How do you open/close/read/write to a file in PHP? Do the C fprintf() and fwrite() family of functions work?

http://ca.php.net/fprintf
http://ca.php.net/fwrite

In short, yes. All fopen(), fclose() are supported as well.

Check the left side of the linked pages for a "directory" of file functions in PHP.


geometry
circumcircle
incenter
geometry proposed

Let a triangle $ABC$ and its circucircle with center $O$ and its incircle with center $I$ . Take any point $K$ of the circumcircle .Draw the tangents from $K$ to the incircle which meet the circumcircle at $L,\ M$ . Prove that the line $LM$ is tangent to the incircle of $ABC$ .

let $ KI$ intersect the circumcircle at$ T$ and $ TO$ intersect circumcircle again at $ T'$ and let $ MK$ be tangent to incircle at $ N$ by similarity of triangles $ KIN$ and $ TT'M$ we have $ IN.TM\equal{}2R.r$
the power of $ I$ wrt circumcircle=$ KI.IT\equal{}R^{2}\minus{}OI^{2}\equal{}2R.r$ hence $ TM\equal{}TI\equal{}TL$ hence $ I$ is incenter of $ KLM$.

That's also my proof . Congratulations Amir S  :D  .Nice work

In fact there is a nicer form (equivalent to that one)
Let two circles $ C_{1},\ C_{2}$ . If there exist a triangle inscibed to one and circumscribed to the other then there exist an infite number of triangles with that property . :wink:

This does have a name: http://www.cut-the-knot.org/Curriculum/Geometry/PonceletChain.shtml


inequalities
algebra
polynomial
Rational Root Theorem
number theory open
number theory

Consider the equation $x^2 + y^2 + z^2 = x^5 + y^5 + z^5$.

How many solutions in integers are there?
Can we find an infinite family of solutions for this equation?
Can we find all solutions?

$x^2(x^3 - 1) + y^2(y^3 - 1) + z^2(z^3 - 1) = 0$

From which we immediately derive the solutions $x = y = z = 0, 1$.  I strongly suspect there are no other integer solutions:

$x^2(x-1)(x^2 + x + 1) + y^2(y-1)(y^2 + y + 1) + z^2(z - 1)(z^2 + z + 1) = 0$

For $x, y, z > 1$ all of the terms are positive and for $x, y, z < 0$ all of the terms are negative.  WLOG, if $x = 0, 1$, then $y < 0$ and $z > 1$, but because the $y-1, z-1$ and the $y^2 + y + 1, z^2 + z + 1$ terms are symmetric along different lines, it seems impossible that there are any other solutions (I realize this isn't a rigorous proof).

Edit:  Okay, so I have shown that any other solutions must have one of the variables $x = 0, 1$.  Taking that:

$y^5 + z^5 = y^2 + z^2$
$y^5 - y^2 = z^2 - z^5$

Substitute $y = -k$ - we now want a solution for $z, k \in \mathbb{N}$.

$k^5 + k^2 = z^5 - z^2$
$z^5 - k^5 = z^2 + k^2$

The left side is minimized when $z, k$ are consecutive fifths - however, for $k > 2$ it is easily shown that the difference between consecutive fifths becomes much larger than $z^2 + k^2$ could be (there should be a way to say this more rigorously, but I never was very good at inequalities) and hence no nontrivial integer solutions are possible.

Edit #2:  Rational root theorem on the above equation!


Let be $ a,b,c \in \mathbb{N}$ and $ f: \mathbb{N}\rightarrow \mathbb{N}\ ,\ f(n)\equal{}an^2\plus{}bn\plus{}c$ .

a)Find $ f(n\plus{}2)\minus{}f(n\plus{}1)\ ,\ n\in \mathbb{N}$;

b)Find $ a,b,c$ such that :

 $ \left[\frac{f(n)}{2}\right]\plus{}\left[\frac{f(n\plus{}1)}{2}\right]\equal{}(n\plus{}1)^2\ ,\ (\forall)n\in \mathbb{N}$

No one wants to try this function ? I think it is very nice  :clap2:

(a) is easy, so I'll solve (b)

Put $ n=2k$, simplify and use integer part properties to get

$ 4ak^2+2(a+b)k+\left[{c\over 2}\right]+\left[{a+b+c\over 2}\right]=4k^2+4k+1$

Equating the coefficients, we get

$ 4a=4\implies a=1$

$ 2(1+b)=4\implies b=1$

$ \left[{c\over 2}\right]+1+\left[{c\over 2}\right]=1\implies \left[{c\over 2}\right]=0\implies c=1$ (since $ c$ is natural)

Therefore, candidate function is $ f(n)=n^2+n+1$. We must check it:

$ \left[{n^2+n+1\over 2}\right]={n^2+n\over 2}$, as $ n^2+n=n(n+1)$ is always even. Hence

\begin{eqnarray*}\left[{f(n)\over 2}\right]+\left[{f(n+1)\over 2}\right]={n^2+n\over 2}+{n^2+2n+1+n+1\over 2}={2n^2+4n+2\over 2}=(n+1)^2\end{eqnarray*}

Thus the desired function is $ f(n)=n^2+n+1$

[hide="Conjecture for part 2"] Solutions are f(n)=n^2+n and f(n)=n^2+n+1.[/hide]


AMC
AIME
analytic geometry
graphing lines
slope
geometry
symmetry

http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_10

How would we go about investigating to get the first solution to the problem in the above? I just don't see the line the problem wants to find and have difficulty believing the solution since I don't know where it came from.

The circles are laid out like this:

[code]
AB
CDE
FGH
[/code]

Find the point where F and G touch, and the point where B and D touch. Connect these two points for the line you're looking for. A-C-F are to the left of the line; E-G-H are to the right, and B-D is split into two regions of equal area.

Hope that helps :)

It just... doesn't seem intuitive to me. Even when I draw it, it doesn't look like it cuts in in half. The fact that three circles are on opposite sides of the line is clear but B and D don't look even to me.

How did you find the solution when you were solving the problem or did it just jump out at you?

How do we use the idea of the slope of 3 to solve the the problem? Should I sketch such a line and move it around convenient points around the figure and then guess and check until I find the right one?

A line through the point of tangency of two congruent circles must divide their area in half because there is rotational symmetry about that point.

[quote="not_trig"][quote="mathnerd314"]The circles are laid out like this:

[code]
AB
CDE
FGH
[/code]

Find the point where F and G touch, and the point where B and D touch. Connect these two points for the line you're looking for. A-C-F are to the left of the line; E-G-H are to the right, and B-D is split into two regions of equal area.

[/quote]

Wait... is G completely to the right of the line? Because then the line would be the common external tangent of F and G and wouldn't go through the point where B,D touch.[/quote]

No - Only A and C are completely to the left of the line, only E and H are completely to the right, and two pairs of circles, B/D and F/G, are split equally.

[quote="not_trig"]Wait... is G completely to the right of the line? Because then the line would be the common external tangent of F and G and wouldn't go through the point where B,D touch.[/quote]

Sorry! My bad. As matt276eagles said, B-D and F-G are both split equally.

And aufha, I looked at the solution in hopes that I could explain it to you. ;)


Are USAMO problems in increasing order of difficulty?

The problems are in increasing difficulty order for each day.

Meaning that the two easiest should be #1 and #4, while the two hardest are #3 and #6.


geometry
geometric transformation
rotation

In a theme park, you walk into a giant can-shaped room that can rotate. When the angular speed of the "room" reaches a certain, value, the bottom "floor" drops out. However, you don't fall down. Instead you "stick" to the wall.

a. Explain why in terms of the forces involved.
b. If the coefficient of static friction between and the side wall of the curved room is 0.30 and the radius of the room is 2.5 m, what is the minimum rotational speed of the "can" so that you can "stick" to the wall of the bottomless room?

My ideas:
For part B, wouldn't you do this:

$ F_{c}\equal{}\frac{mv^{2}}{r}\equal{}\mu N$

How do you continue?

friction always opposes direction of motion, think of what it has to oppose in order to stop falling

[quote="1/(ln x)"]For part B, wouldn't you do this: $ F_c \equal{} \frac {mv^2}{r} \equal{} \mu N$[/quote]

That's not correct. What are the forces applied to the person and their directions? We have the weight, vertical direction and downward, the friction, also vertical but upward, and the reaction force of the wall, with the direction of the center of rotation and centripetal. So we should write

$ \frac {mv^2}{r} \equal{} N$.

But the person don't move in the vertical direction, so we must also have

$ \mu N \equal{} mg \equal{} > N \equal{} \frac {mg}{\mu}$

and now you can find from these two expressions a relation between the angular velocity and the given data.


Why is it true for the following? can someone please answer or explain since I don't understand why it is true:

[b]If v is an eigenvector of a matrix A, then v is an eigenvector of A+cI for all scalars c and  I is the identity matrix[/b].

[quote="kouboy"]if v is an eigen vector of A, then Av=pv for some p. now (A+cI)v=Av+cIv=Av+cv=(p+c)v. hence v is an eigen vector of A+cI but with eigen value p+c[/quote]

but how do you know that p+c actually exists?


[quote="stvs_f"]if:
\[ \log_{12}18 \equal{} a , \log_{24}54 \equal{} b
\]
what is ab+5(a-b) ????[/quote]

If you have
\[ a\log_{24}54
\], then this is equivalent to;
\[ \log_{24}54^a
\]
and if this is equivalent to $ b$, then that means;
$ 24^b \equal{} 54^a$
I don't know what to do from there, but this may be helpful;

\[ log_{x}y \plus{} log_{x}z
\]$ \equal{}$
\[ log_{x}yz
\]

[quote="Wickedestjr"][quote="stvs_f"]if:
\[ \log_{12}18 \equal{} a , \log_{24}54 \equal{} b
\]
what is ab+5(a-b) ????[/quote]

If you have
\[ a\log_{24}54
\]
, then this is equivalent to;
\[ \log_{24}54^a
\]
and if this is equivalent to $ b$, then that means;
$ 24^b \equal{} 54^a$
I don't know what to do from there, but this may be helpful;
\[ log_{x}y \plus{} log_{x}z
\]
$ \equal{}$
\[ log_{x}yz
\]
[/quote]

Have you noticed that there was a comma between the $ a$ and the $ \log_{24}54$ ? ^^

Lol anyways,
[hide]
We know that $ 12^a \equal{} 18$ and $ 24^b \equal{} 54$ and by factoring the $ 12$ and the $ 24$, we have $ (2^2 * 3)^a \equal{} 2^2a * 3^a \equal{} 18$ and $ (2^3 * 3)^b \equal{} 2^{3b} * 3^b \equal{} 54$. Since $ 54 \equal{} 3 * 18$, we can say that $ 3 * (2^{2a} * 3^a) \equal{} 2^{2a} * 3^{a \plus{} 1} \equal{} 54$. Thus we have $ 2^{2a} * 3^{a \plus{} 1} \equal{} 2^{3b} * 3^b$. And You can do the rest.
[/hide]

No one has actually posted the entire solution...

[hide="solution"]
Let $ \ln 2 \equal{} x, \ln 3 \equal{} y$

We have $ a \equal{} \frac{x\plus{}2y}{2x\plus{}y}, b \equal{}\frac{x\plus{}3y}{3x\plus{}y}$

$ a\minus{}b \equal{} \frac{(3x^2 \plus{} 7xy \plus{} 2y^2)\minus{}(2x^2 \plus{} 7xy \plus{} 3y^2)}{(2x\plus{}y)(3x\plus{}y)} \equal{} \frac{x^2 \minus{} y^2}{(2x\plus{}y)(3x\plus{}y)}$

$ ab \equal{} \frac{x^2\plus{}5xy \plus{} 6y^2}{(2x\plus{}y)(3x\plus{}y)}$

$ \Rightarrow ab \plus{} 5(a\minus{}b) \equal{} \frac{6x^2 \plus{} 5xy \plus{} y^2}{(2x\plus{}y)(3x\plus{}y)} \equal{} \frac{(2x\plus{}y)(3x\plus{}y)}{(2x\plus{}y)(3x\plus{}y)} \equal{} 1$

$ ab \plus{} 5(a\minus{}b) \equal{} \boxed{1}$

POST $ 300$!! :surf: [/hide]


If $x$ is a real number such that $x^2-x$ is an integer and for some $n \geq 3$ $x^n-x$ is also an integer prove that $x$ is an integer.

It has been posted several times on the forum. The problem appeared for the first time in romanian national olympiad, proposed by Laurentiu Panaitopol.  :)


Suppose $ D \subset \mathbb{R}^ \plus{}$ is some set. Suppose we are given that some numbers, $ x_1, \ldots, x_n$ in $ D$, and we are given that their sum is $ s$.

1. Find the maximum possible value of $ x_1 x_2 \ldots x_n$.
2. Find the exact values of $ x_i$ that make this possible.

I am aware that when $ D \equal{} \mathbb{R}$, the terms are simply equal, and it is a simple matter of the AM-GM. However, in other sets (say, of integers), the maximum apparently occurs when the elements are "closest together". This problem is essentially asking for us to formalize this.

If there's not enough information, please let me know. I kind of made it up on the spot.

Furthermore, I do not know the solution to this problem myself.

I think it really does depend on exactly what set it is, and if it's finite or infinite. For a finite there's probably no formal proof other than trying out all possibilities.

When $ D \equal{} \mathbb{N}$, show that every $ x_i$ must be equal to either $ \left\lfloor \frac {s}{n} \right\rfloor$ or $ \left\lfloor \frac {s}{n} \right\rfloor \plus{} 1$, and show that this condition uniquely determines the product.  Generally, if $ \frac {s}{n}$ is not a limit point of $ D$ show that every $ x_i$ must be equal to either $ \sup D \cap \left( \minus{} \infty, \frac {s}{n} \right]$ or $ \inf D \cap \left[ \frac {s}{n}, \infty \right)$.  If $ \frac {s}{n}$ is a limit point of $ D$ but not contained in $ D$ then there is no maximum.  Otherwise, $ x_i \equal{} \frac {s}{n}$.

Well, certainly we can consider some special cases, can we? For instance, if D is simply the set of all positive reals, then the product is maximized when each of the terms are equal.

I can see a problem occurring if there actually is no maximum (say the sum were 9, we had 3 terms, and $ D \equal{} \mathbb{R} \minus{} \{3\}$.) Indeed, that would result in a problem, but we could still the supremum of the set of products (which I suppose is what I'm actually asking for.)

I was just intrigued that in integer products, the products were always maximized when the terms were closest together. What if, in particular, D is a subset of the integers? For a particular example, suppose we have 6 numbers whose sum is 30, and D is the set of even integers. I would initially be tempted to guess that 4, 4, 4, 4, 4, 6, 6, 6, 6, 6 would maximize the product, though that's just heuristic reasoning. I never really explored why I would choose this set. If, instead, it happened that the problem was 5 numbers, and the product was 25, what would I do?

And t0rajir0u, what if it's a special subset of N?

I've already answered both of those questions.

Excuse me then; my apologies. I didn't really fully understand what you said, and therefore skimmed over it. :/ Pathetic excuses...

[quote]If, instead, it happened that the problem was 5 numbers, and the product was 25, what would I do? [/quote]
I'm an idiot. Parity argument fail.

I assume, then, that when selecting the terms as you described, we try to get the number of each as close as possible to the other. We still need to show that this definitely yields the smallest possible product (I'll try to reason this out myself.)

I'll think over what you said. Thanks, t0rajir0u.

[quote="Zhero"]we try to get the number of each as close as possible to the other. [/quote]
Nope.  When there are only two distinct possible values, there is at most one choice of the number of each that works.  This is because $ n$ is fixed.


Let $a,b,c$ be positive reals. Prove that for all $n\ge m\ge 0$
\[\frac{a^{n}}{\sqrt{b^{n}c^{n}}}+\frac{b^{n}}{\sqrt{c^{n}a^{n}}}+\frac{c^{n}}{\sqrt{a^{n}b^{n}}}\ge \frac{a^{m}+b^{m}+c^{m}}{\sqrt[3]{a^{m}b^{m}c^{m}}}\]


I don't think we have a section about statistics (do we actually? :huh: ), I reckon we should have one. :D

[quote="Xevarion"]What kind of statistics do you want to see?[/quote]

Just statistics in general, like expectation algebra, estimators etc.

I would second adding "Probability and Statistics" subforum to the College Playground. We've got quite a few problems like that in Calculus and Advanced subfora :)

Could this include things like stochastic calculus? Could that be included in the name of the forum? I don't know if anyone other than myself would really care to discuss stochastic processes or stochastic differential equations, but it seems like an appropriate place...

I think that a subforum about probability, stochastic processes, stochastic differential equations, statistics etc.. would be a good think. I would be greatly interested in such a subforum  :)

so what about this probability section ? any news ?

Seems like a lot of reasonable people requested it, so it will be created. I just need you guys to figure out a proper naming (so people understand what's it about from the start) ... For now I'll just name it Probability and Statistics but as I said that may be changed in the future.

I think just name it "Probability and Statistics", people will certainly understand what it is about. I look forward to this new forum. :)

so when will this probability forum be created ?

How is it that there are topics in it already dated 2006?  :o  Were they moved?


Prove that for every positive integer $ n$ there exists an n_digit number divisible by $ 5^n$ all of whose digits are odd.


We use $ k^{2003}\plus{}(10A\minus{}k)^{2003}\vdots 10$ then easy to find that remainder!

[quote="the future"]
Its $ 200(45) \plus{} 16 \mod 10$ which is still $ 6$, but you had $ 200(55)$...lucky it still ends in 5.[/quote]

 :rotfl: I thought I had a residue class of 10 in modulo 10!

You can use [url=http://en.wikipedia.org/wiki/Euler%27s_theorem]Euler's Theorem[/url], which states that:

$ a^{\phi(n)}\equiv1\mod{n}$

where $ \phi(n)$ is the Euler Phi Function.

Can someone explain it a bit more?  :oops:

[quote="10000th User"][quote="ZzZzZzZzZzZz"]It doesn't have to be so complicated to find the remainers.  Each integer is clearly congruent to only one of $ 1, 2, 3, 4, ..., 10$ (mod 10).  Notice that $ x^{2003}\equiv y^{2003} \text{(mod 10)}$ implies $ x\equiv y \text{(mod 10)}$, so every ten terms will be congruent to different numbers (mod 10), and the sum of the remainders of every ten consecutive terms will be 45.[/quote]Careful with this 'generalization'. Consider odd powers and even powers, ie $ 6^2\equiv4^2\pmod{10}$ but $ 6\not\equiv4\pmod{10}$.[/quote]

I has assumed that would occur, but I fail to see how that proof is sufficient. It makes sense that the unique residue classes would produce another unique residue after being raised to a power, but saying $ x^{2003}\equiv y^{2003}\mod{10}$ does nothing for me.

Can you try to be a bit clearer? I think it's important that I understand.


A compoud contains only carbon, hydrogen, and oxygen.  Combution of 10.68 mg of the compound yields 16.01 mg of CO2(carbon dioxide) and 4.37 mg H2O(water).  The molar mass of the compound is 176.1 g/mol.  What are the empirical and molecular formulas of the compund?

I have no clue how to work this problem.  Can someone [i]please [/i]help me?

from the mass of $CO_2$...you can find the mass of Carbon... and from water, you can get the grams of hydrogen...take those two masses and subtract them from 10.68 and you'll get the mass of oxygen...convert all of them to mols, divide by lowest number of mols and get three whole numbers for your subscripts for your empirical formula

if you're there, you take the molar mass...176.1 g/mol and divide that by the empirical formula mass to get a whole number, which you mulitply that by the subscripts in your empirical formula to get the molecular formula!


Let$ a \plus{} b \plus{} c \equal{} 1,a,b,c\geq 0$.Prove that
$ \frac {1}{2a \minus{} 5b^2} \plus{} \frac {1}{2b \minus{} 5c^2} \plus{} \frac {1}{2c \minus{} 5a^2}\geq \frac {3}{(a^2 \plus{} b^2 \plus{} c^2)^2}$
Ofcourse,with $ a,b,c$ such that All denominators $ > 0$ :lol:

Nice,my problem is really easy,i am sorry,(because,my proof is harder than yours)
wait for a hard problem
[hide="to mitdac123"]anh quên mất cách nay,xin lỗi làm mất time của em[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

Is the set of constant polynomials in $\mathbb{Z}$[x] an ideal? If not, give two cosets, whose product is not a coset.

I think the set of constant polynomials in $\mathbb{Z}$[x] is an ideal... but i cant think of two cosets whose product is not a coset.

please advise

The set of constant polynomials is not an ideal as say Let $c \in \mathbb{Z}$ and $f(X)$ a polynomial in $\mathbb{Z}[X]$ of non zero degree,then their product has non zero degree.
Sorry,i do'nt understand your question on cosets as if the set under question is not an ideal how will u talk about cosets?.

I believe he meant that he thought it was not an ideal.

So now we need to find two cosets of c such that coset 1 * coset 2 is not a coset.  The cosets take the form of:

c + x
c + x^2
c + x^2 + x
...

So for the multiplication of two cosets, it seems that you always come out with some constant term c, and thus, it is a coset.  We need to find two which when multiplied, don't have a constant term on the end.

[quote]Sorry,i do'nt understand your question on cosets as if the set under question is not an ideal how will u talk about cosets?.[/quote]

For the quotient group R/S, S does not have to be an ideal of R.  So even though c is not an ideal of R, you can still talk about it's cosets.

ok sorry i get it,thanks ricky
k so Let $C$ be the set of all polynomials,Consider $\mathbb{Z}[x]/ C$,consider the coset $C+x$,now $x,x+1$ both belong to this coset,but $x * x = x^2 \in C+ x^2,x*(x+1) = x^2 +x \in C+ (x^2 + x)$,so u can't get any coset bu multiplying $C+x$ with itself.


calculus
integration
real analysis
real analysis theorems

I have a few general queries about PDEs
(1)  Is Integral transform method (apart from separation of variables) the only standard method to solve PDEs. Are there any other powerful analytical methods (non-numerical, including Finite element)
(2)  Is there a general guide to tell when should we choose the right type of integral transform for the problem. For example, whether Laplace or Fourier should we use?

(1) There are no methods for solving PDEs. There are methods for solving certain types of PDEs. For some equations it's Fourier transform, for others, conformal mapping or maybe Legendre transform. The PDE book by Evans lists a few.

(2) Is there a general guide to tell when to divide and when to multiply? My calculus students want to know.


1997 #8: Unit cubes are glued together to make a cube several units on each side. Some of the faces of this large cube are painted. When the cube is taken apart, there are exactly 45 unit cubes without any paint. How many unit cubeswere used to create the larger cube?

1998 #10: Two regular square pyramids have all edges 12 cm in length. The pyramids have parallel bases and parallel edges, and each has a vertex at the center of the other pyramid's base. What is the total number of cubic centimeters in the volume of the solid of intersection of the two pyramids? Express answer in simplest radical form.

Answers:

[hide]#8: 125
#10: 72sqrt2[/hide]

[hide]#10...
hard to explain without diagram...
try to imagine what this would look like in your head....
the intersection of the two square pyramids ends up being two square pyramids that are stuck together. The height of each these is 1/2 of the 12 inch pyramids so the sides are 1/2 of the 12 inch pyramids too!!! Cool..
So the intersection is the area of 2 square pyramids with all sides of 6.
You can find that the height is sqrt 18 or 3sqrt2
2((6*6*3sqrt2)/3))=72sqrt2!!!![/hide]

[hide]
8. Try to find different ways that you an multiply 3 numbers to get 45 unit cubes, because if you realize, no matter how you paint it, there is a rectangular prism that doesn't have paint on it's cubes. So you 5*3*3. Since 5 is the biggest, the cube must be 5*5*5 or 125 unit cubes.[/hide]


It's been 86 years. 86 long tough years. Ever since 1918, this has been the moment a whole nation has been waiting for. We've been through so much, both pain and happiness. From Slaughter's mad dash, to Fisk's home run, to Buckner, to Aaron Boone. So many times we've come oh so close, but fallen at the last step.

This one wasn't just for the rings. This was to finally stop the Babe from haunting us. This was so there would never be another 1918 chant in Yankee Stadium. This was a win to finally allow Buckner, Pesky and everyone else breathe a sigh of relief, knowing they'd be forever forgiven in New England and in the fanbase around the world. This was for everybody whose parents or grandparents taught them to be forever loyal. This one's for all the loyal fans who have passed away, wanting to see their beloved team win just once in their life-time.

http://p086.ezboard.com/fsonsofsamhornbostonredsox.showMessageRange?topicID=14279.topic&start=1&stop=20

This topic at another forum really touched me. I admit that I have suffered nothing. I've been a Red Sox fan since I first came to the US at age 5. The first baseball game I ever watched on TV was a Red Sox victory. I've only known how it feels to have your hopes crushed for 9 short years. In 1999, I watched Pedro's great bullpen performance against the Indians, and then another gem pitched against the Yankees. Even with that confidence, we lost in 5 games in the ALCS. Just last year, I sat on my sofa, stunned, watching Aaron Boone's home run fly over the left field bleachers in the 11th inning. Shortly thereafter, I cried. My beloved team had come so close, just to stumble once again. But these are nothing compared to the older generations of Red Sox Nation. They were the ones who watched with so much hope as the Red Sox came within 3 outs of becoming world champions over and over again. In that forum topic, there are so many dedications, to deceased loved ones and friends as well as the living. This championship was for them, so they can all rejoice.

God bless all the players, for giving us memorable moments, like Ortiz's two playoff walkoffs, Schilling's bloody sock game, there's too many to name. Thanks for keeping hope throughout, even down 3-0 with just 3 outs remaining. For riding momentum to a championship. For ending the curse. For giving all of our nation a chance to cry tears of joy for once, instead of tears of sorrow. For letting us walk down the street without having to hear "1918" or "the curse of the bambino" in jeering tones. Thank you for winning, and finally putting an end to all the pain of 86 years.

2004, the year the curse was broken. We've waited and waited. Even with all the suffering that happened, few ever lost faith. This one's for everybody who stayed loyal to our team. For everyone who never gave up, no matter how dire the situation. It's for everyone who could've walked away from all of it a million times, but stayed for the entire journey. This victory's for us Red Sox Nation! We deserved it! And finally, WE HAVE A CHAMPIONSHIP!

hehe red sox rock
my language teacher is a BIG fan of the cardinals
and this morning when he was out of his classroom but the door wasnt locked me and a couple friends went in and wrote SWEPT! and Red Sox Rock on his chalkboard
my lang. teacher said when he was watching the game and when they finally lost he said he pounded his coffee table really hard and "punched a hole in his wall."
when he found out i wrote the stuff on his chalkboard he said its time to enter grades into his gradebook

naughty naughty!
today one of my friends who's a REAL BIG red sox fan wore all red(even though she used to live in new york)
i was supporting the red sox, but i dont feel it necessary to post 5 paragraphs on it (wow, must have taken a LONG time to write all that!)
but, it was fun to watch the match! they SWEPT
<my dad said that it was mostly cuz they had their confidence level up, since they beat the Yankees>


Hello everyone,

I am having some trouble with the following trig equations and would appreciate some help.

Thank you very much!

---

1. Solve $ \tan x \plus{} \sec2x \equal{} 1, [\frac { \minus{} pi}{2}, \frac {pi}{2}]$.

2. Solve $ \cos^2x \minus{} 2\sin x\cos x \minus{} \sin^2x \equal{} 0, 0 \leq 2x \leq \pi$.

---

[b]Rough Work:[/b]

[b]1.[/b]

$ \frac {\sin x}{\cos x} \equal{} 1 \minus{} \frac {1}{\cos 2x}$

$ RS \equal{} \frac {\cos 2x \minus{} 1}{\cos 2x}$

$ RS \equal{} \frac {1 \minus{} 2\sin^2 x \minus{} 1}{\cos 2x}$

$ RS \equal{} \frac { \minus{} 2\sin^2 x}{\cos 2x}$

I tried to manipulate the trigonometric equation, but I did not arrive at anything useful.

[b]2.[/b]

$ 1 \minus{} \sin^2 x \minus{} \sin 2x \minus{} \sin^2 x \equal{} 0$

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

$ (1 \minus{} 2\sin^2 x)^2 \equal{} (\sin 2x)^2 \leftarrow$ I am squaring both sides and being mindful of extraneous solutions.

$ 1 \minus{} 4\sin^2 x \plus{} 4\sin^4 x \equal{} 4\sin^2x(1 \minus{} \sin^2x)$

$ 8\sin^4x \minus{} 8\sin^2x \plus{} 1 \equal{} 0$

Let $ u \equal{} \sin x$:

$ 8u^4 \minus{} 8u^2 \plus{} 1 \equal{} 0$

$ \therefore u \equal{} \sin x \equal{} \minus{} 0.383, \minus{} 0.924, 0.383, 0.924$

$ \Rightarrow x \equal{} \minus{} 0.393, 3.53, \minus{} 1.18, 4.32, 0.393, 2.75, 1.18, 1.96$

When I checked for the validity of these solutions by subsitution in the original equation, I found that $ x \equal{} 0.393, 1.96$ both gave values close to zero.

However, the answer provided only lists [hide]$ 0.393$[/hide] as the solution.

P.S.: Also, if anyone has a more efficient way to solve this equation without getting a quartic equation, please show me! Thanks!

#2 should be instantly recognizable as $ \cos 2x \equal{} \sin 2x$.  What can you do from here?  (You'll also note that if $ x \equal{} 1.96$ then $ 2x$ is outside of the desired range.)

Thanks for your response, t0rajir0u.

I worked through the equation and solved x to be $ \frac{\pi}{8}$.

However, in this process, I divided $ \cos 2x \equal{} \sin2x$ by $ \cos 2x$. Would there be any other method to solve this equation, because I am thinking that dividing by $ \cos 2x$ may be a mathematically perilous move?

You can use the [url=http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/sumprod.htm]sum-to-product[/url] formulas if you really want, but it's a waste of time.  Just note that if $ \cos 2x \equal{} 0$, then $ \sin 2x \equal{} \pm 1$ and the two cannot be equal.


inequalities
function
algebra proposed
algebra

Does there exist a function $ f: \mathbb{Q}\to\mathbb{Q}$ such that $ f(x)f(y)\le|x \minus{} y|$ holds for any $ x,y\in \mathbb{Q},$ $ x\ne y,$ and for each $ x\in \mathbb{Q}$ the set $ \bigl\{y\in \mathbb{Q}\,|\, f(x)f(y) \equal{} |x \minus{} y|\bigr\}$ is infinite? (V. Brayman)

[quote="xxp2000"]Let $ x \equal{} \frac mn\ne0$ with $ (m,n) \equal{} 1$, then $ f(x) \equal{} \frac1m$, $ f(0) \equal{} 1$. This definition should work.[/quote]

I think you are wrong. Take $ x\equal{}\frac{1}{3}$ and  $ y\equal{}\frac{2}{3}$ then $ f(x)f(y)\equal{}\frac{1}{2} > \frac{1}{3}\equal{}y\minus{}x$

I proved that there is no such function but my solution  is a little bit longer.I will post it soon.

It was a typo. $ f(x)\equal{}\frac1n$ not $ f(x)\equal{}\frac1m$.

Thanks a lot.

[quote="xxp2000"]It was a typo. $ f(x) \equal{} \frac1n$ not $ f(x) \equal{} \frac1m$.

Thanks a lot.[/quote]

You are right. I made wrong assumption while solving the problem. Thanks a lot...


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a non-trivial group such that every normal subgroup of $ G$ is finitely generated. Prove that there is no non-trivial normal subgroup $ N$ of $ G$ such that $ G\cong\dfrac{G}{N}$.

Suppose that we have a surjective homomorphism $ f: G\to G$ with $ \ker f\equal{}N$.

Let $ f^{(i)}$ denote $ f$ applied $ i$ times, let $ M_i\equal{}\ker f^{(i)}$, and let $ M\equal{}\cup_{i\equal{}1}^{\infty} M_i$, then we claim that $ M$ is normal and not finitely generated.

First, we note that $ M_i \subsetneq M_{i\plus{}1}$, due to the surjectivity of $ f$ and the nontriviality of $ N$, so $ M$ is not equal to any $ M_i$.  But if $ M$ was finitely generated, then some $ M_i$ would contain each generator and we would have $ M\equal{}M_i$.

Finally, $ M$ is normal because each $ M_i$ is normal.


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 12 B
AMC 10

What are your goals for the AMC?

I want to get at least 130.

Read my signature. It's 150 but I'll make a bunch of stupid mistakes if I'm not careful...

[quote="Iversonfan2007"]to get above the national average[/quote]

To get below the national average :P.

I don't care what I get as long as I make AIME qualifying scores on both tests I take, then make USAMO. Though I really can't prepare more this year, I will be studying hard to make MOP next year.

Remember, chess64, that we had a deal that we would both make MOP so we could play chess in real life. I think we'll have to move that back to next year, ok? MOP qualification rules for 8th graders are asinine.

[quote="now a ranger"]I'll be happy just making AIME.[/quote]
Honestly, I'd be happy making the AMC, there's a pretty high chance that I won't be able to take it this year.

[quote="MysticTerminator"][quote="bpms"]Info:
7th grade:
Last year: AMC-100
AIME-15
USAMO-17

This year I'm going for:
AMC-150
AIME-9
USAMO-40
IMO-20[/quote]

dude I suspect you are not serious

this is because you do not actually exist[/quote]
Yes, I am the opposite of Yahoo! Slurp who is a human that's IP adress makes AoPS think he is a bot.

aiming for one fifty.

go now, run and hide.

[quote="MysticTerminator"][quote="bpms"]Info:
7th grade:
Last year: AMC-100
AIME-15
USAMO-17

This year I'm going for:
AMC-150
AIME-9
USAMO-40
IMO-20[/quote]

dude I suspect you are not serious

this is because you do not actually exist[/quote]
I don't understand.  :huh:

Well, I used to get around 12 on old practice tests last year, which translated into a 9 on the real thing because you're bound to make stupider mistakes. It's some law of mathematical entropy that makes you do worse during the real thing.

[quote="chess64"]Yeah, like I used to get 8-9's and then I only got a 5...[/quote]

Which is why I got no higher than 5s on practices last year and a 7 on the real AIME?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131535]Qualifying for USAMO is now a more reasonable goal[/url] for this year.

Good luck, everyone.

[quote="1337h4x"]Does anyone have thoughts on how this will affect the minimum selection index required to take USAMO?[/quote]
The index'll probably go down about 5 or so points I think.

I predict this year's index score will be no lower than last year's.

AMC 12: 130
AIME: 10

so hopefully USAMO.

Dude I'm terrified of the AMC  :|

AMC 10: Not taking it
AMC 12A: 135
AMC 12B: 140+

I usually make ~11s on practice AIMEs, so...

AIME: 11

USAMO: 5+

[quote="mysmartmouth"][quote="Pakman2012"]I'm aiming for a x on the AMC
and a y on the AIME
and a z on the USAMO.[/quote]

Where $x+y+z = 207$[/quote]
Interestingly, the perfect scores on them seem to resemble 2007, the current year.

For more strange coincidences, Google "Lincoln and Kennedy".

Why are the AMCs harder than the AIMEs to do well on for me? That's really strange. I'm horrible under time constraints. Oh, the horror, the horror.

some people are better under time pressure. Others are slow workers. Therefore, AIME and AMC are suitable for different types of problem solvers.

Goals:

AMC 10A: 126
AMC 12B: 102
AIME: 2

My goals are

AMC 12: 130
AIME: 12
USAMO: 16+

Hmm...I suppose goals are supposed to be somewhat unattainable...

AMC 12: 135+
AIME: 9+ (all I actually want to do is qualify for USAMO...)
USAMO: 7+

 :whistling:

Oh- this is a good question. Getting a perfect on the 12 is much much harder than on the 10- this is the first time I haven't felt like I had a viable shot at acing something....
But
140+ on the 12? 135? As long as I go through to AIME with a solid score for my index....
enough on the AIME to qualify for USAMO
enough on USAMO to qualify for MOP
I think that posting actual numbers for goals isn't that productive- There have been years where a 7 made USAMO (pre-expansion, if I remember right) and there have been years where a higher score then that wouldn't have.


trigonometry
inequalities
inequalities unsolved

It is a problem from beginner's corner.But i think it quite hard for beginners :( ,so i send it here:

If $ x$, $ y$, $ z$ are nonnegative reals satisfying $ yz+zx+xy=1$, then prove that

$ \sum\frac{\left(1-x^{2}\right)^{2}}{1+x^{2}}\geq 1$,

where the $ \sum$ sign stands for a cyclic sum.

In fact, this one can be solved by trigonometry if we notice that x= tg(A/2),....., ABC is a triangle. After replacing, we have an easy inequality. :D.

Treegoner,if we have x=tgA/2,then that inequality will become:
Sum(cosA/cos(A/2)) 2  \geq 1.
But is it really too easy?I have had solution but it is too long.
Can you make it shorter? :?

Either I am very tired or stupid at this moment or the problem is very hard. Look what I did, with no success:
  Reduce it to  \sum x^2+4 \sum 1/(1+x 2 ) \geq 10. Put x=a/ \sqrt ab+bc+ca  and so on and reduce it to:
    \sum a 2 / \sum ab+ 8abc/(a+b)(b+c)(c+a) \geq 2. With some tricks we can get to  \sum a^3+3abc \geq  \sum a 2 (b+c)+abc/(ab+bc+ca)*( \sum a 2 - \sum ab), which is STRONGER THAN SCHUR! I made some courage and found the equivalent form  \sum a^4(b+c)+2abc(ab+bc+ca) \geq  \sum a^3(b+c) 2 . Can anyone prove this?

Well, if a,b,c are sides of a triangle, then the last inequality can be written  \sum ( b+c-a)*(b-c) 2 /a \geq 0, true. So, we have to deal with the case a>b+c.

The last inequality you writed is right. And this is my solution for it.
Let a \geq c \geq b and m=b-c;n=c-a then it can be writed:
(b+c-a)*m 2 /a+(a+c-b)n 2 /b+(a+b-c)(m+n) 2 /c \geq 0
(bc+ab+(c-a) 2 )m 2 /ac+(ac+ab+(b-c) 2 )n 2 /bc+2mn(a+b-c)/c \geq 0.
But notice that if a \geq c \geq b ,2mn(a+b-c)/c \geq 0.
So we have the result we need.
 :)

Very nive solution for it. I also found a solution in the case when, for example b>=a+c (and thus the inequality is proved, since when a,b,c are sides of a triangle, I proved it). In this case the inequality can be written:
  (a-b) 2 ( a/c+a/b+b/c+c/b-2)+(b-c) 2( c/a+c/b+a/b+b/a-2) \geq 2(c-b)(b-a)(1-a/b-c/b). Obviously, the LHS is at least a(b+c)/bc(a-b) 2 +c(a+b)/ab(b-c) 2  \geq 2|a-b||b-c|* \sqrt (a+b)(c+b) /b. All we need is that (a+b)(a+c) \geq (a+c-b) 2 , or 3b(a+c) \geq a 2 +c 2 +ac, which foloows from b \geq a+c.

Wow, in fact I'm wrong at first when I thought one was easy. In fact, this is JACK GARFULKEL inequality. I have found a very nice solution for another JACK GARFULKEL. The solution is in Inequality _ Open question.
   I know how to prove this one but it's very long and I have another direction for this one. It finally give a tighter inequality that I don't know how to prove it. But I'm sure it's right. I'm looking forward to your reply to this one. :D

I finally have a very short solution to this one. What a luck! :D
 We have to prove that
    [ \sum a^2/ \sum bc] + 8abc/prod(b+c)>=2.
  <=> [ (\sum a)^2/ \sum bc] +8abc/prod(b+c) >=4.
      It 's easy to show that there always exists a triangle ABC such that
        sinA =  \sqrt( (ab+ca+bc) /(b+a)(a+c)),....
     Hence this inequality for every x,y,z we have  \sum xy(sinC)^2 <=1/4(x+y+z)^2, we have
        [ (\sum a)^2/ \sum bc] >= 4* \sum ab/[(a+c)(b+c)] =K
     But interestingly, K + 8abc/prod(a+b) = 4 and we finally get what we need.  :).
    And we also have the solution to JG inequality.

Treegoner, I see that you apply this inequality with sinA, sinB, sinC virtuously. Congratulations!


You are given a piece of paper. You can cut the paper into $8$ or $12$ pieces. Then you can do so for any of the new pieces or let them uncut and so on.
Can you get exactly $60$ pieces¿ Show that you can get every number of pieces greater than $60$.

[hide]We start with $1$ piece.  Each cut adds $7$ or $11$ pieces.  Then we can get any number of pieces that can be expressed as $1+7n+11m$, where $n$ and $m$ are positive integers.  Suppose $1+7n+11m=60$.  Then $7n+11m=59$.  We have $n\le 8$ and $m\le 5$.  Looking $\pmod{7}$, we see that $4m\equiv 3\pmod{7}$, which has no solutions for $1\le m\le 5$.  Therefore, it is impossible to get $60$ pieces.

We can get any number of pieces from $61$ to $67$:

$61=1+1\cdot 11+7\cdot 7$
$62=1+3\cdot 11+4\cdot 7$
$63=1+5\cdot 11+1\cdot 7$
$64=1+0\cdot 11+9\cdot 7$
$65=1+2\cdot 11+6\cdot 7$
$66=1+4\cdot 11+3\cdot 7$
$67=1+6\cdot 11+0\cdot 7$.

We can get any number of pieces $n$ greater than $67$ by taking the $k\in[61,67]$ such that $n\equiv k\pmod{7}$ and adding $7$ until we get the desired number of pieces.

In general, if we can make cuts into either $n$ or $m$ pieces, then the lowest number of pieces that we can not get is $1+(n-1)(m-1)-(n-1)-(m-1)$.[/hide]

[hide]
You can't make 60 from chicken mcnugget formula. 1+7*11-7-11=60.

Every cut will increase the total number of pieces by 7 or 11.
Base:
61=1+7*7+11
62=1+4*7+3*11
63=1+7+5*11
64=1+9*7
65=1+6*7+2*11
66=1+3*7+4*11
67=1+6*11
Inductive step: If you can make n number of pieces, then you can make n+7 number of pieces.
So, we can make any number of pieces bigger than 60.[/hide]


A bag contains blue and green marbles. If 9 green marbles are removed from the bag, the probabbiliy of drawing a green marble from the remaining marbles would be 9/16 . If instead 7 blue marbles are added to the bag, the probability of drawing a blue marble would be 7/16 . What was the number of blue marbles in the bag before any changes were made?

[hide="ANSWER"]7[/hide]

MY SOLUTION:
[hide]I used a calculator (This is a team round problem.) to generate a table of values after getting the equation g = (9b + 63)/7.  I got many values that work.  Can anyone explain why 7 is the correct answer?  Thanks.[/hide]

You just set it up with equations==>let b be the number of blue marbles and g be the number of green marbles.  Hence, b+g are the number of marbles in the bag.  So, we get:
(g-9)/(g+b-9)=9/16 and
(b+7)/(g+b+7)=7/16
Now, All you have to do is solve.

I was gonna post the same thing Z-coli said, but then he already posted it.

Hint: Try avoiding using "b" as a variable because when you solve lots of times you get confused over "b" and "6"

I DID solve.  However, I found many solutions.

For example:

A few ordered pairs that work...

28, 45
42, 63

.
.
.

[quote="Z-coli"]There are two equations.  Unless they give the same information about the variables, there should only be 1 solution.[/quote]

Actually, both equations yield $ g \equal{} \frac{9b \plus{} 63}{7}$ after cross-multiplying both equations.


Area of CBM=9/3=3. Since the length (CB) is 3, the base (BM) is 2, so $ CM\equal{}\sqrt{2^2\plus{}3^2}\equal{}\sqrt{13}$


Let $ a,b,c\in\mathbb{R}_\plus{}$. Prove that $ 11(a^6\plus{}b^6\plus{}c^6)\plus{}40abc(ab^2\plus{}bc^2\plus{}ca^2)\geq 51abc(a^2b\plus{}b^2c\plus{}c^2a)$.

[quote="arqady"][quote="fjwxcsl"]
\[ (a^6 \plus{} b^6 \plus{} c^6) \plus{} 50abc(ab^2 \plus{} bc^2 \plus{} ca^2)\geq 51abc(a^2b \plus{} b^2c \plus{} c^2a).
\]
[/quote]
Are you know a poof of  this inequality without computer?  :wink:[/quote]
I haven't even seen a proof with computer. :wink: But the problem I gave should be easier.

[quote="MeKnowsNothing"]Let $ a,b,c\in\mathbb{R}_ \plus{}$. Prove that $ 11(a^6 \plus{} b^6 \plus{} c^6) \plus{} 40abc(ab^2 \plus{} bc^2 \plus{} ca^2)\geq 51abc(a^2b \plus{} b^2c \plus{} c^2a)$.[/quote]
Assum $ c\equal{}max${$ a;b;c$}

$ 11(a^6 \plus{} b^6 \plus{} c^6) \plus{} 40abc(ab^2 \plus{} bc^2 \plus{} ca^2)\geq 51abc(a^2b \plus{} b^2c \plus{} c^2a)$

$ \leftrightarrow M(a \minus{} b)^2 \plus{} N(a \minus{} c)(b \minus{} c) \ge\ 0$

With $ M \equal{} 11(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b)^2 \minus{} 11abc^2 \ge\ 0$

$ N \equal{} 11(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} c)(b \plus{} c) \plus{} 40ab^2c \minus{} 51a^2bc \ge\ 0$

We have done!

Why $ N\geq 0$ and why $ M(a\minus{}b)^2\plus{}N(a\minus{}c)(b\minus{}c)\geq 0$? We might have that $ N(a\minus{}c)(b\minus{}c)\leq 0$.

[quote="MeKnowsNothing"]Why $ N\geq 0$ and why $ M(a \minus{} b)^2 \plus{} N(a \minus{} c)(b \minus{} c)\geq 0$? We might have that $ N(a \minus{} c)(b \minus{} c)\leq 0$.[/quote]

$ N \equal{} 11(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} c)(b \plus{} c) \plus{} 40ab^2c \minus{} 51a^2bc$

 $ \equal{} 11(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca \plus{} c^2) \plus{} 40ab^2c \minus{} 51a^2bc \ge\ 0$

 $ \equal{} 11(a^2 \plus{} b^2)(ab \plus{} ca \plus{} c^2) \plus{} 11b^3c \plus{} 40ab^2c \plus{} [11(c^3b \plus{} c^3a \plus{} c^4 \plus{} a^2bc \plus{} c^2ab) \minus{} 51a^2bc]$

$ \rightarrow N \ge\ 0$ with $ c\equal{} max${$ a;b;c$} :wink:

[quote="MeKnowsNothing"]Let $ a,b,c\geq0$. Prove that

$ 11\left(a^6 \plus{} b^6 \plus{} c^6\right) \plus{} 40abc\left(ab^2 \plus{} bc^2 \plus{} ca^2\right)\geq 51abc\left(a^2b \plus{} b^2c \plus{} c^2a\right).$

I haven't even seen a proof with computer.[/quote]$ 11\sum_{sym}{a^6} \plus{} 40abc\sum_{cyc}{ab^2} \minus{} 51abc\sum_{cyc}{a^2b}$

$ \equal{} \sum_{cyc}{\frac {(b \minus{} c)^2}{6}\left[66a^2(a \minus{} c)^2 \plus{} 3abc(11a \plus{} 23b \plus{} 21c) \plus{} 22\left(2b^4 \plus{} 4b^3c \plus{} 2bc^3 \plus{} c^4\right)\right]}$

$ \geq 0.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=154225

[quote="fjwxcsl"]
\[ (a^6 \plus{} b^6 \plus{} c^6) \plus{} 50abc(ab^2 \plus{} bc^2 \plus{} ca^2)\geq 51abc(a^2b \plus{} b^2c \plus{} c^2a).
\]
[/quote]

We also have
\[ (a^6 \plus{} b^6 \plus{} c^6) \plus{} 51abc(ab^2 \plus{} bc^2 \plus{} ca^2)\geq 52abc(a^2b \plus{} b^2c \plus{} c^2a).
\]


The bigger mass weighs $ 12kg$, while the smaller mass weighs $ 5kg$.
Find individual accelerations. Pulleys are massless, strings are inextensible.
[hide="My solution:"]
I get the answer, with $ a_m \equal{} 2a_M$. Since, if $ M$ accelerates with $ a$, the two strings will be pulled by $ a$. $ m$ will get an acceleration $ a$ due to $ 1^{st}$rope, and another acceleration $ a$ due to the pulley which is pulled by $ a$.

To verify, I argued:
"Consider the smaller pulley as your reference. Now, you are sitting on it observing $ m$. It appears to accelerate with $ a$ towards you. You are attached to another string. You too are accelerating with $ a$. So with respect to a stationary observer, $ m$ accelerates by $ 2a$"

But a friend of mine argues the acceleration relations are wrong. I'd like to know your opinion.
[/hide]

hello
I m getting $ a_m\equal{}3a_M$ :(
thank u

[quote="kabi"]hello
I m getting $ a_m \equal{} 3a_M$[/quote]
That was my first conclusion, but if you try to verify, like I have mentioned in the main question, I believe you get the true values.

[quote="kabi"]hello
I m getting $ a_m \equal{} 3a_M$ :(
thank u[/quote]I think kabi is right.
I point $ A$ moves $ x$, $ B$ moves $ x$, $ C$ moves $ 2x \plus{} x$.
Hence, $ a_m\equal{}3a_M$. Correct.
But then what is wrong with my analysis: the verification?

hello
Here is my method::
bigger pulley be P and smaller be Q.
since the pulley P is fixed,accelerations of two ends of string suppported by it should be equal and opp. .
thus, if block $ M$ moves downwards with $ a_M$ pulley Q moves upwards with acc. $ a_M$.
Now if we consider d string supported by Q.
we will get $ a_M\plus{}a_M\plus{}a_M\minus{}a_m\equal{}0$

[quote="kabi"]hello
Here is my method::
bigger pulley be P and smaller be Q.
since the pulley P is fixed,accelerations of two ends of string suppported by it should be equal and opp. .
thus, if block $ M$ moves downwards with $ a_M$ pulley Q moves upwards with acc. $ a_M$.
Now if we consider d string supported by Q.
we will get $ a_M \plus{} a_M \plus{} a_M \minus{} a_m \equal{} 0$[/quote]
Ok. Different people have different ways. Nice method (didn't understand the last part though! I find my way more convenient, I'll stick to it :rotfl: )

[quote="Rajiv"]
Ok. Different people have different ways. Nice method (didn't understand the last part though! I find my way more convenient, I'll stick to it :rotfl: )[/quote]
hello
It will better to stick to ur method as now there is no time to learn new methods.
thank u

[quote="aadil"]kabi how did you find that the acceleration of the smaller pulley is thrice the acceleration of the larger one[/quote]
i wrote in my II post.
Kindly see it.

We have got $ a_m = 3a_M$, alright! But which block moves up and which one down? I noticed an anomaly when I entered the the relations of $ T_1, T_2 \; and \; a$ into a linear program solver.
See figure:
Let us find the mass $ m$ that can balance $ M$.
We get:
$ \begin{align*}
& 2T_2 = T_1
\\
& 3T = 12g
\\
& T = mg
\\
& \therefore mg = 4g
& \therefore \boxed {m \ge 4kg}$

Now find relations of $ T_1, T_2, \; and \; a$ for $ 5kg$, you get only two equations, for the third equation, use conservation of energy and $ s_m = 3s_M, \; s \to \; displacement$ and $ v_m = 3v_M$.

[quote="Rajiv"]
"Consider the smaller pulley as your reference. Now, you are sitting on it observing $ m$. It appears to accelerate with $ a$ towards you. You are attached to another string. [b]You too are accelerating with $ a$[/b]. [/quote]yy?bcoz they r connected by same string??--NOOO

Here is the simplest solution
Let $ T$ in rope thru larger pulley
Then tension in each arm of smaller pulley =$ \frac{T}{2}$
Now ,the father of all --Constraint relation
$ T_1 a_1 \equal{} T_2 a_2$
$ T_1 \equal{}\frac{T}{2} \plus{} T \equal{} \frac{3T}{2}$
$ T_2 \equal{}\frac{T}{2}$
which gives $ a_2 \equal{} 3a_1$

do i need 2 tell nething else???

[hide]though kabi's method rely's on the same, this 1 is easier 2 interpret,i hope :) [/hide]

[quote="rituraj007"]
do i need 2 tell nething else???
[/quote]
No! Thanks! I got it! Thanks a lot! :D
well I use the concept of displacement of the strings which I find is easier to do, I hope it is foolproof... :maybe:


modular arithmetic
induction
algebra unsolved
algebra

Let $M$ be a natural number greater than $9$,written in decimal system with digits $1;3;7;9$.Prove that $M$ has at least a prime divisor not less than $11$

$2\nmid M$ and $5\nmid M$ because of the last digit of $M$.
So assume $M=3^{m}7^{n}$ for some $m,n\in\mathbb{N}_{0}$.
But for $a\in\{1,3,7,9\}$ we have $3a\equiv 1,3,7,9\pmod{20}$ and $7a\equiv 1,3,7,9\pmod{20}$.
So by induction $3^{m}7^{n}\equiv 1,3,7,9\pmod{20}$ for all $m,n\in\mathbb{N}_{0}$
and the second last digit of $M$ is even. Contradiction!

[quote="chien than"]$\nmid$
What is it?
Is it $\neq$[/quote]
No, it means "does not divide".
$2,3,5,7$ are the only primes less than $11$ and $M$ must be a product of primes.
If $2|M$, then last digit of $M$ is even.
If $5|M$, then last digit of $M$ is $0$ or $5$.

[quote="olorin"]But for $a\in\{1,3,7,9\}$ we have $3a\equiv 1,3,7,9\pmod{20}$ and $7a\equiv 1,3,7,9\pmod{20}$.
So by induction $3^{m}7^{n}\equiv 1,3,7,9\pmod{20}$ for all $m,n\in\mathbb{N}_{0}$
and the second last digit of $M$ is even. Contradiction![/quote]
What induction? For example $7*73=511$ had digit 5.

[quote="chien than"]CAn you prove that: $3^{m}7^{n}\equiv 1,3,7,9\pmod{20}$ for all $m,n\in\mathbb{N}_{0}$[/quote]
[quote="Rust"]What induction? For example $7*73=511$ had digit 5.[/quote]
What is so hard to understand?
If $a\equiv 1,3,7,9\pmod{20}$ and $b\equiv 1,3,7,9\pmod{20}$, then also $ab\equiv 1,3,7,9\pmod{20}$.
So the set $X: =\{a\in\mathbb{Z}: a\equiv 1,3,7,9\pmod{20}\}$ is closed under multiplication. ($a,b\in X\Rightarrow ab\in X$)
And since $1,3,7\in X$ we must have $3^{m}7^{n}\in X$ for all $m,n\in\mathbb{N}_{0}$.
(This can be done by induction on $m$ and $n$ [b]Rust[/b], a trivial one)

And now if $M$ is a counterexample to the problem, then $2\nmid M$ and $5\nmid M$,
so $M$ has only prime factors $3$ and $7$.
And therefore $M=3^{m}7^{n}\in X$ for some $m,n\in\mathbb{N}_{0}$,
and the second last digit of $M$ is even. Contradiction!

Please explain what you don't understand.
I can't see it. :(

I think Rust read it too quickly and got confused... it made sense to me and I can't find any errors. But I am most likely wrong...

the reason why the Rust example doesn't work is because $73\not = 3 \mod{20}$ so it is not closed.


function
integration
linear algebra
matrix
number theory
prime factorization
college contests

I dont have too much time at hand now, and these computers are not really fun to use either, so can anyone start the topics for me? Well, here are the problems from Category II. At least, I hope these are the problems - I am writing them down from memory, so I may have forgotten something.

[b]Problem 1.[/b]
Let $ E$ be the set of all continuously differentiable functions $ f: [0,1]\to\mathbb{R}$ satisfying $ f(0) = 0$ and $ f(1) = 1$. Find the maximum of $ \int_0^1(1 + x^2)(f^{\prime}(x))^2dx$ for $ f\in E$, and find at least one function $ f\in E$ for which this maximum is obtained.

[b]Problem 2.[/b]
Given positive integers $ p$, $ q$, $ r$ such that $ r$ is odd and $ p^2 = q^2 + r^2$, and some matrix $ A\in\mathbb{Z}^{n\times n}$ satisfying $ p^2A^{p^2} = q^2A^{q^2} + r^2I_n$, where $ I_n$ is the $ n\times n$ unity matrix. Prove that $ |\det A| = 1$.

[b]Problem 3.[/b]
[discuss at http://www.mathlinks.ro/viewtopic.php?t=268467 please]
Let $ a$ be a positive integer. A permutation $ \pi$ of the set $ \{1,2,...,n\}$ will be called [i]vojtechnical[/i] if for every divisor $ d$ of $ n$ and for arbitrary $ i$ and $ j$ from the set $ \{1,2,...,n\}$ satisfying $ d\mid i - j$, we have $ d\mid\pi(i) - \pi(j)$.

For every $ n$, we denote by $ v(n)$ the number of all vojtechnical permutations of the set $ \{1,2,...,n\}$.

Find the prime factorization of $ v(2520)$.

[b]Problem 4.[/b]
Given three integers $ n$, $ m$ and $ k$ satisfying $ 1\leq m\leq n$. Let $ A_1$, $ A_2$, ..., $ A_k$ be $ m$-element subsets of $ \{1,2,...,n\}$ such that for every $ i\in\{1,2,...,k\}$, there exists a partition $ S = S_{i,1}\cup S_{i,2}\cup ...\cup S_{i,m}$ of $ S$ into $ m$ pairwisely disjoint subsets such that $ |A_i\cap S_{i,r}| = 1$ for every $ r\in\{1,2,...,m\}$, but for every $ j\in\{1,2,...,k\}$ there exists some $ r\in\{1,2,...,m\}$ such that $ A_j\cap S_{i,r}| = \emptyset$.
Prove that $ k\leq\binom{n - 1}{m - 1}$.

[b]Problem 5.[/b]
A sequence $ (b_0,b_1,...,b_{s - 1})$ of integers is called [i]nauseating[/i] if for every $ i\in\{0,1,...,s - 1\}$, the number $ b_i$ satisfies $ b_i < s$ and is equal to the number of $ j\in\{0,1,...,s - 1\}$ satisfying $ b_j = i$.
Find all integers $ s$ such that there exists a nauseating sequence $ (b_0,b_1,...,b_{s - 1})$ of $ s$ integers. Prove that every nauseating sequence $ (b_0,b_1,...,b_{s - 1})$ satisfies $ b_{s - 1} = 0$.

I don't have the Category I problems with me now, but it seems like they were much better (what is not that much of an assertion, anyway).

  darij

The problems for both categories from 2009 are available here:
[url=http://vjimc.osu.cz/j19/j19problems1.pdf]Category 1[/url],
[url=http://vjimc.osu.cz/j19/j19problems2.pdf]Category 2[/url].

darij, there were only 4 problems per category :wink:

Well, yeah, the quick appearance of the correct problem sheet and the lack of interest about the VJ on MathLinks have turned my little April fool troll void... I was hoping for someone to dismiss problem 5 as a joke, noticing that [url=http://www.mathlinks.ro/viewtopic.php?t=17457]it has already been too ugly for IMO 2001[/url], and to accept problem 3 instead, which has nothing to do with Vojtech Jarnik (it is my own problem, inspired by [url=http://www.mathlinks.ro/viewtopic.php?t=265566]a posting from t0rajir0u[/url]).

Anyway, as we already know the Category I problems: can anyone solve #3 combinatorially? It is easily seen to be equivalent to
http://www.mathlinks.ro/Forum/viewtopic.php?t=3798
http://www.mathlinks.ro/Forum/viewtopic.php?t=5684
but I have never seen a combinatorial solution of that one (though books claim it exists).

  darij


What is the remainder when $ 6 000 002$ is divided by 9?

Note that $ 6 000 003$ has digit sum 9, so that number is divisible by 9. Since $ 6 000 002$ is just 1 less than $ 6 000 003$, it has remainder $ \boxed{8}$.

ill answer this in a series of steps

[hide="THIS IS MY EXPLANATION"]1.adding all the digits of $ 6000003 $ is 9

2.adding all the digits of $ 6000002 $ is 8 and the 2#'s are 1 apart so the answer is

3.[hide="ANSWER"]8[/hide][/hide]

[hide="."][size=200][color=#0080BF]PLEASE RATE MY POST THANK YOU SO MUCH!!![/color][/size][/hide]

[size=50][b][i][u]please be fair when you're rating my post and please PM if there is a better way to explain this thank you so much!![/u][/i][/b][/size]


A collection F of subsets of a set X is said to have the finite intersection property if the intersection of any finite subcollection {F1, F2, .. Fn} of F is not empty. Show that a topological space X is compact if and only if for every collection {F_(∝ ): ∝∈A} of closed subsets of X which has the finite intersection property, we have the intersection of all F_(∝ ) is not empty for all ∝∈A. A is the set of index.

Yes. But I am still not clear how to do that.

Did you try to pass to the complements and to figure out what the condition means for [b]open[/b] sets? ;)

I think you would need three things.

1. If P then Q <=> If -Q then -P.
2. De Morgan law
3. The fact that only countable intersection/countable union works for open/close sets.

So I guess you'll need to use De Morgan law to change the statement of finite cover property into language of closed sets, then use the reversed inverse to prove it, and use 3. I remember myself proved in this way. Sorry I don't have time to write a detailed proof.

Something like 3. is not needed here.

Anyway, it seems that 3. claims that, say, the countable intersection of open sets is open. But this is not true in general.

You are right. An easy counter example is $ \displaystyle\cap_{i \equal{} 1}^{\infty}(0 \minus{} \frac {1}{i},1 \plus{} \frac {1}{i})$ is $ [0,1]$. I mistook finite with countable. Sorry for the mistake.


I am learning the following pieces:

Bach Chaconne in D-minor
Concerto 2 in D minor by Wieniawski
Paganini's Caprices
Hewbrew Melody

$\text{My list}$

Personally, I love the Mendelssohn Violin Concerto 3rd Mvt.
It is so animated and the theme is very nice. It's really great in the fact that it is very technical. I'm trying to play it (it's not turning out real well.)

Also, the Kabalevsky Violin Concerto Opus 48.
This isn't a really popularly played piece, but this piece sounds difficult to play if you play at tempo, and it really is. There is much to be considered when playing this piece. The theme is really nice for Kabalevsky. Kabalevsky got better as time passed (see Piano Concerto No. 3...it's remarkable.)

The Tchaikovsky Violin Concerto.
In my opinion, the king of all violin concertoes. Phenominal. The technique is beyond my wildest dreams, but the final result of days of practice is great. I'm working on it, but, as with the Mendelssohn, much is left to be desired.

The Brahms Violin Concerto.
My example for an "impossible piece." Something I will never achieve in my life. The chords are overwhelming, though the piece, once put together, is realy nice.

Praeludium and Allegro, by Fritz Kreisler.
Fabulous piece. I've mastered it, as my private teacher suggested it for me. I played it for NYSMMA, my local musical evaluation festival, and got a 98/100, 2 points deducted for sightreading. I got full score on the piece!!! Anyways, the overall expressiveness required to master such a piece is great. I don't understand how people can play this without feeling...you should definitely give this piece a try, no matter what your ability!!!

Two other impossible pieces are the Dance of the Goblins by Bazzini and the Ernst violin Concerto in F# minor. The latter piece is obsolete because it is so hard. I spend time learning it. It did not turn out well.

I am going to learn my favorite piece of all time next week: The Mendelssohn violin concerto.

[quote="fireninja0"]
Caprice No.24, by Paganini
     My favorite piece, this caprice is tricky. The variation with the tenths is particularly hard. Does anyone have any tricks for tenths?[/quote]

The only trick I know is to practice various scales in tenths because the "tenths" variation is simply variations of the A-minor scale.

To start, practice the A melodic, harmonic, and natural minor scales slowly in tenths. I recommed playing chord by chord to ensure proper intonation. Also, try some harder scales such as f# minor and E-minor. In addition, keep practicing thirds and sixths.

Here is another suggestion that my private teacher tells me.
First learn a piece, and then put it away for one year. Next year, the piece will be much easier and sound much better because you will have played more pieces and learned more techniques in the mean time.

Hope this advice helps!

Note: A good recording of the Paganini 24 caprice is the Heifetz recording. Don't be disappointed if you can't play as well as he does. This caprice really takes quite a few years.

[quote="The Sigma"]
Bach Chaconne in D-minor
[/quote]

Oh oh ... good luck. This one is hard like hell to master; I don't know for the violin, but as for the classical guitar, this is extremely hard.

[quote="Rhapsody in Green"]Oh, I'm playing the Wieniawski Polonaise in D too! It's a really nice piece, although not very many people seem to be familiar with it.
I'm also playing:

Bach Presto from Sonata No. 1

Lalo Symphonie Espagnole (first movement)[/quote]
My teacher thought it was too much for me, so he made me drop the Polonaise.  :(  :(
But now he gave me the Zigeunerweisen (Gypsy Airs) by Sarasate! Go figure!  :rotfl:  :huh:
The Gypsy Airs and Paganini Caprice 9 are now my "fun pieces," stuff that I practice a little bit to save for later.

My competition pieces are (in preferred order to play for judges):
Concerto No. 3: Saint-Saens
Perpetuum Mobile: Novacek
Partita No. 1, Sarabande: Bach
Banjo and Fiddle: Kroll

My teacher loves the way I play the Perpetuum Mobile, so it's kind of my "ace in the hole." Whenever I need a contrasting piece, he tells me to bring this piece back. :D

Preludium & Allegro...of course.

The orchestration is nice for the Tchaikovsky Concerto, but I don't like the composition as a whole.

Some of the other small (in length) Kreisler pieces (i.e., Tambourin Chennois).

Bruch Concerto...Sibelius...maybe...

My favorite has to be the Mendelssohn Concerto. No doubt.

But, overall, I'm much more into chamber music (sorry, virtuosi)!

No! Not the Praeludium and Allegro!  :maybe:
I didn't really like that song...maybe it's because I lost a competition with that song...

I think the ultimate program would be:

Poéme by Chausson
Carmen Fantasy by Sarasate

That, my friends, would be the best program ever. Only problem: People don't view non-concerto-y type pieces as "standard" kind of repertoire, even though these two pieces are even harder than  some concertos, so they're not good for audition! :mad:

Why not the bach double?
that always sounds really nice

Heh...I'm the beginner here...

Just finished Sicilienne and Rigaudon...that piece was getting on my nerves...

Finishing up this other piece as well (called "Hejre Kati"...a small piece)

Start recently on Vitali Chaconne (love it).  Will probably start Banjo and Fiddle in a couple weeks, and then supposed to do Paganini Cantabile at some point after that...

Ack. I'm not a fan of Bach Double.

Sarasate Carmen Fantasie
Prokofiev Violin Concerto 1

for me right now.

And now I'm playing Wieniawski F sharp minor.

[quote="fireninja0"]@Funicia: Do you mean the Bach for two violins? Yeah, that's a really nice piece!

@K81o7: Wow, the Vitali Chaconne! That really is a cool piece; I saw a video of Heifetz playing it once! The Banjo and Fiddle has a bit of a steep learing curve (unless you don't do spicatto) and the slow part is really good. You're lucky, you get to play the Cantabile! Weh, I never get pieces like that! :([/quote]
oh yes. asmallgoat plays it all the time.

No i do not! I hate it :wallbash_red: it's SOSOSOSOSO stupid!

It's so overplayed these days... :sleeping:

My current pieces:

Wieniawski Violin Concerto No. 2 in D Minor
Beethoven Violin Concerto in D Major
Bruch Violin Concerto No. 1 in G Minor

Of these three, I think the Bruch is the best. I really love the theme and it's very much like the Tchaikovsky Violin Concerto in that it is strong and very engaging.
But it's hard also. Getting all the chords in tune is quite difficult :(

[quote="speedcuber"]hmm...
im in seventh grade...
i havent been to any competitions.

right now, im playing saint-saens concerto number 3 too!
i just started though, so im on that part with all the sixteenth notes near the bottom of the second page and on the third.[/quote]

I am currently practicing on American Rhapsody, Farandole, Poet and Peasant Overture, Collections from Corpse Bride, The Juggler, and this one other piece that I do not know the name of... :lol:

[quote="hwenterprise"]$ \text{My list}$

Personally, I love the Mendelssohn Violin Concerto 3rd Mvt.
It is so animated and the theme is very nice. It's really great in the fact that it is very technical. I'm trying to play it (it's not turning out real well.)

Also, the Kabalevsky Violin Concerto Opus 48.
This isn't a really popularly played piece, but this piece sounds difficult to play if you play at tempo, and it really is. There is much to be considered when playing this piece. The theme is really nice for Kabalevsky. Kabalevsky got better as time passed (see Piano Concerto No. 3...it's remarkable.)

The Tchaikovsky Violin Concerto.
In my opinion, the king of all violin concertoes. Phenominal. The technique is beyond my wildest dreams, but the final result of days of practice is great. I'm working on it, but, as with the Mendelssohn, much is left to be desired.

The Brahms Violin Concerto.
My example for an "impossible piece." Something I will never achieve in my life. The chords are overwhelming, though the piece, once put together, is realy nice.

Praeludium and Allegro, by Fritz Kreisler.
Fabulous piece. I've mastered it, as my private teacher suggested it for me. I played it for NYSMMA, my local musical evaluation festival, and got a 98/100, 2 points deducted for sightreading. I got full score on the piece!!! Anyways, the overall expressiveness required to master such a piece is great. I don't understand how people can play this without feeling...you should definitely give this piece a try, no matter what your ability!!![/quote]

I like the first movement for Mendelssohn. I recently finished it. Actaully I don't know which you're talking about, since there are many.

I'm working on Sibelius and Bach "Preludio" for contest, but I don't know why I'm even bothering. My hand shakes so hard every audition/competition. It's like all that hard work just disappears. :(

for violin:
Concerto in D Minor by JS Bach

and La Travieta by i forgot who :P

however I wish I could play the violin part for Celine Dion's song To Love You More

I'm playing Wieniawski's 2nd Concerto right now along with Meditation by Tchaikovsky for my audition. Afterwards, I'll either play Zigeunerweisen or Carmen Fantasy by Sarasate.

Mendelsohn's On Wings Of Song. it's so beautiful! plus a bunch of concertos by frederich seitz.

My main solo is Rondo Capriccioso by Saint-Saens. Fairly long and hard, especially the arpeggios :wink: .

Other pieces are Mendelssohn's Concerto in E Minor, working on Mvt. 2 right now. It sounds great and isn't too hard, except for the double stops which I still need practice on.

Bach A Minor 3rd Mvt. (Allegro Vivace), which is actually harder than it seems when played with an orchestra. I plan on writing a cadenza for this before the recapitulation.

Paganini's Motto Perpetuo, even though I haven't worked on that since I "passed it". Great piece for showing off fingerwork if you play it fast though.

My future goal is...Vitali Chaconne?

I have to agree that tarantella looks pretty intimidating.

I also want to learn Zigeunerweisen someday.

Well, I'm playing a bunch:

A Wieniawski Mazurka and a Concerto

A Pierre Rode Concerto

A Ludwig Schpor Concerto

Zigeunerweisen.
(This one is pretty darn difficult... ;])

Mendelssohn Concerto in E minor is pretty nice; currently playing that

I'm also playing Lalo Symphonie Espagnole and Paganini Concerto No.1 and a lot more random songs

But I feel my teacher might drop paganini soon... since i'm not exactly playing it well

wahh u guys r all so good!!

current piece- bach concerto in a minor 3rd movement.
next piece- mozart no.3 in who-knows-what major/minor  :D

I play a Tartini (have you heard of him?? :P) and a Vivaldi. The latter is pretty dificult...  :(

Right now I'm working on Wieniawski's Concerto No. 2. :D It is probably the first full concerto I've worked on. (For example, I learned the first and third movements of the Kabalevsky concerto but not the second.) My favorite movement is the third one (which I'm learning right now) because of the part with all the spiffy double stops :D:D:D

Anyone have any tips for up-bow staccato? I'm working on the [i]molto moderato[/i] section with a bunch of up-bow staccato, and I fudge it up every time xD

I reallyreallyREALLY want to learn Zigeunerweisen (sheesh, I had to look up the spelling xD). And the Carmen Fantasy. And the Lalo Symphonie Espagnole. And... uh, I'll stop there. lol.

Zigeunerweisen. Left-hand pizzicato is a pain though.

[quote="Squarerootofisoscelestria"][quote="fireninja0"]This is where you can post any violin pieces that you are currently playing. Comment on your pieces!

My list is...

Polonaise in D, by Henri Wieniawski
     This piece is tricky because of the G-string and E-string string-crossings. I like the middle section because of the expressivo on the G string.

Banjo and the Fiddle, by Kroll
     This piece is also tricky, but fun. I like plucking the pizzicato sections, but I have trouble transitioning from pizzicato to arco.

Concerto No. 3, by Saint-Saens
     Another great piece. Again, I like the expressivo on the G-string. This piece was written for cello, then was transposed to violin.

Caprice No.24, by Paganini
     My favorite piece, this caprice is tricky. The variation with the tenths is particularly hard. Does anyone have any tricks for tenths?[/quote]

Oh a trick for the tenths...
Well you kinda got to stretch your hand toward the direction of your pinky.  Keep your pinky real straight and try not to tighten your index finger.
It gets kind of easier if you do scales with tenths.  My teacher made me do it  :mad: It takes some time.

I'm working on Paganini Caprices #1,6,and 15. (Finished 5,11,13,14,16)
Caprice #1 is driving me mad :cursing: .  Any suggestions on the ricochet?[/quote]

For 10ths, I stretch out my hand all the way and just hope it's long enough  :(

I want to learn Zigeunerweisen. Just finishing Bruch Concerto 3rd movement and gonna go into Mendelssohn soon.


[quote="N.T.TUAN"]What will occur if $(x+a)^{4}+(x+b)^{4}=c$? :D[/quote]
Denote $y=x+\frac{a+b}{2}$ :)

[quote=Possible][hide=Don't Do This]Bash using quartic formula[/hide][/quote]

[hide = Don't Do This] Bash using WolframAlpha[/hide]

[quote=ilovepizza2020][quote=Possible][hide=Don't Do This]Bash using quartic formula[/hide][/quote]

[hide = Do This] Bash using WolframAlpha[/hide][/quote]

FTFY :P

It can be written as $$2 \left(x^2-2 x-1\right) \left(x^2-2 x+9\right)=0$$

[quote=4everwise]Find all real solutions to $x^{4}+(2-x)^{4}=34.$

I really liked this problem so I thought I'd post it. :P
[hide="Approach"]Let $x=y+1.$ Equation becomes $(y+1)^{4}+(y-1)^{4}=34,$ and we're left with a quadratic in $y^{2}$! :D

I believe this works for any such quartic, where the two constant terms differ by an even number.  :) [/hide][/quote]

You found this in Intermediate Algebra Last Chapter Symmetry?

(S)he posted it before Intermediate Algebra released.

Oh, then @OP may have found it in HMMT, which is the source of a similar problem.

[quote=OlympusHero][quote=4everwise]Find all real solutions to $x^{4}+(2-x)^{4}=34.$

I really liked this problem so I thought I'd post it. :P
[hide="Approach"]Let $x=y+1.$ Equation becomes $(y+1)^{4}+(y-1)^{4}=34,$ and we're left with a quadratic in $y^{2}$! :D

I believe this works for any such quartic, where the two constant terms differ by an even number.  :) [/hide][/quote]

We would like to get some symmetry. We note that if we let $x=y+1$, we get lots of symmetry! We have $(y-1)^4+(y+1)^4=2y^4+12y^2+2$, or $y^4+6y-16=0=(y^2+8)(y^2-2)$. We are working in the reals, so $y = \pm \sqrt{2}$. Therefore, $\boxed{x = \pm \sqrt{2} + 1}$.[/quote]

How did you go from 2y^4+12y^2+2$, to $y^4+6y-16?

@above They divided both sides by 2 and then subtracted 17 from both sides


trigonometry
limit
calculus
calculus computations

$ \frac{\text{sinc}(2x)}{\text{sinc}\, x}=\frac{\frac{\sin(2x)}{2x}}{\frac{\sin x}{x}}=\frac{\sin(2x)}{2 \sin x}=\cos x$
$ \Rightarrow \prod_{k=1}^{n}\cos\left(\frac{x}{2^{k}}\right) =\prod_{k=1}^{n}\frac{\text{sinc}\left(\frac{x}{2^{k-1}}\right)}{\text{sinc}\left(\frac{x}{2^{k}}\right)}=\frac{\text{sinc}\left(\frac{x}{2^{0}}\right)}{\text{sinc}\left(\frac{x}{2^{n}}\right)}\to \text{sinc}\, x$


prove that :
$ n $ is a prime number if and only if  $ \sum ^{n} _{i=1}[\frac ni]-\sum ^{n-1} _{i=1}[\frac {n-1}{i}]=2$

Try to show that [code]\sum ^{n} _{i=1}[\frac ni]-\sum ^{n-1} _{i=1}[\frac {n-1}{i}] [/code]   equals the numberof divisors of n.it is so easy. :D


calculus
derivative
integration
geometry
vector
Gauss
linear algebra

you know how college calculus is based on 4 quarters and in high school you can go over 3 (AB/BC)
i know AB is derivative and basic integration
and     BC is adv integration, polar/parematric, arclength, surface area, and series
what about D (4th quarter)
is there any E, F G etc?

and for second year, what do most people take for math? is there like a calculus II or do you go to linear algebvra?

There's vector calculus: lines, planes, vectors, dot and cross products (which isn't really calculus but is conventionally included); partial derivatives, mutli-variable chain rules, and Lagrange multipliers; multiple integrals; parametric equations revisited in vector terminology, including curvature; line and surface integrals; theorems of Green, Gauss, and Stokes.

Look at the back third of the full-length version of a popular calculus textbook (Stewart, Edwards & Penney, etc.) to get the idea.

American campuses typically think of courses by term (semester or quarter) rather than by year. It's typical for a semester campus (like ours) to put this material as a third semester, called Calculus III. (AP Calculus BC is Calculus I and Calculus II.) Quarter system campuses often have two quarters of vector calculus.

Linear algebra and differential equations are also typical second-year courses, often indepentent (in prerequisite terms) of the vector calculus course; some places bundle these together as one course. Vector calculus, linear algebra, and differential equations is a common package of courses for a wide variety of technical majors. Computer science and math majors are often also directed to courses that are varying mixtures of "discrete structures" and "transition" ("how to prove things") courses.

The most selective and highly prestigious institutions are the most likely to have ideosyncratic courses that only they offer.

a lot of high schools dont of bc calc b/c they dont have teachers that can teach, or they dont have enough students to fill up a whole class.  so for further semisters i dont know if high schools could do it, also college board would have to offer exams for those latter sections, which i dont think they do.


[quote="frt"]Solve the system for positive reals $x$ and $y$:

$y^3=x^2+x-1$
$x^3=y^2+y-1$[/quote]

Substract second from first and we have:
$(y-x)(y^2+xy+x^2)=(x-y)(x+y)+(x-y)$
$(y-x)(y^2+xy+x^2+x+y+1)=0$

We have two cases:
1) $x=y$ and 2) $y^2+xy+x^2+x+y+1=0$

1) $x^3=x^2+x-1$
$x^2(x-1)-(x-1)=0$
$(x-1)(x^2-1)=0$
$x=1, x=-1$

Since $x$ is positive only $x=1$ satisfy conditions.

2) $x^2+x(y+1)+y^2+y+1=0$
$D=(y+1)^2-4(y^2+y+1)=-(3y^2+2y+3)$

Since $D$ must be $\geq0$ so $3y^2+2y+3\leq0$ but this inequality doesn't have solutions.

So the only solution is $(x,y)=(1,1)$

Is it true that for all functions that are not odd, intersections with its inverse only occur along the line $y=x$?

If it is, we can just realize that $f(x)=\sqrt[3]{x^2+x-1}$ is not odd, thus $x=y=k$ so the system becomes $k^3=k^2+k-1 \iff k(k^2-1)-(k^2-1)=0 \iff (k-1)(k^2-1)=0$, so $k= \pm 1$.Checking, both of these work so our solutions are $\boxed{ (\pm 1, \pm 1)}$.

[Edit: Minor changes... for which functions does the first statement not apply?]

[quote="cincodemayo5590"][quote="t0rajir0u"][quote="cincodemayo5590"]Is it true that for all functions that are not odd, intersections with its inverse only occur along the line $y=x$?[/quote]

This only holds for functions that are either strictly convex or strictly concave, I think.[/quote]

are you sure? because it seemed to hold for the function $f(x)=\sqrt[3]{x^2+x-1}$, which isnt strictly concave/convex, as i look at my grapher...[/quote]

I mean in general.  It may hold in specific for other functions, but you can only automatically show that it holds for strictly convex and strictly concave functions.  :)

Edit:  Okay, I suppose it would make more sense to change the criterion:

Iff $f(x) - x \le 0 \forall x$ or $f(x) - x \ge 0 \forall x$ then no intersection points off of $y=x$ exist.

let $f(x)=\sqrt[3]{x^2+x-1}$

then our system is

$x=f(y)$ and $y=f(x)$

--
Lemma 1: for $x\ge -1$, we have that $x\ge f(x)$

$x^3-x^2-x+1\ge 0$
$x^2(x-1)-(x-1)\ge 0$
$(x+1)(x-1)^2\ge 0$ which is true
--

--
Lemma 2: for $x\ge 0$, we have $f(x)\ge-1$
$x^2-x-1\ge1$
$x(x-1)>0$ which is true using the domain
--

looking at the system, $x=f(f(x))$, and by Lemma 1,

$x\ge f(x)$

by Lemma 2, $f(x)\ge -1$ for $x\ge 0$, so we can apply Lemma 1 on $f(x)$ to get:

$f(x)\ge f(f(x))$

but then $x\le f(x)$ by substitution of $x=f(f(x))$

so combining those, $x=f(x)$, then implying $x=y$ using the original system...so

$x^3-x^2-x+1=0$, $x=\pm1$, then $(x,y)=(-1,1), (1,1)$

p.s. frt great problem, where do you find such nice problems? :)

[quote="Altheman"]p.s. frt great problem, where do you find such nice problems? :)[/quote]
Thanks! Actually, I made this problem :lol:
I basically tried to make a "nice" system, and I made some adjustments until the results turned out nicely :)
(My original solution was similar to eLeM's solution.)

The problems that I post in GS/Intermediate are from various resources (math competitions, Internet, some Japanese textbooks, etc...), but some of them are my original.


A spherical glass globe is inscribed in a cylinder. Latitude and longtitude lines are painted on the globe. A light source at the center of the globe projects an image of the lines onto the cylinder. If the radius of the globe is 12 cm, find the area (on the cylinder) of the image of the region bounded by the lines of 0 and 45 degrees latitude and 0 and 30 degrees longtitude.

IF you draw it, you'll find that the height of this region is 12 cm.  The length (of the rectangle, if you flattened it out) is $2\pi$, being 1/12 the radius.  Therefore, the area of this region is $24\pi$.


Show that for $ n\geq 6$ a square can be dissected into n smaller squares, not necessarily all of the same size.

[hide="Almost full solution"]
If we can dissected a square into $ n$ smaller squares, then we can dissected into $ n+3$ smaller squares (then we must dissected one square into $ 4$  identical parts).
And now we must find solution for e.g. $ n=4,6,8$ (and it isn't hard ;) ) then we'll show that what we want :) [/hide]


function
analytic geometry
trigonometry
algebra
polynomial

Just wondering, is there a way to write the polar function $r=a\theta$ as a function in the Cartesian coordinate system? I got as far a $y=x\tan(a^2+x^2+y^2)$ and got sidetracked. or maybe that isn't right at all. I don't know...

Wait a sec, that's not right, it's $y=x\tan^2(\frac{x^2+y^2}{a^2})$. Either way, it's not fully simplified. And it's probably wrong.

[quote="Moubinool"]The curve
Arc length and polynomial Legendre

http://mathworld.wolfram.com/ArchimedesSpiral.html[/quote]

Umm, yes, but that's not what I'm looking for. I'm looking for a Cartesian function in the form $f(x)=...$ (For example, the kind of function you can graph with graphing calculators)

My derivation goes like this:

$r=a\theta$
$\sqrt{x^2+y^2}=a\tan^{-1}(\frac{y}{x})$
$x^2+y^2=a^2\tan^{-2}(\frac{y}{x})$
$\frac{x^2+y^2}{a^2}=\tan^{-2}(\frac{y}{x})$
$y=x\tan^{2}(\frac{x^2+y^2}{a^2})$

From there you can use various identities and eventually get an equation of the form $y\tan^{2}{y}=...$, but the question is, can you then simplify it to an equation of the form $y=...$?

Please note that my method may be right, wrong, or somewhere in between.

I believe you got confused by the meaning of $\tan^{-1} \theta$.  What that is is an inverse function, not the inverse of a number.  hehe

By the way, I doubt you can have a "y=" thing because there are infinitely many values of y for any given finite x.  You could use your calculator's polar graphing capabilities...

probability1.01 is right.

$(\tan^{-1}{x})^2 \not = \tan^{-2}{x}$.

That's probably why we have $\cot{x}$ to represent the reciprocal of $\tan{x}$.


function
number theory solved
number theory

how duz one do this:
Define the sequence a_1, a_2, a_3, ... by a_1 = 1,
a_n = a_(n-1) + a_[n/2]. Does the sequence contain infinitely many multiples of 7?

Wow! This is really tough. I'm sure the answer is "yes", because I investigated this for about 400 numbers using Mathematica, but nothing I've tried so far worked. Aren't the solutions available somewhere?

This looks like a very promising source for an obsession :).

First note that a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 5 and a_5 = 7. Then there exists i such that a_i = 0 mod[7].

Suppose, for a contradiction, that there is a finite (but nonzero) number of i such that a_i = 0 mod[7].
Let k be the greatest subscript such that a_k = 0 mod[7]. Note that k > 1.
Then, from the recurrence formula, we easily see that
a_(2k+1) = a_(2k) = a_(2k-1) mod[7]. Let x be this common value mod[7]. Note that x =/= 0 mod[7].
Let y = a_(4k-3).
Then, we easily see that :
a_(4k-2) = y + x mod[7]
a_(4k-1) = a_(4k-2) + x = y+2x mod[7]
a_(4k) = y+3x mod[7]
a_(4k+1) = y+4x mod[7]
a_(4k+2) = y+5x mod[7]
and a_(4k+3) = y+6x mod[7].
Since x =/= 0 mod[7], and 7 is a prime, we deduce that the numbers 0,x,2x,3x,4x,5x,6x give each of the class modulo 7. Thus, one of the a_i's, where i = 4k-3, 4k-2,...,4k+3 is divisible by 7. Since for such a i we have i \geq  4k-3 > k, this contradicts the maximality of k.

Pierre.

Thanks! I guess you've spared me the inconvenience of an obsession! :D:D


What is the probability that a randomly chosen partition of the number $50$ has a prime number in it.
A partition is a collection of one or more positive integers with repetitions allowed. Eg. partition of 5 is $\{2,3\}$, $\{1,1,3\}$ etc.
Generalization:

What is the probability that a randomly chosen partition of $n$ has a prime number in it.

[hide]Well, there are only about 204226 partitions.

NOw all we need to do is write 'em all out and then find the number, n, then the answer is

$\frac{n}{204226}$, and simplify if necessary.[/hide]

Now that was an enlightening post. \sarcasm...

Could anyone expalain how mathgeniuse^ln(x) found that there were 204226 partitions?

He used either
1) Google
or
2) [url=http://mathworld.wolfram.com/PartitionFunctionP.html]These[/url] formulas.[/url]

Generating functions to solve the problem although it needs a computer

I rememberd that 50 had 204226 partitions, so yeah.

Let's see...

The generating function for partitions is $\prod_{n\in\mathbb{N}}\frac{1}{1-x^{n}}$

We'd subtract each prime less than 50 from 50, take the partitions of those, and add them together to get the numerator.

The generating function for that is

$\sum_{p\in\mathbb{P}}\prod_{n\in\mathbb{N}}\frac{x^{p}}{1-x^{n}}$

So then you just take the coefficient of $x^{n}$ in the second over the coefficient of $x^{n}$ in the first.

Is there a nice way to do this?

There is no formula for it, so I don't think there is a nice way. :maybe:

[quote="JBL"]Hamster, I think what you suggested double-counts a large number of things that you remove, doesn't it?  The correct generating function for the no-primes-partitions would be $\prod_{n \in S}\frac{1}{1-x^{n}}$ where $S$ is the set of positive integers excluding the primes.

The sequence is at http://www.research.att.com/~njas/sequences/A002095[/quote]
I was thinking of something along those lines.
Does #H34N1 want to comment about if he had a solution in mind or just pulled this out of thin air?

[quote="JBL"]Hamster, I think what you suggested double-counts a large number of things that you remove, doesn't it?  The correct generating function for the no-primes-partitions would be $\prod_{n \in S}\frac{1}{1-x^{n}}$ where $S$ is the set of positive integers excluding the primes.

The sequence is at http://www.research.att.com/~njas/sequences/A002095[/quote]

Ahh you're right. I double counted everything with 2 primes, triple with 3, etc. Thanks for pointing that out.


calculus
integration
function
geometry
rectangle
real analysis
real analysis unsolved

Give an example of a function $ f : A \times B \to \mathbb{R}$, where $ A$ is a rectangle in $ \mathbb{R}^k$ and $ B$ is a rectangle in $ \mathbb{R}^n$, such that $ \int_{A \times B} f$ exists but one of the integrals
\[ \int_{x \in A} \int_{y \in B} f(x,y)\]
and
\[ \int_{y \in B} \int_{x \in A} f(x,y)\]
exists and the other does not.

I've tried variations on the modified Dirichlet (i.e. "popcorn") function to no avail.

Okay, we can't have this as stated; if $ \int_{A\times B} f$ converges absolutely, the iterated integrals exist and are equal to each other and to the double integral. That double integral just doesn't make sense without absolute convergence.

On the other hand, we can drop that condition. Can you find a (reasonably nice) function $ f$ on a product of intervals $ A\times B$ so that $ \int_A\int_B f(x,y)\,dy\,dx$ exists and $ \int_B\int_A f(x,y)\,dx\,dy$ doesn't?

[quote="jmerry"]...a (reasonably nice) function [/quote]
Just don't take "reasonably nice" to restrict you to a single formula. Piecewise defined functions may be easier to manipulate.

My favorite example is almost a step function. Chop the unit square up into non-overlapping rectangles in a recursive way, with the rectangles alternating between a $ 1\times 2$ shape and a square shape, getting smaller and smaller as you approach one corner of the unit square. Assign constant values for the function on each sub-rectangle, with all of the values being $ \pm$ some power of 2, half plus, half minus. Fiddle with the numbers until you make it work the way you want it to.

This isn't technically a step function, or even a simple function, because it takes on infinitely many values. But each cross-section (except for the very edges) is a step function. Computing the interated integrals is just a matter of careful bookkeeping.


It was a dark and stormy night. All of the AoPS mafia moderators were huddled up in a single room in Alpine, CA, situated at the AoPS headquarters trying to figure out the flavor for the next game in the progression, 46. It turned out that james4l was going to mod this game.

“Hmmm...” mused Brut3forc3, who happened to mod Game 38. “Why don’t we play a meatworld game with all the moderators?”
“What?” cried Peter. “I am not going to allow james4l to mod another 30 player game! He is extremely inconsistent! He didn’t even show the townie PMs! And seriously now... meatworld with 30 people?”
“Wait, wait, wait. What if townie PMs were not a problem anymore?”, james4l offered.
Just then, a knock came in the door. It was locked – both inside and outside. A shout which later was identified as Valentin Vornicu shouted, “GREAT NEWS! WE’VE INSTALLED A NEW ARTIFICIAL INTELLIGENCE IN EACH TOPIC! THE TOPICS NOW THINK FOR THEMSELVES!”
It was almost unheard due to the webmaster’s raved shouting, but just then, a bolt of lightning hit the building, wiping out all of AoPS’s servers, except twenty mafia games, each imbued with the artificial intelligence module.
Even the navigational interface was unresponsive. The thirty-plus moderators could only wait, as the door locking system was controlled by the disks themselves. Only time would tell as to which side would reign the disk at the end.
Day 1 Starts Now.

RULES RULES RULES (Please read)
[list]
    [*] Votes must be in bold. If you do not bold your vote, it will not be counted.
    [*] Once a player has reached lynching state, his pleas are useless and any attempts to unvote will be unheeded. You may vote: no lynch – majority votes of this kind are necessary to end the day without a death.
    [*] The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
    [*] Once your death scene has been posted, you’re dead. Stop typing. The only thing you can type is a "bah" post.
    [*] Don’t edit/delete previously submitted posts. If you must, post again with EBWOP (Edit by way of post).
    [*] Don’t quote any PMs from me. Neither true, nor false. Seriously. To ensure rule-reading, post a different word with two "k" s.
    [*] If you have a night choice to make, it is due by the posted deadline. I will not wait for you – if you do not submit a choice to me, sorry - [b]no choice or a random choice will be made, depending on what is applicable.[/b]
    [*] Treat this game as a commitment. Be considerate – don’t leave us hanging. [b]If the moderator considers this to disrupt game flow, YOU MAY BE REPLACED.[/b]
    [*] Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.
[*] You win if your group is the last group standing.[/list]

In bold are rule changes / additions.

[u]Still Alive (20)[/u]
1) AIME15
2) PowerOfPi
3) jjfun1
4) spaceguy524
5) mathfreak123
6) isabella2296
7) SuperNerd123
8) AceOfDiamonds
9) mathelete
10) cf249
11) Bugi
12)monkeygirl13
13)15!!!
14)pythag011
15) meewhee009
16) dragon96
17) zapi2007
18) binettheorem
19) ernie
20) randomguy64

[u]Deleted From Registry (0)[/u]

First Confirmer.

You should have started the game like:

"On a dark and stormy night Snoopy suddenly saw a flash of lightning. Ten seconds later he heard the sound of thunder. The speed of sound is 1088 feet per second and one mile is 5280 feet. Estimate, to the nearest half-mile, how far Snoopy was from the flash of lightning."

The answer is 2, which I believe is a (D).

Confirm. "a different word with two "k" s" (Am I supposed to post the phrase, or a word? In the latter case, "kick".)

Addition after preview: Darn, dragon beat me.

now, it's a little too obvious

*knocks on wood*

confirm!

[quote="binettheorem"]Confirm. What are you guys doing?[/quote]

Please read instructions, thank you.

Am I allowed to edit before the game even starts?

EDIT: I didn't actually edit this post.

I confirm kookily.

This game has the potential to be the best mafia game ever.  I want to be blown away by awesomeness!

[quote="dragon96"]Oops. Please don't send any kamikaze suiciders after me, or else I will make a kraken eat you up. Or don't make me wear a kamelaukion which was sent by Kinko's. Whatever you do, do not make me perform a kathakali. If you do, I will call you a Kallikak. (pwn) However, I am fine if you ask Kriskovitch - the teacher that always wears khaki pants - to make me run 3 cross countries, since I do that every week anyways. KWinK![/quote]

Ow. Mod vs mod? :P

[quote="pythag011"]/confirm.

I would like to know everyone's thoughts on a NUMBER CLAIM.[/quote]

Read the first posts. (and the rest too) kthanks.

Konfirmk. So we are mafia games... Hm...

I despise kicking soccer balls.

[b]Confirm.[/b]

I knew it...

I just knew you were scum, AIME...but then I let it go :P

Well played, Mafia.

[url=http://www.quicktopic.com/43/H/rp4J3k2dwrxY]Here[/url] is the Mafia QT.

aejrblekarutyapeiurtartlawrgajevndbfnvacajehpaoehparthyjkahkjzzj rkjgbdngbdznrbglkejrhtlakjethlykjrhtlkjsrtsjjk?

Really?

Wow...

Notes on PR:

Vigenere Cipher: A simple numeric substitution would have easily bypassed this restriction.
Element Cipher: Fortunately a player ingame invented the cipher? :P
Wordle: I can think of putting messages in the filename, the metatags if jpg, or photo comments.

Oh wow, I should have thought of that.

What, like 1=a, 2=b, ...? :wallbash_red:

I was a pretty bad sk.
I targeted the town too often and should have been looking for more balanced sides...

oh well.

Well, you targeted me which was good.

Why? Because I was Hated and I messed up the PR too much.

Every time I broke the PR, I would take one less vote to lynch; I think I would have been auto-lynched at some point. Such as during lylo.

Although it didn't really help :P

GOOD JOB TOWN!!!

Instead of listening to AoD and 15 (and later me), you left mathfreak.
mg13 was always suspicious...

@AIME15: hate you :P you gave the idea of giving me PR
wow people why me???

And also, if you managed to kill pythag on time (N1), my doc abilities would stop it hahahaha

[quote="monkeygirl13"]Wordle pr was a great idea! pythag stopped talking.
We where lucky someone else killed pythag.
I'm surprised that you guys let me live so long. I was scummy from day 1.[/quote]

PRs in general are not good ideas, especially when they give people who stop caring about the game due to lack of scumhunting an excuse to get away with it.

I don't understand why nobody noticed my "Vote: AIME15, Unvote"

I did  :lol: See the [url=http://www.quicktopic.com/43/H/rp4J3k2dwrxY]Mafia QT[/url].

Actually that was the reason I wanted to put PR on you the next day also.

When is the next Mafia game going to open? That way I know when to sign up.

Check the sign-ups thread. Make sure not to make the same mistakes as in this game (hiding text, etc)

lastpost (this is used to preserve order of locked topic)


How to prove by contradication that  $ a^{2} \plus{} b^{2} \neq c^{2}$ where a, b and c are prime

Pythagorean equation

Suppose there is a triple
$ a^2\equal{}(c\minus{}b)(c\plus{}b)$

So we have only one possibility: $ c\minus{}b\equal{}1,c\plus{}b\equal{}a^2$

But the only primes that differ by 1 are 2 and 3, and then $ a\equal{}\sqrt{13}$, contradiction

$ a^{2} \plus{} b^{2} \equal{} c^{2}$

If $ a$ and $ b$ are odd, then so are $ a^{2}$ and $ b^{2}$, so $ c^{2}$ is even, so $ c$ is even. So all three can't be odd.

Now if $ c$ is even, $ c^{2} \equiv 0 (mod 4)$ and thus so are $ a^{2}$ and $ b^{2}$

If $ a$ is even but both $ b$ and $ c$ are odd, then $ b \equal{} 2m \plus{} 1$ and $ c \equal{} 2n \plus{} 1$.

$ a^{2} \equal{} c^{2} \minus{} b^{2}$

$ \equal{} 4n^2 \plus{} 4n \minus{} (4b^{2} \plus{} 4b)$

$ \equal{} 4c(c \plus{} 1) \minus{} 4b(b \plus{} 1)$

Therefore $ 8$ is a factor of $ a^{2}$, so $ 4$ is a factor of $ a$.

So $ a$ cannot be $ 2$.

Oops, just seen Bugi's neat proof.


DB and DC are two tangents to the circumcircle of the triangle ABC.line L  is tangent to the circle in point A.and line L'

is paralell to L,and goes through D.prove that:

if L' intersects AB,CD in pointsE,F,then D will be the midpoint of EF.

Let the intersectiong of $L$ and $DB$, $DC$ be $M$ and $N$ respectively. Since $L || L'$, so:
\[ \dfrac{ED}{MA}=\dfrac{DB}{BM} \text{and: } \dfrac{NA}{DF}=\dfrac{NC}{DC} \]
Multiply them togather, and notice that $DB=DB$, $CN=NA$, $MA=MB$, we get $\dfrac{ED}{DF}=1$.

thanks to all u posts .im glad because u solve the problems in a very easily way.but i cant solve them easily.

i should use pole and polar and harmonic and orthogonal circles and conjugates points and .... :|

I found this proof because I knows nothing about pole and polar and harmonic and orthogonal circles and conjugate points...


$ ABC$ is a triangle and $ A^\prime$, $ B^\prime$, and $ C^\prime$ are points on $ BC$, $ CA$, and $ AB$, respectively. Given that:

$ AA^\prime, BB^\prime, CC^\prime$ concur at $ O$, and

$ \dfrac{AO}{A^\prime O} \plus{} \dfrac{BO}{B^\prime O} \plus{} \dfrac{CO}{C^\prime O} \equal{} n$

Find, with proof, the value of:

$ \dfrac{AO}{A^\prime O}  \times \dfrac{BO}{B^\prime O}  \times \dfrac{CO}{C^\prime O}$

:noo:

[hide=" :noo: "]

Assign masses of $ a$, $ b$, and $ c$ to the points $ A, B, C$, respectively.

Then $ O$ has a mass of $ a \plus{} b \plus{} c$.

By the given,

$ \dfrac{AO}{A^\prime O} \plus{} \dfrac{BO}{B^\prime O} \plus{} \dfrac{CO}{C^\prime O} \equal{} n$

$ \Rightarrow \dfrac{b\plus{}c}{a} \plus{} \dfrac{c\plus{}a}{b} \plus{} \dfrac{a\plus{}b}{c} \equal{} n$

$ \Rightarrow \dfrac{b^2c \plus{} bc^2 \plus{} c^2a \plus{} ca^2 \plus{} a^2b \plus{} ab^2}{abc} \equal{} n$

$ \Rightarrow \dfrac{b^2c \plus{} bc^2 \plus{} c^2a \plus{} ca^2 \plus{} a^2b \plus{} ab^2}{abc} \plus{} 2 \equal{} n \plus{} 2$

$ \Rightarrow \dfrac{b^2c \plus{} bc^2 \plus{} c^2a \plus{} ca^2 \plus{} a^2b \plus{} ab^2 \plus{} 2abc}{abc} \equal{} n \plus{} 2$

$ \Rightarrow \dfrac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc} \equal{} n \plus{} 2$

$ \Rightarrow \dfrac{b\plus{}c}{a} \times \dfrac{c\plus{}a}{b} \times \dfrac{a\plus{}b}{c} \equal{} n \plus{} 2$

$ \Rightarrow \dfrac{AO}{A^\prime O}  \times \dfrac{BO}{B^\prime O}  \times \dfrac{CO}{C^\prime O} \equal{} n \plus{} 2$

[/hide]

:noo:

[hide="Solution"]Let $ AC'\equal{}a$
$ BC'\equal{}b$
$ BA'\equal{}c$
$ CA'\equal{}d$
$ CB'\equal{}e$
$ B'A\equal{}f$

By Menelaus on $ AC'BCA'O$, we have that $ \frac{AC'}{C'B}*\frac{BC}{CA'}*\frac{A'O}{AO}\equal{}1$
Similarly, $ \frac{BA'}{A'C}*\frac{CA}{AB'}*\frac{B'O}{OB}\equal{}1$
and $ \frac{CB'}{B'A}*\frac{AB}{BC'}*\frac{C'O}{OC}\equal{}1$

In terms of our notations: $ \frac{a}{b}*\frac{c\plus{}d}{d}*\frac{A'O}{OA}\equal{}1$, etc.

First, we find the values of $ \frac{AO}{A'O}$  and the other ratios using these relations.

$ \frac{AO}{A'O}\equal{}\frac{a}{b}*\frac{c\plus{}d}{d}$

$ \frac{BO}{B'O}\equal{}\frac{c}{d}*\frac{e\plus{}f}{f}$

$ \frac{CO}{C'O}\equal{}\frac{e}{f}*\frac{a\plus{}b}{b}$

We now wish to find the product of these three, which is $ \frac{a}{b}*\frac{c}{d}*\frac{e}{f}*\frac{a\plus{}b}{b}*\frac{c\plus{}d}{d}*\frac{e\plus{}f}{f}$

However, we know by Ceva's on triangle ABC that $ \frac{a}{b}*\frac{c}{d}*\frac{e}{f}\equal{}1$

Thus, we need to find $ \frac{a\plus{}b}{b}*\frac{c\plus{}d}{d}*\frac{e\plus{}f}{f}\equal{}\frac{acf\plus{}bce\plus{}bcf\plus{}ade\plus{}adf\plus{}bde\plus{}bdf\plus{}ace}{bdf}$

Since we know the sum of the three ratios is $ n$,

$ \frac{a}{b}*\frac{c\plus{}d}{d}\plus{}\frac{c}{d}*\frac{e\plus{}f}{f}\plus{}\frac{e}{f}*\frac{a\plus{}b}{b}\equal{}\frac{acf\plus{}adf\plus{}bce\plus{}bcf\plus{}ade\plus{}bde}{bdf}\equal{}n$ (by combining into one fraction)

Note that this differs from the product by $ \frac{ace\plus{}bdf}{bdf}$

But by Ceva's, we know that $ ace\equal{}bdf$, so $ \frac{ace\plus{}bdf}{bdf}\equal{}2$, and our final answer is $ \boxed{n\plus{}2}$[/hide]

Hmm forgot to mention that my solution was hidden in the second :noo: . It's quite similar to yours (@resurrection) except the

[hide="Spoiler 1"] mass points [/hide]

mask/simplify all the

[hide="Spoiler 2"] Menelaus [/hide]


The arithmetic mean of nine numbers is 54. If two numbers $ u$
and $ v$ are added to the list, the mean of the eleven-member list
becomes 66. What is the mean of $ u$ and $ v$?

If the arithmetic mean of nine numbers is fifty-four, then the sum is 486. If the mean of 11 numbers is 66, then the sum is 726. Therefore, the mean of $ u$ and $ v$ is $ \frac{726 \minus{} 486}{2} \equal{} 120$.