Find the distance between a point $(x_{1},y_{1})$ and the line $Ax+By+C=0$.

Vector solution? :ninja:

do u want a vector solution or a normal solution..because thats not to complicated..

let $(x_{0},y_{0})$ be a point on the line

note that $(A,B)\cdot (x-\frac{C}{2A},y-\frac{C}{2B})=0$ is the line, so the vector $(A,B)$ is perpendicular to the line

now we have that $(A,B)\cdot (x_{1}-x_{0},y_{1},y_{0})=|(A,B)|* |(x_{1}-x_{2},y_{1}-y_{2})|\cos\theta$

looking at the right triangle, we see that $|(x_{1}-x_{0},y_{1}-y_{0})|\cos\theta$ is our desired distance, isolating that, we get

$\frac{|Ax_{1}-Ax_{0}+By_{1}-By_{0}|}{\sqrt{A^{2}+B^{2}}}=\frac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$

because $(x_{0},y_{0})$ is on the line, it satisfies $-Ax_{0}-By_{0}=C$

You can generalize the technique to find the distance between a point and a plane.

And with linear algebra you can extend this to the case where your finding the distance between a point in $R^{n}$ and a subspace in $R^{n}$ using the orthogonal decomposition theorem.
 :thumbup:

Yeah, that is right for a line.

A plane is just

for plane $Ax+By+Cz+D=0$, and the point $(x_{0},y_{0},z_{0})$

The answer is $\frac{|Ax_{0}+By_{0}+Cz_{0}+D|}{\sqrt{A^{2}+B^{2}+C^{2}}}$.

[quote="Altheman"][quote="SM4RT"]So for a point in fourth dimension:
$Aw+Bx+Cy+Dz+E=0$, and the point $(w_{0},x_{0},y_{0},z_{0})$

The answer is $\frac{|Aw_{0}+Bx_{0}+Cy_{0}+Dz_{0}+E|}{\sqrt{A^{2}+B^{2}+C^{2}+D^{2}}}$.[/quote]

the algebra is the same, but it looses meaning...what does $Aw+Bx+Cy+Dz+E=0$ describe precisely? (other than analytically, set of points perpendicular to (A,B,C,D)

a certain type of space?[/quote]

true. but maybe you can get some type of artificial geometric interpretation if you're doing something in physics.


1) Prove that $p$ divides the numerator of $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{p-1}$ in simplest form

2) Prove that $p^{2}$ divides the numerator of $1+\frac{1}{2}+...+\frac{1}{p-1}$ in simplest form.

[hide] for 1), this equals $(1+2+...+p-1)$ in $Z[p]$. For 2)...[/hide]

You mean for $p$ to be an odd prime, right?

[hide="1"]Since $p$ was obviously never canceled, we may as well consider the unsimplified numerator. Note that $\frac{1}{k}+\frac{1}{p-k}=\frac{p}{k(p-k)}$. We may add the terms in pairs like these. There are clearly a even number of terms, since p is odd, and each contributes a multiple of $p$ to the numerator, hence the numerator is divisible by p.[/hide]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147369]this thread[/url] for the second problem.

Can somebody tell me what does this mean?
 for 1), this equals $(1+2+...+p-1)$ in $Z[p]$. For 2)...

I supposed $p$ is a prime greater than 2 for 1) and greater than 3 for 2)

$P=1+\frac{1}{2}+...+\frac{1}{p-1}=\sum_{k=1}^{p}\frac{1}{k}=\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k}+\frac{1}{p-k}}$
$=\sum_{k=1}^{\frac{p-1}{2}}{\frac{p-k+k}{k(p-k)}}=\sum_{k=1}^{\frac{p-1}{2}}{\frac{p}{k(p-k)}}=p\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k(p-k)}}$

note that no factor on the denominator of this sum will be greater than $p-1$,
and as $p$ is prime "it won't cancel" then $p$ divides the numerator of $P$. 1) is proved.

For 2) we only need to prove $p$ divides the numerator of $\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k(p-k)}}$
Let's take $\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k(p-k)}}$ in mod p(proving that this is valid is an easy exercise)
$\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k(p-k)}}=\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k(-k)}}=-\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k(k)}}=-\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k^{2}}}$  **
 notice that in mod p  we have ${\frac{p+1}{2}}$(counting cero) distinct cuadratic residues
and since $k<{\frac{p-1}{2}}$ we have in ** all cuadratic residues so we can rewrite it as follows:
$-\sum_{k=1}^{\frac{p-1}{2}}{\frac{1}{k^{2}}}=-\sum_{k=1}^{\frac{p-1}{2}}{k^{2}}$
$=-\frac{(\frac{p-1}{2})(\frac{p-1}{2}+1)(2(\frac{p-1}{2})+1)}{6}=-\frac{(\frac{p-1}{2})(\frac{p-1}{2}+1)(p)}{6}$
wich is divisble by $p$, so $p^{2}$ divides $P$

and we're done.  :)  :lol:  :!:
(jiji for when I finished writing this two solutions were already posted jeje)
________________________________________________________________

Thanks Phelpedo :) Do you know Chuck Norris has counted to infinite twice?

[hide="@ElChapin"]

For 1), I considered the sum in the field $Z[p]$, aka mod p. Then it's just a permutation of $1+2+...(p-1)$[/hide]


geometry
inequalities
inequalities solved

Let a,b,c, S be sides and area of ABC, m,n,p are positive number satisfying X= 2(mn+np+pm)-m^2-n^2-p^2 >0.
       Prove that
    \sum m*a^2 >=4 \sqrt X *S + \sum m*(b-c)^2. :D

This is definitely nice, but it's the first and the last trigonometric inequality I will start. Let us write it as  \sum m(a^2-b^2-c^2+2bc) \geq 4S \sqrt X. This can also be written  \sum mbc sin 2 A/2 \geq S \sqrt X, or  \sum m* tgA/2 \geq  \sqrt X, or  \sum (m/cosA/2)^2 \geq 2 \sum m/cosA/2*n/cosB/2*sinC/2. And now we put m/cosA/2=x and so on and use  \sum xycosA' \leq 1/2* \sum x 2 in the triangle with angles ( \pi -A)/2,....

That 's exactly my direction. Congratulations! :)  :)  :)


search
number theory proposed
number theory

The integers $a$ and $b$ have the property that for every nonnegative integer $n$ the number of $2^na+b$ is the square of an integer. Show that $a = 0$.

Masoud Zargar

Simple search for "2^na+b" brought up two (the second one from the spanish community):
http://www.mathlinks.ro/Forum/viewtopic.php?t=23626
http://www.mathlinks.ro/Forum/viewtopic.php?t=61390


function MATHCOUNTS algebra polynomial AMC AIME \/closed

We plan to schedule a Math Jam for 11:00 PM ET (8:00 PM PT) on Friday, June 10.

This Math Jam will take place at a late hour due to the unique opportunity that we have to demonstrate the Math Jam to an audience of mostly mathematics, science, and technology professionals.

The topic of the Math Jam will be a basic level of generating functions.  The Math Jam should be accessible to strong MATHCOUNTS/middle school students and above.

We want to make sure that there will be enough students (at least as many as in a typical AoPS class) in order to run the Math Jam.  Please post below if you would like to participate.  If you do not post, you can still participate.

SUre, I'd love to participate.  But would there be a lot of pressure or something.  Would professional mathematicians be watching, and if none of us figured out the problem would they think less of your site?  I'm shaking already...

Also, how long would it run?  11:00 isn't that late, of course.  but if it ran until 2, for example...

I don't think I would be able to make it unfortunately.

I'm not sure, but the impression I get is that the real objective of the math jam is to show the classroom environment and how the students interact with the instructors.  Just like in a regular class, if things start to lag, the instructor can push the conversation along by giving hints and such.  Basically, as long as a good number of people show up, I don't think you'd have to worry about everyone being dumbfounded by the problems.

As for the time, it will probably go for about an hour maybe a little longer.  But you can always leave at any time...it's not like you are committed to stay until the very end.

[quote="bubala"]SUre, I'd love to participate.  But would there be a lot of pressure or something.  Would professional mathematicians be watching, and if none of us figured out the problem would they think less of your site?  I'm shaking already...[/quote]You haven't participated to any other MathJams before, have you? :)

It's not at all a pressure situation.  These people aren't scouting for college or anything like that.  It's a group of technologically minded people who discuss all kinds of technological issues.  I've been to talks with the group ranging from AI to biotech.  They are friends of the education world.  They simply want a glimpse at the potential of technology for education.

I can participate. I live in California, so it's only at 8:00 for me  :) .

can we come late, as usual? then i can make it around 9:30 PT.

[quote="xxreddevilzxx"]basic level of generating function: from strong MATHCOUNTS/Middle school students and on. I hope that helped. (It was on the first post of the thread)[/quote]

Yeah, I know. I read that. But that only says where the level begins at--it doesn't say how challenging it gets.  :P

I think that most of the problems could be followed by a good MATHCOUNTS student if they remain sharp throughout the Math Jam and don't get hung up on the initial idea of using polynomials to help count things.

The hardest problem will be no harder than a medium AIME problem so long as the basic concept is understood.

I would probably come.  Generating functions sound interesting.  Are the other "technology minded people" going to be participating or just watching?

[quote="themonster"] Are the other "technology minded people" going to be participating or just watching?[/quote]

Watching and discussing.  The truth is that they will probably pay less attention to the class as to the fact that the subject matter is unusual and advanced and taught through the internet.  Essentially, what matters is that they see that teaching this way is an educational advancement for students who will one day be the ones generating new technology.

Otherwise, it's just a regular Math Jam -- an opportunity to learn along with intellectual peers.


First 4 numbers are:
0 1 -1 -5

a.3
b.0
c.-3
d.-5

and explain why you choose????????/ (a/b/c/d)Thankss

all options are correct.There are infinite answers
without specifying any property of the sequence, any option may fit into the sequence.
for example, considering the sequence to be $t_{n+1} = t_n + 2.(-1)^{[\frac{n}{2}] +[\frac{n-1}{2}] +[\frac{n-2}{2}] +...}$ the next term comes out to be 3.
here $[x]$ refers to greatest integer $\leq x$

The trick is to consider differences of each consecutive pair and try to form a sequence with them or again see the consecutive differences until only two terms are left as two terms are always in AP or GP hence by recurrence of $a_{n+1} - a_n$ we can find a different sequence (maybe complex...).

Terrible problem.  There isn't a particular reason to pick one type of sequence over another.  I could propose the following solutions:

1.  The first differences, $1, -2, -4$, continue $1, -2, -4, 8, -16, -32, 64, ...$, making the next number $\boxed{3}$.

2.  The sequence is a third-degree polynomial.  The next number is $\boxed{-10}$.

3.  The sequence is a polynomial of degree $\ge 4$.  The next number is $\boxed{arbitrary}$.

4.  The sequence begins with $0$, then a pentagonal number, then the negative, then the negative of the next pentagonal number, then the original positive one, and so forth.  Then the next term is $\boxed{5}$.

Do you see what I'm getting at?  Where is this problem from?

hey!!! t0rajir0u
i was only a minute faster than you!!!
well have qritten the same thing as i have!!


Geometry. The volume of the box is represented by (x^2 + 5x + 6)(x + 5). Find
the polynomial that represents the area of the bottom of the box. (The box had x + 2 on the side of it.)

[hide]The height of the box is $x+2$.

The polynomial $\left(x^{2}+5x+6\right)\left(x+5\right)$ can be factored into $(x+2)(x+3)(x+5)$.

The area of a base is $(x+3)(x+5)=\boxed{x^{2}+8x+15}$.[/hide]

[quote="jamesstine"]Can 4x^2-9 be factored?[/quote]

Yea as a general rule, a difference of two squares is in the form of $ax^{2}-b$ if and ONLY if $a$ and $b$ are perfect squares.


If $ a,b,c,d \in \mathbb{Z}$, then $ (a\minus{}b)(a\minus{}c)(a\minus{}d)(b\minus{}c)(b\minus{}d)(c\minus{}d)$ is divisible by $ 12$.

[quote="moldovan"]If $ a,b,c,d \in \mathbb{Z}$, then $ (a \minus{} b)(a \minus{} c)(a \minus{} d)(b \minus{} c)(b \minus{} d)(c \minus{} d)$ is divisible by $ 12$.[/quote]

[hide="Hint"]
[color=blue]
Consider residue class of $ \ 3,\ \{0,\ 1,\ 2\}$ and residue class of $ \ 2,\ \{0,\ 1\}$.
[/color]
[/hide]

[hide="Complete Solution"]
[color=blue]
Atleast two of the numbers $ \ a,\ b,\ c,\ d$ belong to same residue class of $ \ 3$,their difference is divisible by $ \ 3\ .$ Since we have multiplied all possible differences of the numbers $ \ a,\ b,\ c,\ d$ , product $ \ (a \minus{} b)(a \minus{} c)(a \minus{} d)(b \minus{} c)(b \minus{} d)(c \minus{} d)$ is divisible by $ \ 3$.

Now Consider following cases with respect to residue class of $ \ 2$:

Case 1 : Out of the numbers $ \ a,\ b,\ c,\ d$, two numbers are even and two are odd, their respective difference is divisible by $ \ 2$, product $ \ (a \minus{} b)(a \minus{} c)(a \minus{} d)(b \minus{} c)(b \minus{} d)(c \minus{} d)$ is divisible by $ \ 4$.

Case 2 : Out of the numbers $ \ a,\ b,\ c,\ d$, one numbers belong to one residue class of $ \ 2$ and three belong to other residue class of $ \ 2$. Consider these three numbers of same residue class, difference of any two of them is divisible by $ \ 2$. Since in the product they will give three terms in between them, product $ \ (a \minus{} b)(a \minus{} c)(a \minus{} d)(b \minus{} c)(b \minus{} d)(c \minus{} d)$ is divisible by $ \ 8$.

$ \implies$ In any case product is divisible by atleast $ \ 4$.

Hence $ \ (a \minus{} b)(a \minus{} c)(a \minus{} d)(b \minus{} c)(b \minus{} d)(c \minus{} d)$ is divisible by $ \ 3\times 4\equal{}12$.
[/color]
[/hide]


Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $ 7$ songs, which was more than any other girl, and Mary sang $ 4$ songs, which was fewer than any other girl. How many songs did these trios sing?

$ \textbf{(A)}\ 7 \qquad
\textbf{(B)}\ 8 \qquad
\textbf{(C)}\ 9 \qquad
\textbf{(D)}\ 10 \qquad
\textbf{(E)}\ 11$

[hide="Click for solution"]
Tiny and Alina each sang either $ 5$ or $ 6$ times.  If $ N$ denotes the number of songs sung by trios, then $ 3N\equal{}4\plus{}5\plus{}5\plus{}7\equal{}21$ or $ 3N\equal{}4\plus{}5\plus{}6\plus{}7\equal{}22$ or $ 3N\equal{}4\plus{}6\plus{}6\plus{}7\equal{}23$.  Since the girls sans as trios, the total must be a multiple of $ 3$, so only $ 21$ works.  Thus, $ N\equal{}\frac{21}{3}\equal{}7$, which is the number of songs the trios sang, or $ \boxed{\textbf{A}}$.
[/hide]

Challenge: Show that this can be attained.

Can be attained:
HTM,HTM,HAM,HAM,HTA,HTA,HTA


modular arithmetic
number theory
prime factorization

For how many integers $ n$ is $ \frac{n}{100\minus{}n}$ also an integer?

If $ \frac{n}{100\minus{}n}\equal{}k$, then $ 100k\minus{}kn\equal{}n$, or $ kn\minus{}100k\plus{}n\equal{}0$, or $ (k\plus{}1)(n\minus{}100)\equal{}\minus{}100$.  Thus $ n\minus{}100$ ranges over the divisors of $ \minus{}100$, of which $ 9$ are positive and the same amount are negative.  Thus, there are $ 18$ solutions for $ n$.

[hide="Another Method"]
This is equivalent to finding all n such that:

$ 100 \minus{} n \equiv 0 \pmod n \implies 100 \equiv n \pmod n \implies 100 \equiv 0 \pmod n \implies n|100$

Therefore, we simply must find all factors, negative or positive, of 100.

Since there are 9 positive factors (because its prime factorization is $ 2^25^2$), there are 9 negative factors. This gives an answer of $ \boxed{18}$.
[/hide]


Hi. I have some problem. It's about functional Equation.
Please help me.  :D

Let f be function so that
(i) f(0) is not equal 0
(ii) f(xy) = f(x)f(y) for all real x , y

what is the value of f(2548). ?  :blush:

:rotfl:  :rotfl:  :rotfl:

it is so easy:
let $x=0$
then:
$f(0)=f(0)f(y)$
$f(0)(f(y)-1)=0$
$f(y)=1$

Yo. Thanks.  :D

But Do you think it has only one  solution.?

Can $f(2548) = 2548$. ?

[quote="kueh"]No! because $f(2548) - 1 = 0$![/quote]

How do you know that $f(2548) = 2$ ?

Please tell me.  :blush:


[b]Thank for your's reply. But I hope you don't give me a similar exercise, please give me exact solved of my prolems. Because I am not much intelligent.[/b]

[color=blue]Prolem 1:[/color]
Let ABCD is a quadrangle, AC and BD meet at O. Let I, J be the midpoints of AB and CD. Let H and K be the orthocentrics of OAD triangle and OBC triangle. Prove that the line IJ is perpendicular to the line HK.

[color=blue]Prolem 2:[/color]
Let ABC be a triangle inscribed in (O; R) circle and let (I; r) be a circle inscribed in ABC triangle. Prove that: $ OI^{2}\equal{} R^{2}\minus{}2Rr.$

[color=blue]Prolem 3:[/color]
Let ABC is a triangle and let point O. Three lines through O and perpendicular to the lines OA, OB, OC sequence meet BC, CA, AB at M, N, P.
Prove that these point M, N, P are in a straight line.

[b]---------------------------------------------------------------------
Thank for reply - You are a good friend - I wish you good luck.[/b]

Because I don't know how to use vector symbols.So I only give you my hint to these problem.
You are Vietnamese ,if now you are in grade 10 (or prepare).I think these folowing hints are useful for you
Pr1. Please prove that vector$ IJ$ $ \times$ vector$ HK$ $ \equal{}0$
by   2.vector$ IJ$ =vector$ AD$ +vector$ BC$  and vector$ HK$=vector$ HO$ +vector$ OK$
Pr2. Note that $ I \equal{}[\frac{A}{a},\frac{B}{b},\frac{C}{c}]$
 $ (a\plus{}b\plus{}c)$.vector$ OI$ =$ a$vector$ OA$ +$ b$vector$ OB$ + $ c$vector$ OC$
Pr3.There is a very nice solution by vector method.But I don't remember.
 suppose that $ O\equal{}[\frac{A}{x},\frac{B}{y},\frac{C}{z}]$ ... :!:

Why do you repost the problems? I didn't give you "similar exercises", but [b]exactly the same exercises[/b], partly with different notations. And the respective threads contained exact solutions of the problems. Please ask in these threads (to which I linked) what you don't understand about the solutions, rather than reposting the problems.

  Darij


probability
combinatorics unsolved
combinatorics

 hi i have this math problem that i have. i did the entire problem but i got it incorrect can you help me as to where i went wrong. i even have the correct answer but i dont know how to get it. thanks

Problem
use a sum or diffenrece identity to find an exact value:
Problem   tan 5 pie/12
the identity that you are to use is tan(u +v)=sinu cosv+cosu sinv
                                                                -----------------------
                                                                 cousu cosv -sinu sinv
or the identity tan(u-v) =sinu cosv -cous u sinv
                                   --------------------------
                                    cos u cos v=sinu sinv

my attempt:
i multiplyed the 5 pie/12 with 180/pie to convert it to degrees as it is easier for me to work with degrees
that answer equlled 75 degrees
then i did tank(45 degrees+30 degrees) so that the 45 equalls u and 30 equalls v and i pluged it into the equation

tan 45+ tan 30
______________
1-tan 45 tan 30

1+1/ square root of 3
________________
1- 1X 1/the square root of 3

simplyfying to 1+ quare root 3/3
                     ________________
                    1- square root 3/3
=1
 but the answer in the text book is 2 + the square root of 3

Where did i go wrong.....
thanks

This is not at all the right place for your question.  Regardless, the issue is in the last step.  $ \frac{\sqrt{3} \plus{} 1}{\sqrt{3} \minus{} 1}$ is clearly not equal to $ 1$; in fact, it's equal to exactly what the textbook claims.

99% probability that this is the same poster as chocolatelover, who repeatedly posts homework problems in completely incorrect forums.  I recommend not responding until behavior improves.

I AM [b]NOT[/b] chocolatelover. i am sorry i dont know where to post this problem. this is the first time i have used this system and the first post i have made. would you be so kind as to tell me where would be the correct location to post this.


geometry
3D geometry
modular arithmetic
number theory proposed
number theory

[quote="Tiks"][quote="pluricomplex"][quote="Fermat -Euler"][size=150]Find all $5$-digit numbers which equal the cube of the sum of their digits[/size][/quote]

huh sounds simple like as : $\overline{a_1a_2a_3a_4a_5}=\sum_{k=1}^5a_k^3$ with $a_1\neq0$. Fact we have $a_1\cdot 10^4\leq RH=LH\leq 5\times9^3$ and i think it's wellknown $5\times9^3<10^4$ :D so $a_1=0$ , contract ![/quote]
$\overline{a_1a_2a_3a_4a_5}=\sum_{k=1}^5a_k^3$ or   $\overline{a_1a_2a_3a_4a_5}=(\sum_{k=1}^5a_k)^3$??????????????  :?: :?[/quote]

ah thi's my bad English:
If that put $a=a_1+\cdots+a_5$ then $a^3\equiv a\pmod 9$ then $9\mid a$ or $9\mid a^2-1$ then so $9\mid a-1$ or $9\mid a$ or $9\mid a+1$. Use some simple calculating with note that $1\leq a\leq45$ we will get all results.

[quote="pluricomplex"][/quote][quote="Tiks"][quote="pluricomplex"][quote="Fermat -Euler"][size=150]Find all $5$-digit numbers which equal the cube of the sum of their digits[/size][/quote]

huh sounds simple like as : $\overline{a_1a_2a_3a_4a_5}=\sum_{k=1}^5a_k^3$ with $a_1\neq0$. Fact we have $a_1\cdot 10^4\leq RH=LH\leq 5\times9^3$ and i think it's wellknown $5\times9^3<10^4$ :D so $a_1=0$ , contract ![/quote]
$\overline{a_1a_2a_3a_4a_5}=\sum_{k=1}^5a_k^3$ or   $\overline{a_1a_2a_3a_4a_5}=(\sum_{k=1}^5a_k)^3$??????????????  :?: :?[/quote][quote="pluricomplex"]

ah thi's my bad English:
If that put $a=a_1+\cdots+a_5$ then $a^3\equiv a\pmod 9$ then $9\mid a$ or $9\mid a^2-1$ then so $9\mid a-1$ or $9\mid a$ or $9\mid a+1$. Use some simple calculating with note that $1\leq a\leq45$ we will get all results.[/quote]
pluricomplex I don't think that this is corectly solution,becous we have very many variants :roll:

what ? it's ugly but not fail ! For exm. $a=9,18,27,36,45$ and $a+1,a-1$ ... about $5\times 3=15$ cases at most :D  .
$a=9$ then $\overline{a_1\cdots a_5}=a^3=9^3=729$ , cotract!

..... after 14 these nice works uhu? :D


On a cube, what is the sum of the number of vertices, edges, and faces?

A.    18	  B.20	C.   22	D.  24	E.   26

[hide=" click to reveal hidden content"]

$ V\plus{}E\plus{}F\equal{}8\plus{}12\plus{}6 \equal{} 26$

[/hide]

[hide]There are 8 vertices, 12 edges, and 6 faces, so the sum is 8+12+6=26 => [size=200][color=red]E.[/color][/size][/hide]


In a chess competition there were 2 student's from  7-th grade and some other students from 8-th grade.
Both the 2 students from grade 7 took 8 points. The students from grade 8 took equal number of point with each other.

THe rules:
1) THe students $ x_1$ and $ x_2$ should play only 1 game with each other.
2)Who wins take 2 point
3)In the draw each student take 1 point.

How students in grade 8 were in  the competition?

Well im not quite sure if this is right

[hide]Well since there are only 2 7th graders and each person can only pay one game against eachother that means there are 2 different possibilities 1. that they tied and they each ended up with 1 point after their match. 2. that one seventh grader one. Making the score 2points to a seventh grader and 0 points to the other seventh grader.

Even though i just explained 2 different possibilities there are (i think) infineteley many possibilities.[/hide]

this is how i think it would be solved: :maybe:
 :?: so there are two 7th graders, and they both have 8 points
if the played each other once/ one would get i and i if it was a draw assume they played 7 8th graders all getting a draw since they all have to receive an equal number of point,
so most easily every time the game was played it was a draw so there are 9 people and 9-2 7th graders 7 8th graders
but then again there other ways they could have won ,too, so to figure those out wouldn't be so easy, but this is a logical simple solution :blush:
at least i tried  :surf:


linear algebra
matrix
ratio
geometry
geometric transformation
rotation
algebra

We have a system of three masses in a row, joined by two identical springs. The two outer objects have masses of 16 (units) each, while the middle object has a mass of 12 (units).

Assume that all motion takes place in one dimension, the direction the masses are lined up in. Assume there are no frictions and resistances of any kind.

[Does this system resemble a carbon dioxide molecule? Perhaps.]

Write the equations of motion as

$ \mathbf{y}'' \equal{} KA\mathbf{y}$

where $ \mathbf{y} \equal{} \begin{bmatrix}y_1 \\
y_2 \\
y_3\end{bmatrix}$ are the net displacements of each mass from a "home" position, $ A$ is a $ 3\times 3$ matrix, and $ K$ is some constant. Adjust the constant so that $ A$ is an integer matrix.

Find the eigenvalues and eigenvectors of $ A.$ Interpret these results physically. In particular, what is the ratio between the frequency of the "asymmetrical stretch" oscillation and the frequency of the "symmetrical stretch" oscillation?

We need to consider the free-body diagram of each mass.  The left most mass experiences its $ k_{1}x_{1}$ from the first spring and $ k_{2}x_{2}$ from the right spring. The middle mass experiences $ k_{2}x_{2}$ on the left and $ k_{3}x_{3}$ on the left, and the final mass experiences $ k_{3}x_{3}$ on the left and $ k_{4}x_{4}$ on the right.  We also need to consider the external forces on each mass.

For simplicity, the two mass system is given by
$ \frac{d^{2}x_{1}}{dt^{2}}\equal{}\minus{}(k_{1}\plus{}k_{2})x_{1}\plus{}k_{2}x_{2}\plus{}F_{1}(t)$

$ \frac{d^{x}x_{2}}{dt^{2}}\equal{}k_{2}x_{1}\minus{}(k_{2}\plus{}k_{3})x_{2}\plus{}F_{2}(t)$

Assume no external forces. In other words, set $ F_1(t)\equal{}F_2(t)\equal{}0.$

However, you do not seem to have allowed for the differing masses. And there are three masses, not two. (We will get a $ 3\times 3$ matrix.)

Also: there are only two springs. There is no spring to the left of the left mass or to the right of the right mass. And the two springs in between the three masses are identical.

In my expression, I forgot to add in the $ m_{1}$ and $ m_{2}$ terms so yes, those should be there.  In your setup, we need only adjust the free-body diagrams so that the two end springs would experience fewer forces.

My intent here is this specific concrete problem. I want to see a very specific $ 3\times 3$ matrix (preferably with integer entries) and I want to see it diagonalized.

Here's the setup:

$ 16y_1''\equal{}k(\minus{}y_1\plus{}y_2)$
$ 12y_2''\equal{}k(y_1\minus{}2y_2\plus{}y_3)$
$ 16y_3''\equal{}k(y_2\minus{}y_3)$

Multiplying by the appropriate constants to get the masses up to a common multiple, we wind up with this:

$ \frac{d^2}{dt^2}\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}\equal{}\frac{k}{48} \begin{bmatrix}\minus{}3&3&0\\4&\minus{}8&4\\0&3&\minus{}3\end{bmatrix} \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}$

The eigenvalues and eigenvectors of that matrix:

$ \lambda\equal{}0,\ \begin{bmatrix}1\\1\\1\end{bmatrix}:$ Not an oscillation at all. Just uniform center of mass motion.

$ \lambda\equal{}\minus{}3,\ \begin{bmatrix}1\\0\\\minus{}1\end{bmatrix}:$ The symmetric stretch.

$ \lambda\equal{}\minus{}11,\ \begin{bmatrix}3\\\minus{}8\\3\end{bmatrix}:$ The asymmetric stretch.

Hence, the vibrational frequency associated to the asymmetric stretch is $ \sqrt{\frac{11}3}$ times the vibrational frequency associated to the symmetric stretch, at least for the most common isotopes (carbon-12 and oxygen-16).

Physical considerations:

Since the asymmetric stretch changes the dipole moment of the molecule while the symmetric stretch does not, the asymmetric stretch should be a much stronger absorber or emitter of electromagnetic radiation (in the infrared range).

By confining this to one dimension, I overlooked several other possible motions that can be seen by looking at it in three dimensions. One is rotations (with frequencies in the microwave range, serving to spread the infrared absorptions into bands rather than lines). Another is the out-of-line bending. This simple exercise has no way of knowing the relative strengths of the restoring forces for stretching versus bending, so cannot be used to predict the relative frequencies.

I do use this in a linear algebra class as a lecture example. I also don't ever compute the characteristic polynomial - I find the eigenvalues by other means.


geometry
circumcircle
trigonometry
cyclic quadrilateral
IMO Shortlist

$ ABC$ is an acute-angled triangle. $ M$ is the midpoint of $ BC$ and $ P$ is the point on $ AM$ such that $ MB \equal{} MP$. $ H$ is the foot of the perpendicular from $ P$ to $ BC$. The lines through $ H$ perpendicular to $ PB$, $ PC$ meet $ AB, AC$ respectively at $ Q, R$. Show that $ BC$ is tangent to the circle through $ Q, H, R$ at $ H$.

[i]Original Formulation: [/i]

For an acute triangle $ ABC, M$ is the midpoint of the segment $ BC, P$ is a point on the segment $ AM$ such that $ PM \equal{} BM, H$ is the foot of the perpendicular line from $ P$ to $ BC, Q$ is the point of intersection of segment $ AB$ and the line passing through $ H$ that is perpendicular to $ PB,$ and finally, $ R$ is the point of intersection of the segment $ AC$ and the line passing through $ H$ that is perpendicular to $ PC.$ Show that the circumcircle of $ QHR$ is tangent to the side $ BC$ at point $ H.$

Let $HQ,HR$ cut $BP,CP$ at $D,E$ respectively. Also let $CP, BP$ cut $AB,AC$ at $X,Y$ respectively. By Ceva, $XY//BC$, so $XP:PC=YP:PB$. As $MP=MB=MC$, so $\angle BPC=90^{\circ}$. Therefore $QD:DH=XP:PC=YP:PB=RE:EH$, and hence $QR//DE$. Since $PDHE$ is a cyclic quadrilateral, so $\angle HQR=\angle HDE = \angle HPE = 90^{\circ}-\angle PCH = \angle EHC$. Therefore $BC$ is tangent to the circle through $Q,H,R$ at $H$.

This is a problem be impli from perpendicular circle

[u][b]Notations[/b][/u]$HQ \cap BP=K,HR \cap PC=L$

From $CM=BM=PM$ it is easy to see that $\angle{BPC}=90^{\circ} \Rightarrow KPLH$ is a rectangle.Nice proof of $KL \parallel QR$ by yptsoi.I proved it using trigonometry letting $\angle{BHQ}=x$.After that it is just $\angle{BHK}=\angle{BPH}=\angle{KPH}=\angle{KLH}=\angle{QRH} \Rightarrow BC$ is tangent to $\odot{QRH}$ at $H$.


A bag contains 10 marbles numbered 1-10. A marble is removed from the bag, and without replacing it, a second marble is drawn. What is the probability that both numbers are odd? Express your answer as a common fraction.


[quote="ckck"]A trapezoid is inscribed and circumscribed about two circles. The bases of the trapezoid are 4 and 8. What is the area of the trapezoid?[/quote]
[hide]Ok, we know the trap is cyclic, so the legs are congruent. I can use my voodoo magic to tell you that the legs both have length 6. We can use ptolomey's to find that the diagonals are length $2\sqrt{17}$. Now we can use Heron's to find that the area of the triangle is $16\sqrt{2}$. The area of the other triangle is half of this since the two share an altitude but the base is half, so the area of that is $8\sqrt{2}$ so the area of the trap is $24\sqrt{2}$[/hide]


[quote="diophantient"][quote="Art of Owna"]In regular hexagon ABCDEF with side length 6, what is the value of $(FD)^{2}-(ED)(AD)$[/quote]
[hide]
$108-6 \cdot 12=36$[/hide][/quote]
ya thats right, I just notcied you could solve this by just finding every value..instaed of just doing a trick and not having to find every value.


$ K$ takes $ 30$ minutes less time than $ M$ to travel a distance of $ 30$ miles. $ K$ travels $ \frac {1}{3}$ mile per hour faster than $ M$. If $ x$ is $ K$'s rate of speed in miles per hours, then $ K$'s time for the distance is:

$ \textbf{(A)}\ \dfrac{x \plus{} \frac {1}{3}}{30} \qquad\textbf{(B)}\ \dfrac{x \minus{} \frac {1}{3}}{30} \qquad\textbf{(C)}\ \dfrac{30}{x \plus{} \frac {1}{3}} \qquad\textbf{(D)}\ \frac {30}{x} \qquad\textbf{(E)}\ \frac {x}{30}$

[hide="A bit easy?"]
Time equals distance over rate, so $ t\equal{}\frac{30}{x}$ or $ \boxed{\textbf{(D)}}$.
[/hide]

No, I do not see how it seems to be that easy. Please elaborate for us lesser bright minds. What happened to all of the other variables presented in the problem?

Filler information is extra information in a problem that you don't need to solve it.


Two different numbers are randomly selected from the set
$ S \equal{} \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. The probability that their
sum is 12 would be greater if the number $ n$ had first been
removed from set $ S$. What is the value of $ n$?

every number in the set EXCEPT FOR 6 can combine with another number in the set to make a 12. thus, the probability would be greater if 6 was removed.

answer: $ \boxed{6}$

[hide=Quick Solution]We can abuse symmetry to solve this problem.

Suppose that the answer is $x$. If $x$ was removed, then this would just have the same probability as when you remove $12-x$. This would mean that $12-x$ is also an answer to this question.

Since the problem can only have one answer, this only means that $x=12-x\implies x=\boxed6$.

[hide = Quicker Solution]12/2 = 6 it always works for these problems[/hide]

[hide=Sol]Each number has a pair: $1+11, 2+10, 3+9, 4+8, 5+7$ except for $6$, so if $6$ is $1$ of the randomly choosen numbers, we can be [b]sure[/b] that the sum will not equal $12$. Once one of the other numbers (not $6$) have been choosen there is at a $\frac{1}{10}$ probability that the other number will be $12$'s pair and therefore work. So removing $6$ would increase the probability the most.[/hide]

[hide=oh look 895 posts]
Note that $1+11=12,$ and $2+10=12 \cdots,$ all the way to $6+6=12.$ However, we cannot choose $6$ twice so removing $\boxed{6}$ will increase the probability the most.
[/hide]

[hide=Solution (to basically 2 problems???)]Well, first of all, there are two copies of this problem but just written in two ways. You just remove the number without any sets of 12.[/hide]


Prove that has infinite (a,b) such that $a^{2006}+1$ divise by b and $b^{2006}+1$ divise by a

now i will write solution: $(a,b)=(F_{2n-1},F_{2n+1})$


limit
real analysis
real analysis unsolved

some body asked me for a reason could any body help??:
we know that $\lim_{x\to\infty}\frac{x}{x+1}=1$
we also know that $\lim_{x\to\infty}[\frac{x}{x+1}]=0$
now what is the answer for $[\lim_{x\to\infty}\frac{x}{x+1}]=0 or 1$


[quote="Zenox"][hide="Solution"]To find the mean of an arithmetic sequence, add the first and the last terms and divide by two. 17+99 / 2 = 58. To increase the mean to 59, you would have to remove 17, as doing that would make the mean 19+99 / 2 , which = 59.

So, 17.[/hide][/quote]

Thanks!

[quote="7h3.D3m0n.117"]What integer should be removed from the list of odd integers from $17$ through $99$, inclusive, such that the average of the remaining odd integers is $59$?

I tried this problem then stopped because my way was too tedious and confusing...any ideas?[/quote]
[hide]
the current average is 58. (the average of an arithmetic sequence is half the sum of the first and last numbers) to increase the average by 1, remove the lowest number $17$.
[/hide]
[quote="Ubemaya"]Dumb solution:

There are 42 numbers that sum to 2436.
We want 41 numbers with an average of 59, so they sum to 2419.

So you need to remove $17$.

The problem with your solution is that it replies too much on luck, since it just happened that the first term is removed. If a middle term is removed, then you can't just add the first and last terms.[/quote]
most of the times mathcounts problems are tricky and have a "simple" way of doing it that is way easier than some random problme from nowhere. you would have to trust your luck on problems like this to do it fast

[quote="Zenox"][quote="Ubemaya"]
The problem with your solution is that it replies too much on luck, since it just happened that the first term is removed. If a middle term is removed, then you can't just add the first and last terms.[/quote]

Actually, no, it doesn't. 17 is the only possible way to get an integer value when removing 1 number from the sequence. Removing 119, which would give a mean of 57, doesn't work, as 119 isn't in the sequence. Removing 58, which would keep the mean at 58, doesn't work, as 58 is an even number and isn't in the sequence. Anything smaller than 17 wouldn't work either, as it wouldn't be in the sequence.[/quote]

Soo... I don't get it. How did you do this problem? Did you just say "let's see what happens when you remove 17..."? Because that's what it seems like you're doing...

they figured out the mean and since it is all odd number noticed that they need to take that out if it was some other odd number they were still on their way to your solution because you have to do that anyway

Although the Zenox trick is hard to miss, if you did miss it then the formal way without any guess whatsoever (or observation) is just using weighted averages:

42 numbers average 58 ... to raise average by one number to 59, drop the number corresponding to $59-42=17$

if this method seems too intuitive, here is proof:
shift number system so that 59 corresponds to zero. then we have a series which averages -1, with 42 member, so that its sum is -42. by removing a term from the series, we want to bring the average to zero, so sum of all terms is zero. Thus, $sum_{initial}=-42, sum_{final}=0\implies$ that the term to remove is $-42$, which corresponds to $17$.

Thanks... although it's still kinda confusing to me.


Given $n$ lines in plane, determinate all possible numbers of intersection points!
For $n = 3$, one can get $0$, $1$, $2$ or $3$ points (when all lines are parallel, all intesecret in one point, two are parallel, and every two have different intersection).

alexilic

If all lines are in general position i.e. no two lines are parallel and no three meet at a point then number of intersections is n[b]C[/b]2=n(n-1)/2

Hence maximum number of inter section points is n(n-1)/2

We can obtain any number of points between 0 and n(n-1)/2
(I'll try to find a proof for this)

It's not possible to have two intersection points of four lines...


inequalities
real analysis
real analysis unsolved

N(x) is norm of x ; prove that
 N(x) is smaller than Max[N(x-y) , N(x+y)]


at least i think its wrong...

http://www.artofproblemsolving.com/Wiki/index.php/1990_AIME_Problems/Problem_6

I agree; it seems the correct solution should be
\[ \frac{3}{42}\equal{}\frac{45}{\frac{3}{4}x}\Rightarrow x\equal{}840\]


calculus
integration
trigonometry
logarithms
calculus computations


so did i.
who are you all?
or mainly the first poster.
i figure charizard is dkang.

nice eli
i'm gonna guess that the guy up there is a sixth grader

why do you have like 5 aops accounts, daniel?
-_____-

I'm a mysterious person jessica. You don't know where I'll strike next.

uhhmmmmm sure. haha.
no one ever posts in the florida forum. |:

post.
The message is too small. Please make the message longer before submitting.

yeah, i noticed one of the last few posts are like from 2007... o.O

hey matthew, sorry to break this to you (if you ever see this message)
but it's been long-established that the florida aops forum is dead. it was killed by california.

[quote="mathlete5"]matthew, are you matthew kang???[/quote]

who are you?

jimmy who?
deerlake i'm guessing?

and yeah, i'm pretty sure unappropriated3 is matt kang.

the only Jimmy that would know Matthew that I know of is Deerlake Jimmy... but why would he make a thread that says
"Fairview Middle School"
"If you go to fairview, post here because I might know you!!"

[quote="resurrection"]i go to fairview. i think[/quote]

go away lyndon D:

Okay, now what's with all these various high schoolers becoming random forum spammers...*cough* resurrection and ra5249... :maybe: This happened in the IL forum as well as perhaps 500 other forums I could think of...

gauss i think you're stalking me and ra... you always manage to find our random spammy posts. lol.

and jessica. no. i dun wanna go away. anyways you belong in indiana remember =P

Like 15 people in my grade have an AoPS account


[url]http://zetamac.com/arithmetic/[/url]

Is this a great speed drill for mathcounts countdown prep?

I tried it and got a record score of 59 so far after 4 plays. is this good? What would be a typical good score to be considered good at countdown, respectively or something?

Arithmetic won't help you much in countdown; it's mostly faster thinking and tricks.

I agree with AIME15 on this. Usually(95%) the cd problem with involve some easy trick. But yes, some question do use basic arithmetic so learning how to be fast on them is useful.

argh i can do arthimetic fast but dont know any tricks.....
anyone know where i can get them...what do u people just research.....

Old Countdowns seem to be very useful, i've actually seen lots of almost exact problems repeat. So yea, FTW is a good place to find old cds.

Also, unless you think you can make National CD, I would be focusing more on sprints and targets.

[quote="not_trig"]what? mathcounts countdown is totally about fast arithmetic at least 30% of the time

it pays to be a fast calculator in MC

later on not so much[/quote]

I totally agree with this. Yes, mathcounts does involve tricks, but most of them require arithmatic anyway.

It's usually only arithmetic that requires you to know addition with numbers less than 10, or multiplication less than 10, which you should all know :wink:


For all real number $x$ consider the family $F(x)$ of all sequences $(a_{n})_{n\geq 0}$ satisfying the equation \[a_{n+1}=x-\frac{1}{a_{n}}\quad (n\geq 0).\] A positive integer $p$ is called a [i]minimal period[/i] of the family $F(x)$ if

(a) each sequence $\left(a_{n}\right)\in F(x)$ is periodic with the period $p$,

(b) for each $0<q<p$ there exists $\left(a_{n}\right)\in F(x)$ such that $q$ is not a period of $\left(a_{n}\right)$.

Prove or disprove that for each positive integer $P$ there exists a real number $x=x(P)$ such that the family $F(x)$ has the minimal period $p>P$.


inequalities
geometry
incenter
circumcircle
geometry proposed

Let $ H,I,O$ be orthocentre, incenter and circumcircle of triangle $ ABC$ , respectively . Prove that

$ OH \ge IH.\sqrt{2}$

if $ Q$ be the midpoint of $ OH$. we konw that $ Q$ is the center of nine point circle. we have $ OH^2 \plus{} 4QI^2 \equal{} 2OI^2 \plus{} 2HI^2$ and by feuerbach theorem we have $ 2QI \equal{} R \minus{} 2r$ and $ OI^2 \equal{} R^ 2\minus{} 2Rr$ and so $ (R)\geq(2r)$ so $ OH^2 \equal{} (2HI^2 \plus{} R^2 \minus{} 4r^2)\geq(2HI^2)$ :)


prove that the coefficient $a$ in
\[\left(\sum_{i=0}^{n-1}x^{2^{i}}\right)^{n}\ =\ \dots+ax^{2^{n}}+\dots \]
is equal to the number of sequences $(k_{1},\dots,k_{n})$ of nonnegative integers such that $\sum 2^{n-k_{i}}=2^{n}$.

I have tried with Newton's formula but I don't know how to sum these factorials :(

[quote="t0rajir0u"]Er... the way you have your summation written, $a = 0$.  What did you mean?[/quote]

He/She might have meant to raise the summation to some power. (That's my guess, at least, because of the parentheses.)

you're right, there was a mistake, there should be $n$-th power
now it's ok
please help me to solve this task

[hide="hint"] Find a bijection between each sequence $(k_{1}, k_{2}, \ldots, k_{n})$ and each way that we can choose $n$ terms (one from each product) to get a  $x^{2^{n}}$ term. [/hide]

n(1,1,...,1)-(k_1,k_2,...)=(a_1,a_2,...) the sum of powers that gives \sum 2^a_i=2^n

this is what your generating function gives


i have no clue how hard this is

every prime that can be expressed as 4n-1 can never be the sum of two squares and that every prime that can be expressed as 4n+1 can always be

and

prove that any integer is the sum of 4 squares

These are non trivial results, which you can find from looking on the internet.  For the last question, there is a complicated factorization which trivializes the problem.

see here:

http://mcraefamily.com/MathHelp/BasicNumberSquareSums.htm

For the case x^2 + y^2 = 4k+3, however, it is easy - it is impossible because all quadratic residues mod 4 are either 0 or 1.


inequalities
LaTeX
rearrangement inequality
inequalities unsolved

Let $a$, $b$ and $c$ be the sides of a triangle. Prove that $a^{2}b\left(a-b\right)+b^{2}c\left(b-c\right)+c^{2}a\left(c-a\right)\geq 0$.

The following more general ineq. is valid for $r \ge 2$ and $a,b,c$ length-sides of an triangle:

$a^{r}b(a-b)+b^{r}c(b-c)+c^{r}a(c-a) \ge 0$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=149630 .

[quote="probability1.01"]Isn't this just Schur, with $b \ge a-c$ and same with others?[/quote]
don't think so. the sign of the bracket cannot be confirmed

[quote="spoudyal"]Hmmm ... No one used the rearrangement inequality? That is the method I had used and I thought it was the title of the topic but now I see it is "Old IMO inequality."[/quote]
in what way are you going to use it?

More difficult is the following statement:
If $0 \le r \le 1$ and $a,b,c$ are the length-sides of an triangle, then:

$a^{2}b(a^{r}-b^{r})+b^{2}c(b^{r}-c^{r})+c^{2}a(c^{r}-a^{r}) \ge 0$.

[quote]
spoudyal wrote:
Hmmm ... No one used the rearrangement inequality? That is the method I had used and I thought it was the title of the topic but now I see it is "Old IMO inequality."

in what way are you going to use it?
[/quote]

I will post solution with or without LATEX soon, but there are a few fractions that will look weird without LATEX.


calculus
integration
ratio
arithmetic sequence

For the arithmetic sequence a_1,a_2,...,a_16 it is known that a_7+a_9=a_16. Find each subsequence of three terms that forms a geometric sequence.

I'm assuming the three terms of the subsequence must be distinct, otherwise we can have $ \{a_{1},a_{1},a_{1}\}$, etc.

[hide]We can let $ a_{1}=a$ and $ a_{k+1}-a_{k}=d$ for all positive integers $ k$. We have $ (a+6d)+(a+8d)=(a+15d)$, so $ a=d$. So our sequence is $ a_{k}=ka$ for all positive integral $ k$, and a fixed $ a$. We find that the following triples form geometric sequences, which we systematically get by considering the common ratio: $ \{a_{1},a_{2},a_{4}\},\{a_{2},a_{4},a_{8}\},\{a_{3},a_{6},a_{12}\},\{a_{4},a_{8},a_{16}\},\{a_{1},a_{3},a_{9}\},\{a_{1},a_{4},a_{16}\}$.[/hide]


A rectangular frame measuring 10'' by 15'' has five square holes for photos.  Four of the holes are each 2'' by 2'', and one hole is 3'' by 3''.  What portion of the frame's area was not cut away for the photo holes?  Express your answer as a common fraction.

[asy]filldraw((0,0)--(24,0)--(24,15)--(0,15)--cycle,gray);
filldraw((9.4,4.9)--(14.6,4.9)--(14.6,10.1)--(9.4,10.1)--cycle,white);
filldraw((3,3)--(3,6)--(6,6)--(6,3)--cycle,white);
filldraw((3,9)--(6,9)--(6,12)--(3,12)--cycle,white);
filldraw((18,3)--(21,3)--(21,6)--(18,6)--cycle,white);
filldraw((18,9)--(21,9)--(21,12)--(18,12)--cycle,white);[/asy]

3000th post.

The area of the entire rectangular frame is $ 150$.  There are four holes with area $ 4$, and one hole with area $ 9$.  This is just $ \frac{150\minus{}4(4)\minus{}9}{150} \implies \frac{150\minus{}16\minus{}9}{150} \implies \frac{150\minus{}25}{150} \implies \frac{125}{150} \implies \boxed{\frac56}$.


If $ x,y,z>0$, then

$ 2(x\plus{}y\plus{}z)\plus{}3\sqrt[3]{xyz}\minus{}4(\sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{zx})\geq0$

[quote="cnyd"]

I think $ 2(x \plus{} y \plus{} z) \plus{} 3\sqrt [3]{xyz} \minus{} 3(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx})\geq0$

because ; $ 2[6,0,0] \plus{} [2,2,2]\geq 3[3,3,0]$

if we use Schur inequality $ [6,0,0]\geq 2[4,2,0] \minus{} [2,2,2]$

$ \implies$  $ [6,0,0] \plus{} 2[4,2,0]\geq 3[3,3,0]$  which is obviously true  :wink:[/quote]

$ \Longleftrightarrow  2\sum(x^4(x\minus{}y)(x\minus{}z))\plus{}2\sum(x^3(y\plus{}z)(x\minus{}y)(x\minus{}z))\plus{}2(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2$ $ \plus{}\sum(y^2z^2(x\minus{}y)(x\minus{}z))$

$ \plus{}6xyz\sum(x(x\minus{}y)(x\minus{}z))\plus{}xyz\sum((y\plus{}z)(x\minus{}y)(x\minus{}z))\geq 0$

[quote="cnyd"]$ x \equal{} a^{6},y \equal{} b^{6},z \equal{} c^{6}$  $ \implies$

$ [6,0,0] \plus{} \frac {[2,2,2]}{2} \minus{} 2[3,3,0]\geq 0$

$ 2[6,0,0] \plus{} [2,2,2]\geq 4[3,3,0]$

But I think it's not true

for $ x \equal{} y \equal{} z \equal{} 1$  $ \implies$  $ 9 \minus{} 12\geq 0$

I think $ 2(x \plus{} y \plus{} z) \plus{} 3\sqrt [3]{xyz} \minus{} 3(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx})\geq0$

because ; $ 2[6,0,0] \plus{} [2,2,2]\geq 3[3,3,0]$

if we use Schur inequality $ [6,0,0]\geq 2[4,2,0] \minus{} [2,2,2]$

$ \implies$  $ [6,0,0] \plus{} 2[4,2,0]\geq 3[3,3,0]$  which is obviously true  :wink:[/quote]
i'm stupid, i can't understand what you write,  explain it for me please  :blush:


İn the P(x) polynom the greatest degree is $ n$.
Such that for $ x\equal{}1,2,...,n,n\plus{}1$
$ P(x)\equal{}2^x$
$ P(n\plus{}2)\equal{}?$

Posted before many times/ $ P(n\plus{}2)\equal{}2^{n\plus{}2}\minus{}2$.


Hi,

We all know Harvard is famous for its Math 55, and MIT for its course 18. Does anyone know if there is a similar type of track at Princeton?

Also

I'm thinking about taking MAT 215 - Analysis of a single variable as a freshman. I plan to be a math major and scored well on cnotests like USAMO. Is this the track I want to get on or is there something more or less advanced that's right for me?

To clarify, Course 18 at MIT refers to the whole mathematics department.  Saying you're a Course 18 major doesn't say too much about your track except that you're majoring in mathematics in some way.

Also,

[url=http://www.math.princeton.edu/undergraduate/Placement.html]Placement Information for Undergraduates[/url]

Yes, Princeton tries to serve up a very challenging set of math courses for well prepared students.

Princeton has a very high quality math undergraduate program, but I believe that the number of math majors is much smaller than Harvard or MIT.  And of course, there's no analog of Math 55.  If you're seriously interested in learning more about their curriculum than what's available online, you should talk to Xevarion.


[i]Mou tin edose enas filos sto sxoleio kai me parakalese na tou grapso ti lysi gia na tin pei ston kathigiti tou (sto frontistirio) o opoios kai tou eixe dosei i askisi.
Yparxoun dyo lyseis-i mia me Stewart kai Cosine Law einai profanis-i geometriki pou ekana eina omorfi!
Episis i askisi kati mou thimize apo tin arxi,pragma pou paei na pei oti einai gnosti (isos na yparxei kai se vivlio toy kyriou Stergiu).[/i]

Sas ti dino ligo allagmeni:

[b]Dinetai isopleuro trigono ABC kai simeio D tis BC.Proekteinoume tin AC kata tmima CE=BD kai tin AB kata tmima BZ=CD.An gnorizoume ti gonia ZAD, na vreite ti gonia AEZ.[/b]
[hide]Fysika o kathigitis den pistepse to filo mou...[/hide]

Ωραία ασκησούλα Νίκο!

Προεκτείνω την $BZ$ κατά τμήμα $IZ$ έτσι ώστε $BI=BD$. Έστω $M$ το σημείο τομής της $ID$ με την $AE$. Θα έχουμε το τρίγωνο $AIE$ ισόπλευρο και το τρίγωνο $AIM$ ορθογώνιο. Επομένως $DM$ μεσοκάθετος της $AE$ και έτσι $AD=DE$, άρα το τρίγωνο $ADE$ είναι ισοσκελές με $\angle{DAE}=\angle{AED}\quad(1)$. Επίσης το τρίγωνο $ZDE$ ισοσκελές και $\angle{DEZ}=30^\circ$. Επομένως $\angle{AEZ}=30^\circ+\angle{DAE}$ από $(1)$. Όμως $\angle{DAE}=60^\circ-\angle{ZAD}$ έτσι θα έχουμε τελικά: $\angle{AEZ}=30^\circ+60^\circ-\angle{ZAD}\implies\angle{AEZ}=90^\circ-\angle{ZAD}$

PS: Μερικές αποδείξεις λείπουν καθώς είναι στοιχειώδεις και μπορείτε να τις βρείτε εύκολα...

PS2: (Είναι εύκολο) Δείξτε ότι $D$ είναι το περίκεντρο του τριγώνου $AZE$

isws ton pistepsei etsi (αν είναι συμμαθητής σου, δε θυμάμαι αν δίνεται στα δεδομένα... :) )
θεωρούμεμε ένα ορθοκανονικό σύστημα συντεταγμένων με Α ανήκει στον y'y και y'y κάθετος στην BC και BC πάνω στον x'x. τότε ξέρουμε τα Α,Β,C και βρίσκουμε το D γνωρίζοντας ότι ανήκει στον x'x και τη γωνία ΖΑD.  βρίσκουμε επίσης το Z γνωρίζοντας ότι ανήκει στην ΑΒ και ότι (ΖΒ)=(DC) που το ξέρουμε. βρίσκουμε το Ε ομοίως και τώρα βρίσκουμε τη ζητούμενη γωνία.

Poly oraia kai oi dyo.Den ksero telika isos na to ekana pio eukolo to provlima.To arxiko elege ndo AD=DE.

H lysi mou:
a)Theoro to A' pou prokyptei apo to A me anaklasi stin BC
b)To A'BC einai isopleuro
c)To DA'E einai isopleuro
kai telos.
An de sas arkoun auta pou egrapsa gia na diamorfosete ti lysi tote boro na sas steilo ki ena arxeio pou tin exo katharograpsei!(apla tora den to vrisko...)

afou exoume vrei ola ta simeia vriskoume ta AD,DE kai diapistwnoume oti einai isa

Γειά σας και από μένα. Θα ήθελα να προτείνω και εγώ μια λύση αυτής της ωραίας άσκησης. Προεκτείνω την BC κατά CL=BD και την CB κατά BK=CD. Τότε τα τρίγωνα KBZ και CLE είναι ισόπλευρα. Επίσης τα τρίγωνα ABD και DLE είναι ίσα καθώς και τα τρίγωνα ADC και KDZ επίσης ίσα. Άρα AD=ZD=DE. Άρα αν (πλέον αναφέρομαι σε γωνίες) ZAD=θ τότε και AZD=θ και DAE=DEA=60-θ. Επίσης από τις προηγούμενες ισότητες τριγώνων προκύπτουν: EDC=θ και BDZ=60-θ. Άρα ZDE=120 μοίρες, άρα ZED=30 μοίρες (από το ισοσκελές ZDE). Άρα AEZ=AED+DEZ=90-θ.
Να είστε όλοι καλά και να ασχολείστε με τα μαθηματικά. Παρά το αξίωμα της μη πληρότητας νομίζω ότι στα μαθηματικά βρίσκει κανείς την αίσθησης της πληρότητας.
Κώστας Καπένης


geometry
parallelogram
trigonometry
perimeter
trig identities
Law of Cosines

Find the Circumference and surface of the triangle AHG if  angle alpha (or angle BAC) is 60 scale

I'm taking CDBA to be a parallelogram (at least the file name is paralelogrami.jpg  :rotfl:).

Law of cosines on triangle AHB to get AH.
Law of cosines on triangle CAG to get AG.
Law of cosines on triangles AHB and CAG to get angle GAH.

Finding the perimeter and the area (if that's what you meant) of triangle AHG should be straightforward.

By the way, how to create a picture?

sure that CDBA  is a parallelogram   :P
yes i did mean for the perimeter and area , but i don't understand what is "straightforward" " i'm bad in English  and how to solve  "straightforward"?
Do you want to know how i made the picture?

LOL  "Straightforward" is an adjactive meaning easy, simple, and/or direct in this context... (there are other meanings too - check off a dictionary)  :P


geometry
perimeter
conics
hyperbola
parallelogram
rectangle
perpendicular bisector

Let $ABC$ a triangle and $P$ a point such that the triangles $PAB, PBC, PCA$ have the same area and the same perimeter. Prove that if:
a) $P$ is in the interior of the triangle $ABC$ then $ABC$ is equilateral.
b) $P$ is in the exterior of the triangle $ABC$ then $ABC$ is right angled triangle.

The locus of points such that the areas $S(\triangle PAB) = S(\triangle PCA)$ are equal is the median line AD of the triangle $\triangle ABC$, where D is the midpoint of the side a = BC. Denote the remaining sides b = CA, c = AB and WLOG, assume $b \ge c$. From equality of the perimeters $p(\triangle PAB) = p(\triangle PCA)$, we get

$AB + BP + PA = p(\triangle PAB) = p(\triangle PCA) = AC + CP + PA$

$BP - CP = AC - AB = b - c \ge 0$

This means that the points A, P lie on the opposite branches of a hyperbola with the focal points B, C. If b > c, this hyperbola intersects the median line AD at 2 points A, P. The 2nd intersection P is symmetrical to A with respect to the midpoint D, hence, it lies outside of the triangle $\triangle ABC$. Thus we see that for the specified point P to be inside of the triangle $\triangle ABC$, this hyperbola has to degenerate into the perpendicular bisector of the segment BC, which means that b - c = 0 and the triangle $\triangle ABC$ is isosceles with $AB = c = b = CA$. Similarly, it can be shown that it must be BC = a = c = AB, i.e., that the triangle $\triangle ABC$ must be equilateral. The point P then obviously is the intersection of the medians, i.e., the centroid of the equilateral triangle $\triangle ABC$.

Let b > c and let P be the other intersection of our hyperbola with the median line AD. The quadrilateral ABPC is a parallelogram, because its diagonals BC, AP cut each other in half. The area of the triangle $\triangle PBC$ is equal to the same areas of the triangles $\triangle PAB, \triangle PCA$, because both the paralleogram diagonals BC, AP cut its area in half, hence, $S(\triangle PBC) = S(\triangle PAB)$. But their perimeters can be equal only if the diagonals of the parallelogram ABPC are equal, because

$BP + PC + CB = p(\triangle PBC) = p(\triangle PAB) = AB + BP + PA$

implies PA = CB. But this means that the parallelogram ABPC has to be a rectangle and the angle $\angle CAB = 90^\circ$ right, which is what we are supposed to show.


How many whole numbers between 200 and 999 do not contain the digits 3, 5, or 7?
How many whole numbers between 2000 and 9999 do not contain the digits 3, 5, or 7?
Generalize this. (How many whole numbers between 2*10x and (x number of 9) do not contain the digits 3, 5, or 7?

[hide="Part 1"]
There are $5$ choices for the hundred's place (2,4,6,8,9).
There are $7$ choices for the ten's place (0,1,2,4,6,8,9)
There are $7$ choices for the one's place (0,1,2,4,6,8,9)
So there are $5*7*7=245$ total possibilities.
But you must subtract $2$; don't count 200 and 999.
So there are $245-2=243$ possibilities.
[/hide]

[hide="Part 2"]
There are $5$ choices for the thousand's place (2,4,6,8,9).
There are $7$ choices for the hundred's place (0,1,2,4,6,8,9)
There are $7$ choices for the ten's place (0,1,2,4,6,8,9)
There are $7$ choices for the one's place (0,1,2,4,6,8,9)
So there are $5*7*7*7=1715$ total possibilities.
But you must subtract $2$; don't count 2000 and 9999.
So there are $1715-2=1713$ possibilities.
[/hide]

[hide="Part 3"]
Looking at the pattern, the total number of possibilities is
\[5*7^{x}-2 \]

[/hide]

wow, im so tired i was doing 49 * 7 - 2

sorry

jorian

Please don't revive old posts. The question is from 2003.

[quote="rd5493"]Woah, sry, i didnt realize that.  i saw this one at the bottom of page 1, so i decided to answer it.  i guess someone else answered it recently and i saw it at the bottom.  woah, look at the dates.  it took almost 4 yrs for someone to reply to his/her message :rotfl:[/quote]
Rule of thumb, if Tare posted in it, you shouldn't be :)

[quote="rd5493"]For the first one
[hide]The numbers we can use for the hundreds place are $1,2,4,6,8,9$
The numbers we can use for the tens place are $1,2,4,6,8,9,0$
The numbers we can use for the ones place are $1,2,4,6,8,9,0$

So, there are $6$ for the hundreds place, and $7$ for both the tens and ones place

$6*7*7=294$[/hide]

For the second one
[hide]The numbers we can use for the thousands, hundreds and tens digits are the same we can use for the hundreds, tens, and ones in the previous problem.  Plus, for the ones digit of this problem, you have $7$ choices.

So, $6*7*7*7=2058$[/hide]
I'll leave the 3rd one for someone else[/quote]
For the first one you can't have 1 as a hundreds digit because it says 200-999, so one isn't a possibility there.

oops, change my answers to

[hide]245 [/hide]
and

[hide]1715[/hide]


calculus limit function trigonometry algebra polynomial articles

This might serve as a preview of calculus for students who haven't yet reached that level of math, or just a review (as it is for me) for those who have.

Let's try to keep this concise, and not stray too far from calculus.

Anyway the first tidbit (let's start basic here):

[size=150][b]Concept of a limit[/b][/size]

[b]If $ f(x)$ becomes arbitrarily close to a [i]single number $ L$[/i] as $ x$ approaches $ c$ [i]from either side,[/i] the limit of $ f(x)$ as $ x$ approaches $ c$ is $ L$.[/b]

For example, let's say we want to find the limit of $ f(x) = \cfrac{x^3 - 1}{x - 1}$, provided $ x\neq{1}$, as $ x$ approaches $ 0$.

The notation:
\[ \lim\limits_{x\to{0}}\cfrac{x^3 - 1}{x - 1}
\]
For this, we can just use [b]direct substitution:[/b]
\[ \lim\limits_{x\to{0}}\cfrac{0^3 - 1}{0 - 1} = \cfrac{ - 1}{ - 1} = \boxed{1}
\]
[hide="Another example"]
\begin{align*} & \lim\limits_{x\to{1}}\cfrac{x^2 + x - 2}{(x - 1)(x + 3)},x\neq{1,3} \\
& = \lim\limits_{x\to{1}}\cfrac{(x + 2)(x - 1)}{(x - 1)(x + 3)} \\
& = \lim\limits_{x\to{1}}\cfrac{x + 2}{x + 3}\qquad\qquad\text{We can now use direct substitution} \\
& = \cfrac{1 + 2}{1 + 3} \\
& = \boxed{\cfrac{3}{4}} \end{align*}
I'm going to guess that you've gotten the hang of this direct substitution thing by now. It's not that hard. :D
[/hide]
[hide="Another"]
\[ \lim\limits_{x\to{0}}\cfrac{|x|}{x}
\]
The graph:
[asy]import graph;
size(200);

Label f;
f.p=fontsize(10);
xaxis(-4,4,LeftTicks(f,5));
yaxis(-4,4,RightTicks(f,5));

real f(real x) {return -1;}
D(graph(f,-4,0));
real f(real x) {return 1;}
D(graph(f,0,4));[/asy]
We can see that the function gives two numbers when approached from the left and from the right. Therefore,
\[ \lim\limits_{x\to{0}}\cfrac{|x|}{x} = DNE\text{, or Does Not Exist.}
\]
[/hide]
[hide="Examples enrich the mind XD"]
\[ \lim\limits_{x\to{0}}\cfrac{2}{x^2} = DNE
\]
because it gives no numerical value (we'll talk about infinite limits later).
[/hide]
[hide="The last example"]
\[ \lim\limits_{x\to{0}}\sin\cfrac{1}{x} = DNE
\]
This function shows an oscillating behavior, which does not approach any value.
[/hide]

you don't need to graph the abs value thing for the abs(x)/x thing since its not continuous at 0 anyway

1. Not everybody is as smart as you.
2. It's easier to see visually. :)

EDIT: actually, that demonstrates one sorta-important part of this - it doesn't matter what actually happens at $ x \equal{} 0$. What's important is what the [b]rest of the graph[/b] does as it approaches $ x \equal{} 0$. That's the limit. I'll get to the discontinuities later, but what we really want to see now is that the function approaches two different values from the left and the right sides, so we don't know what the limit really is. The limit does not exist (DNE).

I know a fair bit of calculus myself (thank you, Mr. Watson!).

Here are a few suggestions (I'm not trying to force anything on you):
And, I think that you should first explain WHY direct substitution "works" in finding the limits of rational functions or polynomial functions.
I also think that, although you defined the limit well, you should add more detail, including the fact that it doesn't matter whether $ x\equal{}c$ exists or not; just what the value that is being approached is.  And perhaps you should include a short, non-rigorous explanation of what "arbitrarily close" means.
Finally, I think that you should start with intuition, as traditionally done in teaching calculus.  Include more graphs at the beginning.  For example, you could draw $ f(x)\equal{}x^2$ and graphically show why the limit is $ 1$ as $ x$ approaches $ 1$ (both the right and the left of the graph seem to "close in" on the value $ y\equal{}1$).  Tables could also be included.  For example, let's say you make up some function and want to find the limit as $ x$ approaches $ 5$.  Then you can make a table of $ f(x)$ and $ x$, with values of $ x$ that are approaching $ 5$, such as $ x\equal{}4$, $ x\equal{}6$, $ x\equal{}4.5$, $ x\equal{}5.5$, $ x\equal{}4.9$, $ x\equal{}5.1$, etc. and show that the values of $ f(x)$ are approaching a certain value.

I know that's a lot of criticism, and I have plenty more.  But I do admit that I doubt I could make an article like this that is perfect, and I can continue to offer ideas if you like.

heh criticisms are welcome. :D

Soooo in response to your suggestions:

It's true that making a table, or looking at the graph of a function is a more accurate way of determining whether or not a limit actually exists. But they're tedious, especially if you don't have an aid like a graphing calculator.

So why does direct substitution work? Actually, it doesn't always work. As I've mentioned, in a case such as when $ f(c)\neq\lim\limits_{x\to{c}}f(x)$, direct substitution is totally inaccurate. It's only a shortcut that works [i]most[/i] of the time, especially for polynomials (this is where the topic of continuity might come in) and trig functions (sine, cosine, though not tangent). So even though it's not the Swiss army knife of limits, this is one of the first things you should try in the process of evaluating a limit, and it's always good to double-check. :)

I wish I could be more detailed than I currently am, but I don't exactly have the time lol.. These are based mostly off of my notes, and I'm not teaching a full course or anything anyway. :) But please, if you think some things are lacking (which will probably happen quite a bit), feel free to add. :P

[quote="undefined117"][hide="Common Sense"]
[u]Basic Limits[/u]
Let $ b$ and $ c$ be real numbers and let $ n$ be a positive integer. Then

$ \lim\limits_{x\to{c}}b = b\qquad\lim\limits_{x\to{c}}x = c\qquad\lim\limits_{x\to{c}}x^n = c^n.$

[u]Properties of Limits[/u]

Scalar multiple: $ \lim\limits_{x\to{c}}[b\cdot{}f(x)] = b\lim\limits_{x\to{c}}f(x)$

Sum/Difference: $ \lim\limits_{x\to{c}}[f(x)\pm{}g(x)] = \lim\limits_{x\to{c}}f(x)\pm\lim\limits_{x\to{c}}g(x)$

Product: $ \lim\limits_{x\to{c}}[f(x)\cdot{}g(x)] = \lim\limits_{x\to{c}}f(x)\lim\limits_{x\to{c}}g(x)$

Quotient: $ \lim\limits_{x\to{c}}\cfrac{f(x)}{g(x)} = \cfrac{\lim\limits_{x\to{c}}f(x)}{\lim\limits_{x\to{c}}g(x)}\qquad,\lim\limits_{x\to{c}}g(x)\neq{0}$

Power: $ \lim\limits_{x\to{c}}[f(x)]^n = \left[\lim\limits_{x\to{c}}f(x)\right]^n$[/hide]
What do you make of $ \lim\limits_{\theta\to{0}}\cfrac{\sin\theta}{\theta} = \cfrac{0}{0}$? To evaluate this properly, we need

[size=150]The Squeeze Theorem[/size]

(Thanks, Mrs. Gates :))

If $ h(x)\leq{f(x)}\leq{g(x)}$ for all $ x$ in an open interval containing $ c$, except possibly at $ c$, and if $ \lim\limits_{x\to{c}}h(x) = L = \lim\limits_{x\to{c}}g(x)$, then $ \lim\limits_{x\to{c}}f(x)$ exists and is equal to $ L$.

[hide="Application"][asy]size(200);
pair O=D("O",(0,0),SW), A=D("A (1,0)",(1,0),SE), B=D("B (\cos\theta,\sin\theta)",IP(A..dir(30)..(0,1),O--2*dir(35)),W), C=D("C (1,\tan\theta)",IP(A--(1,1),O--2*dir(35)),NW);
D((-.2,0)--(1.2,0),black); D((0,-.2)--(0,1.2),black);
D(A..dir(30)..(0,1),red); D(O--1.5*dir(35)); D(B--A--C);
MP("\theta",O,5*dir(17));[/asy]
$ A_{\triangle{OBA}}\leq{A_{\text{sector }OBA}\leq{A_{\triangle{OCA}}}}$

$ \cfrac{1}{2}\sin\theta\leq\cfrac{\theta}{2}\leq\cfrac{1}{2}\tan\theta$

$ \sin\theta\leq\theta\leq\cfrac{\sin\theta}{\cos\theta}$

$ 1\leq\cfrac{\theta}{\sin\theta}\leq\cfrac{1}{\cos\theta}$

$ 1\geq\cfrac{\sin\theta}{\theta}\geq\cos\theta$

$ \lim\limits_{\theta\to{0}}1\geq\lim\limits_{\theta\to{0}}\cfrac{\sin\theta}{\theta}\geq\lim\limits_{\theta\to{0}}\cos\theta$

$ 1\geq\lim\limits_{\theta\to{0}}\cfrac{\sin\theta}{\theta}\geq{1}$

$ 1\geq{1}\geq{1}$

so $ \boxed{\lim\limits_{\theta\to{0}}\cfrac{\sin\theta}{\theta} = 1}$[/hide]
[hide="And from that"]
$ \lim\limits_{\theta\to{0}}\cfrac{1 - \cos\theta}{\theta} = \cfrac{0}{0}$

hmm.

Rewritten as

$ \lim\limits_{\theta\to{0}}\cfrac{(1 - \cos\theta)(1 + \cos\theta)}{\theta(1 + \cos\theta)} = \lim\limits_{\theta\to{0}}\cfrac{1 - \cos^2\theta}{\theta(1 + \cos\theta)}$

$ = \lim\limits_{\theta\to{0}}\cfrac{\sin\theta}{\theta}\cdot\cfrac{\sin\theta}{1 + \cos\theta} = \lim\limits_{\theta\to{0}}\cfrac{\sin\theta}{\theta}\cdot\lim\limits_{\theta\to{0}}\cfrac{\sin\theta}{1 + \cos\theta} = 1\cdot\cfrac{0}{1 + 1} = \boxed{0}$
[/hide]

[hide="Practice"]
1. $ \lim\limits_{\theta\to{0}}\cfrac{\cos\theta\tan\theta}{\theta}$

2. $ \lim\limits_{x\to{0}}\cfrac{\sin{x}}{6x}$

3. $ \lim\limits_{x\to{0}}\cfrac{\tan^2x}{x}$[/hide][/quote]

1. $ 1$ Simplify $ \cos{x}\tan{x}$ into $ \sin{x}$ Or just use l'hospital's rule.
2. $ 1/6$ Use $ \frac {\sin{x}}{x}$
3. $ 0$

You are beastly at latex

Should we get in the derivative rules?
$ f(x) \equal{} x^r$
where ''r'' is any real number, then
$ f'(x) \equal{} rx^{r\minus{}1}$
$ \frac {d}{dx}e^x \equal{} e^x$
$ \frac {d}{dx}a^x \equal{} \ln(a)a^x$
$ \frac {d}{dx}\ln(x) \equal{} 1/x,\qquad x > 0$
$ \frac {d}{dx}\log_a(x) \equal{} \frac {1}{x\ln(a)}$
$ \frac {d}{dx}\sin(x) \equal{} \cos(x).$
$ \frac {d}{dx}\cos(x) \equal{} \minus{} \sin(x).$
$ \frac {d}{dx}\tan(x) \equal{} \sec^2(x).$

[quote="undefined117"]yep.

Now, on to the subject of

[size=150]Continuity[/size]

If $ \lim\limits_{x\to{C}}f(x) \equal{} L$ for any constant $ C$, observe the following graph of $ f$:

[asy]import graph;
size(200);

Label f;
f.p=fontsize(10);
xaxis(-4,4);
yaxis(-4,4);

real a(real x) {return 1/(x+1);}
D(graph(a,-4,-3.1)); D(graph(a,-2.9,-1.25)); D(graph(a,-.75,.9)); D(graph(a,2.6,4));
D(circle((-3,a(-3)),.1)); fill(circle((-3,1),.1),blue); D('a',(-3,0),SE); D((-3,a(-3))--(-3,1),dashed);
D((-1,-4)--(-1,4),dashed); D('b',(-1,0),SW);
D((1.1,-1.5)--(2.4,-1)); D(circle((1,-1.55),.1)); D(circle((1,a(1)),.1)); D((1,-1.55)--(1,a(1)),dashed); D('c',(1,0),SW);
fill(circle((2.5,-.95),.1),blue); D(circle((2.5,a(2.5)),.1)); D((2.5,-.95)--(2.5,a(2.5)),dashed); D('d',(2.5,0),SE);[/asy]

[u]Discontinuities[/u]

Removable: There is a [b]hole[/b] at $ a$, and $ \lim\limits_{x\to{a}}f(x) \equal{} L\neq{f(a)}$. One can remove such removable discontinuities by using methods such as cancellation.

Nonremovable: There is a [b]gap[/b] at $ b$, and $ \lim\limits_{x\to{b}}f(x) \equal{} DNE$.

Non removable: There is a [b]jump[/b] at $ c$ and $ d$, and the limit of $ f$ at both $ x$ values $ DNE$.

However, the limit of $ f$ as $ x\to{c}$ from the left exists but is not equal to the limit of $ f$ as $ x\to{c}$ from the right. This can be denoted as $ \lim\limits_{x\to{c^ \minus{} }}f(x)\neq\lim\limits_{x\to{c^ \plus{} }}f(x)$. Also, $ \lim\limits_{x\to{d^ \minus{} }}f(x) \equal{} f(d)\neq\lim\limits_{x\to{d^ \plus{} }}f(x)$.

[b]Continuity[/b] - a function $ f$ is continuous at $ c$ iff $ \lim\limits_{x\to{c}}f(x) \equal{} f(c)$.
[hide="More common sense"]
If $ b$ is a real number and $ f$ and $ g$ are continuous at $ x \equal{} c$, then the following functions are also continuous at $ x \equal{} c$:

Scalar multiple - $ b\cdot{f(x)}$

Sum/difference - $ (f\pm{g})(x)$

Product - $ (fg)(x)$

Quotient - $ (\cfrac{f}{g})(x),g(c)\neq{0}$

and

[b]Composite Functions[/b] - for $ f(g(x))$, if $ g$ is continuous at $ c$, and $ f$ is continuous at $ g(c)$, then $ f(g(x))$ is continuous at $ c$.
[/hide]

[b]Intermediate Value Theorem[/b] - If $ f$ is continuous on $ [a,b]$, and $ k$ is any number between $ f(a)$ and $ f(b)$, then there is at least one number $ c$ in $ [a,b]$ such that $ f(c) \equal{} k$.

For a continuous function, if $ x$ takes on all values on $ [a,b]$, then $ f(x)$ must take on all values in $ [f(a),f(b)]$.

EDIT: faulty latex - forgot an underscore.[/quote]

how did you learn asymptote.

Thanks moogra! :lol:

@cognos: by experimenting with examples from [url=http://piprim.tuxfamily.org/asymptote/index.html]here.[/url] :)

[size=150]Infinite Limits[/size]
[hide="okay I`m going to skip some stuff here cause I find them pretty much self-explanatory and boring.."]

[b]Definition of a Vertical Asymptote:[/b] If $ f(x)$ approaches $ \infty$ or $ \minus{}\infty$ as $ x$ approaches $ c$ from the right or the left, then the line $ x\equal{}c$ is a vertical asymptote of the graph of $ f$.

A zero denominator sometimes gives a vertical asypmtote:

$ f_1(x)\equal{}\cfrac{x^2\plus{}9}{x^2\minus{}9}$, V.A. at $ x\equal{}\minus{}3$ and $ x\equal{}3$

$ f_2(x)\equal{}\tan{x}$, V.A. at $ x\equal{}\cfrac{\pi}{2}\plus{}\pi{k}$

but not always:

$ f_3(x)\equal{}\cfrac{x^2\plus{}5x\plus{}6}{x^2\minus{}4}\equal{}\cfrac{(x\plus{}2)(x\plus{}3)}{(x\plus{}2)(x\minus{}2)}\equal{}\cfrac{x\plus{}3}{x\minus{}2}$, V.A. at $ x\equal{}2$, but as you can see with $ f_3(\minus{}2)\equal{}\cfrac{(0)(1)}{(0)(\minus{}4)}\equal{}\cfrac{0}{0}$, an [b]indeterminate form[/b], there is a hole at $ x\equal{}\minus{}2$.

$ f_4(0)\equal{}\cfrac{\sin{0}}{0}\equal{}\cfrac{0}{0}$, an indeterminate form again. There is a hole at $ x\equal{}0$, not a vertical asymptote.

Let $ f$ and $ g$ be continuous on an open interval containing $ c$. If $ f(c)\neq{0}$, $ g(c)\equal{}0$, and there exists an open interval containing $ c$ such that $ g(x)\neq{0}$ for all $ x\neq{c}$ in the interval, then the graph of the function $ h(c)\equal{}\cfrac{f(x)}{g(x)}$ has a V.A. at $ x\equal{}c$. Visualize:

[asy]import cse5;
import graph;
size(200);

xaxis(-3,6);
yaxis(-5,5);

real f(real x) {return x^2/4+1;}
real g(real x) {return x-1;}
real h(real x) {return f(x)/g(x);}

D(graph(f,-3,4)); MP('f(x)',(-2,f(-2)),NE);
D(graph(g,-3,6),green); MP('g(x)',(5,g(5)),.5SE);
D(graph(h,-3,.7),red); D(graph(h,1.3,6),red); MP('h(x)',(4,h(4)));
D((1,-5)--D('c',(1,0),2SE)--(1,5),dashed);[/asy][/hide]

Definition of a derivative
$ \lim\limits_{h\to{0}}\frac{f(x\plus{}h)\minus{}f(x)}{h}$

Elaborating on that:

[asy]import graph;
size(250);

xaxis(-1,7);
yaxis(-2,6);
real a(real x) {return x^2/8+1;}
real b(real x) {return .875x-.25;}
D(graph(a,-1,6.5));
D(graph(b,-1,7),purple);
D(D((2,a(2)))--MP('x',(2,0)),dashed); D(D((5,a(5)))--MP('x+h',(5,0)),dashed);
D((2,-1)--(5,-1),black,Bars); MP('h',(3.5,-1));
D((2,a(2))--(6,a(2)),dashed); D((5,a(5))--(6,a(5)),dashed);
D((6,a(2))--(6,a(5)),black,Bars); MP('f(x+h)-f(x)',(6,a(2)+(a(5)-a(2))/2),E);[/asy]

The [b]derivative[/b] is the slope of a curve at a point, aka the [b]instantaneous rate.[/b]

The slope of a secant line through a curve like the one shown above will equal $ \cfrac{f(x \plus{} h) \minus{} f(x)}{h}$, the [b]difference quotient.[/b] Now imagine the value of $ h$ becoming infinitely smaller:
\[ \lim\limits_{h\to{0}}\cfrac{f(x \plus{} h) \minus{} f(x)}{h} \equal{} f'(x)
\]
which is what moogra got. :lol:

It's also like the slope at any point
$ \lim\limits_{h\to{0}}\frac {f(x \plus{} h) \minus{} f(x)}{h}$
$ f(x) \equal{} ax^b$
$ f'(x) \equal{} a*bx^{b \minus{} 1}$
ex:
$ f(x) \equal{} 3x^{21}$
$ f'(x) \equal{} 21*3x^{20}$


Hello friends,

I mean help me out friends. I tried my best to understand the algorithm for 8-queen problem given in Brassard and Bratley (but could not). Earlier I had tried on my own. But my efforts were fruitless. Please suggest me a good website [b](i.e a website with some powerpoint presentations)[/b] to read about this problem (which also contains a good intro to backtacking).

Thanks in advance...

Do you know what's wikipedia and google?

I know very well what these sites are but they return too many pages when searched despite the specifications I give.
Infact, I have tried advanced search as well but I did not find results satisfying me. Sorry, if that bothers you....But help me out anyway.

Thank You

That doesn't bother me. That was the answer ;-)

This is taken from USACO. I hope it helps...

Search Techniques
Sample Problem: n Queens [Traditional]

Place n queens on an n x n chess board so that no queen is attacked by another queen.
Depth First Search (DFS)

The most obvious solution to code is to add queens recursively to the board one by one, trying all possible queen placements. It is easy to exploit the fact that there must be exactly one queen in each column: at each step in the recursion, just choose where in the current column to put the queen.

 1 search(col)
 2     if filled all columns
 3         print solution and exit

 4   for each row
 5       if board(row, col) is not attacked
 6            place queen at (row, col)
 7            search(col+1)
 8            remove queen at (row, col)

Calling
search(0) begins the search. This runs quickly, since there are relatively few choices at each step: once a few queens are on the board, the number of non-attacked squares goes down dramatically.

This is an example of depth first search, because the algorithm iterates down to the bottom of the search tree as quickly as possible: once k queens are placed on the board, boards with even more queens are examined before examining other possible boards with only k queens. This is okay but sometimes it is desirable to find the simplest solutions before trying more complex ones.

Depth first search checks each node in a search tree for some property. The search tree might look like this:

The algorithm searches the tree by going down as far as possible and then backtracking when necessary, making a sort of outline of the tree as the nodes are visited. Pictorially, the tree is traversed in the following manner:

Complexity

Suppose there are d decisions that must be made. (In this case d=n, the number of columns we must fill.) Suppose further that there are C choices for each decision. (In this case c=n also, since any of the rows could potentially be chosen.) Then the entire search will take time proportional to cd, i.e., an exponential amount of time. This scheme requires little space, though: since it only keeps track of as many decisions as there are to make, it requires only O(d) space.


Prove that $ (1\minus{}a\plus{}a^2)(1\minus{}b\plus{}b^2)(1\minus{}c\plus{}c^2)\geq abc$ , with $ a,b,c$ real numbers

[hide="Hidden"]We have
$ 1 \minus{} a \plus{} a^2 > a$ if $ a\ge 0$ and
$ 1 \minus{} a \plus{} a^2 > \minus{} a$ if $ a < 0$, so we can say $ 1 \minus{} a \plus{} a^2 > |a| > 0$.
So
$ (1 \minus{} a \plus{} a^2)(1 \minus{} b \plus{} b^2)(1 \minus{} c \plus{} c^2) > |a|(1 \minus{} b \plus{} b^2)(1 \minus{} c \plus{} c^2) > |a\parallel{}b|(1 \minus{} c \plus{} c^2) > |a\parallel{}b\parallel{}c| \equal{} |abc| > abc$.[/hide]

[quote="Armenia"][hide="Hidden by mod"]We have
$ 1 \minus{} a \plus{} a^2 > a$ if $ a\ge 0$ and
$ 1 \minus{} a \plus{} a^2 > \minus{} a$ if $ a < 0$, so we can say $ 1 \minus{} a \plus{} a^2 > |a| > 0$.
So
$ (1 \minus{} a \plus{} a^2)(1 \minus{} b \plus{} b^2)(1 \minus{} c \plus{} c^2) > |a|(1 \minus{} b \plus{} b^2)(1 \minus{} c \plus{} c^2) > |a\parallel{}b|(1 \minus{} c \plus{} c^2) > |a\parallel{}b\parallel{}c| \equal{} |abc| > abc$.[/hide][/quote][hide="ignore this"]$ 1 \minus{} a \plus{} a^2 > \minus{} a$ does not mean that $ 1 \minus{} a \plus{} a^2 > |a|$. $ 1 \minus{} a \plus{} a^2 < \minus{} a$ does.[/hide]$ 1 \minus{} a \plus{} a^2 \equal{} a$ if $ a \equal{} 1$.

[quote="towersfreak2006"]If $ a < 0$, then $ |a| \equal{} \minus{} a$, so $ 1 \minus{} a \plus{} a^2 > \minus{} a$ does imply $ 1 \minus{} a \plus{} a^2 > |a|$.
However, those strict inequalities should be non-strict, since there is equality at $ 1$.[/quote]Oh yeah, you're right.


absolute value
number theory unsolved
number theory

Prove or disprove:there exist infinitely many pairs of distinct integers $ (x,y,z,w)$
satisfies $ x^3\plus{}y^3\plus{}z^3\plus{}w^3\equal{}0$

I think it would be more interesting if $ x,y,z,w$ are pairwise distinct to absolute value.

[quote="xxp2000"]$ (a,6a,8a, \minus{} 9a)$[/quote]
You are right!Thank you very much :)
Could you prove the existence of infinitely many $ x,y,z,w$ suit the equation ,$ (x,y,z,w) \equal{} 1$ with their absolute value are distict?


calculus
derivative
integration
function
trigonometry
limit
real analysis

i know there are 2pi radians in 360 degrees but thats it

are there any formulas or tricks i need to know that can help me(basics please)

why are they used opposed to degrees?

Well, radians are usually measured in terms of $ \pi$, and almost everything dealing with circles have to do with $ \pi$. For example, the circumference is $ 2\pi r$.

So, using radians fits nicely in calculations. No conversions are needed from degrees.

Are you searchin for this :

 Radian is  $ rad \equal{} \frac {180}{\pi}$ so $ 2(\pi)\frac{180}{\pi} \equal{} 360degree$

The reasons to use them are all about calculus. Derivatives and integrals of trigonometric functions work out nicely in radians.

The basic proportion for converting degrees to radians or vice versa is $ \frac{radians}{2\pi}\equal{}\frac{degrees}{360}$.

7h3.D3m0n.117 is right. After working with radians a bit, you should be able to know certain angles on the spot.

there are certain limits that come out very nicely in radians

For example:
$ \lim_{x\to 0} \frac{\sin x}{x} \equal{} 1$

You can try it on your calculator. Plug in a very small value for x and see what you get.
Another way to see it is to graph the two functions $ y\equal{}x$ and $ y\equal{}\sin x$
When you get near zero, they overlap almost exactly, but only in radians
Try both of those again in degrees and you will see that it won't work.

These limits are the basis of trig in calculus, so radians are essential here.

Here is the way I do quick conversions
I know that $ \pi radians \equal{} 180 degrees$ (or a half circle)

When I see something like $ \frac{2\pi}{3}$, I split the 180 into thirds (60), and then count two of them (120).

When I see something like $ \frac{5\pi}{6}$ I split it into sixths (30) and count 5 (150).

Going the other way is a little tougher, but most everything you see will come in multiples of thirty or 45

[quote="chickendude"]When I see something like $ \frac {2\pi}{3}$, I split the 180 into thirds (60), and then count two of them (120).

When I see something like $ \frac {5\pi}{6}$ I split it into sixths (30) and count 5 (150).[/quote]

That's the way I do it  :lol:  rather than just blankly memorize random ones. It makes you think what you are actually doing.

Also with the use of radians, make sure to label everything correctly, even if it's on scratch paper. A $ \sin 1$ could be either radian or degrees, and could be the fatal mistake in a competition if not labeled correctly.

[quote="chickendude"]there are certain limits that come out very nicely in radians

For example:
$ \lim_{x\to 0} \frac {\sin x}{x} \equal{} 1$
[/quote]

Er, sorry? That limit works just as well in degrees. Different units of measure generally don't change limits to zero or infinity.

Nope.  In degrees, $ \lim_{x \to 0} \frac {\sin x}{x} \equal{} \frac {\pi}{180}$.  The reason is that when $ \sin x$ is considered as a radian function, the "degree version" of $ \sin x$ is given by $ \sin \left( \frac {\pi}{180} x \right)$.  You can verify for yourself on your calculator that $ \sin 1^{\circ} \approx \frac {\pi}{180} \ll 1$.

In any case, I'm surprised no one's brought this up yet, but the fundamental reason that radians make things work out nicely is that the length of an arc given by

$ L \equal{} r \theta$

where $ r$ is the radius is only valid in radians.  As [b]123456789[/b] pointed out, when $ \theta \equal{} 2 \pi$ (a full circle) the corresponding arc is the circumference of the circle, which is $ 2 \pi r$ as expected.

[quote="7h3.D3m0n.117"]Make sure you can quickly identify simple radians and degrees. For example, $ \frac {\pi}{3} \equal{} 60^{\circ}$ and vice versa. This will save you [i]a lot[/i] of time in higher math classes such as trig or calculus.[/quote]

For this, a unit circle is nice thing to remember.

http://upload.wikimedia.org/wikipedia/commons/c/c9/Unit_circle_angles.svg

The trigonometric functions are different for degrees? I didn't know that...

In (real) analysis, the trigonometric functions are treated as functions $ \mathbb{R} \to \mathbb{R}$; that is, their inputs are dimensionless (so we prefer to work in radians).  In other words, although $ \frac {\pi}{3}$ and $ 60^{\circ}$ represent the same angle, they are not the same real number (when viewed as dimensionless inputs to a trigonometric function).  Thus, the function $ \sin_{rad}(x)$ such that

$ \sin_{rad} \frac {\pi}{3} \equal{} \frac {1}{2}$

and the function $ \sin_{deg}(x)$ such that

$ \sin_{deg} 60 \equal{} \frac {1}{2}$

are not the same function!  In fact, as I have said, $ \sin_{deg} x \equal{} \sin_{rad} \left( \frac {\pi}{180} x \right)$.


At 7:40 p.m., Bob passed mile marker 134. At 8:20 p.m., he passed mile marker 176. What is the number of miles per hour in his average speed?


calculus
integration
trigonometry
function
calculus computations

source : Stewart 5e Calculus Chapter 8.2

 integral of 1/(cos x- 1)

Thank you for helping me

From trigonometry $\cos x -1 = - 2 \sin^{2} \left( \frac{x}{2}\right)$

so the integral is $\int \frac{1}{ \cos x -1} \, dx = \cot \left( \frac{x}{2}\right)$

 :)

An alternate method.

Inspiration: any trig function can be viewed as a square root. If you've got an expression containing a square root in the denominator, you have the algebraic option of rationalizing the denominator.

$\frac1{\cos x-1}=\frac{-1}{1-\cos x}=\frac{-1}{1-\cos x}\cdot \frac{1+\cos x}{1+\cos x}$

$=\frac{-1-\cos x}{\sin^2x}=-\csc^2x-\csc x\cot x$

$\int -\csc^2x-\csc x\cot x\,dx=\cot x+\csc x+C=\frac{1+\cos x}{\sin x}+C$

Note: I am not disagreeing with Ramanujan. For a further exercise in trigonometric identities, try showing that our answers agree.

Another approach, I think it works.
[hide="hint"] Let $t=\tan \frac{x}{2}$...[/hide]


\[ \left\{ \begin{array}{l}
 2\sqrt {x \plus{} y}  \minus{} \sqrt {5x \minus{} 3y}  \equal{} 2 \\
 \sqrt {x \plus{} 2}  \plus{} x \minus{} y \equal{} 2 \\
 \end{array} \right.\]

hello, from the second equation we have $ y\equal{}\sqrt{x\plus{}2}\plus{}x\minus{}2$, inserting this in the first one we have
$ 2\sqrt{2x\minus{}2\plus{}\sqrt{x\plus{}2}}\equal{}2\plus{}\sqrt{2x\plus{}6\minus{}3\sqrt{x\plus{}2}}$. Squaring this and simplifying again we have
$ 6x\minus{}18\plus{}7\sqrt{x\plus{}2}\equal{}4\sqrt{2x\plus{}6\minus{}3\sqrt{x\plus{}2}}$.Squaring gives $ 3x^2\minus{}199x\plus{}326\equal{}\minus{}12\sqrt{x\plus{}2}(7x\minus{}17)$.
Hence $ (x\minus{}2)(1296x^3\minus{}18792x^2\plus{}45649x\minus{}11522)\equal{}0$. The only solution of our equation is $ x\equal{}2$ and $ y\equal{}2$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, from the second equation we have $ y \equal{} \sqrt {x \plus{} 2} \plus{} x \minus{} 2$, inserting this in the first one we have
$ 2\sqrt {2x \minus{} 2 \plus{} \sqrt {x \plus{} 2}} \equal{} 2 \plus{} \sqrt {2x \plus{} 6 \minus{} 3\sqrt {x \plus{} 2}}$. Squaring this and simplifying again we have
$ 6x \minus{} 18 \plus{} 7\sqrt {x \plus{} 2} \equal{} 4\sqrt {2x \plus{} 6 \minus{} 3\sqrt {x \plus{} 2}}$.Squaring gives $ 3x^2 \minus{} 199x \plus{} 326 \equal{} \minus{} 12\sqrt {x \plus{} 2}(7x \minus{} 17)$.
Hence $ (x \minus{} 2)(1296x^3 \minus{} 18792x^2 \plus{} 45649x \minus{} 11522) \equal{} 0$. The only solution of our equation is $ x \equal{} 2$ and $ y \equal{} 2$.
Sonnhard.[/quote]
I think the following reasoning is better.
Let $ \sqrt {x \plus{} 2} \equal{} t.$ Hence, $ 2\sqrt {2x \minus{} 2 \plus{} \sqrt {x \plus{} 2}} \equal{} 2 \plus{} \sqrt {2x \plus{} 6 \minus{} 3\sqrt {x \plus{} 2}}\Leftrightarrow$
$ \Leftrightarrow2\sqrt {2t^2 \plus{} t \minus{} 6} \equal{} 2 \plus{} \sqrt {2t^2 \minus{} 3t \plus{} 2}\Leftrightarrow$
$ \Leftrightarrow2\sqrt {2t^2 \plus{} t \minus{} 6} \minus{} t \minus{} 2 \equal{} \sqrt {2t^2 \minus{} 3t \plus{} 2} \minus{} t\Leftrightarrow$
$ \Leftrightarrow\frac {7(t \plus{} 2)(t \minus{} 2)}{2\sqrt {2t^2 \plus{} t \minus{} 6} \plus{} t \plus{} 2} \equal{} \frac {(t \minus{} 1)(t \minus{} 2)}{\sqrt {2t^2 \minus{} 3t \plus{} 2} \plus{} t}\Leftrightarrow t \equal{} 2$ since,
$ \frac {7(t \plus{} 2)}{2\sqrt {2t^2 \plus{} t \minus{} 6} \plus{} t \plus{} 2} > 1 > \frac {t \minus{} 1}{\sqrt {2t^2 \minus{} 3t \plus{} 2} \plus{} t}.$

[quote="Dr Sonnhard Graubner"]hello, from the second equation we have $ y \equal{} \sqrt {x \plus{} 2} \plus{} x \minus{} 2$, inserting this in the first one we have
$ 2\sqrt {2x \minus{} 2 \plus{} \sqrt {x \plus{} 2}} \equal{} 2 \plus{} \sqrt {2x \plus{} 6 \minus{} 3\sqrt {x \plus{} 2}}$. Squaring this and simplifying again we have
$ 6x \minus{} 18 \plus{} 7\sqrt {x \plus{} 2} \equal{} 4\sqrt {2x \plus{} 6 \minus{} 3\sqrt {x \plus{} 2}}$.Squaring gives $ 3x^2 \minus{} 199x \plus{} 326 \equal{} \minus{} 12\sqrt {x \plus{} 2}(7x \minus{} 17)$.
Hence $ (x \minus{} 2)(1296x^3 \minus{} 18792x^2 \plus{} 45649x \minus{} 11522) \equal{} 0$. The only solution of our equation is $ x \equal{} 2$ and $ y \equal{} 2$.
Sonnhard.[/quote]
Hello ,we can  by put \[ \left\{ \begin{array}{l}
 u \equal{} \sqrt {x \plus{} y}  \\
 v \equal{} \sqrt {5x \minus{} 3y}  \\
 \end{array} \right.\]


function
real analysis
real analysis unsolved

Let $f:[0,2] \rightarrow R$ be a derivable function such that $f((x^2-2x) \cdot f'(x)+x)=5$
for all $x \in [0,2]$

Find f.  :)

I don't know all $f$ satisfy that conditon, but for a glance $f(x)=5$ seems working :D

This is the answer but make an attempt to prove it, too! :)  :lol:

Consider this "a bit" general problem:
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a derivable function and let $h(x)$ be a function that has two different zeros $a<b$. If
\[
f(h(x)f'(x)+x)=k
\]
Then $f([a,b])=k.$

Clearly $f(a)=f(b)=k$. By Mean value Theorem, there exists $c\in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}=0$. It follows that $f(c)=k$. Let $x\in (a,b)$. If $x=c$ then $f(x)=k$, if not then $x\in (a,c)$ or $x\in (c,b)$. WLOG, $x\in (a,c)$. Using the same argument, there is $c_2\in (a,c_1)$ (here  $c_1:=c$) such that $f(c_2)=k$. Using this argument repeatedly, we have a sequence $(c_n)$ converges to $x$. Since $f$ is continuous, then $k=f(c_n)\rightarrow f(x)$. Hence $f(x)=k$.

Let me know if I made some stupid mistake.


inequalities
floor function
number theory unsolved
number theory

Prove for every natural $n>4$ we have :
$n\sqrt{n+4}\le \left\lfloor{n\sqrt{n+3}}\right\rfloor+1$

It is not true when n is big number. For example $8*\sqrt{12}>[8*\sqrt{11}]+1$.


IF

2+1+3+4 = 40
5+4+2+4 = 90
3+2+5+8 = 126
4+3+6+7 =  180

then 7+8+1+10 = ????

That is the whole info i have . Nothing more is provided .

Kindly provide me the solutions asap.

I think the reason nobody is replying, including me, is that the puzzle doesnt make sense. Does it specify anything about the "+" signs?


Prove for all $x,y,z>0$ with $xy+yz+zx=1$ that
\[\frac{27}{4}(x+y)(y+z)(z+x)\geq \left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^{2}\geq 6\sqrt 3 \]

Posted again see here http://www.mathlinks.ro/Forum/viewtopic.php?p=508186&sid=65c34b3dd83f6dddd7f06a20c8d95956#p508186


calculus
integration
trigonometry
limit
absolute value
real analysis
real analysis unsolved

problem: lim(x->0,x=/=0) (1/(x)^2)*[integral(0 to x)tsin(1/t)]?

solution:-|t|<=tsin(1/t)<|t|=> -|t|<=integral(0 to x)tsin(1/t)<=|t|=> -1/|x|<=integral(0 to x)tsin(1/t)<=1/|x| and when x->0 it follows the limit  of the integral is zero.

1)lim(a->0)integral(0 to a)sin(1/x)dx=?

2)f :[0,1]->reals, continuous and a_n=[integral(0 to 1)|f(x)|^ndx]^(1/n). find lim a_n.

3)a_n=integral(0 to +infinity)1/(1+x^n)dx. a)prove {a_n}(n=2 to +infinity) is convergent b)guess its limit c)justify it

Here is the solution to problem number 1.

The limit is 0.

$I(a)=\int_{0}^{a}\sin\frac{1}{x}dx=\int_{1/a}^{\infty}\frac{\sin y}{y^{2}}dy$.
Thus,

$|I(a)|\leq \int_{1/a}^{\infty}\frac{dy}{y^{2}}=\frac{-1}{y}\biggm|_{1/a}^{\infty}=a$

It follows that

$\lim_{a\rightarrow 0}I(a)=0$.

[quote="martin1985"]problem: lim(x->0,x=/=0) (1/(x)^2)*[integral(0 to x)tsin(1/t)]?

solution:-|t|<=tsin(1/t)<|t|=> -|t|<=integral(0 to x)tsin(1/t)<=|t|=> -1/|x|<=integral(0 to x)tsin(1/t)<=1/|x| and when x->0 it follows the limit  of the integral is zero.[/quote]
It is true that $\lim_{x\to 0}\frac1{x^{2}}\int_{0}^{x}t\sin(1/t)dt = 0$, but your solution is not correct.

[quote]2)f :[0,1]->reals, continuous and a_n=[integral(0 to 1)|f(x)|^ndx]^(1/n). find lim a_n.[/quote]
Let $M$ be the maximum value of $|f(x)|$.  Note that $a_{n}\leq M$ for all $n$.

Since $f$ is continuous, for any $\epsilon>0$ you can pick an interval $I\subset [0,1]$ of length $\delta>0$ on which $|f(x)|>M-\epsilon$.  Then

$a_{n}\geq \left(\int_{I}|f(x)|^{n}dx\right)^{1/n}\geq \delta^{1/n}(M-\epsilon).$

So $\varliminf a_{n}\geq \lim \delta^{1/n}(M-\epsilon) = M-\epsilon$.  Since this holds for all $\epsilon>0$, it follows $\varliminf a_{n}\geq M$, and therefore $\lim a_{n}=M$.

[quote]3)a_n=integral(0 to +infinity)1/(1+x^n)dx. a)prove {a_n}(n=2 to +infinity) is convergent b)guess its limit c)justify it[/quote]
Show that $\int_{1}^\infty \frac{dx}{1+x^{n}}\to 0$, and that $\int_{0}^{1}\frac{dx}{1+x^{n}}\to 1$.

why the solution of "problem" isn't correct?

[quote="martin1985"]solution:-|t|<=tsin(1/t)<|t|=> -|t|<=integral(0 to x)tsin(1/t)<=|t|=> -1/|x|<=integral(0 to x)tsin(1/t)<=1/|x| and when x->0 it follows the limit  of the integral is zero.[/quote]
$|t\sin(1/t)|\leq |t|$ is correct, but the rest of it isn't.  You will need to use more than the fact that $|\sin(1/t)|\leq 1$.  Let

$F(x)=\frac1{x^{2}}\int_{0}^{x}t\sin(1/t)dt = \frac1{x^{2}}\int_{1/x}^\infty \frac{\sin t}{t^{3}}dt$,

and use the fact that for $k\in\mathbb{Z}$, $k>0$,

$0\leq\int_{2k\pi}^{(2k+2)\pi}\frac1{t^{3}}\sin tdt \leq \frac2{(2k\pi)^{3}}-\frac2{((2k+2)\pi)^{3}}$

to conclude $0\leq F(1/(2k\pi)) \leq 1/(\pi k).$

A simpler approach: integrate by parts.

$\int_{0}^{x}t\sin\left(\frac1t\right)\,dt=\int_{0}^{x}t^{3}\cdot\frac1{t^{2}}\sin\left(\frac1t\right)\,dt$

$=\left.-t^{3}\cos\left(\frac1t\right)\right|_{0}^{x}+\int_{0}^{x}3t^{2}\cos\left(\frac1t\right)\,dt.$

The absolute value of the first term (the boundary terms) is no larger than $x^{3}.$ The absolute value of the second term is bounded by

$\int_{0}^{x}3t^{2}\,dt=x^{3}.$

Hence both terms are $O(x^{3})$ as $x\to0.$ The desired limit is $\lim\frac{O(x^{3})}{x^{2}}=0.$

Here is a generalization of problem number 3.

Let $k\geq 1$ be a natural number. Calculate:

$\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{dx}{(1+x^{n})(1+x^{2n})\cdots(1+x^{kn})}$.


maybe a picture might help, but this is pretty simple

I'm not sure about this, but you could also try making a v vs. t graph and finding the area under it (this is equal to position).

[quote="mdk"]A 7200 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.30m/s^2 and feels no appreciable air resistance. When it has reached a height of 505m , its engines suddenly fail so that the only force acting on it is now gravity.

How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?

How fast will it be moving just before it crashes?[/quote]

1) Till h=505m
v2-u2=2ah

Solve for v

2) Now, We know v, a=-g (-9.8 or -10), h=-505

h=ut+.5at2

Solve for t,I think you will get one negative value which you can neglect.

3) Velocity at bottom
Solve by energy conservation principles or by v2-u2=2ah.

that works. Now look at the other problem.

[quote="mdk"]A 7200 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.30m/s^2 and feels no appreciable air resistance. When it has reached a height of 505m , its engines suddenly fail so that the only force acting on it is now gravity.

How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?

How fast will it be moving just before it crashes?[/quote]

First, solve $ v_{h}^{2} \equal{} 2ah$ for $ v_h$.

Second, solve $ h \plus{}v_h t \minus{} \frac{1}{2} g t^2 \equal{} 0$ for $ t$.

Finally, solve $ v\left(t\right) \equal{} v_h \minus{} g t$ for $ v\left(t\right)$.


geometry
analytic geometry
calculus
calculus computations

Let P ( t , f ( t ) ) be any point on the curve f(x) = x / ((1 + X^2)^2)

and let Q be the point at which the y-axis intersects the line that is tangent to the given curve at the point P . Suppose that A ( t ) denotes the area of the triangle OPQ , where O = ( 0 , 0 ) . Is there an absolute maximum value for the area A ( t ) ? If so, find this absolute maximum value and determine the coordinates of all the points P for which that maximum is reached.

kenn, you are given $ f(x) \equal{} \frac{x}{(1\plus{}x^2)^2}$.

Your area equation is wrong.  If we let $ b\equal{}P_y \minus{} Q_y$ the we have $ b\equal{}f(t) \minus{} (f(t)\minus{}tf'(t)) \equal{} tf'(t)$ and $ h\equal{}t$ so $ A(t) \equal{} \frac{1}{2}bh \equal{} \frac{1}{2}t^2 f'(t)$.

So now:
$ A(t) \equal{} \frac{1}{2}t^2 \bigg(\frac{1\minus{}3t^2}{(1\plus{}t^2)^3}\bigg)$
$ A(t) \equal{} \frac{t^2\minus{}3t^4}{2(1\plus{}t^2)^3}$
$ A'(t) \equal{} \frac{(1\plus{}t^2)^3 (2t\minus{}12t^3) \minus{} 3(1\plus{}t^2)^2(2t)(t^2\minus{}3t^4)}{2(1\plus{}t^2)^6}$
$ A'(t) \equal{} \frac{t[(1\plus{}t^2) (1\minus{}6t^2) \minus{} 3(t^2\minus{}3t^4)]}{(1\plus{}t^2)^4}$
$ A'(t) \equal{} \frac{t[1\minus{}8t^2\plus{}3t^4]}{(1\plus{}t^2)^4}$

so $ t\equal{}0$ or $ t^2 \equal{} \frac{8 \pm \sqrt{52}}{3} \equal{} \frac{4 \pm \sqrt{13}}{3}$ so $ t \equal{} \pm \sqrt{ \frac{4 \pm \sqrt{13}}{3}}$ which gives $ t \approx \pm 1.592226039$ or $ t \approx \pm .36260572$, which give $ A(0)\equal{}0$, $ A(\pm 1.592226039) \approx \minus{}.1895189$ and $ A(\pm .36260572) \approx .02748187$.  So the maxiumum area (area is always positive) is approximately $ .1895189$ given by $ t \equal{} \pm \sqrt{ \frac{4 \plus{} \sqrt{13}}{3}}$


[quote][color=darkred]Let $ ABCD$ be a cyclic convex quadrilateral. Denote the intersection $ E\in AC\cap BD$ .

Consider two mobile points $ M\in (AD)$ and $ N\in (BC)$ so that $ E\in MN$ .

Ascertain the minimum of the sum $ \frac {MA}{MD} \plus{} \frac {NB}{NC}$ and when it is touched.

What happens when $ ABCD$ is [b]no[/b] cyclically ?[/color][/quote]

hmm.. :maybe:

$ \frac {MA}{MD}\plus{}\frac {NB}{NC}\equal{}\frac {AE\sin (\angle AEM)}{ED\sin (\angle DEM)}\plus{}\frac {BE\sin (\angle DEM)}{EC\sin (\angle AEM)} \ge \frac {2AE}{DE}$

Minimum when AE=DE

[quote="Virgil Nicula"][quote="hell_ever"]$ \frac {MA}{MD} \plus{} \frac {NB}{NC} \equal{} \frac {AE\sin (\angle AEM)}{ED\sin (\angle DEM)} \plus{} \frac {BE\sin (\angle DEM)}{EC\sin (\angle AEM)} \ge \frac {2AE}{DE}$ Minimum when AE=DE[/quote]
why ?! Explain please larger ..[/quote]

$ \frac {MA}{MD} \plus{} \frac {NB}{NC} \equal{} \frac {AE\sin (\angle AEM)}{ED\sin (\angle DEM)} \plus{} \frac {BE\sin (\angle DEM)}{EC\sin (\angle AEM)}$
$ \equal{}\frac {AE.EC \sin^2 (\angle AEM)\plus{} BE.ED \sin^2 (\angle DEM)}{DE.EC\sin (\angle AEM) \sin (\angle DEM)}$

Now, $ AE.EC\equal{}BE.ED$ and $ \sin^2 (\angle AEM)\plus{} \sin^2 (\angle DEM)\ge 2\sin (\angle AEM) \sin (\angle DEM)$ by AM-GM inequality.

$ LHS \ge \frac {2AE.EC}{DE.EC}\equal{} \frac {2AE}{DE}$

Again why minimum is touched when $ AE \equal{} DE$ ?!

And what happens when $ ABCD$ is no cyclically ? See the enunciation.


AMC
AIME
number theory
relatively prime
AIME I

Can anyone help me with the solutions to these problems?

6. For how many ordered pairs (x,y) of integers is it true that 0 < x < y < 10^6 and that the arithmetic mean of x and y is exactly 2 more than the geometric mean of x and y?

7. Suppose that x, y, and z are three positive numbers that satisfy the equations xyz = 1, x + 1/z = 5, and y + 1/x = 29.  Then z + 1/y = m/n, where m and n are relatively prime positive integers.  Find m + n.

If you are looking for the solutions, then the wiki has many of them:
[list]
[*][url=http://www.artofproblemsolving.com/Wiki/index.php/2000_AIME_I_Problems/Problem_6]Problem 6[/url]
[*][url=http://www.artofproblemsolving.com/Wiki/index.php/2000_AIME_I_Problems/Problem_7]Problem 7[/url][/list]


Let  x,y,z>0. Show  that
Σ{x/[x+sqrt(x+y)(x+z)]} \leq 1

By Cauchy, (x+y)(x+z)>=(x+ \sqrt yz)^2,

And GM>=HM,  \sqrt yz>=2yz/(y+z).

So

x/(x+ \sqrt (x+y)(x+z) )<=x/(x+ (x+ \sqrt yz))

=x/(2x+ \sqrt yz)<=x/(2x+ 2yz/(y+z) )=(xy+xz)/2(xy+yz+zx)

Therefore  \sum x/(x+ \sqrt (x+y)(x+z) )<= \sum (xy+xz)/2(xy+yz+zx) =1, as desired.

Thanks.

You may easily prove that there exists a triangle ABCsuch that
    \sqrt ((x+y)(x+z)/x^2) = 1/cosA,......
and you just have to prove this easy inequality
    \sum 1/(1+1/cosA) <=1.
 :D  :D