jack and joe went to duck hunting. Both of them has 1/2 chance to hit the shooted duck. If jack shoots 12 ducks, and joe shoots 13 ducks, what would be the probability of Joe's hit ducks are more than Jack'S??

sorry for english mistakes cause translated (un)successfully by me

(that's why i selected math  :D)

[hide]It's $1/2$, after $12$ shoots there is a probability of $p$ that they have shoot the same number of ducks and $\frac{1-p}{2}$ for each of them to have shoot more ducks. Then for the shoot n°$13$, joe will win with probability $p\times\frac{1}{2}$ (he shoots one duck) + $\frac{1-p}{2}\times (\frac{1}{2}+\frac{1}{2}$) (he was winning before the shoot $13$ so if he miss or hit he wins) then probability is $\frac{p}{2}+\frac{1-p}{2}=\frac{1}{2}$[/hide]

great job, how i missed that..
by the way my answer was 13/25 :blush:


function
algebra
polynomial
complex numbers
projective geometry
complex analysis

Let $ u(z)$ be harmonc in all of $ \mathbb{C}$.  Suppose $ |u(z)|\le c|z|^n$ for some positive constant $ c$.  Prove that $ u$ is the real part of a complex of degree at most $ n$.

I believe you're asking the following:

Let $ f(z)$ be an entire function. Suppose there exist constants $ M,R_0,k$ such that: $ |f(z)|<M|z|^k$ when $ |z|=R>R_0$, then $ f(z)$ must be a polynomial.

If so then:

If $ f(z)$ is entire, then it can be written as a Taylor series with center at the origin:
\[ f(z)=\sum_{k=0}^{\infty}a_n z^n\]
with each coefficient given by Laurent's Theorem:
\[ a_n=\frac{1}{2\pi i}\mathop\oint\limits_{|z|=R}\frac{f(z)}{z^{n+1}}dz\]
then:
\begin{align*}
|a_n|=\left|\frac{1}{2\pi i}\oint\limits_{|z|=R}\frac{f(z)}{z^{n+1}}dz\right|&\leq\frac{M}{2\pi}\oint\limits_{|z|=R}\frac{|z^k|}{|z^{n+1}|}dz\\ \vspace{12pt}
&\leq \frac{M}{2\pi}\frac{R^k}{R^{n+1}}2\pi R\\
&\leq \frac{M R^k}{R^n}
\end{align*}
Since this expression is true for all $ R>R_0$, we can let $ R\to\infty$ and thus $ a_n=0$ for $ n>k$. Thus $ f(z)$ must be a polynomial of degree not greater than k.

it seems that initially, you only have the bound $ u \equal{} O(z^k)$ instead of $ f \equal{} O(z^k)$ so to finish the argument, you must also prove $ v \equal{} O(z^k)$ for $ v$ the harmonic conjugate of $ u$, I think ?  is this in fact true ?


superior algebra
superior algebra solved

Is the number pi+e rational or irational?

take a look at http://mathforum.org/library/drmath/view/51617.html


geometry
3D geometry
sphere
geometry solved

Each point of the 3-dimensional space is coloured with exactly one of the colours red, green and blue. Let $R$, $G$ and $B$, respectively, be the sets of the lengths of those segments in space whose both endpoints have the same colour (which means that both are red, both are green and both are blue, respectively). Prove that at least one of these three sets includes all non-negative reals.

I think I saw this one In Engel book, anyway here is a solution:

Supoose non of the colors realises all distances, assume r1,g1,b1 are the smallest distances such that the red, green, blue points dont realise.

now suposse r1>g1>b1. take a red point and draw a sphere of radius r1, then the sphere is made of blue and green points. If tha sphere is only of one color, g1, or b1 is realised, so assume it is not. take a green point on the sphere and draw another sphere of radius g1, then the intersection of these 2 spheres is blue but it has 2 points at a distance b1, contradiction, so one of the sets must have all distances.

There are few details I didn't include, like the case when the space is of two colors only.....but In fact if tha plane is colored in two colors, the same result holds......!

Something is not clear in the solution above.

First, why do you take $r_1$, $g_1$, $b_1$ to be the smallest distances not realised? Those values may not exist in some configurations, and however the proof works fine also if $r_1$, $g_1$, $b_1$ are taken arbitrarily among the distances not realised. The important thing is just to suppose that $r_1\geq g_1\geq b_1$.

Second, you didn't prove that the common circle of the two spheres is large enough to have two points at distance $b_1$. This is easily done by saying that $r_1\geq g_1$ implies that the diameter of our circle is at least $\sqrt 3 g_1>b_1$.

[quote="pascual2004"]I think I saw this one In Engel book, anyway here is a solution: [/quote] BTW Darij, how does the BWM problem selection takes place ? Slowly I am beginning to believe that they really like to take problems from the Engel book for the competition...  :D

[quote="orl"]BTW Darij, how does the BWM problem selection takes place ? Slowly I am beginning to believe that they really like to take problems from the Engel book for the competition...  :D[/quote]

In fact, the problem we are speaking about was a BWM 1985 one, long before Engel's book was written, but in 2003 two BWM problems were in Engel (one in the same form as proposed in the competition, but without solution, the other one with solution but with slightly different coefficients). I really don't know how the problems are selected for the BWM, but Arthur Engel belongs to the problem selection comitee!

  Darij


function
factorial
algebra
polynomial
modular arithmetic
quadratics
AMC

Find the smallest two values of the integer n for which n^2-n+41 is composite

$n=41,42$ since the function $f(k)=k^{2}+k+41$ gives primes for $k=1,2,\cdots 39$.

How can u be sure that $f(k)$ will give primes for $k=1,2,\cdots ,39$ ?
You didn't check one by one, didl u ?  :lol:

$n^{2}-n=(n-1)^{2}+(n-1)$

So $n^{2}-n+41$ = $(n-1)^{2}+(n-1)+41 = m^{2}+m+41$ ($m=n-1$)
and $n=40$ is the largest number for which it is still prime.

And the fact that $n^{2}+n+41$ is prime for $1,2,\cdots39$ is a "well-known" trivia fact.  :wink: (Or not.)

[quote="Albanian Eagle"]you don't have to check all values
you should just check k=0,1,2,3  :wink:[/quote]

Can you tell me why is that ?
I'm sorry but I didn't understand your reason... :maybe:

Don't worry it happens!  :D
remember the famous http://www.mathlinks.ro/Forum/viewtopic.php?p=366550#p366550   :wink:

The function $f(k)=k^{2}+k+17$ also gives primes for $k=1,2,\cdots15$ :D

If you want more of those, you may like to know that the ring of integers of $\mathbb{Q}(i \sqrt{d})$ is principal, so factorial, for d = 1, 2, 3, 7, 11, 19, 43, 67 and 163.
(now compute the discriminant of those two examples given in the thread :) )

I guess there are infinitely many negative integers $n$ such that $n^{2}+n+41$ is composite.

[quote="fermat3"]therise result general of your problem.here you are: prove that doesnt exist a polynom who containt infinitly many primes.
in fact  supose that $f(x_{0})=p$ then we get easly $f(x_{0}+kp)$ is composite. :wink:[/quote]

Yes, you're right, but how can we proceed to prove it from the fact that $f(x_{0}+kp)$is composite ? Thx  :)
[Sorry, but I thought you prove that the polynomial contains infinite number of composites..]

He probably means composite instead of prime. Espcially, his statement is false for all polynomials of type $ax+b$, $\gcd(a,b)=1$. It's likely to be false for $x^{2}+1$, too (open problem).

i meant that : doesnt exists a polynom who contain primes for every natural number.
lets prove that:
supose that such polynom exist and chose a $n_{0}$ such $f\left(n_{0}\right)=p$. with a easy calcul we find that exist a polynom $g$ such $f\left(n_{0}+kp\right)=p(1+g\left(t\right))$. and we are done. :wink:

The way I solved the problem that was posted was that I considered what primes could possibly divide $n^{2}-n+41$.  If we consider the primes less than 41 we are basically solving for $n^{2}-n+41 \equiv 0 \pmod p$ for some prime $p < 41$.  But this is easy since we can use the quadratic formula and get that we have solutions iff there is some sort of $\sqrt{-163}$ i mod $p$.  We could then check the primes  $2,3,5,7, \dots, 37$ and conclude that they have no solutions by quadratic reciprocity and other such methods to compute the quadratic residues and conclude that these primes will never divide $n^{2}-n+41$.  Thus $n^{2}-n+41$ can only be composite if it is at least as big as $41^{2}$, so then $n \ge 41$, where we quickly check that 41 and 42 work.

Now I have a question about this well-known fact about why you can produce a lot of primes.  Fundamentally, it seems the only issue is that some quadratic residue doesn't appear until a large prime is reached.  So for -163, the first prime is 41.  It seems like certain other numbers are special too.  I just had a recent homework problem that asked me to prove that -163 has class number 1, and I also know that it is in some sense the last one that has class number 1.  Is there some connection here? and why would it be?  It might be obvious, but I haven't thought about it much. Just looking at my proof for $h(-163) = 1$, it doesn't seem obvious the relevant connection to quadratic residues.  I know that the point made by mathmanman is also another cool property of this class number business.  Is there some correlation between how often a number is a quadratic residue and the size of it's class number?

ZetaX, yeah $x^{2}+1$ is a funny one.  I think a lot of people tried to use the even stronger conjecture that we get infinitely primes for any irreducible primitive polynomial on USAMO number 3 last year :).


quadratics
AMC
AIME
algebra
quadratic formula

[hide="Answer!!!"]$\frac{x+y}{2}=2\sqrt{xy}$, then $x+y=\sqrt{16xy}$, then $x^{2}+2xy+y^{2}=16xy$. So $x^{2}+y^{2}=14xy$, so $x^{2}-14xy+y^{2}=0$. Quadratic formula and you get $7\pm4\sqrt{3}$.[/hide]
Sorry if my earlier and WRONG post confused you.

[quote="bigboy82892"][hide="Answer!!!"]None! $\frac{x+y}{2}=2\sqrt{xy}$, then $x+y=\sqrt{2xy}$, then $x^{2}+2xy+y^{2}=2xy$. So $x^{2}+y^{2}=0$ and $x^{2}=-y^{2}$, which is impossible.[/hide][/quote]I didn't read the problems when posting, so I posted a misleading title.  :blush:

It's AM is 2 greater than GM.  :D

[quote="4everwise"]For how many ordered pairs $(x,y)$ of integers is it true that $0<x<y<10^{6}$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y?$[/quote]
[hide]$\frac{x+y}{2}-2=\sqrt{xy}$
$x+y-4=2\sqrt{xy}$
$x^{2}+y^{2}+16-8x-8y+2xy=4xy$
$x^{2}+y^{2}-8x-8y-2xy+16=0$
$x=\frac{8+2y \pm \sqrt{4y^{2}+32y+64-4y^{2}+32y-64}}{2}$
$x=\frac{8+2y \pm 8\sqrt{y}}{2}$
$x=4+y\pm 4\sqrt{y}$
y has to be a perfect square. There are 999 perfect squares between 0 and $10^{6}$ noninclusive. Clearly x and y have to be positive. We know that $4+y>4\sqrt{y}$ for secondary values. This is only true for y=1. So we have 1000 ordered pairs. [/hide]
I hope this isn't an AIME question...

[hide]
Sorry for the nonlatex

(x+y)/2=2+sqrt(xy)
x+y=4+2sqrt(xy)
x-2sqrt(xy)+y=4
(sqrt(x)-sqrt(y))^2=4
Simplifies to
sqrt(y)-sqrt(x)=2

Y is greater than 0 and x so the min value of y is 3 and max is one below 10^3 so numbers between 3 and 999 inclusive is [b]997[/b]

I think?

[/hide]

[hide]
$(x+y)/2-2=\sqrt(xy)$
$x+y-4=2\sqrt(xy)$
$x-2\sqrt(xy)+y=4$
$(\sqrt(x)-\sqrt(y))^{2}=4$

From plugging in small numbers, we see that the pairs $(1,9)$ $(4,16)$ and $(9,25)$ all satisfy this equation. Let us conjecture that $x$ and $y$ satisfy this equation if and only if $(x,y)$ are of the form $(n^{2}, (n+2)^{2})$ with $n>0$.
Because our solutions $x$ and $y$ must satisfy $x<y<10^{6}$, and because $(10^{3})^{2}=10^{6}$,
$y\le(10^{3}-1)^{2}$.

Since it is a given that $x$ and $y$ are both positive, it follows from our conjecture that $x$ and $y$ can only take on $10^{3}-1$ different values, since this follows from the upper bound we set for $y$.  Furthurmore, because our solutions take the form $(n^{2}, (n+2)^{2})$, we must disregard the cases where $y=1^{2}$ or $y=2^{2}$.  This leaves us $10^{3}-1-2=997$ possible values.
To make this rigorous, let us now prove that $(x,y)$ is a solution if and only if $(x,y)$ is of the form $(n^{2}, (n+2)^{2})$.
First, we will show that $x$ must be a square.
$(\sqrt(x)-\sqrt(y))^{2}=4$
Because $x<y$,
$\sqrt(y)-\sqrt(x)=2$
$\sqrt(y)=2+\sqrt(x)$
$y=4+4\sqrt(x)+x$
$(y-4-x)/4=\sqrt(x)$
Thus, $\sqrt(x)$ is rational. Because square roots of integers are either integers or irrational, $\sqrt(x)$ is an integer and $x$ is a square.  Let $n$ be the positive square  root of $x$.  Now we will show that if $x=n^{2}$, then $y=(n+2)^{2}$
We know that $\sqrt(y)=2+\sqrt(x)$, which can now be rewritten as $\sqrt(y)=2\pm(n)$ because $x=n^{2}$.
However, we know that $y>x$.  Thus $\sqrt(y)=2-n$ is impossible.  Thus we have  $\sqrt(y)=2+n$
Squaring both sides gives us $y=(n+2)^{2}$, which is what we needed to show.  So $997$ is the solution to this problem.[/hide]

Wow, here's what I did
It simplifies to $ (x \minus{} y)^2 \plus{}16\equal{} 8(x \plus{} y)$.
Then let $ x \minus{} 4 \equal{} 4n$, then let $ x \plus{} y \equal{} 2k.$
We have $ n^2 \plus{} 1 \equal{} k$
Also $ x \equal{} 2n \plus{} k$ and $ y \equal{} k \minus{} 2n$ from the system.
Substituting, $ y \equal{} n^2 \minus{} 2n \plus{} 1 \equal{} (n \minus{} 1)^2$ and $ x \equal{} n^2 \plus{} 2n \plus{} 1 \equal{} (n \plus{} 1)^2$
Substituting in the original, we find these work. So y can range from 3 to 999, so the answer is 997.

is this an AIME question?
if so, from which year?

@above: you have totally unnecessary manipulation.
[hide="Quicker"]
We have $x + y - 2\sqrt{xy} = 4$ so square rooting gives $\sqrt{y} - \sqrt{x} = 2$ since $y > x$. So $x = (\sqrt{y} - 2)^2$. And so $y + 4 - x = 4\sqrt{y}$ so $y$ is a square. And then $997$ values.
[/hide]

Video Solution: [url]https://www.youtube.com/watch?v=--IsG-iWMVA[/url]
Hope you enjoy!

We can identify a lower bound for $x$ and an upper bound for $y$, which gives an interval for $y$, with $997$ elements!

$\color{blue} \boxed{\textbf{SOLUTION}}$

$$\frac{x+y}{2} - 2 = \sqrt{xy} \implies x+y-2\sqrt{xy}=4 \implies (\sqrt{y}-\sqrt{x})^2=4$$

$$ y > x \implies \sqrt{y}-\sqrt{x}=2 \implies (\sqrt{y} -2)^2=x \implies y \in (3^2,4^2,...999^2) $$

So, there are $997$ ordered pair of $(x,y)$


farz konid $G_{k}$ yek graphe dobakhshi bashad ke raashayash zirmajmoohaye $k$ ozvi va $k+1$ ozvi az $\{1,2,3,\dots,2k+1 \}$ bashand , va do ras dar $G_{k}$ mojaverand agar yeki az anha ba afzoodan ya hazfe tanha yek onsor az digari motefavet shavad .. ba dar nazar gereftane yek rase $X=\{ x_{1},x_{2},\dots,x_{k}\}$ va $0=x_{0}\ < x_{1}\ < \dots\ < x_{k}$ , farz konid $t$ bozorgtarin $i$ bashad ke be ezaye an $2i-x_{i}$ max shavad.. sabet konid ke majmooe yalhaye tashkil shode ba peivastane $X$ be $X\cup\{ x_{t}+1 \}$ yek tataboghe kamel az $G_{k}$ ast. ( masalan agar $X=\{ 1,3,4,7\}$ , darim $t=3$ va $X$ ba $\{ 1,3,4,5,7\}$ joor mishavaad)

no idea???? :(     ye zarre fekr konid khob dge


let $ n$ points in the plane $ (n \geq 5)$ such that there aren't among them 3 aligned .
prove that there is at least $ {n\minus{}3 \choose 2}$ convex quadrilateral  such that their corners are amoung the n points .


What is the general form of the sequence {1, 11, 111, 1111, 11111, ...}?

Hint
[hide]
What happens when you multiply the $n$th term of the sequence by 9 and add 1?[/hide]

Look at the repeating number 0.111..... or 1/9 as we know it.

let  a = 1/9

We know this holds our sequence, but how? Every time we multiply by ten,  we get:

a = 1.11....  10a = 1.11..  100a = 11.11..  1000a = 111.11...

We find that our desired sequence and 10xa are only seperated by 0.11111... or a

since [(10x)/9] - [1/9]  = (10x - 1)/ 9

so the function for this sequence is:

   f[x]: (10x - 1)/9

[quote="rzsolt"]
so the function for this sequence is:

   f[x]: (10x - 1)/9[/quote]

I don't think this makes sense, unless you mean $\dfrac{10^x - 1}{9}$.

The simplest way to do this:

Take the number N = 11...11 with n digits 1.

Multiply it by one: 9N = 99...99 with n digits 9.

Add 1: 9N + 1 = 100...00 with n zeroes.

So: $9N + 1 = 10^n$ or $N = \dfrac{10^n - 1}{9}$.

[quote="frt"]Here's a similar one- What is the general form of the sequence $1$, $12$, $123$, $1234$, ...?[/quote]
[quote="Altheman"]hmm, i think you could define it recursively, then just transfer it over to a closed form (too bad i don't know how to do that with this kind of recursion)
$a_{n} = 10a_{n-1} + n$ with $a_1=1$[/quote]
[hide="hint"]Let $a_{n}$ be the $n$th term of the sequence $1$, $12$, $123$, ... and $b_{n}$ be the $n$ term of the sequence $1$, $11$, $111$, ...

Then...
[hide]$a_n=\sum\limits_{k=1}^{n} b_k$[/hide][/hide]

[quote="Altheman"]the first term is 1, then 10+1, 100+10+1, etc, notice that to find each term, you make a geometric sequence ratio of 10 and initial term of 1[/quote]

The $n$ th term of $a_n$ is $a_n=1+10+10^2+10^3+\cdots+10^{n-1}=\dfrac{1-10^n}{1-10}=\dfrac{10^n-1}{9}$

[quote="Altheman"]hmm, i think you could define it recursively, then just transfer it over to a closed form (too bad i don't know how to do that with this kind of recursion)
$a_{n} = 10a_{n-1} + n$ with $a_1=1$[/quote]

$a_n=10a_{n-1}+n\Longleftrightarrow a_n+\dfrac{1}{9}n+\dfrac{10}{81}=10\left\{a_{n-1}+\dfrac{1}{9}(n-1)+\dfrac{10}{81}\right\}$

$\Longleftrightarrow a_n+\dfrac{1}{9}n+\dfrac{10}{81}=10^{n-1}\left\{a_1+\dfrac{1}{9}(2-1)+\dfrac{10}{81}\right\}=\dfrac{1}{81}\cdot 10^{n+1}$,yielding

$a_n=\dfrac{1}{81}(10^{n+1}-9n-10)\ (n=1,2,\cdots)$

[quote="frt"]Interesting. Here's a similar one- What is the general form of the sequence $1$, $12$, $123$, $1234$, ...?[/quote]
Does this even make sense? I mean, what happens when you reach 123456789? What is the next term?  :?  :D

Ah, good point, I didn't think about that :oops:  Maybe I should have said "What is the general form of the $n$th term of the sequence $1$, $12$, $123$, ... ,$123456789$  ($1\le n\le 9$)?"

By the way, kunny's answer (the second one) is correct :)


dANG COULDN"T WE REST!!

wait 5 mins pl0xx

Um, sorry to burst your bubble, but you cheated.

um. first of all, pop added 5 each time last time

secondly, ace messed up subtraction, so i pretended she minused two instead of wutever # she wrote and then i minused three, eventually getting down to zero..... TWICE

pl0x check back

I was posting one more than I did the last time every time, because if you posted, then it would just go up one and it would be right. When ace posted, it was supposed to go up two, but I didn't have time to change it, so I posted one less, which was good for the girls.
When you posted after each other, I went up two extra, and the problem was there.


Prove that for any positive integers $n$ and $k$ there exist positive integers $a>b>c>d>e>k$ such that
\[n=\binom{a}{3}\pm\binom{b}{3}\pm\binom{c}{3}\pm\binom{d}{3}\pm\binom{e}{3}\]

[i]Radu Ignat[/i]

Set $ a\equal{}x,b\equal{}x\minus{}1,c\equal{}x\minus{}2,d\equal{}x\minus{}3,e\equal{}y$ and use signs $ \minus{}\minus{}\plus{}\plus{}$. Now check again what you want to archieve.


Let be given real matrix $ A = (a_{ij})_{nxn} $ satisfying
$ a_{ij} = 0 $  if  $ i = j  $ and $ = \pm 1 $ if  $ i \ne j $.
Prove that:
a) If n = 3 then there exist such matrix $A$ for which $ det(A) = 0 $;
b) If n = 4 then we have $ det(A) \ne 0$ for all such A;

[quote="Namdung"]a) If n = 3 then there exist such matrix $A$ for which $ det(A) $;[/quote]
Huh?

[quote="Namdung"]
a) If n = 3 then there exist such matrix $A$ for which $ det(A) = 0 $;
b) If n = 3 then we have $ det(A) \ne 0$ for all such A;[/quote]

Do you by any chance mean

b) If $n\neq 3$ then $det(A)\neq 0$ for all such $A$ ?

a)

A=
(0  1  1)
(1  0  -1)
(1  1  0)

det(A)=0

b) for n=4 we work in Z/(2Z) any matrix with the condition become in M(4;Z/(2Z))

(0  1  1  1)
(1  0  1  1)
(1  1  0  1)
(1  1  1  0)

its determinant is class(-3)=class(1) (modulo 2)


geometry
incenter
vector
parallelogram
geometry proposed

Let $P$ be an inceter. If $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}=\overrightarrow{0},$ then prove that this triangle is a regular triangle.

Let $AP$ meet $BC$ at $A'$.  The vector equation implies that if we draw the parallelogram $BPCX$, then $B, P, X$ are collinear. $BP$ must then pass through the midpoint of $BC$, which means $AB$ = $AC$. Then since $2PA' = PX$, the vector equation implies $2PA' = PA$. Now drop a perpendicular $PY$ to $BA$. Since $PA' \perp BC$ and $P$ is the incenter, $PA' = PY$. Then $2PY = PA$. Triangle $PYA$ is thus a 30-60-90 triangle, and $ABC$ is obviously equilateral.

lest it be known...

It is very easy to prove that point  P  is the barycenter of the triangle(just take the midpoint M  of BC ect). But if the incenter and the barycenter coincide , the triagle is regular.

[u] babis[/u]


Find sum: $ S\equal{}\sin{x}\plus{}4\sin{2x}\plus{}9\sin{3x}\plus{}...\plus{}n^2.\sin{(nx)}$

Let $ f(x) \equal{} \sum_{k\equal{}1}^n \minus{} sin kx$. We have $ f''(x) \equal{} \sum_{k\equal{}1}^n k^2sin kx$. Also,

$ \sum sen kx \equal{} \frac{1}{2} \sum e^{ikx} \minus{} e^{\minus{}ikx}$.

hello, i have derived
$ \sum_{i\equal{}1}^n i^2\sin(ix)\equal{}$
$ \frac{1}{8} \csc ^3\left(\frac{x}{2}\right) \left(n^2 \cos \left(n x\plus{}\frac{3 x}{2}\right)\plus{}\left(n^2\plus{}2 n\plus{}1\right) \cos \left(\frac{x}{2}\minus{}n x\right)\plus{}\left(\minus{}2 n^2\minus{}2
   n\plus{}1\right) \cos \left(n x\plus{}\frac{x}{2}\right)\minus{}2 \cos \left(\frac{x}{2}\right)\right)$
Sonnhard.


inequalities
geometry
perimeter
inequalities theorems

In some cases when I try to use Cauchy or Holder I get the reverse inequality. So it becomes impossible to use those inequalities there.
For example,
If $ a,b,c$ are the side lengths of a triangle with perimeter $ 3$, prove that, \[ L\equal{}\sum \frac{1}{\sqrt{a\plus{}b\minus{}c}} \geq \frac{9}{\sum ab}\] [i](Pham Kim Hung)[/i]
Using Holder I get, \[ (a\plus{}b\plus{}c)L^2 \geq 27 \iff L \geq 3\]
Finally, $ \sum ab \geq 3$. Here the inequality is wrong and precisely the sign is reversed. (This inequality has a cute solution using AM-GM).

Another Example: Let $ a,b,c,d$ be positive real numbers such that $ \sum a^2\equal{}4$, prove that, \[ \sum \frac{a^2}{b} \geq 4\]
Here If we use cauchy directly we get, $ \sum \frac{a^2}{b} \geq a\plus{}b\plus{}c\plus{}d$. We need to prove that $ a\plus{}b\plus{}c\plus{}d\geq 4$
But AM-QM gives, $ \left (\frac{a\plus{}b\plus{}c\plus{}d}{4} \right)^2 \leq \frac{\left (\sum a^2 \right )}{4}\equal{}1 \iff \sum a \leq 4$!!!

So what should I do in these cases? I know one method that we need to multiply the fraction with something (so that the degree is higher) which makes the expression "bigger". Is there any more method of this type for these inequalities :?:

Any kind of help is appreciated. Thanks in advance. :)

When you can't solve an inequality that way it just means that the inequality to be proven is stronger than the inequality you are trying to apply (or the way of applying your inequality). Then obviously you will have to apply [i]some stronger inequality/technique that will make it stronger[/i] in order to solve it. This is what solving (inequality) problems is all about; it wouldn't be any fun otherwise, would it? :wink:

The second inequality is one of arqady's and it was discussed [url=http://www.mathlinks.ro/viewtopic.php?t=212248]here[/url].

Hmm...so I [i]strongly[/i] need to find [i]stronger techniques[/i] to solve [i]stronger[/i] inequalities.;)

BTW what's the definition of strong inequality. I mean when I find that some inequality $ P\geq Q$ is stronger than $ P \geq R$ or something like this?

Thanks a lot! :)

There is no definition of a strong inequality. It is just a relative idea. If we are dealing with some inequality $ a\ge b$, and we find some $ c$ so that $ a\ge c\ge b$ we simply say that the inequality $ a\ge c$ is [i]stronger[/i] than $ a\ge b$. (Similarly $ b\le c$ is stronger than $ b\le a$.)


function
LaTeX
real analysis
real analysis unsolved

Let $ f,g: \mathbb{N}\rightarrow\mathbb{N}^*$ be two functions with the following properties
[list][*] g is surjective
      [*] it exists a natural so that $ f(a)\leq a$
      [*] $ f[g(n)\plus{}m]\geq f(m)\cdot f(n) \forall m,n \in\mathbb{N}$[/list]
Prof that it exists $ n_0\in\mathbb{N}$ so that $ f(n_0)\equal{}g(n_0).$

Just put dollar signs around your $ \text{\LaTeX}$ code; that gets it interpreted in math mode. And there's an edit button, so please go back and try that on your original post.

Can you tell me what $ \mathbb{N}^*$ is?

$ \mathbb{N}^*\equal{}\mathbb{N}\setminus\{0\}$ is the set of all not null natural numbers

See [url=http://www.mathlinks.ro/viewtopic.php?t=172663]here[/url] the same problem :)


OK so last night I was watching Terminator 2, and one of the evil robots looks [b]exactly[/b] like Mr. Lindhorst (not the hair, just the face).

Even if this is in the Missouri forum, you have to remember than Missouri doesn't have a population of 20.  Most people in Missouri probably have no idea who Mr. Lindhorst is.

Yeah but everyone who posts in this forum does (whos form MO)

What if JRav is actually Mr. Lindhorst's Aops account..????? :o  :o

[quote="tinytim"]What if JRav is actually Mr. Lindhorst's Aops account..?????[/quote]

Nah, don't worry.  JRav is a portion of my actual name shortened.  And I live in Long Island, NY and I'm currently in Baltimore, MD so don't worry, I'm not stalking anyone in the MO forums?

Or am I? :police:


A, B, C are digits. The 3-digit number ACB is divisible by 3, BAC is divisible by 4, BCA is divisible by 5, and CBA has an odd number of factors. Find ABC.

[hide="Hint"]$ BCA$ is divisible by $ 5$ $ \Rightarrow$ $ A \equal{} 5$.
$ BAC$ is divisible by $ 4$ $ \Rightarrow$ $ 10A \plus{} C \equal{} 52 \ \text{or} \ 56$.
$ ACB$ is divisible by $ 3$ $ \Rightarrow$ $ A \plus{} B \plus{} C \equal{} 7 \plus{} B \ \text{or} \ 11 \plus{} B$ is divisible by $ 3$.[/hide]


function
calculus
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domain
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real analysis

When there is the Riemann-integral for the characteristic function?

In general, a function is Riemann integrable if and only if it is continuous almost everywhere (on some bounded set), so the characteristic function would have to be continuous almost everywhere.

Or if it has a countable number of points, where she is not-continuous...

That's what continuous a.e. means... and I didn't know functions had gender. :)

Continuous a.e. means that a function is continuous on the entire domain except possibly on a set of measure 0.  Any countable set has measure zero so if a function is continuous except at a countable number of points, it is continuous a.e.

The second comment was a joke, based on your statement that "...where she is not continuous."

JRAV, the joke was understood :P
However to "prove" the joke : Functions are curvy ;)

[quote="Christiano89"]The problem is that i don't know measure theory... But I can understand sth...[/quote]
To put it really simply, the measure of a set is basically its size from a length, area, or volume point of view.  The exact definitions are a bit more rigorous, but to say that a set has measure 0 essentially says that it has no length (in $ \mathbb{R}$; area would be $ \mathbb{R}^{2}$ and so on).  As a an example, $ \mathbb{Q}$ has measure 0 and perhaps more curiously, the Cantor set has measure 0.

Now I can understand it. So, the characteristic function has Riemann Integral if the set A, where we define the characteristic function, has measure 0?

Not quite.  The Riemann integral of $ \chi_{A}$ exists if and only if $ \chi_{A}$ is continuous a.e.  This means that $ \chi_{A}$ can't be have too stupidly.  As a quick example, consider $ \chi_{\mathbb{Q}}$.  This function is 0 a.e. (the irrationals) and 1 on a set of measure 0 (the rationals).  But in fact, $ \chi_{\mathbb{Q}}$ is continuous nowhere (can you prove this?) so it is as far from Riemann integral as you can get.  In fact, if you try to compute the upper and lower sums, the upper sum is always 1 while the lower sum is always 0, so the integral couldn't possibly exist.

Roughly speaking, you need the set $ A$ to be "pretty connected."  By that, I mean that it can't look like $ \mathbb{Q}$ or something similar.  Recall that a function is Riemann integrable if and only if you can force the upper and lower sums to converge.  If the $ \chi_{A}$ bounces between 0 and 1 too much, you can't make the two sums converge, and hence it can't be integrable.


calculus function factorial algebra domain

I'm thinking of putting this one on the tournament I'm writting. I need some confirmation for my answer.  ;)

Let $x^{!}=x!(x-1)!(x-2)!\ldots 1!$. Find the smallest $x$ such that $2007^{2007}$ divides $x^!$.

[hide]x=3? or u mean $2007^{2007}$ divide x?[/hide]

yeah, I mean $2007^{2007}$ divides $x^!$.

by the way, the way I had originally written it would give you $x=1$ since 1 divides $2007^{2007}$

[hide]$2007=3^2\cdot233$, so we need 233 appearing 2007 times.that means $a-233+1=2007,a=2239$ and 3 we have enough, so $x=2239$[/hide]

But you forgot that when we hit x=446 we ge two 223's in the ppf of x^!, and when we hit x=669 we get 3, etc.
Also, hide your next answer

[hide]
$2007=9 \cdot 233$. In $233^!$ the number 233 appears just 1 time. In $466^!$ number 233 appears 232 times (from 234! to 465!) plus one time in 233! and 2 times in 466!. In general, in the number $(233 \cdot k)^!$, $k \in \mathbb{N}$ the number of 233s is given by $232 \cdot \frac{k(k-1)}{2}+\frac{k(k+1)}{2}$. Following this reasoning and doing some calculus I obtained as result 1554...[/hide]

Question related: If I have $2$ numbers, $x$ and $y$, and $x!=y!$, or $x!!=y!!$, or the $x$ and the $y$ with the same number of $!$'s, must $x$ be equal to $y$? (without considering the solution $(0;1)$ or $(1;0)$).

Everyone's answers so far are too large.  The function hydro gave works up until $k = 233$, when it stops taking into account the higher powers of 233 that get introduced.  It didn't affect his answer, because we aren't working with numbers that large, but something else did -- his answer is too big.  An easy way to see this is simply to look at $k = 5$, which gives a sufficiently large value in his function and so gives us an upper bound of $x = 5\cdot233 = 1165$, smaller than his 1554.

The answer, for those who are interested, is [hide]1084[/hide]

Wait you guys, everyone made a senseless error. $233$ is not the prime we're looking at, its $223$.
My answer was close to JBL's, but I used $223$ instead of $233$.

Jose, I think that $x$ does equal $y$
To see this let $f_n(x)=x!!\ldots!$ ($n$ factorial signs) for $n\geq 1$ and $f_o(x)=x$.
Then, ignoring the cases where $f_i(x)=0$ or $f_i(x)=1$,
$f_n(x)=f_n(y)$ can never be true if $f_{n-1}(x) \not= f_{n-1}(y)$ since the factorial function always takes on different values for different inputs (once again, ignoring $x=0$ and $x=1$).

Maybe a better way to show it would be to show that $f(x)=x!$ is one-to-one when we restrict the domain.

Ignoring $x=0$, suppose that $f(x)$ is not one-to-one and let $f(x)=f(y)$, where, wlog, $x=y+k$ where $k$ is a positive integer. This means $x(x-1)\ldots1=y(y-1)\ldots1$, or $(y+k)(y+k-1)\dots(y+1)=1$, a contradiction if $y,k\geq1$. So $f(x)$ is one-to-one (with altered domain) so the existance of an inverse function is guaranteed. That is, ignoring $x=0$ or $y=0$, $f(x)=f(y)$ implies $x=y$.

Indeed.  That's what I get for not doing the arithmetic myself.  In any event, I get 1059 as the answer.

I agree with JBL. Here's my explanation.
I'm going to go slow just in case there is a mistake (there are lots of places for small counting errors). We get one 223 from $223^!$ and another for $224^!$ etc. up to $(223+222)^!=445!$. So that's $245-223+1=223$ 223's so far. We get another two 223's starting at $446^!$. This goes up to $(446+222)^!=668^!$. From this group we have $2(668-446+1)=2(223)$ 223's. We get three 223's starting at $669^!$ and this continues up to $(669+222)^!=891^!$. From this group we hae $3(223)$ 223's. Up to this point we have a total of $223(1+2+3)=1338$ 223's. So we need $2007-1338=669$ more. Starting in the next set (at $892^!$), we get another four 223's. Let $k$ be how much we increase from 892 (e.g., k=1 means we go to 893).  The $k$ we seek is the smallest $k$ such that $4k>669$. Thus $k=168$. With $k=168$ we are at $891+168=1059$


Let $\{a_{n}\}_{n \ge 1}$ be a sequence of positive integers such that \[0 < a_{n+1}-a_{n}\le 2001 \;\; \text{for all}\;\; n \in \mathbb{N}.\] Show that there are infinitely many pairs $(p, q)$ of positive integers such that $p>q$ and $a_{q}\; \vert \; a_{p}$.

An lovely problem :
Let consider the sequence as follow :
[b]1[/b]$ x_1,..,x_{2001}$ where $ 0<|x_i-x_j|\leq 2001$ and $ x_i\in \{a_n\}$
[b]2[/b]$ (x_1)!+x_1,...,x_{2002}!+x_2002$
...
...

[b]2002[/b]${ \prod_{i=1}^{2001}{x_{n_i}}+x_{n_i}}$
From this relative we has $ x_{ki}|x_{mi}\forall m\geq k$
Apply pegeonhole we has exist ${ (x_{ki},x_{kj})\in\{a_n}$
So we has $ x_{ki}|x_{mi}$

How exactly are you defining that sequence? I don't see the pattern.

is there somobody who know the solution?


Hi,

So no one knows how to do it?
Thank you anyway for reading.

[quote="Matharith"]
2m! = (m+1)^2m-1 + ....D(2m!)... ((m-1)!)^2
[/quote]

should be
\[ 2m! \equal{} (m \plus{} 1)((m \plus{} 1)^{2m \minus{} 2} \plus{} ....D(2m!)... ((m \minus{} 1)!)^2)
\]
Then you are very bright to discover this!

So get more study in Combinatorics to solve your problem!

It is a $ \prod$ sign.

Thank you.
What I'm looking for is to expand it as sum of three distincts parts not as product.
I'm looking just for the expansion of D(2m!).

2m! = A + B + C
where
A= (m+1)^2m-1
B=...D(2m!)...
C= ((m-1)!)^2

I know there is a formula to express it. I tried to find it on internet without success.

I still do not understand why no one has answered.
Since the expansion is in the range of high school it will be easy to point to some reference somwhere on internet or to write the expansion.
So thank you anyway.

Introductory Combinatorics,3E
By Richard A.Brualdi
is a further text.

The answer to the question is fairly trivial.
It it more important to learn more.

[quote="Matharith"]I still do not understand why no one has answered.[/quote]
That's because there really isn't much to say about this problem.  We have
$ \prod_{i\equal{}1}^{k} (t \minus{} r_i) \equal{} \sum_{j\equal{}0}^{k} t^j e_{k\minus{}j}(r_1, ... r_k)$

where $ e_n$ are the [url=http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial]elementary symmetric polynomials[/url].  Then what you're essentially asking for are the elementary symmetric polynomials of $ 1, 2, ... n^2$.  There isn't a particularly nice way of writing them all.

Thank you for help.
I have found another way to develop some specific factorials.
Thank you for your references.


Let n be a positive integer >5 and m be the greatest positive integer such that $\frac{m(m-3)}{2}<n$
Prove that there exists a m-element subset S of {1,2...n}such that no two
sums a+b/$a,b\in{S}$,${a}\neq{b}$ are equal  :)

This appears to be incorrect;  for n = 6, any 5-element subset will have two different pairs with the same sum.  For small values, the answer appears to follow the Fibonacci sequence;  the smallest value of n for which you can find a suitable m-element subset is $F_{m+1}$.


function
algebra
polynomial
rational function
complex analysis
real analysis
real analysis unsolved

[quote="Cezar Lupu"][color=blue] [b] Problem. [/b]

  Let $s\geq 2$ be a positive integer. Prove that there is no rational function $R(x)$ such that

                                   \[ \frac{1}{2^{s}}+\frac{1}{3^{s}}+\ldots+\frac{1}{n^{s}}=R(x),  \]
  for all positive integers $n\geq 2$.
[/color][/quote]
$n = x$?

Using complex analysis:

We have $R(x)-R(x-1)=\frac1{x^s}$ at the integers. If $R$ is a rational function, $S(x)=R(x)-R(x-1)$ is as well. $S(\frac1x)$ is therefore a meromorphic function; since we know its values on a convergent sequence, we know them everywhere and $S(\frac1x)=x^s$, $S(x)=\frac1{x^s}$. Now we can see that $R(z)$ has a pole at $z=0$, a pole at $z=-1$, and so on- a pole of order $s$ at each nonpositive integer. A rational function cannot have infinitely many poles, and we are done.

Using algebra:

Suppose $R(x)=\frac{p(x)}{q(x)}$ with $p$ and $q$ relatively prime. We have $R(x)-R(x-1)=\frac{p(x)}{q(x)}-\frac{p(x-1)}{q(x-1)}= \frac{p(x)q(x-1)-p(x-1)q(x)}{q(x)q(x-1)}$. If $r$ is an irreducible factor of $q$, then both $r(x)$ and $r(x-1)$ divide the denominator. If $r(x)$ divides the numerator, $r(x)$ divides $q(x-1)$ and $r(x+1)$ divides $q(x)$. Since $q$ has finitely many irreducible factors, there is some $n\ge0$ for which $r(x+n)$ divides $q(x)$ but $r(x+n+1)$ does not. Similarly, there is some $m\le0$ for which $r(x+m)$ divides $q(x)$ but $r(x+n-1)$ does not. Then both $r(x+n)$ and $r(x+m-1)$ divide the denominator, and neither divides the numerator.
Since $R(x)-R(x-1)=\frac1{x^s}$, we have a contradiction. This denominator has only one irreducible factor, and we have shown that it must have 2 if $R$ is a non-polynomial rational function; if $R$ is a polynomial, the denominator would have no irreducible factors.

One place I don't see any solutions: real-variable calculus.


calculus derivative function algebra polynomial factorial integration

I'm a newbie to Calculus, though I've been doing highschool contest math for a while. I just started teaching myself calculus, and I understand the concept of a derivative pretty well.

I realize that we can take Nth derivatives of functions fairly easily, but how about fractional derivatives? How would I go about finding one, if one existed?

I have an idea of how do to it with the gamma function, for polynomial functions.

Basically you extend the idea of the Nth derivative with factorial to the gamma function, and plug in real numbers (N > 0) and evaluate.

Is this along the right track, or is my idea completely false?

[quote="DiscreetFourierTransform"]how about fractional derivatives? [/quote]

I cannot, offhand, think of a reason why this concept would be meaningful.  The function $ f^{(k)}(x)$ as a function of $ k$ returns a function of $ x$, rather than a value, so it would be difficult to make the concept of a fractional derivative sensible (to my mind).

One can make the notion make sense given the full tools of Fourier analysis (including tempered distributions.)

Suppose that a function $ f(x)$ has as its Fourier transform

$ \widehat{f}(\xi)\equal{}\int_{\minus{}\infty}^{\infty}f(x)e^{\minus{}2\pi i\xi x}\,dx.$

Then (assuming the derivatives make sense in some sense),

$ \widehat{f'}(\xi)\equal{}2\pi i\xi\widehat{f}(\xi)$ and $ \widehat{f^{[n]}}(\xi)\equal{}(2\pi i\xi)^{n}\widehat{f}(\xi).$

That gives us the following idea: to take the $ \alpha$ derivative of $ f,$ where $ \alpha$ is not necessarily an integer, take the inverse Fourier transform of $ (2\pi i\xi)^{\alpha}\widehat{f}(\xi).$

But even that has certain problems, as we are forced to make an arbitrary choice in choosing a branch of $ (i\xi)^{\alpha}.$

Is there an intuitive way to look at it as a graph transformation?

(i.e the 1/2th derivative of a function is some combination of the graph of the derivative and the original function)?

Interesting.  Apparently, the concept does exist:

http://mathworld.wolfram.com/FractionalDerivative.html
http://mathworld.wolfram.com/FractionalIntegral.html


function
algebra
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complex analysis unsolved

Find all pairs $(n,m)$ of of positive integers $\geq 2$ such that there is a pair $f,g$ of nonconstant entire functions satisfying the equation $f^{m}+g^{n}= 1$.

The fact that the only such pair is $(2,2)$ follows from the first part of this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119070]other problem[/url] you've posted:

Suppose, for example, that $n>2$. We know that $f^{m}-1$ is a non-constant $n$'th power of an entire function. We can factor $f^{m}-1$ as $\prod_{i=0}^{m-1}(f-\zeta^{i})$ where $\zeta$ is a primitive $m$'th root of unity. Because the factors in this decomposition are mutually coprime, each one of them must be a non-constant $n$'th power, and hence the constant function $\zeta-1=(f-1)-(f-\zeta)$ is a sum of non-constant $n$'th powers. Because $n>2$, this contradicts the result in that other thread.

And a slightly more difficult problem: find all triples $(n,m,k)$ of positive integers $\geq 2$ for which there are nonconstant entire functions $f,g,h$ with $f^{m}+g^{n}+h^{k}= 0$.   :roll: This generalizes both threads...

Hmm, for polynomials this is not an easy problem, there are some really nasty huge polynomials that satisfy this relation for suitable $k,m,n$.

I agree it is not an easy problem. But it only asks for all possible $(n,m,k)$ for which there are solutions; and for polynomials, immediate application of the $ABC$-theorem renders only the possibilities $(2,2,k), (2,3,3), (2,3,4), (2,3,5)$, for which of which there is an example (and yes, some examples have nasty coefficients :)). But the question doesn't ask for a description of all such examples.

However, for the entire function problem, $(3,3,3)$ is a solution also (which isn't for the polynomial problem).

I thought I had a solution to the general problem showing that $(2,2,k), (2,3,3), (2,3,4), (2,3,5)$, and $(3,3,3)$ are the only solutions, but now I see that it only works for $m=n=k$... So it is possible that there are other solutions...

Just an idea: I know that there is an $abc$-theorem for meromorphic functions (in Nevanlinna theory), but I'm not familiar with these results. Maybe it would help...


Yes, me again  :blush:

My current equation is attached to this post as an image.

We can use either the elimination or substitution method for this equation. Now, I know how to use both methods. I am unsure of how to go about using one of these methods on this type of equation. Should I make a common denominator, eliminate the fractions, or what?

This is definitely not an olympiad problem, thus totally wrong here.
But the worst is to add it as picture: typ it down, when possible in $\LaTeX$.

Yes.

Whenever you have a system of equations, manipulate one of them so that it has the same coefficient for one of the terms.  Then you can add or subtract equations to get rid of a variable :)

once you have a common denominator, you can multiply both sides by it to simplify.


lets see if I learned form last time :)

[hide="answer"]
$\log_{2}{\left(\frac{9x+5}{x^{2}-1}\right)}= 2$
$\log_{2}{\left(\frac{9x+5}{x^{2}-1}\right)}= \log_{2}{4}$
$\Downarrow$
$4x^{2}-9x-9 = 0$
$\Downarrow$
$x_{1,2}= \frac{9 \pm 15}{8}= 3,-0.75$

$-0.75$ is extraneous to logs so it is discarded which leaves as the only value of x, $\framebox{3}$.
[/hide]


[quote="shyong"]4 positive integers $p,q,r,s$ such that $p>q>r>s$ .

If $ps=qr$ , and $p+s=2^{m}$ , $q+r=2^{n}$ with $m,n\in\mathbb{N}$ , show that

$s=1$[/quote]

It looks like an old IMO problem, isn't it? :)

Look here:[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=165811#p165811[/url] :wink:

thanks  :) . I dont know that it is an old IMO problem .


I know this sounds totally lame and makes me look as if I have WAY too much free time, which I do. Let's start a math jokes marathon. It's basically you post as many math jokes that have not already been posted and you win. Marathon lasts forever. :D

I have an awesome example :first:
Why was 6 afraid of 7?

Because seven ate (eight :rotfl: ) nine

 :whistling:

[hide="What did 0 say to 8?"]
Nice belt.
[/hide]

What is the integral of one over cabin d cabin?

Answer: Log cabin plus sea. House boat!

[Editor's note: Log here is the natural log]

A biologist has several snakes that he plans to study. However, after several weeks, he noticed they haven't reproduced at all. He wonders out loud, "Why haven't you reproduced yet?". To his surprise, one of the snakes replies "Get us some trees and we will be able to."

So hte biologist gets several trees to put in with the snakes. Soon, the snakes have many baby snakes. The biologist asks, "Why did you need the trees?". One of the snakes replies, "Well you see, we're adders, and we need logs to multiply."

A biologist, a physicist, and a mathematician are watching a house. Two people enter. Soon afterwards, three people leave.

The biologist concludes: "They reproduced."
The physicist concludes: "Our measurements were inaccurate."
The mathematician concludes: "If one person enters the house the house will be empty again."

=================

A engineer, a physicist, and a mathematician (specifically, a topologist) are given a survival test. They are locked in a room with a can of tuna, but no can opener. After three days the people conducting the study check on the engineer's room. There are many dents in the wall, but the engineer is eating out of the tuna can. Then they go into the physicist's room. There are many notes written on scrap paper, one single dent in the wall, and the physicist is eating out of the tuna can. Then they go into the mathematician's room. There are many notes written on scrap paper, no dents in the wall, but the mathematician is nowhere to be seen! They open the can of tuna and out comes the mathematician.

"Dammit, I got a sign wrong."

================

there are a ton here.

[url]http://www.math.utah.edu/~cherk/mathjokes.html[/url]

Mathematician, Physicist, Engineer walking through a field come upon a farmer.

The farmer asks what is the best way to construct a fence that will contain his livestock (ie., most area for least perimeter).

The engineer states that it is a circle.

The physicist points out that if the fence goes in a direction infinitely, there will be an infinite area.

The mathematician just sits there for a while and eventually stands up, puts a small fence around himself and says "I declare myself to be outside".

Salary Theorem (or, as i like to call it, the "Paris Hilton Theorem")
The less you know, the more you make.
Proof:

Postulate 1: Knowledge is Power.
Postulate 2: Time is Money.

As every engineer knows: Power = Work / Time
And since Knowledge = Power and Time = Money
It is therefore true that Knowledge = Work / Money .
Solving for Money, we get:
Money = Work / Knowledge
Thus, as Knowledge approaches zero, Money approaches infinity, regardless of the amount of Work done.

Interesting Theorem:
All positive integers are interesting.
Proof:
Assume the contrary. Then there is a lowest non-interesting positive integer. But, hey, that's pretty interesting! A contradiction.

[quote="daermon"]Paris Hilton Theorem[/quote]

 :rotfl:

A counter-proof to your next one:

All integers are boring.
Proof:
Assume the contrary.  So there exists one non-boring integer.  But who cares?

Terrorists capture a mathematician, a physicist, and an engineer. They bring them up on the roof of a sky-scraper and tell them that there is a 1m by 1m piece of cloth stretched at the bottom of the sky-scraper that is strong enough to balance the shock from the fall from the top of the building. Each of the three people must jump off. If he lands on the cloth safely, he can walk free. Otherwise, he dies...
So, the engineer is first. He does a bit of calculations, jumps off, and... dies.
The physicist is next. He does a bit more calculations, jumps off, and... falls safely on the piece of cloth.

Then, it's the mathematician's turn. He keeps calculating and writing formulas for several hours, then finally jumps off,
[hide="and..."]starts flying up, and shouting in horror:

"Wrong sign!!!"[/hide]

This one is just bad:

People get sinus infections, but what about cosinus infections?

"Do you love your math more than me?"
"Of course not, dear - I love you much more."
"Then prove it!"
"OK... Let R be the set of all lovable objects..."

Congratulations for reviving an 8-year-old thread.

[quote=wlm2]What happened to the plant in math class?
It grew square roots...[/quote]

No. Roots are 3D, so it grew cube roots.

and then it was grpahed on a 3d plane: complex roots

3 year bump!

1, 2, 3, 4, and 5 were in a classroom. Why was 3 kicked out?

[hide=Answer]Because 3 was being the mean.[/hide]


A piece of wire 72 cm long is cut into two equal pieces and each is formed into a circle. What is the sum, in square centimeters, of the areas of these circles?

Each wire is 36cm long. This makes a circle with circumference 36, radius $ \frac{18}{\pi}$ and area of $ 18^2\equal{}324*2\equal{}\boxed{648}$

Two equal pieces implies the circumference is 36.  Hence $ 36\equal{}2\pi r\Rightarrow r\equal{}\frac{18}{\pi}$.  Then the area is just $ \frac{324}{\pi}$.  Both circles have this area, so the sum is just $ \frac{648}{\pi}\approx 206.3$.


calculus
derivative
algebra
polynomial
integration
linear algebra
linear algebra unsolved

Let $C^{\infty}\{\mathbb{R}\}= \{f: \mathbb{R}\longrightarrow\mathbb{R}\}$ such that $f$ has derivatives of all orders.

$C^{\infty}\{\mathbb{R}\}$ includes polynomials, exponential, sin, cos, etc.

Let $T: C^{\infty}\{\mathbb{R}\}\longrightarrow C^{\infty}\{\mathbb{R}\}$ be defined by:
$f \longrightarrow f^{'}=T(f)$

Determine the eigenvalues and eigenvectors of T: $T(f)=f^{'}=\lambda f$

Thanks for the help!

$\frac{dy}{dx}=\lambda y$ is separable differential equation: $\int \frac{dy}{y}=\int\lambda dx$, hence $y=Ce^{\lambda x}$.


Find $xy+zy+xz$ if $x^2+y^2+z^2=1$, and x,y,z are positive reals.

I don't think an exact value is possible...  :huh:

That makes sense.  What's wrong with what I did though? It seems I found the maximum instead.

[quote="K81o7"]Why maximum?
We know $xy+yz+zx \le x^2+y^2+z^2$ with equalith iff $x=y=z$.  Hence, the maximum would HAVE to be the value of $x^2+y^2+z^2$, which is 1.[/quote]
[quote="K81o7"]Well we know $xy+yz+xz \le x^2+y^2+z^2$.
We know that $xy+yz+zx = \frac{(x+y+z)^2-1}{2}$ in this case.  Thus, we should minimize $x+y+z$.  This occurs, technically, as $x$ goes to 1 and $y$ and $z$ go to 0, making $x+y+z$ hone in on 1.  Of course, we're dealing in positive reals here.  However, we can go as close to zero as we want.  Hence, $0 < xy+yz+zx \le 1$.[/quote]
[quote="pkothari13"]If you do AM-GM on $xy+yz+zx$ and $x^2+y^2+z^2$, you will note the RHS is the same for each:

$\frac{x^2+y^2+z^2}{3} \ge \sqrt[3]{x^2y^2y^2}$

and

$\frac{xy+yz+zx}{3} \ge \sqrt[3]{x^2y^2y^2}$.

Because we know that the maximum value for $\sqrt[3]{x^2y^2y^2}$ is $\frac{1}{3}$, we can conclude that the minimum value of $xy+yz+zx$ is $1$ with equality occuring when $x=y=z = \frac{\sqrt{3}}{3}.$

I hope this reasoning is correct lol.[/quote]
[quote="Altheman"][hide]say we have (0,0,1), then the expression is 0, those are not positive, but instead, it could be 10^{-100}, and it approaches 0

i think that it should be maximum...then the am-gm is the way to go, but to make pkothari13's argument a little better, just subtract the 2 inequalities to get

$x^2+y^2+z^2-( xy+yz+zx) \ge0$
$1 \ge xy+yz+zx$

where the maximum of $xy+yz+zx=1$ is attained when $x=y=z=\frac{1}{\sqrt3}$

another thing that might work (i am not sure if i am using this correctly, but)

by the rearrangement inequality,

$x^2+y^2\ge xy+yx$

summing those with x,y,z taken pairwise gives us the same result[/hide][/quote]
Um...no
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz$
Because the LHS is a square
$0\leq x^2+y^2+z^2+2xy+2yz+2xz$
$-(x^2+y^2+z^2)\leq 2xy+2yz+2xz$
$-1\leq 2xy+2yz+2xz$
$-1/2\leq xy+yz+xz$
With min -1/2.

 :P

$x,y,z$ are positive real numbers. So $xy+yz+zx\geq 0$

[quote="deej21"][quote="leepakhin"]When any two of $x,y,z$ tends to zero, $xy+yz+zx$ tends to zero also. So
$0<xy+yz+zx\leq 1$.
(Maybe you are asking for the maximum? :? )[/quote]

True, but x y and z don't tend to zero.  Remember we have $x^2 + y^2 + z^2 = 1$

[hide="maybe..."]
We know by AM-GM that $xy + yz + zx \ge 3 \sqrt[3]{x^2y^2z^2}$... Now this doesn't help us, but we know the minimum occurs when there is equality.  If there is equality, all variables are equal, and the partial product is $1$.
[/hide][/quote]
But I said "any two of $x,y,z$", not all of them.

[quote="PenguinIntegral"]
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz$
Because the LHS is a square
$0\leq x^2+y^2+z^2+2xy+2yz+2xz$
$-(x^2+y^2+z^2)\leq 2xy+2yz+2xz$
$-1\leq 2xy+2yz+2xz$
$-1/2\leq xy+yz+xz$
With min -1/2.

 :P[/quote]

Of course it is true that $xy+yz+zx\ge -1/2$ but that does not make it a the best bound for the minimum.  As others have pointed out, 0 is the best minimum possible.  (And why would you say that $-1/2$ is a minimum value for something that is clearly positive :?: ...that's just redundant).


quadratics
geometry
3D geometry
trigonometry
algebra
difference of squares
special factorizations

[hide]
If the average of the two is $7$, that means their sum is equal to $14$. Also, we know that their product is $45$. So we get:

$x+y=14$
and
$xy=45$

Now all you have to do is solve the system.
[/hide]

I guess solve the system algebraically?

Say that $s$ is the sum and $p$ is the product.
$x+y=s$
$xy=p$

$x=p/y$

$p/y+y=s$

$p+y^2=sy$

$y^2-sy+p=0$

So basically you solve the quadratic with $a=1$ $b=-s$ and $c=p$

[quote="mathnerd314"]Yeah, but is there a more clever way to do it without finding x and y individually?

For example, there's a "more clever way" to solve this problem:

$x+y=16$
$xy=30$
Find $x^2+y^2$

The solution would be to square $x+y$ to get $x^2+2xy+y^2=256$. Plug $xy=30$ in and get $x^2+y^2=256-60=196$.[/quote]

That's one of those slick solutions that'll determine whether you win stuff like countdown or not. Start trying to solve the system and you'll get x=8-rt34.y=8+rt34.  Not fun.

It's not exactly guessing and checking, but when you see 45, the first thing that comes to my mind is 5. Then you see 9 pairs up with that quite nicely. 1/45 and 3/15 don't come to mind at all unless I actually think harder.

well as someone said..

First thing comes to mind is 5*9 and it JUST WORKS =)

Well, some things won't just "come to your mind".

For example, try this problem:
(2001 nationals)

If $a+b=1$, and $a^2+b^2=2$, then what is $a^3+b^3$?

Solving for a and b will result in radicals and a fraction. Cube them if you like...

And what if there was trig involved instead of a and b? (it was on a high school competition and the problem [b]said[/b] something like: hint: do not solve for the variable)

CRAP nvm edit, miscubed (a+b)

CRAP double edit -- realized I only had a small mistake and it was easy to fix...

Well the answer is [hide]$5/2$[/hide]

Take my word for it, ugh... i typed it all out then deleted it!

[quote="mathnerd314"]The arithmetic mean of two numbers is 7, and their product is 45. What is their difference?

(Without guessing and checking)

In general, if the sum of two numbers is a, and the product is b, how do I find the difference?[/quote]

Do they mean integers? then it's 5 and 9. It was faster to do guess & check on this problem for me, took about 3 seconds. Difference is $\boxed{4}$

[hide="big hint"] Make a and b roots of a quadratic.  Then you have the sum, and product of the roots, how can you write out a quadratic with those and solve them?[/hide]

i dont think anyone here has mentioned that if you say $x+y=a$, $xy=b$, and assume $x>y$, then $x-y = \sqrt{a^2-4b}$...can anyone figure out why?

its a good problem solving strategy to know.

also, note that $a^3+b^3 = (a+b)(a^2-ab+b^2)$.

that should help a lot on the problem where you are given $a+b$ and $a^2+b^2$, and are asked to find $a^3+b^3$ - it should be very easy to find.

[quote="mathnerd314"]The arithmetic mean of two numbers is 7, and their product is 45. What is their difference?

(Without guessing and checking)

In general, if the sum of two numbers is a, and the product is b, how do I find the difference?[/quote]
[hide="Without Guess and Check"]
If we write the "difference of squares" factorization, we have:
$a^2-b^2=(a-b)(a+b)$

We let $(a-b)$ be the smaller of the two numbers.
And we let $(a+b)$ be the larger of the two numbers.
We seek the difference of the two numbers, which is $2b.$
They gave us the average of the two numbers, $a=7.$
They gave us the product of the two numbers $(a-b)(a+b)=45.$
We just plug them in to the handy formula, and presto:

$\begin{eqnarray*} a^2-b^2 &=& 45\\ 7^2-b^2 &=& 45\\ 49-b^2 &=& 45\\ b &=& 2$

And the difference of the two numbers is $2b=4.$

With lots of practice, you can do many "difference of squares" problems in your head at lightning speed.  Think:  $7,\hbox{ squared is }49,\hbox{ minus }45\hbox{ is }4,\hbox{ rooted is }2,\hbox{ doubled is }4.$
[/hide]

it wouldnt be so easy if we said the mean was 6.5 ... in that case, the method i mentioned above would be faster.


1. Is there a setting such that starting a new line in the editor makes a new line in the compiler without the new line symbol? ex:

asdfasd
asdfasdfa

If I type this in the editor, it shows up as one line separated by a space in the compiled version. Is it possible to type what I did above and have it show the two words on two different lines by adding something to the preamble or something and not using the new line character?

2. Is there a Latex editor that builds what you're typing as you write, like in real time without pressing the build file button.

On the first: There shouldn't be, because it defeats the principles of the system. The same code should produce the same output in any instance of $ \text{\LaTeX}$, and those .tex files in the editor are just plain text.

Incidentally, fully skipping a line in the code does get you to a new line in the output, and is probably what you should use for text paragraph breaks.


Does anybody have good distance and speed problems or ratio problems? I need help mostly on the milk / paint mixing ratio problems and all problems concerning distance and speed and time, etc. All problems concerning anything remotely to do with these topics are welcome.

A nut mix is 99% by weight almonds. 1000 grams of this nut mix is taken. What is the weight, in grams, of the almonds that must be removed in order to make the new mixture 98% by weight almonds?


AMC
AIME
MATHCOUNTS
algebra
polynomial
function
AIME I

Is there a difference in difficulty between the AIME and the alternative AIME? Are we allowed to choose which one to take? My math teacher seems to think we can choose, in which case I would MUCH rather have the later date (which is what last year's date was around) since it would give me spring break to prepare a little bit.

They encourage that you take the first one unless you have a conflict.  And I believe your school doesn't have t pay for the first one whereas if you take the alternate they have to pay a fee.

not that i actually have any clue...but from my experience, it seems like the A test is easier...just a little bit

edit: I meant "AIME I," not "A"...

there shouldn't be a difference generally or it wouldn't be fair.

However you do 16 extra days to practice.. so I'm not sure whether I should take the AIME I or AIME II.

It will for me at least since I have almost no prep in AIME and I need to build the stamina slash skill for it. I am sure that spring break prep time will help like crazy.

Unless you have never seen an AIME before and are going to use those 16 days to acquaint yourself with them I doubt the extra time helps a whole lot.  For someone like Feiqi, I don't see much happening in those 16 days.  When you have trained all year and pretty much know all the techniques, theorems, etc. that you need for the AIME the only thing left after practicing old tests is mentally preparing.  The AIME is not like the SAT or AP tests.  Last minute cramming will probably hurt more than help you.

If you think that those 16 days are going to make the world of difference for you, then go ahead and take the AIME II.  But as for which one you will score better on it's just a coin flip, imo.  The AIME II might be easier for you than the AIME I.

[quote="Magnara"]16 days won't make a difference.[/quote]
Do you really believe this?  If so, are you also suggesting that this site's one-day AIME seminar is of no value?  [I strongly disagree, and I suspect Richard and Mathew would also disagree.]

The AIME has a very novel test-format.  A student who has never seen an AIME question will benefit just by getting used to the style of questions and answers asked on the AIME.  Summing ${p\hbox{, }q\hbox{, and }r}$ in the expression $p+q\sqrt{r}$, for example.

I think a student who hasn't been practicing AIME-level questions regularly -- such as Vinny -- could see a significant increase by practicing old tests over the span of a week.

that is definitely not true.

In 7th grade for mathcounts,

I finished ninth in chapter round, and after a month of INTENSE training, I tied for fourth in the state, but finished fifth after tie breakers and barely missed nationals.

a short intense workout does help.  Practicing 3 hours a day for 16 days is about the same as 1 hour a day for 1 month and half.

Now that the AMC forum has been re-opened, I've moved this thread where it belongs.

The thing about mathcounts is that you can learn how to do certain types of problems and you are guaranteed that you will see them on the competition.  Their questions are much more straightforward than AIME questions.  The questions on AIME are much more unpredictable.

Depending on your current status the extra 16 days will produce different results for different ppl.  It also depends on how much you are able to study during that time.  Nonetheless, I still don't see 16 days making a ton of difference for ppl already at the level of getting 5 or more uestions right.

16 days won't dramatically increase your math ability, but....I don't really know... maybe 16 days can prepare me better mentally

[quote="juggle7"]Can you take both the AIME I and the AIME II?[/quote]

I wish, but no, you can only take one

[quote="juggle7"]Can you take both the AIME I and the AIME II?[/quote]No.

You generally take the AIME on the regular date (March 8th) if possible.  If you need to take it on the alternate date (March 22), it costs your school extra, so most schools naturally try to discourage this.

Roughly 80% of students take the AIME I.  The two tests are designed to be roughly the same level of difficulty (last year, the average score on the AIME I was 2.25 and the average score on the AIME II was 1.96)

[quote="Ravi B"]I don't think it is meant to be a choice.  If you are available to take the AIME 1, then you are supposed to.  Only if you have a conflict are you supposed to take the AIME 2.  At least that's how I understand the rules.[/quote]My understanding of the rules is that you're not required or even necessarily encouraged to take the AIME I.  But administering the AIME II necessitates the school paying extra.

I think the extra time does matter some. I will be taking the AIME-II only because my school has spring break from March 2-21. Because of a lot of schoolwork, I am rarely able to study too much during February and have been restricted to 2 hours during the week and about 4 hours each weekend. Spring break will be the main phase of my training for the AIME and I hope it'll help. So it probably helps also if one has not trained enough for it.

That's strange.  For some reason I have always heard that the AIME II is harder.

Well, I doubt that I'm imagining things when I say this, but whoever says that AIME problems are "unpredictable" is not wholly correct.  I've noticed several recurring problem types.  Specifically, problems dealing with the representation of numbers by their base-ten digits, problems like "add all n for which so and so polynomial function of n is a perfect square," problems involving congruences, and problems involving functions of the roots and coeficients of polynomials, appear alot.  So it is extremely helpful to do alot of practice problems for the AIME--the more you do, the more prepared you'll be for whatever they throw at you on this year's test.

I wouldn't take the alternate. Looking at last years scores, the average on the alternate was much lower (which may be because most people took the first one, but still) so I would just stick with the regular one despite having a bit more time.

[quote="vinny919"]So do you guys think it is a smart idea to take the alernate AIME? Last year I only got 2 right, but this year I am able to get 3 correct in about an hour (I'm about to go finish the remaining 2 hours of the test). I think I can improve a lot, but if the Alt test is harder, will it balance out and should I just take the regular?[/quote]

Plus... if your school budget's drained like it is here in Glastonbury, then ya you shouldn't take the alternates, because the schools have to pay quite a bit, (considering they can't buy even a glue stick here :? ). But if you've never seen the AIMEs, then maybe it'll help you to take the Alternates and practice in those 20 or so days. BTW guys, this is my hundreth post! (I felt that I should mark the occasion...)


[hide]'Tis C

I can explain it, but non-latex, so bear with me.

Ok, so draw two concentric circles. the radius of the smaller circle is "r" and the radius of the larger is "R"
so we know that pi(R^2) - pi(r^2) = 12.5pi   (Area of larger circle - area of smaller circle = annulus which we are told is 12.5pi)
Take away the pi's and we get R^2  - r^2 = 25/2
Now, looking at the two concentric circles, we can draw the radius of the smaller circle and the tangent that is perpendicular to it. We can now draw the larger circle's radius, creating a right triangle, where R is the hypotenuse, r is one of the legs and 1/2 of the tangent (t) is the other leg.
Now we know (t/2)^2 + r^2 = R^2
We can rewrite this (t/2)^2 = R^2 - r^2
Since we know R^2  - r^2 = 25/2,
we can say (t/2)^2 = 25/2
so t/2 = sqrt (25/2) = 5/(sqrt  2)
Multiply 2 to both sides to get the length of the tangent and we get 10/(sqrt 2)
Rationalize the denominater by multiplying (sqrt 2)/(sqrt 2)
and you end up with 5(sqrt 2)
[/hide]

[quote="MCrawford"]The area of the ring between two concentric circles is $12 \frac{1}{2} \pi$ square inches.  The length of a chord of the larger circle tangent to the smaller circle, in inches, is:
\[ \text{(A)}\ \frac{5}{\sqrt{2}}\ \ \ \ \text{(B)}\ 5\ \ \ \ \text{(C)}\ 5\sqrt{2}\ \ \ \ \text{(D)}\ 10\ \ \ \ \text{(E)}\ 10\sqrt{2} \][/quote]

[hide]Ah, a fun conventional problem.

Notice that the area of the ring can be easily deduced if we simply reduce the radius of the smaller concentric circle to 0. Hence, the radius of the larger circle is now the radius of the once annulus/ring. The area is $\frac{25}{2} \pi$, so the radius r can be calculated:

$\pi r^2 = \frac{25}{2} \pi \\ \Rightarrow r^2 = \frac{25}{2} \\ \Rightarrow r = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$.

Since we want the chord/diameter times this by 2.

Our answer is indeed $5\sqrt{2}$. (C)[/hide]

aren't they looking for the diameter then though? A chord connects two sides of a circle.

Edit- posted before I realized the author above had changed answer... anyways..

[hide]
R=radius of big circle; r=radius of small circle
R^2*pi-r^2*pi=25*pi/2
factor out and cancel the pis (R^2-r^2=25/2)
draw in the tangent chord and u can see that it forms a right triangle with the chord/2  and r being radii and R being the hypotenuse. and since
c^2-b^2=a^2, we know that the chord/2= the square root of 25/2 which is 5/sqrt2
since that is onloy half the length of the chord, the chord is 10/sqrt2 or [b]5rt2[/b]

(i really need to learn Latex)[/hide]


geometry
geometric transformation
reflection
parallelogram
similar triangles

Let $ABC$ be a triangle. $D$ is the midpoint of $BC$, $E$ is a point on $AC$ such that $BE=2AD$. Let $BE$ and $AD$ intersect at $F$. If $\angle FAE=60^{\circ}$, find $\angle FEA$.

[hide]
Reflect $A$ over the point $D$ to $X$, so $ABXC$ is a parallelogram and $\overline{AE}\parallel \overline{BX}$. This makes the angles $\angle EAF$ and $\angle FXB$ equal. Also, $\angle AFE=\angle XFB$. It follows that triangles $\triangle AFE$ and $\triangle XFB$ are similar.

The similar triangles give $\frac{XF}{AF}=\frac{BF}{FE}$, or $\frac{AX}{AF}=\frac{BE}{FE}$. But $BE=2AD=AX$, so $AF=FE$ and $\angle FEA=\angle FAE$, which in this case is $\boxed{60^\circ}$.
[/hide]

Let the parallel through $B$ at $AD$ intersect $AC$ at $P$, then $BE=BP, E\neq P$. Since $AF\parallel BP\implies AF=EF$; with $\angle FAC=60^\circ$ the triangle $AFE$ is equilateral.

Best regards,
sunken rock


[quote="towersfreak2006"]If a cell is $10 \text{ }\mu\text{m}$ in diameter, what is its approximate mass?
A) $7.1*10^{-14}\text{ kg}$
B) $3.3*10^{-10}\text{ g}$
C) $5.2*10^{-4}\text{ }\mu\text{g}$
D) $2.5*10^{-9}\text{ g}$
E) $9.7*10^{-3}\text{ }\mu\text{g}$[/quote]

Logic tells me it is A.

Are we supposed to know this, or just enough for process of elimination?

Well, if there are billions and billions of cells in each person, and people (babies excluded)  weigh $\text{something}\cdot10$kg, i'd have to pick A as well.


Let P(x) be an irreducible polynomial of an odd degree with rational coefficients.Rational polynomials Q(x) and R(x) are such that $P(x)$ divides $Q(x)^{2}+Q(x)R(x)+R(x)^{2}$.
Prove that:
$P(x)^{2}$ divides $Q(x)^{2}+Q(x)R(x)+R(x)^{2}$

Write $Q^{2}+QR+R^{2}=H.P$. Assume $x_{0}$ is a real root of $P$ , then $Q(x_{0})=R(x_{0})=0$, so $H(x_{0})=0$, therefore $P|H$ because $P$ is minimal polynomial of $x_{0}$, and we're done!


Solve for natural $ n,a$ if

$ a^{1008} \equal{} n^6 \plus{} 3 n^5 \minus{} 5 n^4 \minus{} 15 n^3 \plus{} 4 n^2 \plus{} 12n \plus{} 3$

No solutions.
Easily prove that the RHS is not a perfect square for $ n$ being a natural while the LHS is a perfect square.

[quote="harekrishna"]Agr_94_Math is right. You can easily prove that RHS always ends in 3.
This is because the RHS is $ (n \minus{} 2)(n \minus{} 1)(n)(n \plus{} 1)(n \plus{} 2)(n \plus{} 3) \plus{} 3$

Now here is another problem.

Solve for natural $ x,y$ if

$ (x!)^{12343210^{y}} \equal{} 10072007300740075007.......10000000007$[/quote]
Well, $ x!$ is only odd for $ x\equal{}1$, so there are no solutions. Nice (but absurd) problem statement! :P

Brut3Forc3 is right. Also observe that the LHS is a perfect square but the RHS isn't.


geometry
incenter
geometric transformation
reflection
circumcircle
search
geometry unsolved

Problem 1:
Let $ ABC$ be a triangle, $ I$ its incentre; $ A_{1}, B_{1}, C_{1}$ be the reflections of $ I$ in $ BC, CA, AB$ respectively. Suppose the circumcircle of triangle $ A_{1}B_{1}C_{1}$ passes through $ A$. Prove that $ B_{1}, C_{1}, I, I_{1}$ are concyclic,
where $ I_{1}$ is the incentre of triangle $ A_{1}B_{1}C_{1}$.

something goes wrong... $ AA_1B_1C_1$ are concyclics!maybe i'm missing something...

We have $ \angle C_1IB_1 \equal{} 180^o \minus{} \angle BAC$
$ \angle C_1I_1B_1 \equal{} 90^o \plus{} \frac {1}{2}\angle C_1A_1B_1 \equal{} 90^o \plus{} 90^o \minus{} \frac {1}{2}\angle C_1AB_1 \equal{} 180^o \minus{} \angle BAC$
So $ \angle C_1IB_1 \equal{} \angle C_1I_1B_1 \Rightarrow C_1I_1IB_1$ is cyclic quadrilateral.

[b]Happy the Independence Day of Vietnam![/b]  :thumbup:


Princeton
college
real analysis
complex analysis

Can someone recommend a good introductory book on complex analysis? Many thanks.

Fundamentals of Complex Analysis (with Applications to Engineering and Science), 3rd. edition. by E. B. Saff & A. D. Snider. Prentice Hall, 2003.

This is the book that my university uses for the undergraduate level complex analysis course.

To what degree do you want a complex analysis book to help you learn applications and computational techniques, versus a complex analysis book that will focus on theory and analysis techniques?

[quote="Phelpedo"]I've heard that the [i]Princeton Lectures in Analysis [/i] (volume ii is complex analysis) is excellent. I'd like to hear from someone who used it.[/quote]

it's really really good, especially if you have a prior course (say, at the level of blue rudin) in real analysis.  this is the book that i recommend to everyone, and the book i wish that i had used to learn complex analysis from (i used ahlfors)


[url=http://www.mathlinks.ro/viewtopic.php?t=80127&ppage=200]Sos - sum of squares[/url]


quadratics
Vieta
algebra
system of equations
quadratic formula

Resolve into factors :
1)  x^2 +2(a^2 + b^2) + 3ax - b(3x + 5a)
2) x^4 + 6x^3 + 4x^2 - 15x + 6
Thanking you
Regards
Sreedhar

Can you post me the complete solution
Sreedhar

I think you just guess that it factors into at least two quadratics and write it as $ (x^2\plus{}ax\plus{}b)(x^2\plus{}cx\plus{}d)$ and then set up a system of equations for the coefficients using Vieta's or whatever.

[hide="Solution to 1"]
$ x^2 \plus{} 2(a^2 \plus{} b^2) \plus{} 3ax \minus{} b(3x \plus{} 5a) \equal{}$
$ x^{2} \plus{} (3a \minus{} 3b)x \plus{} (2a^{2} \plus{} 2b^{2} \minus{} 5ab)$

Using the quadratic formula, we get:

$ x \equal{} \frac {(3b \minus{} 3a)\pm \sqrt {9a^{2} \minus{} 18ab \plus{} 9b^{2} \minus{} 8a^{2} \minus{} 8b^{2} \plus{} 20ab}}{2} \equal{} \frac {(3b \minus{} 3a)\pm (a \plus{} b)}{2}$
$ x \equal{} 2b \minus{} a, b \minus{} 2a$

So, the factors of x are:
$ (x \minus{} (2b \minus{} a))(x \minus{} (b \minus{} 2a)) \equal{} (x \minus{} 2b \plus{} a)(x \minus{} b \plus{} 2a)$[/hide]

Edited: Changed the mistake caught by Arrange your tan...

[quote="mismatchtea"][hide="Solution to 1"]
$ x^2 \plus{} 2(a^2 \plus{} b^2) \plus{} 3ax \minus{} b(3x \plus{} 5a) \equal{}$
$ x^{2} \plus{} (3a \minus{} 3b)x \plus{} (2a^{2} \plus{} 2b^{2} \minus{} 5ab)$

Using the quadratic formula, we get:

$ x \equal{} \frac {(3b \minus{} 3a)\pm \sqrt {9a^{2} \minus{} 18ab \plus{} 9b^{2} \minus{} 8a^{2} \minus{} 8b^{2} \plus{} 20ab}}{2a} \equal{} \frac {(3b \minus{} 3a)\pm (a \plus{} b)}{2a}$
$ x \equal{} \frac {2b \minus{} a}{a}, \frac {b \minus{} 2a}{a}$

So, the factors of x are:
$ (x \minus{} (\frac {2b \minus{} a}{a}))(x \minus{} (\frac {b \minus{} 2a}{a})) \equal{} (ax \minus{} 2b \plus{} a)(ax \minus{} b \plus{} 2a)$[/hide][/quote]

No, the coefficient on $ x$ is 1, not $ a$, so the denominator in the quadratic formula fraction is $ 2$,
not $ 2a$.

Then the factors are $ (x \minus{} (2b \minus{} a))(x \minus{} (b \minus{} 2a)) \equal{} (x \minus{} 2b \plus{} a)(x \minus{} b \plus{}2a)$.

[quote="Arrange your tan"][quote="mismatchtea"][hide="Solution to 1"]
$ x^2 \plus{} 2(a^2 \plus{} b^2) \plus{} 3ax \minus{} b(3x \plus{} 5a) \equal{}$
$ x^{2} \plus{} (3a \minus{} 3b)x \plus{} (2a^{2} \plus{} 2b^{2} \minus{} 5ab)$

Using the quadratic formula, we get:

$ x \equal{} \frac {(3b \minus{} 3a)\pm \sqrt {9a^{2} \minus{} 18ab \plus{} 9b^{2} \minus{} 8a^{2} \minus{} 8b^{2} \plus{} 20ab}}{2a} \equal{} \frac {(3b \minus{} 3a)\pm (a \plus{} b)}{2a}$
$ x \equal{} \frac {2b \minus{} a}{a}, \frac {b \minus{} 2a}{a}$

So, the factors of x are:
$ (x \minus{} (\frac {2b \minus{} a}{a}))(x \minus{} (\frac {b \minus{} 2a}{a})) \equal{} (ax \minus{} 2b \plus{} a)(ax \minus{} b \plus{} 2a)$[/hide][/quote]

No, the coefficient on $ x$ is 1, not $ a$, so the denominator in the quadratic formula fraction is $ 2$,
not $ 2a$.

Then the factors are $ (x \minus{} (2b \minus{} a))(x \minus{} (b \minus{} 2a)) \equal{} (x \minus{} 2b \plus{} a)(x \minus{} b \plus{} 2a)$.[/quote]

 :blush: Thanks.


We could use the same tehnic as in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=210516]here[/url].
So , from Holder we have :
\[ \left(\sum\frac{a}{\sqrt[3]{a^{3}\plus{}(b\plus{}c)^{3}}}\right)^{3}\left(\sum a(a^{3}\plus{}(b\plus{}c)^{3})\right) \geq \left(a\plus{}b\plus{}c\right)^{4}\]
It is enough to prove that \[ (a\plus{}b\plus{}c)^{4}\geq \sum a(a^{3}\plus{}(b\plus{}c)^{3})\equal{}(a\plus{}b\plus{}c)(\sum a^{3} \plus{}6abc)\]
After some simplifications, the following is equivalent to $ \sum ab(a\plus{}b) \geq 0$ .
Equality occurs for $ (a,b,c) \sim (k,0,0)$ where $ k>0$.


function
real analysis
algebra unsolved
algebra

is there bijection from R^R to the power set of R?

More generally, if $A$ is an infinite set, then there is a bijection between $A^{A}$ and $2^{A}$.

By [url=http://planetmath.org/encyclopedia/SchroederBernsteinTheorem.html]Cantor-Bernstein[/url] and the trivial observation that there is an injection from $2^{A}$ to $A^{A}$, it suffices to prove that there is an injection from $A^{A}$ into $2^{A}$.

Because $|A\times A|=|A|$ (proven on the forum before; the most recent topic is in the "Real Analysis" section, I believe), we can partition $A$ into $|A|$ subsets $(A_{x})_{x\in A}$. Let $\phi: A\to 2^{A}$ be an injection. For every function $f: A\to A$ (an element of $A^{A}$, that is), consider the function $\Phi(f): A\to\{0,1\}$ (i.e. an element of $2^{A}$) defined as follows: $\Phi(f)|_{A_{x}}=\phi(f(x))$. In other words, the restriction of $\Phi(f)$ to the subset $A_{x}$ of $A$ is equal (after identifying $A_{x}$ to $A$ through a bijection, but that would only make things more complicated :)) to the element $\phi(f(x))$ of $2^{A}$.

Clearly, $\Phi: A^{A}\to 2^{A}$ is an injection, and we're done.


trigonometry
ratio
geometry
circumcircle
inequalities
trig identities
Law of Cosines

Guys I dont know when to use law of cosines and when to use law of sines. Can someone explain it to me and give me practice examples?

Law of Cosines: $ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \cos C$

Law of Sines: $ \frac{a}{\sin A} \equal{} \frac{b}{\sin B} \equal{} \frac{c}{\sin C}$

Notice that it depends on the information given and what is needed. I don't have any examples though.

Hey Nate hows it going?

Alrite so, we use Law of Cosines when we have 2 sides and one included angle. Law of sines is used when we have 2 sides and 2 angles.

Try these:
In triangle RST, r=3√2, s=1, and the measure of angle T is 135. Find t.

(This one is harder)
The vertices of a triangle inscribed in a circle separate the circle into arcs whose measures are in the ratio 2:3:4. If the measure of the radius of the circle is 10. Find the longest side of the triangle.

Law of Sines:
In triangle CAR, a=24. Angle A is 27. And measure of C is 83. Find c to the nearest integer.

In triangle RST, Sin r is 0.4, and the measure of angle S is 30. Find the ratio of r:s.

Oh and I forgot how do you find the area of a triangle using trig? And hey Navdeep

And can you give a hint to how to start the 2nd problem?
I'll give the answers at once

There are many, many, many, many ways. But most are derived from $ \frac {ab}{2} \sin A$.

Hint for second problem: The measures of the arcs add up to 360. Once you find that, notice that the measure of an inscribed angle if half of the arc it opens up to. One last thing, the FULL law of sines is

$ \frac{a}{\sin A} \equal{} \frac{b}{\sin B} \equal{} \frac{c}{\sin C} \equal{} 2R$,

where $ R$ is the radius of the circumcircle of the triangle.

Uhm. Draw a diagram. Label the arcs in the ratio 2:3:4. So Find each arc. A central angle is congruent to it's intercepted arc. Law of cosines.

i got the answer 20 but if its 20, it cant be a triangle. (triangle inequality rules) did I do something wrong?

And thanks a lot Marc. I get it now :)

Consider [url=http://www.themathpage.com/atrig/law-of-sines.htm]this link[/url] and [url=http://www.themathpage.com/atrig/law-of-cosines.htm]this link[/url] for the laws of sines and cosines.

Actually no, thats the answer. I think the exact answer was like 19.8... something of that sort. But 20 is right. Its like when you first learned about triangles and the teacher labeled it a 1,2,3 triangle. It's perfectly fine. Do the Law of Sines now.  :D


hey friends
wanna hear your smart opinion
i love playing [url=http://casino-on-net.com]casino games[/url].
is there  a connetction between some of the games to math?
is there a math subject i can learn in order to improve my gambling skills?

The subject of probability has many historical links to gambling; most of the problems that spurred its development were on gambling.

If you understand some probability, you'll know that the house always has an advantage; the expected value of the money you get back is smaller than what you bet. Wild things can happen in a few bets, but over a large number the house is almost certain to make money. If you gamble, do it for the fun and thrill, and don't risk anything you can't afford to lose.

Hi,
A bit of knowledge of game theory may help you to increase your gambling skills. There is a documentary on gambling and mathematics, "Break Vegas". It ought to be pretty informative.
Sanjith


What is the remainder when you divide 123,456 by 9?

A number is divisible by 9 if the sum of its digits is divisible by 9... So for 123456, the sum of the digits is 21... The last number divisible by 9 is 18... so the remainder is 21 - 18 = 3... :)

Don't you mean that the last number before 123456 divisible by 3 is 123453? The concept in this problem is that If a number's sum exceeds a multiple of 9 by x, then then it is divided by nine, the remainder is x.

[hide="Solution"]$1+2+3+4+5+6 = 21$

$21 \equiv 3 (mod 9)$

Thus the answer is $\boxed{3}$[/hide]


geometry
perimeter
circumcircle
inequalities
inequalities solved

I sure hope I haven't made any mistakes...

We know that N,G and I are collinear and NI=3GI (you can prove this by using baricentric coordinates). So we want to prove that 3GI*HI<=2OI^2.
Now, I,m not so sure about this, but how about trying to prove that angle GIH >= 90 ? I'm going to assume it's true, because I think I can actually remember a problem in which you had to prove this. Could you also try to prove it (or maybe prove it's wrong)? I think that you can use the fact that in any triangle a bisector is between the corresponding altitude and median.

Anyway, assuming that GIH>= 90 degrees , here goes:
We know that G,O,H are collinear and OH=3OG, so we apply Stewart's thm in triangle OIH, and then we divide by OG, getting:2*OI^2=6*OG^2+3*GI^2-HI^2=3/2*GH^2+3*GI^2-HI^2>=3/2*GI^2+
3/2*HI^2+3*GI^2-HI^2(because GIH>=90 so GH^2>=GI^2+HI^2)=9/2*GI^2+1/2*HI^2>=3*GI*HI so we get the conclusion.

Let ABC be a triangle. Prove that

NI*HI<=2OI^2(where N is Nagel's point)

 What is Nagel's point? Can anyone explain what is NI,HI and OI?

[quote="Maverick"]Let $ ABC$ be a triangle. Prove that

$ NI\ HI\leq 2\ OI^2$,

where $ N$ is Nagel point, $ H$ is orthocentre, $ O$ is circumcentre, $ I$ is incentre.[/quote]Denote $ s$ is the semi-perimeter of $ \triangle ABC$, $ R$ and $ r$ are the radii of circumcircle and incircle, respectively.

$ NI^2\equal{}s^2\minus{}16Rr\plus{}5r^2$,

$ HI^2\equal{}4R^2\plus{}4Rr\plus{}3r^2\minus{}s^2$,

$ OI^2\equal{}R(R\minus{}2r)$

go for

$ 4\ OI^2\minus{}NI^2\minus{}HI^2\equal{}4r(R\minus{}2r)\geq 0\Longrightarrow$

$ 2\ OI^2\geq \frac{NI^2\plus{}HI^2}{2} \geq NI\ HI$.

By the way, the the following reverse inequality hlods :

$ 2\ OI \leq NI \plus{} HI$.


superior algebra
superior algebra unsolved

let $ R$ be an associative and not commutative ring and let $ n > 2$ be fixed integer.
Prove that $ x^n \equal{} x$ for all $ x\in R$ implies $ xy^{n \minus{} 1} \equal{} y^{n \minus{} 1}x$ for all $ x,y\in R$.

in a ring with no nilpotent elements (except $ 0$), the idempotent elements are central (because if $ e$ is such an idempotent and $ x$ another element, we calculate $ (ex\minus{}exe)^2\equal{}0, (xe\minus{}exe)^2\equal{}0$, thus $ ex\equal{}exe\equal{}xe$). in the exercise, the powers $ y^{n\minus{}1}$ are clearly idempotent and there are no nilpotent elements.

[quote="-oo-"]in a ring with no nilpotent elements (except $ 0$), the idempotent elements are central (because if $ e$ is such an idempotent and $ x$ another element, we calculate $ (ex \minus{} exe)^2 \equal{} 0, (xe \minus{} exe)^2 \equal{} 0$, thus $ ex \equal{} exe \equal{} xe$). in the exercise, the powers $ y^{n \minus{} 1}$ are clearly idempotent and there are no nilpotent elements.[/quote]
1)Nice solution. What exactly was you motivation? Was it just this theorem that you knew?

2)What are etomoligies for the nilpotent and indempotent?

1) reading in ml or other communities ;)
2) I'll try it: nilpotent elements produce nothing, and idempotent elements just produce themselves ;). lat. nil = nothing, lat. potentia = power, ability, lat. idem = the same
by the way, it's [i]etymologies[/i].

Just for the record: This is from the Vojtech Jarnik competition 2001, category 2, problem 4.


[quote="GameBot"]Each good worker can paint my new house alone in 12 hours.  Each bad worker can paint my house alone in 36 hours.  I need my house painted in 3 hours.  If I can only find 3 good workers, how many bad workers must I also find in order to have my house painted on time?[/quote]
3 bad workers = 1 good worker.  hence  3 bad +3 good paint ur house in 3hours.
Answer:  3 bad

One good worker is the same as 3 bad workers.

One good worker and one bad worker working together is the same as 4 bad workers working together, who will take 36/4 = 9 hours to paint my dream house. Therefore, I need 9/3 good and bad workers working together to paint the house that I need so urgently.


function
calculus
derivative
complex analysis
real analysis
real analysis unsolved

Construct a function $ f(x) \in C^{\infty}( \minus{} \infty, \plus{} \infty)$ that satisfy $ f^{(2n \plus{} 1)}(0) \equal{} n$,and $ f^{(2n)}(0) \equal{} 0$,for all $ n\geq0$.

Furthermore,show that there are infinity $ f(x)$ satisfy the condition above

[color=red](thanks mornik, Because my careless,I made some mistake in my notation :huh: )[/color]

I'm not sure if I got this correctly: in your notation, would $ f^3$ mean the third derivative of $ f$ ($ f'''$) or something else (for example, $ (f(x))^3$ or $ f(f(f(x)))$)?

If it's the first one (derivative), take $ n \equal{} 0$ and we have $ f'(x) \equal{} 0$. Thus, $ f(x)$ is a constant function. However, $ f'''(x)\neq 1$, so there would not be any such functions.

If it's the second one (power), again, take $ n \equal{} 0$ and we have $ f^{2\cdot 0 \plus{} 1}(x) \equal{} (f(x))^1 \equal{} 0$. However, again, $ (f(x))^3\neq 1$, so there still do not exist any such functions.

If it's the third option (composition), again take $ n \equal{} 0$ and you get $ f(x) \equal{} 0$, so $ 0 \equal{} f(f(f(x)))\neq 1$ and there still aren't any solutions.

Or did I get something wrong? :oops:

The power series $ f(x) \equal{} \sum_{k \equal{} 0}^{\infty}\frac {kx^{2k \plus{} 1}}{(2k \plus{} 1)!}$ gives an entire function which satisfies these condition.

That's the only globally analytic solution. To get other solutions, consider $ f(x) \plus{} g(x)$ where $ g$ is a $ C^{\infty}$ function that vanishes identically at $ 0.$ One such example is $ g(x) \equal{} \begin{cases}e^{ \minus{} 1/x} & x > 0 \\
0 & x\le 0\end{cases}.$ There are infinitely many more examples like that.

That function can also be written as $ f(x) \equal{} \frac {1}{2} (x \cosh x \minus{} \sinh x) \equal{} \frac {1}{4}\left( (x \minus{} 1)e^x \plus{} (x \plus{} 1)e^{ \minus{} x}\right)$.


algebra
polynomial
Vieta
product of roots

Find $ x$, $ y$, and $ z$ if

\begin{align*}
x+y+z &= 7\\
xy+yz+zx &= -14\\
xys &= -120
\end{align*}

--

[hide]The solution says that $ x$, $ y$, and $ z$ are the roots of the polynomial $ t^3-7t^2-14t+120=0$, but how do you get this polynomial?  Thanks![/hide]

[quote="Zed_darkx"]Because it has 3 roots , it is a cubic polynomial in the form
ax^3-bx^2+cx-d=0

here -b/a= sum of the roots.
c/a = sum of the product of roots taken two at a time.
-d/a= product of the roots.
This is Vieta's formula or general called Theory Of Equations.[/quote]

Ok, I got it now.  Thanks!

THe main part is noticing that the symmetric sums are related to Vieta's.

[quote]\begin{align*}
x+y+z &= 7\\
xy+yz+zx &= -14\\
xys &= -120
\end{align*}[/quote]

I believe you mean

\begin{align*}
x+y+z &= 7\\
xy+yz+zx &= -14\\
xyz &= -120
\end{align*}

[quote="314.math"]Find $ x$, $ y$, and $ z$ if

\begin{align*} x + y + z & = 7 \\
xy + yz + zx & = - 14 \\
xys & = - 120 \end{align*}

--

[hide]The solution says that $ x$, $ y$, and $ z$ are the roots of the polynomial $ t^3 - 7t^2 - 14t + 120 = 0$, but how do you get this polynomial?  Thanks![/hide][/quote]

[hide]
$ t^3 - 7t^2 - 14t + 120 = 0 \implies (t-5)(t+4)(t-6)=0$
$ \therefore x,y,z\in\left\{ -4,5,6\right\}$
[/hide]
 :)


Do you think calculators should be used on contests?

ugh. you should have specified what contest. so im not gonna answer

Overall, I think that calculators should not be [i]needed[/i] for competitions.  Of course, there are competitions that require speed and or massive calculations, which require the use of a calculator.  But that would just mean that it's not a very good math competition since extreme arithmetic is a pretty low level of problem making.

-interesting_move


Let ABC be an isosceles triangle with |AB|=|AC|.Let take a D point on [AB] line segment that |AD|=1. If |BC|=1 and |CD|=$\sqrt2$ find "A" angle.
Please use synthetic(pure geometrical) ways.

http://www.mathlinks.ro/Forum/viewtopic.php?t=22849 .

  darij

$BD=x,\ 0<x<1,\;\ u=m(\widehat {ABC})\ (AC=1+x,\ B=C=u,\ A=\pi -2u)$.

$CD^2\cdot AB+DA\cdot DB\cdot AB=CA^2\cdot DB+CB^2\cdot DA\Longrightarrow x^3+x^2-2x-1=0\ \ (1)$.

$CD^2=BC^2+BD^2-2\cdot BC\cdot BD\cdot\cos u\Longrightarrow x^2-2x\cos u -1=0\ \ (2)$.

From the elimination of the variable $x$ between the relations $(1),\ (2)$  we obtain

$8\cos ^3 u -4\cos^2 u -4\cos u +1=0\ \ (3)$

For the solving this equation see http://www.mathlinks.ro/Forum/viewtopic.php?t=59595


Βάζω και μια όμορφη διοφαντική, που δεν με ικανοποίησε αρκετά η λύση που είδα, αλλά έχει πολύ απλή εκφώνηση :)

Να λύσετε στους θετικούς ακέραιους:

$ x^2 \plus{} y^2 \plus{} x \plus{} y \plus{} 1 \equal{} xyz$

Καλησπέρα Δημήτρη (και φίλοι αναγνώστες),

Λοιπόν η λύση που προτείνω είναι η εξής:

Υποθέτουμε χωρίς βλάβη της γενικότητας ότι $ x \geq y$.

Τότε $ x^2 \plus{} y^2 \plus{} x \plus{} y \plus{} 1 \geq y^2 \plus{} y^2 \plus{} y \plus{} y \plus{} 1$ δηλαδή $ xyz \geq 2y^2 \plus{} 2y \plus{} 1$, άρα $ 2y^2 \plus{} (2 \minus{} xz)y \plus{} 1 \leq 0$ οπότε για τη διακρίνουσα $ \Delta_1$ του τελευταίου ως προς $ y$ τριωνύμου ισχύει $ \Delta_1 \leq 0 \Leftrightarrow (xz)^2 \minus{} 2(xz) \minus{} 4 \leq 0$ απ' όπου πρέπει $ 1 \minus{} \sqrt {5} \leq xz \leq 1 \plus{} \sqrt {5}$ κι έτσι (αφού οι $ x,z$ είναι θετικοί ακέραιοι - δεν βρίσκω το λόγο γιατί, ίσως γιατί ο συγγραφέας θέλει να την κάνει ευκολότερη ή ίσως γιατί με τον τρόπο που έχει στο μυαλό του να μη βγαίνει εύκολα αν τη λύσεις στους ακεραίους) $ xz \equal{} 1$ ή $ 2$ ή $ 3$.

Από δω και πέρα η κατάσταση είναι εύκολη καθώς με μερικές περιπτώσεις παίρνουμε τα $ x,z$ και αντικαθιστώντας στην αρχική παίρνουμε το $ y$.

π.χ. αν $ xz \equal{} 3$ τότε $ (x,z) \equal{} (1,3)$ ή $ (3,1)$ και δεν παίρνουμε ακέραιες λύσεις για το $ y$.

Ελπίζω να ήταν πιο κομψή από τη λύση που είδες.
Αλέξανδρος

Αγαπητέ Αλέξανδρε νομίζω ότι έχεις λάθος. Η εξίσωση είναι αρκετά πιο απαιτητική απ' ότι φαίνεται. Η ιδέα είναι η άπειρη κάθοδος (απ' όπου προκύπτει ότι $ z\equal{}5$) και οι γενικευμένες εξισώσεις Pell, οι οποίες είναι ο λόγος για τον οποίο έλεγε μάλλον ο Δημήτρης ότι δεν του άρεσε η λύση. Δημήτρη, πάντως το πολύ πολύ για να αποφύγεις τον γενικό τύπο της Pell μπορείς να τη λύσεις κατασκευαστικά (όπως περίπου έλεγε ο Σιλουανός για το πρόβλημα που είχαμε φέτος στο tst, κάτι σαν "άπειρη άνοδος") ή αναδρομικά, αν και κατά τη γνώμη μου και οι τρεις μέθοδοι είναι παρεμφερείς.

Ηλίας

Η λύση του βιβλίου είναι ΠΟΛΥ διαφορετική αλλά διαφορετικό είναι και το αποτέλεσμα :P

:maybe: Με μπέρδεψε αρκετά η λύση σου γιατί δεν έβρισκα το λάθος... Ομως νομίζω το κατάλαβα. Στην αρχή υποθέτεις ότι $ x,y$ είναι ένα ζεύγος που ικανοποιεί την εξίσωση (σωστά?). Μετά όταν φτάνεις στο τριώνυμο $ 2y^2\plus{}(2\minus{}xz)y\plus{}1\leq 0$ είναι λάθος που λες ότι η διακρίνουσα είναι μικρότερη του 0 γιατί δεν ισχύει για όλα τα $ y$.

Ηλία αυτή ακριβώς είναι η ιδέα!

Αγαπητοί Ηλία και Δημήτρη,

έχετε απόλυτο δίκιο για την άσκηση και σας ευχαριστώ πολύ! Φυσικά και δεν είναι σωστή! Η εξάρτηση του y από το x χαλάει τα πράγματα και ο δρόμος αυτός δεν έχει επιστροφή!

Μια κι αναφέρατε την άσκηση του tst, αναφέρω το link με τη λύση στο 2ο ερώτημα (το πρώτο βγαίνει πράγματι με άπειρη κάθοδο) για να τη θυμηθεί κάθε ενδιαφερόμενος.

http://www.mathlinks.ro/viewtopic.php?p=1102294#1102294

Αλέξανδρος


geometry
3D geometry
tetrahedron
geometry unsolved

The faces of a tetrahedron are formed by congruent triangles.  If $\alpha$ is the angle between a pair of opposite edges of the tetrahedron, show that

$cos\alpha=\frac{\sin(B-C)}{\sin(B+C)}$

where B and C are angles adjacent to one of these edges in a face of the tetrahedron.

Two questions.

Could someone try to explain a visualtization of the angle $\alpha$?

Also, what exactly do they mean by angles adjacent to one of these edges in a face of the tetrahedron?

Thanks

Could someone try to explain a visualtization of the angle \alpha?

Possibly... but I could be mistaken:

If you take two rods and hold them a distance apart and at an angle[ alpha] to one another, the four end-points will form, in space, a tetrahedron.  I think they are referring to the angle mentioned here.  Viewed full center, you would see an X, with the two lines of the X being one further away than the other.

The second angle mentioned is, I think, one between one of those rods [lines of the X] and a line joining two end-points, so forming a face of the tetrahedron.

Yeah, the first part would make sense, for the solution used something like that sort repeatedly, but I don't know for sure.

It is pretty hard to visualize.

And the second one is much easier to notice.

Thanks


From the set $ \{1,2,3,\ldots,2009\}$ we choose $ 1005$ numbers, such that sum of any $ 2$ numbers isn't neither $ 2009$ nor $ 2010$. Find all ways on we can choose these $ 1005$ numbers.

[hide="Solution"]

Divide the numbers into groups $ (1,2009), (2,2008),\ldots ,(1005)$. We see that we can take a maximum of 1 number per group, as the sum is 2010 in every group, but as we take 1005 numbers, there will be exactly 1 number per every group.

We take 1005, so we can't take 1004. Now we have to take 1006 (exactly 1 number per group), so we can't take 1003, so we must take 1007 and so on.

We see that if we take $ n$, we can't take $ 2009\minus{}n$, so we must take $ n\plus{}1$. Now from this we see that we take all the number from 1005 to 2009, exactly 1005 numbers.

Answer: 1 way, all the numbers from 1005 to 2009 inclusive [/hide]

i think you are wrong
see
 the set is of 1,2,3, ....2005
if we divide it into four parts such that

1,2,3,....1004 : 1005 : 1006,1007,....2009
futher into five parts
1--  1,2,3,...502
2--  503,504,...1004
3--  1005
4--  1006,....1507
5--  1508,....2009

if i form 2010 or 2009
from any two (502) numbers .+1 total(1005)
it contradicts the question and then the answer would be zero

select 502 numbers from 1 group and 502 from 5
n remaining one 1005 from group 3

then we will have 1 number which doesn't add up with anything to form 2009 or 2010

similarly

we can take group 2 and group 4
and repeat it here also we need to take 1005 also
in this also 1005 will noit match

...
..
hence the answer will 0 according to me

It's impossible to be 0, since the set {1005,1006,...,2009} is already a solution (verifiable). My solution is the same as Bugi's solution.

sry i juzt focused on the first and the last number of a particular set

Why not numbers from 1 to 1005, every other sum will be greater then 1006+1007=2013


$  \begin{array}{c}z=100-(x+y)\\ 6x+20y+z=200 \end{array}\}\Rightarrow 5x+19y=100\Rightarrow
\{ \begin{array}{c} y=5k \\
x=20-19k \\ z=80+14k \end{array} ; k\in \mathbb{Z}.$

Thank you so much!
I still have a question, though.

How did you come to the solution that z=80+14k?
I understand everything else.
Thank you again!

$ 5x\plus{}19y\equal{}100\Rightarrow 19y\equal{}5.(20\minus{}x)\vdots 5\Rightarrow y\vdots 5\Rightarrow y\equal{}5k; k\in\mathbb{Z}\Rightarrow x\equal{}20\minus{}19.k\Rightarrow\\
\Rightarrow z\equal{}100\minus{}(x\plus{}y)\equal{}100\minus{}\left[(20\minus{}19k)\plus{}5k\right]\equal{}80\plus{}14k.$


[quote="GameBot"]In the game of sliding sleds, there are only two ways to score points. A ram is worth 4 points and a jam is worth 5 points. What is the largest score that cannot be attained in playing the game?[/quote]

Using the Chicken McNugget Theorem(use the AoPSWiki if you don't know what that is), our answer is $ 4 \times 5 \minus{} (4 \plus{} 5) \equal{} 11$.