[quote="thanhnam2902"]Let $ ABC$ is a equilateral triangle. Let $ M$ is a point  in pland show that $ MA, MB, MC$ are lenght of three sides of another right triangle. Find locus of point $ M$.[/quote]
it means
$ MA^2=MB^2+MC^2$and all cyclic
now,with O is a point such that $ OA->=OB->+OC->$(-> means vector,sorry for my poor $ LaTex$ :blush: )
and now we have $ OA^2-OB^2-OC^2=2OM^2$,now after checking somes case we find locus of M is circle $ (O,R)$ where $ R=\sqrt {\frac{OA^2-OB^2-OC^2}{2}}$
I see I didnt use $ ABC$is a equilateral triangle ,may be I am wrong


Solve

$ 3x\plus{}3y\plus{}3z\equal{}13$

$ \frac{3}{x}\plus{}\frac{3}{y}\plus{}\frac{3}{z}\equal{}13$

$ xyz\equal{}1$.

Hi Sonnhard. I got the same six solutions as follows:

Rearranging the given equations,

$ 3(x\plus{}y\plus{}z)\equal{}13$

$ 3(xy\plus{}yz\plus{}zx)\equal{}13$

$ xyz\equal{}1$.

So $ x, y, z$ are the roots of the cubic equation

$ w^{3}\minus{}\frac{13}{3}w^{2}\plus{}\frac{13}{3}w\minus{}1\equal{}0$

$ 3w^{3}\minus{}13w^{2} \plus{}13w\minus{}3\equal{}0$

$ (w\minus{}1)(3w^{2}\minus{}10w\plus{}3)\equal{}0$

$ w\equal{}1, 3$ or $ \frac{1}{3}$.

The six permutaions of these give the same six solutions as you have.


Which is bigger 2^3000 or 3^2000?

Yes, I know that I can use logs for this but this was a non-calculator problem. And I really suck at these types of problems.

The answer is:

[hide]3^2000[/hide]

I got it right because I guessed, but I don't know the concept of how to do it. What if it were comparing 5^1234 and 2^4321? 3^10293 and 283^234?

I don't have the log values of every number memorized =P.

~G.L.

I think the question was more on the lines of an approach. As you said log is generally a handy tool for this if not
what you could do is try and express both the l.h.s and the r.h.s in terms of a common exponent (well try to) and then compare the bases to identify which one's greater

So for example if you are comparing -

5^1234 and 2^4321

We can see that 4321 is approximately 3.5 times 1234

so we have,

5^1234 and (~11)^1234

Now you know whats greater :)

Note: A lot of approximations is done here, which is a handy tool for such problems as the magnitude of numbers are so big

That's probably not your most effective example $ 2^{4321} > 2^{3702} \equal{} 8^{1234} > 5^{1234}$

Suppose you had $ 2^{19}$ and $ 3^{12}$. Now you can't do a long string of large inequalities. Let's rewrite them as $ 8^6 \cdot 2$ and $ 9^6$.

So now, divide by $ 8^6$ and now you're comparing $ 2$ and $ (1 \plus{} 1/8)^6$

We can expand the RHS for a few terms with the binomial theorem: the first few terms are:
$ \blacktriangleright 1^6 \equal{} 1$ << we need $ 1$ more to break a total of $ 2$
$ \blacktriangleright 6 \cdot 1^5 \cdot \frac {1}{8} \equal{} \frac {3}{4}$ << we need $ 3/4$ more to break a total of $ 2$
$ \blacktriangleright 15 \cdot 1^4 \cdot \frac {1}{8^2} \equal{} \frac {15}{64}$ << we need $ 1/64$ more to break a total of $ 2$
$ \blacktriangleright 20 \cdot 1^3 \cdot \frac {1}{8 \cdot 8^2} \equal{} \frac {20}{8 \cdot 64} > \frac {1}{64}$ << we broke $ 2$!

We don't even need to calculate out the rest of the terms. We know that $ 3^{12}$ is greater than $ 2^{19}$

By the way, the two numbers here are 531441 and 524228 -- they don't differ by orders of magnitude.


If you are given $\log{8} = .9031$ and $\log{9} = .9542$, then the only logarithm that cannot be found without the use of tables is:
\[ \text{(A)}\ \log{17} \qquad \text{(B)}\ \log{(5/4)} \qquad \text{(C)}\ \log{15} \qquad \text{(D)}\ \log{600} \qquad \text{(E)}\ \log{.4} \]

[hide]
(A) Maybe?

(B)
This is equal to
$(\log9 - \log8) - (\log(9) - \log(10))$
So this one can be found

(C)
This is
$log5 + log3$
log of 5 is equal to
$(\log9 - \log8) - (\log9 - \log40)$
$(\log9 - \log8) - (\log9 - (\log4 + \log10))$
$(\log9 - \log8) - (\log9 - ((2/3)\log8 + \log10))$
log3 is the same as
$(1/2)\log9$
So this works

(D)
This is the same as
log(8) + log(5) + log(15)
Which all can be found so its not this one

(E)
Same as log(2/5)
which is equal to
log2 - log5
(1/3)log8 - log5

So I suppose it must be A
[/hide]


function topology calculus derivative algebra polynomial induction

If somebody solves this problem, it will be possible to present a full proof of the famous Denjoy theorem stating that there exists a differentiable function $f : \mathbb R\to \mathbb R$ such that $f'$ changes sign on each open interval. I know that grobber, harazi and some other people were once quite interested in such things, so I hope this problem won't stay unsolved for long. :)

So here is the question: suppose that $P$ and $Z$ are two disjoint countable dense subsets of $\mathbb R$. Construct an increasing diffeomorphism $\varphi:\mathbb R\to \mathbb R$ such that $\varphi(P)\subset Z$ and $\varphi(Z)\subset P$.

An idea.

$\phi$ a homeomorphism is easy to arrange, by taking an bi-infinite sequence $p_i \in P$ with distance between adjacent points bounded above and below, taking "midpoints" of the intervals $z_i \in (p_i,p_{i+1}) \cup Z$ and sending $p_i \to z_i$.  Then repeat the construction until all points are exhausted.
So the problem seems to be differentiability.

I think differentiability of $\phi$ can be enforced in this construction [i]if[/i] $P$ and $Z$ can be exhausted in such a way that, at stage $n$, the set $S_n$ of points (from $P \cup Z$) already used has a positive lower bound on the distance between adjacent points.  (By constructing a smooth positive function $f_n$ with zeros at $S_n$ and $|f_n| + |f'_{n}| < 1$ on all $\mathbb{R}$, and making sure that new points at stage $n+1$ of the construction give derivative close to $c_n f_n'$ at the points of $P$ whose $\phi$-values are determined at this stage, where $c_n \to \infty$ goes to zero very rapidly).

I am not certain whether such an exhaustion is always possible (it seems so).

Yes, this idea eventually leads to the result, though it is easier to add points one by one rather than by whole progressions. Now how about a reasonably complete solution everyone would be able to read and understand? Can anybody do it?

I will try to give a solution which is understandable ( I hope also that it is correct ).
Let us prove the following statement:
If $ a_{1},a_{2}, \ldots \in \mathbb{R} $ is a sequence of distinct numbers and
$ A_{1},A_{2}, \ldots \subset \mathbb{R} $ is a sequence of dense subsets in $ \mathbb{R} $,
then there exists an entire function $ f : \mathbb{C} \rightarrow \mathbb{C} $,
such that $ f( \mathbb{R} )  = \mathbb{R} $, $ f |_{ \mathbb{R} }  $ is an
increasing diffeomorphism $ \mathbb{R} \rightarrow \mathbb{R} $ and for every
$ n \geqslant 1 $, $ f(a_{n}) \in A_{n} $. The question easily follows from this,
because we can enumerate elements of $ Z \cup P $ as a sequence
$ a_{1},a_{2}, \ldots $, such that elements with odd index are in $ Z $ and with even in
$ P $, and take $ A_{2k+1} = P $ and $ A_{2k} = Z $.
Let us prove this statement. First of all, take some $ b \in \mathbb{R} $, such that
$ b \neq a_{1},a_{2}, \ldots $, and define a sequence of polynomials
$ p_{1}(x) = (x-b) $, $ p_{2}(x) = (x-a_{1}) $, $ p_{3}(x) = (x-b)(x-a_{1})(x-a_{2}) $,
$ p_{4}(x) = (x-a_{1})(x-a_{2})(x-a_{3}) $, and so on, so
$ p_{2k}(x) = (x-a_{1})(x-a_{2}) \ldots (x-a_{2k-1}) $ and $ p_{2k+1}(x) = (x-b)(x-a_{1})
\ldots (x-a_{2k}) $. So we see that all $ p_{m}(x) $ are real polynomials of
odd degree, thet $ p_{m}(a_{m}) \neq 0 $ and $ p_{m+1}(a_{m}) = p_{m+2}(a_{m}) =
p_{m+3}(a_{m}) = \ldots = 0 $. Now for every $ m $, define $ M_{m} $ to be the maximal
value of $ | p_{m}(z) | $, where $ | z | \leqslant m $ runs over all complex numbers
with absolute value less or equal to $ m $. Also we know that $ p_{m}(x) $ is of odd degree,
so $ p_{m}'(x) $ is of even degree, so there exists $ C_{m} := min_{ x \in \mathbb{R} }
p_{m}'(x) $. Now for every $ m \in \mathbb{N} $, take $ \varepsilon_{m} > 0 $
small enough, such that we will have $ \varepsilon_{m}M_{m} \leqslant (1/2^{m}) $
and $ \varepsilon_{m}C_{m} \geqslant (-1/2^{m+1}) $. Now by induction we will find
real numbers $ c_{1},c_{2}, \ldots \in \mathbb{R} $. Suppose we have found
$ c_{1},c_{2}, \ldots , c_{m} $. Take $ f_{m}(x) = x + c_{1}p_{1}(x) + c_{2}p_{2}(x) +
\ldots + c_{m}p_{m}(x) $. We know that $ p_{m+1}(a_{m+1}) \neq 0 $, so because $ A_{m+1} $
is dense in $ \mathbb{R} $, we can find some $ c_{m+1} \in \mathbb{R} $, such that
$ 0 <  c_{m+1}  \leqslant \varepsilon_{m+1} $ and $ f_{m}(a_{m+1}) + c_{m+1}p_{m+1}(a_{m+1})
\in A_{m+1} $.
 So by induction we have found a sequence $ c_{1},c_{2}, \ldots \in \mathbb{R} $, such that
$ | c_{m}p_{m}(z) | \leqslant (1/2^{m}) $ for $ |z| \leqslant m $, for every $ m $,
so the series $ f(z) = z + c_{1}p_{1}(z) + c_{2}p_{2}(z) +  c_{3}p_{3}(z) + \ldots $
converges uniformly on compact subsets of $ \mathbb{C} $ to a holomorphic function
on $ \mathbb{C} $, i.e. to an entire function. As we have built $ c_{1},c_{2}, \ldots $,
we see that they are real, so $ f( \mathbb{R} )  \subseteq \mathbb{R} $.
Also $ f(a_{m}) \in A_{m} $ for every $ m $. Now, because we have uniform convergence
of holomorphic functions $ f(z) = z + c_{1}p_{1}(z) + c_{2}p_{2}(z) +  c_{3}p_{3}(z) + \ldots $,
then we can take a derivative and we have that
$ f'(z) = 1 + c_{1}p_{1}'(z) + c_{2}p_{2}'(z) +  c_{3}p_{3}'(z) + \ldots $ for every
$ z \in \mathbb{C} $. Now if we take $ z = x \in \mathbb{R} $, the we have
$ f'(x) = 1 + c_{1}p_{1}'(x) + c_{2}p_{2}'(x) +  c_{3}p_{3}'(x) + \ldots $,
but we know that $ c_{m}p_{m}'(x) \geqslant (-1/2^{m+1}) $, so
$ c_{1}p_{1}'(x) + c_{2}p_{2}'(x) +  c_{3}p_{3}'(x) + \ldots \geqslant -( 1/4 + 1/8 + \ldots ) =
-1/2 $, so $ f'(x) \geqslant 1/2 $ for every $ x \in \mathbb{R} $.
Therefore it is easy to see that $ f |_{ \mathbb{R} } : \mathbb{R} \rightarrow \mathbb{R} $
is an increasing diffeomorphism.

I hope it is correct.

Thanks, Leva! Now the promised proof of the Denjoy theorem.

Take the function $g$ constructed in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2822[/url]. Let $Z$ be a dense countable subset of the (dense) set of points where $g'=0$. Sinse $g$ is strictly increasing, the set of points where $g'>0$ is also dense and we can choose some dense countable subset $P$ of it. Now let $\varphi$ be the increasing diffeomorphism that maps $P$ into $Z$ and $Z$ into $P$. Put $f(x)=g(x)-g(\varphi(x))$. Then $f'(x)=g'(x)>0$ on $P$ and $f'(x)=-g'(\varphi(x))\varphi'(x)<0$ on $Z$. The end.


Prove that $ \ \frac{1}{x\plus{}y} \plus{} \frac{1}{y\plus{}z} \plus{} \frac{1}{z\plus{}x} > \frac{3}{x\plus{}y\plus{}z}$, for all positive numbers $ x, y$ and $ z$.

[hide="Proof"]$ x\plus{}y < x\plus{}y\plus{}z$, so $ \displaystyle\frac{1}{x\plus{}y} > \displaystyle\frac{1}{x\plus{}y\plus{}z}$. By similar logic, since $ x$, $ y$, and $ z$ are all positive, $ \displaystyle\frac{1}{x\plus{}z} > \displaystyle\frac{1}{x\plus{}y\plus{}z}$ and $ \displaystyle\frac{1}{y\plus{}z} > \displaystyle\frac{1}{x\plus{}y\plus{}z}$. Therefore, $ \displaystyle\frac{1}{x\plus{}y} \plus{} \displaystyle\frac{1}{x\plus{}z} \plus{} \displaystyle\frac{1}{y\plus{}z} > \displaystyle\frac{1}{x\plus{}y\plus{}z} \plus{} \displaystyle\frac{1}{x\plus{}y\plus{}z} \plus{} \displaystyle\frac{1}{x\plus{}y\plus{}z}$, or $ \displaystyle\frac{1}{x\plus{}y} \plus{} \displaystyle\frac{1}{y\plus{}z} \plus{} \displaystyle\frac{1}{z\plus{}x} > \displaystyle\frac{3}{x\plus{}y\plus{}z}$.[/hide]

[hide="Another Solution"]
Multiply by $ 2(x \plus{} y \plus{} z)$ on both sides,

$ ((x \plus{} y) \plus{} (y \plus{} z) \plus{} (z \plus{} x))\left(\frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x}\right) > 6$

But by Cauchy-Schwarz we have

$ ((x \plus{} y) \plus{} (y \plus{} z) \plus{} (z \plus{} x))\left(\frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x}\right)\ge 9 > 6$ :ninja:

And we are done.
[/hide]

[hide="Probably simpliest solution"] Easily from Andreescu Inequality,

$ \frac{1}{x\plus{}y} \plus{} \frac{1}{y\plus{}z}\plus{}\frac{1}{z\plus{}x} \geq \frac{9}{2(x\plus{}y\plus{}z)} > \frac{6}{2(x\plus{}y\plus{}z)}\equal{}\frac{3}{x\plus{}y\plus{}z}$ [/hide]


Look at the following, for example:

$n^{(\frac{1}{n}})$

The powers become illegibly small. How do I make them larger? Thank you.

How about putting the power in a box? n^{\mbox{\left(\frac{1}{n}\right)}} gives
$n^{\mbox{\left(\frac{1}{n}\right)}}$


Solve equation

$ \sqrt[5]{(x\minus{}2)(x\minus{}32)} \minus{}\sqrt[4]{(x\minus{}1)(x\minus{}33)}\equal{}1$

[hide]$ u: \equal{}\sqrt[5]{(x\minus{}2)(x\minus{}32)},v: \equal{}\sqrt[4]{(x\minus{}1)(x\minus{}33)}$. Then

$ u\minus{}v\equal{}1\land u^5\minus{}v^4\equal{}31$

So we get $ u^5\minus{}(u\minus{}1)^4\minus{}31\equal{}0\iff u^5\minus{}u^4\plus{}4u^3\minus{}6u^2\plus{}4u\minus{}32\equal{}0$

Write this as

$ u^5\minus{}2u^4\plus{}u^4\minus{}2u^3\plus{}6u^3\minus{}12u^2\plus{}6u^2\minus{}12u\plus{}16u\minus{}32\equal{}0$

which yields

$ (u\minus{}2)(u^4\plus{}u^3\plus{}6u^2\plus{}6u\plus{}16)\equal{}0\quad(*)$

The second factor can be written as $ u^2(u^2\plus{}u\plus{}3)\plus{}3u^2\plus{}6u\plus{}16$, and is obviously always positive since both the quadratic trinomials involved are always positive. Hence the only solution to $ (*)$ is $ u\equal{}2\iff x^2\minus{}34x\plus{}32\equal{}0\iff x\equal{}17\pm\sqrt{257}$[/hide]


function
algorithm
modular arithmetic
number theory
relatively prime
Euclidean algorithm
Diophantine equation

I know that for any integer $x$ such that $x\neq{mp}$ for some prime $p$ and integer $m$, there is a $y\in\{1,2,\dots,p-1\}$ such that $xy\equiv{1}\bmod p$.
after some messing around i noticed that if $x$ is not a multiple of $n$, not necessarily prime, then there are $\phi(n)$ numbers $y$ in the set $\{1,2,\dots,n-1\}$ such that $xy\equiv{1}\bmod n$. I also noticed that the only $y$ and $x$ for which this works are those such that $x$ and $y$ are both relatively prime to $n$. could someone please provide me with a proof of this observation.

The proof of this is straightforward. $\varphi(n)$ is the number of integers $1\leq x \leq n$ such that $(x,n) = 1$. The most obvious set of size $\varphi(n)$ is the integers $1 \leq x \leq n$ relatively prime to $n$.

First let us prove that for all $x$ such that $(x,n) > 1$, there is no inverse. Suppose there is an inverse. Then for some $y, k$, $xy = kn+1$ Suppose $a = (x,n)$. $a\mid x \implies a\mid xy$ and $a\mid n \implies a\mid kn \implies a\nmid kn+1$ Contradiction.

Now we simply have to prove that every $x$ where $(x,n) = 1$ has an inverse modulo $n$. I will offer a constructive proof.

We use the Euclidean algorithm to get $(x,n)$. e.g.

$(25,17)$

$25 = 1\cdot17+8$
$17 = 2\cdot 8+1$
$8 = 8\cdot1+0$ and we are done.

Now we work our way backwards through this to find a linear combination of 25 and 17 that makes 1.

$1 = 17-2\cdot 8$
$1 = 17-2(25-17) = 3\cdot 17-2\cdot 25$

Take this expression modulo 25 and we get $3\cdot 17 = 1$, so 3 is the inverse of 17 modulo 25. This always works and can be made both general and rigorous, but I don't know how to phrase it correctly. Maybe someone else can help?

I see what you're saying, but your formulation is incorrect.  IF $gcd(x, n) \neq 1$ then there do not exist [b]any[/b] numbers $y$ such that $xy \equiv 1 \bmod n$.

What your statement should be is this:

There exist $\varphi(n)$ numbers $x$ in the set $\{ 1, 2, 3, ... n-1 \}$ such that there exists $y$ such that $xy \equiv 1 \bmod n$.

[hide="Discussion"] It's easy to see that there is at most $\varphi(n)$ such numbers, since if $gcd(x, n) \neq 1$ then $gcd(xy, n) \neq 1$ and so $xy \not \equiv 1 \bmod n$.

So we are looking at the set of numbers relatively prime to and less than $n$.  This is the set of [b]units[/b] $\bmod n$, which has size $\varphi(n)$.

It turns out that this set is a multiplicative group ($\bmod n$) - in other words:

- The product of two members of this set is a member of this set.
- Every member has a multiplicative inverse.  (This is the essence of your question.)

The question of how to find a multiplicative inverse is better phrased this way:

Given $a$ such that $gcd(a, b) = 1$, how do we find $a^{-1}= x$, an integer $\bmod b$ such that

$ax \equiv 1 \bmod b$

?  Well, this is better written as

$ax = 1-by$
$ax+by = 1$

Which is a linear Diophantine equation, the familiar subject of Bezout's lemma.  Can you work out the rest of the proof? [/hide]

$\varphi(x)$ is Euler's totient function (not torrent).

If $\gcd(x,n)=d>1$, then for $m,y\in\mathbb{Z}$, there are no solutions to $xy=1+mn$.
Proof. Suppose there exists such a solution. Then note that $d|xy-mn$, so $d|1$, but $d>1$. Contradiction.

If $\gcd(x,n)=1$, then there exists $y$ for which $xy\equiv 1\pmod{n}$.
Proof. By a corollary to the Euclidean algorithm, you can find $y,m\in\mathbb{Z}$ such that $xy+nm=\gcd(x,n)=1$. Then $xy\equiv 1\pmod{n}$. (If $y\not\in\{1,2,\ldots,n-1\}$, reduce it $\mod n$.)

[quote="Bictor717"]If $\gcd(x,n)=1$, then there exists $y$ for which $xy\equiv 1\pmod{n}$.[/quote]

Consider the numbers $0 \cdot x, 1 \cdot x, \ldots, (n-1) \cdot x$ (modulo $n$). If we suppose that two of them are equal, i.e. $i \cdot x \equiv j \cdot x \left( \bmod \, n \right)$, then we get that $(i-j) \cdot x \equiv 0 \left( \bmod \, n \right) \, \, \Longleftrightarrow i \equiv j \left( \bmod \, n \right)$ (since $\gcd (x, n) = 1$), i.e. $i = j$. Thus, the numbers are pairwise distinct, so they give exactly $n$ distinct values. But each number $k \cdot x$ can take one of the following values $\left\{ 0, 1, \ldots, n-1 \right\}$ ($n$ values), so $1$ must be "reached".


integration
function
real analysis
real analysis unsolved

Check:
Let's $ u(x)$ is harmonic function and
$ \int \int \int _{\mathbb R^{3}} \frac {u^{2}(x)}{(1 \plus{} |x|)^{3}}dx < \infty$ then $ u \equal{} const$

We have for every $ r > 0$ and $ a$
${ |u(a)| \leq \frac {1}{V(B(a,r))}\iiint_{B(a,r)}(1 + |x|)^{3/2}\frac {|u(x)|}{(1 + |x|)^{3/2}} \, dx}$
${{ \leq \frac {1}{V(B(a,r))}( \iiint_{\mathbb{R}^3}\frac {u^2(x)}{(1 + |x|)^3} \, dx} )^{1/2} ( \iiint_{B(a,r)}(1 + |x|)^3 \, dx} )^{1/2}$
For a fixed $ a$ the $ \limsup_{r\to \infty}$ is a constant, independent of a. So it follows that $ u$ is constant by Liouville's theorem.

I guessed a little bit before, but I put things into Maple and got
$ \iiint_{B(a,r)}(1 \plus{} |x|)^3 \, dx \leq \iiint_{B(0,r \plus{} |a|)}(1 \plus{} |x|)^3 \, dx \equal{}$
$ \frac23\,(r \plus{} |a|)^{6}\pi \plus{} {\frac {12}{5}}\,(r \plus{} |a|)^{5}\pi \plus{} 3\,(r \plus{} |a|)^{4}\pi \plus{} \frac43\,(r \plus{} |a|)^ {3}\pi$
Since $ V(B(a,r)) \equal{} \frac {4\pi r^3}{3}$ my claim is justified (well the $ \limsup$ is less than a constant independent of $ a$).


solve equation :
 $ sin^{2008}x \minus{}cos^{2007}x \plus{} ... sin^2x\minus{} cosx \equal{}1004$

Is the answer $ {x}{\equal{}}{pi/2}$ and $ {x}{\equal{}}{pi}$???

[quote="bboypa"]$ \sum_{i \equal{} 1}^{1004}{\sin^{2i}{x}} \minus{} \sum_{i \equal{} 1}^{1004}{\cos^{2i \minus{} 1}{x}} \equal{} 1004 \minus{} \sum_{i \equal{} 1}^{2008}{\cos^{i}{x}} \equal{} 1004$
so, $ \sum_{i \equal{} 1}^{2008}{\cos^{i}{x}} \equal{} \frac {\cos{x}(1 \minus{} \cos^{2008}{x})}{1 \minus{} \cos{x}} \equal{} 0$ from which, in $ [0,2\pi]$:
$ x \equal{} \pi, x \equal{} \pm \frac {\pi}{2}$

or equivalently, $ x \equal{} \frac {k\pi}{2}, \forall k \in N, e 4 \nmid k$

bye
paolo[/quote]
thanks bboypa this solution  is clear  and right.
   again thanks


Suppose $A$ is a complex number and $n$ is a positive integer such that $A^{n}=1$ and $(A+1)^{n}=1$. Prove that $n$ is divisible by 6 and that $A^{3}=1$.

Let $A=a+bi$.  Now, since $A^{n}=1$, we have $|A|=1$, and similiarly, $|A+1|=1$.  This implies

$a^{2}+b^{2}=1=(a+1)^{2}+b^{2}$, which means $a=\frac{-1}{2}$.

Now, by De Mouivre's Theorem, $\exists k_{1}\in\{0,1,\dots,n-1\}$ such that $A=cis\left (\frac{2\pi k_{1}}{n}\right )$.
Then $cos\left (\frac{2\pi k_{1}}{n}\right )=\frac{-1}{2}$ which means $\frac{k_{1}}{n}=\frac{1}{3},\frac{2}{3}$.  Hence $A^{3}=1$.

Now, remember that $A+1$ must also be an n-th root of unity, and by some simple calculations, $A+1=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}=cis\left (\frac{2\pi k_{2}}{n}\right )$ for some $k_{2}\in\{0,1,\dots,n-1\}$.  This means $\frac{k_{2}}{n}=\frac{1}{6},\frac{5}{6}$, and hence $6|n$.

[hide="interesting geometric solution"]
consider the unit circle. let the given point $A$ lie on the unit circle, somewhere
we are given that a horizontal line through A hits the unit circle at a distance of 1 (to the right) from the original point. But 1 is the radius of the circle; therefore connecting the radii to thes two vertices yields equilateral triangles, from which the result follows...[/hide]

[quote="Phelpedo"]Suppose that $z\in C$ and $n\in N^{*}$ so that $z^{n}=1$ and $(z+1)^{n}=1$. Prove that $6\ |\ n$ and $z^{3}=1$.[/quote]
$|\begin{array}{ccccc}z^{n}=1 & \Longrightarrow & |z|=1 & \Longrightarrow & z\overline z=1\\\\ (z+1)^{n}=1 & \Longrightarrow & |z+1|=1 & \Longrightarrow & (z+1)(\overline z+1)=1\end{array}$ $\Longrightarrow$ $|\begin{array}{c}z\overline z=1\\\\ z+\overline z=-1\end{array}$ $\Longrightarrow$ $t^{2}+t+1=0\begin{array}{c}\nearrow z\\\\ \searrow \overline z\end{array}$ $\Longrightarrow$ $\boxed{\ \begin{array}{c}z^{3}=1\\\ z^{2}+z+1=0\end{array}\ }\ .$

$|\begin{array}{c}n=3k+r\ ,\ 0\le r\le 2\ ;\ z^{3}=1\end{array}$ $\implies$ $1=z^{n}=(z^{3})^{k}\cdot z^{r}\Longrightarrow r=0\Longrightarrow 3\ |\ n\\\\ 1=z^{n}\cdot (z+1)^{n}=(z^{2}+z)^{n}=(-1)^{n}\Longrightarrow (-1)^{n}=1 \Longrightarrow 2\ |\ n \end{array}$ $\Longrightarrow \boxed{\ 6\ |\ n\ }\ .$

me@home

How did you get the horizontal line stuff? I am sort of confused.

In the complex plane the x axis represents all reals... adding by a real shifts a point in its horizontal line...
Very nice solution Virgil Nicula!!! Very aesthetic, good job!!

the center, $O$, $A$, and $A+1$ form an equilateral triangle!

[hide="result follows..."]so 60 = 360k/n, 6|n,
equilateral triangle has altitude as x=axis, then arg(Z)=-30, 210, A^3=1 follows[/hide]


induction
algebra
polynomial
arithmetic sequence
number theory unsolved
number theory

Let $\, p \,$ be an odd prime.  The sequence $(a_n)_{n \geq 0}$ is defined as follows: $\, a_0 = 0,$ $a_1 = 1, \, \ldots, \, a_{p-2} = p-2 \,$ and, for all $\, n \geq p-1, \,$ $\, a_n \,$ is the least positive integer that does not form an arithmetic sequence of length $\, p \,$ with any of the preceding terms. Prove that, for all $\, n, \,$ $\, a_n \,$ is the number obtained by writing $\, n \,$ in base $\, p-1 \,$ and reading the result in base $\, p$.

[hide]Let a 'good number' denote a non-negative integer with no digit of $p-1$ when expressed in base $p$.

Note that we have a 1-1 correspondence between the integers and the good numbers, i.e. we can express each integer in base $p-1$ and read it in base $p$, producing a good number, and conversely we can read each good number's digits in base $p$ as if they were in base $p-1$ to produce an integer. Thus, it suffices to simply show that $a_{n}$ is the $n$th (counting from $n=0$) good number.

Note that for $a_{0}$ through $a_{p-2}$, $a_{n}$ is automatically the $n$th good number. We shall now proceed by induction, i.e. we shall show that if $a_{0}$ through $a_{k}$ are the first $k+1$ good numbers, then $a_{k+1}$ is the next good number.

So assume that $a_{0}$ through $a_{k}$ are the first $k+1$ good numbers for some $k \ge p-2$. We now prove the following lemma:

-------------------------------------------------------
[b]Lemma 1[/b] Any arithmetic sequence of length $p$ contains a number with a digit of $p-1$ in base $p$

[u]Proof[/u]: Consider an arithmetic sequence with first term $x$ and common difference $d$ given by \[x+ad \] for integer $0\le a \le p-1$.
Let $d_{m}$ denote the rightmost nonzero digit of $d$, and say that this is in the $m$th place (i.e. the coefficient of $d^{m}$). Also, let $x_{m}$ denote the $m$th digit of $x$. Now, when we form the arithmetic sequence, the $m$th digit will be formed by successively adding $d_{m}$ to $x_{m}$ and taking that value $\mod{p}$, as a digit must be between $0$ and $p-1$ and any excess multiples of $p$ carry over to higher digit places. We also note that because all the digits to the right of $d_{m}$ (if any) are zero, none of them will carry over to the $m$th place. Because $p$ is prime, $d_{m}$ is nonzero by definition, and $a$ ranges over all the residues $\mod{p}$, the digit in this place must sometime be $\equiv p-1 \mod{p}$ and thus $= p-1$, and in particular this occurs when $a \equiv d_{m}^{-1}(-x_{m}-1) \mod{p}$ (which exists because $d_{m}$ is nonzero and thus has an inverse in this prime modulus). Therefore, a number with a digit of $p-1$ must appear somewhere in a sequence of length $p$, and the lemma is proven.
-------------------------------------------------------

This now means that the next good number does not form any arithmetic sequences of length $p$ with $a_{0}$ through $a_{k}$ because none of them have digits of $p-1$. If we can show that each of the integers between $a_{k}$ and the next good number do form arithmetic sequences of length $p$ with (some of) the numbers $a_{0}$ through $a_{k}$, then the next good number is the lowest number that does not and we are done.

Consider an integer, call it $s$, that is between $a_{k}$ and the next good number. By definition of the 'next' good number, it is not good and has a digit of $p-1$ somewhere. Now form a sequence by replacing each $p-1$ digit of $s$ with each of the integers $0$ through $p-2$ successively, i.e. replace each $p-1$ digit with 0 and this is the first number in the sequence, replace them all with 1 and this is the second number in the sequence, etc., and finally place $s$ at the end of the sequence. This forms an arithmetic sequence of length $p$ whose common difference is a number with a digit of 1 at each place where $s$ had a digit of $p-1$ in base $p$. Because all the terms before $s$ in this sequence have no digits of $p-1$ and are thus good numbers and we assumed that $a_{0}$ through $a_{k}$ consisted of all the good numbers less than $s$, all of the members of this sequence besides $s$ are members of $\{a_{0},a_{1},...,a_{k}\}$, meaning that $s$ does form an arithmetic sequence of length $p$ with (some of) $a_{0}$ through $a_{k}$. Therefore, the next good number is the lowest number that does not and therefore equals $a_{k+1}$, completing the induction, so $a_{n}$ is always the $n$th good number or the number formed by reading $n$ in base $p-1$ and reading the result in base $p$.

QED[/hide]

Seems too easy, do we really need $p$ is an odd prime? I suspect there might be something wrong with this...

[hide]Consider $a_{p-1}$. The preceding $p-1$ terms clearly form the arithmetic sequence $\{0,1,\ldots,p-2\}$, hence we cannot have $a_{p-1}=p-1$, meaning $a_{p-1}=p$. After this, we can have $a_{p}=p+1$, $a_{p+1}=p+2$, etc., until we have another $p-1$ terms that form an arithmetic sequence, in which we will need $a_{2(p-1)}=2p$, instead of $2p-1$, as it would complete an arithmetic sequence of length $p$. Then, we can verify by induction that $a_{k(p-1)}=kp$, and also $a_{k(p-1)+c}=kp+c$, where $c<p-1$.

Using $a_{k(p-1)+c}=kp+c$, we find that $n$, when represented in base $p-1$, consists of the digits of the number $k$ in base $p-1$, followed by the single digit $c$, as $c<p-1$. Finally, $kp+c$ in base $p$ looks the same, beginning with the digits of $k$, followed by the digit $c$, as $c<p-1<p$.

Therefore, we conclude that $a_{n}$ is the number obtained by writing $n$ in base $p-1$ and reading in base $p$.  $\Box$[/hide]

That doesn't work - try $ n \equal{} (p\minus{}1)$

[hide="Solution"]Define a "set number" as a number that can be written in base $p$ without the digit $p-1$. Starting with zero, these are $g_0, g_1, \cdots$.
Define a “non-set number” as all other nonnegative integers.
We must prove that $a_i=g_i$ for all nonnegative $i$.

We do this by induction:
Base case: $a_0, a_1, \cdots, a_{p-2}$ are all set numbers because $a_k=k$ for $0 \leq k \leq p-2$.
Assume that $a_0, a_1, \cdots, a_k$ are the first $k+1$ set numbers. Thus, $a_i=g_i$ for $0 \leq i \leq k$.

Lemma 1: $g_{k+1}$ can be added onto $a_0, a_1, \cdots, a_k$ without creating arithmetic sequences of length $p$ with $a_0, a_1, \cdots, a_k$.
Proof: we prove this by showing that all arithmetic sequences of length $p$ have at least one non-set number.
All terms of the sequence are of the form $y+md$, for $0 \leq m \leq p-1$.
Let $y=Q(p)=c_np^n+\cdots+c_0p^0$ such that $0 \leq c_i \leq p-1$ for $0 \leq i \leq n$.
Each term of the sequence has a unique representation as this type of polynomial, because each coefficient represents a digit base $p$.
Similarly, let $d=H(p)= z_np^n+\cdots+z_0p^0$ such that $0 \leq z_i \leq p-1$ for $0 \leq i \leq n$.
Choose distinct $m$ and $j$ between $0$ and $p-1$.
So $y+md=Q(p)+mH(p)$ and $y+jd=Q(p)+jH(p)$.
Choose $r$ between $0$ and $n$.
We will show that for no value of $r$ are the coefficients of $p^r$ in $Q(p)+mH(p)$ and $Q(p)+jH(p)$ congruent modulo $p$ (for $m \neq j$). Since there are $p$ possible values of $m$ and $j$ and $p$ possible values modulo $p$, there must be at least one coefficient of $p^r$ that is $p-1$ modulo $p$ for some term in the sequence. If this is true, then that term will be $k_np^n+\cdots+k_{r+1}p^{r+1}+k_rp^r+\cdots+k_0p^0$. $k_r=pl+(p-1)$, so the term is $k_np^n+\cdots+(k_{r+1}+l)p^{r+1}+(p-1)p^r+\cdots+k_0p^0$, which is a non-set number.
The coefficient of $p^r$ in $Q(p)+mH(p)$ is $c_r+mz_r$, and $c_r+jz_r$ in $Q(p)+jH(p)$.
Assume $c_r+mz_r \equiv c_r+jz_r$ (mod $p$).
Then $m \equiv j$ (mod $p$) since $p$ and $z_r$ are relatively prime.
But $m$ and $j$ are between $0$ and $p-1$ and $m \neq j$, so they are distinct modulo $p$.
Contradiction.
This completes the lemma.

Lemma 2: $g_{k+1}$ is the smallest term that can be added to $a_0, a_1, \cdots, a_k$ without creating arithmetic sequences of length $p$ with $a_0, a_1, \cdots, a_k$.
Proof: we prove this by showing that all numbers between $a_k=g_k$ and $g_{k+1}$ form arithmetic sequences of length $p$ with $a_0, a_1, \cdots, a_k$.
Let $b=b_np^n+\cdots+b_0p^0$ such that $g_k<b<g_{k+1}$. This means that $b$ is a non-set number.
Let $B(p)=(p-1)(p^{e_r}+\cdots+p^{e_0})$ and let $G(p)=z_{f_s}p^{f_s}+\cdots+z_{f_0}p^{f_0}$ such that $0 \leq z_{f_i} \leq p-2$ for $0 \leq i \leq s$, $\{f_0, f_1, \cdots, f_s\} \cap \{e_0, e_1, \cdots, e_r\} = \{{\o}\}$, $\{f_0, f_1, \cdots, f_s\} \cup \{e_0, e_1, \cdots, e_r\} = \{0,1,\cdots,n\}$, and $b=B(p)+G(p)$.
Since $b$ is a non-set number, $B(p) \neq 0$.
Let $d= p^{e_r}+\cdots+p^{e_0}$.
We see that $b$ is the $p-1$th term of an arithmetic sequence of length $p$, first term $G(p)$, common difference $d=\frac{B(p)}{p-1}$.
All the other terms in this sequence are $G(p)+m\frac{B(p)}{p-1}$ for $0 \leq m \leq p-1$.
Since $G(p)$ has no coefficients equal to $p-1$, $G(p)+m\frac{B(p)}{p-1}$ has no coefficients equal to $p-1$ for $0 \leq m \leq p-2$. Thus, all other terms in the sequence are set numbers less than $b$, so they are contained within $a_0, a_1, \cdots, a_k$.
This completes the lemma.

Thus, the smallest term that can be added to $a_0, a_1, \cdots, a_k$ that will not create any arithmetic sequences of length $p$ with any of the preceding terms is $g_{k+1}$. Thus, $a_j=g_j$ for all nonnegative $j$, by strong induction.

$Q.E.D.$[/hide]


If $f : \mathbb{R} \to \mathbb{R}$ is infinitely differentiable and $f(x)$ is is rational for all rational $x$, must $f'(x)$ be rational for all rational $x$?

Hi Myth,

I found them on Gabriel Carroll's website. He was a former USA IMO team member, and has some good problems on his website. It is [url=http://www.people.fas.harvard.edu/~gcarroll/math/index.htm]here[/url].

But where exactly on that page? There's a lot of stuff to look at there...

[quote="prime_lover"]I found them on Gabriel Carroll's website. He was a former USA IMO team member, and has some good problems on his website. It is [url=http://www.people.fas.harvard.edu/~gcarroll/math/index.htm]here[/url].[/quote]
As I remember, Gabriel published many problems whose solutions he doesn't know  :?

On Gabriel's page, go to the document for "problems previously labeled as unsolved".  There you can find a solution by David Speyer.  The answer is "no".


limit
real analysis
real analysis unsolved

Let f(0)>0 and f(n)=f([n/2])+f([n/3])+n/6.
 a) Prove that  \lim f(n)/n=1;
 b) Prove that for all x>ln (5/2)/ln 3 we have  \lim (f(n)-n)/n^x=0.

I don't think this is so ugly or difficult. I would really like to see different solutions.


Is there a way to enclose my text in a box?  Could I color that box in?

i think you type \boxed{type stuff here}
$\boxed{test}$
$\boxed{[color=red: d132744085]test[/color: d132744085]}$

that doesn't work? is that a glitch?

I think make the box a single row/column tabular environment and use the colortbl package, but I can't figure out how to use the thing.

Oops.  I spelt the package name wrong.  That turned out well.  I can't figure out how to put the latex in here, but here's the code(make sure to install colortbl package):

\begin{tabular}{
   >{\color{white}
     \columncolor[gray]{.2}}r r
 }
\textsc{Lecture Ticket 1} & \textsc{Heat}  \\
\hline
\end{tabular}

Color isn't available on the forum which is why you couldn't show the effect here.


calculus
integration
function
trigonometry
calculus computations

We know there are some elementary functions which primitives are not elementary (such as $ \int\frac {\sin x}x dx$). However, how can we determine the primitives of a elementary function are elementary or not?

Besides, if we know the primitives of a function are elementary, is it sure that we can intergate the function by the methods in text (change variable and integration by parts, with some basic formulas)?

Hi Ifai.  Here's a link describing the phenomenon and I'm quick to admit I don't understand it.  Just giving you the reference:

http://web.archive.org/web/20010731204011/http://pauillac.inria.fr/algo/bsolve/constant/itrexp/wiener.html

Or look [url=http://math.hunter.cuny.edu/ksda/papers/churchill.pdf]here[/url] for more detailed exposition.


since most of dem over here have taken fiitjee open test let me put forth my question

i've recieved a mail from fiitjee statin u've got scholarship from fiitjee 4 your rank in open  test but u can exercise it only in case u join their courses gmp and rtpf  :|  :huh:
now i feel dat almost al've got dis scholarship(no matter wat's their score)
but just wanna clarify it and also know if joinin those courses is of any use at al?

12th for sure

if u had enrolled in fiitjee den u would've known dat gmp and rtpf is for 12th only :D

i think there are tests for eleventh too.
neways what scholarship can they offer you, jee is hardly a month and a half away :rotfl:

200 Rs Discount in GMP is what I got  :rotfl:  :rotfl:  :rotfl:

I bet anit would get free GMP

I got about Rs.350/- for GMP and Rs.175/- for RTPF and then Rs.375/- for both. :rotfl:  :rotfl:  :rotfl:

all of u havent ordered for gmp and irtpf i feel
i have got no mail whatsoever from fiitjee

king nerds dont require any mr anit sahu, they are directly gonna get the material  :D   :rotfl: . I AM REFERRING TO YOU. That was fer those mentally damaged poor souls who believe that i am a nerd  :P

Probably they want to offer u the entire package for free for the expected "AIR 1" in IIT-JEE. :rotfl:

Same thoughts da rock  :rotfl:  :rotfl:   :D
P.S: Tell in the other thread of cbse how u have done da rock

hey chemrock ur typing is very bad :lol:
u forgot 4 zeros after 1 in air 1 in iitjee 2008
i m not even sure if i will qualify in jee because i m pretty weak in phy and chem
i fear that i wont be able to clear the cutoff

Yes 100 in physics is a pretty lousy score eh? :P  :rotfl:  :D

anit you'll soon die of me hands i swear  :D
All create havoc   :D  Pardesi wont be here for a month and i wish we create enough havoc to occupy his holidays in mod editing  :rotfl:

[quote="anitsahu"]all of u havent ordered for gmp and irtpf i feel
[/quote] What do you mean by that statement??

Once you do get the mail, tell me how much discount you are getting.

And probably fiitjee feels that all India toppers dont require GMP.

[quote="madness"]king nerds dont require any mr anit sahu, they are directly gonna get the material  :D   :rotfl: . :P[/quote]

so master madness wat do u mean by they r directly gonna get d material :?:

FIITJEE is the stupidest of them all. I get FT5 scorecard after I get open test scorecard.  :rotfl:  :rotfl:  :rotfl:


Putnam
Diophantine equation
number theory solved
number theory

Prove that there exists infinitely many positive integers $n$ such that $n, n+1$, and $n+2$ can be written as the sum of two perfect squares.

By Pell equation : x^2 - 2y^2 =1 has infinitely many solutions
then let a=2y^2 = y^2 +y^2
we have  a+1 = x^2 + 0^2 ;   a+2 = x^2 + 1^2

Both of your solutions coincide with some of those given by the contestants taking part!

This was actually originally a putnam problem.

Here is mi solution:
Let $n=(2x)^2=4x^2$
$\implies n+1=4x^2+1^2$
$\implies n+2=4x^2+2=(2x-1)^2+(4x+1)$
$\implies 4x+1$ it's a perfect square
From here it is easy to see that $x = {6,12,20,30, ...}$
We will show that these comply:
[i]Case examples:[/i] $x=6 \implies 4x+1=25=5^2$
[i]Case Hypothesis[/i]: We will assume that they are fulfilled until a certain $k$
[i]Last case[/i]: $n=k+1 \implies 4(k+1)(k+2)+1=4k^2+12k+9=(2k+3)^2$
And that's it, we already show that they satisfy $x = {6,12,20,30, ...}$ with $n=4x^2$


l  :maybe: et a,b,c be positive real numbers such that  a+b+c=2
Prove the inequality
$ \frac{1}{2}(a^4\plus{}b^4\plus{}c^4)\plus{}1\geq a^3\plus{}b^3\plus{}c^3$
 :roll:

[quote="xin007"]l  :maybe: et a,b,c be positive real numbers such that  a+b+c=2
Prove the inequality
$ \frac {1}{2}(a^4 \plus{} b^4 \plus{} c^4) \plus{} 1\geq a^3 \plus{} b^3 \plus{} c^3$
 :roll:[/quote]
It's wrong. Try $ a\equal{}\frac{5}{3}$ and $ b\equal{}c\equal{}\frac{1}{6}.$ :wink:

[quote="xin007"][quote="arqady"][quote="xin007"]l  :maybe: et a,b,c be positive real numbers such that  a+b+c=2
Prove the inequality
$ \frac {1}{2}(a^4 \plus{} b^4 \plus{} c^4) \plus{} 1\geq a^3 \plus{} b^3 \plus{} c^3$
 :roll:[/quote]
It's wrong. Try $ a \equal{} \frac {5}{3}$ and $ b \equal{} c \equal{} \frac {1}{6}.$ :wink:[/quote]

 you make a mistake :(
 the inequality is true  :D[/quote]
Yes, you are right. Here is my proof:
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$
Hence, $ \frac {1}{2}(a^4 \plus{} b^4 \plus{} c^4) \plus{} 1\geq a^3 \plus{} b^3 \plus{} c^3\Leftrightarrow(a\plus{}b\plus{}c)^4\geq8\sum_{cyc}(a^3b\plus{}a^3c)\Leftrightarrow f\left(w^3\right)\geq0,$
where $ f$ is a linear function. Hence, $ f$ gets a minimal value, when $ w^3$ gets an extremal value,
which happens, when two numbers from $ \{a,b,c\}$ are equal or when $ w^3\equal{}0.$
In these cases we obtain that your inequality is true.

[quote="xin007"]l  :maybe: et a,b,c be positive real numbers such that  a+b+c=2
Prove the inequality
$ \frac {1}{2}(a^4 \plus{} b^4 \plus{} c^4) \plus{} 1\geq a^3 \plus{} b^3 \plus{} c^3$
 :roll:[/quote]

we prove that ineq holds for any $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 2$:

$ \frac {1}{2}(a^4 \plus{} b^4 \plus{} c^4) \plus{} 1\geq a^3 \plus{} b^3 \plus{} c^3$

$ \iff 8(a^4 \plus{} b^4 \plus{} c^4) \plus{} (a \plus{} b \plus{} c)^4 \ge 8(a \plus{} b \plus{} c)(a^3 \plus{} b^3 \plus{} c^3)$

$ \iff 27u^4 \minus{} 48v^2 \minus{} 72u^2v^2 \plus{} 96v^4 \plus{} 8uw^3 \ge 0$

fix $ u$ and $ v$, just need to prove that when $ w$ get min, i.e. just need to prove when $ c \equal{} 0$ and $ a \equal{} b$ ineq holds.

(1), when $ c \equal{} 0$, ineq becomes $ (a \minus{} b)^4 \ge 0$, clear true;

(2), when $ a \equal{} b$, we setting $ a \equal{} b \equal{} tc,(t > 0)$, ineq becomes $ c^4(16t^3 \plus{} 8t^2 \plus{} (1 \minus{} 4t)^2) \ge 0$, clear true.

so, ineq holds.


Are all the prizes within a position (i.e. gold, silver, etc) the same? That is, does someone who gets a 80 get the same recognition/prizes that someone who gets a 90 does? Are there further rankings within these categories?

I ask because I'm wondering if it's worth skipping my homework and studying to do USAMTS again.

We haven't yet decided exactly how we will do the prizes, except that students will get to choose what prizes they receive from a list of available prizes.  This will be done on a first come, first serve basis with higher scores getting to choose earlier.  For example, all perfect scores will be able to start choosing on Date X.  On Date X+(some small number), the rest of the gold winners will choose.  And so on.  We probably will make small date differences within the wide silver and bronze ranges.


quadratics algebra function domain conics parabola

How do you find the inverse of a generic quadratic equation with a domain restriction?
Say the equation is Ax^2+Bx+C=0, then the inverse of that with x=>-B/2A is _______.

Thank you very much in advance.

Are you asking about the inverse of $ y\equal{}ax^2\plus{}bx\plus{}c$ for $ x \ge \minus{}\frac{b}{2a}$.

Switch $ x$ and $ y$:
$ x\equal{}ay^2\plus{}by\plus{}c$

Then solve for $ y$:
$ ay^2\plus{}by\plus{}(c\minus{}x)\equal{}0$, so $ y \equal{} \frac{\minus{}b \pm \sqrt{b^2\minus{}4a(c\minus{}x)}}{2a}$,

Since $ x \ge \minus{}\frac{b}{2a}$ in the original equation, $ y \ge \minus{}\frac{b}{2a}$ in the inverse.

Thus, the inverse is $ y \equal{} \frac{\minus{}b \plus{} \sqrt{b^2\minus{}4a(c\minus{}x)}}{2a}$

So, for x=> -b/2a, the inverse of y=ax^2+bx+c,
the inverse is
y' = (-b + sqrt(b^2-4a(c-x)))/(2a)

with 4ax+b^2-4ac=>0

Right?

That's why x must be greater than or equal to -b/2a, I think. Because if the expression under the radical is negative, then the answer is nonreal.

x=> -b/2a so that it is only half of the parabola that you're looking for the inverse of. If not, there would be no inverse, at least the inverse would not be a one-on-one function.

I don't get the
"with 4ax+b^2-4ac=>0"
part of the question.

I don't see why it needs to have that constraint.

BTW, this is from an old math competition. I can't remember the name and there's no cover page to it, so.......

I think I deleted half of my earlier post.

Yeah, the -b/2a rule is so that the inverse is a function.

The answer to your other question lies under the radical in the equation of the inverse.

The expression is b^2-4a(c-x), which distributes and simplifies to 4ax+b^2-4ac. If this value is negative, then taking the square root will give you a nonreal value. We don't want (or graph) imaginaries, so that is why there is a constraint on that.


I need to learn how to construct a graph using the equilateral method.
i am not given any mesurements. and i dont have any nubers to plug into any equations.
aparently this is possible so please, please help me!

PLEASE PLEASE.
if you know anything please help.
its so fusterating.

Please don't post the same thing twice (or three times). http://www.artofproblemsolving.com/Forum/viewtopic.php?t=183820 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=183824


uhh i did it a differnet way? and got a different answer?

[hide]
$f(x) = ax^3 + bx^2 + cx + d$
$f(\displaystyle\frac{1}{2}) = \frac{1}{8}a + \frac{1}{4}b + \frac{1}{2}c+d$
$f(-\displaystyle\frac{1}{2}) = -\frac{1}{8}a + \frac{1}{4}b -\frac{1}{2}c + d$
$f(0) = d$

so when you add $f(\displaystyle\frac{1}{2})$ and $f(-\displaystyle\frac{1}{2})$ and divide by $f(0)$ you get:

$\displaystyle\frac{\frac {1}{2}b + 2d}{d} = 997 \Rightarrow \frac{b+4d}{d} = 1994 \Rightarrow b=1990d$

you know from vieta's formula (i think) that $c = r_1r_2 + r_1r_3 + r_2r_3$ and you want to find the sum of the reciprocals of that

so to find the polynomial with roots $\displaystyle\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$, you reverse the coefficients

so your new polynomial is $p(x) = dx^3 + cx^2 + bx + a$, and to find what you want, you get $\displaystyle\frac {b}{d}$ which is equal to $\fbox {1990}$

[/hide]

its the same i messed up i forgot negative sign on denominator.


is 0/0 0, undefined, or 1??? please reply asap

$ \frac{0}{0}$ is undefined because division by $ 0$ is undefined.

Undefined, and please read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=189641]this.[/url]

EDIT: Darnit t0rajir0u beat me to it.

You seem to have gotten those last two mixed up.  $ \frac{0}{x} \equal{} 0$ if $ x \neq 0$ and $ \frac{1}{0}$ is also undefined.

$ \frac {0}{0}$ can take any value
Does one of this value be $ \sqrt{\minus{}1}$ or this can take only real values?

[quote="Elio (n)"]$ \frac {0}{0}$ can take any value[/quote]

It [i]can't[/i] take on any value because division by zero is [i]not defined.[/i]

[b]Definition:[/b]  We say that $ \frac {1}{b} \equal{} c$ if and only if $ b \cdot c \equal{} 1$.  $ c$ is then called the [i]multiplicative inverse[/i] of $ b$.

[b]Definition:[/b]  $ \frac {a}{b} \equal{} a \cdot \frac {1}{b}$.  (This is the proper definition of division.)

Clearly there exists no number $ c$ such that $ 0 \cdot c \equal{} 1$.  We conclude that $ 0$ does not have a multiplicative inverse and therefore that division by $ 0$ is not defined.

[quote="stevenmeow"]f 0/0=x than multiplying by 0 on both sides (i know, i know)
u get 0=0.
this also means that 0/0 can be any number[/quote]

This proof is written backwards; you are assuming the result you want in order to prove it.  This proof can't be written forwards because the operation of multiplying by zero is not reversible.  You could say this about [i]any[/i] statement.  If I wanted to prove that $ 2 \equal{} 1$, I could do it by noting that

$ 2 \cdot 0 \equal{} 1 \cdot 0$

according to your logic, which is ridiculous.

[quote="stevenmeow"]Actually the TECHNICALLY 1/0 (LaTeX newbie) is infinite because, going back to the second grade definition, infinite 0's are "needed" to fill up a "1"

and here is my fairly feeble proox of 0/0:

since 0/0 translates into how many zeros can fill up a 0, then we can start by filling up a 0...
But the zero is ALREADY FULL!
But the zero can also be filled with 1 zero!
or it can be filled with ANY NUMBER OF 0's
this is why it is undefined...the expression can take the value of any number.

Also: if 0/0=x than multiplying by 0 on both sides (i know, i know)
u get 0=0.
this also means that 0/0 can be any number

This concludes my feeble proof of 0/0[/quote]
Using the end result to prve the end result is called [b]circular reasoning[/b]. It is important to watch out for it.

Let $ a$ be a nonzero real number. Then, if division by $ 0$ is defined, $ \frac{a}{0}\equal{}b$, where $ b$ is a real number. This means that if you multiply $ b$ by $ 0$, you will obtain the nonzero real number $ a$. However, this is absurd, because $ b \cdot 0$ is clearly $ 0$.

Take any number. 1, for example. Now keep adding 0 to it until you get to 1. This is the principle behind multiplication, only it results in a non-existent answer here. Also 0/0 is undefined because there is an infinite number of times you can add 0 to get 0.

And btw, Elio (n), the square root of -1 is an entirely different concept. We define this as [i]i[/i], and this is the basic unit for a branch of numbers called imaginary numbers (numbers involving [i]i[/i]), and complex numbers (numbers in the form a+b[i]i[/i]).

[u]Examples[/u]

Imaginary:

sqrt(36)=sqrt(36)*sqrt(-1)=6[i]i[/i]

Complex numbers: (All numbers are complex numbers, see below)

3+0[i]i[/i]=3

0+5[i]i[/i]=5[i]i[/i]

3+5[i]i[/i]

Hope this was helpful!


hallo guys please try this problem:
http://www.mathlinks.ro/viewtopic.php?p=1424731#1424731

Why are you linking us to problems that appear to have already been discussed?

it was discussed after i linked. but the point is: i,in my second post in the discussion pointed out that the solutions posted are wrong  :) and told the reason. then i myself posted a solution. i want you people to verify it. :(


[quote="dinoman"]If xy=c, 9xy=a, and 2a-5c=kxy, then k= ?[/quote]

[hide="this is my answer"]9c=a
18c-5c=kc
13c=kc
k=13[/hide]

[quote]If xy=c, 9xy=a, and 2a-5c=kxy, then k= ?[/quote]

[hide="no looky except dinoman"]18xy - 5xy = kxy
it means

18 - 5 = k

k = 13[/hide]

[quote="dinoman"]If xy=c, 9xy=a, and 2a-5c=kxy, then k= ?[/quote]
[hide="solution"]
c = xy
5c = 5xy
a = 9xy
2a = 18xy

18xy - 5xy = 13xy

k = 13
[/hide]

[quote]Well, you all did what the question expected of you, correctly. I have a question, though: can anyone think of a way to make k something different than the answer you've given and still have all of the equations be correct?[/quote]

how do you do that? then wouldn't the equation be false when you insert the new answer to the equation to check your answer? :?

Well, in general, yes -- it is true that if you have a solution and you change it, you probably won't still have a solution.  I'm asking if you can, within the framework of the problem, find a circumstance where all the equations are true and k is not equal to 13.  It's tricky.  You might want to think about what other freedom you have.  (That is, you no longer have the freedom to set k = 13.  Is there any other way to affect values of expressions in the question?)

That's not too hard...
[hide="answer"]
0

Set everything to 0 and see what the value of k is.
[/hide]

[hide]In fact you can set either $x$ or $y$ to 0.[/hide]


calculate pH of following solution

a) 40ml of 0.050M  Na2CO3 + 50ml of 0.040 M HCl

b)40ml of 0.020M Na3PO4 + 40 ml of 0.040 M HCl

Try to view each one of those as titrations. For example, in (a) you added equal chemical quantities of Na2CO3 and HCl, so what will happen first?


Is there any 7x7 matrix, name $ A$ such that null(A)=3,null($ A^2$)=4 and null($ A^3$)=7, here null($ X$) is defined as the dimension of kernel of matrix $ X$.
Thanks in advance.

The answer is no because necessarily $ dim(ker(A^3))\minus{}dim(ker(A^2))\leq{dim}(ker(A^2))\minus{}dim(ker(A))$.
cf Jordan's theory.

[quote="loup blanc"]The answer is no because necessarily $ dim(ker(A^3)) \minus{} dim(ker(A^2))\leq{dim}(ker(A^2)) \minus{} dim(ker(A))$.
cf Jordan's theory.[/quote]

How to prove it?

We may assume $ A\equal{}diag(J_{n_1},\cdots,J_{n_k},U)$ where $ J_{n_i}$ is the nilpotent Jordan's block of dimension $ n_i\geq{1}$ and $ U$ is an invertible matrix.
 $ dim(ker(A))\equal{}k,dim(ker(A^2))\equal{}k\plus{}card(i;n_i\geq{2}),dim(ker(A^k))\equal{}dim(ker(A^{k\minus{}1}))\plus{}card(i;n_i\geq{k})$.


calculus
geometry
analytic geometry
graphing lines
slope

Source: 360 PMC
Let $M$ be a point on the hypotenuse $BC$ of a right triangle $ABC$ and let points $N$,$P$ be the feet of the perpendiculars from $M$ to $AB$ and $AC$. Find the position of $M$ such that the length $NP$ is minimal.

I don't kno how to do this geometrically. I did it analytically, and then by the co-ordinates I found, I realized (this was the hardest thing, actually, and it took the longest) the geometric significance.

And that would be:
[hide]M is always located on the altitude of the hypotenuse.[/hide]
Am I right?

You kno, it's an interesting result, for me.

By the way, what's "360 PMC"?

Yes you are right. To solve the problem geometrically, you should notice that $NP \equiv AM$. This is because ANMP is a rectangle.  And thus NP takes minimum if and only if AM takes minimum. Hence M is the projection of A to BC.

Oh wow, that's right. I can't believe I didn't notice that.

I will blame it on studying calculus for the past few months, while geometry got left behind after I finished the geometry course in my high school in the first semester.  Wow, I really made it a pain in the ass for me.  lol, This is not an interesting result at all.

[quote="Astronum"]If $A(0,0)$ is at the origin of the referencial, B has coordinates $(0,y)$ and $C(x,0)$, then the position of $M$ corresponds to the minimum value of:

$x^{2}(a^{2}+1)+b(2ax+b)$, where $a$ is the slope of $BC$ and $b$ the ordinate of $B$[/quote]ya lol i tried that, it worked, and it's a pain in the ass..

much easier to think about it geometrically, u kno?

[quote="The Master"]
By the way, what's "360 PMC"?[/quote]
360 Problems for Math Contests ..

[quote="The Master"][quote="Astronum"]If $A(0,0)$ is at the origin of the referencial, B has coordinates $(0,y)$ and $C(x,0)$, then the position of $M$ corresponds to the minimum value of:

$x^{2}(a^{2}+1)+b(2ax+b)$, where $a$ is the slope of $BC$ and $b$ the ordinate of $B$[/quote]ya lol i tried that, it worked, and it's a pain in the -..

much easier to think about it geometrically, u kno?[/quote]

Yes, much easier. I read your thoughts.  :lol:


modular arithmetic
logarithms
combinatorics unsolved
combinatorics

Determine the number of subsets $ A$ of the set $ T\equal{}\{1,2,...,2005\}$ such that the sum of elements of $ A$ is divisible by $ |A|$.

$ |A|$ : the number of elements of set $ A$.

That's an interesting argument zhoraster.  I think you answered a different question to the one posed though.  You worked out the number of subsets of any size which add to a multiple of 2005, wheras the question asked was to find all subsets that add to a multiple of the size of the subset..

Oh yes indeed. I saw $ |T|$ not $ |A|$  :blush:

But it is possible to modify the argument and we will get something like the Catalan number. I will post soon.

The original problem is nothing like easy.

[quote="JBL"]I think this is it: http://www.research.att.com/~njas/sequences/A114702[/quote]
I don't think so.

In fact, I computed almost what is needed. Namely, there is a bijection $ \varphi$ between non-empty subsets of $ \overline{1,n}$ with sum divisible by $ n$ and subsets containing $ n$ with sum divisible by $ |A|$.

The bijection is as follows.

If $ A \equal{} \{a_1 < \dots < a_k\}$ and $ \sum_{j \equal{} 1}^k a_k\equal{} rn$ then $ \varphi(A) \equal{} \{a_{k \minus{} r \plus{} 1} \minus{} n \plus{} s,\dots, a_k \minus{} n \plus{} s,a_1 \plus{} s,\dots,a_{k \minus{} r} \plus{} s\}$, where $ s \equal{} n \minus{} a_{k \minus{} r}$.

So the number of $ A\subset \{1,\dots,n\}$ with sum divisible by $ |A|$ is
\[ m_n \equal{} \sum_{m \equal{} 1}^n(a_m \minus{} 1) \equal{} \sum_{m \equal{} 1}^n \frac {1}{m}\sum_{d|n, d\text{ odd}}\varphi(d)2^{m/d} \minus{} n.\]
The GF of $ m_n \plus{} n$ is
\[ \frac {1}{1 \minus{} z} \sum_{d\text{ odd}} \frac {\varphi(d)}{d}\log \frac {1}{1 \minus{} 2z^d}.\]
This resembles two-coloured neckplaces.

By the way, for odd $ n$ the number $ a_n$ of $ A\subset \{1,\dots,n\}$ with sum divisible by $ n$ is same as the number of two-coloured necklaces (non-free, i.e. symmetric necklaces are different) of length $ n$. Is there a bijection?


[quote="quykhtn-qa1"]Let $ a,b,c,d$ be positive numbers.Prove that:
 $ \frac {(a \minus{} b)(a \minus{} c)}{a \plus{} b \plus{} c} \plus{} \frac {(b \minus{} c)(b \minus{} d)}{b \plus{} c \plus{} d} \plus{} \frac {(c \minus{} d)(c \minus{} a)}{c \plus{} d \plus{} a} \plus{} \frac {(d \minus{} a)(d \minus{} b)}{d \plus{} a \plus{} b} \ge \frac {(a \plus{} b \minus{} c \minus{} d)^2}{4(a \plus{} b \plus{} c \plus{} d)}$[/quote]
Dear quykhtn,
I think this is a nice and hard inequality. My solution used the same way at the solution which I posted at mathvn forum, but it is too stuffy. May I know your solution? Thanks.

[quote="can_hang2007"][quote="quykhtn-qa1"]Let $ a,b,c,d$ be positive numbers.Prove that:
 $ \frac {(a - b)(a - c)}{a + b + c} + \frac {(b - c)(b - d)}{b + c + d} + \frac {(c - d)(c - a)}{c + d + a} + \frac {(d - a)(d - b)}{d + a + b} \ge \frac {(a + b - c - d)^2}{4(a + b + c + d)}$[/quote]
Dear quykhtn,
I think this is a nice and hard inequality. My solution used the same way at the solution which I posted at mathvn forum, but it is too stuffy. May I know your solution? Thanks.[/quote]
Sorry,it must be: $ LHS \ge \frac{(a+b-c-d)^2}{8(a+b+c+d)}$
Here is my Solution: :).If it 's not true,I am very sorry
First,we have:
$ \frac{a-b}{a+b+c}-\frac{c-d}{c+d+a}=\frac{a^2-c^2+2ad-2bc-ab+cd}{(a+b+c)(c+d+a)}$
$ =\frac{a^2-c^2+ad-bc-2ab+2cd+(a-c)(b+d)}{(a+b+c)(c+d+a)}$
So,$ 2(\frac{a-b}{a+b+c}-\frac{c-d}{c+d+a})=\frac{(a-c)(2a+2c+b+d)+3(a+c)(d-b)}{(a+b+c)(c+d+a)}$
Similar,we have:
$ 2(\frac{b-c}{b+c+d}-\frac{d-a}{d+a+b})=\frac{(b-d)(2b+2d+a+c)+3(b+d)(a-c)}{(b+c+d)(d+a+b)}$
Then,
$ 2.LHS =\frac{(a-c)^2(2a+2c+b+d)}{(a+b+c)(c+d+a)}+\frac{(b-d)^2(2b+2d+a+c)}{(b+c+d)(d+a+b)}+3(a-c)(b-d)(\frac{b+d}{(b+c+d)(d+a+b)}-{\frac{a+c}{(a+b+c)(c+d+a)})}$
$ -$ $ If$ $ (a-c)(b-d) \ge 0$,we have:
$ (a+b+c)(c+d+a) \le \frac{(2a+2c+b+d)^2}{4}$
So,$ \frac{2a+2c+b+d}{(a+b+c)(c+d+a)} \ge \frac{4}{2a+2c+b+d}>\frac{2}{a+b+c+d}$
Similar,we have:$ \frac{2b+2d+a+c}{(b+c+d)(d+a+b)}>\frac{2}{a+b+c+d}$
On the other hand $ (a+b+c)(c+d+a)=(c+a)(a+b+c+d)+bd>(a+c)(a+b+c+d)$
So,$ \frac{a+c}{(a+b+c)(c+d+a)} <\frac{1}{a+b+c+d}$
Then,we have:
$ 2.LHS  \ge 2\frac{(a-c)^2+(b-d)^2}{a+b+c+d}-3\frac{(a-c)(b-d)}{(a+b+c+d)}$
$ \ge \frac{(a-c)^2+(b-d)^2}{2(a+b+c+d)} \ge \frac{(a+b-c-d)^2}{4(a+b+c+d)}$
we have Q.e.D
$ -$ $ if$ $ (a-c)(b-d) \le 0$,similar,we have Q.E.D
Equality holds if and only if $ a=c;b=d$

@ Can_hang2007:
 em rat kham phuc anh,mong duoc anh giup do nhieu   :)

[quote="can_hang2007"][quote="quykhtn-qa1"]Let $ a,b,c,d$ be positive numbers.Prove that:
 $ \frac {(a \minus{} b)(a \minus{} c)}{a \plus{} b \plus{} c} \plus{} \frac {(b \minus{} c)(b \minus{} d)}{b \plus{} c \plus{} d} \plus{} \frac {(c \minus{} d)(c \minus{} a)}{c \plus{} d \plus{} a} \plus{} \frac {(d \minus{} a)(d \minus{} b)}{d \plus{} a \plus{} b} \ge \frac {(a \plus{} b \minus{} c \minus{} d)^2}{4(a \plus{} b \plus{} c \plus{} d)}$[/quote]
This inequality is true, too, dear quykhtn-qa1. You can try on it. :-)[/quote]
Thank you very much.But we have an nice inequality than stronger: :)
$ \frac{(a\minus{}b)(a\minus{}c)}{a\plus{}b\plus{}c}\plus{}\frac{(b\minus{}c)(b\minus{}d)}{b\plus{}c\plus{}d}\plus{}\frac{(c\minus{}d)(c\minus{}a)}{c\plus{}d\plus{}a}\plus{}\frac{(d\minus{}a)(d\minus{}b)}{d\plus{}a\plus{}b} \ge \frac{17(a\plus{}b\minus{}c\minus{}d)^2}{40(a\plus{}b\plus{}c\plus{}d)}$
 :)


glycerol on treatment with oxalic acid gives acrolein. is it pinacol pinacolone rearrangement foll by reduction or is there some other mechanism altogether

No, that's not the mechanism. Here's the correct one:

Here the answer also depends on the reaction conditions.

At higher temperatures about 120 degrees only it eliminates and that too forming allyl alcohol and not acrolein. It will first form glyceryl dioxalate which then on heating will evolve CO2 to form allyl alcohol.

At lower temperatures only a monooxalate is formed which will expell HCOOH and revert back to glycerol. This is what I have of heard of and have been thought.

I don't agree with that, since acrolein formation is very favourable. Anyway, the question was not on the product but on the mechanism. Also, having in mind the usual and very important application of this reaction (the in situ dehydration of glicerol in acid to acrolein is one of the principal synthetic methods for the preparation of quinolines) I don't think its relevant at all (and that should not be the purpose any related problem) to consider the importance of such possible side reactions.

isnt the above mechanism d same as pinacol pinacolone rearrangement foll by dehydration. but thanks neways. btw what chemrock says seems to be rite coz at higher temp, around 220c, allyl alcohol is the major product. at least dat is wat my book sez.

No, the mechanism above is not a rearrangement.


Hey. I was wondering if anyone knows the dimensions of an Olympic archery target. I've been training very hard and tend to get about 25% to 35% yellow and the rest red. I'm not sure if my target is smaller or bigger than the standard target.

Also, I compete in some fairly regional competitions, but I want to know if there are any multi-state competitions that anyone knows about that is illegible for someone under 18.

[url]http://en.wikipedia.org/wiki/Image:Archery_Target_80cm.svg[/url]
maybe this helps?

About the multi-state competition, I'm sorry but I can't help : I have no idea what country you live in.  It could be Mongolia... but Liberia just as well. :|

i live in gondor jk jk

i live in the USA


1)  If $ v_1, v_2 ... v_j$ are vectors such that $ v_i \cdot v_j \equal{} 0$ (the dot product) where $ i$ does not equal $ j$, show that they are linearly independent

2)  Let V and W be subspaces in $ R^n$ such that dim(V) + dim(W) = n (I think this refers to the amount of columns in matrix A).  Show there exists a matrix such that N(A) = V and C(A) = W.

This question seemed a bit too easy.  The column space and null space of a matrix A is always a subspace so V and W are subspaces.  By rank-nullity, the dimension of N(A) + dimension of C(A) = n, and since N(A) = V and C(A) = W then dim(V) + dim(U) = n.  Is this right?

3)  If V lies in $ R^n$ and is a subspace, does there exist A such that N(A) = V?

This one seems pretty intuitive.  The null space of a matrix A is always a subspace.  I think the point of the question is to ask can the null space be any subspace V?  I'm not sure how I can show that the nullspace of a matrix can be anything.

Thank you for your help

A useful characterization: the null space of a matrix is the orthogonal complement of the row space. (This can be a little weird outside of $ \mathbb{R}$, but there's still a symmetric bilinear form there)

Given a desired null space, can you find row vectors that span its orthogonal complement?

On the other hand, #2 is strictly stronger, and you haven't answered it.

On #1, one must also state as a condition that none of the $ v_i$ is the zero vector. In fact, if one leaves $ \mathbb{R}^n$ for stranger and more general vector spaces, one should state that as $ v_i\cdot v_i\ne 0$ for each $ i.$ If you don't have this condition, then the statement is false.

The proof is simple. Suppose $ \sum_i\alpha_iv_i\equal{}0.$ Take the inner product with $ v_j$ for some specific $ j.$ This then becomes

$ \sum_i\alpha_iv_i\cdot v_j\equal{}0\cdot v_j$

$ \alpha_jv_j\cdot v_j\plus{}\sum_{i\ne j}\alpha_iv_i\cdot v_j\equal{}0$

$ \alpha_jv_j\cdot v_j\equal{}0.$

Then, with the condition that $ v_j\cdot v_j\ne 0,$ we have that $ \alpha_j\equal{}0.$ We can repeat for each $ j.$ Hence if a linear combination of the $ v_i$ is zero, then each coefficient is zero. That's independence.


find a, b , c , d , e  if abcde X 4 =edcba


a,b,c > 0. Prove that:
sqrt [a/(8b+c)] + sqrt [b/(8c+a)] + sqrt [c/ (8a+b)] is greater or equal than 1.

I think, with weighted Jensen, that $ \sum_{cyc}\frac {\sqrt {a}}{\sqrt {15b \plus{} c}} \ge 1$: as it is homogeneous, let $ \sum_{cyc}\sqrt {a} \equal{} 1$. Then let $ f(x) \equal{} \frac {1}{\sqrt {x}}$ and apply Jensen. Then you get LHS $ \ge$ $ \frac {1}{4\sqrt {\sum_{cyc}{ab}}}$. But $ x \plus{} y \plus{} z \equal{} 1 \implies \sum_{cyc}x^2y^2 \le \frac {1}{16}$.

I don't think you applied correctly Jensen,check your proof again.

[hide="wrong again, ignore please"]Oh. Sorry. I think it should be LHS $ \ge$
\[ \frac {1}{\sqrt {\sum_{cyc}{(8a\sqrt {a} \plus{} a\sqrt {b})}}}
\]
which is not pretty anymore. Hmm. But by Rearrangement, the denominator is at most $ \sqrt {9\sum_{cyc}a\sqrt {a}}$. By Power Mean, $ \sum_{cyc}a\sqrt {a}\ge \frac {1}{9}\sum_{cyc}\sqrt {a} \equal{} \frac {1}{9}$, so the proof is complete.

EDIT: argh, power mean goes the wrong way.

EDIT ARGH THIS IS ALL WRONG.[/hide]

OK how about this: Let $ abc\equal{}1$. apply weighted AM-GM with weights that are the denominators. Then, LHS $ \ge$ $ abc^k$ for some positive $ k$, so $ LHS\ge 1$. (I am still not sure about this :( )


trigonometry
real analysis
real analysis unsolved

OK, last question of the day.....  :D

for $ x>0$, prove

$ e^{x}\cdot \sin x\;=\;\sum_{k=1}^\infty\;\frac{2^{\frac{k}{2}}\cdot \sin \left( \frac{k\pi}{4}\right)}{k!}\;\;x^{k}$

Expand the RHS of $ e^{x}\sin x = e^{x}\frac{e^{ix}-e^{-ix}}{2i}$ will give the answer.

Or let $ I_{n}= 2^{-\frac{n}{2}}e^{x}\sin\left( x-\frac{n\pi}{4}\right)$, then $ I_{0}= e^{x}\sin x$, $ \frac{dI_{n+1}}{dx}= I_{n}$ and $ I_{n+8}= \frac{1}{16}I_{n}$. Integrate each side of $ -e^{x}\leq I_{0}\leq e^{x}$ 8$ n$ times, product $ 2^{4n}$ to each side, and take $ n\to\infty$.


geometry
geometric transformation
dilation

Would someone mind briefly explaining the differences and similarity between electrolysis of saturated salt solution and dilute salt solution please? The equations for the Anode and Cathode will be much appreciated.

do u mean concentrated salt solution?

The difference is the preference of ions to be discharged.

For example is NaCl solution.

1) When it's concentrated solution, you will get Hydrogen gas @ the cathode and Chlorine gas @ the anode.

2) When it's dilute solution, you will get Hydrogen gas @ the cathode and oxygen gas and water @ the anode.

In the 1st case, the concentration of chloride ions is higher than the the hydroxide ions in thesolution. Thus, chloride ions are discharged although chlorine is less reactive in term of electonegativity than hydroxide ions.

In the second case, as the concentration of OH- ions is the same or higher, OH- ions will be discharged, producing oxygen gas and water. This's so b'coz OH- ions are more reactive in term of electronegativity.

That's what I know. Sorry if it's confusing  :P


Hey,

Newbie to here :)

Anyways,

Every 3rd FN is divisible by 2 (f(3)), and every 4th is divisible by 3 (f(4)), and every 5th FN is divisible by 5 (f(5)), Right?

How to prove that it works forever?

(I hope I make sense, forgive me if I dont)

You can give me hints at the beggining, and If I still do not get it, the answer can be revealed.

Please, kindly forgive my shortcomings.

What can you say about $ F_{n\plus{}k} \pmod{F_n}$ and $ F_k \pmod{F_n}$? Note that $ F_0 \equiv F_n \equiv 0 \pmod{F_n}$; try to use the recurrence relation from there.

What does Mod mean?

I am starting AP calculus, I dont know waht that means yet.

Is it possible that you post the solution inside a spoiler?

If $ a \equiv b \mod m$, it simply means that $ a$ and $ b$ have the same remainder upon division by $ m$.

There is a related statement in a HMMT Guts Round that $ \gcd (F_m, F_n) \equal{} F_{\gcd (m,n)}$. How do you prove this?

If possible, can someone put the proof in a spoiler?


integration
inequalities
inequalities proposed

?

take $f(x)=x^2$. you get ${\sf Osc}(f;[0,1])=1$ whereas $|f^{\prime\prime\prime}|_2=0$, thus the inequality is trivially false.

Peter

[quote="flip2004"]... prove that
\[\mbox{\sf Osc}(f;[0,1])  \le \frac{1}{2\sqrt{30}}\parallel f'''\parallel_2  \hsp .\]  [/quote]
I proved in  $ C_0^{3}[0,1] $ following inequality
\[\mbox{\sf Osc}(f;[0,1])  \le \frac{4}{15} \parallel f'''\parallel_2  \hsp .\]
Though my estimates are quite brute force.

Hi Peter and Myth,
many thanks. I change the proposed question . Regards, Alex


trigonometry
rational
point set
algebra
roots of unity
IMO
IMO 1975

Can there be drawn on a circle of radius $1$ a number of $1975$ distinct points, so that the distance (measured on the chord) between any two points (from the considered points) is a rational number?

Hmm... I think the answer is yes and this will go by induction.

Assume we have such a structure for $n$, and consider a chord [a,b] on the circle. Then if there is a point with rational distance to a and to b, by ptolemy's theorem it has rational distance to every other point in the structure. (since the quadrilateral is cyclic)

But now we have to prove that there is a chord for which there are an infinity of points having rational distance to both points...

Wlog assume one point is $(1,0)$, and represent a second point as $[\cos(t),\sin(t)]$. Then we find that $\sqrt{2-2\cos(t)}=2\sin\left(\frac t2\right)$ has to be rational, so there are an infinity of points at rational distance of that one point, more specifically all points with rational y-ordinate... but then I miss the rigorous argument to complete the proof... anyone able to finish?

Start with the two points A=(-1,0) and B=(1,0).

Another point (cos t, sin t) has rational distance to both if both cos(t/2) and sin(t/2) are rational.
An infinite number of these points can easily be found by either using the formula for Pythagorean triples or intersecting a line through (-1,0) with rational slope with the circle.

Ptomely's theorem ensures that any two points X,Y among them have also rational distance as XY is the only unknown distance in ABXY.

Since there are infinitely many primitive Pythagorean triples, there are infinitely many angles $\theta$ s.t. $sin\theta ,cos\theta$ are both rational. Call such angles good. By angle-sum formulas, if $a,b$ are good, then $a+b,a-b$ are also good.

For points $A,B$ on the circle $\omega$, let $\angle AB$ be the angle subtended by $AB$. Now inductively construct points on $\omega$ s.t. all angles formed by them are good; for 1,2 take any good angle. If there are $n$ points chosen, pick a good angle $\theta$ and a marked point $A$ s.t. the point $B$ on $\omega$ with $\angle AB=\theta$ is distinct from the $n$ points. Since there are infinitely many good angles but finitely many marked points, such $\theta$ exists. For a previously marked point $P$ we have $\angle BP=\pm \angle AP\pm \angle AB$ for suitable choices for the two $\pm$. Since $\angle AP ,\angle AB$ are both good, it follows that $\angle BP$ is good, which finishes induction by adding $B$.

Observe that these points for $n=1975$ work: since $AB=2sin\angle AB$ for $A,B$ on the circle, it follows that $AB$ is rational, and so we're done.

Here are 2 very nice solutions:

1) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2016134#p2016134

2) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2182412#p2182412

[hide=yes]Since there are infinitely rational points on the unit circle.\\

we choose such a point $z_{A}$ s.t. $z_{A}=\cos{\alpha_{A}}+i\sin{\alpha_{A}}$ s.t. $(\cos{\alpha_{A}},\sin{\alpha_{A}})\in \mathbb{Q}$\\
so we get $z_{A}^2=\cos{2\alpha_{A}}+i\sin{2\alpha_{A}}$\\

similarly we choose $z_{B}^2=\cos{2\alpha_{B}}+i\sin{2\alpha_{B}}$ s.t. $\alpha_{A}\neq \alpha_{B}$ and $(\cos{\alpha_{B}},\sin{\alpha_{B}})\in \mathbb{Q}$\\

so we get |$z_{A}^2-z_{B}^2|=\sqrt{2-2\cos{2(\alpha_{A}-\alpha_{B})}}=2|\sin{(\alpha_{A}-\alpha_{B})}|=2|\sin{\alpha_{A}}\cdot \cos{\alpha_{B}}-\cos{\alpha_{A}}\cdot \sin{\alpha_{B}}| \in \mathbb{Q}$\\

since choice of $z_{A}$ and $z_{B}$ was arbitrary , so for any no. of points we can have that distance between any two points is rational\\

hence it works for $n=1975$ also as desired $\blacksquare$ [/hide]

[quote=kunterbunt]Ptomely's theorem ensures that any two points X,Y among them have also rational distance as XY is the only unknown distance in ABXY.[/quote]
Your proof is very nice but maybe it is 'Ptolemy' and not 'Ptomely'?


When I look at my normal report, it shows 1 "normal" topic, 1 "passed" topic, and the rest "mastered".  However, when I click to view my badges, it shows over 30 topics in the "passed" section.  How can I get these synced again?

If you've mastered a topic, you've also passed a topic, thus you have those badges as well.

How many badges do you see in your mastered badges section?

Ahh, whoops - you're right.  The different colors mixed me up :lol:

Yeah, I have the same question, why are the "Passed Subjects" and "Mastered Subjects" badges different?


Solve in integers the equation:

   $5x^2+9y=8$ ;)

We have $5x^2\equiv 8(9)$ then multiplying by $2$, $x^2\equiv 7(9)$ and $x\equiv 4$ or $5 (9)$


Does Mathcounts publish its tiebreaker system?

I believe the first tiebreaker is # of questions correct in the last 10 questions of the Sprint Round. Then, it goes to predetermined questions in the Sprint Round.

With individuals I think it is sprint score first. Then it is the difficulty of the problems answered.

With team it is the team score first and then something with the individual scores.

It is sprint score first, then number of total in the last 10 of the sprint round, then comparison of problems in that subset, and then I am not surew after that

The system is as follows:

individual overall score
questions answered from 30-21 correctly. (did this person get #30 right compared to this person)
questions answered from 8-5 on the target correctly.
questions answered from 20-11 correctly
questions answered from 4-1 correctly on the target
questions answered from 10-1 correctly.

If the people have the exact same score, and answered the same questions correctly, then a specific tiebreaker will be made at the countdown round.

are you sure?? I thought it was your sprint score first :?  :?

Whoops!  :blush:
This is the official procedure: (FROM 2004 OFFICIAL MATHCOUNTS HANDBOOK)

Sprint round score.
The total number correct of a subset of sprint round problems Use problems 30, 29, 28, 27, 26, 25, 24, 23, 22, and 21.
Specific problems from the first sprint round tiebreaker subset compared in the order from 30-21.
The problems from the target round compared in the following order:8, 7, 6, 5, 4, 3, 2, 1.
The total number correct of a subset of sprint round problems 20-11
Specific problems compared from the second sprint round tiebreaker subset compared in the order from 20-11
Specific problems compared from the third sprint round tiebreaker subset compared in the order from 10-1
If tie still exists, go to the Tiebreaker round booklet for further instructions. (Huh.  I guess they have a tiebreaker round.  :? )

For team ties:
Sum of team member's sprint round scores
The total number correct of a subset of team round problems:10, 9, 8, 7, 6
Specific problems compared in the subset:10, 9, 8, 7, 6
Specific problems compared in the subset: 5, 4, 3, 2, 1
If tie still exists, go to the Tiebreaker round booklet for further instructions.

[quote="Klebian"]Yeah, they do have a tiebreaker round.
Everyone gets 2 problems, and a calculator.
You have up to 6 minutes to work on them.
Once you finish, raise your hand, and someone'll record what time people answered.
After raising hand, you can't work on problems.
Then the system for that is:
Number of problems answered correctly
If tie, order buzzed in.[/quote]
So kind of like countdown and target mixed.


Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits.  Find the sum of the elements of $\mathcal{S}$.

[hide="my friend thought 10 was a digit"]
If the digit $n$ is in one of the positions $a,b,c$, then there are $9\cdot 8$ ways of filling the other two positions. So, each digit appears in each position $72$ times. The sum of all digits is $1+2+...+9=45$, and $45\cdot 72=3240$.

Note that the repeating decimal is equal to $\frac{100\cdot a+10\cdot b + c}{999}$. The sum of all such fractions is

$\frac{324000 + 32400 + 3240}{999} = 360$.[/hide]

[hide="The way I did it"]First we count the number of permutations of 10 numbers into three slots to be $10\cdot9\cdot8=720$ total decimals. Now, since no digit is special in any way, each of $a,b,c$ will be any of the digits an equal number of times. So the average value for $a$, $b$, or $c$ will be $\frac{0+1+\cdots+9}{10}=4.5$. So the average of all of the decimals will be $\frac{450+45+4.5}{999}=\frac{499.5}{999}=\frac12$. $720$ numbers averaging $1/2$ must sum to $720\cdot1/2=\fbox{360}$.[/hide]

[hide]The total number of permutations is 10 C 3 = 720. For each abc, there is a unique number (9-a)(9-b)(9-c) as well. Each of these pairs sum to 1. So 720/2 = 360.[/hide]

[hide="Solution that I screwed up on the real test"]
$\mathcal{S} = \left\{ \frac{012}{999},\frac{013}{999},\cdots,\frac{986}{999},\frac{987}{999}\right\}$. There are $10(9)(8)=720$ total elements of $\mathcal{S}$ and the $k$th and $(1000-k)$th elements sum to $1$. This means that there are 360 pairs of numbers that each sum to 1, so the total is then $360(1) = \boxed {360}$.
[/hide]

I did this problem a little differently: first, figure out the number of different repeating decimal numbers there are, and that is 10*9*8, since u cant have two of the same digit. then, the sum of the highest and lowest value for the repeating digits is 12/999 + 987/999, which is 1. from that, u can tell that adding up all the numbers will get exactly one half the numbers u added together, thus 720/2 =360

[hide]Basically, I found the number of ways to get three unique digits $(10*9*8=720)$ and divided by two, realizing that there is a counterpart to every solution, that is, since every set can be represented as $\frac{x}{999}$, there is also a number $\frac{999-x}{999}$ that works as a solution. Since $\frac{999-x+x}{999}=1$, the sum has to be $\frac{720}{2}=\fbox{360}$.[/hide]

[quote="salahodeen"]I did this problem a little differently[/quote]

No you didn't :D Read my solution again.

Yeah, I guess that would be the easiest way.  I made a really careless error on this and came up with 500 :(.

Yeah, that's what I got the first time I did it because I didn't see "distinct." :roll:

[quote=ethancui0529][quote=RedFireTruck][hide=Solution]
There are $10\cdot9\cdot8=720$ numbers of that form.

The mean of the possible digits is $4.5$ so the average number will be $.\overline{111}\cdot 4.5=\frac{4.5}{9}=\frac12$.

Multiplying the mean by the total number of numbers gives us the sum of $720\cdot \frac12=\boxed{360}$
[/hide][/quote]
15 year bump yikes[/quote]

this thread is 3 years older than me :o

[quote=ethancui0529]lol same, @RedFireTruck[/quote]

DoUbLe AgE rEvEaL pOgGeRs

[quote=judgefan99][quote=RedFireTruck][hide=Solution]
There are $10\cdot9\cdot8=720$ numbers of that form.

The mean of the possible digits is $4.5$ so the average number will be $.\overline{111}\cdot 4.5=\frac{4.5}{9}=\frac12$.

Multiplying the mean by the total number of numbers gives us the sum of $720\cdot \frac12=\boxed{360}$
[/hide][/quote]

nice solution :)[/quote]

thank you :)

Rft can you help me here?

[url]https://artofproblemsolving.com/community/c63h2520651[/url]

Solved with [b]amuthup[/b], [b]js104[/b], [b]crazyeyemoody907[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLsclesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b].

Clearly, if $x \in S$ then $1-x \in S$. Thus, the answer is simply $\frac{\left| S\right|}{2}$. However, $S$ has $10 \cdot 9 \cdot 8 = 720$ elements, for an answer of $\fbox{360}$.


function
integration
calculus
real analysis
real analysis solved

let [tex] f(x,y) = 0 [/tex] if x or y is irrational and [tex] f(x,y) = 1/q [/tex] if x is rational and y = p/q in lowest terms.

show that f is integrable on A = [0,1] x [0,1] and find [tex] \int_A{f} [/tex]

so far, i got:
let L(f,P) be lower sum, U(f,P) be upper sum, glb and lub be greatest lower bound and least upper bound.

L(f,P) = 0 since the greatest lower bound of each partition is 0.  similarly U(f,P) = 1/q.

we know 0<1/q < 1.
also, L(f,P) < U(f,P)

so
0< lub {L(f,P)} <  glb {U(f,P)} < 1

i'm stuck here.  usually, if we have something of the form
c < lub{L(f,P)} < glb {U(f,P)} < c, we can say
lub{L(f,P)} = glb{U(f,p)} by the squeeze theorem.  but how do i get them to be equal in this question?  is the integral supposed to be 0?  if so, how would i arrive at this?

The lower sums are all zero, no matter what the partition. That's the easy side. What needs work is the upper sums.

Fix some $\epsilon>0.$ For how many values of $y$ is $f(x,y)>\epsilon$? For that to be true, we would have to write $y=\frac pq$ for some $p<q<\frac1{\epsilon}.$ This happens only a fininte number of times.

Find a partition of the square into horizontal strips. Isolate each of the finitely many values of $y$ in very thin strips. Then

upper sum $<1\cdot$(sum of width of thin strips) $+\epsilon\cdot 1.$

This can be made arbitrarily small. I've glossed over several grubby details, but this is the spirit of the argument.

This is essentially a one-dimensional proof despite the integral in two dimensions.


In my new language, I have only 5 different letters - A,B,C,D,E.  Every word in my language has only 4 letters, and every single word has at least one repeated letter (the repeated letters need not be adjacent, so ABCA is a word, but ABCD is not).  How many different words can be formed in my language?

How exactly did you get that??

and If [b]at least one[/b] letter must be repeated then more than one can be repeated like AAAA or AABB, right?

Oops...I didn't see the "at least 1 repeated letter"...With that, I think it is:

(5^4)-(5*4*3*2)=505.

Don't just give the answer, give an explanation!  (Incidentally, I didn't try to figure out if your answer is correct.)


limit
logarithms
calculus
calculus computations

$=+\infty$ because $\lim_{x\to 0^{+}}\ln{x}=-\infty$.


Assume that $ f(x)\in C[0,+\infty)$ (which means $ f(x)$ is continuous when $ x\in[0,+\infty)$)  s.t. for $ \forall l\in\mathbb{R}$, there's a limited number of roots of $ f(x)=l$  ($ x\in [0,+\infty$)  (including the case that there is no root of $ f(x)=l$)

Then prove,

1) if $ f(x)$ is bounded, then $ \lim_{x\to+\infty}{f(x)}=A$  (here $ -\infty<A<+\infty$)

2) if $ f(x)$ is unbounded, then ${ \lim_{x\to+\infty}{f(x)}=+\infty}$

this isn't exactly formal, but it should give u an idea...

1. since $ f(x)$ is bounded, there exist constants $ a$ and $ b$ such that $ a\leq f(x) \leq b$ for all $ x\geq 0$ (in fact, we may take $ a\equal{}\inf\{f(x)\}_{x\geq 0}$ and $ b\equal{}\sup\{f(x)\}_{x\geq 0}$). so if $ f(x)$ does have a limit as $ x\rightarrow \infty$, it is a real number. if $ \lim_{x\rightarrow \infty} f(x)$ does not exist, then for EVERY real number $ r\in [a,b]$, and for every natural $ n$ and $ \epsilon >0$, there exists $ x_n>n$ s.t. $ f(x_n)\not\in (r\minus{}\epsilon, r\plus{}\epsilon)$. now, by the given condition, we must have that for some natural number $ N$, either $ f(x_n)\geq r\plus{}\epsilon$ for all $ n\geq N$ or $ f(x_n)\leq r\minus{}\epsilon$ for all $ n\geq N$. (otherwise, by the IVT, $ f(x)\equal{}r$ would have unbounded roots. assume WLOG that $ f(x_n)\geq r\plus{}\epsilon$ for all $ n\geq N$, and pick $ r\equal{}\frac{a\plus{}b}{2}$ (remember, the above statement is true for EVERY $ r$). by continuity, there must in fact be a number $ M$ such that $ f([M,\infty)\subseteq [\frac{a\plus{}b}{2},b]$ (or else, again by the IVT, $ f(x)\equal{}\frac{a\plus{}b}{2}$ would have unbounded roots. repeating this procedure, we cut the interval length of the image of $ f$ in half at each step. since these interval lenghts clearly go to 0, $ f(x)$ must in fact have a limit, contradicting our initial assumption.

(2) this goes pretty much the same way.

1) The bounded sequence $ f(1), f(2), ...$ contains a subsequence converging to some $ A \in R$. Let $ \epsilon > 0$. You know $ f(x) \equal{} A \plus{} \epsilon$ cannot happen if $ x$ is large enough. But you also know $ f(n) < A \plus{} \epsilon$ for infinitely many $ n$. The intermediate value theorem will then give $ f(x) < A \plus{} \epsilon$ for all $ x$ large. Etc. The proof of 2) is nearly identical.


all i need is the equation and i can get it from here

The store has 10kg of mixed cashews and pecans,which sell for 8.40 per kg. cashews alone sell for 8.00 per kg and pecans sell for 9.00 per kg. how many kgs of each are in the mix?

10kg of mixed cashews and pecans cost 10*8,4=84 dollars.

x kilos of cashews taken
y kilos of pecans taken

Equation system:
x+y=10
8x+9y=84
---------------

x=10-y
8(10-y)+9y=84
y=4
x=6

Answer: 6 kilos of cashews and 4 kilos of pecans.

Also if you want to do quickly, say u have n pounds of the 9 dollar one. The average price will be 8.n0 so n=4


Given 4 natural numbers. We know that sum of squares of arbitrary 2 of them is divisible bym multiply of two others. Prove that At least three of them are equal. Can be this problem generalised in some way?

WLOG, Let $ a_1\le a_2\le a_3\le a_4$ be the four natural numbers and   $ gcd(a_1,a_2,a_3,a_4) \equal{} 1$
Then $ a_3 a_4|{a_1}^2 \plus{} {a_2}^2$   but   $ 2a_3 a_4 \ge ({a_1}^2 \plus{} {a_2}^2)$ and equality holds only when $ a_1 \equal{} a_2 \equal{} a_3 \equal{} a_4$
Then we get $ a_1 \equal{} a_2 \equal{} a_3 \equal{} a_4 \; or \;a_3 a_4 \equal{} {a_1}^2 \plus{} {a_2}^2\dots \dots 1$
Similarly from   $ a_3 a_4|{a_1}^2 \plus{} {a_2}^2$ and $ 2a_2 a_4 \ge ({a_1}^2 \plus{} {a_3}^2)$ we get
$ a_4 \equal{} a_3, a_2 \equal{} a_1$(in this case it's easy to check that all of them are equal)  or
$ a_2 a_4 \equal{} {a_1}^2 \plus{} {a_3}^2\dots \dots 2$
$ 1 \minus{} 2$ and we get $ a_2 \equal{} a_3 \equal{} k$, where k is a positive number, then $ ka_4|({a_1}^2 \plus{} k^2)$, so $ ka_4 \equal{} ({a_1}^2 \plus{} k^2)$  by similar claim.Now if $ a_4$ and $ a_1$ has a common divisor $ m$, then $ m$ divides $ k$ also, which leads to contradiction to our assumption at the beginning.So $ gcd(a_1,a_4) \equal{} 1$.Also $ a_1 a_4|2k^2 \equal{} 2(ka_4 \minus{} a_1^2)$ , $ a_4|2a_1^2$ .As $ gcd(a_1,a_4) \equal{} 1$ , $ a_4 \equal{} 2$(then at least 3 equal) or $ a_4 \equal{} 1$(then all of them are equal).Done.


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Find all continuous functions $f\colon [0;1] \to R$ such that
\[\int_{0}^{1}f(x)dx = 2\int_{0}^{1}(f(x^{4}))^{2}dx+\frac{2}{7}\]

There are infinetely many solutions. For example we can find solutions $f(x)=a+bx^{c}$. It give $a+\frac{b}{c+1}=2a^{2}+\frac{4ab}{4c+1}+\frac{2b^{2}}{8c+1}+\frac{2}{7}$.


probability
limit
inequalities
function
advanced fields
advanced fields unsolved

Suppose n, m are non-negative integers, for real-valued probability W(n, m) (between 0 and 1), if n<=m(m>0), then W(n, m)=1 and W(n, 0)=0. We have the following recursive equation (r>d):
W(n, m)= W(n-r, m)*n/(n+m) + W(n-r+d, m-d)*m/(n+m).

I conjecture that if n<<m^(r/d), then lim_{m->Inf} W(n, m)=1 (Inf means Infinity). And if n>>m^(r/d), then lim_{m->Inf} W(n, m)=0.

But I do not know how to prove, thanks very much.

Here is the easy half ($ \lim\equal{}0$). Choose $ p,q>0$ very small such that $ p/q\equal{}r/d$ and put $ V(x,y)\equal{}\frac{y^p}{(x\plus{}y\plus{}r)^q}$ for $ y>0$ and $ V(x,y)\equal{}0$ for $ y\le 0$ (sorry, I still prefer $ x$ to $ n$ and $ y$ to $ m$ :P). All we need to check to show that $ V$ dominates $ W$ (possibly with some large constant factor) is the inequality
\[ V(x\minus{}r,y)\frac x{x\plus{}y}\plus{}V(x\minus{}r\plus{}d,y\minus{}d)\frac y{x\plus{}y}\le V(x,y)\]
for large $ x$ and $ y\ge 1$ (plus, of course, that $ V$ dominates $ 1$ for $ y\ge x$, but this part is obvious).
Now notice that $ (y\minus{}a)_\plus{}^p\minus{}y^p\le \minus{} pay^p/y$ and
$ z^{\minus{}q}\minus{}{z\plus{}b}^{\minus{}q}\le qb z^{\minus{}q}/z$, provided that $ y,z\ge 1$, $ 0<p,q<1$, $ a,b>0$ and $ pa<1$. Using these inequalities with $ a\equal{}d$, $ b\equal{}r$, $ z\equal{}x\plus{}y$, we immediately get
$ V(x\minus{}r,y)\frac x{x\plus{}y}\plus{}V(x\minus{}r\plus{}d,y\minus{}d)\frac y{x\plus{}y}\le \frac{y^p}{(x\plus{}y)^q}\minus{}
pd\frac{y^p}{(x\plus{}y)^{q\plus{}1}}$
$ \le V(x,y)\minus{}pd\frac{y^p}{(x\plus{}y)^{q\plus{}1}}\plus{}rq
\frac{y^p}{(x\plus{}y)^{q\plus{}1}}\equal{}V(x,y).$
I still have some trouble with the other part if $ r/d>2$.

OK, here is the other half. It took me quite a while to bring the argument to a postable form :).

We'll do it in 2 steps. First of all, we'll show that the limit is $ 1$ if $ y\ge x^\gamma$ with $ \gamma>\frac dr$. To this end, choose $ \alpha,\beta>0$ such that $ \alpha-\beta=1$ and $ \frac \alpha\beta<\frac rd$ but otherwise arbitrary; let $ z=x+y$, $ t=y+2d$, and consider functions of the form $ \varphi\left(\frac{t^\alpha}{z^\beta}\right)$. Then, proceeding almost exactly as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185627]here[/url], we see that the gain in the first order terms is about $ \varphi'(\tau)\frac\tau z$ where $ \tau=\frac{t^\alpha}{z^\beta}$ and the possible loss in the second order terms is only $ |\varphi''(\tau)|\frac{\tau^2}{zt}$ (we used that $ t$ is dominated by $ z$ here). But $ \frac\tau t=\left(\frac tz\right)^\beta=O(1)$, and we can finish as before.

Now comes the second step. We know now that $ W(x,y)\ge 1-\delta$ if $ y\ge x^\gamma$ and $ x$ is large enough. So, it suffices to check the recurrence inequality for the minorant only for $ t\le z^\gamma$. We shall do the same as before but consider now the functions of the form $ \varphi \left(\frac{t^\alpha}{z^\beta}\xi(z)\right)$ where $ \alpha-\beta=1$, $ \frac \alpha\beta=\frac rd$ and $ \xi$ is a positive bounded away from both $ 0$ and $ \infty$ decreasing convex function.
The first order terms gain now reduces to $ \varphi' \tau|\xi'|$ but the second order loss now is about $ \varphi'\frac\tau z|\xi'|+|\varphi''|\tau^2(t^{-1}z^{-1}+|\xi'|^2)$. The first term here is definitely dominated by the gain term; also, we will assume that $ \xi'=O(z^{-1})$. So, it is enough to find $ \xi$ such that $ \frac{\tau}{zt}\le|\xi'|$. But $ \frac{\tau}{zt}=z\left(\frac tz\right)^\beta\le z^{-1-\beta(1-\gamma)}$ and we see that it will suffice to take $ \xi(z)=1+z^{-\kappa}$ with sufficiently small $ \kappa>0$.

The end.


Duke college calculus AMC USA(J)MO USAMO geometry

[quote="MysticTerminator"]SUCKS[/quote]
You really put a lot of thought into that MysticTerminator. :rotfl:

Their math courses are definitely better than high schools, but they're pretty silly and useless in college, and YOU CANT TAKE COURSES AT DUKE!!! Also, the heavy work load and lack of freedom always makes people who go there get worse (at competitive math). On the other hand, almost everybody there is smart, and they have interesting and challenging courses.

I might go  :D

Arnav thinks everyone sucks  :wink:

No offense to him.

what

I didn't say anyone sucked

I was addressing the issue raised in the first post

YAY YOU ARE USING THE AWESOME ACRONYM FOR AMP

NCSSM sucks, or the math courses?

Why is it so bad?

[quote="MysticTerminator"][quote="b-flat"]What do you think about the North Carolina School of Science and Math?  How do their math courses compare to those of other high schools?  Also, for the people that chose not to go to there, why did you choose not to go there?[/quote]

SUCKS!! [/quote]
It should be clear by now what he belives sucks. :rotfl:

yoooo NCSSM wtf

this is basically the entirety of my opinion on the matter

also yeah it is really junk that they just like send their school to the state math contest

but they provide the site so maybe they get special consideration for that

who knows

like it doesn't actually affect me I guess

but IT'S STILL KINDA JUNK

[quote="MysticTerminator"]yoooo NCSSM wtf[/quote]

[quote="MysticTerminator"]but IT'S STILL KINDA JUNK[/quote]

(VERY) Well said.

but considering that former ARMLers (and sometimes former MOPpers i think) don't qualify (or barely scrape in - yasha) through other regions, whereas everybody and their mother gets to qualify from NCSSM, that's lame.   :noo: really lame, guys.  i don't give a s* if you provide space, that doesn't mean you can fill it with the kind of crap that you do.

[quote="Teki-Teki"]whereas everybody and their mother gets to qualify from NCSSM[/quote]

ahem...was that a slight on my previous comment?

And...NCSSM solicits a lot... :maybe:

not really, i was talking about the state math contest.  in which case my comment should make perfect sense.

[quote="jb05"]With all that said, NCSSM is a WONDERFUL place for those who are very successful but not amazing in math (and I assume the sciences are the same way). It is also a wonderful place for people in very small cities who have no access to comparable peers, universities, etc.[/quote]
I attended NCSSM 1982-1984 and agree with this assessment.  I lived on a cattle farm in Claremont, North Carolina, population 800, a real life version of Mayberry from "The Andy Griffith Show."  I ran across this forum post while doing a Google search for "NCSSM sucks" because I became curious about what malcontents, if any, had to say about the school.  I am currently reading [i]The Teenage Liberation Handbook[/i] by Grace Llewellyn and it sparked the question in my mind about malcontents.  I definitely agree that the lack of freedom would stifle those accustomed to much more, but in my case NCSSM offered [i]more[/i] freedom than I had before that time.  In retrospect, I can see that I could still have made it to my current job as a mechanical engineer at NASA without NCSSM, but whether I would have made it remains another question entirely.

I would suggest that those who already enjoy a great deal of personal liberty and excellent educational prospects without NCSSM stay put.  Those whose lives basically suck due to a lack of readily visible educational and social opportunities should by all means apply.  But do give the Llewellyn book a read just to explore other possibilities.

This thread has inspired me to write and to publish "Advice for Those Considering NCSSM" at

http://tinyurl.com/3dbl9l

Feel free to register and to post your thoughts using the "Discuss this Article" link at the bottom of the article.

Hey,

I'm '05, and I think NCSSM weakened my math skills when I was there. On one hand, my statecon rank dropped from 8 to 11 to 30, but I also increased AIME score, so it's hard to say what happened. Basically, I think what happened is that it's hard to train for math stuff when you're at SM because there's a pressure to be well-rounded, but if you're pretty math-heavy, you can load it up in your free time. Lots of normal people go to SM and feel motivated to train for math because of the math people around, but for the top math people, you're not really motivated, because the school doesn't put much emphasis on the contests.

[quote="kimmy16"]I disagree with the competitive math gets worse at NCSSM, too. The Ligon MathCounts coach, Mr. Rohan Lewis, placed pretty well at Comprehensive in all his years there; you can check out '98 & '99 Comprehensive finals. And he still remembers all his math, but then again, he's not that old.[/quote]

Yeah, but how good of an indicator is the Comprehensive Contest of math ability?  I find it far more significant that Mr. Lewis's only USAMO qualification came in 10th grade, BEFORE he went to NCSSM.  I think that might just be good testimony to the fact that the best math students' competitive skills, although they may not get worse at NCSSM, do not increase at the same rate as the skills of students not at NCSSM.

Oh, by the way, that USAMO qualification was in 1997, in case you care.

[quote]With all that said, NCSSM is a WONDERFUL place for those who are very successful but not amazing in math (and I assume the sciences are the same way). It is also a wonderful place for people in very small cities who have no access to comparable peers, universities, etc.[/quote]

^
 That'd be me, I believe in both respects. I'm in CD 3 and I hope to go, but I'm a freshman in pre-cal, and the said thing is there are probably 1-5 people in the entire county who is a freshman and in pre-cal.

I have high grades (in 6 honors and 1 AP class), literally about 2 pages of ECs, and extenuating circumstances. My school thinks I'm pretty much in, but I'm a worrier.

Oh, and I now I feel like a bad math student looking at the stuff on this site. The only math competition I've been in was the county one, and I didn't place. (3 weeks in Algebra II competing with Cal BC students on an Algebra-Cal test)

@aigiqinf: Don't worry.  If you think you suck, you probably don't.  But if you do suck, that's OK because NCSSM will still take you.  I was talking to a teacher about recommendations after I got him to write me one for RSI and he said that a lot of the kids who got in from Enloe and had asked him for recommendations were not the ones he wrote the best recommendations for.

Another problem in approach: if we look at the concentration of AoPSers, it's probably > 90% in < 8 counties.  (Wake, Durham, Orange, whatever Myers Park is in, Mecklenburg, and New Hanover, plus maybe one other I'm forgetting.)  Somebody from, say, Alamance County would have a [i]radically[/i]different perspective than anybody in the 7 metro counties, just because of the comparative advantage of going to NCSSM versus staying at their current school.

hmm my NCSSM discovery day is... like next week

I will probably not go... but I may... bleh that is like saying nothing

"this may or may not be a right angle"

I believe I have said something rather similar to that once

also guys let's do math instead of talking about NCSSM

for example

given two points and a ratio, how do you find the center of the circle of Apollonius of the two points?

bleh that was a trivial problem

hmm another less trivial problem

find all polynomials $ P$ with real coefficients such that $ P(x) \in \mathbb{Q}$ iff $ x\in\mathbb{Q}$

I think I am pretty close on that one... maybe I should finish it

1. Am I actually allowed to answer that, or are you supposed to do it for something important?

2. Post math problems in another thread. This is definitely an NCSSM thread. Unless you have math problems directly pertaining to NCSSM?

uh the first is from a finished thing

and the second is from APMO '01

and third of all math is more important than ncssm

find the magnitude of the suckiness of ncssm.

answer: dunno.  my meter broke.

I don't know.
It's a darn lot better than the high
schools in my county, so. I'll probably
go there in 11th and 12th grade.
if I make it.  :D