induction
inequalities
quadratics
LaTeX
ratio
algebra
system of equations

I'm at it again. For the first one I think you're supposed to assume that the statement is true for all $n \le p$, and then show that the inequality is true for $n = p+1$. However, I'm pretty much clueless at how to go past showing the statement to be true for $n=1$ in both problems.

1. Let $p_n$ be the $n$:th prime number, $n = 1,2,3, ...$. Show that $p_n \le 2^{2^n}$

2. A sequence $\{x_n\}^\infty_0$ is defined recursively through
\[x_0 = A \quad x_1 = B\]
\[x_{n+2} = \alpha x_{n+1} + \beta x_n \quad n = 0,1,2, ...\]
A, B, $\alpha$ and $\beta$ are given numbers and $\alpha$,$\beta$ chosen so that the quadratic equation $r^2 = \alpha r + \beta$ has two real roots $r_1$ and $r_2$. Prove by induction that
\[x_n = Cr^n_1 + Dr^n_2\]
where the constants C and D are the solutions to the system of equations:
\[C + D = A\]
\[r_1C + r_2D = B\]

How do I display a system of equations in LaTeX?

For #1, do you mean $p \le 2^{2^n}$?  Because as it is I'm not sure it's true.

For #2,

the base case(s) are the hardest part.  After that, just see that

$x_n = \alpha C r_1^{n-1} + \alpha D r_2^{n-1} + \beta C r_1^{n-2} + \beta D r_2^{n-2} = C r_1^n + D r_2^n$

by plugging in the quadratic in r.

For #1, yes, you are right.

For #2, I still don't get it. Is it possible that you could show the solution more elaborately? I'm not even sure I CAN show the base cases.

[quote="Kalle"][quote="Magnara"]I have to say, my solution to #1 amuses me.

[hide]Note first that using the method of Euclid's proof of the infinitude of the primes, we have $p_{n+1}\leq{p_1p_2...p_n+1}$.

The base case is obvious, it just says that $2<4$.  Therefore, assume that for all integers $1\leq{k}\leq{n}$, we have $p_n\leq{2^{2^n}}$.  Then, as noted above, we have $p_{n+1}\leq{p_1p_2...p_n+1}\leq{2^{2^1}2^{2^2}...2^{2^n}=2^{2^1+2^2+...+2^n}+1=2^{2^{n+1}-1}+1\leq{2^{2^{n+1}}}}$.[/hide][/quote]
This is probably obvious, but how do we get that $2^1 + 2^2 +...+2^n = 2^{n+1} - 1$?[/quote]
Let's consider a more general case. Find
$S=a+a^2+a^3+\cdots +a^{n-1}+a^n$
Multiply this by $a$,
$aS=a^2+a^3+a^4+\cdots +a^n+a^{n+1}$
Subtracting the first one from the second one,
$(a-1)S=(a^2+a^3+a^4+\cdots +a^n+a^{n+1})-(a+a^2+a^3+\cdots +a^{n-1}+a^n)=a^{n+1}-a$
Therefore $S=\frac{a^{n+1}-a}{a-1}$
(In fact this result is very obvious if you know the factorization $a^n-b^n=(a-b)(a^{n-1}+a^{n-1}b+\cdots +ab^{n-2}+b^{n-1})$.)

PS Why has Kalle's post disappeared?  :?:  :?  :huh:

[quote]This is probably obvious, but how do we get that ?[/quote]
it is a geometric sequence...with a ratio of 2

[quote="leepakhin"]
PS Why has Kalle's post disappeared?  :?:  :?  :huh:[/quote]
I deleted it right after posting it (or so I thought I did) because I saw that it was a geometric sum. I was tired :O

We'll prove that $p_n \leq 2^n$ (which is obviously less or equal $2^{2^n}$. Obviously bases (for $n=1$ and $n=2$) hold ($2 \leq 2$, as well as $3 \leq 4$). Assume that we know that $p_n \leq 2^n$, and now we have to prove that $p_{n+1} \leq 2^{n+1}$. Since $p_{n+1} \leq 2p_n$ (by Bertrand, we have prime number between $n$ and $2n$ for every $1 < n \in N$, so inductively we actually proved strict inequality for all $n\geq2$).


Prove that $ (p\minus{}2)!\equal{}1 mod p$

with $ p$ prime number

This is a direct consequence of [url=http://mathworld.wolfram.com/WilsonsTheorem.html]Wilson's theorem[/url].  Similarly, we can show $ (p\minus{}3)! \equiv \frac{p\minus{}1}{2} \bmod p$.

[quote="t0rajir0u"]This is a direct consequence of [url=http://mathworld.wolfram.com/WilsonsTheorem.html]Wilson's theorem[/url].  Similarly, we can show $ (p \minus{} 3)! \equiv \frac {p \minus{} 1}{2} \bmod p$.[/quote]

Hello

we have

$ (p\minus{}2)!\equal{}1mod (p)\implies (p\minus{}2)!\cdot (p\minus{}1)\equal{}(p\minus{}1)mod (p)\implies (p\minus{}1)!\equal{}\minus{}1mod(p)$

But we assume that $ (p\minus{}2)!\equal{}1mod p$ is true before prove it...and it's not correct..

Multiplicative inverses exist mod a prime, so all of those steps are reversible.

By Wilson's Theorem:

$ (p\minus{}1)! \equiv \minus{}1 (mod p) \iff (p\minus{}1)(p\minus{}2)! \equiv \minus{}1 (mod p)$

Since $ p\minus{}1 \equiv \minus{}1 (mod p)$

Hence $ (p\minus{}2)! \equiv 1 (mod p)$


inequalities
trigonometry
inequalities unsolved

[quote="gemath"]Given $x,y,z>1$ such that $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}+2=xyz$
prove that
$3(xy+yz+zx)\ge 4(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}+3)$  :lol:[/quote]
I think the inequality isn't neccesarily true.

Let $x=y=x=2$. Then we have $x,y,z>1$ and $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}+2=2+2+2+2=8=2\cdot 2\cdot 2=xyz$

Now let us se if the inequality holds in this specific case.

$3(xy+yz+zx)=3(4+4+4)=36\le 44=4(2+2+2+3)=4(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}+3)$

Apparently it doesn't. Or have I missed something?

oh Alfred $2+2+2+3=9$ it isn't equal $11$ :D  :D

Use the inequality:
$\sum \frac{\cos A \cos B}{\cos C}\ge \frac{3}{ 2}$

And let:
$\frac 1{xy}=\cos^{2}A$

You can prove the stronger one.

I don't understand so much, because if you let $\frac{1}{yz}=cos^{2}A$ then
$x=\frac{cosA}{cosBcosC}$ improbable $x>1$ moreover do $x,y,z$ satisfy the equal $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}+2=xyz$   :)  :)

I don't think that $x,y,z>1$ is necessary.($x,y,z>0$ is enough.)
with $\frac{xy}z+\frac{zx}y+\frac{yz}x+2=xyz$
we get:
$\frac{1}{yz}+\frac{1}{xy}+\frac{1}{zx}+\frac{2}{xyz}=1$
which means
$(\sqrt{\frac{1}{yz}})^{2}+(\sqrt{\frac{1}{xy}})^{2}+(\sqrt{\frac{1}{zx}})^{2}+2\sqrt{\frac{1}{yz}}\sqrt{\frac{1}{zx}}\sqrt{\frac{1}{xy}}=1$
so we can let:
$\sqrt{\frac{1}{yz}}=\cos A$ and similar.$A,B,C$ are three angles of a acute triangle.
Then the inequality is equal to:
$3\sum \frac{1}{\cos^{2}A}\ge 4+\frac 4{\cos A\cos B \cos C}$
but:
we have:$\sum \frac{\cos A \cos B}{\cos C}\ge \frac{3}{2}$(You can find it in the forum,And it is not hard.)
This means:
$\sum \frac{1}{\cos^{2}A}\ge \frac{3}{2\cos A\cos B \cos C}$
which is stronger than your inequality. :)

[quote="gemath"]I think $x,y,z>0$ isn't enough because if you let $\sqrt{\frac{1}{yz}}=cosA$
$\Rightarrow 0<\frac{1}{yz}<1\Rightarrow yz>1$ thus you should need $x,y,z>1$[/quote]
but the condition:
$\sum \frac{xy}{z}+2=xyz$
has imply $0<\frac{1}{yz}<1$...

ok thank for a good solution!, we can infer if $x,y,z>0$ such that
$x^{2}+y^{2}+z^{2}+2xyz=1$ then we can put
$x=cosA,y=cosB,z=cosC$ :)  :)

[quote="RDeepMath91"][quote="zhaobin"]I don't think that $x,y,z>1$ is necessary.($x,y,z>0$ is enough.)
with $\frac{xy}z+\frac{zx}y+\frac{yz}x+2=xyz$
we get:
$\frac{1}{yz}+\frac{1}{xy}+\frac{1}{zx}+\frac{2}{xyz}=1$
....[/quote]

Are u sure ??
Isn't it $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{xyz}=1$ ??  :huh:[/quote]
Sorry,That's my fault.
But If it is,I think the inequality is wrong.
Because the inequality is $equal$ to
$3(\cos A+\cos B+\cos C) \ge 4+4\cos A \cos B \cos C$
But it is obvious wrong.

Thank RDeepMath91 for your remarks, but it is really true inequality, can zhaobin show me cleanly why it is flase ? :)


is my alcumus score supposed to decrease even if i havnt solved any problems to cause it to do so?

i don't think so...[hide="though"]
my score has never dropped....[/hide]


[hide]9 times 1/9 = 1

and so on.. :-)  :D

like i like pie said[/hide]

[hide="Another way(can also help you turn fractions into repeating decimals)"]Assume that $0.\overline{9}=x$. Then $10x=9.\overline{9}$. Then, you can do this: $10x-x=9$. Then $9x=9$. So, $x=1$. I learned it in 3rd grade from my very smart tutor (who I stopped seeing 5 years ago :oops: ), Mrs. Stephanie Kim.[/hide]

[hide="Another example of how to use this method to get fractions from decimals."]I'm going to take a random decimal $0.\overline{483}$. Call it $x$. Multiply by $1000$. $1000x=483.\overline{483}$. $999x=483$. $x=\frac{483}{999}$.[/hide]
Oh yeah, and by the way, I already asked this in the Rubix forum, I know that QED stands for some weird Latin thingy but [u][b]what does it mean[/b][/u] and [b][u]when is it used[/u][/b]? :?:

Q.E.D. is an abbreviation of the Latin phrase "quod erat demonstrandum" (literally, "which was to be demonstrated"). In simple terms, the use of this Latin phrase is to indicate that something has been definitively proven.

QED, may be written at the end of mathematical proofs to show that the result required for the proof to be complete has been obtained.

Don't copy straight from wikipedia... :roll:

[quote="8parks11"]is copiyng from wikipedia suddenly illegal?
im just giving him a definition from an encyclopedia lol.
you could be redundant and post your own definition if you want to
 :wink:[/quote]
I didn't say it was illegal... I just meant you shouldn't plagiarize other people's work but instead try to put it in your own words.

[quote="Necrowarrio0"]UH? WHy is 0.9 recurring 1???[/quote]
Read my proof.

1/3=0.333333333333 repeated (do long division)
1/3*3=1
0.333333 repeated=1 because of the multiplication property of equality (multiply both sides of 1/3=0.333333333 by 3, you get two equal things, namely 1 and 0.9999999 repeated)
0.9999 repeated=1

[quote="Dracomage94"]how do u do the
"click to show solution" thing?[/quote]
Click the "Hide" button (next to the formatting buttons) and enter the text you want hidden between the tags. Same for Code, Quote, Image, and URL.

Here's a pattern:

$\frac{1}{3}= 0.33333333333...$

$\frac{2}{3}= 0.66666666666...$

$\frac{3}{3}= 0.99999999999...$

That's a neat pattern!

Here's the general formula:

If a decimal has a repeating block $Q$, then the fraction is $\frac{q}{10^{k+1}-1}$, where $k$ is the number of digits in $Q$.

Don't be intimidated by this. Here's an example:

$0.234234234234... = \frac{234}{999}= \frac{26}{111}$

This is an application of the geometric sum

$\frac{1}{10^{k}}+\frac{1}{10^{2k}}+... = \frac{1}{10^{k}-1}$

Generally, for $r > 1$ we have the identity

$\frac{1}{r}+\frac{1}{r^{2}}+... = \frac{1}{r-1}$

Any two distinct real numbers have an arithmetic mean unequal to either of the two. So, if 0.999... were not equal to 1, there would have to be a real number between 0.99... and 1. Clearly this is not the case.

this problem bothers me....first, to define this problem rigirously, we mean find:

$S=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}9*10^{-k}$

second, the statement $S=1$ is not the same as when we say $1=1$, when we say that $1=1$, we mean that they are actually equal, when we say that $S=1$ we mean as $n$ gets arbitrarily large, $S$ gets arbitrarily close to $1$

[quote="Altheman"]this problem bothers me...[/quote]

Intuitively, we would like for each real to have a unique representation in each base.  Unfortunately, any real with a terminating base-$b$ representation does not have a unique base-$b$ representation precisely because $0.999... = 1$.

Certain proofs showing that a bijection exists between, say, $\mathbb{R}$ and $\mathbb{R}^{2}$ rely on "mixing the digits" in the coordinates of a member of $\mathbb{R}^{2}$ to produce a unique member of $\mathbb{R}$.  The basic concept is sound, but this particular scheme fails precisely because of the lack of completely unique base-$b$ representation of some reals.  (Can you figure out a way to modify it so that it does work?)


calculus
integration
geometry
rectangle
analytic geometry
function
calculus computations

Computer the integral of f(x,y,z) = xyz over the rectangle with vertices (1,0,1), (2,0,0), (1,1,1), and (2,1,0).

Can someone please help? I'm pretty sure I get everything, but I don't know how to start it off. How do I replace the z?

(I graphed it on a xy-plane and I get a square. Now what?  :? )

If you're using $x$ and $y$ as the primary coordinates, you need $z$ as a function of $x$ and $y$. It should be a linear (affine) function, since the region is meant to be a quadrilateral.
Any three vertices should be enough to determine the function; just set up an appropriate linear system for the coefficients and solve.

Once you have $z$, you can find the function and the area multiplier in the integral.


The rails on a railroad are $ 30$ feet long. As the train passes over the point where the rails are joined, there is an audible click. The speed of the train in miles per hour is approximately the number of clicks heard in:
$ \textbf{(A)}\ 20\text{ seconds} \qquad\textbf{(B)}\ 2\text{ minutes} \qquad\textbf{(C)}\ 1\frac{1}{2}\text{ minutes} \qquad\textbf{(D)}\ 5\text{ minutes}\\
\textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
$ 30$ fps corresponds to $ 1$ click per second.  In miles per hour, $ \frac{30 \cdot 60^2}{5280}\equal{}\frac{225}{11}$ mph corresponds to $ 1$ click per second.  $ 1$ mph thus corresponds to $ \frac{11}{225}$ clicks per second, which is about $ 1$ click per 20 seconds.  So $ x$ miles per hour corresponds to about $ x$ clicks in $ 20$ seconds, or $ \boxed{\textbf{(A)}}$.
[/hide]


ok so the riddle goes like this:

100, 121, 144, 169, number thats not square, number thats not square, ..., _?

you could think that _ is a perfect square, but it's not.

if the number is x, then the first term is x expressed by base n, next term is base (n-1), next term is base (n-2)...

but i can't seem to remember x or n.

what is this riddle?

I think n=13, x=169. For higher n, the next number is 196, a square. For lower n, 169 must be a number in a base lower than 10, which it is not.


[color=green]Prove that if the Miquel point is on the circumcircle of triangle ABC, than the 3 Menelaus points on the sides are collinear.[/color]

Thanks very much guys!!!! :lol:

Sorry for asking, but what did you mean with "3 Menelaus points"?

Let $P_B,P_C$ be any point on $AC,AB$ respectively. Let $M$ be the intersection of the circumcircle of $\triangle ABC$ and the circumcircle of $\triangle AP_{B}P_{C}$. Let $P_A$ be the intersection of $BC$ and $P_{B}P_{C}$.

It is equivalent to show that $C,P_{A},M,P_{B}$ and $B,P_{A},M,P_{C}$ are concyclic. Indeed it is obvious as $\angle P_{C}P_{B}M = \angle P_{C}AM = \angle BCM$ and $\angle MP_{C}A = \angle MP_{B}A = \angle MP_{A}B$.

[quote="suedenation"]Do you know Menelaus theorem?[/quote]

I do know Menelaus Theorem, but I've just understood your problem by seeing on the figure by yptsoi. Thank's to him :)

If I were you, I will write the problem as:
Let $P_a,P_b,P_c$ be a point on (or the extension of) the side  $BC,CA,AB$ on triangle $ABS$. If circumcircle of $ABC$, $AP_bP_c$, $BP_cP_a$, and $CP_aP_b$ are concurrent, then $P_a$,$P_b$,$P_c$ are collinear.

Forgive me for this. I think I have to learn English more.


function
calculus
derivative
real analysis
calculus computations

prove that $ 0\equal{}y(0)\equal{}y'(0)\equal{}y''(0)\equal{}...$ does not have a unique solution

$ y$ can be any [url=http://en.wikipedia.org/wiki/Bump_function]bump function[/url] centered suitably far away from zero.

Well, we can surly say that it exists at $ x\equal{}0$, so we can do the Taylor series around $ 0$ and we would get $ f(x)\equiv 0$. I don't understand the "bump function" answer though...

There is a reason why this argument doesn't work, milin.  Suppose $ f : \mathbb{R} \to \mathbb{R}$ is infinitely differentiable. The problem is, $ f$ does not need to correspond to it's power series in a point in any interval, as is clear from the bump function examples (at the points where $ f$ smoothly goes from being constant zero to non-zero). This happens when the remainder terms in the taylor expansions do not tend to zero in a small interval.

If $ f$ does correspond to it's taylor series in some small intervals, then indeed $ f$ is zero as can be seen from the fact that the set of points where all derivatives vanish is open (by the taylor series). But also due to continuity, the complement is open. It's impossible to partition $ \mathbb{R}$ into two non-empty open sets, so the result follows.

Analytic functions, that is functions that are complex differentiable, do always correspond to their taylor series.


A leaf is a vertice which has only one edge emanating from it.

[quote="Xixas"]How many labeled trees on $n$ vertices having $k$ leaves are there?[/quote]

it's the answer for $n\geq 3$
maybe... if $f(k)$ is the answer (fixed n) and $g(k)= nCk (n-k)^{n-2}$
$g(k)= sum(i=k...n-1)_{i}C_{k}f(i).$
we can know $f(i)$ from above

For example, when $n=3$, $f(1)=0,f(2)=12,f(3)=4$
$g(2)=24=f(2)+3C2 f(3)$
$g(3)=4 =f(3)$
it maybe works well.


I was looking at the Mid term challenge for the intermediate algebra course and founed that the filename of it was MidTerms0lv3r.pdf.  Notice that that is midterm followed by "solver" in 1337.  Is this a coincidence?

So long as it's been mentioned, I've always wondered: what in heavens' name is the point of '1337'?

1337 is a whole dialect. No, a language. No, a society! It's so unbelievably awesome, it can only be described as 1337.

Actually, it is a language, often used in computer multiplayer games, like warcraft, starcraft, etc. It's just numbers taking the place of some letters. 1337 by itself means "elite" which got shortened to leet and then to 1337.

No no, I know full well what it is.  I just don't get why it exists.

Why does it exist? Why does english exist? Why does math exist? WHY DO WE EXIST???????

Whew, sorry about that. It's just a fun "language" made up by some people and now, it's been spread all over the internet games. See if you can figure it out. It's cool, since there's only a couple letters that you can't make with it.

[quote="nr1337"][quote="yif man12"]Why does it exist? Why does english exist? Why does math exist? WHY DO WE EXIST???????
[/quote]

42

It's the answer to life the universe and everything.[/quote]

Heh, that's exactly what my friend said right when I showed him my post.

[quote="ComplexZeta"]Douglas Adams made a mistake. The meaning of life, the universe, and everything is $\frac{1}{2\pi i}$.[/quote]

How did you get that?

It's very important for complex integrals.

Ah, yes, I suppose so.

A lot of functions are important for complex integrals.  I've seen the Beta function and Gamma function used a lot...

[quote="nr1337"][quote="yif man12"]Why does it exist? Why does english exist? Why does math exist? WHY DO WE EXIST???????
[/quote]

42

It's the answer to life the universe and everything.[/quote]

Isn't the question supossed to be 9*6 in base 13?

We don't exist because existence is a state of mind. We do not exist, but we just believe that we exist. Hehe. I am messed up...


trigonometry
calculus
calculus computations

Find the range of $ k$ for which on the $ xy$ plane, two curves $ y\equal{}k(x\minus{}x^3),\ x\equal{}k(y\minus{}y^3)\ (k>0)$ have the intersection points $ (\alpha ,\ \beta)$ such that $ \alpha \neq \beta$ in the first quadrant.

let $ f(x) \equal{} k(x \minus{} x^3)$. solving the equation $ f(x) \equal{} x$ for $ x > 0$, we have $ kx \minus{} kx^3 \equal{} x \Rightarrow kx^2 \equal{} k \minus{} 1 \Rightarrow x \equal{} \sqrt {\frac {k \minus{} 1}{k}}$. denote $ \sqrt {\frac {k \minus{} 1}{k}} \equal{} r$. note that if $ k\leq 1$, $ f(x) \equal{} x$ has no solution $ x > 0$, and in that case the graphs have no intersection off the line $ y \equal{} x$ (since they're symmetric across $ y \equal{} x$. so from here on assume $ k > 1$ so that $ r$ exists. now since $ f$ is concave, and the two graphs in question are symmetric across $ y \equal{} x$, it's not hard to see that they'll intersect at a point off the line $ y \equal{} x$ if and only if $ f'(r) < \minus{} 1$. (drawing the tangent line to $ f$ at $ (r,r)$ will make this clear.) we have $ f'(r) < \minus{} 1 \Leftrightarrow k \minus{} 3kr^2 < \minus{} 1 \Leftrightarrow k \minus{} 3(k \minus{} 1) < \minus{} 1 \Leftrightarrow$ $ k > 2$, and this is the required range.

[hide]
If an intersection point is $ (p,q)$ where $ p,q$ positive and distinct, then $ q \equal{} k(p \minus{} p^3),p \equal{} k(p \minus{} p^3)$. Subtracting and dividing by the nonzero $ p \minus{} q$,
\[ \minus{} 1 \equal{} k(1 \minus{} p^2 \minus{} pq \minus{} q^2)\Rightarrow k \equal{} \frac {1}{p^2 \plus{} pq \plus{} q^2 \minus{} 1}
\]
Plugging into $ q \equal{} k(p \minus{} p^3)$ gives
\[ q(p^2 \plus{} pq \plus{} q^2 \minus{} 1) \equal{} p \minus{} p^3\Rightarrow 0 \equal{} p^3 \plus{} p^2q \plus{} pq^2 \plus{} q^3 \minus{} p \minus{} q \equal{} (p \plus{} q)(p^2 \plus{} q^2 \minus{} 1)
\]
Since $ p,q$ positive, $ p \plus{} q$ nonzero, so $ p^2 \plus{} q^2 \equal{} 1$. We then get $ 1<p^2 \plus{} pq \plus{} q^2 < \frac {3}{2}$, so $ k > 2$.

To see that all such $ k$ work, each $ k > 2$ can be expressed in the form $ 2\csc 2\theta$ where $ 0 < \theta < \frac {\pi}{4}$. Then $ p \equal{} \cos \theta, q \equal{} \sin \theta$ satisfies the system.
[/hide]

pleurestique, scorpius119, your answers are correct. :)

My solution is almost same as scorpius119's.

After getting $ p^2 \plus{} q^2 \equal{} 1\ (p\neq q,\ p > 0,\ q > 0)$ in scorpius119's, we have $ k \equal{} \frac {1}{pq} > 2$ by a well-known inequality.

How about this problem?

[color=blue]Find the range of $ k$ for which on the $ xy$ plane, two curves $ y \equal{} k(x \minus{} x^5),\ x \equal{} k(y \minus{} y^5)\ (k > 0)$ have the intersection points $ (\alpha ,\ \beta)$ such that $ \alpha \neq \beta$ in the first quadrant.[/color]

the same approach i gave above resolves this problem for any $ f(x) \equal{} k(x\minus{}x^r)$, $ k>0$, $ r>1$. $ f$ has the fixed point $ p \equal{} \sqrt[r\minus{}1]{\frac{k\minus{}1}{k}}$ in the first quadrant if $ k>1$. again, $ f$ is concave, so as before, we're looking for $ k$ such that $ f'(p) < \minus{}1$:

$ f'(p) < \minus{}1 \Leftrightarrow k\minus{}krp^{r\minus{}1} < \minus{}1 \Leftrightarrow k\minus{}r(k\minus{}1) < \minus{}1 \Leftrightarrow  r\plus{}1 < k(r\minus{}1) \Leftrightarrow \frac{r\plus{}1}{r\minus{}1} < k$

Thank you, pleurestique. :)

That was  my final purpose.

kunny


At King Arthur's Court, $ 2n$ knights gathered at the Round Table. Each has at most $ n \minus{} 1$ enemies among the others. Prove that Merlin the Wizard can devise a seating arrangement such that no knight will be next to any of his enemies.

Pierre.

Again, I'll have to bug you with a detail, and I am sorry for this :): is the "enemy" relationship mutual?

Consider the relationship of "not being an enemy". The corresponding graph has $2n$ vertices and the minimal degree is $\ge n=\frac{2n}2$. It's a well-known theorem (Dirac's, maybe:  :? ) that in this case there's a hamiltonian cycle.

Yes, it is well known now. But imagine school students, who didn't know it, of course.


function
algebra
functional equation
topology
real analysis
real analysis theorems

Prove that any measurable function f:R->R that satisfies Cauchy's functional equation f(x+y)=f(x)+f(y) for any x,y is automatically continuous.
Hint:Use Lusin's theorem.

Notice that $f(x+h)=f(x)+f(h)$, it suffices to show the continuity at $x=0$

From Luzin's Theorem, there exist a bounded closed set $F$ such that $f(x)$ is continuous on $F$. Thus $f$ is uniformly continuous on $F$. That is, given $\epsilon>0$, there exists $\delta_1$ such that for any $x,y\in F$ satisfying $|x-y|<\delta_1$, it holds that $|f(x)-f(y)|<\epsilon$.

Now we need a powerful theorem(Steinhaus, 1920):
If $mE>0$ then $E-E$ contains $B(0,\delta)$ for some $\delta$.

Apply this theorem to $F-F$, there exists $\delta_2$ such that $[-\delta_2, \delta_2]\subset F-F$. Let $\delta=\min\{\delta_1,\delta_2\}$. For any $z\in [-\delta, \delta]$, there exists $x,y\in F$ such that $x-y=z$. Thus $|f(z)|=|f(x-y)|=|f(x)-f(y)|<\epsilon$, whence it follows the continuity at $x=0$.


[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135796[/url]

Same rules, same concept as VM 2.

Inserts homework.

Oh boy, not this again.

Receives F.  Inserts failure HW.

Receives a bunch of paper shreds.

Inserts a vending machine.

Receives a pine tree... which grows apples.

Inserts  :huh: .

Receives a green ninja. Who's smiling.

Inserts  :yoda:

Receives very bad grammar. Indeed.

Inferts a fettler, a warrior, a merchant, and a dipolmat.

There are no typos in this poft. Brownie points for gueffing what I'm making a fpoof of correctly.

ur little sister

inserts sister
recieves a lemon

recieves: mad confuzzled parents
inserts: dark chocolate

receives white chocolate

inserts Cheez-it. :D

Receives anti-Cheez-it.

Inserts Open Office.

Receives empty can
OH ARE YOU SRS??
Inserts that worthless can

Recieves a recycling bin.

Inserts badd speelinge

recives an English paper with a 0 on it.

Inserts this:
 :stink:

Recieves an F.

Inserts a waterbender from the Southern Water Tribe.

Receives  :stretcher:

Inserts a random AoPS user.

receives a random AoPS user.

inserts evilhamster

Receives hamsterevil.
 :|
Inserts  [hide="this thing"][img]http://roflrazzi.files.wordpress.com/2009/03/celebrity-pictures-bela-lugosi-vampires-sparkle.jpg[/img][/hide]

Receives a photo of a mirror.

Inserts a bouncy ball w/ slime

Receives a very slimy tennis racket.

Inserts a picture of a picture of a picture of a picture...

Receives a report card.

Inserts a tarantula.

recieves a penguin

inserts AAAAGGGGGHHHHHH and UUUGGHHHH!!!!

Receives the best thing in the world.

Inserts the USA.

receives the world.

inserts a F1 rocket engine. :D


How did you do at the recent Grissom Tournament?

Could Grissom have messed up any more? At least this time there weren't as many protests on the tests, but all those screw-ups at the award ceremony more than made up for it. Sigh now I have another identical tiger-head trophy that looks more for hunting than math.

Last competition of the year for us

what place did you/your team get (in what catergory(pre-alg, alg, geom...))?

do you go to randolph?

Lets see our massive team of 4 got 4th in the large school division even though Randolph is small school. And this was in Geometry, where I got 22nd with a 74.2. so yea thats about it

they count private schools as big schools (or at least thats wat my coach said).

i knew sid knew you......

22nd with a 74.2??? WOWZAS

I was the person to go up to get the first graphing calculator/6th alg THANKS TO YOU!!!!! jk,jk. How did you know my name before that though?

[quote="lifesapain"]NO they dont... All private schools except maybe 2 or 3 are placed in the small school division. But they knew Randolph would pwn again as usual if competing with other small schools. So they bumped us up just like at Vestavia and Hoover[/quote]

alright....

you keep avoiding me asking HOW YOU KNOW MY NAME!!!! :wink:

Grisshom? Why does nobody tell me about these things? :dry:

Is it like... a by invite competition? That would explain it because they don't care about Mobile  :harhar:

Yup Mobile's just that isolated part of Alabama that nobody cares about. And i think for the most part it is an invite competition, but you can probably register yourself anyway. Haha theres gonna be less and less schools going to Grissom's every year

If you all need information on the Grissom tournament, go to http://www.geocities.com/grissom_math

And to address one question, it is not an invite-only competition, and teachers can register on that website.

Furthermore: In general, it is probably not such a good idea to publicly bash a tournament under any circumstances. On the other hand, suggestions for improving any tournament for the future would be welcome in this forum.

oh and by the way, there were a couple of questions that were dropped due to protests (on the geom tests) that i know of

21,22,18, and they were considering dropping 11, but i dont know if they did. + i dont understand why they would, i got it right and it wasnt a hard or incorrect problem..
anyway, those are the one that they were considering/dropped
(which killed my score)


For how many different values of $ x$ does $ x^5 \equal{} x^4 \plus{} 72x^3$?

One value is 0.
If x isn't 0 then:
$ x^2\equal{}x\plus{}72$ which has 2 solutions, so 3 values.

Note that it only asks for the NUMBER of roots, not the actual roots themselves.

However, akhil's method of actually finding all solutions is much more rigorous. For example, what if the polynomial was $ x^5\equal{}18x^4\plus{}81x^3$? Also a problem that readily becomes a quadratic after dividing out $ x^3$, $ x^2\equal{}18x\plus{}81$, but this quadratic has a repeated solution! ($ 9$) It's much safer to make sure you have all different solutions.


If $ a \nabla b \equal{} 3a \minus{} b^2$ find $ 2 \nabla (3 \nabla 1)$.

i got that too, but the answer says -75. Fishy.

wait no. answer is -58, i typed -75. heh


[hide="idea"]
use the rule for divisibility of 9 and 11... and probably some other stuff too
[/hide]

Consider the prime factorization of 37!
We have 37! = 2^n × 5^8 × m, where n  >= 8 and m is an integer not divisible by 5.
Hence 37! is divisible by 2^8 × 5^8 = 10^8, but not by 10^9.

Therefore 37! ends in precisely eight zeroes, and so h = 0.

Determine g

Next we calculate g by means of the following observation: g = 37! / 10^8  (mod 10.)
We may divide 37! by 10^8 = 2^8 × 5^8 by dividing each multiple of 5 by all of its factors of 5, as well as removing an additional factor of 2^8 by striking out 2 = 2^1, 8 = 2^3, and 16 = 2^4.
Hence g 	= 	3 × 4 × 1 × 6 × 7 × 9 × 2 × 11 × 12 × 13 × 14 × 3 × 17 × 18 × 19 × 4 × 21 × 22 × 23 × 24 × 1 × 26 × 27 × 28 × 29 × 6 × 31 × 32 × 33 × 34 × 7 × 36 × 37  (mod 10.)
 	= 	3 × 4 × 1 × 6 × 7 × 9 × 2 × 1 × 2 × 3 × 4 × 3 × 7 × 8 × 9 × 4 × 1 × 2 × 3 × 4 × 1 × 6 × 7 × 8 × 9 × 6 × 1 × 2 × 3 × 4 × 7 × 6 × 7  (mod 10.)
 	= 	3 × 4 × 6 × 7 × 9 × 2 × 2 × 3 × 4 × 3 × 7 × 8 × 9 × 4 × 2 × 3 × 4 × 6 × 7 × 8 × 9 × 6 × 2 × 3 × 4 × 7 × 6 × 7  (mod 10.)

We may now multiply term-by-term, reducing mod 10 as we go.
We begin: 3 × 4 = 2 (mod 10),  2 × 6 = 2 (mod 10),  2 × 7 = 4 (mod 10), and so on.

After 27 such multiplications (mod 10), we obtain g = 4.

Determine a, b, c, d, e, and f
We now have 37! 	= 13763753091226345046315979581abcdef400000000
 	= 13,763,753,091,226,345,046,315,979,581,abc,def,400,000,000

Note that 1001 = 7 × 11 × 13 and 999 = 3^3 × 37.
Hence 999999 = 1001 × 999 = 3^3 × 7 × 11 × 13 × 37, and so 37! is divisible by 999999.

Further, since (10^6)n = 1^n = 1  (mod 999999), the sum of the digits of 37!, grouped six at a time beginning with the units digit, is congruent to 0  (mod 999999.)
(If we consider each group of six digits as a single digit, this is the base 10^6 equivalent of the familiar base 10 test for divisibility by 9.)

So we have 000000 + (100000d + 10000e + 1000f + 400) + (581000 + 100a + 10b + c) + 315979 + 345046 + 091226 + 763753 + 13 = 0  (mod 999999.)
Hence 100000d + 10000e + 1000f + 100a + 10b + c = 902580  (mod 999999.)

Since 0  <= a, b, c, d, e, f  <= 9, it follows that d = 9, e = 0, f = 2, a = 5, b = 8, c = 0.

Putting the results together, we have a = 5, b = 8, c = 0, d = 9, e = 0, f = 2, g = 4, h = 0.


This is a Function problem that was taken out of my Algebra 2 book.

Problem: If g(x)=x², then g(x+1)= ?

Answer from the book: x²+2x+1

How do you get x²+2x+1 from g(x+1)= (x+1)²  ??


Hello, this is my problem:
When i type greek characters outside the dollar symbols they appear ok in pdf file, but the characters between  dollar symbols appears in english...How can i fix that? Thank you.

Use the commands \alpha, \beta, \gamma etc to get $ \alpha, \beta, \gamma, \dots$. See [[LaTeX:Symbols#Greek_Letters]]

Hello, well it works fine....but do you think i can use my greek keyboard to insert greek characters between the dollar symbols? Thanks again.

You may need to use Omega - see [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/49562dc7e08d79dd/6a8b1896094203fa]Using Omega in MiKTeX[/url]. But it may be better to consult the experts who use TeX with a Greek keyboard. See [url=http://www.eutypon.gr/eft/index.html]Greek TeX Friends Group[/url]. I don't have a Greek keybard so I think that they will be better able to advise you.

Look here:

http://tobi.oetiker.ch/lshort/lshort.pdf

at page 32


Consider 64 objects, each with a different weight. What is the minimum number of pair wise weight comparisons needed to determine the heaviest and 2nd heaviest objects?

[hide]
There are 64 objects, $ a_1,a_2...a_{64}$
First select an object at random,  for instance $ a_1$, and weight it with the remaining $ 63$ objects
From that you can determine the the two heviest pairwise weights - suppose it's $ a_1 \plus{} a_2$ and $ a_1 \plus{} a_3$.
Now you have to weigh $ a_2 \plus{} a_3$ to check whether $ a_1 < \{a_2,a_3\}$

So you need $ 64$ pariwise weight comparisons [/hide]


I failed to make USAPhO even though I got 1a, 2,3,4 right(no joke) on the semifinalist. However, my solutions got there late probably due to the delay of a snowstorm. My coach went to post office on Tuesday and mailed the solution to the AAPT. The mail should arrive by Thursday, but the solution is delivered on Saturday. According to what my coach told me about the phone conversation between Annette Coleman and him, AAPT checked the mailbox on Friday and Monday, and therefore AAPT got my solution packet on Monday instead. Because all the physicists left on Sunday, nobody was available to grade my paper, along with the papers from four other schools.  :mad: Now .. great.. I can't even take the semifinal despite doing considerably better than my friends who made the semis.

I think AAPT should use the system MAA use during USAMO. AAPT should let coaches to fax the solutions in so that there will be NO delay in the delivery...

Help.. What should I do? I don't think I can take the semifinal, at least the chance is very very slim. But any advices are appreciated..  :maybe:

I think you are making too big of a deal about this.  So you missed USAPhO, it's not the end of the world.  There isn't really anything you can do except hope that they read your solutions and accept you.  But yes, perhaps they should reform the solution submission system (say by fax as you suggested, or possibly an online test).

JRav, keep in mind that for some students, physics is their one big thing that they've prepared for all year, and it's a really big deal. Think about how a former MOPer would feel if he or she got disqualified on the AIME because of lost mail. But, yes, you're right; there's probably nothing that can be done at this point (the semis have already begun and will end on the 19th).

Allan Z, I would recommend that your teacher use at least one-day mail or even next-day FedEx. I don't know how that would be affected by a snowstorm, but at least there are [i]some[/i] guarantees. In the case of a snowstorm, I doubt FedEx would delay more than one or two days past the "one-day" deadline.

This seems pretty unfair.  Reminds me of a time when my mom got a late charge on a bill due on Sunday (why??!)  because her check arrived on Sat and they didn't open it until Monday.

It would probably be worth making calls to see if they will grade your test (and you should make it then) and you might be able to take the semi before the window closes.  You have nothing to lose by calling, I think.

[quote="Allan Z"]Thank you all for your suggestions. I guess I am going to call AAPT and ask them whether I can take the test provisionally and let them grade my quarterfinal later. If I don't qualify for the semifinal, AAPT can just discard my results.[/quote]

Allan, are they letting you do this? Because my physics teacher told me yesterday that AAPT sent him semifinal tests for my school to take, but he didnt really tell me anything else. Neither me nor the other kid at my school who took the quarterfinals test are on the AAPT website's list of semifinalists, and i was confused as to what is going on.


Can anyone please help? Can anyone either post or pm to me the answer key for this test, or does anyone know where I can find online without downloading potentially risky files? Thanks.

Unless I'm mistaken, don't you get a score report with an answer key provided? Or are you just trying to figure out your score before you get it?

I would just be patient and wait for the PSAT score to come out. And I doubt that anybody would have the answer key, let alone upload it, since it hasn't even been sent out yet.

Gah, I know, PSAT takes forever to come out and builds way too much suspense  :(

Does anyone know when the score reports will come out?

Last year, (in my county), they came out during the second week of January, if I remember correctly.

Most nerve-wracking wait ever up to that point. (College apps are more nerve-wracking...  :maybe: )

well, we already got ours, but I'm not sure when they usually come out.

Already got yours? That means other places will get score reports soon, too?

If it's too much for one person to take the time to post all the answers, could you maybe post them from just one section - list the section, and the correct answer choices - ie: Section 1) 1a,2c, 3e,... etc. and maybe someone else could kindly post the answers to another section? Thank you for anyone who can post or pm me if you'd prefer.

[quote="Stokes93"]I'm not sure... we haven't gotten them officially handed out yet, but my guidance counselor looked at my scores for me - so maybe your school already has them, but hasn't handed them out?[/quote]

Possibly. Does CollegeBoard send out all score reports at once? If so, I might just ask my guidance counselor if I can see my score.

[quote="buzzer11"]Possibly. Does CollegeBoard send out all score reports at once? If so, I might just ask my guidance counselor if I can see my score.[/quote]

I'd like to think so, but I recall that people in the same county as me received their official score reports earlier than I did.

I actually got my score report two weeks ago. Did you take the Wednesday test or the Saturday test? I have the answers for the Wednesday test. Also, if you head over to college confidential you'll undoubtedly find some answers somewhere.

I took the Wednesday test but unfortunately do not remember most of the questions/answers... I guess I'll have to wait.
But over two weeks difference, that's weird.

I heard from someone about PSAT scores today. I don't think score reports were released though.

Score reports have not be released? How about the people saying they already have theirs?

Well, there was a senior complaining this morning about a junior who supposedly earned a 237 on his PSAT, according to his guidance counselor. I don't think the score reports were distributed, because our school schedule would need to be revamped in order to get score reports distributed to the ~1200 students who took it and no such disturbance took place. It's much like the actual SAT: you get your scores but you don't get your actual score report for a few more weeks.

As for people having their score reports... maybe it is nonuniform across the nation.

Oh, you meant in your school specifically. I thought you meant in general.  :blush:
So I guess counselors have them now. Maybe I'll just wait it out :)

I think our school obtained score reports, although they tried to keep it quiet. I talked to a junior who got a 223 and was really mad because he messed up some easy math problem.

I hope the minimum score for MD remains constant so he becomes a semifinalist... then our school would have at least two semifinalists.  :)

got score report today... and got 211 and i'm in 8th gd.

By now schools should have already recieved the score reports. If you haven't gotten yours yet, it means that the counselors haven't released them for whatever reason (I say laziness, I know mine haven't given them yet).

My counselor said we would get them in January. D:

Some of my friends have gotten them in the mail. :)
that means mine should be coming soon?

240, well done.
I'm not awesome at the writing section which keeps me from getting perfect.
did you send your RSI app yet? relatedly what did u put as your second research interest after math? they can't be two diff fields of math right?

haha, thanks.
I haven't really thought too much about the RSI app... I really should start finishing it up though. I don't think I'll make it in, so that hasn't been at the top of my priorities list. :)
I don't think they can be two different fields of math, but I'm not sure... what are you thinking about putting as your research interests?

No they cant be two fields of math :(
math then economics.
yea im not going to make it either idt i will even apply its hopeless unless you know like algebraic topology or won the olympiad.

Don't be so pessimistic!
But then again, can you say a true statement and be pessimistic at the same time? :P

238 - I'm in 9th grade.

I made a stupid mistake in the reading section.

Why was this year's PSAT writing curve so brutal? 220-Junior, evidently missed NJ merit scholarship cutoff by 1 point. Sigh.

Also, does anyone here know when the December SAT II "official" score reports will be mailed home? It would help if I could actually put my SAT II scores on that RSI application...

Darjn. You guys are beasts. Stop talking down yourselves like this which only makes us feel inferior.

227, 11th grade. I was halfway through bubbling the second-to-last question in writing when time was called, so I thought two questions would be omitted, but I actually got the question correct.

I don't think colleges really care about your exact PSAT score as long as it's above 220-225, so everyone on here has done really well.

more to the point, colleges don't SEE your PSAT score. They only know if you get semifinalist-- so anything above 220 will suffice.


The area of a circle inside of a triangle. the answer is defined as the largest possible circle that fits entirely within the triangle formed by those three points

It should be noted that Point A and Point B are 240 kilometres apart, Point B and Point C are 330 kilometres apart, and Point A and Point C are 270 kilometres apart. The Maraquan Circle, according to legend, is defined as the largest possible circle that fits entirely within the triangle formed by those three islands.

What is the area of the Circle, in square kilometres?

[hide="Hint"] $A=rs$, $A$ is the area (from Heron), $r$ is inradius, $s$ is semiperimeter. [/hide]


A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits ($ \spadesuit$, $ \heartsuit$, $ \diamondsuit$, and $ \clubsuit$), such that there is exactly one card for any given rank and suit.  Two of the suits ($ \spadesuit$ and $ \clubsuit$) are black and the other two suits ($ \heartsuit$ and $ \diamondsuit$) are red.  The deck is randomly arranged. What is the probability that the top card is a face card (a Jack, Queen, or King)?

Each of the $ 13$ ranks are equally likely to be the top card; thus the probability is $ \boxed{\frac3{13}}$.

Why in the world does it say so much about the suits and colors?

Some people may not be familiarized with a normal deck of cards.

It's not that hard if you read the problem. :)

[hide=Smartgoodness' submission]Since there are $3$ face cards (Jack, Queen, and King) and $4$ suits for each one ($\diamondsuit~\heartsuit~\clubsuit~\spadesuit$), there are $4\cdot 3=12$ face cards. That means the probability of randomly choosing a face card is $\frac{12}{52}=\boxed{tfrac{3}{13}}$.

@bove you forgot the backslash, it should be [hide=this.]
$\frac{12}{52}=\boxed{\tfrac{3}{13}}$
[/hide]


[quote="chien than"]Find $x$ such that:
$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}=\frac{x-1}{x}$[/quote]
Let $y^{2}=\frac{x-1}{x}$. If $-1\le x<0$, then $y>1$ and $\sqrt{x+1}=1+y$ contradition. Therefore $x\ge 1,0\le y<1$.
One solution $y=0$ or $x=1$. For other solutions $y^{4}+2y^{3}-y^{2}-2y+1=0.$ It give 2 roots $\frac{1}{2}<y_{1}<y_{2}<1$ or $x_{i}=\frac{1}{1-y_{i}^{2}}.$

$\sqrt{x-\frac{1}{x}}$ impliks $x \in [-1,0) \cup [1,\infty)$
$\sqrt{1-\frac{1}{x}}$ impliks $x \in (-\infty,1]$
so the set of solution is $[-1,0) \cup{1}$.
so the $RHS$ is posif impliks that $\sqrt{x-\frac{1}{x}}\geq\sqrt{1-\frac{1}{x}}$ wich givs that $x \geq 1$ so the only solution is $x=1$.

as u said it .[quote]One solution $y=0$ or $x=1$[/quote]

A hint : Put $\frac{1}{x}=\cos\phi$.
[quote="chien than"]Find $x$ such that:
$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}=\frac{x-1}{x}$[/quote]

[hide]
\[\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}=\frac{x-1}{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}\]
So either $x=1$ or $\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}=x$.

In the second case, adding to the original equation gives
\[2\sqrt{x-\frac{1}{x}}=\frac{x-1}{x}+x=\left( x-\frac{1}{x}\right)+1\]
Letting $y=\sqrt{x-\frac{1}{x}}$, we have $2y=y^{2}+1$ so $y=1$. This results in $x^{2}-x-1=0$, so $x=\frac{1+\sqrt{5}}{2}$ (this works in the original equation, the negative one clearly doesn't satisfy the second-case equation) in this case.

Therefore, the solutions are $\boxed{1,\frac{1+\sqrt{5}}{2}}$.
[/hide]

A similar problem: Solve equation
 $2x+\frac{x-1}{x}-\sqrt{1-\frac{1}{x}}-3\sqrt{x-\frac{1}{x}}=0$

rezult: $x=\frac{1+\sqrt{5}}{2}$


trigonometry
function
ARML
geometry
3D geometry
algebra
polynomial

From the 2001 Georgia ARML tryout: help, please!

12. What is the angle formed by the diagonals of a cube?

a. arcsin 2/3    b. arccos 2/3   c. arctan 1/3   d. arcsin 1/3   e. arccos 1/3

23. A certiain rational function r(x)= f(x)/g(x), where f is a polynomial of degree 2 and g is a polynomial of degree 3. Which of the following statements most completely describes the graph of r?

a. r has the x axis as a horizontal asymptote.
b. r has a horizontal asymptote and at least one vertical asymptote.
c. r has at least one zero.
d. r has 2 zeros and 3 vertical asymptotes.
e. r has one oblique asymptote and at least one vertical asymptote.

26. (x-6)^2 + (y-7)^2 is less than or equal to 25. The objective function (i.e. profit) is 3x+4y. What point of the feasible region maximizes the objective function?

a. (9,11)   b. (10,4)   c. (2,10)   d. (3,3)   e. (10,10)

21. Two miles upstream from his starting point, a canoeist passed a log floating in the river's current. After paddling upstream for one more hour, he paddled back and reached his starting point just as the log arrived. Find the speed of the current.

a. 1/2 mph   b. 1 mph    c. 1 1/2 mph    d. 2 mph    e. 2 1/2 mph

[hide="23"]As $x\rightarrow\infty$, the function $r\left(x\right)\rightarrow 0$, so you have a horizontal asymptote.  Setting $r\left(x\right)=0$ is equivalent to setting $f\left(x\right)=0$.  $f$ isn't guaranteed to have a zero so neither is $r$ (although it is possible).  This narrows it down to choices $\left(a\right)$ and $\left(b\right)$.  Since $g\left(x\right)$ is of degree $3$ with real coefficients, it will be equal to $0$ for some $x$.  This means there will be a vertical asymptote, so your answer is $\boxed{\left(b\right)}$.[/hide]

[hide="26"]You could do it the long way, but since you are given numbers you might as well just check them by hand.  The first choice $\left(a\right)$ maximizes the profit function and it is certainly within the constraint (actually, it is a point on the perimeter of the disk).[/hide]

[hide="21"]Let the rate at which he paddles be $r$ and the rate of the current be $x$.  Then you have
\[ d=r-x  \]
\[ t=\frac{2}{x}=1+\frac{d+2}{r+x}\implies 2r=2xr\implies x=1\implies\boxed{\left(b\right)}.  \]
Note:  I set this one up really fast so I may have made a mistake in the expansion and collection of terms.  You might want to double-check using the equations I gave.[/hide]

[hide="12"]
Consider the triangle formed by two space diagonals and one side of the cube.
The ratio of side lengths is $1: \frac{\sqrt{3}}{2} : \frac{\sqrt{3}}{2}$.

By Law of Cosines,
$\cos{\theta}=\frac{\left(\frac{\sqrt{3}}{2}\right)^2 +\left(\frac{\sqrt{3}}{2}\right)^2 -1^2}{2\left(\frac{\sqrt{3}}{2}\right)^2} =\frac{1}{3}$

Hence the angle is $\text{arccos}\frac{1}{3}$.
[/hide]


Consider $ \triangle ABC$ and points $ M \in (AB)$, $ N \in (BC)$, $ P \in (CA)$, $ R \in (MN)$, $ S \in (NP)$, $ T \in (PM)$ such that $ \frac {AM}{MB} \equal{} \frac {BN}{NC} \equal{} \frac {CP}{PA} \equal{} k$ and $ \frac {MR}{RN} \equal{} \frac {NS}{SP} \equal{} \frac {PT}{TN} \equal{} 1 \minus{} k$ for some $ k \in (0, 1)$. Prove that $ \triangle STR \sim \triangle ABC$ and, furthermore, determine $ k$ for which the minimum of $ [STR]$ is attained.

[hide="Hint"]I think $ \frac {MR}{RN} \equal{} \frac {NS}{SP} \equal{} \frac {PT}{TM} \equal{} \frac {1}{k}$
We have:
$ \overrightarrow{RS} \equal{} \overrightarrow{RN} \plus{} \overrightarrow{NS} \equal{} \frac {k}{k \plus{} 1}\overrightarrow{MN} \plus{} \frac {1}{k \plus{} 1}\overrightarrow{NP}$
$ \equal{} \frac {k}{k \plus{} 1}\overrightarrow{MB} \plus{} \frac {k}{k \plus{} 1}\overrightarrow{BN} \plus{} \frac {1}{k \plus{} 1}\overrightarrow{NC} \plus{} \frac {1}{k \plus{} 1}\overrightarrow{CP}$
$ \equal{} \frac {k}{(k \plus{} 1)^2}\overrightarrow{AB} \plus{} \frac {k^2}{(k \plus{} 1)^2}\overrightarrow{BC} \plus{} \frac {1}{(k \plus{} 1)^2}\overrightarrow{BC} \plus{} \frac {k}{(k \plus{} 1)^2}\overrightarrow{CA}$
$ \equal{} \frac { k^2 \minus{} k \plus{} 1}{(k \plus{} 1)^2}\overrightarrow{BC}$
Similarly then OK!
Minimum when $ k \equal{} 1$
Sorry if I have a mistake![/hide]

Well, PhilAndrew, can you post the rest of Romanian District 2007?

(That is what I requested on http://www.mathlinks.ro/viewtopic.php?t=181940, but no response so far)


that's correct! you could've also done something like this:3996*2000/2, where 3996 is the sum of the first and last terms and 2000 is the number of terms.


Let p/q = 1 - 1/2 + 1/3 - 1/4 + ... - 1/2k + 1/(2k+1)
If (p,q)=1 is it possible to determine k so that p is divisible by 1997?


The number of diagonals that can be drawn in a polygon of 100 sides is:

$\textbf{(A)}\ 4850 \qquad
\textbf{(B)}\ 4950\qquad
\textbf{(C)}\ 9900 \qquad
\textbf{(D)}\ 98 \qquad
\textbf{(E)}\ 8800$

[hide]$ 100$ sides are formed from $ 100$ vertexes.  Assuming this is a convex polygon, there are $ \binom{100}{2}\equal{}4950$ to choose two vertexes to form line segments (diagonals and sides of the polygon).  Of these, there are $ 100$ sides.  The answer is $ 4850$, A.[/hide]

Isn't the answer 4850. n(n-3)/2 is the number of diagonals.

Yes. Using $ \frac {n(n \minus{} 3)}{2}$ is equivalent to what azn6021023 did.

[hide="Because..."]
Given $ n$ vertices we choose 2 of them to form a line, then subtract away the sides which we don't need.

$ \binom{n}{2} \minus{} n \equal{} \frac {n(n \minus{} 1) \minus{} 2n}{2} \equal{} \frac {n(n \minus{} 3)}{2}$.
[/hide]

[quote="grn_trtle"]$ \binom{n}{2} \minus{} n \equal{} \frac {n(n \minus{} 1) \minus{} 2}{2} \equal{} \frac {n(n \minus{} 3)}{2}$[/quote]
You meant $ \binom{n}{2} \minus{} n \equal{} \frac {n(n \minus{} 1) \minus{} 2n}{2} \equal{} \frac {n(n \minus{} 3)}{2}$.


abstract algebra
algebra proposed
algebra

Let $ z \in \mathbb{C}$ such that for all $ k \in \overline{1, 3}$, $ |z^k \plus{} 1| \le 1$. Prove that $ z \equal{} 0$.

Assume $ z$ has a positive norm. Let $ \theta\equal{}\arg (z)$

If $ \minus{}\frac{\pi}{2}\le \theta\le \frac{\pi}{2}$, let $ x\equal{}z$.
If $ \frac{3\pi}{4}\le theta le \frac{5\pi}{4}$, let $ x\equal{}z^2$.
Otherwise, let $ x\equal{}z^3$.

In each case $ \minus{}\frac{\pi}{2}\le \arg(x) \le \frac{\pi}{2}$ and by our assumption $ x$ has a positve norm, but in which case $ |x\plus{}1|>1$. Contradiction

Somebody has another solution to this problem?
Please, post it .
Thank you


Hi there--I have a copy of the AMC12B from 2002 with the multiple choice blanked that I created in order to make a greater challenge for my students.

However, unfortunately problem 4 doesn't make any sense without the answer choices left in.  (It's one of those "Which of the following statements is not true?" questions.)

My master copy of the questions is back on the AMC CD-ROM at my house--if anyone has a copy of the test and would be willing to type in the choices for answers on # 4 of the 12B for 2002, we'd be grateful!

Thanks!

No need to be concerned about this any longer, thanks.  I found a copy with the choices still in.


This inequality is much easier than previous but it's quite nice I suppose.

For positive reals a,b,c where $a+b+c=1$ prove:

$\frac{(a+b)(b+c)(c+a)}{\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}} < max(a,b,c)$

Well, at least it's pretty easy...

Since $\frac{a+b+c}{3} \leq max(a,b,c)$, it suffices to prove

$9(a+b)(b+c)(c+a) < (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = (a+b+c)^4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$, but

$(a+b+c)^4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) > (a+b)(b+c)(c+a)(a+b+c)\left(\frac{ab+bc+ca}{abc}\right)$

$=(a+b)(b+c)(c+a)\left(\frac{(a+b+c)(ab+bc+ca)}{abc}\right) \geq 9(a+b)(b+c)(c+a)$,

where the last inequality follows from AM-GM.


In how many ways can the word MATHCOUNTS be arranged if the arrangement has 6 letters but no more than 2 vowels?

[hide="Note:"]I do not have an answer, but maybe the person who made this mock mathcounts test will recognize this problem and tell us the answer.  :D [/hide]

ARGH. This problem REALLY bugs me. I'm gonna post my answer, no solution, because I doubt it's correct.
Click the smiley below! :)
[hide=":juggle:"]$9000$?[/hide]

[hide="Note:"]maybe the person who made this mock mathcounts test will recognize this problem and tell us the answer.  :D [/hide][/quote]*cough**cough*
[hide="solution"]
First find out how many subsets of MATHCOUNTS you can make.
Case 1: 1 T
We are leaving out 1 T, so there are $_9P_6=3024$ ways to arrange them
Case 2: 2 Ts:
We already know that there are 2 Ts.
Thus we can choose the other 4 letters is $\binom 84=70$ ways.
We can arrange them in $6!/2=360$
Thus there are 25200 ways with 2Ts
Case 3: no Ts
Since we are leaving out the two Ts, we have $_8P_6=336$ ways to choose and arrange the other letters.
That is a total of 28560 ways.
Now the only non working arrangements are those with 3 vowels.
For 1 T, there are 5 spots left and $\binom 53=10$ ways to choose spots for the vowels and 6 ways to arrange them. Thus there are 60 ways.
For 2 Ts, there are 4 more spots for letters and there are $\binom 43=4$ ways to choose spots for the 3 vowels and 6 ways to arrange them. Thus there are 24 ways to have the 3 vowels.
For 0 Ts, there are 6 more spots so $\binom 63=20$ ways to determine the spots of the three vowels and 6 ways to arrange them. Thus there are 720 ways with 0Ts.
Summing the non working ways, we get $720+24+60=804$
Subtracting from the total number, we get 27756, the answer[/hide]

See, I knew he would recognize it.  :lol:


How do you export a file from texnic center to something less ugly? I have miktex already downloaded, except I somehow could not get the file to load onto miktex, so i only have the raw code.

I don't undertsand your problem. What do you mean by export? What does file load onto MikTeX mean?
Have you read [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Basics]LaTeX:Basics[/url] to see how to write and compile a file in TexnicCenter to produce a pdf?

Thanks for the link. I never actually read it.


The sum of two numbers is 32, and their product is 96. What is the sum of the reciprocals of the two numbers? Express your answer as a common fraction.

Let the two numbers be $ x$ and $ y$. $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{x\plus{}y}{xy}\equal{}\frac{32}{96}\equal{}\boxed{\frac{1}{3}}$


inequalities
trigonometry
inequalities proposed

1)Let $ a_1,a_2,...,a_N$ be positive real numbers then :
           $ \sum_{n = 0}^{N}\sum_{n = 0}^{N}{\frac {a_ma_n}{m + n + 1}} \leq \pi \sum_{n = 0}^{N}{a_{n}^{2}}$
 Here $ N \in \mathbb{Z_{ + }}$
 2)Let $ a_1,a_2,...,a_N$ be  real numbers such that  : $ 0 < \sum_{k = 0}^{ + \infty}{a_k^{2}} < + \infty$ If $ 0 \leq \alpha < 1$
Then :
                $ \sum_{p,q = 0,(i + j)  even}^{ + \infty}{\frac {a_pa_q}{p + q + 1 - \alpha}} \leq \frac {\pi}{2cos(\frac {\pi\alpha}{2})} \sum_{k = 0}^{ + \infty}{a_{k}^{2}}$

[b]the more strict Hardy-Hilbert inequality[/b]:For $ p,q>1$ such that $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}1$,$ a_n,b_n>0$ and $ \sum_{n\equal{}1}^{\plus{}\infty}{a_n^p}<\infty,\sum_{n\equal{}1}^{\plus{}\infty}{b_n^p}<\infty$,then
\[ \sum_{n\equal{}0}^{\infty}{\sum_{m\equal{}0}^{\infty}{\frac{a_mb_n}{m\plus{}n\plus{}1}}}<\frac{\pi}{\sin{\frac{\pi}{p}}}\left(\sum_{n\equal{}0}^{\infty}{a_n^p}\right)^{\frac{1}{p}}\left(\sum_{n\equal{}0}^{\infty}{b_n^q}\right)^{\frac{1}{q}}\]


Prove that the number of odd coefficients in the expansion of $ (1\plus{}x)^n$ is a power of $ 2$ for any $ n\in \mathbb{N}$.

If $ n\equal{}2^{k_1}\plus{}2^{k_2}\plus{}...\plus{}2^{k_m},0\le k_1<k_2<...<k_m$, then $ C_n^k$ odd if and only if $ k\equal{}\sum_i \theta_i2^{k_i}$, were $ \theta_i\equal{}0 \ or \ 1$. Number of suth k is $ 2^m$.


A lattice point is a point in the plane with only integer coordinates. How many non-negative coordinates like on the graph of 13x+15y=273?

If you rearrange it to get 15y=13(21 - x) then, since 13 has no factors in common with 15, it must mean that (21 - x) is a multiple of 15. This can only happen for non-negative x,y if x = 6 or 21. This means that the two solutions are (6,13), (21,0)!


A quadrilateral is inscribed in a circle. If an angle is inscribed into each of the four segments outside the quadrilateral, the sum of these four angles, expressed in degrees, is:

$ \textbf{(A)}\ 1080\qquad
\textbf{(B)}\ 900\qquad
\textbf{(C)}\ 720\qquad
\textbf{(D)}\ 540\qquad
\textbf{(E)}\ 360$

Let's call the quadrilateral $ ABCD$.  Drawing in all of the angles specified in the problem, we form an octagon inscribed in the circle and we're looking for the sum of the measures of one of the sets of alternating angles.  All angles of the octagon are inscribed angles and therefore equal to half the measure of their intercepted arcs.  Adding all these angles, we get

$ \frac{1}{2}((360\minus{}arc AB)\plus{}(360\minus{}arc BC)\plus{}(360\minus{}arc CD)\plus{}(360\minus{}arc DA))$

and simplifying, we get

$ 720\minus{}\frac{1}{2}(arcAB\plus{}arcBC\plus{}arcCD\plus{}arcDE)$.

Since the arcs add up to a complete circle and therefore 360 degrees, the answer is (D) 540.


f: {1,2,...,50} ---> {1,2,...,80}

are they? that simultaneously comply with:

f(1) is even

f(2)=4

f(3)=6

My solution: I think that given only the first condition (f(1) is even) then there are 40, but taking into account the second there are 39, and finally the third 38. Is it correct? Thanks.

How can a function give $ 80$ different outputs given only $ 50$ inputs?

[b]@i_like_pie:[/b] An injective function does not require every element of the "into" set to have a corresponding argument in the "from" set, but the opposite must be true.

Each value of f for arguments ranging from 1 to 50 must be distinct.

The second and third restrictions simply remove two elements from the "into" set, leaving 78 to deal with for 48 inputs.

We had 40 even numbers before, but with 2 taken away, there are 38 even numbers left to pick for f(1).

From there, there are 77 remaining numbers to choose for f(4), then 76 for f(5), etc., 31 choices for f(50).

Hence, the number of functions $ f$ is equal to:

$ 38 \cdot 1 \cdot 1 \cdot 77 \cdot 76 \cdot \ldots \cdot 32 \cdot 31 \equal{} 38 \frac {77!}{30!}$

[color=red]Edit: Fixed! Thanks[/color]

I was way off!!! This is a combinatory problem...

One question: did you mistakenly write that 38 instead of a 31? (the one between 32 and the equal sign)

Yes, I believe that is a typo; it should be $ 31$ instead of $ 38$.


Let $x$ be a prime number such that $x + 1$ is a perfect square.Find the sum of all such prime numbers... :?
[hide]
{As far as I'm concerned , I can find only one prime : 3 that seems to fit the category....Or Am I missing something ??  :? }[/hide]

You should notice that $x+1=k^2 \rightarrow x=k^2-1 \rightarrow x=(k-1)\cdot (k+1) $ But what do you know about x ?

[hide]You're not missing anyhing karthik_k. $x$ can indeed only be $3$
Here's why:
Like Anto told us, $x$ is given by:
\[x = (k-1).(k+1)\]
Since we know that x is a prime number, and we know that k is a natural number, $(k-1)$ or $(k+1)$ must be one, else would x be divideable by 2 integers and thus not be a prime. If we say that $(k+1) = 1$, then $k$ must be $0$. This solution is not acceptable, because $k^{2}$ would not be a perfect square.
If we say that $(k-1)=1$, then $k = 2$, which is indeed the only solution.
So $x=3$[/hide]

there is only one solution karthik_k
because from anto's method x=(k-1)(k+1)
now x=1*x so k has to be 2

[quote="riddler"] k has to be 2[/quote]\
don't you mean 3?
3+1=4
and 4 is a square

$k$ was a different variable we used to help prove that $3$ was the only answer for $x$.  $k=2$, but $x=3$.

[quote="karthik_k"]But is there any mathematical way of proving that there exists no  natural number $x$ such that $x \neq 3 $ and $x + 1$ is a perfect square ?[/quote]

I think Jan's solution pretty much proves it...

[quote="Jan"]This solution is not acceptable, because $k^{2}$ would not be a perfect square.[/quote]
although $k=0$ is not acceptable, but the above statement is self-contradiction, how could $k^2$ not be a perfect square in which $k^2$ itself is a perfect square? my explanation is if $k=0$, $x=0$ and 0 is not a prime number. done.

[quote]Jan wrote:
This solution is not acceptable, because  would not be a perfect square.

although  is not acceptable, but the above statement is self-contradiction, how could  not be a perfect square in which  itself is a perfect square? my explanation is if ,  and 0 is not a prime number. done.[/quote]

That's not true: If $k = 0$, then $x = -1$ because $x = (k-1).(k+1), \text{so we see that:}$ \[x=-1.1=-1\]

But still, you have a point when you say that sentence wasn't 100% correct. Better would indeed be: k=0 is not an acceptable solution, because x=-1 then, and -1 is not a prime number.
i'm wondering... Is 0 a perfect square? or not?

why not? 0 is a perfect square because $0^2=0$; and yea, i was thinking of $x+1$ that led me to $x=0$  :blush:

[quote="tarquin"]I thought perfect squares were only positive integers ..[/quote]

Yeah I think that's generally accepted, on contests at least.

I think Jan's solution shows that there is only one "unique" value of $x$ satisfying the problem. And since it showed that there is only one, it also proves that there is none other than the "unique" value (obviously).


geometry
geometric transformation
dilation
reflection
trapezoid
circumcircle

The Simson Line of a point P with respect to triangle $ABC$ meets BC at L and the altitude AD at K. If H is the orthocentre of triangle $ABC$, prove that PK is parallel to LH.

[hide]
We will prove the useful fact that that the Simson line passes through the midpoint of $\overline{PH}$. [hide="if you treat this as well known, you can just skip past it"]To start, note that a dilation centered at $P$ with factor 2 sends this midpoint to $H$ and the Simson line to a parallel one. Let $P_{2}$ be the image of $L$. Then $P_{2}$ lies on the new line. Showing that $H$ is on the new line is equivalent to proving $\overline{HP_{2}}$ parallel to the Simson line.

Now observe that $P_{2}$ is also the reflection of $P$ over $\overline{BC}$. Let $H_{2}$ be the reflection of $H$ over $\overline{BC}$.Then $PH_{2}HP_{2}$ is an isosceles trapezoid due to symmetry.

On the other hand, if we extend $\overline{PL}$ to meet the circumcircle of $\triangle ABC$ again at $Q$, then $PH_{2}AQ$ is cyclic (some angle chasing reveals $\angle BHC=180^\circ-A$, and it's pretty well-known anyway) and $\overline{AH_{2}}\parallel \overline{PQ}$, so $PH_{2}AQ$ is [b]another[/b] isosceles trapezoid with $\overline{PH_{2}}$ as a leg. This gives us $\overline{AQ}\parallel \overline{HP_{2}}$.

It remains to show $\overline{AQ}$ parallel to Simson line. Let $M$ be the intersection of the Simson line with $\overline{AB}$. Since $AQBP$ is cyclic, $\angle AQP=\angle ABP$.

But also $BPLM$ is cyclic from $\angle BMP=\angle BLP=90^\circ$, so $\angle MLQ=180^\circ-\angle MLP=\angle ABP$. Finally, $\angle AQP=\angle MLQ$, so $\overline{AQ}\parallel \overline{LM}$, as desired.

[/hide]
So letting $X$ be midpoint of $\overline{PH}$, we have $\triangle XPL\cong \triangle XHK$ by ASA congruence. This gives $KH=PL$, so that $KHLP$ is a parallelogram. It follows that $\overline{PK}\parallel \overline{LH}$.
[/hide]


Find the point on the line y= 2x-4 that is closest to the point (1,3)

[hide]A line perpendicular to the line will give us the shortest length.
the slope will be -.5 so y=b-x/2.  Inputing the point (1,3) we find that b=3.5.
To find the line where they intersect we set the two lines equal to each other so 2x-4=(7-x)/2.
Solving for x we get x=3 and inputting we find y=2.  Therefore the point is (3,2).
On a side note the distance between (1,3) and (3,2) is sqrt(5)
[/hide]

formula:
Line: $ax + by + c = 0$
Points: $(x, y)$

Distance: $\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$ (sub in the x-y points in the numerator)

$\frac 5 {\sqrt 5} = \sqrt 5$

If you know the distance is $\sqrt5$ can you find the point (easily)?

well i was thinking just making another equation for the line and using the perpendicular slope. then plugging in (1,3) and solving for the answer. simple algebra, no big equations

GIven what the problem is asking, bobthebuilder, I think your way is more straightforward.


[quote="post233"][quote="Rust"][quote="Gauss_87"]Find all couples $(x,y)$ with $x,y \in Z$ for which:
$x+y$, $x-y$, $x \cdot y$ is a perferct square.[/quote]
Let x+y=n^2,x-y=m^2,xy=(k/2)^2. We have n^4=k^2+m^4. Ferma prove, that this equation has not solution.[/quote]
But Fermat's theorem does not only prove that $n^4=k^4+m^4$ has no positive solutions?[/quote]
Ferma proved that equations x^4+y^4=z^2, and x^4+y^2=z^4 have not solutions.

[quote="Rust"][/quote][quote="post233"][quote="Rust"][quote="Gauss_87"]Find all couples $(x,y)$ with $x,y \in Z$ for which:
$x+y$, $x-y$, $x \cdot y$ is a perferct square.[/quote]
Let x+y=n^2,x-y=m^2,xy=(k/2)^2. We have n^4=k^2+m^4. Ferma prove, that this equation has not solution.[/quote]
But Fermat's theorem does not only prove that $n^4=k^4+m^4$ has no positive solutions?[/quote][quote="Rust"]
Ferma proved that equations x^4+y^4=z^2, and x^4+y^2=z^4 have not solutions.[/quote]
Could you give the solution for this theorem?

[quote="Lovasz"][/quote][quote="Rust"][/quote][quote="post233"][quote="Rust"][quote="Gauss_87"]Find all couples $(x,y)$ with $x,y \in Z$ for which:
$x+y$, $x-y$, $x \cdot y$ is a perferct square.[/quote]
Let x+y=n^2,x-y=m^2,xy=(k/2)^2. We have n^4=k^2+m^4. Ferma prove, that this equation has not solution.[/quote]
But Fermat's theorem does not only prove that $n^4=k^4+m^4$ has no positive solutions?[/quote][quote="Rust"]
Ferma proved that equations x^4+y^4=z^2, and x^4+y^2=z^4 have not solutions.[/quote][quote="Lovasz"]
Could you give the solution for this theorem?[/quote]
Exactly Ferma proved if integers x,y,z satisfyed condition
x^4+y^2=z^4  or  x^4+y^4=z^2
then xy=0.

[quote="Rust"][quote="Gauss_87"]Find all couples $(x,y)$ with $x,y \in Z$ for which:
$x+y$, $x-y$, $x \cdot y$ is a perferct square.[/quote]
Let x+y=n^2,x-y=m^2,xy=(k/2)^2. We have n^4=k^2+m^4. Ferma prove, that this equation has not solution.[/quote]

What are you saying?
$(2k^2,2k^2) \forall k \in Z^+$ is a solution; $(0,n^2),(n^2,0)$ is solution!

I have a simple proof of these but i have not yet a proof they are the unique (if they're)..

I find solution, were x+y,x-y, xy natural (x is not equal y).

Ok but the problem didn't say: $x \neq y$ or $x,y \neq 0$..
But in this way we have solved it together, because it's proved there're NO other solution!  :rotfl:
Thanks!

[quote="Rust"][quote="Gauss_87"]Find all couples $(x,y)$ with $x,y \in Z$ for which:
$x+y$, $x-y$, $x \cdot y$ is a perferct square.[/quote]
Let x+y=n^2,x-y=m^2,xy=(k/2)^2. We have n^4=k^2+m^4. Ferma prove, that this equation has not solution.[/quote]
I think we can solve equation $n^4=k^2+m^4$ by pifagorian substitution.


[img]http://www.nxbgd.com.vn/toanhoctuoitre/tapchi/Images/200712095527_d.JPG[/img]

[img]http://www.nxbgd.com.vn/toanhoctuoitre/tapchi/Images/200712095538_d3.JPG[/img]

who can answer this excercise , please send me to email
anhtuan_kt01@yahoo.com
thank!
[/img]

Please lock this topic because these problems are in use in the Math competition of Mathematical Youth  Magazine.
Please don't send the solutions to [b]anhtuan_kt01[/b].
And he must be to blame for his wrong attitude.


function
integration
trigonometry
calculus
calculus computations

By direct application of Beta function to $ \int_0^{\frac{\pi }
{2}} {\sin ^{2k \plus{} 1} (x)\,dx}$ yields the result $ \frac{{\sqrt \pi   \cdot k!}}
{{2\left( {k \plus{} \cfrac{1}
{2}} \right)!}}.$ Can we turn this last result into an elegant form?

In a nicer closed form, unfortunately no. But, if you want a non closed form, easy to use in computer computations, then you could use:

$ \frac {\pi k!}{2\left(k \plus{} \frac {1}{2}\right)!} \equal{} \sum^k_{i \equal{} 0}\frac {( \minus{} 1)^i}{2i \plus{} 1}\binom{k}{i}$

what does it mean : $ (n\plus{}\frac{1}{2})!$  something like $ \Gamma (n\plus{}\frac{3}{2})$ :maybe:


Let ABC be a triangle where AB = AC and let D be a point on segment AC so that AD = BC. Prove that angle(B) = 3 angle(A) iff BD^2 = 2BC^2.

This equivalent to

sin 2(3 \pi /7) + sin 2(\pi /7) - 2cos(\pi /7)sin(\pi /7)sin(3 \pi /7) = 2sin2(\pi /7)

that easy to be checked.

May be. But what about a pure geometric solution?

I) angle(B) = 3 angle(A) ==>  BD^2 = 2BC^2

Let E be a point on AB so that CE=CB and let M=c(BCE).c(ADE)

then we have that <EMB = <ECB = <EAD, so A, M and D are collinear.  If O is the center of c(BCE) then <OED = 7/2 <A = 90°.

Therefore DE^2 = DM*DA. Given that ABC is similar to AEB then BC^2 = BE*BA, but BE*BA = BM*BD, as they are secant by A to c(ADE). From those relation BM=MD=BD/2 and finally BD^2 = 2BC^2 as claimed.

II) BD^2 = 2BC^2   ==> angle(B) = 3 angle(A)

properly inverting the steps of I)  one can prove this implication.

A direct proof of the fact that

(BD/BC)^2 = 2  ==> angle(B) = 3 angle(A)  ==> <A = Pi/7.

Let E be a point on AB so that CE=CB. Let M the second point were c(BCE) cuts AD. As <EMB = <ECB = <A then EMDA is cyclic. From secant theorem we have that BM*BD = BE*BA = BC^2 = BD^2/2  then BD = 2DM.  Now as DM/AD = AD/BD it results that DAM ~ DBA then <DMA = <BAD which implies that DE=DA=BC=CE and from here the thesis.


basically, it looks like as if you want to find the smallest possible value for the 3 square roots.

the first one:
[(4-b)^2+3^2}^(1/2)
- to make this as small as possible, b must be 0

[(a-b)^2+a^2}^(1/2)
- a must equal b

[(4-a)^2+(3-a)^2}^(1/2)
- a must equal 4, 3

since b=4, a=b, then a=4

plug it in. the answer should be 8, or if you count negative roots: -8

very nice with the asymptote :D

Basically

$ \sum \sqrt{ (a_i)^2\plus{}(b_i)^2}\ge \sqrt{(\sum a_i)^2\plus{}(\sum b_i)^2}$


Recently I've seen a few posts where the poster simply quotes the previous person's post without additional comments. Is this a joke? Does it mean anything? Or is my internet stuffing up and the person actually did post something and I'm just not seeing it?

maybe the person just grouped together some quotes from different places answering a previously asked question
or maybe someone quoted to emphasize a point

or it could be that they were bored and wanted to spam, but who knows

[quote="AIME15"]No, it's supposed to say "your message only contains quote and code tags.  please add some text of your own".[/quote]

Hmmm you're probably right and that it was a very minor variation.

e.g. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=238124

Well I don't see any variations but there probably is one. :P

EDIT:  I just quoted the post above me, and I don't need to add any text myself.  Edited though.

[quote="mewto55555"][quote="dragon96"]I think if you quote, then change what's inside the quote tag, then you don't need any text. But, I could be wrong[/quote]

[b]I added: I used to do that if the message was too short. But sometimes when I quote, I don't need to add any text, like this case.[/b][/quote]

I've changed what's inside the quote tag before, and it works fine.


I copied and pasted the "first document" example from the LaTeX entry in AopsWiki, but after building, it says 0 pages and when I try to view output, it says "Cannot execute the command". What's wrong?!   :mad:

If you are using TexnicCenter then you may have forgotten to tell it where to find LaTeX. Go to Build, Define Outout Profiles, Wizard then click the ... button and navigate to the directory holding latex.exe; it will be something like C:\Program Files\MiKTeX 2.6\miktex\bin. After the Wizard try compiling again.

[quote="erdogankerem123"]That didn't work, but thanks anyways  :( . I'll try uninstalling, and reinstalling, and see if that helps...[/quote]

Do you get any errors/warnings? Are you sure you copied the entire thing?


Let $ a,b,c$ be positive real numbers such that
$ 2(a^2\plus{}b^2\plus{}c^2)\plus{}ab\plus{}bc\plus{}ca \equal{}9.$
Prove that
$ \frac{a^2}{2a\plus{}b}\plus{}\frac{b^2}{2b\plus{}c}\plus{}\frac{c^2}{2c\plus{}a} \ge 1.$

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that
$ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca \equal{} 9.$
Prove that
$ \frac {a^2}{2a \plus{} b} \plus{} \frac {b^2}{2b \plus{} c} \plus{} \frac {c^2}{2c \plus{} a} \ge 1.$[/quote]
Rewrite the inequality as
$ \frac {a^2\plus{}ab\plus{}b^2}{2a \plus{} b} \plus{} \frac {b^2\plus{}bc\plus{}c^2}{2b \plus{} c} \plus{} \frac {c^2\plus{}ca\plus{}a^2}{2c \plus{} a} \ge 3$
And we have
$ 9(a\plus{}b\plus{}c)\equal{}[2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca](a\plus{}b\plus{}c)\equal{}\sum(c^2\plus{}ca\plus{}a^2)(2a\plus{}b)$
I can't continue.Can somebody help me?
[hide="can_hang2007"]Can you post your solution?[/hide]

[quote="leedt26"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that
$ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca \equal{} 9.$
Prove that
$ \frac {a^2}{2a \plus{} b} \plus{} \frac {b^2}{2b \plus{} c} \plus{} \frac {c^2}{2c \plus{} a} \ge 1.$[/quote]
Rewrite the inequality as
$ \frac {a^2 \plus{} ab \plus{} b^2}{2a \plus{} b} \plus{} \frac {b^2 \plus{} bc \plus{} c^2}{2b \plus{} c} \plus{} \frac {c^2 \plus{} ca \plus{} a^2}{2c \plus{} a} \ge 3$
And we have
$ 9(a \plus{} b \plus{} c) \equal{} [2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca](a \plus{} b \plus{} c) \equal{} \sum(c^2 \plus{} ca \plus{} a^2)(2a \plus{} b)$
I can't continue.Can somebody help me?
[hide="can_hang2007"]Can you post your solution?[/hide][/quote]
I think from it we can prove the inequality of nguoivn.
$ {\left( {\frac{a}{{2a \plus{} b}}} \right)^3} \plus{} {\left( {\frac{b}{{2b \plus{} c}}} \right)^3} \plus{} {\left( {\frac{c}{{2c \plus{} a}}} \right)^3} \ge \frac{1}{9}$

[quote="leedt26"][quote="leedt26"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that
$ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca \equal{} 9.$
Prove that
$ \frac {a^2}{2a \plus{} b} \plus{} \frac {b^2}{2b \plus{} c} \plus{} \frac {c^2}{2c \plus{} a} \ge 1.$[/quote]
Rewrite the inequality as
$ \frac {a^2 \plus{} ab \plus{} b^2}{2a \plus{} b} \plus{} \frac {b^2 \plus{} bc \plus{} c^2}{2b \plus{} c} \plus{} \frac {c^2 \plus{} ca \plus{} a^2}{2c \plus{} a} \ge 3$
And we have
$ 9(a \plus{} b \plus{} c) \equal{} [2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca](a \plus{} b \plus{} c) \equal{} \sum(c^2 \plus{} ca \plus{} a^2)(2a \plus{} b)$
I can't continue.Can somebody help me?
[hide="can_hang2007"]Can you post your solution?[/hide][/quote]
I think from it we can prove the inequality of nguoivn.
$ {\left( {\frac {a}{{2a \plus{} b}}} \right)^3} \plus{} {\left( {\frac {b}{{2b \plus{} c}}} \right)^3} \plus{} {\left( {\frac {c}{{2c \plus{} a}}} \right)^3} \ge \frac {1}{9}$[/quote]
Yes, I inspired my inequality from this inequality. Anyway, I will wait for a nice proof of my inequality.

[quote="leedt26"]
I think from it we can prove the inequality of nguoivn.
$ {\left( {\frac {a}{{2a \plus{} b}}} \right)^3} \plus{} {\left( {\frac {b}{{2b \plus{} c}}} \right)^3} \plus{} {\left( {\frac {c}{{2c \plus{} a}}} \right)^3} \ge \frac {1}{9}$[/quote]
It isn't my ineq, leedt26. But I have a nice proof for it by Am-Gm and CS which I mentioned on some forums. And my proof is different with Can_hang's proof :)


$ a,b>1,\gcd(a,b)\equal{}1$, set $ S\equal{}\{ax\plus{}by: x,y\in\mathbf{N}\}$, denote $ f(a,b)\equal{}\max\{n\in\mathbf{N}: n\not\in S\}$, find the expression of $ f(a,b)$.

The value of $ f(a,b)$, commonly known as the [url=http://mathworld.wolfram.com/FrobeniusNumber.html]Frobenius Number[/url] of $ ax \plus{} by$ with $ x,y \in \mathbb{N}$, is $ ab \minus{} a \minus{} b$.

[i]Partial Proof:[/i] Let $ d \equal{} ab \minus{} a \minus{} b$. We will first show by contradiction that $ d \not\in S$. Suppose that $ d \in S$. Then there exist $ x,y \in \mathbb{N}$ such that $ ax \plus{} by \equal{} d$. Note that $ (x_0,y_0) \equal{} (b\minus{}1,\minus{}1)$ is a solution to this equation. Then all solutions can be given by $ (x,y) \equal{} (x_0 \plus{} bt, y_0 \minus{} at) \equal{} (b \minus{} 1 \plus{} bt, \minus{}1 \minus{} at) \equal{} ((t\plus{}1)b \minus{} 1, \minus{}(at\plus{}1))$. Since $ x > 0$, we have that $ t \geq 0$. Also, since $ y > 0$, we have that $ t < 0$. We have reached a contradiction, and thus $ d \not\in S$.

To complete the proof, the remaining step is to show that $ d \geq c$ for any integer $ c \not\in S$. I've been having some trouble with this. Any takers?

That's when $ x, y \in \mathbb{N} \cup \{ 0 \}$.  Here $ f(a, b) \equal{} ab$.

Woops, sorry everyone! Thanks for the correction t0rajir0u.

But what about $ x, y \in \mathbb{N} \cup \{ 0 \}$, it is much more difficult and I wait for a solution.

It's not more difficult, a number $ n\equal{}ax\plus{}by$ can be represented with $ x,y \in \mathbb N_0$ iff $ n\plus{}a\plus{}b \equal{} a(x\plus{}1)\plus{}b(y\plus{}1) \equal{} a x^\prime\plus{}by\prime$ can be represented with $ x^\prime,y^\prime \in \mathbb N$.


induction
combinatorics proposed
combinatorics

The director has found out that six conspiracies have been set up in his department, each of them involving exactly $3$ persons. Prove that the director can split the department in two laboratories so that none of the conspirative groups is entirely in the same laboratory.

Let us notice that $6$ is the best possible - the claim is no longer true for $7$ conspiracies (look at the $7$ "lines" of the Fano projective plane, on $7$ points).

The proof for $6$ is quite convoluted - I do possess one, made of cute little steps, but it's too much to write down, so this is just to certify the claim is correct.

It's quite easy but like mavropnevma said it's a bit too long. If you look at it like a graph we have a number of components. Each component consists of a few linked conspiracies. So if we prove that it is possible to do this in each component then we're done. We use induction on the number of conspiracies linked in one component. When a conspiracy is linked to the component it's done in either of these three ways:
1. By adding 2 new persons
2. By adding 1 new person
3. By adding 0 person to the component and linking two persons that only have one neighbour in common.( :!: We use the number 6 here :!: )

It's not so hard to observe that if $n$ (the number of persons) is less than $7$, the conclusion follows.

If $n \geq 7$, use induction.  Observe that for $n \geq 7$, there are at least two persons who are not involved in the same conspiracy. Let $x_1$ and $x_2$ be those persons. Now, if we consider $(x_1, x_2)$ as  a single person and apply the induction hypothesis, we are done.

Indeed, we must have $n\geq 5$, since $\binom {4} {3} = 3 < 6$. For $n=5$, we have $\binom {5} {3} = 10 > 6$, so we can split into a laboratory of $3$ which is not a conspiracy, and the remaining $2$. For $n=6$, we have $\binom {6} {3} = 20$ possible triplets, i.e. $10$ pairs $(T,\overline{T})$; since there are $6$ conspiracies, there must exist such a pair free of conspiracies.

just a few discussions will do.
it suffices to discuss whether $n\le 5$ or $n\ge 6$
the former one is trivial;the latter one can be proved by Fubini principle.


calculus
integration
function
floor function
derivative
calculus computations

$ \int_{0}^{x}\frac {2t}{2^{[ t ]}}\, dt$    where [ t ] denotes the greatest integer function

this works unless $ x$ is actually meant to be an integer provided that $ x\ge2,$ thus:

$ \int_{0}^{x}{\frac{2t}{2^{\left\lfloor t \right\rfloor }}\,dt}\equal{}\sum\limits_{k\equal{}0}^{x\minus{}1}{\int_{k}^{k\plus{}1}{\frac{2t}{2^{k}}\,dt}}\equal{}\sum\limits_{k\equal{}0}^{x\minus{}1}{\frac{2k\plus{}1}{2^{k}}},$ and the value of that sum will give ya the result.

If $ x$ is not an integer then you must consider the 'left over' bit which is  given by:

$ \int_{\left\lfloor x \right\rfloor}^{x} \frac{2t}{2^{\left\lfloor x \right\rfloor}}\, dt \equal{} \frac{x^2 \minus{} \left\lfloor x \right\rfloor^2}{2^{\left\lfloor x \right\rfloor}}$

@integrability

but how will we find the sum? :huh:

There are a few ways to do it - here's one of them.
I'll ignore the second term because it's just a GP

\begin{eqnarray*} \sum\limits_{k = 0}^{x - 1}{ky^k} & = & y\sum\limits_{k = 0}^{x - 1}{ky^{k - 1}} \\
& = & y\dfrac{d}{dy}\sum\limits_{k = 1}^{x - 1}{y^{k}} \\
& = & y\dfrac{d}{dy}(y \dfrac{y^{x - 1} - 1}{y - 1})
\\&=& y{\dfrac{(x-1)y^x - xy^{x-1} + 1}{(y-1)^2}\end{eqnarray*}

(Use quotient rule for the differentiation), then sub in $ y = \frac {1}{2}$.


Help me Prove : If a +b+c =3 => a^n + b^n +c^n > or= a^k + b^k +c^k with all n and k belong N* , n > k

[hide="proof"]
By power mean inequality
$ \sqrt [n]{\dfrac{a^n \plus{} b^n \plus{} c^n}{3}}\ge\dfrac{a \plus{} b \plus{} c}{3}\equal{} 1, \forall n\in\mathbb{Z}^ \plus{}$

Also, if $ x\ge 1$ and $ y\ge z$ then $ x^y\ge x^z$.

So
$ \sqrt [k]{\dfrac{a^k \plus{} b^k \plus{} c^k}{3}}\ge\sqrt [n]{\dfrac{a^k \plus{} b^k \plus{} c^k}{3}}$

And, by power mean inequality
$ \sqrt [n]{\dfrac{a^n \plus{} b^n \plus{} c^n}{3}}\ge\sqrt [k]{\dfrac{a^k \plus{} b^k \plus{} c^k}{3}}$

Then
$ a^n \plus{} b^n \plus{} c^n\ge a^k \plus{} b^k \plus{} c^k$
[/hide]

:)

Then it is not always true true.  :|
Take for example $ a \equal{} \minus{} 15$, $ b \equal{} c \equal{} 9$.
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} \minus{} 1917$ and $ a^5 \plus{} b^5 \plus{} c^5 \equal{} \minus{} 641277$.

Another proof when $ a, b, c>0$.
[hide]
Cauchy: $ \dfrac{a^{n\plus{}2}\plus{}b^{n\plus{}2}\plus{}c^{n\plus{}2}}{a^{n\plus{}1}\plus{}b^{n\plus{}1}\plus{}c^{n\plus{}1}}\ge\dfrac{a^{n\plus{}1}\plus{}b^{n\plus{}1}\plus{}c^{n\plus{}1}}{a^n\plus{}b^n\plus{}c^n}$
[/hide]
:)


If $m_a, m_b,m_c$ are the medians and $r_a, r_b, r_c$ are the exradii of a triangle (labelled accordingly), prove that
\[
\frac{r_ar_b}{m_am_b} + \frac{r_br_c}{m_bm_c} + \frac{r_cr_a}{m_cm_a} \geq 3
\]

We had this almost a year ago: http://www.mathlinks.ro/Forum/viewtopic.php?t=3485

  Darij


algebra
polynomial
function
linear algebra
matrix
linear algebra unsolved

A rather hard problem:
  Characterize all functions $f$ from $M_n(K)$ to $K$ with the property that they are polynomials with respect to the elements of $A$ (that is, they are polynomials in $n^2$ variables) and $ f(AB)=f(BA)$ for all $A,B$.

I can find them for infinite fields $K$, I think.

We extend $f$ in the obvious way to $\mathcal M_n(\tilde K)$, where $\tilde K$ is the algebraic closure of $K$. Since $f(AB)-f(BA)$is a polynomial function in $2n^2$ variables which is null in all the $2n^2$-uples with elements in $K$, and $K$ is infinite, $f$ must be identically zero on $\tilde K$, so $f(AB)=f(BA)$ in $\mathcal M_n(\tilde K)$ as well.

$f$ is constant on each similarity class, so it suffices to determine it for Jordan matrices. Given a Jordan $A$ matrix with some $1$'s below the main diagonal, all matrices having $\lambda\ne 0$ instead of those $1$'s are similar to it, so $f$ has the same value for all of these, which means that it must have the same value for the matrix we get by making $\lambda=0$, i.e. the diagonal matrix with the exact same eigenvalues as $A$ (including multiplicities). This means that $f$ induces an $n$-varianle polynomial $\bar f$ with $\bar f(\lambda_1,\ldots,\lambda_n)=f(A)$ whenever $(\lambda_i)_i$ is the spectrum of $A$. $\bar f$ is symmetric since $f$ is constant on similarity classes, so it's a polynomial in the coefficients of the characteristic polynomial of the matrices. The same holds when $f$ is viewedas the old function, defined on $\mathcal M_n(K)$. Conversely, every such $f$ satisfies the requirements, so this is the answer: when $K$ is infinite, at least, $f$ is a polynomial in the coefficients of the characteristic polynomial of its argument.


All are questions from the path to combinatorics for undergraduates:

1)Determine the no. of functions f: {1,2,3,...,1999) -> {2000,2001,2002,2003} satisfying the condition that f(1) + f(2) + f(3) ....+ f(1999) is odd.

[hide]My answer :$ 1000. 2^{1999}$[/hide]

2) In how many ways can COMPUTER be spelled by moving either down or diagonally to the right in fig:

C
O O
M M M
P  P  P  P
U  U  U  U  U
T  T  T  T  T  T
E  E  E   E  E  E  E
R  R  R   R  R  R  R  R

[hide]I tried generating function[/hide]

3) A basketball team consists of 12 pairs of twin brothers. On the first day of training, all 24 players stand in a circle in such a way that all pairs of twin brothers are neighbors. In how many ways can this be done?

[hide]My answer :$ 11! . 2^{12}$[/hide]

[quote="Conjurer"]All are questions from the path to combinatorics for undergraduates:

1)Determine the no. of functions f: {1,2,3,...,1999) -> {2000,2001,2002,2003} satisfying the condition that f(1) + f(2) + f(3) ....+ f(1999) is odd.

[hide]My answer :$ 1000. 2^{1999}$[/hide][/quote]
Hmm, I get
[hide]Give any random configuration for f(1) to f(1998). Then f(1999) is fixed as even or odd to make the sum of all the f(i) odd. Therefore, only half of all possible configurations work, and there are $ 2\cdot4^{1998}$ possible ways.[/hide]

[quote="Conjurer"]2) In how many ways can COMPUTER be spelled by moving either down or diagonally to the right in fig:

C
O O
M M M
P  P  P  P
U  U  U  U  U
T  T  T  T  T  T
E  E  E   E  E  E  E
R  R  R   R  R  R  R  R

[hide]I tried generating function[/hide][/quote]
Well, you don't need generating functions.  :wink:
[hide]Note that no matter which row you are in (and what paths you have already chosen), there are always 2 choices of where to go next. Therefore there are $ 2^7\equal{}128$ choices total.[/hide]

But for number two, is a move down one and two to the right considered diagonally down?

C
[b]O[/b]O
MM[b]M[/b]
...

[quote="Conjurer"]I dont think you can jump like that.

Can anyone try the third?[/quote]
Oh, I skipped that one because I think your answer is right. :P
[hide="My solution to confirm"]Consider moving around the pairs of twins each as one "person". Fixing one group (since it is a circle), you can arrange the pairs in $ 11!$ ways. Then within each pair, there are $ 2$ ways to arrange it, giving $ 11!\cdot2^{12}$ as the final answer.[/hide]


function
algebra
polynomial
Vieta
floor function

Let $ m,n$ be integers such that $ m>n>3$. Prove that the roots $ x_1,x_2$ of the equation $ x^{2}\minus{}mx\plus{}n\equal{}0$ are integers if and only if the number $ \left[mx_{1}\right]\plus{}\left[mx_{2}\right]$ is a perfect square.

is the question worded correctly?
beucase $ [mx_1] \plus{} [mx_2] \equal{} m^2$

Not necessarily. I think he means the floor function by brackets.

Anyways, $ [mx_1] \plus{} [mx_2] \leq mx_1 \plus{} mx_2 \equal{} m^2$ (Vieta) and $ [mx_1] \plus{} [mx_2] > mx_1 \plus{} mx_2 \minus{} 2 \equal{} m^2 \minus{} 2$. Since $ m > 3$, we have $ m^2\minus{}2 > (m\minus{}1)^2$, so the only possible perfect square $ [mx_1] \plus{} [mx_2]$ can be is $ m^2$. This occurs when both $ x_1, x_2 \in Z$, since then the floors become equal to their insides. Also, if either is not an integer than $ [mx_1] \plus{} [mx_2] < m^2$ cannot be a perfect square, so we are done.


[quote="hydro"]$f(f(0))=f(0) \rightarrow f(0)=0 \vee f(x)=x$[/quote]

This implication is clearly false; for instance, any constant function satisfies the first condition but neither of the second.   Your second deduction has the same problem.  It is possible to derive $f(0)=0$, but it is not so straight-forward.

[quote="hydro"]$x=y=0 \rightarrow f(f(0))=f(0) \rightarrow f(0)=0 \vee f(x)=x$, but this second case doesn't work

$y=0 \rightarrow f(x^2)=f(f(x)) \rightarrow f(x)=x^2$

... then?[/quote]

The second implication doesn't work either.  You're assuming $f^{-1} (x)$ exists.

[quote="violetcraze"]Find all functions $f: R \to R$ which satisfys:

$f(x^2 + y) + f(f(x) - y) = 2f(f(x)) + 2y^2$[/quote]

$y = -x^2 \implies f(0) + f(f(x) + x^2) = 2f(f(x)) + 2x^4$
$y = f(x) \implies f(0) + f(f(x) + x^2) = 2f(f(x)) + 2f(x)^2$

Hence

$f(x)^2 = x^4$

Clearly then $f(0) = 0$, so setting $x = 0$, we have $f(y) + f(-y) = 2y^2$.

Suppose $f(c) = -c^2$. Then we have $f(-c) = 2c^2 - f(c) = 3c^2$. But then we have $f(-c)^2 = 9c^4 \ne c^4$ unless $c = 0$. Hence $f(x) = x^2$ for all $x$.


HI GUYS
i am a new member to this site i think india is going to do well in the current tournament. i am undergraduate studying in singapore but fully in touch with cricket and hope that india does well this time around.Anyways hope to see you guys around.

If anyone is intrested in writing articles on cricket than contact me on vvmehta2@hotmail.com
have fun

i think we will make it to the finals and lose to west indies or australia....depends on the Ind VS WI match...I am a grrr888 cricket fan so i watch each match and i think india HAS a chance of doing well only if they beat WI

Did anyone see India bt england (gd game) england got bowled out for 125! India will definately get into the finals I think. Just wish Pakistan would do to, so India can win this time. :lol:

That match was too good...but now Australia has beaten england...though an INDO-PAK match is much more exciting and the victory more sweet, i think Aus will do sumthing 2 upset us in the IND-AUS match...

This is kind of stupid.. but I reckon Pakistan will beat australia which will make it easier for us to win!

I hope so!! but when is PAK playing AUS?
Do u think IND will succumb to psychological pressure and lose to WI or Aus? i hope not...Rudi Webster should be of SOME use.
i hope atleast this time Dhoni,Kaif,Yuvraj,Raina etc score big...bcoz of them the batting is collapsing each time!

india lost!!  :mad:  :(
is this alvida (bye-bye) to the Champions Trophy???

:( India is indeed out of the Champions Trophy...

:(  :mad:   HOW could we lose so badly to Australia??? bye-bye champions trophy....
I think the Aussies will break the jinx and win the trpohy this time...then hopefully they'll become overconfident and lose the Ashes!! :D

On the lines of cricket...another topic 4 discussion-
Do u think India has what it takes to win the World Cup 2007?
What wid no 2 digit scores other than the bowler's card...i say NO. What about u'll?

no chance for this yrs,australia is on a roll...anyway india made 91 against South africa,what more to say...thnk god i'm not an indian  :roll:

The South Africa match was painful to watch. Australia is so going to win the ahses, when you beat the auzzies once you can't do it again because they come back twice as strong.

India got drubbed 5-0 in the ODIs against SA.
We won the 1st test by 123 runs and lost the 2nd one by 175 runs.
We are really pathetic.
Sehwag and Jaffer need to play county cricket for atleast a year, the way they're going. Wonder why Dravid is not scoring...no-one had any hopes of Ganguly scoring anyway. Sreesanth and Kumble are the only good players left in the team.. :(
England became over-confident and the unstoppable Ozzies won the Ashes...  :(
Looks like they'll be adding another World Cup to their tally next year...

Hello,
    Jaffer and Sehwag,I am not too sure if a county would sign them up.They ought to come back and play our domestic cricket.And England,well,they simply got overwhelmed.Yes,Aussies are well on their way to the World Cup.
Sathej

[quote="ashwinkrishnan"]If its someone who is going to win the world cup , it has got to be INDIA.they are just messing uo with things now and thus seem to be in disorder.If they just get rid of sehwag and bring ig Uthappa , Pathan,Kumble,sreesanth and laxmam back all willl be well .[/quote]
EXACTLY!!!
But yesterday, Jaffer scored a century, the Lil'Master made 64, Kaarthick and Ganguly made 63...that was expected.. but Sehwag actually made 40!!!
India reached 414!! :clap:  :winner_first:
You know...we just MAY have a chance to win the series...but if Pathan, Kumble, Uthappa and Sreesanth don't figure in India's world cup plans then we deserve to lose.. :)

[quote="ashwinkrishnan"]If its someone who is going to win the world cup , it has got to be INDIA.they are just messing uo with things now and thus seem to be in disorder.If they just get rid of sehwag and bring ig Uthappa , Pathan,Kumble,sreesanth and laxmam back all willl be well .[/quote]
I don't know, they definitely should bring back pathan, but I've never been a fan of Laxman, maybe it's after that one day I woke up at two (I live across the world) hoping to see great batting on the fifth day of the test match and out came VVS, bunting almost every ball. I was so mad that I lost my sleep just to watch VVS. Once Dhoni comes back, India will be doing a good amount better I think. Dhoni=pwnser.

[quote="Padmavathi"][quote="bpms"] Once Dhoni comes back, India will be doing a good amount better I think. Dhoni=pwnser.[/quote]
I totally disagree... :D Dhoni's become to arrogant after his ads and stardom. Tell me...after that 183 has he made a SINGLE century?? no...but look at how Dinesh Kaarthick is performing...simple...quiet...willing to bat at any position..even the jinxed opening one...always on the sidelines...not exactly in the spotlight...but yet he scores so much more than M.S. Dhoni....Dhoni is ok with not performing because he's sure he CAN'T lose his place...but if he gets a little shock treatment like what they did to poor Irfan...then maybe he'd straighten out... :D[/quote]
He didn't make a single century, yes, however he has had 5 50s since then. I'll admit that it was 32 matches, and also that I don't live in India so I don't know about stardom and ads and arrogance. I just watch the matches.

Hmm, right now watching the ODI, just watched Bravo get injured. Right now India is actually doing pretty good, just hope that they can keep it up.
To quote 3 K G: India pe koi bharosa nahi kersakte.

This should be an interesting series, I hope India can snap back into shape for the world cup after their shameful loss in South Africa.
EDIT: I've don't really like Ganguly too much just because he's seemed arrogant, but with the way he's batting now my opinion might get changed. He just smacked a SIX!
EDIT 2: OMG another 6 to bring up 50 for Ganguly!!!
EDIT 3: Gambhir out :(
EDIT 4: OMG another 6 from Ganguly!!!
EDIT 5:Over 47: 2 sixes!!! Over 48: 2 more sixes!!! 36 runs from those two overs!!!
338 Runs!!!

AT LAST, the moment of triumph has come for india. this was very uch needed to boost confidence in the team AND TO SHUT LARA'S BRAGGING.
saw the way how Dada batted and.... dhoni and ...sachin and.... dravid,..... augurs well for the team

I think this is the first time I've seen Dravid smack it. This just shows that he can bat in any setting.

A man of dravids calibre just needed the right situation and he got it yesterday


hi im new to the whole programming thing so my sister let me borrow a java book and when i did the basic stuff like making the hello world program, imaginary rectangle, and input output thing it worked out great  :lol:
but then i moved on to the next chapter and it gave me a command to use in a programm

generator.nextInt(6)

so i can make a lottery thing but then my eclipse platofrm said it was wrong and made some suggestions like

Random generator.nextInt(6)

which was still wrong somehow and i never got it to work?
then i did an output think with color and for the program it just had to insert pink and i got a color however another excercise on the page said make it green by adjusting the amount of red, green, and blue components?
how do u do that

Did you make sure to declare the variable 'generator' first?
[code]Random generator = new Random ();
System.out.println (generator.nextInt (6));[/code]
As for your second question, I think what you're looking for is that green is (r, g, b) = (0, 255, 0); that is, full green and no red or blue. If this doesn't answer your question, could you be more specific?

thx that makes perfect sense buit wut command do u put before the
(r,g,b)?

If you're being asked to adjust a colour in a program that already exists, then the method should already be there. I highly doubt the book hasn't already told you what method to use.

one: the book did not give me the exact command
two: i typed in the command u gave e and there are no red lines now yet when i try to run this specific program there is no server to run it on but when i run the other programs there es a server


1. If gcd(a,b) = 1, prove that
(a) gcd(a-b,a+b)  <= 2
(b) gcd(a-b,a+b,ab) = 1
(c) gcd(a2-ab+b2,a+b) <= 3

2. Prove that (a+b) / (c+d) is irreducible if ad-bc=1

Hi firefox,

My attempts :

[hide]

(1) (a)

If d = gcd(a-b,a+b), then d divides (a+b)+(a-b) = 2a, and also (a+b)-(a-b) = 2b.

Since, gcd(a,b)=1, so only common divisors of 2a and 2b are 1 and 2.

Thus, d = gcda,b) = 1 or 2 which is < 2.

(b)

From part (a), gcd(a-b,a+b) = 1 or 2. So, d = gcd(a-b,a+b,ab) = 1 or 2. Let d = 2.

Then, d divides ab implies, one of a or b is even. If it is so, then the other is also even, since (a+b) is even too. Thus, if d = 2, then 2 divides a and b both (contradiction).

So, gcd(a-b,a+b,ab) = 1.

(c)

a 2 -ab+b 2 = (a+b)2 - 3ab.

So, any common divisor of (a 2 -ab + b 2) and (a+b) will divide 3ab too.

So, d = gcd(a 2 -ab + b 2, a+b) divides 3ab,

which implies that d can be 1, 3 or any divisor of a or b. Now, if d is a divisor of a, then since d divides (a+b) too, so d also divides b (contradiction), and similarly for b.

Thus, d = 1 or 3.

[/hide]

PS : Firefox, I guess you won't get 2 as the gcd in (c). It can be 1 or 3 only.

(Thanks Peter for the correction)

Sambit

you're right about the 1 or 3, sambitnayak, but I think you made a typo in the second one.. the gcd should be 1 there ;)

Hi Firefox,

Attempt at (2):

[hide]Assume (a+b) and (c+d) have a common divisor n (> 1). So, a is congruent to (-b) mod n, and c is congruent to (-d) mod n.

Thus, ad - bc modulo n = (-b)*d - b*(-d) mod n = 0 mod n; i.e., n (which is greater than 1) divides (ad-bc).

So, we proved that if (a+b)/(c+d) is reducible, then ad-bc has a divisor greater than 1 and hence it is not equal to 1.

Thus, contrapositive says that if (ad-bc) = 1, then (a+b)/(c+d) is irreducible.[/hide]

Sambit