summer program
MathPath
MATHCOUNTS
algebra
polynomial
analytic geometry
floor function

[quote="akalra1"]how do you solve that... and how many people get into this MathPath thing anyway![/quote]
This site might help.

http://mathpath.org/apply.htm

I got both of those but decided not to go to mathpath

[quote="robinhe"][quote="Knuth's_check"]Here are two problems from this year's MathPath quiz, to which I strongly doubt my solutions.

#3
Suppose that a,b,c,d are real numbers such that
$a^2+ b^2= 1$
$c^2+ d^2= 1$
$ac + bd = 0.$
Show that
$a^2+ c^2= 1$
$b^2+ d^2= 1$
$ab + cd = 0$.

AND

Let Z(n) denote the number of zeros at the end of n!. For example, Z(11) = 2, since
11! = 39916800 ends with 2 zeros. Compute Z(2006), without a calculator or computer.[/quote]

Those might be a little hard for the mathcounts forum (at least the first one).

[hide="Hint for the first one?"]I can't quite remember how I did this or if I got it right, but...

Try adding the first two statemenets and twice the third statement in the given.[/hide]

[hide="Hint for number 2"]$10=5\times 2$.  Since there are always more $2's$ than $5's$, try counting how many powers of $5$ are in $2006!$.[/hide][/quote]

[hide]Actually  I believe you have to multiply the two polynomials not add them.  Then find a way to use abcd to help.

The second one, yeah the same hint.[/hide]

:showoff:  :thumbup:
use coordinate grid
[hide]the first 2 equations are 2 points on a circle,
the 3rd is just perp. lines[/hide]

[quote="mathgeniuse^ln(x)"]

[hide]Actually  I believe you have to multiply the two polynomials not add them.  Then find a way to use abcd to help.

The second one, yeah the same hint.[/hide][/quote]

Thanks. I actually thought of multiplying them, but I dont't think I got any closer to proving the thing.

[quote="robinhe"]
Those might be a little hard for the mathcounts forum (at least the first one).
[/quote]

Yeah, sorry. I didn't know where else to post them, but a moderator can move them to a more appropriate forum. :?

for the first one I had two solutions: one used algebra to prove it and the other just proved that a = d and b = c.

[quote="Knuth's_check"]Here are two problems from this year's MathPath quiz, to which I strongly doubt my solutions.

#3
Suppose that a,b,c,d are real numbers such that
$a^2+ b^2= 1$
$c^2+ d^2= 1$
$ac + bd = 0.$
Show that
$a^2+ c^2= 1$
$b^2+ d^2= 1$
$ab + cd = 0$.

AND

Let Z(n) denote the number of zeros at the end of n!. For example, Z(11) = 2, since
11! = 39916800 ends with 2 zeros. Compute Z(2006), without a calculator or computer.[/quote]

I think these are the solutions, if I am not allowed to post it just tell me or the moderator and he/she will delete it.

[hide="Number 3"]1.  Multiplying the first two equations of the first system we get
$a^{2}c^{2}+b^{2}d^{2}+b^{2}c^{2}+a^{2}d^{2}=1$.

Squaring the third equation of the first system we get
$(ac+bd)^{2}=a^{2}c^{2}+b^{2}d^{2}+2abcd=0$.

We well now subtract the bottom equation from the top equation to get $b^{2}c^{2}-2abcd+a^{2}d^{2}=1$.

Now we move to the second system of equations

First, square the third equation of the second system to get
$a^{2}b^{2}+c^{2}d^{2}+2abcd=0$.

Now add this equation to the last equation that we achieved before the above equation.  We get

$a^{2}b^{2}+a^{2}d^{2}+b^{2}c^{2}+c^{2}d^{2}=1$.

Now we multiply the first and second equations of the second system to get

$a^{2}b^{2}+a^{2}d^{2}+b^{2}c^{2}+c^{2}d^{2}=1$.

This is the same result as what we were trying to show assuming that this was true.  Hence the systems are good.

I don't know if this would get any credit since I assumed that what we were trying to prove is true and then proving it is true, but this is a way of proving it.[/hide]

[hide="Number 6"]For the second question
Notice how if you want to have a zero at the end of the number, a ten must be a factor of that number.  10=5*2, so our goal is to get 2s and 5s from 2006! in order to be able to find the number of zeroes at the end of the number.  But also notice that more 2s are in 2006! than 5.  So we really need to find how many 5s are in 2006! to get the answer.

It is basically
$\lfloor\frac{2006}{5}\rfloor+\lfloor\frac{2006}{25}\rfloor+\lfloor\frac{2006}{125}\rfloor+\lfloor\frac{2006}{625}\rfloor$
This equals
$401+80+16+3=500$

So there are 500 zeroes at the end of 2006![/hide]

[quote="tyy"][hide]
Note that the problem didn't specify what a,b,c,d have to be as long as their satisfy.

[/hide][/quote]

I am like positive that you will get little or no credit by finding a quadruple that satisfies these conditions and showing that this holds true.  That is why you should probably prove the general case like I did, though I don't think I did great in explanation.

But what is the quadruple that you found that works.

Note that there have to be infinitely many triples to this problem I believe.

[hide="Number 3 Solution 1"]
So for the first equation subtract $b^2$ from both sides to get $a^2 = 1 - b^2$. Do the same for $c^2$ to get $c^2 = 1 - d^2$.

Multiply both equations to get $a^2c^2 = 1 - b^2 - d^2 + b^2d^2$.

Do the same for $b^2$ and $d^2$ and multiply together to get $b^2d^2 = 1 - a^2 - c^2 + a^2c^2$

Now for the 3rd equation, subtract $bd$ from both sides and square to get $a^2c^2 = b^2d^2$. Subsitute in $1 - b^2 - d^2 + b^2d^2$ for $a^2c^2$. Simplifying the equation gives $1 = b^2 + d^2$. Go back to $a^2c^2 = b^2d^2$ and subsitute $1 - a^2 - c^2 + a^2c^2$ for $b^2d^2$ and simplify to get $1 = a^2 + c^2$.

So using these last 2 equations we have $a^2 = 1 - c^2$ and $b^2 = 1 - d^2$. Multiply together to get $a^2b^2 = 1 - c^2 - d^2 + c^2d^2$. We know that $c^2 + d^2 = 1$ so we now have $a^2b^2 = 1 - 1 + c^2d^2$. Simplify and take the square root to get $ab = -cd$, and add $cd$ to both sides to get $ab + cd = 0$.
[/hide]

[hide="Number 3 Solution 2"]
Take the 3rd equation and subtract $bd$ from both sides to get $ac = -bd$. We can ignore the negative sign since we'll be dealing with squares. Divide both sides by $bc$ to get $\frac{a}{b} = \frac{c}{d}$. There $a = dk$ and $b = ck$ where k is a constant. But since $a^2 + b^2 = c^2 + d^2$, $k$ is $1$. Therefore $a = d$ and $b = c$.
[/hide]

[quote="Phelpedo"]I suppose (0,1,0,1) works.

Again, merely finding a triplet that works will probably receive no credit; the problem asks you to prove that [i]all[/i] triplets work.[/quote]

actually I think it is a [i]quadruplet[/i].  But whatever.

[quote="Phelpedo"]I suppose (0,1,0,1) works.

Again, merely finding a triplet that works will probably receive no credit; the problem asks you to prove that [i]all[/i] triplets work.[/quote]

Actually that quadruplet doesn't work.  I'll let you figure out why.

Also you can probably figure out what you really meant.

[size=59][color=white](0,1,1,0) could be it.[/color][/size]

[quote="Knuth's_check"]
Let Z(n) denote the number of zeros at the end of n!. For example, Z(11) = 2, since
11! = 39916800 ends with 2 zeros. Compute Z(2006), without a calculator or computer.[/quote]

Just to generalize, the formula for the number of 0's at the end of $n!$ is:

$Z(n)=\sum_{p=1}^{\lfloor\log_5 n\rfloor}\lfloor\frac{n}{5^{p}}\rfloor$

[quote="lotrgreengrapes7926"][quote="mathgeniuse^ln(x)"]yeah, plus since there are 3 equations and 4 variables I believe there should be infinitely many solutions.[/quote]
This doesn't always have to be true. You could have 2 equations with 3 variables describing 2 tangent spheres, which only has 1 solution.[/quote]

Oh, duh. :blush:

I meant to say usually.  So, it is possible that there are a finite number of solutions, but I think it will be infinitely many.

So, I think my solution would do.

This is obviously true if any two members are zero. Now, if they aren't, then

$$ac=-bd$$
$$\dfrac{a}{d}=\dfrac{-b}{c}$$

$$\dfrac{a^2}{d^2}=\dfrac{1-b^2}{1-c^{2}}=\dfrac{b^2}{c^2}$$

$$\dfrac{1-b^{2}-1+c^2}{1-c^2}=\dfrac{b^2-c^2}{c^2}$$

$$-c^2=1-c^2$$ [If $b^2-c^2 \neq0$]

But this is contradiction and hence we conclude that

 $$b= \pm c$$
And $$a= \mp d$$

And the required result follows.

$$Q.E.D.$$


probability
symmetry
analytic geometry
graphing lines
slope
number theory
relatively prime

Twenty-five points are arranged in a five by five grid. If two distinct points are chosen at random from the grid, then what is the probability that the segment connecting these two points will not pass through any of the other points?

Tons of ugly counting yields $\frac{13}{21}$. Count the number of possible (not going through another point) segments for each point (symmetry is helpful to speed it up) and then take the fraction of possibilities over the total for each one (24) and add them all.

I'm not sure, but wouldn't you have to do something like...

For two distinct points $(x_1,y_1), (x_2,y_2)$, with $x_1, y_1, x_2, y_2$ all between 0 and 5, find out how many different ways you can have $x_2-x_1$ relatively prime with $y_2-y_1$. Then it'll be $1-\frac{that number}{total number of possibilities}$.

Because the line'll pass through a point if it's slope (given by $\frac{y_2-y_1}{x_2-x_1}$) is reducible. Right?

I dunno.

I'm bad at this whole "math" thing.

-BLt

EDIT: Forgot to take into account horizontal and vertical lines. Bah. Oh well, I'm wrong.

I hate these types of problems.  :mad:

[hide]After mad counting, I get the nice fraction 2/3. [/hide]


analytic geometry
geometry
3D geometry
sphere
trigonometry
vector

Explain why the equation $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$ represents the surface of a sphere, centre the origin.
Find the equations of the planes through the point (2, 0, 0) which are tangential to the sphere and parallel to the [i]y[/i] axis.

The equation is just the distance formula applied at the origin. It describes all points $ 1$ unit away from the origin, which is the surface of a sphere.

123... thanks. Any help with the second part?

The plane parallel to the y-axis that goes through $ (2, 0, 0)$ is the plane $ x \equal{} 2$, which doesn't intersect the sphere at all.

No, there are an infinity of planes through that point that are parallel to the $ y$-axis.

If we invert the point in the circle, we get (1/2,0,0), the plane x=1/2 intersects the sphere at y^2+z^2=1/2; x=1/2

Then the tangency point can be written as $ (1/2,\frac {\cos\theta}{\sqrt {2}},\frac {\sin\theta}{\sqrt {2}})$...

We have that $ (0,1,0)$ is on the plane.

Since it is a sphere, the vector from the center to the point of contact is normal to the plane...so we have

$ (1/2,\frac {\cos\theta}{\sqrt {2}},\frac {\sin\theta}{\sqrt {2}})$ is parallel to  $ (0,1,0) \, x \, [ (1/2,\frac {\cos\theta}{\sqrt {2}},\frac {\sin\theta}{\sqrt {2}}) \minus{} (2,0,0)]$

Since the cross product of two vectors in a plane is parallel to the normal vector.

Then you can solve for $ \theta$ or whatever...

Note that this is sort of overkill...It is easy to see that the tangency point must be in the $ xz$ plane. i.e. $ y\equal{}0$, so $ z\equal{}\pm \frac{1}{\sqrt{2}}$


Cybil and Ronda are sisters. The 10 letters from their names are placed on identical cards so that each of 10 cards contains one letter. Without replacement, two cards are selected at random from the 10 cards. What is the probability that one letter is from each sister's name? Express your answer as a common fraction.

Much simpler solution:

It doesn't matter what the first letter is.  The second must be from the other sister's name, so there is a 5/9 chance of that.

Related reading: Introduction to Counting & Probability Chapter 9: Think about it!

[hide=Bogus Solution]
We can multiply both probabilities to get $5/18$.[/hide]
[hide=Solution] We need to add $2$ cases.
$$\text{Case~1:~C,R}$$
$$1/2*5/9=5/18$$
$$\text{Case~1:~R,C}$$
$$1/2*5/9=5/18$$
Now we add the $2$ cases:
$$5/18+5/18=\boxed{5/9}$$

Just do 10 choose 2 as total possibilities and 5*5 as the ones that work because 5 choices for Cybil and 5 choices for Ronda.
25/45 = 5/9 !


summer program
MathPath
AwesomeMath
PROMYS

The new MathPath Qualifying Quiz is now available at [url]http://www.mathpath.org[/url].  MathPath also has a new scholarship structure, offering 1 in 4 students $\$$1000 in financial aid (more info on the scholarships is available [url=http://www.mathpath.org/fees/schol.htm]here[/url]).  We will be hosting a Math Jam for MathPath in March.

We will have a Math Jam next week (Tuesday evening) for Awesome Math ([url]http://www.awesomemath.org[/url]), a new camp from Zuming Feng and Titu Andreescu.

[url=http://www.mathpath.org]MathPath[/url] is a great program and I highly recommend it. There are lots of AoPSers there.

are all programs as expensive as this. i don't think i could pay this much for a summer program no matter how much i wanted to go.

[quote="fireemblem13"]are all programs as expensive as this. [/quote]

No, not all summer programs are as expensive as MathPath.  (For a list of some summer programs, click on Resources in the left sidebar, then the Internet Guide tab and the Summer Programs button.)  However, MathPath is the only national-level mathematics summer program I know of which is geared toward middle-school students---someone please post a rebuttal if I'm misinformed.

I know that at least 1 8th grader has been accepted to PROMYS, which is a national level math camp (i think it's geared mostly towards number theory). I myself went to mathpath, i thought it was alotta fun, I highly recommend it!

I hope lots of AoPSers are applying to MathPath this year. It was a great experience for the people who attended last year.


vector
linear algebra
matrix
linear algebra solved

You may assume that they are all real matrices.

$A_{m\times n}$, $B_{n\times s}$ satisfy that $rank(A)=rank(AB)$
Show that there exists $C_{s\times n}$ such that $A=ABC$.

Let $v_1,v_2,\ldots,v_n$ be the column vectors of $A$.

We know that the spaces spanned by $\{v_1,\ldots,v_n\}$ and $\{b_{11}v_1+\ldots+b_{n1}v_n,\ldots,b_{1s}v_1+\ldots+b_{ns}v_n\}$ have the same dimension. Denote $\sum_i b_{ik}v_i$ by $w_k$, where $k=\overline{1,s}$. The hypothesis tells us that all the $v_i$'s can be written as linear combinations of the $w_j$'s. After writing down the linear combinations we can gather the coefficients to form the matrix $C$.

For a problem like this, why should we care what the field is? It's obviously a general linear property which holds over arbitrary fields, and every reasonable proof works over arbitrary fields.


Given triangle $ABC,P$ is an arbitrary point inside it $d_{a},d_{b},d_{c}$ are distance from $P$ to $BC,CA,AB$ reps prove that
$\frac{MA}{d_{a}}+\frac{MB}{d_{b}}+\frac{MC}{d_{c}}\ge 6$

Sorry I mean $M$ is $P$ it must be
Given triangle $ABC,P$ is an arbitrary point inside it $d_{a},d_{b},d_{c}$ are distance from $P$ to $BC,CA,AB$ reps prove that
$\frac{PA}{d_{a}}+\frac{PB}{d_{b}}+\frac{PC}{d_{c}}\ge 6$

Hi gemath. I am not good in inequalities but we can easy to prove it, based on another well known, coming from the past.

So, let $\bigtriangleup DEF$ the cevian triangle of $P,$ with respect to $\bigtriangleup ABC.$

Based on [b][size=100]Van Aubel’s[/size][/b] theorem, states that $\frac{AP}{PD}= \frac{AE}{EC}+\frac{AF}{FB}$ and similarly $\frac{BP}{PE}= \frac{BD}{DC}+\frac{BF}{FA},$ and $\frac{CP}{PF}= \frac{CE}{EA}+\frac{CD}{DB},$

we have that $\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}\geq 6,$ $,(1)$ because of $\frac{AE}{EC}+\frac{CE}{EA}\geq 2$ etc.

So, because of $PD\geq d_{a}$ and $PE\geq d_{b}$ and $PF\geq d_{c},$ from $(1)$ your inequality it follows.

[b][size=100]NOTES.[/size][/b]

[b][size=100]1)[/size][/b] –  Based also on [b][size=100]Van Aubel’s[/size][/b] theorem, we have also the well known inequality $\frac{AP}{PD}\cdot \frac{BP}{PE}\cdot \frac{CP}{PF}\geq 8$ $,(2)$

[b][size=100]2)[/size][/b] –  Also the not well known (I think) equality $\frac{AP}{PD}\cdot \frac{BP}{PE}\cdot \frac{CP}{PF}-\left( \frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}\right) = 2$ $,(3)$ for any arbitrary point $P,$ inwardly to $\bigtriangleup ABC$ ( easy to prove ).

[b][size=100]3)[/size][/b] – For any two points $P,$ $P',$ isotomic with respect to a given triangle $\bigtriangleup ABC$ inwardly to it, we have the ( easy to prove based on $(3)$ ) equalities:

$\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}= \frac{AP'}{P'D'}+\frac{BP'}{P'E'}+\frac{CP'}{P'F'}$ $,(4)$ and $\frac{AP}{PD}\cdot \frac{BP}{PE}\cdot \frac{CP}{PF}= \frac{AP'}{P'D'}\cdot \frac{BP'}{P'E'}\cdot \frac{CP'}{P'F'}$ $,(5)$

where $\bigtriangleup DEF,$ $\bigtriangleup D'E'F',$ are the cevian triangles of $P,$ $P'$ respectively wrt $\bigtriangleup ABC.$

Kostas Vittas.

Thank you vittasko, that's good solution, it is source that I find out this problem, too. Thus we had $\frac{PA}{d_{a}}+\frac{PB}{d_{b}}+\frac{PC}{d_{c}}\ge 6$ and $\frac{PA+PB+PC}{d_{a}+d_{b}+d_{c}}\ge 2$ and I hope the following is true
$\frac{PA+PB}{d_{a}+d_{b}}+\frac{PB+PC}{d_{b}+d_{c}}+\frac{PC+PA}{d_{c}+d_{a}}\ge 12$ :)

Sorry my conjecture is $\frac{PA+PB}{d_{a}+d_{b}}+\frac{PB+PC}{d_{b}+d_{c}}+\frac{PC+PA}{d_{c}+d_{a}}\ge 6$


The number of ordered pairs $ (x , y)$, where $ x$ and $ y$ are integers satisfying the equation $ x^3 \plus{} y^3 \equal{} x^2y \plus{} xy^2 \plus{}1$ is

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

[hide]$ (x\plus{}y)\equal{}\frac{1}{(x\minus{}y)^2}$

therefore only $ (0,1)$ and $ (1,0)$[/hide]

$ x^3\plus{}y^3\minus{}x^2y\minus{}xy^2\equal{}1\Leftrightarrow(x\plus{}y)(x\minus{}y)^2\equal{}1$

$ \left\{\begin{array}{ll}x\plus{}y\equal{}1\\x\minus{}y\equal{}1\end{array}\right.\Rightarrow x\equal{}1, \ y\equal{}0$

$ \left\{\begin{array}{ll}x\plus{}y\equal{}1\\x\minus{}y\equal{}\minus{}1\end{array}\right.\Rightarrow x\equal{}0, \ y\equal{}1$

There are two pairs who satisfy the equation.


function
real analysis
real analysis solved

Show that there exists an integrable $f:[0,1] \rightarrow R$ such that:

a) $f([0,1])=[0,1] \cap Q$;
b) $f([0,1])=[0,1]-Q$.

I think, it is too easy for Lebesque's integral.
Do you assume integrable in Lebesque's or Rieman's mean?

By "integrable" I mean "Riemann integrable". I don't really know what Lebesgue integrable means :).

a) Let q1, q2, q3,... be sequence of all rational numbers from [0,1].
Define f(x)=qn if 1/(n+1) < x <= 1/n and f(0)=0.

b) Define f(x)=x+R(x)/:sqrt: 2 where R(x) is Riemann's function (if x is irrational then R(x)=0 and if x=m/n then R(x)=1/n, R(0)=1)

In both cases a) and b) function f(x) is integrable on [0,1] in Riemann's mean due to Lebesgue's criteria.

I did something very similar, with one extra observation: do you know for sure that $x+\frac{R(x)}{\sqrt{2}}\leq 1$ ? I think we should define $t(x)$ to be irrational, $>1$, s.t. $x+\frac{R(x)}{t(x)}\leq 1$. Actually, it's enough to say this, because $t$ obviously exists. We don't really need to construct it.

If x is irrational then f(x)=x and if x=m/n (m<n) then f(x)=m/n+1/n:sqrt:2 < (m+1)/n <= 1. And also define f(1)=1/:sqrt:2.

Oh, you're right. Sory for not realizing it was Ok.


Let be a triangle $ ABC$ and $ M\in BC$ such that $ MB\equal{}MC\ ,\ N\in AC$ such that $ \widehat{ANB}\equal{}\widehat{CBN}\ ,\ P\in AB$ such that $ CP\perp AB$ . Let be $ MA\equal{}m_a\ ,\  NB\equal{}b_b\ ,\ CP\equal{}h_c$. Prove that :

$ m_a^2\plus{}b_b^2\plus{}h_c^2\equal{}m_a\cdot b_b\plus{}b_b\cdot h_c\plus{}h_c\cdot m_a\Longleftrightarrow AB\equal{}BC\equal{}CA$ .

$ m_a^2\plus{}b_b^2\plus{}h_c^2\equal{}m_a\cdot b_b\plus{}b_b\cdot h_c\plus{}h_c\cdot m_a\Longleftrightarrow  m_a \equal{} b_b \equal{} h_c$
if $ ABC$ equilateral then $ m_a \equal{} b_b \equal{} h_c$
if $ m_a \equal{} b_b \equal{} h_c$
$ sin\widehat{MAB} \equal{} \frac{a sinB }{2m_a} \equal{} \frac{a sinB}{2h_c} \equal{} \frac{1 }{2}$   then  $ sin\widehat{BMA} \equal{} \frac{2c sin\widehat{MAB}}{a} \equal{} \frac{c }{a} (1)$
$ sin\widehat{CAM}\equal{} \frac{a sinC }{2m_a} \equal{}\frac{a sinC }{2asinB} \equal{} \frac{c}{2b}$  thus  $ sin\widehat{AMC} \equal{} \frac{2b sin\widehat{CAM}}{a} \equal{} \frac{c }{a} (2)$
$ (1)$ and $ (2)$  implies $ \widehat{BMA} \equal{} \widehat{AMC}$  therefore $ ABC$ isoceles
since $ sin\widehat{MAB} \equal{} \frac{1 }{2}$ hence $ \widehat{MAB} \equal{} \frac{\pi }{6}$ so  $ ABC$ equilateral.


Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{ab}{a^2\plus{}5b^2}\plus{}\frac{bc}{b^2\plus{}5c^2}\plus{}\frac{ca}{c^2\plus{}5a^2} \le \frac{1}{2}\]

We can prove this inequality using Cauchy Schwarz. :)

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac {ab}{a^2 \plus{} 5b^2} \plus{} \frac {bc}{b^2 \plus{} 5c^2} \plus{} \frac {ca}{c^2 \plus{} 5a^2} \le \frac {1}{2}
\]
We can prove this inequality using Cauchy Schwarz. :)[/quote]
x*y/(x^2+k*y^2)+z*y/(y^2+k*z^2)+x*z/(z^2+k*x^2) <= 3/(k+1)
k^4-59*k^3+285*k^2-59*k+1=0

The inequality is equivalent to:
$ \sum{\frac{1}{5\frac{a}{b} + \frac{b}{a}} \leq \frac{1}{2}}$
Let $ \frac{a}{b}=x,\frac{b}{c}=y,\frac{c}{a}=z$ then the ineq becomes:
$ \sum{\frac{x}{x^2+5}} \leq \frac{1}{2}$
WLOG, we assume that $ x \leq y \leq z => yz \geq 1$
Let $ f(x,y,z)= \sum{\frac{x}{x^2+5}}$
It is not hard to show that: $ f(x,y,z) \leq f(x,\sqrt{yz},\sqrt{yz}) \leq 1/2$
@can_hang2007: Can you post your Cauchy-Schwarz solution?  :lol:

I have an ugly proof letting $ a \equal{} 1$, $ b \ge 1$ $ c\ge 1$, then expanding, using AM-GM to get rid of odd exponents, then proving the discriminant less than 0, I will post it later).

[quote="can_hang2007"][quote="setoptnk"]Let $ f(x,y,z) \equal{} \sum{\frac {x}{x^2 \plus{} 5}}$
It is not hard to show that: $ f(x,y,z) \leq f(x,\sqrt {yz},\sqrt {yz}) \leq 1/2$
[/quote]
Can you please show the "easy work" as you said? ;)[/quote]
Oops...Sorry for my stupidity, i have a mistake in expanding. I will try it again :(


Fin equation of the tangent line at the given value of x on the curve.
2xy - 2x + y = -14, x = 2
This is what i did.==>
2y + 2xdy - 2 + dy = 0
2*2y-2*2+y = -14
5y=-10
y=-2
2*(-2) +2*2dy -2 + dy = 0
5dy = 6
dy=5/6
But this isnt answer what i need, what i did wrong?

Its a very small mistake. Check the last two lines of your algebra.
 :wink:


I know what f(x) means but I have no clue on where to start or what they mean when I see problems like these.  Could someone [b]detailedly[/b] explain the steps involved and what these equations means?

1. The function f is defined for all real values of x and it satisfies the equality [tex]f(x) + xf(1-x) = 3+x.[/tex] Find the value of [tex]f(\frac{5}{3})[/tex].

2. Let f be a function from the positive reals to the positive reals. If [f(xy)] :^2:  = x[f(y)] :^2: for all positive number x and y and f(2) = 6, find f(50).  (I posted this question before in the intermediate forum but i wanted a more thorough solution.)

Robert0201 wrote:
1. The function f is defined for all real values of x and it satisfies the equality [tex]f(x) + xf(1-x) = 3+x.[/tex] Find the value of [tex]f(\frac{5}{3})[/tex].

[hide]OK what you might want to do here is to plug in 5/3, and then treat f(5/3) and f(-2/3) as variables.  Try plugging in -2/3 = x as well.  Don't know of any easier ways...[/hide]

For, the second one I did it like this:

[hide]f(50)=f(25*2)

so (f(25*2)) :^2: =25(f(2)) :^2:

Plug in f(2)=6 to get 25*36

we want the square root of this which is 5*6or 30.[/hide]

[hide]The answer would be b.  Since on the interval 3 :le: x :le: 7 it is increasing its lowest point will be when x=3 and it will increase in some way all the way up to x=7.  Since f(-x)=-f(x) f(-3)=-5.  Going from -7 to -3 you can see that it will increase and -5 will be the maximum.  I can't really explain why but you have to visualize it to understand.[/hide]

[hide]This might be a clearer explanation.  When it's between 3 and 7, everything is one sign.  When we make everything negative, we're just doing the same thing in another direction.  So now we call the most extreme in the "forward" direction the minimum and the "backward" direction the maximum.  OK so this is less clear perhaps.  I don't know, but hopefully you understand one way or another.[/hide]

For your last question, try drawing a few pictures.

I just don't get what f(something) means.  Is f(xy) equal to f(x)f(y)?  If so, why?  Is f(x+y) equal to f(x) + f(y)?  Why? I don't know how I can know a lot other things about math and don't know this simple concept.  Maybe its just that I've never really been taught this stuff or just not taught it well or maybe its just me.

f(x) just means a function of some defined variable.

Let's say we have f(x) = x^2. This means that whatever is put into f(x), you would replace in the x^2 portion as well. For instance, f(y)=y^2, f(3)=3^2, f(x+y)=(x+y)^2. You just plug in for the x in both places.

As for number 1. This might be a bit over your head if you are just trying to grasp the idea of a function.

Spoiler:[hide]

Consider the circular nature of 1-x, plugging in 1-x to x gives 1-x, now plugging in 1-x to 1-x will return x, and the patter repeats.

SO, we have:

[tex]f(x)+xf(1-x)=3+x[/tex]

Plug in 1-x for x:

[tex]f(1-x)+f(x)-xf(x)=4-x[/tex]

[tex]f(1-x)=-f(x)+xf(x)+4-x[/tex]

and from the first equation:

[tex]\displaystyle{f(1-x)=\frac{3+x-f(x)}{x}}[/tex]

Equate the two:

[tex]\displaystyle{\frac{3+x-f(x)}{x}=-f(x)+xf(x)+4-x}[/tex]

[tex]3+x-f(x)=4x-x^2-xf(x)+x^2f(x)[/tex]

[tex]3-3x+x^2=f(x)-xf(x)+x^2f(x)[/tex]

[tex]x^2-3x+3=f(x)(x^2-x+1)[/tex]

[tex]\displaystyle{f(x)=\frac{x^2-3x+3}{x^2-x+1}}[/tex]

[tex]\displaystyle{f\left(\frac{5}{3}\right)=\frac{7}{19}[/tex]

[/hide]


algorithm
modular arithmetic
Miscellaneous Problems

it is the same exercise http://www.mathlinks.ro/Forum/viewtopic.php?t=151005  :)

bye
paolo

Actually, this number is substantially larger

i know but the "algoritm" is the same :)
we find a,b s.t. $ 5^a \parallel{} n!$ and $ 2^b \parallel{} n!$, so bye $ (1*3*7*9)^{2m}\equiv 1 \pmod {10}$, we have that the last nonzero digit is k s.t. $ k \equiv 2^{b \minus{} a} \pmod {10}$ and is trivial.

[quote=Peter]What is the rightmost nonzero digit of $1000000!$?[/quote]

$\sum_{k=1}^{\infty}[\frac{1000000}{5^k}]=249998$.

We have to get $x\equiv \frac{1000000!}{2^{249998}\cdot 5^{249998}}$ $(mod 10)$.

Apparently $x$ is even, so $x=0,2,4,6,8$.

On the other hand, $\frac{1000000!}{5^{249998}}\equiv (1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 8\cdot 9)^{100000}\equiv 6^{100000}\equiv 6$ $(mod 10)$.

Then $2^{249998}\cdot x\equiv 6\Leftrightarrow x\equiv 4,9$ $(mod 10)$.

Therefore $x$ is 4.

The answer is $\boxed 4$.$\blacksquare$


induction
superior algebra
superior algebra unsolved

Any help at all with this one would be appreciated...

If S is a finite set and f is a 1-1 mapping of S into itself, show that f must be surjective.

[u]Attept to solve[/u]:
For every $ s \in S$, there exists f (s1) such that f(s1) = f(s2) such that s1 = s2 and therefore every s is mapped such that y=f(s) is onto.

a rigorous proof has to use induction on $ |S|$.

I think first you need to define what [url=http://en.wikipedia.org/wiki/Finite_set]finite[/url] means.
There are many equivalent ways to do that, and your problem is one of them.


Non-negative real numbers $ p, q$ and $ r$ satisfy $ p\plus{}q\plus{}r\equal{}1$.

Prove that

$ 7(pq\plus{}qr\plus{}rp) \le 2\plus{}9pqr$.

$ \Leftrightarrow \sum pq(p\plus{}q) \le \sum (p^3\plus{}q^3) \Leftrightarrow 0 \le \sum (p\minus{}q)^2(p\plus{}q)$

We can use Schur's inequality:
[b]Solution:[/b]
The inequality is equivalent to
\[ 7(pq \plus{} qr \plus{} rp) \le 2(p \plus{} q \plus{} r)^2  \plus{} 9pqr\]

It's enough to prove that \[ 2(pq \plus{} qr \plus{} rp) \le p^2  \plus{} q^2  \plus{} r^2  \plus{} 9pqr\]
But which is true from Schur's inequality


Alice buys three burgers and two sodas for $ \$3.20$, and Bill buys two burgers and a soda for $ \$2.00$. How many cents does a burger cost?

We have this system of equations, where $ x$= the cost, in cents, of a burger, and $ y$= the cost, in cents, of a soda.

$ 3x\plus{}2y\equal{}320$
$ 2x\plus{}y\equal{}200$

Multiplying the bottom expression by $ \minus{}2$ and adding yields

$ \minus{}x\equal{}\minus{}80 \implies x\equal{}\boxed{80}$


Find all numbers such that $\frac{(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}}{2}$ is a perfect square or prove that none exist, where n is odd.

[hide="Cheap Beginning"]Well, we know that it works for some number (or numbers?) because $n=1$ automatically yields $1$, a perfect square  :D [/hide]

[hide]
Hmm... well if we let $a_{n}$ be the expression, we can derive the relation

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

because $x^{n+1}+y^{n+1}= (x+y)(x^{n}+y^{n})-xy(x^{n-1}+y^{n-1})$. I'm not sure how to find out when this recursion produces squares though.[/hide]

You could always do binomial expansion, and sort through the termswith combinatorics to reach a big expression with powers of 2 andother things, but that seems harder still...

This is the generating pell equation (due to paladin's recursion)

Thus all numbers in this form must solve a^2-2b^2=-1, since we are considering odd coefficients.
so assume a is a square. Then a must obviously be odd. Assume a is a square. Then a=4x^2+4x+1.

Thus (4x^2+4x+1)^2=1=2b^2, then 8x^4+16x^3+12x^2+4x+1 =b^2
(2x^2+2x)^2+(2x^2+2x+1)^2 =b^2
Now we can apply primitive pythagorean indentity (since 2x^2+2x and 2x^2+2x+1 are relatively prime). 2x^2+2x is even, 2x^2+2x is in the form 2uv.  and 2x^2+2x+1=u^2-v^2

By pythagoreas u^2-v^2-1=2uv. (u-v)^2-2v^2=1. Which nicely forms a pell equation with infinitely many solutions.

Thus there are infinitely many solutions such that our expression is a square.

edit---- oh wait x could still be irrational...... oops

Ok...I'm an idiot. Care to generate a few for us?

[quote="chess64"][hide]$\sum_{k=0}^{(n-1)/2}2^{k}\cdot\binom{n}{2k}$[/hide]

maybe[/quote]I got the exact same thing, but can't think of what to do from there.

I tried using the same method that you use to derive geometric sequence sums but I suck at combinatorics...

[quote="Scrambled"][quote="Phelpedo"][quote="Scrambled"]This is the generating pell equation (due to paladin's recursion)

Thus all numbers in this form must solve $a^{2}-2b^{2}=-1$, since we are considering odd coefficients.
so assume a is a square. Then a must obviously be odd. Assume a is a square. Then $a=4x^{2}+4x+1.$

Thus $(4x^{2}+4x+1)^{2}=1=2b^{2},$ then $8x^{4}+16x^{3}+12x^{2}+4x+1 =b^{2}$
$(2x^{2}+2x)^{2}+(2x^{2}+2x+1)^{2}=b^{2}$
Now we can apply primitive pythagorean indentity (since $2x^{2}+2x$ and $2x^{2}+2x+1$ are relatively prime). $2x^{2}+2x$ is even, $2x^{2}+2x$ is in the form $2uv$.  and $2x^{2}+2x+1=u^{2}-v^{2}$

By pythagoreas $u^{2}-v^{2}-1=2uv. (u-v)^{2}-2v^{2}=1.$ Which nicely forms a pell equation with infinitely many solutions.

Thus there are infinitely many solutions such that our expression is a square.[/quote] (latex added for convenience)

$(2x^{2}+2x+1)^{2}=4x^{4}+8x^{3}+8x^{2}+4x+1$, so the expression doesn't add up correctly.[/quote]
yes it does...[/quote]

Ok...you're correct in that there are infinitely many solutions for the last equation. However, just because $u$ and $v$ are integers doesn't mean that $x$ is an integer. In fact, I'm still pretty convinced that there are no solutions.


The results for the November 1 MML competition were (mostly) posted yesterday.  Results from Division II haven't reached the statistician yet.

November report: http://www.geocities.com/mmlleague/November.doc
(The revised report, when the Div II results are received, will have the same URL.)

League website: http://www.geocities.com/mmlleague/reports

The revised report for the November 1 meet was put online this afternoon (I received notice at 5:42 PM EST).  Same link as shown above.


geometry
incenter
angle bisector
geometry solved

ABC is a triangle. D is a point on the side BC. A is equidistant from the incenter of ABD and the excenter of ABC which lies on the internal angle bisector of B. Show that AC = AD

[quote="amir2"]which lies on the internal angle bisector of B[/quote]

Actually, I think you mean the internal angle bisector of $\angle C$, and, in this case, it's just an angle chase. Let the incenter of $ABD$ be $I_d$, and the mentioned excenter be $I_c$. We know that $\angle AI_dB=\pi-\angle AI_cB$. Now express those angles in terms of the angles of $ABC$ and $\angle ADB$ to get the desired result.

[quote="grobber"][quote="amir2"]which lies on the internal angle bisector of B[/quote]

Actually, I think you mean the internal angle bisector of $\angle C$,[/quote]

Why? I think Amir2's problem is correct, and your solution works pretty well with $I_c$ being the excenter of triangle ABC lying on the internal angle bisector of the angle B (rather than C).

[b]EDIT:[/b] This problem was recently discussed again on http://www.mathlinks.ro/Forum/viewtopic.php?t=30389 .

  Darij


For $ a,b, c,d$ such that $ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2\equal{}5$. Prove that:

$ (5 \minus{} a \minus{} 2b)(5 \minus{} c \minus{} 2d)(5 \minus{} ac \minus{} bd) \leq (\frac {15}{2})^3$

 :oops:

[quote="Honey_S"][quote="vipCD_A1"]For $ a,b, c,d$ such that $ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2$. Prove that:

$ (5 \minus{} a \minus{} 2b)(5 \minus{} c \minus{} 2d)(5 \minus{} ac \minus{} bd) \leq (\frac {15}{2})^3$

 :oops:[/quote]

Your inequality is not true vipCD, let try $ a \equal{} d \equal{} ...$ and $ b \equal{} c \equal{} 0$  :)[/quote]

I have edit,


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi
i would help me ,  i try solved it ;
If H is a subgroup of G , [G:H]=n ,then there is homomorphism from G to S.n with kerf is a Subgroup of H

Define: $ \phi: G \to S_{n}: \phi(a) \equal{} \l_{a}$, where $ \l_{a}$ acts on the left cosets by $ \l_{a}(gH) \equal{} agH$.
Each $ \l_{a}$ permutes the left cosets of $ H$, and we have:
$ \l_{ab}(gH) \equal{} (ab)gH \equal{} a(b(gH)) \equal{} (\l_{a} \circ \l_{b} ) (gH) \implies \phi(ab) \equal{} \phi(a) \circ \phi(b)$.
Hence $ \phi$ is a homomorphism.

Show now that the $ ker \phi$ is a subset of H by showing that $ x \in ker \phi \iff$ every conjugate of x is in H.


function algebra polynomial quadratics algorithm integration calculus

If told to use decomposition to integrate a fraction like (in factored form):

1/x^3

Is the form:

(Ax^2 + B + C)/(x^3)

or

(A)/(x) + (Bx + C)/(x^2) + (Dx^2 + Ex + F)/(x^3)

Thanks!
-Matt

We're talking about rational functions, which is to say polynomials divided by polynomials.

Step 1: using long division if necessary, make sure you're dealing with a proper fraction, whose numerator has a lower degree than the denominator.

Step 2: factor the denominator completely into linear and quadratic factors. (Of course, this is the step for which no true algorithm exists.)

What next? The technique of partial fractions is only needed if the fraction you're dealing with has at least two [i]different[/i] factors in the denominator. If that's not the case, then we don't need the technique.

Here's an example:

$ \int\frac{x^{2}}{(x+2)^{3}}\,dx$

My first step will be the substitution $ u=x+2,$ so that $ du=dx$ and $ x=u-2$

$ \int\frac{(u-2)^{2}}{u^{3}}\,du= \int\frac{u^{2}-4u+4}{u^{3}}\,du$

$ =\int\frac1u-\frac4{u^{2}}+\frac4{u^{3}}\,du= \ln|u|+\frac4u-\frac2{u^{2}}+C$

$ =\ln|x+2|+\frac4{x+2}=\frac2{(x+2)^{2}}+C$

That may not have answered what you thought you were asking, but when  you ask a question as badly as you did, there's not much we can do.

In terms of general partial fraction decomposition (without the motivation of integrals), what you should be looking at is either

$ \frac{Ax^{2}+Bx+C}{x^{3}}$

Or

$ \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}$

The two forms are equivalent, so it's a matter of which one is algebraically easier to work with (which would probably be the first).


I took some Aops Classes over the summer and I saved all the transcripts. However, I didn't save the message board, because in the e-mail it said that the boards would be archived. Does anyone know where in the forum the classes have been saved?

Send an email to classes@artofproblemsolving.com; include your username and the class name.

Well, I've found that if you go to your "Previous posts" thing on your profile page, you can access the course homepage by going to your respective posts (Assuming that you posted at least once on the class forum). Does this work for everyone?

I also use 'Previous Posts', since I can't find the 'Archive' section.  Does anyone know where it is?

[quote="rrusczyk"]You should be able to find the class forum in the "Archive" section of the "Online School" portion of the Forum.  (Sony, I assume you mean the class forum, not the class homepage.)[/quote]

 :D  I guess I was referring to the Forum, but on the forum, you can see the "hierarchy (like this: Home » Online School » Classes Information) and it says Archive there. Is that where you can access it?

The archive should be available from the Forum main page now.  If you're still having trouble, send me a message.


In $CHCl_{3}$ there exists two types of bond angles, $108.0^{\circ}$ and $110.1^{\circ}$. Explain why this is so in terms of bond polarity and indicate where each bonding angles occur.

There is going to be an angle between the $Cl$ molecules, and an angle between $H$ and $Cl$.

I don't recall much of my electronegativity charts, but from what I remember, $Cl$ is more electronegative than $C$, which is in turn more electronegative than $H$.  This means that $Cl$ in the $C-Cl$ bond is negative, while $H$ in the $C-H$ bond is positive.

Then the $Cl$ atoms should repel each other while they should be slightly attracted to the $H$ atoms ($Cl$ also has more electron levels, and so is a "fatter" atom), which means the larger angle should be between the $Cl$ atoms.

I think.  :D


function
real analysis
real analysis unsolved

Does there exist a surjective function $ f$, $ f: R\to R$, such that the expression
                                   $ f(x \plus{} y) \minus{} f(x) \minus{} f(y)$
takes exactly two values $ 0$ and $ 1$ for various real $ x$ and $ y$.

Today i have solved this problem: $ f(x)\equal{}\frac{1}{2}([x]\minus{}\{x\})$, where $ [x]$ is the greatest integer not exceeding $ x$, and $ \{x\}\equal{}x\minus{}[x]$, satisfies the required condition.


probability
geometry
geometric transformation
rotation
counting
distinguishability

[quote="md45"]is it 104/216 for Q8[/quote]

Umm... that wouldn't be the correct answer anyway simply because it's not in simplest form!

[hide="Hint"]$ \frac {104}{216} \equal{} \frac {13}{27}$.[/hide]

[quote="Jongao"]8.  Each of 3 standard 6-sided dice is rolled once.  What is the probability that the 3 numbers rolled are the lengths of the sides of an obtuse triangle?

10.  A circular tabletop is divided into 4 congruent sectors by 2 diameters that are perpendicular to each other.  Each sector is to be painted with one of 4 colors.  How many distinct ways can the table be painted? (A color may be used on more than 1 sector, but paintings that are the same after a rotation are not considered distinct.)[/quote]

10. We split this into cases:

1. All the colors are the same (4 choices). There is 1 way to

put this on a table in a distinguishable way, so 4*1=4 for this case

2. 3 of the same, 1 different (12 choices). There are 4

distinguishable arrangements, so 12*4=48.

3. 2 of the same, 2 other of the same:

4*3*4*2=96.

4+48+96=148.


geometry
rectangle
algebra
system of equations

[hide]The system of equations is

3w=2L+3
4L=2(w+L)+12

where w is the width and L is the length,

we get L=21, w=15.
[/hide]

[quote="ryan22"]Three times the with of a certain rectangle exceeds twice its length by three inches and four times its length is twelve more than its perimeter. Finds its dimensions of its rectangle. This is why i love algebra more that boring useless geometry!! Have fun

A) 31,14 B) 21,32 C) 56,48 D) 21,15[/quote]

[hide]
I believe this is a popular theme for Mathcounts. :)

$3W=2L+3$

$4L=2L+2W+12$ which becomes $2W=2L-12$

Subtract the second equation from the first:

$W=15$

Substitute back in:

$3(15)=2L+3$

$45=2L+3$

$2L=42$

$L=21$
[/hide]


Discussion and solutions on Bay Area Math Olympiad that took place today.

[quote][b]Problems[/b]

1.  An integer is called [i]formidable [/i]if it can be written as a sum of distinct powers of 4, and [i]successful [/i]if it can be written as a sum of distinct powers of 6.  Can 2005 be written as a sum of a formidable number and a successful number?  Prove your answer.

2.  Prove that if two medians in a triangle are equal in length, then the triangle is isosceles.  (Note:  A median in a triangle is a segment which connects a vertex of the triangle to the midpoint of the opposite side of the triangle.)

3.  Let $n \ge 12$ be an integer, and let $P_1, P_2, \ldots, P_n, Q$ be distinct points in the plane.  Prove that for some $i$, at least $n/6-1$ of the distances
\[P_1 P_i, P_2 P_i, \ldots, P_{i-1} P_i, P_{i+1} P_i, \ldots, P_n P_i\]
are less than $P_i Q$.

4.  There are 1000 cities in the country of Euleria, and some pairs of cities are linked by dirt roads.  It is possible to get from any city to any other city by traveling along these dirt roads.  Prove that the government of Euleria may pave some of these dirt roads so that every city will have an odd number of paved roads leading out of it.

5.  Let $D$ be a regular dodecahedron which is inscribed in a sphere of radius $R$.  Let $I$ be a regular icosahedron which is also inscribed in a sphere of radius $R$.  Which has the greater volume?  Prove your answer.
Note:  A [i]regular polyhedron[/i] is a geometric solid, all of whose faces are congruent regular polygons, in which the same number of polygons meet at each vertex.  A regular [i]dodecahedron[/i] is a polyhedron with 12 faces which are regular pentagons and a regular [i]icosahedron[/i] is a polyhedron with 20 faces which are equilateral triangles.  A polyhedron is [i]inscribed in a sphere[/i] if all its vertices lie on the surface of the sphere.[/quote]

There they are, word for word, but without the pictures of the icosahedron and the dodecahedron.

What a shame. I've seen #5 in a book before.

How is number 5 done?  I couldn't get it.


pigeonhole principle
ceiling function
combinatorics unsolved
combinatorics

There are 1987 students who came from 6 different classes.Give them 1987 numbers 1,2,...,1987 ,for every student get only one number.For any positive integer N,  the student who get N and the student who get 2N are not classmates.Please prove that there exists a student ,his number is the sum of his two classmates' numbers.

Suppose the contrary -- i.e., there is no such student -- and let the classes be called $ A, B, C, D, E, F$. By Pigeonhole, one of the classes, say $ A$, has at least $ \lceil \frac{1987}{6}\rceil \equal{} 332$ students. Let their numbers be $ a_1<a_2<\cdots<a_{332}$. by our assumption, the $ 331$ differences $ a_{332}\minus{}a_{1}, ..., a_{332}\minus{}a_{331}$ do not occur in $ A$. so they occur among classes $ B, C, D, E, F$. one of these classes, say $ B$, must have at least $ \lceil \frac{331}{5}\rceil \equal{}67$ of these differences. call these differences $ b_1 < b_2 < ... < b_{67}$. then the 66 differences $ b_{67}\minus{}b_{1}, ..., b_{67}\minus{}b_{66}$ cannot occur in $ B$, and they cannot occur in $ A$ because any difference of $ b_i$'s is a difference of $ a_i$'s, and no difference of $ a_i$'s occurs in $ A$. thus one of $ C, D, E, F$ must contain at least $ \lceil \frac{66}{4}\rceil \equal{} 17$ of these differences. continuing this reasoning, we get that $ 16$ differences aren't in any of $ A$ through $ C$, $ \lceil \frac{16}{3} \rceil \minus{}1 \equal{} 5$ differences aren't in $ A$ through $ D$, $ \lceil \frac{5}{2} \rceil \minus{} 1 \equal{} 2$ differences aren't in $ A$ through $ E$, and finally $ \lceil \frac{2}{1} \rceil \minus{} 1 \equal{} 1$ difference isn't in any of $ A$ through $ F$ - contradiction.


Could we have a table BBCode?

e.g.

[table][row][cell][/cell][cell][/cell][/row][row][cell][/cell][/row][/table]

You can makes tables with $ \text{\LaTeX}$...

\begin{tabular}{|c|c|}\hline a & b\\ \hline c & d\\ \hline\end{tabular}

Yes, but things like links for each individual cell or images/BBCode contained in the cells is impossible with LaTeX.

[quote="Temperal"]Darn. Is there any way to make LaTeX into a link?[/quote]

You can try,

[code][url=http://www.artofproblemsolving.com]$\text{\LaTeX}$[/url][/code]

which gives

[url=http://www.artofproblemsolving.com]$ \text{\LaTeX}$[/url]

But this is probably useless in your problem with LaTeX tables.

One more thing: is there any way to include a graphic in the forum latex?
[code]
$\includegraphic{http://i19.tinypic.com/61thy09.gif}$[/code]

doesn't work.

[quote="Valentin Vornicu"][quote="Temperal"]One more thing: is there any way to include a graphic in the forum latex?
doesn't work.[/quote]Why would you need that? You have the [img ] bbcode tag.[/quote]

I'm guessing he wants to put the images in the tables, which can only be made through LaTeX.

Wouldn't it just be more efficient to make an image of the table then post the image


It's equivalent of asking how many 5's divide 75!
It's $ \displaystyle [\frac{75}5]\plus{}[\frac{75}{25}]\equal{}18$

Hi;

Or you can go here:

http://www.wolframalpha.com/input/?i=75!

And count them!

[quote="chipmunk88"] :maybe: Wait how does that work, 75/5=25 and 25/5=5[/quote]

In order to make a zero appear at the end of a number, the number must be divisible by $ 10$. Now $ 10$ has th eprime factorization $ 2(5)$. Therefore, in order to "make" a zero, we need a factor of $ 2$ and a factor of $ 5$. Clearly, there are more factors of $ 2$ than factors of $ 5$. So, we count the number of $ 5$ (because we need one of both, so we count the lesser one.) Also, in case you don't know, $ 75!\equal{}75(74)(73)\dots(3)(2)(1)$. So the number of $ 5$s in $ 75!$ is $ \frac{75}{5}\equal{}15$ (since there are $ 15$ positive integers less than or equal to $ 75$ which have $ 5$ as a factor). However, $ 25$, $ 50$ and $ 75$ produce an additional factor of $ 5$ since they have a $ 5^2$ in their prime factorization. We're interested in counting the number of factors of five, not the number of integers which have five as a factor. That is why those cases must be included. Thus, the answer is $ 15\plus{}3\equal{}18$.

Feel free to ask any questions. :)


geometry
circumcircle
perpendicular bisector
power of a point
radical axis
geometry unsolved

In the triangle $ ABC$, denote by $ A_1, B_1,$ and $ C_1$ the feet of altitudes. Let $ P$ be the perpendicular foot $ C_1$ on line $ A_1B_1$, and let $ Q$ be that point of line $ A_1B_1$ for which $ AQ \equal{} BQ$. Show that $ \angle PAQ \equal{} \angle PBQ$, and find the size of these two angles in terms of angles $ A,B,C$.

Let $ X$ be the intersection of $ A_1C_1$ with $ AB.$ If $ M$ is the midpoint of $AB,$ then $ QM$ is the perpendicular bisector of $ AB.$ Note that $ PQMC_1$ is cyclic $\Longrightarrow$ $ XP \cdot XQ \equal{} XM \cdot XC_1,$ but $ X$ lies on the orthic axis of $\triangle ABC,$ i.e. the radical axis of the circumcircle and the 9-point circle $\odot( A_1B_1C_1)$ $\Longrightarrow$ $ XA \cdot XB \equal{} XM \cdot XC_1$ $ \Longrightarrow$ $ XP \cdot XQ \equal{} XA \cdot XB$ $ \Longrightarrow$ $ PQAB$ is cyclic. Since $ (X,A,C_1,B)=-1$ and $ PC_1 \perp PX,$ it follows that $ PC_1$ bisects $ \angle APB.$ As a result, $ PC$ and  the perpendicular bisector $ CM$ meet at $ Y$ lying on $\odot( PQAB).$ Hence, $ \angle PAQ \equal{}|  \angle A \minus{} \angle B|.$


One gallon of paint covers 400 square feet. If paint is only sold
in whole gallons, how many gallons of paint do you have to
buy to paint a surface of 925 square feet?

2 gallons are not enough, because they only cover 800 sq.ft.
You need 3 gallons (1200 sq.ft) to cover all 925 sq.ft.

$ \lceil \frac {925}{400} \rceil \equal{} \boxed{3}$

The proper way. :P

Minor correction:

$ \lceil{\frac{925}{400}} \rceil \equal{} \boxed{3}$

The [i]true[/i] proper way!  :P

$ \left\lceil{\frac{925}{400}}\right\rceil =\boxed{3} $

The [i]true true[/i] proper way! :D


Very nice problem...

Show that there are arbitrarily long strings of consecutive integers, each divisible by the square of some other integer.

It is directly result of Chinese reminder Theorem:
There exist a soluito of the equations:
$x\equiv-i(\mod p_{i}^{2})$where $p_{i}$be distinct primes
then $x+1,\dots,x+n$ satisfy your condtion

It can be extended, show that for every $m$ there exist $n$ such that $n+1,n+2,...,n+m$ are all divided with some perfect square, but $n+m+1$ is square free.

There are also many simmilar problems like this...

It is a well-known problem and it is a nice application of chinese remainder theorem,also.


How do you solve the maximum flow problem when there are vertex constraints?  That is, the flow into a vertex cannot exceed a number.

This should be sent off to either the Computer Science or Combinatorics forum.

Just replace each vertex with two vertices 'in' and 'out'. All edges into the vertex now go into 'in'; all edges leaving the vertex now leave 'out', and you connect in and out with an edge of weight (maximum flow for this vertex). That way, the graph is identical other than the fact that only a fixed flow can pass through the vertex.


Simplify $ 6 \plus{} 3(18 \minus{} 4^2) \div 2$

Remember the order of operations! $ 18 \minus{} (4^2) \equal{} 18 \minus{} 16 \equal{} 2 \rightarrow 3 \cdot 2 \equal{} 6 \rightarrow 6/2 \equal{} 3 \rightarrow 3 \plus{} 6 \equal{} \boxed{9}$

Use \cdot instead of *, for getting that dot for the multiplication.


Let $ ABC$ be a triangle with inradius $ 2$, and let $ P$ be any point. Prove that $ \max \{AP,BP,CP\} >3$

I will prove a stronger result!
From $ (\overrightarrow{AP} \plus{} \overrightarrow{BP} \plus{} \overrightarrow{CP})^2\geq 0$ then:
$ AP^2 \plus{} BP^2 \plus{} CP^2 \plus{} 2\overrightarrow{AP}.\overrightarrow{BP} \plus{} 2\overrightarrow{BP}.\overrightarrow{CP} \plus{} 2\overrightarrow{CP}.\overrightarrow{AP}\geq 0$
$ \Leftrightarrow AP^2 \plus{} BP^2 \plus{} CP^2 \plus{} (AP^2 \plus{} BP^2 \minus{} AB^2) \plus{} (BP^2 \plus{} CP^2 \minus{} BC^2) \plus{} (CP^2 \plus{} AP^2 \minus{} CA^2)\geq 0$
$ \Leftrightarrow 3(AP^2 \plus{} BP^2 \plus{} CP^2)\geq AB^2 \plus{} BC^2 \plus{} CA^2 \equal{} a^2 \plus{} b^2 \plus{} c^2$
But it is very easy and well-known to check that $ a^2 \plus{} b^2 \plus{} c^2\geq 36r^2$ (in fact $ ab \plus{} bc \plus{} ca\geq 36r^2$) then:
$ 3(AP^2 \plus{} BP^2 \plus{} CP^2)\geq 144$
Then $ max(AP,BP,CP)\geq 4$
Equal when $ a \equal{} b \equal{} c$ and $ P$ be the center of triangle $ ABC$.
Sorry if I post a mistake solution!


is there $ S\subset R^2$, ($ |S|>1$ ) such that $ S$ is connected and for every $ x$ and $ y$( $ x\neq y$) there is no path in $ S$ betwen $ x$ and $ y$?

Wonder what you exactly mean by

[quote="hamidreza"] ($ |S| > 1$ )...[/quote]

In any case, you may want to take a look at [url=http://en.wikipedia.org/wiki/Knaster-Kuratowski_fan]this[/url] site. :)

Best of luck,

[quote="coquitao"]Wonder what you exactly mean by

[quote="hamidreza"] ($ |S| > 1$ )...[/quote]

In any case, you may want to take a look at [url=http://en.wikipedia.org/wiki/Cantor_set]this[/url] site. :)

Best of luck,[/quote]

it means  that $ S$ is not  trivial (singltons satisfy the desired property )

Then, the [i]set[/i] in the link does [b]serve[/b] your purposes, right?

Thanks in advance
But I can't find the answer of my question
because The Sierpinski carpet is path connected.

[quote="hamidreza"] The Sierpinski carpet is path connected...[/quote]

I wasn't talking about Sierpinski's carpet, [b]hamidreza[/b]. I was referring to [b]Cantor's leaky tent[/b] instead.

You may want to take a look at pages 31-32 of the following text now:

[url=http://books.google.com.mx/books?id=DkEuGkOtSrUC&dq=counterexamples+in+topology&printsec=frontcover&source=bn&hl=es&ei=AtuuSaeIGYjHnQf7-7nABg&sa=X&oi=book_result&resnum=4&ct=result#PPA31,M1][color=BLUE]Counterexamples in Topology[/color].[/url]

[b][color=red]BTW[/color][/b], I've just [b]edited[/b] the link in my first post of the thread.

Regards,  :wink:

Also look at my construction [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246172]here[/url] (Probably, it is pretty similar to the one coquitao referred to).


function
limit
integration
inequalities
geometry
3D geometry
logarithms

Let $ f$ be a continuous function on the unit interval $ [0,1]$. Show that \[ \lim_{n \rightarrow \infty}  \int_0^1... \int_0^1f(\frac{x_1+...+x_n}{n})dx_1...dx_n=f(\frac12)\] and \[ \lim_{n \rightarrow \infty}  \int_0^1... \int_0^1f (\sqrt[n]{x_1...x_n})dx_1...dx_n=f(\frac1e).\]

A couple of approaches have been given here:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=226389]link[/url]

Let me try to post full solution above problems  :) .
for the first:
some observation:
since $ f \in C[0,1]$ then for all $ \epsilon > 0$ one can choose such $ \delta$such as if $ |x - y| < \delta$ then $ |f(x) - f(y)| < \epsilon$
fix some small delta and consider the set
\[ E_{n}(\delta) = \{ (x_{1},x_{2},...,x_{n})\jn [0,1]^{n} : \left| \frac {x_{1} + x_{2} + ... + x_{n}}{n} - \frac {1}{2} \right| > \delta \}
\]
consider function :
\[ F(x) = \left( \sum_{1\leq k \leq n}x_{k} \right) - \frac {n}{2}
\]
then Chebushev inequality implies that  if $ F \in L^{0}(B)$ then :
\[ \lambda (B(|F| > t)) \leq \frac {1}{t^{2}}\int_{B} F^{2}
\]
thus we have
\[ \lambda (E_{n}(\delta)) \leq \frac {1}{(n\delta)^{2}}\int_{[0,1]^{n}}\left( \sum_{1\leq k \leq n} x_{k} - \frac {1}{2} \right) ^{2} =
\]

\[ = \frac {1}{(n\delta)^{2}}\int_{[0,1]^{n}} \sum_{1\leq k \leq n} (x_{k} - \frac {1}{2} ) ^{2} \approx \frac {Const}{n}
\]
since the system of functions $ x_{k} - 1/2$ on the unit cube is orthogonal.
thus measure of $ E_{n}(\delta)$ is very small , hence :
\[ \int_{[0,1]^{n}} \left| f(\frac {\sum x_{k}}{n}) - f(\frac {1}{2}) \right| \leq \int_{E_{n}(\delta)} | ... | + \int_{[0,1]^{n} - E_{n}(\delta)} | ... | \leq \frac {Const}{n} + \epsilon
\]
we are done .
 similary for the second problem :
we need to consider the set :
\[ E_{n}(\delta) = \{ (x_{1},...,x_{n}) \in [0,1]^{n} : \left| \frac {1}{n} \sum \ln x_{k} + 1 \right| > \delta \}
\]
note that the system of function $ \ln x_{k} + 1$ is orthogonal since $ \int_{[0,1]} \ln x dx = - 1$ and also note that :
$ (x_{1}x_{2}...x_{n})^{1/n} = \exp(1/n \sum \ln x_{k})$ and use continious of our function . done


MATHCOUNTS
trigonometry
geometry
geometric transformation
reflection
trig identities
Law of Sines

[quote="bpms"]If a right triangle has a 15 degree angle and the hypotenuse has length 10, then what is the length of the shortest leg?

The hard part:
Since this is the mathcounts forum, NO TRIG AT ALL.[/quote]
how do we do it without some trig???
*confused* does the cosine formula count as trig?

all things dealing with sines and cosines
and other things derived from these 2 such as tangent
can be counted as trig

i'm confundled  :huh:

is it possible ?

this is just a highly educated guess, but is the answer[hide]6 (w/o trig)[/hide]???
-jorian

edit: or is it[hide]~2.5881904510252074[/hide]???

do you have any explanations?
or just random guesses.

There might be some way to do it by doubling the angle and using 30-60-90 identities, but I can't get anywhere doing that...

perhaps we can drop millions of altitudes until we have a 30-60-90 or a  45-45-90

[quote="daermon"]its pretty easy using trig, but I don't know how to do it without...
[hide="trig solution"]by the law of sines, $10=\frac{x}{\sin{15}}=\frac{4x}{\sqrt6-\sqrt2}$ so
$x=\frac{5(\sqrt6-\sqrt2)}{2}$
jhredsox, u are right:  the answer is 2.58819...[/hide][/quote]

Do not post trig solutions, they are near worthless to the thread if they are in the MATHCOUNTS forum.

[hide="Idea"]Reflect the triangle over the hypotenuse so you have a happy 30 degree angle.  [/hide]

This problem is very easy. My friend showed me a solution.

[hide]
From the 75 degree angle, draw a line hitting the longer leg so that it divides the 75 degree angle into a 60 degree angle and a 15 degree angle. Now we have a 30-60-90 triangle and an isosceles triangle.

30-60-90 triangle facts can be easily proven using geometry, so I won't bother.

Call the short leg x. The long leg of the 30-60-90 triangle is $x\sqrt{3}$. The hypotenuse of the 30-60-90 triangle is $2x$, and since the 15-15-150 triangle is isosceles the other leg of the isosceles triangle is also 2x. Now by Pythagoras we have $x^{2}+x^{2}(2+\sqrt{3})^{2}= 100$ Simplifying, we get $x = 5\sqrt{2-\sqrt{3}}$.
[/hide]

Last night I solved it using kyyuanmathcount's idea. It was pretty long, though. What I did was take the 30-75-75 triangle, then from one vertex I drew a right angle and extended the other leg to make a 30-60-90 triangle. Then I did a bunch more stuff, then I made similar triangles and solved it. If anyone wants to see my exact solution I'll be happy to type it up.

EDIT: I just read daermon's solution and realized that $\sqrt{2-\sqrt{3}}$ can be simplified to $\frac{\sqrt{2}+\sqrt{6}}{4}$.

If you reflect it you can use cosines(not that it isn't trig.)
But you can...
i mean if you make it so you have a 30-75-75
I don't know what you are getting at though because I can't see it....

I'll post my solution sometime tonight, when my parents give me internet for an hour or so. (I'm typing this at a friend's house).

wow i am doing school
umm yeah i wont spam but just to tell everyone
i am amazed at how early some of you get out
because i have a lot of school to do during the holiday break

and...you can do trig at the comp. right? there is no way they can see you did it?

It's not like trig is not allowed, it's just that middle schoolers aren't supposed to know it.

And I'm 99.9% sure that a problem like this will never appear on a MathCounts test.

This topic will be subject to mass deletions (or lock if I'm feeling lazy) if we continue the trig talk.  We don't CARE if you know a solution w/ trig.  Mathcounts requires no trig.  And...use common sense Funcia, the testgraders won't shoot you if they see the word "cosine" on your paper.

[quote="bpms"]If a right triangle has a 15 degree angle and the hypotenuse has length 10, then what is the length of the shortest leg?

The hard part:
Since this is the mathcounts forum, NO TRIG AT ALL.[/quote]

Extend the shorter leg.

Then, draw a fifteen degree angle on the opposite side of the hypotenuse.

Then, figure a way of using Angle Bisector Theorem and 30-60-90 ratios.

[quote="daermon"]its pretty easy using trig, but I don't know how to do it without...
[hide="trig solution"]by the law of sines, $10=\frac{x}{\sin{15}}=\frac{4x}{\sqrt6-\sqrt2}$ so
$x=\frac{5(\sqrt6-\sqrt2)}{2}$
jhredsox, u are right:  the answer is 2.58819...[/hide][/quote]
Um, just wondering how is it the law of sines when it is a right triangle?
To ubemaya: for future reference always check if nested radicals can be simplified, just a suggestion.
To mathgenius: That is the way I did it :)

Yay, I feel smart, knowing that is the same way bpms did it.

Wait, how does Neil's method work? I don't get it...

I would think that you need Steward's since the length given is the line instead.

[hide]Let the triangle be ABC where angle A=15 and angle C=90.
Flip the triangle over to get a 30-75-75 triangle and point B lands on Point D.

Drop the altitude from B to AD (it meets at E) and that altitude is 5 because of the 30-60-90 triangle.  Let BC=x.  ABC is simila to BDE.

AC=$\sqrt{100-x^{2}}$
$\frac{10}{2x}=\frac{\sqrt{100-x^{2}}}{5}$
$x^{4}-100x^{2}+625=0$

$x=\boxed{2.58881}$

checking this with the calculator, it is also 2.588

[color=blue]kyyuanmathcount: hide tags please ;)[/color]
[/hide]


There are two point charges at distance r (r is unknown) from each other.First one is $q_{1}$ with plus sign,and the second charge is $q_{2}$ with minus sign.Electric field leaves $q_{1}$ with the angle $\alpha$ to the line joining the charges and reaches the second charge $q_{2}$ with the angle $\beta$.Find the angle $\beta$.

just conserve the number of electric lines of force in the region u have specified
${q_{1}}(1-cos{\alpha}) ={q_{2}}(1-cos{\beta})$

using which we can get the value of ${\beta}$.Her take only the absolute value of the charges.

well this problem is really to find the equation of the electric field lines
(it requires quite a bit of calculation so let me tell u the underlying process)
now say $ y \equal{} f(x)$ represnts the equation of the electric fild line take the line joining $ \plus{} q_{1}$ and $ \minus{} q_{2}$ as the $ X$ $ \minus{}$ $ axis$.
now the rest part is just applying columb's law
laso use the fact that the electric field at a point $ (x,y)$ can be made into
$ E_{x}$ and $ E_{y}$ where
$ \frac {E_{x}}{E_{y}}$ $ \equal{}$ $ \frac {dy}{dx}$.
then convert the variables like distance of point from  charges into polar coordinates.
then you will get a beautiful equation
which gives you the answer

$ \sin{\frac {\beta}{2}}$ $ \equal{}$ $ \sqrt {\frac {q_{1}}{q_{2}}}\sin{\frac {\alpha}{2}}$

sorry to bump....but that;s almost one year since i last posted :o ...actually that was the best time i suppose with Djole,Namiqfizik,Thaakisfox all here i miss them all :rotfl:  :rotfl:  :rotfl:  :rotfl:
why aren't they coming :rotfl:  :rotfl:  :rotfl:  :(  :(

[quote="pardesi"]sorry to bump....but that;s almost one year since i last posted :o ...actually that was the best time i suppose with Djole,Namiqfizik,Thaakisfox all here i miss them all  :rotfl:
why aren't they coming ::(  :([/quote]

Ans
[hide]Because they are all scared of me.  :rotfl: [/hide]

And you forgot Rushil

no da he wasn't active then....i am talking of feb march april may and june 2007  :)


probability
number theory unsolved
number theory

Find the integer solution from equation $[ m\sqrt{2}]=[(2+\sqrt{2})n]$,which $[x]$ denotes the integer part from $x$

$n=0\to m=0$, If n<0, then m<0 too and (-m,-n) - solution too.
Therefore sufficiently find all positive integer solutions.
Let $n\in N$, then $[(1+\sqrt 2 )n]\le m\le [(1+\sqrt 2 )n+1]$.
For any n we have one solution with probability $\frac{1}{\sqrt 2 }$.


Let $k$ be inscribed circle in triangle $ABC$. Is corect: For every point P on circle kdistinct than touching ones is true $'sin \measuredangle PAC \cdot sin \measuredangle PBA \cdot sin \measuredangle PCB=1$

Are you sure???  :?:

It would be correct if all these angles were $90^0$

every sin is $\leq 1$, so your product is $\leq 1$

(in fact, they can't be all $90^0$, so you'll have a product $<1$)


inequalities
floor function
quadratics
algebra
quadratic formula
binomial theorem

If $x+\frac1x=a$ Then what is $x^b+\frac1{x^b}$ in terms of $a$ and $b$

[quote="tetrahedr0n"]It's called the $b$th Chebyshev Polynomial. To show that it is possible to do, we use this recursive formula:

$x^b + 1/x^b = (x+1/x)(x^{b-1}+1/x^{b-1})-(x^{b-2}+1/x^{b-2})$[/quote]
Is there a way to put it in summation form, instead of recursive and only in terms of $a$ and $b$

[quote="Ravi B"]\[
  x^b + \frac{1}{x^b}
    = \frac{1}{2^{b-1}}
         \sum_{j = 0}^{\lfloor b/2 \rfloor} \binom{b}{2j} {(a^2-4)}^j a^{b - 2j} \, .
\][/quote]
wow, how did you derive that?  :?:
no, thats not right because if you plug in $a=2$, then you will always get $0$, when it should always be $1$. And if you plug in $b=1$, you should get $a$, but your formula shows $a^2-4$... I think it needs some fixing

I'll try to post the derivation tomorrow.

The formula is correct.  When $a = 2$, it will always give 2; don't forget the $j =0$ term, which is nonzero because $0^0 = 1$.  When $b = 1$, it will always give $a$; there will be only one term in the summation ($j = 0$).

Nevermind, that works, I didn't see the $^j$ part and when a=2, you will have a $0^0$, so you have to say, when $a=2$, $\sum=2$

Okay, here is a derivation of the formula.  By the quadratic formula, one solution of $x + 1/x = a$ is
\[
  x = \frac{a + \sqrt{a^2 - 4}}{2} \, .
\]
The other solution is the reciprocal
\[
  \frac{1}{x} = \frac{a - \sqrt{a^2 - 4}}{2} \, .
\]
Raising both to the $b$th power, we get the formula
\[
  x^b + \frac{1}{x^b}
    = \frac{1}{2^b}
         \bigl[ {(a + \sqrt{a^2-4})}^b + {(a - \sqrt{a^2-4})}^b \bigr] \, .
\]
This formula can already be considered a solution to your original problem, although you might object to the square roots.  To get rid of them, we can apply the binomial theorem to both parts.  Fortunately, half the terms, the ones involving an odd power of $\sqrt{a^2 - 4}$, will cancel out.  We are left with terms involving an even power of $\sqrt{a^2 - 4}$, which gets rid of the square roots.  Each such term will appear twice, once in each part.

Ok, I understand all that, but how in your formula id you have a $\frac1{2^{b-1}}$ instead of $\frac1{2^b}$?

That's because each of the even-powered terms appears twice, once from ${(a + \sqrt{a^2-4})}^b$ and once from ${(a-\sqrt{a^2-4})}^b$.  That gives us a constant factor of $\frac{2}{2^b}$, which is $\frac{1}{2^{b-1}}$.


Find all pairs $(x, y)$ of rational numbers such that $y^2 =x^3 -3x+2$.

$y^{2}=(x-1)^{2}(x+2)$
$y=(x-1)\sqrt{x+2}$
So we need x+2 to be a perfect square.
(2,2)
(7,18 ) (this was a smiley earlier, hence the space)
(14,52)
(23,110)
There are infinitely many.

[quote="Moonmathpi496"]You found integer solution.But what about rational solutions? Example: 1/4,7/9 are some solutions.[/quote]
[quote="Thangtonghop"]I think to change to Rational roots, we find p/q+2= perfect rational square. It   's easy because (p,q)=1 so p+2q, q=perfect squares.  We solve [code] x^2- 2y^2= p [/code] , which is Pell equation[/quote]

[quote="TTsphn"][quote="bpms"]$ y^{2} \equal{} (x \minus{} 1)^{2}(x \plus{} 2)$
$ y \equal{} (x \minus{} 1)\sqrt {x \plus{} 2}$
So we need x+2 to be a perfect square.
(2,2)
(7,18 ) (this was a smiley earlier, hence the space)
(14,52)
(23,110)
There are infinitely many.[/quote]
I think solution :
$ x \equal{} m^2 \minus{} 2,m\in Q$
$ |y| \equal{} |(m^2 \minus{} 3)m|$[/quote]

I think , when x=1 , x+2 need not a perfect square.
and x=1 => y=0 is solution. but x=1=m^2-2 , m=3^(1/2) is not rational number.

so, m is also include 3^(1/2)

my english is very poor. I do not well english. sorry.
it is first time I write reply in ML :blush:

\[ y\equal{}(x\minus{}1)\sqrt{x\plus{}2}\]
is rational. Thus, either $ \sqrt{x\plus{}2}$ is rational or $ x\minus{}1$ is zero. Thus the solution set is:
\[ (x,y)\equal{}(1,0)\qquad (x,y)\equal{}\left(m^2\minus{}2,m(m^2\minus{}3)\right),\qquad m\in\mathbb{Q}\]


A quadrilateral $ABCD$ has an inscribed circle with center $O$. Let $E$ and $F$ be the points of intersection of the line $BD$ with this inscribed circle. Let $H$ be the foot of the perpendicular from the point $O$ to the line $AC$. Prove that $\measuredangle BHE=\measuredangle DHF$.

Hint: try to prove that HC is the angle bisector of angles <BHD and <EHF.

sprmnt21,this problem is not easy as you thought! :mad:

Both of you(James potter and sprmnt21) are right and none of you are left.
yes we have to prove that$\angle EHC=\angle CHD$
and this isn't very easy I think.

Let $NL\cap MK=X$ ,the polars of $B,D$ pass through $X$ and we know that $MN\cap LK=P'$ is on pole of $X$.therefore $B,D,P'$ are collinear.

like this we have $AC$ is polar of $P'$ hence $OP'\perp AC$ so $P\equiv P'$.

now we have $(PQ,EF)=-1\ \ (\star)$

Notice that $A(PS,MN)=-1$ hence $(PQ,BD)=-1\ \ (\star \star)$ .

from $(\star)(\star \star)$ we have $\angle BHQ =\angle QHD \ ,\ \angle EHQ =\angle QHF$ Cause $OH\perp AC$

Now we are at home .

Hi guys,I solve  this problem  without using polars,but I used more not easy theorems :( ,and also I get to the realy interesnting results :!:.
[b]Nice problem James,thank you[/b] ;) .
Also I want to know,wher you found it? :roll:

[quote]Also I want to know,wher you found it?[/quote]
I found this problem in [color=red][b]Mathematics and Youth[/b][/color],a magazines from [color=red][b]Vietnam[/b][/color](I have a friend in Vietnam! :) )
I post this problem in this section because the solution in that magazines is too long. They use two lemma
1)Let be given $\Delta ABC$ and two point $M,N$ on the side $BC$.Prove that:$\frac{BM.BN}{CM.CN}=\frac{AB^2}{AC^2}\longleftrightarrow\widehat{BAM}=\widehat{CAN}$
2)Let be gien a circle $(O)$ and a point $A$ outside $(O)$.A line through $a$ intersects $(O)$ at $B$ and $C$.Two tangent of $(O)$ at $B$ and$C$ intersect each other at $K$.$AE,AE$ are two tangents of $(O)$($E,F$ are tangent point).Prove that $K,E,F$ are collinear.
So I want to find a better one,but a syntheticals one!

I used second lemma too,and I think that is well-know lemma(at lest for me :lol:),and can we faind that magazine in internet?(Mathematics and Youth)

[quote="James Potter"][quote]And can we faind that magazine in internet?(Mathematics and Youth) [/quote]
I'm sorry you cannot.My friend scan this magazines and post it to me everymonth.If you want I can post you but [b][u]Can you Read and Speak Vietnamese[/u][/b]?(example:[b]Ban co biet noi va doc tieng viet khong[/b]?)[/quote]
I am afreid,I don't know  [b]Vietnamese[/b],but [b]thenk you[/b].

[hide="To amir2,"]
 Your proof is so wonderful.
 [/hide]
 My proof is also simple. I use the following fact (I don't have much time right now, so I just tell you about the fact I use. Note that my solution also uses properties of pole and polar.)
 [i]Fact[/i] Let $X$ be a point and $(O)$ be a circle. Suppose $X'$ is the inversion of $P$ wrt $(O)$. Through $X$, we draw two rays such that they intersect $(O)$ at $M$, $Q$ and $N$, $P$ ($M$ lies between $X$ and $Q$, $N$ lies between $X$ and $P$). Then $\measuredangle{MX'N} = \measuredangle{PX'Q}$

[quote="James Potter"]
2)Let be gien a circle $(O)$ and a point $A$ outside $(O)$.A line through $a$ intersects $(O)$ at $B$ and $C$.Two tangent of $(O)$ at $B$ and$C$ intersect each other at $K$.$AE,AE$ are two tangents of $(O)$($E,F$ are tangent point).Prove that $K,E,F$ are collinear.
[/quote]

Can you post the proof of this lemma?
Thx..


[quote="Virgil Nicula"]Let $a,b,c$ be positive numbers. Prove that $\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)\ge 9abc+13\cdot \sum a(b-c)^{2}$.[/quote]
I think, your inequality is wrong. Try $a=b=1$ and $c=2.$
$\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)\ge 9abc+2\cdot \sum a(b-c)^{2}$ is true already.

Thank you, Arkady ! You are right. I corrected above.

[quote="Virgil Nicula"][color=darkblue]Let $a,b,c$ be positive numbers. Prove that $\boxed{\ \left(a^{2}+b^{2}+c^{2}\right)(a+b+c)\ge 9abc+2\cdot \sum a(b-c)^{2}\ }$.
[b][u]Remark.[/u][/b] This inequality is equivalently with the following :
$1\blacktriangleright$ $a^{3}+b^{3}+c^{3}\ge 3abc+\sum a(b-c)^{2}$.
$2\blacktriangleright$ $(a+b+c)^{3}\ge 27abc+4\cdot \sum a(b-c)^{2}$.
$3\blacktriangleright$ $(a+b+c)\cdot\sum (b+c-2a)^{2}\ge 6\sum a(b-c)^{2}$.
$4\blacktriangleright$ $\sum (b+c-a)(b+c-2a)^{2}\le 4\cdot\sum a(b-c)^{2}$.[/color][/quote]
[color=darkred]Three interesting identities :[/color] $\left\{\begin{array}{c}\sum a(b+c-2a)^{2}=4\sum a(a-b)(a-c)+\sum a(b-c)^{2}\\\\ a^{3}+b^{3}+c^{3}-3abc=\sum a(a-b)(a-c)+\sum a(b-c)^{2}\\\\ \sum (b+c-a)^{2}(b+c-2a)=4\sum a(b-c)^{2}\end{array}\right\|$ $\implies$ $\boxed{\left\|\begin{array}{c}\sum a(b+c-a)(b+c-2a)\ge 0\\\\a^{3}+b^{3}+c^{3}\le 3abc+\sum a(b+c-2a)^{2}\\\\ (a+b+c)^{3}\ge 27abc+\sum (b+c-a)^{2}(b+c-2a)\end{array}\right\|}$.


Prove that for all natural $n$: $9^n(9^n+1)$ is divisible by $3^n(3^n+1)$.

for n=1, 9(9+1)=90 which is not divisible by 3(3+1)=12

the given statement is only true for n=0 since $gcd(9^n,3^n+1)=1$ and $gcd(9^n+1,3^n+1)=gcd((3^n+1)(3^n-1)+2,3^n+1)=2$,so $9^n(9^n+1)$ is not divisible by $3^n+1$ if $9^n(9^n+1)>2$. if we replace $9^n(9^n+1)$ by $9^n(9^n-1)$ the statement is trivially true. maybe this was meant.


calculus
integration
trigonometry
calculus computations

Compute
\[
\displaystyle\int_1^{\sqrt{3}}\displaystyle\frac{x^2-1}{x^4+x^3+3x^2+x+1}\,dx
\]

$I = \int_1^{\sqrt 3} \frac{x^2 - 1}{x^4 + x^3 + 3x^2 + x + 1}\text dx = \int_1^{\sqrt 3} \frac{1}{x^2 + x + 3 + \frac 1 x + \frac{1}{x^2}}  \frac{x^2-1}{x^2} \text dx$

Substitute $y = x + \frac 1 x$, [color=white].[/color] $y^2 = x^2 + 2 + \frac{1}{x^2}$, [color=white].[/color] $\text dy = \left( 1 - \frac{1}{x^2} \right) \text dx= \frac{x^2 - 1}{x^2} \text dx$

$I = \int_2^{\frac{4 \sqrt 3}{3}} \frac{\text dy}{y^2 + y + 1}$

Substitute $z = y + \frac 1 2$, [color=white].[/color] $z^2 = y^2 + y + \frac 1 4$, [color=white].[/color] $\text dz = \text dy$

$I = \int_{\frac 5 2}^{\frac{3 + 8 \sqrt 3}{6}} \frac{\text dz}{z^2 + \frac 3 4}$

Substitute $v = \frac{2}{\sqrt 3} z$, [color=white].[/color] $v^2 = \frac 4 3 z^2$, [color=white].[/color] $\text dv = \frac{2}{\sqrt 3} \text dz$

$I = \frac{2\sqrt 3}{3} \int_{\frac{5 \sqrt 3}{3}}^{\frac{8 + \sqrt{3}}{3}} \frac{\text dv}{v^2 + 1} = \left[ \frac{2 \sqrt 3}{3} \arctan v \right]_{\frac{5 \sqrt 3}{3}}^{\frac{8 + \sqrt{3}}{3}} = \frac{2 \sqrt 3}{3} \left( \arctan{\frac{8 + \sqrt 3}{3} - \arctan{\frac{5 \sqrt 3}{3}} \right)}$

Let $\alpha = \arctan{\frac{8 + \sqrt 3}{3}}$, [color=white].[/color] $\beta = \arctan{\frac{5 \sqrt 3}{3}}$.

$\tan{(\alpha - \beta)} = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{3(8 + \sqrt 3 - 5\sqrt 3)}{9 + 5 \sqrt 3 (8 + \sqrt 3)} = \frac 3 2\ \frac{2 - \sqrt 3}{3 + 5 \sqrt 3} = \frac{13 \sqrt 3 - 21}{44}$

$I = \frac{2 \sqrt 3}{3} \arctan{\frac{13 \sqrt 3 - 21}{44}}$

yetti has some serious integrating skills.


limit
logarithms
integration
function
real analysis
algebra
rational function

Is the following true?

$ \lim_{n \to \infty} \frac{\sum_{k \equal{} 2}^{n} k^{b\minus{}1} \ln{k}}{n^{b} \ln{n}} \equal{} \frac{1}{b}$ for $ b \geq 1$

I tried to generalize this after observing a pattern with specific values....  how would I go about showing this if it is true?

Write as $ A \equal{} \lim_{n \to \infty} \sum_{k\equal{}2}^{n} \left( \frac{k}{n} \right)^{b\minus{}1} \frac{\ln k}{\ln n} \cdot \frac{1}{n}$.  Let $ B \equal{} \lim_{n \to \infty} \sum_{k\equal{}2}^{n} \left( \frac{k}{n} \right)^{b\minus{}1} \frac{\ln n}{\ln n} \cdot \frac{1}{n} \equal{} \int_{0}^{1} x^{b\minus{}1} \, dx \equal{} \frac{1}{b}$; then

$ A \minus{} B \equal{} \lim_{n \to \infty} \frac{1}{\ln n} \sum_{k\equal{}2}^{n} \left( \frac{k}{n} \right)^{b\minus{}1} \ln \left( \frac{k}{n} \right) \cdot \frac{1}{n} \equal{} \lim_{n \to \infty} \frac{1}{\ln n} \left( \int_{0}^{1} x^{b\minus{}1} \ln x \, dx \right) \equal{} 0$

so $ A \equal{} B$ as desired.  Strictly speaking, that last part should involve a bounding argument, but I hope you understand the point that the Riemann sum converges to a finite limit when $ b \ge 1$ whereas $ \frac{1}{\ln n}$ converges to $ 0$.

For $ b>1$, the bounding argument is trivial; the function can be extended to be continuous on the closed interval. For $ b\equal{}1$, the sum is actually closer to zero than the integral.

[quote="t0rajir0u"]Write as $ A \equal{} \lim_{n \to \infty} \sum_{k \equal{} 2}^{n} \left( \frac {k}{n} \right)^{b \minus{} 1} \frac {\ln k}{\ln n} \cdot \frac {1}{n}$.  Let $ B \equal{} \lim_{n \to \infty} \sum_{k \equal{} 2}^{n} \left( \frac {k}{n} \right)^{b \minus{} 1} \frac {\ln n}{\ln n} \cdot \frac {1}{n} \equal{} \int_{0}^{1} x^{b \minus{} 1} \, dx \equal{} \frac {1}{b}$[/quote]

Yes, your method makes sense.  Just wondering: how did you think of writing it like that?  A bit of luck, experience, or was there a pattern?  Like, how did you think of using a Riemann Sum?  (Just curious in case something similar works in the future for other expressions.)

Thanks!!!

An awful lot of infinite sums can be written in the form $ \lim_{n \to \infty} \sum_{k\equal{}k_0}^{n} f \left( \frac{k}{n} \right) \frac{1}{n}$ (see, for example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191805]this recent one[/url]).  The tricky part is that when $ f(x)$ has particular algebraic properties (such as being a rational function or being multiplicative like the logarithm) this form can get obscured.  Just keep in mind that this is one of the techniques available for dealing with sums involving both $ k$ and $ n$ and look to apply it wherever possible.  :)


Studiati functia : $f(x)=\frac{1+|\sin(x)|}{|\cos(x)|}$ din punctul de vedere al domeniului de definitie si al graficului.

Domeniul de definitie e: $\mathbb{R}-\{\frac{\pi}{2}+2k\pi, \frac{3\pi}{2}+2k\pi \}, k\in\mathbb{Z}$. Cum functia e periodica cu perioada $2k\pi$ e suficient sa o reprezentam pe intervalul $(\frac{\pi}{2},\frac{3\pi}{2})$. Dupa explicitare $f$ are forma:

$f(x)=\{\begin{array}{cccc} \frac{1+\sin(x)}{-\cos(x)}, \ x\in(\frac{\pi}{2},\pi] \\ \frac{1-\sin(x)}{-\cos(x)}, \ x\in(\pi,\frac{3\pi}{2}) \\ \frac{1-\sin(x)}{\cos(x)}, \ x\in(\frac{3\pi}{2},2\pi] \\ \frac{1+\sin(x)}{\cos(x)}, \ x\in(2\pi,\frac{5\pi}{2}) \end{array}$

Dupa derivare ramurile 1 si 3 sunt negative pt valorile pe care le ia $x$ shi ptentru celelalte 2 ramuri functia e pozitiva (de aici ne rezulta monotonia). Se mai fac limitele in punctele ${\{\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}}\}$ shi dau toate infinit, adica avem 3 asimptote veticale (prin pct mentionate). Cum $f>0$ peste tot graficul va fi peste OX. Mai dai cateva valori($\pi, 2\pi$) shi iese graficul de forma unui W interupt :).

[color=darkblue][b][u]Etapa 1.[/u] Domeniul de definitie $D_{f}$ ([u]de lucru[/u]).[/b]
$1.1.\ f$ - periodica $(T=\pi )$ $\Longrightarrow$ $f: \left (-\frac{\pi}{2},\frac{\pi}{2}\right)\rightarrow R$, unde $f(x)=\frac{1+|\sin x |}{\cos x}\ .$
$1.2.\ f$ - para $\Longrightarrow$ $f: \left[0,\frac{\pi}{2}\right )\rightarrow R$, unde $f(x)=\frac{1+\sin x}{\cos x}\ .$
$1.3\ f(x)=\frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}=\frac{\\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\cos\frac{x}{2}}=\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}$ $\Longrightarrow$ $\boxed{\ f: D_{f}\equiv \left[0,\frac{\pi}{2}\right)\rightarrow R\ ,\ f(x)=\tan\left(\frac{x}{2}+\frac{\pi}{4}\right)\ }\ .$

[b][u]Etapa 2.[/u] Intersectia cu axele de coordonate si precizarea capetelor inchise ale graficului[/b] $G_{f}$. De exemplu, $(0.1)\in G_{f}\ .$

[b][u]Etapa 3.[/u] Comportarea functiei pe inchiderea $\overline{D_{f}}$ si stabilirea $D_{\mathrm{cont\ f}}$ - domeniul de continuitate).[/b]
$D_{\mathrm{cont\ f}}=D_{f}$ si $f\left(\frac{\pi}{2}-0\right)=\infty$, adica dreapta de ecuatie $y=\frac{\pi}{2}$ este asimptota verticala de la stanga.

[b][u]Etapa 4.[/u] Studiul primei derivate (monotonia functiei, puncte de extrem, puncte de nonderivabilitate si interpretarea lor geometrica etc).[/b]
Domeniul de derivabilitate $D_{f'}=D_{f}$ si $f'(x)=\frac{1+\sin x}{\cos^{2}x}=\frac{1}{1-\sin x}>0$, adica functia $f$este strict crescatoare : $f\s\nearrow\ .$

[b][u]Etapa 5.[/u] Studiul derivatei a doua $f''$ (concavitate, convexitate, puncte de inflexiune etc.).[/b]
$D_{f''}=D_{f}$ si $f''(x)=\frac{\cos x}{(1-\sin x)^{2}}>0$, adica functia $f$ este convexa.

[b][u]Etapa 6.[/u] Tabelul de variatie a functiei (tabelul "expert").[/b]

[b][u]Etapa 7.[/u] Reprezentarea propriu zisa a graficului functiei[/b] $f\ .$

[b][u]Etapa 8.[/u] Concluzii generale si specifice (semnul functiei; $Im\ f$- imaginea functiei; [u]extinderea rezultatelor obtinute de la domeniul de lucru[/u] $D_{f}$ [u]la domeniul maxim de definitie[/u] $D_{\mathrm{max\ f}}\ ;$ intuirea unor proprietati geometrice care nu au fost sesizate anterior si eventuala demonstratie analitica ale acestora; se raspunde la problemele : injectie, surjectie, bijectie).[/b]

[b][u][size=134]Tema pentru vacanta.[/size][/u] Reprezentati graficele urmatoarelor functii [/b]$f_{k}\ ,\ k\in \overline{1,15}\ :$

$\begin{array}{ccc}f_{1}(x)=x\cdot\sqrt{\frac{x+1}{x+2}}\ \ \ \ \ & f_{2}(x)=|x|\cdot \sqrt{\frac{x}{x+1}}\ \ \ \ \ & f_{3}(x)=\sqrt [3]{x^{2}(x-3)}\ \\\\ f_{4}(x)=\sqrt{-\ln|\sin x|}\ \ \ \ \ & f_{5}(x)=\frac{\ln |x|}{\ln |x|-1}\ \ \ \ \ & f_{6}(x)=\ln\left(1+\frac{1}{|x-1|}\right)\ \\\\ f_{7}(x)=\arccos (3x-4x^{3})\ \ \ \ \ & f_{8}(x)=\frac{\sin 2x}{\sin x-\cos 2x}\ \ \ \ \ & f_{9}(x)=\max \{x-\arctan x\ ,\ \ln (1+x^{2})\}\ \\\\ f_{10}(x)=\frac{x^{2}}{x+1}e^{\frac{1}{x}}\ \ \ \ \ & f_{11}(x)=\left|\arctan\frac{x+2}{2x-1}{+\arctan\frac{1}{x}\right|\ \ \ \ \ }& f_{12}(x)=(x-1)\cdot\arcsin \frac{|x|}{\sqrt{1+x^{2}}}\ \\\\ f_{13}(x)=\frac{x}{1+xe^{|x-1|}}\ & f_{14}(x)=|x-1|\cdot e^{-\frac{1}{|x|}}\ \ \ \ \ & f_{15}(x)=x\cdot\arcsin\frac{x-1}{\sqrt{2(1+x^{2})}}\ \end{array}$[/color]


calculus integration geometry perimeter AMC

How many scalene triangles have all sides of integral lengths and perimeter less than 13?

[hide]Since the perimeter's less than 13, it is less than or equal to 12, so 5 is the maximum length for a side of a triangle. There are 5 choices and 3 sides, the answer is 8.[/hide]

Sorry Tare, nope.
Try again
try listing out the possibilites, and u'll see ur flaw

Oh I see, a+b>c, I forgot. So the answer's [hide]3[/hide], si?

meh... list 'em is easiest...

[hide]

2 3 4

3 4 5

2 4 5

So 3.[/hide]

Am I right?  Am I right?  Eh, I'll just see what Tare got in a moment...


Solve the equation:

$ \minus{}3|x\minus{}|x\plus{}2\parallel{} \equal{} \minus{}9(x\plus{}4)$

[hide]$ |x \minus{} |x \plus{} 2|\,| \equal{} 3(x \plus{} 4)$

The RHS must be non-negative, hence $ x\geqslant \minus{} 4$.

The critical point of $ |x \plus{} 2|$ is $ x \equal{} \minus{} 2$, so we split the argument into two cases.

[b]Case 1.[/b] $ \minus{} 4\leqslant x < \minus{} 2$.

Then $ |x \plus{} 2| \equal{} \minus{} x \minus{} 2$, so the equation becomes

$ |2x \plus{} 2| \equal{} 3x \plus{} 12\iff 2|x\plus{}1|\equal{}3x\plus{}12$. Since every $ x$ in the observed interval is smaller than $ \minus{} 1$, we get $ |x \plus{} 1| \equal{} \minus{} x \minus{} 1$, hence the equation becomes $ \minus{} 2x \minus{} 2 \equal{} 3x \plus{} 12\iff x \equal{} \minus{} {14\over 5}$. As that value lies within the boundaries, we accept it as a solution.

[b]Case 2.[/b] $ x\geqslant \minus{} 2$

Then $ |x \plus{} 2| \equal{} x \plus{} 2$, so the equation becomes

$ | \minus{} 2| \equal{} 3x \plus{} 12\iff 3x \plus{} 10 \equal{} 0\iff x \equal{} \minus{} {10\over 3}$. However, that value is smaller than $ \minus{} 2$, so we don't accept it as a solution.

[b]Conclusion.[/b] The only solution to the given equation is $ x \equal{} \minus{} {14\over 5}$[/hide]


Find infinitely many pairs of narutal numbers$(a,b)$ that $a,b$ shares the same prime factors and also $a+1,b+1$ shares the same prime factors .

Well... one possibility is $a=6k+3, b=6l+3$... or more generally, $a=m(2k+1),b=m(2l+1)$ where $m$ is odd (though it's possible I misunderstood the problem).

They are supposed to share all prime factors, not just the factor $3$ or $2$.

Yes, this was found by Erdös, who asked if is there others than those above.
Makowski found $m=3\cdot 5^{2}$ and $n=3^{5}\cdot 5$. I don't know if there are general patterns.

Pierre.


inequalities
trigonometry
geometry
analytic geometry
function
Pythagorean Theorem
similar triangles

MOVED FROM INEQUALITIES POST:

While trying to make inequalities, I came across a rather funny proof of sin(x) :le: 1...

By Cauchy-Schwarz

[tex] \displaystyle \\ \Bigg(\sin^2(\frac{x}{2}) + \cos^2(\frac{x}{2})\Bigg)\Bigg(\cos^2(\frac{x}{2}) + \sin^2(\frac{x}{2})\Bigg) \ge 2\sin(\frac{x}{2})\cos({\frac{x}{2}). [/tex]

And therefore

[tex] \displaystyle\\ (1)(1) \ge 2\sin(\frac{x}{2})\cos({\frac{x}{2}). [/tex]

[tex] \displaystyle\\ 1 \ge \sin(x) [/tex]

:-p I am The Win.

Since you're already relying on the fact that sin2 x + cos2 x = 1, I think that |sin x| :le: 1 follows directly from there by the trivial inequality.

The harder thing to prove is that  sin2 x + cos2 x = 1.

[quote="Osiris"]Wouldn't proving sin2(x) + cos2(x) = 1 be the same as proving the Pythagoras Theorem?

I have a proof of Pythagoras that does not involve areas of squares, which I think to be quite elegant. Available upon request.[/quote]

Could it possibly have to do with Baramgupta (I can't spell, but it's a person from India)?

depends on ur definition of sin and cos...  i was taught that (cos(x),sin(x)) is a point on the unit circle (but it still boils down to the definition of a circle as a set of points equildistant to center blah blah)

You can easily prove the Pythagorean Theorem by drawing an altitude to the hypotenuse and then using similar triangles.

Label the legs a and b and the hypotenuse c. The altitude divides c into x and y, where x is adjacent to a and y is adjacent to b. Using similar triangles, a/x=(x+y)/a and therefore a^2 = x^2 + xy. Similarily, b/y=(x+y)/b and thus b^2=y^2+xy. Thus a^2+b^2=x^2+2xy+y^2=(x+y)^2=c^2. Q.E.D.

Right. It depends on your definition of cosine and sine. Initially we all learn them as the x and y coordinates of the points on a unit circle; but these "geometric" definitions are not so great. For one thing, they make it impossible to actually compute values of sine and cosine. They also don't allow us to extend the functions to complex values.

A better definition for cosine and sine is through power series (or equivalently as the real and imaginary parts of eix). Then it's a little more work to show that sin2 x + cos2 x = 1.

[quote="Osiris"]Proof. Suppose that triangle ABC is right, with AB > AC > BC. Construct circle with centre at B and radius BC. Extend AB to intersect the circle at another point D. Then, by Power of a Point,

(c - r) (c + r) = a2
c2 - r2 = a2
c2 = a2 + r2

Since r = b, we arrive at the required theorem.[/quote]

How is Power of a Point proved?

I'm not really good with trig or anything...I just came to advance forum for no reason at all...

but I'm wondering something

[tex] \displaystyle \\ \Bigg(\sin^2(\frac{x}{2}) + \cos^2(\frac{x}{2})}[/tex]

Shouldn't that not equal to 1 because...

[tex]sin^2x + cos^2x =1[/tex]

The identity applies for all x, so that means dividing x by 2, we still have a valid angle for the identity to hold.

[quote="kool_dudy"]So would this mean this would even work if x was divided by any real number?[/quote]

Yes. You can put anything you want in the parentheses, as long as it's the same thing in both parentheses, and it will still be true. You can put in 4 or x/pi or a^2-4 or 123.456 if you wanted and it will still be true.


calculus derivative algebra function domain Support integration

Some tricky questions: If we are working with H = L^2(R) and we consider the differentiation operator Af = df/dx what is the adjoint and the domain of the adjoint of A in each case below:

a) Domain(A) = C^(infinity) function with compact support on R.
b) same as above AND g(0)=0.
c) g is now a distribution and g and g'

now H is L^2( (0, inf)) and we ask the same question if the domain is

d) same as a) but now the support is in (0, infinity)
e) g are C^(infinity) and we have both g and g' in L^2((0, infinity))

MY guess for a): integration by parts gives A* as (-d/dx) but then you need integration by parts to work so...how do I know how to choose the domain and how do you know you are picking the smallest? Is it unique?

Lastly: can a compact operator satisfy some sort of relation P(A) = 0 where P(x) is a polynomial? My guess is yes provided the unit ball of the range is finite dimensional. If you look at T^2 = T this can happened if T if of finite rank. So for example T(x) = <x,e>e works where here T is acting on a Hilbert space. Is there anything more we can say?

What could go wrong when you integrate by parts? Functions might not have a nice derivative, and boundary terms might not be zero. Now, what can we say about existence of derivatives and limits at $ \infty$ for $ C^{\infty}$ functions with compact support?


[quote="Slippy"]I can't seem to figure out what the simplified form of sqroot(x+8)=10sqroot(2)[/quote]
Simplified form?  Do you mean that you want to find the value of x?
[hide="Solution"] We have that $ \sqrt{x\plus{}8}\equal{}10\sqrt{2}$.  So, $ (\sqrt{x\plus{}8})^2\equal{}(10\sqrt{2})^2$.  Thus, $ x\plus{}8\equal{}200$.  Therefore, we can say that $ x\equal{}\boxed{192}$. [/hide]


Wish I could help you, there may be a "Set options to default" option somewhere, but I don't know for sure.  My help file is missing, for some reason, so you might want to look up "margins" with the Search tab under the Help window.  Happy hunting.

I have Word 2000, and for me, it's under the File -> Page Setup dialog box.  Bottom left corner, there's a button that says default, and it will change the current settings to the default settings.  You can do the same thing with fonts.

Silly me.  I missed that button entirely...


induction
combinatorics unsolved
combinatorics

Given positive integer $n > 2004$. Numbers 1, 2, 3, …, $n^2$ were filled in the squares of an $n\times n$ checkerboard such that every square had just one number. Square $A$ was called "good" if

[b](i)[/b] The number filled in $A$ is larger than at least 2004 numbers in the squares at the same row of $A$.
[b](ii)[/b] The number written in $A$ is larger than at least 2004 numbers in the squares at the same column of $A$.

Find the greatest number of "good" squares in $n\times n$ checkerboard.

A square is good if i and ii are simultaneously fullfilled or just one of them at least?

I believe the answer is $n(n-2004)$.

This is a maximum, because only $n-2004$ of each column can fulfil the condition, and there are n columns.

This can also be constructed by taking: "first $n-2004$ squares of each column, then stagger by one for each column after that (ie for the second column, first $n-2003$ less the first square)" and fill them with the biggest numbers.

Hopefully I haven't misread or anything because it seems too easy.

Isn't it the same as Vietnam 3 in project 2002?

Pierre.

Bah, just a question of point of view... :D

Pierre.

I don't know what you mean, bet vietnamese problem asks to find the smallest number of good squares, while the current problem asks to find the greatest number of good squares.
In my opinion, these are distinct problems.

I think we can use induction on n as the offical solution of the vietnamese problem


geometry
3D geometry
sphere
AMC
AIME
ratio
percent

AIME15 has a sphere of cheese of radius 1. Around it is an infinitesimally thin layer of wax. buzzer11 wants some cheese, so AIME takes a circle cookie cutter of radius 1/4 and cuts exactly down the center of his cheese sphere. He gives the cut piece to buzzer11 and takes the outer piece for himself.

a) Find the ratio of the volume of AIME15's cheese to buzzer11's cheese.
b) What percent of the wax layer did AIME15 give away?

You're on the right track. Maybe a little more specific.


topology
geometry
3D geometry
sphere
vector
group theory
abstract algebra

let $ M$ be manifold which is compact,  no boundary

Find all $ M$ such that every continuous map $ f : M\to M$ has a fixed point ?

I am getting stuck on this problem,  :(  any idea is highly appreciated .  Thanks

Not a complete answer, but here's some spaces with that property: $ \mathbb{R}P_n$ for even $ n$, by the Hairy Ball theorem.

Maps with no fixed points exist in: any Lie group, any sphere, any product $ f\times g$ if $ g$ has a map with no fixed point, ...

jmerry

I know the sphere ([url]http://en.wikipedia.org/wiki/Hairy_ball_theorem[/url]) is one of them but I have difficulty to know why $ \mathbb{R}P^n$ for $ n$ even is one of them, could you give me more details? thanks

No. The sphere is not one of them- mapping each point to its antipode is a counterexample.

Now, what about those projective spaces? $ \mathbb{R}P^n$ is $ S^n/\{\pm 1\}$, so if we have a map $ f$ from $ \mathbb{R}P^n$ to itself, it extends (by undoing the quotient) to a map $ g$ from $ S^n$ to $ \mathbb{R}P^n$ that sends antipodes to the same thing, and then lifts to a map $ h$ from $ S^n$ to itself such that either $ h(x)\equal{}\minus{}h(\minus{}x)$ or $ h(x)\equal{}h(\minus{}x)$ everywhere. Project the vector $ h(x)\minus{}x$ onto the tangent plane to the sphere at $ x$; this forms a vector field, and it's zero at $ x$ iff either $ h(x)\equal{}x$ or $ h(x)\equal{}\minus{}x$. By the Hairy Ball theorem, it must be zero somewhere if $ n$ is even. That zero quotients back to a fixed point of $ f$ on the projective space.


If AB be divided at O in such a manner that l.AO= m.OB, and if P be any point, Prove that l.AP2 + m.BP2 = (l+m)OP2 + l.AO2 + mBO2.

If a,b,c be length of sides of triangle ABC then find the locus of P such that a.PA2 + b.PB2 + c.PC2 is a constant. Where l,m are constants and A,B are fixed points and a,b,c are the lengths of the sides of the triangle.

[quote="kalyanram_a"]If AB be divided at O in such a manner that l.AO= m.OB, and if P be any point, Prove that l.AP2 + m.BP2 = (l+m)OP2 + l.AO2 + mBO2.[/quote]

This is known as the [b]Stewart Theorem[/b].

[quote="kalyanram_a"]If a,b,c be length of sides of triangle ABC then find the locus of P such that a.PA2 + b.PB2 + c.PC2 is a constant. Where l,m are constants and A,B are fixed points and a,b,c are the lengths of the sides of the triangle.[/quote]

The locus is a circle with center at the incenter of triangle ABC.
This is a special case of Proposition 2 in

[url]http://forumgeom.fau.edu/FG2001volume1/FG200109.pdf[/url]

  Sincerely,
  Darij Grinberg