function
calculus
derivative
inequalities unsolved
inequalities

If $a,b,c$ are real numbers $a\geq 1$ and $a+b+c=0$ then find the minimum value of $a^{4}+b^{4}+c^{4}-3abc$.

we have that

${a+b+c=0}$
${c=-(a+b)}$

Creating a function $f$ and substituing the value of c

${f(a,b)=a^{4}-3 b c a+b^{4}+c^{4}}$
${f(a,b)=a^{4}-3 (-a-b) b a+(-a-b)^{4}+b^{4}}$
${f(a,b)=a^{4}+3 b (a+b) a+b^{4}+(a+b)^{4}}$

For the minimum the partial derivatives must be equal to zero and/or inspetionate the box of the function will be ploted

${\frac{\partial f(a,b)}{\partial a}=0}$
${4 a^{3}+3 b a+4 (a+b)^{3}+3 b (a+b)=0}$
${(2 a+b) \left(3 b+4 \left(a^{2}+b a+b^{2}\right)\right)=0}$

${\frac{\partial f(a,b)}{\partial b}=0}$
${4 b^{3}+3 a b+4 (a+b)^{3}+3 a (a+b)=0}$
${(a+2 b) \left(4 b^{2}+4 a b+a (4 a+3)\right)=0}$

Solving this system

${\left\{ \begin{array}{l}(2 a+b) \left(3 b+4 \left(a^{2}+b a+b^{2}\right)\right)=0 \\ (a+2 b) \left(4 b^{2}+4 a b+(4 a+3) a\right)=0 \end{array}\right.}$

We get the following ordered pairs

${\left( \begin{array}{ll}-\frac{1}{4}&-\frac{1}{4}\end{array}\right)\text{ }\left( \begin{array}{ll}0 & 0 \end{array}\right)\text{ }\left( \begin{array}{ll}\frac{1}{2}&-\frac{1}{4}\end{array}\right)\text{ }\left( \begin{array}{ll}-\frac{1}{4}& \frac{1}{2}\end{array}\right)}$

But remembering

${a\geq 1}$

So this test of derivative fail to find the minimum. But now we will test the value ${a = 1}$ because it is the smallest value of a. It define de Box of the funcion ${f}$

So ${a}$ must be ${1}$
Using this on the function

${f(1,b)=b (b+1) (2 b (b+1)+7)+2}$

Deriving in function of ${b}$ and igualing to zero.

${\frac{\partial f(1,b)}{\partial b}=0}$
${4 b^{3}+3 b+4 (b+1)^{3}+3 (b+1)=0}$
${(2 b+1) (4 b (b+1)+7)=0}$

The minimum value of b is ${b=-\frac{1}{2}}$

Finishing ${a = 1}$ and ${b=-\frac{1}{2}}$

So:

${f\left(1,-\frac{1}{2}\right)=\frac{3}{8}}$

That is the minimum

[quote="me@home"]If $a,b,c$ are real numbers $a\geq 1$ and $a+b+c=0$ then find the minimum value of $a^{4}+b^{4}+c^{4}-3abc$ [/quote]
We can easily check that
\[a^{4}+b^{4}+c^{4}-3abc \ge a^{4}+\frac{(b+c)^{4}}{8}-\frac{3a(b+c)^{2}}{4}\\ =\frac{9a^{4}}{8}-\frac{3a^{3}}{4}\ge \frac{3}{8}\]
:)

Nice solutions, both of you :)

Now how did you come upon that identity? And did it require previous knowledge of the answer or was it farily easy to see?

[quote="toanhocmuonmau"]
\[a^{4}+b^{4}+c^{4}-3abc \ge a^{4}+\frac{(b+c)^{4}}{8}-\frac{3a(b+c)^{2}}{4}\\ =\frac{9a^{4}}{8}-\frac{3a^{3}}{4}\ge \frac{3}{8}\]
:)[/quote]

OK I see what you mean:
$b^{4}+c^{4}\geq (b+c)^{4}/8$ is Am-Gm for fourth degree
$(b+c)^{2}/4>bc$
The second is true as follows: let $x=b+c/2$
then, $x^{2}>(x-x')(x+x')\iff x'^{2}>0$, and we pick suitable $x'$ which proves $(b+c)^{2}/4>bc$

Then, we use $a\geq 1$ to prove $\frac{9a^{4}}{8}-\frac{3a^{3}}{4}\ge \frac{3}{8}$.

[quote="me@home"]

$(b+c)^{2}/4>bc$
The second is true as follows: let $x=b+c/2$
then, $x^{2}>(x-x')(x+x')\iff x'^{2}>0$, and we pick suitable $x'$ which proves $(b+c)^{2}/4>bc$

[/quote]

can't we just say $\frac{(b+c)^{2}}{4}\geq bc \Leftrightarrow (b-c)^{2}\geq 0$?

why is there $x$ and $x'$?


inequalities
triangle inequality
complex analysis
real analysis
complex analysis theorems

suppose that {bn} converges to b and b/=0 and bn/=0 for any n.  Show that there is a number k>0 such that lbnl>k for every n.  you may assume that bn>0 to simplify the write up.

does anybody have a good proof for this (not too complicated)

By the definition of limit, using $\frac{|b|}2$ for $\epsilon,$ we have that there exists $N$ such that $|b_n-b|<\frac{|b|}2$ for $n>N.$ You can use the triangle inequality to rearrange that to

$|b_n|>|b|-\frac{|b|}2=\frac{|b|}2.$

Then let $k=\min\left\{|b_1|,|b_2|,\dots,|b_N|,\frac{|b|}2\right\}.$

This has very little to do with complex analysis - it's a basic real analysis argument, even when applied to complex-valued sequences.


algebra
polynomial
inequalities
algebra proposed

Assume that  $+\infty>{\mathbf x}_{\mathbf 1}\ge x_{2}\ge x_{3}\ge{\mathbf x}_{\mathbf 4}\ge x_{5}>-\infty$   are the roots of equation
\[(*)\; \; \; \; x^{5}-x^{4}-x^{3}+Ax^{2}+Bx+C = 0\; \; ,\; \; x\in{\mathbb R}. \]
[b]Assuming ${\mathbf x}_{\mathbf 1}-{\mathbf x}_{\mathbf 4}\ge \sqrt{4.2}\; ,\; \;$ solve (if possible) the equation $(*)$ [/b].

acctually there is no algebraic solution for a pollynomial of 5th degree.

If you set A,B,C as a numbers.

because your condition for the roots, only says that the roots are reals.

[quote="yagaron"]acctually there is no algebraic solution for a pollynomial of 5th degree.
because [b]your condition for the roots, only says that the roots are reals.[/b][/quote]
Thanks for interest ! However condition $x_{1}-x_{4}\ge \sqrt{4.2}$ enables us to solve the problem !

i'm interested on the solution

but i dont know how to solve  :blush:

Nice [b][color=blue]"pco" [/color][/b] !
Please see :
http://www.mathlinks.ro/Forum/viewtopic.php?p=856053#856053


Answer

If we look at the problem, we could guess a result that holds for all $ n$ that could be proved by induction. My first guess was the $ x_{n+1}> y_{n}$ but this didn't give an obvious path to $ x_{n+2}> y_{n+1}$ in the induction step. However, once I realised why it didn't work, I just used a "simple" adaption.

Conjecture: $ x_{n+1}> 2y_{n}\, \forall n \in N$

Lets see if conjecture holds for $ n = 1$:

$ LHS = x_{2}= 3^{x_{1}}= 3^{3^{x}_{0}}= 3^{3^{3}}= 3^{27}$

$ RHS = 2y_{1}= 2\cdot 4^{y_{0}}= 2\cdot 4^{4}= 2\cdot 2^{8}= 2^{9}$

$ LHS > RHS \Rightarrow$ Conjecture holds for $ n = 1$

Lets assume the conjecture holds for $ n = k$:

$ \Rightarrow x_{k+1}> 2y_{k}$

Lets consider $ x_{k+2}$:

$ x_{k+2}= 3^{x_{k+1}}$

$ \Rightarrow x_{k+2}> 3^{2y_{k}}$

$ \Rightarrow x_{k+2}> 9^{y_{k}}$

We now have to show that if $ m \geq 1$ then $ 9^{m}> 2\cdot 4^{m}$

Lets assume that $ 9^{m}\leq 2\cdot 4^{m}$ for $ m \geq 1$

$ \Rightarrow \log{9^{m}}\leq \log{2\cdot 4^{m}}$

$ \Rightarrow m\log{9}\leq \log{2}+m\log{4}$

$ \Rightarrow m(\log{9}-\log{4}) \leq \log{2}$

$ \Rightarrow m\log{2.25}\leq \log{2}$

$ \Rightarrow m \leq \frac{\log{2}}{\log{2.25}}$

$ \Rightarrow m < 1$

This contradicts the fact that $ m \geq 1$

Hence if $ m \geq 1$ then $ 9^{m}> 2\cdot 4^{m}$

$ \Rightarrow x_{k+2}> 2\cdot 4^{y_{k}}$

$ \Rightarrow x_{k+2}> 2y_{k+1}$

This completes the inducton step. Since the conjecture holds for all $ n \geq 0$, then it will certainly hold for $ n = 2007$.

$ \Rightarrow x_{2008}> 2y_{2007}$

$ \Rightarrow x_{2008}> y_{2007}$ QED


[b]http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12A_Problems/Problem_15[/b]

I don't understand the explanations

Here is the relevant thread for this particular problem: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=424232]Click here[/url]

I like the solution by [b]joml88[/b].

The AoPS Wiki is a handy first-reference, however it is also helpful to see the discussion threads for these problems, which can be found in the [b]"Contests"[/b] section of the website [i](see the horizontal bluish navigation bar that you use to log in and check private messages)[/i]

Welcome to AoPS  :)


hey all,

I have this problem i have been trying to solve, for personal interests, but anything i try does not work.  Maybe I am forgetting something simple.

Given:
[img]http://filexoom.com/files/12152/pyramid.JPG[/img]

The numbers on the left are the rows, and the numbers on the right are the total squares?

Question: How many rows do i need to get to 1,000,000 total squares.

Help?

p.s. sorry if this was posted in the wrong section, i'm new to the site.

[hide]The numbers on the right are triangular numbers.

$\frac{n(n+1)}2\ge1000000$

$n=\boxed{1414}$[/hide]

why is it called exponential regression? there is no exponential, and there is no regression...

Two of the many questions left by NXD, who left 1 post with a strange diagram only to disappear into obvlivion :spidy:  :noo:


Suppose $\alpha$,$\beta$,$\gamma$ are the roots of the cubic equation

$x^3+2x^2+3x-5=0$

Show that the cubic equation with roots $\alpha\beta-\gamma^2$,$\alpha\gamma-\beta^2$,$\beta\gamma-\alpha^2$ is

 $x^3-5x^2+15x-67=0$

[quote="shyong"][quote="Farenhajt"]Is there a quicker way than ugly and very lengthy application of Newton's sums and Vieta's formulas?[/quote]

well , the quicker way I saw is using the identity $\alpha\beta-\gamma^2=2\gamma+3$ to compute . Other than that im not sure .[/quote]

You are right. I could made a confirmation.

The desired cubic equation is $\left(\frac{x-3}{2}\right)^3+2\left(\frac{x-3}{2}\right)^2+3\cdot \frac{x-3}{2}-5=0,$ yielding $x^3-5x^2+15x-67=0.$

[quote="arqady"][quote="Farenhajt"]... and Vieta's formulas?[/quote]
Farenhajt, sorry, why not Viet's formulas? Thank you.[/quote]

He was born Italian, so he's known by both Italian and French versions of his name: Francesco Vieta and François Viette.

I misspelt it: In French it should be François Viète.

And I must correct myself - he was French, but the Latin (not Italian!) version of his name was Franciscus Vieta (in the meantime I checked the data on the Net)

[quote="Altheman"][quote="shyong"]...using the identity $\alpha\beta-\gamma^2=2\gamma+3$ to compute . Other than that im not sure .[/quote]

how do you show that?[/quote]

From $\alpha \beta\gamma=5,$ we have $\alpha \beta -\gamma ^2=\frac{5}{\gamma}-\gamma ^2=\frac{5-\gamma ^3}{\gamma}=\frac{2\gamma ^2+3\gamma}{\gamma}=2\gamma +3.$


geometry
3D geometry
analytic geometry
sphere

n is a positive integer. $S$ is the set of all triples $(a, b, c)$ such that 1 ≤ a, b, c, ≤ n. What is the smallest subset X of triples such that for every member of S one can find a member of X which differs in only one position. [For example, for n = 2, one could take X = { (1, 1, 1), (2, 2, 2) }.]

duplicates can occur if it's something like (a,b,#) and (@,b,c) where # = c and @ = a. So I don't think your logic works.

[quote="agolsme"]I might be misunderstanding you, but I don't see how you get rid of the problem I mentioned before.

Let's say you have an array (1,2,x) where x is free to shift. and let's say you have an array (x,2,3) where x is free to shift. These two arrays will intercept. So you can't just take some random $n^{2}/(3n-2)$ 3-tuples and hope it'll work.[/quote]

:blush:

To be honest, I misread the question, I thought the question was "determine the size of X", not "determine the elements in X". So you're right, my method lacks an explanation of *how* to choose these elements (I'll rework my solution taking this into account; I've spotted another mistake in it as well).

[quote="agolsme"][For example, for n = 2, one could take X = { (1, 1, 1), (2, 2, 2) }.][/quote]

This looks wrong to me.  But what exactly do you mean by "differs in only one position"?  Do you mean, for example, $(5, 1, 2), (3, 1, 2)$?  (Does order matter?)

And if so, the problem is trivial.  For $n \ge 2$ take $X = \{ (1, 1, 1), (1, 1, 2) \}$.

yes, order matters.

and no {(1,1,1), (1,1,2)} doesn't capture everything...

note that it would not get $(3,2,4)$ and a lot of others because it differs in more than one coordinate.

Oh, for every member of [b]S[/b].  Sorry, I completely misread that part  :blush:

Edit 2:  Nevermind.  =/

I am pretty sure it's $n^{2}/2$ but I don't know how to prove that it is.

When you say "only" one position, do you mean exactly one position?

yes so if you think of this as a covering problem - $(1,2,3)$ covers $(1,2,k); (k,2,3); (1,k,3)$ for $k = 1,2,\ldots,n$

If you think of this as a cube of $nxnxn$ dimensions, you are trying to find the most efficient way to cover the cube by picking unit cubes. Every unit cube you pick colors/covers all the rows, column and depths that it's on.

Is it possible to arrange 40 points in a $10\times10\times10$ cube so that there are 4 in each each plane parallel to the coordinate planes?  If so, it "might be" possible to do it with those 40 points, as follows:  in any fixed plane, the 4 points in the plane knock out $19+17+15+13 = 64$ vertices, and we have 36 points left over to take out the remaining 36 vertices.

I think this is commonly known as a covering problem. See sphere packings and coverings. I don't recall whether or not there's a specific formula, but it's easier to work with just 0's and 1's initially. Rather than ≤n. And as this is an optimization problem, I don't know if a formula exists, but I guess there is one as it's being asked for in the problem...

Also consider the following problem: what's the maximal number of ordered n-tuples $(a_{1}, \cdots , a_{n})$ (first work with $a_{i}\in \{ 0, 1 \}$) such that no two cover a similar n-tuple. (This problem is also computationally difficult, and I doubt there is a formula.)

One last thing, you can represent it nicely with geometry (not efficient though). Make an n-dimensional unit cube and put spheres with radius 1 centered at vertices. (The spheres can overlap.) Once you have all the points touching a sphere (adjacent to, or inside) your covering is complete.

(Edit: didn't see that it had already been noted as a covering problem...)


\[ \frac{a^{2}+bc}{a(b+c)}+\frac{b^{2}+ca}{b(c+a)}+\frac{c^{2}+ab}{c(a+b)}\geq 3 \Leftrightarrow M_a(b-c)^2+M_b(c-a)^2+M_c(a-b)^2\ge0  \]
$M_a=a^2b^2+a^2c^2+abc(a+b+c)-b^2c^2$
Let $a\ge b\ge c$
It's easy to prove that
$M_b\ge0,M_b+M_a\ge0,M_b+M_c\ge0$

[quote="zhaobin"]another solution:
$\frac{bc}{a(b+c)}+\frac{ca}{b(c+a)}+\frac{ab}{c(a+b)}\geq \frac 3 2$.
and
$\frac{a^{2}}{a(b+c)}+\frac{b^{2}}{b(c+a)}+\frac{c^{2}}{c(a+b)}\geq \frac 3 2$.[/quote]
Very nice solution :)

[quote=sqing][quote=Cezar Lupu][color=blue] Let $a, b, c$ three positive real numbers. Prove that the following inequality holds true:
      \[ \frac{a^{2}+bc}{a(b+c)}+\frac{b^{2}+ca}{b(c+a)}+\frac{c^{2}+ab}{c(a+b)}\geq 3.  \]
 [/color][/quote]
Let    $a,b,c>0$   , [url=https://artofproblemsolving.com/community/c6h1756749p11472823]prove that[/url]: $$\frac{a^2+bc}{a(b+c)}+\frac{b^2+ca}{b(c+a)}+\frac{c^2+ab}{c(a+b)}\ge \frac{a+b+c}{\sqrt[3]{abc}}$$[/quote]

I think SOS still helps here.

[quote=sqing]
Let    $a,b,c>0$   , [url=https://artofproblemsolving.com/community/c6h1756749p11472823]prove that[/url]: $$\frac{a^2+bc}{a(b+c)}+\frac{b^2+ca}{b(c+a)}+\frac{c^2+ab}{c(a+b)}\ge \frac{a+b+c}{\sqrt[3]{abc}}$$[/quote]
Let    $a,b,c>0.$ For $n\in N^+$ , prove or disprove  $$\frac{a^2+bc}{a^{n-1}(b+c)}+\frac{b^2+ca}{b^{n-1}(c+a)}+\frac{c^2+ab}{c^{n-1}(a+b)}\ge \frac{a+b+c}{\sqrt[3]{(abc)^{n-1}}}$$

Let $a,b,c$ be positive real numbers . For $n\in N^+$ , [url=https://artofproblemsolving.com/community/c6h1627441p10206241]prove that[/url]$$\frac{a^{n-1}(a^2+bc)}{b+c}+\frac{b^{n-1}(b^2+ca)}{c+a}+\frac{c^{n-1}(c^2+ab)}{a+b}\geq a^n+b^n+c^n .$$


For each diagram, prove the points $P,Q,R$ colinear.

please explain 1st diagram a bit. i see it as just a cluster of arbitary lines.

The vertices of triangles $ABC$ and $DEF$ are joined and meet at one point. Their corresponding sides are extended and meet at 3 points: $P,Q,R$. Prove $P,Q,R$ colinear.

For the circles diagram, all lines except $L$ are each tangent to 2 circles. Prove $P,Q,R$ colinear.

Please write down the assumptions and the claim to be proven. I think those problems has something to do with Desargues' theorem and Monge's theorem.


Let $ m$ and $ n$ be natural numbers with $ m > n$. Prove that
$ 2(m \minus{} n)^2(m^2 \minus{} n^2 \plus{} 1)\geq 2m^2 \minus{} 2mn \plus{} 1.$

$ 2(m\minus{}n)^2(m^2\minus{}n^2\plus{}1)\geqslant 2m^2\minus{}2mn\plus{}1 \\ \iff (m\minus{}n)^2(m^2\minus{}n^2\plus{}1)>m(m\minus{}n) \\ \iff (m\minus{}n)^2(m\plus{}n)\plus{}m\minus{}n>m \\ \iff m(m\minus{}n)^2\plus{}n(m\minus{}n)^2>n$
The last ineqality is obvious, becouse $ m(m\minus{}n)^2>0$ and $ n(m\minus{}n)^2\geqslant n$


given a disc divided into 6 sectors, those sectors are labelled clockwise 0, 1, 0, 1, 0, 0. the following move is allowed: choose any two adjacent sectors and add 1 to each entry. question is, using such moves, can you get all the entries to be equal?

See the diagram attached. The sum of all of the numbers in one color field must equal the sum of all the numbers in the other color field. Since blue starts off at 2, and green starts off at 0, and their difference is invariant (1 is added to each color each time) they can never be equal.


The set of values of $m$ for which $x^2+3xy+x+my-m$ has two factors, with integer coefficients, which are linear in $x$ and $y$, is precisely:
(A) $0,12,-12$
(B) $0,12$
(C) $12,-12$
(D) $12$
(E) $0$

[hide="my answer"] Set $x^2+3xy+x+my-m=0$. Solve this quadratic equation and let the Discriminant be a square number,
I got the answer (B) $0,12$ [/hide]


inequalities
induction
inequalities unsolved


geometry
rectangle
rotation
geometric transformation
reflection
combinatorics unsolved
combinatorics

A $ 2n \times 2n$ board must be covered (without overlays and without passing the limits of the board) with things like

$ \Box \Box$ (a $ 2\times1$ rectangle)

or$ \Box \Box$
$ \Box \Box$ (another tetris peace; this is not a square: I used the "or" to make the piece be in the correct form).

You can rotate or reflect the pieces. Prove that you are going to use at least $ n \plus{} 1$ of the first kind.

Are we supposed to cover a board? What's the size of the board?

Sorry... It was not well organized (the information was in the title, but it was wrong). Now I've edited. I hope it is clear now.

Could you please post the complete test?

Is it $ n\plus{}1$ the minimum? do you have an example for small cases such as $ n\equal{}2, 3$ or $ 4$ ?

No... I haven't found examples for $ n > 1$. But this bound can be obtained just for odd $ n$. The number of squares in the board is divisible by $ 4$. Thus, the number of rectangles must be even (the number of squares they fullfil is multiple of $ 4$). For even $ n$, it can't be $ n \plus{} 1$. However, I can show an example for $ n \plus{} 2$.


:) it is given a n-gon with no angle greater than 180 degrees. If a1>=a2>=...>=an prove that a1=...=an.

How about dividing a rectangle in two cuadrilateral parts by a segment?

i cannot understant you. please describe what you want to say in more details.

Maybe I misunderstood the problem, here is my cuadrilateral $ A_1A_2A_3A_4$ with the angles $ 100$, $ 90$, $ 90$ and $ 80$.

I'm sorry all angles should be equal ... the error here is mine. If some of the moderators may edit the first post it will be very good.


algebra
polynomial
function
algebra unsolved

Consider those polynomials $p(x)=x^4+ax^3+bx^2+cx+1$ whose coefficients are all positive numbers less than 3, and which do not have any real root. Find the maximum value of the product $abc$.

[quote="chess64"]Why do you say that $p(-1)>0$?[/quote]
In fact, $p(x)>0$ for every $x$, because $p$ is continuous, $p(0)=1>0$, and we're told $p$ doesn't have any real root.
(If there was a value of $x$ with $p(x)\leq 0$, either $x$ is a root, or by the intermediate value theorem for continuous functions, there would be a real root between $1$ and $x$).

[quote="sunchips"]Note: $p(-1)>0$, so $5\geq b+2>a+c$, thus $3*2.5*2.5>abc$ This can be obtained, with $b=3$ and $a$ and $c$ as close to $2.5$ as possible[/quote]
Yes, that's the basic idea.
From my previous post, we have $p(x)>0$ for every $x$.
In particulatr, $p(-1)>0$ gives $b+2>a+c$.
If $a,b,c<3$, we have strictly $5>b+2>a+c$.
Thanks to AM-GM, for every fixed $a+c$ ($<b+2$), the maximum $ac$ occurs at $a=c$. So, the greater $b$ becomes, the greater $abc$ can become.
Therefore, in this case, $2.5*3*2.5$ is the lowest upper bound for $abc$.
I think we can't call it a maximum, because it's strictly unreachable.

Oh right, I forgot that there are no real roots.


Find all functions R-R satisfying for all $x$ and $y$.
$xf(y)+yf(x)=(x+y)f(x)f(y)$

$x = y$ gives $x \cdot f(x) \cdot (f(x)-1) = 0$

Take any subset $E$ of $\mathbb{R}$, $f(x) = 1$ if $x \in E$, $f(x) = 0$ elsewhere, and $f(0)=0$
This make all the solutions.


Let $a$ be a number with $n$ digits $(n>1)$. A number $b$ has $2n$ digits and we get it by wirting two copies of $a$ one after the other. If $\frac{b}{a^2}$ is an integer number $k$, find all possible values of $k$.

That's what I got, but I can't find which are all the possible values.

Hint:  Once you have what you have above, the answer is not hard ...
[hide]Can  a factor of $10^n+1$  be 2? [hide]or 3?  [hide]or 5? .... Also  now only a few (one) prime number left before $k$ becomes too large.. (remember k can not be > 9  :)[/hide] [/hide][/hide]

I see what you're getting at, but somehow it isn't working when I try different $n$ (which is making me think I made a mistake in my above post).  Consider $n=2$.  Then we have $a$ is a two-digit number and

\[ k=\frac{101}{a}. \]

But $101$ is prime so the result of dividing it by a two-digit number won't be another integer.

??

The fact that $a$ has $n$ digits, I think, is why.

[quote="Gyan"]How about n=3 :)
[hide]Try 1001, and k=7  and 1001/7 = 143 [/hide]
But the main point is  you are asked  [b]possible value(s)  of k only [/b]..  and need not worry about  possible values of n.
[hide]and need not worry that not only 1001/7 is integer but  also 1000000001/7  etc ... would work too  but k is always 7 :)
[hide]BTW possible values of n would be 3, or 9  or 15 or of the form 6m+3 but you need not worry  about that...[/hide]
[/hide][/quote]
I see that, but the original problem said $n>1$, so I figured it meant [b]all[/b] $n>1$.

[quote] see that, but the original problem said $n>1$, so I figured it meant all $n>1$  .[/quote]
Yes, for all $n>1$ WHEN   k is integer, there is only  one possible    [hide] 7 [/hide]  value for k.
..(It does [b]not[/b] mean that for all values of n>1 , k would come  out as an integer..)
Hope that helps .

Oh yes, IF $\frac{b}{a^{2}}$ is an integer.  For some reason I thought it was required that $\frac{b}{a^{2}}$ is an integer for all $n\in\mathbb{N}$.  Thanks for clearing that up!


greetings people of aops. i will be co-moderator on this part of the forum along with tare (and MysticTerminator as of September 9th). if any of you have questions that need to be answered, or any interesting challenges, go ahead. im going to try my best to keep discussions on track, teach interesting concepts, and bring up cool bean challenges. mostly, ill be getting questions from mathcounts (chapter, state, nationals), and amc (8, 10, and maybe 12). i may post some "hard" questions that would suit better on the amc thread, but they may teach good concepts. im not going to limit myself to mathcounts and amc questions though. ill keep it on that level though. any questions or suggestions go ahead. and please dont get offended if i delete any of your posts. ill try my best not to, but if they're really off topic, ill go ahead. other than that, ill try to be as helpful as my limits go. and we all will benefit from this. HAPPY SOLVING!

Also, a word from our beloved Administrator...
[quote="Mr. Rusczyk"]We will periodically post practice problems in our MATHCOUNTS section of the message board - we ask that experienced students (i.e. those beyond 8th grade) leave these problems for the less experienced users of the site to discuss.

If you have a problem that would be good for beginning students, please post it here!

Thanks much![/quote]

This is the 2nd post ever! The first post is locked :(


[hide]
12 squares, 8 hexagons, 6 octagons
Because every point is shared by 3 polygons, the number of vertices is $\frac{12(4)+8(6)+6(8)}{3}=48$
Because every edge is shared by 2 polygons, the number of edges is $\frac{12(4)+8(6)+6(8)}{2}=72$
Because $12+8+6=26$, there are 26 faces.
Euler's: $V+F=E+2$ check

In other words theyre asking for a space diagonal, so we take the total number ways to connect 48 points in space and subtract face diagonals and edges.

The total number of space diagonals, face diagonals, and edges: $\binom{48}{2}=1128$
Face diagonals: Every point is shared by a square, a hexagon, and an octagon. There are 15 points around this point to which we could extend a diagonal, but of them 3 are edges, so there are really only 9. If we do this for all 48 vertices, we have to divide by 2 to account for diagonals being counted twice...
$\frac{48(9)}{2}=216$=face diagonals
there are 72 edges

$1128-216-72=\boxed{840}$

Sorry for the long solution

[/hide]

[hide]Use Eulers formula $V + F = E + 2$
Since $2$ sides of each face make an edge, we sum the sides of the polygons and divide by 2. $(4*12 + 8* 6 + 6*8)/2 = 72 = E$

Sum the faces so $F = 26$. Now solve for the number of vertices, $V= E - F + 2 = 48$
So now we find the number of total segments. Notice that each segment is the segment between 2 vertices. So there are $\binom{48}{2}$ total segments = $24*47$

Now we find how many segments are not along an edge or a face. There are $72$ edges but we still need to account for the overcounting of segments along faces.

For each square, there are $\binom{4}{2} - 4 = 2$ segments along each face.(subtract 4 to account for the edges, you could have not counted edges and just use choose for finding the segments along faces but this is just how i did the problem).

For each hexagon, there are $\binom{6}{2} - 6 = 9$.

For each octagon, there are $\binom{8}{2} - 8 = 20$

So now find the total amount of segments along faces, which is $2*12 + 9*8 + 20*6 = 216$

Now subtrac the number of total segments - edges - segments along faces = $24*47 - 72 - 216 = \boxed{840}$.[/hide]

[quote="nsato"]A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons.  At each vertex of the polyhedron one square, one hexagon, and one octagon meet.  How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?[/quote]

[hide]
Every edge is counted twice which means there are $\frac{12 * 4 + 8 * 6 + 6 * 8}{2} = 72$ edges total. And we know that $v+ F - 2= e$ $V = E + 2 - F = 72 - 26 + 2 = 48$ There are then $\binom{48}{2}$ possible ways of chosing a segment from 2 vertice. We need to subtract all the edges and all the face diagonals. There are $12 * 2 + 20 * 6 + 9*8 = 216$ diagonals due to faces alone so we must subtract this and the edges from $\binom{48}{2}$ and we end up with $840$[/hide]

V-E+F = 2
so F is given as 26.

Here is where my solution is different from the others because I solved for vertices instead. Since they told you that every single vertex contains the meeting of the 3 shapes, you know that each vertex has 3 edges coming from it. However, this counts every edge twice. So E = 3V/2. Now you can use V-E+F = 2 knowing the F and having E in terms of V to get V = 48. Then with E = 3V/2, E = 72. Then I used the same method as previously posted: (48c2) - (Edges) - (Sum of Diagonals for the different shapes), achieving 840.

Honestly, all the solutions presented look essentially the 'same', imho, except for the difference in preferences on what to find first (edges, vertices, faces, etc) or how to find the other quantities ($ V\equal{}E\minus{}F\plus{}2$, $ 2E\equal{}3V$)...

In other words, there are many ways of starting to do this problem but they all converge to the same final combinatorial argument. I think there is more than enough sufficient information given by the problem that allows [b]seemingly[/b] different ways of tackling this problem, which is exactly how a lot of AIME problems are designed to be.

[quote="10000th User"][quote="cdymdcool"][quote="minsoens"][hide]
12 squares, 8 hexagons, 6 octagons
Because every point is shared by 3 polygons, the number of vertices is $ \frac {12(4) \plus{} 8(6) \plus{} 6(8)}{3} \equal{} 48$
Because every edge is shared by 2 polygons, the number of edges is $ \frac {12(4) \plus{} 8(6) \plus{} 6(8)}{2} \equal{} 72$
Because $ 12 \plus{} 8 \plus{} 6 \equal{} 26$, there are 26 faces.
Euler's: $ V \plus{} F \equal{} E \plus{} 2$ check

In other words theyre asking for a space diagonal, so we take the total number ways to connect 48 points in space and subtract face diagonals and edges.

The total number of space diagonals, face diagonals, and edges: $ \binom{48}{2} \equal{} 1128$
Face diagonals: Every point is shared by a square, a hexagon, and an octagon. There are 15 points around this point to which we could extend a diagonal, but of them 3 are edges, so there are really only 9. If we do this for all 48 vertices, we have to divide by 2 to account for diagonals being counted twice...
$ \frac {48(9)}{2} \equal{} 216$=face diagonals
there are 72 edges

$ 1128 \minus{} 216 \minus{} 72 \equal{} \boxed{840}$

Sorry for the long solution

[/hide][/quote]

I don't think we should subtract 6 from the edges, because we are using the 1128 total edges to subtract those edges which are on not interior.  So should 1128 - 15*48/2 ?[/quote]No. minsoens is correct but may have not explained it correctly his idea. There are 4+6+8-1=15 points to be considered but he forgot that there are only 12 points that surround this point because two faces share an edge, and hence also the vertex at the other end of the edge. There are 3 edges and so there are 9 points that would create face diagonals.

Another way of looking at it is to see that there are n-3 diagonals originating from a given vertex. Then, 1+3+5=9 face diagonals.[/quote]
i understand that there are only 12 ponts surround this point but why do we have to subtract 3 edges away?  I though we are considering the case of when the edges which are not interior, so it should include the 3 edges?

[quote="minsoens"]$ \binom{48}{2}$ = # of line segments joining every pair of vertices on the polyhedron = (# of space diagonals) + (# of face diagonals) + (# of edges)
In my solution, I considered my face diagonals and edges separately. When I was calculating the number of face diagonals, I excluded the edges.. and later subtracted the number of edges from 1128.[/quote]That's right. That was the basis of minsoens solution. Even if the edges were included,

$ \binom{48}{2}\minus{}\frac{48\cdot12}{2}\equal{}840$.

Sorry for the revive, but I found a fast solution:
[hide=Solution]In any polyhedron with each vertex having such a well-defined structure finding, [u]"How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?"[/u] is rather easy in that we don't need to do casework. The number of points not on any face that vertex is on including itself joined with that vertex is valid. By Euler's formula there are $48$ vertices, so there are $35$ possible segments for every vertex but divide by $2$ so you don't overcount and you get $35 \cdot 24 \implies \boxed{840}$.[/hide]

We know that $V-E+F=2$, where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number of faces in the given polyhedron. It becomes apparent that the number of faces is $12+8+6=26$, so $V-E+26=2 \implies V-E=-24$.
Note that each angle in a regular hexagon is $120$ degrees, in an octagon $135$ degrees, and in a square, $90$ degrees. Take the net of the given polyhedra and note that from each vertex, the "degree" is $3$ (*) because protruding from the vertex, there are either one edge from a hexagon and two from a square or two edges from an octagon and one from a square. Given that the degree is $3$, we know that $E=\frac{3V}{2}$. Plugging this into $V-E=-24$, we have that $E=72$ and $V=48$ (**). We now conclude that the total number of diagonals is $\binom{48}{2} = 1128$. However, this includes the diagonals of the faces, which needs to be subtracted. Since, there are $\binom{n}{2}-n$ diagonals in an $n$-gon (excluding the edges), there are $12 \bigg(\binom{4}{2}-4\bigg)+8\bigg(\binom{6}{2}-6\bigg)+6\bigg(\binom{8}{2}-8\bigg) =$ $12 \cdot 2 + 8 \cdot 9 + 6 \cdot 20 =$ $24 + 72 + 120 = 216$ non-edge diagonals and $E=72$ edge-diagonals. Hence, there are $1128-216-72=\boxed{840}$ diagonals that do not lie on the interior of a face or an edge.

(*) When I say "degree", I mean that number of edges protruding from the given vertex.
(**) We could have immediately noted that $V=12 \cdot 4=8 \cdot 6=6 \cdot 8=48$.

We'll find the total amount of lines connecting two vertices and subtract off face diagonals + edges. There are $\frac{4 \cdot 12 + 6 \cdot 8 + 8 \cdot 6}{3} = 48$ vertices, so $\binom{48}{2}$ total ways. Next, a square has $\binom{4}{2}-4=2$ face diagonals, a hexagon has $\binom{6}{2}-6=9$ face diagonals, and an octagon has $\binom{8}{2}-8=20$ total face diagonals. This gives $2 \cdot 12 + 8 \cdot 9 + 6 \cdot 20 = 216$ face diagonals. There are also $26$ faces, so $48+26=E+2 \implies E = 72$ and hence there are $72$ edges. The answer is $\binom{48}{2}-216-72 = \boxed{840}$.

Wait I’m getting $\frac{48 \cdot (47-13)}{2}$, why is it $48-13$ and not $47-13$ because we exclude the vertices itself, right?

Perhaps I’m just missing something


This problem was given as a homework at our IMO training (2nd homework problem set of the IMO training 2004, geometry, problem 4). It is rather simple; as somebody pointed out, the most difficult part of the problem was to draw a good figure. Anyway, I was delighted to see that the majority of the participants had exactly the same figure as I had  :)

Let the lines BD and CE intersect at a point X. Then, since < AEC = < ADB, we have < AEX = < ADX, and thus, the points A, D, E and X lie on one circle. Hence, < AXE = < ADE, and since we have given < ABC = < ADE, we thus obtain < AXE = < ABC. Thus, < AXC = 180° - < AXE = 180° - < ABC, what implies that the points A, B, C and X lie on one circle. Now, everything is simple:

  < BAC = < BXC       (since the points A, B, C and X lie on one circle)
       = < DXE = < DAE        (since the points A, D, E and X lie on one circle).

  Darij

[quote="grobber"]In some convex pentagon $ ABCDE$, we have $ \angle ABC \equal{} \angle ADE$ and $ \angle AEC \equal{} \angle ADB$. Show that $ \angle BAC \equal{} \angle DAE$.[/quote]

If X = DB^EC then AXDE is cyclic. Therefore <ABC = <ADE = <AXE.THis means that ABCX is cyclic too. From here <BAC = <BXC = <DXE = <DAE = <QED  ;) .

let $ O$ be the intesection  of $ EC$  and $ BD$ .since $ \angle AEO\equal{}\angle ADO$  we have $ AEDO$ is cyclic so $ \angle EAD\equal{}\angle EOD\equal{}\angle BOC$.
and $ \angle ABC\equal{}\angle EDA\equal{}\angle AOE\equal{}180\minus{}\angle BOC$ so $ ABCO$ is cyclic  so $ \angle BOC\equal{}\angle BAC.$

[hide="Solution"]
Denote the intersection between $ EC$ and $ BD$ to be $ P$.  Hence, $ \angle PEA \equal{} \angle ADP$, so $ APDE$ is cyclic.  Thus, $ \angle ABC \equal{} \angle ADE \equal{} \angle APE$, which means that $ APCB$ is cyclic.  Hence, $ \angle BAC \equal{} \angle BPC \equal{} \angle EPD \equal{} \angle DAE$.  [/hide]

[hide]Let intersection of BD and CE be F.

Since angle AEC = angle ADB, AFDE is cyclic. Thus, angle ADE = angle AFE, and DAE = angle DFE = angle BFC. Since angle ADE + angle DAE + angle AED = 180, from triangle ADE, angle AFB = angle AED. Thus in quadrilateral ABCF, angle ABC + angle AFC = 180, and thus it is cyclic. Thus, angle BFC = angle BAC. Since angle DAE = angle BFC, angle DAE = BAC, and we're done.[/hide]


logarithms algebra polynomial calculus derivative

Hello People,

Can somebody proof if $\log ((\log n)^{5}) \in O(\log n)$ ?

$\log((\log n)^{5}) = 5 \log \log n$

Now suppose that $\log \log n$ grows faster than $\log n$. Then $n$ grows faster than $e^{n}$, but it can be proven that $e^{n}$ grows faster than any polynomial by continuously taking derivatives, and $n$ is a polynomial, so we have a contradiction, and

$\log((\log n)^{5}) \in O(\log n)$


integration
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calculus computations

$ \int log_3x dx$

$ \int_{0}^{pi}sin^2xdx$

$ \int(x^2)/(x^2 \minus{} 4x \plus{} 5)dx$

In no.1, you may use the change of base, either of base 10 or base $ e$. In no.2, consider $ \sin^2 x \equal{} \dfrac{1 \minus{}  \cos 2x}{2}$. In no.3, you may use partial fractions.

[quote="corps0507"]In no.2, consider $ \sin^2 x \equal{} \dfrac{1 \minus{} \cos 2x}{2}$.[/quote]thanks a lot!

$ \int \dfrac{1 \minus{} \cos 2x}{2} \equal{} (1/2)x \minus{} 1/4\int2cos(2x)dx \equal{} (1/2)x \minus{} 1/4sin(2x) \plus{} c$

$ log_3x \equal{} (lnx)/(ln3)$ so $ \int log_3xdx \equal{} 1/(ln3)\int(lnx)dx\equal{}x/(ln3)ln(x)\minus{}x\plus{}C$
it should be ok, right?

but I need an help to start the  num 3  :blush:

hello, for number one, it should be
$ \displaystyle{\int \log_3 xdx\equal{}\frac{1}{\ln(3)}\int \ln(x)dx\equal{}\frac{1}{\ln(3)}\int 1\cdot\ln(x)dx\equal{}\frac{1}{\ln(3)}\left(x\ln(x)\minus{}\int dx\right)\equal{}\frac{1}{\ln(3)}\left(x\ln(x)\minus{}x\right)\plus{}C}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, for number one, it should be
$ \displaystyle{\int \log_3 xdx \equal{} \frac {1}{\ln(3)}\int \ln(x)dx \equal{} \frac {1}{\ln(3)}\int 1\cdot\ln(x)dx \equal{} \frac {1}{\ln(3)}\left(x\ln(x) \minus{} \int dx\right) \equal{} \frac {1}{\ln(3)}\left(x\ln(x) \minus{} x\right) \plus{} C}$
Sonnhard.[/quote]yea thanks I did a mistake in my solution.

I'm stuck on this one...

$ \int(x^2)/(x^2 \minus{} 4x \plus{} 5)dx$

For 3 you can make use of this equality:
\[ \frac {x^2}{x^2 \minus{} 4x \plus{} 5} \equal{} 1 \plus{} 2\cdot \frac {2x \minus{} 4}{x^2 \minus{} 4x \plus{} 5} \plus{} \frac {3}{(x \minus{} 2)^2 \plus{} 1}
\]
This gives you three integrals; the first integrand is constant, the second can be done with the substitution $ u \equal{} x^2 \minus{} 4x \plus{} 5$, and the third is the derivative of $ 3\arctan (x\minus{}2)$.

hello, for the second one you have
$ \displaystyle{\int_{0}^{\pi}\left(\frac{1}{2}\minus{}\frac{1}{2}\cos(2x)\right)dx\equal{}\left[\frac{1}{2}x\minus{}\frac{1}{4}\sin(2x)\right]_{0}^{\pi}\equal{}\frac{\pi}{2}\minus{}\frac{\sin(2\pi)}{4}\minus{}\left(\frac{0}{2}\minus{}\frac{sin(2\cdot0)}{4}\right)}$
Sonnhard.

hello, for the last integral we have
$ \int\frac{x^2}{x^2\minus{}4x\plus{}5}dx\equal{}\int 1dx\minus{}\int \frac{4x\minus{}5}{x^2\minus{}4x\plus{}5}dx$
We compute the second integral
$ \int \frac{4x\minus{}5}{(x\minus{}2)^2\plus{}1}dx$. By setting $ t\equal{}x\minus{}2$ we have
$ \int \frac{4t\plus{}3}{t^2\plus{}1}dt\equal{}2\int \frac{2t}{t^2\plus{}1}dt\plus{}3\int\frac{dt}{t^2\plus{}1}dt$
Sonnhard.


When cyclopentanecarbaldehyde is treated with phenylhydrazine in acetic acid a product B is obtained with 90% yield. Propose a structure for B.


calculus
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real analysis unsolved

1. f is continuous in [0,1].
---what change in variables we do to prove that integral(0 to pi)xf(sinx)dx=(pi/2)integral(0 to pi)f(sinx)dx
---prove:F(x)=integral(0 to 1)f(x+t)dt is differentiable and F'(x)=f(x+1)-f(x)

2. sinx+sin2x+...+sinnx?
note:i put t(x)=cosx+isinx and=>t^k=coskx+isinkx and thought about taking Im[t+t^2+...t^n] but it doesn't work. any easy way?

3. f defined and m times differentiable in [0,1] and f^(k) is never zero in [0,1],k=0,1,..,m-1(f^(k)=kth derivative of f), and |f^(m)(x))|>=M for all x in [0,1].prove max(x in [0,1])|f(x)|>=M/m!

10x

First, please learn how to use Latex -- it makes reading what you write so much easier.

For the first part of the first one, if the integral on the left is $I$ and the integral on the right is $J$ then taking $x \mapsto \pi-x$ on the left gives $I = 2J-I$, and hence $I = J$.

For the second one, you're just wrong that it doesn't work: sum the geometric series, multiply numerator and denominator by the conjugate of the denominator and then take the imaginary part.  (This has appeared on the forum before.)

I strongly suspect the third has also appeared on the forum before.


Are you sure?  I don't think that this problem is true.  (I drew a couple of figures and saw that the distance from $ O$ to $ MN$ can be both bigger and smaller than the distance to $ PQ$.)

[quote="tmy1018"]let $ \odot O_{1}$ $ \odot O_{2}$ meet in two points $ P$ ,$ Q$. point $ O$ is the midpoint of $ O_{1}O_{2}$. let $ A$ ,$ C$ be two points on $ \odot O_{1}$, and $ B$, $ D$ be two points on $ \odot O_{2}$. line $ AB$ and line $ CD$ both pass through $ P$. $ M$ is the midpoint of $ AB$, $ N$ is the midpoint of $ CD$.
prove $ O$ is closer to line $ MN$ than it is to line $ PQ$.
(by closer,  it means the distance is smaller)[/quote]

-------------------
Also I don't think that this problem is true. I have made some tests
 :wink:


Let $ P\equal{}(2\minus{}3\minus{}4\plus{}7)^{2347}$ and $ Q\equal{}(\minus{}2\plus{}3\plus{}4\minus{}7)^{2347}$.  What is the value of $ (2\plus{}3\plus{}4\plus{}7)^{(P\plus{}Q)}$?

[hide="Solution"]
We can simplify what is inside the parentheses and get:

$ P\equal{}2^{2347}$
$ Q\equal{}(\minus{}2)^{2347}$

Since in $ Q$ the power is odd, you can pull out the negative sign.

$ Q\equal{}\minus{}(2^{2347})$

Now add:

$ P\plus{}Q\equal{}(2^{2347})\minus{}(2^{2347})\equal{}0$

Now we compute the final value:

$ (2\plus{}3\plus{}4\plus{}7)^{(P\plus{}Q)}\equal{}16^0\equal{}1$

$ Answer\equal{}1$
[/hide]


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Let f be Riemann integrable in [a,b] and positive on [a,b] (f>0). Show that the Riemann integral of f from a to b is also positive. It is trivial to show that it is non-negative but in fact, it also can't be zero. It is also trivial if f is continuous in the interval but the statement is still true even if it isn't.

No measure theory tools/techniques allowed.

This is a little hard to do directly from the definition of Riemann-integrability. The standard argument would be to let $E_{k}= \{x \in [a,b] \; | \; f(x) \geq \frac{1}{k}\}$, observe that $[a,b] = \cup E_{k}$ and hence one $E_{k}$ must have positive measure. For this particular problem one could say that there is a closed subset of positive measure where $f$ is continuous. However, I can't think of a way to show this without referring to measures.

However, the first argument can be modified. By the Baire Category Theorem at least one of the $E_{k}$'s is not nowhere dense in $[a,b]$, say $E_{k_{0}}$.  Hence there is an interval $(x_{0}-r, x_{0}+r)$ with $x_{0}\in [a,b]$ such that  $(x_{0}-r, x_{0}+r) \cap [a,b] \subset \bar{E_{k_{0}}}$. Hence for any $s \in (x_{0}-r, x_{0}+r) \cap [a,b]$ there is a sequence $(s_{n})$ in $[a,b]$ converging to $s$  such that $f(s_{n}) \geq \frac{1}{k_{0}}$. Hence any upper sum must be larger than the length of $(x_{0}-r, x_{0}+r) \cap [a,b]$ times $1/k_{0}$

A little hard or impossible? The distinction is quite important to me.

Basically, if there is even just one point in which the function is continuous, then there is a neighborhood in which it is positive and in that neighborhood the integral of the function will be positive.

Therefore, one point of looking at it, is to try and prove that a function that is nowhere continuous is not Riemann integrable. The argument is trivial if one knows that a function is Riemann integrable iff the set of its discontinuities has measure zero. The question is: Can it be done without resorting to measure theory?

I did provide an argument that proves the original statement. And it doesn't use any measure theory. I thought it was a good argument  :P

If you are set on doing it with continuity we could do it this way (but it will only be more complicated):
Assume $f$ is nowhere continuous. Let
$E_{k}= \{ x \in [a,b] \; | \; \inf_{\delta > 0}\left( \sup_{|y-x|<\delta \;, |z-x| < \delta}|f(y)-f(z)|\right) > 1/k\}$
The assumption is that $[a,b] = \cup E_{k}$. Again using the Baire Category Theorem in more or less the same way you can derive that $f$ is not Riemann integrable.

For another approach see [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1727409521&t=37648]jmerry's post[/url].

Note that jmerry also uses the Baire Category Theorem he just proves it instead of referring to it.

I absolutely love jmerry's approach. The problem I had with the previous proofs is that I have no idea what the BCT is about and my knowledge regarding measure theory is quite rudimentary. Kalle, can you explain in a few words what the BCT is about and how it's relevant to jmerry's proof?

Thanks!

A version of the Baire theorem: An intersection of countably many dense open sets in a complete metric space is dense.

I didn't think it was relevant to my proof, so I didn't mention it. That argument stands on its own, and now is the first time I've seen anyone say Baire category is involved.
It does use some of the same techniques for the proof; the proof in my analysis text for the theorem I quoted uses a sequence of nested balls to construct a point of intersection. The details are different.

I might be confused but this seems simple.  :maybe:

Don't you merely consider that the integral is a supremum to lower Stieltjes sums which, by hypothesis, are [b]always[/b] positive?

Non  :?:

It's clear that the sums (I call them Riemann) are non-negative, but it's not that obvious that they are $>0$. What if $f$ would be such that $f \left( \frac{m}{n}\right) = \frac1{n}$ for all $\gcd(m,n) = 1$? You wouldn't be able "to stick boxes under the graph" then. Of course, any such function would have to be non-integrable, but you do need a way to avoid such anomalies.

I still don't see what the problem is.  :(

The fact is that [i]any [/i]integrable function satisfies that property, regardless of how 'well-behaved' it is. Unless my Mathematical Analysis book taught me an incorrect theory of integration?  :huh:

Well, do you at least agree that for any Riemann integrable function, the integral is defined to be the supremum of lower sums?

jmerry, maybe you are not using exactly Baire's theorem, I don't feel like comparing carefully just for argument's sake. Anyway, the idea does feel to be about the same, and I don't think that supplying a proof of Baire's theorem will make my first argument be any harder than yours. I like it, it avoids showing that there is a point of continuity.

Thunder22: The Baire theorem is a theorem in topology which is very powerful and important. The version I used is that in a complete metric space a countable union of nowhere dense sets has empty interior. A nowhere dense set has closure with empty interior. In many applications it is used in the way I have - you write an open set (often the whole space) as a countable union of well chosen sets, so that you may conclude that one of them is not nowhere dense and use whatever implication this has. I used it more or less to mimic measure theoretical proofs - measures respect countable unions while Riemann integration in generally does not. So I replaced measure theoretic considerations with topological using Baire's theorem. Baire's theorem is not particularly difficult to show, about the same caliber as jmerrys argument that there is a point of continuity.

SouthStar: The problem is to show that the infimum over some interval is larger than 0. It is not obvious. For me, Baire sets is a natural tool to do it.


Brouwer's Fixed Point theorem can be deduced from the fact that a closed ball does not retract to its boundary. For, if a map from the closed disk to itself has no fixed points, by sending each $ x$ to the intersection of the ray $ f(x) \to x$ with the boundary we obtain such a retraction. The following, then, generalizes on this point.

[color=darkred][b]Fact.[/b][/color] [color=brown]A closed ball does not retract to a manifold.[/color]

In particular, the ball does not retract to its boundary, or to any sphere for that matter. (We do not specifically state "the manifold is a subset of the ball," for it must be so if it were to be its retract.) The fact that the composition \[ \mathcal H_n (M) \stackrel{i_{\ast}}{\longrightarrow} 0 \stackrel{r_{\ast}}{\longrightarrow} \mathcal H_n (M)\] is the identity reduces the fact to manifolds whose homology groups are all trivial; perhaps this somehow makes it obvious, though there is a different approach that is also quite brief itself.

That does make it trivial, together with the observation that any retract of a compact and connected set is compact and connected. I would modify the statement to exclude zero-manifolds; we can retract to a point.

Now, what was the point of this thread? Is there a question?


Hello,
Suppose P is a linear map in L(V).(sorry, I don't know how to express in term of matrix) Does the condition PPv=Pv for any v in V imply that P is an orthogonal projection? Or we need some other conditions?

[quote="t0rajir0u"]Every idempotent linear map is a projection, but not necessarily an orthogonal one.  A projection can also be oblique.[/quote]

Could you explain a bit more about oblique potent and give an example when P can be not orthogonal?

[quote="t0rajir0u"]Wikipedia gives the example $ \left[ \begin{array}{cc} 0 & 0 \\
\alpha & 1 \end{array} \right]$.  It is definitely idempotent, but it "skews" the vector it projects.[/quote]

O.K, but still I think for some base your matrix is orthogonal projection(not the same base as the base of this matrix).

There must have been something wrong with what I thought, here I show my prove, please show me where the mistake is.

Suppose PPv=Pv for any v in V, where P is in L(V):

let U be the null space of P and W be the range of P, obviously, the sum of dimensions of U and W equals the dimension of V.

Now we show that V is the direct sum of U and W. otherwise, there must exists a non-zero vector x that appears both in U and W. Since x is in W(range of P), this indicates that there is y that x=Py. so x=Py=PPy(by condition)=Px which suggests that Px is nonzero and thus x can't be in U.

So V is the direct sum of U and W. It is obvious that P is the orthogonal projection on W. Because for any z in W we have z=Pa=PPa=Pz; for any z in U we have Pz=0. Q.E.D

It's not enough that $ W$ and $ U$ be disjoint and sum to $ V$; they also need to be orthogonal subspaces.  And the problem with your first comment is that the basis you need is not itself orthonormal.


In Call of Duty 4, there are many weapons and perks available for a player. The primary weapons are the 7 assualt rifles, 5 submachine guns, 3 light machine guns, 2 shotguns, and 5 sniper rifles. There are 5 different side arms. There are also 3 different types of grenades. There are three perk slots - 7 perks in the first and second, while 8 in the third. Players are able to choose only one perk per slot.

Questions -

1.) A player chooses a primary weapon and a side arm, along with three perks and one of the three grenade types. How many different combinations are possible?

2.) A player chooses two primary weapons (no side arms), along with two additional perks and one of the three grenade types. How many different combinations are possible? (You cannot choose the same weapon for both primaries).

:mad: there are 22 primary weapons.
ans 1.)
multiply them all
=22*5*(7+7+8)*3
=330(22)
=7260       , if my calc. are right

ans. 2)
if we choose a category of weapon that has 'n' number of weapons ,
then permutation=22*(22-n) for primary weapons
two perks are chosen , so permutation=(8+7) or (7+7)
so here ans =22*(22-n)*3*15
                  or
                 =22*(22-n)*3*14
 since there are 5 different primaries , so no. of solutions =5*2
=10 :wink:

      do give reply :)


Do you have a pet? If so take teh poll.
I have a kitty cat who is so cute and can kinda count to 2 (we have an alarm and one bleep means open door/ just closed door).

Our family has one rabbit, three fish, and one gerbil.

our family has 2 extremely ridiclously fat guinea pigs that want food everytime you walk by.
they smell like poo too.
i have no idea what is under that fur of theirs, and i don't wanna know either.

one humongous stinky dog. and a very cute ferret. do chipmunks count?

I thought that ferrets were supposed to stink....

i don't have a pet, unless you count my little brother!  :rotfl:  :rotfl:  :rotfl:


geometry
inradius
perimeter
perpendicular bisector
geometry proposed

The vertices of a square sheet of paper are $ A$, $ B$, $ C$, $ D$. The sheet is folded in a way that the point $ D$ is mapped to the point $ D'$ on the side $ BC$. Let $ A'$ be the image of $ A$ after the folding, and let $ E$ be the intersection point of $ AB$ and $ A'D'$. Let $ r$ be the inradius of the triangle $ EBD'$. Prove that $ r\equal{}A'E$.

Take D’ a variable point on (BC); the perpendicular bisector of DD’ intersects AD at P. D’P is the image of AD when folding to bring D on (BC). Take A’ on D’P so that A’D’ = AD. The triangle PDD’ is isosceles and, since its altitude from D’ equals the side of the square, so will the one from D, hence we conclude that PD’ is the tangent from D’ to the circle (D,DC).
If M is its tangency point to this circle we see that D’M = D’C and EM =EA, so the perimeter of the triangle BD’E is 2a (a – the side of the square) and, being right angled at B, its inradius will be r = p - D’E; but p= AD = A’D’, hence r = A’D’ - D’E = A’E.

Best regards,
sunken rock

This Sangaku problem is known as "The problem of the coin and the sheet". See pages 29 and 98 in the below attachment.


geometry
rectangle
symmetry
ratio
summer program
Mathcamp
function

Is it possible to partition a (1*sqrt(2)) rectangle into a finite number of squares?

Since 1 can not be partitioned into the sum of a finite number of irrational numbers we wish to see if $\sqrt{2}$ can be partitioned into a finite sum of rational numbers. However, since $\sqrt{2}=\frac{1}{1+\frac{1}{1+\frac{1}{....}}}$ which is an infinite fraction, I believe the answer is that a $1$ x $\sqrt{2}$ rectangle can not be partitioned into a finite number of squares. I could be wrong though, this is a quick intuition  ;)

Imagine that the $1 \times \sqrt{2}$ square is made out of a resistive material. Line the top of and bottom of the rectangle with a conductive material so all points at the top of the rectangle are at the same potential, and likewise for the bottom of the rectangle. Now suppose that we can cut the rectangle into squares. It is clear by symmetry that no current will flow sideways through the square. Therefore, we can line the top and bottom of each of our squares with conductive material without changing total resistivity of the rectangle.

It is easy to confirm that the resistance between the top and the bottom of a rectangle whose sides are in ratio r:1 is independent of the size of the rectangle, and is equal to r. It follows that all squares have the same resistance, let's call it 1, and the $1 \times \sqrt{2}$ has resistance $1/\sqrt{2}$ (or $\sqrt{2}$, depending whether you view the side with length $1$ or $\sqrt{2}$ as the "top"). The resulting network is a series-parallel network, and it follows from the series-parallel resistance rule that the total resistance of any rectangle that can be dissected into squares must be rational, QED.

We didn't really use any physics, so this argument could probably be made mathematically rigorous (but why bother!) I learned the really neat technique given here from Julian Gibley at Mathcamp '04 in his talk on dissecting the square into squares.

By the way, you're intuition was totally correct, DPopov, that this problem has to do with continued fractions. The total total resistance of a complicated series-parallel circuit is a sort of continued fraction related to its components, as I learned solving [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36527]this[/url]
problem. If you wanted to fill a $1 \times \sqrt{2}$ rectangle with an infinite number of squares, the easiest way to do it would be to first put in one $1 \times 1$ square, then two $\sqrt{2}-1 \times \sqrt{2}-1$ squares, etc., the number of squares you put in each time is the next number you read off in the continued fraction expansion of $\sqrt{2}$

Ah, that is awesome, great way of thinking about it!!! :lol:

Here's another cool technique (this is a slightly modified proof, but basically the same as that one, of a problem discussed a long time ago in the Olympiad forum, and solved by lhvietbao):

Let $f:\mathbb R\to\mathbb R$ be an additive function which takes the value $0$ in $x$ iff $x$ is rational (such functions can easily be constructed using the axiom of choice; see the last few lines). Assume a rectangle is tiled with squares and one of its sides is $1$. Define the function $\varphi$ on the set of rectangles with sides parallel to the ones of the initial rectangle by setting $\varphi(\mathcal R)=f(a)f(b)$, where $a,b$ are the lengths of the sides of the rectangle $\mathcal R$.

It's easy to see that $\varphi$ is additive, in the sense that if a rectangle $\mathcal R$ is tiled with $\mathcal R_1,\ldots,\mathcal R_n$, then $\varphi(\mathcal R)=\varphi(\mathcal R_1)+\ldots+\varphi(\mathcal R_n)$. $\varphi$ takes the value $f(a)^2$ on all squares of side $a$, so if at least one of the squares paving $\mathcal R$ had an irrational side, $\varphi(\mathcal R)$ would be $>0$ (because it's the sum of the squares of $f(a)$ for all sides $a$ of the squares paving $\mathcal R$). However, $\varphi(\mathcal R)=0$, because if its sides are $1$ and $t$, then $\varphi(\mathcal R)=f(1)f(t)=0$, since $f(1)=0$.

The above means that all the squares have rational sides, so, in particular, $t$ must also be rational. However, note that a lot more has been proved, namely that all the squares tiling a rectangle with a rational side have rational sides.

For the construction of $f$, consider a basis $(a_i)_{i\in I}$ of $\mathbb R$ as a vector space over $\mathbb Q$ which contains $1$ (assume $a_{i_0}=1$), and set $f(\sum\alpha_ia_i)=\sum_{i\ne i_0}\alpha_ia_i$ (for each real  $\sum\alpha_ia_i$, only finitely many of the $\alpha_i$'s are non-null).


I don't understand the rules and strategies of playing mancala. Can anyone help me?

Google has answers.

[url]http://boardgames.about.com/cs/mancala/ht/play_mancala.htm[/url]

Actually, I would say that the strategy is that you calculate before you make a move..

This game is fun.


function
trigonometry
calculus
calculus computations

Write an expression for the function, f(x), with f '(x) = sin^2(x) + sin(x) and f(pi/2) = 5.

HELP PLEASE!

Let's try to be more helpful in our answers.

$f ' (x) = \sin ^2 x + \sin x$

$f ' (x) = \frac{1-\cos (2x)}{2} + \sin x \rightarrow f (x) = \frac{x}{2} -\frac{\sin (2x)}{4} - \cos x + C$

Since $f (\frac{\pi}{2}) = \frac{\pi}{4} + C = 5$

$f(x) = \frac{x}{2} - \frac{\sin (2x)}{4} - \cos x + 5 - \frac{\pi}{4}$


Your condition equavalent to $f(x)+f(x^{2})=0$ (consider $x\to x^{2^{m}}, m=0,1,...,n$).
If $g(y)=f(e^{y})$, then $g(y)+g(2y)=0$, if $g(\pm 2^{t})=s_{\pm}(t)$, then $s_{\pm}(t+1)=-s_{\pm}(t)$ ($s_{\pm}(t)$ periodic with period 2).
For all $s_{\pm}(t), \ t\in [0,1)$ we get solution $f(x)$.

[quote="N.T.TUAN"]Given $n \in\mathbb{N}$, ﬁnd all continuous functions $f : \mathbb{R}\to \mathbb{R}$ such that for all $x\in\mathbb{R},$
\[\sum_{k=0}^{n}\binom{n}{k}f(x^{2^{k}})=0. \]
[/quote]

$\sum_{k=0}^{n}\binom{n}{k}f(x^{2^{k}})=0$ $\Leftrightarrow$ $\sum_{k=0}^{n-1}\binom{n-1}{k}f(x^{2^{k}})+\sum_{k=0}^{n-1}\binom{n-1}{k}f(x^{2^{k+1}})=0$.
If I call $g(x)=\sum_{k=0}^{n-1}\binom{n-1}{k}f(x^{2^{k}})$, I have $g(x^{2})=-g(x)$ and so $g(x)=g(x^{4^{-p}})$ for $x\geq 0$ and so, with continuiity (when p$\rightarrow+\infty$) $g(x)=c$ for any $x\geq 0$ and, since $g(x)=-g(x^{2})$ for any x, we have $g(x)=c$ for any $x\geq 0$x, and $g(x)=-c$ for any $x\leq 0$ and so $g(x)=0$ for any x

So $\sum_{k=0}^{n-1}\binom{n-1}{k}f(x^{2^{k}})=0$

So any solution for rank n is also solution for rank $n-1$
The only solution at rank 0 is $f(x)=0$

So the only solution at rank n is $f(x)=0$.

--
Patrick

[quote="Rust"]Why $f(x)+f(x^{2})$ give $f(x)\equiv 0$?[/quote]

As I said : $g(x)+g(x^{2})=0$ $\Rightarrow $ $g(x^{2})=-g(x)$ $\Rightarrow $ $g(x)=-g(x^{\frac{1}{2}})$ for $x\geq 0$ $\Rightarrow $ $g(x)=g(x^{\frac{1}{4}})$ for any $x\geq 0$
Hence $g(x)=g(x^{4^{-p}})$ for any $x\geq 0$
When $p\rightarrow+\infty$ and because g is continuous, $g(x)=g(1)$ for any $x>0$
Since $g(x)=-g(x^{2})$ for any x, we have then $g(x)=-g(1)$ for any $x<0$
With the continuity at 0, the result is $g(x)=0$

--
Patrick


analytic geometry
graphing lines
slope
geometry
function

How can you prove that from the formula $ a\equal{}\frac{v\minus{}v_0}{t\minus{}t_0}$ we can find that $ l\equal{}v_0t\plus{}\frac{at^{2}}{2}$  :)

There is no need to integrate.

1) Consider a graphic representation v = v(t) for a constant acceleration motion (which will be a line with a slope given by $ a$).

2) By definition of velocity, the distance traveled $ l$ between $ t_0$ and $ t$ is given by the area under the line v = v(t).

3) Then write, using the area, $ l$ as a function of $ v_o$, $ v$, $ t$, and $ t_o$. Don't forget that, for simplicity and without loss of generality, we can take $ t_o \equal{} 0$.

4) Finally use the expression $ a \equal{} \frac {v \minus{} v_o}{t \minus{} t_o}$.

Who are first and second derivation  of   $ t$ ?


topology
advanced fields
advanced fields unsolved

Show that the torus minus one point is homotopy equivalent to the figure eight.
I have no idea to prove such facts.please solve this and give an advice to proof or disproof such statement.

The torus is the quotient space of $ I \times I$ where the top/bottom edges and left/right edges are identified. Remove
the point from the center; then you can deformation retract to the edges of the square, getting a figure-eight.


prove that the equation $y!-y^{2}-y+19=0$ has no solution for natural numbers.

[hide]
for $y=1,2,3$ there are no solutions. Let $y \ge 4$. We want to prove that $y! > y^{2}+y \rightarrow (y-1)! > y+1$. This is our inductive hypotesis.
So we have to prove that $y! > y+2 \rightarrow y((y-1)!-1) > 2$.
Since $(y-1)! > y+1$, $(y-1)!-1 > y$, so $y((y-1)!-1) > y^{2}> 2$, that is obviously true cause $y \ge 4$ [/hide]

well... $y^{2}+y$ has to be even, and y! has to be even when y>2. so the difference of two even numbers cannot come out odd..

The whole thing is divisible by $y$ so y has to be 1 or 19.

you show really nice proofs! thanks  :) . my way
$y!-1=(y+5)(y-4)$ so in the right side we have to numbers that their difference is $9$. so if mod it by $9$ they have the same residue. on the right side when $y\ge9$ the residue is $8$. so we check $1^{2},2^{2}...8^{2}mod 9$ and none is 8.


how to solve $ ax^3 \plus{} bx^2 \plus{} cx \plus{} d \equal{} 0$

i have read some notes but still dont understand   :P [/code]

Check here
http://en.wikipedia.org/wiki/Cubic_equation
I think it's useful

Daniel


trigonometry
quadratics
geometry
circumcircle
algebra
similar triangles
geometry unsolved

Construct a triangle $\triangle ABC$, given the side $a$, the angle $\measuredangle A$ and the segment $s = mb+nc$, where $m, n$ are given reals.

Let $R$ be the triangle circumradius. Then

$a = 2R \sin{\widehat A}$, $b = 2R \sin{\widehat B}$, $c = 2R \sin{\widehat C}$

Since $\measuredangle C = 180^{o}-(\measuredangle A+\measuredangle B)$

$\sin{\widehat C}= \sin{(\widehat A+\widehat B)}$.

$s = mb+nc = 2R (m \sin{\widehat B}+n \sin{\widehat C}) = 2R [m \sin{\widehat B}+n \sin{(\widehat A+\widehat B)}] =$

$= 2R [(m+n \cos{\widehat A}) \sin{\widehat B}+n \sin{\widehat A}\cos{\widehat B}]$

Rearranging this equation and squaring it, in order to get rid of $\cos{\widehat B}$, yields

$[s-2R (m+n \cos{\widehat A}) \sin{\widehat B}]^{2}= (2R n \sin{\widehat A})^{2}(1-\sin^{2}{\widehat B})$

Denoting $x = \sin{\widehat B}$, we get the following quadratic equation for $x$:

$4R^{2}[(m+n \cos{\widehat A})^{2}+(n \sin{\widehat A})^{2}] x^{2}-4sR(m+n \cos{\widehat A}) x+s^{2}-4R^{2}(n \sin{\widehat A})^{2}= 0$

$4R^{2}(m^{2}+n^{2}+2mn \cos{\widehat A}) x^{2}-4sR(m+n \cos{\widehat A}) x+s^{2}-(na)^{2}= 0$

The triangle circumradius $R$ and the segment $R \cos \alpha$ can be easily constructed and so can be the seqment $q = 2(mR+nR \cos{\widehat A})$. The segment $r = \sqrt{s^{2}-(na)^{2}}$ is one leg of a right angle triangle with hypotenuse $s$ and the other leg $na$. Let $\triangle MNP$ be a triangle with sides $PM = 2mR, PN = 2nR$ and an angle $\measuredangle MPN = 180^{o}-\measuredangle A$. This triangle can be easily constructed. By cosine theorem, the side $p = MN$ is equal to

$p = MN = 2R \sqrt{m^{2}+n^{2}+2mn \cos{\widehat A}}$

The quadratic equation for $x = \sin{\widehat B}$ then becomes

$p^{2}x^{2}-2sqx+r^{2}= 0$

${x = \sin{\widehat B}= \frac{1}{ }p [\frac{sq}{p}\pm \sqrt{(\frac{sq}{p})^{2}-r^{2}}}]$

The segment $t = \frac{sq}{p}$ is constructed with the help of similar triangles, one with sides $p, q$, the other with sides $s, t$, and with equal angles $\widehat{st}= \widehat{pq}$. After this, the segment $u = \sqrt{t^{2}-r^{2}}$ is constructed as one leg of a right angle triangle with the hypotenuse $t$ and the other leg $r$. Finally, $x = \sin{\widehat B}= \frac{t \pm u}{p}$ is constructed as the angle against the leg $t \pm u$ of a right angle triangle with hypotenuse $p$.

[size=134][hide="Introduction"][/size]Let $S = mb+nc$. This product reminds me the Ptolemy's Theorem.

Suppose that we have constructed the triangle $ABC$. Let $w$ be its circumcircle. Let $E$ be a mobile point on the arc $BC$ which not contains the point $A$.

Then from Ptolemy in $ABCD$ we have $EA\cdot a = EB\cdot b+EC \cdot c$

So, one could think that we should choose the position of $E$ such that $\frac{EB}{m}=\frac{EC}{n}$.

Now I think it's easy
$\hline$[/hide]
[size=134][hide="Construction"][/size]
[color=brown][b]1.a)[/b][/color] Start with the segment $BC=a$

[b][color=brown]1.b)[/color][/b] Construct the locus of points $P$ which has the property $\angle BPC=A$. Namely, this locus is an arc $w_{1}$ of a circle which passes through $B,C$. Clearly $A\in w_{1}$

[b][color=brown]1.c)[/color][/b] Let $w_{2}$ be an arc such $w=w_{1}\cup w_{2}$ is a circle

[b][color=blue]2.a)[/color][/b] Construct an angle $\angle xDy=180-A$

[color=blue][b]2.b)[/b][/color] Take the points $M\in Dx, \ N\in Dy$ such that $DM=m, \ DN=n$.

[b][color=blue]2.c)[/color][/b] On the arc $w_{2}$ construct a point $E$ such $\triangle BCE \sim \triangle MND$

Then we have $k = \frac{BC}{MN}= \frac{BE}{DM}= \frac{CE}{DN}\Rightarrow$

$BC = k\cdot MN$
$BE = k\cdot m$
$CE = k \cdot n$

From Ptolemy in $ABCD:$

$AE \cdot BC = b \cdot BE+c \cdot CE$

$AE \cdot a = b \cdot k \cdot m+c \cdot k \cdot n$

$AE \cdot a = k\cdot (b \cdot m+c\cdot n)$

$AE \cdot a = k\cdot S$

$AE = \frac{kS}{a}= \frac{S}{MN}$

The point $E$ is constructed. What we need now is to construct a circle $q=(E,AE)$. Then $A \in q \cap w$

$\hline$
[/hide]
[size=134][hide="Discussion"][/size]Let $R$ be the radius of $w$.

If $AE<2R$ then we have two solutions
If $AE=2R$ then we have one solution
If $AE>2R$ then we have no solutions[/hide]


probability
combinatorics unsolved
combinatorics

X,Y and Z are playing a game and they have the same probability to win each set of the game.Who first wins 3 sets will be the winner. Know that X won the first set, find the probability that Y will be the winner.

Obviously, it's $\frac{5}{16}=31.25\%$.

There are only 10 variants of the game (I don't write the first winning of X):

1. XX --- X wins
2. XYX --- X wins
3. XYYX --- X wins
4. XYYY --- Y wins
5. YXX --- X wins
6. YXYX --- X wins
7. YXYY --- Y wins
8. YYXX --- X wins
9. YYXY --- Y wins
10. YYY --- Y wins

So, the desired probability is $\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{8}=\frac{5}{16}$.

I think you forget that Z also takes part in the game.

Indeed. I'm sorry.

There are 192 variants of the game.

XX
XYX
XYYX
XYYY
XYYZX
XYYZY
XYYZZX
XYYZZY
XYYZZZ
XYZX
XYZYX
XYZYY
XYZYZX
XYZYZY
XYZYZZ
XYZZX
XYZZYX
XYZZYY
XYZZYZ
XYZZZ
XZX
XZYX
XZYYX
XZYYY
XZYYZX
XZYYZY
XZYYZZ
XZYZX
XZYZYX
XZYZYY
XZYZYZ
XZYZZ
XZZX
XZZYX
XZZYYX
XZZYYY
XZZYYZ
XZZYZ
XZZZ
YXX
YXYX
YXYY
YXYZX
YXYZY
YXYZZX
YXYZZY
YXYZZZ
YXZX
YXZYX
YXZYY
YXZYZX
YXZYZY
YXZYZZ
YXZZX
YXZZYX
YXZZYY
YXZZYZ
YXZZZ
YYXX
YYXY
YYXZX
YYXZY
YYXZZX
YYXZZY
YYXZZZ
YYY
YYZXX
YYZXY
YYZXZX
YYZXZY
YYZXZZ
YYZY
YYZZXX
YYZZXY
YYZZXZ
YYZZY
YYZZZ
YZXX
YZXYX
YZXYY
YZXYZX
YZXYZY
YZXYZZ
YZXZX
YZXZYX
YZXZYY
YZXZYZ
YZXZZ
YZYXX
YZYXY
YZYXZX
YZYXZY
YZYXZZ
YZYY
YZYZXX
YZYZXY
YZYZXZ
YZZXX
YZZXYX
YZZXYY
YZZXYZ
YZZXZ
YZZYXX
YZZYXY
YZZYXZ
YZZYY
YZZYZ
YZZZ
ZXX
ZXYX
ZXYYX
ZXYYY
ZXYYZX
ZXYYZY
ZXYYZZ
ZXYYZX
ZXYYZY
ZXYYZZ
ZXYZX
ZXYZYX
ZXYZYY
ZXYZYZ
ZXYZZ
ZXZX
ZXZYX
ZXZYYX
ZXZYYY
ZXZYYZ
ZXZYZ
ZXZZ
ZYXX
ZYXYX
ZYXYY
ZYXYZX
ZYXYZY
ZYXYZZ
ZYXZX
ZYXZYX
ZYXZYY
ZYXZYZ
ZYYXX
ZYYXY
ZYYXZX
ZYYXZY
ZYYXZZ
ZYYY
ZYYZXX
ZYYZXY
ZYYZXZ
ZYYZY
ZYYZZ
ZYZXX
ZYZXYX
ZYZXYY
ZYZXYZ
ZYZXZ
ZYZYXX
ZYZYXY
ZYZYXZ
ZYZYY
ZYZYZ
ZYZZ
ZZXX
ZZXYX
ZZXYYX
ZZXYYY
ZZXYYZ
ZZXYZ
ZZXZ
ZZYXX
ZZYXYX
ZZYXYY
ZZYXYZ
ZZYXZ
ZZYYXX
ZZYYXY
ZZYYXZ
ZZYYY
ZZYYZ
ZZYZ
ZZZ

The desired probability is $\frac{153}{729}$.

P.S. I didn't use computer. So, I could be mistaken.

Dear Remike,

Thanks a lot for your solution. But I hope you keep on finding another solution for this problem.

Hoang

[quote="DusT"]I think I have something a little less calculatory than the one above ! :D
Let every lattice point in the XYZ plane represent a state of the games, that is how many times did X, Y and Z win will be reprezented by the x, y and z coordinates. Now, what we need to know is the number of the paths from $(1,0,0)$ to the plane $(x,3,z)$, with $x,z<3$. Take all the pairs of points and calculatate, they will $3\cdot 3$ , and the formula is $\binom{x+z}{x}$ I think!
Now, the entire number of paths can be calculated in the same way :)[/quote]

This will work, but the paths are to $(x,2,z)$ because the last step is a Y.

There is a nice answer (which I do not remember) for the number of games of
length $N$ where the number of victories satisfies $n(X) > n(Y) \geq n(Z)$ at all times.

you can cut short your work even further by noticing that $P(Y$ wins)=$P(Z$ wins) (for every game that Y wins just swap Y and Z and that will give a game that Z wins and vice-versa).Also,
  $P(X$ wins)+$P(Y$ wins)+$P(Z$ wins) = 1.so it just suffices to calculate $P(X$ wins) and that reduces your work even further.

[quote="fleeting_guest"][quote="DusT"]I think I have something a little less calculatory than the one above ! :D
Let every lattice point in the XYZ plane represent a state of the games, that is how many times did X, Y and Z win will be reprezented by the x, y and z coordinates. Now, what we need to know is the number of the paths from $(1,0,0)$ to the plane $(x,3,z)$, with $x,z<3$. Take all the pairs of points and calculatate, they will $3\cdot 3$ , and the formula is $\binom{x+z}{x}$ I think!
Now, the entire number of paths can be calculated in the same way :)[/quote]

This will work, but the paths are to $(x,2,z)$ because the last step is a Y.
[/quote]
Actually, it was ok because [quote="DusT"] Now, what we need to know is the number of the paths from $(0,1,0)$ to the plane $(x,3,z)$, with $x,z<3$. Take all the pairs of points and calculatate, they will $3\cdot 3$ , and the formula is $\binom{x+z}{x}$ I think!
[/quote]
I said we start from $(0,1,0)$, but it is actually Ok. Also, seshadri's observation (I made it, but I forgot to mention it) helps even more .....

[quote="DusT"][quote="fleeting_guest"] This will work, but the paths are to $(x,2,z)$ because the last step is a Y. [/quote]
Actually, it was ok because [quote="DusT"] Now, what we need to know is the number of the paths from $(0,1,0)$ to the plane $(x,3,z)$, with $x,z<3$. Take all the pairs of points and calculatate, they will $3\cdot 3$ , and the formula is $\binom{x+z}{x}$ I think!
[/quote]
I said we start from $(0,1,0)$, but it is actually Ok.[/quote]

Your posting said we start from $(1,0,0)$, which is correct because X wins the first game.
It is also given that Y wins the last game, so the last move is from $(x,2,z)$ to $(x,3,z)$.
So the correct thing to count is paths from $(1,0,0) \to (x,2,z)$ with $x,z \leq 2$.


trigonometry
inequalities
inequalities proposed

[i]   Let $ a , b, c$ be positive real numbers such that

 $ | sina \plus{} sinb \plus{} sinc | \geq \sqrt {2}$

  Prove that

  $ | cosa \plus{} cosb \plus{} cosc | \leq \sqrt {5}$[/i]

   [i]    Argady , Do you like this problem? :lol: [/i]

 [i] Love An Forever

 Huỳnh Võ Phương An[/i]

Let $ a \equal{} b \equal{} c \equal{} \alpha$ such that $ \sin \alpha  \equal{} \frac{{\sqrt 2 }}{3},\cos \alpha  \equal{} \frac{{\sqrt 7 }}{3}$

We have

$ \sin a \plus{} \sin b \plus{} \sin c \equal{} 3\sin \alpha  \equal{} \sqrt 2$

$ \cos a \plus{} \cos b \plus{} \cos c \equal{} 3\cos \alpha  \equal{} \sqrt 7  > \sqrt 5$

So your inequality is wrong!


geometry
3D geometry
sphere
vector
analytic geometry
pyramid
geometry unsolved

Prove that among any five distinct rays $ Ox$, $ Oy$, $ Oz$, $ Ot$, $ Or$ in space there exist two which form an angle less than or equal to $ 90^{\circ}$.

We assume for the sake of contradiction that all angles determined by any two of the five rays are greater than $ 90^\circ$.

Consider a sphere centered at $ O$, and let $ Ox,Oy,Oz,Ot,Or$ intersect the sphere at $ x, y, z, t, r$ respectively.
The plane perpendicular to $ Ox$ at $ O$ cuts the sphere into two hemispheres.
Let $ A$ be the hemisphere containing $ x$. $ y$ cannot lie on $ A$, because if it did, the angle $ \angle xOy$ would be less than or equal to $ 90^\circ$.
By the same argument $ z, t, r$ cannot lie on $ A$. Define hemispheres $ B$, $ C$, $ D$, and $ E$ similarly, to contain $ y, z, t, r$, respectively and have similar properties.

Each of the five hemispheres has a measure of $ 2 \pi$ steradians. Because of the assumption, any two hemispheres must overlap at a solid angle of less than $ \pi$ steradians. Let the intersection of $ A$ and $ B$ have measure $ \theta < \pi$.
The union of $ A$ and $ B$ must have measure $ 2\pi \plus{} 2\pi \minus{} \theta \equal{} 4\pi \minus{} \theta$, and a complement $ G$ of measure $ 4\pi \minus{} (4\pi \minus{} \theta) \equal{} \theta$, and be bounded by two semi-great-circles that meet at two points, $ p$ and $ q$. The complement $ G$ is contained within what it would be if its measure were its upper bound of $ \pi$.

The intersection of $ G$ and $ C$ has a minimum of $ \dfrac{\theta}{2}$ as $ z$ approaches either $ p$ or $ q$, so it is greater than $ \dfrac{\theta}{2}$. Thus the union of $ A, B, C$ has measure greater than $ (4\pi \minus{} \theta) \plus{} \dfrac{\theta}{2} \equal{} 4\pi \minus{} \dfrac{\theta}{2}$, so the complement $ H$ of this union has measure less than $ \dfrac{\theta}{2} < \dfrac{\pi}{2}$ and is contained within what it would be if $ z$ were at $ p$ or $ q$.

$ H$ is contained within what it would be if $ z$ were at $ p$ or $ q$ and, since $ H$ is a subset of $ G$, within what $ H$ would be if $ \theta$ equalled $ \pi$. This would be a $ 90^\circ$-$ 90^\circ$-$ 90^\circ$ spherical triangle, so $ H$ is a subset of a $ 90^\circ$-$ 90^\circ$-$ 90^\circ$ spherical triangle. Since $ t$ and $ r$ must lie on $ H$, and therefore a $ 90^\circ$-$ 90^\circ$-$ 90^\circ$ spherical triangle, $ \angle tOr \le 90^\circ$. We have contradicted our assumption, so we are done.

This is my solution.

Let $\vec{x}, \vec{y}, \vec{z}, \vec{t}, \vec{r}$ be the the vector in the rays' directions.
We can set up our coordinate system such that, we may assume that all rays start at the origin $O$ and we may assume that $\vec{x} = \vec{k}$, that is pointing straight upwards. So that means the other four vectors must have a negative $z$ component, that is, pointing at the bottom half-space.

Let $P$ be the plane $z = -1$ and let $Y,Z,R,T$ be the intersection of $O_y, O_z, O_r, O_t$ with $P$. These points must exist because all those four rays are pointing "downwards". We now have a pyramid $O.YZRT$ and let $S$ be the point $(0,0,-1)$, that is, the projection of point $O$ onto $S$.

Examine the 2D figure on plane $P$. We have a quadrilateral $YZRT$, and a point $S$. Draw the following four rays: $\vec{SY}, \vec{SZ}, \vec{SR}, \vec{ST}$ all originating from $S$.  It's easy to see that two of those rays must form an angle less than or equal to 90 degrees, for the sum of all four angles must equal to 360 degrees.

WLOG, suppose the angle between $\vec{SY}$ and $\vec{SZ}$ is $\theta \leq 90^o$. But not that $SO$ is perpendicular to plane $P$, and hence perpendicular to all vectors on plane $P$.
$\vec{OY} . \vec{OZ} = (\vec {SY} - \vec{SO})(\vec{SZ} - \vec{SO}) = \vec{SY} . \vec{SZ} + SO^2$ because $SO \perp SY, SZ$

But since the angle between $\vec{SY}$ and $\vec{SZ}$ is $\leq 90^o$ then $\vec{SY} . \vec{SZ} \geq 0$ which means $\vec{OY} . \vec{OZ} > 0$, which means that the rays $OY$ and $OZ$ form an acute angle.


Asymptote
quadratics
Vieta
algebra
quadratic formula

3. How many vertical asymptotes does $ f(x)\equal{} /frac{x^2\minus{}6x\minus{}16} {x^3\minus{}5x^2\minus{}2x\plus{}24}$ have?

I don't know how you would factor the denominator.

10. If a,b, and c are roots of the equation $ f(x)\equal{}x^3\plus{}2x^2\plus{}5x\minus{}9$, find $ a\plus{}bc\plus{}c\plus{}ac\plus{}abc\plus{}b\plus{}ab$

I don't know how to find roots of equations that aren't quadratic formula possible on this one. Is there a good place to learn about this?

Haha explanation would be nice  :lol:

Do you know the Rational Roots Theorem yet? That's one way to factor... Anyway,
[hide="3"]
I'm guessing you meant to type $ \dfrac{x^2-6x-16}{x^3-5x^2-2x+24}$. Factor the top, getting $ \dfrac{(x+2)(x-8)}{x^3-5x^2-2x+24}$. Now we use my conjecture that teachers repeat a root on the top and the bottom to "trick" you, and check for $ (x+2)$ and $ (x-8)$. The first one factors out. Then we have:
$ \dfrac{(x+2)(x-8)}{x^3-5x^2-2x+24}=\dfrac{(x+2)(x-8)}{(x+2)(x^2-7x+12)} = \dfrac{(x+2)(x-8)}{(x+2)(x-3)(x-4)}$. The $ x+2$ makes a hole at $ x=-2$, and the $ (x-3)(x-4)$ makes asymptotes at $ x=3$ and $ x=4$.
[/hide]
[hide="10"]
Using Vieta's Law, we get:
\begin{eqnarray*}
a+b+c&=&-2\\
ab+bc+ac&=&5\\
abc&=&-(-9)
\end{eqnarray*}
Add the three equations to get $ a+bc+c+ac+abc+b+ab=12$.
[/hide]


In the USAMTS solutions, I used "\begin{align} ....3 lines of equations.... \end{align}" Later, I used it again, though it started numbering the equations at 4. How can I make it begin numbering at 1?

Sorry, if this has been posted before, I couldn't find it. :lol:

Yes that's the default for equation numbering so that every equation has a unique number you can refer to. But if you'd prefer to reset the counter so it starts at one again put \setcounter{equation}{0} before the next \begin{align}.

Thanks! Are equations usually numbered continuously, even going on to different problems?

I don't know about USAMTS itself (maybe someone else can comment?) but in mathematical papers it's usual for every equation to have a different number; this is sometimes can be achieved by using numbering systems like 1.1, 1.2, ... , 2.1, 2.2,... where the paper (unlike USAMTS) has sections.

Could other people who have submitted solutions say what they do?

I usually write, for example, 3.1 for my first labeled equation in problem 3.


calculus
probability
integration
function
derivative
calculus computations

I'm trying to review some calculus for a probability class I am taking. There are a lot of integral's that require me to integrate exponentials. I have come across an 'easy' question and I am having problems with it.

I have included a small picture of my work. It is wrong and I am not sure why. From what I can tell, it is like another example in the book except it is -2x rather then x^2 for e. Thanks for any help.

[img]http://www.tempestseason.com/images/math.jpg[/img]

[quote="dayrinni"]I'm trying to review some calculus for a probability class I am taking. There are a lot of integral's that require me to integrate exponentials. I have come across an 'easy' question and I am having problems with it.

I have included a small picture of my work. It is wrong and I am not sure why. From what I can tell, it is like another example in the book except it is -2x rather then x^2 for e. Thanks for any help.
[/quote]

You can't just substitute in du for dx in this case since there is a constant factor involved.  Once you introduce the constant factor, you can find algebraically that your answer is the same as that in the book.  BTW, I hate the format of the book's answer, there's nothing wrong with negative exponents.

[quote="dayrinni"]Thanks for your reply. I changed it around to:
Int[  e^x/2 ] but I still wasn't able to get anywhere.[/quote]
What haoye meant was this: $\int_{0}^{1}e^{-2x}\, dx =-\frac12 e^{-2x}\Big|_{0}^{1}$.

So I forgot to mulitple each side by 1/2:
u = -2x du: -2
and, the -1/2 comes from the mulitplication to make sure it doesn't change, ie: -2/2=1.
-1/2 Int(e^(-2x))

My calc book does not go into this very well. I have what appears to be a very important question now!

When do I multiple each side by 1/x? Where x = d/du. When there is a constant in e's exponent?

I also am wondering how the book got the answer it did. I can see how e can be on the bottom because of the negative exponent, but it does not explain how it stays on top.

Sorry, its been 2 years since I've even looked at my calc book.

THANK YOU!!

The substitution rule for definite integrals:
$\int_{a}^{b}f(g(x))\cdot g'(x)\,dx=\int_{x=a}^{x=b}f(u)\,du= \int_{g(a)}^{g(b)}f(u)\,du$
by the substitution $u=g(x),du=g'(x)\,dx$.
This is essentially the chain rule written backwards.

The most common mistakes students make applying this rule all result from transforming only part of the integral, while ignoring other parts. In your case, you left out the constant factor required to transform $dx$ into $du$. You also neglected the limits of integration; your second line is not a true equation.

[quote="dayrinni"]When do I multiple each side by 1/x? Where x = d/du. When there is a constant in e's exponent?[/quote]Don't think like this.
When making a substitution, convert [b]everything[/b] into the new variable. The function and the limits transform by ordinary composition; the $d$(whatever) transforms by the derivative of the transformation. Because of this, it is standard to write the substitution in the two parts $u=g(x),du=g'(x)\,dx$ or $x=h(u),dx=h'(u)\,du$.
The $d$(whatever) is part of the integral sign. If you write down an integral without one, you risk many kinds of mistakes.

Your example, reworked carefully:
$\int_{0}^{1}e^{-2x}\,dx$ ($u=-2x,du=-2\,dx$)
$=-\frac12\int_{0}^{1}e^{-2x}\cdot-2\,dx=-\frac12\int_{0}^{-2}e^{u}\,du$
$=\.-\frac12e^{u}|_{0}^{-2}=-\frac12e^{-2}+\frac12= \frac12(1-e^{-2})$
I would leave the answer in this form. Your book prefers to treat $e^{-2}$ as $\frac1{e^{2}}$, and its form of the answer is the result of finding a common denominator for the two terms.

[By the way, the example you refer to with an integral of $e^{-x^{2}}$ is entirely different. The integral of that function over any finite interval can only be found using numerical methods.]


inequalities
Cauchy Inequality
inequalities proposed
133

Actually, the following inequalities hold (chapter 7):

(a) $\sum \frac 1{y^{2}+ryz+z^{2}}\geq \frac{2r+5}{(r+2)(xy+yz+zx)}$ for $-1 \leq r \leq 2$;

(b) $\sum \frac{xy+pyz+zx}{y^{2}+z^{2}}\geq 2+p/2$ for $-1 \leq p \leq 4$.

For $r=2$, from (a) we get Iran Inequality.

Thank you Vasc and argady.
Can you give another solution (shorter) ?
Another inequality :
If $a,b,c$ are positive numbers and $k$ is a real number in $[-2;2]$, then we have
$\frac{1}{x^{2}+kxy+y^{2}}+\frac{1}{y^{2}+kyz+z^{2}}+\frac{1}{z^{2}+kzx+x^{2}}\ge \frac{9}{(2-k)(x^{2}+y^{2}+z^{2})+2k(xy+yz+zx)}$
For $k=2$, we get Iran 96 inequality  :lol:
For $k=1$, we get a famous inequality (in Old and New Inequalities).

[quote="ehku"]Can you, Vasc or argady? It's not too hard.[/quote]
may you give a hint please?

In Algebraic Inequalities there is a stronger inequality for $-\frac 1{4}\leq k\leq 2$:

$\frac{1}{x^{2}+kxy+y^{2}}+\frac{1}{y^{2}+kyz+z^{2}}+\frac{1}{z^{2}+kzx+x^{2}}\ge \frac{9(5+2k)}{2+k}.\frac{1}{(4-2k)(x^{2}+y^{2}+z^{2})+(1+4k)(xy+yz+zx)}$.

For $k=1$, we get an inequality sharper than my "famous" inequality (in Old and New Inequalities). :lol:

But your inequality is nicer. :oops:
We may prove it using the inequality
$\frac{(2-k)\sum{a^{2}}+2k\sum{bc}}{b^{2}+kbc+c^{2}}\geq 4\frac{ab+bc+ca}{(b+c)^{2}}$,
which is equivalent to $(2-k)[(b+c)a-b^{2}-c^{2}]^{2}\geq 0$.

[quote="ehku"]Another inequality :

If $ x,y,z$ are positive numbers and $ k\in [ \minus{} 2, 2],$ then

$ \frac {1}{y^{2} \plus{} kyz \plus{} z^{2}} \plus{} \frac {1}{z^{2} \plus{} kzx \plus{} x^{2}} \plus{} \frac {1}{x^{2} \plus{} kxy \plus{} y^{2}}$

$ \ge \frac {9}{(2 \minus{} k)(x^{2} \plus{} y^{2} \plus{} z^{2}) \plus{} 2k(yz \plus{} zx \plus{} xy)}.$

It's not too hard.[/quote]Let $ x,y,z$ are nonnegative numbers such that $ yz \plus{} zx \plus{} xy > 0$.

For $ k\geq \minus{} 2$, we have

$ \frac {1}{y^2 \plus{} kyz \plus{} z^2} \plus{} \frac {1}{z^2 \plus{} kzx \plus{} x^2} \plus{} \frac {1}{x^2 \plus{} kxy \plus{} y^2}$

$ \geq \frac {9}{2\left(x^2 \plus{} y^2 \plus{} z^2\right) \plus{} k(yz \plus{} zx \plus{} xy)},$

with equality if $ x \equal{} \minus{} 1 \minus{} k,y \equal{} z \equal{} 1$ when $ k\in [ \minus{} 2, \minus{} 1].$
[hide="Proof."]Using Cauchy inequality,

$ \sum{\left(y^2 \plus{} kyz \plus{} z^2\right)}\sum{\frac {1}{y^2 \plus{} kyz \plus{} z^2}}\geq9.$[/hide]
For $ \minus{} 1\leq k\leq 2$, we have

$ \frac {2 \plus{} k}{y^2 \plus{} kyz \plus{} z^2} \plus{} \frac {2 \plus{} k}{z^2 \plus{} kzx \plus{} x^2} \plus{} \frac {2 \plus{} k}{x^2 \plus{} kxy \plus{} y^2}$

$ \geq\frac {9(5 \plus{} 2k)}{2(2 \minus{} k)\left(x^2 \plus{} y^2 \plus{} z^2\right) \plus{} (1 \plus{} 4k)(yz \plus{} zx \plus{} xy)},$

with equality if $ x \equal{} 0,y \equal{} z \equal{} 1.$
[hide="Proof."]$ \sum{\frac {2 \plus{} k}{y^2 \plus{} kyz \plus{} z^2}} \minus{} \frac {9(5 \plus{} 2k)}{2(2 \minus{} k)\left(x^2 \plus{} y^2 \plus{} z^2\right) \plus{} (1 \plus{} 4k)(yz \plus{} zx \plus{} xy)}$

$ \equiv F(x,y,z,k) \equal{} F\left(x,y,z,\frac {2p \minus{} 1}{p \plus{} 1}\right)$

$ \equiv\frac {G(x,y,z,p)}{\left[2\sum{x^2} \plus{} (3p \minus{} 1)\sum{yz}\right]\prod\left[(p \plus{} 1)\left(y^2 \plus{} z^2\right) \plus{} (2p \minus{} 1)yz\right]}.$

$ G(x,y,z,p) \equal{} G(x,x \plus{} s,x \plus{} s \plus{} t,p)$

$ \equal{} 2(4p \plus{} 1)^2(3p^2 \plus{} 7p \plus{} 1)(s^2 \plus{} st \plus{} t^2)x^4$

$ \plus{} (4p \plus{} 1)^2(2s \plus{} t)[(6p^2 \plus{} 13p \plus{} 1)s^2 \plus{} (6p^2 \plus{} 13p \plus{} 1)st \plus{} 6(2p^2 \plus{} 5p \plus{} 1)t^2]x^3$

$ \plus{} (4p \plus{} 1)[2(15p^3 \plus{} 37p^2 \plus{} 14p \plus{} 1)s^4 \plus{} 4(15p^3 \plus{} 37p^2 \plus{} 14p \plus{} 1)s^3t$

$ \plus{} 9(18p^3 \plus{} 52p^2 \plus{} 25p \plus{} 3)s^2t^2 \plus{} (132p^3 \plus{} 394p^2 \plus{} 197p \plus{} 25)st^3$

$ \plus{} 2(15p^3 \plus{} 49p^2 \plus{} 29p \plus{} 4)t^4]x^2$

$ \plus{} (4p \plus{} 1)(2s \plus{} t)[3p(p \plus{} 2)(p \plus{} 1)s^4 \plus{} 6p(p \plus{} 2)(p \plus{} 1)s^3t$

$ \plus{} (36p^3 \plus{} 122p^2 \plus{} 79p \plus{} 11)s^2t^2 \plus{} (33p^3 \plus{} 113p^2 \plus{} 73p \plus{} 11)st^3$

$ \plus{} 6(p \plus{} 1)(p^2 \plus{} 4p \plus{} 1)t^4]x$

$ \plus{} (p \plus{} 1)t^2[3(15p^3 \plus{} 45p^2 \plus{} 20p \plus{} 2)s^4 \plus{} 6(15p^3 \plus{} 45p^2 \plus{} 20p \plus{} 2)s^3t$

$ \plus{} 3(19p^3 \plus{} 70p^2 \plus{} 38p \plus{} 5)s^2t^2 \plus{} 3(4p \plus{} 1)(p^2 \plus{} 6p \plus{} 3)st^3 \plus{} 2(p \plus{} 1)(4p \plus{} 1)t^4]$

$ \geq 0,$

which is clearly true for $ x\leq y\leq z.$[/hide]
[quote="Vasc"]Actually, the following inequality holds (chapter 7):

$ \sum \frac {2 \plus{} r}{y^{2} \plus{} ryz \plus{} z^{2}}\geq \frac {5 \plus{} 2r}{yz \plus{} zx \plus{} xy}$ for $ \minus{} 1 \leq r \leq 2,$

with equality if $ x \equal{} 0,y \equal{} z \equal{} 1.$[/quote]For $ r \geq \minus{} 2$, we have $ \sum \frac {1}{y^{2} \plus{} ryz \plus{} z^{2}}\geq \frac {2 \minus{} r}{yz \plus{} zx \plus{} xy},$

with equality if $ x \equal{} 0,y \equal{} 2,z \equal{} 1 \minus{} r \minus{} \sqrt {(r \plus{} 1)(r \minus{} 3)}$ when $ r\in [ \minus{} 2, \minus{} 1].$

See : http://www.mathlinks.ro/Forum/viewtopic.php?t=184883


Meh.

$ x^7\minus{}1\equal{}0\implies{x_1\equal{}1}$

$ x^6\plus{}x^5\plus{}\ldots\plus{}1\equal{}0$

Since $ x$ is non-zero, we can divide by $ x^3$ to get

$ x^3\plus{}x^2\plus{}x\plus{}1\plus{}\frac1{x}\plus{}\frac1{x^2}\plus{}\frac1{x^3}\equal{}0$

Now let $ z\equal{}x\plus{}\frac1x$, equate and pwn.

Hm I was thinking more along the lines of DeMoivre's Theorom. Let's see if your way works though.

We have

$ x\plus{}\frac1x\equal{}z$

$ x^2\plus{}\frac1x^2\equal{}z^2\minus{}2$

$ x^3\plus{}\frac1x^3\equal{}z^3\minus{}3z$

Thus, we have $ z^3\plus{}z^2\minus{}2z\minus{}1\equal{}0$.

That's actually a really good method. OF course, you'd have to cubic formula that and then solve for x given z, so that might end up pretty nasty.

[hide="Hint"]Think about the reverse of DeMoivres Theorem[/hide]

So...$ \left(\cos x\plus{}i\sin x\right)^n\equal{}\cos\left(nx\right)\plus{}i\sin\left(nx\right)$.
The $ n$th roots of unity can be expressed as an $ n$-gon on the complex plane. Too lazy.

$ \left(\cos x \plus{} i\sin x\right)^n \equal{} \cos\left(nx\right) \plus{} i\sin\left(nx\right)$.

Actually that's incorrect. That's assuming on the complex plane it lies on the unit circle.

I suck at trig. I will proceed with my old solution.

$ z^3 \plus{} z^2 \minus{} 2z \minus{} 1 \equal{} 0$.

Let $ z \equal{} y \minus{} \frac13$.

$ y^3 \minus{} y^2 \plus{} y \minus{} \frac1{27} \plus{} y^2 \minus{} \frac23y \plus{} \frac19 \minus{} 2y \plus{} \frac23 \minus{} 1 \equal{} 0$

$ \implies{y^3 \minus{} \frac53y \minus{} \frac7{27} \equal{} 0}$

Now let $ y \equal{} w \plus{} \frac5{9w}$ to get

$ w^3 \plus{} \frac {125}{729w^3} \minus{} \frac7{27} \equal{} 0\implies{w^6 \minus{} \frac7{27}w^3 \plus{} \frac {125}{729} \equal{} 0}$.

This is solvable by normal means.

Alternatively,

$ \cos{\frac{2k\pi}{7}}\plus{}i\sin{\frac{2k\pi}{7}}$ where k is an integer

Meh, what if we don't want $ \operatorname{cis}$? Eh...that only leads to the polynomials above.

Meh. The seven roots of unity are
$ {1, \minus{} 0.900968867\pm0.4338837391i, \minus{} 0.222520934\pm0.9749279122i,0.6234898019\pm0.7818314825i}$. Just for future reference; I can just search them on AoPS. :wink:


geometry
3D geometry
tetrahedron
geometry proposed

Let $ ABCD.A'B'C'D'$ be a parallelepiped and let $ S_1;S_2;S_3$ denote the areas of the sides $ ABCD;ABB'A';ADD'A'$,respectively.Given that the sum of squares of areas of all sides of tetrahedron $ AB'CD'$ equals 3,find the smallest possible value of
$ T\equal{}2(\frac{1}{S_1}\plus{}\frac{1}{S_2}\plus{}\frac{1}{S_3})\plus{}3(S_1\plus{}S_2\plus{}S_3)$

To mod: Pls lock this topic because that problem is on Mathematical and Youth Magazine last month.
To "I love math very much": You are bad  :mad:


geometry
geometric transformation
rotation

why does earth spin?what is angular speed depending on?

For the first question consult
[url]http://en.wikipedia.org/wiki/Planetary_formation[/url]
[url]http://en.wikipedia.org/wiki/Earth_rotation[/url]
its more a atrohistorical question. Propably beacause the clouds which formed
earth had slight movement and by conservation of angular momentum.
Also check out:

 [url]http://en.wikipedia.org/wiki/Angular_momentum[/url]
[url]http://en.wikipedia.org/wiki/Angular_velocity[/url]
[url]http://en.wikipedia.org/wiki/Angular_frequency[/url]

In the long run, tidal effects break the "local" conservation of momentum and slow down (or speed up) rotation of planetary bodies. In the case of the earth, these tidal effects are pretty small, but the day is about two hours longer than it was 600 million years ago. For the earth's moon, the tidal effects have been big enough to lock together the period of its revolution around the earth and its own rotation.


trigonometry
inequalities
function
calculus
inequalities proposed

Prove that for every triangle $ABC$
\[\Bigl(\sin\frac A2+\sin \frac B2+\sin \frac C2\Bigl)\Bigl(\tan\frac A2+\tan \frac B2+\tan \frac C2\Bigl)\geq \tfrac{3\sqrt3}2.\]

I think this inequlity isn't true because we have $\tan \frac A2+\tan \frac B2+\tan \frac C2 \geq \sqrt {3}$ and $\sin \frac A2+\sin \frac B2+\sin \frac C2 \leq \frac 32$ but this inequality .... :?

Oh, Ehsan2004 believe me, this inequality is true, I think. Just your inequalities don't work here. ;)

I think this inequality should be in this form :

$(\sum \sin^2\frac A2)(\sum \tan \frac A2)\geq \frac {3\sqrt {3}}{2}$ :huh:

[quote="cuong"]i don't have time now but this kind of problem is very well_known

just use formular to change it into three sides of a triangle

and a=x+y , b=y+z, c=z+x; :rotfl:  :ninja:[/quote]
  Don't try to do that . It's useless ! :lol:

[quote="Megus"]I cannot prove the inequality above but I obtained quite nice equivalent form:

With $x+y+z=1$ prove:

$\frac{x}{\sqrt{x+yz}}+\frac{y}{\sqrt{y+zx}}+\frac{z}{\sqrt{z+xy}}\geq \frac{3\sqrt3}{2}\sqrt{xy+yz+zx}$

So it will be nice if the above inequality is true  :)[/quote]

I have a solution to the given inequality using calculus. It is too ugly and complicated to post here.

Why, calculations, when Cauchy solves it imediately? ;)

Cezar, believe it or not, here we post from time to time solutions to problems. Were you someone who proved that he has solutions for his comentaries, I would have believed what you say. But I quite remember some remarks of the same kind to difficult problems, which turned out to be just non-sense. So, please, feel free to post solutions from time to time.

Vasc, can you please show exactly how you apply Jensen's inequality? I tried something like this:

$\sum \frac{x}{x+y+z}f(x+yz) \geq f(\sum x^2+ 3xyz)$

but now I don't know how to prove nicely (I didn't finish my calculations because they were too messy, but it seems like it follows from Muirhead):

$f(\sum x^2 + 3xyz) \geq f(\frac{4}{27(xy+yz+zx)})$

let x=tgA/2, y=tgB/2,z=tgC/2;

so xy+yz+zx=1

we must prove:

$\sum\frac{x}{\sqrt{x^2+1}}\geq\frac{3\sqrt{3}}{2(x+y+z)}$

using cauchy schwart we get: :P

$lhs\geq\frac{(x+y+z)^2}{\sum x\sqrt{x^2+1}}$

using cauchy schwart again we gain:

$lhs\geq\frac{(x+y+z)^2}{\sqrt{(x+y+z)(x^3+y^3+z^3+x+y+z)}}$

we only have to prove :

$\frac{(x+y+z)\sqrt{x+y+z}}{(x^3+y^3+z^3+x+y+z)}\geq\frac{3\sqrt{3}}{2(x+y+z)}$

our ineq is equivalent to:

$(x+y+z)^5\geq\frac{27}{4}(x^3+y^3+z^3+x+y+z)$

this ineq is solved completely by nttu in topic"prove or disprove"

the corporation between two student of"le hong phong high school for the gifted" :rotfl:  :rotfl:

[quote="Megus"]Vasc, can you please show exactly how you apply Jensen's inequality? I tried something like this:

$\sum \frac{x}{x+y+z}f(x+yz) \geq f(\sum x^2+ 3xyz)$

but now I don't know how to prove nicely. [/quote] Megus, we have to show that
$(x^2+y^2+z^2+3xyz)(xy+yz+zx)\le \frac{4}{27}$
for $x+y+z=1$.
Setting $A=xy+yz+zx$, the inequality becomes
$(1-2A+3xyz)A \le \frac{4}{27}$.
From $(x+y+z)^2 \ge 3(xy+yz+zx)$ and $(xy+yz+zx)^2 \ge 3xyz(x+y+z)$ we get $A \le \frac{1}{3}$ and $3xyz \le A^2$.
Thus, it suffices to show that $(1-2A+A^2)A \le \frac{4}{27}$; that is
$(3A-1)^2(3A-4) \le 0$. This is clearly true.

[quote="Vasc"][quote="Megus"]I cannot prove the inequality above but I obtained quite nice equivalent form:

With $x+y+z=1$ prove:

$\frac{x}{\sqrt{x+yz}}+\frac{y}{\sqrt{y+zx}}+\frac{z}{\sqrt{z+xy}}\geq \frac{3\sqrt3}{2}\sqrt{xy+yz+zx}$[/quote]
This inequality follows applying Jensen's Inequality to the convex function $f(t)=\frac{1}{\sqrt t}$.[/quote]

I do not need Jensen, or any calculus, just AM-GM and Cauchy. Here is a solution:

Let $x=\tan (A/2)$,... then $xy+yz+zx=1$ and the desired inqe is transformed into
\[(x+y+z)\Bigl(\frac{x}{\sqrt{1+x^2}}+\frac y{\sqrt{1+y^2}}+\frac z{\sqrt{1+z^2}}\Bigl)\geq\frac{3\sqrt3}2.\]

We have $\sqrt{1+x^2}=\sqrt{(x+y)(x+z)}\leq (x+y+x+z)/2$ by AM-GM, so that we have
\[\frac{x}{\sqrt{1+x^2}}\geq \frac{2x}{2x+y+z}.\]

Two other similars and using Cauchy gives the desired inequality. We are done.

[quote="silouan"][quote="pvthuan"][Two other similars and using Cauchy gives the desired inequality. We are done.[/quote]
Could anyone said to me how we get the desired ineq because I cannot find the way by Cauchy, Please someone finish this ineq.[/quote]

It remains to show that
\[\frac{x}{2x+y+z}+\frac y{2y+z+x}+\frac z{2z+x+y}\geq\frac{3\sqrt3}{4(x+y+z)}.\]
Now we use $x/a+y/b+z/c\geq \frac{(x+y+z)^2}{ax+by+cz}$ and $xy+yz+zx=1$ to prove the above. I think this hint is enough for you to proceed, Silouan.

[quote="silouan"]For me it is not enough. Please write it with details.
in the other hand i have another slution for this but could you pst your solution please?[/quote]
OK, By C-S we have
\[\frac x{2x+y+z}+\frac y{2y+x+z}+\frac z{2z+x+y}\geq\frac{(x+y+z)^2}{2(x+y+z)^2-2}.\]
Now we need to prove that
\[\frac{(x+y+z)^2}{2(x+y+z)^2-2}\geq\frac{3\sqrt3}{4(x+y+z)}\]
Put $t=x+y+z$, $t\geq\sqrt3$, then we have to prove $t^3/(t^2-1)\geq\tfrac{3\sqrt3}2$. This can be factored into $(t^2-3)^2(4t^2-3)\geq0$, which is obvious for $t\geq \sqrt3$. (We have $t\geq\sqrt3$ since $(x+y+z)^2\geq3(xy+yz+zx)=3$.)

[quote="silouan"]For me it is not enough. Please write it with details.
in the other hand i have another slution for this but could you pst your solution please?[/quote]

I have posted my proof, please give yours, Silouan

[quote="silouan"]For me it is not enough. Please write it with details.
in the other hand i have another slution for this but could you pst your solution please?[/quote]
I posted my solution, what about yours, Silouan.

My solution :
write $(x+y+z)(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y})\geq\dfrac{3\sqrt 3}{4}\sqrt{\sum xy}$ as $\sum\dfrac{x}{x+y+z}\sum\dfrac{\dfrac{x}{x+y+z}}{1+\dfrac{x}{x+y+z}}\ge\dfrac{3\sqrt 3}{4}\sqrt{\sum(\dfrac{x}{x+y+z}\dfrac{y}{x+y+z})}$.
Then homogenise with $x+y+z=2$

And a solution by mecrazywong
[quote="mecrazywong"]Homogenalize the inequality,\[ (x+y+z)(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y})\geq\dfrac{3\sqrt 3}{4}\sqrt{\sum xy} \]
Dehomogenalize with x+y+z=1, we have to prove\[ \sum\dfrac{x}{1+x}\ge\dfrac{3\sqrt 3}{4}\sqrt{\sum xy} \]
By Cauchy Schwarz,\[ \sum\dfrac{x}{1+x}\ge\dfrac{1}{\sum x(1+x)}=\dfrac{1}{1+\sum x^2} \]The result follows from $\sum xy(1+\sum x^2)^2=2(2\sum xy)(1-\sum xy)^2\le2(\dfrac{2\sum xy+1-\sum xy+1-\sum xy}{3})^3=\dfrac{16}{27}$.[/quote]

If $a,b,c$ are the sides of a  triangle $ABC$, then

$\sum\sin\frac{A}{2}\sum\tan\frac{A}{2}\geq \frac{2}{\sqrt3}\sum\frac{a+b+c}{2a+b+c}\geq \frac{3}{2}\sqrt{3}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=119112


Prove in the shortest way that the Napoleon`s inner triangle has area $\triangle '=\frac{\triangle}{2}-\frac{1}{24}\sqrt{3}(a^{2}+b^{2}+c^{2})$, where $\triangle$ is the area of the original triangle.

Look at the picture for names of points. The largest angle of the triangle is $\alpha = \angle BAC \geq 60^{\circ}$.

Cosine rule in $\Delta DEA$

$\Delta' = \frac{\sqrt{3}}{4}DE^{2}= \frac{\sqrt{3}}{4}\left(\frac{b^{2}+c^{2}}{3}-\frac{2}{3}bc\cdot\cos{\alpha-60^{\circ}}\right)$
Which rewrites to
\[\Delta' =-\frac{bc. \sin{\alpha}}{4}+\sqrt{3}\cdot\frac{2b^{2}+2c^{2}-2bc\cos{\alpha}}{24}\]

So we are done.

Is there a shorter way?  :maybe:


[hide="Solution"]
So most of this stuff is basic side finding based on the given conditions. Then, we do some angle chasing to find that $ \triangle QAB\sim \triangle B'DC$, which means that $ \frac68\equal{}\frac{B'D}{16}\implies B'D\equal{}12\implies AD\equal{}20$, so the perimeter is $ 2(20\plus{}16)\equal{}\boxed{72}$.
[/hide]