this seems real challenging (atleast to me)
given a number n(n<=1000) find a minimum number which is a multiple of n with the sum of digits equal to n..
like for
10 its 190
11 209
12 48
20 3980
etc
the output will definitely not fit into the long long datatype for n>100 it seems .take care abt that
please help me with this
best regards
vasu

I hope this shows up well... this code is in C++ and should help no matter which language you are using.  There may be one or two mistakes, and it's not perfect, but it's a good guideline.  It's not completed; I'll tell you where it needs to be worked on via a comment.

[code]#include <iostream>
using namespace std;

int tentothe(int y);

int main()
{
  int n, x = 0, y, runningtotal, helper, current, solution;

  cout << "Enter your number: ";
  cin >> n;

  for (solution = 0; solution == 0; x++)
  {
    current = n * x;  //set the current number you're trying

    for (y = 1; tentothe(y) <= current; y++);

    for (helper = 0 ; y >= 0; --y)
    {
      runningtotal = (current / tentothe(y)) - helper;
      helper = runningtotal;  //Not sure about this...
      //NOT FINISHED!!! PLEASE FINISH IF YOU SEE WHAT I'M GETTING AT. : )
    }
  }
}

int tentothe(int y)
{
// make 10 go to the yth power.  There should be an easier way to do this than a
//recursive function, but I'm blanking on it
}[/code]

To .cpp: I'm not sure exactly what you are trying to do, but it looks like you are going through every single multiple of n, finding the sum of digits, and seeing if it works. That isn't going to work at all here; not only because the multiple of n won't fit into a 'long long' variable, but even if it did or you use some arbitrary precision arithmetic, it will take far too long to find a solution.

To galois: this looks very similar to a problem from the recent icpc world finals. Checking every multiple will be too slow; but I'm guessing that checking every single number whose digits add to n will run fast enough. This shouldn't be too hard to do (just store powers of 10 mod n, and repeatedly check for length ceil(n/9), ceil(n/9)+1, ..). I'd expect it to run fast enough as well, since the number of possibilities is increasing very fast, so you should end up finding the correct multiple reasonably quickly.

hey stephen
a doubt here.... i was trying sumthing similar so to say...but what's the point in computing powers of 10 mod n whel all u r checking for are numbers  with digital sum n and on finding one check if it is divisible by n..if it does we are done else continue..
why wud we require the modulo values for the powers..??

kind regards
vasu

If you are using arbitrary-precision arithmetic you won't have to; I was just assuming you would use, say, an integer array of digits, in which case you'd need to check if 1*last digit + 10 * secondtolast + 100 * thirdtolast + ... is divisible by n, and thus can just use powers mod n to reduce the calculations to fit inside an int.

[edit] Actually, you don't even need that; using an array, you just keep multiplying by 10, adding the next digit, then modding by n.

[edit again]Although, depending on how you are producing partitions of n, you're probably just changing a couple of digits at a time, so it would be faster to just add/subtract a power of 10 or so from a running total.

Sorry about that messy code, I had to turn off the computer more quickly than I expected and had to stop writing without a completed program or many comments.  I think that I had an ok idea, but TripleM's [i]does[/i] seem faster once the program can fill in the array with integers... I might work on that

TripleM: When you speak of a 'long long integer', do you just mean the long int type?  That should be enough for standard purposes... And unless you do a lot of programming using an array, it'll be the same limit either way...

galois: Did you mean me when you said Stephen?  Anyways, I used the powers of 10 in an attempt to isolate digits... 1234 / 1000 = 1, 1234 / 100 = 12, etc.  I was trying to use that to my advantage.  I might keep working on this problem later, it seems pretty interesting...

well long long means 64-bit...
and i see some pattern in the patterns dude
for example n=28;
minimum number will be 4 digits
1999
next is 2899,2989,2998,
then 3799,3889,3898,3979,3988,3997
then 4699,4789,4798,4879,4888,4897,4969,4978,4987,4996
then ...
do u notice the patterns the difference are all of the form 9.10^k... except in certain cases where it's to figure out as well
i guess this help in generating the new number in the sequence as strings...

so u c .cpp the long long wont help here .u might have answers running into more than 20 odd digits easily..
and 64-bit allows maximum 18 digits...

best regards
vasu

looks like something wrong with the above post...guys pay little heed to the above for the moment

Hmm...which programming language are you using?

Well, I've tried coding this using my method, and while it works on most inputs <=1000, there are still several where it takes too long.
I've also tried a dynamic programming approach; the time limit at SPOJ is 7 seconds, so this way is working for me up to about input size 600 or so (I'm using java, so it will be faster for you), but 1000 is still too slow.

I have one more idea of how to speed this up, but I'll have to think about it for a while.


geometry
rectangle
trigonometry
trig identities
Law of Sines
AMC

[quote="mathfanatic"]Source: MOCP January 2001, #57

Let ABCD be a rectangle and let E be a point on the diagonal BD with angle DAE = 15:^o:.  Let F be a point in AB with EF perpendicular to AB.  It is known that EF = 1/2*AB and AD=a.  Find the measure of angle EAC and the length of segment EC (in terms of a).[/quote]

Do you  mean F is [i]on[/i] AB?

mathfanatic wrote:Source: MOCP January 2001, #57

Let ABCD be a rectangle and let E be a point on the diagonal BD with angle DAE = 15:^o:.  Let F be a point in AB with EF perpendicular to AB.  It is known that EF = 1/2*AB and AD=a.  Find the measure of angle EAC and the length of segment EC (in terms of a).

[hide]Alright. From right triangle EAF, we have  cos 15 = EF/AE, or 2 cos 15 = AB/AE. Now, using the law of sines in triangle ABE, we fine that equivalent to sin AEB/sin ABD. Ok, now let EAC be x. Then x+15 =CAD=ADB, and AEB=15+FEB=15+ADB=30+x. So, our equation is equivalent to 2 cos 15= sin (30+x)/sin(15+x). Expanding, we have the RHS equal to (sin 30 cos x +cos 30 sin x)/(sin 15 cos x + cos 15 sin x). Using the double angle formulas in the numerator and multiplying the denominator to the other side, we get 2 sin 15 cos 15 cos x + 2 cos :^2: 15 sin x = 2 sin 15 cos 15 cos x +

 2 cos :^2: 15 sin x - sin x. Cancelling like terms, we get sin x =0, or E lies on AC. Thus, angle EAC = 0, and using right triangle ACD, we get EC= AC/2 = a/(2 cos 15). [/hide]

Thanks for your help ComplexZeta. I think I may actually have solved a geometry problem!

So that's where that trig problem you gave me came from.


Who is going to ARML 2009?
I know it early...
Anyone from Oklahoma???

ehh, wrong forum.

you should post this in ARML section.


ratio
geometry
circumcircle
geometry unsolved

In triangle $ ABC$, $ \angle A \equal{} 90^{\circ}$. $ M$ is the midpoint of $ BC$. Choose $ D$ on $ AC$ such that $ AD \equal{} AM$. The
circumcircles of triangles $ AMC$ and $ BDC$ intersect at $ C$ and at a point $ P$.
What is the ratio of angles: $ \frac{\angle ACB}{\angle PCB}$? (Find with complete proof)

It is surprising that no one sent a solution to this problem...(Probably it was quite easy for this section... :huh: )
OK...here is my solution.
We have $ CDPB$ cyclic. So, $ \angle ADP \equal{} \angle MBP$. Also, $ CMPA$ cyclic and $ \angle CAP \equal{} \angle PMB$. And $ AD \equal{} CM \equal{} MB$.
So $ \triangle APD \cong \triangle MPB$. So, $ DP \equal{} BP \iff \text{arc } DP \equal{} \text{arc } BP$ and the result follows.

Oh...I did not answer correctly :P
So...$ \angle ACB \equal{} 2\angle PCB$ and the ratio is certainly $ 2$ :wink:


inequalities
geometry
circumcircle
geometry proposed

Given are $\triangle ABC$ with any $M$ inside triangle, circumradius $R$ prove that
$\frac{MBMC}{a}+\frac{MCMA}{b}+\frac{MAMB}{c}\ge \frac{\sqrt{3}(R^{2}-OM^{2})}{R}$
$\frac{a}{MBMC}+\frac{b}{MCMA}+\frac{c}{MAMB}\le \frac{3\sqrt{3}R}{R^{2}-OM^{2}}$ :)  :)


Given two distinct numbers $ b_1$ and $ b_2$, their product can be formed in two ways: $ b_1 \times b_2$ and $ b_2 \times b_1.$ Given three distinct numbers, $ b_1, b_2, b_3,$ their product can be formed in twelve ways:

$ b_1\times(b_2 \times b_3);$ $ (b_1 \times b_2) \times b_3;$ $ b_1 \times (b_3 \times b_2);$ $ (b_1 \times b_3) \times b_2;$  $ b_2 \times (b_1 \times b_3);$ $ (b_2 \times b_1) \times b_3;$ $ b_2 \times(b_3 \times b_1);$ $ (b_2 \times b_3)\times b_1;$ $ b_3 \times(b_1 \times b_2);$ $ (b_3 \times b_1)\times b_2;$ $ b_3 \times(b_2 \times b_1);$ $ (b_3 \times b_2) \times b_1.$

In how many ways can the product of $ n$ distinct letters be formed?

If I haven't miss understood the question

If $ a_n$ is the number of ways to multiply $ n$ numbers then

$ a_n \equal{} 2^{n\minus{}2} n! \quad n>1$ and $ a_1\equal{}1$

[b]Proof[/b]

$ a_n \equal{} 2 \times n \times a_{n\minus{}1}$ - reasoning, there are two ways to order multiplying by k by the rest and n ways to choose k


inequalities
function
real analysis
real analysis unsolved

given d(x,y) is a metric on X
show that d'(x,y) = d(x,y) / [1+d(x,y)]
also define a metric.

I am struggling in the details of triangular inequility....can someone help me out????

What you want to show is the following: If $a \le b+c,$ then $\frac{a}{1+a}\le \frac{b}{1+b}+\frac{c}{1+c}.$ Here $a,\ b,\ c \ge 0.$

The simplest way to show the above (without fancy inequalities) is by multiplying all the terms, getting rid of the denominators and cancelling terms on either side of the resulting inequality. This actually turns out to be quite easy.

Hi,
I think this method should work:

Please observe that the function $f(x)=\frac{x}{1+x}$ is an increasing function on $(0,\infty)$.
Thus we get , via $d(x,y)\leq d(x,z)+d(z,y)$, that

$\frac{d(x,y)}{1+d(x,y)}\leq \frac{d(x,z)+d(y,z)}{1+d(x,z)+d(y,z)}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(y,z)}{1+d(x,z)+d(z,y)}$

$\leq \frac{d(x,z)}{1+d(x,z)}+\frac{d(y,z)}{1+d(y,z)}$,
and you are done.
Didi


find the velocity v in the figure assuming that all the pulleys and ropes are of negligible mass in the file attached to the post..

$ v \equal{} \frac{4}{3}\;m/s$

Attach images directly, you need not compress them. If they are too large, save in .gif format.

Consider the displacement of the point on different parts of the string.


geometry
circumcircle
power of a point
geometry proposed

Given are three circles $\omega_{i}$ ($i=1,2,3$) with equal radius, who have a common point $P$. Of these three circles $PA, PB$ and $PC$ are diameters of $\omega_{1}, \omega_{2}$ and $\omega_{3}$ respectively.

Draw tangents $l_{1}, l_{2}$ and $l_{3}$ to the three circles, through the points $A,B$ and $C$ respectively. Call $l_{1}\cap l_{2}= \left\{G\right\}, l_{1}\cap l_{3}= \left\{H\right\}$ and $l_{2}\cap l_{3}= \left\{I\right\}$.

Call $X$ and $Y$ the midpoints of $GH$ and $GI$ respectively. Now, call $PX \cap \omega_{1}= \left\{J\right\}$ and $PY \cap \omega_{2}= \left\{K\right\}$.

(a) Prove that $YKJX$ is a cyclic quadrilateral.

If we call $L$ the midpoint of $XY$, then the circle through $L,K$ and $Y$ intersects $PC$ again in $M$.

(b) Prove that $PA$ is tangent to the circumcircle of $\Delta ALM$

Again, this is a self-made problem. I don't think it is very nice, because it is too transparant. If anybody has any comments or suggestions to improve my problems or any other remarks, they would be really appreciated. Please tell me (by posting or PM) what you think, even if you think this is a lousy problem (Especially if you think it is lousy!)
The same goes for http://www.mathlinks.ro/Forum/viewtopic.php?t=124552  :)

[quote="Jan"]
(a) Prove that $YKJX$ is a cyclic quadrilateral.
[/quote]

Let $\omega$ be the circle with centre $P$ and radius $r : = PA = PB = PC$.

Then circle $\omega_{1}$ is the inverse of line $GH$ with respect to $\omega$, as $\omega_{2}$ is the inverse of line $GI$.  In particular, $J$ is the inverse of $X$, and $K$ is the inverse of $Y$.  Hence, $PX \cdot PJ = PY \cdot PK = r^{2}$.  Then by power of a point, quadrilateral $XJKY$ is cyclic.

[quote="Jan"]
If we call $L$ the midpoint of $XY$, then the circle through $L,K$ and $Y$ intersects $PC$ again in $M$.
[/quote]

Are the variables correct here?  The circumcircle of triangle $LKY$ appears to intersect $PC$ twice.

Oh I'm so sorry nsato, that is indeed a typo. $L$ is not the midpoint of $XY$, but the midpoint of the circle segment formed by XY, ie the intersection point of $PC$ and $XY$ is called $L$.

Sorry for the mistake, now it should be correct.
I fear you will find part (b) too easy, too transparent.

Just a note: Another statement, equivalent to (b) could be: Prove that the circumcentre of $\Delta ALM$ lies on the line $GH$.


floor function
number theory
prime factorization
number theory unsolved

Show that for all prime numbers $p$, \[Q(p) = \prod^{p-1}_{k=1}k^{2k-p-1}\] is an integer.

[hide]For any prime $q$, let $x_{q}=\lfloor{\frac{p-1}{q}}\rfloor$
Then the power of $q$ in the prime factorization of this number is $\sum_{i=1}^{x_{q}}{2qi-p-1}=x_{q}(q(x_{q}+1)-p-1)$
But $qx_{q}\ge p-q+1$, because were it p-q, that would mean $q|p$, a contradiction.  Then the exponent of $q$ is nonnegative, as desired...[/hide]


If you walk for $ 45$ minutes at a rate of $ 4$ mph and then run for $ 30$ minutes at a rate of $ 10$ mph, how many miles have you gone at the end of one hour and $ 15$ minutes?

\[ \textbf{(A)}\ 3.5 \text{ miles} \qquad
\textbf{(B)}\ 8 \text{ miles} \qquad
\textbf{(C)}\ 9 \text{ miles} \qquad
\textbf{(D)}\ 25 \frac{1}{3} \text{ miles} \qquad
\textbf{(E)}\ 480 \text{ miles}
\]

[hide]You walked $ \frac {3}{4} \times 4$ or $ 3$ miles.  You ran $ \frac12 \times 10$ or $ 5$ miles.  Thus, the answer is $ \boxed{\textbf{(B)}\ 8 \text{ miles}}$.[/hide]


I am looking for a good book with great MC problems and easy to understand and well-written solutions and i am stuck between the handbooks and the all time greatest mathcounts hits. Please help soon thank you

By "stuck between", do you mean you can't decide between to two of them? If so, I'd get the all-time greatest MathCounts problems.

thank you izzy
has anyone read both the books and has an opinion about this?

Personally, I believe that you shouldn't get either of these books. Really, doing problems is the best way to improve. I'm assuming your primary goal is to do well in MC. You can't really get good at it by doing thousands and thousands of MC problems. There just aren't enough hard problems compared to the loads of easy problems. Of course, doing MC problems will increase your speed, but you shouldn't worry about that until the competitions actually start. Basically, you should just do many hard problems that challenge you, and there really are many problems out there (AMCs, Purple Comet, etc.) that can do that and prepare you for the hardest MC problems. So that's my suggestion: first challenge yourself, and then work on actual MC problems.

Edit: Why neither of the books are really that good: the handbooks CAN be found on Google if you are sophisticated with the search function and the "Greatest Problems" book really doesn't have such great problems.

i would say handbook

but try to get all the past tests, and if you can solve all the mc problems fairly quickly, move on to harder stuff

if i could have only one or the other, i would take the all time greatest problems. i have the handbook and most of the problems are school level, while the greatest problems are state to national problems-with solutions. so i guess it depends on whether you are striving to make your school team, or to qualify for nationals.

goodluck with either(or both) :D


function
algebra
polynomial
ratio
integration
calculus
abstract algebra

Suppose $ f$ is entire and that $ |Ref|\leq M|z|^k$ for sufficiently large $ |z|$. Show that $ f$ is a polynomial of degree at most $ k$.

It seems like you can replace a part of the hypothesis of lots of standard problems with $ Ref$ instead of $ f$ and still get the same strength conclusion.

for me, i would think about $ e^{f}$, but I am not sure

finish using the Hadamard factorization theorem.

[quote="Albanian Eagle"] the Hadamard factorization theorem.[/quote]

what is that??

That's a big overkill. ;) Just notice that the gradient of a harmonic function is not much greater than the function itself and that the conjugate harmonic functions $ \text{Re\,}f$ and $ \text{Im\,}f$ have essentially the same gradients. So, any reasonable growth restriction on the real part automatically transfers to the imaginary part and, thereby, to $ |f|$.

The alternative is just to use the [url=http://en.wikipedia.org/wiki/Schwarz_formula]Schwarz formula[/url].

I've been using the Schwarz formula on problems like these, but it seems cheating. I was looking to see if anyone had a way of doing it without just recovering $ f$ as the first step.

Thanks. I'll try to figure out the details of that gradient method.

My concern with relating growth of a harmonic function with its harmonic conjugate was that you can define an analytic branch $ f(z)\equal{}\log(1\minus{}z)$ on $ |z\minus{}1|<1$. Then as I approach 1 along the real axis $ |Im(f)|\equal{}|arg(1\minus{}z)| \equiv 0$ but its conjugate $ |Re(f)|\equal{}|log|1\minus{}z\parallel{}\to \infty$ blows up.

I guess this has to do with continuity on the boundary or something?

Yes, that's about the boundary. That logarithmic singularity shows up as a jump discontinuity in the imaginary part.

With the Schwarz formula, you're not using it at values anywhere near the boundary; you want to apply it at some fixed ratio between the point of interest's radius and the outer circle for this application.

Since I just built this for another question, I'll post my version, which doesn't [i]exactly[/i] recover $ f$:

Let $ g$ be the real part of $ f$. It's harmonic, so we have the Poisson formula $ g(z)\equal{}\frac1{2\pi R}\int_{|w|\equal{}R}g(w)\frac{R^2\minus{}|z|^2}{|z\minus{}w|^2}\,d|w|$. Now, $ g(z)\equal{}\frac12f(z)\plus{}\frac12\overline{f(z)}$, so $ f'(z)\equal{}2\frac{\partial g}{\partial z}$. We differentiate that integral formula:
$ 2g(z)\equal{}\frac1{\pi R}\int_{|w|\equal{}R}g(w)\frac{R^2\minus{}z\overline{z}}{(z\minus{}w)(\overline{z}\minus{}\overline{w})}\,d|w|$
$ 2\frac{\partial g}{\partial z}\equal{}\frac1{\pi R}\int_{|w|\equal{}R}g(w)\left(\frac{\minus{}\overline{z}}{(z\minus{}w)(\overline{z}\minus{}\overline{w})}\minus{}\frac{R^2\minus{}z\overline{z}}{(z\minus{}w)^2(\overline{z}\minus{}\overline{w})}\right)\,d|w|$
$ f'(z)\equal{}\frac1{\pi R}\int_{|w|\equal{}R}g(w)\frac{\minus{}z\overline{z}\plus{}w\overline{z}\minus{}R^2\plus{}z\overline{z}}{(z\minus{}w)^2(\overline{z}\minus{}\overline{w})}\,d|w|$
$ f'(z)\equal{}\frac1{\pi R}\int_{|w|\equal{}R}g(w)\frac{w\overline{z}\minus{}R^2}{|z\minus{}w|^2(z\minus{}w)}\,d|w|$

Now, choose $ |z|\equal{}r\equal{}\frac12R$. We have $ |w\overline{z}|\equal{}2r^2$ and $ |z\minus{}w|\ge r$, so $ \left|\frac{w\overline{z}\minus{}R^2}{|z\minus{}w|^2(z\minus{}w)}\right|\le \frac{6r^2}{r^3}\equal{}\frac{6}{r}$ and
$ |f'(z)|\le\frac1{2\pi r}\int_0^{2\pi}2^kMr^k\cdot\frac{6}{r}\cdot 2r\,d\theta\equal{}6M\cdot 2^kr^{k\minus{}1}$
Therefore $ f'$ must be polynomial of degree $ k\minus{}1$ or less; integrating, $ f$ is polynomial of degree $ k$ or less.

As a side note, if we merely start with a one-sided bound $ \text{Re}(f(z))\le M|z|^k$, we can derive a two-sided bound $ |\text{Re}(f(z))|\le M'|z|^k\plus{}C$ of the same order. Away from the boundary, the Poisson kernel doesn't vary that much.


calculus
integration
function
limit
logarithms
topology
real analysis

$h : [0,\infty) \to [0,\infty)$ is a nonnegative measurable function with $\lim_{x\to\infty}\frac{h(x)}{x}= \infty$.
Prove that for each $\epsilon > 0$ there exists a nonnegative continuous function $f: (0,1) \to [0,\infty)$ satisfying :
\[\int_{0}^{1}f = \epsilon \ ,\quad \int_{0}^{1}h \circ f = \infty \]

Inspired by the problem of finding a function $f \ge 1$ satisfying :
\[\int_{0}^{1}f < \infty \ ,\quad \int_{0}^{1}f \cdot \log{f}= \infty \]

Some hint.

We can divide the problem into three.

Step 1. solve easier problem:
Strengthen the hypothesis part by assuming that h(x) = x g(x) for all x with g(x) a continuous non-negative strictly-increasing function. Weaken the conclusion part by replacing the requirement $\int_{0}^{1}f = \epsilon$ with the easier requirement $\int_{0}^{1}f < \infty$.

Step 2. Weaken the hypothesis back.

Step 3. Strengthen the conclusion part back.

[b]NOTATION[/b]
1. $\int f$ = $\int_{0}^{1}f(t)\,dt$. (this is for convenience)
2. $f: (a,b)\uparrow(c,d)$ means that $f$ is a continuous strictly-increasing function with $f(a+) = c$ and $f(b-)=d$, i.e. $f$ is a orientation-preserving homeomorphism from $(a,b)$ onto $(c,d)$.

[b]LEMMA[/b]
Let $N>0$ and $\ell: (N,\infty)\uparrow(0,\infty)$. Then there is a function $f: (0,1)\uparrow(N,\infty)$ with $\int f < \infty$ and $\int f \cdot (\ell\circ f) = \infty$.

[b]PROOF OF LEMMA[/b]
It is easy to find a continuous positive function $w$ defined on $(0,1)$ and a function $g: (0,1)\uparrow(N,\infty)$ satisfying the following two equations:
\[\int w \cdot g < \infty \]

\[\int w \cdot g \cdot (\ell\circ g) = \infty \]
(Hint: try $w = 1/g$).
We may assume $\int w = 1$ and define $W: (0,1)\uparrow(0,1)$ to be an antiderivative of $w$.
Let $f = g\circ W^{-1}$. Then change of variable shows:
\[\int f = \int w \cdot g \]

\[\int f \cdot (\ell\circ f) = \int w \cdot g \cdot (\ell\circ g) \]
[b]THEOREM[/b]
Let $h: [0,\infty)\to[0,\infty)$ be a measurable function with $\lim_{t\to\infty}h(t)/t = \infty$.
For any $\epsilon >0$ there is $f: (0,1)\uparrow(0,\infty)$ with $\int f = \epsilon$ and $\int h \circ f = \infty$.

[b]PROOF OF THEOREM[/b]
It is easy to find a positive real $N>0$ and a function $\ell: (N,\infty)\uparrow(0,\infty)$ such that $\ell(t) \le h(t)/t$ for all $t \in (N,\infty)$.
By the LEMMA, there is $g: (0,1)\uparrow(N,\infty)$ with $\int g < \infty$ and $\int g \cdot (\ell\circ g) = \infty$.
We may assume $\int g$ is large enough so that it is greater than the given $\epsilon$.
Then we can find $b\in(0,1)$ and $f: (0,1)\uparrow(0,\infty)$ such that $f \le g$, $\int f = \epsilon$, and that $f=g$ on $(b,1)$.
We have $\int_{0}^{1}h\circ f \ge \int_{b}^{1}h\circ g \ge \int_{b}^{1}g \cdot (\ell\circ g) = \infty$.

I'm still interested in other's solution. Also what about a discrete version of the theorem (sequence version)?

Discrete version: $h\colon [0,\infty) \to [0,\infty)$ is a nonnegative measurable function with $\lim_{x\to 0}\frac{h(x)}{x}= \infty$. For each $\epsilon > 0$ there exists a  series with positive terms $a_{n}$ such that $\sum_{n}a_{n}\le \epsilon$ but $\sum_{n}h(a_{n})$ is divergent.

Proof: for $k\ge 1$ find an integer $M_{k}$ such that $h(x_{k})/x_{k}\ge 2^{k}$ where $x_{k}=\epsilon 2^{-k}/M_{k}$. Consider the sum of $M_{k}$ equal terms $\epsilon 2^{-k}/M_{k}+\dots+\epsilon 2^{-k}/M_{k}$. This sum is equal to $\epsilon 2^{-k}$. Combining such blocks of $M_{k}$ elements $(k\ge 1)$ into an infinite series, we get a series with $\sum_{n}a_{n}=\epsilon$. On the other hand, the $k$th block of the series $\sum_{n}h(a_{n})$ sums up to at least $\epsilon/M_{k}+\dots+\epsilon/M_{k}=\epsilon$. Hence $\sum_{n}h(a_{n})$ diverges.

P.S. I speculate that the lack of responses to the original problem has something to do with somewhat annoying requirements ($f$ must be continuous, $\int f=\epsilon$) that add little, if anything, to the substance of the problem.


inequalities
algebra
polynomial
trigonometry
induction
triangle inequality
inequalities unsolved

Let $n$ be a positive integer, and  $a_j$, for $j=1,2,\ldots,n$ are complex numbers. Suppose $I$ is an arbitrary nonempty subset of $\{1,2,\ldots,n\}$, the inequality $\left|-1+ \prod_{j\in I} (1+a_j) \right| \leq \frac 12$ always holds.

Prove that $\sum_{j=1}^n |a_j| \leq 3$.

[quote="Singular"]The trick is in figuring out how to handle $|ab + a + b| \leq 1/2$.. I still haven't figured it out yet.

However, one can write $\sum |a_i| \leq \Bigg| \sum a_i \Bigg| \leq \frac{1}{2} + Z$, and this looks like it could reasonably do it, if good enough bounds can be found for the symmetric polynomials.

(where Z is the modulus of the 2nd symmetric polynomial of the $a_i$, plus the 3rd, 4th, .. to the nth.)[/quote]

did you reverse the sign of triangle inequality?

Yes, maybe someone has the official solution ?

sorry.I can't find the office solution...
and I 'm very sorry for breaking one's promise :(  :(

It doesnot mind but there are so many Chinese in the forum, noone has the solution?

Fortunately,I find the offical solution now.And I think this problem is hard.Some problem is easy if you know the solution,but I think this one not competely is.
now let me post the offical solution.
let $a_j+1=r_j e^{i \theta_j},|\theta_j| \leq \pi$ for $j=1,2,\cdots,n$
with the condition we have
$|\prod_{j \in I} r_j e^{i \sum_{j \in I} \theta_j} -1 |\leq \frac 1 2$   $(*)$
we will prove  a lemma first.
$lemma$,if $r,\theta$ are two real number,$r>0,|\theta| \leq \pi$ and $|r e^{i \theta}-1|<\frac 1 2$
then we have :
(1)$\frac 1 2 \leq r \leq \frac 3 2$
(2)$|\theta| \leq \frac{\pi}6$
(3)$|re^{i \theta}-1| \leq |r-1|+|\theta|$
proof of lemma  (1),(2) can be proved in a graph easily(also you can prove it by algebra)
(3) is because
$|re^{i \theta}-1=|r \cos \theta +r\sin \theta-1|=|(r-1)(\cos \theta+i \sin \theta)+(\cos \theta-1+i \sin \theta)| \leq |r-1|+\sqrt{(\cos \theta-1)^2+\sin^2 \theta}=|r-1|+2|\sin \frac{\theta}{2}| \leq |r-1|+|\theta|$
so the lemma is proved.
return to the orginal problem:
with the lemma we have:
$\frac 1 2 \leq |\prod_{j \in I} r_j | \leq \frac 3 2$ ,$|\sum_{j \in I}| \leq \frac{\pi}6$
and $|a_j|=|r_j e^{i \theta_j}-1| \leq |r_j-1|+|\theta_j|$,
so $\sum^n_{j =1}|a_j| \leq \sum^n_{j =1}|r_j-1|+\sum^n_{j=1}|\theta_j| =\sum_{r_j \geq 1}|r_j-1|+\sum_{r_j < 1}|r_j-1|+\sum_{\theta_j \geq 0}|\theta_j|+\sum_{\theta_j < 0}|\theta_j|$
but we have
$\sum_{r_j \geq 1}|r_j-1| =\sum_{r_j \geq 1}(r_j-1) \leq \prod_{r_j \geq 1}(1+r_j-1)-1(**) \leq \frac 3 2 -1=\frac 1 2$
$\sum_{r_j <1 }|r_j-1| =\sum_{r_j \geq 1}(1-r_j) \leq \prod_{r_j < 1}(1-(1-r_j))^(-1)-1 \leq(***) \frac 3 2 -1 \leq 2-1=1$
$\sum^n_{j=1}|\theta_j-1| =\sum_{\theta_j \geq 0}|\theta_j| -\sum_{\theta_j < 0}|\theta_j| \leq \frac{\pi} 6 + \frac{\pi} 6 = \frac{\pi} 3$
so $\sum^n_{j =1}|a_j| \leq \frac 1 2 +1+\frac{\pi} 3<3$
$(**)$ and $(***)$ can be proved by induction easily.
so the problem are proved.
maybe this is my longest post in mathlink. :lol:
and I hope it is clear for you. :)

[quote="zhaobin"]$\sum_{r_j <1 }|r_j-1| =\sum_{r_j \geq 1}(1-r_j) \leq \prod_{r_j < 1}(1-(1-r_j))^(-1)-1 \leq(***) \frac 3 2 -1 \leq 2-1=1$[/quote]

Zhaobin, I don't how to prove this inequality. (i'm refering to this one : $\sum \ldots \leq \prod_{r_j<1} \ldots$) It doesn't seem obvious by induction. Can you help me?

ok,with pleasure. :)
but an another nice proof is coming now :)
 just notice $\frac 1 {1-x} \ge 1+x$ for $0<x<1$
so $\prod_{r_j < 1}(1-(1-r_j))^{-1}-1 \ge \prod_{r_j < 1}(1+(1-r_j))^{-1}-1 \ge \sum_{r_j \geq 1}(1-r_j)$
is it clear?
and I think induction also works.

thanks. now it is clear. my misunderstanding came from that $-1$. i didn't realize it was the exponent


logarithms
calculus
calculus computations

hi
I have a question
find the series expansion for ln(x) centered at point a=1 using the series for
1/(1-u) .where does it converge and why ?/

i tried but i didnt get the right answer . also am not sure if the center has affect .
for example what if the center was a=2 or a=3 .

Hi zidbatos. You are asking for the Taylor expansion of $\log (1-x)$. The easiest way is to follow dheardie's hint.
As to different centers, e.g. $\log (a-x)$, where $a >0$, there is just an additional constant term and some adjustment to $x$ in the expansion. Try it!


$ \lim_{x\to\infty}2^{x}\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{3}}}}}}$

When I solve it, I get $ 2^{x}*0$. Then that becomes $ \infty*0$ which is indeterminable. I don't know the graph and don't want to apply l'hospital's rule. Is there an alternative way to do this problem?

No it's just $ 2^{x}$ times the square root part

Which doesn't answer t0r's question.  How many square roots are there meant to be in "the square root part" -- presumably, some function of $ x$, yes?  (Also, this question would have been much better placed in the Calculus Computations section.)

If there are always an infinite number of squareroots, the big squareroot term converges to 0. Since it is a constant, you evaluate it first, and then take the limit in terms of x, so it's just 0. Just like $ \lim_{x \to \infty} (x \cdot 0) \equal{} 0$, even though x increases without bound.

If there are x surd symbols, and the squareroot term [u]is approaching[/u] (as apposed to "equal to") 0 while 2^x is increasing, you have a much more interesting problem. Let the nth term be f(n). Then:

$ f(n\plus{}1) \equal{} 2^{n\plus{}1} \sqrt{2 \minus{} \sqrt{4\minus{} \left( \frac{f(n)}{2^n} \right) ^2}}$

Letting $ g(x) \equal{} f(x)^2$, we can square and rearrange to get:

$ g(x\plus{}1)  \minus{} g(x) \equal{} \frac{1}{4} \left( \frac{g(x\plus{}1)}{2^{x\plus{}1}} \right) ^2$

So, it's clear by telescoping that the limit is positive, if it exists.

[quote="The Zuton Force"]So, it's clear by telescoping that the limit is positive, if it exists.[/quote]

You could also arrive at this conclusion by noting that $ 2^x>0\forall x$ and $ \sqrt{y}\geq0\forall y$, so their product is $ \geq0$.

I meant that it does not converge to 0  :wink:

The answer depends on whether the first square root outside the 3 and/or the final square root are considered as part of the $ x$ square roots.

[hide="you can use"]
\[ 2\cos \frac{\theta}{2}\equal{}\sqrt{2\plus{}2\cos \theta}\]
[/hide]


Let $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ be real numbers such that \[2 \leq a_i/b_i \leq 3\textrm{ for }1 \leq i \leq n.\]
Determine the maximal value of \[\frac{\left(\sum a_i^2\right) \left(\sum b_i^2\right)}{\left(\sum a_ib_i\right)^2}.\]

Let's see if this works.

We have [tex](a_i-2b_i)(a_i-3b_i)\le0[/tex], i.e. [tex]a_i^2+6b_i^2\le5a_ib_i[/tex]. Sum this to get [tex]\sum a_i^2+6\sum b_i^2\le5\sum a_ib_i[/tex]. Next by AM-GM,

[tex]2\sqrt{6\sum a_i^2\sum b_i^2}\le\sum a_i^2+6\sum b_i^2\le5\sum a_ib_i[/tex], i.e.

[tex]\displaystyle \frac{(\sum a_i^2)(\sum b_i^2)}{(\sum a_ib_i)^2}\le\frac{25}{24}[/tex].

I'm not sure if we actually can get 25/24, though.

(sorry, its 2 in the morning... i'll take a second look when i get up)

Ok, it's 3 in the morning, I've showered, and I now know that we can get 25/24. Set

[tex]a_1=\sqrt{6(n-1)}[/tex],
[tex]a_2=\cdots=a_n=3[/tex],
[tex]b_1=\sqrt{3(n-1)/2}[/tex], and
[tex]b_2=\cdots=b_n=1[/tex].

Of course this only works if [tex]n\ge2[/tex]. If n=1 then the whole thing equals 1 anyway, so I suppose I should be more specific and say that the maximum value of 25/24 is attainable iff n>1.

Moreover we can prove the following generalisation:

Let $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ be real numbers such that \[0 < m \leq a_i/b_i \leq M\textrm{ for }1 \leq i \leq n.\]
Then

 \[1 \leq \frac{\left(\sum a_i^2\right) \left(\sum b_i^2\right)}{\left(\sum a_ib_i\right)^2} \leq \frac{(M+ m)^2}{4Mm}.\]

We could for example prove that the trinomial \[ f(x) = \left(m \sum a_i^2\right) x^2 - \left( (m + M) \sum a_ib_i \right) x + M \sum b_i^2\] has at least one zero, because then its discriminant is $\geq 0$.
(In fact, $f(1/m) \leq 0$ and $f(1/M) \geq 0$.)


$120^4 + 272^4 + 315^4 + n^4 = 353^4$
Find n.

Answer:
[hide]
I got the answer 900 using Chinese Remainder theorem. But are there any better methods? Mine was kinda tedious.
[/hide]

[quote="amcavoy"]Do you mean $n^{2}=900\implies n=30$?[/quote]

oops, sorry, yup. typo.  :blush:

Well, $n=-30$ is also a solution.

Masoud Zargar

[quote="boxedexe"]Well in that case, there are infinite solutions since we could have $\pm 30i^k$, where $k\in\mathbb{N}$. :P

Masoud Zargar[/quote]

those are not all distinct solutions...consider the peridicity of powers of $i$, also this argument is like

x=1 has infinite solutions since

$1,1^2,1^3,1^4,...$ are all solutions

Solutions are only $\pm 30$ I think.... Why do you say that there are more??

[quote="José"][quote="t0rajir0u"]As a fourth-degree polynomial in $n$ there exist(s):

One solution $\in \mathbb{N}$,
Two solutions $\in \mathbb{Z}$,
Two solutions $\in \mathbb{R}$,
Four solutions $\in \mathbb{C}$,

Okay?  :D[/quote]

Yes, but they can be the same sometimes[/quote]

For example, I don't know, $x^4-1=0$

$x^4-1=0$ has four distinct solutions...... right?

$1$
$-1$
$i$, $-i$

An example of where there are not four [u]distinct[/u] solutions is:

$x^4-4x^3+6x^2-4x+1=0$


geometry
rhombus
perimeter
rotation
3D geometry
AMC
AIME

EXPLAIN HOW YOU SOLVED THE PROBLEM PLEASE.

1.In a plane, 10 lines intersect such that no 3 lines meet at the same point. What is the maximum number of regions created by the lines?

2.The lengths of the diagonals of a rhombus are 6 inches and 8 inches. If a circle is inscribed in the rhombus, how many inches are in its circumference? Use 3.14 as an approximation for pi, and express your answer as a decimal to the nearest tenths.

5.The areas of two adjacent squares are 256 square inches and 16 square inches, respectively, and their bases lie on the same line.  What is the number of inches in the length of the segment that joins the centers of the two inscribed circles?  Express your answer as a decimal to the nearest tenth.

6.A sequence of numbers a1, a2, a3... is defined by a1=7, a2=-6 and a, =an-1-an-2 for n>2. What is the sum of the first 2000 terms of the sequence?

7. Two runners began running in the same direction o a quarter-mile track. The first runner began five feet ahead of the second runner. The first runner's  pace is 6 minutes per mile, while the second runner's pace is 8 minutes per mile. Both runners continue at their respective paces for 119 minutes. How many times will the first runner pass the second runner?

9. Regular hexagons are placed side-by-side in a continuous pattern. What is the  maximum number of congruent hexagons that can be placed sid-by-side such that the perimeter of the resulting figure is less than 100cm?    Side to side is 3cm

10. Right AABC has legsmeasuring 8cm and 15cm. The triangle is rotated about one of its legs. What is the number of cubic centimeters in the maximum possible volume of the resulting solid? Express your answer in terms of pi.

#1: It's one more than a 10th triangular number.
#2: Rhombus is orthodiagnal. So find the right triangle.
#5: Draw a triangle connecting both of the radii.

Rest, I am not sure of..

what year is this and what round (chapter, state, nats)? YOu might be able to find some solution on the internet..

For number 10, the formula for a cone is πr²h/3, where r is the radius of the base and h is the height of the cone. So find a way to maximize the radius of the base.

5.

There are two squares, 256 sq. units and 16 sq. units

It says they are adjacent, so they are touching. Also, their bases lie on the same line.

Kinda like this.

|---------|
|           |_____
|           |      |
|_______|____|
Let's call the 256 sq. unit square A and the 16 unit square B
so find the side length of each square. square a is 16 units and square B is 4 units

So to get the center of the inscribed circle, find the center of the square.

So the center of square B is half of 4 off of each side, or 2.

The center of square A is half of each side, or 8.

Then make a right triangle.
The center of square A is 10 to the left of the center of square B

The center of square A is 6 above the center of square B

10^2 + 6^2 = 136 using a calculator to sq. root it you get 11.7 when u round it

[quote="abacadaea"]9th triangluar number. take 1 for example. it obviously have no intersection point[/quote]True, but go beyond it. It always gives 1 more than the number of line you can draw. But I don't think it's 9th one.

[quote="darit28"]how is a cone related[/quote]
The question said: "The triangle is rotated about one of its legs."
I think the radius becomes 15, because we know that 8,15,17 is a pythagorean triple, but we were told that the leg was rotated..

Oh, I think someone mentioned this before, but the area of a cone is just $ \frac{\pi r^2 \cdot h}{3}$

For #9,I will assume that the side of a hexagon is 3.
 I think it's going to be $ 7$.

The first and last hexagons will have 5 sides out and one in between have 4.
So I think the equation is $ (3)5\plus{}(4\cdot3)n\plus{}(3)5<100$
$ n\equal{}5$ and we count the first and last to get $ \boxed{7}$

[quote="darit28"]how is a cone related[/quote]
After rotating, you get a cone. I seem to remember an AIME problem similar to this. Anyway, you just find the area.


I'm incredibly stuck on two proofs:

[i]1. Prove that the altitudes of any triangle are concurrent.

2. Prove that the angle bisectors of any triangle are concurrent.[/i]

Does anyone know how to do these?

[hide="Hint on 2"] Draw the incircle. [/hide]

[hide="Hint on #1"]Connect the feet of the altitudes and prove that the orthocenter is actually the incenter of the smaller triangle.[/hide]

I've been able to prove #1 but am still very stuck on #2 (I tried the hint)

The angle bisector from $A$ and the angle bisector from $B$ meet at a point $I$, such that ....
Can you see how this tells you that the bisector from $C$ goes through this point?

(Remember, an angle bisector of an angle happens to be the locus of points that are [b]equidistant from the two lines that make up the angle[/b])

I understand that this is true.  And I believe that it works.  But I can't for the life of me come up with a rigorous proof.

A rigourous proof: (?)

Let the angle bisector from the vertex $A$ and the angle bisector from the vertex $B$ meet at a point $I$. (They cannot be parallel, just by considering that the sum of the angles in a triangle is $180^{o}$).

$I$ lies on the bisector of $\angle A$ and hence the perpendicular distance from $I$ to the sides $AB$ and $AC$ are equal.
Similarly, since it lies on the bisector of $\angle B$, the perpendicular distance from $I$ to the sides $AB$ and $BC$ are equal.

From these two, the perpendicular distance from $I$ to the sides $AC$ and $BC$ are equal- hence $I$ lies on the bisector from $\angle C$. We have shown that the angle bisectors from $\angle A,\angle B$ and $\angle C$ all pass thorugh a point $I$: they are concurrent.

[hide="2"]Use bysector theorem and then Ceva reciprocal :)[/hide]


geometry
circumcircle
cyclic quadrilateral
geometry proposed

Let $ A_1A_2...A_n $ be a n-equilateral gons inscribed in (O) . $ M \in $ the arc $ A_nA_1 $ (M is different from $ A_1 , A_n $) . Prove that :
 a) $ \frac{1}{MA_1.MA_2} + \frac{1}{MA_2.MA_3} + ... + \frac{1}{MA_{n-1}.MA_n} = \frac{1}{MA_n.MA_1} $
 b) If n is an odd number then : $ MA_1 + MA_3 + ... + MA_n = MA_2 + MA_4 + ... + MA_{n-1} $

This is problem 253 from the last chapter of Geometric Transformations by I. M. Yaglom (Circular Transformations), which has never been translated to English. The first three chapters has been published by MAA as Geometric Transformations I (Isometries), Geometric Transformations II (Similarities) and Geometric Transformations III (Projectivities).

Denote $a = A_1A_2 = A_2A_3 = ... = A_nA_1$ the side of the regular n-gon $A_1A_2...A_n$. Let $d_1 = MA_1$, $d_2 = MA_2$, ... $d_n = MA_n$ be the distances of the point $M$ on the arc $A_nA_1$ of its circumcircle $(O)$ from the vertices $A_1, A_2, ..., A_n$. We have to show that

$\frac{1}{d_1d_2} + \frac{1}{d_2d_3} + ... + \frac{1}{d_{n-1}d_n} = \frac{1}{d_nd_1}$ [color=white]...[/color] (for any n)

$d_1 + d_3 + ... + d_{n-2} + d_n = d_2 + d_4 + ... + d_{n-1}$ [color=white]...[/color] (for an odd n)

Invert the n_gon in a circle centered at the point $M$ and an arbitrary radius $r$. The circumcircle $(O)$ is carried into a line $o$ and the vertices $A_1, A_2, ..., A_n$ into points $A_1', A_2', ..., A_n'$ on this line. The distance of the images $A_i', A_j'$ of points $A_i, A_j$ in an inversion with center $M$ and power $k = r^2$ is

$A_i'A_j' = A_iA_j \frac{r^2}{MA_i \cdot MA_j}$

Hence, the distances $a_{12} = A_1'A_2'$, $a_{23} = A_2'A_3'$, ..., $a_{n1} = A_n'A_1'$ are equal to

$a_{12} = a \frac{r^2}{d_1d_2}$, $a_{23} = a \frac{r^2}{d_2d_3}$, ..., $a_{n1} = a \frac{r^2}{d_nd_1}$

Since the inverted vertices $A_1', A_2', ..., A_n'$ are collinear and arranged on the line $o$ in this order, it is clear that

$a_{12} + a_{23} + ... + a_{n-1,n} = a_{n1}$

Substituting the above expressions for $a_{12}, a_{23}, ..., a_{n-1,n}, a_{n1}$ and reducing by $ar^2$, we get the desired result:

$\frac{1}{d_1d_2} + \frac{1}{d_2d_3} + ... + \frac{1}{d_{n-1}d_n} = \frac{1}{d_nd_1}$

Let $b = A_1A_3 = A_2A_4 = ... = A_{n-1}A_1 = A_nA_2$ be the shortest diagonal of the regular n-gon $A_1A_2...A_n$, where n is odd.

The distances $b_{13} = A_1'A_3'$, $b_{24} = A_2'A_4'$, ..., $b_{n-1,1} = A_{n-1}'A_1'$, $b_{n2} = A_n'A_2'$ are equal to

$b_{13} = b \frac{r^2}{d_1d_3}$, $b_{24} = b \frac{r^2}{d_2d_4}$, ..., $b_{n-1,1} = b \frac{r^2}{d_{n-1}d_1}$, $b_{n2} = b \frac{r^2}{d_nd_2}$

Since the inverted vertices $A_1', A_2', ..., A_n'$ are collinear and arranged on the line $o$ in this order, it is clear that

$a_{12} + a_{23} = b_{13}$

$a_{23} + a_{34} = b_{24}$
...

$a_{n-2,n-1} + a_{n-1,n} = b_{n-2,n}$

Since the inversion center $M$ is carried into the point at infinity, the last 2 equations are different:

$a_{n-1,n} + b_{n-1,1} = a_{n1}$

$b_{n2} + a_{12} = a_{n1}$

Substituting the above expressions for $a_{12}, a_{23}, ..., a_{n-1,n}, a_{n1}$,  $b_{13}, b_{24}, ..., b_{n-1,1}, b_{n2}$ and reducing by $r^2$, we get:

$a\frac{1}{d_1d_2} + a\frac{1}{d_2d_3} = b\frac{1}{d_1d_3}$

$a\frac{1}{d_2d_3} + a\frac{1}{d_3d_4} = b\frac{1}{d_2d_4}$
...

$a\frac{1}{d_{n-2}d_{n-1}} + a\frac{1}{d_{n-1}d_n} = b\frac{1}{d_{n-2}d_n}$

$a\frac{1}{d_{n-1}d_n} + b\frac{1}{d_{n-1}d_1} = a\frac{1}{d_nd_1}$

$b\frac{1}{d_nd_2} + a\frac{1}{d_1d_2} = a\frac{1}{d_nd_1}$

Removing the fractions, rearranging and indexing the equations by $bd_i$,

[color=blue][b](1)[/b][/color] [color=white].....[/color] $bd_1 = ad_2 - ad_n$

[color=blue][b](2)[/b][/color] [color=white].....[/color] $bd_2 = ad_3 + ad_1$

[color=blue][b](3)[/b][/color] [color=white].....[/color] $bd_3 = ad_4 + ad_2$
...

[color=blue][b](n-2)[/b][/color] [color=white]..[/color] $bd_{n-2} = ad_{n-1} + ad_{n-3}$

[color=blue][b](n-1)[/b][/color] [color=white]..[/color] $bd_{n-1} = ad_n + ad_{n-2}$

[color=blue][b](n)[/b][/color] [color=white].....[/color] $bd_n = ad_{n-1} - ad_1$

Adding separately the odd equations (1), (3), ... (n-2), (n) and the even equations (2), (4), ... (n-1), we get

$b(d_1 + d_3 + ... + d_{n-2} + d_n) = 2a(d_2 + d_4 + ... + d_{n-1}) - a(d_n + d_1)$

$b(d_2 + d_4 + ... + d_{n-1}) = a(d_1 + d_n) + 2a(d_3 + d_5 + ...  + d_{n-2})$

Subtracting the 2nd equation from the 1st one, rearranging the result and reducing by $2a + b$, we get the desired result

$(2a + b)(d_1 + d_3 + ... + d_{n-2} + d_n) = (2a + b)(d_2 + d_4 + ... + d_{n-1})$

$d_1 + d_3 + ... + d_{n-2} + d_n = d_2 + d_4 + ... + d_{n-1}$

For n = 3 (equilateral triangle $\triangle A_1A_2A_3$), both results reduce to

$d_1 + d_3 = d_2$

where $d_1, d_2, d_3$ are the distances of the vertices $A_1, A_2, A_3$ from a point $M$ on the arc $A_3A_1$ of its circumcircle. This directly follows from Ptolemy's theorem for the cyclic quadrilateral $A_1A_2A_3M$:

$A_1A_2 \cdot MA_3 + A_2A_3 \cdot MA_1 = A_3A_1 \cdot MA_2$

Since $A_1A_2 = A_2A_3 = A_3A_1 = a$,

$a d_3 + a d_1 = a d_2$

$d_3 + d_1 = d_2$


Find the smallest positive integer $ n$ such that $ 107n$ has the same last two digits as $ n$.

[hide="Solution"]We are looking for the smallest positive integer $ n$ such that $ 107n\equiv n\pmod{100}$.

$ 107n\equiv 7n\pmod{100}\quad\Longrightarrow\quad7n\equiv n\pmod{100}$

Subtracting $ n$ from both sides, we get $ 6n\equiv 0\pmod{100}$.

The smallest positive integer $ n$ such that $ 100\mid 6n$ is $ \boxed{50}$.[/hide]


I am going to post a short tutorial on combinations and permutations.

First I will go through an example before I define these terms:

How many ways can you choose a shortstop, a second baseman, and a catcher from a team of 9 players?

First we consider how many choices we have for the shortstop... there are 9.  Now that we have a shortstop how many choices do we have for the second baseban... we have 8 choices(any of the 8 remaining players).  Finally how many choices do we have for the catcher... 7.  We multiply these numbers together to find the total number of ways we can pick these three positions 9*8*7=504.
(note: it doesn't matter which position we pick first, each way will give us the same answer)

Now let's say that we wanted to find out how many ways we can choose two of these players to play catch.

Using the same logic as last time we find 9*8=72.  But wait we have overcounted... Let's say that two players are A and B.  If we just do 9*8 we will count A playing catch with B as different than B playing catch with A.  So we need to find out how many ways we can arrange A and B or any of the two players.  We can arrange them in 2!=2 ways.  Since we have overcounted by a factor of two we must divide 72 by 2 to get 36 different ways.

If we wanted to find out how many ways to pick groups of three we would do 9*8*7/(3*2*1)=84.

Ok the first example demonstrated the idea of permutations.  A permutation is an arrangement of some objects when order DOES matter.  It is often denoted by $P(n,r)$ or $_nP_r$  where you have a large group of n things and you want to find how many ways you can arrange r of them(so that order does matter).

The second example was of combinations.  A combination is an arrangement of some objects where the order of them DOES NOT matter.  There are a few ways it is denoted:  $_nC_r= ^nC_r=C^n_r=C(n,r)={n\choose r}$  I personally like the last one but the second to last one is easiest to use if you don't have $\LaTeX$  ;) .

Problems:
(please leave these problems to those who are new to this)

1.  How many arrangements are there of the word "math"?

2.  How many arrangements are there of the word "Mississippi"?

3.  How many 5 letter words are there( a word is any arrangement of letter i.e. it doesn't have to be an actual "word")?

4.  How many 5 letter words are there if you can only use a letter once?

5.   How many diagonals does an n-sided polygon have?

6.  How many ways can six people sit around a table?

7.  How many ways can six people sit around a table if Ben and Jamie wont sit next to each other?

8.  How many distinct arrangements of 1001011110 are there?

Feel free to post more problems if you want.

In how many ways can you split up a group of n persons into $k \leq n$ teams (a team must have at least 1 person)?

1.  [hide]

We have 4 diffrent letters and positions so we have 4!=24 diffrent possipilitys [/hide]

3.[hide]

We have 5 positions 26 letters thus for every position are 26 diffrent possibilitys that means 26*26*26*26*26=26^5=11881376 diffrent ways

[/hide]

4.[hide]

It's like problem 3 ecept we have 26letters for position 1 25letters for position 2 etc. so there are 26*25*24*23*22=7893600 possibilitys

[/hide]

Is that correct so far?

For joml's #1, and other similar to it, don't you mean "How many arangements are there when putting the letter in the work 'math' together"?  Looking at 1, I would say 1, the word "math".  It isn't really that clear to me what I'm supossed to be solving (my questions isn't much better though).

By arrangements I mean like tham, and athm...

6.

[hide]The persons ABCDEF are sitting around the table. A can change 5 times his position with one of the others to get a new order. B has already changed possition with A so he can still change with CDEF. Then C has changed with AB so there remain DEF etc. Thus we can use the sum formula

n(n-1)/2=6*5/2=15[/hide]

i have another problem.  how many ways can you draw n-2 triangles on an n-gon?


I haven't proved the problem :
      "Proved that have no k,n,m,a >=2 in N satisfy:
                 (m-k+1)....m=a^n.
with k<m.

This is a very difficult problem but the result holds :
Erds and Selfridge have proved that the product of at least two consecutive positive integers is never a (non trivial) power.
The proof uses 10 pages and some direct computations, so I will clearly omit it here and only give you the reference. Note that the proof also uses graph theory...

Reference :
P.Erds, J.L.Selfridge, 'the product of consecutive integers is never a power', Illinois Journal of Math., 19 (1975), p.292-301.

Pierre.

[quote="pbornsztein"]so I will clearly omit it here and only give you the reference.[/quote]

Maybe somebody could scan the relevant pages and upload it onto MathLinks? This proof has been mentioned a couple of times, but clearly the length has scared anybody off typing it.


integration
trigonometry
algebra
partial fractions
calculus
calculus computations

$ t=\tan \frac{x}{2}\Longrightarrow \cos x=\frac{1-t^{2}}{1+t^{2}}$.

[quote="BanishedTraitor"][quote="IrinaSharova"]$ \Rightarrow\;I=\frac{2}{\sqrt{a^{2}-1}}\int\frac{du}{1-u^{2}}$
$ \Rightarrow\;I=\frac{2}{\sqrt{a^{2}-1}}arctanh\;u+C$
$ \Rightarrow\;I=\frac{2}{\sqrt{a^{2}-1}}arctanh\;(\sqrt{\frac{a-1}{a+1}}\tan{\frac{x}{2}})+C$[/quote]

Or $ I=\frac{2}{\sqrt{a^{2}-1}}\int\frac{du}{1-u^{2}}$ which gives $ I = \frac{2}{\sqrt{a^{2}-1}}\int \frac{du}{(1+u)(1-u)}$ and use partial fractions:

$ I = \frac{1}{\sqrt{a^{2}-1}}\int (\frac{1}{1-u}+\frac{1}{1+u}).du$

So: $ I = \frac{1}{\sqrt{a^{2}-1}}ln | \frac{1+u}{1-u}|$ and then re-substitute the variables.[/quote]

Yes, either way will do, :)  they are equivalent.


Do they have to be strictly increasing/decreasing? e.i. Does the number $ 2344$ count? What about $ 2222$ ?

Same question for "Help 2 :( " .

[quote="marcopolo"]How many of the 4 digit integers ( from 1,000 to 9,999) have 4 distinct digits that are either increasing or decreasing?[/quote]

Just saying, if you want a better response, you may want to try to name your threads better. We understand that you need help, say something like: increasing/decreasing digits combo
Not something like: combinations (there's thousands of problems involving combinations) or help!


How to prof that sin( sqrt(x) ) < sqrt ( sin (x) ),  0 < x < Pi/2 ???

ARO 2006 : http://www.mathlinks.ro/Forum/viewtopic.php?t=86557

Thank you very very much!!!! It's not very difficult problem, i don't know why i couldn't to solve it.


calculus
integration
function
calculus computations

[quote="kunny"]A function $f(x)$ satisfies $f(x)=f\left(\frac{c}{x}\right)$  for some real number $c(>1)$ and all positive number $x$.

If $\int_1^{\sqrt{c}} \frac{f(x)}{x} dx=3$, evaluate $\int_1^c \frac{f(x)}{x} dx$[/quote]
I guess the answer is $6$ :P

Let $u=\frac{c}{x}$, so $du=\frac{-c}{x^{2}}dx$, so $\int_1^{\sqrt{c}} \frac{f(x)}{x} dx= \int_c^{\sqrt{c}}\frac{u f(u)}{c}(\frac{-x^2}{c})du=\int_{\sqrt{c}}^c \frac{f(u)}{u} du$.

Therefore, $\int_1^c \frac{f(x)}{x} dx=\int_1^{\sqrt{c}} \frac{f(x)}{x} dx+\int_{\sqrt{c}}^c \frac{f(u)}{u} du=3+3=6$,  $\mathbb{QED}.$


Express 7 as a linear combination of 15 and 4, or prove that it is impossible.

there is something wrong with the questions above. It was given by my teacher for revision. He says there's an error inside and u got to show it why it is wrong.

I so far know that gcd (15,4) = 1 hence it contradicts to 7???

I got lost from there as if I don't know to show that there's an error ( or I could just fail to explain correctly that there's an error)

My apologies cuz this is my first time trying such questions as well as first time studying elementary no. theory.

I would guess that the question is wrong since it didn't specify whether the combinations of 15 or 7 could be negative? (Which is required for the problem to be solved correctly...)

But I didn't see any other errors.


inequalities
LaTeX
inequalities unsolved

much easier to read it latex: $\sqrt{e^e\cdot\pi^{\pi}} > e^{\pi}$.


Let $ ABC$ be a triangle with angles $ \pi/4 < \alpha, \beta, \gamma < \pi/2$.

a) Prove that it is possible to put, inside this triangle, three squares with this four properties:

i) the three squares have the same area

ii) they all have a common vertex $ K$

iii) $ K$ is the only point which belongs to two different squares

iv) each square has two [b]opposite[/b] vertex on the perimeter of $ ABC$.

b) Consider the square (external to $ ABC$) whose side is $ BC$, and let O be its centre. Let $ r_A$ be the symmetric of the line $ AK$ with respect to the bisector of $ \alpha$. Prove that $ O \in r_A$.

The problem should not be so difficult. Drawing equal antiparallel MN, PQ and RS to BC, CA and AB respectively, we know the hexagon MNQPSR is cyclic, let's K being its' excircle's center. All we need, basically, is to find the lenght of MN such as the distance from K to it is half of its lenght, then to show K is the isogonal conjugate of intersection of OA and similar ones from B and C w.r.t. ABC.
However I do not have time for such a search.

Best regards,
sunken rock


How many different ways can the letters of the world FOUR be scrambled so that the first letter is a vowel and the last letter is a consonant?

There are 2 vowels and 2 consonants.
The first letter is a vowel, so there are $ 2$ choices.
The last letter is a consonant, so there are again $ 2$ choices.
The 2 middle letters are 1 consonant and 1 vowel, in any order.
There is only 1 vowel left and 1 consonant left, but there are $ 2$ possible orders
So the answer is $ 2*2*2 \equal{} \boxed{8}$


function
calculus
derivative
analytic geometry
graphing lines
slope
inequalities

find if this function is increasing or decreasing
$ f(x)\equal{}ln(x^2\plus{}1)\minus{}e^{\minus{}x}\plus{}1$

EDIT : oops, sry. its $ f(x_1)$ and $ f(x_2)$, not $ f'(x_1)$ and $ f'(x_2)$.

when $ x_1 < x_2$, if $ f(x_1)\le f(x_2)$,a function is increasing.

as you said, $ f(x) = ln(x)$ is an increasing function in $ (0,\infty)$

therefore, x can't be negative.

$ f(0) = 0$
$ f(\infty) = \infty$

but it could possibly be decreasing at some point.

we should find the derivative of the function.

$ f'(x) = \frac {2x}{x^2 + 1} + e^x$

in a graph, if a slope of a point is 0, that point is the graph's maximum or minimum.

so we are looking for a value of x that $ \frac {2x}{x^2 + 1} + e^x = 0$.

if $ 0\lex$, the value of expression is larger than 0.

and x can't be negative.

therefore, there are no maximum or minimum in this function.

so the function is continuously increasing.

"When $ x_1<x_2$ ,if  $ f'(x_1)\le f'(x_2)$ then the function is increasing"
is that right?
Because I'm thinking the function
$ f(x)\equal{}lnx$
which all we know that is increasing function in $ (0,\plus{}oo)$
but the derivative of this function is
$ f'(x)\equal{}\frac{1}{x}$
if we choose $ x_1\equal{}1 , x_2\equal{}e$
is not right that $ f'(x_1)\le f'(x_2)$
if understanding right...?!

sorry for wrong solution, spinos.

i editted the solution.

[hide="Partial Solution"]$ f(x) \equal{} \ln(x^2 \plus{} 1) \minus{} e^{ \minus{} x} \plus{} 1$

Obviously the +1 doesn't change increasing or decreasing.

$ f(x) \equal{} \ln(x^2 \plus{} 1) \minus{} e^{ \minus{} x}$

Differentiating

$ f'(x)\equal{} \frac{2x}{x^2\plus{}1} \plus{} \frac{1}{e^x}$

When $ x\geq{0}$ you can see both terms are greater than 0.

When $ x<0$ for the slope to be positive (and hence for f(x) to be monotone increasing),

$ \frac{1}{e^x}>\frac{\minus{}2x}{x^2\plus{}1}$

Since $ x^2\plus{}1$ is always positive

$ \frac{x^2\plus{}1}{e^x}>\minus{}2x$

Adding $ 2x$ to both sides,

$ \frac{x^2\plus{}1\plus{}2xe^x}{e^x}>0$

Since e^x is positive,

$ x^2\plus{}1\plus{}2xe^x>0$

I can't figure out where to go from there
[/hide]

Little error [b]mewto55555[/b]
[hide="Correction"]
In the case considering, $ f' \left( \minus{} x \right) \equal{} e^{x} \minus{} \frac {2x}{x^{2} \plus{} 1}$.
Following the inequality with the correction reaches the conclusion.
$ e^{x} > \frac { \minus{} 2x}{x^{2} \plus{} 1} \implies f' \left( x \right) > 0 \ \forall \ x \Leftrightarrow f \left( x \right)$ is an increasing function.
$ \left( e^{x} \equal{} \sum_{n \equal{} 0}^{\infty} \frac {x^{n}}{n!} \right)$.
[/hide]


I think it is here :

http://www.mathlinks.ro/Forum/viewtopic.php?t=82512


geometry
percent
AMC
USA(J)MO
USAMO
trigonometry
area of a triangle

Is it generally easy to change courses once you get there? Has anyone had any experience doing so?

Are the two geometry courses [i]Computational Geometry[/i] and [i]Geometric Proofs[/i] independent of each other? Does one require the knowledge of the other course? (I ask this because I can hardly recognize any of the terms under the description of "Computational Geometry", but I feel that I understand the description of "Geometric Proofs".) How difficult is "Geometric Proofs"?

I think the most appropriate courses for me in each topic would be Alg. 2.5, Combinatorial Arguments, Computational Geometry, and Number Theory. However, these are all afternoon session classes  :( .

I am also very worried that I have overestimated my own skill...

[quote="Meggy-Meg"]In my experience, it's not that hard to change courses even when you've already started. The teachers didn't take roll last year, so it was possible to just ditch one class and go to another.

Also, don't worry about overestimating your own skill. You are the person who knows best what's good for you, and if you did overestimate, then you will end up learning more. Don't worry about it![/quote]

So if we underestimate our skill, we can change classes?

Would it be bad if you took Geo Proofs but didn't understand the material of Computational Geometry? I need to find a course for the morning....

I don't think that's a very good idea though...since you'll probably just fall behind

For me, I'm thinking about Geo Proofs and Alg 2.5, since I understand basically everything in Computational Geo, and I need an intermediate afternoon course.

But the problem is that Geo and Alg are probably my best 2 subjects, so I'm thinking I might want to take Combo or NT, since I'm pretty bad at those.

So should I do my first choice, or should I do Computational Geo and Modular Arithmetic/Counting Strategies, since I don't want to take any beginner classes.

Yeah, I have a similar problem as Carbon57's.... I don't want to take any beginner classes, and I think Combs 2 and NumThe 2 might be too easy... but then that leaves all my classes in the afternoon session...

I was really disappointed when I got the course selection sheet for two reasons:
-I thought we were taking 4 courses
-Certain courses have conflicting schedules

 :(  Oh well I'm sure it'll be fun anyways

Last year I wasn't sure about the whole idea of taking only two courses, but having been at the camp, I think that it is best. Four classes is too much information to absorb in such a short time. There is a difference between learning a technique, and absorbing it into your repertoire of techniques that you are comfortable with.

I think that each class has a pretty wide range of difficulty. I remember in the problem sets, there were some people who solved very few problems alone, some that could do say half, I could do like 95 percent of them and pretty much nobody could solve everything. Also you can always get more difficult problems from your teachers if you think its too easy. But if that doesn't help, I don't think that it should be that hard to switch classes.

And I don't think that the counting strategies class should be that easy. Topics like recursion or bijections can be easy in some contexts, but they can be very difficult in other contexts. Apparently this class also covers some entry level USAMO problems...

[quote="leoxnlin"] Are the two geometry courses [i]Computational Geometry[/i] and [i]Geometric Proofs[/i] independent of each other? Does one require the knowledge of the other course? (I ask this because I can hardly recognize any of the terms under the description of "Computational Geometry", but I feel that I understand the description of "Geometric Proofs".) How difficult is "Geometric Proofs"?[/quote]

Hello, I will be part of the team teaching the "Geometric Proofs" class. I can give you my thoughts. Of course we plan to teach the class at a level after assessing the nature of the students in the class. But you should probably review those theorems (It shouldn't take more than 5 minutes to look up say Ceva or Menelaus so you know the statement at least). The goal of the class is to teach how to use these techniques more than to introduce them. I plan to prove a number of those theorems (like ptolemy). The law of sines is useful. Stewart's, law of cosines, heron's formula are unlikely to come up because they are mainly computational.

Altheman, in your opinion is it good to take both olympiad geometry from AOPS and Geometric Proofs from AMSP?

leoxnlin. dont wry bout ur skill. ur rly talented :) u'll be fine in the classes u mentioned.
comp. geo was a cool class [ i liked it --- even tho it was prettyy like 'wth am i doing' for me..] but u'll be fine XD

Check the topic post date, shelly32494.  :wink:

I am fairly certain that the courses change from year to year.
I am desperately hoping for combinatorics and number theory.


linear algebra
matrix
integration
Putnam
superior algebra
superior algebra solved

We have the matrices A,B in M_n(R) which verify the relations: A^(2002)=A^(2003) and A+B=In. Prove that In-AB is invertible!

Cheers! :D :D

are you sure this is the right question? maybe A2002= B2003 ?!

This problem was given in GM 3/2003 and the text is like I gave it .. perhaps it's a typing error there?! but I have found no erata!

cheers! :D :D

in that case from (1-x)2002(1-1+x) = 0 => (1-x)2002x=0 which means that the proper values of B can only be 1 and 0 (1).

but In - AB = In-B+B2 = (B+ eIn)(B + e2In) where  e3=1,  e <> 1 and we obviously cannot have e or e2 as proper values of B (because of (1)), so det In -AB >0 (!).

[quote="Lagrangia"]A,B in M_n(R) which verify the relations: A^(2002)=A^(2003) and A+B=In. Prove that In-AB is invertible!
[/quote]

I-AB=I-(A-A)

First case if A=A projection matrix, I-AB=I is invertible

Second case if A-A=/=0

lemma 1: (A-A)^2002=0
prof: (A-A)^2002=A^(2002)(A-I)(A-I)^2001=(A^2003-A^2002)(A-I)^2001=0

lemma 2: If M \in Mn(R) such that there exist q \in N \{0;1}, M^q=0 then
I-M is invertible, its inverse is I+M+...+M^(q-1)

prof: (I-M)(I+M+...+M^(q-1))=
I+M+...+M^(q-1)-M-M-...-M^(q-1) - M^q = I
---------------------------------------------------------------------
lemma 2 with M=A-A gives the inverse of I-AB=I-(A-A) is
I+(A-A)+....+(A-A)^2001

Remark: If M is nilpotent then det(I-M)=1
Indeed nilpotent matrix M is similar to triangular superior with diagonal
(0,...,0), under the diagonal only 0
si I-M is similar to

(1 * * .........................................* *)
( :D  1 *......................................* *)
(....................................................*)
(....................................................*)
( :D   :D   :D  ........................... :D 1)
the determinant of this matrix is 1

Let me guess this is 12th grade problem in GM, isn't it ?
Problems in G :D M Gazeta  :D  Matematica  :D  is very nice  :D
 :D  \cup \beta  \int n :D  :D  \lambda

well ... that's about what I proved :) nice soln though mo \cup bi :)

This one is actually an 11th grade problem,moubi! :D :D

This is alos another 11th grade problem:

Consider the matrices A,B of order n with real entries such that A^3=B^3 and A^2B=B^2A.
Prove that at least one of the matrices A^2+B^2, A-B, A+B is invertible..

this shouldn't be too hard! :D

cheers! :D :D

Your second problem proves to be false if A=B=O_n. What's the correct text?

maybe both A and B are different from 0n ?!

Like the other problem, I took it from the GM, and the text is correct how I posted it! At least it is the same like in the gazzette. I also think that A,B <>On!

cheers! :D :D

Well then just take A=B such that det(A)=0. :D

Hmm.. I see what you mean andrei.. :D :D
Lets try to solve the problem if A and be are both different than On!

cheers! :D :D

This is an older Putnam problem or something, from 1991. It's in the book with olympiads for grades 11-12 , 2003 from Gil Publishing House at the "Chosen Problems in Algebra" section.

Cheers ;)

There's another thing : A<>B. The soln: (A^2+B^2)(A-B)=On (calculate it) so if A^2+B^2 is invertible then A-B=On, which is false.

Okay! I'll move this post to solved problems because the last problem is totally flawed. (If A=B and detA=0 then ...)


[quote="Garry Denke"]Where are h's equations published?

[b]h = e^2 * z0 / (2 alpha)
h = e^2 / (2 alpha) * e0 * c
h = e^2 * u0 * c / (2 alpha)[/b]

I can't find h anywhere.

Thanks![/quote]

i really could not understad by "h's equations " and please specify what ur variables are :rotfl:

$\alpha$ is the fine structure constant, $h$ is Planck's constant, $e_{0}$ (rather $\varepsilon_{0}$) is the vacuum electric permeability, $c$ is the speed of light, $e$ is the electron charge... The middle equation is correct, I don't know what $z_{0}$ and $u_{0}$ mean. Can you clarify your question? Do you want to confirm that these equations are correct?

In sci.physics Sue found the second,

http://physics.nist.gov/cuu/Images/alphaeq.gif
http://physics.nist.gov/cuu/Constants/alpha.html

but the first and the last I can't find.

http://physics.nist.gov/cgi-bin/cuu/Value?z0
http://physics.nist.gov/cgi-bin/cuu/Value?mu0

Yes, thanks!


Find the sum:

$ \sum\limits_{n\equal{}1}^{\infty}\left(n\ln \left(\frac{2n\plus{}1}{2n\minus{}1}\right)\minus{}1\right)$.

It can be shown using a series expansion for ln (1+x) that the series converges. the sum? well, mathematica gives it as ___.  :roll:

Maybe someone can find where the alternating harmonic series is.

$ \sum_{n\equal{}1}^m n\ln(2n\plus{}1)\minus{}\sum_{n\equal{}1}^m n \ln(2n\minus{}1)\minus{}m$

$ \equal{}\sum_{n\equal{}0}^m n\ln(2n\plus{}1)\minus{}\sum_{n\equal{}0}^m (n\plus{}1) \ln(2n\plus{}1)\plus{}(m\plus{}1)\ln(2m\plus{}1)\minus{}m$

$ \equal{}\underbrace{\minus{}\sum_{n\equal{}0}^m \ln(2n\plus{}1)}_{\ln\frac{m!\, 2^m}{(2m\plus{}1)!}}\plus{}(m\plus{}1)\ln(2m\plus{}1)\minus{}m$

For $ m\to\infty$ use the stirling approximation and you get $ \frac{1\minus{}\ln 2}{2}$ .


function
modular arithmetic
floor function
number theory
relatively prime
number theory unsolved

[quote="FelixD"]Let $ N$ be a positive integer. How many non-negative integers $ n \le N$ are there that have an integer multiple, that only uses the digits $ 2$ and $ 6$ in decimal representation?[/quote]
For any $ p$ which is relative prime to 10 , we will show that there exist an integer $ k$ such that it divisible by $ p$ .Let's consider number
$ k\equal{}\frac{6.10^n(10^m\minus{}1)}{9}\plus{}\frac{2.(10^n\minus{}1)}{9}\equal{}\frac{6.10^{m\plus{}n}\minus{}4.10^n\minus{}2}{9}$
Only need to chose $ m,n$ such that both of $ m\plus{}n$ and $ n$ are multiple of $ \varphi(9p)$ ,by using Euler's function we can easy to show that $ k$ is a multiple of $ p$
Consider a number $ k$ only use the digit $ 2,6$ then it is not hard to show that $ 5$ is not a divisor of
$ k$ and $ \frac{k}{2}$ only uses the digit $ 1,3$ , it means that $ 2\parallel{}k$
Therefore n satisfy condition if and only if $ n$ is not  a multiple of $ 4$ or $ 5$ . The number of positive integer satisfy condition is :
\[ N\minus{}[\frac{N}{4}]\minus{}[\frac{N}{5}]\plus{}[\frac{N}{20}]
\]

It is necessary that $ n$ is not divisible by both $ 4$ and $ 5$,because $ 22,26,62,66$ are all not divisible by $ 4$,and units digit of multiple of $ 5$ never be $ 2$ or $ 6$.
If even $ n$ satisfies the condition,then so does $ \frac {n}{2}$,
therefore we prove there exists such multiple of $ n$ that satisfies the condition when $ n$ is even and not divisible by $ 4$ or $ 5$.

Let $ n \equal{} 2m$,where $ m\in \mathbb{N}$ is relatively prime to $ 10$.
we know there exists a positive integer $ a$ such that
$ 10^a\equiv 1\pmod{m}$.
then $ ma$-digit integer $ b \equal{} \underbrace{111\cdots 11}_{ma}\equiv \underbrace{\underbrace{11\cdots 1}_{a} \plus{} \underbrace{11\cdots 1}_{a} \plus{} \cdots \underbrace{11\cdots 1}_{a}}_{m}\equiv 0\pmod{m}$,

which means $ 2b\equal{}\underbrace{222\cdots 22}_{ma}$ is a multiple of $ n$,and we get the answer $ N \minus{} \left\lfloor \frac {N}{4}\right\rfloor \minus{} \left\lfloor \frac {N}{5}\right\rfloor \plus{} \left\lfloor \frac {N}{20}\right\rfloor$.


algebra
polynomial
superior algebra
superior algebra theorems

I want to be able to create a multiplication table of Zmod2[x]/<x^3+x>.  I just don't even know what to google.  Can someone please provide the name of the topic that would cover this technique of polynomial multiplication over a polynomial modulus?  Thanks!

The Chinese remainder theorem for rings may be most useful here, as it reduces your case to a very simple one.
The topic you'd be looking for here would be ideals and modulo in rings. In this case, chinese remainder theorem for rings.
=Uncool-


we know $\prod_{cyc}(a+b-c) \leq abc$
so the given inequality is $\equiv$ to proving
$\sum_{cyc}a^{2}\leq 2\prod_{cyc}(a+b-c)+1$ ------------*

set,
$a+b-c=u$
$b+c-a=v$
$c+a-b=w$
so,
the  inequality * $\equiv$ $8uvw+4$ $\geq$ $\sum_{cyc}(u+v)^{2}$
where $0 \leq u,v,w \leq 1$
set ,$f(x)=(x+u)^{2}+(x+v)^{2}+(u+v)^{2}-8uvx-4$
$f^{''}(x)=4>0$
so $f$ attains maximum in it's boundary .so we have to check just $4$ values now
namely $(d,e,f)$ where $d,e,f$ are either 0 or 1.
so we have by checking $max[f(u,v,w)]=f(1,1,1)=0$
and we are done.

[quote="RDeepMath91"]If $f''(x) > 0$, then I think $f$ attains its minimum at its boundary[/quote]

Yes that's true . since the variables are in [0,1] , (1,1,1) is a boundary  :wink:

[quote="RDeepMath91"]Hm.. If $f''(x) > 0$, then I think $f$ attains its minimum at its boundary..  :maybe:[/quote]
no that's not true it attains it's maximum in the boundary .

[quote="SOUMYASHANT NAYAK"]let $a,b,c \geq 0$
if
$b+c \leq 1+a$ ,
$a+c \leq 1+b$,
$a+b \leq 1+c$.
Prove that
$\sum_{cyc}a^{2}$ $\leq$ $2abc+1$[/quote]

let be
$A=1-a$
$B=1-b$
$C=1-c$

the conditions become
$A,B,C\leq 1$
$A+B\geq C$
$B+C\geq A$
$C+A\geq B$

adding the first two we have $B\geq 0$ and the same for $A$ and $C$, so $A,B$ and $C$ are sides of a triangle, and exist $x,y,z\geq 0$ such that
$x+y=C$
$y+z=A$
$x+z=B$

(we have $x,y,z\leq 1$ too)

the inequality becomes
$ABC+A^{2}+B^{2}+C^{2}\leq2AB+2BC+2CA$

$4xy+4yz+4zx\geq x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)+2xyz$

but we have

$4xy+4yz+4zx=2xy+2yz+2zx+x(y+z)+y(z+x)+z(x+y)\geq 2xyz+x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)$
for $1\geq x,y,z \geq 0$.

[quote="shalex"]This is actually an old problem, and I have such kind of idea.
[/quote]
do you know the source of this problem

[quote="shalex"]I'm sorry I don't know the exact source it comes from,
but I saw this problem in a document before this competition was held.[/quote]
they have copied almost all the problems i suppose.... :mad:


I could say nothing but to confess it's to hard :D


algebra
polynomial
analytic geometry
function
domain
conics
parabola

when solving polynomials, for example $ P(x) \equal{} x^2 \minus{} 3x \plus{} 2$
we say $ p(x) \equal{} 0$ then when we find the roots right?
we say $ p(x) \equal{} 0$ because the roots of that polynomial is when it the polynomial crosses the x-axis, the $ y$ coordinate equals to $ 0$ therefore $ P(x) \equal{} 0$ right ?

when it comes to  imaginary roots we still do the same thing, eventhough they dont pass the x-axis at all, why?
for example we say $ x^2 \plus{} 1 \equal{} 0$ then we solve it. and it gives $ \pm{i}$ i am confused. because it doesnt pass the x-axis, why do we asume $ P(x) \equal{} 0$,and we then solve, it doesnt pass the x-axis so $ p(x)\neq0$

can someone explain this please.

I think it crosses the "x-axis" on the imaginary plane.

I might be being really dumb right now, but I'm pretty sure that's not true, 7h3.D3m0n.117.

When you talk about crossing the x-axis on the coordinate plane, you're only considering a domain of real numbers, not complex numbers. So not passing the x-axis does not imply that there are no roots (real or non-real) to the equation.

[quote="azjps"]I might be being really dumb right now, but I'm pretty sure that's not true, 7h3.D3m0n.117.[/quote]

Yeah, my illogical way of thinking got in the way of my reasoning; even I'm not sure anymore.

But wouldn't it make sense in the sense of the complex plane though?

[quote] you're only considering a domain of real numbers, not complex numbers.[/quote]

how can you say the complex numbers domain exists , when the first equation that found the imaginary number $ x^2 \plus{} 1$ doesnt have imaginary root, (when no imaginary number exists how can you say that imaginary domain exists, any proof?)

Once we start using words like "complex plane" in the context of solving polynomials, we have to be very careful to sort out what we mean by "plane."

Imaginary roots cannot be depicted on a [b]real[/b] Cartesian plane at all.  You are correct that $ x^2 \plus{} 1 \equal{} 0$ has no real solutions - in other words, there is no real point at which the graph crosses the [b]real[/b] $ x$-axis.

It's possible to graph the "complete" graph of a parabola by letting both $ x$ and $ y$ take complex values.  Then the graph of $ y \equal{} x^2 \plus{} 1$ will "cross" the [b]complex[/b] $ x$-axis at $ (\pm i, 0)$.  (In terms of real dimensions, the parabola $ y \equal{} x^2 \plus{} 1$ over the complex numbers is actually a surface (and the complex $ x$-axis is a plane), so our notion of "crossing" must be modified somewhat.)

The problem with visualizing this graph is that $ x$ now has two (real) dimensions and so does $ y$, so the resulting graph is [b]four-dimensional.[/b]

Like all numbers, it's just an abstract term defining something observed. Why can't we picture a 4th dimension based on 3 dimensional space?

Complex numbers are not "proven" to exist. They are [b]defined[/b] in a logically consistent way that we want them to be.

EDIT: Henry beat me...

Post conflict like 3x; Hmm, that's what I thought  :) , but I didn't realize that the complex x-axis wasn't referring to a line; thanks.

I was going to say that complex numbers are defined etc but it seems that a bunch of people have beaten me to it.

[quote="binomial_4eva"]when the existence of imaginary number is not proven, how can we say that imaginary or complex axis exists?  any proof? or is it just an assumption?[/quote]

How can we say that the real axis exists?  There are actually some mathematicians who think that the infinite precision of a real number has no sensible meaning.

so its just an assumption that its there?  i thought in mathematics things are proven first then accepted

Remember even though it's mathematics, math is still a science. Science can be overwhelmingly and [seemingly] true, but all it takes is one counterexample to shatter what we think is true now and send scientists back to the drawing boards.

[quote]How can we say that the real axis exists? [/quote]
this is a good question. thanks for reminding me i wanted to ask, i am looking for an answer for that my self

[quote]based up on nothing but observation[/quote]
when have we observed the complex axis, have you seen the complex axis?

[quote="Temperal"]Mathematics is based on an axiomatic system, which basically means that there are some things that we just have to accept as true based upon nothing but observation.[/quote]

That's not what an axiomatic system is at all.  Mathematics is a tool we use to study patterns that we observe.  To formalize those patterns, we [b]define[/b] the terms we use - line, point, real number - and write out [b]axioms[/b] regarding those terms - statements that we hold to be true - and then prove things from those definitions and axioms.  The resulting patterns may or may not describe the observations we make, but the point is that [b]from our definitions[/b], the results are irrevocably true.

The real axis exists because we have axiomatically defined the real numbers (for example, using [url=http://mathworld.wolfram.com/DedekindCut.html]Dedekind cuts[/url]).  The complex numbers exist because we can axiomatically define operations on them (or we can define the notion of a field extension or an algebraic closure and work from there).  We use those systems because they are [b]useful[/b].

no but you said [quote]based up on nothing but observation[/quote] so i am asking when have u observed the complex-axis, how does it look like ?

[quote="binomial_4eva"]so i am asking when have u observed the complex-axis, how does it look like ?[/quote]

Keyword is [i]when[/i]. When mankind develops a sensible visual for the 4th dimension, the microscopic world, and the imaginary axis, you'll be the first to know.

The complex $ x$-axis is the set of points $ (x, 0)$ where $ x \in \mathbb{C}$.  As such it, it looks like an [url=http://mathworld.wolfram.com/ArgandDiagram.html]Argand plane[/url] (which I prefer to the use of the term "complex plane," since when we do not distinguish between real and complex dimensions the term is ambiguous).

"Observation" is a tricky word when it comes to mathematics.  Observations are the basis for our axiomatic systems, but they do not comprise them.  Mathematics is fundamentally different from other branches of science.  The other branches of science [b]use mathematics as a tool[/b] (and I do not know how often I can say this) to analyze patterns in physical observations.  Mathematics itself is fundamentally different.  Mathematical objects are [b]formal[/b] - they exist because we say they exist.  Whether we can visualize four-dimensional space is [b]irrelevant[/b] to whether we can define it mathematically.

Sorry, perhaps not on observation, then. But the basis of the post is that some things are not proven in an axiomatic system.

Not since the 1800s.  What are you talking about?

Starting (I believe) in the late 1800s, mathematics underwent a crisis of formality.  Mathematicians realized that they did not actually know what their intuitive notions (say, of a set) meant and sought to rigorize all of mathematics (see for example [url=http://en.wikipedia.org/wiki/Principia_Mathematica]Principia Mathematica[/url]).  All of modern mathematics now takes place in the context of axioms.

So when you say that some things (by which I take it you mean "mathematical statements") aren't proven in an axiomatic system, what things are you referring to?

[url=http://mathworld.wolfram.com/Axiom.html]Axioms[/url] don't need to be proven.  The way in which you've worded your statements has behind it the implicit assumption that there is a possibility that a given axiom might be wrong.

The actual word we want to discuss here is [url=http://mathworld.wolfram.com/Consistency.html]consistency[/url].  Interestingly enough, it's not actually possible (in some sense) to prove that the axioms of mathematics are consistent.  See [url=http://mathworld.wolfram.com/GoedelsIncompletenessTheorem.html]Godel's incompleteness theorem[/url], and look up a few things about the philosophy of mathematics if you really want to get into this kind of stuff :)


Daca (G; *) grup necomutativ infinit, atunci rezulta ca exista H, subgrup propriu, necomutaiv si infinit ?

English version :
If (G; *) is an infinite nonabelian group, implies the existence of an infinite, nonabelian proper subgroup H ?

Cautam pe net chestii care nu prea aveau legatura cu asta, si am gasit din greseala definitia [url=http://planetmath.org/encyclopedia/InfiniteDihedralGroup.html]grupului diedral generalizat[/url]. Dupa aia a fost usor :).

Se pare ca raspunsul e "nu". Adica exista grupuri infinite neabeliene care nu au subgrupuri proprii infinite neabeliene. Fie $N=\mathbb Z_{p^\infty}$ grupul radacinilor unitatii de ordin putere a lui $p$, unde $p$ e un numar prim. Asta e un exemplu binecunoscut de grup abelian infinit care are numai subgrupuri proprii finite. Acum fie $G$ produsul semidirect $N\rtimes\mathbb Z_2$, unde elementul netrivial din $\mathbb Z_2$ actioneaza pe $N$ prin conjugare trimitand fiecare element in inversul sau.

Fie acum $H\le G$ un subgrup. $H\cap N$ e subgrup normal in $H$, de indice $1$ sau $2$. Daca e $1$, atunci $H\le N$, si $H$ e abelian. Daca e $2$, atunci $H$ e, la randul sau, izomorf cu produsul semidirect $(H\cap N)\rtimes\mathbb Z_2$. Daca $H\cap N$ e infinit, atunci $H\cap N=N$, si deci $H=G$. Am aratat deci ca avem una din trei variante: $H$ e ori abelian, ori finit, ori egal cu $G$.

Sper ca nu am gresit pe undeva.


Let $F_{i}$ represent the ith fibonacci number. Let $F_{a}$, $F_{b}$, $F_{c}$, and $F_{d}$ be the sides of a convex (nondegenerate) quadrilateral, with $a<b<c<d$. Find the greatest possible value for $d-b$.

[quote="tim1234133"][hide="solution"]

The 'quadrilateral inequality' is that for any non-degenarate quarilateral with side lengths a,b,c,d  a+b+c >d

We prove that $\frac32 \cdot F_{i}\leq F_{i+1}\leq 2F_{i}$

Clearly $F_{i-1}\leq F_{i}$, so ${F_{i+1}= F_{i}+F_{i-1}\leq 2F_{i}}$
This establishes the second part of the inequality.

$F_{i}\leq 2 F_{i-1}$ so $F_{i+1}= F_{i}+F_{i-1}\geq \frac32 F_{i}$
which establishes the first part of the inequality.

Now assume that c<d-1

Then $F_{d}\geq \frac94 F_{c}$

$F_{b}\leq \frac23 F_{c}$, and $F_{a}\leq \frac23 F_{b}$ so $F_{a}\leq \frac49 F_{c}$

So $F_{a}+F_{b}+F_{c}\leq \frac{19}{9}F_{c}< F_{d}$
However this contradicts the 'quadrilateral inequality'.

So d=c+1

Suppose b=c-2

$F_{c-1}=F_{b}+F_{a}$
$F_{a}+F_{b}+F_{c}= F_{c-1}+F_{c}= F_{c+1}= F_{d}$
Again by the 'quarilateral inequality', the quadrilateral degenerates.
Clearly if $b<c-2$, $F_{a}+F_{b}+F_{c}< F_{d}$, which again doesn't work

So the only possibility is d-b = 2 which is attainable, for example if
a=1 b=2 c=3 d=4 [/hide][/quote]
Thats a very nice solution, but its very long. Do you have a shorter solution?

[quote]Do you have a shorter solution?[/quote]

[hide="My revised (=shorter) solution"]

Suppose $b \leq d-3$
Then $a \leq d-4$

So $F_{a}+F_{b}\leq F_{d-2}$
and since $c \leq d-1$

$F_{a}+F_{b}+F_{c}\leq F_{d-2}+F_{d-1}= F_{d}$
and we have a contradiction by the 'quadrilateral inequality' as before

[/hide]

[quote]Also, is there an "n-gon inequality"? [/quote]

Yes; it is an extension to the triangle inequality. To prove it I suppose you can just cut the n-gon into triangles or something

this isn't anything like my solution, but we both used the same elements. thanks tim1234133. :)

[quote="vishalarul"]Is it possible to do this problem for an arbitrary $n-gon$?

Also, is there an "n-gon inequality"? :maybe:
Please prove the "n-gon inequality" if it exists. If not, please provide a counterexample.[/quote]

The "n-gon inequality" is very weak; it's just the repeated application of the triangle inequality. So if $a_{1}, a_{2}, \ldots, a_{n}$ are the sides, then

$a_{i}< \sum_{j \neq i}a_{j}$.

An easy way to prove this is by induction on $n$ (what happens when you add a side?).


Prove or disprove: for any $ n \in \mathbb{N}$, there exist positive integers $ x_{1}, x_{2}, \ldots, x_{n}$ such that
\[ \phi(x_{1}x_{2}\ldots x_{n}) = x_{1}x_{2}\ldots x_{n}\cdot \left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}\right). \]
Here as in fashion, $ \phi(n) = |\{a \in \mathbb{N}: \gcd(a,n) = 1\;\text{and}\; 1\leq a\leq n\}|$ and $ \mathbb{N}= \{1, 2, \ldots\}$. :)

[quote="s.tringali"]Here as in fashion, $ \phi(n) = |\{a \in \mathbb{N}: \gcd(a,n) = 1\}|$ and $ \mathbb{N}= \{1, 2, \ldots\}$. :)[/quote]
Hmm, It must is $ 1\leq a\leq n$.

Ehm... :oops: Yes, it must! So please edit my post (I can't any more) and fix it - thank you. :)


The electric field of an electromagnetic (EM) wave is $\mathbf{E}=E_0\mathbf{x}_0\exp{i(kZ-\omega t)}$, where $E_0$ is a real constant, and $\omega=ck/\widetilde{n}$. Here $\omega$ is real, $c$ is the speed of light in vacuum, and $\widetilde{n}$ is the complex dielectric constant of the medium.
(a) Briefly discuss what will happen to the EM wave amplitude as it propagates in the medium if $\widetilde{n}$ is real, imaginary, or complex.
(b) Find the magnetic field $\mathbf{E}$, and the time-averaged (over one period) Poynting’s vector $\langle\mathbf{S}=\frac{1}{\mu_0}{\mathbf{E}\times\mathbf{B})\rangle}$.
(c) The quantity $q=\frac{d\langle\mathbf{S}\rangle}{dz}$ describes the loss of EM wave energy to the medium. Calculate $q$ and briefly discuss the physical meanings of the results if $\widetilde{n}$ is real, imaginary, or complex.
(d) With reference to the results above, does an EM wave that decreases in amplitude while propagating always loose energy to the medium?
--From [i]Pan Pearl River Delta Physics Olympiad 2005[/i]

(a) if n is real ,nothing is happend , the norme of E still constant .
if n is complex, its only depends on the imaginary part and the direction of the wave(z>0 or <0 ),for exemple if z>0 and Im(n)<0 the norme increase ...
(b) B=E/c , ||<S>||=(1/2µc)E² .
(c)i dont understand the statement
i'm i reight ?


[quote="Lovasz"]Find all the functions $f: R \to R$ satisfy:
                    $(f(x))'.\int{f(x)dx}=f(x)$[/quote]
What do you mean $\int{f(x)dx}$???
Is it a set of $\{\int{f(x)dx}+C\}$? Or this is concrete integral?

For that matter, $(f(x))'$ is terrible notation; $f(x)$ is a real number that can't be differentiated, unless there's another variable somewhere.

If $\int$ indicates an antiderivative, that's a lot of trouble too, because it doesn't have a unique value anywhere.


modular arithmetic
induction
number theory unsolved
number theory

This one I think should be an easy one, but I can't solve it.

Show that if $ a$ is an odd integer, then for any $ n \ge 1$

\[ a^{2^n} \equiv 1 \pmod{2^{n\plus{}2}}\]

[b]Lemma:[/b]  The greatest power of $ 2$ dividing $ {2^n \choose k}$ is $ n$ minus the greatest power of $ 2$ dividing $ k$.

[i]Proof.[/i]  Apply [url=http://planetmath.org/encyclopedia/KummersTheorem.html]Kummer's theorem[/url].

Now write $ a \equal{} 2k \plus{} 1$.  Then binomial expansion gives

$ a^{2^n} \equiv 1 \plus{} 2^{n \plus{} 1} k \plus{} {2^n \choose 2} (2k)^2 \bmod 2^{n \plus{} 2}$

where the other terms vanish $ \bmod 2^{n \plus{} 2}$ by the lemma.  This is congruent to $ 1 \plus{} 2^{n \plus{} 1} \left( k \minus{} k^2 \right) \bmod 2^{n \plus{} 2}$ and $ k \minus{} k^2$ is even.

There's also a totally straightforward proof by induction, but where's the fun in that  :P  The binomial argument lets you prove something stronger (as I recall):  namely, that $ 5$ has maximal order.


vector
linear algebra
matrix
geometry
linear algebra unsolved

Non-singular symetric bilinear form $ g$ acts on $ E \times E$, where $ E$ is a finite dimentional vector space.

Define a symetric bilinear form $ g_{r}$ on $ E^{r}\times E^{r}$ by:
$ <g_{r}, (h_{1},...,h_{r}) \times (k_{1},...,k_{r})=det(<g, h_{i}\times k_{j}>)$.

If $ g$ is positive definite, is $ g_{r}$ also positive definite?

[quote="mlok"]This is essentially the [url=http://en.wikipedia.org/wiki/Gramian_matrix]Gram matrix[/url] of a system of vectors, except that the inner product is not the standard one. Let $ A$ be the matrix of $ g$ (in your notation, $ \langle g, x\times y\rangle =y^{*}A x$). Let $ B_{h}$ be the matrix with $ r$ columns, where the columns are vectors $ h_{1},\dots , h_{r}$. Define $ B_{k}$ similarly. You should be able to check that the value of $ g_{r}$ is $ \det (B_{k}^{*}AB_{h})$. When $ k=h$, the matrix $ B_{h}^{*}AB_{h}$ is positive semidefinite, because $ A$ is [url=http://en.wikipedia.org/wiki/Positive-definite_matrix]positive definite[/url]. It is positive definite if and only if the matrix $ B_{h}$ has rank $ r$. For instance, if $ h_{1}=\dots =h_{r}$, then $ B_{h}$ and $ B_{h}^{*}AB_{h}$ have rank $ 1$, and therefore $ g_{r}$ vanishes. So, the form $ g_{r}$ is not positive definite.[/quote]

Thanks, now I totaly understand.

Also the question was about r-dimentional area in a Riemann structure on a pure manifold. The correct defination of $ g_{r}$ should be:

Define a symetric bilinear form $ g_{r}$ on $ \bigwedge^{r}E \bigotimes \bigwedge^{r}E$ by:
$ <g_{r}, (h_{1}\wedge... \wedge h_{r}) \otimes (k_{1}\wedge...\wedge k_{r}){>}=det(<g, h_{i}\times k_{j}>)$.

So $ B_{h}$ in your post is either rank $ r$ or $ h_{1}\wedge ... \wedge h_{r}=0$ by defination of
exterior product. So $ g_{r}$ is positive definite.


trigonometry
Euler
Pythagorean Theorem
geometry
algebra
binomial theorem

Solve the following equation:
$\cos^{n}x-\sin^{n}x = 1$,
where n is a positive integer and x is a real number.

nice ! that's exactly the question  :P
the second equation u wrote says only it exists x such that a = cos x and b = sin x, then replace it into the first one...
what did you try with the complex forms ??

[quote="Thales418"]${\cos^{n}(x)-\sin^{n}(x)=1}$

${a=\cos (x)}$
${b=\sin (x)}$

${a^{n}=b^{n}+1}$
${a^{2}+b^{2}=1}$

Just Need to solve this system[/quote]

You could have said this immediately, without all that intermediate work.

Yea, i don't think either that u need that Euler-stuff:
Given a right triangle a, b, 1 with 1 as the hypotenuse,
and x as the angle opposed to side b, we have:
$cos(x) = a; sin(x) = b$
Substituting&simplifying:
$I: a^{n}-b^{n}= 1$
And because of the Pythagorean Theorem:
$II: a^{2}+b^{2}= 1$

Now we can square the first equation and take the 2nd one to the power of n and set them equal in order to get:

$(a^{n}-b^{n})^{2}= (a^{2}+b^{2})^{n}$

which is, using the Binomial Theorem:

$a^{2n}-2a^{n}b^{n}+b^{2n}= a^{2n}+\sum_{k=1}^{n-1}{n-1 \choose k}a^{2(n-1-k)}b^{2k}+b^{2n}$

(note that the sum must always be positive for $a,b > 0$)

$0 = \sum_{k=1}^{n-1}{n-1 \choose k}a^{2(n-1-k)}b^{2k}+2a^{n}b^{n}$

As the sum and $2a^{n}b^{n}$ are both positive for $a, b > 0$, it is obvious that there is no 2-tupel $(a, b)$ for $a, b > 0$ solving this equation.

Because $a, b$ are the legs of the right-angled triangle, we now know that there exists no such triangle such that,
$a^{n}-b^{n}= 1$, for $n \in \mathbb{N}$.
Substituting back, this yields that there is also no real x such that, $cos^{n}(x)-sin^{n}(x) = 1$.

Lol, this must be wrong, it was too easy. sry if it is, im a beginner!

EDIT:
lol gawd im stupid^^ ya its pretty ovious what i proved and it doesn't bring us any further^^ i should have looked at the problem more closely

i know that dont need euler..

but when i was doing i thougth that if i put this it will simplicate, but not. but i didnt want to erase the "euler" so....

and TearOfMercury  you did not find a explicit formula for x

[quote="TearOfMercury"]Lol, this must be wrong, it was too easy. sry if it is, im a beginner![/quote]
You correctly showed there are no solutions in the case that $\cos x$ and $\sin x$ are the legs of a non-degenerate right triangle with hypotenuse 1, i.e. when $0 < \cos x, \sin x < 1$.  (For these situations, you also did far more work than was necessary.)  The tricky part is what happens when we are outside of this range, particularly when $\cos x$ is still positive but $\sin x$ is negative and $n$ is odd.  Those solutions will be ugly, I imagine.  (There are also trivial solutions when $\sin x = 0$.)

$\boxed{n\ge 2\Longrightarrow 1=\cos^{n}x-\sin^{n}x\le |\cos x |^{n}+|\sin x |^{n}\le \cos^{2}x+\sin^{2}x=1}$

a.s.o.

[quote]$\boxed{n\ge 2\Longrightarrow 1=\cos^{n}x-\sin^{n}x\le |\cos x |^{n}+|\sin x |^{n}\le \cos^{2}x+\sin^{2}x=1}$[/quote]
so is this supposed to be the solution?  :huh: if it is could s.o. explain it to me? otherwise would s.o. please post a solution  :roll:

$a\le b\le a\Longrightarrow a=b\ .$
When $a-b=|a|+|b|$ ?
When for $0\le a\le 1$, $0\le b\le 1$, $n\ge 2$ we have $a^{n}+b^{n}=a^{2}+b^{2}$ ?


I had guessed who you are from your PM... :P

I had guessed who you are when pardesi told me...

[quote="Euclidean Geometer"]Well,
Yes, now lemme formally introduce myself,
I am Raghavendra from Bangalore.
Da same one who pissed shre off... <sorry though da >
Same class as Abhimanyu (alter ego)

Well how s the board mugging goin on all...
For me its not started as yet. I really like da last minute slogging. Under all pressure...
You know the thrill one can obtain from dat is itself  'thani'
LOADSSS of thrill  :D[/quote]
hello raghavendra, and
 :welcome: welcome to this forum :welcomeani:

hi da...chumma formality ke... :D ....but anyways it has been real fun knowing u  :)


[quote="mdk"]show that for n greater than or equal to 2, (n^3+(n+2)^3)/4 is a composite integer.[/quote]
[hide="Hint"]Consider parity.
If $ n$ is even,
$ \frac{n^{3}+(n+2)^{3}}{4}= (n+1)\cdot \frac{3(n(n+2))+4}{4}$
The second factor must be an integer (since n(n+2) has 2 factors of 2).
I'll leave the odd case to you.
[/hide]


linear algebra
matrix
limit
linear algebra unsolved

It is readily checked that the eigenvalues of $A_{1}$ are ${\frac{1}{2},\frac{1}{3},\frac{1}{5}}$.
Using these, the matrix $A_{1}$ is diagonizable because they form an basis for $R^{3}$. Hence $A_{1}=P^{-1}BP$ where $B$ har the eigenvalues on the diagonal.
When $n-\to infinity$, $B$ will turn towards the nullmatrix, leaving your answer the nullmatrix.


USAMTS
geometry
AMC
AIME
LaTeX
analytic geometry
graphing lines

[size=150][b]DO NOT[/b][/size] POST  ANSWERS, SOLUTIONS, COMMENTS, ETC. IN THIS THREAD UNTIL WEDNESDAY OCTOBER 8!!!

THIS THREAD IS ONLY HERE TO ELIMINATE THE NEED FOR 500 THREADS ON ROUND 1, AND TO SERVE AS A DISCLAIMER AGAINST POSTING [b]ANYTHING[/b] ON ROUND 1 UNTIL OCTOBER 8!!

Mmm, good idea. :)
I'll just link to the [url=http://www.nsa.gov/programs/mepp/usamts.html]USAMTS website[/url] and the [url=http://www.nsa.gov/programs/mepp/usamts/problems/problems15R1.pdf]problem[/url].

P.S. We can start discussing these problems tomorrow since according to the official website's [url=http://www.nsa.gov/programs/mepp/usamts/mail.html]page about deadlines[/url] "Solutions need to be postmarked (or faxed) [i]before [/i]the deadline date of the round."

Also, you can't mail anything out on a Sunday.  But, actually, you can fax in answers I believe.  (Or am I just making that up?)  So we should wait until after Sunday.

without discussing particular problems, how did you guys do?

My justification to one of the problems was about half a solution from which the answer came out of nowhere, but I think the rest were OK... at any rate, I'm 99% sure the numerical answers were right...

Finding the answers themselves was not hard; proving them was. I found #3 the easiest and #4 the hardest.

#2, I believe, is impossible to "prove", since prime numbers have no patterns whatsoever.

All of them can be proved. I found #1 the hardest.

Question: how soon do you get your results back. Is it like they say, 2 months, or do you USUALLY get it back before the next round starts.

You usually get the results after the next round has started but before the deadline, if I remember correctly.

It's a free public service.  If you want them to go faster, I'm sure they'd accept payment for additional graders.

Or stop giving so much money away for tax-exemption (USAMTS is sponcered by the government) :wink:

Which problem do you guys think is the easiest?  Don't explain how, just which....

Number three was, then 5, 4, 2, and 1. i did not find a solution to one  :cry: .

definately 3 is easiest, then 2, then 1 , 4 , 5. seems like a lot of people thought 1 was hard.... it thought it was ok.

Right, but a lot more work than that must be shown to get any credit.  Particularly for #1.  Some insightful discovery (although patterns work, I suppose) is probably preferred.

The [url=http://www.nsa.gov/programs/mepp/usamts/problems/credits15R1.pdf]credits and quick answers[/url] are posted at http://www.nsa.gov/programs/mepp/usamts/problems/credits15R1.pdf

yep, 20 for 20. i thought that the graders would be tougher...some of mine were a bit verbose.

:lol:, you're absolutely right, 5 a piece, 5 problems, 4 times. 100 points total.

am i the only one who hasn't gotten miine yet?

Bah.  No.  I haven't gotten mine yet, either.  It's probably because I live significantly farther from the NSA headquarters than some of the other people on this forum...but I guess that means there's a lesser chance of NSA hitmen gunning me down. </jk>  :D

Phht, they'll get you anywhere on Earth even if you were hiding in a bunker in A********** (which the government successfully hushed the media because of political reasons)

Got my results in the mail today...yay.  24/25.

I too thought they would be much harsher in grading...especially with my sorta-sticky-did-at-the-last-minute #1, which was fairly ugly, or my let's-work-with-50-pairs-of-similar-triangles #5, which ended up with a commendation.  But I don't mind being graded fairly loosely, now that I think about it.

Just how good is a commendation?

I got 25 and two commendations, both of which are extremely unexpected.

The first problem was a sort of put together some BS last second proof that made no sense, yet I got it commended.

In the fifth problem, I only did superimposition onto coordinate plane and got commended.

Did any of you get commendations that were unexpected?

cool!! you asked them to post the answers on my birthday! wow! cool!

Well, my proof for #5 was even simpler, probably - I just constructed some similar triangles - and I got commended on that...

Maybe it's the best x% of the proofs for such-and-such a problem...or no, maybe not.

No, it's totally subjective -- whenever a grader likes it for whatever reason, it gets commended.

I got a 20, with 1 commendation.  I completely skipped 4.

Anyone else find 4 difficult?

Hm.  #4 was the second one I got (after #3).  I found #1 the hardest, just because the gaps were kinda abstract.