[quote="Allnames"]-Find all the functions from $ \mathbb R^ \plus{} \to \mathbb R^ \plus{}$ that satisfy:
$ x^2(f(x) \plus{} f(y)) \equal{} (x \plus{} y)(f(yf(x)))$[/quote]
From condition we have :
$ \frac {f(yf(x))}{x^2} \equal{} \frac {f(xf(y))}{y^2}$(1)
Now take $ x\to y$ then we have :
\[ f(xf(x)) \equal{} xf(x)
\]
Therefore f has a fixed point $ a$ (i.e $ f(a) \equal{} a$)
From $ f(xf(x)) \equal{} xf(x)$ we have $ f(a^n) \equal{} a^n,\forall n\in N$
Take $ x \equal{} a^m,y \equal{} a^n$ into (1) we have $ a^{2m} \equal{} a^{2n}\Rightarrow a \equal{} 1$
It means that f has only a fixed point is 1 . So $ xf(x) \equal{} 1$ ,it gives $ f(x) \equal{} \frac {1}{x}$
Easy to check that this function satisfy condition .

Nice,TT sphn,and here is my proof
If $ f$ is const we have $ f(x)\equal{}0$,wrong
If not,Fix $ y$ we can easy find $ f(x_1)\equal{}f(x_2)$<=>$ x_1\equal{}x_2$
Let $ x\equal{}y\equal{}1$ we have $ f(f(1))\equal{}f(1)$ thus$ f(1)\equal{}1$
Let $ x\equal{}1$ we have $ yf(y)\equal{}1$ hence $ f(y)\equal{}\frac{1}{y}$done  :lol:

Suppose $f(a)=f(b)$ for distinct $a,b$. we have $\frac{a^2}{a+y}=\frac{b^2}{b+y}$, we get $ab=-y(a+b)$ impossible, $a,b,y>0$, so $a=b$, so f is injective.Take $x=y=1$ and we have $f(1)=f(f(1)),$ so$ f(1)=1$ Taking $x=1$,we have $yf(y)=1$ so the conclusion follows.

[color=#f00]Ans:[/color] $f(x)=\frac{1}{x}$.
Sol: Let $P(x,y)$ be the given assertion, \[P(x,x)\implies f(xf(x))=xf(x).\] Let $Q=xf(x)$, then \[P(Q,1)\implies Q^3+Q^2f(1)=(Q+1)Q \implies Q^2+(f(1)-1)Q-1=0\] this is a quadratic equation in $Q$ and since $Q\ne 0$, we also have the discriminant being positive, so $Q$ must be constant which means \[xf(x)=C \implies f(x)=\frac{C}{x}\] for some positive real $C$. Plugging into our original equation yields $f(x)=\frac{1}{x}$ and it is easy to see that this satisfies our equation.$\square$

@above: just notice.
$$f(xf(x))=xf(x)$$
Now plug $x\to xf(x)$ and we have
\begin{align*}
f(xf(x)f(xf(x)))&=xf(x)f(xf(x))\\
f(xf(x)xf(x))&=xf(x)xf(x)\\
f((xf(x))^2)&=(xf(x))^2
\end{align*}
So by induction letting $a=xf(x)$ we have $f(a^k)=a^k$.

Now we take $(m,k)\in\mathbb R^2$ for $m\neq k$ s.t $x=a^m$ and $y=a^k$. Using the relation we get $a^{2m}=a^{2k}$. Now say $k$ is fixed value.

Consider $g(x)=a^x$. So $g(2m)=g(2k)\forall m$ so $g$ is constant which can be only possible for $a=1$. Hence $a=1$ and $xf(x)=a=1$.

We plug in $x=y$ from here we get that $xf(x) = f(xf(x))$.

We define set $S$ to be $S= \{ xf(x) \mid \forall x \in \mathbb{R}^+ \}$, from the above we have that if $x \in S$ then we have that $f(x)=x$.

Now we plug in $x,y$ such that they are in $S$ this gives us $x^2=f(yf(x))=f(yx)=f(xf(y))=y^2$.
This gives us that $x=y$, which means that $xf(x)$ is a constant value. By direct computation we get that the only solution is $f(x)=\frac{1}{x}$

Nice problem.

Let $P(x,y)$ be the assertion of $x^2(f(x)+f(y))=(x+y)f(yf(x))$.

$P(x,y)\implies x^2(f(x)+f(y))=(x+y)f(yf(x))$
$P(y,x)\implies y^2(f(x)+f(y))=(x+y)f(xf(y))$
Thus, $f(yf(x))\cdot y^2=f(xf(y))\cdot x^2$
Setting $y=1$, we have that $f(f(x))=f(xf(1))\cdot x^2$.
$P(1,1)\implies f(1)=f(f(1))$
$P(1,x)\implies f(x)+f(1)=(x+1)f(xf(1))\quad \heartsuit$
$P(x,1)\implies x^2(f(x)+f(1))=(x+1)f(f(x))=(x+1)f(xf(1))\cdot x^2$
Thus, $f(x)+f(1)=(x+1)f(xf(1))\quad \diamondsuit$
$f(1)=f(f(f(1)))=f(f(1)^2)\cdot f(1)^2\implies f(f(1)^2)\cdot f(1)=1$.
Substituting $x=f(1)$ to $\diamondsuit$, we get that $f(f(1))+f(1)=(f(1)+1)f(f(1)^2)\implies 2f(1)^2=f(1)+1\implies  2f(1)^2-f(1)-1=0$
Factorizing, we get that $(f(1)-1)(2f(1)+1)=0$, thus $f(1)=1$.
Thus by $\heartsuit$, $f(x)+1=(x+1)f(x)\implies f(x)=\frac{1}{x}$.
This does work.

Answer. $f(x)=\frac{1}{x}\,\forall x\in \mathbb R^+$.

[quote=Allnames]-Find all the functions from $ \mathbb R^\plus{} \to \mathbb R^\plus{}$ that satisfy:
$ x^2(f(x)\plus{}f(y))\equal{}(x\plus{}y)(f(yf(x)))$[/quote]

Nice problem.
$P(x,x)\implies f(xf(x))=xf(x)$
$P(xf(x),yf(y))\implies f(xyf(x)f(y))=(xf(x))^2$
$P(yf(y),xf(x))\implies f(xyf(x)f(y))=(yf(y))^2$
Then $(xf(x))^2=(yf(y))^2\implies f(x)=\frac{c}{x},c>0$
$f(xf(x))=xf(x)\implies \frac{c}{x\cdot \frac{c}{x}}=x\cdot \frac{c}{x}\implies c=1$
Then
$f(x)=\frac{1}{x}$

Let $P(x,y)$ be the assertion $x^2f(x)+x^2f(y)=xf(yf(x))+yf(yf(x))$.
$P(1,1)\Rightarrow f(f(1))=f(1)$
$P(f(1),1)\Rightarrow f(1)=1$
Switching $x,y$ we have:
$\frac{P(y,x)}{P(x,y)}\Rightarrow x^2f(xf(y))=y^2f(yf(x))$
Taking $y=1$, it's $f(f(x))=x^2f(x)$.
$P(x,1)\Rightarrow\boxed{f(x)=\frac1x}$ which works.

Let $P(x,y)$ be the assertion of the problem

*Claim:* there are two positive real numbers $a,b$ such that $f(a)=f(b)=c$

$P(a,y): a^2(c+f(y))=(a+y)(f(yc))\implies \dfrac {c+f(y)}{f(yc)}=\dfrac {a+y}{a^2}$
$P(b,y): \dfrac {c+f(y)}{f(yc)}=\dfrac {b+y}{b^2}$
So $\dfrac {b+y}{b^2}=\dfrac {a+y}{a^2}$ hence $a=b$ so $f$ is injective
$P(1,1): f(f(1))=f(1)\implies f(1)=1$
$P(1,x): 1+f(x)=(1+x)f(x)$
Finally $\boxed {f(x)=\dfrac {1}{x}  \forall x\in \mathbb R^+}$

Denote the assertion by $P(x,y).$
If $f(u)=f(v)$ then $P(u,z)$ and $P(v,z)$ gives
$u^2(v+z)=v^2(u+z)$ so $u=v.$
$P(1,1)\implies f(1)=1.$
$P(1,x)\implies f(x)=x^{-1}$ this fits.


suppose $A\in M_{n}(\mathbb{R})$ and $tr(A^{k})=0$ for $1\le k\le n$. prove that all eigenvalues of $A$ are $0$.

This is true over any field of characteristic zero, not just $\mathbb{R}$. We do need to divide by all integers $\le n$, so it can fail in other fields.

You don't even need to hypothesize that eigenvalues exist- the condition is enough to ensure that $X$ is nilpotent. Newton's relations work even if you don't have eigenvalues; the $a_{i}$ are defined using the characteristic polynomial of $X$, and the $s_{i}$ are defined using the traces of powers of $X$. (By the way- that last equation has a missing coefficient)

Sorry,  I did not realize that missing $n$. Fixed it now.


When Dave walks to school, he averages 90 steps per minute, each of his steps 75cm long. It takes him 16 minutes to get to school. His brother, Jack, going to the same school by the same route, averages 100 steps per minute, but his steps are only 60 cm long. How long does it take Jack to get to school?

$\textbf{(A) }14 \frac{2}{9}\qquad
\textbf{(B) }15\qquad
\textbf{(C) }18\qquad
\textbf{(D) }20\qquad
\textbf{(E) }22 \frac{2}{9}$

[hide]$S=90\,\frac{\textrm{steps}}{\textrm{min}}\cdot 75\,\textrm{cm} \cdot 16\, \textrm{min}$

$S=100\,\frac{\textrm{steps}}{\textrm{min}}\cdot 60\,\textrm{cm} \cdot t\, \textrm{min}$

$100\,\frac{\textrm{steps}}{\textrm{min}}\cdot 60\,\textrm{cm} \cdot t\, \textrm{min}=90\,\frac{\textrm{steps}}{\textrm{min}}\cdot 75\,\textrm{cm} \cdot 16\, \textrm{min}$

$t=\frac{90\cdot75\cdot16}{6000}=18\,\textrm{min}$[/hide]

[quote="riddler"][hide="ans"]
$\boxed{B}$
[/hide][/quote]

Your answer is incorrect and moreover, you SHOULD NEVER just post the thread to put a letter. If you repeatedly make post does not contain any work and only letter, you'll get a warning from me and eventually be banned from this forum.

I know I'm posting a series of problems and you may not be the first one to post. But still, post your work, okay?

BTW, you're not the only one who is doing it. So don't worry if this sounds too harsh  :)

Sorry, I guess Lobo solitaire beat me. The answer I meant was 18 but I think I wrote the no. wrong :lol:

I found this problem quite weird for the level of AHMSE...

[hide="sorta hint"]you don't have to mulitply the entire thing[/hide]

[quote="Silverfalcon"]When Dave walks to school, he averages 90 steps per minute, each of his steps 75cm long. It takes him 16 minutes to get to school. His brother, Jack, going to the same school by the same route, averages 100 steps per minute, but his steps are only 60 cm long. How long does it take Jack to get to school?

A. $14 \frac{2}{9}$
B. 15
C. 18
D. 20
E. $22 \frac{2}{9}$[/quote]

[hide]Let $x$ be the time Jack needs to go to school.

We have: $75*90*16=60*100*x$

$=\Rightarrow x=18$

$\boxed{C}$[/hide]

[hide]Dave goes at 6750cm/min, while Jack goes at 6000cm/min. The ratio of Dave to Jack is $\frac{6750}{6000}=\frac{9}{8}$.

So the time ratio is the same, $\frac{9}{8}\cdot16=18$. [/hide]


LaTeX
inequalities proposed
inequalities

$ a,b,c>0$ $ \frac {2}{b(a \plus{} b)} \plus{} \frac {2}{c(c \plus{} b)} \plus{} \frac {2}{a(a \plus{} c)}\ge \frac {27}{(a \plus{} b \plus{} c)^2}$
[hide="My solution"]  we could use $ a \plus{} b \plus{} c \equal{} 1$ so $ LHS \equal{} 2(\frac {(\frac {1}{\sqrt {b}})^2}{a \plus{} b} \plus{} \frac {(\frac {1}{\sqrt {c}})^2}{c \plus{} b} \plus{} \frac {(\frac {1}{\sqrt {a}})^2}{a \plus{} c})\ge (\frac {1}{\sqrt {a}} \plus{} \frac {1}{\sqrt {b}} \plus{} \frac {1}{\sqrt {c}})^2$ by Angel and
 $ (\frac {1}{\sqrt {a}} \plus{} \frac {1}{\sqrt {b}} \plus{} \frac {1}{\sqrt {c}})^2\ge\frac {81}{(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2}\ge 27$ because by CS $ (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2\ge 3(a \plus{} b \plus{} c)^2 \equal{} 3$  [/hide]

Maybe $ a;b;c>0$
Because, try $ a\equal{}\minus{}2;b\equal{}1;c\equal{}3$

[quote="kunny"][quote="dima ukraine"]$ \frac {2}{b(a \plus{} b)} \plus{} \frac {2}{c(c \plus{} b)} \plus{} \frac {2}{a(a \plus{} c)}\ge \frac {27}{(a \plus{} b \plus{} c)^2}$
[hide="My solution"]  we could use $ a \plus{} b \plus{} c \equal{} 1$ so $ LHS \equal{} 2(\frac {(\frac {1}{\sqrt {b}})^2}{a \plus{} b} \plus{} \frac {(\frac {1}{\sqrt {c}})^2}{c \plus{} b} \plus{} \frac {(\frac {1}{\sqrt {a}})^2}{a \plus{} c})\ge (\frac {1}{\sqrt {a}} \plus{} \frac {1}{\sqrt {b}} \plus{} \frac {1}{\sqrt {c}})^2$ by Angel and
 $ (\frac {1}{\sqrt {a}} \plus{} \frac {1}{\sqrt {b}} \plus{} \frac {1}{\sqrt {c}})^2\ge\frac {81}{(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2}\ge 27$ because by CS $ (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2\ge 3(a \plus{} b \plus{} c)^2 \equal{} 3$  [/hide][/quote]

Why can you define $ \sqrt {a}$ e.t.c. :?:[/quote]
Thanks its a typo I edit it

it is easy with am gm inequality.
first take the AM GM of the LHS and then again use the fact
$ (a\plus{}b\plus{}c)/3>\equal{}(abc)^1/3$
and use AM GM for $ 2(a\plus{}b\plus{}c)\equal{}(a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a)>\equal{}.....$
i cant use latex...so sorry for exponents

[hide="Latex tips"]\frac{a+b+c}{3} gives $ \frac{a\plus{}b\plus{}c}{3}$ when within dollar signs.
(abc)^{1/3} gives $ (abc)^{1/3}$ and possibly more preferable:
(abc)^{\frac{1}{3}} gives $ (abc)^{\frac{1}{3}}$
\ge gives $ \ge$
Tip: quote ppl's posts to see the latex codes they have used. (you need not post that. Just cancel out when you are done. the quote button appears in the bottom right corner of a post.)[/hide]


Let X and Y be sets and T be a σ-algebra of subsets of Y. Let f: X→ Y be a function. Show that {f-1[F]:F element of T} is a σ-algebra of subsets of X.

I know this isn't very difficult but I am having trouble understanding sigma algebras and Borel sets.

Any help would be appreciated

A $\sigma$-algebra is a collection of sets $T\subseteq Y$ such that
1) If $A\in T$, then $Y\setminus A\in T$;
2) If $A_{1},A_{2},\dots \in T$, then $\bigcup_{n=1}^{\infty}A_{n}\in T$.

We must prove that $\{f^{-1}(A)\colon A\in T\}$ satisfies 1) and 2). The key fact is that unions, complements (and intersections) are well-behaved under preimages. Specifically,
A) $X\setminus f^{-1}(A)=f^{-1}(Y\setminus A)A\in T$;
B) $f^{-1}\left(\bigcup_{n=1}^{\infty}A_{n}\right)=\bigcup_{n=1}^{\infty}f^{-1}(A_{n})$.

Use A) and B) to prove the properties 1) and 2) for $\{f^{-1}(A)\colon A\in T\}$.


x,r,a,b are positive integer. (x+1)^3-x^3=r^2  prove this equation r=(a^2+b^2)

Hmm...this is pretty interesting. I think you can make that claim stronger by saying that if $ r^2\equal{}(x\plus{}1)^3\minus{}x^3$, then $ r\equal{}a^2\plus{}(a\plus{}1)^2$ for some positive integer $ a$.

I noticed that the first couple of solutions were $ (x, r)$:
$ (0, 1) \rightarrow 1\equal{}0^2\plus{}1^2$
$ (7, 13) \rightarrow 13\equal{}2^2\plus{}3^2$
$ (104, 181) \rightarrow 181\equal{}9^2\plus{}10^2$
$ (1455, 2521) \rightarrow 2521\equal{}35^2\plus{}36^2$
$ (20272, 35113) \rightarrow 35113\equal{}132^2\plus{}133^2$.

Also I noticed that the first few $ r^{2}$ are the first couple terms of  [url=http://www.research.att.com/~njas/sequences/A006051]A006051[/url]

[quote="1/(ln x)"]I think you can make that claim stronger by saying that if $ r^2 \equal{} (x \plus{} 1)^3 \minus{} x^3$, then $ r \equal{} a^2 \plus{} (a \plus{} 1)^2$ for some positive integer $ a$.[/quote]
You are right. This problem had already been posted here, and I have written a solution in that topic. You may check it at: http://www.mathlinks.ro/viewtopic.php?t=236014.


function
number theory
number theory unsolved

let a, b be integers such that gcd(a,m) = gcd(b,m) = 1, let n, m be positive integers, let g be a primitive root modulo m.

Find the necessary and sufficient condition on the index of a (mod m) (denoted I(a) ) and the index of b (mod m) (denoted I(b) ) for the congruence equation ax^n congruent to b (mod m) to have integer solutions.

I have:

ax^n congruent to b mod m iff

I(ax^n) = I(b), which is true iff

I(ax^n) congruent to I(b) mod (phi(m)) (Phi(m) refers to euler's phi function)

i.e. I(a) + nI(x) congruent to I(b) mod (phi(m))

i.e. nI(x) congruent to I(b) - I(a) mod (phi(m))

which has solutions iff gcd(n, phi(m)) divides ( I(b) - I(a) )

is this correct, or is there some further conclusion I can draw? Thanks!

looks correct to me.  btw, I believe a similar problem is treated in NSato's number theory article.  look in the AOPS resource section for it; I believe the part I'm thinking of is in the chapter about primitive roots.


can you send me iran national  maths olymp questions  2004 and 2005
mail murat2030@gmail.com

Iran 2005 TST: [url]http://www.mathlinks.ro/Forum/topic-34210.html[/url]
Iran 2005 pre-TST: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=30004[/url]


I wanna start memorizing digits of pi, so I was wondering to how many digits of it do you know?

Topics such as this are better off in the Games & Fun Factory.

Moved.

lol i don't know any....... let's see....

3.14159265.......that's it =] the crazy guy on mathmaticious knows a lot  :P

3.14  :D

The message is too small. Please make the message longer before submitting.

Poincare, how many digits of pi do YOU know?

[quote="pythag011"]ln(-1) / sqrt(-1) = pi.

I know infinitely many.

BTW, ln(-1) / sqrt(-1) actually does = pi if you extend ln to negative integers.

Proof:

e^ipi = -1
i*pi = ln(-1)
pi = ln(-1) / sqrt{-1}

BTW, my calculator actually tells me that it equals pi...[/quote]

So the definition of ln(-1) = pi*i

Interesting. Can we extend this to all other negative numbers? Yes. Pythag, please check that my math is correct. (It's likely to be wrong.)

ln(-1)+ln(x) = pi*i+ln(x)
ln(-x)=pi*i+ln(x)

Also:

e^ipi = -1
e^(i*pi/2) = -1^(1/2) = (+/-) i

i*pi/2= ln(i) = ln(-i) = pi*i+ln(i)= pi*i+pi*i/2
Or, pi*i=0

Therefore, we cannot make a logical extension for logarithms of imaginary/complex numbers.

EPIC FAIL

[url=http://en.wikipedia.org/wiki/Complex_logarithm]Complex logarithms[/url]...
It's like how if you take a square root, you get only the positive value. So... just defined to be a function giving the prinicpal value?

um ajain, i am pretty sure he was joking. i really don't bother trying to memorize pi, but here goes 3.14159265358979323846264

"Sometimes 'pi = 3.14' is (a) infinitely faster than the 'correct' answer and (b) the difference between the 'correct' and the 'wrong' answer is meaningless. And this is why I get upset when somebody dismisses performance issues based on 'correctness'. The thing is, some specious value of 'correctness' is often irrelevant because it doesn't matter. While performance almost _always_ matters. And I absolutely _detest_ the fact that people so often dismiss performance concerns so readily."

-Linus Torvalds

I disagree with Linus, actually...

Meh, I suck.

3.141592653589732346264338

I haven't tried to memorize $ \pi$ in a long time...
When I was trying, I easily made 5 digits a day with only a few minutes of practice

No, I memorized all that. Do you really think I would be able to memorize a million digits of pi? I copy-pasted that. lol. Yes I fail.


geometry
rotation
rhombus
Asymptote
geometric transformation

A plane contains points $ A$ and $ B$ with $ AB \equal{} 1$. Let $ S$ be the union of all disks of radius $ 1$ in the plane that cover $ \overline{AB}$. What is the area of $ S$?

$ \textbf{(A)}\  2\pi \plus{} \sqrt3 \qquad \textbf{(B)}\  \frac {8\pi}{3} \qquad \textbf{(C)}\  3\pi \minus{} \frac {\sqrt3}{2} \qquad \textbf{(D)}\  \frac {10\pi}{3} \minus{} \sqrt3 \qquad \textbf{(E)}\ 4\pi \minus{} 2\sqrt3$

Does anyone have a different solution than those offered in the previous discussion?  I cannot seem to find a good solution to this problem on the internet...

you don't like my solution :P , i agree that it is not that good, but it is understandable if you draw a picture and work through the solution

Sorry about that.  I meant that I can't follow your solution very well.  I just need to write it out and look at a diagram with this one I guess.  Let me see if I can try it...

Thank you.

Umm... Altheman, your solution led to choice (D), but the answer is (C). You might want to check your work.

ya.. hather got the solution i wanan post. straight forward i think.. with that diagram  :D

[hide="Solution"]
The left most circle and right most circle that cover $AB$ are the ones with $AB$ for a radius.  These two circles intersect at two points, $C$ and $D$.  The diameter through $B$ in the right one intersects the circle at $E$ and the diameter through $A$ in the right one intersects the circle at $F$.

The highest point in $S$ is the circle that has $AB$ as a chord.  The radii drawn to the endpoints of this chord form an equilateral triangle with side 1, so we know that this ceneter is $C$.  The lowest point, naturally, will have center $D$.

Now, we can rotate point $F$ about $A$ onto a point on circle C, diameterically opposite $A$.  If we let this point be $F'$, then we can draw a sixth of a circle centered at $C$ through $F$ and $F'$.  This sector, of radius 2, will contain points in $S$, since we can rotate Circle B as we did Point F onto Circle C, creating circles that are contained in $S$ while in motion.  We can also rotate Circle C downwards onto Circle D, so we have a $120^\circ$ sector with radius 2 on the right.

We can do the same thing on the left, creating two sectors with area $\frac{4\pi}{3}$.  We also have two sixths of circles C and D that we haven't counted.

These each have area $\frac{\pi}{6}$.

The two $120^\circ$ sectors overlap, so we must subtract that overlap, which comes in the form of Rhombus $ACBD$.  This has twice the area of equilateral $ABC$.  Therefore, we must subtract $\frac{\sqrt{3}}{2}$.

The area of $S$ is $2\left(\frac{4\pi}{3}\right)+2\left(\frac{\pi}{6}\right)-\frac{\sqrt{3}}{2}=\boxed{3\pi-\frac{\sqrt{3}}{2}}$; $\boxed{\mathbb C}$.
[/hide]

[quote="negativebplusorminus"]
Now, we can rotate point $F$ about $A$ onto a point on circle C, diameterically opposite $A$.  If we let this point be $F'$, then we can draw a sixth of a circle centered at $C$ through $F$ and $F'$.[/quote]
I don't quite understand how you do this.  I'm assuming that this is the diagram that you have constructed:
[asy]/* File unicodetex not found. */

 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(7.52cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -2.64, xmax = 4.87, ymin = -2.24, ymax = 6.47;  /* image dimensions */

 /* draw figures */
draw((-0.04,2.1)--(2.16,2.12));
draw(circle((-0.04,2.1), 2.2));
draw(circle((2.16,2.12), 2.2));
draw(circle((1.04,4.02), 2.2));
draw(circle((1.08,0.2), 2.2));
draw((xmin, 0.01*xmin + 2.1)--(xmax, 0.01*xmax + 2.1)); /* line */
 /* dots and labels */
dot((-0.04,2.1),dotstyle);
label("$A$", (0.09,2.27), NE * labelscalefactor);
dot((2.16,2.12),dotstyle);
label("$B$", (1.77,2.36), NE * labelscalefactor);
dot((1.04,4.02),dotstyle);
label("$C$", (0.91,4.26), NE * labelscalefactor);
dot((1.08,0.2),dotstyle);
label("$D$", (0.98,-0.31), NE * labelscalefactor);
dot((4.36,2.14),dotstyle);
label("$E$", (4.5,2.3), NE * labelscalefactor);
dot((-2.24,2.08),dotstyle);
label("$F$", (-2.58,2.33), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */[/asy]

If you rotate $F$ about $A$, then $F'$ must be on circle $A$ as well, and from your solution, it must also be on circle $C$, so the only possible points for $F'$ to be are the two intersections of circle $A$ and circle $C$.  However, I don't see how that is diametrically opposite $A$ or a circle can be drawn with center $C$ and radius $2$ that passes through $F$ and $F'$, in fact, $CF=\neq 2$.  I think I am constructing your diagram incorrectly.

[quote="AlphaBetaTheta"]I don't understand the problem and solution fully. AB doesn't need to covered fully correct? How do you know which region to find the area of?[/quote]
The problem statement says "Let $S$ be the union of all disks of radius $1$ in the plane that cover $\overline{AB}$."  Although it does not specifically say "...that [i]entirely[/i] $\overline{AB}$", I think it should be assumed that that is the case.  Or, you can try the other interpretation and see that the answer you get is not in the list of choices. :P

[asy]
/* File unicodetex not found. */
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(7.52cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -2.64, xmax = 4.87, ymin = -2.24, ymax = 6.47; /* image dimensions */
/* draw figures */
draw((-0.04,2.1)--(2.16,2.12));
draw(circle((-0.04,2.1), 2.2));
draw(circle((2.16,2.12), 2.2));
draw(circle((1.04,4.02), 2.2));
draw(circle((1.08,0.2), 2.2));
draw((xmin, 0.01*xmin + 2.1)--(xmax, 0.01*xmax + 2.1)); /* line */
/* dots and labels */
dot((-0.04,2.1),dotstyle);
label("$A$", (0.09,2.27), NE * labelscalefactor);
dot((2.16,2.12),dotstyle);
label("$B$", (1.77,2.36), NE * labelscalefactor);
dot((1.04,4.02),dotstyle);
label("$C$", (0.91,4.26), NE * labelscalefactor);
dot((1.08,0.2),dotstyle);
label("$D$", (0.98,-0.31), NE * labelscalefactor);
dot((4.36,2.14),dotstyle);
label("$E$", (4.5,2.3), NE * labelscalefactor);
dot((-2.24,2.08),dotstyle);
label("$F$", (-2.58,2.33), NE * labelscalefactor);
draw((-0.04,2.1)--(1.04,4.02)--(2.16,2.12)--(1.08,0.2)--cycle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]
Can someone tell me why my solution is incorrect?  I can't figure out why.

[hide="Flawed Solution"]If we add the area of the sectors of the circle bounded by $CBDE$, $CADF$, $CAB$ (larger sector), and $ADB$ (larger sector), plus the area of the rhombus, we get $\frac{2}{3}\pi+\frac23\pi+\frac56\pi+\frac56\pi+\frac{\sqrt3}{2}=3\pi+\frac{\sqrt3}{2}$.  We counted the 4 areas that are identical to the one bounded by C, B, and the intersection of circles C and B twice, so we have to subtract out each of their areas once, or just the areas of the intersections of any two circles twice:
[asy]/* File unicodetex not found. */

 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(5.62cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -2.08, xmax = 3.54, ymin = -1.04, ymax = 3.24;  /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39,0);

draw((-0.12,1.28)--(0.74,2.7)--(1.54,1.24)--(0.68,-0.18)--cycle, zzttqq);
draw(arc((-0.12,1.28),0.6,-61.38,58.62)--(-0.12,1.28)--cycle, qqwuqq);
draw(arc((1.54,1.24),0.6,118.62,238.62)--(1.54,1.24)--cycle, qqwuqq);
draw(arc((0.68,-0.18),0.6,58.62,118.62)--(0.68,-0.18)--cycle, qqwuqq);
draw(arc((0.74,2.7),0.6,-121.38,-61.38)--(0.74,2.7)--cycle, qqwuqq);
 /* draw figures */
draw(circle((-0.12,1.28), 1.66));
draw(circle((1.54,1.24), 1.66));
draw((-0.12,1.28)--(0.74,2.7), zzttqq);
draw((0.74,2.7)--(1.54,1.24), zzttqq);
draw((1.54,1.24)--(0.68,-0.18), zzttqq);
draw((0.68,-0.18)--(-0.12,1.28), zzttqq);
 /* dots and labels */
dot((-0.12,1.28),dotstyle);
label("$W$", (-0.48,1.12), NE * labelscalefactor);
dot((1.54,1.24),dotstyle);
label("$X$", (1.7,1.16), NE * labelscalefactor);
dot((0.74,2.7),dotstyle);
label("$Y$", (0.72,2.84), NE * labelscalefactor);
dot((0.68,-0.18),dotstyle);
label("$Z$", (0.62,-0.56), NE * labelscalefactor);
label("$120^\circ$", (-0.06,1.14), NE * labelscalefactor,qqwuqq);
label("$120^\circ$", (0.98,1.14), NE * labelscalefactor,qqwuqq);
label("$60^\circ$", (0.54,0.1), NE * labelscalefactor,qqwuqq);
label("$60^\circ$", (0.56,2.2), NE * labelscalefactor,qqwuqq);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */[/asy]
The area of the curved figure WYXZ is the area of the sector of circle X bounded by YWZ plus the area of the sector of circle W bounded by YXZ minus the area of the rhombus to correct for overcounting, which is $\frac13\pi+\frac13-\frac{\sqrt{3}}{2}=\frac23\pi-\frac{\sqrt{3}}{2}$.  Since we have to subtract two of those, we have $3\pi+\frac{\sqrt3}{2}-2\left(\frac23\pi-\frac{\sqrt{3}}{2}\right)=\boxed{\frac{5\pi}{3}+\frac{3\sqrt3}{2}}$ which is none of the choices, so it's clearly incorrect.  But I don't see a flaw in my PIE argument, and cannot find a flaw in my calculations, so I think there's something wrong with my method.  Help please?[/hide]

[asy]
size(300);
pair A, B, C, D,E,F,G,H;
A = (0,0);
B = (1,0);
C = (1/2,sqrt(3)/2);
D = (1/2,0-sqrt(3)/2);
F = (2,0);
E = (-1,0);
G = (1,sqrt(3));
H = (1,-sqrt(3));
draw(A--B);
draw(A--G);
draw(A--H);
draw(Circle(A,1));
draw(Circle(B,1));
draw(Circle(A,2));
draw(Circle(C,1));
draw(Circle(D,1));
draw(Circle(B,2));
dot("$A$", A, W);
dot("$B$", B, SE);
dot("$C$",C,N);
dot("$D$",D,S);
dot("$E$",E,E);
dot("$F$",F,SE);
dot("$F'$",G,N);
dot("$F''$",H,N);[/asy]
There's the diagram I used.  If you look at my solution, you'll see that I said that $F$ was the intersection point on the right, not the left.

The flaw in your solution is in your diagram.  You found the area of the wrong thing.  $S$ also has to include the area swept out by the rotation of $F$ described in my solution.

Basically, it's the intersection of the two biggest circles in my diagram, minus the little awkward-looking sections on the top and bottom (so that the top and bottom boundaries are the edges of $\odot C$ and $\odot D$ in my diagram).

Sorry to revive, but in the wiki solution it says to subtract the area of the two equilateral triangles. Why do we need to do this? Aren't they included in the disks?

This is probably a dumb question :P
EDIT: Oh, you're just correcting for overcounting.

[quote="genesis2"]Wait, can someone clarify what the "disks" mean in this problem?[/quote]
Closed disk is the region inside a circle INCLUDING the boundary of the circle.

"circle" by itelf just refers to the boundary ("circumference").

Open disk refers to the region within the circle EXCLUDING the boundary.

Hope that clears it up.


Here's a problem that I don't know how to solve. Help would be appreciated:

1. If there are 6 boxes and there are 12 identical balls, how many ways can the balls be placed such that there is at least 1 ball in each box?
2. If there is no minimum for the above question, then how many ways can the balls be placed into the boxes?

Please show your answer and your method. I know the answer to the first one, but I need to know whether you get it right to know if your method is correct.

I also need some practice probability problems. Thank you in advance.

[hide]Balls and urns pwns this. We need the sum of the balls in each of the 6 to sum to 12. Since each must get atleast one, let's take 6 away from the 12 by giving one to each box. Now you just apply balls and urns.

For the second, it's just the same except the 6 just add to 12.[/hide]

1

[hide]put 1 ball in each box, you have 6 balls left

put 5 dividers among the 6 balls

it becomes from 11 chosse 5, or 11C5[/hide]

2

[hide]put 5 divider among 12 balls

17C5[/hide]


AMC USA(J)MO USAMO inequalities calculus integration function

Show that for any non-negative reals x, y, $[5x]+[5y] \ge [3x+y]+[x+3y]$. Hence or otherwise show that $\frac{(5a)! (5b)!}{( a! b! (3a+b)! (a+3b)! )}$ is integral for any positive integers a, b.

$[5x]+[5y] \geq [3x+y]+[x+3y]+[x]+[y]$

maybe? Just at a glance, i think this solves the problem, but at the moment im a bit to busy to see if its true. Ill go for it later. I dont see how the previous condition can solve the problem, but i might be wrong

The first part of the problem is common sense.

[3x+y]+[x+3y] can be at most 3x+y+x+3y+1+1 = 4x+4y+2.

For any non-negative real, [x][u]>[/u]1, and the same for y.

Hence, 4x+4y+2 [u]<[/u]5x+5y[u]<[/u][5x] + [5y]

[quote="neelnal"][quote="Farenhajt"][quote="neelnal"]For any non-negative real, [x][u]>[/u]1[/quote]

For [i]non-negative[/i] reals $[x]\geqslant 0$[/quote]

I thought it is greatest integer function, so even .00000001 would become 1?[/quote]

$[x]$ is the greatest integer [i]less than or equal to[/i] x.

Hey Jeff!

So my suggestion for proving the first part is:

let $x = x_{0}+x_{1}$, where $x_{1}< 1$, therefore $[x] = x_{0}$, similar notation for $y$ as well.

Then the problem is $x_{0}+y_{0}+[5 x_{1}]+[5 y_{1}] \ge [3 x_{1}+y_{1}]+[3 y_{1}+x_{1}]$ Therefore the most constraining evaluation of this occurs when $x_{0}= y_{0}= 0$.

Now here is the hard part, I made a graph in $x$ and $y$, such that the range of the graph was $0 \le x < 1$ and $0 \le y < 1$. By drawing the region bounds, which were of the form $y = \frac{-1}{3}x+\frac{k}{3}$ and $y =-3x+k$, where $k \in{1,2,3}$. Then in each bounded region, I wrote the value of the RHS sum and checked to make sure that tracing the interior of a given boundary near to the edges didn't invalidate the claim.

I saw you at the the last TCHS, although you coded the first problem faster than I did.  GG.  On a more relevant note, how do you use the inequality to prove the combinatorial statement? Hopefully there is a better way to solve the inequality than the way I did.


cho minh hoi .trong Mathlink co phan nao trao doi tai lieu ko

[url=http://www.mathlinks.ro/Forum/index.php?f=121]ONE[/url]
[url=http://www.mathlinks.ro/Forum/resources.php]TWO[/url]
[url=http://www.mathlinks.ro/Forum/index.php?f=177]THREE[/url]
[url=http://www.mathlinks.ro/Forum/index.php?f=81]FOUR[/url]
I think it is not enough for u,but I gave u all I know. ;)  ;)


Stronger Inequality; $ a,b,c > 0$ prove that
$ \sum_{cyc}\frac{2a}{2a\plus{}b\plus{}c} \leq \frac{3}{2}$

[quote="soruz"]If $ a,b,c > 0$ then prove:
$ \frac {1}{2} \le \frac {2a}{2a \plus{} b \plus{} c} \plus{} \frac {2b}{a \plus{} 2b \plus{} c} \plus{} \frac {2c}{a \plus{} b \plus{} 2c} \le \frac {3}{2}$.[/quote]

$ p \equal{} a \plus{} b \plus{} c$  ---->
$ \frac {2a}{2a \plus{} b \plus{} c} \plus{} \frac {2b}{a \plus{} 2b \plus{} c} \plus{} \frac {2c}{a \plus{} b \plus{} 2c} \equal{} \frac {2a}{p \plus{} a} \plus{} \frac {2b}{p \plus{} b} \plus{} \frac {2c}{p \plus{} c} \equal{} 6 \minus{} 2(\frac {p}{p \plus{} a} \plus{} \frac {p}{p \plus{} b} \plus{} \frac {p}{p \plus{} c})\leq 6 \minus{} 2(3\sqrt [3]{\frac {p^{3}}{(p \plus{} a)(p \plus{} b)(p \plus{} c)}}) \equal{} 6 \minus{} 2(\frac {3p}{\sqrt [3]{(p \plus{} a)(p \plus{} b)(p \plus{} c)}})\leq6 \minus{} 2(\frac {3p}{\frac {p \plus{} a \plus{} p \plus{} b \plus{} p \plus{} c}{3}}) \equal{} 6 \minus{} 2(\frac {9p}{4p})\equal{}\frac{3}{2}$
$ a \equal{} b \equal{} c$

[quote="apollo"]Stronger Inequality; $ a,b,c > 0$ prove that
$ \sum_{cyc}\frac {2a}{2a \plus{} b \plus{} c} \leq \frac {3}{2}$[/quote]

$ \Leftrightarrow$ $ \sum \left( 1\minus{} \frac{b\plus{}c}{2a\plus{}b\plus{}c} \right) \leq \frac{3}{2}$ $ \Leftrightarrow$ $ \sum \frac{b\plus{}c}{2a\plus{}b\plus{}c} \geq \frac{3}{2}$
Bur from Cauchy we have:

$ \sum \frac{(b\plus{}c)^2}{2a(b\plus{}c)\plus{}(b\plus{}c)^2}\geq \frac{4(a\plus{}b\plus{}c)^2}{6(ab\plus{}bc\plus{}ca)\plus{}2(a^2\plus{}b^2\plus{}c^2)}\geq \frac{3}{2}$ $ \Leftrightarrow$ $ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ca$

Another solution: We can assume that $ a\plus{}b\plus{}c\equal{}1$,by jensen $ f(x)\equal{}\frac{2x}{x\plus{}1}$ is concave
therefore $ f(\frac{1}{3}) \geq \frac{f(a)\plus{}f(b)\plus{}f(c)}{3} \rightarrow L.H.S. \leq \frac{3}{2}$

$ \sum\limits_{cycl}\frac{2a}{2a\plus{}b\plus{}c}> 1$
and we can't replace $ 1$ by any number greater than $ 1$.

Let $a,b,c\geq 0$ and $ab+bc+ca\neq 0$. Prove that
$$ \frac{a}{2a+b+c}+\frac{b}{2b+c+a}+\frac{c}{2c+a+b}  \geq\frac{1}{2} $$
[quote=zaya_yc]$ \frac{2a}{2a\plus{}b\plus{}c}\plus{}\frac{2b}{a\plus{}2b\plus{}c}\plus{}\frac{2c}{a\plus{}b\plus{}2c}\geq\frac{2a}{2a\plus{}2b\plus{}2c}\plus{}\frac{2b}{2a\plus{}2b\plus{}2c}\plus{}\frac{2c}{2a\plus{}2b\plus{}2c}\equal{}1$[/quote]
[url=https://artofproblemsolving.com/community/c6h1750587p15736646]h[/url]


In the figure, angle AFB is 82 degrees and the central angle AOB (arc AB) is 92 degrees. Chord AD is congruent to chord BC. What is the measure of the central angle AOD.

Insufficient information to find out the answer. Just change the position of D along the circumference and change the position of C accordingly.

congruence of angles ADF and BCF are insufficient conditions i think; are angles C and D right angled? Triangle OCD is obviously an isosceles but not equilateral . . . any hint to solve this problem?

Totally no fixed position for F, no condition stated for AD. I think this question does not provide sufficient information to solve.  :maybe:

How could you possibly get an answer?
Points D and C can be moved around freely...

[quote="geotrig"]Some how I got, as my answer, 346/3 or 115 1/3 degrees.

Is there an actual answer or did you make this up???[/quote]

It's a problem given to me by my colleague. I try to solve it for 2 days and got no result, so I thought of posting it here. Yeah, it seems the information is insufficient, but I am not sure though.

I believe I did it right this time!

I figured out that I added the wrong variables for angles :oops_sign:

This time I got 124.5...I'll check my answer again though.

stephecheng and phulam146 said the conditions are insufficient. GEOTRIG, how could you extract an answer?  :?:


function
topology
real analysis
real analysis unsolved

Let $X$ be a metric space,$A$ a closed set of $X$ and $f: A\to R$ a continous function.Show that we can find a continous function $g: X\to R$ such that f is its restriction to $A$ and both functions have the same upper and lower bounds($sup_Xg=sup_Xf$, $inf_Xg=inf_Xf$).
And since it's not that easy to came with a function like that,here's the function:
[hide]
$g(x)=f(x)$ if $x \in A$
$g(x)=\frac{inf_{y\in A}(d(y,x)f(y))}{d(x,A)}$,when $x\in X-A$.Now show that this function verifies those properties.
[/hide]

Tietze's stronger theorem states that every continuous function defined on a closed subset of a normal space with values in $[a,b]$ (or $\mathbb R$) can be extended to a continuous function defined on the whole space with values in the same set. We can apply that here to the function $f: A\to [\inf_A f,\sup_A f]$ and we're done (I know there are simpler solutions for metric spaces, but I like this approach more :)).


a,b,c positive reals

prove

sigma cyclic a+3[b+c]/b+c>= 9/2

This seems to be: for $ a, b, c >0$, prove that

$ a\plus{}\frac{3(b\plus{}c)}{b} \plus{} c \plus{} \frac{3(c\plus{}a)}{c} \plus{} a \plus{} c\plus{} \frac{3(a\plus{}b)}{a} \plus{} b \ge \frac{9}{2}$,

that is,

$ 2(a\plus{}b\plus{}c) \plus{} 9 \plus{} 3(\frac{c}{b} \plus{} \frac{a}{c} \plus{} \frac{b}{a}) \ge \frac{9}{2}$.

But $ a, b, c >0$!


Does there exist a function $f: \mathbb{N} \rightarrow \mathbb{N}$ such that \[f^{\left(f(n)\right)}(n) = n + 1,\;\forall n \in \mathbb{N},\] where $f^{(k)}(n) = f\left(f^{(k - 1)}(n)\right),\ \forall k \in \mathbb{N}$?

[quote="Arne"]Does there exist a function $ f: \mathbb{N} \rightarrow \mathbb{N}$ such that
\[ f^{\left(f(n)\right)}(n) \equal{} n \plus{} 1,\;\forall n \in \mathbb{N},
\]
where $ f^{(k)}(n) \equal{} f\left(f^{(k \minus{} 1)}(n)\right),\ \forall k \in \mathbb{N}$?[/quote]

I suppose that, as generally in this forum, $ 0\notin\mathbb{N}$

As a consequence, it does not exist any $ n_0\in\mathbb{N}$ such that $ f(n_0)\equal{}1$. Else we would have $ n_0\plus{}1\equal{}f^{f(n_0)}(n_0)\equal{}f(n_0)\equal{}1$ and so $ n_0\equal{}0$

$ 1)$ We know that $ f(1)\neq 1$. Assume then $ f(1)\equal{}2$. Then $ f^{f(1)}(1)\equal{}2$ and so $ f^2(1)\equal{}2$ and so $ f(f(1))\equal{}2$ and $ f(2)\equal{}2$.
But this is impossible since we would have then $ f^k(2)\equal{}2$ $ \forall k$ and $ f^{f(2)}(2)\equal{}3$ would be impossible.
So $ f(1)>2$.

$ 2)$ $ f(x)$ is injective : $ f(a)\equal{}f(b)>1$ implies $ a\plus{}1\equal{}f^{f(a)}(a)\equal{}f^{f(b)}(a)$ $ \equal{}f^{f(b)\minus{}1}(f(a))\equal{}f^{f(b)\minus{}1}(f(b))$ $ \equal{}f^{f(b)}(b)\equal{}b\plus{}1$ and so $ a\equal{}b$

$ 3)$ $ f^{f(1)}(1)\equal{}2$, then $ f^{f(1)\plus{}f(2)}(1)\equal{}f^{f(2)}(2)\equal{}3$ and so on : So $ \boxed{f^{\sum_{k\equal{}1}^{n}f(k)}(1)\equal{}n\plus{}1\: \: \forall n>0}$

$ 4)$ But, since $ f(1)>2$ and so $ f^{\sum_{k\equal{}1}^{f(1)\minus{}1}f(k)}(1)\equal{}f(1)$ and so $ f(f^{(\sum_{k\equal{}1}^{f(1)\minus{}1}f(k))\minus{}1}(1))\equal{}f(1)$ and, since $ f(x)$ is injective :

$ f(f^{(\sum_{k\equal{}1}^{f(1)\minus{}1}f(k))\minus{}2}(1))\equal{}1$ which is impossible (consider that $ f(1)>2$ implies that $ (\sum_{k\equal{}1}^{f(1)\minus{}1}f(k))\minus{}2>0$

So, such function $ f(x)$ does not exist.


[quote="muks"]Of all the triangles on the same base and of equal areas, prove that the isosceles has the least perimeter.

Thanks in advance. :lol:[/quote]

[geogebra]2bdb3cb682fcafaf50c8c79e38afc84141be0548[/geogebra]

$ PR \plus{} RQ \equal{} P'RQ < P'R' \plus{} R'Q \equal{} PR' \plus{} R'Q$

How did you arrive at the conclusion that $ P'R'\equal{}PR'$ ?Can you please explain?

Thank you.

[quote="muks"]How did you arrive at the conclusion that $ P'R' \equal{} PR'$ ?Can you please explain?

Thank you.[/quote]

In the graph above, $ P',Q'$ are formed by reflecting $ P,Q$ along the straight line (which is parallel to $ PQ$) in the middle.


I'm sure nearly everyone is familar with Grahm's Gas Law
$\frac{v_a}{v_b}=\sqrt{\frac{M_b}{M_a}}$ where M and v are the molar masses and velocities respectively of gases a and b.
If $v_a\sqrt{M_a}=v_b\sqrt{M_b}$ would it not be true that $v_a\sqrt{M_a}=v_x\sqrt{M_x}$?  If so, would that not imply that there exists some constant C such that $C=v_x\sqrt{M_x}$ where x is a general gas?
I do not have accurate enough methods of testing my hypothesis and finding this constant if one exists, nor have I ever heard of one mentioned.  Either I hypothesized correctly or someone can help me to catch my mistake.  Thanks in advance.

You are right about that constant.  It can be shown using kinetics and stuff like that that:

avg. speed of a gas molecule=[tex]\sqrt{\frac{3RT}{M}}[/tex], where R is the gas constant, T is Kelvin temperature, and M is the molar mass of the gas.  Note you could easily derive Grahm's law from here.

To answer your question: multiply each side by [tex]\sqrt M[/tex].  You then get that the constant you were wondering about is

[tex]\sqrt{3RT}[/tex]


Let $ (O)$ be a semicircle with diameter $ AB \equal{} 2$ and $ C,D\in (O)$ such that the distance between the two incentres of $ \triangle ABC$ and $ \triangle ABD$ is $ \sqrt {2}$. Calculate $ CD$.

Draw the circle of diameter AB and note M the middle of the semicircle which does not contain C and D.
It is well known that the incenters of the triangles ACB and ADB lay on the circle C(M,MA) and, as its radius is sqrt (2), therefore the angle between the angle bisectors of C and D is 60 deg, which shows that the distance CD is sqt(3).

Best regards,
sunken rock


[quote="#H34N1"]A square sheet of paper has area $6 \text{ cm}^{2}$. The front is white and the back is black. When the sheet is folded so point $A$ rests on the diagonal as shown, the visible black area is equal to the visible white area. How many centimeters is $A$ from its original starting position? Express your answer in simplest radical form.

[img]7231[/img][/quote]
[hide]Black area is forming an isosceles right triangle. Let $a$ be the length of its leg.  So the black area is $\frac{a^{2}}{2}$. The visible white area is $6-a^{2}$. Since $\frac{a^{2}}{2}= 6-a^{2}$, $a = 2$. Hence $A$ is $\boxed{2\sqrt{2}}$ cms from its original position.[/hide]

[hide]
The area of the area of the black region is the height of the triangle (the one that lies on the diagonal of the square) squared. Therefore

$x^{2}=6-2x^{2}$
$3x^{2}=6$
$x^{2}=2$
$x=\sqrt{2}$
$2x$ (the desired distance) = $2\sqrt{2}$
[/hide]


Nα λυθει στο $\mathbb{R}$ το σύστημα

$x=\sqrt{\mid \frac{1-y}{1+y} \mid}$

$y=\sqrt{\frac{1+z}{2}}$

$z=\sqrt{\frac{1}{1+x^2}}$

Cheers :)

[quote="silouan"]Νομίζω ότι είναι απλή χρήση του θεωρήματος εδώ http://www.mathlinks.ro/Forum/viewtopic.php?p=447041#p447041

Κάνω λάθος ?[/quote]

ποια είναι η συνάρτηση??

Ο cmad βρήκε μια ωραία λύση........
σιλουανε τι ακριβώς εννοείς??

[quote="jj-the-king"]θέτεις χ = εφu προκύπτει με πράξεις ότι συν2u=συνu/2 KAI προκύπτει ένα πολυώνυμο τετάρτου βαθμού με ρίζες
1,-1/2, (-1+ sqrt(5))/2 , (-1- sqrt(5))/2 από τις οποίες γίνεται δεκτή μόνο η 1 και άρα y=1,x=0,z=1[/quote]

ωραιος.  :first:
αλλα απλα θα μπορούσες να λύσεις την συν2u=συνu/2 στο [0,$\frac{\pi}{2}$].....

Το σύστημα λύνεται και παίρνοντας περιπτώσεις για το y 0<y<1, y=1 και y>1 ως εξής:

 i)y=1. Τότε χ=0, y=1 και z=1
 ii)y<1. Τότε x^2=(1-y)/(1+y) οπότε από τις άλλες δύο παίρνουμε:
  y^2=(1+z)/2 και z^2=(1+y)/2 και αφαιρώντας κατά μέλη y=z οπότε τελικά y=1 άτοπο.
 iii)y>1. Kατ' ανάλογο τρόπο έχουμε τελικά 8y^3(y^2-1)+y-1=0 που είναι επίσης άτοπο αφού y>1.

Άρα μοναδική λύση είναι η x=0,y=1,z=1.


search
analytic geometry
conics
hyperbola
function
Asymptote
algebra

Find all pairs $ (m,n)$ of integers such that $ 3m\plus{}4n\equal{}5mn$.

[quote="Altheman"]Find all pairs $ (m,n)$ of integers such that $ 3m + 4n = 5mn$.[/quote]

[hide="Solution"]

\begin{align*}
5mn &= 3m + 4n \\
5mn - 3m - 4n &= 0 \\
25mn - 15m - 20n &= 0 \\
(5m - 4)(5n-3) &= - 12
\end{align*}

Clearly we only need to search for factors of 12

Hence $ \boxed{(m, n) = (1,3)}$
[/hide]

[quote="SimonM"]

\begin{align*} 5mn & = 3m + 4n \\
5mn - 3m - 4n & = 0 \\
25mn - 15m - 20n & = 0 \\
(5m - 4)(5n - 3) & = - 12 \end{align*}

Clearly we only need to search for factors of 12

Hence $ \boxed{(m, n) = (1,3)}$
[/hide][/quote]

The flaw is that when adding twelve to one side, you also ADD 12 to the other side...

[hide]m=4n/(5n-3). For |n|>3, 0<4n/(5n-3)<1, so m isn't an integer. So it suffices to check n=-3,-2,-1,0,1,2,3. We find that we have the solutions (0,0),(2,1),(1,3).[/hide]

[quote="mathcrazed"]
The flaw is that when adding twelve to one side, you also ADD 12 to the other side...[/quote]

Whoops. I was more trying to give the general idea rather than giving a solution

Can somebody actually type up a full solution, in latex.

[hide="Solution?"]
Using Algebra:

\begin{align*} 5mn & = 3m + 4n \\
5mn - 3m - 4n & = 0 \\
25mn - 15m - 20n & = 0 \\
25mn - 15m - 20n + 12 & = 12 \\
(5m - 4)(5n - 3) & = 12 \\
\end{align*}

Now we find all integer solutions (if any) for six cases: $ 5m - 4 = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ ; $ 5n - 3 = \pm 12, \pm 6, \pm 4, \pm 3, \pm 2, \pm 1$. Note that both $ m$ and $ n$ must be integers.

(In the following cases, the “negatives” will refer to the negative values of $ 5m - 4$ and $ 5n - 3$.)

[b]Case 1[/b]:
$ 5m - 4 = 1 \Rightarrow 5m = 5 \Rightarrow m = 1$
$ 5n - 3 = 12 \Rightarrow 5n = 15 \Rightarrow n = 3$.

This gives us our first pair, $ \boxed {(1, 3)}$. The negatives are not considered, since $ 5m - 4 = - 1 \Rightarrow m = \frac {3}{5}$.

[b]Case 2[/b]:
$ 5m - 4 = 2 \Rightarrow 5m = 6 \Rightarrow m = \frac {6}{5}$, so there is no integer solution here. $ 5m - 4 = - 2$ gives us $ m = \frac {2}{5}$, so we can dismiss the negatives.

[b]Case 3[/b]:
$ 5m - 4 = 3 \Rightarrow 5m = 7 \Rightarrow m = \frac {7}{5}$ — no integer solution.
$ 5m - 4 = - 3 \Rightarrow m = \frac {1}{5}$.

[b]Case 4[/b]:
$ 5m - 4 = 4 \Rightarrow 5m = 8 \Rightarrow m = \frac {8}{5}$ — again, no int. solution. $ 5m - 4 = - 4$, however, gives us $ m = 0$, and $ 5n - 3 = - 3$ gives us $ n = 0$.

Our second pair: $ \boxed {(0,0)}$.

[b]Case 5[/b]:
$ 5m - 4 = 6 \Rightarrow 5m = 10 \Rightarrow m = 2$
$ 5n - 3 = 2 \Rightarrow 5n = 5 \Rightarrow n = 1$.

And our third pair is $ \boxed {(2,1)}$. $ 5m - 4 = - 6 \Rightarrow m = - \frac {2}{5}$, so we dismiss it.

[b]Case 6[/b]:
$ 5m - 4 = 12 \Rightarrow 5m = 16 \Rightarrow m = \frac {16}{5}$ — no integers here, either. The negatives can be dismissed, as $ 5m - 4 = - 12 \Rightarrow m = - \frac {8}{5}$

[b]Answer[/b]: Our pairs are $ \boxed {(0,0), (1, 3), (2, 1)}$. [/hide]
Any corrections would be appreciated. :P

A different approach that I used. Not word for word what I wrote, but the same idea.

[hide="My way"]
\[ 3m + 4n = 5mn \\
5mn - 4n = 3m \\
n(5m - 4) = 3m \\
n = \frac {3m}{5m-4}
\]

(Eqnarray is not working for some reason?) By graphing the above equation on a coordinate plane with an m-axis and n-axis, we get a hyperbola with a parent function of $ Y = x$. Because \[ 5m-4 = 0\] and \[ \frac{3m}{5m} = \frac{3}{5},\] we know the hyperbola must have a vertical asymptote at \[ m = \frac{4}{5}\] and a horizontal asymptote at \[ n = \frac{3}{5}.\] This graph has a root when the numerator is 0, so clearly, the root occurs at (0,0), which is one solution.

As m becomes negative and farther away from 0, the curve converges toward $ \frac{3}{5}$ from "below", just like its parent function. However, there are no integers strictly between 0 and $ \frac{3}{5}$, we know that there must be no solutions where m is negative. Therefore, all other solutions, if any, must have a positive value for $ m$.

When m grows larger from $ \frac{4}{5}$ (the vertical asymptote), n drops down and approaches $ \frac{3}{5}$ from the postive side, again similar to the parent function. Therefore, there exists a point (k,1) on this hyperbola, and there are no solutions with an m-value greater than k. We try natural number values for m to find the solutions (1,3) and (2,1), so we know that $ k = 2$, and that all integer solutions have been found. Thus, the solutions so this system are [b](0,0), (1,3) and (2,1).[/b][/hide]

[hide="Formalized I guess"] Rearranging gives $ m(5n \minus{} 3) \equal{} 4n\Leftarrow m \equal{} \dfrac{4n}{5n \minus{} 3}$.

Case 1: $ n > 3$.
We have $ m \equal{} \dfrac{4n}{5n \minus{} 3} > 0$, since $ 4n,5n \minus{} 3 > 0$. Also, note that $ n > 3\Leftarrow 4n < 5n \minus{} 3\Rightarrow\dfrac{4n}{5n \minus{} 3} \equal{} m < 1$, so $ 0 < m < 1$. Therefore, $ m$ cannot be an integer and we have no solutions here.

Case 2: $ n < 0$.
Again, $ m > 0$, since $ 4n,5n \minus{} 3 < 0$. Now, Since $ n < 0$, we have $ n < 3\Leftarrow 4n > 5n \minus{} 3\Rightarrow m < 1$, so $ 0 < m < 1$. Again, $ m$ cannot be an integer, so we have no solutions.

Case 3: $ 0\le n\le 3$.
$ n \equal{} 0\Rightarrow m \equal{} 0$, so we have the solution $ (m,n) \equal{} (0,0)$.
$ n \equal{} 1\Rightarrow m \equal{} 2$, so we have the solution $ (m,n) \equal{} (2,1)$.
$ n \equal{} 2\Rightarrow m \equal{} \dfrac{8}{7}$, which is not an integer.
$ n \equal{} 3\Rightarrow m \equal{} 1$, so we have the solution $ (m,n) \equal{} (1,3)$.

We have exhausted all cases, and thus $ \boxed{(m,n) \equal{} (0,0),(2,1),(1,3)}$.[/hide]

@metafor: nice solution. One thing that you can consider is writing a table (certainly if you were writing this on a contest, it wouldn't be a pain to format). In other words something like

$ \begin{tabular}{cccc}
5m\minus{}4 & 5n\minus{}3 & m& n\\
1 & 12 & 1 & 3 \\
2 &  . & 7/5 & .\\
\end{tabular}$

more later...

[hide="@zephyredx"]I understand what you are saying but I feel like there are places that your solution gets 'hand wavy'.

What do you mean by 'parent function' of a hyperbola? I'm not familiar with this. Where did $ Y$ and $ x$ come from?

Your analysis of the behavior of the function is correct I suppose, but you didn't really prove it.

The basic idea that you used was that: say $ y \equal{} \frac {3x}{5x \minus{} 4}$. I'd also think that you should break up your casework more clearly:

Case 1: $ ( \minus{} \infty,0]$, the function is strictly decreasing, and between $ [0,\frac {3}{5}]$...

Case 2: $ (0,\frac {4}{5})$ $ x$ is not an integer

Case 3:...[/hide]

[hide="@CatalystOfNostalgia"]

Good, some arrows seem switched. I think sentences should be in the form "X & Y => Z" instead of "Z because X & Y"...but thats not really important. [/hide]

Sorry I actually meant $ Y \equal{} \frac{1}{x}$.  :oops:  Typoed? Not sure.

Well I guess the parent function part wasn't that useful. Only explanation for the shape of the graph I could think of at that time though. Sorry I'm not really familiar with the parenthetical notations for domains and ranges...is () strict or non-strict?

[quote="Altheman"] Good, some arrows seem switched. I think sentences should be in the form "X & Y => Z" instead of "Z because X & Y"...but thats not really important.[/quote]

Yeah I realized that...I think that was just a LaTeX error. I switched all of my arrows, and when I edited the post, apparently I left some out.

The interval: $ x\in [3,5)$ means $ 3\le x<5$. That is to say, $ [$ means you include the endpoint, $ ($ means you leave out the endpoint. Note that a parenthesis always goes with infinity: $ (\minus{}\infty,\infty)$ or whatever.


A game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door." You begin by pointing to door number 1. The host shows you that door number 3 has a goat. What should you do (assuming you want the best odds of winning the car)?

Switch to 2: the chances of picking on the first try are still 1 in 3, so the chances of #2 is 2 in 3.

pick a door. if it has a goat switch the door so u have 66 2/3% of winning the car

A very similar problem can be found in AoPS volume 2 (and beyond). And yes, it's based on Monty Hall.

[quote="lingomaniac88"]Read this:

http://en.wikipedia.org/wiki/Monty_Hall_problem[/quote]

yeah hes right

I read the Wiki Article, but I still don't get it....Wouldn't they be the same probability?

[quote="1=2"]What if there were $ 10^10$ doors?[/quote]

Do you mean $ 10^{10}$?

yeh

teh massage is too long. please make da massage shorter before sub-sitting.

No matter how many doors there are, you still should switch.

yes. because, this is what 1=2 is trying to say:

Pretend, instead of 3 doors, there are 1000000000000000000000000000000000000 doors. you pick one. I declare
99999999999999999999999999999999999 of them have a goat in it. Will you stick with your original choice or pick the one i "just so happned" to skip over?


inequalities
logarithms
function
inequalities theorems

for $x>y>0$
$\frac{1}{x}<\frac{\ln x-\ln y}{x-y}<\frac{1}{y}$
(corrected)

You have reversed the sign.
If $x>y>0$, then $\frac{1}{x}<\frac{\ln x-\ln y}{x-y}<\frac{1}{y}$.
In fact, this is just an immediate conclusion from the mean-value theorem, applied for the function $f(t)=\ln t$.


rotation
geometry
parallelogram
3D geometry

The vertices of quadrilateral $ ABCD$ are at $ (0,0),(9,3),(6,0),$ and $ (9,-3)$.  This dart-shaped quadrilateral and its interior are rotated about the y-axis (yes, the y-axis) to form a solid figure.  What is the volume, in cubic units and terms of $ \pi$, of this bizarre solid?

I don't understand how this could be concave.  I got a parallelogram, but I could be wrong. :maybe:   The question sounds pretty interesting though. :)

Well you start at $ A(0,0)$ then draw to $ B(9,3)$ then $ C(6,0)$ and finally to $ D(9,-3)$ and then back to $ A$.
I should have made that clear, because there are many different ways to draw a concave quadrilateral with just four points given.

But then the side lengths would cross each other and make this "hourglass" shape.  Could it be that you made a typo.  Of course, I could be being reaaaaaaaaaaaaaaaaaaaaally stupid, but I wouldn't know. :P

No, I'm being reaa...aally stupid.  I meant $ C(6,0)$, not $ C(0,6)$. :blush:

[hide="HINT"]Just focus on triangle $ ABC$ for now.  Then multiply by $ 2$ at the end.[/hide]

um draw a line from (9,3) to the y axis, from there draw a line through (9.3) and (6,0) and extend that to the y-axis also this gives a right triangle with lenghts 9,9 9rt2...now rotate this over the y-axis giving a volume of $ 243\pi$, the volume of the part above the x-axis will be $ 1-(\frac{2}{3})^{3}=\frac{19}{27}=171\pi$..... the volume of the cone formed by the smaller right triangle between $ ABC$ and the y-axis is $ \frac{1}{3}\pi(9^{2})3=81\pi$ subract to get $ 90\pi$ formed by rotating $ ABC$ over the y-axis so multiply by 2 to get $ 180\pi$?


Yes, i am. Before I posted the first, I didn't see computations and Tutorials. After i saw it, I think I posted in the wrong place. So I re-posted it. Sorry about it. [quote="fredbel6"]http://www.mathlinks.ro/Forum/viewtopic.php?t=64274
are you double posting this?[/quote]


induction
number theory unsolved
number theory

prove:

if $\sum_{i=0}^{k}{(2^n)^i}$ is prime then $n$ must be a power of $k+1$

[hide]
i couldnt apply induction correctly here... ;) [/hide]


i hope the expert inequality solvers can explain me the exact rationale behind assuming things and normalizing etc..
i am unsure why it is done. someone help plz!

Look at my posts at
http://www.mathlinks.ro/viewtopic.php?t=252146
http://www.mathlinks.ro/viewtopic.php?search_id=429332617&t=252980


geometry
circumcircle
inequalities
geometry proposed

:lol: I desighned this one 2 years ago :!:   I agree its not a nice problem :wallbash_red:

Let $ABC$ be an acute-angled triangle such that $AC<BC$ and $\angle C=60^{\circ}$. $D$ is a point on the side $AC$( near to $A$) such that $DC=BC$ and $M$ is the midpoint of side $BC$. $C_{1}$ and $C_{2}$ with centers $O_{1}$ and $O_{2}$ respectively are the circumcircles of the triangles $ABC$ and $ADM$ respectively. $C_{1}$ and $C_{2}$ intersect at a point $P$(other than $A$). Prove that:

$a) S_{ABC}= \frac{\sqrt{3}}{2}BC^{2}-4S_{ADM}$

$b) (AM-O_{1}M)(O_{1}P+O_{1}O_{2}) < AB^{2}+AC^{2}$

ok here's the solution of part $a)$ : Let $BC$ meet circle $C_{2}$ at $N$. Now because $\Delta MND$ is right angled, $DN$ is a diameter of circle $C_{2}$ . Notice that: $\angle DCN = 60^{\circ}=\frac{DN-AM}{2}\Longrightarrow AM=60^{\circ}$ (here $DN$ and $AM$ were arcs)
As a consequence $AM = O_{2}M = O_{2}A = O_{2}D = \frac{DN}{2}$ so $AM$ is equal to the radius of circle $C_{2}$ , thus : $AM = \frac{AM.AD.MD}{4S_{ADM}}\Rightarrow 4S_{ADM}= AD.MD$,
we have $MD = \frac{\sqrt{3}}{2}BC$ and $AD = CD-AC = BC-AC$, thus:
$4S_{ADM}= ( BC-AC )\frac{\sqrt{3}}{2}BC =\frac{\sqrt{3}}{2}BC^{2}-\frac{\sqrt{3}}{2}AC.BC$
 also we know that $S_{ABC}= AC.BC.sin60^{\circ}$
  As a consequence: $S_{ABC}= \frac{\sqrt{3}}{2}BC^{2}-4S_{ADM}$
 ( because $S_{ABC}> 0$ , we can also say that the inequality: $BC^{2}> \frac{8}{\sqrt{3}}S_{ADM}$ is correct too )
 ok, can anyone now solve part $b)$?


ceiling function
induction
combinatorics unsolved
combinatorics

Let $n \neq 4$ be a positive integer. Consider a set $S \subseteq \{1,2,\ldots,n\}$ with $|S| > \big\lceil \frac{n}{2} \big\rceil$. Prove that there exist $x,y,z \in S$ with $x+y=3z$.

You may find it in :
R.L. Graham, J. Nesetril, 'The mathematics of Paul Erdös', vol.I, p.218-227.

Pierre.

Some of us don't have this book..
I think I found a quite ugly solution:
for $n\leq 15$ we can verify directly (sorry I didn't make something nicer)..
Now we prove for larger values of $n$.
Let $A$ be our set with $|A|>\frac n2$. Suppose, for sake of contradiction that the conclusion fails.
Firstly, note that for $n>k\geq 5$, we have $|A\bigcap \{1,\cdots,k\}|\leq \frac k2$ (otherwise apply the induction hypothesis.
Next, we prove that we can find $t \in A$ with $\frac{n+5}3<t\leq \frac{2n}3$.
Suppose not.
Let $B=A\bigcap [\frac{2n+1}3;n]$, $C=A \bigcap [1;\frac{n+5}3]$. So $B\bigcup C=A$
Now we look at $n$ mod $3$.
If $n=1$ $(mod$ $3)$
We see that $|B|\leq \frac{n+5}6$ by induction hypothesis. But of couse $|C|\leq \frac{n+2}3$.
Since $|A|\geq \frac{n+1}2$ we have two cases:
Case 1: $|C|=\frac{n+2}3$. Then $|B|\geq \frac{n+1}6$. Now if we take $a,c$ the least elements from $B$ and $b$ the biggest one, then we see that $b\geq \frac{n-4}3$ by induction hypothesis. Hence one of $3b-a,3b-c$ lies in $C$ and we have contradiction.
Case 2:$|B|=\frac{n+5}6, |C|\geq \frac{n-1}3$. Then if we take $a,b,c$ the smallest $3$ elements from $B$ and $d$ the biggest one, then $d\geq \frac{n-1}3$ by induction. Now one of
$3d-a,3d-b,3d-c$ lies in $C$ and again we get contradiction, finishing the case $n=1($ $mod$ $3)$.
The other cases are handled analogously. So we've proven the existence of our $t$.
Now by induction hypothesis $3t-n\geq 6$ so the set $\{1,\cdots ,3t-n-1\}$ has at most $\frac{3t-n-1}2$ elements from $A$ hence more than half of the elements of $\{3t-n, \cdots n\}$ are from $A$. Now if we group them into pairs $(3t-n,n),(3t-n+1,n-1), \cdots$ all having sum $3t$ since $A$ has more than half of the elements, $A$ contains at least one pair $(x,y)$. Then $x+y=3t$
QED


What are the specific rules for saying whether a triangle is "formed" or not? Are we are supposed to count triangles which have lines going through them, or are we only looking at the smallest possible triangles?

E.g., if we hypothetically had just a single triangle which had a line going through it, forming a smaller triangle and a trapezoid, does that count as 2 or 1 triangles? (both the larger and smaller triangles having at least one point on the circle as a vertex)

um...i'd answer but we're not allowed to discuss it...

I know that there are a lot of things we aren't allowed to discuss, but I don't see how clarifying the problem's statement is really help. This isn't something like "what's the formula for ____ " or "is this proof of ____  valid", something that would give away a hint of how to do the problem. Its a question about what the problem is in the first place.

Well, I don't know either way, maybe this really is something people aren't allowed to discuss. Sorry, if it is - I didn't mean to cause trouble. But if it is, wouldn't people to be forced to guess the meaning of (potentially) ambiguous problems?

I guess one of the USAMTS staff will resolve this issue.

Use your own judgement; this is not likely to be answered.

You will need to interpret this on your own as best you can.  If you think that there's some ambiguity in the problem statement, then clearly state your assumptions in your solutions.

Beyond that, we cannot discuss the question while the round is in progress.


Say I have a polynomial x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1.  How would I find the roots?

Multiplying by $(x-1)$ gives you $x^{9}-1 = 0$.

Thus, $x^{9}= 1$, and you know that $x \neq 1$.

So, you conclude that the solutions for $x$ are the other 8 complex roots of $1$.

So, $x = e^{2 n \pi i / 9}$, for $n = 1, 2, 3, 4, 5, 6, 7, 8$.

Oh my god...I can't believe I'm so stupid >_<  Thanks. *Seppuku*


If $b>1$, $x>0$ and $(2x)^{\log_b 2}-(3x)^{\log_b 3}=0$, then $x$ is

$\text{(A)}\ \frac{1}{216} \qquad \text{(B)}\ \frac{1}{6} \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ \text{not uniquely determined}$

[hide]
As $(2x)^{\log_b 2}-(3x)^{\log_b 3}=0$, we know that $(2x)^{\log_b 2}=(3x)^{\log_b 3}$. So, taking logs on both sides: $(\log_b 2)\cdot(\log 2x) = (\log_b 3)\cdot(\log 3x)$. Multiplying both sides with $\log b$, yields $(\log 2)\cdot(\log 2 + \log x) = (\log 3)\cdot(\log 3 + \log x)$
After a little work, we get that \[ \begin{aligned} \log x & = \frac{\log^2 2 - \log^2 3}{\log 3 - \log 2} \\ & = \frac{-(\log 3 + \log 2)(\log 3 - \log 2)}{\log 3 - \log 2} \\ & = - (\log 3 + \log 2) = - \log 6 = \log \frac{1}{6}\end{aligned} \]So $\boxed{x=\frac{1}{6}}$.
[/hide]

[quote]

(2x)^{\log_b 2}=(3x)^{\log_b 3}. So, taking logs on both sides: \left(\log_b 2\right)\cdot\left(\log 2x\right) = \left(\log_b 3\right)\cdot\left(\log 3x\right).
[/quote]

i am new at logs and i have just started but i don't understand how this works?

sorry dont worry, lol i understand now..

>< im so slow..><  :blush:

[quote=Kurt Gödel] So, taking logs on both sides: $(\log_b 2)\cdot(\log 2x) = (\log_b 3)\cdot(\log 3x)$. Multiplying both sides with $\log b$, yields $(\log 2)\cdot(\log 2 + \log x) = (\log 3)\cdot(\log 3 + \log x)$
After a little work, we get that \[ \begin{aligned} \log x & = \frac{\log^2 2 - \log^2 3}{\log 3 - \log 2} \\ & = \frac{-(\log 3 + \log 2)(\log 3 - \log 2)}{\log 3 - \log 2} \\ & = - (\log 3 + \log 2) = - \log 6 = \log \frac{1}{6}\end{aligned} \]So $\boxed{x=\frac{1}{6}}$.
[/quote]

can someone explain this entire part? i'm lost

they moved 1 part to the other side took logs ( undoing exponents) multiplied by the log of the algebraic base to eliminate it ...   after a bit of work they used difference of squares,  then simplified and got a negative of a log, which is the log of the reciprocal ...


The ratio of domestic stamps to foreign stamps in John’s stamp collection is 3:1. If John sold thirty of his domestic stamps, the ratio of domestic stamps to foreign stamps would be 1:2. How many foreign stamps does John have in his collection?

Thanks for answer in advance

[hide="Solution"]Let the number of domestic stamps be d and the number of foreign stamps be f; d+f is the total amount of stamps. Then we have

$ \dfrac{d}{d\plus{}f}\equal{}\dfrac{3}{4}$ and

$ \dfrac{d\minus{}30}{d\plus{}f\minus{}30}\equal{}\dfrac{1}{3}$.

Cross-multiplying,

$ 4d\equal{}3d\plus{}3f \Longrightarrow d\equal{}3f$

$ 3d\minus{}90\equal{}d\plus{}f\minus{}30 \Longrightarrow 2d\equal{}f\plus{}60$. We want to solve for $ f$, so we substitute:

$ 2d\equal{}2(3f)\equal{}6f\equal{}f\plus{}60 \Longrightarrow 5f\equal{}60 \Longrightarrow \boxed{f\equal{}12}$.[/hide]

[hide="Another way to do it"]
You could also multiply the first ratio by 2 to get $ 6: 2$.  Then you could conclude that 30 domestic stamps are equivalent to $ \frac{5}{8}$ of John's total stamps (because the of the ratio's change).  Then you divide $ \frac{30}{5}$ to get 6.  So 6 is $ \frac{1}{8}$ of John's total collection.  Multiply by 2 to get the number of foreign stamps: \[ \boxed{12}\].
[/hide]


probability
inequalities
triangle inequality

A strip of paper is available for each whole number length from 1 inch to 15 inches. What is the probability that if a strip is randomly selected, it can be used as the third side of a triangle with two sides of 5 and 7 inches? Express your answer as a common fraction.

By the triangle inequality, the number can't be greater than 5+7-1 is 11, or any less than 7-5+1 is 3. There are 11-3+1=9 possible choices, and 15 total strips, so our answer is $ \boxed{\frac{3}{5}}$.


just notice $3x^4+y^4 \geq 2\sqrt{3}x^2y^2$
so $\frac{xy}{\sqrt{(x^2+y^2)(3x^2+y^2)}}\leq \frac{xy}{(2\sqrt{3}+4)x^2y^2)}$
so the $k$ we have to find is $\sqrt{3}+1$

[quote="hardsoul"][quote="nickolas"]Find the maximum  value  of  $k$  such that for all  positive numbers $x,y$  the  following inequality  holds  $\frac{xy}{\sqrt{(x^2+y^2)(3x^2+y^2)}}\leq \frac{1}{k}$.[/quote]

The inequality is homogenous so take $x=1$
After simplifying we get $y^4+y^2(4-k^2)+3=(y^2-\sqrt{3})^2+y^2(2\sqrt{3}+4-k^2)\ge 0$ for all $y$ hence at $y^2=\sqrt{3}$ gives the first item <0 so we must have $k\leq \sqrt{4+2\sqrt{3}}$.
Now putting $k=\sqrt{4+2\sqrt{3}}$ in the inequality yields to $y^4+y^2(4-k^2)+3=(y^2-\sqrt{3})^2\ge 0$
So $k=\sqrt{4+2\sqrt{3}}$
Sorry I didn't see your proof, zhaobin... :roll:[/quote]
why??and $\sqrt{4+2\sqrt{3}}=\sqrt3 +1$...


calculus
integration
function
combinatorics unsolved
combinatorics

Find the number of non-negative integral solutions to this Diophatine equation
$ x_1 \plus{} x_2 \plus{} ... \plus{} x_k \equal{} n$
I tried to use a recurrence relation to slove this problem, but it became pretty hard to solve after a few steps, can anyone help me? (Maybe by using another better method?) Thanks.
I'm really sorry to have left something important when I first posted this.
There's also the condition that $ x_1\geq x_2\geq ... \geq x_n \geq 0$

use the bars and shorts method aka the sticks and stones method aka the boxes and balls method

This is just the number of partitions of $ n$ into $ k$ or fewer parts.  See, for example, [url=http://www.research.att.com/~njas/sequences/A008284]A008284[/url] or [url=http://www.research.att.com/~njas/sequences/A106254]A106254[/url] in the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url].  Nice generating functions, but not nice formulas.

whoa, that be an addendum to reckon with!


algebra
function
domain
analytic geometry
conics
parabola
logarithms

I need a clear and baby-like definition of the words:

[size=150]Domain and Range.[/size]

What exactly is a function's domain and range?

Please, in simple words.

I'll give a simple definition and anybody who wishes to expand/ clarify can do so.

The domain is all the possible x values of the function, that is, if you were looking at the coordinate grid it would be the horizon. The range is all the possible values of y that correspond to those x values, if you were looking at the coordinate grid this would be what is stemming vertically from the horizon, like the sky (trying to make it poetic??  :P)

So, basically we can write the domain and range in terms of the point (x, y) this way:

(domain, range), right?

usually the domain and range will be more than one value, such as for... $y = x^{2}$ the domain is all the real numbers, but the range is $y\geq 0$

For example, say x = - 4, then y = x^2 becomes y = 16.

Since it is POSITIVE 16, then the range is y = +16 or 0, right?

The domain is - 4, right?

Well the domain for the graph $y=f(x)=x^{2}$ is $\boxed{x\in\mathbb{R}}$ this means that for $x$ can be any real value for $f(x)$..

The range however is $\boxed{y\in\mathbb{R}\text{ such that }y\ge0}$ and if you draw the parabola you see $f(x)\ge0$

these specific examples can't really be applied to all functions.

domain and range are words we use with functions: they're things about the function that tell us how its graph might work.

you might know that the definition of a function is that it's basically a rule which takes a number (x), does something to it, and turns out a number (y). [b]x can never give two or more ys as an output: only one (or none). [/b]

the set of all possible numbers that are allowed to go in the function (all the xs) is called the domain. the set of all possible numbers that are able to come out (all the ys) is called the range. [b]every number in the domain is assigned to one but never two or more numbers in the range.[/b] the function, or the rule, describes how to find which xs are assigned to which ys.

you may have noticed the bolded parts above are basically the same, but there's a difference: in the first I said x can output 1 or 0 ys. in the second I said the domain is assigned to exactly one number in the range, never zero.  that's because if it doesn't have an output, it can't really be in the domain.

also note about the bold parts: the same y can be assigned to more than 1 x, but the same x can never be assigned to more than 1 y. in the function $f(x)=x^{2}$, if we put -5 in for x, the output will be 25. it will never be any other number. but if we put in 5 for x, the output is also 25. that's okay. each x has only 1 y, even if 1 y has 2 xs.

that's sort of demanding conditions. what if your function doesn't fit (for every x, at most one y)? it ain't a function.

If you have $f(x)=x$, the domain is basically all the possible values that $x$ can be and the range is all the possible values that $f(x)$ can be.

Here, try some examples and we'll help you out if something goes astray:

(a) $f(x)=x^{2}+10$

(b) $g(x)=\sin(x)$

(c) $h(x)=\sqrt{16-x^{4}}$

(d) $p(x)=\log ( x^{2}-9)$

A couple trickier ones...

(e) $q(x)=\frac{1}{x^{2}+3x+2}$

(f) $r(x)=\tan(x)+\cot(x)$

Remember, a function's domain is only where it is defined; for example you cannot take the square root of a negative number (and remain in the set of real numbers), or divide by zero. The range is the set of possible numbers which the function "maps" its domain to; that is, the numbers which could come out of the function from plugging in those numbers in the domain.


A bead of mass m is fitted on to a rod and can move on it without friction.At the initial moment the bead
 is in the middle of the rod .The rod moves translationally in a horizontal plane with an acceleration A in
 the direction forming an angle $ \alpha$  with the rod.Find the acceleration of the bead relative to the rod.


Is it just me or,

do you hope that capcom will finally tell us what happened between megaman 8 and megaman x?

This is their last chance!!!!!


Colleen ranks 50th in her class of 250. What is her percentile rank?

Alg. teacher: 20th percentile.
Me: 80th percentile.

Who is right, and why?

I think percentile is the percentage you are better than, which makes the answer 80%

I think your teacher is just looking for the part of the 250 they are, making it not necessarily the 80%. In a way, it could be 20%.

...even though they're wrong. The answer is, then, 80%.

Hm. Makes sense, Alex. Thanks.

Percentile is % that you do better than. Got it! :)

[quote="alexhhmun"]IF it was what % did better than you, then it would be 49/250 or 79.8, 19.6%...

IF it was who did better or equal to you, then it's 20%

IF it is who you did better than, it is 80%.

IF it is who you did better/equal to, it is 80.4%[/quote]

Neither makes max percentile 99th.

There is no accepted standard for a definition of percentile.  The answer should be around 80% by any definition.

Max percentile is 99th because 100th percentile implies you did better than everybody else - including yourself.  This is more a convention than anything else; if you only allot two spots to print someone's percentile, 100th percentile doesn't fit.


I haven't seen it, but I really want to. Do you reccomend it?

it was hilarious, but only if you have watched both of these movies.

willy wonka and the choclate factory (or something like that)
narnia

those are the two main ones

It was OK, not the funniest movie I've ever seen, but moderately funny.  I'd recommend The Wedding Crashers for a good laugh ;)

[quote="SoxFan34"]Good. i've seen both the Chronicles of Narnia and Willy wonka and the Chocolate Factory. I thought Narnia was good, but Willy Wonka was pitiful. I cant wait to see it.[/quote]

lol.  willy wonka was really wierd.

narnia was OK but tey tried to put all that emotional crap in it like saying "for Narnia!!!".  also they forgot to put blood on the sword when Peter (i think) killed that one dude (prolly to keep it pg  :D )

Well, I saw it. I thought, for the most part, it was absolutely awesome. VERY FUNNY.
[hide="WHY?"] Why, God, why was Mr.Tumnus gay with the beaver? WHY?!?![/hide]

It was pretty inapropriate, but oh well, ive seen worse

It blew every single type of [censored] imaginable. Even fish [censored].


It's a special case of the Finsler--Hadviger inequality.

But Why ???

Finsler Hadviger is
\[ \left( {\sum {a^2 } } \right) \ge 4\sqrt 3 S + \sum {(a - b)^2 }  \]

How you can prove that
\[ (\sum {R_a } )^2 \ge \sum {(a^2 - \left( {b - c} \right)^2 )}  \]

 :lol:

[quote="Hawk Tiger"][quote="vardanch"]:pilot:

Prove that for every P point inside triangle
\[ \left( {R_a + R_b + R_c } \right)^2 \ge 4\sqrt 3 S  \]

(\[ PA=R _a  \]) .[/quote]
the minium value of the expression
$R_a + R_b + R_c$ is$sqrt{a^2+b^2+c^2+4sqrt{3}S}$when when each angle of the triangle ABC is less than $\frac{2\pi}{3}$[/quote]

But why???

No.

If the minium value of the expression
$R_a + R_b + R_c$ is$sqrt{a^2+b^2+c^2+4sqrt{3}S}$ hense
$(R_a + R_b + R_c)^2$>=$a^2+b^2+c^2+4sqrt{3}S$

but $a^2=R_a^2+R_b^2-2*R_a*R_b*cosl1$

 :lol:

Maybe I lost a constant coeffient which should be multipied to the right side.You can find it.
But I am sure about the form of the minium value.


algebra
polynomial
calculus
induction
derivative
complex analysis

[b]Find the first few terms in the Laurent expansion of $ \frac {1}{e^z \minus{} 1}$ around $ z \equal{} 0$

Hint: Show that since $ \frac {1}{e^z \minus{} 1}$ has a simple pole,
$ \frac {1}{e^z \minus{} 1} \equal{} \frac {b_1}{z} \plus{} a_0 \plus{} {a_1}z \plus{} {a_2}z^2 \plus{} ...$
Then cross-multiply (using worked example 4.2) and solve for $ {b_1},{a_0},{a_1},{a_2}$[/b]

Worked example 4.2:

[i]Suppose that $ \sum{a_n{z^n}}$ and $ \sum{b_n{z^n}}$ have a radii of convergence $ \geq r_0$. Define $ c_n \equal{} \sum_{k \equal{} 0}^{n}a_k{b_{n \minus{} k}}$. Prove that $ \sum{c_n}{z^n}$ has radius of convergence $ \geq r_0$ and that inside this circle of radius $ r_0$ we have
$ \sum_{n \equal{} 0}^{\infty}c_n{z^n} \equal{} (\sum{a_n{z^n}})(\sum{b_n{z^n}})$

Solution: This way of multiplying out two power series is a generalization of the manner in which polynomials are multiplied. A direct proof can be given but would be somewhat lengthy. If we use Taylor's theorem, the proof is fairly simple. Let $ f(z) \equal{} \sum_{n \equal{} 0}^{\infty}a_n{z^n}, g(z) \equal{} \sum_{n \equal{} 0}^{\infty}b_n{z^n}$ and let A={z such that $ |z| < r_0$} Then f and g are analytic on A, so fg is also analytic on A. By Taylor's theorem, we can write:
$ (f.g)(z) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {(f.g)^{(n)}(0)}{n!}{z^n}$
for all z in A. It is a simple exercise(as in calculus) to show by induction that the nth derivative of the product $ f(z)g(z)$ is given by:
$ (f.g)^{(n)}(z) \equal{} \sum_{k \equal{} 0}^{n}\frac {n!}{k!(n \minus{} k)!}f^{(k)}(z)g^{n \minus{} k}(z)$
Hence
$ \frac {(f.g)^{(n)}(0)}{n!} \equal{} \sum_{k \equal{} 0}^{n}\frac {1}{k!(n \minus{} k)!}f^{(k)}(0)g^{n \minus{} k}(0) \equal{} \sum_{k \equal{} 0}^{n}a_k{b_{n \minus{} k}}$
Thus $ \sum_{n \equal{} 0}^{\infty}c_n{z^n}$ converges on A(and therefore, by Taylor's theorem, the radius of convergence is $ \geq r_0$) and on A

$ \sum_{n \equal{} 0}^{\infty}c_n{z^n} \equal{} (f.g)(z) \equal{} (\sum{a_n{z^n}})(\sum{b_n{z^n}})$[/i]

This is basically telling you exactly what you must do.  $ e^{z}\minus{}1$ is an easy Laurent series so just multiply and match coefficients.  Just set $ 1\equal{}e^{z}\sum_{i\equal{}\minus{}\infty}^{\infty}a_{i}z^{i}$ and solve.


Prove that the sides a,b,c of any triangle suck that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$   satisfy the inequality
$ \sum\frac{a}{a^2\plus{}b\plus{}c}<\equal{}1$


Sorry for such a trivial question, but is it necessary to fill in all empty digits on the scantron with zeros?  For example, which of these answers be correct?

_ denotes a blank circle

007
_07
__7

just to be safe, bubble in 007. It'll be correct if you bubble in 007.

Yes, you need to write 007.

http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/TM-AIME,2008.pdf


Normalization is one choice for this. That is, let $a+b+c=3$.

[quote="neelnal"]What is normalization?[/quote]
It's assuming such constraints as $a+b+c=3$, $abc=1$, etc. in homogenous inequalities. See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=54263[/url].


algebra
system of equations
algebra proposed

$x=y=z=1$ (by inspection :D). It remains to prove uniqueness I guess ;)

[quote="A cute angle"]Hawk Tiger,

you left off the required parentheses.  It must be

ln(x^2 - x + 1).[/quote]
Thank you,it's a typo.I have edited.


given $ f(z)\equal{}exp(z\plus{}\frac{1}{z})$

Show that $ Res[f,0]\equal{}\sum_{k\equal{}0}^{\infty}\frac{1}{k!(k\plus{}1)!}$

Expand about $ 0$ and look for the coefficient on the $ \frac{1}{z}$ term.

\begin{eqnarray*}
f(z) & = & \sum_{n=0}^\infty\frac{(z+1/z)^n}{n!}\\
 & = & \sum_{n=0}^\infty\sum_{k=0}^n\frac{{n \choose k } z^k(1/z)^{n-k}}{n!}\\
 & = & \sum_{n=0}^\infty\sum_{k=0}^n\frac{1}{k!(n-k)!}\frac{z^k}{z^{n-k}}\\
\end{eqnarray*}

So the $ \frac{1}{z}$ term occurs when $ n=2k+1$, so plug that in and the coefficient is the sum $ \sum_{k=0}^\infty\frac{1}{k!(k+1)!}$.


For $a,b,c$ positive reals such that $ab+bc+ca =3$ prove that

$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{1}{\sqrt{1+c^2}}$

Myth said to me that

$LHS \geq 1$

if $ab \leq 3$ and he never makes mistakes  ;) , so it is true that $LHS \geq 1$

but I cannot prove it

Perhaps it is better to write :)

$\sqrt{\frac{1}{1+a^2}}+\sqrt{\frac{1}{1+b^2}} \geq 1$

for $ab \leq 3$

it is my proof here. let$\frac{1}{\sqrt{1+a^{2}}}=x,\frac{1}{\sqrt{1+b^{2}}}=y$
we can get $(\frac{1}{x^{2}}-1)(\frac{1}{y^{2}}-1)\leq9$,then if $x+y<1$
 we can have the contradiction

[quote="manlio"]Thank you very much , zhaobin.

It is correct and very nice. :)[/quote]
My thought is from 41st-imo.
if I rememeber correctly. :)

I have an easier way :
 The ineq is equivalent to
 $ \sqrt{a^2+1} + \sqrt{b^2+1} \geq \sqrt{a^2+1}.\sqrt{b^2+1} $
 We have :
 $ \sqrt{a^2+1} + \sqrt{b^2+1} \geq \sqrt{(a+b)^2 +4} $
 We will prove
 $ \sqrt{(a+b)^2 +4} \geq \sqrt{a^2+1}.\sqrt{b^2+1} $ (1)
  Fortunately , $ (1) <==> (ab-3) (ab+1) \leq 0 $


geometry
rectangle
combinatorics unsolved
combinatorics

Let A and B be two rectangles such that it is possible to get rectangle similar to A by putting together rectangles equal to B. Show that it is possible to get rectangle similar to B by putting together rectangles equal to A.

Hmm, such a problem was disscused in this section, see a post by Billzhao, something with rectangles...

well it is not the same problem, but an idea that solves it


The roots of the equation x^5 - 40x^4 + Px^3 + Qx^2 + Rx + S = 0 are in geometric progression. The sum of their reciprocals is 10. Compute the numerical value of S.

If you don't mind, would you please identify the source of the problem?  Thanks.

A bit simple for Advanced.

[hide] Let the roots be a, ak, ... , ak^4. We know, from the polynomial, that their sum is 40. So a(k^5-1)/(k-1)=40. Next, we have that the reciprocals of the roots are 1/a, 1/ak, ... , 1/ak^4 sum to 10. Or (1/k^5 - 1)/(1/k - 1) = 10. From a little algebra this is 1/k^4((k^5-1)/(k-1))=10a, which implies 1/k^4 * 40/a = 10 a, or 4 = a^2k^4. What we want to find out is the negative of the product of the roots, which is -(a^5*k^10) = -4^(5/2) =  - 32.[/hide]

Hopefully I didn't make too many dumb mistakes

[quote="tetrahedr0n"]Hopefully I didn't make too many dumb mistakes[/quote]
Nothing Major here. Why not ak^2 = -2?

I think it's actually a bit nicer if [hide]you let the roots be a/r^2, a/r, a, ar, ar^2.[/hide]


analytic geometry
topology
function
calculus
calculus computations

Hi,

I was wondering how one would prove:

div (F X G) = G * ( curl F) - F * (curl G)

without using brute force.

Thanks!

Here's the same formula, in radically different language: $ d(F\wedge G)\equal{}dF\wedge G\minus{}F\wedge dG$ when $ F$ and $ G$ are 1-forms on $ \mathbb{R}^3$.
$ d$ here is the divergence on 2-forms and the curl on 1-forms, while the wedge is either the cross product or the dot product in various coordinate systems (This part is special to $ \mathbb{R}^3$). In that form, this formula generalizes to $ d(F\wedge G)\equal{}dF\wedge G\plus{}(\minus{}1)^kF\wedge dG$ when $ F$ is a $ k$-form and $ G$ is an $ l$-form. It's just another facet of the product rule.

Of course, building everything needed to understand that terminology is far more complicated than just running this calculation.

I'll run a piece of the calculation in this manifold language:
$ d((f\,dx)\wedge (g\,dy))\equal{}d(fg\,dx\wedge dy)\equal{}d(fg)\wedge dx\wedge dy\equal{}(g\,df\plus{}f\,dg)\wedge dx\wedge dy$
($ d$ on 0-forms is the gradient, and it satisfies the usual product rule)
$ d((f\,dx)\wedge (g\,dy))\equal{}g\,df\wedge dx\wedge dy\plus{}f\,dg\wedge dx\wedge dy$
$ \equal{} (df\wedge dx)\wedge (g\,dy)\plus{}dg\wedge (f\,dx)\wedge dy\equal{}(df\wedge dx)\wedge (g\,dy)\minus{}(f\,dx)\wedge (dg\wedge dy)$
$ \equal{}d(f\,dx)\wedge (g\,dy)\minus{}(f\,dx)\wedge d(g\,dy)$
That last move uses the antisymmetry of the wedge product to move $ dg$ one step farther right. Multiplying by a function is commutative, so that part can move freely where we need it. Every 1-form is a linear combination of $ dx$, $ dy$, and $ dz$ terms, so we can do this for all combinations and combine them by bilinearity.


integration
logarithms
calculus
calculus computations

[quote="Carcul"]Compute $ \int \frac {x^{17}}{x^{24} \plus{} 1}\ \mathrm {dx}$.[/quote]
[b][u]An easy extension.[/u][/b] For $ \alpha\in \mathcal R^*_\plus{}$ , $ \int\frac {x^{3\alpha \minus{}1}}{x^{4\alpha}\plus{}1}\ \mathrm {dx}\equal{}$ $ \frac {1}{\alpha}\cdot\int\frac {\left(x^{\alpha}\right)^2}{\left(x^{\alpha}\right)^4\plus{}1}\cdot\left(x^{\alpha}\right)'\ \mathrm {dx}\equal{}$

$ \frac {1}{\alpha}\cdot F\left(x^{\alpha}\right)\plus{}\mathcal C$, where $ F(t)\equal{}\int\frac {t^2}{t^4\plus{}1}\ \mathrm {dt}$. Denote $ G(t)\equal{}\int\frac {1}{t^4\plus{}1}\ \mathrm {dt}\ .$ Thus,

$ \left\|\begin{array}{c}
F(t)\plus{}G(t)\equal{}\int\frac {\left(t\minus{}\frac 1t\right)'}{\left(t\minus{}\frac 1t\right)^2\plus{}2}\ \mathrm {dt}\\\\
F(t)\minus{}G(t)\equal{}\int\frac {\left(t\plus{}\frac 1t\right)'}{\left(t\plus{}\frac 1t\right)^2\minus{}2}\ \mathrm {dt}\end{array}\right\|$ $ \Longrightarrow$ a.s.o.

Here's the detailed answer:

[hide="Generalization"]$ P_n \equal{} \int \frac{x^{3n\minus{}1}}{x^{4n}\plus{}1}\,dx \equal{} \int \frac{x^{n\minus{}1}}{x^{2n}\plus{} \frac{1}{x^{2n}}}\,dx \equal{} \frac{1}{2} \left(\int \frac{x^{n\minus{}1}}{x^{2n} \minus{} 2 \plus{} \frac{1}{x^{2n}} \plus{} 2}\,dx \plus{} \int \frac{x^{n\minus{}1}}{x^{2n} \plus{} 2 \plus{} \frac{1}{x^{2n}} \minus{} 2}\,dx\right)$

$ \equal{} \frac{1}{2}\left(\int \frac{x^{n\minus{}1} \plus{} \frac{1}{x^{n\plus{}1}}}{\left(x^n \minus{} \frac{1}{x^n}\right)^2 \plus{} 2}\,dx \plus{} \int \frac{x^{n\minus{}1} \minus{} \frac{1}{x^{n\plus{}1}}}{\left(x^n \plus{} \frac{1}{x^n}\right)^2 \minus{} 2}\,dx\right)$

$ \equal{} \frac{1}{2n}\left(\int \frac{n\left(x^{n\minus{}1} \plus{} \frac{1}{x^{n\plus{}1}}\right)}{\left(x^n \minus{} \frac{1}{x^n}\right)^2 \plus{} 2}\,dx \plus{} \int \frac{n\left(x^{n\minus{}1} \minus{} \frac{1}{x^{n\plus{}1}}\right)}{\left(x^n \plus{} \frac{1}{x^n}\right)^2 \minus{} 2}\,dx\right)$

Now make $ x^n \minus{} \frac{1}{x^n} \equal{} t$ and $ x^n \plus{} \frac{1}{x^n} \equal{} y$, and we get

$ P_n \equal{} \frac{1}{2n}\left(\int \frac{1}{2\plus{}t^2}\,dt \plus{} \int \frac{1}{y^2\minus{}2}\,dy\right) \equal{} \frac{1}{4n \sqrt{2}}\left[2 \arctan\left(\frac{t}{\sqrt{2}}\right) \plus{} \ln \left|\frac{y \minus{} \sqrt{2}}{y \plus{} \sqrt{2}} \right| \right]$

and so

$ P_n \equal{} \frac{1}{4n \sqrt{2}}\left[2 \arctan\left(\frac{x^{2n}\minus{}1}{x^n \sqrt{2}}\right) \plus{} \ln \left|\frac{x^{2n} \minus{} x^n \sqrt{2} \plus{} 1}{x^{2n} \plus{} x^n \sqrt{2} \plus{} 1} \right| \right] \plus{} C$.[/hide]


The areas of three of the faces of a right, rectangular prism are $ 24 \hspace{.6mm} \mathrm{cm}^2$, $ 32 \hspace{.6mm} \mathrm{cm}^2$, and $ 48 \hspace{.6mm} \mathrm{cm}^2$. What is the volume of the prism, in cubic centimeters?

If x, y, and z are the dimensions of the rectangular prism,

xy = 24, xz = 32, yz = 48

x = 4, y = 6, z= 8

$ Volume \equal{} 4 \times 6 \times 8 \equal{} \boxed{192}$

You don't need $ (x,y,z)$ multiplying $ xz=32,xy=24,yz=48$ together gives $ (xyz)^2=24\cdot32\cdot48$ taking $ \sqrt$ of this gives $ xyz=192$


I'm wondering who AoPSers think will win the NCAA Basketball Tournament. I'm going with Duke!!!

For sure it is not going to be [img]http://dept.gatech.edu/adam/buzz/buzz.gif[/img] They haven't done much lately!

Best,

Duke easy, i live near their campus 2!!!!

wowww why are so many people picking duke...? they do have the slight problem that they're just not good this year

I'm rooting for UNC but have to give credit to Illinois.

[quote="yif man12"]They did play Wake Forest pretty tough though.[/quote]

worst part about it was that they (GT) were leading at the half, but halftime can swing momentum in the opposite direction...

i havent followed college basketball this year (not really ever, actually) but i have to go withthe deamon deacons (W.F) just cuz tim duncan used to go there and Chris Paul is awesome! :D

yeah, why are the majority dukies? Illini are the best team.......

the time on this thing is way off...

For the people who said Duke.... The team is dangerously thin.... If something happens to Reddick, I wouldn't be surprised if Duke is bounced in the second round! Therefore my pick has to go to some team that is deep. My choice is either North Carolina, Illinois, or Wake Forest. These teams have great talent and is deep on the bench. I would like to see all three teams get #1 seedings and make it to the Final Four. That would be amazing!

lol, duke was pretty upsetting sun. oh well...the better team won. UNC used their strength, bigger bodies, to pound the glass. felton just threw the ball at the rim cause he knew his guys would crash the boards. luckily, duke went up in the AP rankings  :first:

check out ESPN guru's preseason picks. the guy who chose Notre Dame has NO idea what he's thinking :rotfl:

http://sports.espn.go.com/ncb/preview2004/news/story?id=1923021&nocache

i have no clue how good any of the teams are this year, so i'm going w/ kansas just for fun.

[quote="baseballfreak2"]I'm from CA, and I support Stanford all the way. [hide] But they might not even make the NCAA's........ :blush: [/hide][/quote]

dude, Stanford is the best!!! But i gotta support Duke cause they're local. Stanford beat No.10 Washington over the weekend. They're in for sure. They might not last round 2, however...

[quote="MysticTerminator"]wasn't Stoudamire sort of sucking the whole game?[/quote]

2-13 FG made, 9 points in 39 min. but all arizona needed was a basket, and stoundamire is still (or should be) the primary choice.

If Reddick was 2 for 1653 in a game, if I were Coach K, I'd still have him taking the final shot.

[quote="MysticTerminator"][quote="plokoon51"][quote="mcalderbank"]Er, wow, that made me dizzy! Hmmm, I pick Illinois.[/quote] ?????????????...huh????

anyways, i picked Illinois (cause i'm all for Duke) and i thought Illinois' outside shooting was good enough for a win. their outside shooting was miserable in the 1st half...

congrats to UNC. they used the Shaq-like May to dominate the inside paint. To all u UNC fans, better watch out for Duke next year. Our freshman class will be a great asset and a ton of your (UNC) players will be leaving. :D[/quote]

umm. yeah. ton meaning what? 1 or 2? (other than the seniors)[/quote]

The ton includes McCants, and some dude that i cant remember at the moment.

[quote="MysticTerminator"][quote="plokoon51"][quote="mcalderbank"]Er, wow, that made me dizzy! Hmmm, I pick Illinois.[/quote] ?????????????...huh????

anyways, i picked Illinois (cause i'm all for Duke) and i thought Illinois' outside shooting was good enough for a win. their outside shooting was miserable in the 1st half...

congrats to UNC. they used the Shaq-like May to dominate the inside paint. To all u UNC fans, better watch out for Duke next year. Our freshman class will be a great asset and a ton of your (UNC) players will be leaving. :D[/quote]

umm. yeah. ton meaning what? 1 or 2? (other than the seniors)[/quote]

non-seniors? well, 1 or 2 maybe 3. McCants is going. Felton is probably going. Marvin Williams is thinking of going. May is staying put, however. (darn :P )

[quote="plokoon51"][/quote][quote="MysticTerminator"][quote="plokoon51"][quote="mcalderbank"]Er, wow, that made me dizzy! Hmmm, I pick Illinois.[/quote] ?????????????...huh????

anyways, i picked Illinois (cause i'm all for Duke) and i thought Illinois' outside shooting was good enough for a win. their outside shooting was miserable in the 1st half...

congrats to UNC. they used the Shaq-like May to dominate the inside paint. To all u UNC fans, better watch out for Duke next year. Our freshman class will be a great asset and a ton of your (UNC) players will be leaving. :D[/quote]

umm. yeah. ton meaning what? 1 or 2? (other than the seniors)[/quote][quote="plokoon51"]

non-seniors? well, 1 or 2 maybe 3. McCants is going. Felton is probably going. Marvin Williams is thinking of going. May is staying put, however. (darn :P )[/quote]

Thats what i was gonna say!! Felton!! I cant believe i forgot his name :-(

[quote="juicybooty911"]did y'll see how Augustine fouled out (9min, 0pt)[/quote]

its really funny when players just come into the game, pick up a foul (sometimes a bad call) and then they have to sit back down. although augustine didn't do a thing, his backup (forgot his name) did excellent.

Illinois had 3 big men - I think ingram and smith.

Well, I hear both the Huskies and Zags both have good chances next year, so I'm hoping they do well!

it's never too early to think about next year

http://sports.espn.go.com/ncb/news/story?id=2030071

yea it is. let us bask in the glory of a TAR HEELS NATIONAL CHAMPIONSHIP


is optical isomerism and chirality  defined for a carbon atom that is doubly bonded. I know that they exhibit cis trans isomerism but recently i read about optical isomers and chirality of haloarenes.

Sure it's defined.  It's just not a chirality center, as the plane containing the double bond is a mirror plane.  Of course, however, you can have chiral alkenes: you could have some other stereocenter in the molecule somewhere.

if you refer the book on stereoisomerism by p.s. kalsi it is clearly written that unsymmetrical alkenes  exhibit stereoisomerism.the cis trans structures are diastereomers.also allenes of odd carbon atoms are another example.


For those of you who don't know, I ran a contest from June-July on this very forum called the MATHCOUNTS challenge. Now, you may remember it having very hard, yet creative and tricky problems, and the problems were hard because the thrill of nationals was just winding down.

I'm thinking of running another one in the fall, and the problems will be easier, to prepare you for school and chapter, but will still be tricky. Every now and then on Mathcounts you run into a problem that you don't know the shortcut or trick for, and it might not even have a trick, and this contest will prepare you for that.

Also, I have problem writing credentials :P  C'mon, two of my problems are in this year's handbook and I finished 14th at nationals last year.

The contest may have a different format than last year, possibly eliminating people who fail to score above a predetermined score, using cummulative results, having the same style as last time, or something completely different I have in mind :D

[b]Now, I don't feel like putting a ton of effort into a contest that few people will participate in, so, I want to know if you would participate. It's not that much of a time effort. Feel free to reply with comments or questions as well. [/b]

[quote="ccy"]I looked up the last competition and it looks hard.  :(  I'll try anyway.[/quote]

Like I said, [u]this one will be much easier[/u], but will still prepare you for the times in Mathcounts when there's a problem you haven't seen before.

Also, [b]this is by no means a sign up list[/b], if I decide the challenge is happening, I'll probably post one a week before the competition.

The only reason I dropped out of the previous one was because I thought I wouldn't have much computer access at MathPath. But I was wrong.

It was a fun competition, and I'm happy I have another chance.

Actually, I was thinking of starting my own competition, but I was too lazy to look up problems, so I was thinking of a system where the competitors submit their own problems and the system is rigged to weed out impossible or extremely difficult problems (i.e. a bonus for every person who answers your problem correctly).

If there is one, when approximately will it be?

It'll most likely be October-November, when people have finished the handbook and are looking for problems to do.

Hmm. This looks fun. If i have the time i'll participate.

This is by no means a sign-up list, as sirorange pointed out.

And treething, weren't you in 8th grade last year? I'm probably wrong, though...

Oh. Whoa. False assumption there.

Do you like my competition idea? (Although it kind of sprang from your last round...)

Well, I like it, except for the fact that you'll find it'll be hard to solve every hard problem people send in on time, and that people might run out of problems quickly.

I was thinking about starting a chain competition, where this is a school, chapter, state and nationals competition. The winner of each competition (well, some rules apply  :P ) gets to host the next level of competition (i'll probably start the school competiiton, the winner of that holds the chapter one).

I want at least 20 participants in the next MATHCOUNTS challenge  :D  And guys, you want to sign up when I post the sign up, because this one's gonna be REALLY cool.

When will that be, tentatively speaking?

May i join the competition? I have so many things to do, but i will try. :ninja:

I remember you Kevin, I'm Noah, I was 14th :)

[quote="uberl33tmage"]OOO! I'd like to join to!! I'm Kevin, #17 from last year's nationals.[/quote]

Oh, and you're posting in the Basics?  :P  just kidding..

sirorange, what do you mean by really cool?  :?

I mean really fun and interesting :D

EDIT: Meh, looks like it won't be happening.


Here is a problem I want to solve:

Fresh strawberries contain $ a$% weight percent solids and $ b$% weight percent water.  To make strawberry jam, fresh strawberries are crushed and then mixed with sugar in a $ c: d$ mass ratio (i.e., $ c$ lbs of strawberries for every $ d$ lbs of sugar).  This mixture is next cooked in a vessel to evaporate enough water so that the remaining jam contains $ k$% water by mass.  This is then cooled and bottled.

How many lbs of fresh strawberries and how many lbs of sugar are needed to make $ n$ lbs of jam?

Now, thnis is probably really simple, but how would I do this?

I figured that since there are $ kn$% lbs of water in $ n$ lbs of jam and $ bx$% lbs of water in fresh strawberries (where $ x$ is the number of lbs of fresh strawberries to be used), I could set $ bx \equal{} kn$... and thus $ x \equal{} \frac {kn}{b}$.  However, I'm not sure if this is correct.  Once I get the number of lbs of fresh strawberries, how would I find the number of lbs of sugar?

with the quantities $ n$, $ a$, $ b$, $ c$,$ d$ and $ k$ in lb for example:
$ J$ means jam
$ W$ means water
$ Su$ means sugar
$ St$ means strawberries

$ St: n J(\frac{k W}{100 J})(\frac{100 St}{b W})\equal{}\frac{nk}{b}St$

$ Su: \frac{nk}{b}St(\frac{d Su}{c St})\equal{}\frac{dnk}{bc}Su$

 :)


Can some one tell me a good reason as to why the moon/sun is larger when it is at the horizon and also why it looks more reddish .

The moon (or sun) isn't larger at the horizon - it's the same angular width as it is anywhere else in the sky. This is an illusion, an artifact of the human brain's processing of visual information.

When the moon is high in the sky, you have no visual clues suggesting distance; as far as your eyes can tell, it could be fairly close. But when it's near the horizon, your visual field also picks up trees, buildings, hills in the distance - all the clutter that makes up the horizon. Compared to a tree or a building or a distant hill, the moon is obviously much further away, and so your brain (as part of the unconscious processing of the visual image) interprets it as much bigger.

There are some other illusions that have to do with the perception of distance and size. Stand a few kilometers away from a major airport and watch the airplanes in the landing glidepath. A Boeing 747 appears to move much slower than a small coprorate jet, even though they must actually be travelling at very nearly the same speed.

But the redness at the horizon is real. Shorter wavelength light (the blue end of the visual spectrum) suffers more scattering in the atmosphere than does longer wavelength (red) light. The rays of the sun near the horizon travel a longer path in the atmosphere than do the rays of the sun high in the sky, and hence suffer more scattering, appearing both dimmer and redder.

"Why is the sky blue?" is a related question.

In the area about an airport, most planes have to fly the same speed, 250 knots, because it's the speed limit (everyone's on radar all the time).  Otherwise the air traffic controllers would have to space the planes at great distances.  (Only the really little ones that can't go 250, or the big blimps go slower.)

You have the A answer before this one.  However, there is some distortion of the image from top to bottom, but it is not significant and tends to make the sun/moon squashed or actually a bit smaller up and down than side to side.

[quote="Kent Merryfield"]

But the redness at the horizon is real. Shorter wavelength light (the blue end of the visual spectrum) suffers more scattering in the atmosphere than does longer wavelength (red) light. The rays of the sun near the horizon travel a longer path in the atmosphere than do the rays of the sun high in the sky, and hence suffer more scattering, appearing both dimmer and redder.

"Why is the sky blue?" is a related question.[/quote]

That's rayleigh's law if I'm not mistaken