limit
real analysis
real analysis theorems

[quote="romano"]Let $ (x_n) , (y_n) $ be two sequences satisfying :
 i) $ (x_n) $ strictly increase to infinity
 ii) $ \lim_{n\to \infty}\frac{x_n-x_{n-1}}{y_n-y_{n-1}}=g $
  Prove that : $ \lim_{n\to \infty}\frac{x_n}{y_n} = g $
 (do not use Toeplitz theorem !)[/quote]
Do you mean $y_n \nearrow +\infty$?
And what's Toeplitz's Theorem?

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=4215

[quote="romano"]Let $ (x_n) , (y_n) $ be two sequences satisfying :
 i) $ (x_n) $ strictly increase to infinity
 ii) $ \lim_{n\to \infty}\frac{x_n-x_{n-1}}{y_n-y_{n-1}}=g $
  Prove that : $ \lim_{n\to \infty}\frac{x_n}{y_n} = g $
 (do not use Toeplitz theorem !)[/quote]

actually Stolz theorem says that $(y_n)_n$ must be a strictly increasing(decreasing) sequence which tends to infinity (zero)...

Oh , you don't know Toeplitz theorem ???? Maybe , in your country , it's called different name .
   I have a question : Does there exist the inversion of Stolz theorem ??

yes.. but it's not exactly [b]reciproc[/b].. you need some more condions...
$(x_n)_n$ and $(y_n)_n$ are two sequences of real numbers with the properties:
(i) $(y_n)_n$ strictly increasing and unbounded
(ii) exists $\lim_{n\to\infty}\frac{x_n}{y_n}=l\in\overline{\mathbb{R}}$
(iii) exists ${\lim\frac{y_n}{y_{n+1}}=y\in\mathbb{R}_+\{1}$

Then there exists $\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=l$.

It's $y_n$ which should be increasing and unbounded (or decreasing and convergent to 0). And the reciprocal includes the condition that $\lim \frac {b_n}{b_{n+1}}\neq 1$.

There is a likely theorem.
It is:Let the sequence $x_{n}$ monotinically tends to $0$.Let $y_{n}$ is a sequence
s.t. $y_{n}{\rightarrow}{0}$ and $\lim_{n{\rightarrow}\infty}{\frac{y_{n+1}-y_{n}}{x_{n+1}-x_{n}}}=a$,then
$\lim_{n{\rightarrow}\infty}{\frac{y_{n}}{x_{n}}}=a$.
The Cesaro-Stolz theorem and the theorem above are are similiar to l'Hospitale rules for functions.


How many grams of Cadium (II) Chloride will form when 24.6 grms of Amonium Chloride reacts with Cadium (II) nitrate?

This is the same stuff. And by the way it should be "Cadmium". Start by writing the balanced chemical equation:

$2NH_{4}Cl+Cd(NO_{3})_{2}\longrightarrow CdCl_{2}+2NH_{4}NO_{3}$.

Now convert each mass of reagent in number of moles by using the molar mass of each. Then decide which is the limiting reagent. Finally, use the stoichiometry of the equation and do your calculations using the limiting reagent.


[geogebra]7458c5f6df2388e538d61c17f8247b4b7d053b6e[/geogebra]

AA_1, BB_1, CC_1 is perpendicular to BC,CA,AB, respectively.

A_2 , B_2 , C_2 is midpoints of AA_1 , BB_2 , CC_2, respectively.

Prove that ∠B_2 A_1 C_2 + ∠A_2 B_1 C_2 + ∠ B_2 C_1 A_2 = 180.[/code]

Let L,M,N be midpoints of BC,CA,AB, then ∠BLB₂=∠C=∠BHA₁, hence circumcircle of HLA₁ involves B₂,and similiar way C₂ also be involved that.
Therefore ∠B₂A₁C₂=∠B₂LC₂=∠A


[quote="Yuriy Solovyov"]Let $ q_1 \equal{} x > 1$ and $ q_{n \plus{} 1} \equal{} 2q_{n}^2 \minus{} 1$ for $ n\geq 1$. Find
\[ \prod_{k \equal{} 1}^{\infty}\left(1 \plus{} \frac {1}{q_k}\right)
\]
[hide]
If $ x \equal{} 2$ our product is equal to $ \sqrt {3}$.
[/hide][/quote]
Don't know if it helps at all....
$ \prod_{k \equal{} 1}^{n}\left(1 \plus{} \frac {1}{q_k}\right) \equal{} \frac{x\plus{}1}{x} \cdot \prod_{k \equal{} 1}^{n\minus{}1}\left(\frac {q_{k\plus{}1} \plus{} 1}{q_{k\plus{}1}}\right) \equal{} \prod_{k \equal{} 1}^{n\minus{}1}\left(\frac {2q_k^2}{q_{k\plus{}1}}\right) \cdot \frac{x\plus{}1}{x} \equal{} 2^{n\minus{}1} \prod_{k\equal{}1}^{n\minus{}1} \left ( q_k \right ) \cdot \frac{x\plus{}1}{x} \cdot \frac{q_1}{q_n} \equal{} \left ( 2^{n\minus{}1} \prod_{k\equal{}1}^{n\minus{}1} q_k \right ) \cdot \frac{x\plus{}1}{q_n}$

I don't know if the substitution $ x \equal{} \csc \theta$ could be helpfull?

Let $ \varphi \equal{} \varphi_1$ and $ \varphi_n \equal{} \text{arccosh} \, q_n$.

Then $ \cosh \varphi_{n\plus{}1} \equal{} q_{n\plus{}1} \equal{} 2q_{n}^{2} \minus{} 1 \equal{} \cosh(2 \varphi_n)$ and hence $ \varphi_n \equal{} 2^{n\minus{}1} \varphi$.

This shows that $ 2^{n} \sinh \varphi \prod_{k\equal{}1}^{n} q_k \equal{} \sinh (2^{n} \varphi)$.

Then referring to Mathias_DK's calculation,

$ \prod_{k \equal{} 1}^{n}\left(1 \plus{} \frac {1}{q_k}\right) \equal{} \sqrt{\frac{x\plus{}1}{x\minus{}1}} \tanh (2^{n\minus{}1} \varphi) \to \sqrt{\frac{x\plus{}1}{x\minus{}1}}$

as $ n \to \infty$.

Ok, this was not so hard because of the formula $ \cosh(2\theta) \equal{} 2\cosh^2\theta \minus{} 1$. Look at [url=http://www.mathlinks.ro/viewtopic.php?t=259993]this[/url]  :)


inequalities
geometry
3D geometry
algebra
polynomial
factorization
sum of cubes

Give x,y,z satisfying $ x^8+y^8+z^8 =3 $.
Prove that:
[tex] x^3+y^3+z^3 \leq 3xyz +6 [/tex]  :)

And what isnot easy in this inequality? We obviously have $ xyz\geq -1$ from AM-GM. Thus, it is enough to prove that $ x^3+y^3+z^3\leq 3 $. But $ x^3+y^3+z^3\leq |x|^3+|y|^3+|z|^3$ and then apply Power-Mean inequality.

You are right,Harazi this is really easy one. :( But what about:
[tex] x^3+y^3+z^3 < 3xyz + \sqrt 27 [/tex] :cool:

Well, this one is hard. I couldn't find any smart solution, so I will try later. Maybe I will be more lucky. Anyway, this is a nice problem. I think the maximum of the difference between the sum of cubes and three times the product is 4, but I'm not sure.

I still haven't found any smart solution, so if you have a smart solution, not with ugly computations, please post it. My idea was to reduce the inequality to:
 $ x^3+\sqrt{27}\geq y^3+z^3+3xyz$ for $ x,y,z\geq 0$ such that $ x^8+y^8+z^8=3$ and then to use the fact that $ y^3+z^3\leq 2\cdot\sqrt[8]{(\frac{3-x^8}{2})^3} $ and a similar one for $ yz$ and so to reduce everything to an inequality in x. But this is the hell on earth.

I think I've got it. As far as I remember I have proved that $ x^3+y^3+z^3-3xyz\leq ( x^2+y^2+z^2)\cdot\sqrt{x^2+y^2+z^2} $ for any reals. From here it follows.

Victory shall be ours! It's done! Just observe that $ (x^3+y^3+z^3-3xyz)^2=(a+2b)(a^2-2ab+b^2)$ where $ a=x^2+y^2+z^2, b=xy+yz+zx$ so, its enough to prove that $ (a+2b)(a^2-2ab+b^2)\leq a^3$ which comes down to $ b^2(3a-2b)\geq 0$, obvious.

Congratulation,that is exact what I did :) .In fact,my friend asked me about the least value of the number but $ \sqrt 27 $ is the smallest number I found.Maybe that is 4 but I don't have any solution . :(


$x;y;z>0 and x+y+z=1$
find k bestest such that : $xy+yz+zx>\frac{kxyz}{2+xyz}$

:?:   What you mean ??? Maybe you have mistake ???
Let $x,y\to 1/2$ $z\to 0$ $\Rightarrow$ $xyz\to 0$ $\Rightarrow$ $kxyz\to 0$
$(\sum{xy})(2+xyz)>2xy>1/2>kxyz$ for very big $k$

oh, but for the k big the inequality can't be true

The best possible value of $k$ is:

$1485\sqrt{3}= 2572.09545...$

@ guest : can you post solution ? thanks

but ì ưe have n number
how do find max k :o


superior algebra
superior algebra theorems

Let $R^{**}$ denote that set of $R$ excluding $-1$ and the operation * defined by $a+b+ab$.  Prove that $R^{**}$ is a group. I have proved that it is associative and found its identity, but I can't find its inverse.

You want to show that for every $a\not=-1$, there exists $b\not=-1$ such that $ab+a+b=e$

On one hand, $e*e=e$ for all $a$, so that $2e+e^{2}=e$, whence $e=0$ (since obv $e\not=-1$).

Next we want to find $a^{-1}$.  Fix $a$ and let $b=a^{-1}$.  We have $ab=0$, so that $a+b+ab=0\implies b(1+a) =-a$, or
\[b=-\frac{a}{1+a}\]
as desired.


Prove that the first thousand digits after the decimal point in the value of $(6+\sqrt{35})^{1980}$ are all $9$

A related well known result, from which the problem can be easily derived:

We show that for any natural $m$ and any $n \geq 1$, $(m+\sqrt{m^2-1})^n=k+\sqrt{k^2-1}$ for some natural $k$.

For $n=1$ it is obvious; now assume it true for $n$ and look at $(m+\sqrt{m^2-1})^{n+1}=(m+\sqrt{m^2-1})^n(m+\sqrt{m^2-1})$. It's easy to see by expanding $(m+\sqrt{m^2-1})^n$ that we will get something of the form $a+b\sqrt{m^2-1}$. By hypothesis, $b^2(m^2-1)+1=a^2$. Now we have

\[(a+b\sqrt{m^2-1})(m+\sqrt{m^2-1})=am+(m^2-1)b+(a+bm)\sqrt{m^2-1}.\]
Here $(am+b(m^2-1))^2-(m^2-1)(a+bm)^2=a^2-b^2(m^2-1)=1$ by hypothesis, so the induction is done.


topology
real analysis
real analysis unsolved

Prove that there is no group structure on $\mathbb{RP}^2$ with the usual topology making it into a topological group.

This is a part of this problem:
Find the closed (compact, connected and with no boundry) surfaces that are topological groups with the usual topology given by the surface structure.

Now you're just teasing us, after having solved the one about $\mathbb S^2$ :).

It's rather easy, given a topological group and a covering space of this group (assuming any nice connectedness properties that you may need: path connectedness and local path connectedness, both for the space and its covering space), to endow the covering space with a topological group structure. Now, $\mathbb S^2$ is a covering space of $\mathbb {RP}^2$ through the quotient map which identifies two antipodal points, so if $\mathbb {RP}^2$ was a topological group, $\mathbb S^2$ would also be one, and there's a nice proof on the forum saying that it's not :).

Actually I didn't do/finished this part about $\mathbb{RP}^2$ in exam time  :blush:
But the proof on the forum about $S^2$ is indeed magnificent  :P


trigonometry calculus function derivative algebra polynomial

$2.9 - \displaystyle\frac{2.9^3}{3!} + \displaystyle\frac{2.9^5}{5!} - + ... = \sin 2.9 = 0.239...$

So the thousandth digit is 9.  (I'm assuming you're allowed to use a calculator because this problem is a ridiculous exercise in division by hand.)

[url=http://www.ucl.ac.uk/Mathematics/geomath/level2/series/ser121.html]Look here[/url] to find a short explanation about the Taylor Series of $\sin x$ (I am assuming you know calculus; by your other post you seem to be taking it, so there you go).

Essentially, differentiating a function $f(x)$ n number of times, and plugging x = 0 into each of those derivatives, allows you to construct an infinite polynomial (formally known as a Taylor Series, or Maclaurin Series) that has the same values at x = 0 for every derivative of $f(x)$, and it can be shown that this implies that the series and the function are equal (on some interval).

$\sin x = x - \displaystyle\frac{x^3}{3!} + \displaystyle\frac{x^5}{5!} -+ ... = \displaystyle\sum_{k=0}^{\infty} \left(-1 \right)^k \displaystyle\frac{x^{2k+1}}{(2k+1)!}$

Um.. aren't Maclaurin and Taylor Series different ?
From what I've learnt, Taylor series expansion are more general than Maclaurin...

Just asking  :)


Hello fellow mathematicians,

I'm new here and already have a problem which I could use some help with...

How can I prove the following:

Let (A_n) be a stricly monotone sequence of sigma-Algebras over X. Then the infinitely countable union of all the A_n's is not a sigma-Algebra.

Thanks for your input.

P.S: Sorry, for the inconvenience, but I just installed a new system and haven't jet installed LateX.

Best,
Tom

Your prerequisites read that we have sigma algebras $ \sigma_{1}\subset\sigma_{2},\ldots$ and we construct $ \sigma_{sup}\equal{}\bigcup_{i}\sigma_{i}$. Now choose $ X_{1}\in\sigma_{1}$ and for $ i>1$ select $ X_{i}\in\sigma_{i}\setminus\sigma_{i\minus{}1}$, then the countable union $ X\equal{}\bigcup_{i}X_{i}\not\in\sigma_{sup}$, since $ X\not\in\sigma_{k}$ for any $ k$


A regular hexagon has a perimeter of 36 centimeters. Calculate the number of centimeters in the sum of the lengths of all diagonals of the hexagon. Express your answer in simplest radical form.

EDIT: lol, i only did easy ones

since the perimeter is 36, the side length of the hexagon is 36/6=6.

and the length of a diagonal AD, BE, FC is 12.

so its 12*3=36

and the length of other diagonals is $ 6\sqrt{3}$.

lets take triangle ABD as an example. AD is twice the length of AB, so its 30-60-90 triangle. Therefore, BD becomes $ 6\sqrt{3}$

and since there are 6 such diagonals, its $ 36\sqrt{3}$

answer : $ 36\plus{}36\sqrt{3}$

thanks for pointing out, leoxlin. :D

:o FantasyLover made a mistake!

Hint:

[hide]A hexagon has two distinct types of diagonals. One of them is easy to find the length of, and the other is not as easy.

Fact: in hexagon $ ABCDEF$, $ AB \perp BD$.  :) [/hide]

But I don't know how to find the length of the hard one!

Also, Do I find the length of those two diagonals and add them up, or am I supposed to find the length of the long one, multiply it by three, then find the length of the short one, multiply it by six, and add them together?

First we find the side length, which is $ 36 \div 6\equal{}6$ cm. Now, it would help to know where all the diagonals are.

Drawing it out, or using logic tells us that: The hexagon has 3 diagonals that connect opposite vertices. The lengths of these are $ 2s$, where s is the side length (made apparent if you draw it out). So, $ 3 \cdot 2 \cdot 6\equal{}36$ cm total for these diagonals.

The hexagon also has $ 6$ diagonals that do not connect opposite vertices. These connect every vertex with one 2 vertices away. Looking at the hexagon, we could take one of those diagonals and remove the section it splits the hexagon, an isosceles triangle. This is a $ 120\minus{}30\minus{}30$ triangle. We draw the altitude, or height, to the triangle, to create two 30-69-90 triangles

[asy]draw ((-1.732,0)--(0,1)--(1.732,0)--cycle);
draw ((0,1)--(0,0));[/asy]

By the Pythagorean Theorem, the base of this triangle, which is the diagonal, is $ \sqrt{3} s$. The side length is 6 and we have 6 of these diagonals, so we multiply $ \sqrt3 \cdot 6 \cdot 6\equal{}36\sqrt3$ cm. Add that to the 36 cm we had for the other diagonals and we get the answer of $ \boxed{36\sqrt3\plus{}36}$ or $ \boxed{36(1\plus{}\sqrt3)}$.

Phew! :D


function
topology
calculus
calculus computations

Let $I \subset R$ be a nonempty open set and
$f: I \to R$ be a function with the property that $f^{2}: =f \cdot f$ and $f^{3}: =f\cdot f\cdot f$ are in $C^{\infty}$.Where $C^{\infty}$ is the class of infinitely differentiable functions.Prove or disprove that then $f \in C^{\infty}$, too.

[quote="spider_boy"][quote="Kent Merryfield"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122583]Recently posted.[/url] And not suitable for the "computations and tutorials" forum.[/quote]

Can you please give me direct link containing the solution? I can not find it myself.This was my homework problem and I would be very happy if i see the solution. :)[/quote]

someone help? :(

[quote="harazi"]Man, is it so difficult to understand that this "homework problem" is extremely difficult and the solution is too long and complicated to write it here? If your teacher is such a great genius to ask this in homework, then why don't you ask him for a solution? I don't know I have the feeling he will give you the short and wrong one.  :wink:[/quote]

No man.This was homework problem, but it was not mandatory.It was 2 star challenge problem.So we didnot have to solve this problem.But I am interested in the solution.As you say I also guessed that this was very hard, because if you saw our normal mandatory hmw problems you could also guess that 2 star problem must be hard. :lol:  :D


[quote]No pencil whipping here - you are our conscience.
[/quote]

Does anyone know what this means?  Thanks!

Some context would be nice.  A quick internet search reveals the following:

[quote]After looking at the Ford Th!nk site and remembering crticism that I had recently read, the phrase “pencil-whipping” came to mind, i.e., a [b]false accountability[/b] — checking items on a list to go through with the show but the action has no substance.[/quote]

It probably says that on a form or application you have to fill out - it's telling you to fill out the form honestly and not say that something was done when it wasn't, and to uphold your moral values.

It's funny how stupid I sound sometimes.

In team games where one player keeps score for both his team and the opposing team (such as dominoes), the phrase "pencil whipping" is used when the scorekeeper attempts to give his team extra points without the other team noticing.


[quote="Labiliau"]Let two matrix $A \in R^{m \times n}$ and $B \in R^{m \times n}$. Show that there exists a regular matrix $S \in R^{m \times m}$ such that $B = SA$ if and only if $A$ and $B$ have the same Gauss-Jordan matrix.[/quote]

There is Gauss-Jordan Elimination

http://mathworld.wolfram.com/Gauss-JordanElimination.html

It's method. There is square Jordan matrix. What is it, Gauss-Jordan matrix, first? Solution follows from determination.

Details. If n<m, we can find a matrix $S$ for any $A$, $B$. In other case, not.  $B = SA$ says $A$ and $B$ tiedly connected, but Jordan matrices of them they will be different, naturally.


Extremal combinatorics
graph theory
combinatorics
Extremal Graph Theory
IMO Shortlist

Translation into graph language:
A complete graph contains 10 vertices and all its edges colored either in one of the two possible colors,red or black,or in both colors.
We need to prove that one can choose edges of either red or black color,such that this subgraph contains two connected components,each with odd number of vertices.
If got it correctly,the needed subgraph may not contain all vertices,right?Can edge be colored in both colors?
Does round trip means that all vertices can be connected by using only one color,or it just means that edges are not ordered?

Maybe one should note if there is a direct service from $ x$ to $ y$ and $ x \neq y,$ then this airline also offers this direct service from location $ y$ to $ x.$ Try to prove the following statement:

Colour the edges of a complete graph $ K_m$ with $ n$ colours and if we have that $ m > 2^n, n \geq 1,$ then within $ K_m$ there exists a monochromatic circle of odd length. A subgraph $ G'$ of some graph $ G$ is called monochromatic if all its edges have the same colour. Use induction.

[hide="Wikipedia: circle/loop"]http://en.wikipedia.org/wiki/Graph_(mathematics)#Loop[/hide]

By way of contradiction, suppose any two odd cycles of the same color intersect.

Since $10>R(3,3)$ there is a monochromatic triangle, let us say it is black and call it $T$.

Consider the remaining $7$ vertices, there is no odd black cycle in here, so the black part of the graph is bipartite and therefore there are at least two complete red subgraphs, the possibilites for the sizes are: $(1,6),(2,5),(3,4)$ in all of the cases except $2,5$ It is clear that there is going to be two disjoint red triangles.

So we only have to check the case in which we have a black triangle $T$, a red $K_2$ and a red $K_5$ (and they are all mutually disjoint). Pick vertices $t_1$ and $t_2$ in $T$, notice that if any of them has at least two red edges towards the red $K_5$ then we can obtain two disjoint red triangles. So each of them has at east $4$ black edges towards the $K_5$. Therefore there must be a vertex $x$ in the $K_5$ so that $t_1,t_2,x$ forms a black triangle.

Now let $t_3$ be the remaining vertex in $T$. It must send at least one black edge toward the $K_2$. Call the other endpoint of this edge $y$. As before, notice at least $4$ of the edges from $y$ to the $K_5$ and from $t_3$ to the $K_5$ must be black. Therefore there is a vertex $z$ in the $K_5$ (other than $x$) so that $t_3,y,z$ forms another black triangle.

Can someone remove the part about $1990>2^{10}$ in the title? It kinda gives the problem away :(

How does it give the problem away, what does $1990$ have to do with the problem statement? Also, I thought what we had to prove is that there are two disjoint odd cycles of the same color. Is this not the case?

Oops. The title made me think of a similar problem:

1990 cities are served by ten airlines, such that between any two cities some airline offers a two-way flight between them. Show that one of the airlines can offer a round-trip with an odd number of cities.

oh yeah, that one is very nice. I think Ori also confused it with this one :)

[quote=orl]Ten localities are served by two international airlines such that there exists a direct service (without stops) between any two of these localities and all airline schedules offer round-trip service between the cities they serve. Prove that at least one of the airlines can offer two disjoint round trips each containing an odd number of landings.[/quote]

Graph theoretic interpretation : Take a $K_10$ graph (10 vertices and each pair of vertices has an edge between them) color each edge either blue or red (these represent the airlines respectively). We wish to show that there exist 2 disjoint odd monochromatic cycles.

[b][u][i]Solution: [/i][/u][/b] (slightly similar to @Gamamal) We know that since $10>R(3,3)=6$ (where $R$ refers to ramsey numbers), there must exist a monochromatic triangle. WLOG let it be blue colored. Now, delete these three vertices from the graph. We now wish to show that there exists either an odd blue cycle or $2$ disjoint odd red cycles on the remaining $K_7$ graph. If there exists a blue odd cycle, we are done, so assume not. Now we know that $K_7$ is not bipartite so there must exist an odd cycle and since $7>R(3,3)=6$, There infact exist multiple red triangles . Now delete $3$ vertices of a red cycle such that in the resulting graph all the blue edges are incident to a single vertex, which is clearly possible otherwise if in all the sub-graphs of $4$ vertices there are two blue edges that are not adjacent, it is possible to show that we must have multiple edges that are blue and that there does exist a blue triangle, which is a contradiction to our assumptions.
[hide="Proof"] Consider 4 vertices $A,B,C,D$ and WLOG let the (atleast) 2 blue non-adjacent edges be crossing i.e. $AC,BD$ (like the diagonals of a square). Now consider another pair of vertices $E,F$ and the subgraph $CDEF$ which must have blue $EF$ since $CD$ is blue. But now consider $CBEF$ and $ABEF$ in which $CB$ and $AB$ must also be blue. But this implies that $ABC$ is a blue triangle, contradiction
 $\blacksquare$  [/hide]
  Finally we now have a $K_4$ graph which must have an odd cycle (not necessarily monochromatic) since complete graphs are non-bipartite. But note that the number of blue edges must be $\leq 3$ by Turan's theorem since there doesn't exist any blue triangle here (by assumption). Now there are three cases (number of blue edges $=1,2,3$) and in each one it is easy to verify (by constructing edges WLOG) that there infact does exist a red triangle and we are done.


calculus
integration
trigonometry
real analysis
real analysis unsolved

I know there is a tricky way to calculate that but I can't find it.

$ \int_0^\pi \frac{xsinx}{1\plus{}cos^2x}$

Yeah, there is one:
\[ \int_0^{\pi} x f(\sin x) \, \mathbf{d}x \equal{} \frac {\pi}{2} \int_0^{\pi} f(\sin x) \, \mathbf{d}x
\]
By the way - have a look at this topic http://www.mathlinks.ro/viewtopic.php?t=180448


inequalities
function
inequalities proposed

[quote="weiquan"]I have created two interesting inequalities, and I would like to know what methods can be employed to solve them:

$ 2 \sqrt {\frac {x^2 + y^2 + z^2}{3} } + 3 \sqrt [3]{xyz} \leq 5 \left( \frac {x + y + z}{3} \right)$
$ 4 \sqrt {\frac {x^2 + y^2 + z^2}{3} } + (\sqrt {3} - 1) \sqrt [3]{xyz} \geq (3 + \sqrt {3}) \left( \frac {x + y + z}{3} \right)$

I am also curious as to what the best constants are as well as how to find and prove them.[/quote]

A very interesting problem! I can only tell you what I did - which didn't work out. :(

Consider the function:
\[ f_{x,y,z} (p) = \left( \frac {x^p + y^p + z^p} 3 \right)^{1/p},
\]
which is defined on all real $ p$ (at $ p = 0$, the function is the G.M. of $ x,y,z$). We know that the function is increasing - thanks to Power Mean inequality. Now, if the function were concave, our job would be done. Unfortunately, it's not.

In fact, I tried to look for $ c_1, c_2 > 0$ such that:
\[ c_1 f(2) + c_2 f(0) \le (c_1 + c_2) f(1).
\]
It turns out that there's a critical value for $ c_2 / c_1$ for this inequality to hold. When $ c_2 / c_1 >  a : = 1.10874112...$, the inequality holds; otherwise it does not. [ Note that if $ f$ were concave it ought to hold for $ c_1 = c_2$. ]

I tried inputting the above constant into an inverse symbolic calculator - didn't get anything interesting.

First I think that your inequality should have condition $ a,b,c \ge 0$.
For example, if we let $ c\equal{}0,a\equal{}\minus{}b$ then that e is wrong. So I try proving this with $ a,b,c\ge0$.
If $ abc\equal{}0$ then the e becomes
$ 2\sqrt {\frac{{a^2  \plus{} b^2 }}{3}}  \le \frac{{5\left( {a \plus{} b} \right)}}{3}$
$ 2\sqrt {\frac{{a^2  \plus{} b^2 }}{3}}  \le \frac{{2\left( {a \plus{} b} \right)}}{{\sqrt 3 }} \le \frac{{5\left( {a \plus{} b} \right)}}{3}$
Which is true.
If $ abc>0$ then we assume $ abc\equal{}1$, then we have to prove that
$ 2\sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}}  \plus{} 3 \le \frac{{5\left( {a \plus{} b \plus{} c} \right)}}{3}$
$ \Leftrightarrow 5\left( {a \plus{} b \plus{} c} \right) \minus{} 2\sqrt {3\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)}  \ge 9$
Setting $ f\left( {a,b,c} \right) \equal{} 5\left( {a \plus{} b \plus{} c} \right) \minus{} 2\sqrt {3\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)}$ and $ c \equal{} \max \left\{ {a,b,c} \right\}$ we have
$ f\left( {a,b,c} \right) \minus{} f\left( {\sqrt {ab} ,\sqrt {ab} ,c} \right) \equal{} \left( {\sqrt a  \minus{} \sqrt b } \right)^2 \left( {5 \minus{} \frac{{2\sqrt 3 \left( {\sqrt x  \plus{} \sqrt y } \right)^2 }}{{\sqrt {2xy \plus{} z^2 }  \plus{} \sqrt {x^2  \plus{} y^2  \plus{} z^2 } }}} \right) \ge 0$
Finally we need to prove $ f\left( {t,t,c} \right) \ge 9$ with $ t^2 c \equal{} 1$
Which is true.

[quote="Honey_S"]First I think that your inequality should have condition $ a,b,c \ge 0$.
For example, if we let $ c \equal{} 0,a \equal{} \minus{} b$ then that e is wrong. So I try proving this with $ a,b,c\ge0$.[/quote]
[quote]Setting $ f\left( {a,b,c} \right) \equal{} 5\left( {a \plus{} b \plus{} c} \right) \minus{} 2\sqrt {3\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}$ and $ c \equal{} \max \left\{ {a,b,c} \right\}$ we have
$ f\left( {a,b,c} \right) \minus{} f\left( {\sqrt {ab} ,\sqrt {ab} ,c} \right) \equal{} \left( {\sqrt a \minus{} \sqrt b } \right)^2 \left( {5 \minus{} \frac {{2\sqrt 3 \left( {\sqrt x \plus{} \sqrt y } \right)^2 }}{{\sqrt {2xy \plus{} z^2 } \plus{} \sqrt {x^2 \plus{} y^2 \plus{} z^2 } }}} \right) \ge 0$
Finally we need to prove $ f\left( {t,t,c} \right) \ge 9$ with $ t^2 c \equal{} 1$
Which is true.[/quote]

Yes, sorry, I have edited my post. Should it be
$ f\left( {a,b,c} \right) \minus{} f\left( {\sqrt {ab} ,\sqrt {ab} ,c} \right) \equal{} \left( {\sqrt a \minus{} \sqrt b } \right)^2 \left( {5 \minus{} \frac {{2\sqrt 3 \left( {\sqrt a \plus{} \sqrt b } \right)^2 }}{{\sqrt {2ab \plus{} c^2 } \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2 } }}} \right) \ge 0$?
But nice proof! How about the other one? :P

I think SOS can give the best constants, but I don't know how.. anyone can help?

[quote="weiquan"]I have created two interesting inequalities, and I would like to know what methods can be employed to solve them:

$ 2 \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2}{3} } \plus{} 3 \sqrt [3]{xyz} \leq 5 \left( \frac {x \plus{} y \plus{} z}{3} \right)$
$ 4 \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2}{3} } \plus{} (\sqrt {3} \minus{} 1) \sqrt [3]{xyz} \geq (3 \plus{} \sqrt {3}) \left( \frac {x \plus{} y \plus{} z}{3} \right)$
for all non-negative reals $ x$,$ y$,$ z$

I am also curious as to what the best constants are as well as how to find and prove them.[/quote]
It suffices for you to consider two cases (it's enough :) ): $ a\equal{}b\equal{}1$ or $ c\equal{}0$
It follows from EV theorem. :)

[quote="can_hang2007"]
It suffices for you to consider two cases (it's enough :) ): $ a \equal{} b \equal{} 1$ or $ c \equal{} 0$
It follows from EV theorem. :)[/quote]

Can you please explain? I don't see how EV theorem can be used here. Are the best constants as:
$ c \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2}{3} } \plus{} (1 \minus{} c) \sqrt [3]{xyz} \leq \frac {x \plus{} y \plus{} z}{3}$
is true for $ c \leq .4742...$ (I don't know what the skew equality case might be)
$ c \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2}{3} } \plus{} (1 \minus{} c) \sqrt [3]{xyz} \geq \frac {x \plus{} y \plus{} z}{3}$
is true for $ c \geq \frac {2\sqrt{3}}{3\sqrt{2}}$
?


By AM-GM, $ (abc)^2 \plus{} 1 \plus{} 1 \ge 3(abc)^{2/3}$. So we have to show
$ 3(abc)^{2/3} \plus{} a^2 \plus{} b^2 \plus{} c^2 \ge 2(ab \plus{} bc \plus{} ca)$
Let $ a^{1/3} \equal{} p$, $ b^{1/3} \equal{} q$, $ c^{1/3} \equal{} r$. Then we have to show
$ 3p^2q^2r^2 \plus{} p^6 \plus{} q^6 \plus{} r^6 \ge 2(p^3q^3 \plus{} q^3r^3 \plus{} r^3p^3)$
Which is true by Schur and Muirhead.

Whoever posted the solution using the discriminant of the quadratic.
Why did you delete your post?

[quote="Dimitris X"]It may be a piece of cake for you but i think its nice.
If $ a,b,c$ positive reals show that
$ (abc)^2 \plus{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2\geq2(ab \plus{} bc \plus{} ca)$[/quote]
Consider $ (b^2c^2 \plus{} 1)a^2 \minus{} 2(b \plus{} c)a \plus{} b^2 \plus{} c^2 \plus{} 2 \minus{} 2bc$ as a quadratic. Then the discriminant is $ \minus{} 4b^2c^2(b \minus{} c)^2 \minus{} 8(bc \minus{} 1)^2$, which makes it obvious.
[quote="dgreenb801"]
Whoever posted the solution using the discriminant of the quadratic.
Why did you delete your post?[/quote]
I miscalculated and thought I was wrong. Now the right result should be there :)

Btw. It is a easy if you know this ineq created by Darij Grinberg.
It says $ 2abc \plus{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 1 \ge 2(ab \plus{} bc \plus{} ca)$. And since $ a^2b^2c^2 \plus{} 1 \ge 2abc$ the result follows.


floor function
logarithms
function
topology
geometry
3D geometry
sphere

suppose that $f: \mathbb{R}^{n}\to \mathbb{R}^{n}$ satisfies $\|f(y)-y\| \leq \frac{\|y\|}{3}$. Is it true that $f$ is surjective ?

Are we also assuming continuity? Because without that, I can do something silly like

$f(x)=\text{sgn}(x)\text{exp}\left(1.1\left\lfloor\frac{\ln |x|}{1.1}\right\rfloor\right)$

oh yes, functions are continuous  :blush:

Then the answer is yes. Pick a point $a$ and consider the straight-line homotopy $f_{t}(x)=tf(x)+(1-t)x$ on a sphere of radius $2\parallel a\parallel $. Since $\parallel f_{t}(x)\parallel \ge 4\parallel a\parallel /3>\parallel a\parallel $ for any $x$ on the sphere, it follows that $f$ has degree $1$ about $a$.

There is no way to prove this result with on calculus level. Indeed, it implies Brower's theorem: a sphere is not a retract of a ball.

I was figuring there had to be some algebraic topology in there - including that it's simple (just the intermediate value theorem) in the $n=1$ case.


algebra
polynomial
number theory unsolved
number theory

Let $f(x)=x^{3}+a\in\mathbb{Z}[x]$ be a cubic polynomial with integer coefficients (for example $f(x)=x^{3}+1$).

when is the set $\{f(n): n\in\mathbb{Z}$ and $f(n)$ is prime $\}$ infinite?

It is unsolved problem.
If $f(n)$ is irreducible (in Q) and gcd(a_i)=1, and for any p<n=deg(f(x)) exist x, suth that $p\not |f(x)$ must be infinite primes f(n) (Shensel's conjecture).


CEMC
function
email
geometry
3D geometry
Mafia
summer program

What's the usual cutoff for the Seminar on Fermat?

(i know.. Some of you probably know why I'm posting this, but I really want to go back!!  :D )

lol hey, the Twins have got make a comeback this year again.

Hey was just thinking, considering how they've been promoting girls at the seminar like last year, you're a given, lol.  You don't even need an invite, come on!

[quote="YY"]Haha~evil twins!!!!!
That's one of the best parts I remembered from the seminar...And I still have no idea what your little war with Ian was about...

(Screaming) I REALLY hope I can make it in this year again!
Sigh...[/quote]

Neither do we.  It was just fun, hehehe.  We gotta make a comeback this year, Ian almost beat us!  Actually, I think he did, but I'm not saying that.

Well as long as we did decent on the Fermat, then there's hope.  Especially if they're after girls, then yiyi you shouldn't have much of a problem, considering you've went once already

[quote="jpark"][quote]Actually, I think he did, but I'm not saying that. [/quote]

No, we beat him by half a point!!  :D[/quote]

Hehehe ok good.

Well the whole point is that I really screwed up on Fermat, only 138 or something...And I really don't think I'll ever do good on Euclid.
Ah well...

And that war of yours...was that about who's more evil?

hey, don't forget Hypatia is left too, hehehe.

The war, yeah pretty much who was worse.  He cheated though!

Is Waterloo Seminar the same as or related to the CMS Awards Banquet?

And does this, [quote]Almost 5,000 students wrote the 2003 Canadian Open Mathematics Challenge and the winners of the 2003 Open and the 2004 CMO will be honoured at the CMS Awards Banquet which will take place at the University of Waterloo on Tuesday, June 15, 2004 from 5:30 p.m. to 8:00 p.m. [/quote] still hold true this year?  (from [url]http://www.cms.math.ca/MediaReleases/2004/cmo.html[/url])

They didn't award COMC winners last year.

They haven't awarded COMC for 4 years.  To my knowledge, the Invites are not based on COMC at all.

What exactly goes on during the award ceremony and the seminar in general?

and also, why is the canadian computing competition not included? it is part of the CEMC like all the math contests that are awarded.

[quote="jpark"]The award ceremony is for the students who receive some sort of prize, I think.

As for the CCC, isn't there another camp for the computers only? Probably that's why..  :D[/quote]

Yeah, there's a computer camp I'm pretty sure.

Seminar, basically a week of great lectures, (some not so great)  fun team competitions, and a lot of fun overall.  You meet a lot of great people there

[quote="TheDreamer"]Hey Valentin, could you?  Just curious to know :)[/quote]Romanian math camps are no longer what they used to be :) So I can't give you any details ;)

For all you chinese people, I'll probably bring military chess (Jun Qi). have fun..

lol good for you, but I suck at it....
Also, I will bring Chinese chess, but still, I suck at it too.....
 :wacko:

You might have to wait a while for a reply. I'm pretty sure the experienced ones knowledgable about the Waterloo Seminar are currently attending it.

[quote="YooSam"]How many percent do we need to get invited to the seminar basically?

Do we need to pay for attending there?[/quote]

I am not sure what you mean by "how many percent"

From my understanding, to get invited to the seminar you either have to win an award in either the Cayley, Fermat, Euclid, COMC, CMO, APMO or be in grade 10, 11 and do well in the COMC AND the Euclid.

The only thing you pay for is the transportation [color=red][mod: plus 150 dollars for the seminar fee][/color], and it is worth it to attend the seminar.

[quote="mak"]
From my understanding, to get invited to the seminar you either have to win an award in either the Cayley, Fermat, Euclid, COMC, CMO, APMO or be in grade 10, 11 and do well in the COMC AND the Euclid.

The only thing you pay for is the transportation [color=red][mod: plus 150 dollars for the seminar fee][/color], and it is worth it to attend the seminar.[/quote]

Not quite.  Though all those contests are mentioned there, they will only invite on the Waterloo ones.  So Cayley, Fermat, Euclid, and sometimes the Hypathia (and the gr 10 equivalent which I can't remember)

Heh, was great to see you again mak!

Hey mak, you didn't show up on the last night...we missed yah...so i couldn't get your msn :(

What is your MSN? :)

Thanks, it was great meeting all you guys too!

Yufei is right, you do have to pay $150 for seminar fee, but you can make your school pay for that!  ;)

Enjoy your summer! NO REALLY, enjoy your summer while you can now!!! (I miss summer breaks...ha ha ha  :(  )

[quote="lightrhee"]This year's Waterloo Seminar was great :) . I loved every single hour of the seminar![/quote]I think the time has come to update your signature then  :lol:

[quote="cheeZZcake"]I got my invitation last Friday,
and I'm fascinated that Waterloo actually sent me one b/c I didn't even do well on Fermat :0

(random passer-by -- found this forum thru Google!)[/quote]

I also found AoPS-MathLinks forum through Google search engine; what a weird coincidence.  :ninja:

I think many new people in mathlinks , particularly our Canadian Community find it through google search engine (including myself!) BTW, we were playing chinese chess at the seminar this year. Daniel is very good! And we had a lot of fun with it! Are you up to a match Douglas? Then, I'll see you next year in the seminar, if both of us are lucky enough! (I doubt I'll) :lol:

Our mighty mod :omighty: , billzhao, has asked us to discuss Waterloo Seminar [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41476]here[/url].
So we should stop posting here.


geometry
circumcircle
combinatorics proposed
combinatorics

Let S be a set of points in the plane satisfying the following conditions:
(a) there are seven points in S that form a convex heptagon; and
(b) for any five points in S, if they form a convex pentagon, then there is a point of S that lies in the
interior of the pentagon.
Determine the minimum possible number of elements in S.

[hide="I think..."]...it is 8. Seven points forming a regular heptagon and its geometrical centre.

[u]Proof:[/u]If you don't want the geometrical centre to be inside one of the pentagons you draw using the vertices of the heptagon, you have to choose $ 5$ adjacent vertices of the heptagon.

The new line you are drawing to complete the pentagon divides the heptagon into two parts one with four sides and the other with three sides and you want to draw a pentagon with the four vertices on one side. Clearly by length of arcs of the circumcircle , the angle subtended by the major arc at the centre is reflex and hence the geometrical centre lies inside the pentagon. The same argument shows that you can't draw a pentagon with the geometrical centre as one of the vertices otherwise the pentagon would be concave which is false.

We proved that $ 8$ points work. Suppose $ 7$ points which form a regular heptagon work. Draw a pentagon and it encloses the area within the heptagon which doesn't contain the remaining vertices.

Thus proved[/hide]


Let $m$ and $n$ be positive integers such that $n \leq m$.  Prove that
\[ 2^n n! \leq \frac{(m+n)!}{(m-n)!} \leq (m^2 + m)^n \]

A stronger inequality is $\binom{m+n}{m-n} \geq 2^{n}\big(n!\big)^2$.

a solution using induction on $n$ :
when $n=1$ the case is clear
so suppose the inequality holds for $n$ we prove it for $n+1$
$2^{n+1}(n+1)!\le\frac{(m+n+1)!}{(m-n-1)!}\le(m^2+m)^{n+1}\Longleftrightarrow\\
2^nn!(2)(n+1)\le\frac{(m+n)!}{(m-n)!}(m+n+1)(m-n)\le(m^2+m)^n(m^2+m)$
applying the induction it suffices to prove :
$2(n+1)\le (m+n+1)(m-n)\le m^2+m$
which is easy after expanding and using $m\ge n+1$

I like the task, pure calculations :ddr: . Let's start by proving the left inequality. We transform our equation (factorial division) into:[hide]
$(!)(m-n+1)(m-n+2)\dots m(m+1)\dots(m+n) \geq 2^n$
Since $m\geq n$, the left equation is greater than $2n!$ (just substitute $n$ instead of $m$). By Induction, we easily prove that $(2n!) \geq 2^n*n!$.
Let's prove the right inequality. Note that our product is of the form:
$\prod(m-n+k)(m+n+1-k) k\in [1:n]$
Before that, it is easy to guess if you multiply the first and last, second and penultimate, etc. particles. If we prove that all these paired products are less than or equal to $(m^2+m)$ then the problem is solved. Well, let's compare them:
$(m-n+k)(m+n+1-k)|(m^2+m) \Rightarrow n^2+k^2+n|2nk+k \Rightarrow n\geq k, n^2+k^2\geq 2nk (AM \geq GM)$. So $\prod(m-n+k)(m+n+1-k) \leq(m^2+m)^n.$[/hide]


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Prove that
Let V and W be [url=http://planetmath.org/encyclopedia/GModule.html]G-modules[/url] with V [url=http://en.wikipedia.org/wiki/Group_representation#Reducibility]irreducible[/url]. Then dim Hom(V, W) is the multiplicity of V in W.

If W contains no submodule isomorphic to V, dim Hom(V, W) = 0 by Schur's lemma.
For a general W, can anyone explain me why dim Hom(V, W) is the multiplicity of V in W?

Thanks.

[quote="algtop"]Let V and W be G-modules with V irreducible. Then dim Hom(V, W) is the multiplicity of V in W.[/quote]

This is not true in general.  For it to be true, the modules need to be over an algebraically closed field of characteristic 0.

In that case W is a direct sum W' ⊕ V^n for submodule W' with no composition factor isomorphic to V and some non-negative integer n; the submodules W' and V^n are uniquely determined, not just up to isomorphism.  Every homomorphism from V to W has its image entirely in V^n, so is just a homomorphism from V to V^n.  In other words, Hom(V,W) = Hom(V,V^n), and over an algebraically closed field K with V irreducible, Hom(V,V^n)=Hom(V,V)^n = K^n has dimension n.

Over a general (unital, associative, commutative) ring R, if V is an irreducible G-module, then V is also a G-module over R/M=K for some uniquely determined maximal ideal M of R. Again over a general ring R, Hom(V,W) = Hom(V,Soc(W)) and Soc(W) = W' ⊕ V^n as before.  Then Hom(V,W) = Hom(V,V^n) = Hom(V,V)^n = E^n where E is a division ring finite dimensional over the "base field" K, and so E^n has dimension a fixed multiple of the multiplicity of V in the socle of W.

[quote="JackSchmidt"]
This is not true in general.  For it to be true, the modules need to be over an algebraically closed field of characteristic 0.

In that case W is a direct sum W' ⊕ V^n for submodule W' with no composition factor isomorphic to V and some non-negative integer n; the submodules W' and V^n are uniquely determined, not just up to isomorphism.  Every homomorphism from V to W has its image entirely in V^n, so is just a homomorphism from V to V^n.  In other words, Hom(V,W) = Hom(V,V^n), and over an algebraically closed field K with V irreducible, Hom(V,V^n)=Hom(V,V)^n = K^n has dimension n.

Over a general (unital, associative, commutative) ring R, if V is an irreducible G-module, then V is also a G-module over R/M=K for some uniquely determined maximal ideal M of R. Again over a general ring R, Hom(V,W) = Hom(V,Soc(W)) and Soc(W) = W' ⊕ V^n as before.  Then Hom(V,W) = Hom(V,V^n) = Hom(V,V)^n = E^n where E is a division ring finite dimensional over the "base field" K, and so E^n has dimension a fixed multiple of the multiplicity of V in the socle of W.[/quote]

Thanks for your reply. Now I understand the first part of your explanation, but I am still having difficulty understanding some key concepts of the second part of your explanation. I think the second part has something to do with Maschke's theorem.

From wiki, Maschke's theorem states that
"Let G be a finite group and K a field whose characteristic does not divide the order of G. Then KG, the group algebra of G, is a semisimple algebra."

I am trying to find a counterexample if K is a field whose characteristic divides |G|. If K is a field whose characteristic divides |G|, is KG still an algebra? Can you plz show me an example or counterexample of this one?

Thanks.

[quote="algtop"]Let V and W be G-modules with V irreducible. Then dim Hom(V, W) is the multiplicity of V in W.[/quote]
(indicated some hypotheses are needed, such as good characteristic)

[quote="algtop"]
I am trying to find a counterexample if K is a field whose characteristic divides |G|. If K is a field whose characteristic divides |G|, is KG still an algebra? Can you plz show me an example or counterexample of this one?[/quote]

Yes, KG is still an algebra.  The easiest example to think about is G cyclic of order 2 and K an algebraically closed field of characteristic 2.  KG is the very familiar ring K[x]/(xx).  If G is generated by the element g of order 2, then taking x=1+g, xx=(1+g)^2 = 1^2 + g^2 = 1 + 1 = 0.  KG has exactly one isomorphism class of simple module, KG/(1+g) = KG/(x) ≅ K, where x acts as 0 and g acts as 1.

Now, the regular module KG has dimension 2, so it must have two composition factors in any composition series, so the multiplicity of K in KG is 2.  However, Hom(K,KG) = Hom(K, xKG ), since if f in Hom(K,KG), then x*f(k) = f(xk) = f(0) = 0, so that the image of f is always contained in { f in KG : x*f = 0 } = xKG, the ideal of KG generated by x.  It is not hard to check that xKG ≅ K is the unique simple submodule of KG and also of course the unique up to isomorphism simple KG module.  In particular, Hom(K,KG) = Hom(K,xKG) = Hom(K,K) = K has dimension 1.  The multiplicity of K in KG is 2, but the dimension of Hom(K,K) is 1.

If the group ring part is weird, then switch to the commutative ring k[x]/(x^2).  It is a local, artinian ring whose only ideals are 0, (x), and the whole ring, so very easy to work with.


limit
function
complex numbers
complex analysis
complex analysis unsolved

If anyone understands how to approach these problems Please offer any help possible

Let $f$:$\mathbb{C}$ $\to$ $\mathbb{C}$ be defined by

f(z) = $\{$$z^{2}$           if   Im(z) < Re(z)
   -- and $\{$ $2i(Re(z))^{2}$        if Im(z) $\ge$ Re(z)

Show that
a) f is continuous throughout $\mathbb{C}$

b) f is not differentiable for[b] Im(z) > Re(z)[/b]

Thanks in advance for any help.

$f(z)=\{ \begin{array}{rcl}z^{2}& \mbox{for}& Im(z)<Re(z) \\ 2i(Re(z))^{2}& \mbox{for}& Im(z)\geq Re(z) \end{array}$

For a) need to show f(z) exists,  limit exits along the diagonal line Re(z)=Im(z), and that limit is equal to $2i(Re(z))^{2}$

Maybe:

$\mathop\lim_{y\to x}(x+iy)^{2}$

Although that might not be sufficiently rigorous.

For part b), consider the Cauchy-Riemann equations for a complex-valued function which has a pure imaginary value.


:| Can someone explain these three things: Sin Cos and Tan

$ \sin\equal{}\frac{\text{opposite}}{\text{hypotenuse}}$, $ \cos\equal{}\frac{\text{adjacent}}{\text{hypotenuse}}$, $ \tan\equal{}\frac{\text{opposite}}{\text{adjacent}}$

http://en.wikipedia.org/wiki/Trigonometry

I imagine them as functions of base angles of a right triangle. But there are other ways to think of them (e.g. the unit circle).


Prove the intermediate value theorem using uniform continuity.

They say the hint is to divide [a, b] into sufficiently many equal-
length subintervals.

[I don't know how the hint helps to prove it.]

If we can find a sequence of points $ x_n$ converging to some $ x$, with $ f(x_n)$ converging to $ y$, we have $ f(x)\equal{}y$.

WLOG, the value we're trying to hit is zero, and $ f(a)<0<f(b)$. Choose $ \epsilon>0$. Next, choose $ \delta$ so that $ |f(x)\minus{}f(y)|<\epsilon$ whenever $ |x\minus{}y|<\delta$ and $ x,y\in I\equal{}[a,b]$. Cut $ I$ into subintervals of length $ \le\delta$, and look at the values at the endpoints. Some will be positive, and some will be negative; find a subinterval $ [a_k,b_k]$ on which it changes sign (from negative to positive). We have $ \minus{}\epsilon<f(a_k)\le 0\le f(b_k)<\epsilon$.
Repeat with smaller $ \epsilon$ to get a sequence of points $ a_k$ with $ f(a_k)\to 0$. With just a bit more work, you can extract a convergent subsequence (Bolzano-Weierstrass, or just choose it monotone).

The uniform continuity is basically superfluous- the whole thing works with the successive bisection algorithm. We do need completeness of $ \mathbb{R}$, in the form that monotone sequences converge. Without uniform continuity, we have $ a_k$ increasing and $ b_k$ decreasing, $ |a_k\minus{}b_k|\to 0$, and $ f(a_k)<0<f(b_k)$. Both sequences converge to some $ c$, so $ 0\ge \lim f(a_k)\equal{}f(c)\equal{}\lim f(b_k)\ge 0$.


linear algebra
matrix
induction
skew symmetric

given $ A \in M_n(K)$ such that $ A^{T}\equal{}\minus{}A$.Prove that $ rank(A)$ is even.

What's $ K$? This is clearly false if $ K$ is a field of characteristic $ 2.$ (Just take the matrix with $ 1$ as the upper left entry and zeros everywhere else.)

Actually $ K$ doesn't need to be $ \mathbb{C}$, the result is true as long as $ K$ doesn't have characteristic 2.
Proceed by induction on $ n$.
If $ n\equal{}1$, it's obvious.
Assume the result is true for $ n$ and consider $ A \in \mathcal{M}_{n\plus{}1}(K)$ with $ A\equal{}\minus{}A^T$.
Write $ A\equal{}
\begin{pmatrix}
B & c \\
\minus{}c^T & 0
\end{pmatrix}$ where $ B \in \mathcal{M}_{n}(K)$ with $ B\equal{}\minus{}B^T$.
If $ c\in \mbox{Im}(B)$, then $ \mbox{rank}(A)\equal{}\mbox{rank}(B)$.
If $ c\notin \mbox{Im}(B)$, then $ \mbox{rank}(A)\equal{}\mbox{rank}
\begin{pmatrix}
B \\
\minus{}c^T
\end{pmatrix} \plus{}1 \equal{}\mbox{rank}(B) \plus{}2$.


geometry
incenter
inequalities unsolved
inequalities

Let $ABC$ be a triangle, $I$ the incentre, and $A'$, $B'$, $C'$ the intersection of $AI$, $BI$, $CI$ with the circumcircle. Then
$0\leq \sum IA'-\sum IA \leq 2(R-2r)$.

[quote="Kalimdor"]I can find a veryeasyway to prove $\ \sum IA'-\sum IA \geq 0$
since by calculating angles, we can prove: $\ IA'=A'B=A'C, IB' = AB'=B'C, IC'=AC'=BC'$
then $\ \sum IA'-\sum IA$ is equivalent to this one: $\ \frac{A'B+A'C+B'C+AB'+AC'+BC'-2IA'-2IB'-2IC'}{2}$
which is equal to $\ \frac{A'B+A'C+B'C+AB'+AC'+BC'-(IA'+IB')-(IA'+IC')-(IB'+IC')}{2}$
but the $\ 2(R-2r)$ part I can't find it. Can some one pls help?
Frank Morgan[/quote]
Maybe you mean
$\sum IA'-\sum IA =\frac{A'B+A'C+B'C+AB'+AC'+BC'-(IA+IB)-(IA+IC)-(IB+IC)}{2}$
and because $A'B+A'C-(IB+IC) \geq 0$
so $\ \sum IA'-\sum IA \geq 0$
does that you mean? :)

Just take this one: prove $\ AA'+BB'+CC'\leq \ A'B+A'C+B'C+AB'+AC'+BC'$
Then extend CB', pick point X on it which makes$\ B'X=BA$, do the same thing on BA' and AC'(Get points $\ Y,Z$, connect$\ AX,CY,BZ$.)


analytic geometry
function
complex analysis
complex analysis theorems

The map $ e^z$ defines a map from an infinite cylinder to the punctured complex plane.

Say we have a coordinate system with a coordinate $ h$ of height on the cylinder, and a coordinate $ \theta$ of angle around on the cylinder, where $ h \in (\minus{}\infty,\infty)$ and $ \theta \in \[0,2\pi)$.

Then $ e^z$, for $ z$ on the cylinder with coordinates $ h$ and $ \theta$, is defined as the point in the complex plane with angle $ \theta$ and radius $ f(h) \equal{} e^h$. This of course defines a conformal mapping, as it is holomorphic.

I wrote the function for radius in the plane in terms of height on the cylinder as "$ f(h)$" since I'm wondering what other functions $ f$ would give a conformal mapping from the cylinder to the punctured plane. Is there a nice answer to this?

If $ F_1$ and $ F_2$ are two conformal maps from the cylinder to punctured plane, then $ F_1\circ F_2^{\minus{}1}$ is a conformal automorphism of the punctured plane. And there are not many of those: just $ z\mapsto az$ and $ z\mapsto a/z$, where $ a\in\mathbb C\setminus\{0\}$.


[tex]

      a)  find the sum
                                sinx  +  sin3x  +  sin5x  +.......+  sin99x

      b) find the sum
                               cos2x + cos4x + cos6x +......+ cos2nx  [/tex]

[hide]The easiest way is to  sum geometic series   of the type e^ix  and then  find the real (or imaginary) part as the case may be:

For example  if   z=e^(ix) then the second series is nothing but

Real part of (z^2+z^4+......)

[/hide]

Treething wrote:This probably doesn't help at all, but...

[hide]
sin(2x) = 2sinxcosx
sin(3x) = (2sinxcosx)cosx + sinx(cos2x - sin2x)

You'd probably want to look for a pattern and keep going down the list all the way to sin(99x).
[/hide]

Although, I think the , well known, method I hinted above is the easiest but... looking at the pattern may help too.. specially if:

If you remember that   2 sin A sin B =  cos((A-B )  - cos (A+B)

and    2  sin A cos B =  sin (A+B) + sin (A-B)

Final expression can be as simple as  (sin  (50x))^2  / (sin x)

Hidden TextYou write    sin ((2n+1) x) as   2  ( sin((2n+1) x))* (sin x ) / (2 siin x)
= (1/(2*sin x) ( cos (2nx) - cos ((2n+2)x)

and you start seeing things like:
(1/(2sinx)( 1 - cos 2x + cos 2x - cos 4x + cos 4x - ....

(1/(2sinx) (1 -  cos (100)x()  (All  middle terms cancel out...)
Similary the other series
Use  2 sin A cos B = sin (A+B) + sin (A-B)

and  (1/(2sin x) ( 2 sinx cos 2x + 2 sin x cos 4x + 2 sin x cose 6 x ..)

can again be reduced to (1/2sinx) ( sin 3x  - sin x + sin 5x -sin 3x ...)

All but the first and last term  will cancle out nicely ..


hey!Είμαι καινούρια στο forum! Μπραβο πολύ καλή προσπάθεια!
Έχω μία ερώτηση για σας..πως θα αποδεικνύατε ότι υπάρχει μοναδική έκφραση μιας δύναμης του k ως γραμμικός συνδιασμός παραγοντικών πολυωνύμων?thanxxxxxxxxxxx.

Χρησιμοποίησε τον τύπο του Newton. Στην προκειμένη περίπτωση, οι συντελεστές είναι οι περίφημοι αριθμοί του Stirling δευτέρου είδους. Υπάρχει συνδυαστική καθώς και αλγεβρική απόδειξη. Βλ. οποιοδήποτε καλό βιβλίο Διακριτών Μαθηματικών ή Συνδυαστικής.

Lance


[quote="eryaman"]Have you guys decided whether or not this will be offered in the winter?  :?:[/quote]

Unless we get a significant flood of enrollments in the AIME B, we will not be scheduling the AIME A.


Let $ a,b,c>0$, prove that $ \sqrt{1\plus{}a^2\plus{}b^2}\plus{}\sqrt{1\plus{}b^2\plus{}c^2}\plus{}\sqrt{1\plus{}c^2\plus{}a^2}\geq \sqrt{9\plus{}2(a\plus{}b\plus{}c)^2}$

well it is just minkowsky,isnt it? :| $ LHS\geq\sqrt{3^2\plus{}(a\plus{}b\plus{}c)^2\plus{}(a\plus{}b\plus{}c)^2}\equal{}RHS$

Yes, it's easy  :lol:
And what about this problem?
Let a,b,c>0. Prove that:
$ P\equal{}\frac{1}{2(a\plus{}b)}\plus{}\frac{1}{3(b\plus{}c)}\plus{}\frac{1}{6(c\plus{}a)}\geq \frac{6}{4a\plus{}5b\plus{}3c}$

well(andreescu)
$ LHS\equal{} \frac{9}{18(a\plus{}b)}\plus{}\frac{4}{12(b\plus{}c)}\plus{}\frac{1}{6(c\plus{}a)}\geq\frac{36}{24a\plus{}30b\plus{}18c}\equal{}RHS$

QED

I will add one :
if $ ABC$ is a triangle , $ AB \equal{} a,BC \equal{} b,AB \equal{} c$
prove that :
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}\le 2$
sorry for bad english
I'm still learning

[quote="mathson"]I will add one :
if $ ABC$ is a triangle , $ AB \equal{} a,BC \equal{} b,AB \equal{} c$
prove that :
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}\le 2$
[/quote]
$ \sum_{cyc}\frac{a}{b\plus{}c}<\sum_{cyc}\frac{2a}{a\plus{}b\plus{}c}\equal{}2.$


probability
geometry
analytic geometry
inequalities
trapezoid
ratio
integration

[quote="fizsle"]Bender orders two robot painters to paint a 1000 meter long circular fence. One robot painter is ordered to paint a randomly selected 720 meter stretch of the fence while the other isordered to paint a randomly selected 750 meter stretch of the fence. Find the probability that when the robot painters are done, the entire fence has been painted.

my answer is 1/250 .... is this right???[/quote]
[hide="I got a different answer"]$\frac{2}{280\cdot250}=\frac{1}{35000}$[/hide]

[hide="I got"]47/100. Lets say that one robot paints 750 meters of the fence. Next, the other robot paints 720 meters. For the second robot to paint the extra 250 meters of fenced area, it could start painting anywhere from $720-250=470$ meters away from the empty space to right at the empty space, so it could start from 470 of the 1000 starting places.[/hide]

or it could start from the other direction... so it should be 47/50

No, because if they go in the other direction, the probability of the whole fence getting painted is also 47/100, and you would average the probabilities or something like that, to get 47/100.

i think it is 1/251 .....since there are 251 outcomes and only 1 is favorable... minimum length that will get painted is 750m...i.e incase they paint on top of each other...and maximum is 1000m so there are 251 possible outcomes...and we want only 1 out of this...

You keep disregarding the fact that the coordinates need not be integer - that is, if we put some zero point anywhere on the circle, then one robot can paint the stretch between, let's say, $0.4$ and $720.4$, and the other one between, lets say, $13+\sqrt{2}$ and $763+\sqrt{2}$. Therefore, the number of both favorable and possible outcomes is infinite and you can't apply the counting method.

[quote="Dr. No"][quote="fizsle"]Bender orders two robot painters to paint a 1000 meter long circular fence. One robot painter is ordered to paint a randomly selected 720 meter stretch of the fence while the other isordered to paint a randomly selected 750 meter stretch of the fence. Find the probability that when the robot painters are done, the entire fence has been painted.

my answer is 1/250 .... is this right???[/quote]
[hide]
No, but a good try nonetheless.

The probability that our 750 meter painter leaves a given sector unpainted is $\frac{1000-750}{1000}= 0.25$

Likewise, the probability that our 720 meter painter leaves a given sector unpainted is $\frac{1000-720}{1000}= 0.28$

The joint probability that both robots leave the same sector unpainted is $0.25\times 0.28 = 0.07$

Therefore, the probability that the entire fence is painted is:
$P(\text{painted}) = 1-P(\text{unpainted}) = 1-0.07 = 0.93$
[/hide][/quote]

The answer is $\frac{47}{100}$ - think of it this way, it doesn't matter where the first painter paints.

I agree with Treething.
Once the first robot painter has finished, we can choose the origin so that the interval painted is $(0,750)$.
The other painter will paint $(x, 720+x)$ ($\mod 1000$ in the circular sense).
The limits of x so that all the fence is painted are when $x=750$ or $720+x=1000$.
The interval of successful x measures $750+720-1000$.
So the probability of success is $\frac{750+720-1000}{1000}\equiv 47\%$.
In general it's $\frac{L_{1}+L_{2}-L}{L}$ if $L\leq L1+L2\leq 2L$.

[quote="Dr. No"][quote="Treething"]The answer is $\frac{47}{100}$ - think of it this way, it doesn't matter where the first painter paints.[/quote]
You've failed to take that fact into account in your supposed solution.

Your answer, 47%, is the proportion of fence that will recieve two coats if one robot begins exactly where the other leaves off. This number has nothing to do with the chances that the entire fence will be painted!

The robot with 750 meters of paint will not paint 250 meters of fence. Likewise the robot with 720 meters of paint will not paint 280 meters of fence. The probability that a given point lies within the 250 meter interval unpainted by Robot_750 is 0.25. The probability that a given point lies within the 280 meter interval unpainted by Robot_720 is 0.28.  These events are independent such that the probability that a given point lies within the intersection of the two unpainted intervals -- the definition of what it means to leave some part of the fence unpainted -- is $0.25 \times 0.28 \equiv 0.07$

$P(\text{all painted}) = 1-P(\text{some unpainted}) = 1-0.07\equiv 0.93$.[/quote]

$470$ is the length of the arc within which the second painter can begin his painting and end up painting the entire fence.

Change the problem a bit. Say one person only paints $100$ meters; the other only paints $200$ meters. According to your method, the probability that the entire fence is painted is equal to $1-\frac{9}{10}\times \frac{8}{10}= \frac{7}{25}$, which is clearly false.

[quote="Dr. No"][quote="Fraxia"]
$470$ is the length of the arc within which the second painter can begin his painting and end up painting the entire fence.[/quote]True, but that does not answer the question.
[/quote]

Actually, I think that it does answer the question. The second painter can start his painting in $470/1000$ of the fence, so the probability that the painter paints the right bit is $47/100$

[quote="Treething"]well use common sense, it's obviously not 0.93 or $\frac{1}{3113513947210378109}$ or whatever the other people said...[/quote]
That doesn't answer my question.

For shortness and for keeping the generality of the solution, put $a=\frac{720}{1000}=0.72$ and $b=\frac{750}{1000}=0.75$.

Let's put some angular coordinates. Choose an arbitrary ray from the center of the circle as a starting point, and let robot $A$ (one which paints $a$-th part of the fence) start at angle $x$, and the robot $B$ (one which paints $b$-th part of the fence) start at angle $y$. Both angles are measured on the interval between $0^\circ$ and $360^\circ$.

If $x\geqslant y$, then we must have $x\leqslant y+b\cdot 360^\circ$ (that is, the first robot mustn't start its painting past the endpoint of the second robot's stretch) and $x+a\cdot 360^\circ\geqslant y+360^\circ$ (that is, when the first robot paints its part, he must at least reach the starting point of the second robot). If we graph these inequalities, we find that they describe a trapezoid whose area is $A_{1}=\left({a^{2}\over 2}-{(1-b)^{2}\over 2}\right)\cdot 360^\circ^{2}$

If $x\leqslant y$, then we must have $y\leqslant x+a\cdot 360^\circ$ (that is, the second robot mustn't start its painting past the endpoint of the first robot's stretch) and $y+b\cdot 360^\circ\geqslant x+360^\circ$ (that is, when the second robot, paints its part, he must at least reach the starting point of the first robot). Similarly, we find that the area described by these conditions is $A_{2}=\left({b^{2}\over 2}-{(1-a)^{2}\over 2}\right)\cdot 360^\circ^{2}$.

Adding up those areas and dividing by $360^\circ^{2}$, which is the area of all the possible outcomes, we find (after simplifying) that $P=a+b-1=0.47$, which analytically proves that ${47\over 100}$ is the correct answer.

[quote="i_like_pie"]There have been several different assumptions. What is the correct answer?[/quote]

Either of the painters may start at any point on the circle, i.e. the starting points are random.

Let $A$ be the event that a particular given point is painted by the 750 meter painter; likewise let $B$ be the event that said point is painted by the 720 meter painter.

The probability that said point is painted by one painter, or the other, or both is $P(A\cup B)$.

$P(A\cup B) = P(A)+P(B\cap A^{c})$

$P(B\cap A^{c}) = P(B)-P(A\cap B)$

$\Rightarrow P(A\cup B) = P(A)+P(B)-P(A\cap B)$.

$P(A) = 0.75$.

$P(B) = 0.72$.

Since $A$ and $B$ are independent $P(A\cap B) = P(A)\cdot P(B) = 0.75\times 0.72 = 0.54$.

$\therefore P(A\cup B) = 0.75+0.72-0.54 = 0.93$.

$\boxed{\text{The probability that the entire fence is painted is }0.93.}$

Dr No, you shouldn't work with isolated points (each point is not independent), but with intervals (only the two starting points are independent).

[quote="Dr. No"][quote="Fraxia"]
$470$ is the length of the arc within which the second painter can begin his painting and end up painting the entire fence.[/quote]True, but that does not answer the question.

[quote="Fraxia"]Change the problem a bit. Say one person only paints $100$ meters; the other only paints $200$ meters. According to your method, the probability that the entire fence is painted is equal to $1-\frac{9}{10}\times \frac{8}{10}= \frac{7}{25}$, which is clearly false.[/quote]
You've constructed a straw man fallacy. In the problem as stated together both robots use an adequate amount of paint to completely cover the entire fence. In your variation they do not.[/quote]

I would call it a straw man fallacy if your solution was created under the assumption that an adequate amount was painted, and actually used that fact. But, your solution doesn't distinguish between whether or not there is an adequate amount and you can't just arbitrarily create these limitations.

Your method doesn't work because those two probabilities aren't independent because we're talking about location. For example, in the case where the arcs that the two painters do not pint overlap, if a given point is in one of the unpainted arcs, it can't be in the other at the same time. The location of a point in one arc has an impact on whether or not it's in a different arc.

[quote="fizsle"]I have 2 questions now....
1)why is it not 1/251
2) Why is everyone concerned about the circle??? the length is 1000m so why bother if it is a circle or square....or am i missing out something???[/quote]

1) It is not $\frac{1}{251}$ because you cannot calculate the required probability by using a discrete probability distribution, as you seem to have tried to do.

2) No, you're not missing anything. The fact that the interval is a circle is a spoiler thrown into the problem to lead you astray. The problem is an ideal example of a uniform probability distribution. A uniform probability distribution is one in which the probability of an event occuring is proportional to the length of the interval in which it occurs. The intervals in question here are merely laid out as arcs on a circle.

[quote]I would call it a straw man fallacy if your solution was created under the assumption that an adequate amount was painted, and actually used that fact.[/quote]
[color=blue]It was and it did. Between the two robot painters there 470 meters more paint than is required to paint the entire fence.[/color]

[quote]But, your solution doesn't distinguish between whether or not there is an adequate amount and you can't just arbitrarily create these limitations.[/quote]
[color=blue]I didn't find it necessary to explicitly state that assumption since it is part of the problem statement. For your example, what you are actually calculating is the conditional probability that an arbitrary point on the fence is painted given an insufficient amount of paint to cover the entire fence, and this is in fact 7/25, just as you calculated.[/color]

[quote]Your method doesn't work because those two probabilities aren't independent because we're talking about location. For example, in the case where the arcs that the two painters do not pint overlap, if a given point is in one of the unpainted arcs, it can't be in the other at the same time. The location of a point in one arc has an impact on whether or not it's in a different arc.[/quote]

[color=blue]They are independent. Where one robot starts to paint has no influence on where the other robot starts to paint. The choices of starting positions are random; it says so in the problem statement.[/color]

ok guys... i was bored and made a program to do this thing repeated times during study hall

i repeated it 1337 times and it worked a little more than 600 times... i'd say that's close to $\frac{47}{100}$

[quote="K81o7"]Wow...this has become a huge argument over something so small..

It's simply the probability that a randomly chosen real number $x$ in the interval $[0, 1000)$ satifies $x \ge 280, x \le 750$...so the probability is just $\frac{47}{100}$...(ie x is the robot's starting point.

Now, Dr. No, you calculated that the probability that a given point will be painted is $0.93$.  But that's not the probability that the whole fence will be painted...[/quote]

I can’t believe it’s taken me so long to figure this out.

We have two uniformly distributed random variables, $X_{1}$ and $X_{2}$.

$X_{1}\le 720$ and $X_{2}\le 750$.

$P(X_{1}+X_{2}\le 1470) = 1$.

$P(X1+X_{2}\le 1000) = \frac{1000}{1470}$

We want $P(X_{1}+X_{2}> 1000)$.

$P(X_{1}+X_{2}> 1000) = 1-P(X_{1}+X_{2}\le 1000) = 1-\frac{1000}{1470}= \frac{470}{1470}$

$\boxed{\therefore P(X_{1}+X_{2}> 1000) = \frac{47}{147}}$

[quote="fizsle"]i think we require to find P( x1+ x2 >= 1000) as 1000 will also satisfy our requirement. Also another thing i'm confused about is that anything >1000 will create a duplicate sample space...for eg 1010 metres would be the same as 10 metres (since it is a circular fence) which is already included in the sample space... i'm not too sure but there seems some flaw in the logic... :oops:[/quote]
fizsle,

If the sum $X_{1}+X_{2}> 1000$ then the entire fence is painted. Some points will be painted more than once since it is a circular fence but the goal will be met.

Further the '=' part of the inequality is always given as "$\le$" rather than "$\ge$" for reasons of definition. $Y$ is a random variable if and only if $\lbrace \xi\colon Y(\xi)\le t\in \mathbb{R}\rbrace$ is an event. Also $P(Y = a) = 0$, where $a$ is an arbitrary point in the range of $Y$. In other words, the probability that $X_{1}+X_{2}$ will equal exactly 1000 is zero.

[quote="lordWings"]Sorry, Dr. No. But I think that your interpretation is not what the problem asks.
The only uniformly distributed random variables in this problem are the starting point of each robot, not the meters painted.
Can you define your variables $X_{1}$ and $X_{2}$?
Check the solutions with final result $470/1000$.[/quote]
The result $470/1000$ is incorrect.

The probability that the entire fence is painted must equal the probability that A or B or both painted an arbitrary section of fence, $P(A\cup B)$.

Since $A\subset (A\cup B)$, $P(A)\le P(A\cup B)$, and since $B\subset (A\cup B)$, $P(B)\le P(A\cup B)$.

If according to you $P(A\cup B) = 0.47$, while $P(A) = 0.72$ and $P(B) = 0.75$ then $0.72 \le 0.47$ and $0.75\le 0.47$. Clearly neither is correct. Therefore, by [i] modus tollens[/i] it is not the case that $P(A\cup B) = 0.47$.

By contrast the question I answered was:

[i]Given that it is possible to paint a strip 1470 meters long, what is the probability that such a strip is at least 1000 meters long?[/i].

I assume that:
- Robot number $1$ paints a continuous extension of $720$m of the circular fence.
- Robot number $2$ paints a continuous extension of $750$m of the circular fence.

If the minimum distance between the center points of those two extensions is $x\in (0,500)$, it can be observed that:
- when $x\in (0,15)$ there is a total overlapping and $250$m of the fence aren't painted.
- when $x\in (15,265)$ there is just a partial overlapping and the non-painted extension of the fence decreases linearly from $250$m to $0$m.
- when $x\in (265,500)$ there is partial overlapping on both sides, so the whole circular fence is painted.

If $x\in (0,500)$ is a uniformly distributed random variable, the probabilities of the three cases are $3\%$, $50\%$ and $47\%$ respectively.

I think Dr. No works with other assumptions, but I'm not sure if I understand them.
I think he assumes that each one of his robots have a can of paint enough to paint $720$m and $750$m respectively, and then they paint independent points randomly until they have finished their cans.

[quote="Dr. No"]
We have two uniformly distributed random variables, $X_{1}$ and $X_{2}$.

$X_{1}\le 720$ and $X_{2}\le 750$.

$\boxed{P(X1+X_{2}\le 1000) = \frac{1000}{1470}}$
[/quote]

The boxed is wrong. Since $X_{1}$ and $X_{2}$ are independent and continuous, we must plot the graph and use geometric probabilities, which show that $P(X_{1}+X_{2}\le 1000) = 1-\frac{47^{2}}{2\cdot 75\cdot 72}$

Also, take a good look what your variables are. You work with the presumption that the robots paint ANYWHERE between $0$ and $720$ or $0$ and $750$ respectively, and that's simply not the case. Hence you work NOT with the starting points, but with the segment LENGTHS, which steers you off the course.

[quote="Farenhajt"][quote="Dr. No"]
We have two uniformly distributed random variables, $X_{1}$ and $X_{2}$.

$X_{1}\le 720$ and $X_{2}\le 750$.

$\boxed{P(X1+X_{2}\le 1000) = \frac{1000}{1470}}$
[/quote]

The boxed is wrong. ...[/quote]
No, it's not wrong.

There are two random variables in this problem, $X_{1}\text{ and }X_{2}$, the lengths painted by the first and second robot respectively. Adding these together, $X_{1}+X_{2}\le 1470\text{ meters}$.

Now paint a strip 1470 meters long in a straight line along the ground. Next cut the fence vertically and lay it out flat parallel to the strip that you painted, with one end flush with an end of your painted strip. These flush ends you can consider the starting point of the first painter. If the ending point of the second painter does not fall beyond the far end of the fence, i.e. $X_{1}+X_{2}\le 1000$, then the painters will have failed to paint the entire fence. The probability that this the case is the ratio of length of the fence to the length of the strip painted on the ground, $P\left(X_{1}+X_{2}\le 1000\right) = \frac{1000}{1470}$.

The desired result is the complementary probability that the painters did not fail to paint the entire fence, $P\left(X_{1}+X_{2}> 1000\right) = 1-P\left(X_{1}+X_{2}\le 1000\right) = 1-\frac{1000}{1470}= \frac{470}{1470}$.

Since the fence is circular any distance greater than 1000 meters will result in sections already painted being painted again; nevertheless, 1000 meters is the success criterion, not the sample space.

[quote="perfect_radio"][quote="Dr. No"]There are two random variables in this problem, $X_{1}\text{ and }X_{2}$, the lengths painted by the first and second robot respectively. Adding these together, $X_{1}+X_{2}\le 1470\text{ meters}$.[/quote]

versus

[quote="fizsle"]Bender orders two robot painters to paint a 1000 meter long circular fence. [size=125][color=blue][b]One robot painter is ordered to paint a randomly selected 720 meter stretch of the fence while the other is ordered to paint a randomly selected 750 meter stretch of the fence.[/b][/color][/size] Find the probability that when the robot painters are done, the entire fence has been painted.

my answer is 1/250 .... is this right???[/quote]

PS: If your interpretation was true, then the answer would be $\frac1{720}\cdot \int_{0}^{720}\frac{\min \left\{ 1000-x, 750 \right\}}{750}\, dx$, which gives the same numerical value Farenhajt obtained.[/quote]
It is unclear what distribution you're using.

Assuming $X_{1}\text{ and }X_{2}$ are uniformly distributed, subject to the constraint that each robot must use all of his paint (the section of fence painted is at least 750 meters long), what we want is actually: $P\left(1000 < X_{1}+X_{2}\le 1470| X_{2}= 750\right) =\frac{1}{720}\int_{250}^{720}dt = \frac{(720-250)}{720}=\frac{470}{720}= \frac{47}{72}$.

What we really want is this: $\frac1{1000}\cdot \int_{0}^{1000}\left[ x \geq 280, \, x \leq 750 \right] \, dx$, where $[P(x)]$ is $1$ if $P(x)$ is true and $0$ otherwise.
See K81o7's solution for more details.
[quote="K81o7"]Wow...this has become a huge argument over something so small..

It's simply the probability that a randomly chosen real number $x$ in the interval $[0, 1000)$ satifies $x \ge 280, x \le 750$...so the probability is just $\frac{47}{100}$...(ie x is the robot's starting point.

Now, Dr. No, you calculated that the probability that a given point will be painted is $0.93$.  But that's not the probability that the whole fence will be painted...[/quote]
Do we agree now?

I got yet another different answer(this agrees with above). If one painter can paint 720 and the other 750, then together, if both of them have to paint, the minimum distance around the arc that they paint is 750(they paint over each other as much as possible). Next, if they happen to paint the max., they can go to 1470. Clearly, this is over 1000, but is nonetheless sucessful. It is a possible outcome. Next, since there are a total of 1470-750 total distance possible. Since only distances 1000 and up work then there are 1470-1000 sucessful outcomes that work. Therefore P(even)= 47/72. Such logic can be easily obtained from the "three types of probability" in the articles section.


Find all complex numbers $ z$ such that $ 4z^2 \plus{} 8|z|^2 \equal{} 8$.
My solution
$ 4z^2 \plus{} 8|z|^2 \equal{} 8$
$ \Leftrightarrow z^2 \plus{} 2|z|^2 \minus{} 2 \equal{} 0$
$ \Leftrightarrow |z|^2 \equal{} \frac {2 \minus{} z^2}{2}$
Since the LHS is real the RHS has to real
It is only real if $ z \equal{} yi$ or $ z \equal{} x$
Case 1: $ z \equal{} yi$
$ y^2 \equal{} \frac {2 \plus{} y^2}{2}$

$ y \equal{} \pm\sqrt {2}$
$ z \equal{} \pm\sqrt {2}i$

Case 2: $ z \equal{} x$
$ x^2 \equal{} \frac {2 \minus{} x^2}{2}$

$ x \equal{} \pm\frac {\sqrt {6}}{3}$

$ z \equal{} \pm\frac {\sqrt {6}}{3}$
$ \therefore z \equal{} \pm\frac {\sqrt {6}}{3}$ or$ z \equal{} \pm\sqrt {2}i$

The part i am doubtful is about is there any other $ z$ such that $ z^2$ is real

Let $ z \equal{} a \plus{} bi$. So, $ z^2 \equal{} a^2 \plus{} 2abi \minus{} b^2 \Rightarrow 2abi \equal{} 0$. Hence, either $ a \equal{} 0$ or $ b \equal{} 0$. You're fine.


function
induction
combinatorics unsolved
combinatorics

Define a sequence of strings as follows: Let $R_{0}$ = 1 , and let $R_{n}$ be the string obtained by replacing each instance of  $i$ in $R_{n}-1$ with $123...i$ and finally adding $n+1$ at the end. So, (for example) $R_{1}= 12$, $R_{2}= 1123$, $R_{3}= 11121234$. Show that if we write $R_{n}$ down and $R_{n}$ backwards under that, then each column will contain exactly one $1$ .

I see why it works (sort of), but I can't formulate a rigorous proof. Obviously you want to use induction. A technique may be to instead of using all integers, just use primes, then multiply numbers in the same column, and show that you only get primes (or 1). For instance...

1123
3211
____
3223 (only primes)

You may have trouble when you get to double digits though. It seems though that your sequence loses a lot of its splendor when you get to double digits. I would keep 10 as one 'digit' just to preserve the beauty.

Sorry for my clumpsy unintelligible formalization.

Call $S$ the set of all finite strings of natural numbers, $L(\alpha)$ the length of the string $\alpha$ and $\alpha(i)$ the i'st element in $\alpha$. Given $\alpha,\beta\in S$, define the relation $Q(\alpha,\beta) \leftrightarrow [ L(\alpha)=L(\beta)=L \wedge [\alpha(i)=1\leftrightarrow\beta(L-i+1)\neq1 ]]$. Note that $[Q(\alpha,\chi) \wedge Q(\beta,\delta)] \rightarrow Q(\alpha\beta,\delta\chi)$. More in general, if we define the "great-yuxtaposition operator"  $\Gamma_{i=k}^{l}[g(i)]=g(k)g(k+1)g(k+2)...g(l)$  where the $g(i)$ are strings, we have that  $[\forall i|0\leq i\leq (l-k)\; (Q(g(k+i),h(l-i))]\rightarrow Q(\Gamma_{i=k}^{l}[g(i)],\Gamma_{i=k}^{l}[h(i)])$.
What we need to show is that $Q(R_{n},R_{n})\;\forall n \in N_{0}$.
Let $f(\alpha)$ be the string obtained by replacing in the string $\alpha$ each $\alpha(i)$ ($i\leq L(\alpha)$) by $12...\alpha(i)$. Note that $f(\alpha\beta)=f(\alpha)f(\beta)$. Using this notation the recursion generating our sequence can be written as $R_{n}=f(R_{n-1})n$
$R_{0}=1=f^{0}(1)=\Gamma_{i=0}^{0}[f^{0-i}(i+1)]$ and $R_{n-1}=\Gamma_{i=0}^{n-1}[f^{n-1-i}(i+1)] \Rightarrow$
$R_{n}=f[\Gamma_{i=0}^{n-1}[f^{n-1-i}(i+1)]]n=[\Gamma_{i=0}^{n-1}[f(f^{n-1-i}(i+1))]]n=$
$=[\Gamma_{i=0}^{n-1}[f^{n-i}(i+1)]]n=\Gamma_{i=0}^{n}[f^{n-i}(i+1)]$, (where the exponents indicate the number of iterations of the function $f$ and the argument below is the 1-number string to which we start applying the function) so the formula holds by induction $\forall n \in N_{0}$.
Let's now analize $g(n,m)=f^{n}(m+1)$. $g(n,0)=f^{n}(1)=1$ and $g(0,n)=f^{0}(n+1)=n+1$, so $Q(g(n,0),g(0,n))$ $\forall n \in N$.
Also,$g(n,m)=f^{n-1}(f(m+1))=f^{n-1}(12...m(m+1))=$
$=f^{n-1}(12...m)f^{n-1}(m+1)=f^{n-1}(f(m))f^{n-1}(m+1)=$
$=f^{n}(m)f^{n-1}(m+1)=g(n,m-1)g(n-1,m)$. If we suppose that $Q(g(a,b),g(b,a))$ holds  $\forall a,b|a+b<m+n$, then, having, $g(m,n)=g(m-1,n)g(m,n-1)$  and  $g(n,m)=g(n-1,m)g(m-1,n)$,  $Q(g(m,n),g(n,m))$  by the observation made at the beginning, so the statement holds $\forall m,n\in N|m+n>0$.
Now, we had that  $R_{n}=\Gamma_{i=0}^{n}[g(n-i,i)]$, and from this, $Q(g(n-i,i),g(i,n-i))$ and the generalization at the beginning it follows that $Q(R_{n},R_{n})$  $\forall n \in N_{0}$,  which is what we wanted to prove.

[ I have considered strings of numbers, not digits, because otherwise $R_{10}$ would begin with 11 and end with 10, which contraddicts it fulfilling the property, as gopherhole112 said. ]


Provide a syntesis of 2-Bromo-3-phenylpropanoic acid from diethyl malonate.

1. EtONa to abstract the proton
2. C6H5CH2Br
3. Hydrolysis of the compd to get 3 phenyl propanoic acid
4. $ \alpha$ bromination
 :)

Yes, that's right.

And how about 2-bromo-2-phenylacetic acid?

1. EtONa in ethanol
2. 1,5 dibromopentane
3. EtONa in ethanol
4. dil. NaOH, H+
5. Heat
6. Ardnt Eitstert
7. Pd/Se/C (aromatisation, my iit prof told me this to convert cyclohexane into benzene)
8. $ \alpha$ bromination
 :)

well here is a long but a method which i beleive carcul gave as a question long long long ago
first do a free radical bromination of cyclohexane
then dehydrhalogenation
then $ 2 eq NBS$
then dehydrobromination

[quote="Madness"]Pd/Se/C (aromatisation, my iit prof told me this to convert cyclohexane into benzene)[/quote]

Ha ha. That's a joke right? And exactly who would do that in the laboratory (except your teacher)? And what would happen to the rest of the molecule? Come on, give it a real thought ok.

[quote="Pardesi"]well here is a long but a method which i beleive carcul gave as a question long long long ago[/quote]

Again, what would happen to the rest of the molecule? And wouldn't it be sympler to start already with a benzene derivative??

OK  :D  By what pardesi said (for aromatisation) is my method right?  :D

iit books have strange quetsions..strange reactions and strange names...
i wouldb't be surprised if Mozingo is a hypothetical guy :P


geometry
geometric transformation
homothety
geometry proposed

Let $A_1A_2A_3A_4$ --- convex quadrilateral, point $P$ inside it. $S_i$ --- incircles of $A_iPA_{i+1}$
($A_5=A_1$) with centers $O_i$. $S_{i+1}$ touch $S_i$ and $S_{i+2}$($i=1,2,3,4, S_5=S_1, S_6=S_2$). Proof, that lines $O_1O_2$, $O_3O_4$, $A_1A_3$ are concurent or parallel.

Since $S_i$ tangents $S_{i+1}$, we have $A_2P+A_3P-A_2A_3=A_3P+A_4P-A_3A_4=A_4P+A_1P-A_4A_1=A_1P+A_2P-A_1A_2$. Then the quadrilateral $A_1A_2A_3P$ is curcumscribed. Let $\omega_2$ be its incircle. There is a homothety $H_{12}$ with center $A_1$ and positive coefficient s.t. $H_{12}(S_1)=\omega_2$; and homothety $H_{32}$ with center $A_3$ and positive coefficient s.t. $H_{32}(\omega_2)=S_2$. Then the homothety $H_{2}=H_{12} \circ H_{32}$ with positice coefficient and some center $C_{2}$, that lies at $A_1A_3$, maps $S_1$ to $S_2$. Hence the points $C_2$, $O_1$ and $O_2$ are collinear. So, $C_2=O_1O_2 \cap A_1A_3$. Similarly, $C_4=O_3O_4 \cap A_1A_3$, where $C_4$ is a center of the homothety that maps $S_3$ to $S_4$.

Let $S_1$ thangents $A_1P$ at $X$ and $S_2$ touches $A_3P$ at $Y$. Let $X'=H_{2}(X)$ and $P'=H_{32}(P)$. Then $XP=PY$ and $X'P'=P'Y$. Therefore the points $X$, $X'$ and $Y$ are collinear. It means that the points $X$, $Y$ and $C_2$ are collinear. But it means that $C_2=XY \cap A_1A_3$. Similarly, $C_4=XY \cap A_1A_3$. Hence $C_2=C_4$ and we are done.

[quote] Then $XP=PY$ and $X'P'=P'Y$ . [/quote]
Why?
[quote] Therefore the points $X, X'$  and $Y$  are collinear.[/quote]
$Y\in XP$???

Oh, there was a typo. I'm sorry.

Instead of "$X'=H_{12}(X)$" I had to write "$X'=H_2(X)$". I have just edited my previous post.

The points $X$, $X'$ and $Y$ are collinear since $XP=PY$, $X'P'=P'Y$ and $XP \parallel X'P$ (it is easy to see that $H_2(XP)=X'P'$, where I mean not the segments but the lines $XP$ and $X'P'$).

If you have any more questions, please ask.

[quote="fagot"][quote]But it means that $C_2=XY\cap A_1A_3$ .[/quote]
If $A_1, P, A_3$ are collinear?[/quote]
No. I don't understand why they must be collinear.

We have shown that $X$, $X'$ and $Y$ are collinear. But $X'=H_2(X)$ Therefore, $C_2$, $X$ and $X'$ are collinear. From the last two collinearities we have that the points $C_2$, $X$ and $Y$ are collinear, that is $C_2 \in XY$. Earlier we have proved that $C_2 \in A_1A_3$. It means that $C_2=XY \cap A_1A_3$.

If $A_1, P, A_3$ are collinear, then the line $XY$ the same, that  $A_1A_3$. Then maybe $C_2, C_4\in XY$,
but $C_2\ne C_4$.

Yes, you are right. This is the special case. But it is rather obvious. Indeed, it is easy to see that $H_2(X)=Y$ (since $X'=Y$) and $H_4(X)=Y$. Then $C_2=C_4$, since they both map $X$ to $Y$.

[quote="fagot"][quote]Then , $C_2=C_4$ since they both maps $X$ to $Y$.[/quote]
Why? Exist many homoteties, which maps $X$ to $Y$.[/quote]
Ok. By definition, $H_2=H_{12} \circ H_{32}$. The homothety $H_{12}$ has center $A_1$ and it maps $X$ to $P$ (since the quadrilateral $A_1A_2A_3P$ is circumscribed and it is a triangle indeed, so the point $P$ must be a tangency point of incircle $\omega_2$ to the side $A_1A_3$). And the homothety $H_{32}$ has center $A_3$ and it maps $P$ to $Y$.

Similarly, $H_4=H_{14} \circ H_{34}$, where $H_{14}$ has center $A_1$ and maps $X$ to $P$, and $H_{34}$ has center $A_3$ and maps $P$ to $Y$.

But there is only one homothety with center $A_1$ that maps $X$ to $P$. Therefore $H_{12}=H_{14}$. Similarly, $H_{32}=H_{34}$. Consequently $H_2=H_{12} \circ H_{32}=H_{14} \circ H_{34}=H_4$. Hence $C_2=C_4$ and we are done.


AwesomeMath
summer program
geometry
email
pigeonhole principle

Solutions to Admission Test A were posted earlier this week. They're available on the AwesomeMath website here:

http://www.awesomemath.org/tests/TestA(2007)sol.pdf
(Copy and paste into address bar - I'm posting from a machine that I believe to be powered by a Super Nintendo, so I'm not sure if my BBCode is working. I'll fix it later.)

Most of the solutions in that file are from submitted papers, but we included a few of our own as well (particularly on the harder geometry problems.)

[quote="roadnottaken"]Are you sure that your computer was powered by a super nintendo?  I think you should check to make sure it doesn't have vaccuum tubes :P :roll: .[/quote]
Okay, that may have been a little harsh on my part.  But I would think that if you typed in the bbcode manually, it would still work (this coming from someone who is not at all well versed in how computers work :maybe: ).

[quote="zacacox"][quote="Aegor"]Very nice solutions, and so much more concise than mine! :P  What is the next step?  For those of us who got accepted early, the money is due on March 1st, but do we receive more information after that?  Thanks for any enlightenment.[/quote]

You'll receive a packet with a lot of fun and exciting forms, packing information, rules, and pre-camp problems sometime towards the end of the school year/beginning of the summer. We'll probably send these out all at once, so it'll be after the application process is done with.

-Zac[/quote]
When do we find out if our payment was recieved?

[quote="Aegor"][quote="bpms"]
When do we find out if our payment was recieved?[/quote]

If you applied early and sent in your check, they send an email (they did to me).  Also, I recently received a packet with numerous problems concerning the Pigeonhole Principle and other theorems.[/quote]
Eh, I just sent out the check on the 2nd.
That's when it needs to be sent by right?

[quote="bpms"]
Eh, I just sent out the check on the 2nd.
That's when it needs to be sent by right?[/quote]

From the site:  "Payment Receipt Deadline (Tuition)" March 2nd.  So I don't know whether they need it by then or it needs to be postmarked by then.


For what values of $ x$ does the inequality hold?

$ \log_{10}{\left(0.4\right)}\,\cdot\,\log_{2}{\left(0.5^{x \minus{} 5}\right)}\ \le\ \log_{0.4}{\left(x \plus{} 2\right)}$

[color=green]Mod edit: Changed commas to decimal points.[/color]

I'm pretty sure log is not multivariable...

hello, defining $ f(x)=\frac{\ln(x+2)}{\ln(\frac{2}{5})}-\frac{\ln(\frac{2}{5})\ln(\left(\frac{1}{2})^{x-5}\right)}{\ln(10)\ln(2)}$, then we get $ f^'(x)=\frac{1}{(x+2)\ln(\frac{2}{5})}+\frac{\ln(\frac{2}{5})}{\ln(10)}<0$ for $ x>-2$, further $ \lim_{x \to -2+0}f(x)=+\infty$, $ \lim_{x\to \infty}f(x)=-\infty$, so $ f(x)$ has an uniquely determined intersection point with the $ x$-axes $ x_0$ ($ \approx 1,536120$) and for $ -\infty<x\le x_0$ we have $ f(x)\geq0$.
Sonnhard.


[quote="joejia"]The circumference of a circle is $24\pi$. What is the number of square centimeters in the area of the largest square that can fit inside the circle?[/quote]
[hide="clicky here"]24 is the diameter so the are of the square would be 576sq.un. nevermind this is wrong...[/hide]

[hide]24 is the diameter,so, 12 is the radius. the largest possible square that can fit has side length 12*(2^0.5), so, the largest area is 288cm^2.[/hide]


Let P be a polynomial that belongs to $R[x]$, $P(x)=x^3+ax^2+bx+c$, a<>0, with the solutions $x_1, x_2, x_3$ which are distinct 2 by 2.
a) $\frac{x_1}{P'(x_1)} + \frac{x_2}{P'(x_2)} +\frac{x_3}{P'(x_3)}$ is equal to ?
b) Let $Q$ be a polynomial of $degre\; 1$ that belongs to $R[x]$, what is the value of $\frac{Q(x_1)}{P'(x_1)} + \frac{Q(x_2)}{P'(x_2)} +\frac{Q(x_3)}{P'(x_3)}$

Moderators : I think that i put this topic in a wrong forum, but i can not delete it.


A couple of you asked last night during the Math Jam about the Mandelbrot contest.  Info on Mandelbrot can be found at http://www.mandelbrot.org.  It's a mail in contest primarily for high schoolers.  I was one of the founders (after my first year of college), though am not directly involved in the contest anymore.  Sam Vandervelde (IMO 89, multiple time grader at MOP, all around swell guy) runs Mandelbrot.  The contest has a team and an individual portion; the team portion is typically a series of proofs that are great practice in mathematical writing and the type of thinking that goes into tackling really advanced problems like the USAMO/IMO (though the problems on the Team Test are not quite that difficult, primarily because they are broken into bite-size pieces - without these guiding steps, some of these would easily be IMO-level problems).

If you have any more questions about the contest, or any experience with it, you can post them here.

Can I get old copies of the test from the website??? Thanx

There are a few sample texts at the Mandelbrot web site; just go to http://www.mandelbrot.org and visit the "Resources" page, then follow the "Samples" link.  A whole lot more can be purchased directly from the resources page.


AMC
AIME
geometry
trigonometry
trig identities
Law of Sines

Distinct points $A$ and $B$ are on a semicircle with a diamter $MN$ and center $C$. The Point $P$ is on $CN$ and $\angle CAP = \angle CBP = 10^{\circ}$

If $\widehat{MA}=40^{\circ}$, then $\widehat{BN}$ equals?

from the 1983 AHSME.

Wow, this was only from the AHSME? Felt harder than most of the AIME problems from the same year. We went over a similar problem in Olympiad Geometry:

$\angle MCA=40^\circ$, so $\angle ACP=180^\circ-40^\circ=140^\circ$.
Since $\angle CAP=\angle CBP$, $ABPC$ is a cyclic quadrilateral. (Draw this in.)
$\angle ABP=180^\circ-\angle ACP=40^\circ$, since opposite angles are supplementary.
$\angle ABP=\angle CBA+\angle CBP$.
$\angle CBA=\angle ABP-\angle CBP=40^\circ-10^\circ=30^\circ$.
$\angle CPA=\angle CBA=30^\circ$ since they intercept the same arc.
Since $AC$ and $BC$ are both radii of the large circle, $AC=BC$ and $ABC$ is isoceles.
Thus $\angle CAB=\angle CBA=30^\circ$.
$\angle BAP=\angle BAC-\angle CAP=30^\circ-10^\circ=20^\circ$.
$\angle APB=180^\circ-\angle PBA-\angle BAP=180^\circ-40^\circ-20^\circ=120^\circ$ as three angles in $\bigtriangleup ABP$.
$\angle BPN=180^\circ-\angle MPA-\angle APB=180^\circ-30^\circ-120^\circ=30^\circ$. Thus $\widehat{BN}=30^\circ$.

Lots of room for error there...hope everything was clear and correct!

Well... once you see that $\angle BAC=30^\circ$ and $\angle ABC=30^\circ$, then $\angle ACB=120^\circ$, so $\angle BCN=180^\circ-40^\circ-120^\circ=20^\circ$, right?

I think everything you had was correct towersfreak2006 except for the fact that $\widehat{BN}$ is subtended by $\angle BCN$ rather than $\angle BPN$

How did you establish that $ABPC$ was a cyclic quadralateral?

This was the last problem on that year's ahsmc.

[hide="A way to do it with Trig"]All angle measures are in degrees.

Since an arc is equal to its central angle, $\angle ACM=40$.  It is clear that $\angle ACP=140$ and $\angle CPA=30$.
Without loss of generality, let the radius of circle C be 1.  Then $CM=CA=CB=CN=1$.

By the Law of Sines on $\triangle ACP$, we have
\[\frac{\sin{30}}{1}=\frac{\sin{10}}{CP}\]

and, since $\sin{30}=\frac{1}{2}$, $CP=2\sin{10}$.

By the Law of Sines on $\triangle BCP$,
\[\frac{\sin{10}}{2\sin{10}}=\frac{\sin{\angle BPC}}{1}\]

and thus $\sin{\angle BPC}=\frac{1}{2}$.  Looking at the diagram, $\angle BPC$ is obtuse, so $\angle BPC=150$. (We can also look at the answer choices and see that $\angle BPC=30$ does not give any of them).  I'm not sure how to prove this rigorously, but, since this is the AMC 12, I don't need to.

It is now clear that central angle $\angle BCN=180-150-10=20$, and thus
\[\text{arc}\; BN=\angle BCN=\boxed{20^\circ}\]
[/hide]


If $ a,b,c>0$ and $ abc\equal{}1$, then

$ \sum\frac{a(1\plus{}c\plus{}a\plus{}ab)\plus{}2bc}{(a\plus{}2b)^2}\plus{}\frac{(a^2\plus{}b^2\plus{}c^2)^3}{(ab\plus{}bc\plus{}ca)^3}\geq3$

[quote="hedeng123"][quote="filomen"]If $ a,b,c > 0$ and $ abc \equal{} 1$, then

$ \sum\frac {a(1 \plus{} c \plus{} a \plus{} ab) \plus{} 2bc}{(a \plus{} 2b)^2} \plus{} \frac {(a^2 \plus{} b^2 \plus{} c^2)^3}{(ab \plus{} bc \plus{} ca)^3}\geq3$[/quote]

only prove

$ \sum\frac {a(1 \plus{} c \plus{} a \plus{} ab) \plus{} 2bc}{(a \plus{} 2b)^2}\geq2$[/quote]

which the resolution?  :oops:


The arithmetic mean between $ \frac {x \plus{} a}{x}$ and $ \frac {x \minus{} a}{x}$, when $ x \not \equal{} 0$, is:

$ \textbf{(A)}\ {2}\text{, if }{a \not \equal{} 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a \equal{} 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x$

Sorry about that, the problem had a typo.

We have $ \frac{1}{2}\cdot \left(\frac{x \plus{} a}{x} \plus{} \frac{x \minus{} a}{x}\right) \equal{} \frac{2}{2} \equal{} \boxed{\text{(B) }{1}}$.

Other alternate way is by simply looking at each fraction as 1 + a/x and 1 - a/x. Adding them will give 2, and dividing by 2 will give 1. What sunehra did is a standard method, but for this problem, this is something you can also do (basically same but quicker).


I LOVE classical. Mine is Mozart. I also like Beethoven and Vivaldi.

My favorite is definitely Fryderyk Chopin! Have you heard his fourth ballade? It's an epic work of utmost genius that is almost definitely the most touching piece of piano music ever written, in my own and many others' opinions.

Franz Schubert would probably be second.

I also love the music of J.S.Bach, Beethoven, Rachmaninoff, Prokofiev, and Shostakovitch.

[quote="winner2009"]for those who appreciate classical music, who is your favorite CLASSICAL composer? mine is Chopin. original, with deep but subtle tones.[/quote]
should this be moved to the Art & Music section?

[quote="mz94"]rock & kpop all the way[/quote]

What is your favorite CLASSICAL composer?


The squares of a rectangular chessboard with 10 rows and 14 columns are colored alternatingly black and white in the usual manner. Some stones
are placed the board (possibly more than one on the same square) so that there are an odd number of stones in each row and each column.
Show that the total number of stones on black squares is even.

There is an odd number of stones on each column and there are 14 columns
Odd•14=even
Therefore total stones=even
Let a be the set of columns where the number of stones on black squares is odd
Let c be the set of rows where the number of stones on black squares is odd
Let d be the set of columns where the number of stones on black squares is even

Let the intersection of a and c have the number of stones on black squares be even
Then the intersection of a and d has an odd number of stones on black squares

Whenever there is an odd number of stones on black squares in a row or column , there is an even number of stones on white squares

Then the intersection of a and c have the number of stones on white squares be odd
Then the intersection of a and d has an even number of stones on white squares

Now I will find the parity of the number of stones in a
N stones in a= Odd + even + odd +even
= even
Each column has an odd number of stones, so since a is even
Number of columns in a is even
Therefore n black stones in a is even, as it is the set of all columns with an odd number of stones on black
N stones on black squares in b is even
Therefore, total black stones = even +even
=even

Repeat for when the intersection of a and c have the number of stones on black squares being odd

QED