articles
probability
geometry
expected value

Regarding Problem 7 in the article about this year's featured topic (Expected Value), how would one calculate the expected value of the number of special points covered by a randomly placed washer if (for example) the probability of special points falling in certian regions is higher than others?

So for example, if the expected value of the number of special points falling within the region covered by a concentric circle of radius 8 (concentric to the circle of radius 16) is $326$ instead of $650/3$.

[i]Would we be able to assume that there are indeed 326 special points inside this circle of radius 8?[/i] (Essentially calculating expected value on expected value)

Could we then say that the total expected value is $5/11^2*326$ (expected value in the higher probability region) plus $5/(19^2-5^2)*324$ (expected value in the lower probability region)?

But that comes out to be around 18, which is much higher than the expected value for the original problem. I guess intuitively that does make sense, since we are in essense making the total area "smaller" - more precisely, restricting the area of the special points.

Additionally, for the definition, do the events or values x1, x2, etc. have to be mutually exclusive? Problem 5 would indicate no, but is that only because we are adding up the total number of times that BG/GB occurs?

As I understand it, you're asking if the distribution of the points matters.  It depends on what you are finding the expected value of.  For example, suppose we are told that all the special points are within 2 units of the center of the 16 unit circle.  If we want

E[washer placed with center inside 19 unit circle]

then our computation is exactly as shown in the article.  We end up with placements near the center with lots of special points averaged with many, many zeroes for placements far from the center.

However, if we change what we are computing to

E[washer placed with center within 5 units of the center of the 16 unit circle],

then our new distribution gives a very different answer.

[quote="rrusczyk"]For example, suppose we are told that all the special points are within 2 units of the center of the 16 unit circle. If we want

E[washer placed with center inside 19 unit circle]

then our computation is exactly as shown in the article.  We end up with placements near the center with lots of special points averaged with many, many zeroes for placements far from the center.[/quote]

Yes, you answered my question. I was asking basically whether or not further restrictions made a difference to the Expected value. From your answer, the condition, "suppose we are told that 99% of the special points are within 2 units of the center of the 16 unit circle" would yield again the same "E[washer placed with center inside 19 unit circle]".

It's kind of strange to think that whether or not this randomly placed washer covers a point [i]other than[/i] G (whatever we define that to be doesn't matter, since there is no requirement that G must fulfill) is irrelevant. Similarly, in the Boy/Girl problem, it was irrelevant whether the rest of the arrangement contained another BG/GB, all we cared about was that there was $2*18!/(6!12!)$ arrangements such that the first two places contain people of the opposite gender. My lack of understanding of this was what caused me to ask the question in the first place.

Thanks


[url]http://usamts.org/About/U_Contact.php[/url] has a link under "Message Board" that's supposed to link to this forum but doesn't.

We've fixed the link.  Thanks for pointing it out!


Show that a number is rational if and only if its decimal representation eventually becomes periodic.

[quote="mathnerd314"][hide="First part"]The "if" part:

If the decimal becomes periodic, then we can use an algorithm to transform it to $\frac{p}{q}$.

It's kind of hard to explain so I'll give an example:

$x=0.12333333...$
$100x=12.333333...$
$100x-x=12.21$
$99x=12.21$
$x=\frac{1221}{9900}$

Not quite sure how to do the "only if" part.[/hide][/quote]

Is this the same as saying, if the decimal is periodic, then it is rational?

Cause whatever mathnerd3.14 is doing is the right approach.

mathnerd314, just use variables instead of numbers to generalize. :)

prove this: for all $q\in\mathbb{N}$ with $\gcd(q,10)=1$, there exists $n\in\mathbb{N}$ such that $q|(10^{n}-1)$

or in MCCCXXXVII symbols,
prove: $\forall q_{0}\in \{q|q\in\mathbb{N}, \gcd(q,10)=1\}, \exists n\in\mathbb{N}$ s.t. $q_{0}|(10^{n}-1)$

[quote="boxedexe"][hide]Fermat's Little Theorem[/hide][/quote]

$q$ is not always a prime...although this is easily fixed with a stronger theorem (that i know you know...)

Oops...I thought $q$ was a given prime.  Yes, it's quite trivial.  I'll leave the solution to others. :)

[quote="mathnerd314"][quote="mathgeniuse^ln(x)"]Is this the same as saying, if the decimal is periodic, then it is rational?

Cause whatever mathnerd3.14 is doing is the right approach.[/quote]

Yep, now all we have to do is prove that if a number is rational, then it has a periodic representation.

[hide="Flash of insight"]The "only if":

If a number can be represented $\frac{p}{q}$, then there will be at most q remainders when you do long division. Thus, the decimal representation will eventuallly repeat.[/hide]

Q.E.D.[/quote]

If I understand you right, I don't see how this would proves the desired conclusion.


Meredith has a red hat, a blue hat, and a white hat. She also has three sweaters-one red, one blue, and one white. She has three pairs of jeans -one red,one blue, and one white. She wants to wear a red, white, and blue outfit consisting of one hat, one sweater, and one pair of jeans. How many different possibilities does she have?
I did: 3*3*3=27, but it's wrong. Why? I think my reasoning was right. There are three choices for red -red hat,red sweater and red jeans; three choices for white -hat,sweater and jeans, thus three choices for jeans. So Can anyone tell me why I can't use this method? Thank you.

[quote="MIA6"]There are three choices for red -red hat,red sweater and red jeans; three choices for white -hat,sweater and jeans[/quote]
An outfit consisting of a red hat, a white hat, and a blue hat would be very interesting to see :)

[quote="brconterato"]If I understood the problem, she wants to use one of each color of clothes...
so, the number of possibilities are 3*2*1 = 6...
Am I right??[/quote]

Oh, I think you were right.


MATHCOUNTS
ratio
percent
geometry
Pythagorean Theorem

What's the most elegant solution for the following:

TV sets are manufactured so that their length-to-width ratio is a constant.  Their  sizes represent the length of the diagonal of the screen.  By what percent is the  viewing area increased if you buy a 42-inch TV rather than a 37-inch TV?  Express  your answer to the nearest tenth.

[For some reason, this posted twice using the Safari browser.  Admin, please delete the duplicate.  Thanks.]

[hide="Solution"]Let the length and width of the TV set with the 37-inch diagonal be $ l$ and $ w$ respectively. Then, since the ratio of the length and the width must be kept constant, the dimensions of the TV set with the 42-inch diagonal are $ cl$ and $ cw$ for some constant $ c$.

Now, by the Pythagorean Theorem, we have
\[ l^2 \plus{} w^2 \equal{} 37^2\]
and
\[ c^2l^2 \plus{} c^2w^2 \equal{} 42^2.\]
Factoring the second equation, we get
\[ c^2(l^2 \plus{} w^2) \equal{} 42^2.\]
Finally, we plug the first equation into the second to get
\[ c^2(37^2) \equal{} 42^2\Rightarrow{c^2 \equal{} 42^2/37^2}\]
Next, we know the viewing area of the original TV set is, by the formula for area, $ lw$ and the viewing area of the bigger TV set is $ \tfrac{42^2}{37^2}lw$. The percentage of increase is $ \tfrac{42^2 \minus{} 37^2}{37^2}\times100\approx\boxed{22.39}$.[/hide]

[hide="easier"]If you know corresponding measurements $ a$ and $ b$ of two similar two-dimensional objects, then the ratio of their areas is $ (a/b)^2$ ... if they were three-dimensional, the ratio of their volumes would be $ (a/b)^3$.[/hide]

You still need to determine the ratio of length of legs, because this is not always true: $ \frac{\sqrt{l^2\plus{}w^2}}{\sqrt{a^2\plus{}b^2}}\equal{}\frac{l^2}{a^2}$

If plane figures are similar in a ratio a:b, then the ratio of their areas will always be $ a^2/b^2$.

For example... in this seemingly much more difficult problem:
In concave heptagon ABCDEFG, the distance from A to D is 7cm and the area of the pentagon is $ 10cm^2$.
In similar concave heptagon HIJKLMN, the distance from H to K is 8cm.  What is its area?

The area is just 64/49 of the area of the smaller heptagon $ (640/49cm^2)$ ... you don't need to know anything else except that they are similar and you know a corresponding dimension ... same with the televisions (in which case you know the diagonal lengths).

I am not sure why you would need the ratio of the leg lengths, or even how you could possibly know this ratio in the given problem ... now that I look at the problem, I believe Pytheagle made a computational mistake at the end, it should be [hide="this"]$ 28.85%
$[/hide].

Also (Toad) ... you need to either square the left side or square-root the right side of your equation.


Let $ a,b,c \in R$ such that $ a\plus{}b\plus{}c\equal{}1$ prove that
$ 15(a^3\plus{}b^3\plus{}c^3\plus{}ab\plus{}bc\plus{}ca)\plus{}9abc \geq 7$
:)

[quote="devil_maths"]Let $ a,b,c \in R$ such that $ a \plus{} b \plus{} c \equal{} 1$ prove that
$ 15(a^3 \plus{} b^3 \plus{} c^3 \plus{} ab \plus{} bc \plus{} ca) \plus{} 9abc \geq 7$
:)[/quote]
After homogenizing it is:
$ \sum_{cyc} 15 a^3 \plus{} 15a^2b \plus{} 15ab^2 \plus{} 18abc \ge \sum_{cyc} 7a^3 \plus{} 21a^2b \plus{} 21ab^2 \plus{} 14abc \iff$
$ \sum_{cyc} 8a^3 \minus{} 6a^2b \minus{} 6ab^2 \plus{} 4abc \ge 6 \sum_{cyc}a^3 \minus{} a^2 \minus{} ab^2 \plus{} abc \ge 0$
The first is true by AM-GM, and the last by schur.

Edit: I assumed $ a,b,c \in \mathbb{R}_\plus{}$. I doesn't hold when $ a,b,c \in \mathbb{R}$. For instance: $ a\equal{}\minus{}1,b\equal{}c\equal{}1$


Dear Valentin,

You can consider adding the CMI(Chennai Mathematics Insitute) and the ISI(Indian Statistical Insititute) admission tests into the Indian competitions page at the Resources section.

Both of this two tests are comparable to the INMO in its degree of sophistication.I hope then the students aspiring to read in these institutes can benefit too!

[quote="orl"]Every resource manager like Darij Grinberg, ZetaX can do it. Same procedure. You need to post the problems to the national olympiad forum with the URLs to the topic discussion. Then it can be added.[/quote]

Thanks I will do that!


integration
calculus
real analysis
real analysis unsolved

For which values of nonnegative $ \alpha\in \mathbb{R}$ the sequence \[ x_n\equal{}1\plus{}\frac{1}{2^{\alpha}}\plus{}\ldots\plus{}\frac{1}{n^{\alpha}}\minus{}n^{1\minus{}\alpha}lnn\]
is convergent?

For $ \alpha > 1$, the series converges and the trailing term goes to zero, so the sequence converges.  For $ 0 < a \leq 1$, draw a picture to see that $ f(n)\equal{}  \int_1^n \frac{1}{x^a} \, dx \minus{} \sum_{i \equal{} 2}^n \frac{1}{i^a}$ is increasing and bounded between $ 0$ and $ 1$ for all $ n$, so converges.  For $ a \equal{} 1$, this gives us that $ x_n$ converges, while for $ a < 1$ it shows that it diverges (the integral doesn't have the extra log factor).  Similarly, for $ a \leq 0$ we have the series portion growing asymptoticly to $ n^{1 \minus{} a}$, and the extra log factor kills us.  So this converges for $ a \geq 1$ and diverges to $ \minus{}\infty$ otherwise.


integration
calculus
trigonometry
logarithms
calculus computations

$\int ln(x)sin^{2}(3x)\,dx$

i keep doing something wrong and the problem keeps getting bigger with integration by parts. i dont know what im missing because it seems it should be easy. i started it a few different ways from the integration by parts formula and used the power reduction formula also. if someone wants to help id appriciate it.

im gonna see if i can post back with some of my work in a few minutes that shows my integration by parts expansion. itll take me a bit to find out how to use the special text thing on this forum since im new.

If you write $\sin^{2}(3x)=\frac12(1-\cos(6x))$, you'll be able to integrate $\ln x$ times $\frac12$ but, taking the second integral by parts, you'll get $\int\frac{\sin 6x}{12x}\,dx$, which is well-known to be not an elementary function.

ok so i get the following by using power reduction and distributing the ln(x) then using integration by parts.

$\frac{1}{2}\left[(xln(x)-x)-\left(\frac{ln(x)sin(6x)}{6}-\int \frac{sin(6x)}{6x}\,dx \right) \right]$

i also tried using integration by parts from the beginning and i got this.

$ln(x) \left( \frac{6x-sin(6x)}{12}\right)-\int \frac{6x-sin(6x)}{12x}\,dx$

ok thank you for the explanation. i must have been posting as you typed.


Can somebody prove that $ det(I\minus{}A^{t})\equal{}\minus{}det(A\minus{}I)$ ?

Thanks

It's false, and orthogonal matrices are irrelevant.
If those are $ n\times n$ matrices, you need $ (\minus{}1)^n$ as the factor, because you're essentially multiplying the whole matrix by $ \minus{}1$. Transposes don't change determinants.


number theory
relatively prime
number theory unsolved

$ x, y$ are natural numbers.

Find all solutions of $ 3y^2\equal{}x^4\plus{}x$.

[quote="Stephen"]$ x, y$ are natural numbers.

Find all solutions of $ 3y^2 \equal{} x^4 \plus{} x$.[/quote]

Nobody wants to help me to solve this interesting problem...

Pleaseee, I really need a solution.

[quote="ridgers"]x is or 0 or 2 in mod 3[/quote]

Hmmm...so what is the solution?

but you can factorize like this
$ 3y^{2}\equal{}x^{4}\plus{}x$
$ 3y^{2}\equal{}x(x^{3}\plus{}1)$
$ 3y^{2}\equal{}x(x\plus{}1)(x^{2}\minus{}x\plus{}1)$

and we also know that y is even and 4 divides $ x(x\plus{}1)(x^{2}\minus{}x\plus{}1)$ and also that $ x\plus{}1$ and $ x$ are coprime (relavitely prime with each other).

And somebody should help us.

Firstly x =0 as been stated as a solution but unfortunately 0 is not a natural number.
Then substitute z = x^2.
3*y^2 = z^2 + √z .
Thus z must be a perfect square and it must be divisible by 3 .
The possible values would be 3^2 * k^2 ( where k is any natural number).
substituting the above value for z we get
y^2 = 3^3 * k^4 + k .
Thus your problem is does there exists a natural number k such that k + 27 K^4 is a perfect square.
I will let u know if there exists one as of now i am not very sure on how to go on further.

[quote="Stephen"]$ x, y$ are natural numbers.

Find all solutions of $ 3y^2 \equal{} x^4 \plus{} x$.[/quote]

As above, factorize the RHS as $ x(x \plus{} 1)(x^2 \minus{} x \plus{} 1)$. These are pairwise relatively prime, thus two of them are squares and the third is three times a square.

Clearly $ x$ and $ x \plus{} 1$ can't both be squares, so one of them is the number which is three times a square. That means that $ x^2 \minus{} x \plus{} 1$ must be a square. However, $ (x \minus{} 1)^2 < x^2 \minus{} x \plus{} 1 < x^2$. So no solutions.

EDIT: This solution I think actually can easily be extended to prove slightly stronger statement: $ p^ky^2\equal{}x^4\plus{}x$ has no natural solutions for a prime $ p$ and a positive integer $ k$.

[quote="caffeineboy"][quote="Stephen"]$ x, y$ are natural numbers.

Find all solutions of $ 3y^2 \equal{} x^4 \plus{} x$.[/quote]

As above, factorize the RHS as $ x(x \plus{} 1)(x^2 \minus{} x \plus{} 1)$. These are pairwise relatively prime, thus two of them are squares and the third is three times a square.

Clearly $ x$ and $ x \plus{} 1$ can't both be squares, so one of them is the number which is three times a square. That means that $ x^2 \minus{} x \plus{} 1$ must be a square. However, $ (x \minus{} 1)^2 < x^2 \minus{} x \plus{} 1 < x^2$. So no solutions.

EDIT: This solution I think actually can easily be extended to prove slightly stronger statement: $ p^ky^2 \equal{} x^4 \plus{} x$ has no natural solutions for a prime $ p$ and a positive integer $ k$.[/quote]

Thanks.

[quote="caffeineboy"][quote="Stephen"]$ x, y$ are natural numbers.

Find all solutions of $ 3y^2 \equal{} x^4 \plus{} x$.[/quote]

As above, factorize the RHS as $ x(x \plus{} 1)(x^2 \minus{} x \plus{} 1)$. These are pairwise relatively prime, thus two of them are squares and the third is three times a square.

Clearly $ x$ and $ x \plus{} 1$ can't both be squares, so one of them is the number which is three times a square. That means that $ x^2 \minus{} x \plus{} 1$ must be a square. However, $ (x \minus{} 1)^2 < x^2 \minus{} x \plus{} 1 < x^2$. So no solutions.

EDIT: This solution I think actually can easily be extended to prove slightly stronger statement: $ p^ky^2 \equal{} x^4 \plus{} x$ has no natural solutions for a prime $ p$ and a positive integer $ k$.[/quote]

I think you forgot also the cases where $ x(x \plus{} 1)(x^2 \minus{} x \plus{} 1)$ one of this is three and the other perfect squares but also this cases do not bring solution I tried them.
And i have also a quesstion : How can you show that $ x\plus{}1$ , $ x$ and $ x^2 \minus{} x \plus{} 1$ are relatively prime. For $ x$ and $ x\plus{}1$ it is clear but for $ x^2 \minus{} x \plus{} 1$ ??

[quote="aauqen"][quote="ridgers"]I think you forgot also the cases where $ x(x \plus{} 1)(x^2 \minus{} x \plus{} 1)$ one of this is three and the other perfect squares but also this cases do not bring solution I tried them.
And i have also a quesstion : How can you show that $ x \plus{} 1$ , $ x$ and $ x^2 \minus{} x \plus{} 1$ are relatively prime. For $ x$ and $ x \plus{} 1$ it is clear but for $ x^2 \minus{} x \plus{} 1$ ??[/quote]

$ gcd(x, x^2 \minus{} x \plus{} 1) \equal{} gcd(x, x(x \minus{} 1) \plus{} 1) \equal{} 1$
$ gcd(x \plus{} 1, x^2 \minus{} x \plus{} 1) \equal{} gcd(x \plus{} 1, (x \plus{} 1)(x \minus{} 2) \plus{} 3) \equal{} gcd(x \plus{} 1, 3)$

So we only have to additionally check the case where $ x \equiv 2\ (mod\ 3)$.[/quote]

So what can we do when $ x\equiv 2\ (mod\ 3)$ ???

Obviosly $ x\equal{}2\mod 3$. In these case $ 3y^2\equal{}x(x\plus{}1)(x^2\minus{}x\plus{}1)$ we had $ 3|x^2\minus{}x\plus{}1$. But $ 9|x^2\minus{}x\plus{}1\to 9|(2x\minus{}1)^2\plus{}3$ -contradition. Obviosly $ (x,x^2\minus{}x\plus{}1)\equal{}1, (x\plus{}1,x^2\minus{}x\plus{}1)\equal{}(x\plus{}1,3)\equal{}3$.
Therefore must be $ x^2\minus{}x\plus{}1\equal{}3z_1^2, x\equal{}z_2^2,x\plus{}1\equal{}z_3^2\to x\equal{}y\equal{}0$.


Hi,
I'd appreciate any help with the following problem. Let f(z) be holomorphic in D(0,1)/{0}, and assume that |f|^2 is integrable over D(0,1). Show that z=0 is a removable singularity.

It is clear to me how to show that we cannot have a pole of any order, but the essential singularity case I have no idea how to approach.

Thank you in advance!

Here's a nice trick, based on some of the same ideas as Fourier series. Write a Laurent series for $f$: $f(z)=\sum_{n=-\infty}^\infty a_nz^n$. For any $n$ and any fixed radius $r$, $2\pi a_n=\int_0^{2\pi}f(re^{i\theta})e^{-in\theta}$. Choose $n$ and let $g(z)=f(z)-a_nz^n$.

Now $\int\int_{0<|x+iy|<1}|f(x+iy)|^2\,dx\,dy= \int_0^1\int_0^{2\pi}r|f(re^{i\theta}|^2\,d\theta\,dr$
$=\int_0^1\int_0^{2\pi}r|g(re^{i\theta}+a_nr^ne^{in\theta}|^2\,d\theta\,dr$
$=\int_0^1\int_0^{2\pi}r\left(|g(re^{i\theta}|^2+|a_nr^ne^{in\theta}|^2 +2\text{Re }a_nr^ng(re^{i\theta})e^{-in\theta}\right)\,d\theta\,dr$
Since $g$ has its $n$-th Laurent coefficient zero, the third term integrates to zero and the integral becomes
$\int\int_{0<|x+iy|<1}|g(x+iy)|^2\,dx\,dy+\int_0^1\int_0^{2\pi}r^{2n+1}|a_n|^2\, d\theta\,dr$
$=\int\int_{0<|x+iy|<1}|g(x+iy)|^2\,dx\,dy+2\pi|a_n|^2\int_0^1r^{2n+1}\,dr$
The second integral diverges to $\infty$ whenever $n<0$ and $a_n$ is nonzero; since the first integral is positive, the original integral of $|f|^2$ diverges whenever any of the terms of its Laurent series with negative index is nonzero. For the integral to converge, the singularity must be removable.


geometry
parallelogram
analytic geometry
parameterization
probability and stats

Let's suppose that we have bivariate data that is randomly distributed about a line. We know that the data is not normally distributed about the line, but it is instead uniformly distributed. Let's also assume that the uniform distribution is evenly wide throughout. The probabilities are then uniformly distributed on a parallelogram in an x-y coordinate system.

Given we take a simple random sample of a few data from this population (say, 30), how should the equation of the line be estimated and how should the width of the uniform distribution be estimated?

This is my initial thoughts to get the ball rolling.  Perhaps the best way to estimate the width of the uniform distribution will be the min and max of hte sample.
Additionally, if you will be doing repeated samples.  I would suggest the min of the min and the max of the max if that makes since.

For the equation of the line, you can still do a least squares regression, your only issue will be that any inference of the parameters will by void unless you apply some law of large numbers to your coefficient estimates - because the main assumption of linear regression hypothesis testing is that your error term is going to be normal (and your parameter estimates are essentially a linear combination of the error term).   But I'm not exactly sure that a law of large numbers will help over come.  hope this helps

I have one question about the setup. The distribution you describe has more than two parameters. In addition to the two parameters that describe the line, there's the interval in $ x$ over which this line segment is defined (you do describe it as a line segment) and there's the width (in $ y$) of the uniform distribution in your description.

Do you know these additional parameters [i]a priori[/i] or is your knowledge of them limited to what you can see in your experimental data?

An additional question about your points $ (x_i,y_i):$ are you choosing the ordered pair at random in your parallelogram or are you choosing the $ x_i$ according to some fixed pattern (equally spaced or whatever)?

Here's one interpretation:
The points are given by $ y\equal{}p(x)\plus{}q$, where $ p(x)$ is our unknown (linear) model and $ q$ is uniform on $ [\minus{}a,a]$ for some unknown $ a$.
By the nature of uniform distributions, the only meaningful distinction is whether a model is possible. We want our model to be possible with the widest possible range of $ a$ values, which means that we want to minimize $ \max \{y\minus{}p(x)\}.$
For a method to do that, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=238484]here[/url].


[b]Does Friction between two surfaces depend on the area of contact between them?[/b]

And if not, how can we expain that friction between the two surfaces is dependent on the relative velocity of one surface with respect to another when the velocity is quite large (friction decreases)?

According to me, when the velocity is quite large, there is relative vibration between the two surfaces and the area of contact between them becomes less than before, and so there is lesser amount of clinging between the atoms of the two surfaces.

Definitely area of exposure has an affect.  Imagine an ice skater.  The blade is really thin so as to minimize friction.  If the blade was, say, 3 times as thick, the skater could not generate the same speeds.


I want to know how to calculate this sum:

[img]http://img370.imageshack.us/img370/3658/cot2zi2.jpg[/img]

Thx

Hello

By De Moivre's Theorem, we have

$ \cos{(2n\plus{}1)x}\plus{}i\sin{(2n\plus{}1)x}\equal{}(\cos{x}\plus{}i\sin{x})^{2n\plus{}1}$

So comparing imaginary parts, we get that

$ \sin{(2n\plus{}1)x}\equal{}\binom{2n\plus{}1}{1}(\cos^{2n}{x} \sin{x}\minus{}\binom{2n\plus{}1}{3}\cos^{2n\minus{}2}{x}\sin^{3}{x}...\plus{}(\minus{}1)^{n}\binom{2n\plus{}1}{2n\plus{}1}\sin^{2n\plus{}1}{x}$

This can be written as $ \sin^{2n\plus{}1}{x}(\binom{2n\plus{}1}{1}\cot^{2n}{x}\minus{}\binom{2n\plus{}1}{3}\cot^{2n\minus{}2}{x}\plus{}...\plus{}(\minus{}1)^{n}\binom{2n\plus{}1}{2n\plus{}1})$.

Observe that some roots of $ \binom{2n\plus{}1}{1}\cot^{2n}{x}\minus{}\binom{2n\plus{}1}{3}\cot^{2n\minus{}2}{x}\plus{}...\plus{}(\minus{}1)^{n}\binom{2n\plus{}1}{2n\plus{}1}$ are $ x\equal{}\frac{l\pi}{2n\plus{}1}$, where $ l\equal{}1,2,3,...,n$, as those are roots of $ \sin{(2n\plus{}1)x}\equal{}0$

Now, if we make the substituion $ a\equal{}\cot^{2}{x}$, we get $ \binom{2n\plus{}1}{1} a^{n}\minus{}\binom{2n\plus{}1}{3} a^{n\minus{}1}\plus{}...\plus{}(\minus{}1)^{n}\binom{2n\plus{}1}{2n\plus{}1}$, which therefore has roots of $ a\equal{}\cot^{2}{\frac{l\pi}{2n\plus{}1}}$, where $ l\equal{}1,2,...,n$.

So, we see that the desired sum, by Vieta's Formula is $ \frac{\binom{2n\plus{}1}{3}}{\binom{2n\plus{}1}{1}}\equal{}\frac{n(2n\minus{}1)}{3}$.

[hide="Fail"]Note that
\begin{align*} \sin[(2n + 1)x] &= \text{Im}((\cos x + i\sin x)^{2n + 1})\\ & = ( - 1)^n {2n + 1 \choose 1} \cos^{2n} x \sin x + ( - 1)^{n + 1}{2n + 1 \choose 3} \cos^{2n - 2} x \sin^3 x + \cdots \\
& = ( - 1)^n\sin^{2n + 1} x \left[{2n + 1 \choose 1} \cot^{2n} x - {2n + 1 \choose 3} \cot^{2n - 2} x + \cdots\right]\end{align*}
which is a $ 2n$ degree polynomial in $ \cot x$, and roots $ x = \theta_k, \quad k = 1,2 \ldots 2n$, where $ \theta_k = \frac {k\pi}{2n + 1}$. As $ \sum_{i = 1}^{2n} \cot^2 \theta_i = \left(\sum_{i = 1}^{2n} \cot \theta_i \right)^2 - 2\left(\sum_{1 \le i < j \le 2n} \cot \theta_i \cot \theta_j \right)$; by Vieta’s, $ \sum_{i = 1}^{2n} \cot \theta_i = 0, \sum_{1 \le i < j \le 2n} \cot \theta_i \cot \theta_j = - {2n + 1 \choose 3}\left/\right. {2n + 1 \choose 1} = - \frac {n(2n - 1)}{3}$.

Finally, from $ \cot(\pi - x) = \cot x$, we have $ \sum_{i = 1}^{n} \cot^2 \theta_i = \frac 12 \sum_{i = 1}^{2n} \cot^2 \theta_i = \frac {1}{2}\left[ - 2 \left(\frac {n(2n - 1)}{3}\right)\right] = \frac {n(2n - 1)}{3}$.[/hide]


From UIL 2006 Regional Mathematics:

numbers 1-10 are in a bag. 5 numbers are taken w/o replacement. What is the probability that the smallest number taken is 4?

is it possible to use graphing calculator for permutation and combination? if so, how?

For a TI 83, 86, 89, etc... go to math, and over to PRB, the probablity menu.  You will see nPr and nCr, the combination permutation functions.

For example:  If you wanted to find the combination of "7 choose 4", hit $\boxed{7}$, math-->PRB->$\boxed{nCr}$, $\boxed{4}$.  Hit enter to compute.

[quote="Farenhajt"][hide="Solution"]In order for $4$ to be the smallest number, the remaning four numbers must be chosen among $5,6,7,8,9,10$. Hence the probability is $\frac{{6\choose 4}}{{10\choose 5}}={15\over 252}={5\over 84}$[/hide][/quote]

You have to choose 4 as well..

[quote="Farenhajt"][quote="Ignite168"]You have to choose 4 as well..[/quote]

No, I NEED to have it, so it's fixed.[/quote]

Yeah but do you have the probability in there of you getting 4?

Maybe you should review the concept of probability - it's the ratio of the number of FAVORABLE outcomes to the number of ALL THE POSSIBLE outcomes. Hence first I simply count 5-member subsets which have 4 as the least member, and then divide that by the number of all the possible 5-member subsets.

Is it clear now?


If $a,b,c$ are the sides of the triangle then show that $(a+3b-c)(b+3c-a)(c+3a-b)\ge 27(a+b-c)(b+c-a)(c+a-b)$.
Well I solved it opening brackets.(it took me 10 minutes.) :D

It's worth noting that letting $x=a+b-c$ and so on induces the exact same inequality (that's my substitution of choice)


calculus integration derivative function geometry logarithms trigonometry

I've just finished integration (at least, the basics) in my calculus class, and we've moved on to the differentiation and integration of special functions (logarithmic, for example).
Note: When I refer to area, the area in question can be positive and negative (though in real-life calculus application problems, the restriction usually is that area can never be negative).
I was thinking of certain definite integrals, and I found that $ \int^1_0\ln x\,dx \equal{} \minus{} 1$.  The more interesting fact to me is that $ \int^a_0\ln x\,dx \equal{} c$, where $ a\ge 0$ and $ a,c$ are constants.  I intuitively thought that the area had to be constant because as $ x\rightarrow 0^ \plus{}$ in the function $ f(x) \equal{} \ln x$, the graph of the function approaches negative infinity but is sort of "bounded."  In other words, the area that is being accumulated in $ \int^k_a\ln x\,dx$ (with $ a > 0$) as $ k\rightarrow 0^ \plus{}$ is getting smaller until it reaches $ 0$ as $ k$ reaches $ 0$, so the area reaches a limit value.
But then I found that $ \int^{\frac {\pi}{2}}_0\tan x\,dx \equal{} \plus{} \infty$ and $ \int^1_0\frac {1}{x}\,dx \equal{} \plus{} \infty$!  If I used a similar argument as in the above paragraph for those two definite integrals, then the values of the integrals should be constant as well!  I don't understand why those two areas are unbounded, while the area $ \int^1_0\ln x\,dx$ IS bounded.
This is what I want an intuitive explanation for (notice that, in this post, I haven't been rigorous but instead intuitive).  Is there such an explanation?

Some functions grow faster than others, so some improper integrals converge while others diverge. Logarithms in particular are small compared to positive powers of $ x$ near $ \infty$, and small compared to negative powers of $ x$ near zero.


Find the sum of the digits in the decimal representation of (654,000,003,987,000)^2.

this is simply the sum of the digits of 654^2 + the sum of the units digits of 2*654*3987 + the sum opf the units digits of 3987^2.

This equation is derived off the sum of squares, and there will be 18 zeroes at the end of (654000000000)^2(we eliminate the last 3 zeroes) there will be 16 digits in 2*654000000000*3987.  There are 9 zeroes at the end of that and there are 8 digits in 3987^2.

Keep in mind that it is not the sum of the units digits of (654^2+2*654*3987 + 3987^2).

A hint, if you want to see how Scrambled got to that step:  write the number you're squaring as x + y, where it's relatively easy to square x, to square y, and to compute xy.

for example square 17 you could write it as (10+7)(10+7) which is not neccesarly easier here, but for 10000007^2 it would be..


In Caddlebot's warehouse, everybody makes the same amount of money, except for Caddlebot. If Caddlebot makes $800,000 a year and Bobert (who works at Caddlebot's warehouse) makes $20,000 a year, how many people work at Caddlebot's warehouse (including Caddlebot himself) if the average salary of all the workers is $68750?

[quote="Iron Fist"]In Caddlebot's warehouse, everybody makes the same amount of money, except for Caddlebot. If Caddlebot makes $800,000$ a year and Bobert (who works at Caddlebot's warehouse) makes$20,000$ a year, how many people work at Caddlebot's warehouse (including Caddlebot himself) if the average salary of all the workers is $68750$?[/quote]

[hide]$\frac{800000+20000x}{x+1}=68750$, $x=15$, 16 people[/hide]


let $ D(a,b,c,d) \equal{} \left|\begin{array}{cccc}1\plus{}a&1&1&1\\ 1&1\plus{}b&1&1\\ 1&1&1\plus{}c&1\\ 1&1&1&1\plus{}d\end{array}\right|$
then $ D(a,b,c,d)\equal{}
 \left|\begin{array}{cccc}1 \plus{} a & 1 & 1 & 1 \\
1 & 1 \plus{} b & 1 & 1 \\
1 & 1 & 1 \plus{} c & 1 \\
1 & 1 & 1 & 1\end{array}\right| \plus{} \left|\begin{array}{cccc}1 \plus{} a & 1 & 1 & 0 \\
1 & 1 \plus{} b & 1 & 0 \\
1 & 1 & 1 \plus{} c & 0 \\
1 & 1 & 1 & d\end{array}\right|$.
in the first determinant we remplace  $ C_i$ by $ C_i \minus{} C_4$ for $ i \equal{} 1,2,3$ then it equal to $ abc$.in the second determinant after developping by $ C_4$
 we find $ D(a,b,c,d) \equal{} abc \plus{} dD(a,b,c)$, with the same way $ D(a,b,c) \equal{} ab \plus{} cD(a,b)$ and so $ D(a,b,c,d) \equal{} abc \plus{} abd \plus{} bcd \plus{} acd \plus{} abcd$.


Given $ a, b, c > 0$. Prove that: $ a \plus{} b \plus{} c \plus{} \frac{1}{abc} \geq\ \frac {2(ab \plus{} bc \plus{} ca \plus{} 1)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ a + b + c + \frac {1}{abc} \geq\ \frac {2(ab + bc + ca + 1)^2}{(a + b)(b + c)(c + a)}$
 :)[/quote]

Try:${ a = b = c = \frac{1}{2}}$
It isn't true  :P

[quote="hophinhan"][quote="caubetoanhoc94"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ a \plus{} b \plus{} c \plus{} \frac {1}{abc} \geq\ \frac {2(ab \plus{} bc \plus{} ca \plus{} 1)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
 :)[/quote]

Try:$ a \equal{} b \equal{} c \equal{} \frac {1}{2}$
It isn't true  :P[/quote]

It true :|[/quote]

I'm sorry  :( .Thanks;hophinhan  :(.It isn't easy  :(

[quote="caubetoanhoc94"][/quote][quote="hophinhan"][quote="caubetoanhoc94"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ a + b + c + \frac {1}{abc} \geq\ \frac {2(ab + bc + ca + 1)^2}{(a + b)(b + c)(c + a)}$
 :)[/quote]

Try:$ a = b = c = \frac {1}{2}$
It isn't true  :P[/quote]

It true :|[/quote][quote="caubetoanhoc94"]

I'm sorry  :( .Thanks;hophinhan  :(.It isn't easy  :([/quote]

[b]
Solution:[/b]

$ (a + b)(b + c)(c + a) \ge \frac {8}{9}(a + b + c)(ab + bc + ca)$

We need to prove :  $ a + b + c + \frac {1}{abc} \geq\ \frac {2(ab + bc + ca + 1)^2}{\frac {8}{9}(a + b + c)(ab + bc + ca)}$

$ < = > \ p + \frac {1}{r} \ge \frac {9(q + 1)^2}{4pq}$

$ < = > \ 4pq(pr + 1) \ge 9r(q + 1)^2$

$ < = > \ 4pq \ge r(9(q + 1)^2 - 4p^2q)$

+ $ 9(q + 1)^2 - 4p^2q \le 0$ ...It's obvious

+ $ 9(q + 1)^2 - 4p^2q \ge 0$ ...By AM_GM :

$ r(9(q + 1)^2 - 4p^2q) \le \frac {pq}{9}(9(q + 1)^2 - 12q^2) = \frac {pq}{9}(36 - 3(q - 3)^2) \le 4pq$

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ a \plus{} b \plus{} c \plus{} \frac {1}{abc} \geq\ \frac {2(ab \plus{} bc \plus{} ca \plus{} 1)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
 :)[/quote]
$ \Leftrightarrow (a \plus{} b \plus{} c)((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc) \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}\geq 2(ab \plus{} bc \plus{} ca)^{2} \plus{} 4(ab \plus{} bc \plus{} ca) \plus{} 2$
$ \Leftrightarrow (a \plus{} b \plus{} c)^{2}(ab \plus{} bc \plus{} ca) \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}\geq (a \plus{} b \plus{} c)abc \plus{} 4(ab \plus{} bc \plus{} ca) \plus{} 2$
using $ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}\geq 8$
and $ a^{2} \plus{} b^{2} \plus{} c^{2}\geq ab \plus{} bc \plus{} ca$
we need prove :
$ (ab \plus{} bc \plus{} ca)^{2} \plus{} 6 \geq (a \plus{} b \plus{} c)abc \plus{} 4(ab \plus{} bc \plus{} ca)$
because
$ \frac {1}{3}(ab \plus{} bc \plus{} ca)^{2} \geq (a \plus{} b \plus{} c)abc$
and $ \frac {2}{3}(ab \plus{} bc \plus{} ca)^{2} \plus{} 6 \geq 4(ab \plus{} bc \plus{} ca)$
Done !  :D

[quote="nguoivn"][quote="iqtimo"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ a \plus{} b \plus{} c \plus{} \frac {1}{abc} \geq\ \frac {2(ab \plus{} bc \plus{} ca \plus{} 1)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
 :)[/quote]
$ \Leftrightarrow (a \plus{} b \plus{} c)((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc) \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}\geq 2(a \plus{} b \plus{} c)^{2} \plus{} 4(ab \plus{} bc \plus{} ca) \plus{} 2$
[/quote]
Are you sure? Notice that $ RHS \equal{} 2(ab \plus{} bc \plus{} ca \plus{} 1)^2$  :maybe:
[quote="hophinhan"]
[b]
Solution:[/b]

$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge \frac {8}{9}(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$

We need to prove :  $ a \plus{} b \plus{} c \plus{} \frac {1}{abc} \geq\ \frac {2(ab \plus{} bc \plus{} ca \plus{} 1)^2}{\frac {8}{9}(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$

$ < \equal{} > \ p \plus{} \frac {1}{r} \ge \frac {9(q \plus{} 1)^2}{4pq}$

$ < \equal{} > \ 4pq(pr \plus{} 1) \ge 9r(q \plus{} 1)^2$

$ < \equal{} > \ 4pq \ge r(9(q \plus{} 1)^2 \minus{} 4p^2q)$

+ $ 9(q \plus{} 1)^2 \minus{} 4p^2q \le 0$ ...It's obvious

+ $ 9(q \plus{} 1)^2 \minus{} 4p^2q \ge 0$ ...By AM_GM :

$ r(9(q \plus{} 1)^2 \minus{} 4p^2q) \le \frac {pq}{9}(9(q \plus{} 1)^2 \minus{} 12q^2) \equal{} \frac {pq}{9}(36 \minus{} 3(q \minus{} 3)^2) \le 4pq$[/quote]
Nice proof, hophinhan  :)[/quote]
thank, i edited it

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ a \plus{} b \plus{} c \plus{} \frac {1}{abc} \geq\ \frac {2(ab \plus{} bc \plus{} ca \plus{} 1)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
 :)[/quote]
Applying the Cauchy Schwarz Inequality, we have
$ LHS\cdot (ab^2\plus{}bc^2\plus{}ca^2\plus{}abc) \ge (ab\plus{}bc\plus{}ca\plus{}1)^2,$
and
$ LHS \cdot (ac^2\plus{}ba^2\plus{}cb^2\plus{}abc) \ge (ca\plus{}ab\plus{}bc\plus{}1)^2.$
It follows that
$ LHS \ge \frac{(ab\plus{}bc\plus{}ca\plus{}1)^2}{ab^2\plus{}bc^2\plus{}ca^2\plus{}abc},$
and
$ LHS \ge \frac{(ab\plus{}bc\plus{}ca\plus{}1)^2}{a^2b\plus{}b^2c\plus{}c^2a\plus{}abc}.$
Therefore
$ 2LHS \ge (ab\plus{}bc\plus{}ca\plus{}1)^2\left( \frac{1}{a^2b\plus{}b^2c\plus{}c^2a\plus{}abc}\plus{}\frac{1}{ab^2\plus{}bc^2\plus{}ca^2\plus{}abc}\right) \ge \frac{4
(ab\plus{}bc\plus{}ca\plus{}1)^2}{(a^2b\plus{}b^2c\plus{}c^2a\plus{}abc)\plus{}(ab^2\plus{}bc^2\plus{}ca^2\plus{}abc)}\equal{}\frac{4(ab\plus{}bc\plus{}ca\plus{}1)^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}.$
So, by dividing both sides for $ 2$, we obtain the result. :)

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ a \plus{} b \plus{} c \plus{} \frac {1}{abc} \geq\ \frac {2(ab \plus{} bc \plus{} ca \plus{} 1)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
 :)[/quote]
Applying the Cauchy Schwarz Inequality, we have
$ LHS\cdot (ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc) \ge (ab \plus{} bc \plus{} ca \plus{} 1)^2,$
and
$ LHS \cdot (ac^2 \plus{} ba^2 \plus{} cb^2 \plus{} abc) \ge (ca \plus{} ab \plus{} bc \plus{} 1)^2.$
It follows that
$ LHS \ge \frac {(ab \plus{} bc \plus{} ca \plus{} 1)^2}{ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc},$
and
$ LHS \ge \frac {(ab \plus{} bc \plus{} ca \plus{} 1)^2}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc}.$
Therefore
$ 2LHS \ge (ab \plus{} bc \plus{} ca \plus{} 1)^2\left( \frac {1}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \plus{} \frac {1}{ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc}\right) \ge \frac {4 (ab \plus{} bc \plus{} ca \plus{} 1)^2}{(a^2b \plus{} b^2c \plus{} c^2a \plus{} abc) \plus{} (ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc)} \equal{} \frac {4(ab \plus{} bc \plus{} ca \plus{} 1)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}.$
So, by dividing both sides for $ 2$, we obtain the result. :)[/quote]

Very very nice  :blush: .

[hide="phục quá"]Cần gì làm nấy  :D ...[/hide]


Prove that, when $p$ is prime, for a certain $n$, that $n^{2}\equiv-1\pmod{p}$, if and only if $p\equiv1\pmod{4}$ and that such an $n$ does not exist if and only if $p\equiv3\pmod{4}$.
[hide="hint"]wilson's theorem[/hide]

is this too easy for olympiad? :maybe:
if so please move it

[quote="daermon"]$n\equiv-1\pmod{p}$[/quote]
Must be $n^{2}\equiv-1\pmod{p}$.

Well known, I remember posting it myself some time ago.

The nicest solution in my opinion that was given is the following:
$n^{2}\equiv-1 \pmod p \Rightarrow (n^{2})^{\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}$
So by Fermat's theorem, $(-1)^{\frac{p-1}{2}}=1$ so $\frac{p-1}{2}$ is even. Hence, $p \equiv 1 \pmod 4$

[u]Please post your solution using Wilson's theorem daermon![/u]

Jan - Nice work but one still has to prove "if and only if" part, that is  given ,$(-1)^{(p-1)/2}\equiv 1 \implies$ that there exist a n such that $n^{2}\equiv-1$ .. No big deal but may require a little more detail.

One interesting practical question: given $p$ can you find $n$ - in a reasonable time .. for example, $5882353$ is prime and 1 mod 4, find $n$ such that $n^{2}\equiv-1 \bmod 5882353$ (You can use calculator but no computer program)

[hide="wilsons theorem solution"]
first of all, if $p\equiv3\pmod{4}$ then $n^{2}\equiv2\pmod{4}$, a contradiction.
now, we just need to prove that we can always find a solution to $n^{2}\equiv1\pmod{p}$.  Here is where we use wilson's theorem.  We have $(p-1)!\equiv-1\pmod{p}$.  We can write this as $1\cdot2\cdot...\cdot(p-2)\cdot(p-1)\equiv-1\pmod{p}$.  However, we can also write this as $1\cdot2\cdot...\cdot(-2)\cdot(-1)\equiv-1\pmod{p}$ because $p-n\equiv-n\pmod{p}$.  Pairing the first and last, second and second to last, and all the other terms accordingly gives $(-1^{2})(-2^{2})(-3^{2})...(-\frac{p-1}{2}^{2})\equiv-1\pmod{p}$.  It is easy to prove that, if $p\equiv1\pmod{4}$ then there will be an even amount of negatives on the left side and thus they cancel out and we have $(\frac{p-1}{2})!^{2}\equiv-1\pmod{p}$ and we are finished.
also, if $p\equiv3\pmod{p}$ then the negatives will not cancel out and we have $(\frac{p-1}{2})!^{2}\equiv1\pmod{p}$ which doesn't help us with this problem.[/hide]

gyan, if you had a reeeeealy powerful computer then you could use my method  :P


geometry
rectangle
search
rotation
combinatorics unsolved
combinatorics

A 7.7 square is covered by sixteen 3.1 and one 1.1 tiles. What are the pemissible positions of the 1.1 tile?

[quote]A $7 \times 7$ square is covered by sixteen $3 \times 1$ and one $1 \times 1$ tile. What are the pemissible positions of the $1 \times 1$ tile?[/quote]

This can be extended for any $(3m+1) \times (3n+1)$ rectangle:

We show a trivial construction for all solutions $(x,y)$ with $x,y \equiv 1 \,\, mod \, 3$
Except for the row and column containing cell $(x,y)$ the rest of the cells form $4$ rectangles with sides multiples of $3$
The number of cells in either row or column is a multiple of $3$ as well

Colour the rectangle with $3$ colours such that cell $(i,j)$ has the same colour as cell $(k,l)$ iff $(i+j) \equiv (k+l) \,\, mod \, 3$
Each $3 \times 1$ tile covers $1$ square of each colour
Since there exists a solution with the $1 \times 1$ tile in cell $(1,1)$,
There is exactly one more square of the colour of cell $(1,1)$
Thus for any solution $(x,y)$, $x+y \equiv 1+1 \,\, mod \, 3$
Similarly for the colouring such that cell $(i,j)$ has the same colour as cell $(k,l)$ iff $(i-j) \equiv (k-l) \,\, mod \, 3$,
We obtain $x-y \equiv 1-1 \,\, mod \, 3$

Therefore $x,y \equiv 1 \,\, mod \, 3$

[quote="weiquan"][quote]quote]

Colour the rectangle with $3$ colours such that cell $(i,j)$ has the same colour as cell $(k,l)$ iff $(i+j) \equiv (k+l) \,\, mod \, 3$
[/quote][/quote]
Why? I don't understand. Can you exprees your solution more clearly?

[quote="HUY VAN"][quote="weiquan"][quote]quote]

Colour the rectangle with $3$ colours such that cell $(i,j)$ has the same colour as cell $(k,l)$ iff $(i+j) \equiv (k+l) \,\, mod \, 3$
[/quote]
Why? I don't understand. Can you exprees your solution more clearly?[/quote][/quote]

It is actually already clear enough: Cell $(i,j)$ is the cell in row/column $i$ and column/row $j$. Cells $(i,j)$ and $(k,l)$ have the same colour if and only if $(i+j) \equiv (k+l) \,\, mod \, 3$. By this condition alone you will have to use exactly $3$ colours.

This has surely appeared elsewhere on the forums since this old post (it was recently reposted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=517502 ), but I don't know what to search for. Here's another (more transparent?) way to see what's going on: color the board checkerboard with 1, 0 and -1 (ignore the dots):
[code]
. 1  0 -1  1  0 -1  1
. 0 -1  1  0 -1  1  0
.-1  1  0 -1  1  0 -1
. 1  0 -1  1  0 -1  1
. 0 -1  1  0 -1  1  0
.-1  1  0 -1  1  0 -1
. 1  0 -1  1  0 -1  1
[/code]
Every 3x1 gives sum 0 while the whole board has sum 1, so the 1x1 must cover a square numbered 1.  Now rotate the coloring and the same principle applies.

Problem 19 from here
https://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=165148&p=920922
I just found this link


calculus
derivative
function
trigonometry
calculus computations

Find the derivative of the given functions using the difference quotient.

1) f(x) = (sin(PI/2 + t) - 1)/t

2) f(x) = (cosx + 1)/(x-3PI)

If we have a function $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x)-v'(x)u(x)}{v^2(x)}$, so for the first problem, we have

$u(x) = \sin\left(\frac{\pi}{2}+x\right)-1 = \cos x -1 \Rightarrow u'(x) = -\sin x$

$v(x) = x \Rightarrow v'(x) = 1$

so $f'(x) = \frac{1-x\sin x-\cos x}{x^2}$


analytic geometry
graphing lines
slope
trigonometry

A book with a mass of $ m$ is held in equilibrium on a board with a slope of $ \theta$ degrees by a horizontal force. What is the Normal Force exerted on the book?

[quote="Carcul"]What doubt do you have in a problem like this?

[hide="Answer"]From equilibrium in the perpendicular direction to the board we get $ N \equal{} P \cos \theta \plus{} F \sin \theta$. From equilibrium parallel to the board we have $ P \sin \theta \equal{} F \cos \theta$. Now just combine both expressions.[/hide][/quote]

I have no doubt, I just think it's an interesting problem because in another forum the poster considered the normal force to be $ N\equal{}mg$. So I guess I was wondering if people here would make the same mistake...

[hide]By the way the full answer for anyone who cares is $ mg\sin^2 \theta \sec \theta \;\plus{}\; mg\cos \theta$[/hide]


calculus
integration
geometry
trapezoid
function
algebra solved
algebra

Is there a way to find the exact value of the following integral?

integration of (x^-0.5)(e^-x) dx. Sry for the horrible format. :blush:

The question is for trapezium rule but I was wondering if it were possible to find the exact value.

no elementary primitive.

what's the range of the integral? the integration from 0 to $\infty$ is $\sqrt\pi$.

just wondering, is there a trick to quickly find that there's no elementary primitive?

or is it just if you don't find any? :)

you reduce the integral to a error function (erf)...

How did you get sqrt pi? And how do you reduce it to an error function? (what's an error function anyway?)

Suppose $t = x^{\frac{1}{2}}$
thus $dx = 2tdt$

\[\int_0^\infty x^{-\frac{1}{2}} e^{-x} dx = 2\int_0^\infty e^{-t^2}dt\]

Error function : [tex]\hbox{erf}(z) = \frac{2}{\sqrt\pi}\int_0^z e^{-t^2}dt[/tex]

we are going to evaluate $I = \int\limits_0^\infty e^{-t^2}dt$.

[tex]I^2 = \int\limits_0^\infty e^{-x^2}dx \int\limits_0^\infty e^{-y^2}dy = \int\limits_0^\infty\!\int\limits_0^\infty e^{-(x^2+y^2)}dxdy = \int\limits_0^{\frac{\pi}{2}}\!\int\limits_0^\infty e^{-r^2} r dr d\theta = \frac{\pi}{2} \cdot  \int\limits_0^\infty e^{-r^2} d(\frac{1}{2}r^2) = \frac{\pi}{4} [/tex]

thus [tex]I = \frac{\sqrt\pi}{2}[/tex].


I am new to latex , i have made my first file but i dont know how to insert that file in Aops' posts ...

do i need to attach it ...

$ \text{\LaTeX}$ on AOPS is not based on making something and attaching it. Just put the math mode code directly into your post, and let the forum software do the rest.


Given are$\triangle ABC$ with centroid $G$,denote $d_{a}$ is distance from $G$ to internal bisector of $A$ simila define $d_{b},d_{c}$ prove that one of three number $\frac{d_{a}}{sin\frac{A}{2}},\frac{d_{b}}{sin\frac{B}{2}},\frac{d_{c}}{sin\frac{C}{2}}$ is sum of two others!

We have

$\frac{d_{a}}{sin \frac{A}{2}}= \frac{3d_{M/AI}}{2sin \frac{A}{2}}$

$= \frac{3|d_{B/AI}-d_{C/AI}|}{4sin \frac{A}{2}}= \frac{3|b-c|}{4}$.

From here, it is easy to see the result.


This has been posted about on the Wiki by azjps, but a search doesn't seem to turn up anything on the forums, and wiki discussion is never noticed.

Would the admins turn on monobook.css on the AoPSWiki? I realize there's a security hole in monobook.js, but I can't really see how CSS could do that... as far as I know, one simple variable setting in LocalSettings.php does the trick.

I really don't understand what you want :)

http://en.wikipedia.org/wiki/Help:Monobook.css

That explains it.

I first asked Temperal about this and he said that monobook.css had to be turned on

So I used the search and found this thread. Could you please turn it on, Mr. Vornicu?


vector
analytic geometry
function
integration

A vector field $ \vec{E}$ is given by the following expressions (in spherical coordinates):

$ \vec{E} =  \{ \begin{array}{c}\,\,\, kr\,\hat{e}_r,\,\,\,0 < r < a \\ k\frac{a^3}{r^2}\, \hat{e}_r,\,\,\,a < r < b \\ 0, \,\,\, r > b \end{array}$.

(a) Show that $ \vec{E}$ can represent an electrical field and find the charge distribution responsible for it.

(b) Compute the electrostatic energy corresponding to that charge distribution.

(c) Find the potential associated to this field.

You'll  find many of this kind of problem in a Brazilian book: Problemas de Física - Eletricidade - Eletrostática , by Eduardo Wilner. I guess that the language will not be a problem to you, isn't it so?
But let's look to this one.

[hide]
a) To represent an eletrostatic field the condition is that the vector field be conservative or $ \; \nabla \times \vec{E} \equal{} 0 \;$ and that is satisfied since $ \; \vec{E} \;$ is radial and  the spherical coordinates expression of the curl  would be, in this case, $ \; {1 \over r}\; ({1\over {sin \theta}}{{\partial E} \over{\partial \phi}} \hat {e}_\theta \minus{} {{\partial E} \over {\partial \theta}} \hat {e}_\phi ) \equal{} 0 \;$.

For the charge distribution, using, $ \; \rho \equal{} \epsilon_0 \;\;\nabla\cdot\vec{E} \;$ (assuming permissivity of vacuum), we'll have for $ \; 0 < r < a,\;\; \rho \equal{} 3 k \epsilon_0 \;$ and a surface distribution at $ \; r \equal{} b, \;\;\sigma \equal{} \minus{} k \epsilon_0 {{a^3} \over{b^2}} \;$.

b) For the electrostatic energy we could use the density as a function of the electric field (  $ \; {1\over2} \epsilon_0 E^2 \;$)  an integrate in the region 0 < r < a , plus that in  a < r <b, getting
\[ W \equal{} \pi \epsilon_0 k^2 a^5 ({{12 }\over 5} \minus{} 2 {a \over b}) \;
\]
.

c) For the potential, choosing  $ V \equal{} 0$ for $ r \geq b \;\;$ (till  $ \infty\;\;$ as usual), and integrating,
\[ \;\;V \equal{} {\int_r}^b \; Edr \; \equal{} k a^3( {1\over r} \minus{} {1 \over b}), \;\; a \leq r \leq b
\]
and
\[ \;\; V \equal{} {\int_r}^{a} Edr \; \plus{} ka^3 {{b \minus{} a} \over{ab}} \equal{} k ( {3 \over 2} a^2 \minus{} {{a^3}\over b} \minus{} {{r^2}\over 2}) ,\;\; 0 \leq r \leq a\;\;\;.
\]
[/hide]

I didn't post the problem because I don't know how to solve it, but because I thought it could be interesting. No, the language wouldn't be a problem, although I don't like to read books written in brazilian.
Finally, your answer to (a) is incomplete: there is one more condition that the vector field must fulfill in order to represent an electric field.

I don't think there is such a language as "brazilian", rather portuguese.... That is why Immanuel suggested the book for you, since you live in portugal.... :wink:

And I dont see why, the answer is incomplete. A vector field (ok, a physicists) is fully defined via its curl and divergence..., that is the electric field is given by the first two Maxwell equations for the electrostatic case:

\[ \text{div}\mathbf{E}(\mathbf{r})\equal{}\frac{\rho(\mathbf{r})}{\varepsilon}\]

\[ \text{curl}\mathbf{E}\equal{}0\]

And here we assumed the linear response of the homogenous isotropic surrounding...

Of course there is no language called "Brazilian": I obviously meant "Portuguese spoken and written in Brazil". Do you have enough knowledge of Portuguese in order to discuss this subject? If you have, then you should know that there is a striking difference between spoken and written Portuguese in Portugal and Brazil (and other Portuguese speaking communities, notably Azores and Madeira). This difference is perhaps similar to English spoken in England and in the USA, but more pronounced. And it just happen that I (and many other people) simply "don't like" the way Portuguese is written in Brazil. So that I prefer with no doubt to read a book in English (which is what I am accostumated to) than in Portuguese from Brazil. This is hard to explain to someone who is not from Portugal, so it might looks awkards.

What condition must an electric field obbey in surfaces?

Right Carcul!!

I thought that (not being a portuguese or brazilian national) the "brazilian portuguese" was similar to the "portuguese portuguese".

Thanks for the enlightenment. :D  :D


Need help with this problem

Question:

Find all incongruent solutions modulo 35 of the following system
7x + 5y = 13 (mod 35)
11x + 13y = 4 (mod 35)

Not sure what to do or how to get the solutions

Thanks

[hide="sketch of solution"]
Break up the congruences into (mod 5) and (mod 7). When working modulo 5, the y-term drops and you can find x (mod 5), then substitute into the second equation. A similar thing happens with (mod 7). (However, most systems aren't this nice; for the general case, you can take linear combinations (i.e 2 times one congruence, plus 3 times the other) to eliminate a variable.)

Now that you have x,y modulo 7 and modulo 5, you can find x,y modulo 35.
[/hide]


inequalities AMC AIME induction geometry trigonometry USA(J)MO \/closed

AoPS is now offering a course on inequalities and optimization.  This course will include problem solving techniques relevant to the AMC (advanced problems), AIME, and Olympiad problem solving levels.

Material covered will include classical inequalities, other famous inequalities, optimization methods, and a host of problem solving techniques including a variety of algebraic manipulations, induction, and utilization of the properties of functions.

Information about the course is located in the Online Classes area of the site.

If you have any questions about this course you may post them here or PM me.

[quote="answerseeker95"]Can you please name some famous inequalities covered in the course?[/quote]

RMS-AM-GM-HM inequality chain
Rearrangement
Cauchy-Schwartz
Schur's
Triangle
Minkowski
Jensen
Power Mean
and others

Much of the course however will involve various algebraic techniques that can knock out many inequalities without invoking famous inequalities and even prove the famous inequalities.

Not every inequality or technique will be covered equally and we won't cover the derivations of some of them, though perhaps students can tackle those on the board.

The course will be taught such that students who do not understand how to apply the more advanced inequalities will still learn a lot (so long as they have a good grip on algebra).

There will also be some trigonometry, but it will be mostly packaged into one lesson.

[quote="zabelman"]Im going to sign up for the class today or tomorrow, but I don't know if I should do audited or evaluated, since I've never done a class like this before. (I'm currently in the geometry class, but it didn't have an audited/evaluated option). Any advice for how I should decide?[/quote]

If cost is not an issue then I would suggest taking the course as evaluated (unless you are extremely confident in your proof-writing skills).  For the evaluated students I will be focusing my comments as much on guiding presentation of work as on problem solving techniques.  I think this will be very beneficial to any students who will at any point take the USAMO, take higher level mathematics courses in college, or plan to one day work in a quantitative field.


topology
geometry
geometric transformation
reflection
invariant
advanced fields
advanced fields unsolved

I can't seem to find this written down anywhere, but it would be a very nice to be able to use. Is the following true?

If $ X$ is connected and $ Y$ is connected, then any two maps $ f: X\to Y$ and $ g: X\to Y$ are homotopic. (I don't mean path homotopic, that would definitely be false.)

I'm not sure what "path homotopy" means when $ X$ is something other than $ [a,b]$.

I think that homotopy theory would be very boring if the above statement were true. Fortunately, it is false. For example, consider $ X\equal{}Y\equal{}S^1$, $ f$ is the identity, $ g$ is the reflection about some diameter.

Right. That was dumb. That "theorem" would essentially imply that every connected space is simply connected.

I guess here is my real problem. I don't seem to see how to prove that two maps are not homotopic. If there exists a homotopy, then you can write it down. If there doesn't, then it seems as if you just say, well, there clearly can't be a homotopy.

Take for example $ X\equal{}\{(x,y) : x^2\plus{}y^2\equal{}1\}$ and $ Y\equal{}\{(x,y) : (x\minus{}1)^2\plus{}y^2\equal{}1\}\cup\{(x,y): (x\plus{}1)^2\plus{}y^2\equal{}1\}$.

Now take the mappings $ f(x,y)\equal{}(x\minus{}1, y)$ and $ g(x,y)\equal{}(x\plus{}1,y)$.

It seems as if there cannot be a homotopy, but I don't see how one could prove that.

You can consider the fundamental group.
That of a circle is Z (say generated by c), and that of the wedge of two circles is Z * Z, the two generators being the left circle (say a), and the right circle (say b) respectively.
If the two maps you mentioned are homotopic, it induces the same map on the fundamental group. Yet, one corresponds to the imbedding with c -> a, another corresponds to c -> b, which are not equal.

I should have been more clear. I do know some techniques that require knowledge of things built off of the definition of homotopy (i.e. fundamental group). Is there any way to do this using just the definition?

Suppose this occurs as a problem in a text before fundamental group is defined as an exercise to learn how to deal with homotopy. Or would this problem not occur in that situation?

For subsets of the plane you can look at [url=http://en.wikipedia.org/wiki/Winding_number]winding numbers[/url] - about $ \minus{} 1$ and $ 1$ in your example.

When one wants to show that two "objects" are not "equivalent", the usual thing to do is to invent an "invariant" that is preserved by "equivalences", and is not the same for the two objects. This is done not just for homotopies, but also for homeomorphisms, isomorphisms, and other -morphisms.

OK, just to see if I understand. Homotopic maps preserve winding number, so if $ f$ and $ g$ are homotopic, then the winding number about $ p\equal{}(\minus{}1,0)$ should be the same. But the winding number about $ p$ for $ f$ is $ 1$, but for $ g$ is $ 0$. Thus they cannot be homotopic.

Right. The winding number about $ ( \minus{} 1,0)$ is the same for all maps $ f_t$ in the homotopy ($ f_0 \equal{} f$, $ f_1 \equal{} g$) because $ f_t$ is not allowed to pass through $ ( \minus{} 1,0)$.


Ροδόλφε σε ευχαριστούμε πολύ για τα υψηλού επιπέδου θέματα που μας στέλνεις.
  Σκέφτομαι να ανοίξω από τη νέα χρονιά ένα χώρο που θα βάζουμε όλα τα θέματα για τη Γ' Λυκείου. Γιατί όπως ξέρεις οι περισσότεροι εδώ ενδιαφέρονται κυρίως για ολυμπιακά θέματα.
 Καλή Ανάσταση !!!

 Μπάμπης

Πολύ καλή σκέψη Μπάμπη. Καλό Πάσχα σε όλους!!


Consider a standard deck of playing cards. A group of three people play by evenly distributing the 51 cards and discarding the last card. They then attempt to make three of a kind triads from the cards they have got . What is the probability that a player will be able to form the maximum 5 triads ? If that be the case, what are the chances of his opponents to get no triad ?

PS: A triad is three cards of the same number/figure regadrless of the suite, eg. three aces,three queens etc.

No one has an opinion  :maybe:

Else, Mods please move this to Olympiad section

Isn't the answer to the first question just $ \binom{13}5\binom43\binom{52 \minus{} 15}2$ divided by $ \binom{52}{17}$?

I haven't looked into this much, but I'm inclined to say that you overcounted. Specifically, you choose 5 triads, and then 2 more cards -- however, if one  or both of the two chosen cards happens to make a four-of-a-kind with the cards you already have, then that case is over-counted.

I would suggest that you break it down as:

Case 1: 5 different triads and 2 entirely different singles
Case 2: 1 four-of-a-kind, 4 triads, one single
Case 3: 2 four-of-a-kinds, 3 triads

to avoid the chance of overcounting.


Is there a book like the "Art of Problem Solving" for finding tricky ways to solve physics problems? Or is there really no need because the clever techniques to solve math problems often apply to physics problems? I've had a lot of experience with math contests, and little experience with physics so I wouldn't know...Thanks!

Try 200 Puzzling Problems in Physics -Gnadig et. al. -Cambridge University Press.
I have not read the AOPS book so I can't compare exactly, but this contains some very nice problems and equally nice solutions.


[quote="Ignite168"]Bunny hops 20 times a minute and SUPERBunny hops 20 times a second and the length of both their hops is 1 feet long, after 1 hour, how much farther will SUPERBunny have traveled? Express your answer as a rational number.[/quote]

[hide]SUPERBunny haps $20*60-20=1180$ feet farther

Therefore, the answer is $1180*60=\boxed{70800}$[/hide]

[hide]After one hour, 'bunny' has hopped $20\cdot 60=1200$ feet; SUPERBunny has hopped $20*3600=72000$ feet. Therefore SUPERBunny has gone [b]70800[/b] feet further[/hide]

Where can I get a SUPERBunny?

[hide]In one hour, a bunny can hop $20\times60=1200$ feet and a SUPERbunny can hop $20\times60\times60=72,000$ feet.

Thus, the SUPERbunny hops $72,000-1200=70,800$ feet farther.[/hide]

[quote="b-flat"][quote="mtms5467"][hide]$20\cdot60^{2}-20\cdot60=70800$ $feet$[/hide][/quote]

Tip:  use \mbox{feet} or \text{feet} to make text in $\LaTeX$.  However, don't use too much text because the administrators don't like it when you do.  :wink:[/quote]
I've always been wondering why not, people keep saying not to, does it use up space or something? (Just wondering)

[quote="Klebian"]I've always been wondering why not, people keep saying not to, does it use up space or something? (Just wondering)[/quote]
If you have a slow computer (like me), it will take longer to load the page.
And I don't know if it's just me, but if the page isn't done loading and you press a hide, the page doesn't finish loading.

[quote="Ignite168"]Bunny hops 20 times a minute and SUPERBunny hops 20 times a second and the length of both their hops is 1 feet long, after 1 hour, how much farther will SUPERBunny have traveled? Express your answer as a rational number.[/quote]
[hide]
Well, u first get the distance Bunny and SUPERBunny can travel in one hour
$\text{Bunny}=1,200 \text{feet}/ \text{hour}$
$\text{SUPERBunny}=72,000 \text{feet}/ \text{hour}$
then u subtract that
$72,000-1,200=\boxed{70,800}\text{feet}$[/hide]


Four friends shared a bottle of water. The first drank $ \frac14$ of the water. The second drank $ \frac13$ of the remainder. The third drank $ \frac12$ of what was left. The fourth friend drank the last six ounces. How many ounces of water were originally in the bottle?

Okay, let's call the friends A, B, C, and D.  D drank the last 6 ounces.  C drank half of what was left at the time, so before C drank the water, there were 12 ounces.  B drank one-third of what was left at that time, so there were 18 ounces.  Finally, A drank one-fourth of the original amount, so there were originally $ \boxed{24}$ ounces.

$ C \equal{} \frac12(C \plus{} 6) \implies C \equal{} \frac{C}2 \plus{} 3 \implies \frac{C}2 \equal{} 3 \implies C \equal{} 6$

$ C \plus{} D \implies 6 \plus{} 6 \implies 12$

$ B \equal{} \frac13(B \plus{} 12) \implies B \equal{} \frac{B}{3} \plus{} 4 \implies \frac{2B}{3} \equal{} 4 \implies B \equal{} 6$

$ B \plus{} (C \plus{} D) \implies 6 \plus{} 12 \implies 18$

$ A \equal{} \frac14(A \plus{} 18) \implies A \equal{} \frac{A}{4} \plus{} \frac92 \implies \frac{3B}{4} \equal{} \frac92 \implies A \equal{} 6$

$ A \plus{} (B\plus{}C\plus{}D) \implies 6 \plus{} 18 \implies \boxed{24}$


group theory
abstract algebra
induction
Galois Theory
superior algebra
superior algebra unsolved

One can show that
1. A group $G$ of order $pq$, where $p$ and $q$ are different primes, is solvable.
2. A group $G$ of order $pqr$, where $p$, $q$ and $r$ are different primes, is solvable.

3. Is a group $G$ of order $pqrs$, where $p$, $q$, $r$ and $s$ are different primes, solvable?
4. If it is, how far can we generalize?

Hi.

You can generalize this to arbitrary many primes: Every finite group of squarefree order is solvable. You can say even more: Every finite group of squarefree order is a semidirect product of two cyclic groups.

The fact that groups of squarefree order are solvable follows from Burnside's Normal Complement Theorem, which states that if a Sylow subgroup of a finite group is contained in the center of its normalizer, then it has a normal complement.

If our group $G$ has order $p_{1}\cdot\ldots\cdot p_{k}$ with $p_{1}<\ldots<p_{k}$, then we apply induction on $k$ as follows: notice that a Sylow $p_{1}$-subgroup $H$ of $G$ is contained in the center of its normalizer. This is because the normalizer acts by conjugation on $H$, and its elements, regarded as automorphisms of $H$, have order dividing $p_{2}\cdot\ldots\cdot p_{k}\ (*)$. The automorphism group of $H$, however, has order $p_{1}-1$, which has no prime factors in common with $(*)$. This means that the action of the normalizer of $H$ on $H$ by conjugation is trivial, i.e. $H$ is contained in the center of this normalizer. Now apply the theorem cited above to conclude that $G$ has a normal subgroup $K$ of index $p_{1}$. By the induction hypothesis, $K$ is solvable, and since $G/K$ is cyclic and hence also solvable, $G$ itself must be solvable.

Now, using the solvability of $G$, we can prove the second assertion. There is a proof (well, an attempt, at least; there may be some mistakes) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=metacyclic&t=47370]here[/url] of the fact that $G$ is metacyclic, i.e. $G'$ and $G/G'$ are both cyclic. Because $|G'|$ and $[G: G']$ are coprime, by a theorem of Hall (stating that a solvable group of order $ab,\ (a,b)=1$ has a subgroup of order $a$) there is a subgroup $L$ of $G$ of order $[G: G']$, which must be isomorphic to $G/G'$ (it maps injectively into and hence onto $G/G'$ through the projection from $G$ to $G/G'$). This means that $G$ is the semidirect product of $G'$ by $L$, both of which are cyclic.

[quote="Gockel"] Every finite group of squarefree order is a semidirect product of two cyclic groups.[/quote]
Fantastic result! Do you have a reference for a proof?


Find all primes p such that p divides the numerator of the value $ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{(p \minus{} 1)/2}$ expressed in its lowest term as a fraction

Let $ m \equal{} p \minus{} 1, S_m \equal{} \sum_{x \equal{} 1}^{p \minus{} 1}x^m$, then
\[ (2^m \minus{} 1)S_m \equal{} \sum_{x \equal{} (p \plus{} 1)/2}^{p \minus{} 1}\sum_{k \equal{} 1}^mC_m^kp^k(2x \minus{} p)^{m \minus{} k} \equal{} mp2^{m \minus{} 1}\sum_{x \equal{} 1}^{(p \minus{} 1)/2}x^{m \minus{} 1} \plus{} O(p^2).
\]
Therefore $ S(p)\equal{}\sum_{k \equal{} 1}^{(p \minus{} 1)/2}\frac 1k \equal{} \frac { \minus{} 2(2^{p \minus{} 1} \minus{} 1)}{p}\mod p.$
For example $ p\equal{}3$ $ S(3)\equal{}1\mod 3$, $ p\equal{}5$ $ S(5)\equal{}4\mod 5$.
$ p|S(p)$ if and only if $ p^2|2^{p\minus{}1}\minus{}1$. Rnown only 2 primes $ 1093,3511$ satisfyed these condition.


calculus
integration
function
geometry
3D geometry
sphere
limit

Assume that $ \Omega$ is a function defined on the unit sphere $ \mathbb{S}^1 \subset \mathbb{R}^2$ and is integrable with mean value zero. Define the tempered distribution $ W_{\Omega}$ by setting $ W_{\Omega}(\varphi) \equal{} \lim_{\epsilon \to 0} \int_{|x| > \epsilon} \frac{\Omega(x/|x|)}{|x|^2}\varphi(x)\,dx$ for Schwartz functions $ \varphi$. Let us study the singular integral operators $ T_{\Omega}\colon f \mapsto W_{\Omega} * f$.

Suppose that we fix some Schwartz (or a smooth compactly supported) function $ f$ and consider the operator $ H_f\colon \Omega \mapsto T_{\Omega}f$ which instead acts on the kernels $ \Omega$. Also, suppose that $ 1 < p < \infty$. Let me ask the following, regrettably not too well thought through, question:
Is it possible to obtain a bound of the form $ \|H_f(\Omega)\|_{L^p(\mathbb{R}^2)} \le C(f,p)\|\Omega\|_{L^1(\mathbb{S}^1)}$? (We can actually assume a little bit more of the kernels $ \Omega$ and modify the RHS accordingly if it is relevant.)

Here's a suggestion for an example to work through. What happens if you let $ \Omega(\theta) \equal{} \cos n\theta?$

Note that as a family of examples, these $ \Omega$ all have the same $ L^1$ norms; they differ in the size of their derivatives. What do the operator norms of the singular integrals generated by these $ \Omega$ look like?


college
Putnam
calculus
college contests

i'm a senior in high school coming into math competitions very late in the game.  i took the AMC this year and did miserably, scoring a 97.  although i'm a very inexperienced problem solver, i've been excelling in math with solid grades; i've taken multivariable calc at berkeley and am currently taking differential equations.

i'm eager to take the putnam exam in college but i'confused as to how to prepare for it.
i was wondering if it's worth the time going over art of problem solving vol. 1 & 2, or whether i should go straight to books such as the art and craft of problem solving and problem-solving through strategies.

Putnam is much more like the AMC/AIME than the kind of math that is likely to be taught in schools, so if problem-solving is your main deficiency, it's the main skill you'll need to bulk up on to do well.

Problem-Solving Strategies is more of an Olympiad-level problem solving text than it is a college-level problem-solving text, so it wouldn't help much unless you were already scoring much higher on the AMC and AIME.

ACoPS has more college-level problems, but it's in the same vein.  I'd suggest you build up your basics with AoPS first.  They're definitely worth your time.

Any of the books you mentioned suffer from the weakness that they don't have the Putnam in mind as a target.  You should make sure that you familiarize yourself with the test and the format, and ideally do a bunch of practice problems, before you take it.  The standard for excellence on the Putnam is quite different than on the AMC or other high school contests.  One incomplete solution typically gets you into the top half of competitors, one full solution into the top quarter, and completing half of the contest gets you Honorable Mention, depending on the year, of course.  As a result, spending time wisely and finding the one or two questions on each half that you'll be able to solve are really important, and those are skills that come from working similar types of problems.

Does your college-to-be have a course devoted to studying for the Putnam, or a club or study group? Many do. It's worth your effort to find out.

yeah, berkeley has an honors course devoted to problem solving and the putnam exam.  however, after looking at the syllabus, the demands for the course seem high and not very receptive to beginners.

(I went to Bekeley as an undergrad. We came in fourth in the Putnam two years in a row, just ahead of Stanford.)

Don't worry about it. Take H90. Enthusiasm counts for a lot, and you can learn problem-solving skills.

The real expectations of the class are that you will try the problems and present some solutions. You are not expected to solve all of the problems- nobody is.
Also, you would be expected to take the Putnam at the end.

If you're asked to write up solutions, consider that practice in proof-writing skills. You've probably never written serious proofs before, and communication is an essential skill. If nobody else understands it, it's not a solution.

[quote="JBL"]Can you post the syllabus/requirements or give me a link?[/quote]

Joe, the course is called Math H 90. Look [url=http://www.cs.berkeley.edu/~oholtz/H90/]here[/url] or [url=http://www.cs.berkeley.edu/~wkahan/MathH90/index.htm]there.[/url]

Thanks.  I say take the class -- it looks good, and it doesn't look like it assumes anything you shouldn't have met at some point, especially if you go through any of the books that've been suggested.

Spending time on AoPS (both or either volume) certainly won't hurt.  I only started them after I began studying calculus and there's a lot of stuff that just isn't taught as part of a school's normal curriculum.  They've really helped a lot.   ACoPS is much more readable if you already have a background in the topics it covers, in my opinion.

Mind if i ask, are international students studying in the US elligible for the putnam competition?   :)

[b]violetcraze[/b], the answer to your question above is [b]YES[/b]. I am an international student myself, and I took the test last year.


Find a formula for $sin(2^{n}x)$, for $n \in \mathbb{N}$.


1. For what values of the constant $ k$ will the following function not be its own inverse.
     $ \textbf{f}(x) \equal{} \dfrac{x \minus{} k}{x \minus{} 1}$

[hide][Question 2]
2.Graph the set of points satisfying the following conditions for the ordered pairs(x,y)

lx+1l<4   ,    $ 2 \le y < 5$.

All I need are the set of points that satisfy the conditions. I'm not sure how to do it. I think you need to find the domain and the range and then graph it from there, but again, i'm not sure.[/hide]

$ y \equal{} \frac {x \minus{} k}{x \minus{} 1}$
inv.
$ x \equal{} \frac {y \minus{} k}{y \minus{} 1}$
Solving for y:
$ xy \minus{} x \equal{} y \minus{} k$
$ xy \minus{} y \equal{} x \minus{} k$
$ y(x \minus{} 1) \equal{} (x \minus{} k)$
$ y \equal{} \frac {x \minus{} k}{x \minus{} 1}$

So we see that it is it's own inverse for all values of K.
Is this right?
I think I might be missing a potential division by zero, but it's late.


calculus
real analysis
function
algebra
rational function
Measure theory

the sequence $ x_0, x_1, x_2, ... $ is defined by $ x_0 = 2 $ and $ x_{n+1} = x_n - \frac 1{x_n} $ for $ n \geq 0 $.  is the sequence bounded?

First of all, I am not sure this really belongs to Algebra or Olimpiad level. In my humble opinion, this should go to Calculus or even Advanced fields. [color=red][Moved accordingly. -mathmanman][/color]

While I do not know what happens for $ x_{0}=2$, let me show that for almost all $ x_{0}$ the sequence is unbounded. Nothing will change if we consider the mapping $ T\,: \,x\mapsto \left|x-\frac1x\right|$ on $ [0,+\infty]$ instead. The key observation is that $ T$ preserves the Lebesgue measure on $ [0,+\infty]$, i.e., $ \mu(T^{-1}E)=\mu(E)$ where $ T^{-1}E$ is the full pre-image of the set $ E$. Let now $ E_{n}$ be the set of all points whose orbit stays on the interval $ [0,n]$ forever, i.e., the points $ x$ such that $ x, Tx, T^{2}x, T^{3}x,\ldots\in[0,n]$. It is a bounded measurable set. Our aim is to show that its measure is $ 0$. Note that for any $ m$, $ E_{n}\subset T^{-m}(E_{n})\cap[0,n]$ (the right set is the set of the points $ x\in[0,n]$ such that $ T^{m}x, T^{m+1}x,\dots\in[0,n]$, which is, clearly, a milder restriction on $ x$. But $ \mu(T^{-m}E_{n})=\mu(E_{n})$ and, unless $ \mu(E_{n})=0$, $ \mu(T^{-m}E_{n})\cap[n,+\infty]>0$ (just consider the preimages $ >1$ during each step). Thus, if $ \mu(E_{n})>0$, we get a set of larger measure contained in a set of smaller measure, which is impossible.

I wonder whether $ E_{n}$ is countable or a Cantor-type set for large $ n$. Maybe all orbits are either eventually periodic (it is not very hard to see that every periodic orbit is repelling) or unbounded but that will require more careful analysis.

It seems like the following recent article can shed some light on this problem and may be worth reading anyway:

[i]Iteration of the rational function z-1/z and a Hausdorff moment sequence.[/i] by Christian Berga, and Antonio J. Duránb in Expositiones Mathematicae Volume 26, Issue 4, 2008, Pages 375-385


I had a minor argument with my stats teacher over this question. (this was last year btw) I didn't really bother to keep going though, because it was an extra credit problem and I already had way over a 100 average.

----

You bet $\$2$ on this following game: You continue to flip a fair coin until you flip tails. You are paid $\$1$ for every heads you get. Is this game fair or not?

[hide="Solution"] The expected value is $\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+... = 1$.  So... no. [/hide]

This game is fair in statistics wise. Every flip is independent to each other. So the expected earn of two flips is $1$ dollar since there will be one heads in two flips.

In statistics as $n$ increases the number of heads and tails will be approximately equal. If you flip it $500,000$ times, there is a good chance that there will be nearly $250,000$ heads and $250,000$ tails.

you will pay $2$ dollars for every two flips. As the flips $n$ increases, the data will turn approximately normal. So it is fair in my opinion.

Edit: If the numbers were switched it would be fair. If you pay $1$ dollar for every flip and get $2$ dollars for a tail, it would be fair. In this case,I think it's unfair!

Oh crap, I remembered the question wrong >< You actually pay $\$1$ to play this game. :oops:

so then by t0rajir0u's argument its fair?

Well, my stats teacher said something like

"Well the sum will never actually reach 1, just arbitrarily close to it, so it's not fair..."

Look at $t0rajir0u$'s solution.
The expected value reaches 1.
Your teacher saids it will never actually reach 1.

Maybe use a random digit table and find out.

Do a test.
Do a t-proportion-test to see how significant the result is.
(to conclude if its fair or not.)
It will result in a fair experiment if you follow the conditions

($np>30,nq>30$ large enough sample size, $SRS$)

[quote="randomdragoon"]"Well the sum will never actually reach 1, just arbitrarily close to it, so it's not fair..."[/quote]

Put it this way:  Expected value isn't all that good a measure of the value of a game.  What if it were $2^{n}$ dollars for $n$ heads instead of $n$ dollars?  Then the expected value would tend to $\infty$, but you still wouldn't play the game for more than, say, $5$ dollars a pop.


The question is to show that $ {n}^5\minus{}{n}$ is always divisible by 15.
My solution[/hide]the expression must be divisible by 3 and 5.
$ {n}^5\minus{}{n}\equal{}{n}{(n\minus{}1)}{(n\plus{}1)}{(n^2\plus{}1)}$ which contains the product of three consecutive number.
To make it divisible by 5 i transformed the expression to fermats little theorem $ {n}^5\equal{}{n}mod5$. My question is i want to see anyother proof and is mine absolutely correct for all n

In fact it is not only divisible by 15, but by 30. And for odd n or n which are multiples of 8 it is divisible by 120.

Also, please change your hide tags so your solution is actually hidden, or it might ruin the problem for those who wish to attempt it.

$ n^2\plus{}1\equiv n^2\minus{}4\equal{}(n\minus{}2)(n\plus{}2)\mod 5$

So $ n^5\minus{}n\equiv (n\minus{}2)(n\minus{}1)(n)(n\plus{}1)(n\plus{}2)\mod 5$


$a,b,c$ are non-negative real such that $a+b+c=2$ , prove that

$a^2bc\leq \frac{1}{4}$

Find also the maximum of $a^nbc$ for any natural $n$ .

 :)

[quote="snagg"]maybe I'am wrong but that way you've proved that $a^22b2c\leq \frac{1}{4}$ which is not the same thing[/quote]

He had proved that $a^2.2b.2c\leq1$

so what do you think about the maximum of

$a^nbc$

for any natural $n$ ? :)

Use the same as before
$a^nn\cdot bn\cdot c\leq (\frac{na+nb+nc}{n+2})^{n+2}=(\frac{2n}{n+2})^{n+2}$

this inequality can be generalised as such:

let $\alpha_1 + \alpha_2 + \cdots + \alpha_k = n$

$\alpha_1^{\beta_1} \cdot \alpha_2^{\beta_2} \cdots \alpha_k^{\beta_k} \leq (\frac{n}{\beta_1 + \beta_2 + \cdots + \beta_k})^{\beta_1 + \beta_2 + \cdots + \beta_k} \cdot \beta_1^{\beta_1} \cdot \beta_2^{\beta_2} \cdots \beta_k^{\beta_k}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Having trouble with this one for some reason:

Let $ H$ be a normal subgroup of $ G$, and let $ m \equal{} [G: H]$. Then $ a^{m} \in H$, $ \forall a \in G.$

I want to try something such as homomorphism $ \phi : G \rightarrow G/H$ and somehow showing that $ a^{m}$ maps to the trivial coset $ H$. Is this good direction in which to proceed? Or is there a simpler way?

It's a good idea. Now use Lagrange's/Fermat's theorem.

$ a$ is contained in the coset $ aH$, $ a^i$ is contained in the coset $ a^iH$, if $ b$ is in the coset of $ a^j$ and $ c$ in that of $ a^k$ then $ bc$ is in the coset of $ a^{k \plus{} j}$. (just use $ H$ is normal)  Since $ H$ is normal, the cosets form a group. The property described for the cosets of $ a$ means that they are also a (cyclic) subgroup of the group of cosets. Then the order of this group divides $ m$ meaning that every coset powered to the $ m$ is $ H$. Then every element powered to the $ m$ is in $ H$

Let $ a\in G$. Then each element of $ G/H$ is of the form $ aH$.
Then $ H \equal{} (aH)^m \equal{} a^mH \implies a^m \in H$.
So all you need to do is showing $ G/H$ is a group. But almost every textbook would do this.


Let a triangle ABC . $ AL $ is the bisector of $ <BAC $ . two lines $ l_1 , l_2 $ are parallel together . $ l_1 $ passes  by $ B $ . $ l_2 $ passes by C . $ d(A,l_1)=d(A,l_2) $ . $ M \in l_1 $, $ N \in l_2 $ satisfy : $ LM $ intersects AB at the midpoint of $ LM $ and $ LN $ intersects $ AC $ at the midpoint of $ LN $ . Prove that : $ LM= LN $

[b]Lemma.[/b] Let a triangle $ \triangle ABC$ , $ AL $ is the bisector of $ \angle BAC $. Two lines $ l_1 , l_2 $ are parallel together . $ l_1 $ passes  by $B$ . $ l_2 $ passes by C with $ d(A,l_1)=d(A,l_2) $ . Let $l$ be the perpendicular to $AL$, $M=l \cap l_1, N=l \cap l_2, D=AB \cap ML, E= AC \cap LN$. Then $MD=DL$ and $LE=EN$.

[b]Proof.[/b]
Let $C'$ be the symmetric of $C$ wrt $A$. From the angle bisector theorem we have:
\[\frac{{BL}}{{LC}} = \frac{{BA}}{{AC}} = \frac{{BA}}{{AC'}} = \frac{{BM}}{{MC'}}\]
Thus the triangles $ \triangle MBL$ and $ \triangle C'BC$ are similar (SAS similarity theorem). Hence
\[\frac{{MD}}{{DL}} = \frac{{C'A}}{{CA}}{\rm{  }} \Rightarrow {\rm{  }}MD = DL\]
Similarly can be proved that $LE=EN$.  $\blacksquare$

[b]Solution of problem.[/b]
The triangles $\triangle MDA$ and $\triangle AEN$ are congruent (SAA) since
[b](1)[/b] $MA=AN$
[b](2)[/b] $ML \| AC$ (by previous lemma)  $ \Rightarrow \angle DMA=\angle EAN$
[b](3)[/b] $MD=DL, MA=AN \Rightarrow DA \| LN \Rightarrow \angle MAD= \angle ANE$
Consequently $MD=EN \Rightarrow ML=LN. \blacksquare$


My thunder comes before the lightning;
My lightning comes before the clouds;
My rain dries all the land that it touches.

Who or what is the “I” referred to in the puzzle?

Source: Puzzle Periodical

i dont think so.

is it a forest fire that started?

hmm i think a volcano is a good answer (good going daine!)


And since this is on the topic of precalculus, does any one know when the precalculus book is coming out??????????

[quote="Blazer123"]Does anyone know when the next precalculus  class will be offered after the one starting on December 4th?

Thanks.[/quote]

Possibly spring, 2010.  No later than summer, 2010.

I hope to have the book out by the end of this year.

If the class is offered during the spring, will it be offered in the summer?

Because I'm thinking about taking Alg. 3 in the spring and precalculus in the summer...

Thanks.

We're not sure yet.  It will depend on the enrollment level of the Algebra 3 classes in the fall and spring.


prove that$ (cos^2A\plus{}1)(cos^2B\plus{}1)(cos^2C\plus{}1) \ge \frac{225}{64}$

Is the right hand side supposed to be $ \frac{125}{64}$?

hello, how did you get $ \frac{125}{64}$? If the angles $ A\equal{}B\equal{}C\equal{}\frac{\pi}{6}$ we get
$ (\cos(\frac{\pi}{6})^2\plus{}1)^3\equal{}\frac{343}{64}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, how did you get $ \frac {125}{64}$? If the angles $ A \equal{} B \equal{} C \equal{} \frac {\pi}{6}$ we get
$ (\cos(\frac {\pi}{6})^2 \plus{} 1)^3 \equal{} \frac {343}{64}$
Sonnhard.[/quote]

Errr... $ A \equal{} B \equal{} C \equal{} \frac {\pi}{3}$?

$ \left(\cos\left(\frac {\pi}{3}\right)^2 \plus{} 1\right)^3 \equal{} \left(\frac 54\right)^3 \equal{} \frac {125}{64}$

hello, yes ok, this was my misstake, sorry.
Sonnhard.


Let $ 1,4,\cdots$ and $ 9,16,\cdots$ be two arithmetic progressions. The set $ S$ is the union of the first $ 2004$ terms of each sequence. How many distinct numbers are in $ S$?

$ \textbf{(A)}\ 3722\qquad \textbf{(B)}\ 3732\qquad \textbf{(C)}\ 3914\qquad \textbf{(D)}\ 3924\qquad \textbf{(E)}\ 4007$

Overcounting, we find that there are $ 2004 \cdot 2\equal{}4008$ total terms. We can easily find that the equations for the arithmetic progressions are $ y\equal{}3a\minus{}2$ and $ y\equal{}7b\plus{}2$. We find that the two equal when $ 3a\minus{}2\equal{}7b\plus{}2 \implies 3a\minus{}4\equal{}7b$. This means that
\[ 3a\minus{}4 \equiv 0 \pmod{7} \implies 3a \equiv \minus{}3 \pmod{7} \implies a \equiv \minus{}1 \equiv 6 \pmod{7}.\]
Since the greatest value of $ a$ that satisfies $ 7a\plus{}6 < 2004$ is $ 285$, there are $ 4008\minus{}\left(285\plus{}1\right)\equal{}3722 \rightarrow \boxed{\textbf{A}}$ distinct numbers in $ S$.

Another old revive, but it is for a question. [quote="123456789"]$ 4008\minus{}\left(285\plus{}1\right)$[/quote] Why do we do this and not just $4008-285$?

the terms of the first progression are: $a_k=1+3k$ where $0 \le k \le 2003$
the first common term of both progressions is $16$ and all common terms satisfy $c_n=16+21n$. Obviously the numbers in the first progression grow slower then the numbers in the second. so to find the common terms we need to find when $16+21n \le 1+3 \times 2003=6010$, so $21n \le 5994 \Rightarrow n \le 285$.
The first common term is $c_0=16$ and the last is $c_{285}=6001$ so there are $286$ common terrms.


The sequence $ (a_n)$ is given by $ a_1\equal{}1,a_2\equal{}0$ and:

$ a_{2k\plus{}1}\equal{}a_k\plus{}a_{k\plus{}1}, a_{2k\plus{}2}\equal{}2a_{k\plus{}1}$ for $ k \in \mathbb{N}.$

Find $ a_m$ for $ m\equal{}2^{19}\plus{}91.$

After analyzing the the terms of this sequence, I found that

$ a_{2^{n}\plus{}k}\equal{}k$ or $ 2^{n}\minus{}k$, whichever was smaller $ (0 \leq k \leq 2^{n})$

In words, $ a_{m}$ is the distance of $ m$ from its nearest power of 2.

[hide="Possible approach for a solution"]
We could probably prove the hypothesis by induction.
[/hide]

Hence the answer is 91.


please, can someone help me..

In a big serial 96% of products satisfy to standards. Rude control good product proclaims for ''good'' with probability 0,98 and bad product proclaims for ''good'' with probability 0,05. Find the probability that a chosen product is ''good'' .

Thank you very much

let $ P(G| \plus{} )$ be probability that a chosen product is ''good'' ( that product proclaimed for "good" is really satysfies to standart)
$ P( \plus{} |G) \equal{} 0,98$ is probability that good product is proclaimed for  "good"
$ P( \plus{} |B) \equal{} 0,05$ is probability that  bad product is proclaimed for  "good"
$ P(G) \equal{} 1 \minus{} P(B) \equal{} 0,96$   is probability  that product satisfies to standards

so

$ P(G| \plus{} ) \equal{} \frac {P( \plus{} |G) \cdot P(G) }{P( \plus{} |G) \cdot P(G) \plus{} P( \plus{} |B) \cdot P(B) } \equal{} \frac {0,98 \cdot 0,96}{0,98 \cdot 0,96 \plus{} 0,05\cdot 0,04}$$ \equal{} \frac {2352}{2357} \equal{}$~$ 0,9979$

 :roll:

thank you..i've solved problem but differently, maybe more complicated but Bayes is the better solution..tnx very much


did anyone notice that #10 on the repechage:

Find all positive integers a and b that satisfy the equation: $ |3^a\minus{}2^b|\equal{}1$.

is the exact same problem in Mathematical Olympiad Challenges by andreescu problem 3.6.1  :wink:

if only i had started working on this book; i would've saved 3 hours :)

well... I dont have the book... but I solved it pretty easily anyways...

I guess they used it because its just a good problem....  :maybe:  :maybe:

[quote="max_tm"]Could anyone give me a hint on this one? My NT is a little rusty...  :blush:. Thanks![/quote]

split the cases to 1 and -1

Then take modulo 3, and notice something

then use basic factoring techniques

YEA...

I JUST FOUND IT WHILE READING THE BOOK...

if i worked on the book i could have solved the question T.T

(but i qualified for olympiad anyways lol)


this is for the countdown cup results.

i will post the matchups either 2day or 2morrow.

good luck.

[b]Round 1[/b]

14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1
4 FantasyLover VS 29 vamathletes
1 superllamel VS 32 blue1421
10 RomanianGenius VS 23 isabella2296
13 Brut3Forc3 VS 20 bob einstein
6 yongyi781 VS 27 PWNER Express
17 hurdler DEFEATS 16 budi713, 3-1
5 alexhhmun VS 28 mathmagician1729
3 ra5249 VS 30 Meta Knight
12 lotsofmath VS 21 ash-Pokemon
9 ernie VS 24 AIME15
7 vallon22 DEFEATS 26 nikeballa96, 3-0
2 mewto55555 DEFEATS 31 Mind Storm, 3-0
15 kl2836 VS 18 violin321
11 BOGTRO VS 22 thdanh90
8 Jongao VS 25 kookamunga

[b]Round 2[/b]

skaldjfhrulez VS 4/29
1/32 VS 10/23
13/20 VS 6/27
(16)hurdler VS 5/28
3/30 VS 12/21
9/24 VS 7 vallon22
2 mewto55555 VS 15/18
11/22 VS 8/25

oops, ignore my question on the other thread about the pairings....

DO NOT SPAM ON THIS THREAD!  :bomb: :bomb: :bomb:

Since i lost hurdler is now 16th seed?
 :(  :(

SUPERLLAMEL (1) DEFEATS BLUE1421 (32)

S(4-0)
S(4-1)
S(4-0)

Blue can confirm this.

ROMANIANGENIUS(10) DEFEATS ISABELLA2296(23) 3-0

R(4-1)
R(4-0)
R(4-1)

[b]Round 1[/b]

14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1
4 FantasyLover VS 29 vamathletes
1 superllamel DEFEATS 32 blue1421, 3-0
10 RomanianGenius DEFEATS 23 isabella2296, 3-0
13 Brut3Forc3 VS 20 bob einstein
6 yongyi781 DEFEATS 27 PWNER Express, 3-0
17 hurdler DEFEATS 16 budi713, 3-1
5 alexhhmun VS 28 mathmagician1729
3 ra5249 VS 30 Meta Knight
12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1
9 ernie VS 24 AIME15
7 vallon22 DEFEATS 26 nikeballa96, 3-0
2 mewto55555 DEFEATS 31 Mind Storm, 3-0
15 kl2836 DEFEATS 18 violin321, 3-2
11 BOGTRO VS 22 thdanh90
8 Jongao VS 25 kookamunga

[b]Round 2[/b]

14 skaldjfhrulez VS 4 FantasyLover/29 vamathletes
1 superllamel VS 10 RomanianGenius
13 Brut3Forc3/20 bob einstein VS yongyi781
(16)hurdler VS 5 alexhhmun/28 mathemagician1729
3 ra5249/30 Meta Knight VS l12 otsofmath
9 ernie /24 AIME15 VS 7 vallon22
2 mewto55555 DEFEATS 15 kl2836, 3-0
11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga

[b]Quarter-Finals[/b]

????? VS 1 superllamel/10 RomanianGenius
????? VS ?????
????? VS ?????
2 mewto55555 VS ?????

[b]Semi-Finals[/b]

????? VS ?????
????? VS ?????

[b]Finals[/b]

????? VS ?????

@budi713- no, the parenthesis indicate that he upset that seed[code][hide=Index][/hide][/code]

UGH MEWTO WENT TO QUARTER FINALS BUT I DIDNT EVEN FINISH ROUND 1

WHERE IS VAMATHLETES???

ok im gonna pm him

FANTASYLOVER PWNS VAMATHLETES
4-3 FL WINS
4-0 FL PWNS
4-1 FL PWNS

Round 1

14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1
4 FantasyLover DEFEATS 29 vamathletes, 3-0
1 superllamel DEFEATS 32 blue1421, 3-0
10 RomanianGenius DEFEATS 23 isabella2296, 3-0
13 Brut3Forc3 VS 20 bob einstein
6 yongyi781 DEFEATS 27 PWNER Express, 3-0
17 hurdler DEFEATS 16 budi713, 3-1
5 alexhhmun VS 28 mathmagician1729
3 ra5249 VS 30 Meta Knight
12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1
9 ernie VS 24 AIME15
7 vallon22 DEFEATS 26 nikeballa96, 3-0
2 mewto55555 DEFEATS 31 Mind Storm, 3-0
15 kl2836 DEFEATS 18 violin321, 3-2
11 BOGTRO VS 22 thdanh90
8 Jongao VS 25 kookamunga

Round 2

14 skaldjfhrulez VS 4 FantasyLover
1 superllamel VS 10 RomanianGenius
13 Brut3Forc3/20 bob einstein VS yongyi781
(16)hurdler VS 5 alexhhmun/28 mathemagician1729
3 ra5249/30 Meta Knight VS 12 lotsofmath
9 ernie /24 AIME15 VS 7 vallon22
2 mewto55555 DEFEATS 15 kl2836, 3-0
11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga

Quarter-Finals

14 skaldjfhrulez/4 FantasyLover VS 1 superllamel/10 RomanianGenius
????? VS ?????
????? VS ?????
2 mewto55555 VS ?????

Semi-Finals

????? VS ?????
????? VS ?????

Finals

????? VS ?????

You against winner of 13v20 game yongyi.

[quote="ernie"]Round 1

14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1
4 FantasyLover DEFEATS 29 vamathletes, 3-0
1 superllamel DEFEATS 32 blue1421, 3-0
10 RomanianGenius DEFEATS 23 isabella2296, 3-0
13 Brut3Forc3 VS 20 bob einstein
6 yongyi781 DEFEATS 27 PWNER Express, 3-0
17 hurdler DEFEATS 16 budi713, 3-1
5 alexhhmun VS 28 mathmagician1729
3 ra5249 VS 30 Meta Knight
12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1
9 ernie VS 24 AIME15
7 vallon22 DEFEATS 26 nikeballa96, 3-0
2 mewto55555 DEFEATS 31 Mind Storm, 3-0
15 kl2836 DEFEATS 18 violin321, 3-2
11 BOGTRO VS 22 thdanh90
8 Jongao VS 25 kookamunga

Round 2

4 FantasyLover DEFEATS 14 skaldjfhrulez, 3-0
1 superllamel VS 10 RomanianGenius
13 Brut3Forc3/20 bob einstein VS yongyi781
(16)hurdler VS 5 alexhhmun/28 mathemagician1729
3 ra5249/30 Meta Knight VS 12 lotsofmath
9 ernie /24 AIME15 VS 7 vallon22
2 mewto55555 DEFEATS 15 kl2836, 3-0
11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga

Quarter-Finals

4 FantasyLover VS 1 superllamel/10 RomanianGenius
????? VS ?????
????? VS ?????
2 mewto55555 VS ?????

Semi-Finals

????? VS ?????
????? VS ?????

Finals

????? VS ?????

...now that i found vamathletes and played with him...

I NEED TO FIND SUPER!!!!

WHERE IS SUPER??? I HAVE 2 TOURNEYS WITH HIM!!!

WHAT AN UPSET!!!!!!!!!!!!!!!!! OMGGGGGGGGGGG!!!!!!

ROMANIANGENIUS(10) DEFEATS SUPERLLAMEL(1)

R(4-2)
S(4-2)
R(4-1)
R(4-0)

I HAVE SUCCEEDED WHERE MY BROTHER HAS FAILED! BRING IT ON FANTASY!

AHHH! I've been inactive! Sorry!
When can I play bob einstein?

14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1
4 FantasyLover DEFEATS 29 vamathletes, 3-0
1 superllamel DEFEATS 32 blue1421, 3-0
10 RomanianGenius DEFEATS 23 isabella2296, 3-0
13 Brut3Forc3 VS 20 bob einstein
6 yongyi781 DEFEATS 27 PWNER Express, 3-0
17 hurdler DEFEATS 16 budi713, 3-1
5 alexhhmun VS 28 mathmagician1729
3 ra5249 VS 30 Meta Knight
12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1
9 ernie DEFEATS 24 AIME15, 3-0
7 vallon22 DEFEATS 26 nikeballa96, 3-0
2 mewto55555 DEFEATS 31 Mind Storm, 3-0
15 kl2836 DEFEATS 18 violin321, 3-2
11 BOGTRO VS 22 thdanh90
8 Jongao VS 25 kookamunga

Round 2

4 FantasyLover DEFEATS 14 skaldjfhrulez, 3-0
10 RomanianGenius DEFEATS 1 superllamel, 3-1
13 Brut3Forc3/20 bob einstein VS yongyi781
(16)hurdler VS 5 alexhhmun/28 mathemagician1729
3 ra5249/30 Meta Knight VS 12 lotsofmath
9 ernie VS 7 vallon22
2 mewto55555 DEFEATS 15 kl2836, 3-0
11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga

Quarter-Finals

4 FantasyLover VS (1)RomanianGenius
????? VS ?????
????? VS 9 ernie/7 vallon22
2 mewto55555 VS ?????

Semi-Finals

????? VS ?????
????? VS ?????

Finals

????? VS ?????

...then im going to semi-finals?

eh that was fast...

[quote="RomanianGenius"]FANTASYLOVER(3) DEFEATS ROMANIANGENIUS(10)

R(4-1)
F(4-2)
F(4-2)
F(4-2)[/quote]

only fantasy is seeded #4  :P

____________________________________________________________________________________________

14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1
4 FantasyLover DEFEATS 29 vamathletes, 3-0
1 superllamel DEFEATS 32 blue1421, 3-0
10 RomanianGenius DEFEATS 23 isabella2296, 3-0
13 Brut3Forc3 VS 20 bob einstein
6 yongyi781 DEFEATS 27 PWNER Express, 3-0
17 hurdler DEFEATS 16 budi713, 3-1
5 alexhhmun VS 28 mathmagician1729
3 ra5249 VS 30 Meta Knight
12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1
9 ernie DEFEATS 24 AIME15, 3-0
7 vallon22 DEFEATS 26 nikeballa96, 3-0
2 mewto55555 DEFEATS 31 Mind Storm, 3-0
15 kl2836 DEFEATS 18 violin321, 3-2
11 BOGTRO VS 22 thdanh90
8 Jongao VS 25 kookamunga

Round 2

4 FantasyLover DEFEATS 14 skaldjfhrulez, 3-0
10 RomanianGenius DEFEATS 1 superllamel, 3-1
13 Brut3Forc3/20 bob einstein VS yongyi781
(16)hurdler VS 5 alexhhmun/28 mathemagician1729
3 ra5249/30 Meta Knight VS 12 lotsofmath
9 ernie VS 7 vallon22
2 mewto55555 DEFEATS 15 kl2836, 3-0
11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga

Quarter-Finals

4 FantasyLover DEFEATS (1)RomanianGenius, 3-1
????? VS ?????
????? VS 9 ernie/7 vallon22
2 mewto55555 VS ?????

Semi-Finals

((1)) FantasyLover VS ?????
????? VS ?????

Finals

????? VS ?????
____________________________________________________________________________________________

Games that can be played currently:

13 Brut3Forc3 VS 20 bob einstein
5 alexhhmun VS 28 mathemagician1729
3 ra5249 VS 30 Meta Knight
11 BOGTRO VS 22 thdanh90
8 Jongao VS 25 kookamunga
7 vallon22 VS 9 ernie

Vallon22 beats Ernie

4-2 E
4-2 V
4-1 V
4-2 V

I will be out of town until july 13th, starting tomorrow, so i forfeit to alexhmun or mathemagician, (whoever should face me next, unless they wish to wait for me.......)

[quote="ernie"][quote="AIME_is_hard"]is it too late for me to join?[/quote]

yeah its too late, AIME

vallon, lemme tell u that i typoed 6 TIMES

HOLY COW!  :lol:[/quote]

u wish! :)

bump...

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OMG!!!
it's spam!!! :spam:  :spam:  :spam:

ITS MORE SPAM!

:spam: :spam: :spam:

Too many smilies in your post: 2359.
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it's DDR!!! :ddr:
and yoda!!! :yoda:
and a sick guy being carried off on a stretcher!!!  :stretcher:

I believe you are spamming. :spam: :spam: :spam:

The second one (:harhar:) is one of my favorites.

DId this just die? Have the new tournaments made everyone forget this one? Ernie you should pm those who need to play.

No, I have cancelled this tourney, as I have to Ernie's Tourney (the new one).

Yay and my tournament is alomst over. I'll keep PPT up though, as it won't start for probably a month.

[i]or[/i] we could just stop posting in this thread

True, but an instinct tells me we'll never stop posting on this thread.  :D