[b]Round 1 has begun. Submissions due Sunday, September 30 at 7:00 PM EST[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1098575318&t=149694]Game 1 & Rules[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1098575318&t=151415]Game 2[/url]

Klebian, if you could help moderate it again, that'd be great.

[u]Sign-up list (10/10 alive)[/u]
Nerd_of_the_Ages
AstroPhys
perfect628
CheeseIsGood
mnmath
mz94
Klebian
xpmath
shaggy75
temperal

The one I modded pretty much died.  So I called it off.

Why not? well, actually, I can think of several thousand reasons why not, but... I'll join.

Klebian, I can no longer edit the original post.  Could you edit in Temperal (and any other sign-ups)?

10 players is good.  Let's call it a game.

[b][u]ROUND 1 SUBMISSIONS[/b][/u]
[u]Format:[/u]
Title: "HTA Rd. 1 - Username - Number"  (And no, don't just type "HTA Rd. 1 - Username - Number".)
Body: Comments/Ideas/Thoughts

[u]Submit by:[/u]
Sunday, September 30 at 7:00 PM EST (UTC - 4).
Expect results by 8:00 at the latest.

Remember:  If anybody fails to submit by the due date, they will be marked as submitting 100.  If you choose to submit 100, please submit 100 and don't just not submit.

You may begin.

I chose a COMPLETELY random number this time.

Oh, dang, I totally forgot about this game, lol.
srry... guess I'll try to remember next game

With all two submissions in, I can end the round (and game  :lol: ) early.
[b][size=200]ROUND 5 RESULTS:[/size][/b]
[hide="Submissions"]Shaggy: 50
xpmath: 19[/hide][hide="Half the Average"]Half the Average is [b]17[/b][/hide][hide="Below Half the Average"]nobody[/hide][hide="Scores"]Shaggy: 105
xpmath: 98[/hide][hide="Above 100"]Shaggy[/hide][hide="Winner"]xpmath is the winner![/hide][hide="Notes"]Shaggy, if you had strategically chosen 100 this time, you would have knocked xpmath out easily.
xpmath, you were cutting it very close with 19, but it seemed to work out.[/hide][hide="Game 5?"]If people want me to start a Game 5, I will.[/hide]

And I didn't even turn in round 1 :roll:
If we have a game 5, I'll play.

[quote="miyomiyo"]
xpmath, you were cutting it very close with 19, but it seemed to work out.[/hide][hide="Game 5?"]If people want me to start a Game 5, I will.[/hide][/quote]

Do so!

Okay, I'll make a Game 5.

Funny that Shaggy hasn't posted about his loss yet.

First of all, Congrats xpmath!!! :lol:

Second of all, i would also like a 5th game

Finally, i didnt post about my loss, b/c i didnt think i was going to win anyway.

[quote="xpmath"]Thanks :)
But why wouldn't you think you were going to win? You had a 24 point lead on me.[/quote]

oops...... :blush:

this is embarrasing but i forgot that the highest score under 100 wins..... :oops:

.....i thought it was the lowest....

[quote="miyomiyo"]No, it's the last person to remain under 100.[/quote]

i really need to listen to directions more. :wink:


1=a+b+c=abc=a-b-c=a/b/c ,solve for a, b, c.

If I understand that, it'll be a:
$ \{\begin{array}{l}1=a+b+c \\ 1=abc \\ 1=a-b-c \\ 1=\frac{a}{\frac{b}{c}}=\frac{ac}{b}\end{array}$
So $ b\neq 0\wedge c\neq 0$
(1)+(3):
$ a=1$, thereafter:
$ \{\begin{array}{l}b=-c \\ bc=1 \\ b=-c \\ 1=\frac{c}{b}\end{array}$
(4): $ b=c$, then in (1) we have $ c=-c$, so $ c=0$, but it's impossible.

from $ a=1$
$ b+c=0$ then $ c=-b$
$ bc=1$ then $ b=i$ and $ c=-i$

so the only solution for the first 3 equations is $ (1,i,-i)$
which is incompatible to the last one?!


Hello all,

I'm using TeXnicCenter and creating a PDF document.  I'd like to add hyperlinks.

I tried the \url and \urladdr command and that only prints the link but doesn't make it a hyperlink.  I realize I can use the URL menu item in this forum but how do I incorporate this into a PDF document outside of the forum?

$ \url{http: //www.artofproblemsolving.com/Forum/viewtopic.php?t=44533}$

$ %Error. "urladdr" is a bad command.
{http: //www.artofproblemsolving.com/Forum/viewtopic.php?t=44533}$

Thanks,
Shaw

Use the [url=http://www.tug.org/applications/hyperref/manual.html]hyperref package[/url] then use \href{URL}{text} or \url{URL} if you don't want any text (see section 4 of the manual). The manual will tell you how to change the link colours and lots of other options.

Thanks a bunch Steve.  The MikTex Browse Package makes it a breeze to install the packages and I like that they are nicely documented.  Works fine now. :)


[i]Το νεότερο και δυσκολότερο ίσως μέρος των Ολυμπιάδων είναι η συνδυαστική.Δεν είναι τυχαίο που ο κλάδος αυτός αναπτύσσεται ραγδαία τα τελευταία σχετικά χρόνια και κυρίως με την ανάπτυξη των υπολογιστών. Στο χώρο αυτό θα βάζουμε ασκήσεις συνδυαστικής που συναντήσαμε και μας άρεσανείτε  , στα ελληνικά είτε σε άλλες γλώσσες , ώστε να έχουν σιγά σιγά όλοι ένα ελάχιστο πεδίο ασκήσεων. Προτείνω να μην ανοίγουμε πολλά topic και χανόμαστε.[/i]

 Αρχίζω με ένα πρόβλημα από διαγωνισμού Πανεπιστημίου στις ΗΠΑ. Βάζω και τη λύση του , ώστε να αρχίσουμε την προθέρμανση καλά !

[b] [color=red]ΑΣΚΗΣΗ - 1 [/color][/b]

 ΔΕΣ το συνημμένο αρχείο

[b][color=red]ΑΣΚΗΣΗ 2 [/color][/b]

[i] Την άσκηση τη δίνω κατευθείαν με τη λύση της για να την έχουμε στο αρχείο. Όποιος τη μελετήσει και έχει χρόνο , ας τη στείλλει στα ελληνικά , κάνοντας Quote. Τη δυνατότητα  αποστολής  άσκησης με εικόνα  τη χρωστάω στον [color=blue]Pontios[/color] , που μου έδειξε μερικά κόλπα. Τον ευχαριστώ πολύ και από αυτή τη θέση .

Μπάμπης[/i]

H askhsh 2 genikeuetai eukola: Kathe shmeio tou epipedou xrwmatizetai me ena apo $n$ xrwmata. Na deixtei oti uparxei orthogwnio pou oles oi korufes tou einai tou idiou xrwmatos. Kai mia endiaferousa texnikh:

[b][color=red]ΑΣΚΗΣΗ 3 [/color][/b]

An $a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1}=0$ me $a^{2}_{i}=1$ gia $1\leq i\leq n$, na deixtei oti $4/n$.

askisi 3

$a_{i}^{2}=1 \Rightarrow a_{i}=1 \ h \ a_{i}=-1$

$S=a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1}=0$ stoxos einai

na deixoume oti to $S'$ to opoio prokiptei apo to $S$ otan

parembenoume me ton metasximatismo $a_{i}\rightarrow-a_{i}$.

paramenei "ametablito" $\mod 4$

Enas tropos, einai efarmosoume ton metasximatismo $\forall a_{j}, \ me \ a_{j}=-1$  tote tha exoume $S=S' (\mod 4) \Rightarrow 0=n (\mod 4)$.

endeiktika pernume $b_{1}=a_{j}a_{j+1}, \ b_{2}=a_{j+1}a_{j+2}$ (me epikentro to $a_{j+1}$)

anigun 3 periptosis $(x.b.g) b_{1}=1, b_{2}=-1, \ b_{1}=1, b_{2}=1 \ b_{1}=-1, b_{2}=-1$

se kathe periptosi blepume oti $b_{1}+b_{2}=b_{1}'+b_{2}' \mod 4$.

analitkotera trexe sto biblio toy engel 1 kefaleo (pou oloi pleon exoume :) )

Poli oraia , ti lete gia auto :

Kathe simeio tou Z^2 einai eite Mauro eite Kokkino. Iparxei tetragono (me pleures paralliles stous aksones) me korifes tou idiou xromatos ?

Mporeite na to genikeusete ? Aplopoiisete ?


[b]Prove that [/b]$\left\{\begin{array}{c}|z|\le 1\\\\ \left|z^{2}-2z+2\right|\le 1\\\\ \left|z^{2}-3z+3\right|\le 1\end{array}\right\|\Longleftrightarrow z=1\ .$

Are you wrong?  :wink:
We have: $\left|z^{2}-2z+2\right|\le 1\Longleftrightarrow\left| (z-1)^{2}+1\right|\le 1\Longleftrightarrow (z-1)^{2}=0\Longleftrightarrow z=1$

[quote="April"]Are you wrong?  :wink:
We have: $\left| (z-1)^{2}+1\right|\le 1\Longleftrightarrow (z-1)^{2}=0$[/quote]Why $|u+1|\le 1\Longrightarrow u=0$ ?

[quote="April"]We have: $(z-1)^{2}+1\ge 1$  :D[/quote]Why, here $z$ is complex number !

just consider the loci [which are circles] graphed on the complex plane


Let $ ABCDEF$ be a convex hexagon that has no pair of parallel sides. It is known  that, for every point $ P$ inside the hexagon, the sum: \[ \text{Area}[ABP]\plus{}\text{Area}[CDP]\plus{}\text{Area}[EFP]\] has a constant value. Prove that the triangles  $ ACE$ and $ BDF$ have the same barycentre.

_____________________________________

This problem was proposed by Israel Diaz.

$ Tipe$

[code]Hola amigo Jorge, aqui te envio una prueba del ejercicio, espero este correcta.[/code]


Hello,

I am trying to find the value of $ tan\frac{pi}{2}$ and $ cot\frac{pi}{2}$ but my calculator is giving an ERROR for the closing parenthesis of both trigonometric functions. I am in RADIAN mode and am using a TI-83 plus.

Thank you.

*Note: Moderators, please move topic if this isn't the right forum for graphing calculator discussions.

[quote="AstroPhys"]Tangent at $ \frac{\pi}{2}$ is undefined and cotangent is $ 0$.
$ \tan\frac{\pi}{2}$
\tan \frac{\pi}{2}
$ \cot\frac{\pi}{2}$
\cot \frac{\pi}{2}.[/quote]

Thanks for your reply, AstroPhys.

How would I find the value of $ \cot\frac{\pi}{2}$ on a graphing calculator? I cannot type in the inverse of $ \tan\frac{\pi}{2}$ because I will get an ERROR then.


logarithms
real analysis
real analysis unsolved

Using l'Hopital rull compute:

lim(x-->oo) [ln^k(x+1)-ln^k(x)] for any n natural, n>=1.

have fun!

cheers! :D :D :cool:

is $\ln^k(x)$ supposed to be $(\ln x)^k$ or $\underbrace{\ln \ln \cdots \ln}_{k \text{ times}} x$?

i think it is $(\ln x)^k$.
Thus the limit is zero.

I didn't even notice that someone answered this post! I am sorry!

it is compute the limit of:
$ \ln^{k} (x+1)- \ln^{k} (x)$ $\forall k \geq 1$

And how did you actually used L'Hopital so solve it?


Ring Theory
superior algebra
superior algebra solved

This is the last last one for tonight before I go to bed ...

Show that every proper ideal $I$ of an unital ring $R$ is contained in a maximal ideal of $R$.

It's exactly the same as the proof that there are maximal ideals at all- we just restrict attention to ideals containing $I$.

The union of a nested chain of ideals is an ideal. so every chain (by inclusion) of proper ideals containing $I$ has an upper bound. This upper bound is a proper ideal, since it doesn't contain 1.
These conditions allow us to apply Zorn's lemma: the set of proper ideals containing $I$ has a maximal element, which is a maximal ideal containing $I$.

if you already know that every nontrivial unital ring has a maximal ideal, just apply this to the factor ring $R/I$.


Find the number of positive integers $n$ for which

(i) $n \leq 1991$;

(ii) 6 is a factor of $(n^2 + 3n +2)$.

Consider $n^2 + 3n + 2$ mod 6.

If n = 0, 3 mod 6 it doesnt work.  It works for n = 1,2,4,5 mod 6.

So in total, it works 331*4 + 4 times = 332*4 times.

We can split this up into four cases:

$1. 2|n+1, 3|n+2$
$2. 3|n+1, 2|n+2$
$3. 6|n+1$
$4. 6|n+2$

When counting the first case, we can see that $1, 7, ..., 1987$ work. This yields $332$ numbers.

When counting the second case, we can see that $2, 8, ..., 1988$ work. This yields $332$ numbers.

When counting the third case, we can see that $5, 11, ..., 1991$ work. This yields $332$ numbers.

When counting the fourth case, we can see that $4, 10, ..., 1990$ work. This yields $332$ numbers.

Adding them up, we get $332*4=1328$ numbers.

We have $n^{2}+3n+2=(n+1)(n+2)$.  This is always divisible by 2, and is divisible by 3 iff $n\not\equiv 0\pmod{3}$.  Of the first 1991 positive integers, $\frac{1989}{3}=663$ are divisible by 3.  Therefore, the answer is $1991-663=\boxed{1328}$.

Simply (n+1)(n+2) is a multiple of 3 means that n is not a multiple of 3. So, from 1 to 1991, number of non-multiples of 3 is 1991 - 663 = 1328.

I think this was probably the easiest question asked in INMO.

We want to get that $n^2+3n+2\equiv(n+1)(n+2)\equiv0\pmod6$.

We see that $(n+1)(n+2)\equiv0\pmod2$.

We therefore want $(n+1)(n+2)\equiv0\pmod3$ so $n\not\equiv0\pmod3$.

We know that there are $\lfloor\frac{1991}{3}\rfloor=663$ multiples of $3$ less than or equal to $1991$.

We calculate our answer to be $1991-663=\boxed{1328}$

We see that, $(n+1)$ and $(n+2)$ are consecutive integers which imply that one of them must be even i.e divisible by $2$ which satisfies one of our condition for any $n \leq 1991$. Now for the divisibility of 3, it is pretty easy to see that $\forall n$ that are divisible by 3, the expression isn't divisible by 3 cause then the expression becomes $(3\mathbb{Z}+1)(3\mathbb{Z}+2)$ which is not divisible by 3. So all n of the form $(3\mathbb{Z}+1)$ and $(3\mathbb{Z}+2)$ satisfies the condition. Hence $n \in \{1,2,4,5,7,8,10,11,...,1990,1991\}$. Now by trivial PIE, number of such $n$ are $1991-\lfloor\dfrac{1991}{3}\rfloor=1328$.


Columbia
algebra
polynomial
trigonometry
quadratics
geometry
3D geometry

Hello Everyone,

The next Columbia Math Circle will take place on Saturday, October 3rd at Hammond School in Columbia. This will be a continuation of a planned series of lectures and activities leading up to a presentation of the famous Riemann Hypothesis. I hope that many of you will be able to attend, as I think it will be an enlightening and exciting meeting. The location will again be at Hammond School, 854 Galway Lane, in Columbia from 1-5pm. If you know of someone who would be interested in attending, please pass the message along.

Note that we are also starting a Middle School Math Circle beginning on Oct. 3rd from 9am-12pm. If you know of any students who would be interested in attending, please pass this message along to them as well.

Happy Problem Solving,

Marcus Neal and Patrick Rybarczyk

I hope everyone that came out to the Math Circle today had a good time. Here are solutions to the problems that were given out by Patrick and myself. Feel free to comment, ask questions, or post your own solutions.

Problem Set 2:

1.) $ \boxed{E}$ $ (3 + 8i)( - 2i) = - 6i - 16i^2 = 16 - 6i$

2.) $ \boxed{B}$ $ \frac {1 - 3i}{1 + 3i} = \frac {1 - 3i}{1 + 3i}\frac {1 - 3i}{1 - 3i} = \frac { - 8 - 6i}{1 + 9} = \frac { - 8}{10} - \frac {6}{10}i = - 0.8 - 0.6i$

3.) $ \boxed{E}$ $ P(x) + iP(1 - x) = \sqrt {2}x^5$.
Let $ P(x) = ax^5 + bx^4 + \ldots$, then $ P(x) + iP(1 - x) = (ax^5 + bx^4 + \ldots) + i(a(1 - x)^5 + b(1 - x)^4 + \ldots)$.
The coefficient of $ x^5$ will be $ a + i( - a) = a(1 - i)$. This must be equal to $ \sqrt {2}$.
Therefore, $ a = \frac {\sqrt {2}}{1 - i} = \frac {\sqrt {2}(1 + i)}{(1 - i)(1 + i)} = \frac {1 + i}{\sqrt {2}}$

4.) $ \boxed{A}$ $ \left(\frac {\sqrt {2}}{2} + i \frac {\sqrt {2}}{2}\right)^{41} = (e^{i \pi/4})^{41} = e^{i(10\pi + \pi/4)} = e^{i \pi/4} = \frac {\sqrt {2}}{2} + i \frac {\sqrt {2}}{2}$

5.) $ \boxed{C}$ $ e^{ - i \pi/4}(1 + i) = e^{ - i \pi/4} \cdot (\sqrt {2}e^{i \pi/4}) = \sqrt {2}$

6.) $ \boxed{A}$ If $ 1 + i$ is a root of a polynomial with real coefficients, then $ 1 - i$ is also be a root. The coefficient of $ x^2$ is the negative of the sum of the roots (see Viete's Theorem). So the coefficient of $ x^2$ is $ - (3 + 1 + i + 1 - i) = - 5$.

7.) $ \boxed{B}$ $ \left(\cos{\frac {3\pi}{10} + i \sin{\frac {3\pi}{10}\right)^{10} = (e^{i 3\pi/10})^{10} = e^{3\pi i} = - 1}}$

8.) $ \boxed{C}$ $ \left(\cos{\frac {\pi}{6} + i \sin{\frac {\pi}{6}\right)^{4} = (e^{i \pi/6})^{4} = e^{2\pi i/3} = \cos{\frac {2\pi}{3}} + i \sin{\frac {2\pi}{3}} = \frac { - 1}{2} + i\frac {\sqrt {3}}{2}}}$

9.) See #3

10.) $ \boxed{e}$ If $ 2i$ is a root of the polynomial, then $ - 2i$ is a root. Therefore, $ (x - 2i)$ and $ (x + 2i)$ are factors, meaning that $ (x^2 + 4)$ is a factor.
Performing long division of the polynomial by $ x^2 + 4$, we can find that $ x^2 + 7x + 9$ is the other factor. The quadratic formula gives choice $ \boxed{e}$ as a root of this polynomial.

As an alternative to performing long division, you could also notice that the sum of all 4 roots of the polynomial is -7 and the product of all 4 roots is 36 (by Viete's Theorem). Since the sum of the two roots we already know is 0, and their product is 4; the sum of the remaining two roots is -7, and their product is 9. By Viete's Theorem again, these roots satisfy the quadratic equation $ x^2 + 7x + 9 = 0$. Solving, again, gives choice $ e$.

More to come later....

11.) $ \boxed{A}$ $ (\sin{3^o} \plus{} i\cos{3^o})^{60} \equal{} (i(\cos{3^o} \plus{} i\sin{3^o}))^{60} \equal{} i^{60}(e^{ \minus{} i3^o})^{60} \equal{} (1)e^{ \minus{} i180^o} \equal{} \minus{} 1$

12.) $ \boxed{A}$ Multiply $ x^2 \minus{} x \plus{} 1 \equal{} 0$ by $ (x \plus{} 1)$ to get $ x^3 \plus{} 1 \equal{} 0$. Thus, $ x^3 \equal{} \minus{} 1$ (i.e. $ x$ is one of the two non-real cube roots of $ \minus{} 1$).

13.) $ \boxed{C}$ $ x^2 \plus{} x \plus{} 1 \equal{} 0$ implies $ x^3 \minus{} 1 \equal{} 0$, and that $ x$ is one of the two non-real cube roots of 1.
Let $ x$ be such that $ x \plus{} \frac {1}{x} \equal{} 2\cos \left(\frac {2\pi}{3} \right)$. Then $ x^k \plus{} \frac {1}{x^k} \equal{} 2\cos \left(\frac {2\pi k}{3} \right)$.

So $ \left(x \plus{} \frac {1}{x}\right)^2 \plus{} \left(x^2 \plus{} \frac {1}{x^2}\right)^2 \plus{} \left(x^3 \plus{} \frac {1}{x^3}\right)^2 \plus{} \cdots \plus{} \left(x^{27} \plus{} \frac {1}{x^{27}}\right)^2$
$ \equal{} 4\cos^2\left(\frac {2\pi}{3}\right) \plus{} 4\cos^2\left(\frac {4\pi}{3}\right) \plus{} 4\cos^2\left(\frac {6\pi}{3}\right) \plus{} \ldots \plus{} 4\cos^2\left(27 \cdot \frac {2\pi}{3}\right)$
$ \equal{} 9 \left(4\cos^2\left(\frac {2\pi}{3}\right) \plus{} 4\cos^2\left(\frac {4\pi}{3}\right) \plus{} 4\cos^2\left(2\pi\right)\right)$
$ \equal{} 9(1 \plus{} 1 \plus{} 4) \equal{} 54$

14.) $ \boxed{E}$ $ x \plus{} \frac {1}{x} \equal{} 2\cos \left(\frac {\pi}{1001} \right)$ implies that $ x \equal{} e^{i \pi/1001}$. So $ \alpha^{2002} \equal{} e^{i 2\pi} \equal{} 1$, and $ \alpha^{2002} \plus{} \frac {1}{\alpha^{2002}} \equal{} 2$

15.) $ \boxed{D}$ $ x \plus{} \frac {1}{x} \equal{} 2\cos{12^o}$ implies that $ x \equal{} e^{i 12^o}$. Therefore $ x^5 \equal{} (e^{i 12^o})^5 \equal{} e^{i 60^o}$, and $ x^5 \plus{} \frac {1}{x^5} \equal{} 2\cos{60^o} \equal{} 1$

16.) $ \boxed{(5/8,10)}$ There are 6 possible orientations. When connecting the three given points, a right angle of the square can be located at either $ z$, $ z^2$, or $ z^3$, with the other two points having two possible orientations with respect to that vertex.

If the right angle is at $ z$, then $ z^3 \minus{} z \equal{} \pm i (z \minus{} z^2)$, the sign depending on the orientation of the points. Solving for $ z$, gives $ z \equal{} \minus{} 1 \minus{} i$ or $ \minus{} 1 \plus{} i$. So $ z^2 \equal{} 2i$ or $ \minus{} 2i$ respectively, and $ z^3 \equal{} 2 \minus{} 2i$ or $ 2 \plus{} 2i$ respectively. The area of the square is $ |z^3 \minus{} z|^2 \equal{} |3 \pm i|^2 \equal{} 10$ in both cases.

If the right angle is at $ z^2$, then $ z^3 \minus{} z^2 \equal{} \pm i (z^2 \minus{} z)$, the sign depending again on the orientation of the points. Solving for $ z$, gives $ z \equal{} i$ or $ \minus{} i$. So $ z^2 \equal{} \minus{} 1$ in both cases, and $ z^3 \equal{} \minus{} i$ or $ i$ respectively. The area of the square is $ |z^2 \minus{} z|^2 \equal{} | \minus{} 1 \pm i|^2 \equal{} 2$ in both cases.

If the right angle is at $ z^3$, then $ z^3 \minus{} z \equal{} \pm i (z \minus{} z^2)$, the sign depending on the orientation of the points. Solving for $ z$, gives $ z \equal{} \frac { \minus{} 1 \plus{} i}{2}$ or $ \frac { \minus{} 1 \minus{} i}{2}$. So $ z^2 \equal{} \minus{} i/2$ or $ i/2$ respectively, and $ z^3 \equal{} \frac {1 \plus{} i}{4}$ or $ \frac {1 \minus{} i}{4}$ respectively. The area of the square is $ |z^3 \minus{} z|^2 \equal{} \left|\frac {3 \pm i}{4}\right|^2 \equal{} \frac {5}{8}$ in both cases.


Let $S=\{1,2,3,\ldots,n\}$, $n$ an odd number. Find the parity of number of permutations $\sigma : S \Rightarrow S$ such that the sequence defined by \[a(i)=|\sigma(i)-i|\] is monotonous.

consider the following map $M$, every good permutation $F$ will be mapped to the permutation $G$ such that $G(n+1-i)=n+1-F(i)$, so we get that for $G$, the defined sequence is traslated the same so it is still monotonous.

Now we would like to count for how many  $G$ is  $F$ itself! the pair $(k,F(k))$ appears in $F$, and in $G$ it is $(k,n+1-F(n+1-k))$, so we would get:

$F(k)=n+1-F(n+1-k)$ now, it is easy to see $a(1)=a(\frac{n+1}{2})=a(n)=0$ since $f(\frac{n+1}{2})=\frac{n+1}{2}$, so the only good permutation remaining fixed after the map is the identity.

So sumarising we get an odd number of such permutations!

of course the argument doest works for $1$ :)


geometry
geometric transformation
dilation

Q: Will the pH value increase or decrease when water is added to dilute sodium hydroxide solution?

It is obvious that the pH value will decrease.

However, as water is added....
[H+] decreases -> (pH increases)
[OH-] decreases -> (pOH increases)
at the same time

What confuses me is that how we can determine the final pH change when both of them decreases?

Besides, as both [H+] and [OH-] fall, both pH and pOH rise.
How can pH + pOH still equals 14??

Thanks

i would expect the NaOH to break up into Na+ and OH- and give a basic aqueous soln...giving a pH>7...........plus, NaOH is a strong base....and when you add water, water is neutral, so the NaOH will break up in the solution...basic solution

Water dissociates on its own.

[url]http://www.chembuddy.com/?left=pH-calculation&right=introduction-acid-base-equilibrium#eq1.4[/url]

The best way to think of this problem is to consider the amount of base in the diluted solution.  The water dissociation should be considered after you have the base added.  Dilution willl essentially spread out the base, equivalent to removing some of the base by reaction, say as a percipiate.  If one dilutes with as much water as was initially there, it is the same as removing half of the initial base.

I hope this way of looking at this problem helps.

Your original assumption that "[H+] decreases" is incorrect.  As Borek noted, water donates hydrogens on its own.  So actually, [H+] increases.

How much is the pH of 1M $\text{H}_2\text{SO}_4$? or 4M $\text{HCl}$?


integration
logarithms
advanced fields
advanced fields unsolved

part i. In the Poincare disk model, show that the locus of points with given Poincare distance $d$ from the origin is represented by a Euclidean circle of radius$<d$.

part ii. Use a suitable inversion to show that the locus of points with given Poincare distance $d$ from any given point $P$ is represented by a Euclidean circle.

part iii. Let $C$ be a Euclidean circle with center $(0,s)$ and diameter $s$. It represents a Poincare circle as above. Find its Poincare radius and Poincare center. Is the Poincare radius bigger or smaller than $s$?

Consider the point $z=re^{i\theta}$ in the unit disc.  In the Poincare model, $\gamma(t)=te^{i\theta}$ from $\gamma(0)=0$ to $\gamma(r)=z$ is a geodesic, with length

$d(0,z) =\int_{0}^{r}\frac{2dt}{1-t^{2}}=\log\frac{1+r}{1-r}.$

That is, the Poincare distance $d$ from $0$ to $z$ is related to the Euclidean distance $r$ by $d=\log\frac{1+r}{1-r}$, or $r = \tanh(d/2) \leq d$, with equality iff $r=d=0$.

[quote="countdownmath"]
part iii. Let $C$ be a Euclidean circle with center $(0,s)$ and diameter $s$. It represents a Poincare circle as above. Find its Poincare radius and Poincare center. Is the Poincare radius bigger or smaller than $s$?[/quote]

The circle at $(0,s)$ and diameter $s$ [b]does not[/b] represent a Poincare circle. Something is wrong.


Construct quadrilateral ABCD given lengths AB, BC, CD, DA and the angle AED, where E is the intersection of AB and CD.


induction algebra polynomial function FTW

Consider the sequence 3,5,8,12,17,23,...If we assume that the points belong to a polynomial function, we can use the method of finite differences to determine the degree of the polynomial, and then find the coefficients by plugging in points from the sequence: (1,3), (2,5), (3,8), etc. Find the formula for thr nth term of the sequence.

This should be moved to High School Basics. This forum is specifically for FTW problems.

Well it seems to me that the sequence can be expressed as:

$ 2 \plus{} 1,2 \plus{} (1 \plus{} 2),2 \plus{} (1 \plus{} 2 \plus{} 3),\cdots,2 \plus{} (1 \plus{} 2 \plus{} \cdots \plus{} n),\cdots$

So the $ n$th term should be, $ 2 \plus{} (1 \plus{} 2 \plus{} \cdots \plus{} n) \equal{} 2 \plus{} \dfrac{(n)(n \plus{} 1)}{2}$.


Given $1000000$ points in space, show that the set of pairwise distances has at least $79$ elements.


function
calculus
integration
geometry
geometric transformation
calculus computations

A fancy notepad has edges that follow the graphs of y=0.5sin(x)-2 and y=-0.5sin(x)+2. The notepad is 12 inches long and 1 inch thick.  Explain how you could use cross sections or slices cut perpendicular to the x-axis to find the volume of this notepad

[quote="JRav"]Well what kind of volume?  Around which axis?  Let's assume we're revolving around x, just for simplicity.  Well we cut our infinitesimally thin slices perpendicular.  Each strip has radius as determined by the value of the function at that point.  When we revolve the strip, we get a circle of radius as determined above.  Then all we do is integrate all of these circles along the entire interval.[/quote]
But is the interval [0,12]?

This is not a solid of revolution. Thickness is constant- finding the volume is equivalent to finding an area.

I may have mis-interpreted it.  Do you then think this is a double integral?

No. The most natural translation is a single integral for area, multiplied by thickness.


R.H.S is positive hence L.H.s must be positive
hence x >1        or                     x<-5
for both x>1 after this the function always give the value greater than 10
and x<-5 also give the value greater than 10.
hence the equation have no solution

It's not true [b] ashwinrk-jain [/b]. You can put e.g. $x=1,1$ and then you have: $(3,1)^2 \cdot 0,1 \cdot 6,1=5,8621<10$.

i am so sorry the above equation have fore solution two are real and two are imaginary.
the real solution are mentioned above. and the imaginary solutions are -2 ± i.

the question is solved by taking x²+4x = t
then this quadratic in t having roots 6,-5.
now equate the x²+4x by 6,-5 for getting solutions


Usually, you have to log in twice. After logging in, did you get to the window that says something like "After logging in, you can access..." ? If so, click "click here to log in" again, then type in your name and password again.

If you have already tried this, I can't answer your question.

[quote="IhearNY"]i forgot my password[/quote]

[quote="IhearNY"]i forgot my password

but even though it said it sent be a password through email

i didn't get any

i tried several times for  few days
but i still didn't get any email from USAMTS for password

what should i do?[/quote]Write to usamts@usamts.org describing the situation.  Be sure to include your user id# and your username.


algebra
polynomial
calculus
derivative
number theory
relatively prime
Galois Theory

I believe you need the condition "monic" (which can possibly be weakened to "primitive"): otherwise, we can multiply $ f$ by a large integer (specifically, an integer larger than the discriminant) without changing the value of the discriminant.

If $ f$ contains repeated roots, the discriminant is zero and every ideal contains zero.  Otherwise, $ f$ is relatively prime to its derivative so the ideal $ (f, f')$ is just $ \mathbb{Z}$ and it remains to show that the discriminant is an integer.  Since the discriminant is by definition fixed by every element of the Galois group of the splitting field of $ f$, it is rational, and moreover it is a product of sums of algebraic integers and therefore itself an algebraic integer, hence an integer.

[quote="t0rajir0u"]$ f$ is relatively prime to its derivative so the ideal $ (f, f')$ is just $ \mathbb{Z}$[/quote]
This would only hold for some special rings (Bezout rings, in fact), but not here. $ x$ and $ 2$ are coprime, but the ideal $ (2,x)$ isn't the whole ring (and neither is it's intersection with $ \mathbb Z$; you seemed to have used the wrong ring here).

[quote="dreammath"]Prove that the discriminant of a polynomial $ f$ is contained in the ideal of $ Z[X]$ , which is generated by $ f$ itself and the derivative $ f'$ .[/quote]

[b]HINT [/b] Use the well-known [b]resultant[/b] formula for the discriminant.

At first  , I thought the same as t0rajir0u did .
[quote]HINT  Use the well-known resultant formula for the discriminant.[/quote]
Can u tell me this formula ?
I knew that the discriminant could be defined as a resultant of two polynomials $ f$ and $ f'$ . I don't know whether or not the resultant of two abirtrary integer coefficients polynomials $ P$ and $ Q$ is contained in the ideal of $ Z[X]$ which is generated by $ P$ and $ Q$ ?


Let $ n$ is a positive integer.Find max value of $ n$ in order to $ n$ is divisible by any positive integer $ m$ satisfies that $ m < \sqrt [3]{n}$.

[quote="SherlockBond"]How I can do that?[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=150461


quadratics
modular arithmetic
number theory proposed
number theory

Indeed, not too difficult:

Let $a^2=3x+4y,b^2=4x+3y$. Then $7|a^2+b^2\Rightarrow 7|a,b\Rightarrow 49|a^2,b^2$. On the other hand, $49|4a^2-3b^2=7y\Rightarrow 7|y$ and we're done.

That is exactly what I've done  :)
The key step is to know that if $7|a^2+b^2$ then $7|a$ and $7|b$.

Pierre.

My solution of problem 3

Let  3x+4y=m 2  and 4x+3y=n 2 	Therefore m 2 +n 2 =7(x+y) --> m 2 +n 2  is divided by 7 --> a 2 +b 2  is divided by 7 where m=7k+a and n=7p+b  so a,b=0,1,2…6.
We can’t find such values of a and b so that the sum of their squares is divisible by 7 except a=0 and b=0 ( for a=1 trying all the possible values of b: 0,1,2..6 we don’t get the sum of the squares divisible by 7, same thing for a=2,..6). This means that m,n are both divisible by 7.

m 2 +n 2  is divisible by 49 --> x+y is divisible by 7 since m 2 +n 2 =7(x+y)
n 2 -m 2 =x-y --> x-y is divisible by 49 -->
 x-y is divisible by 7

So we have x-y, x+y both divisible by 7. therefore the result we are looking for follows.

My solution:

Note that the quadratic residues modulo 7 are 0, 1, 2, 4, and thus we immediately see:

[b]Lemma.[/b] The only residue t modulo 7 such that t and -t are both quadratic residues modulo 7 is t = 0.

Now, we have $3x+4y\equiv 3x-3y=3\left( x-y\right)\ \text{mod}\ 7$, and $4x+3y\equiv -3x+3y=-3\left( x-y\right)\ \text{mod}\ 7$. Therefore, both 3 (x - y) mod 7 and -3 (x - y) mod 7 must be quadratic residues modulo 7. Thus, after the Lemma, we must have $3\left( x-y\right) \equiv 0\ \text{mod}\ 7$; hence, $x-y\equiv 0\ \text{mod}\ 7$, and the number x - y is divisible by 7.

From $x-y\equiv 0\ \text{mod}\ 7$, it follows that $3x+4y\equiv 3x-3y=3\left( x-y\right) \equiv 0\ \text{mod}\ 7$. Therefore, the number 3x + 4y is divisible by 7. Since this number is a perfect square, it must also be divisible by 49. Similarly, 4x + 3y is divisible by 49. Thus, (3x + 4y) + (4x + 3y) = 7x + 7y = 7 (x + y) is also divisible by 49. This means that x + y is divisible by 7. Since x - y is also divisible by 7, the number 2x = (x + y) + (x - y) must be divisible by 7, too. Since 2 and 7 are coprime, this entails that x is divisible by 7. Similarly, y is divisible by 7.

  Darij

BTW the following is true:

Given any  prime $p$ in $4k+3$ form and $a,b$ integers so that $a+b=p$ and $(a,b)=1$. Then any $x,y$ integers which $ax+by,ay+bx$ are both squares then $p|x,y$

7 divides a.a+b.b then  7divides a,b.that means 7 divides x-y.So 7 divides x and y.

let $3x+4y=a^2 .      4x+3y=b^2$. then $7(x+y)=a^2+b^2$. so , 7 divides $a^2+b^2$.checking the Q.R.'s mod7 , we get that 7 divides a , 7 divides b. put a=7m , b=7n. so, $7(x+y)=49(m^2+n^2)$. . so , we get that $x+y=7(m^2+n^2)$. so, 7 divides x+y. also $a^2-b^2=y-x$ , so , 7 divides x-y [ as 7 divides a & b]. so , 7 divides 2x , i.e. 7 divides x & so , 7 divides y. :D

^^Scientist, $7|x-y$ does not imply $7|x, 7|y$.

[hide="Solution"]
Let $3x+4y=a^2, 4x+3y=b^2$. Adding we get $7(x+y)=a^2+b^2$ so $7|a^2+b^2$. The quadratic residues $\pmod{7}$ are $\{0,1,2,4\}$; by inspection we must have $a=b=0 \pmod{7}$. So let $a=7a_1, b=7b_1$.

Then $3x+4y=3(x-y)=0\pmod{7} \to x=y \pmod{7}$. So let $x=7x_1+k, y=7y_1+k$. We now have the system

$3x+4y=3(7x_1+k)+4(7y_1+k)=(7a_1)^2$
$4x+3y=4(7x_1+k)+3(7y_1+k)=(7b_1)^2$

Simplifying and taking $\pmod{7}$ we get

$3(x_1-x_2)+k= 0\mod{7}$
$4(x_1-x_2)+k= 0\mod{7}$

Multiply the first equation by $4$ and the second equation by $3$, then subtract to get $k=0 \pmod{7}$. Thus $x=y=0 \pmod{7}$.[/hide]

[quote=pbornsztein]If the positive integers $x$ and $y$ are such that $3x + 4y$ and $4x + 3y$ are both perfect squares, prove that both $x$ and $y$ are both divisible with $7$.[/quote]

We have $3x + 4y = a^2$ and $3y+4x = b^2$ and we get $7(x+y) = a^2+b^2$,$x-y = a^2-b^2$.
 $a^2$ can be equal $0,1,2,4$ $modulo 7$.
Let $k$ $=$ $a^2$ $(mod 7)$ and $l$ $=$ $b^2$ $(mod 7)$ and $k + l$ $=$ $0$ $(mod 7)$.
      $k = 0$ we get $l = 0$.
      $k = 1$ no solution.
      $k = 2$ no solution.
      $k=4$ no solution.
We get the single solution when $a = b = 0$ $(mod 7)$.$x - y = a^2 - b^2$ $\implies x = y$ $(mod 7)$.
Now $3x + 4y = 7x = a^2$ $(mod 49)$ and we get $x = y = 0$ $(mod 7)$ because $a^2 = 0$ $(mod 49)$.

Let $3x+4y=a^2, 4x+3y=b^2$.
So, $0\equiv 7x+7y\equiv a^2+b^2 \pmod 7$. As $7\equiv 3 \pmod 4$, $7\mid a,b$.
$x-y\equiv a^2-b^2\equiv 0 \pmod {49}$. So, $x\equiv y \pmod {49}$.

$0\equiv a^2 \equiv 3x+4y\equiv 7x\equiv 7y \pmod {49}$. So, $x\equiv y \equiv 0 \pmod{7}$.


calculus
integration
calculus computations

Calculate $ \int \frac{1\minus{}x^2\minus{}x^6\plus{}x^8  }{1\plus{}x^{10}}$$ dx$


I'm pretty sure the deadline was yesterday.  I got a 48 which was a very poor performance for me.

im pretty sure i got 55ish... but im from a competitive section so idk...

well I already qualified for the next round.  Also my proctor didn't allow us to keep the tests and I took it a couple of weeks ago and can't remember anything to even discuss.

The deadline is tomorrow.  This shouldn't be discussed yet.

i guess so cus its definetely passed the deadline.

was NaBr the one that could only act as an oxidizing reagent?

Yea.. Sodium anything is a great oxidizing agent .... (i misbubbled that and put BrCl or something =( )

i made a few more stupid mistakes that im ashamed about.. howd everyone else do?

That was one of the ones I got wrong.  In a mind lapse i mixed up reducing and oxidizing agent and put the permanganate one.  Pretty stupid of me lol.  Overall i got 56 outta 60.  It was pretty borderline though because my sections cutoff was 54.  I'm pretty pleased though overall

Meh, edit time limit.

I had 50/60, which qualified in my section. 50+ was my goal, so I'm pretty happy.

so, what kind of scores on the national exam usually are safe for camp? I know ACS doesn't really release information on this, but there's probably a general consensus around here right?

So I apparently got a 46, which just missed passing (we take 12 and our cutoff was 49).

I totally overestimated the difficulty of the test...its no where near the other qualifying exams like AIME and the bio one.  This makes me regret not studying for it outside of school.  Several people told me (including our ACS coordinator who goes to my church) that I had no chance of passing if I just took what I learned in AP Chem + a month of self studying and applied it to the test.  Most people I know didnt even study a month outside of school and passed; most of the people that actually went to the ACS coordinator's tutoring class outisde of school didnt pass.

Anyhow, congratulations to everyone that qualified and good luck on the national test.

Actually, most of the stuff I needed for the test I learned in AP Chem.  The only things that I had to know which I learned by myself was electrolysis (we hadn't learned that yet at the time) and colors of compounds which isn't covered too in depth in AP Chemistry.  There were also some random organic questions which might be a little tricky without further studying, but nothing too bad I think.

I had a friend two years ago who made camp with a 55 on the local. He said that the scores usually varied from 53ish-60.

Although it is general concensus that if you want to make camp, you should be scoring 50+ on the national MC section... 55+ to be safe.

And i thought that usnco local exam was much easier than AIME...

and by much easier i mean MUCH EASIER.....

the questions are mostly formulaic except for those few problems that ask you for random facts. In general I think that the IChO tests how much knowledge in chemistry one has, where as IMO/IPhO tests ... for lack of a better word... brilliance... Anyone walking into AP calculus knows more than enough "math knowledge" in order to solve usamo/IMO level problems, but the beauty of it lies in just that fact. Similarly, everyone knows that F=ma, but that most of the usapho qualifying exam problems can be answered with just this fact alone seems to show that  one cannot simply memorize a textbook equation and expect to be successful. Chemistry is somewhat like this at higher levels (I.e organic chem reaction mechanisms) where one can express creativity in solving a problem by proposing unique synthesis schemes, however, there will always be those facts one has to simply memorize (flame tests etc.)

USNCO local vs. AIME??!! Is that even a comparison?

As an additional note, there have been some rather challenging IChO problems in the past that did test creativity, but I cannot think of any now. However, these are quite rare; chemistry is one of those subjects in which it is hard to distinguish the top performers without going into very high level material.

Hm it'll be interesting to do chem like that most of the chem Ive done so far hasn't really been of the thought provoking type

yea.......cant really compare math and the chemistry.  all math olympiads require  tons and tons of time and you can never be sure that you master a subject.  for chemistry (and physics and bio etc) you read something, do some practice problems and after a while you know everything there is to know about it.  on every MC test, you know that there will be a questions that asks you to calculate the molar mass or a question asking about the shape of a molecule.

and the competition for math olmympiads is definitely more fierce than any other olympiad

[quote="feynman_wannabe"]Yea.. Sodium anything is a great oxidizing agent .... (i misbubbled that and put BrCl or something =( )
[/quote]

Sodium borohydride begs to differ LOL :rotfl:

Also, one of the 2007 US IChO medalists scored a 48 on the national exam MC that year (it was unusually hard...got a 42 myself, one camper from my section that year did get a 52 though).


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Let $f:[-1,1] \to \mathbb{R}$ be a continuous function. Show that:
a)if $\int_{0}^{1}f(sin(x+a))dx=0$ for every real $a$ then $f(x)=0$ for all $x$.
b)if $\int_{0}^{1} f(sin(nx))dx=0$ for every integer $n$ then $f(x)=0$ for all $x$.

The second one follows rather quickly, by reducing it to the first part:

If $g=f\circ\sin$, we will find $\int^1_0g(t+m)dt=0$ for all naturals $m\in\mathbb N$, which means that $\alpha(x)=\int^1_0g(t+x)dt$, which is a continuous function in $x$, takes the value $0$ for all $x=m+2n\pi,\ m\in\mathbb N,\ n\in\mathbb Z$, i.e. in a dense set, so it must be identically null, thus giving the hypothesis of the first part.

It's easy to prove, in general, that if $h$ is a $T$-periodic continuous fucntion, then $H(x)=\int^x_0hdt$ is the sum of a linear function and a $T$-periodic function. We apply this to $h=g$ to conclude that $G(x)=\int^x_0gdt$ is the sum of a linear function and a $2\pi$-periodic function. On the other hand, the hypothesis tells us that $G$ is $1$-periodic, and thus bounded. This means that the mentioned linear function is null, and $G$ is both $1$ and $2\pi$-periodic, and the conclusion that it must be constant follows immediately.


Ok I got this from MathWorld. With a maximum of h stamps, it appears that the answer is (h+3-b)b-2, where b>a. I don't understand why a isn't part of it though. Probably a=1 is necessary, since otherwise 1 is the smallest.

[quote="ElectricSnowman"]Huh? Why would it offend mathfanatic?  And if you think you're gonna offend someone by posting something, then why do you bother posting it?  Unless you like starting flame wars...[/quote]

Its about another thread, don't worry about it.


limit
logarithms
calculus
calculus computations

$ \lim_{x \to \infty} x^{\left[\frac{\log_e(2)}{1\plus{}\log_e(x)}\right]}$

Evaluate this limit without the use of l'hospital's rule.

Nice problem.  This should be a valid solution.

Since $ x\rightarrow \infty$, we can substitute $ x\equal{}e^x$.

$ \lim_{x\rightarrow \infty}e^{x\frac{\ln2}{1\plus{}x}}$

$ \equal{}e^{\ln2\lim_{x\rightarrow \infty}\frac{x}{1\plus{}x}}\equal{}e^{\ln2}\equal{}2$

[hide="Solution"]$ L\equal{}\lim_{x\to\infty}x^{\frac{\ln 2}{1\plus{}\ln x}}\equal{}\lim_{x\to\infty}e^{\frac{\ln 2\ln x}{1\plus{}\ln x}}$

$ \ln L\equal{}\lim_{x\to\infty}\ln 2\cdot\frac{\ln x}{1\plus{}\ln x}\equal{}\ln 2$

$ L\equal{}\boxed{2}$[/hide]


Find all integers $x,y$ such that

$(x-y)^{2}-5(x+y)+25=0$

[hide="Hint"]Working modulo $5$ and then working modulo $2$[/hide]

[hide="a messier approach that works"]
Substitute $u=x+y$ and $v=x-y$.  Notice that $u/2$ and $v/2$ must be integral, so substitute $\alpha = u/2$ and $\beta = v/2$ with the requirement that $\alpha$ and $\beta$ both be integral.

[/hide]

I like lordWings' way more.

calculuslover800: I think you mean that $u/2$ and $v/2$ must be integral.

I checked my [hide="solution"]
$x-y=0 \mod 5$, but substituting $x-y=5k$, we have
$10x=25(1+k+k^{2})$, which is impossible: $0=1 \mod 2$[/hide]

[quote="lordWings"]calculuslover800: I think you mean that $u/2$ and $v/2$ must be integral.

[/quote]

yes. edited.


geometry
geometric transformation
reflection

Find the number of $ 25$-digit positive integers satifying two conditions:

[color=red]*[/color]The digits of integer consists of exactly $ 20$ copies of '$ 1$' and $ 5$ copies of '$ 2$' .

[color=red]*[/color]If we remove the last $ k$ digits from the intger (where $ k$ is any positive integer less than $ 25$), the resulting integer has at least as many '$ 1$'s as '$ 2$'s.

 :starwars:

i believe all 25-digit numbers satisfying the first condition satisfy the second condition
please correct me if i am misunderstanding the problem
if so, then the answer is $ \binom{25}{5}$

[hide="unless"]
you mean that the second statement is true for all k
then:
the answer is simply $ \binom{25}{5}$ (total) minus the # of numbers that satisfy the first condition but not the second
to find which numbers do or do not satisfy the second condition, plot each 25-digit number as follows:
* start at the origin
* if the first digit is 1, move 1 unit to the right, otherwise, move 1 unit up
* do this for all other digits
the graph corresponding to 2112111211112111112111111 is in the attachment
(the graph should be a path connecting lattice points)
for every number that does not satisfy the second condition, there is a path that crosses (not just intersects) the green line (y=x), which means that at one point, there are more 2s than there are 1s
and for every such path, there is a point where the path first crosses over the line y=x
which means for every such path, there is a point where the path first intersects y=x+1 (the blue line)
reflect the path after that point across that point.
the end of the path will always be at point (4,21) nomatter where the path intersects y=x+1
so for every number that satisfies the first condition but not the second, there is a path from the origin to the point (4,21) only turning at lattice points and going only right or up
but is the converse true?
for every such path, there is a point where the path first crosses y=x+1
reflecting the rest of the path after that point yields a path that goes from the origin to (5,20)
so there is  1-1 correspondence!
obviously, cointing paths to (4,21) is easier
there are $ \binom{25}{4}$ of these
the answer is therefore $ \binom{25}{5} \minus{} \binom{25}{4}$
[/hide]


trigonometry
geometry
rectangle
rotation
algebra
system of equations

A cylinder with radius $ 4$ cm and perpendicular height $ 15$ cm is tilted so that it will just fit inside a $ 12$ cm high box. At what angle must the cylinder be tilted?

We don't really care about the three-dimensional part.  In the x-y plane, we get a rectangle with height 8 and length 15.  Now you can just you triangle geometry.

Come on! Anyone, i cant do it, i need to know. Please...

Here is what the answer in the book says [hide="book answer"]$ 16^\circ 50'$[/hide]

If we do what JRav suggested we get $ 28^\circ 4'$

Any ideas

[hide]Well, say we have the side of length horizontal, and the side of length 8 vertical. Now, slowly, we rotate this rectangle about the lower-left corner by some angle $ \theta$.

We know 12 is the height of the upper vertex. Note that the side of length 15 contributes $ 15 \sin \theta$, and the side of length 8 contributes $ 8 \cos \theta$. We thus have the system of equations:

$ 15s \plus{} 8c \equal{} 12 \qquad s^2 \plus{} c^2 \equal{} 1$

Where, $ s$ and $ c$ represent $ \sin \theta$ and $ \cos \theta$, respectively.

Solving, we find that $ (s,c) \equal{} \left(\frac {180 \pm 8 \sqrt {145} }{289} , \; \frac {96 \mp 15 \sqrt {145}}{289} \right)$

We want the case in which $ \theta < 90$, so $ c$ must be positive. Therefore, the solution is:

$ (s,c) \equal{} \left(\frac {180 \minus{} 8 \sqrt {145} }{289} , \; \frac {96 \plus{} 15 \sqrt {145}}{289}\right)$

The answer is therefore $ \boxed{ \theta \equal{} \arcsin \left(\frac {180 \minus{} 8 \sqrt {145} }{289}\right) }$[/hide]

In this case wouldn't $ \theta$ be negative?

$ 180 \minus{} 8 \sqrt{145} \approx 180 \minus{} 8 \cdot 12 > 0$

$ \text{sign} (\arcsin x) \equal{} \text{sign}(x)$, so we're dealing with a positive angle.

The particular angle is roughly $ 16.82838522012 ^ {\circ}$

Oh yeah, I have no idea why I was approximating $ \sqrt {145}$ to be $ 120$ in my head, maybe I need more sleep...


I dont understand this problem...help!

While in Britain Bobby decides to put his money in the bank that pays 6% simple interest per annum, and he gets a part-time job to cover his expenses. Bobbyy remains in britain for 6 months.

Calculate how much interest he recieves for the six months. which equations would you use?

I know its something like....

a=p(1+r/100)^something lol

I'm not sure if the question makes sense. He gets 6% [url=http://en.wikipedia.org/wiki/Simple_interest#Simple_interest]simple interest[/url] per year, and stays in Britain for only half a year. We don't know what money he has deposited either, so I guess it's asking for a percent rather than a cash value of interest. Presuming that he gains interest every month, despite being given the annual figure, I think the answer would be:

[hide="Answer"]
$ \frac {6}{12} \cdot 6 \equal{} \boxed{3\%}$.

So he gets 3% interest, since, after all, by my assumptions he would be getting $ 0.5\%$ interest per month and nothing is compounded so it's just $ 0.5 \plus{} 0.5 \plus{} 0.5 \plus{} 0.5 \plus{} 0.5 \plus{} 0.5 \equal{} \boxed{ 3\%}$. If it was compound interest then it would be a lot harder to work out and would require some formula.[/hide]

Go, find your textbook.  Find the chapter in your textbook where problems of this sort are introduced.  Read this chapter.  Then go solve the problem yourself.


function
complex analysis
complex analysis unsolved

Let $ f$ be an entire function with an essential singularity at $ \infty$ , Prove f is polynomial.

You want a non-essential singularity at $ \infty$, as otherwise it is wrong.

[quote="ZetaX"]Of what¿
You should show some work...[/quote]

I'm agree with you, it's not true that someone considers ML as an exercise solver.


For  positive  real  numbers  a,b,c  with  a+b+c=abc  show  that
1/sqrt(1+a^2) + 1/sqrt(1+b^2) + 1/sqrt(1+c^2) <=3/2
Please  give  me  another solution  than  x=tan^(-1)a, y=tan^(-1)b , z=tan^(-1)c ....
Thanks

Why? This is a vry nice solution and its classical.

Moreover, the problem is clearly made from this substitution...

Pierre.

Just  wanted  to  see  another  solution...but  that  one  is  enought...
thanks!! :)

Don't get upset, please. Ok, here is another solution:
   Let ab=1/z, bc=1/x and ca=1/y. Then x+y+z=1 and a=sqrT(x/yz) and so on. So we need that sum sqrt((xy)/(xy+z))<=3/2. But sum sqrt((xy)/(xy+z))=sum sqrt( x/(z+x)*y/(z+y))<=1/2 sum (x/(z+x)+y/(z+y))=3/2.

i  am  not  apset  harazi!!!!
and thanks  for  the solution!!!!


If $ g(f(x))\equal{}x^2\minus{}1$ and $ f(g(x))\equal{}a^2\minus{}4a\plus{}5$ find $ f(3)\minus{}g(8)$

[quote="rojmmals"]$ f(g(x)) \equal{} a^2 \minus{} 4a \plus{} 5$[/quote]
Where did $ a$ come from :?:  I think it is not the appropriate forum either. This should go to a Algebra forum :arrow:


real analysis
calculus
calculus computations

$a_{n}$=cardinality of natural numbers $\le n$ which begin with 3.
Find
$\liminf_{n\to \infty}\frac{a_{n}}{n}$
$\limsup_{n\to \infty}\frac{a_{n}}{n}$

The numbers beginning with $3$ appear in intervals $3\cdot 10^{k}\ldots 4\cdot 10^{k}-1$, with $10^{k}$ numbers in each interval.  Thus $\liminf$ is realized by the sequence $n_{k}=3\cdot 10^{k}-1$ and $\limsup$ is realized by the sequence $n_{k}=4\cdot 10^{k}-1$. Unless I messed up the bookkeeping, they are $\frac{1}{27}$ and $\frac{5}{18}$ respectively.


I have a general question on the AMC 8 competition. So I was hunting through some of my old awards and stuff, and found an old award from AMC. It said I got third place. Now, I'm pretty sure I knew no math back then, so I'm getting the feeling this is some sort of fluff award. Could someone therefore enlighten me on the awards structure of the AMC-8?

Thanks,
sran

I believe it's third place in the school; see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=amc+award&t=23052]this topic[/url], in which another person found an old third place award.


Hi,

I was wondering if you could help me out on this problem.

[b]Show that an n*n matrix A is invertible exactly when 0 is NOT an eigenvalue of A[/b]

The fact that $0$ is an eigenvalues implies that $\det(A)=0$ .
Should $A$ be invertible , then $AA^{-1}=I_n$so $\det(A)\det(A^{-1})=1$ so $\det(A)\neq 0$ . Contradiction.


Louisa and Darya each rolled three balls toward lines that are
1, 2 and 3 meters away. A person's point total is determined by
squaring the distance between each ball and the line it lands
closest to and then adding the three squares together. The
lowest score wins. Louisa's balls rolled 0.9, 2.2 and 2.7 meters,
while Darya's rolled 1.2, 1.8 and 2.8 meters. By how many
points did Darya win? Express your answer as a decimal to the
nearest hundredth.

this ones pretty calculator bashable: $ (.1^2 \plus{} .2^2 \plus{} .3^2)/3$ is around .37 and $ (.2^2 \plus{} .2^2 \plus{} .2^2)/3$ is around .35, the difference is .02


Please do not double post ;)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302416
Also, I think you have made a mistake in typing because the LHS is odd, while the RHS is even..... please correct the problem .  :maybe:


\[ a,\ b,\ c>0;\: abc\equal{}1\Longrightarrow\frac{a\plus{}b\plus{}c}{3}\ge\sqrt[5]{\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{3}}\]

[quote="Mîndrilă Claudiu"]
\[ a,\ b,\ c > 0;\: abc \equal{} 1\Longrightarrow\frac {a \plus{} b \plus{} c}{3}\ge\sqrt [5]{\frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{3}}
\]
[/quote]
We should prove: $ \left ( \frac{a\plus{}b\plus{}c}{3} \right ) ^5 \ge abc \left ( \frac{a^2\plus{}b^2\plus{}c^2}{3} \right ) \equal{} abc \left ( 3\left ( \frac{a\plus{}b\plus{}c}{3} \right )^2 \minus{} 2 \left ( \frac{ab\plus{}bc\plus{}ca}{3} \right ) \right ) \iff$

$ \left ( \frac{a\plus{}b\plus{}c}{3} \right )^5 \plus{} 2 abc \left ( \frac{ab\plus{}bc\plus{}ca}{3} \right ) \ge 3abc \left ( \frac{a\plus{}b\plus{}c}{3} \right ) ^2$. By AM-GM:
$ LHS \ge 3 \sqrt[3]{\left ( \frac{a\plus{}b\plus{}c}{3} \right )^5(abc)^2\left ( \frac{ab\plus{}bc\plus{}ca}{3} \right )^2} \ge RHS \iff$
$ \left ( \frac{ab\plus{}bc\plus{}ca}{3} \right )^2 \ge abc \left ( \frac{a\plus{}b\plus{}c}{3} \right ) \iff$
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \ge a^2bc \plus{} ab^2c \plus{} abc^2 \iff$
$ a^2(b\minus{}c)^2 \plus{} b^2(c\minus{}a)^2 \plus{} c^2(a\minus{}b)^2 \ge 0$. QED.


calculus
derivative
function
limit
trigonometry
calculus computations

Is there a real function which has a derivative that exists at a point but is not continuous at that point?

No there isnt
$ \begin{array}{l}
 \exists {x_0} \in {D_f}: f'\left( {{x_o}} \right) \equal{} k \in R \Leftrightarrow \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) \minus{} f\left( {{x_0}} \right)}}{{x \minus{} {x_0}}} \equal{} k \\
 \mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) \minus{} f\left( {{x_0}} \right)} \right] \equal{} \mathop {\lim }\limits_{x \to {x_0}} \left( {\frac{{f\left( x \right) \minus{} f\left( {{x_0}} \right)}}{{x \minus{} {x_0}}} \cdot \left( {x \minus{} {x_0}} \right)} \right) \equal{} k \cdot 0 \equal{} 0 \Leftrightarrow \mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) \equal{} f\left( {{x_0}} \right) \\
 \end{array}$

I meant that the derivative exists at the point but the derivative is not continuous at that point.  I know that the existence of the derivative implies continuity for the original function.

The classic example:

$ f(x) \equal{} \begin{cases}x^2\sin\left(\frac1x\right) & x\ne 0 \\
0 & x \equal{} 0\end{cases}.$

Then $ f'(0) \equal{} 0$ exists but $ f'$ is not continuous. You can tweak the example (say, change $ \frac1x$ to $ \frac1{x^2}$ ) to get a function whose derivative exists everywhere but whose derivative is unbounded in a neighborhood of zero.

Or take $ f$ as I defined it above and let $ g(x) \equal{} f(x) \plus{} \frac {x}4.$ Then $ g$ is differentiable everywhere and $ g'(0) > 0,$ but there is no neighborhood of zero on which $ g$ is an increasing function.


[b]Michael Lee is fouled in the act of shooting a 3-pointer, and will get 3 shots at the free-throw line.  If Lee is a 70 percent free-throw shooter, what is the probability that he will make at least 2 of his 3 free-throws?[/b]
[hide="Are you sure you want to check your answer??"][hide="Well then click here!!"][hide="No!!! Not there, there!"]$ 78.4%
$[/hide][/hide][/hide]

[hide] 4 possibilities: make-make-miss; make-miss-make; miss-make-make; make-make-make.
So $ 3(0.7+0.7+0.3)$(first 3 are the same thing in a different order)$ +0.7^3=78.4%
$[/hide]

[hide]
CASE 1: Makes all 3: clearly $ 0.7^3$.
CASE 2: Makes 2: There 8are 3 ways to choose the one he misses, so $ 3(0.7*0.7*0.3)$.
Just add.[/hide]

I was too lazy to use the code for fractions and just put it in decimals, it's easy to convert.

I agree with math154, the question must specify how you should express your answer otherwise fraction, decimal, and percent would have to be accepted.

[quote="math154"]Technically, percent IS probability. Both answers would have to be accepted, as the form of answer was not specified.[/quote]

percent is NOT really probability

[quote="ajain"][b]Michael Lee is fouled in the act of shooting a 3-pointer, and will get 3 shots at the free-throw line.  If Lee is a 70 percent free-throw shooter, what is the probability that he will make at least 2 of his 3 free-throws?[/b]
[hide="Are you sure you want to check your answer??"]$ 78.4%
$[/hide][/quote]
[hide="answer"]
BINOMIAL DISTRIBUTION!
$ \binom{3}{2}(.7)^{2}(.3)^{1}+\binom{3}{3}(.7)^{3}(.3)^{0}$
$ 3*.49*.3+.343$
$ .784 or 78.4%
$[/hide]

OR $ \frac{98}{125}$, as stated earlier.

[quote="math154"]then what is it?[/quote]

well, you can express probability as fractions or decimals as well

and there's also percent of, percent off, etc.  Those clearly aren't probabilities

[quote="math154"]I'm talking about percent [b]chance[/b], though. THAT's probability, eh?[/quote]

eh, maybe you should specify that next time then

Yeah, so basically 3 ways to make 2 out of 3, and 1 way to make all three.

So $ 3\left (\frac{7}{10}\plus{}\frac{3}{10}\plus{}\frac{3}{10}\right ) \plus{} \left (\frac{7}{10}\right )^3 \equal{} \frac{98}{125} \equal{} 0.784 \equal{} 78.4\%$.

By the way, this is funny since I [b]have[/b] a classmate named Michael Lee! :lol:

:rotfl:  :rotfl:  :rotfl:
Weird. Whatever. Can anyone post a really really hard problem from some nats test?

[quote="stevenmeow"]dont you mean 98/125 or $ \frac {98}{125}$?
the problem asked for probability not percent accuracy
you have to be careful on those[/quote]

Actualy this is a multiple choice problem and it [b]was[/b] asking for percent.


[quote="i_like_pie"][quote="Ignite168"]Wow..

This kind of thing seems unrealistic to most people..

Imagine the military in your country suddenly revolting to overthrow the government. The governing laws, the Constitution, the people such as Parliament and the Prime Minister are now suspended.

It has been bloodless though, and I say they deserve credit for that.

What are your thoughts on this? Do you believe the military was justified in revolting? Or do you think they were not justified?[/quote]
I've heard about this whole Thailand thing, but I don't really get it. Who is doing what and why? Why can't the world be peaceful for a single DAY?[/quote]
no the world can't be peaceful even for a single minute.
I blame Bush

[quote="Ignite168"]Wow..

This kind of thing seems unrealistic to most people..
Imagine the military in your country suddenly revolting to overthrow the government. The governing laws, the Constitution, the people such as Parliament and the Prime Minister are now suspended.

It has been bloodless though, and I say they deserve credit for that.

What are your thoughts on this? Do you believe the military was justified in revolting? Or do you think they were not justified?[/quote]

One would have at least to live in Thailand. It's not a country that gathers much attention. Until now. There are many accusations of corruption pending on the deposed prime minister so it could be justified. The ways of the countries who do not belong' let's say, to the privileged developed sector of the world, are significantly different from those of the first world. This is logical. I think one should not judge events happening in the third world by first world paradigms but by a careful reflection and knowledge of what is happening and what the life is like in those countries. So who knows, a coup d'etat is in principle 180º against democracy but democracy as it is understood results a difficult concept for me to apply globally. In the other hand, the coup might be the solution to thailand's problems. The best thing would be if there were somebody from Thailand to explain.

Also Budapest seems to be on fire though it can be the press exaggerating the facts, and sure we could know something from the hungarian guys here.

Ursula

About the Hungary comment, I don't think there are many Hungarians here (maybe because MathLinks was established by Romanians? :( ).  I think there's just one or maybe two.

Yes, please, do we have Thailand members here?

As far as I know, the coup has been bloodless.  They have been banning political gatherings and restricted media.

What IRKS me is that the first thing Belgian media say as a headline is "Belgians in Thailand didn't notice a thing". What do I care about a handful of Belgian tourists in a country where tens of millions of people live. :ninja:

[quote="fredbel6"]What IRKS me is that the first thing Belgian media say as a headline is "Belgians in Thailand didn't notice a thing". What do I care about a handful of Belgian tourists in a country where tens of millions of people live. :ninja:[/quote]
US media is like that too.


algebra
system of equations
algebra unsolved

Find $ x, y \in R$ of this system of equations: $ \begin {cases} x \plus{} y \equal{} x^3(y^3 \minus{} 5) \\
y^3 \minus{} x^3 \equal{} 7 \end {cases}$.

Let $ z\equal{}y^3$, then $ x\equal{}(z\minus{}7)(z\minus{}5)\minus{}y,z\minus{}7\equal{}(z^2\minus{}12z\plus{}35\minus{}y)^3$ or
$ x\equal{}y^6\minus{}12y^3\minus{}y\plus{}35, y^3\minus{}7\equal{}(y^6\minus{}12y^3\minus{}y\plus{}35)^3$.
These equation of 18 degree had only 2 real roots. Ont of roots is $ y\equal{}2\to x\equal{}1$, second $ 1<y<2$.


How many differently dimensioned rectangular prisms of volume 216 can be made from assembling together unit cubes?

I think its $ 19$..did it mentally..is that right?

Answer

You can reduce the amount of work by making use of the $ \tau$ function.

Let the sides be $ a$, $ b$ and $ c$ with $ a \leq b \leq c$.

$ \Rightarrow a^{3}\leq abc = 216 \Rightarrow a \leq 6$

Note $ 216 = 2^{3}3^{3}$

Let $ a = 1 \Rightarrow bc = 2^{3}3^{3}$

$ \Rightarrow \tau (bc) = 4\cdot 4 = 16$

$ 216$ has $ 16$ distinct factors which can be paired to give 8 possible ways.

Let $ a = 2 \Rightarrow bc = 2^{2}3^{3}$

$ \Rightarrow \tau (bc) = 3\cdot 4 = 12$

One factor of $ 2^{2}3^{3}$ is $ 1$ which has already been counted, so a pair can be removed giving 5 possible ways.

Let $ a = 3 \Rightarrow bc = 2^{3}3^{2}$

$ \Rightarrow \tau (bc) = 4\cdot 3 = 12$

Two factors of $ 2^{3}3^{2}$ have already been counted giving 4 possible ways.

Let $ a = 4 \Rightarrow bc = 2 \cdot 3^{3}$

$ \Rightarrow \tau (bc) = 2\cdot 4 = 8$

3 ways already counted giving 1 possible way

Let $ a = 6 \Rightarrow bc = 2^{2}3^{2}$

$ \Rightarrow \tau (bc) = 3\cdot 3 = 9$

4 ways already counted giving 1 possible way.

Hence there are a total of 19 ways.

Although this method looks long when written down, the number of possible ways for each $ a$ can  be written down in seconds without all this working.


The surface of big lake is 31 above the surface of long lake. long lake is half as deep as big lake and the bottom of long lake is eight feet below the bottom of big lake. How deep is each lake?

[quote="nick42"]uhhhh...could you put the problem into simple english? cuz i don't get any of it, especially the part about one surface area being "above" the other.[/quote]
The reason you did not get it is since you need to draw a picture. It is in simple English

[quote="mz94"][quote="smartytarty"]The reason you did not get it is since you need to draw a picture. It is in simple English[/quote]

it is also in horrible grammar. it is terribly worded because you decided to make life easier for yourself and truncate some sentences. im almost positive thats not how its written. please provide the full question.[/quote]
I did give you the full question and there is nothing wrong with the problem. Its just you do not know how to do it>

[quote="PowerOfPi"]People should stop spamming and answer the question instead of whining.
[quote]Before posting, think hard whether your post is necessary or not. Remember, you do not have to press the "Submit" button.[/quote]
This was by isabella2269.[/quote]
Who are you reffering to me or mz94

I mean the people complaining about the grammar and not trying to solve the problem.

If I percieved said quesiton without errs, it is able to be deducted that by choosing the word 'above', he meant greater than, and by choosing the word 'below' he meant less than, so you just use last 2 equations. One is half as deep as the other, while being 8 feet shallower. So that means deeper one is 16 feet, and shallower one is 8 feet. That junk about 31 and surface in the beginning is extra info


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Let $f: [0,1]\to\mathbb{R}$ a continuous function such that $\int_{x}^{1} f(t)dt \geq\frac{1-x^{2}}{3}.$
 Prove that $\int_{0}^{1} f^{2} (t) dt\geq\frac{1}{3}$ ;)

Aaaa, don't say. Just apply Cauchy and it's over. Julien, everybody could have written that. ;)

That's the case though (unless I made another stupid mistake  :( ). Ok let's see:

Cauchy-Scwharz inequality gives:

$\frac{1-x^{2}}{3} \leq \int_{[x,1]}{f(t)}dt \leq \sqrt{\int_{[x,1]}{f^{2}(t)}dt} \sqrt{1-x}$, whence:
$(\frac{1-x^{2}}{3})^{2} \leq (1-x)\int_{[x,1]}{f^{2}(t)}dt$, and then take $x=0$ and you're done.

Just look who's speaking. Anyway, the real reference for this problem is a very old romanian national olympiad.

Yes, Harazi you were right about me and you are right about the refference. Anyway, it's nice to hear from you again. :lol:

Ok, just saw the mistake ... sorry about that.

I didn't say that. I said that there are problems which are very hard to do. ;)

Yeah, I know, and I do agree with it.

I was just joking around ;)

I don't think it's true.
I have an example:f(x)=(1+x)/3
It satisifies all the hyppothesis but the result is not true/
 :?:

I don't quite think so. You should think again. ;)

nguyenkims is correct. His example does satisfy the required hypotheses and

$\int_0^1[f(x)]^2dx=\frac7{27}$ which is smaller than $\frac13.$

julien_santini's proof in #4 actually shows that

$\int_0^1[f(x)]^2dx\ge \frac19.$

Actually differentiating makes it clear that a sufficient condition for $f$ to satisfy the first inequality is: $f(x) \geq \frac{2}{3}x$. In particular taking $f(x)=\frac{2}{3}x$ gives $\int_{[0,1]}{f^{2}(t)dt} = \frac{4}{27}$ which is of course smaller than  $\frac{1}{3}$  and slightly greater than the  $\frac{1}{9}$  previously announced.


Prove:
( (a^2 + b^2) / 2 ) ^ (1/2) >= (a + b)/2 >= (ab)^(1/2) >= 2 / (1/a + 1/b)

by using Geometry.

i know it's classic...but i do want to see the ideas.
thanks... :)


let $\frac{k}{a_{k}}=b_{k}$, then $3b_{k}=2+b_{k-1}$. The remaining part you can solve it easily.


An odd number n teams take part in a hockey tournament. Each team plays exactly one match with any other team. A team receives 2 points for a win, 1 point for a draw, and 0 point for a loss. After the tournament is over, no two teams have the same number of points. Find the largest possible value of the draws in this tournament.

This isn't too bad -- try the first few small cases and the pattern becomes clear.  Proving it is then a little tricky, but not too much if you describe it well.  I was going to say, "This question gets simpler if you divide it into the cases n even and n odd," but then I noticed that we're only dealing with the odd case, anyhow.  (The even case is pretty similar.)

A hint:

[hide]Suppose every team drew every match. Now, what happens when a draw is replaced by a game that is won? How many of the games must be altered so that the outcome has the desired property?[/hide]


trigonometry
function
LaTeX
complex numbers

What does cis theta mean? What does that have to do with de Moiere theorem? Can someone provide a link for a material which teaches you elementary complex numbers? :oops:

Thnx, any help would be greatly appreciated  :)

$cis \theta = e^{i\theta}= i\sin \theta+\cos\theta$

To prove this identity, calculate the taylor series for $e^{i\theta}$ and $i\sin \theta+\cos\theta$.

Basically, this function $f(t) = e^{it}$ maps the real number line into the unit circle in the complex plane.

DeMoivre's Theorem: \[z^{n}= \left(re^{i\theta}\right)^{n}= r^{n}e^{in\theta}= r^{n}(i\sin n\theta+\cos n\theta)\] This lets you raise any complex number to any power.  Above you can see that a complex number has three forms, rectangular, exponential, and trigonometric.  The $r$ denotes the magnitude of $z$.

Let me know if any of this makes sense or if you have any questions.

Just a few question:

How can you use complex numbers to solve the problem of root of unity also posted in the forum?

and prove:

(1+i)^10=32i

Thax :)

[quote="D.P.L"]Just a few question:

How can you use complex numbers to solve the problem of root of unity also posted in the forum?

and prove:

(1+i)^10=32

Thax :)[/quote]
it is wrong

notice that $(1+i)^{2}=2i$

so $(1+i)^{10}=(2i)^{5}=32i^{5}=32i$

[hide="How to find complex roots of unity"]
Let $z$ be the $n$-th complex root of unity which means $z^{n}=1$. Then it's obvious that $|z|=1$, so we have $z=\cos \alpha+i\sin \alpha$ for some $0\leq \alpha <2\pi$. Then $\cos 0+i\sin 0=1=z^{n}=\cos n\alpha+i\sin n\alpha$, so $z^{n}=1\iff n\alpha=2k\pi\iff \alpha=\frac{2k\pi}{n}$ for some $k\in\mathbb{Z}$. Using $0\leq \alpha<2\pi$ we have $k\geq 0$ and $k<n$. Finally $k\in \{0,..,n-1\}$ and $z=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}$ for any $k\in \mathbb{Z}$, $0\leq k \leq n-1$.
[/hide]

Convert the complex number into Cis form, you get $2*(\cos 5\pi/3+\sin 5\pi/3)$ so therefore, that raised to the 5th power is $2^{5}*(\cos 5*5\pi/3+\sin 5*5\pi/3)$, which equals $32*(\cos \pi/3+\sin \pi/3)$ and so you get the desired result.

[quote="D.P.L"]Say if z=a+bi and r=sqr(a^2+b^2). How do you find the angle for cis if z=a+bi=r=sqr(a^2+b^2)*cis theta? :?:  :maybe:[/quote]
You should try learning $\LaTeX$.  :wink:
For $z=a+bi$, in polar form, $r=\sqrt{a^{2}+b^{2}}$ and $\theta=\tan^{-1}\frac{b}{a}$.


The product of 4 distinct positive integers $ a,b,c,d$ is $ 8!$ . The numbers also satisfy $ ab\plus{}a\plus{}b\plus{}1$ $ \equal{}$ $ 323$, $ bc\plus{}b\plus{}c\plus{}1$ $ \equal{}$ $ 399$.Find $ d$.  :rotfl:

[quote="apratimdefermat"]The product of 4 distinct positive integers $ a,b,c,d$ is $ 8!$ . The numbers also satisfy $ ab \plus{} a \plus{} b \plus{} 1$ $ \equal{}$ $ 323$, $ bc \plus{} b \plus{} c \plus{} 1$ $ \equal{}$ $ 399$.Find $ d$.  :rotfl:[/quote][hide]$ (a\plus{}1)(b\plus{}1)\equal{}323\equal{}17\cdot19, 322 \not| \, 8! \Rightarrow \{a,b\}\equal{}\{16,18\}$. $ (b\plus{}1)(c\plus{}1)\equal{}399\equal{}19\cdot21, 132,398 \not| \, 8! \Rightarrow \{b,c\}\equal{}\{18,20\}$ or $ \{56,6\}$. $ \therefore a\equal{}16, b\equal{}18, c\equal{}20, d\equal{}\frac{8!}{abc}\equal{}7$.[/hide]

$ (bc\plus{}b\plus{}c\plus{}1)\minus{}(ab\plus{}a\plus{}b\plus{}1)\equal{}(b\plus{}1)(c\minus{}a)\equal{}399\minus{}323\equal{}76\equal{}4 \cdot 19$.
So, $ b\equal{}3 \ \text{or} \ 18$.

$ b\equal{}3 \Rightarrow \left\{\begin{array}{l}4c\plus{}4\equal{}399 \\ 4a\plus{}4\equal{}323\end{array}\right. \Rightarrow \left\{\begin{array}{l}c\equal{}\frac{395}{4} \\ a\equal{}\frac{319}{4}\end{array}\right.$
This is incorrect.

$ b\equal{}18 \Rightarrow \left\{\begin{array}{l}19c\plus{}19\equal{}399 \\ 19a\plus{}19\equal{}323\end{array}\right. \Rightarrow \left\{\begin{array}{l}c\equal{}20 \\ a\equal{}16\end{array}\right.$
This is correct.

Thus, $ d\equal{}\frac{8!}{abc}\equal{}\boxed{7}$.


See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344949#p344949]USAMO 1991 Problem 4[/url].


geometry
trapezoid
perpendicular bisector

[img]http://img46.imageshack.us/img46/5919/trapezoidoq2.png[/img]

[b]Given:[/b] $BE$ bisects $\angle AEC;\,BE$ is the $\perp$ bisector of $AC$ and $DF$

[b]Prove:[/b] $\angle AFE\cong\angle CDE$

[hide="Answer"]
Triangles ABE and CBE are obviously congruent, meaning angles BEA and CEA are congruent. That also means that AE and CE are congruent. From the given, FE and DE are congruent (BE bisects FD), meaning by SAS, triangles AEF and CED are congruent.

Finally, by CPCTC, angles AFE and CDE are congruent.
[/hide]

[quote="Treething"]...angles BEA and CEA are congruent...[/quote]
I think that's a typo.

[hide]BE is also the perpendicular bisector of AC, so by ASA congruency, $\triangle ABE \cong \triangle CBE$, so $AE=CE$

Since $\angle AEB=\angle CEB$ and $\angle FEB=\angle DEB=90$, subtracting the first from the latter gives $\angle AEF=\angle CED$.

We already know that $FE=DE$.

From these three pieces of info, by SAS congruency, $\triangle AEF\cong\triangle CED$, so $\angle AFE\cong\angle CDE$.[/hide]


geometry
incenter
trigonometry
geometry unsolved

Let incenter of triangle $ ABC$ is $ I$. If $ AI\equal{}3, BI\equal{}5, CI\equal{}7$ then find minimum value of area of $ ABC$

We have

$ \sin \frac{A}{2}\equal{}\dfrac{r}{3}, \ \ \sin \frac{B}{2}\equal{}\dfrac{r}{5}, \ \ \sin \frac{C}{2}\equal{}\dfrac{r}{7}$

From the identity

$ \sum \sin ^2 \frac{A}{2}\plus{}2\prod\sin \frac{A}{2}\equal{}1$,

we get

$ \dfrac{1891r^2}{105}\plus{}2r^2\equal{}1$.

I tested it with the software and it doesn't has any real solution.


We have a quadrilateral $ABCD$ inscribed in a circle of radius $r$, for which there is a point $P$ on $CD$ such that $CB=BP=PA=AB$.

(a) Prove that there are points $A,B,C,D,P$ which fulfill the above conditions.

(b) Prove that $PD=r$.

[i]Virgil Nicula[/i]

The triangle $\triangle ABP$ is equilateral. Let (B) be a circle centered at B and with radius AB = AP = BP = BC and let E be the other end of its diameter through A. The vertex C is located on the arc PE of the circle (B) opposite to the point A. Let the circumcircle (O) of the triangle $\triangle ABC$ with radius r meet the line CP again at D and denote $\phi = \angle PBC,\ 0 < \phi < 120^\circ.$ The triangle $\triangle BPC$ is isosceles with BP = BC, hence $\angle BCP = \angle BPC = 90^\circ - \frac \phi 2$ and

$\angle DPA = 180^\circ - (\angle APB + \angle BPC) = 180^\circ - \left(60^\circ + 90^\circ - \frac \phi 2\right) = 30^\circ + \frac \phi 2$

Quadrilateral ABCD is cyclic, hence $\angle DAB = 180^\circ - \angle BCD = 90^\circ + \frac \phi 2$ and

$\angle DAP = \angle DAB - \angle PAB = 90^\circ + \frac \phi 2 - 60^\circ = 30^\circ + \frac \phi 2$

Thus the triangle $\triangle DAP$ is isosceles with DA = DP. Let (D) be a circle centered at D and with radius DA = DP. The central angle $\angle ADP$ of the minor arc AP of the circle (D) is

$\angle ADP = 180^\circ - 2\ \left(30^\circ + \frac \phi 2\right) = 120^\circ - \phi$

The angle $\angle CAB$ spanning the minor arc BC of the circle (O) is

$\angle CAB = 90^\circ - \frac{\angle ABC}{2} = 90^\circ - \frac{60^\circ + \phi}{2} = 60^\circ - \frac \phi 2$

Since BC = AP and $\angle CAB = \frac{\angle ADP}{2}$, the circles $(D) \cong (O)$ are congruent and DP = r.


algebra
polynomial
inequalities proposed
inequalities

Let $ a,b,c$ be positive real numbers satisfy: $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Find every real number $ k$ such that $ (k\minus{}ab)(k\minus{}bc)(k\minus{}ca) \geq (k\minus{}1)^3$.

\[ k\le{1\minus{}\frac{\sqrt{3}}{3}}\bigvee{k\ge{k_0}},\]

 where

$ k_0 \equal{}1.595538476\ldots$ is a root of the polynomial :

$ 72k^4\minus{}240k^3\plus{}268k^2\minus{}126k\plus{}27$.

How did you get that? I thought $ k \leq 1\minus{}\frac{\sqrt{3}}{3}$ or $ k \geq 1\plus{}\frac{\sqrt{3}}{3}$

I see. I got $ k \geq 1 \plus{} \frac{\sqrt{3}}{4}$ when letting $ a\equal{}b\equal{}\sqrt{\frac{3}{2}},c\equal{}0$. But how can you got the incredible equation whose solution was $ k_0$? And with that value $ k_0$, how did you prove the iequality??? :maybe:

I'm sorry, can you give me a link to a site, where I can read something about generating polynomials like [b]fjwxcsl[/b] did?

I know nothing about the generating polynomials which [b]fjwxcsl[/b] pointed out. By the way, can anybody give some explanation for it or even solve the problem completely?


The diagram to the right shows several numbers in the complex plane. The circle is the unit circle centered at the origin. One of these numbers is the reciprocal of $F$. Which one?

$\text{(A)} \ A \qquad \text{(B)} \ B \qquad \text{(C)} \ C \qquad \text{(D)} \ D \qquad \text{(E)} \ E$

[hide]
$C$

Let $F=a+bi$ . Clearly $a^2+b^2\geq 1$ since the circle is unit circle . Then
$1/F = 1/(a+bi)=(a-bi)/(a^2+b^2)$ . Since $a,b>0$ $1/F$ liesin the fourth quadrant . Also $|1/F|=1/\sqrt{a^2+b^2}\leq 1$ which is inside the circle .

[/hide]


function
number theory unsolved
number theory

Find all $ f: N\rightarrow N$ such that

$ x^2\plus{}f(y)$ divides $ f(x)^2\plus{}y$ for all $ x,y\in N$

Not knowing whether your $ \mathbb{N}$ includes $ 0$, I assumed not.

$ 1^{2} \plus{} f(1)$ divides $ f(1) ^ {2} \plus{} 1$
$ \longrightarrow f(1) \plus{} 1$ divides $ 2$
$ \longrightarrow f(1) \equal{} 1$

Thus for all natural numbers $ x$, we have:
1 : $ 1 \plus{} f(x)$ divides $ f(1) ^ {2} \plus{} x \longrightarrow f(x) \leq x$
2 : $ x^{2} \plus{} f(1)$ divdes $ f(x)^{2} \plus{} 1 \longrightarrow x \leq f(x)$
Combining, we have $ f(x) \equal{} x$.


Each triangle shown is equilateral. What is the probability that a point chosen at random within the largest triangle will lie within the unshaded area? Express the result as a common fraction.

hmmm...

a pic would be very helpful...  :D  :D  :D


trigonometry
geometry
inradius
MATHCOUNTS
area of a triangle
Heron\u0027s formula
power of a point

5. One side of a triangle is divided into segments of lengths 6 cm and 8 cm by the point of tangency of the inscribed circle. If the radxius of the circle is 4 cm, what is the length, in centimeters, of the shortest side of the triangle?

I know this is pretty easy with trig, and possible using Hero's Formula, but I am almost certain that there's some way to get the answer with pure geometry, without Hero's formulas. Anyone got any ideas?

[color=#F5F8FC][hide="Answer for trig."]Draw the three angle bisectors up to the incenter. Then draw the radii of the incircle to the tangency points. Now, use trig to get the four known angles out of the six at the center. Subtract this from 360, and divide by 2, to get one of the angles that was unknown. Get the tangent of this angle, and multiply by 4 to get the unknown part. Now find the shortest side of this 13-14-15 triangle.[/hide][/color]

[hide]You need to know Heron's formula and the Semiperimeter-Inradius formula.

Herons doenst turn out too bad.  $\sqrt{(14+x)(x)(6)(8)}$.

To get area, you can do "s*r" where s is the semiperimeter and r is the inradius.

This gives you $(14+x)(4)$.

Equate them and simplify.  X=7.  The shortest side is thus 6+7=$\boxed{13}$[/hide]

[quote="robinhe"]
Wait.  How did you get those expressions if you only know one side of the triangle?
 :?[/quote]

Heron's formula: $area=\sqrt{s(s-a)(s-b)(s-c)}$

$s-b=6$
$s-c=8$

($s-b$ is the length of the tangent segment from point $B$ to the incircle)

[quote="JesusFreak197"]I would approach it by drawing similar triangles and using Pythagorean.[/quote]

can you post your solution please?
I think that it is difficult.

[quote="robinhe"][quote="kyyuanmathcount"][hide]$\sqrt{(14+x)(x)(6)(8)}$.

To get area, you can do "s*r" where s is the semiperimeter and r is the inradius.

This gives you $(14+x)(4)$.

[/hide][/quote]

Wait.  How did you get those expressions if you only know one side of the triangle?
 :?[/quote]

By power of a point, the sides are 6+x, 8+x, and 6+8.

Except that's not what you put in the equation.

Or, you can use similar triangles, which everyone in MathCounts should know, rather than Power of a Point, which I would expect is known by very few.

Draw the triangle and inscribed circle, and draw the three radii to the points of tangency. Then draw the segments connecting the center of the incircle with the vertices of the triangle, and you have your similar triangles (each right triangle sharing a hypotenuse with another is similar to that other). However, finding $x$ seems to require trig or Heron's, unless you have the 13-14-15 triangle memorized, in which case you should be able to get the answer fairly readily.

So true isn't it?

Did you know there was such thing as a similar circle, or similar squares?  :D


Is there a rule against taking AMC 12 at 8:30 AM in one high school and AMC 10 at 2:45 PM at another high school (in one day)?

Or only one exam should be taken in one day?

Thank you.

You can't take both the AMC10 and the AMC12 of the same letter. Possible due to the fact that they share many of the same questions.

That was fast! Thank you!

I suspected that.

Yes, only one test per date. Either test for students who are eligible for both.


Solve this equation with x,y are integer numbers:
$\ x^{x+y}=y^{12} (1)$
$\ y^{x+y}=x^3 (2)$ :showoff:

Math is my life :weightlift:

[hide]

okay soo taking the natural log of both sides we get:

$(x+y) \ln x = 12 \ln y$

$(x+y) \ln y = 3 \ln x$

multiplying these together we get:

$(x+y)^2 = 6^2 \implies (x+y) = 6$ or $(x+y) = -6$

the latter gives no integral solutions whereas the former gives:

$(6-y +y)\ln (6-y) = 12 \ln y \implies 6-y = y^2$ which has roots $y = 2$ and $y= -3$

this then gives solutions $(4,2)$ and $(9,-3)$

now we just have incorporate the solutions where $x$ and $y$ equal $\pm1$, and we get $(1,1)$ and $(1,-1)$

[/hide]


Support Asymptote LaTeX AoPSwiki articles AMC AIME

Valentin has been upgrading the Wiki.  You'll find a few changes next time you work with it.  First of all, the Wiki now supports Asymptote.  Enclose your Asymptote code in <asy> </asy> tags.

Second, you should find that you don't have to log in to the Wiki after logging in to the forum.  (You may have to click preferences to change your skin from the MediaWiki default to the AoPS skin the first time you visit; we're still working on getting that set up correctly.)

So basically your Wiki login has been merged into the site login, so there is only one place to login in/out of the site (and wiki). However due to username restrictions applicable on the Wiki software, some of you will be logged in with a different username (though similar in some cases) than on the wiki.

For example I_like_pie would be logged in with I like pie.

If you have a username that has uncorrectable characters, like /,& or other such, then, the first time you try to visit the wiki (logged into the site) you will be redirected to a page where you can choose your wiki username. This cannot be a current forum username, current wiki username or an invalid wiki username. After choosing it you will be redirected back to the wiki and logged in.

Also the new code supports Forum $ \text{\LaTeX}$ tags (and the deprecated <math> tags).

[quote="Temperal"]Um.. what happened to the logo, layout, and several other things? Will they be restored?[/quote]Go to preferences, skin -> AoPS Skin. We're trying to fix it now for the old users (new users of AoPSWiki will not have that problem).

Nice  :lol: .

Sorry to pile on more things, but a couple of things (none of urgency) I noticed while I was testing out the new wiki:

- The $ \$$ math tags don't work in show preview mode (and I tend to use show preview a lot). And some really weird parsing errors occur; for example I noticed that the article [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_12A_Problems/Problem_1]here[/url] had a dollar sign denoting money, which seemed to mess up the rest of the math tags. I edited the article, and then everything became even worse, and even reverting didn't seem to fix it. Since <math> tags are now deprecated, how do we use $ \$$ signs for money? (<nowiki> doesn't work)
- On the monobook skin, could links back to the AoPS site be implemented? I actually prefer the monobook skin because the aops wiki layout causes strange things to occur on my browser (math equations disappear, centered items jump to different locations ..).
- Some lines in headings just strangely disappear when I'm in monobook skin. This seems to happen with almost AIME problem / other contest articles.
- It would be nice if monobook.css / monobook.js could be re-implemented (I've brought this up before, and was told that it might be possible for somebody to hack other's passwords?).

The monobook is not going to be modified/upgraded in any way. It's taken directly from the SVN repository of the mediawiki, and it's there more for debug purposes than anything. So any comments about the monobook style will just be ignored from now on.

I know about the math tag problem and I'm trying to fix it.

[quote="azjps"]
- Some lines in headings just strangely disappear when I'm in monobook skin. This seems to happen with almost AIME problem / other contest articles.[/quote]

This happens in every browser, on many webpages. It results from inadequate handling of some CSS.

Use Firefox 2 and there won't be any problems :)
Anyway I've temporarily disabled the use of dollar tags on the Wiki. So all should be back to math tags right now.

Will there be a Mathlinks theme for the wiki?

Layout troubles:

http://www.artofproblemsolving.com/Wiki/index.php/User:Temperal/Introductory_Proportion

Allright, the $ \text{\LaTeX}$ should work now in the Preview as well as in the <nowiki> tags. I've introduced the dollar tag. <dollar/> creates a simple dollar. Also you can create dollars within no wiki tags (but it's more of a hassle, than just click on the dollar add tag button).

I've put up a blank image for the button in the editor, if anyone would be kind enough to create one with a dollar inside (Fireworks required I guess) I would really appreciate it. Don't do dollars with $ \$$ [code]$\$$[/code]  command because it seems to be buggy.

[quote="azjps"]Wow, great work :lol:

There seems to be an error with uploading images also. When I try to do that, I receive an Internal Error, stating "The upload directory (public) is not writable by the webserver. ".[/quote]Will be fixed shortly.

[quote="Temperal"][img]http://i4.tinypic.com/4mlazq8.png[/img]

You locked the layers for some reason, so it's not very good.  :oops:[/quote]I didn't lock anything :) You can use the other images if you want  ... This will do for now.

Where there are those column links, would it be possible to do something from:

[img]http://mathprob.com/_private/get.php/c6c5992d7d4c0d51827a56e3dff9f310/wiki1.gif[/img]

to

[img]http://mathprob.com/_private/get.php/c6c5992d7d4c0d51827a56e3dff9f310/wiki2.gif[/img]

Check the portal page. This current style will be replaced soon with the new one.

[quote="#H34N1"]Oh. I see. Its the new prosilver default phpbb3 theme :D[/quote]Similar (ie rounded corners and some form ideas were used) but not quite.

This may be the wrong thread to ask in but, can we make science related article in the Wiki, like Astronomy et cetra? :)
EDIT Saw the Science section in the Wiki :) , can Astronomy articles be posted in the Physics then?

Yeah I don't see why not. After all, it's a wiki :D

[quote="#H34N1"]Would it be possible to design a ML theme for the wiki? (like the one on the portal?)[/quote]It's been done for a while now.

Preferences->Skin.

Where it normally is.  :wink:

Oh sorry. I saw it before, but I couldn't select it. I figured it out by logging out and logging back in. :D

Another thing about the ML theme: There are strange spaces in between words sometimes. Why is this?

EDIT: and <ul> bullets don't display.


How many zeros are at the end of 75!
Thank you

Pair all of the 5's and 2's.

In 25! there are 6 fives, so it can only be paired with 6 twos. Therefore, there are 6 zeroes.

In 50! there are 12 zeroes.

Do you see the pattern?

So there are $ \boxed {18}$ zeroes in 75!

Thank you. Why didn't I think of that :P


inequalities
function
email
inequalities theorems

I have never seen a proof for Schur's inequality and i am interested
for any r>0 and ai>0
show that
Sum ((ai)^r Product (j!=i) (ai-aj))>=0

[i][ADMIN EDIT] :: at http://www.mathlinks.ro/viewtopic.php?t=1166&postdays=0&postorder=asc&start=12 is the proof for generalised schur inequality for f convex functions ::[/i]

Here a proof of Schur inequality with three variables a,b,c

For any t \in R, (a,b,c)\in R+3

at(a-b)(a-c) + bt(b-c)(b-a)+
ct(c-a)(c-b) \geq 0

with equality iff a=b=c

Proof: the inequality is symetric in a,b,c

if t \geq 0 we can suppose a \geq b \geq c \geq >0
if t  <0 we can suppose 0<a \leq b \leq c

in the two cases we have

at(a-b)(a-c)  \geq bt(b-c)(a-b) and

ct(c-a)(c-b) \geq 0

the inequality follows

in fact I managed to prove (again it's in my book :D) that Schur holds in the general case

\sum  f(x) (x-y)(x-z)  \geq 0 for all convex functions (so x^t is ok) and for all positive increasing functions.

see if you manage to prove it moubi :)

i didnt know t could be less than 0!!  but anyways i dont see how
a^t (a-b)(a-c)>=b^t (b-c)(b-a)  (should be b-a eh?)

please explain and thank you

as I said above it works for all t \in R :D

first bugz it's  a^t (a-b)(a-c) \geq b^t (b-c )(a-b) not a^t (a-b)(a-c) \geq b^t (b-c )(b-a ).

we have (a-b)( a^t (a-c) - b^t (b-c)) .
obviously we have a-c \geq b-c and a^t \geq b^t for t\geq 0.

To answer your question, Schur's doesn't necessary hold for >3 variables, e.g. a1=1, a2=2, a3=2, a4=2 is a counterexample

Let a \geq b \geq c >0
if f is positive, increasing then

f(a)(a-b)(a-c)+f(b)(b-a)(b-c)+f(c)(c-a)(c-b)  \geq 0

------------------------------------------------------------------------------

a-c \geq b-c \geq 0 , a-b \geq 0

f(a) \geq f(b) so f(a)(a-b)(a-c) \geq f(b)(b-c)(a-b)

f(a)(a-b)(a-c)+f(b)(b-a)(b-c)  \geq 0  (1)

we also have f(c)(c-a)(c-b) \geq 0  (2)

(1) and (2) give the result

Valentin why did you take f convex ?

chers

moubinol

well i guess this website is wrong.
http://planetmath.org/encyclopedia/SchursInequality.html

i'll email them

bugzpodder do send the email so quick on the PlanetMath all the number
are positive, Charlie gave us an example with one negative three positive

moubi, I have generalized this inequality, not only for increasing positive functions, but also for convex functions.

for example, the function x |-> 1/x is convex, but decresing. Again schur is true, that is ...

1/x (x-y)(x-z) + ... \geq 0.

:D:D

If you post your proof here, valentin, then we'll call it Vornicu's extention to Schur's Inequality :D

okay: here goes nothin' :D

we shall treat two cases: first the hard one, that is f(b) \geq f(c). We can obviously suppose that a>b>c, otherwise the inequality becomes 0\geq 0. Because f is convex we have

f(a) x1 + f(c) x2 \geq f(a x1 + cx2) = f(b) where x1 = (b-c) / (a-c) and x2 = (a-b)/(a-c) and x1+ x2 = 1.

The above inequality is equivalent with
f(a)(b-c) + f(c) (a-b) \geq f(b)(a-c) that is f(a) \geq f(b)  (a-c) / (b-c) - f(c) (a-b) / (b-c)  which is f(a) (a-c) \geq f(b) (a-c)2 / (b-c) - f(c) (a-b)(a-c) / (b-c) .

Returning to Schur, we shall try to prove, as Moubinool did above, that f(a) ( a-c) - f(b) (b-c) \geq 0. For that it is enough to prove that

f(b) (a-c)2 (b-c) - f(c) (a-b)(a-c) / (b-c) \geq f(b) (b-c) <=>
f(b)(a-c-b+c ) (a-c+b-c) \geq f(c)(a-b)(a-c) <=>  (f(b)-f(c))(a-c) + f(b)(b-c) \geq 0

for the second case, we have even simpler:
f(b)(b-c)(b-a) + f(c)(c-a)(c-b) = (b-c) (f(c) (a-c) - f(b)(b-c) ) = (b-c)[(b-c)(f(c) - f(b)) + (a-b)f(c)] \geq 0 and as  f(a)(a-b)(a-c) \geq 0 we are done.

Note that you have to assume that f is nonnegative...
A counter-example is f(x) = -sqrt(x), which is convex, and the opposite holds...by Schur's inequality.

You use it in the last inequalities you wrote before to conclude.

Thus, Schur's inequality (or Vornicu's inequality) holds for :
- nonnegative convex functions
- nonnegative monotonic functions (not necessarly increasing).

yes I have mentioned above that f must be positive (or non-negative for that matter) :D

No, I mean a=+1, b=c=d=+2.
Then a 2 (a-b)(a-c)(a-d)+b 2 (b-z)(b-c)(b-d)+c 2 (c-a)(c-b)(c-d)+d 2 (d-a)(d-b)(d-c)=1 2 (-1)(-1)(-1)+0+0+0=-1<0
I'm pretty sure Schur doesn't hold for even n because I spent 2 hours proving another inequality using this until my friend doubted the correctness of Schur when n=4 and we found lots of counterexample  :D

[quote="Valentin Vornicu"]okay: here goes nothin' :D

we shall treat two cases: first the hard one, that is $f(b) \geq f(c)$. We can obviously suppose that $a>b>c$, otherwise the inequality becomes $0\geq 0$. Because $f$ is convex we have \[f(a) x_{1}+f(c) x_{2}\geq f(a x_{1}+cx_{2}) = f(b) \text{ where }x_{1}= \frac{(b-c)}{(a-c)}\text{ and }x_{2}= \frac{(a-b)}{(a-c)}\text{ and }x_{1}+x_{2}= 1\] The above inequality is equivalent with \[f(a)(b-c)+f(c) (a-b) \geq f(b)(a-c)\text{ that is }f(a) \geq f(b) \frac{(a-c)}{(b-c)}-f(c) \frac{(a-b)}{(b-c)}\text{ which is }\\ f(a) (a-c) \geq f(b) \frac{(a-c)^{2}}{(b-c)}-f(c) \frac{(a-b)(a-c)}{(b-c)}\] Returning to Schur, we shall try to prove, as Moubinool did above, that $f(a) ( a-c)-f(b) (b-c) \geq 0$. For that it is enough to prove that \[f(b) (a-c)^{2}(b-c)-f(c) (a-b)(a-c) / (b-c) \geq f(b) (b-c) \Longleftrightarrow \\ \Longleftrightarrow f(b)(a-c-b+c ) (a-c+b-c) \geq f(c)(a-b)(a-c) \Longleftrightarrow (f(b)-f(c))(a-c)+f(b)(b-c) \geq 0\] for the second case, we have even simpler: \[f(b)(b-c)(b-a)+f(c)(c-a)(c-b) = (b-c) (f(c) (a-c)-f(b)(b-c) ) = \\ = (b-c)[(b-c)(f(c)-f(b))+(a-b)f(c)] \geq 0 \text{ and as }f(a)(a-b)(a-c) \geq 0\text{ we are done.}\] [/quote]

[quote="Moubinool"]For any $ t \in R, a,b,c>0,$

$ a^t(a\minus{}b)(a\minus{}c) \plus{} b^t(b\minus{}c)(b\minus{}a)\plus{}c^t(c\minus{}a)(c\minus{}b) \geq 0.$

with equality iff $ a\equal{}b\equal{}c.$

[b]Proof[/b] The inequality is symetric in $ a,b,c$,

if $ t \geq 0$ we can suppose $ a \geq b \geq c >0,$

if $ t  <0$ we can suppose $ 0<a \leq b \leq c$,

in the two cases we have

$ a^t(a\minus{}b)(a\minus{}c)  \geq b^t(b\minus{}c)(a\minus{}b)$ and

$ c^t(c\minus{}a)(c\minus{}b) \geq 0$,

the inequality follows.[/quote]For any $ a,b,c\in R, t>0,$

$ |a|^t(a\minus{}b)(a\minus{}c) \plus{} |b|^t(b\minus{}c)(b\minus{}a)\plus{}|c|^t(c\minus{}a)(c\minus{}b) \geq 0,$

with equality if $ a\equal{}b\minus{}c\equal{}0.$


integration
calculus
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]
Find the sum of the following series

$ \sum_{n\equal{}1}^{\infty}n(n\plus{}1)x^n$

[b]3. The attempt at a solution[/b]

$ x\sum_{n\equal{}1}^{\infty}n(n\plus{}1)x^{n\minus{}1}$

$ x\int_{0}^{x}(\int_{0}^{x}f(t)dt)dt\equal{}x(x^2\plus{}x^3\plus{}x^4\plus{}x^5\plus{}\cdots)\equal{}x\frac{x^2}{1\minus{}x}$

$ x\frac{d^2}{dx^2}(\frac{x^2}{1\minus{}x})\equal{}\frac{2x}{(1\minus{}x)^3}$

Using this expression it can be solved quickly:

$ {\sum^{\infty}_{i\equal{}1}{x^i i^n}\equal{}\frac{1}{(1\minus{}x)^{n\plus{}1}}\sum^{n}_{k\equal{}0}{x^k\sum^{k}_{j\equal{}0}{(\minus{}1)^j (k\minus{}j)^n\binom{n\plus{}1}{j}}}}$

where $ {n\in N}$

I agree with azatkgz's answer, with differing intermediate steps.

$ \frac1{1-x}=1+x+x^2+x^3+\cdots$

$ \frac{d}{dx}\left(\frac1{1-x}\right)=1+2x+3x^2+4x^3+\cdots$

$ x^2\frac{d}{dx}\left(\frac1{1-x}\right)=x^2+2x^3+3x^4+4x^5+\cdots$

$ \frac{d}{dx}\left(x^2\frac{d}{dx}\left(\frac1{1-x}\right)\right)=
2x+2\cdot3x^2+\3\cdot4x^3+\cdots=\sum_{n=1}^{\infty}n(n+1)x^n.$

Worked out, $ \frac{d}{dx}\left(x^2\frac{d}{dx}\left(\frac1{1-x}\right)\right)=\frac{2x}{(1-x)^3}.$


probability
calculus
Gauss
geometry
rectangle
integration
3D geometry

Find the probability that three points chosen from a circle form an acute triangle.

[hide="Hint"]
This is equivalent to asking the probability that the set of points inside the triangle contains the center of the circle.[/hide]

[hide="Solution"]
Construct four equidistant marks around the circumference of the circle, and place the first point on one of these marks. Also construct a dotted line from this point to the opposite point on the circle. Clearly, the other two points must be on different sides of this line.

Define point 1 as the border between quadrants 1 and 4. Then wlog let point 2 be on one of quadrants 3 and 4, and point 3 be on one of quadrants 1 and 2.

-If 2 is in q3 and 3 is in q2 (P=1/4), then it must be true
-If 2 is in q4 and 3 is in q1 (P=1/4), then it must be false
-If 2 is in q3 and 3 is in q1 (P=1/4), it is analogous to the probability that of two random numbers chosen on [0,1], the second is smaller than the first, which is 1/2 by symmetry.
-The last case is symmetric to the third.

$P=\frac{1}{4}(1)+\frac{1}{4}(0)+\frac{1}{4}\cdot \frac{1}{2}+\frac{1}{4}\cdot \frac{1}{2}\implies \boxed{P=\frac{1}{2}}$
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[hide]I believe there is a flaw in your solution Zuton. Your sample space (set of possibilities for points 2 and 3) was overly restricted. For example, it's possible that both points 2 and 3 are in the third quadrant; or one in the third quadrant and the other in the forth quadrant. But your calculation does not consider those possibilities.[/hide]

I am pretty sure that we can fix one of the points...[using polar notation] $A=(1,0)$, $B=(1,2x)$ and $C=(1,2y)$ where $0\le x,y\le \pi$, they are independent variables.

We can say that $x>y$ [but we don't change our sample space], then we multiply by $2$ at the end. [Note I am not changing the fact that they are independent if I keep the probablity space the same for $y$. Think of this as using symmetry...]

$\angle A=x-y$, $\angle B=y$, and $\angle C=\pi-x$.

So we break the cases down further... when is each angle obtuse?

Some calculus might be needed. Note that when taking $x>y$, we have to not change our probablity space...i.e. if $0<x<\frac{\pi}{2}$, then the probablity of that happening for fixed $x$ is $\frac{x}{\pi}$, and we need to integrate that over $x\in[0,\frac{\pi}{2}]$

[note the fraction came from $y$, we need to have $y$ less than $x$, but we don't change the probablity space]

So...the problem is done conceptually...WHO WANTS TO DO THE CASEWORK?...I KNOW I DON'T  :P

gauss, you're right

If I'm correct though, all you really have to do is multiply by 1/2, I believe, which removes the chance that points 2 and 3 are on the same side of the divider line, which would give 1/4.... hmmm, seems low

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[b]Edit: Cleaner solution[/b]

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[b]Definitions:[/b]
Fix A somewhere on the circle. Of the other two points, let's start from A and go counterclockwise; call the first point we see B and the other C.

We want the probability that triangle ABC contains the center.

We know that B must be within 180º of A counterclockwise.

Draw a line from A, through the center of the circle, and call it's projection on the other side A'. Do the same for B to produce B'. Since $C$ must lie on arc $A'B'$, the fraction $\frac{m \angle A'OB'}{360}=\frac{m \angle AOB}{360}$ gives the probability for fixed $B$.

Since a uniform distribution of $B$ along this arc implies that $\angle AOB$ varies linearly, we just want the average value of $\frac{m \angle AOB}{360}$ as $m \angle AOB$ varies from $0$ to $180$, which is $P=\frac{1}{4}$
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Seems low, but is correct.  Calculus is not necessary, either: in fact, a standard geometric probability approach (draw it out as a rectangle, where each variable goes from $0$ to $2\pi$; it's even nicer if you only consider the restriction $x > y$) works very nicely, and the conditions for each angle to be acute are quite simple.

[hide]Graph geometrically for A at 0, $B=x,C=y\in{0,2\pi}$, on xy plane.
Given A,B, for C to make the triangle acute, draw perpendiculars to AB from A and B, and C belongs to the arc subtended by the perpendiculars. The length of that arc is the same as MINOR arc AB, since A is at 0 and B is at x, is x. So for $B=x\in{0,\pi}$, C can be in $x/(2\pi)$ of the circle. So our graph up to x=pi is
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The other half would be the mirror image, since we want the minor arc AB
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It takes up $1/4$ the graph.[/hide]

Okay...yea...I saw the $\int x\, dx$ and was like...omg calculus. But that clearly just makes a triangle, so no calc necessary.

This solution gives the correct answer also, but I'm not sure if it's weighted properly or if it's legitimate. Take a look  :maybe:
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Using these three points, it is possible to create a triangle similar to any possible triangle. We can therefore just look at the set of triangles in general.

Consider the set of ordered triples $(a,b,c) \; : \; a+b+c=180,\;a,b,c>0$. This, graphically, is an equilateral triangle in the 3-D cartesian system. Now consider the subset of this for which $a,b,c<90$. This subset is essentially the section contained within the triangle formed by connecting the midpoints of the original equilateral triangle, and thus has 1/4 the area.
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Anyway,

[b]Continuation:[/b]

Suppose 4 points are randomly selected from the surface of a sphere. Find the probability that the tetrahedron with these verteces contains the center of the sphere.

[i]Edit: [/i]I just realized that this was on a Putnam in the 90's