1$, $ S3$, $ E1$, $ E2$ είναι ομοκυκλικά.

[quote="Protonios"]Σε τρίγωνο $ ABC$ γράφουμε κύκλους με κέντρα τα μέσα $ O_{1}$,$ O_{2}$ και $ O_{3}$ των $ BC$, $ CA$, $ AB$ αντίστοιχα που διέρχονται από το ορθόκεντρο $ H$. Αν ο $ (O_{1})$ τέμνει τη $ BC$ στα $ R_{2}$, $ R_{3}$, ο $ (O_{2})$ την $ CA$ στα $ S_{1}$, $ S_{3}$ και ο $ (O_{3})$ την $ AB$ στα $ E_{1}$, $ E_{2}$, να αποδείξετε ότι τα $ R_{2}$, $ R_{3}$, $ S_{1}$, $ S_{3}$, $ E_{1}$, $ E_{2}$ είναι ομοκυκλικά.[/quote]

Είναι $ O_{2}O_{3}\parallel BC$, άρα ο ριζικός άξονας των $ (O_{2}),(O_{3})$ είναι κάθετος στην $ BC$. Αφού διέρχεται από το $ H$, ριζικός άξονας των δύο κύκλων είναι η ευθεία $ AH$.

Επομένως $ AE_{1}\cdot AE_{2}\equal{} AS_{1}\cdot AS_{3}$, που σημαίνει  ότι τα $ E_{1},E_{2},S_{1}S_{3}$ είναι ομοκυκλικά

Ομοίως βρίσκουμε ότι τα $ S_{1},S_{3},R_{2}R_{3}$ είναι ομοκυκλικά

[color=green]Παρατήρηση
Κέντρο του κύκλου είναι το περίκεντρο του τριγώνου[/color]

Λουκάς

Ακριβώς την ίδια προσέγγιση έκανα και εγώ :)

Όμως κόλλησα σε κάτι που ίσως είναι αυτονόητο αλλά παρακαλώ να μου το εξηγείσετε :blush:

Γιατί από αυτά τα δεδομένα βγαίνει ότι και τα 6 σημεία βρίσκονται στον ίδιο κύκλο? Δηλαδή καταλαβαίνω ότι υπάρχουν 3 τετράδες σημείων ομοκυκλικών αλλά πώς προκύπτει ότι όλα είναι ομοκυκλικά :?: (δεν θα μπορούσε να επρόκειτο για τρεις διαφορετικούς κύκλους?

mim κοίτα . Τα $ E_{1},\ E_{2},\ R_{3},\ R_{2}$ είναι ομοκυκλικά και το κέντρο του κύκλου τους είναι το περίκεντρο $ O$ αφού
οι μεσοκάθετοι των $ R_{2}R_{3},\ E_{1}E_{2}$ τέμνονται στο $ O$ . Όμοια και για το άλλο εγγράψιμο οπότε βρίσκεις ότι όλα τα σημεία ισαπέχουν από το $ O$  :wink:

Το φαντάστηκα ότι θα βγαίνει κι αλλιώς. Η λύση μου είναι απαίσια  :(  :mad: αλλά μάλλον σωστή. Πείτε μου κι εσείς.

ΛΥΣΗ:

Προφανώς οι μεσοκάθετοι των $ R2R3$, $ S1S3$, $ E1E2$ διέρχονται από το περίκεντρο O.

Από Πυθαγόρειο είναι $ OR2^{2}\equal{} OO1^{2}\plus{}O1R2^{2}$
ή ισοδύναμα $ OR2^{2}\equal{} OO1^{2}\plus{}HO1^{2}$ (1)

Από 1ο Διαμέσων είναι $ O1O^{2}\plus{}O1H^{2}\equal{} 2O1V^{2}\plus{}OH^{2}/2$, όπου V το κέντρο του Euler.
(2)
Όμως ως γνωστόν από κύκλο Euler $ O1V \equal{} R/2$ (3)
Επίσης γνωστό από ευθεία Euler $ OH^{2}\equal{} 9OG^{2}$ (4)
Τέλος, γνωστή θεωρείται και η $ OG^{2}\equal{} R^{2}\minus{}(1/9)(a^{2}\plus{}b^{2}\plus{}c^{2})$ (5) (αποδεικνύεται με Stewart για όσους δεν την ξέρουν)

Από 4, 5 παίρνουμε $ OH^{2}\equal{} 9[R^{2}\minus{}(a^{2}\plus{}b^{2}\plus{}c^{2})]$ (6)

Από 1, 2, 6 παίρνουμε $ OR2^{2}\equal{} 5R^{2}\minus{}(1/2)(a^{2}\plus{}b^{2}\plus{}c^{2})$ (7)

Ομοίως κυκλικά τα $ OS3$ και $ OE1$, οπότε όλα τα σημεία είναι ομοκυκλικά σε κύκλο κέντρο Ο και γνωστής ακτίνας.

(Το μόνο θετικό της λύσης είναι ότι βρίσκουμε και ακτίνα :P )

ok silouan thanks

Είναι απλώς κρίμα γιατί παιδευόμουν πολύ ώρα να το αποδείξω αυτό ενώ ήταν τόσο απλό :(  ενώ αμέσως βρήκα τις τετράδες των ομοκυκλικών σημείων...

Θεωρούμε ένα εγγεγραμμένο τετράπλευρο $ ABCD$. Οι $ AB$, $ CD$ τέμνονται στο $ E$ και οι $ AD$, $ BC$ στο $ Z$. Οι διχοτόμοι των γωνιών $ <E$, $ <Z$ τέμνονται στο $ K$. Δείξτε ότι τα $ K$, $ M$, $ N$ είναι συνευθειακά, όπου $ M$, $ N$ τα μέσα των διαγωνίων του τετραπλεύρου.

Εξαιρετική άσκηση . (πο ρε τι μου θύμισες , αν ήταν ο Βαφείδης εδώ θα την έλυνε με ένα καταπληκτικό δικό του θεώρημα για κέντρα βάρους σχημάτων , αλλά αφού δεν είναι εδώ  :(  και δεν θυμάμαι τι ακριβώς λέει το θεώρημά του να η λύση μου  :P )

Έστω ότι η διχοτόμος της Ε τέμνει τις AD,BC στα $ M,\ L$ και  ότι η διχοτόμος της Ζ τέμνει τις AΒ,DC στα $ S,\ N$ .
και $ M_{1},\ M_{2}$ τα μέσα των διαγωνίων $ BD,\ AC$ .
Έστω ότι η $ EL$ τεμνει την $ M_{1}M_{2}$ στο $ K$ kai η $ ZN$ στο $ K'$ . Θδο $ K\equiv\ K'$
Λοιπόν είναι $ \frac{\sin{\angle{M_{1}EB}}}{\sin{\angle{M_{1}ED}}}\equal{}\frac{ED}{EB}$ (1) και

$ \frac{\sin{\angle{M_{2}EC}}}{\sin{\angle{M_{2}EA}}}\equal{}\frac{EA}{EC}$ (2)

(1)=(2) άρα  $ \frac{\sin{\angle{M_{1}EB}}}{\sin{\angle{M_{1}ED}}}\equal{}\frac{\sin{\angle{M_{2}EC}}}{\sin{\angle{M_{2}EA}}}$

και αφού οι γωνίες σε κάθε κλάσμα έχουν ίσο άθροισμα θα είναι $ \angle{M_{1}EB}\equal{}\angle{M_{2}EC}$ και
$ \angle{M_{1}ED}\equal{}\angle{M_{2}EA}$ άρα έχουμε ότι $ EK$ διχοτόμος της $ M_{1}EM_{2}$ αρα

$ \frac{KM_{1}}{KM_{2}}\equal{}\frac{M_{1}E}{M_{2}E}$ (3)
Όμως τα τριγώνα $ EDB$ και $ CAE$ είναι όμοια άρα ο λόγος των διαμέσων θα είναι ο λόγος των αντίσοιχων πλευρών άρα

$ \frac{M_{1}E}{M_{2}E}\equal{}\frac{ED}{EA}\equal{}\frac{MD}{MA}$ (4)
Λόγω των (3),(4) θα είναι $ \frac{KM_{1}}{KM_{2}}\equal{}\frac{MD}{MA}$ (5)
Με εντελώς όμοιο τρόπο βρίσκουμε ότι $ \frac{K'M_{1}}{K'M_{2}}\equal{}\frac{SB}{SA}$(6) Τώρα τα τρίγωνα $ ZAB$ και $ EAD$ έχουν μια γωνία ίση και μια γωνία παραπληρωματική άρα συμπεριφέρονται ως όμοια (από το νόμο ημιτόνων)
άρα $ \frac{ED}{EA}\equal{}\frac{ZB}{ZA}$ (7) όμως από το θεώρημα διχοτόμων

$ \frac{SB}{SA}\equal{}\frac{ZB}{ZA}$ και $ \frac{ED}{EA}\equal{}\frac{MD}{MA}$(8) άρα από (5),(6),(7),(8) θα είναι

$ \frac{K'M_{1}}{K'M_{2}}\equal{}\frac{KM_{1}}{KM_{2}}$ άρα $ K\equiv\ K'$ , το ζητούμενο ... :)

Το σχήμα έχει πολλές ενδιαφέρουσες ιδιότητες πχ $ EK\bot ZK$ , $ SMNL$ ρόμβος κτλ

Να σου πω ρε Κώστα (Παπέλη) . Από πότε ξέρεις τα θέματα των ΙΜΟ ένα χρόνο πριν και δεν μας το λες ... ? ε ?
Προφανώς αστιεύομαι αλλά το πρώτο πρόβλημα στο topic είναι το ΙΜΟ pr 1 2008 αν δεν κάνω λάθος ...  :D  (και η λύση σου μοιάζει αρκετά με την πρώτη επίσημη . Μήπως εσύ ο πρότεινες και μας κάνεις την πάπια ?  :P )

Λοιπον, αν και δεν εχω πολυασχοληθει ακομα με Θεματα Γεωμετριας τετοιου επιπεδου, καταφερα να βρω μια λυση στο 1ο προβλημα που αναφερεται στο τοπικ επηδη εμπνευστηκα απο τη λυση που βρηκα πριν απο λιγες μερες στην Ασκηση του Μηνα της ΕΜΕ (Οι 2 λυσεις βασιζονται στην ιδια ιδεα)

Η Λυση μου στο παραπανω προβλημα ειναι η εξης:
Εστω $ S_1, S_2$ οι τομες του $ (O_1)$ με τη πλευρα $ AB$, $ S_3, S_4$ οι τομες του $ (O_2)$ με τη πλευρα $ BC$, $ S_5, S_6$ οι τομες του $ (O_3)$ με τη πλευρα $ AC$

Θα δειξω οτι τα σημεια $ S_1, S_2, S_3, S_4$ ειναι ομοκυκλικα και οτι ανηκουν σε κυκλο με κεντρο το περικεντρο του τριγωνου $ ABC$ (μετα ομοιως αποδεικνειεται οτι τα $ S_3, S_4, S_5, S_6$ και τα $ S_1, S_2, S_5, S_6$ ειναι επισεις ομοκυκλικα και οτι επισεις ανηκουν σε κυκλους με το ιδιο κεντρο, οποτε τελικα θα εχουμε οτι και τα 6 σημεια ανηκουν στον ιδιο κυκλο με κεντρο το περικεντρο του τριγωνου $ ABC$)

Εστω $ H'$ η 2η τομη των κυκλων $ (O_1)$ και $ (O_2)$ (Σε περιπτωση που οι 2 κυκλοι εφαπτονται, εχουμε $ H \equiv H'$.), τοτε η διακεντρος $ O_1O_2$ ειναι καθετη στην κοινη χορδη τους $ HH'$, ομως ειναι $ O_1O_2 \parallel AC$, επομενος ο φορεας του τμηματος $ HH'$ ειναι καθετος στην πλευρα $ AC$, και επηδη απο το $ H$ διερχεται το υψος απο την κορυφη $ B$ στην $ AC$, απο τη μοναδικοτητα της καθετου απο το $ H$ στην $ AC$ εχω οτι το $ H'$ ανηκει στο υψος απο την κορυφη $ B$ στην πλευρα $ AC$, επομενος εχω $ A, H', H:$ συνευθειακα. Απο τη δυναμη του σημειου $ B$ ως προς τους κυκλους $ (O_1)$ και $ (O_2)$ εχω τωρα: $ BS_2*BS_1 \equal{} BH*BH' \equal{} BS_3*BS_4$, που σημαινει οτι τα σημεια: $ S_1, S_2, S_3, S_4$ ειναι ομοκυκλικα. Ακομα, επηδη τα 4 αυτα σημεια ισαπεχουν απο το κεντρο του κυκλου στον οποιο ανηκουν, το κεντρο αυτο θα βρησκεται πανω στην τομη των μεσοκαθετων των τμηματων $ S_1S_2$ και $ S_3S_4$, τα τμηματα αυτα ομως ανηκουν στις πλευρες $ AB$ και $ BC$ του τριγωνου $ ABC$ αντιστοιχα και τα μεσα τους ταυτιζονται με τα μεσα των τμηματων $ AB$ και $ BC$ αντιστοιχα, αρα και οι μεσοκαθετες των τμηματων $ S_1S_2$ και $ S_3S_4$ ταυτιζονται με αυτες των τμηματων $ AB$ και $ BC$ αντιστοιχα οι οποιες τεμνονται το περικεντρο του τριγωνου $ ABC$, επομενος εχουμε οτι το περικεντρο του τριγωνου $ ABC$ ειναι το κεντρο του κυκλου στον οποιο ανηκουν τα σημεια $ S_1, S_2, S_3, S_4$ και τελειωσαμε.

Edit: Προσθεσα στη λυση μου μια σημαντικη λεπτομερια την οποια ειχα παραλειψει αρχικα.
Ευχαριστω κ. Στεργιου για τη διορθωση

[quote="NickNafplio"]Λοιπον, αν και δεν εχω πολυασχοληθει ακομα με Θεματα Γεωμετριας τετοιου επιπεδου, καταφερα να βρω μια λυση στο 1ο προβλημα που αναφερεται στο τοπικ επηδη εμπνευστηκα απο τη λυση που βρηκα πριν απο λιγες μερες στην Ασκηση του Μηνα της ΕΜΕ (Οι 2 λυσεις βασιζονται στην ιδια ιδεα)

Η Λυση μου στο παραπανω προβλημα ειναι η εξης:
Εστω $ S_1, S_2$ οι τομες του $ (O_1)$ με τη πλευρα $ AB$, $ S_3, S_4$ οι τομες του $ (O_2)$ με τη πλευρα $ BC$, $ S_5, S_6$ οι τομες του $ (O_3)$ με τη πλευρα $ AC$

Θα δειξω οτι τα $ S_1, S_2, S_3, S_4$ ειναι ομοκυκλικα (μετα ομοιως αποδεικνειεται οτι τα
$ S_3, S_4, S_5, S_6$ και τα $ S_1, S_2, S_5, S_6$ ειναι επισεις ομοκυκλικα)

Εστω $ H'$ η 2η τομη των κυκλων $ (O_1)$ και $ (O_2)$ (Σε περιπτωση που οι 2 κυκλοι εφαπτονται, εχουμε $ Η \equiv H'$.), τοτε η διακεντρος $ O_1O_2$ ειναι καθετη στην κοινη χορδη τους $ HH'$, ομως ειναι $ O_1O_2 \parallel AC$, επομενος ο φορεας του τμηματος $ HH'$ ειναι καθετος στην πλευρα $ AC$, και επηδη απο το $ H$ διερχεται το υψος απο την κορυφη $ B$ στην $ AC$, απο τη μοναδικοτητα της καθετου απο το $ H$ στην $ AC$ εχω οτι το $ H$ ανηκει στο υψος απο την κορυφη $ B$ στην πλευρα $ AC$, επομενος εχω $ A, H', H:$ συνευθειακα. Απο τη δυναμη του σημειου $ B$ ως προς τους κυκλους $ (O_1)$ και $ (O_2)$ εχω τωρα: $ BS_2*BS_1 = BH*BH' = BS_3*BS_4$, που σημαινει οτι τα σημεια: $ S_1, S_2, S_3, S_4$ ειναι ομοκυκλικα και τελειωσαμε.[/quote]

Νίκο, να προσέξεις την ερώτηση του Δημήτρη που απάντησε ο Σιλουανός, διότι τα εγγράψιμα από μόνα τους δεν αρκούν !
 Απλά το τονίζω για όσους δεν έχουν διαβάσει όλα τα μηνύματα :) .Την άσκηση έτυχα και γω προχθές να τη μελετήσω, μια και έχει κάτι από δύναμη σημείου και με ενδιαφέρει.

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Το πρώτο πρόβλημα όντως το πρότεινα εγώ φέτος στην Ολυμπιάδα. Χάρηκα πολύ όταν είδα ότι τους άρεσε και εκεί. Απ' ότι έμαθα ήμουν ο πρώτος μαθητής που προτείνει θέμα το οποίο τελικά επιλέγεται (έστω και λίγο παραλλαγμένο). Επειδή το είχα σχεδιάσει καιρό τώρα είχα πει πέρσι να σας προετοιμάσω. Σας άρεσε?

[hide]Ελπίζω να μην το πιστέψατε... :rotfl:  :rotfl: [/hide]

Άντε έχουμε και πρωτοπορία στο ελληνικό φόρουμ . Συμπέρασμα : Για προβλήματα της ΙΜΟ της επόμενης χρονιάς διαβάστε ελληνικά ! :D


I have a pouch with pens and pencils in it. At the moment, the ratio of pens to pencils is 3:5. If Phil gives me 3 pens in exchange for 9 of my pencils, I will have an equal number of pens and pencils. How many pens will that be?

Setting up equations give $ \frac{pens}{pencils}\equal{}\frac{3}{5}$. We all have $ \frac{pens\plus{}3}{pencils\minus{}9}\equal{}1$

Solving gives $ pens\equal{}18$,and $ pencils\equal{}30$. Adding $ 3$ pens gives, $ 21$ pens.


Thank you! I already got the [quote]If $x+y+z=0$ then $x^{3}+y^{3}+z^{3}=3xyz$ and $x^{2}+y^{2}+z^{2}=-2xy-2yz-2zx$[/quote] part, but not the rest. Now it seems easy indeed  :)

ok now this may seem very stupid but i tried it again, and no matter how i use am-gm, cauchy or anything else i know on $54x^{2}y^{2}z^{2}\leq 4(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})(x^{2}+y^{2}+z^{2})$, it still doesnt work

Ok, if $xyz=0$ then your problem is trivial else there are two of $x,y,z$ are same sign. WLOG assume they are $x$ and $y$, then $|z|=|x|+|y|$. Put $a=\frac{|x|}{|z|},b=\frac{|y|}{|z|}$, we need prove $54a^{2}b^{2}\leq 4(a^{2}b^{2}+a^{2}+b^{2})(a^{2}+b^{2}+1)$, where $a,b$ are positive reals and $a+b=1$. Now, i think it is easy,again !  :wink:


[quote="chess64"]If 5 hens can lay 24 eggs in 5 days, how many eggs are needed for 8 hens to lay 20 eggs?[/quote]

[hide="answer"](hens *days)/ eggs= constant

(5*5)/24 = 25/24

(8*d)/20 = 2d/5

2d/5 = 25/24 125=48d  so 125/48= 2 29/48  days
i still think something's wrong...
[/hide]

[quote="nigma"][quote="nonie"](hens *days)/ eggs= constant[/quote]
Because the number of eggs you have is directly proportional to the number of hens and days you have, eggs should be the numerator and the product of hens and days should be the denominator, thus the constant should be $\frac{24}{25}$ not $\frac{25}{24}$.

Hope this helps ?[/quote]
ok ill fix it...  :blush:

wait, is nonie's answer correct now? that looks familiar, I don't have the answer key.

The solution manual concurs with nonie :)