Let $\mathbb{N}_0=\{0,1,2 \cdots \}$. Does there exist a function $f: \mathbb{N}__0 \to \mathbb{N}_0$ such that:
\[ f^{2003}(n)=5n, \forall n \in \mathbb{N}_0  \]
where we define: $f^1(n)=f(n)$ and $f^{k+1}(n)=f(f^k(n))$, $\forall k \in \mathbb{N}_0$?

[quote=shobber]Let $\mathbb{N}_0=\{0,1,2 \cdots \}$. Does there exist a function $f: \mathbb{N}_0 \to \mathbb{N}_0$ such that:
\[ f^{2003}(n)=5n, \forall n \in \mathbb{N}_0  \]
where we define: $f^1(n)=f(n)$ and $f^{k+1}(n)=f(f^k(n))$, $\forall k \in \mathbb{N}_0$?[/quote]

[quote=OGGY_666]Therefore function satisfying the equation are f(n)=n(5)^(1/2003) and f(n)=0.[/quote]
Could you kindly explain the notation f(n)=n(5)^(1/2003) ? (what is the meaning of (5) ?)

And give as example what is the exact value of nonnegative integer $f(2)$ in your example ?

Thanks in advance.

[quote]Could you kindly explain the notation f(n)=n(5)^(1/2003) ? (what is the meaning of (5) ?)[/quote]
This means that f(n)=n times 2003rd root of 5.

[quote]And give as example what is the exact value of nonnegative integer f(2) in your example ?
[/quote]
f(2)=2*5(1/2003) which on calculating gives ~2.001607673.
f(f(2))=2*5(2/2003) which on calculating gives ~2.003216639.
f(f(f(3)))=2*5(3/2003) which on calculating gives ~2.004826898.
.....................
.....................
.....................
f[size=55]2003[/size](n)=2*5(2003/2003) which gives the value 10.

[quote=OGGY_666][quote]Could you kindly explain the notation f(n)=n(5)^(1/2003) ? (what is the meaning of (5) ?)[/quote]
This means that f(n)=n times 2003rd root of 5.

[quote]And give as example what is the exact value of nonnegative integer f(2) in your example ?
[/quote]
f(2)=2*5(1/2003) which on calculating gives ~2.001607673.[/quote]
Which unfortunately does not seem to be a nonnegative integer, as required by the problem statement  :whistling:

[quote=OGGY_666][quote]Which unfortunately does not seem to be a nonnegative integer, as required by the problem statement[/quote]
Isn't 2 a non-negative integer?[/quote]
Are you just joking ?
$2$ is indeed a nonnegative integer.
But $f(2)=~2.001607673$, as you suggested, is not (or at least was not $45$ years ago, when I was a student)

[quote=akhan98]Answer yes. f(0)=0 let's determine for natural number.Let f(5n)=5f(n) and all natural number divide in infinite multitude with 2003 number n1,n2,...,n2003 where CGD(ni,5)=1 and let for each multitude f(n1)=5(n2),f(n2)=n3,f(n3)=n2,...,f(n2003)=n1. So it's easy to see that this function satisfy all codition.[/quote]

I don't quite understand this solution.
Could someone help me out?

Consider the function defined as follows:
Group the naturals indivisible by 5 into groups of size $k$ (example: if $k=3$, the groups are $(1,2,3)$, $(4,6,7)$, $(8,9,11)$, so on) and put the first term of each group in set $S_{1}$, the second term of each group in set $S_{2}$, so on until the last term of each group in set $S_{k}$. Now for notation let $S(n)$ denote the nth term of set S. Clearly $S_{1} \cup S_{2} \cup ... \cup S_{k}$ is the set of naturals indivisible by 5, and the sets are pairwise disjoint, so for any natural $n$ indivisible by $5$ there exists unique $t$ and $m$ such that $n = S_{t}(m)$. Then define $f(n)$ as follows:
If $t<k$ then $f(n) = S_{t+1}(m)$
If $t=k$ then $f(n) = 5S_{1}(m)$
If $5|n$ then $f(n) = 5^{v_{5}(n)}f(\frac{n}{5^{v_{5}(n)}})$
$f(0)=0$

From this definition it becomes clear that $f^{k}(n)=5n$ for all non-negative integers $n$, so therefore it's possible for all positive integers $k$.
Setting $k=2003$ creates a function such that $f^{2003}(n)=5n$ for all non-negative integers $n$, so the answer is that it's possible.

[quote=OGGY_666]Yes, there exist a function. Define a sequence a[size=55]0[/size]=n,a[size=55]1[/size]=f(n)...a[size=55]2003[/size]= f[size=55]2003[/size](n).

Using characterstic equation we get:
a[size=55]k[/size]=C(5)^(k/2003) where C=n.
a[size=55]1[/size]=n(5)^(1/2003)=f(n).

Therefore function satisfying the equation are f(n)=n(5)^(1/2003) and f(n)=0.[/quote]

How do you use a characteristic equation on this sequence?


Given a triangle $ABC$ with $\angle{B}=30$ and $AC$ is the shortest side. Points $D$ and $E$ are on $BA$ and $BC$ respectively such that $DA=EC=AC$. Prove that $OI$ is perpendicular to $DE$, where $O$ and $I$ are the circumcentre and incentre respectively.

http://www.mathlinks.ro/Forum/viewtopic.php?t=14655

  darij


linear algebra
matrix
linear algebra unsolved

I dont know this problem belongs to this forum or not? I'm sure that I posted it somewhere but I dont know to which forum!

Prove that for any even n , this matrix is invertible:
$\displaystyle \left(\frac {1}{ {2i+j-1\choose i} }\right)_{i,j=\overline{1,n}}$

I have computed (in MathCad) determinant for n=4 and obtained 0.

oh! my God  :(  I'm sure that that evidance is reliable ,I don't have software to confirm your assertion are sure

I computed with Mathematica the determinant for n=3, ..., 8, and got zero in all cases. Are you sure you don't mean "[b]not[/b] invertible"?

[quote="sam-n"]oh! at least some one give me an idea for this one :([/quote]
i live in iran but didnt see that ?


Given $a,b,c,d,e\in \mathbb{R}$ and the polynomial $f(x)=x^{8}-4x^{7}+7x^{6}-7x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e$ has $8$ real roots, find $a,b,c,d,e$.

[quote="cosinator"]Do the roots have to be distinct?[/quote]

It's not mentioned in the problem. But I think there might be multiple roots.

[quote="jnApp66"]Is there really only one set {a,b,c,d,e} of real numbers that would make all the roots be real?[/quote]

I am not sure...Actually, the original problem is described as such:
Given $a,b,c,d,e\in \mathbb{R}$, the polynomial $f(x)=x^{8}-4x^{7}+7x^{6}-7x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e$ has $8$ factors of degree 1 with real coefficients, find $a,b,c,d,e$.

I think it's equivalent to what I've posted here.

[quote="jelly119"]There's a hint: expand $(x-\frac12)^{8}$. But I couldn't see how and why this works.[/quote]

[hide="why it works"]
Let $r_{k}$ be the roots, and expand out
\[(2r_{1}-1)^{2}+(2r_{2}-1)^{2}+\cdots+(2r_{8}-1)^{2}\]
Knowing the $x^{5}$ coefficient isn't even necessary.
[/hide]


How many diagonals does a regular octagon have?


Find all positive integers $ (m,n)$, such that

$ 2m^2\plus{}3nm\minus{}2n^2\equal{}71$

$ (2m \minus{} n)(m\plus{}2n)\equal{}71$.

$ 71$ is prime so

1) $ 2m \minus{} n \equal{} 1$
    $ m \plus{} 2n \equal{} 71$
or
2) $ 2m \minus{} n \equal{} 71$
    $ m\plus{}2n \equal{} 1$
or
3) $ 2m \minus{} n \equal{} \minus{}71$
    $ m \plus{} 2n \equal{} \minus{}1$
or
4) $ 2m \minus{} n \equal{} \minus{}1$
    $ m \plus{} 2n \equal{} \minus{}71$
Solving all systems shows no integer solutions exist.

Since $ m,\ n>0$, we have $ 2m\minus{}n>0.$ :wink:

Sorry double post.

The problem is easy, so this should be moved to High School Basics.

I solved it in a different way because I couldnt factor the left side  :D

$ 2m^2\plus{}3mn\minus{}(2n^2\plus{}71) \equal{} 0$

Taking the equation as a cuadratic on $ m$ we have:

$ \Delta \equal{} 9n^2 \plus{} 8 (2n^2\plus{}71) \equal{} 25n^2 \plus{}568$

Solutions have to be integers so:

$ 25n^2\plus{}568 \equal{} b^2 \equal{}> 25n^2\minus{}b^2 \equal{} 568 \equal{}> (5n\plus{}b)(5n\minus{}b) \equal{} 568$

We can divide the problem in several cases:

$ (1) 5n\plus{}b \equal{} 71$, $ 5n\minus{}b \equal{} 8$
$ (2)5n\plus{}b\equal{} 142$ $ , 5n\minus{}b \equal{} 4$
$ (3)5n\plus{}b \equal{} 284$ , $ 5n\minus{}b\equal{}2$
$ (4) 5n\plus{}b \equal{} 568$, $ 5n\minus{}b \equal{} 1$

Let $ 5n\plus{}b \equal{} x$  and $ 5n\minus{}b \equal{} y$ then $ 10n\equal{} x\plus{}y \equal{}> 10 \mid x\plus{}y$

And clearly none of our cases has this property. Therefore there are no solutions.


For $n\geq 2,$ prove that if both $n$ and $n^2+2$ are prime, then  $n$ is limited to the case of $n=3.$

Hi,
The solution is trivial.
If possible let n be a prime other than 3.
Therefore n is congruent to 1 or 2 (modulo 3).
This implies, in both cases n^2 +2 congruent to 0(modulo 3) i.e.
Therefore, n is congruent to 0(modulo 3).
But n is a prime.
Therefore, n=3.
Q.E.D

Sanjith

we see that is obvious that this condition holds when $n=3$. Let us then consider the condtions with $n \equiv 1 \pmod3$ and $n \equiv 2 \pmod3$. We do not need to consider $n \equiv 0 \pmod3$, bc that would imply that $n$ is not prime for $n \ne 3$.

If $n \equiv 1 \pmod3$, then $n^2+2 \equiv 0 \pmod{3}$, so $n^2+2$ is not prime. If $n \equiv 2 \pmod3$, $n^2+2$ will still be divisible by three, so the only $n$ for which $n$ and $n^2+2$ are prime is $3$.

Hi Asif,
Thats exactly the same as my solution. But yours looks more comprehensible because you have used LATEX. I got to learn to use it pretty soon.
Sanjtih

Hi Sathej - Best thing about Latex - you will find it easier than you think  (Most of the time, it is just putting a $\$$ on  both sides :)
and x^2+2 would  look like $x^2+2$ .

Hello, guys. Thank you for your replies. :)
Of course your answers are correct. My solution is  same as yours.
This problem was easy for you, isn't it?

kunny

yeah sanjith, i was writing it up while you were, so i didnt know you had posted before me...

It'ss quite obvious for Fermat, since $n(n^2+2)\equiv n^3+2n\equiv n+2n\equiv 3n\equiv 0 mod 3$

Kunny: It was a very beautiful problem, I didn't post my solution, because another posted it before. However, the trick was that you have to use mod...

ok let generelize it:
$n$ and $n^{p-1}+p-1$ are primes $\Rightarrow n=p$


In convex quadrilateral $PQRS,$ diagonals $\overline{PR}$ and $\overline{QS}$ intersect at $T$, with $PT: TR = 5: 4$ and $QT: TS = 2: 5.$ Point $X$ is chosen between $T$ and $S$ so that $QT = TX$ and $\overline{RX}$ is extended its own length to $Y$. If point $Y$ is [i]outside[/i] the quadrilateral, find the ratio of the area of triangle $PSY$ so that of triangle $QRT$.

this just screams mass points to me  ;) (area ratios also work)

[hide="further hint"]let  $YX\cap PS=Z$, find $\frac{RX}{XZ}$[/hide]

let 4x be the area of QTR, PQT is 5x, TSR is 10x and PST is 25/2 x (PT:TR=5:4, QT:TS=2:5)
since QT=TX, so TXR=4x, XSR=6x
connect PX, since PT: TR= 5:4,  TXR=4x, so PXT=5x, and PAX (A is the point that RY intersects PS) is 25/2 x - 5x=15/2 x
so PAX: PXR= 15/2 x: 9x= 5:6. therefore, XR:AX=5:6.
since SXR=6x, so AXS=5x
since XY=XR, so XYS=6x,then YAS=x and 3/2 x, so PYS=5/2 x, then PYS:QTR= 5:8

[quote="vincent_five"]PAX (A is the point that RY intersects PS) is 25/2 x - 5x=15/2 x[/quote]

I think that gives the area of PSX, not PAX.

[hide="problably overly complicated solution but whatever"]
Let $Z$ be the point of intersection between $\overline{PS}$ and $\overline{RY}$. Since $Z$, $X$, and $R$ are collinear, we can use Menelaus' Theorem (off of $\triangle PST$) to get

\[ \frac{PR}{TR}\cdot \frac{TX}{SX}\cdot \frac{SZ}{PZ}=1 \]

Since $\frac{PR}{TR}=\frac 94$ and $\frac{TX}{SX}=\frac 23$, $\frac{SZ}{PZ}=\frac 23$. Using Menelaus [i]again[/i], this time with collinear points $P$, $T$, and $R$ and $\triangle SZX$,

\[ \frac{SP}{ZP}\cdot \frac{ZR}{XR}\cdot \frac{XT}{ST}=1 \]

and using $\frac{SP}{ZP}=\frac 53$ and $\frac{XT}{ST}=\frac 25$, we get $\frac{ZR}{XR}=\frac 32\Rightarrow RZ=3ZY$.

Now it's time to use area ratios. $\frac{[PST]}{[QRT]}=\frac{PT\cdot ST}{QT\cdot RT}$ (note the vertical angles) $=\frac{25}{8}$; $\frac{[PSR]}{[PST]}=\frac{9}{5}$; $\frac{[PSY]}{[PSR]}=\frac 13$ (since the ratio of altitudes is also the ratio of two cevians that make the same angles with the common base). Multiplying these gives $\frac{[PSY]}{[QRT]}=\boxed{\frac{15}{8}}$.
[/hide]


For a positive integer $n$ we denote by $u(n)$ the largest prime number less than or equal to $n$, and with $v(n)$ the smallest prime number larger than $n$. Prove that  \[ \frac 1 {u(2)v(2)} + \frac 1{u(3)v(3)} + \cdots + \frac 1{ u(2010)v(2010)} = \frac 12 - \frac 1{2011}. \]

[hide]
First few terms of the series:
$(\displaystyle\frac{1}{2\cdot3}), (\displaystyle\frac{1}{3\cdot5}, \displaystyle\frac{1}{3\cdot5}), (\displaystyle\frac{1}{5\cdot7}, \displaystyle\frac{1}{5\cdot7}), (\displaystyle\frac{1}{7\cdot11},\displaystyle\frac{1}{7\cdot11},\displaystyle\frac{1}{7\cdot11},\displaystyle\frac{1}{7\cdot11}), (\displaystyle\frac{1}{11\cdot13}$...
If you denote the ith prime as $p_i$, note that a term of the sequence $a_n=\frac{1}{u(n)v(n)}$ appears $v(n)-u(n)$ times in the sequence. This is because every time n is a prime $p_i$, $u(n)=p_i$ and $v(n)=p_{i+1}$. Every time n is between $p_i$ and $p_{i+1}$, $u(n)=p_i$ and $v(n)=p_{i+1}$ also.

If you denote 2011 as the kth prime, the series simplifies to
$\sum_{i=1}^{k}{\frac{p_{i+1}-p_{i}}{{p_{i+1}p_{i}}}}=\sum_{i=1}^{k}{(\frac{1}{p_i}-\frac{1}{p_{i+1}})}=\frac{1}{p_1}-\frac{1}{p_k}=\frac{1}{2}-\frac{1}{2011}$
quite formal terminology for a weak argument... :(
[/hide]

[hide]2011 is prime.
Let's prove the more general assertion, $\frac{1}{u(2)v(2)}+\frac{1}{u(3)v(3)}+...+\frac{1}{u(p-1)v(p-1)}=1/2-1/p$ where p is a prime.
This is true for the base case of 2.
$\frac{1}{(2)(3)}=1/2-1/3$.
Now, assume it true for a given p-1.
Then, for the the next prime-1,
$1/2-1/p+\frac{1}{u(p)v(p)}+\frac{1}{u(p+1)v(p+1)}+\frac{1}{u(p+2)v(p+2)}+...+\frac{1}{u(p+2n-1)v(p+2n-1)}=1/2-\frac{1}{p+2n}$, where p+2n is the next prime.  After removing the halves, one notices that each expression with $\frac{1}{u(p,p+1,p+2,...p+2n-1)v(p,p+1,p+2,...p+2n-1)}$ is the same since they're between prime p and p+2n.  They each equal $\frac{1}{(p)(p+2n)}$ and there's 2n them.
$\frac{2n}{(p)(p+2n)}-1/p+\frac{1}{p+2n}=0$, which is true.[/hide]

Congrats dengmi, you've just solved http://www.mathlinks.ro/Forum/viewtopic.php?p=451267#p451267 too :P

haha, talk about hitting two birds with one stone while blindfolded... :rotfl:


AwesomeMath
summer program
geometry
MathPath
search

[img]http://www.awesomemath.org/images/header.jpg[/img]

(I have to say I went all out with the banner haha).

Hey everybody, I'd like to announce some (awesome) news.

In addition to the University of Texas at Dallas program, there will be a second AwesomeMath Program at the beautiful University of California, Santa Cruz.

Summer Program Dates:
UTD: July 6 to July 27
UCSC: July 31 to August 21

For more info, see the AwesomeMath Summer Program brochure at view the Brochure at http://www.awesomemath.org/AMSP2009.pdf . See the website at http://www.awesomemath.org/summer.shtml .

I'd like give you a sense of my experience at AMSP. Two summers ago, I was a student at AMSP and I had a great time. So much so that I decided to come back as a counselor last summer. Furthermore, I am planning to come back this summer. I really enjoyed helping with the geometry class!

--Alex (or Altheman...as some of you like to call me in real life haha)

whoa. no way.... a second campus and a diff. time??? interesting...

[quote="Altheman"]

--Alex (or Altheman...as some of you like to call me in real life haha)[/quote]

Hey! It's not my fault that I'm stupid...  :maybe:

Not sure if both camp will have the same instructors or not...

I don't think I'll be going to AwesomeMath, but I have to say, all of you who go to UCSC are super lucky! MathPath (another summer math camp) was there in '06 when I went and the campus was awesome; the food and setting were especially great. So pick that session if you can ;-).

(But watch out for mountain lions  :P )

holy cow now they have 2 campuses

but i live like 30 min from UTD anyways

i am not going this time and probably do better

is it in santa cruz to attract people from CA?

well, seems like it's gonna be pretty fun this year

Leon,u r not stupid and y in california??y not in FL??o,and guess who??

pssttt. its steve -___-;; lol
b/c cali is cooollll. nd the gators suckkk [jkjk]
yea and i agree. ur DEF. NOT stupid. [thats me xP]

cali because cal has amazing weatherr and really pretty sceneries :PP

 esp in OC area :D

Last year I went to CTY at Santa Cruz actually, and I have to say the food was TERRIBLE. Maybe my standards are too high...they do have this nice little bookstore which has a shop with lots of junk food and drinks (including energy drinks!!!) next to it.

The campus looks nice though, yes. It gets very foggy in the morning. The sun gets pretty brutal, if you're not lucky.

[quote="serialk11r"]nice little bookstore .... shop with lots of junk food and drinks (including energy drinks!!!) next to it. [/quote]

lmfao. we had like the exact same thing at UTD. XP bookstore w/ little shop + starbucks. & thats pretty much it lol.

[quote="Carbon57"]Yeah...that store was awesome...I spent like at least 4 dollars everyday...and bought a 20 oz. NOS thing before the last night...There was a starbucks?[/quote]

...not to mention a billion miles of floss  :rotfl:

lol.

yeah there was a "starbucks"... well more like the pub sold starbucks coffeee :D yumm.

Yay at least there's a backup now for not making MOP *gulp*. I heard the AMSP in CA is going to be a bit more advanced.

There is a Facebook group now! Search "AwesomeMath 2009" and join!

[quote="LadyKn1ght"]all of you who go to UCSC are super lucky! the food and setting were especially great.[/quote]

/me screams with joy

hey zephyredx please come to CA!


Δίνεται τρίγωνο $ ABC$ με $ <B\equal{}50$, $ <C\equal{}20$. Το ύψος από το $ A$ και η διχοτόμος από το $ C$ τέμνονται στο $ K$. H $ BC$ και η $ AO$, όπου $ O$ το περίκεντρο του $ ABC$ τέμνονται στο $ R$. Να δειχθεί ότι $ RK$ κάθετη στην $ AB$.

Σκοτώστε την όμορφα. :wink:

[quote="Protonios"]Δίνεται τρίγωνο $ ABC$ με $ < B \equal{} 50$, $ < C \equal{} 20$. Το ύψος από το $ A$ και η διχοτόμος από το $ C$ τέμνονται στο $ K$. H $ BC$ και η $ AO$, όπου $ O$ το περίκεντρο του $ ABC$ τέμνονται στο $ R$. Να δειχθεί ότι $ RK$ κάθετη στην $ AB$.
[/quote]

Αν το ύψος ΑΚ τέμνει την BC στο D,τότε <ΑΟΒ = 2C = 40 (υπόψη το Ο είναι εξωτερικό του τριγώνου διότι ,Α = 110 = αμβλεία). Άρα <ΟΑΒ = <ΑΒΟ = 70, άρα
<ΟΑΒ = 30. Θα δούμε ότι και <ΚΒΔ = 30, που αρκεί για να δούμε οτι ΒΚ κάθετος στην ΑΟ (καθώς ΑD κάθετος στην ΒD).

Είναι tan(<KBD) = KD/BD = (CD tan10)/ (AB cos50) = (AC cos20 tan 10)/(AB cos50). Αλλά από τον νόμο των ημιτόνων AC/AB = (sin 50)/(sin 20).

Τελικά tan(<KBD) = (sin 50 cos20 tan 10)/(sin 20 cos50)

To τελευταίο με λίγο brute force βγαίνει tan30. Οι πράξεις είναι λίγο πολλές για να τις γράψω εδώ. Καλού κακού έκανα έλεγχο με κομπιουτεράκι και πραγματικά η αριθμητική τιμή του δεξιού μέλους είναι $ \frac {\sqrt{3}}{3}$.

Ασφαλώς υπάρχει τρόπος να αποφύγουμε την Τριγωνομετρία γιατί το σχήμα έχει πολλές ωραίες γωνίες, αλλά δεν το δοκίμασα.

Τάκης.

Έχω άλλη λύση όχι τόσο elementary. Την έφτιαξα χρησιμοποιώντας τριγωνομετρικό Ceva και την εγγραψιμότητα του ΑΚΟΒ.

Αλλά η λύση μια χαρά είναι. Οι πράξεις δε χρειάζονται γιατί από τη στιγμή που εντοπίσαμε εύκολα με δοκιμή τη γωνία tan30 όλα προσδιορίζονται αμφιμονοσήμαντα στο σχήμα. :wink:


function
inequalities
Geometric Inequalities

Let $0 \leq a, b, c, d \leq 1$. Find the largest possible value of the sum
\[\frac{a}{bcd+1} + \frac{b}{cda+1} + \frac{c}{dab+1} + \frac{d}{abc+1}\]

The LHS is a convex function in each of the variables, thus it reaches it maximum value at  the endpoints.
It is easy to check that this maximum value is $3$,  reached for $(a,b,c,d) = (0,1,1,1)$ or some permutation of this.

Pierre.


hi!
[i][b][u]prove that (a ² +b ²)/(a² - b²) is not a natural number such that a and b are natural numbers[/u][/b][/i]

[quote="Altheman"]Without loss of generality, suppose that $ (a,b)=1$. Note $ a>b$ [$ a=b$ gives undefined, $ b<a$ gives a negative number]. Let $ c=a+b$ and $ d=a-b$. Note that $ (c,d)=(a+b,a-b)=(a+b,2a)=(a,b)$ or $ 2(a,b)=1$ or $ 2$ and $ d>0$. Then we have:

$ 2cd|(c^{2}+d^{2})$

$ c|d^{2}$; $ d|c^{2}$.

If we let $ p\ne 2$ be a prime, define $ p^\alpha\parallel c$ and $ p^\beta\parallel d$. Then $ \alpha\le 2\beta$ and $ \beta\le 2\alpha$. Since $ (c,d)=1,2$, we have that either $ \alpha$ or $ \beta$ is equal to $ 0$. But that implies that $ \alpha=\beta=0$. Hence $ c$ and $ d$ have no prime factors except for maybe $ 2$'s.

i) $ (c,d)=2$.

But this case only leaves $ \text{min}\, \{\alpha,\beta\}=1$, but $ c\ge d$ and the fact that they have no other prime factors $ \implies \alpha\ge \beta$. Hence $ (a,b)=(2^{k+1},2)$ where $ k=0,1,...$. Plugging this back in, $ 2^{k+3}|(2^{2k+2}+2^{2})$ gives a contradiction.

ii) $ (c,d)=1$.

But then $ c$ and $ d$ have no prime factors; $ c\ge d>0$, so they are both $ 1$. Hence $ (a,b)=(1,0)$. I am assuming that you mean that natural numbers means $ 1,2,...$ so we are done.

That fraction can never be an integer.[/quote]

Could you not just say
$ (a, b)=1 \implies (a, a+b)=1, (a, a-b)=1 \implies (a^{2}, a^{2}-b^{2})=1$, so then because the fraction is $ \frac{2a^{2}}{a^{2}-b^{2}}-1$, we must have $ a^{2}-b^{2}|2a^{2}\implies a^{2}-b^{2}|2$ which cannot be so we are done.
?

Haha that is a lot easier. This is just what my 'algorithm' produces so w/e.

$ (a^{2}-b^{2},a^{2}+b^{2})=(a^{2}-b^{2},2a^{2})=2,1*(a^{2},b^{2})=2,1$

I assume that $ a,b > 0$ (otherwise your exercise is wrong because $ (a,b) \equal{} (k,0)$ works).

Suppose by the way of the contradiction that there exists a natural intenger $ k$ such that $ \frac {a^2 \plus{} b^2}{a^2 \minus{} b^2} \equal{} k$. Then, the equation rewrites like : $ a^2(1 \minus{} k) \plus{} b^2(1 \plus{} k) \equal{} 0$. But $ a\geq 1$ and $ b \geq 1$, so $ a^2(1 \minus{} k) \plus{} b^2(1 \plus{} k) \geq 2 \Leftrightarrow 0 \geq 2$, a contradiction. Conclusion follows.

$ a^2\plus{}b^2 | 2a^2$ and $ a^2\plus{}b^2 | 2b^2$ But one of $ 2a^2$ , $ 2b^2$ is less than $ a^2\plus{}b^2$ so we are done (unless a=b, but then the fraction is undefined).

[quote="Zweii"]$ a^2(1 \minus{} k) \plus{} b^2(1 \plus{} k) \equal{} 0$. But $ a\geq 1$ and $ b \geq 1$, so $ a^2(1 \minus{} k) \plus{} b^2(1 \plus{} k) \geq 2 \Leftrightarrow 0 \geq 2$[/quote]

Your argument, if I understand correctly, is that since $ a^2, b^2 > 1$, then $ a^2 (1 \minus{} k) \plus{} b^2 (1 \plus{} k) > 1 (1 \minus{} k) \plus{} 1 (1 \plus{} k) \equal{} 2$.

However, what if $ (a,b,k) \equal{} (2,1,3) \implies 2^2 \cdot ( \minus{} 2) \plus{} 1 ^2 \cdot (4) \equal{} \minus{} 4 \not \geq 2$.

Basically, $ k$ is negative in one term and positive in the other, and thus if we give more weight to the negative term, we could very well make the overall expression negative. So you need to invoke the relationship between $ a,b,$ and $ k$.


[quote="Anh Cuong"]Should it be:
  [tex] f(f(x-y))=f(x) \cdot f(y)-f(x)+f(y)-xy  [/tex] ? :)[/quote]
i have sloved this problem :my result: $f(x)=-x$ But my solution is long and not nice :( ; i will try to find another solution :)

i have the solution for this problem
the first one:give$f(0)$=a;a in real
give $x=y=0$ then we have $f(a)=a^2$
give x=y then we have $f(a)=f(x)^2-x^2$ for all x in real
so we give $x=a$ then we have $2*a^2=a^4$ so $x=0$ or $x=sqrt(2)$ or $x=-sqrt(2)$
if $a=sqrt2$ then we easy to prove that $f(a^2)=a^3-a^2+a$ so give $x =a^2$ then we have the contradition
it's similar for the case $a=-sqrt(2)$(contradition) so $a=0$ hence $f(x)=x$  or $f(x)=-x$
we esay to prove that two solution doesn't take place atthe same time so $f(x)=x$ fo all x in real or $f(x)=-x$ for all x in real are the solution
but $f(x)=x4 isn't the solution
then we have $f(x)=-x$ for all x in real is the function satisfy the condition  :D  :D  :D

If a 2 =2 we choose y =0 then we can see it's wrong.

that was an easy problem .
 Each guy here who is prepairing to IMO maneged it.

Let $ y \equal{} x$, then we find $ f(f(0)) \equal{} [f(x)]^2 \minus{} x^2$. Let $ f(f(0)) \equal{} C$, then $ f(x) \equal{} \pm \sqrt {x^2 \plus{} C}$. Of course, it cannot be both since f is a function, sometimes it will be plus and sometimes it will be minus.
Let $ f(0) \equal{} d$. Letting $ y \equal{} 0$, we have
$ f(f(x)) \equal{} df(x) \minus{} d \minus{} f(x)$
Letting $ x \equal{} 0$ and $ y \equal{} \minus{} x$, we have
$ f(f(x)) \equal{} df( \minus{} x) \minus{} d \plus{} f( \minus{} x)$
Let $ g \equal{} \frac {d \minus{} 1}{d \plus{} 1}$. Setting both expressions for $ f(f(x))$ equal, we find $ gf(x) \equal{} f( \minus{} x)$. But since $ f(x) \equal{} \pm \sqrt {x^2 \plus{} C}$, we find g must be either 1 or -1. $ g \equal{} 1$ gives no possible value for d, so $ g \equal{} \minus{} 1$ and $ d \equal{} 0$. So $ d\equal{}f(0)\equal{}0$ and $ C\equal{}f(f(0))\equal{}0$.  So f(x)= either x or -x.
But now, how do we prove $ f(x)$ can't be x on some values and -x on others? Nhat said it is easy, but never showed it.

The full solution is too big. So I write some of these lines:
I get $f(f(0))=f(0)^2$,$f(x)+f(-x)=2f(0)$.
After that I got $f(x)^2=x^2+f(0)^2$ ,then $f(x)^2-f(-x)^2=0$, if $f(0)=0$,then we get $f(x)^2=x^2\Rightarrow{f(x)=\pm{x}}$
$f(x)=x$,this do not satisfy the eq. ,$f(x)=-x$ satisfies the eq. .
 If $f(x)=f(-x)$, & ,$f(0)\ne{0}$, then we can get $f(x)=\pm{1}$, which does not satisfy the main eq. Hence there does not exist any such $f$.
Hence  , $f(x)=-x$

:Edited:

[quote="pigfly"]Find all function $ f: \mathbb R\to \mathbb R$ satisfying the condition:
\[ f(f(x \minus{} y)) \equal{} f(x)\cdot f(y) \minus{} f(x) \plus{} f(y) \minus{} xy
\][/quote]
I'm always very impressed by those users claiming "I have the solution but it's too long to be posted here" or "this a very easy problem that everbody managed and I'll not post the solution here :lol:
I'm even more impressed when the claim is wrong (so that we cant point the flaw in the proof of the poster since he/she hides it).

Here is a full solution of this rather simple equation :

Let $P(x,y)$ be the assertion $f(f(x-y))=f(x)f(y)-f(x)+f(y)-xy$

1) $f(x)^2=x^2$ $\forall x$
=============
Subtracting $P(0,0)$ from $P(x,x)$, we get $f(x)^2=x^2+f(0)^2$
An immediate consequence is that $f(-x)=\pm f(x)$ $\forall x$

Let then $x\ne 0$ : $f(f(x))=\pm f(f(-x))$ and :
If $f(f(x))=f(f(-x))$, subtracting $P(0,x)$ from $P(x,0)$ implies $f(x)=f(0)$ and so $f(x)^2=f(0)^2$ and so $x=0$, impossible.
So $f(f(x))=-f(f(-x))$ and then adding $P(0,x)$ to $P(x,0)$ implies $f(x)f(0)=0$ and so $f(0)=0$, else $f(x)=0$ $\forall x\ne 0$, which is not a solution.
Q.E.D

2) $f(x)=-x$ $\forall x>0$
================
We got from 1) above that $\forall x$, either $f(x)=x$, either $f(x)=-x$

Suppose now $\exists a,b>0$ such that $f(a)=a$ and $f(b)=-b$ :
$f(f(a-b))=\pm f(f(b-a))$ and :
If $f(f(a-b))=f(f(b-a))$, subtracting $P(b,a)$ from $P(a,b)$ implies $a=-b$ , impossible since $a,b>0$
So $f(f(a-b))=-f(f(b-a))$ and then adding $P(b,a)$ to $P(a,b)$ implies $ab=0$, impossible since $a,b>0$

So : either $f(x)=x$ $\forall x>0$, either $f(x)=-x$ $\forall x>0$

If $f(x)=x$ $\forall x>0$, $P(x,y)$ with $x>y>0$ implies contradiction.
So $f(x)=-x$ $\forall x>0$
Q.E.D.

3) $f(x)=-x$ $\forall x$
============
We got from 1) above that $\forall x$, either $f(x)=x$, either $f(x)=-x$

Suppose now $\exists a,b<0$ such that $f(a)=a$ and $f(b)=-b$ :
$f(f(a-b))=\pm f(f(b-a))$ and :
If $f(f(a-b))=f(f(b-a))$, subtracting $P(b,a)$ from $P(a,b)$ implies $a=-b$ , impossible since $a,b<0$
So $f(f(a-b))=-f(f(b-a))$ and then adding $P(b,a)$ to $P(a,b)$ implies $ab=0$, impossible since $a,b<0$

So : either $f(x)=x$ $\forall x<0$, either $f(x)=-x$ $\forall x<0$ and so :

Either $f(x)=-x$ $\forall x$, which indeed is a solution
Either $f(x)=-|x|$ $\forall x$ which is not a solution

Q.E.D.

Hence the unique solution $\boxed{f(x)=-x}$ $\forall x$

Denote $f(0) = k$. Substituting $x = y = 0$ into the given equation, we get $f(k) = k^2 (1)$
Then substituting $x = y$ and taking into account $(1)$, we have $f(k) =[f(x)]^2 - x^2$, or
$[f(x)]^2 = x^2 + k^2 (2)$ This shows that $[f(-x)]^2 = [f(x)]^2$, or
$[f(x) + f(-x)] .[f(x) - f(-x)] = 0$, . (3) Assume that there is a= 0 such that $f(a) = f(-a)$. Then substituting
$y = 0$ into the given equation one gets $f(f(x)) = kf(x) - f(x) - k$, ,and substituting $x = 0, y = -x$ one obtains $f(f(x)) = kf(-x) + f(-x) -k,$ which together yields $k[f(-x) -f(x)] + f(-x) + f(x) = 2k, (4)$ from which, by substituting $x = a$, we have $f(a) = k. (5)$ On the other hand, from (2) it follows that if $f(x) = f(y)$ then $x^2 = y^2$.Then by (5), the equality $f(a) = k = f(0)$ shows that $a = 0$, which contradicts to  $a= 0$. Thus $f(-x) = f(x)$ for all $x = 0$. Then (3) gives $k[f(x)-1] = 0$, which implies that $k = 0$ (otherwise $f(x) = 1$, contradicts to $f(-1) = f(1))$.So we arrive to $[f(x)]^2 = x^2.$ Now assume that there is $a = 0$ such that $f(a) = a$. Then $a =f(a) = -f(f(a)) = -f(a) =-a$, which implies that $a = 0$. This is impossible, as
$a=0$. Hence, $f(x) = x$, Then from $[f(x)]^2 = x^2$ it follows that $f(x) = -x,$ Conversely, by direct verification we see that this solution satisfies the requirement of the problem. Thus the answer is $f(x) = -x.$

[quote="pco"]I'm always very impressed by those users claiming "I have the solution but it's too long to be posted here" or "this a very easy problem that everbody managed and I'll not post the solution here :lol:
I'm even more impressed when the claim is wrong (so that we cant point the flaw in the proof of the poster since he/she hides it).[/quote]
  You are right , I overlooked it and I edited it.
Though you are right that I did not post the full solution, these lines that I have wrote is enough, and also my solution is smaller than yours.

[quote="MANMAID"]  You are right , I overlooked it and I edited it.
Though you are right that I did not post the full solution, these lines that I have wrote is enough, and also my solution is smaller than yours.[/quote]
Congrats ! You are very smart

Another solution:

Setting $x=y$, we get $f(f(0)) = f(x)^2 - x^2 \Rightarrow f(x) = \sqrt{x^2 + C_1}$ or $f(x) = -sqrt{x^2 + C_1}$, where $C_1 = f(f(0))$. Note that $C_1 = 0.$ Indeed, let $C_1 > 0.$ Note that $x = 0 \Rightarrow f(f(x)) = C_2 f(x) - f(x) + C_2$, where $C_2 = f(0)$. Further, $y = 0 \Rightarrow f(f(-y)) = C_2 f(y) + f(y) - C_2$. Since $C_1 > 0$, $f$ is an even function, so we have that $f(x) = f(-x) \Rightarrow f(f(x)) = f(f(-x)) \Rightarrow C_2 f(x) - f(x) + C_2 = C_2 f(x) + f(x) - C_2 \Rightarrow f(x) = f(0) = 0$, by simple substitution. This is a contradiction, since $|C_1| > 0 \Rightarrow f(x) \neq 0$. Substituting $C_1 = 0$, we get that $f(x) = x$, $f(x) = -x$, $f(x) = |x|$, or $f(x) = -|x|$. We can easily check that of these, only $f(x) = -x$ works. $\box$


induction
combinatorics unsolved
combinatorics

The number of paths between any two distinct vertices in an [i]even[/i] graph is even.

Actually, I think it should be "even" instead of "odd" (think about a triangle: two vertices are joined by two paths).

(by path you mean elementary path, right?)

Yes. It should be even. How to prove it?

We prove that this is true for the more general class of graphs with even degree at each vertex (i.e. considering the possibility that the graph may be disconnected).

Let $G$ be such a graph. Fix vertices $u, v$ in $G$, and let $N(H)$ denote the number of $u-v$ paths in a subgraph $H$ of $G$. We will prove that $N(G)$ is even by induction on the number $n$ of edge-disjoint simple cycles that $G$ can be decomposed into. (A simple cycle being one whose degree at each vertex is exactly 2).

The claim is clearly true for $n = 0, 1$. Suppose n > 1; let $C_1, \ldots, C_n$ be such a decomposition of $G$. We will count the $u-v$ paths via the subgraphs of $G$ made up of some union of the $C_i$s that contain them. By the Principle of Inclusion-Exclusion:

$N(G) =$ number of $u-v$ Eulerian paths in $G + \sum_{\{a_1, \ldots, a_k \} \subset \{1, \ldots, n\}} (-1)^{n-k} N(C_{a_1} \cup \cdots \cup C_{a_k})$

Now the number of $u-v$ Eulerian paths in $G$ is 0, since $u, v$ are distinct. Each of the terms in the summation is even by induction. So the sum is even.

I think this works regardless of whether you talk about paths which do not 'repeat' a vertex, or include those paths that do (but do not repeat an edge). Grobber, was the former what you meant by 'elementary path'?

I don't think your application of the principle of Inclusion and Exclusion is appropriate: Indeed, take the example of two cycles. Check yourself that your formula goes wrong in that case. Your formula was only true if we had that the union over all $i$ of the paths from $u$ to $v$ in $\left(\cup_{j\neq i} C_j \right)$ was equal to all paths from $u$ to $v$.

Oops, my bad. I had assumed that any non-Eulerian $u-v$ path must avoid passing through some $C_i$

Well, I have changed 'Eulerian' to 'even'.
They're the same in West's book.

ok, here is an 'inspired' proof.(inspired by the proof in Bollobas's book to the lemma stated in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Bollobas&t=45852]this[/url])
firstly we remark that the problem is equivalent to showing that the number of circuits of $G$ containing a fixed edge $e=uv$ is odd. this is so since if this were proved, then the problem in question is solved for the case of adjacent vertices.in case the two vertices are not adjacent, then add two extra edges $e_1,e_2$ between the vertices $u$ and $v$; as all of this carriees through even in the case of graphs with multiple edges, the number of circuits containing $e_1$ is odd,but that also includes the case where the circuit is $e_1\cup e_2$ and so the number of $u-v$ paths is again even.the other way is clearly true.

basically the idea is to create a new graph whose odd vertices are the objects in question and that solves the problem automatically.
  let us create a new graph $\mathcal{H}$ with its vertices being the circuits of $G$ that contain $e$ and paths of $G$ that start at $u$(where $e=uv$). this also includes the simple path $P=e$. two vertices are deemed adjacent if one can be obtained from the other by adding or deleting an edge. then note that in this graph all the circuits are odd vertices(actually they have degree $1$ clearly as the only edge that can be deleted from a circuit is its other edge at $u$). however this does not guarantee that these are the $\textit{only}$ odd vertices of $\mathcal{H}$. In fact all of the maximal paths are also degree $1$ vertices. hence we add a few more edges as follows: if $P_1,P_2$ are maximal paths, both starting at $u$, then we say $P_1\leftrightarrow P_2$ iff $P_1\cup P_2$ is a lollipop graph, rooted at $u$(i.e. this graph consists of a cycle and a path starting from $u$ and the cycle and path share exaclty one vertex). in other words, all we are doing is simply adding all the other edges of the end vertex of the maximal path. clearly since all the vertices are even, all the maximal paths also become even vertices. finally, as for the path $e$, its degree in $\mathcal{H}$ is simply $deg(v)-1$ and hence is odd as well.
the upshot is that the odd vertices of $\mathcal{H}$ are precisely the cycles of $G$ containing $e$ and the path $e$ and that finishes the proof.

P.S: The idea of the lollipop graphs is actually what is used in Bollobas's proof the post mentioned above.

edit:P.P.S: this same proof seems to also appear in a paper of Edmonds and Cameron.

[quote="seshadri"]as all of this carriees through even in the case of graphs with multiple edges, the number of circuits containing $e_1$ is odd,but that also includes the case where the circuit is $e_1\cup e_2$ and so the number of $u-v$ paths is again even.the other way is clearly true.[/quote]
I don't understand this ... Could you please tell more details?

if $u,v$ are not adjacent, then just add a couple of $u-v$ edges(we need two to keep the Eulerian property intact). call them $e_1,e_2$. Now by the previous part(or that is what we intend to prove), the number of circuits that contains the edge $e_1$ is odd. hence the number of $u-v$ paths, other than the edge $e_1$ is odd in number since each circuit containing $e_1$ gives exactly one $u-v$ path. but then one of these paths also happens to be the other extra edge that we have added, namely, $e_2$. hence the number of $u-v$ paths in the original graph has to be even.
in the proof of the statement about the number of odd circuits containing an edge, we never have occasion to worry about the edges being multiple and so really speaking, the same proof carries through there also.
this reduces the problem to the case where it suffices to show that for any edge, the number of circuits containing that edge is odd. this should clarify a little bit.

Well, either I am really confused or there's a mistake in your argument - I don't see why the maximal paths are the only other vertices of odd degree.

Another argument would be the following: I start as you did.

Consider the graph $H$ of all paths which may visit a vertex multiple times, but may only visit every edge once. From the condition of the problem, it is clear that the odd vertices of $H$ are circuits and the trivial path thus showing that the number of circuits is odd.

Now we may pair the circuits which visit a vertex multiple times in the following: Let the vertex that occurs first two times be $A$. Reverse the first circle from $A$ to $A$ to get an involution(In fact, for the path with the reversed circle, $A$ stays the first vertex which is visited at least two times).

Peter


ratio
geometry
combinatorics unsolved
combinatorics

There are $1000n^3$ points in the space,where $n$ is a positive integer.Show that we can find two triangles such that their vertices are among the given points and the ratio of their areas is or larger that $n$.

[quote="jin"]There are $1000n^3$ points in the space,where $n$ is a positive integer.Show that we can find two triangles such that their vertices are among the given points and the ratio of their areas [color=darkred]is or larger that $n$[/color].[/quote]

Does it mean $\leq$ or $\geq$?

Yes, it means there exist two triangles with area $S,S'$ and $\frac{S}{S'}\geq n$.


modular arithmetic
number theory
algebra
binomial theorem

Find the remainder when $2^{1990}$ is divided by $1990.$

I will use the following theorem: if $p$ is a prime and $n$ is a natural number, then:
$n^p\equiv{n}(mod\quad (p)$.
$199$ is a prime number, so
$2^{199}\equiv2(mod\quad199)$
$2^{1990}\equiv2^{10}(mod\quad199)$
$2^{1990}\equiv29(mod\quad199)$
Next we have:
$2^4\equiv1(mod\quad5)$
$2^{4\cdot{497}}\equiv1(mod\quad5)$
$2^{1988}\equiv1(mod\quad5)$
$2^{1990}\equiv4(mod\quad5)$
So the last digit of number $2^{1990}$ is $4$ or $9$. It can't be $9$, because this number is even, so the last number is $4$.
$2^{1990}\equiv4(mod\quad10)$
Let $k=2^{1990}$
$199|k-29$
$10|k-4$
$1990|(k-29)(k-4)$
$1990|(k^2-33k+116$
Let $x$ be a sought number
$x\in[0 ; 1989]$
${k}\equiv{x}(mod\quad1990)$
${k-33}\equiv{x-33}(mod\quad1990)$
${k(k-33)}\equiv{x(x-33)}(mod\quad1990)$

${k^2-33k}\equiv{-116}(mod\quad1990)$
$x^2-33x\equiv{-116}(mod\quad1990)$
$1990|x^2-33x+116$

I don't know what next  :(

dondigo, your solution is not correct. Please check the last paragraph.

[hide]we can look at it mod 2,5,199 and use CRT to find the answer

(i use euler's extension of FLT to simplify)

$2^{1990} \mod{2} \equiv 0\mod{2}$ (1)

$2^{1990} \mod{5} \equiv 2^2\equiv 4\mod{5}$ (2)

$2^{1990} \mod{199} \equiv 2^{10} \equiv (256)4 \equiv 57*4 \equiv 228 \equiv 29\mod{199}$

combining (1) and (2), we get $4 \mod 10$, then combining that with (3)

so we have $4+10n=29+199m$, and want least m and n

$10n=25+199m$ so $5|m$, so let $m=5k$
$2n=199k+5$, k must be odd, so let $k=2a+1$
$n=199a+102$

so $a$ can be anything, so we want the lowest a so m is a natural number, so $a=0$

that makes $n=102$, then $4+10(102)=1024$

so we have $2^{1990} \equiv 1024 \mod{1990}$[/hide]

hello, $ 2^{1990} \equiv 1024 \mod 1990$ is correct.
Sonnhard.

[quote="vivek_n"]This is quite easy if you use binomial theorem

write the expression as $ 2^{1990} \equal{} (1990 \plus{} 58)^{180}.2^{10}$
then you get the remainder as $ 29*58 \equal{} 1682$[/quote]

how did you get $ 29*58$ as the remainder?

The remainder is $ 2^{10}*58^{180}$. and if you ask me, that's pretty hard to calculate by hand.

For problems such as these, it's necessary to use the Chinese Remainder Theorem

we can do the same thing very easily
 by fermat's little theorem you get

$ 2^{199}\equiv2(mod\quad199)$

 now by raising $ 2^{199}$ to the power of  $ 10$ we get

$ 2^{1990}\equiv1024(mod\quad199)$

and thus

$ 2^{1990}\ \equal{} 199n \plus{} 1024$

which implies that
 $ 2^{1990}\ \minus{} 1024 \equal{} 199n$ for some positive number n.

by trial and error we get that $ 2^{1990}$ ends with digit  $ 4$
thus $ 2^{1990}\ \minus{} 1024$ will have  $ 0$at its units place.

hence we get $ 2^{1990}\ \minus{} 1024 \equal{} 10k$ for some positive number $ k$.

thus  $ 2^{1990}\ \minus{} 1024$ is divisible by 10 and 199 both.
and since 10 and 199 are co prime  $ 2^{1990}\ \minus{} 1024$ must be divisible by 1990.

hence  $ 2^{1990}\ \minus{} 1024 \equal{} 1990p$ for some positive integer  $ p$.

hence $ 2^{1990}\equiv1024(mod\quad1990)$

Let $ a\equal{}2^{1990}$
By Little Fermat Theorem we have:

\[ 2^{199}\equiv{2}\mod(199)\]
\[ 2^{1990}\equiv{2^{10}\equiv{29}}\mod(199)\]
By trivial induction we have:
\[ 2^{4n}\equiv{6}\mod(10)\]
\[ 2^{1990}\equal{}2^{1988}*4\equiv{6*4}\equiv{4}\mod(10)\]

Let $ a \equal{} 1990q\plus{}r$ Then $ r\equiv{29}\mod(199)$ and $ r\equiv{4}\mod(10)$ and $ 0<r<1990$
$ r \equal{} 199k\plus{}29$. Reducing $ \mod(10)$ we obtain that $ 9k\plus{}9\equiv{4}\mod(10)$
$ \quad k<10$ because $ r<1990$.
Then the only $ k$ that would satisfy the above congruence is $ 5$ It implies $ r\equal{}199*5\plus{}29\equal{}\boxed{1024}$

let 1990=y+1
USING THE REMAINDER THEOREM,
       2^(y+1)=(y+1)(Quotient)+rem    Lets use P(-1) since it leaves out the remainder.
  then  P(-1)=2^(-1+1)
            P(-1)=2^0=1
therefore the remainder is 1 :D is it a solution

Oops, I made a mistake its not the solution :oops_sign:

I still can't understand..........Can anyone explain to me in a simpler manner, please?

the Carmichael function of a positive integer n, denoted$ \lambda(n)$, is defined as the smallest positive integer m such that
$a^m \equiv 1 \pmod{n}$For a power of an odd prime, twice the power of an odd prime, and for 2 and 4, λ(n) is equal to the Euler totient φ(n); for powers of 2 greater than 4 it is equal to one half of the Euler totient...this is a very useful fact

For all positive integers a and b it holds
$\lambda(\mathrm{lcm}(a,b)) = \mathrm{lcm}( \lambda(a), \lambda(b) )$ .
This is an immediate consequence of the recursive definition of the Carmichael function.

And how exactly do you prove that For a power of an odd prime, twice the power of an odd prime, and for 2 and 4, λ(n) is equal to the Euler totient φ(n); for powers of 2 greater than 4 it is equal to one half of the Euler totient ? ;)

Don't need any of this overkill, now do we? :)

[quote=amar_04]As there is no solution using CRT, so I thought of posting a new solution, please forgive me for bumping this old post. :)

We get $2^5\equiv 2\pmod{10}\implies 2^{1990}\equiv 2^{398}\pmod{10}\equiv 4^{199}\pmod{10}\equiv 6\pmod{10}$.

Now by Euler's Theorm we get $2^{198}\equiv 1\pmod{199}\implies 2^{1980}\equiv 1\pmod{199}$ and $2^{10}\implies 29\pmod{199}$. So, $2^{1990}\equiv 29\pmod{199}$. Now let $x=2^{1990}$. Now, $$x\equiv 6\pmod{10}$$ $$x\equiv 29\pmod{199}$$. Now let $n=199\times 10=1990$ and let $N_1=\frac{n}{10}=199$ and $N_2=\frac{n}{199}=10$. Let $x_1,x_2\in\mathbb Z$ such that $199x_1\equiv 1\pmod{10}$ and $10x_2\equiv\pmod{199}$ we get $x_1=9$ and $x_2=20$. Now as $\gcd(199,10)=1990$, by CRT we get $x\equiv (4\times 199\times 9)+(29\times 10\times 20)\pmod{1990}\equiv 1024\pmod{1990}$.[/quote]

Bhaiya,error is there,that will be 4 (mod 10) in 2nd line.

Fairly straightforward if we use the Carmichael function..
If $\lambda(t)$ denotes the Carmichael arithmetic function , then by properties of the Carmichael function $$2^{\lambda(995)}=2^{lcm (\lambda(5), \lambda(199)}=2^{lcm (\phi(5),\phi (199))} =2^{lcm(4,198)}=2^{396} \equiv 1 (mod \,995)$$
Thus we get $$2^{1980} \equiv 1(mod \, 995) \Leftrightarrow 2^{1990} \equiv 2^{10} (mod \, 995)$$
From this we can conclude that $2^{1990} \equiv 2^{10} (mod \, 1990)$

[quote=OCed]You can also use the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url] for this.
\[\lambda(1990)=\text{lcm}(\lambda(2),\lambda(5),\lambda(199))=\text{lcm}(1,4,198)=396,\] so $2^{396}\equiv1\pmod{1990}\implies2^{1990}\equiv2^{10}\equiv\boxed{1024}\pmod{1990}$.[/quote]

Although you've the correct residue, technically for $a^{\lambda(t)} \equiv 1 (mod \,t)$ to occur , $a$ and $t$ must be relatively prime,

Note that $1990 = 199\times10$. We can find $2^{1990} \pmod{199}$ and $2^{1990} \pmod{10}$.

By FLT, we know that:
$$2^{199} \equiv 2\pmod{199}$$
Raising to the power of $10$, we see that:
$$2^{1990} \equiv 2^{10} \equiv 1024\pmod{199}$$

The pattern of last digits of $2^{n}$ is $2, 4, 6, 8$. As $1990 \equiv 2\pmod{4}$, the last digit of $2^{1990}$ is going to be the second number in the pattern, that is $4$. As the last digit of $1024$ is also $4$, $10|2^{1990} - 1024$, $10, 199|2^{1990} - 1024$. This implies that $1990$ divides $2^{1990} - 1024$. So, that implies our answer is just $\boxed{1024}$.

[hide=Solution]We apply the Chinese Remainder Theorem, obtaining the remainder when $2^{1990}$ is divided by $199$ and the remainder when $2^{1990}$ is divided by $10$. From Fermat's Little Theorem, we get $2^{199} \equiv 2 \pmod {199}$. Raising to the tenth power gives $2^{1990} \equiv 1024 \equiv 29 \pmod {199}$. Next, we consider $\pmod {10}$. It's clear that $2^{1990} \equiv 0 \pmod 2$ and $2^{1990} \equiv 4 \pmod 5$, so $2^{1990} \equiv 4 \pmod {10}$. Therefore, $2^{1990} \equiv \boxed{1024} \pmod {1990}$.[/hide]


The solutions of $x^{3}-2x^{2}+3x+4 = 0$  are a, b, and c.
What is the numerical value of (a+b)(a+c)(b+c)?

[hide]By Vieta, $a+b+c=2$, so we have $a+b=2-c$, $b+c=2-a$, $a+c=2-b$. Then, the desired value becomes $(2-a)(2-b)(2-c)=8-4(a+b+c)+2(ab+bc+ca)-abc=8-4(2)+2(3)-(-4)=10$, by Vieta again.[/hide]

[hide="Slightly Different"]
By Vieta's we have that $\prod_{cyc.}(a+b) = \prod_{cyc.}(2-a)$. Now, since \[(2-x)^{3}-2(2-x)^{2}+3(2-x)+4 =-x^{3}+4x^{2}-7x+10 \] we have that $\prod (2-a) = \boxed{10}$ again by Vieta's. [/hide]

[url=http://mathworld.wolfram.com/VietasFormulas.html]Vieta's Formulas[/url] are extremely useful tools in finding the sum, product, etc of a polynomial's roots.

Remember, ALWAYS check the Wiki if you don't know something.

[url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas]link[/url]

If you still have questions, or can't find anything, then ask.


a 12.0 g bullet is fired horizontally into a 100 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring of constant 150 n/m. If the bullet-block system compresses the spring by a maximum of 80.0 cm, what was the velocity of the bullet at impact with the block?

What i have done so far:

I used hooke's law to calculate the force on the spring when it is compressed

F = -kx = (-150 n/m)(0.8m) = -120 N

and i used the potential energy formula (i'm not even sure you need it for the solution, but i did it anyways just for quick reference)

PE = 1/2kx^2 = 1/2(150n/m)(0.8m)^2 = 48 nm

From what i see i can tell that somewhere in the problem you will have to set -120 N = ma

so a would equal (-120 N)/0.112 kg = -1071.42 m/s^2...which seems way too high

*0.112 kg is the mass of the bullet and the mass of the box combined

so yeah...i know you need to find the initial velocity of the bullet which would probably require time and some sort of conservation of momentum equation but i am REALLY stumped on this one. I can't figure out how to put it all together.

As always, your help is greatly appreciated and possibly life saving

Try modelling the initial impact as an inelastic collision, and imagine that the bullet hits the box, they travel with a certain velocity and [i]then[/i] hit the spring and lose all their combined kinetic energy to the potential energy of the spring.

so in total you will use momentum conservation initially, and then energy conservation.

Wow, that's a nice problem & soln.... however, I find it a little unrealistic that the bullet stops immediately after the impact.

yeah.. I'm not sure what you're saying is wrong, I think it is either of two things, this is my reply (to both interpretations just to make sure):

(a) the bullet doesn't "stop" , it slows down to match the speed of the wood after the "collision" in which they join.

(b) if you mean that the bullet doesn't immediately change speed thats right.. I am not sure if this makes a difference to the total energy stored but yes I think the bullet will exert a decreasing force on the box, and my model where all the momentum is transferred from the bullet to the bullet&box [i]before[/i] the box begins to do work on the spring is not entirely accurate; the bullet will still be losing momentum doing work on the box after the box has begun doing work on the spring. I'm not sure if this affects the result but definately worth noting... Im not sure how I'd factor this into the calculations if it did make a difference, although by considering energy conservation, I'm not sure that it would.

This is a classic physics problem.

Momentum conservation is the solution key.

The bullet doesn't stop instaneously, but it stops (relative to  the wood, long before the spring reaches it's maximum compression.

If one uses momentum conservation, the energy consumed by deforming the bullet, wood and heating both is irrelevant.

One must simply assume that neglible momentum (mass) is released in the form of steam or non-wood absorbed particles.

Have fun.

degreestudent> Yes, right. The question is: does it make any difference? Seems like wizard2378 thinks it doesn't.

wizard2378> So you're saying that we shouldn't use energy at all, is that right? Hm.... interesting. I'll think about it...


A rock falls from a height $H$. During the last second of fall, it covers $\frac{3}{4}H$. Taking $g=10m/s^{2}$ and despiting the air resistence, compute:

a) the time of fall;
b) the heigth $H$.

Consider y = 0 for the ground and y = h for the height from where the rock falls. As we are neglecting the air resistance, then the equation of motion for the rock is

$y = h-\frac{1}{2}gt^{2}$.

For y = h let's consider t = 0. Consider also that, for $y = \frac{3}{4}h, t = t_{1}$, and for $y = 0, t = t_{T}$, where $t_{T}$ is the total time of fall. From the problem we now that

$t_{T}-t_{1}= 1$. So now we have:

$\frac{3}{4}h = h-\frac{1}{2}gt_{1}^{2}$ and $h = \frac{1}{2}gt_{T}^{2}$
$\Rightarrow  \frac{h}{4}= \frac{1}{2}gt_{1}^{2}$ and $h = \frac{1}{2}gt_{T}^{2}$.

Dividing these two equations we get:

$\frac{1}{4}= \left(\frac{t_{1}}{t_{T}}\right)^{2}\Rightarrow  t_{T}= 2t_{1}$.

Combining this last relation with $t_{T}-t_{1}= 1$ we get

$t_{1}= 1s$ and $t_{T}= 2s$.

The height $h$ can then be found from $h = \frac{1}{2}gt_{T}^{2}$, giving $h = 20m$.


Hello All,

 look my new puzzle game "Always 100!" at
[url=http://www.gameintellect.com/flash-games/always100.php]www.gameintellect.com/flash-games/always100.php[/url]
 Description:
 The goal of the test is to insert between the given digits some of the four arithmetic operations (addition, subtraction, multiplication and division) in such a way that each of the ten provided expressions is equal to 100.
 If your result will be less than 10 minutes, it will be added in the World's Best Records.

wow great game i got 9:26 on my first try :D

omg i got 4:10, second time i played BEAT THAT XDXD lololol this game is fun

[quote="rem"]4:29. beat that :D
I think something is wrong with the records thing. :?[/quote]

I have added your result to the records table. Please send me (gi[]gameintellect dot com) your real name and country. Thanks.

Author,
 http://www.gameintellect.com

What's wrong with this game????????????????? I got 2:35 and it doesn't output my name on the results list. :(
EDIT: WTF 1:56 still doen't output my name. :(
ha: 2:26 not as good as before but still good. this game is soo addicting :lol:
EDIT:1:59 why does it record only your high score? :?

haha omg i got 9 of them in 1:30 but then i was stuck on the last one for like 4 minutes omg >.<

o i went around that site and the other games were quite fun too, like the maze ones and stuff  :P

edit: i think i found an error in the game, you said in the instructions that you are allowed to make 100 with a blank as an expression, like 90:9*9+4+6=100. But if you complete it with that, the timer stops but it doesnt submit your score, it just sorta freezes.

EDIT just got 1:36  :P yay

[quote="Chocoboom"]
edit: i think i found an error in the game, you said in the instructions that you are allowed to make 100 with a blank as an expression, like 90:9*9+4+6=100. But if you complete it with that, the timer stops but it doesnt submit your score, it just sorta freezes.

EDIT just got 1:36  :P yay[/quote]
The timer freezes because your score is worse than your best score. and the blank thing doesn't affect anything except I find it easier to solve the question without the blank thing(i don't know why).
just got 1:30. very happy :)
Did u see the champion's score? 1:02. that's insane -- that's 6 sec per question. the guy is probably amaing at mental math.

got 1:22 XD lol man.. stupid Masaki... ill beat him one day...

dam just got 1:28

Are you sure that people aren't using programs to find the answer because that much time is necessary to fill the answers but no one can get all answers so quickly!


function
calculus
integration
derivative
LaTeX
trigonometry
real analysis

1. “f” is a differentiable function and f’’(x)>0.Prove that- (integration sign-with upper limit 2π to lower limit 0)f(x)cosxdx ≥ 0.
2. “f” is a differentiable function, satisfying the condition xf(x) =ln x .Now prove that f ^(n) (1) = (-1)^n+1(1+1/2+1/3+1/4+…….+1/n) n! where f^(k)(x) denotes the k-th derivative
pls. solve them up   .extremely sorry for not having LATEX and hence the integratrion sign calamity

2) it's easy try to find taylor expansion at point 1.
1) it's more interesting and wellknown fact in approximation theory.
if $ f$ is $ 2 \pi$ periodic and convex function, then for all $ \ell$ we have
$ ( \minus{} 1)^{\ell}a_{\ell}(g) \geq 0$ where $ a_{\ell}(g)$ is cosinus Fourier coeficients of $ g$  and equal to $ \frac {1}{\pi}\int_{ \minus{} \pi}^{\pi}g\cos \ell t dt$

Proof:
\[ ( \minus{} 1)^{\ell} \pi a_{\ell}(g) \equal{} ( \minus{} 1)^{\ell}\sum_{j \equal{} 0}^{2\ell \minus{} 1}\int_{\frac {j\pi}{\ell} \minus{} \pi}^{\frac {(j \plus{} 1)\pi}{\ell} \minus{} \pi}f(t)\cos \ell t dt \equal{} calculation ... \equal{} \int_{0}^{\pi/\ell} \sum_{j \equal{} 0}^{2\ell \minus{} 1}( \minus{} 1)^{j}f(t \minus{} \pi \plus{} j\pi/\ell)\cos \ell t dt \equal{} \int_{0}^{\pi /2 \ell}(Q(t) \minus{} Q(\frac {\pi}{\ell} \minus{} t)) \cos \ell t dt\]
where $ Q(t) : \equal{} \sum_{j \equal{} 0}^{2\ell \minus{} 1}( \minus{} 1)^{j}f(t \minus{} \pi \plus{} j\pi/\ell) \equal{} \sum_{k \equal{} 0}^{\ell \minus{} 1}f(t \minus{} \pi \plus{} \frac {2 \pi k}{\ell}) \minus{} f(t \minus{} \pi \plus{} \frac {(2k \plus{} 1)\pi}{\ell}) \geq 0$
since $ f$  is convex and hence his first difference will be increasing function.


If
$ x\plus{}y\equal{}a$
$ xy\equal{}b$
$ x\geq y$
then what is $ x$ in terms of $ a$ and $ b$?

Suppose $ x\neq~0$. $ y\equal{}\frac{b}{x}\equal{}a\minus{}x$ gives $ x^2\minus{}ax\plus{}b\equal{}0$. You can do the rest, i guess.

Thanks!
I was approaching the problem in way-overly-complicated ways.


show that there are exactly four distinct sets of integers which satisfy the equations:

     X1+X2+X3=54
    X1 2 +X2 2 +X3 2 =1406

It is not a combinatorial problem. I will move it to Algebra.

If you don't mind, I will use a, b, c to denote X1, X2, X3 resp.

Now, by AM-RMS we have $\displaystyle\frac{a+b}{2}\le\sqrt{\frac{a^2+b^2}{2}}$, so $\displaystyle\frac{54-c}{2}\le\sqrt{\frac{1406-c^2}{2}}$ and solving this one can conclude that $1\le a,b,c\le35$.  Suppose that c is the largest among a, b, c.  Clearly, $1406=a^2+b^2+c^2\le 3c^2$ and it follows that $22\le c \le 35$.  It is easy to show that a, b are the roots of $x^2-(54-c)x+(c^2-54c+755)=0$, letting c = 22, 23, ..., 35 and solving for a or b, one reaches to the conclusion that the solutions for $\{a,b,c\} = \{1, 26,27\}, \{2,21,31\}, \{5,15,34\}, \{9,10,35\}$.  Hence there are 4 distinct set of integers that satisfy the above equations. $\blacksquare$

Can some one show a nicer solution?


What exactly is this?
My textbook says that it is the unique operator on an inner product space $V$ that satisfies $<T(x),y>=<x,T^*(y)>$ for all $x,y\in V$. I don't know what this means in spaces other than $M_{m\times n}(F)$ (not even for the simple spaces $R^n$).

Are you being confused by the use of the term adjoint for a matrix of cofactors involved in computing determinants?  I prefer to call that the adjugate.  Anyway, it's quite a different thing.

Otherwise, it may be that you have been taught to use adjoint for conjugate transpose of a matrix (often denoted by ${}^*$.  This is the same thing: if you represent a vector in a complex vector space by an element of $C^n$, and the inner product by the usual formula $\langle u,v \rangle = u^* v = \sum_i \bar u_i v_i$, then if a map $T$ from the space to itself (operator) is represented by matrix $A$, the adjoint of that map (by your book's definition) will be be represented by the adjoint (conjugate transpose) matrix $A^*$.


Consider the recursion \[ \left(1 \minus{} a_n^2\right)\left(a_{n \minus{} 1} \plus{} a_{n \plus{} 1}\right) \equal{} na_n.\] Show that there is a unique solution satisfying this recursion relation, $ a_0 \equal{} 1$ and $ 0 < a_n < 1$ for all $ n \geq 1$.

Isn't that true that all recursion has a unique solution, but not the converse?

The value of $ a_1$ will determine the solution uniquely, but how to determine $ a_1$? :)

Hint: think about contractions and fixed points.


If $ x,y,z$ are positive reals with sum $ x \plus{} y \plus{} z \equal{} 3$,
Then Prove that  -
$ \sqrt {\frac {x}{1 \plus{} 2yz}} \plus{} \sqrt {\frac {y}{1 \plus{} 2xz}} \plus{} \sqrt {\frac {z}{1 \plus{} 2yx}} \ge \sqrt {3}$

It's very easy if I'm not wrong.By Holder,we have
\[ \left(\sum{\sqrt{\frac{x}{1\plus{}yz}}}\right)^2(\sum{(x\plus{}2xyz)})\geq(x\plus{}y\plus{}z)^3\equal{}27\]
So we need to prove that
\[ \frac{27}{x\plus{}y\plus{}z\plus{}6xyz}\geq 3\Longleftrightarrow x^2\plus{}y^2\plus{}z^2\plus{}6xyz\leq 9\]
\[ \Longleftrightarrow xy\plus{}yz\plus{}zx\geq 3xyz\]
which is ture by AM-GM .

[quote="Tourish"]It's very easy if I'm not wrong.By Holder,we have
\[ \left(\sum{\sqrt {\frac {x}{1 \plus{} yz}}}\right)^2(\sum{(x \plus{} 2xyz)})\geq(x \plus{} y \plus{} z)^3 \equal{} 27\]
So we need to prove that
\[ \frac {27}{x \plus{} y \plus{} z \plus{} 6xyz}\geq 3\Longleftrightarrow x^2 \plus{} y^2 \plus{} z^2 \plus{} 6xyz\leq 9\]

\[ \Longleftrightarrow xy \plus{} yz \plus{} zx\geq 3xyz\]
which is ture by AM-GM .[/quote]
exactly same as my solution! but thr is a typo wherever u have put x,y,z it is actually the squares..
anyway anyone trying this with cauchy or am-gm?


geometry
trapezoid
rectangle
area of a triangle

Given a square ABCD with AB = 1, what is the maximum possible area of a triangle incribed in the square? In other words, what is the maximum area of a triangle with all its vertices touching the sides of ABCD?

[hide]
$0.5$? ;)
i didn't solve it, thats just a guess...
 [/hide]

The answer is
[hide]
$\frac{1}{\sqrt{3}}$.[/hide]

Where's the fun in an answer without justification?

Making good guess is a kind of activities in problem solving. Why not?  :)
Besides, I think everybody will enjoy sharing the idea and solution once it is found.

i don't know how kunny got that, but my guess  was based on that the maximum area is the area of an isoceles tirangle with one side on the square. and the altitude would be $1$. and so base would be equal to $1$ also  so the area would be $0.5$
also you can think of it this way, the area of the right triangle with two sides on the square. would be half the area of the square.
still that was a guess... ;)

You should be able to make an argument that one vertex is on a vertex of the square.  Then find the minimum value of $\frac{a+b+(1-a)(1-b)}{2} = \frac{1+ab}{2}$ if $a, b < 1$.  Well that seems to be if one of them is zero, so I second amirhtlusa's answer...

[quote="blahblahblah"]I get $\frac {1}{2}$ using determinants, but I have a feeling we can inscribe an equilateral triangle in the square and that gives kunny's result.  I'm just trying to figure out why mine isn't yielding that.[/quote]

I don't see an equilateral triangle with a bigger area (the equilateral triangle with one vertex corresponding with one of the square's vertices has area slightly under $\frac{1}{2}$). I think $\frac{1}{2}$ is right.

If you fix two vertices, the area of the triangle is piecewise linear in the position of the other vertex- the maximum must come when it coincides with a vertex of the square. Starting with any triangle and applying this process three times, there is a triangle with equal or greater area that shares all three of its vertices with the square, and all such triangles have area $\frac12$.

I got the same result as jmerry. My argument consists of three parts:
[b]Step 1[/b]
Using parallel transform, it can be shown that if all three vertices do not coincide with any one of the vertices of the square, the area is not maximum. Therefore, the first conclusion is that the required triangle must have one vertex coincide with that of the square.
[b]Step 2[/b]
Assuming one vetex of the triangle coincides with that of the square. Using parallel transform again, we can prove that another vertex must coincide with a vertex of the square.
[b]Step 3[/b]
After Step 1 and Step 2, as the base of the required triangle is constant, we only need to consider its height and which is of course maximum when the height is equal to the side of the square.

The idea of my elementary method is more of less the same as jmerry's argument, but it is clear that jmerry considered the problem at a higher stand point. ( Maybe the argument can be applied to other convex figure instead of a square?)

As a final remark, I want to know if there is any generalization to this simple problem? Please let me know if you have any idea on that. Thank you very much.

 :)

My proof was this:

If two vertices lie on one side, the maximal area is clearly $1/2$

If all vertices lie on distinct sides, WLOG at $(0,x),(y,0),(1,z)$ we may expand the area in a determinant.

We find $\triangle=\frac {1}{2}|y+zx-xy|$ with $0\leq x,y,z\leq 1$.  Since the area is increasing in $z$, we may set $z=1$.  Then we need only maximize $x+y-xy$ (consider $(x-1)(y-1)\geq 0\Rightarrow xy-x-y\geq -1\Rightarrow x+y-xy\leq 1$) and we're done.

[quote="luimichael"]Maybe the argument can be applied to other convex figure instead of a square?[/quote]
Absolutely. If you want to inscribe a $n$-gon in a specified polygon with maximum area, my argument reduces the problem to finite case-checking, since all of the vertices must be vertices of the specified polygon- they may coincide, giving a degenerate $n$-gon. Of course, if you don't have enough symmetries to work with, that's a lot of cases. Also, convexity is not important- you should probably use signed area in this argument anyway.

Dear jmerry,
Although I can't  understand completely the generalization that you mentioned, some trials on special cases should be worthy.
I have tried two special cases below.

Case 1.
Inscribing a quadrilateral inside a regular pentagon ABCDE.
My answer: [hide]
The trapezium ABCD will have maximum area.[/hide]
Case 2.
Inscribing a quadrilateral inside a regular hexagon ABCDEF.
My answer:
[hide]
The rectangle ACDF will have maximum area.
[/hide]

Please point out my mistake if I mentioned something incorrect.

Thanks a lot.

Michael.


In the plane of any triangle ABC, let D,E,F be the midpoints of BC,CA,AB and X,Y,Z the centers of A-excircle, B-excircle, C-excircle respectively. Show that:

(1) The lines DX, EY, FZ meet in a point, labeled M (M is called the Mittenpunkt in of $\triangle ABC$ ).

(2) M is the symmedian point of triangle XYZ.

[b]P.S.[/b] I know a solution in barycentric coordinates. Is there someone of yours which is able to give a synthetic solution ?

It follows immediately from the fact that $ABC$ is the orthic triangle of $XYZ$, so, for example, the triangles $XYZ,XBC$ are inversely similar (i.e. they are similar but oriented differently).

[quote="Leon"]In the plane of any triangle ABC, let D,E,F be the midpoints of BC,CA,AB and X,Y,Z the centers of A-excircle, B-excircle, C-excircle respectively. Show that:

(1) The lines DX, EY, FZ meet in a point, labeled M (M is called the Mittenpunkt in of $\triangle ABC$ ).

(2) M is the symmedian point of triangle XYZ.[/quote]

The crux of the solution is to note that a symmedian from a vertex of a triangle bisects every segment antiparallel to the opposite side. Since the points X, Y, Z are the excenters of triangle ABC, the triangle ABC is the orthic triangle of triangle XYZ, and thus, the segments BC, CA, AB are antiparallel to the sides YZ, ZX, XY of triangle XYZ, respectively. Consequently, the symmedians of triangle XYZ bisect these segments BC, CA, AB; in other words, the symmedians of triangle XYZ pass through the midpoints D, E, F of the segments BC, CA, AB, respectively. Consequently, the lines DX, EY, FZ are simply the symmedians of triangle XYZ, and thus they concur at its symmedian point. $\blacksquare$

  Darij

Thanks Grobber and Darij for your useful suggestions  :) !

[quote]
The crux of the solution is to note that a symmedian from a vertex of a triangle bisects every segment antiparallel to the opposite side.
[/quote]

This theorem is demostrated in the popular book [i]R.Honsberger- Episodes in twentieth century euclkidean geometry[/i], pag. 60

[b]The symmedians of a triangle bisect the sides of its orthic triangle.[/b]

Leon

Dear Mathlinkers,
for (1) in English
http://perso.orange.fr/jl.ayme  vol. 12 Mittenpunckt  with some developpment
and vol. 3 Cinq théorèmes de nagel, p. 14 for a proof in French
Sincerely
Jean-Louis


algebra polynomial Vieta blogs quadratics

Note that Vieta says that the sum of the roots is $ \frac{\minus{}b}{a}$ and the product is $ \frac{c}{a}$. See if you can figure out how to derive these! (Solutions are in my blog - click my website for the Table of Contents.)

[quote="GameBot"]What is the sum of the two values which will make the following expression undefined?
\[ \frac{x\minus{}2}{x^2\minus{}6x\plus{}8}\][/quote]
How about we try using the quadratic formula.  We know that $x^2-6x+8=0$ because if the expression is undefined, that means that the denominator has to be 0.
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case $a=1$, $b=-6$ and $c=8$

We would have $x=\frac{6\pm\sqrt{36-32}}{2}$
$\sqrt{36-32}=\sqrt{4}=2$

$x=\frac{6\pm 2}{2}$

We have two cases to consider:
$x=\frac{6+2}{2}$ and $x=\frac{6-2}{2}$

$x=\frac{6+2}{2}=\frac{8}{2}=\frac{\cancel{8}}{\cancel{2}}=4$

The other case is: $x=\frac{6-2}{2}=\frac{4}{2}=\frac{\cancel{4}}{\cancel{2}}=2$

$\boxed{4+2=6}$

For the expression to be unidentified, the denominator must be 0. You can factor out x^2-6x+8=0. Think what 2 numbers multiply to 8 and sum to -6? Therefore, you get 4+2=6

Lol, my solution doesn't make sense. 4+2 is not -6 (I facepalmed). The best way is indeed Vieta. The denominator needs to be 0, so the sum of the roots is -(-6/1)=6


In a hypothetical situation...

There is a cart,a string is attached to the cart, a mass is attached to the end of the string. There is a pulley that converts the downward pull of the wegiht at the end of the string to horizontal motion of the cart. The downward pull of the weight at the end of the string is increased indefinitely, while the mass is kept constant.

Would the acceleration of the cart increase at a linear rate even as the force on the string downward becomes huge, such as the size of a building?

Basically, does $ F=m\cdot a$ hold at huge values?

Well, the version of Newton's Second Law that holds for ALL values of mass $ m$ and velocity $ \vec{v}$ is ($ \vec{p} \equal{} m\vec{v}$ is momentum, $ \vec{a} \equal{} \frac{d\vec{v}}{dt}$ is acceleration, $ t$ is time):

$ \sum \vec{F} \equal{} \frac{d\vec{p}}{dt} \equal{} \frac{d(m\vec{v})}{dt} \equal{} m\frac{d\vec{v}}{dt} \plus{} \vec{v}\frac{dm}{dt} \equal{} m\vec{a} \plus{} \vec{v}\frac{dm}{dt}$

If the mass is constant (which is what you are talking about), $ \frac{dm}{dt} \equal{} 0$ and the above expression simplifies to $ \sum \vec{F} \equal{} m\vec{a}$. So... the answer I would say is yes, that does hold for huge values of $ \vec{a}$.


trigonometry
function
integration
calculus
calculus computations

I have to find the fourier series for f(x)=x*pi - x*x
it is given that ak=0, So I have to find only the terms with the sine

Can someone help

A Fourier series represents a function on some interval. Which interval do you have?
My guess is $ [0,\pi]$, because then $ f$ can be extended as an odd function to $ [ \minus{} \pi,\pi]$, in agreement with $ a_k \equal{} 0$. Now $ b_k \equal{} \frac {2}{\pi}\int_0^{\pi} f(x)\sin kx dx$ can be evaluated by integration by parts (twice).

[quote="Abelois"]it is on the interval [-pi,pi]
but can someone also calculate what the outcome of bk is?[/quote]
first off, the question you asked is misleading. what you wanted to ask is this: find the fourier series for $ f(x)\equal{}x(\pi\minus{}x),\;\;\;\quad\,0<x<\pi$

now that you have the question, plot the curve on a piece of paper and you will see it is an inverted U-shaped curve within $ 0<x<\pi$

observe there are 2 flavors of fourier representations possible here. one is if you assign a period of $ \pi$, in which case it will be an [i]even [/i]extension ($ \Rightarrow$ only cosine terms will be present, i.e. all $ b_k$'s will be $ 0$). the other possibility is if you assign a period of $ 2\pi$, in which case you will get an [i]odd [/i]extension ($ \Rightarrow$ only sine terms will be present, i.e. all $ a_k$'s will be $ 0$).

now the constraint you imposed implies you want the latter, and so the interval of the odd extended function (of period $ 2\pi$) becomes $ [\minus{}\pi,\pi]$ (which is what you mentioned in your second post).

so, you need to solve $ b_k\equal{}\frac{2}{\pi}\int_0^\pi\;x(\pi\minus{}x)\cdot\sin\,kx\;\;\;\textbf dx$

eventually, you should get
$ f(x)\equal{}\frac{8}{\pi}\left(\dfrac{\sin\,x}{1^3}\plus{}\dfrac{\sin\,3x}{3^3}\plus{}\dfrac{\sin\,5x}{5^3}\plus{}\dfrac{\sin\,7x}{7^3}\plus{}\;\;\;\ldots\;\;\;\ldots\right)$

PS: had your question asked to find the fourier series for $ f(x)\equal{}x(\pi\minus{}x),\;\;\;\quad\,\minus{}\pi<x<\pi$
then obviously the function could neither be even extended or odd extended, so the series expansion would have [b]both [/b]sines and cosines in it, and the overall representation would look much more complicated.
and that is why i stated earlier that the question you asked was misleading.


modular arithmetic
inequalities
triangle inequality

An astronomer added all the distances between  50 stars that he observed with a telescope. The result was S. Suddely a cloud obscured 25 of the stars. Prove that the sum of the distances between the 25 visible stars is less than S/2.

I don't understand, what if the distance between the first and second star is $ 1000$ and the distances between the rest of the stars is $ 1$?

Then $ S\equal{}1048$ and $ S/2\equal{}524$. But if you cover up the last $ 25$ stars, the sum is $ 1024$...

[quote="grn_trtle"]I don't understand, what if the distance between the first and second star is $ 1000$ and the distances between the rest of the stars is $ 1$?
[/quote]
Have you tried drawing this?

[quote="grn_trtle"]I don't understand, what if the distance between the first and second star is $ 1000$ and the distances between the rest of the stars is $ 1$?

Then $ S \equal{} 1048$ and $ S/2 \equal{} 524$. But if you cover up the last $ 25$ stars, the sum is $ 1024$...[/quote]

I think the total distance includes the sum of the distances between all star pairs, not just "adjacent" ones, so in that scenario, the distance would exceed $ 50000$.

[hide="Solution"]Number the still visible stars $ 0, 1, 2, \ldots 24$ and the hidden stars $ 0', 1', 2', \ldots 24'$. Looking at stars $ a$ and $ b$ (still visible), consider also star $ c'$ where $ c \equiv a\plus{}b \pmod{25}$. By the triangle inequality $ d(a,c') \plus{} d(b,c') \geq d(a,b)$. Note that we will never use $ d(a,c')$ or $ d(b,c')$ again, because if we look at $ a$ and $ d$ and find $ c \equiv a\plus{}d \pmod{25}$, then $ d \equal{} b$. Add up all of these inequalities for all possible pairs $ a,b$ of still visible stars. Let $ T$ be the new sum of distances. Then the left hand side of the inequality is less than $ S \minus{} T$, since we never used segments like $ d(x',y')$, and the right side is $ T$. So $ S > 2T$ and the result follows.[/hide]


I was not sure how to call this here it is.

Given, $\{1,2,...,n\}$ with $n>1$ preform the following operation: For any two number $n$ and $m$ replace them by $nm+n+m$. Each time we get one less number in this set (counting multiplicity). After $n-1$ operation the final result is invariant of the choice of the integers.

[hide="Here is my ugly method"]
To keep things simple lets say $n=3$ and we $\{1,2,3\}$. And let $a_{1},a_{2},a_{3}$ be that set not necessarily in that order. We show that we end up with the same result no matter what $a_{1},a_{2},a_$ are:
Initially we have,
$a_{1},a_{2},a_{3}$
Preforming this operation once on the first two:
$a_{1}a_{2}+a_{1}+a_{2},a_{3}$
Once more,
$(a_{1}a_{2}a_{3})+(a_{1}a_{2}+a_{2}a_{3}+a_{2}a_{3})+(a_{1}+a_{2}+a_{3})$
This is basically the premutation of the roots of a cubic if you know what I mean by that. So $a_{1}a_{2}a_{3}$ is the same no matter how $a_{1},a_{2},a_{3}$ are chosen. Similarly $a_{1}a_{2}+a_{2}a_{3}+a_{2}a_{3}$ and $a_{1}+a_{2}+a_{3}$ are the same no matter how $a_{1},a_{2},a_{3}$ are chosen.
In general given,
$a_{1},a_{2},...,a_{n}$
The final result will be,
$\sum_{cyc}a_{1}...a_{n}+\sum_{cyc}a_{1}a_{2}...a_{n-1}+...+\sum_{cyc}a_{1}= \sum_{cyc}\sum_{cyc}a_{1}$ (the RHS is just a joke, apparently not very funny ignore it).
This can be verified by induction.
By similar reasoning it makes no difference what the numbers $a_{i}$ are chosen in the first place because we working with there premutations.
[/hide]

[hide="Giveaway"] Let the current set we have be given by $S$.  Then

$\prod_{s \in S}(s+1)$

Is invariant under the operation. [/hide]


[quote="xxxxx"]find the sum of coefficients of $(x+4y+2z)^{2}$[/quote]

[hide]
$49$??? I prolly did it wrong i must have distributed wrong or something...
[/hide]

[quote="t0rajir0u"]Correct, although it seems like you didn't do it [hide="the fast way..."] The sum of the coefficients of some polynomial $P(x, y, z, ...)$ is given by $P(1, 1, 1, ...)$ (do you see why?)  Ergo,

$(1+4+2)^{2}= 49$ [/hide][/quote]

ugh and i had to distribute that for nothing... :cursing:

ah that solution is so much easier  :lol:

[quote="xxxxx"]so if I was to find the sum of the coefficients of $(2x+5y+8z)^{2}$, then the answer would be 225?[/quote]

yes, according to t0rajir0u's thoery.

"Theory" is rather a large word for it.  Call it a lemma.

what about cubes? like if it was $(x+y+z)^{3}$ would u do $(1+1+1)^{3}$?

[quote="ckck"]No cubing is different. Look at the fourth line of Pascal's triangle.[/quote]
Cubing is the same way...

Looking at Pascal's wouldn't help here, there are 3 coefficients.

[quote="t0rajir0u"][quote="ragnarok23"]Cubing is the same way...[/quote]

Yes.  When I said "polynomials," I meant polynomials.  Cubed expressions are also polynomials.

Edit:  If you're slick, you might point out that this allows you to prove that the polynomial

$(x+1)^{n}= \sum_{k=0}^{n}{n \choose k}x^{k}$

Has sum of coefficients

$2^{n}= \sum_{k=0}^{n}{n \choose k}$

A very nice identity!   :)  (Can anyone find the combinatorial proof of it?  ;) )[/quote]
The combinatorial proof of it in AoPS vol. 1 is so awesome...
[hide]When looking at the number of sets, an element can either be or not be in the set. So when making an arbitrary subset, each element has two choices of whether or not it is in the set so there are $2^{\text{number of elements}}$ subsets possible.[/hide]

It's really simple.  First, consider some random polynomial:

$x^{2}+3x+8$

If we evaluate it for $x=1$ then that just gives the sum of the coefficients, because the x's will always be 1's.  Now the sum of the coefficients squared will equal the sum of the coefficients of the polynomial squared because those two polynomials are really the same polynomial, just rewritten.

It also kills some very frightening examples, such as:

Find the sum of the coefficients of $(529x-485y+318z-361w)^{711}$


The product of the first 1001 primes is divided by the product of the first 1000 primes. How many factors exist for this quotient?

2, as the result is a prime which has 2 factors.

[color=red]Just as in simplifying factorials:

product of first 1001 primes
---------------------------------  = a prime number ( which has [b]2[/b] factors)
product of first 1000 primes [/color] :rotfl:


John and Jane Doe both were born on 1988 on Feb. 29. What dates of 1989 through 1992 would you celebrate their birthday?

No dates of 1989 through 1991, inclusive, would I attempt to celebrate any birthdays for them because they have no February 29 in those years.   I would celebrate on February 29, in 1992,  which is a leap year four years after 1988, another leap year, being their birth year.

I don't know them, so I would not celebrate their birthday.

Incorrect conclusion.

You cannot state for sure that you do not know them because there is nothing
definitive in the puzzle that makes it so.  The puzzle does not state, for instance, that John and Jane Doe are part of any legal proceedings (from the definitions of these names), which THEN would have justified you stating that you don't know them.  As it stands, you may or you may not know them.  However, you cannot lay claim to definitely not knowing them.

maybe he doesn't know any person who's last name is doe who is born on Feb. 29th....

I don't think that THEY can celebrate their birthday at the age of $1$, since it is highly probable that they wouldn't understand the concept.

The original poster, "1=2" altered the original question somewhat, so I am basing this post on that.

They will celebrate their birthday on February 29, 1992, the next leap year after 1988.

(perfect_radio, I don't know what assumptions you made in your statement about them not being likely to understand their birthday celebration.  So be them.)

When February 29, 1992 comes around, they will have become four years older, just as everything else will have aged four additional years.  I predict they will understand some of the concepts of celebrating of their birthday.

If I were them I'd celebrate on March first each year, except for the leap years, because on March first, they will be a year older. Except when I turned 8, I'd probably have made a big deal about how I was actually 2 to all of my classmates, and been like, I'm the tallest two year old ever.


inequalities
function
inequalities proposed

Here is an incomparable easier ( nice, though) inequality:
  For any positive real numbers $ a_1,...,a_n $ with product 1 we have
  $\sqrt{a_1^2+n^2-1}+...+\sqrt{a_n^2+n^2-1}\leq n(a_1+...+a_n)$.

Believe me or not, but $ne^t-\sqrt{e^{2t}+n^2-1}$ is a convex function.
Though why am I writing that?  :(

Can I know the reason of your second statement? Too bad it works with Jensen.

It is your opinion. I do not consider it vain, but I would have liked the function not to be convex. Anyway, interpret it as you want.


function
blogs
number theory unsolved
number theory

For what $ k\in N$ there exists such function $ f: N\to N$, that
\[ f(f(n))\equal{}n^k\]
for all $ n\in N$?

Should be all positive integers $ k$.

[quote]First, let $ f(1) \equal{} 1$.

Let $ S\subseteq\mathbb{N}$ be the set of numbers that aren't perfect $ k$th powers. Take the two smallest elements of $ S$, and remove them. Call them $ a,b$. Define for all nonnegative integers $ n$, $ f(a^{k^{n}}) \equal{} b^{k^{n}}$, and $ f(b^{k^{n}}) \equal{} a^{k^{n \plus{} 1}}$. Thus $ f(f(a^{k^{n}})) \equal{} a^{k^{n \plus{} 1}}$, and $ f(f(b^{k^{n}})) \equal{} b^{k^{n \plus{} 1}}$ as desired. Continue this procedure of removing the two smallest elements of $ S$ and pairing them up in this way ad infinitum.

By examining the exponents in prime factorizations, we can see that for any two elements $ a,b\in S$ with $ a \neq b$, we have that $ a^{k^{n}}\neq b^{k^{m}}$ regardless of $ n$ and $ m$, so there won't be any multiple outputs for this function. Additionally, $ \{a^{k^{n}}| a\in S, n\in\mathbb{N}\cup\{0\}\} \equal{} \mathbb{N}$, so we have defined the function for all natural numbers.[/quote]
Haven't done a detailed check on this, but I don't think there's anything wrong.

Generalized from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150744]this topic[/url].
A further generalization is in [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=185272]this post[/url] from my blog (case $ a$ for $ k > 1$).


geometry
geometric transformation
dilation

WARNING :[b] NOT FOR JEE ASPIRANTS[/b] as of now

I was reading about special relativity and got some doubts
I wud be glad for someone to clear these

DOUBT 1  :  [b]TIME DILATION[/b]
it is derived       t =  t' / [sqrt]1 - [sq](v/c)[/sq] [/sqrt]        taking  a ray of light moving perpendicular to direction of motion and keepin in mind C is a constant

but when we try doing the same thing  by taking a ray of light moving antiparallel to direction of motion  we get

                          t = t' /(1 - v/c )

why ? isnt it wrong

DOUBT 2 :  [b]TWIN TROUBLE  ---- IS IT LEADING TO PARALLEL WORLDS ??[/b]

Acc to theory moving frames time runs slower

NOW consider trains moving in opposite direction with speed v always in straight line

Twins A and B are on the 2 diff trains

[u]A observes [/u]:   It is getting older , faster than B.

[u]B observes [/u]:   It is getting older , faster than A .
[u]
third person on platform observes [/u]:  A and B grow at same rate !

what does it imply ?  Multiple worlds ?

DOUBT 3 :

Have the above 2 paradoxes already been dealt with as they seem obvious . pls [u]provide links [/u]

DOUBT 1:I do not understand how you can do the same thing by taking a ray of light moving antiparallel to direction of motion.Please Explain.

DOUBT 2: Obviously  it does not lead to parallel world.It implies that there is no meaning of absolute space and absolute time.Both time intervel and spatial displacement is relative to different observers.But Proper time is same to all observer.
DOUBT 3:You can consult with a standard special relativity book like Resnick.
Which discuss the twin paradox in detail.


function
calculus
derivative
real analysis
real analysis unsolved

$f$ is defined on $[a,b)$ and it has right-hand derivative at every point in $[a,b)$. Prove that the right-hand derivative function is measurable.


Prove that for $ a,b $ positive real numbers we have

$ \frac{x^n+y^n}{(x+y)^n} + \frac{\sqrt{xy}}{x+y} \geq \frac{1}{2} + \frac{1}{2^{n-1}} $

iff  $ 2 \leq n \leq 7 $

After six months I am again trying to solve it :(

Has anyone tried to solve it?

Hm... I think it's quite hard [for me at least] - looking at it I don't even know what to do exactly with it   :? I try to solve it anyway...

at least, i established the condition $2\leq n\leq 7$ six months ago; it comes from considering $x=1+\epsilon$, $y=1-\epsilon$ as $\epsilon\to 0$. and those cases can, at least i guess so, be easily established one after another.

Did anyone try computer check? Why are you sure that it is true?


[quote="spider_boy"][quote="Tiks"][quote="nik"]In 2001, 2002, 2003 when I was contestant georgians result was more than much beter.
I think 4  medals are much better than 2 too.

Last 13 years: Georgia-Armenia 9:4

irakli
you know me.[/quote]
Do not forget that  your team had 94point,and our had 90,this year,and  that's why I said that your result isn't much better.
Also I think that Armenia and Georgia is brother counterys,so it isn't important for us to have batter results than you,but I can't say the same words about  [b][size=75]a[/size]zerbaijan[/b],and this year our result was much and much more batter than [size=75]a[/size]zerbaijan's result :P  and this is most important for us :wink: .[/quote]

Stop this!.[b][size=75]a[/size]zerbaijan[/b] :mad: :!:.why do you intentionally write Azerbaijan in this way?.Do you understand what are you doing?  This is an international forum and I can not bare your behaviour,You act as a little 1-year-old-boy.
If you continue this I will ask administrator to act accordingly. :!:[/quote]
 :rotfl: Now you don't like this,yes?
But if you can remember you started this :!: .
About the small leters  see your wrot under your posts,
 [quote]Dear users!please take few minutes to watch this site in order to know more about [b]armenian [/b]: [/quote]
If your English is too bad I can tell you that you have to write [size=150]A[/size]rmenian!!! :P
So befor start anything think a little longer :!:

[quote="Tiks"]
 :rotfl: Now you don't like this,yes?
But if you can remember you started this :!: .
About the small leters  see your wrot under your posts,
 [quote]Dear users!please take few minutes to watch this site in order to know more about [b]armenian [/b]: [/quote]
If your English is too bad I can tell you that you have to write [size=150]A[/size]rmenian!!! :P
So befor start anything think a little longer :!:[/quote]

In fact you are wrong, because in some countries, as in mine, the nationality adjiective usually doesn't start with a capitol letter. And of course he didn't make the first letter smaller, as you did. So stop to be angry without any reason.  :!:

[quote="Andreas"][quote="Tiks"]
 :rotfl: Now you don't like this,yes?
But if you can remember you started this :!: .
About the small leters  see your wrot under your posts,
 [quote]Dear users!please take few minutes to watch this site in order to know more about [b]armenian [/b]: [/quote]
If your English is too bad I can tell you that you have to write [size=150]A[/size]rmenian!!! :P
So befor start anything think a little longer :!:[/quote]

In fact you are wrong, because in some countries, as in mine, the nationality adjiective usually doesn't start with a capitol letter. [/quote]
Maybe in your language,but in English the correct version is "Armenian"!
and I am sure he knew that :!:

[quote="Tiks"][quote="ZetaX"]Continue here please, and in a more serious way.
This is not a funny or humorous theme, so discuss it not like you did till know (meant especially at Tiks).
(sorry when I spelled the countries wrong, I didn't know better)[/quote]
First ZetaX,I think this can't be named as a conflict between Armenia and Azerbaijan!
And this isn't even a conflict,I just wanted to show  spider_boy's mistakes.
Second,what do you mean  saying
 [quote]in a more serious way[/quote]?[/quote]

First of all "[b][i]dear[/i][/b]" Tiks, ZetaX is completely true by saying that "This is not a funny or humorous theme" that I think you assume.

Then you say "First ZetaX,I think this can't be named as a conflict between Armenia and Azerbaijan!
And this isn't even a conflict, I just wanted to show  spider_boy's mistakes."

ZetaX has guessed very accurately.
[b]It is a real conflict[/b] :!:  :!:  And you the armenians(or Armenians .I do know the rules that Andreas told for some popular countries but not for your country, and actually I don't have to know.) are always trying to show the world and in some way us that there is not any conflict between our countries by hiding your real faces. But not many people know that in how unimaginably bad way you have killed our old men, brides, women, and children. Not many people know that you have invaded 20 percent of our country and we have known 1 million refugees. (I want to note that we have 8 million people (i mean population..) Do you imagine what this is? And not many people know that you are still stealing our famous music (composer) works, meals, musical instruments etc... And provide it to the whole world as yours.We will never forget all have you done and everyone in the world will know your real faces.

"And I am sure he knew that"
First of all be really sure then speak. And "I just wanted to show spider_boy's mistakes." A man first must know and correct his mistakes, and then he must correct that of others'.
additional links: http://www.karabakh.org/

don't start this here.
do it on your own forum.

math should connect people (and math does that).

I have seen many political discussion (I started the Lebanese Israeli conflict topic in Round Table and requested info from personally involved people).  It should be allowed in my opinion.

In Dutch we write our language with a capital, the French write "français", anyway, I would say don't write any weird things only because you know it irks the other one, and please don't use difference in points or english grammar/spelling as arguments.

You hear about Israel every day, but the Armenian Azeri conflict is one only few people I know have heard of.  People from both parties are present, there's an actually an opportunity here.

Keep it civil.
Thanks,

I was stunned to know, that the Georgian writing system has no distinction between uppercase and lowercase letters, although ones of adjacent Armenia and Azerbaijan do have  :roll:

Also, it would be nice to hear from the Americans, in which cases do they write first and last names of persons in lowercase, do they write country/nationality names in such way too in any cases?

[quote="nonIMO"]Also, it would be nice to hear from the Americans, in which cases do they write first and last names of persons in lowercase, do they write country/nationality names in such way too in any cases?[/quote]

The names of individuals and nations are proper nouns in English and as such are always capitalized.  (Pronouns are not, except for "I.")  Nationalities are, as well.  (If he is from Georgia, he is Georgian, capital "G.")

[quote="JBL"]
The names of individuals and nations are proper nouns in English and as such are always capitalized.  (Pronouns are not, except for "I.")  Nationalities are, as well.  (If he is from Georgia, he is Georgian, capital "G.")[/quote]

But I've seen names of individuals in lowercase on some of American TV sports channels (I can't recall, which one). Maybe the rule is not too strict? This was interesting, because I've never seen this in the Russian-language media (except blogs and forums, of course), although, Russian words are sometimes written in all capitals in the media.

In Russian language nationalities are always written in lowercase (except in the beginning of the sentence). I think, that the English language has an advantage in this sense.

[quote]But I've seen names of individuals in lowercase on some of American TV sports channels (I can't recall, which one). Maybe the rule is not too strict?[/quote]
No, it's very strict in any vaguely formal setting -- I don't think I've ever seen what you are describing.  In fact, there are a (very small) number of people who have made a political or aesthetic statement by [i]not[/i] capitalizing their names (e.e. cummings, bell hooks).

[quote="fredbel6"]Wouldn't a split be appropriate, I'm sure many people would like to explore this matter, but it distracts from the conflict.[/quote]

And it would be a terrible thing to distract from conflict ...

[quote="fredbel6"]JBL, I know you know what I mean  :maybe:[/quote]

Sorry, I was just teasing.  Anyhow, no one has posted on the initial topic of this thread in more than a week, and I'm perfectly happy to have it die.  So, might as well use this thread for something.

About capitalization of "proper nouns" such as people, places, and organizations:

All "proper nouns" should be capitalized.  However, some organizations puposefully do not capitalize their organization name, to create an interesting "modern" design.  Examples:

Proper Capitalization:
[url=http://www.youtube.com/]YouTube[/url]

Improper Capitalization (used to attract attention):
[url=http://www.lowercasetee.com/]lowercase tees[/url] (the name of an organization)

Anyway, generally, all proper nouns should be capitalized out of respect for proper grammar, UNLESS the "creator" of the proper noun chooses not to capitalize the proper noun (very rare).

[quote="Google Define"]
"[b]Proper Noun[/b]" - a name given uniquely to a person (eg Joseph), place (Jerusalem), event (the judgment), group (the Twelve), organization (the Sanhedrin), position (the King of Israel), etc. English proper nouns are usually written with an initial capital letter.
[/quote]

I think I'm beating a dead horse here, but in company or organization logos or advertisements, sometimes you see all lowercase used for effect, even if the name of the organization is normally capitalized.  For example, my San Francisco Bay Area math team shirt, designed for the American Regions Math League (ARML), says "sfba arml" on it, even though the team is actually "SFBA ARML", and normally written as such.  It just looks cool on the shirt.

Also, on a lighter note, capitalizing every other letter of a word is pretty much never advisable.  You'll look like an immature 10-year-old.  :)

[quote="Sly Si"]
Also, on a lighter note, capitalizing every other letter of a word is pretty much never advisable.  You'll look like an immature 10-year-old.  :)
[/quote]

Such as LaTeX?

[quote="fredbel6"]Well, it's a shame.  Going off topic kills it too, and I don't want it die.  Why should this topic not be discussed?[/quote]

But the thread started from the "capitalization conflict". So, it had this meaning originally. Then, why is it off-topic? I have my opinion on the  conflict between Armenia and Azerbaijan and the subsequent war, I read a lot on this. But  more interesting for me is to discuss the "capitalization question". Is it a shame?

[quote=Irshad]Armenia doesn't even have a right to exist.[/quote]

Man, what a great post. Immediately after I read this post, my opinion changed. I no longer think armenia is a country. Yep, that's right. I fr*cking hate armenia so much I'm not even going to bother properly capitalizing its name! In fact, your post was so good that I pressed the upvote button TWICE. I also excuse you for bumping a post that is 12.5 years old and ignoring what BOGTRO wrote in the sticky:[quote]Replies to threads should contribute substantially to the discussion, and should generally be longer than a few sentences.$\phantom{Also remember to censor your minced oaths! }$[/quote]


[quote="Raja Oktovin"]Find all function $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that
\[ f(x \plus{} y)(f(x) \minus{} y) \equal{} xf(x) \minus{} yf(y)
\]
for all $ x,y \in \mathbb{R}$.[/quote]
$ y \equal{} 0$ gives $ f(x)^2 \equal{} xf(x)$ then $ f(x) \equal{} x$ or $ f(x)\equal{}0$
we suppose that $ f(a)\equal{}a$ for some $ a\neq 0$
if $ x\plus{}y\equal{}a$ we have $ f(x\plus{}y)(f(x)\minus{}y)\equal{}xf(x)\minus{}yf(y)\equal{}f(x\plus{}y)(\minus{}f(y)\plus{}x)$ gives $ f(x)\equal{}x$ and $ f(y)\equal{}y$
then $ f(x)\equal{}x$ for all $ x$
if $ a$ don't exist we will have $ f(x)\equal{}0$


parameterization
geometry
3D geometry
sphere
linear algebra
linear algebra unsolved

I am very bad at "analytic geometry". I just had a test on it and calculus. I overcomplicated an important question and want to know if I'm right (won't get to know in a few weeks otherwise):

We have a plane that cuts a disc out of a ball (I don't know if these are the right words. the plane cuts a "filled" circle out of a "filled" sphere). We also get a line on parameter form (parameter $t$ for future reference). Does the intersection between the plane and the line lie in the "disc"?

The official solution: Just find the intersection point, and make sure that the distance between it and the sphere's centre is smaller than the sphere's radii. Very simple.

My complicated solution: I showed that the line intersected the plane for $t = -\frac{3}{2}$. Then I found the t-values for which the line intersected the ball's shell (which is a sphere), call these $t_1, t_2$. Then I showed that $t_1 < -\frac{3}{2} < t_2$. From this it should follow that the point is in the ball, and of course in the plane, and therefore it should be in the disc.

Is this right? I think it is, but I am so bad at geometry.

edit: The reason this solution was "complicated" is that the problem was not designed to be made this way, so I had to some rather hefty calculations and estimations to do it.

The method sounds fine. It's the hard way, but it certainly works.

You've got the language right, too.


Let $ A$ and $ B$ be $ n
\times n$ matrices such that $ AB \equal{} 0.$  Prove that $ (BA)^2 \equal{} 0.$

Reassociate:

$ (BA)^2\equal{}(BA)(BA)\equal{}B(AB)A\equal{}(B0)A\equal{}0A\equal{}0$


calculus
integration
real analysis
real analysis unsolved

Rewrite the integral as:

$ (-1)^{n}\int_{0}^{\infty}e^{-a x}(1-e^{-x})^{n}\hspace{1 mm}dx$

Now let $ u=1-e^{-x}$ to get:

$ (-1)^{n}\int_{0}^{1}(1-u)^{a-1}u^{n}\hspace{1 mm}du$

$ = (-1)^{n}B(n+1,a)=\frac{(-1)^{n}\Gamma(n+1) \Gamma(a)}{\Gamma(n+a+1)}$

where of course I used the Beta function.