geometry
3D geometry
trigonometry
power of a point

I can do the last two  :D as well as the first
[hide]1. We know that $60^3=2^63^35^3$.  Thus, the number of divisors is $7*4*4=112$ (in general, for any $a_1^{b_1}a_2^{b_2}...a_n^{b_n}$ with each a as a prime, it will be $(b_1+1)(b_2+1)...(b_n+1)$ because each factor can either be not part of the divisor, or one of the b powers)

4. Using Power of a Point,
$2.5(92.5)=10x$
$x=\frac{185}{8}$
Therefore, the length of the chord is $\frac{265}{8}$

5. The radius must be 4 because the distance of the center from the x-axis is 4.  Thus,
$x^2+y^2+6x-8x+9=0$[/hide]

well the firs one is easy: [hide] ok to make things easier, you disregard the power that 60 is raised to. then you prime factorize 60. you should get (2^2)(3)(5).  now you multiply each of the exponents by 3, since this is the power that the  60 is raised to. you should now have (2^6)(3^3)(5^3).  to find the number of positive divisors you add one to all the exponents and then u multiply them. this will result in (7*4*4) which equals 112. and that is your answer.[/hide]

the second one is also easy if u know this trick: [hide] to begin you need to know that sin(x)=cos(90-x) and cos(x)=sin(90-x) and finally that (cosx)^2+(sinx)^2=1.  ok now that this is known you change all the (sinx)^2 for 45<x<89  into their corrosponding cosine measure meants.  you then add these to the isne with the same measurement. all the terms will be the sum of a series of 1 ones. for all the terms above 90 u will need a calculator. sorry  but anyway for the terms below 91 the sum is 44.5 i think someone else has posted shoing a more detailed answer.[/hide]

ok im going to skip to the last one: [hide] the desired circle is just  the equation of a circle with a radius of 4. so the equation is (x+3)^2 + (y-4)^2 = 16. you can expand this out and put it into the  desired form.[/hide]

ok the second to last one:[hide] before i begin there is something that you need to know. ok lets say that you are given a circle. and this circle has 2 chords drawn in it so that they intersect.  lets lable the chords AB and CD (these are the endpoints that are on the circle) and they meet at point E.  what you need to know is this:  (AE)(BE)=(CE)(DE). and you solve for the missing value. i am sure that you are capable of doing this on your own. :lol: [/hide]

well sorry but i have not really given much thought to the other question if i am able to i will return and answer it if i can or if no one  else has already done so. well im off to bed now i how that this was helpful.

[quote="jiangm199104"]Hi, I have some problems, please help me to solve them
Let d (n) denote the number of positive divisors of n, including 1 and n, For example, d(12)=6, since 12 has 6 divisors; 1,2,3,4,6,12. Compute d (60³).

Compute sin²0°+ sin²1°+ sin²2°+ sin²3°+ sin²4°+…+ sin²99°

Each of the following are products of two primes.  Only one of these products can be written as the sum of the cubes of two positive integers. Which one?
   A 104729*8512481779		C. 104761*8242254443
   B.104729*8242254443		D. 104761*11401596337
   E. 104729*11401596337

Two and a half centimeters from the end of a 95 cm diameter of a circle, the diameter intersects a chord and splits the chord into two segments, one of which has length 10cm, how long is the chord?

Write the equation of a circle with center (-3,4) which is tangent to the x-aixs, Put your answer in the form x ²+y²+Dx+Ey+F=0
 :D[/quote]

For question 3 , the answer should be A . Say the two cubes are a^3 and b^3 , factorize it we get (a+b)(a^2 -ab + b^2 ) =(a+b){(a+b)^2 - 3ab)

Since all choices above are product of two primes , it follows that the smaller number is (a+b) and the bigger one is {(a+b)^2 -3ab}

Hence , just use mod 3 on [(a+b)^2]  -  [(a+b)^2 -3ab] and if it is 0 then it is the answer .Like for A , a+b = 104729 , (a+b)^2 -3ab = 8512481779  ,  mod 3 and you will get 1-1 =0 (mod 3) .


Prove that if a; b; c are positive real numbers, then

$\displaystyle \sqrt {a^4+b^4+c^4}+\sqrt {a^2b^2+b^2c^2+c^2a^2} \ge \sqrt{a^3b+b^3c+c^3a}+\sqrt {ab^3+bc^3+ca^3}$

See number 2 of http://www.mathlinks.ro/Forum/viewtopic.php?t=18909 . [[b]EDIT:[/b] Of course, I have forgotten to say that the inequality is wrong in the form you give it, but it is clear what you mean.]

  darij

[quote="ehku"]a=b=c that inequality becomes
 $\sqrt{3}a^2+\sqrt{3}a \ge \sqrt{3}a^2+\sqrt{3}a^2$
or $1 \ge a$[/quote]

 Sorry I've edited it


function
integration
topology
real analysis
real analysis unsolved

Let $C[-1,1]$ be the whole sets of the continuous function on $[-1,1]$, for two elements $f,\ g$,

define the distance $d(f,\ g)=\sqrt{\int_{-1}^{1}(f(t)-g(t))^{2}\ dt}$.

Prove that the sequence of points $(f_{n})_{n\geq 1}$ of the metric space $(C[-1,1],\ d)$, for $n\geq 1$, defined by

\[f_{n}(t)=\left\{ \begin{array}{ll}-1 \left(-1\leq t\leq-\frac{1}{n}\right)&\quad \\ nt \left(-\frac{1}{n}\leq t\leq \frac{1}{n}\right)&\quad \\ 1\ \left(\frac{1}{n}\leq t\leq 1\right)&\quad \end{array}\right.\]

isn't convergent sequence but Cauchy sequence.

[hide]
Hint: The sequence converges to $\mathrm{sgn}(x)$ in $L^{2}[-1,1]$. Therefore, it is a Cauchy sequence with respect to $\|\cdot\|_{2}$ norm. Since its limit in $L^{2}[-1,1]$ is not a continuous function, the sequence does not have a limit in the space $(C[-1,1], \|\cdot\|_{2})$.[/hide]


Listed below are five functions, each donated $ g(x)$ and each involving a real number constant $ c > 1$. If $ {f(x) \equal{} 2^{x}}$, which of the five functions yields the greatest value for $ f(g(x))$, for all $ x > 1$?

A. $ g(x) \equal{} cx$

B. $ g(x) \equal{} \dfrac {c}{x}$

C. $ g(x) \equal{} \dfrac {x}{c}$

D. $ g(x) \equal{} x \minus{} c$

E. $ g(x) \equal{} log_{c}x$

Since $ x>1$, $ f(cx) \equal{} 2^{cx}$ gives the largest values; that is $ g(x) \equal{} cx$.

This can be seen by choosing a suitable value of $ c$, say $ c\equal{}2$ and sketching the graphs.


IN THIS PROBLEM THE LETTERS REPRESENT PIECES

I have the total number of pieces and the total of the pieces * the dollar amounts

I need the total number of each piece that = up to both the dollar amount and the total pieces.

Can anyone help with the problem or are their to many unknowns

n * 0.39 + o * 0.63 + p * 0.89 + q * 1.11 + r * 1.35 + s * 1.59 + t * 1.83 + u * 2.07 + v * 2.31 + w * 2.55 + x * 2.79 + y * 3.03 + z * 3.27

What is it that you are asking?

If you want to simplify the expression, you can't.

You can't solve it because there aren't any equations, only an expression.

Clarification  question: do you have a bunch of coins and want to figure out the number of coins that add up to a certain amount of money?

You can use a generating function in that case, which'll get you a recursive equation.


Okay,
This is a location specific question but does anyone know where in Chennai can one find good books on math(preferably libraries,if not,for sale) and do any libraries stock journals like Crux Mathematicorium or Kvant.If so,please let me know.

Cheers
Thanks
Bhargav

You can find all them on internet! I think so :P What is name of books you need?

Old issues of Crux Mathematicorum are I think available online. I think ISI Kolkata stocks Crux Mathematicorum, maybe IMSC does?
You should probably check it out.As for books - Easwara Bookhouse in West Mambalam has a great collection of Olympiad Math books. Problem sets are generally available online in large amounts though - check out old IMOers homepages maybe that'll help.
   AR

I am looking right now for books by Yaglom,a book on basic combinatorial methods by D.Cohen,and a copy of The IMO Compendium,(I dont think springer have an Indian edition and I dont know if I can get it online :P ).
Plus archives of kvant ,if they are available.
Thanks
Bhargav

This is one of what you want :D

http://kvant.mccme.ru/

I have some books by Yaglom. PM me if you want a photocopy. Check out the IMSc library for Cohen, I'm fairly sure they'll have it.
 Also you really must learn Russian - I can't translate .gif files from the Kvant website into English (!).

Thank you ,Tuan.(Unfortunate that I can't read russian,though).
And I ve PMed you Ashwath,thanks.

Cheers
Bhargav

I have some books on level Olympiad (e-book, of course :D ) Do You want have it? :P

[quote]Unfortunate that I can't read russian,though[/quote]
Yaglom and Yaglom books are available in English translation (I think I got a copy  in late 60's ... before most of the kids here were even born :) )

But learning Russian (specially for finding really inexpensive books in sciences and math) is really priceless.

True,and French too.You find so many more journals in french than in English.

Lanchka,

I sent you Crux 2004, please check your mail  :wink:

Nguyen could you send it to me also?
  AR

[quote="ashwath.rabindranath"]Nguyen could you send it to me also?
  AR[/quote]
Please send PM to Lanchka, because Lanchka so is from India  :wink:

[quote="Lanchka"]Thanks ,Big time! :D[/quote]
And  Now is MOSP. Please check your mail :P

well bhargav... u can  try out matscience that's there near IIT . well don't go there and tell u are there for olympiad maths u should be there to learn maths ...go and meet the director he is a really good person prof. Balasubramaniam if u know prof. ranganathan ur coordinator get an intro from him..get aquainted ...go to the library issue books in his name .. xerox them ...


yeah good luck on AMC 8 this week.

and the 10 and 12 if you make it

and AIME if you make that

and USAMO if you make that  :D

and IMO if you make that  :D  :D

...and oh wait, that's it.

[quote="mathcrazed"]I bet I miss one on the AMC 8... I bet... anything. It's gonna be pathetic[/quote]

don't say that to someone who gets 23's and 24's on school sprints...lol :(

[quote="infinity4ever"][quote="unimpossible"]

don't say that to someone who gets 23's and 24's on school sprints...lol :([/quote]

depression hurts, but you don't have to. but seriously arml is like years beyond my ability...  :(  so don't feel bad[/quote]

lol i did so bad at arml

mathcrazed can tell you :D

it's true... he did well during practices, and failed the real competition ~_-. Really unexpected

[quote="infinity4ever"]depression hurts, but you don't have to[/quote]

awww thanks.  :)

[quote="mathcrazed"]I bet I miss one on the AMC 8... I bet... anything. It's gonna be pathetic[/quote]
I bet I made 2 :wallbash_red:

[quote="DonkeyKong"]Gee... You guys are smart.. I know that I made one calculation error.. :blush:

And have a feeling that I missed 3 or 4 more...[/quote]

words of wisdom:  All math geniuses started off knowing nothing about math.  :D

[quote="unimpossible
words of wisdom:  All math geniuses started off knowing nothing about math.  :D[/quote]

Well, I guess so... Thanks for the encouragement.

[quote="unimpossible"]
words of wisdom:  All math geniuses started off knowing nothing about math.  :D[/quote]

Yeah, um, how would a single cell know anything about math, cause that's how math geniuses and all other people start.

But I know what you mean...

yayy longfellow is on the school honor roll for amc 8  :D


function
algebra
domain
calculus
calculus computations

I just got started on derivatives.For the most part, I did everything right, but there are some that gave different answers:

For 1) and 2), A function f is given along with a number (c) in its domain.
a. find the difference quotient of f.
b. Find f'(c) in its domain by computing the limit of the difference quotient.

1) f(x)=3 at c= -5. ans. a.0  b.0. I plugged 3 in for x since there was no x in f(x). so i got (change of x/change of x). I would have thought it would turn into one,

but somehow it is zero.

2) f(x)= 2-x^2 at c=0. ans is a. -(change of x). b.0. I got as far as getting 2x-(change of x). normally,(i thought) 2x would be the answer but, it is different and I

really do not know why.

For 3)Use the definition to differentiate the functions given.

3)(square root 5x). the ans.(square root 5x)/2x.I have no idea how it is that basically.

The function in 1) does not change with $ x$, as you have observed already. So, the change of this function is $ 0$; and $ 0$ divided by the change of $ x$ is still $ 0$.

2) here $ f(0)=2$. The difference quotient is $ \frac{f(x)-f(0)}{x-0}=\frac{2-x^{2}-2}{x-0}=-x$.

3) requires some nifty algebra. Writing $ h$ for the change of $ x$: $ f(x+h)-f(x)=\sqrt{x+h}-\sqrt{x}=\frac{(x+h)-x}{\sqrt{x+h}+\sqrt{h}}$. The rest is easier.


function
search
geometry
3D geometry
induction
algebra unsolved
algebra

[quote="hien"]Find all function $f: Z\rightarrow Z$ such that:
$f(x^{3}+y^{3}+z^{3})=f^{3}(x)+f^{3}(y)+f^{3}(z)$ for all integers $x,y,z$[/quote]
I am sure,this problem has been already posted,try to search! :wink:

Couldn't find it. What is the idea behind it ? The most I could to is to show that the only solutions are $0$, $id$ and $-id$, under the conjecture that for every positive integer $n>1$, $n^{3}$ can be written as the sum of five cubes of integers whose absolute values are less than $n$.
The conjecture seems reasonable but I cannot prove it.

Thanks, that's more or less what I thought, but I didn't carry on the induction !

[quote="hien"]It's not finished. Please try once?[/quote]
I gave you all to have a complete proof :)

Yeah, the indication provided by Aymane is almost the complete solution ...


geometry
geometric transformation
reflection
Euler
search
geometry unsolved

Consider a triangle ABC with P inside.Denote P' as the conjugate point of P.A' is the reflection of P to the side BC.B',C' is similar defined.
Prove that PP' is parallel to the Euler line of the triangle ABC if and only if  AA',BB',CC' is concurent.

Can anyone help me to solve it?
I'm just waiting. :maybe:

Dear Mathlinkers,
I think that this problem has been posted recently:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=131917132&t=269121Sincerely
Jean-Louis


do you really have a ps2 if you really have one your cool like me
but if you hate me its just ok by the way my name is .........  :ninja:

                      who want to be my friend???

What are you asking about the PS2? If your asking do you have one, I do. I like my X-BOX 360 better.


For questions that ask you how many different ways can you make x cents using quarters, dimes, nickels, and/or pennies, I know the methods to find the solution for 0-49 cents (thank you, mewto55555!) -

0-4 cents: One solution
5-9 cents: Two solutions
10-49 cents: ((n-5)((n-5)-1)/2)+3

Is there a method for finding the number of combinations using quarters, dimes, nickels, and/or pennies for values 50 cents and greater? Or do you have to list the possibilities out manually? Thanks!

What about half-dollars?  Should they be included?  This will change the outcome of the problem.

Listing out is really not that bad...refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=995545458&t=234273]this[/url] (especially my post).

I have found a way to solve the problem easily. I was in chemistry class today and started listing out the combinations for some amounts. I noticed a pattern in the amounts of combinations and formed the following conclusion-method.

[u]How to solve problems of said type[/u]:

First find the greatest multiple of 5 that is less than or equal to the amount you are trying to make.

Now consider this:

All of the possible coin combinations fit into exactly one of these two categories:

a. Includes quarters

b. Does not include quarters

Let's say you are trying to make 80 cents. We can find the amount of combinations in each of the two categories, and then add them together to find the answer. If you start listing the combinations in category b, this is what you get (D represents the number of dimes, N represents the number of nickels, and P represents the number of pennies):

[u]|  D  |  N  |  P   |
|  8  |  0  |  0  |
|  7  |  2  |  0  |
|  7  |  1  |  5  |
|  7  |  0  | 10  |
|  6  |  4  |  0  |
|  6  |  3  |  5  |
|  6  |  2  | 10  |
|  6  |  1  | 15  |
|  6  |  0  | 20  |[/u]

As you can see, there is 1 possible combination using no quarters and 8 dimes, 3 possible combinations using no quarters and 7 dimes, and 5 possible combinations using no quarters and 6 dimes. This pattern continues all the way to 17 possible combinations using no quarters and no dimes. Hmm...this really looks like a list of consecutive odd integers. Therefore, there must be 9^2 = 81 combinations in category b since 17 is the 9th positive odd integer!

We can expand on this idea. A goal of 90 cents would therefore have 100 combinations in category b, following the same logic. Therefore, to find the amount of combinations in category b for any amount ending in a zero, truncate the ones digit, add one to the remaining number, and then square the result.

But what if the goal ends in a 5? Listing the combinations for 85 shows this:

[u]|  D  |  N  |  P   |
|  8  |  1  |  0  |
|  8  |  0  |  5  |
|  7  |  3  |  0  |
|  7  |  2  |  5  |
|  7  |  1  | 10  |
|  7  |  0  | 15  |
|  6  |  5  |  0  |
|  6  |  4  |  5  |
|  6  |  3  | 10  |
|  6  |  2  | 15  |
|  6  |  1  | 20  |
|  6  |  0  | 25  |[/u]

As you can see, there are 2 possible combinations using no quarters and 8 dimes, 4 possible combinations using no quarters and 7 dimes, and 6 possible combinations using no quarters and 6 dimes. This pattern continues all the way to 18 possible combinations using no quarters and no dimes. Hmm...this really looks like a list of consecutive even integers. Therefore, there must be 9^2 + 9 = 90 combinations in category b since 18 is the 9th positive even integer!

We can expand on this idea. A goal of 95 cents would therefore have 110 combinations in category b, following the same logic. Therefore, to find the amount of combinations in category b for any amount ending in a zero, truncate the ones digit, add one to the remaining number, square the result, and then add on the truncated number resulting from the second operation.

If this doesn't make sense, please post and I will try to explain my thinking better.

Now that we can find the number of combinations in category b for any goal amount, we need to find out how to find the amount of combinations in category a for any goal amount. Listing for 80 shows this (Q represents the number of quarters, D represents the number of dimes, N represents the number of nickels, and P represents the number of pennies):

[u]|  Q  |  D  |  N  |  P   |
|  1  |  5  |  1  |  0  |
|  1  |  5  |  0  |  5  |
|  1  |  4  |  3  |  0  |
|  1  |  4  |  2  |  5  |
|  1  |  4  |  1  | 10  |
|  1  |  4  |  0  | 15  |[/u]

But wait! This appears to be listing the amount of combinations total for the number 55, which is 25 less than 80. We can easily find the number of combinations in category b for the number 55, but how will be find out for a? But wait...the amount of combinations in category a for 55 is the same as the amount of combinations in all for 30! WOW! That means that the amount of combinations in category a for 80 is equal to the amount of combinations in category b for 55, plus the number of combinations total for the number 30 (can be found by listing or using the formula)!

We can expand this idea into an algorithm! If the number was 85 and we wanted to find the number of combinations in category a, we would find the number of combinations in category b for 60 (85 - 20) and the number of combinations total for 35 (60 - 25) using the formula, and then sum the two values!

Using my method, the number of combinations total for any amount can be calculated! In fact, I could write a C++ or TI-84 program to do it instantly!

So what do you think of my method that I devised today in chemistry class? Think it is useful? Need me to explain it better? Just post here!

So let's see if I got this right...

Say the goal amount was 122c (with coins).
So, you take 120.
And the number of combinations for 120 is $ b(120)\plus{}b(95)\plus{}b(70)\plus{}b(45)\plus{}b(20)\equal{}169\plus{}110\plus{}64\plus{}30\plus{}9\equal{}382$ ways???

Woah... genius...  :D

That is correct! To check my logic I took numbers above 50 and found the number of combinations using listing and my method. The numbers were equal, supporting my logic.

So, does anybody have an instant "plug it in" formula?  I was reading markkram's post...and I got a little lost :blush:

For half-dollars, would you use the same idea?

The idea between 12markkram34's algorithm is pretty much completely based on making the problem easier

casework based on whether there's a quarter or not eliminates the quarter in one case

in the other case, you just take off the quarter and you get the exact same problem with a smaller number

There also exists the method of generating functions, which can't be discussed in much detail here.  However, learn about it and then you can come back to this problem.

Although for coin problems it is pretty tedious...

Guys...12markkram34 posted almost exactly the same thing as I posted in the thread I mentioned earlier...

that is pretty good,
you did that in chemistry?

In chemistry, when he was bored probably :wink:

So you mean I "re-invented the wheel" and came up with something that already exists?

Well, these listing methods have been used by many people before (i.e. computers use them sometimes (more efficient methods now probably)), but it's good that you came up with it by yourself. You always learn more from it if you teach yourself. :)


Ben has a bag which at the start contains 10 red marbles and 12 white marbles. When he draws out a red marble, he puts the marble back and adds 5 more red marbles. When he draws out a white marble, he puts the marble back and adds three more white marbles. What is the fewest number of rounds of draws/replacements/additions after which the bag could contain exactly 43 marbles?

[hide]Ther bag contains 22 marbles, and needs 21 more to have 43. Each time, either 3 or 5 marbles can be added. 21 can be expressed as $7(3)$, or $3(5)+2(3)$. The second way is shorter, and the answer is $3+2=\boxed5$.[/hide]


inequalities
function
integration
topology
Functional Analysis
real analysis
real analysis unsolved

1. from the post:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=129122[/url]

[quote]If $f \in C^{1}[x,y]$,
cauchy schwarz $|f(x)-f(y)|\leq \sqrt{|x-y|}\sqrt{|\int_{x}^{y}f'^{2}|}$[/quote]

How do we get this inequality? What it has to do with Cauchy-Schwarz in Hilbert space?

2. from post:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=borsuk&t=64994[/url]
[quote="grobber"]
Let $\mathcal T$ be the triangle $\{(a,b)\in[0,1]^{2}\ |\ x\le y\}$. On $\mathcal T$ we define $\varphi(a,b)=\left(\frac{1}{2}-\int_{a}^{b}f,\frac{1}{2}-\int_{a}^{b}g\right)$. Since $\varphi$ is constant on $\ell=\{(x,y)\in\mathcal T\ |\ x=y\}$, it's clear that $\varphi$ induces a continuous map $\tilde\varphi$ from $\mathcal D=\mathcal T/\ell$ to the plane, and that $\mathcal D$ can be regarded as a disk in which the points $(0,y),(y,1)$ from $\mathcal T$ are antipodal.[/quote]

$\mathcal D=\mathcal T/\ell$ means quotient? what is the equivalent classes here? why $\mathcal D$ is a disk? And why $(0,y),(y,1)$ from $\mathcal T$ are antipodal?

1) In a Hilbert space we have $|\langle u,v\rangle|\le \parallel u\parallel \parallel v\parallel $. An example is the space $L^{2}[x,y]$ of square-integrable functions on $[x,y]$, with inner product $\langle u,v\rangle=\int_{x}^{y}u(t)v(t)dt$ (I consider real-valued functions). Use the Cauchy-Schwarz in this space with $u(t)=1$ and $v(t)=f'(t)$.

2) Yes, it's a quotient. The equivalence classes are off-diagonal points ($1$-point classes), and the entire diagonal $y=x$. One side of the triangle collapses into a point, but this does not change its topology. For instance, $(x,y)\mapsto (x(y-x),y(y-x))$ is a homeomorphism of $\mathcal{D}$ onto a topological disk in the plane.

Try "Topology" by Munkres.


calculus
inequalities
trigonometry
function
real analysis
real analysis solved

It has been a long time since I have been in a math class, let alone any classes at all. My son showed me this problem yesterday and it just completely boggles me. The part where it gets tricky is the width of the bar! I just don't understand how to solve this. Would anyone here be kind enough to show me the steps in solving this and what the answer would be? Someone has already told me 8.45 meters or so... But I am interested personally in the steps you would take to get to that conclusion. (I tried to figure out the answer to this all night.. I just want to know how to do it for peace of mind.. lol)

Thank you!
++++++++

You have a steel beam, 25m long.
A corridor, 13m wide.
A second corridor, perpendicular to the first.
How WIDE must the second corridor be to allow the steel beam to pass through?

Actually the problem is incomplete: the height of the corridor matters a lot too (if it is over 25 meters, you can just carry the beam in the vertical position). Or should we assume that the beam is so heavy that we cannot lift it from the floor?

Very intelligent observation.  That was how the problem was presented, so I belive that we are to assume it is too heavy to lift vertical, or that the ceiling is too low. Either way, that is how I am interested in the problem being done.. because it's what i've been looking at.

Thank you for response  :)

Edit: I was also thinking, if there was no ceiling, then their would be no limitation right? You would just pick it up and walk around the corner, it would just have to be big enough for your body. So again, I am interested in the problem being solved with the assumtion that the it cannot be turned vertical!  ;)

[quote="Ravi B"]Similar questions have been solved in at least two topics before, for example
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12563]Moving In[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24520]Pole-Corner Problem[/url].[/quote]

I appreciate the post and resources. Even with those I would have to do this problem a little backwards.. I don't mean to sound too mundane  :?  But could somebody throw me a small bone and walk me through it?

Thank you for anyone willing to offer their. I'm trying to use the equations linked to figure it out.. my brain is just a little fried.

Here is one possible approach. The idea is that you should be able to place the beam at any angle $0\leq\alpha\leq\frac\pi 2$:

|13m|
|\**|
|*\*|
|**\|__________
|***\$\alpha$
|****\  $x$m
|_____ \_______
(Sorry for the poor picture.)
This gives the inequality $\frac x{\sin\alpha}+\frac{13}{\cos\alpha}\geq 25$ for all $\alpha$. The minimum of the left hand side is attained when $\tan\alpha=\sqrt[3]{\frac x{13}}$. Finding $\sin\alpha$ and $\cos\alpha$ from here and plugging them in, you get the inequality $(\sqrt[3]{13}\,x^{2/3}+13)\sqrt{\frac{x^{2/3}}{13^{2/3}}+1}\geq 25$, which is best solved numerically (though, in principle, it is cubic in $x^{2/3}$).

From the previous topics, I believe the equation for the minimum is
\[x^{2/3} + 13^{2/3} = 25^{2/3}.\]
Solving for x gives
\[x = (25^{2/3} - 13^{2/3})^{3/2} \approx 5.251080936.\]

A friend of mine told me.

" the function shall have 2 critical points- one of them is non differentiable- you have founded the max, which is the min? "

Isn't x^{2/3} + 13^{2/3} = 25^{2/3} the min?

He said that it is possible to write the equation in a more elegant and simple way. Does anyone have a clue what he might be talking about? 0.0

what is the target function and not only the first order conditions, how to get there and, the most important thing, What is the basic premise, to say, what's the true problem? If someone could help me with this.. i'd really appreciate it.

http://home.ripway.com/2004-6/130079/DSCF0003.JPG

I am trying to make it more simple.. but I am.. coming out with a differnet answer.

 :blush:

Is there any way someone could help me make this into one simple equation? Like the simplest equation you could think of?

I know no one has posted on this for a day now, would someone be kind enough to help me?


geometry
circumcircle
incenter
vector
geometry unsolved

$ \triangle ABC$ has the circumcenter $ O$ and the incenter $ I$.Suppose that $ AB<BC<CA$.Let $ X,Y$ respectively be the points on the rays $ BA,CA$ such that $ BX\equal{}CY\equal{}BC$.Prove that $ XY \perp OI$.

I tried to solve this problem by vectors,but it seems to be complex and nontrivial.Please help.Thanks.

Hi every Body! Can you give me another solution for this problem, please.

Hi livetolove212. Can you solve this problem and not use vector? Thank you very much.

See the names of points in the figure.
Denote $ M,N$ the midpoints of $ AB, AC$.
$ KJ\perp OI \Leftrightarrow KI^2 \minus{} JI^2 \equal{} OK^2 \minus{} OJ^2$ (Pythagore's theorem)
We have:
$ OK^2 \minus{} OJ^2 \equal{} OK^2 \minus{} OB^2 \plus{} OB^2 \minus{} OJ^2 \equal{} KM^2 \minus{} BM^2 \plus{} CN^2 \minus{} JN^2$
$ \equal{} (a \minus{} \frac {c}{2})^2 \minus{} \frac {c^2}{4} \plus{} \frac {b^2}{4} \minus{} (a \minus{} \frac {b}{2})^2$
$ \equal{} (p \minus{} c)^2 \minus{} (p \minus{} b)^2 \equal{} DC^2 \minus{} DB^2 \equal{} IC^2 \minus{} IB^2 \equal{} IK^2 \minus{} IJ^2$
We are done.


Can some one pm me nat MC problems form 1993-2005, please?

http://rapidshare.de/files/15124157/Mathcounts.rar.html

Credit biffanddoc. Also you need winRAR.

[quote="easyas3.14159..."]This is so off-topic, but do you know ccy tyy?[/quote]  :rotfl:


Set $L$ has $4n$ elements. Sets $A_0, A_1, ..., A_k$ each have $2n$ elements, and $A_i \subset L$, $\forall i = 0,1,...,k$.
Prove that $\exists i,j \in \{1,2,...,n\}$, $n(A_i\cap A_j) \geq (1-\frac{1}{k})n$

Fix k, n.   We have k = 0,1,2  is trivial so we dont consider.

Choose largest s so that k(2n - s) <= (4n-s).  Then it is obvious $s \geq \frac{2nk - 4n}{k-1} = 2n - \frac{2n}{k-1} = 2n(\frac{k-2}{k-1}) $ as well as $2(k-2)(k) \ge (k-1)^2$.  The result follows from $(A_i\cap A_j) \ge s$.


A $\pi^{+}$ meson of negligible initial velocity decays to a $\mu^{+}$ (muon) and a neutrino. With what energy does the neutrino move?

(The rest mass of neutrino can be considered zero. The rest energy of the $\pi^{+}$ meson is 140 MeV, and the rest energy of the muon is 106 MeV.)

Isn't this a simple subtraction? As long as none of the energy goes into kinetic energy of the muon, it should just be 34 MeV.

However, if conservation of momentum must be considered, it would be nice to know a few facts about neutrinos. Or maybe I'm missing the point.

Well, I think like you think. But since this is a 5-point problem in Komal, I'm not so sure.

The conservation of energy:
$m_{\pi}c^{2}=\sqrt{m_{\mu}^{2}c^{4}+p_{\mu}^{2}c^{2}}+p_{\nu}c$
since the conservation of momento:
$p_{\mu}=p_{\pi}$
So the energy of the neutrino:
$E_{\nu}=pc=\frac{m_{\pi}^{2}-m_{\mu}^{2}}{2m_{\pi}}c^{2}$

Why is the energy of the muon $\sqrt{m_{\mu}^{2}c^{4}+p_{\mu}^{2}c^{2}}$

And the information on the momentums isn't given anyway, so the problem wants us to ignore?

Well thats the relativistic energy of the muon:
$E_{rest}+E_{kinetic}$.
Since the momento of the system at start is zero, and momento is conserved (no outer force exerted on the system).


I really want to go to an Ivy League school. As far as I am concerned, I have a small disadvantage(?) of being an international student and having lived in the US for only 2 years. I heard that the competition for international students is much more fierce. Anyway, I would like to know your opinions on which I should for academically, like GPA, SAT I & II, APs... and other factors like extra curricular activities, competition... and what would make one stand out in the competition.

Thanks a lot.

Anh Nguyen

When you say you are an international student, you mean you are not a United States permanent resident ("green card" holder), right?

The competition is tough for international student spots at the top universities, but try to take the toughest courses you can (possibly at a local college) and do well in them. Then you may get into a second-tier school with a good reputation even if you don't have the luck to get into a top-tier school.

I'd say that the most important thing for you to do is to spend time learning English. That gets reflected into a lot of the application: your English grades, SAT score, essays, interview, etc. Of course you should also do extracurriculars and compete in fields you're talented in. Grades are always something that shows dedication (especially maintaining straight A's over long periods of time...like 12 years). AP's can be excellent padding for your application. You also want to do some volunteer work - a few hundred hours worth of it in fact.


I have known this problem for a while but still do not know how to solve it. Here it is:

There are 2 non parellel lines and a point X between these two lines. Find the line through X s.t. the triangle formed by this line and the 2 original lines has

a)the minimum area
b)the minimum perimeter.

The first part is relatively easy but the second part is more difficult(in my opinion, that is).

[quote="lammycakes"]
There are 2 non parellel lines and a point X between these two lines. Find the line through [b] X s.t. [/b] the triangle formed by this line and the 2 original lines has
[/quote]

Huh?

 :huh:


Say I use methyl phenyl ketone and ethanal for an aldol condensation. I am going to get two predominant product. Apparently one dehydrates more readily than the other. Is this true? Which one?

The a- hydrogen is selected from the ketone and not the aldehyde....

This is because assuming the a- Hydrogen to be selected from aldehyde....the next step would be the attack of the carbanion.... but ketones are less reactive....so the a- hydrogen is abstracted from the ketone...

[quote="BanishedTraitor"]methyl phenyl ketone[/quote]

Also known as acetophenone. In your reaction both enolates will form, although the one from ethanal more easily. Now consider the attack of ethanal enolate on acetophenone, where you will get

$ \ce{C6H5\minus{}C(OH)(CH3)CH2CHO}$.

Here it is easy to see that removal of the remaining alpha hydrogen is very favourable since it leads to an extended conjugated system phenyl-double bond-carbonyl.

Consider now the attack of acetophenone enolate on ethanal: you will get

$ \ce{C6H5\minus{}COCH2\minus{}CH(OH)\minus{}CH3}$

Here there is still one alpha proton that can be removed, perhaps not so easily as before, leading also to an extended conjugated system.

So the question is why one dehydrates more easily than another. I really don't know.

The answer can be seen in my previous post...

Can we explain in terms of which Hydrogen is more acidic?


I am reading http://www.imomath.com/tekstkut/projg_ml.pdf , but I don't quite understand the definition of perspectivity:
In the paper it is given as:
[b]Definition 3.[/b] Let each of $ l_1$ and $ l_2$ be either line or circle. Perspectivity with respect to the point $ S$
$ ^S = _{\^}$(don't know how to make the symbol) is the mapping of $ l_1 \to l_2$, such that:
(i) If either $ l_1$ or $ l_2$ is a circle than it contains S;
(ii) every point $ A_1 \in l_1$ is mapped to the point $ A_2 = OA_2 \cap l_2$

In (i): If both $ l_1$ and $ l_2$ are circles, should they then both contain $ S$? In (ii) What if $ OA_2 \cap l_2$ has more than one intersection point, which one should be chosen then? And what is $ O$ - why are we not needing $ S$?

Any alternative definition will be very much appreciated as well. :) I'm really strugling to understand this one.


geometry
trigonometry
Pythagorean Theorem

[quote="ffao"]In the diagram, $ ABCD$ is a square of side 1. If $ EF \equal{} 1$, what is the value of $ DF$?

[asy]pair A=(0,0), B=(0,1), C=(1,1), D=(1, 0), F=(1.88320350591352603, 0), E=(1, 0.4689899435404305);

dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
draw(A--B--C--D--cycle);
draw(B--E--F--D);

label("$A$",A,W); label("$B$",B,W); label("$C$",C,plain.E); label("$D$",D,NW); label("$E$",E,NE); label("$F$",F,plain.E);
label("1",A--B,W); label("1",A--D,S); label("1",E--F,NE);

label("$x$",D--F,S);[/asy]

(Try not to solve a cubic! Also, if this is too easy, please move.)[/quote]

is the answer $ x \equal{} \minus{} \frac {1}{2} \plus{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{2} \sqrt {2 \sqrt {2} \minus{} 1}$    ?

$ \angle{EFD} \equal{} \angle{EBC} \equal{} \theta \Longrightarrow x \equal{} \cos \theta$. $ CE \equal{} BC\tan \theta \Longleftrightarrow 1 \minus{} \sin \theta \equal{} \tan \theta$.

$ BA \equal{} AF\tan \theta \Longleftrightarrow 1 \equal{} (1 \plus{} \cos \theta)\tan \theta \Longleftrightarrow 1 \equal{} (1 \plus{} \cos \theta)\tan \theta$

Elliminating $ \tan \theta$ gives  $ 1 \equal{} (1 \plus{} \cos \theta)(1 \minus{} \sin \theta)\Longleftrightarrow \sin \theta \cos \theta \equal{} \cos \theta \minus{} \sin \theta$. Let $ \cos \theta \minus{} \sin \theta \equal{} t$, we have $ \frac {1 \minus{} t^2}{2} \equal{} t\Longleftrightarrow t^2 \plus{} 2t \minus{} 1 \equal{} 0$,

yielding $ t \equal{} \sqrt {2} \minus{} 1$ which satisfies $ 0 < t < 1$, or $ \cos \theta \minus{} \sin \theta \equal{} \sqrt {2} \minus{} 1\ \cdots [1]$. Use $ (\sin \theta \plus{} \cos \theta)^2 \plus{} (\sin \theta \minus{} \cos \theta)^2 \equal{} 2$ to get

$ \cos \theta \plus{} \sin \theta \equal{} \sqrt{2\sqrt {2} \minus{} 1}>0\ \cdots [2]$. From $ [1],\ [2]$, we have $ x \equal{} \cos \theta \equal{} \boxed{\frac {\sqrt {2} \minus{} 1 \plus{} \sqrt {2\sqrt {2} \minus{} 1}}{2}}$.

Comment: The similarity of $ \triangle{EDF}, \triangle{ECB}$ gives directly $ \sin \theta \cos \theta \equal{} \cos \theta \minus{} \sin \theta$.

Of course, the algebraic solution isn't too bad.

[hide="Solution"]Since $ \triangle DEF\sim\triangle CEB$ and $ BC \equal{} 1$, $ BE \equal{} \frac1x$.

From the Pythagorean Theorem,

$ 1 \plus{} (1 \plus{} x)^2 \equal{} \left(1 \plus{} \frac1x\right)^2\implies2 \plus{} 2x \plus{} x^2 \equal{} 1 \plus{} \frac2x \plus{} \frac1{x^2}$

$ \implies x^4 \plus{} 2x^3 \plus{} x^2 \minus{} 2x \minus{} 1 \equal{} (x^2 \plus{} x \plus{} 1)^2 \minus{} 2(x \plus{} 1)^2 \equal{} 0$

$ \implies(x^2 \plus{} x \plus{} 1)^2 \equal{} 2(x \plus{} 1)^2\implies x^2 \plus{} x \plus{} 1 \equal{} (x \plus{} 1)\sqrt2$

$ \implies x^2 \plus{} (1 \minus{} \sqrt2)x \plus{} (1 \minus{} \sqrt2) \equal{} 0\implies x \equal{} \frac {\sqrt2 \minus{} 1\pm\sqrt {2\sqrt2 \minus{} 1}}2$.

It's easy to check that $ \sqrt {2\sqrt2 \minus{} 1} > \sqrt2 \minus{} 1$, and we want the positive root, so the answer is

$ x \equal{} \boxed{\frac {\sqrt2 \minus{} 1 \plus{} \sqrt {2\sqrt2 \minus{} 1}}2}$.[/hide]

[quote="Arrange your tan"][quote="kurt.math"]

is the answer $ x \equal{} \minus{} \frac {1}{2} \plus{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{2} \sqrt {2 \sqrt {2} \minus{} 1}$    ?[/quote]

This answer is consistent (at least when I used it to a rounded 9 decimal places
to the right of the decimal point).

However, there is no claim that kurt.math worked it out himself, and there is
no supporting work at all for it (shown).

This could be from the answer key from "Geometry Beginner Exercises #31."
At any rate, I would want to see at least a framework of the solution
(hidden as this subsection prefers) if I were a passerby reading.

Also I don't know where ffao, the OP, brought in about not trying to solve a
cubic.  The set-ups I looked at kept on coming back (coming down) to
solving the same quartic equation, not a cubic equation.[/quote]

My solution is based on solving the quartic equation.


function
algebra
polynomial
algebra theorems

What is the Intermediate Value Theorem ?

if $f: I\to\mathbb{R}$ is a continuous function on some interval $I$ and if $a,b\in I$ and $f(a)\leq x\leq f(b)$ or $f(a)\geq x\geq f(b)$, then there is a $a\leq \xi\leq b$ s.t. $f(\xi)=x$.

I have 2 observations for what Peter said:

1) The IVT starts from the following [b]Property[/b] that is the Intermediate Value Property (aka IVP):
A function $f: I \to\mathbb{R}$ has the I.V.P. if and only if for all $a<b$, $a,b \in I$, and for all $\gamma \in (f(a),f(b))$ (if $f(a)>f(b)$ then the interval is $(f(b),f(a))$) there exists an $c \in (a,b)$ such that $f(c) = \gamma$. This is also called (in Romania) the Darboux Property. I'm not sure if this is also an international notation.

This definition obviously implies that all continous functions have the IVP, however there are some (rather complicated) examples of non-continous, but IVP functions.

2) The most times this is used (in olympiad problems in general) is the following: if you have an polynomial $P(X) \in \mathbb{R}[X]$, and for some two reals $a,b$ you find that $P(a)\cdot P(b)<0 $ then the polynomial has a root in the interval $(a,b)$.

[quote="jhaussmann5"]Valentin,
I believe Peter's statement is correct: he is stating the Intermediate Value [b]Theorem[/b], you are stating the Intermediate Value [b]Property[/b].

Thus, the theorem is that all continuous functions satisfy the IVP.  See, e.g. [url=http://www.math.uncc.edu/~hbreiter/m1120/lectures/lect16.htm]www.math.uncc.edu/~hbreiter/m1120/lectures/lect16.htm[/url][/quote]
Auch, I misread the topic's name. Indeed you are right, I really had the impression that we are reffering to the property. I'll edit my post to say that it actually refers to the property, not the theorem (I didn't knew there is a theorem for this corollary of IVP  :blush: ).


geometry
rectangle
number theory
greatest common divisor

[quote="linzk3"]A large rectangle is partitioned into four rectangles by two segments parallel to its sides.  The areas of three of the resulting rectangles are 15(top left), 18(top right), X(bottom left), and 54 (bottom right).  Find the area of the fourth Triange.  Give a second solution.[/quote]

[hide] Let's have the 4 dimension lengths be $a,b,c,d$
[code]
______________________
|          |           |
|          |c          |
|____a_____|_____b____ |
|          |           |
|          |d          |
|__________|__________ |
[/code]

Thus, $ac=15$
$bc=18$
$bd=54$
We're trying to find $ad$

$\frac{bd\cdot ac}{bc}=ad$
$\frac{54\cdot15}{18}=\boxed{45}$ [/hide]

[hide="hint"]Give a letter to each of the four segment inside the rectangle[/hide] [hide="procedure"]so we got

ab=15
bc=18
ad=X
cd=54

Since we need to find ad using ab,bcand cd we figure aout the algebraic solution as follows

ab * cd/bc = ad

Now we just substitute

[hide="answer"](15)(54)/(18)=45

so ad=X=45[/hide][/hide]

[quote="lotrgreengrapes7926"][hide] Also, if you find this on a competition, the fastest way would be to factor 15, 18, and 54.
$3\times5=15$
$3\times6=18$
$9\times6=54$
So, $5\times9=\boxed{45}$.[/hide][/quote]

thanks...i need 2 solutions though...do you know how to find another?

There's the find "ad" method, and lotrgreengrapes' method.

The greatest common divisor (actually the only common divisor)(gcd) of 15 and 18 is 3.  So, they must share a side, c, of length 3.

That means that a=5 and c=6.  Since c=6, d=9, and ad=5*9=45.

This IS a different method, if you're wondering (the other one uses algebra, this one uses number theory)


Αν το τριτοβαθμιο πολυωνυμο $ ax^3\plus{}bx^2\plus{}cx\plus{}d$ εχει τρείς ριζες στο διάστημα $ (0,1]$ να αποδειχθεί οτι $ ac\geq{bd}$
(Ελπιζω να βγήκε καλή)

Estw $ r_1 , r_2, r_3$ oi treis rizes tou polyonymou me $ 0 < r_1 , r_2, r_3 \leq 1$

Apo Viete's relations exoume oti:

                                         $ r_1 \plus{} r_2 \plus{} r_3 \equal{} \minus{} \frac {b}{a}$  <=>   $ b \equal{} \minus{} a(r_1 \plus{} r_2 \plus{} r_3)$

                                         $ r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1 \equal{} \frac {c}{a}$  <=>  $ c \equal{} a(r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1)$

                                         $ r_1r_2r_3 \equal{} \minus{} \frac {d}{a}$  <=>  $ d \equal{} \minus{} ar_1r_2r_3$.

Twra,antikathistwntas tis times twn b,c,d stin anisotita pou theloume na apodeiksoume pernoume oti:

                                        $ ac \geq bd$
                                        $ a^2(r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1) \geq a^2(r_1 \plus{} r_2 \plus{} r_3)r_1r_2r_3$
    Epeidi omws $ a\neq0$ arkei na apodeiksoume tin parakatw anisotita gia $ 0 < r_1, r_2, r_3 \leq 1$:

                                       $ r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1 \geq (r_1 \plus{} r_2 \plus{} r_3)r_1r_2r_3$

Epeidi $ 0 < r_1, r_2, r_3 \leq 1$ epetai oti $ 0 < r_1 \plus{} r_2 \plus{} r_3 \leq 3$ kai oti $ 0 < r_1r_2r_3 \leq 1$  .

Opote arkei na apodeiksoume oti:

                                       $ r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1 \geq 3r_1r_2r_3$

Apo AM-GM gia to deksi melos tis desired anisotitas exoume oti:

                                        $ r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1 \geq 3\sqrt [3]{(r_1r_2r_3)^2}$.

Opote arkei na apodeiksoume oti:

                                        $ 3\sqrt [3]{(r_1r_2r_3)^2} \geq 3r_1r_2r_3$ or
                                        $ (r_1r_2r_3)^2 \geq (r_1r_2r_3)^3$ or
                                        $ r_1r_2r_3 \leq 1$ pou isxyei.

~Kostas

Eνας πιο συντομος (οπως την κατασκευασα)
Eιναι $ \frac{1}{r_1}\plus{}\frac{1}{r_2}\plus{}\frac{1}{r_3}\geq{r_1\plus{}r_2\plus{}r_3}$ δηλαδή $ \frac{r_1r_2\plus{}r_1r_3\plus{}r_3r_2}{r_1r_2r_3}\geq{r_1\plus{}r_2\plus{}r_3}$ οποτε $ \frac{c/a}{\minus{}d/a}\geq{\minus{}b/a}$ αρα $ ac\geq{bd}$

Poly pio wraia/kompsi/grigori i lysi sou :)
Btw,theorw pws i askisi afti mporei na genikeftei(den eimai sigouros omws) ws eksis:

        An to polyonymo $ P(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ...a_0$ exei $ n$ rizes sto $ (0,1]$ tote na apodeixtei oti:

            $ a_na_{n \minus{} 2}...a_1 \geq a_{n \minus{} 1}a_{n \minus{} 3}...a_0$

Isws i parapanw na isxyei(epagwgika na vgenei kati isws?  :huh: ) mono se perittou vathmou polyonyma.Tespa,tha to koitaksw an isxyei kati tetoio...

~Kostas


function
limit
calculus
calculus computations

Let $f: \mathbb{R}\rightarrow\mathbb{R}$ be a continuous function (not necessarily differentiable), with $f'_{+}(x)>0$ for all $x$. Does this imply that $f$ is ascending?

Verifying your notation: $f'_{+}(x)=\lim_{h\to 0^{+}}\frac{f(x+h)-f(x)}{h}$, right?
Also, is "ascending" meant to be "increasing"?

For any $t\in\mathbb R$ the set $\{x\colon f(x)>t\}$ is open, hence a disjoint union of open intervals. If $(a,b)$ is such an interval, then $b=+\infty$, for otherwise $f(b)=t$ and $f_{+}'(b)\le 0$. Thus $\{x\colon f(x)>t\}=(a,+\infty)$.


If $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{z^{2}}$ and $x,y,z\in\mathbb{N}$, find the smallest possible value of $x+y+z$.

[hide]Is the answer $47$? If so, I'll post a solution. :maybe: [/hide]

OK.

[hide]After getting a common denominator on the left side and taking the reciprocal of both sides, we get:

$\frac{x^{2}y^{2}}{x^{2}+y^{2}}=z^{2}\implies z=\frac{xy}{\sqrt{x^{2}+y^{2}}}$

Since $x,y,z$ are all natural numbers, the left side must be a perfect square.  Since $x^{2}y^{2}$ is $(xy)^{2}$, $x^{2}+y^{2}$ must be one as well.  That makes $x$ and $y$ the two legs of a Pythagorean triple.

We have to make $x$ and $y$ as small as possible but still making $\frac{xy}{\sqrt{x^{2}+y^{2}}}$ natural.  From here it's trial and error, and we get $x=15$, $y=20$, and $z=12$, which makes the sum $47$.[/hide]

ya u also know this too

[hide]
$\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{z^{2}}$
$\Rightarrow\left(xz\right)^{2}+\left(yz\right)^{2}=\left(xy\right)^{2}$

EDIT: ya sorry about that i've seen this before but this is what i meant

$3^{2}+4^{2}= 5^{2}$ divide by $\left(3\cdot4\cdot5\right)^{2}$
$\Rightarrow\frac{1}{20^{2}}+\frac{1}{15^{2}}=\frac{1}{12^{2}}$

EDIT: thats what i meant with every pyth. triple works..sorry for the inconvenience..thx [b]t0rajir0u][/b]
[/hide]

[quote="SM4RT"]ya u also know this too

[hide]
$\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{z^{2}}$
$\Rightarrow\left(xz\right)^{2}+\left(yz\right)^{2}=\left(xy\right)^{2}$

All pythagorean triples work with this equation..substituting in the smallest one: $z=3,y=4,x=5$..you get your smallest value...
[/hide][/quote]

Clearly untrue.  The LHS is $369$ while the RHS is $400$.


I recommended AoPS to one of my friends (you can thank me later). Is there any way she can join Intro. to Geometry soon? (She wants to take what I'm taking.) Or will she have to wait until the Spring?
Also, I don't know what classes are right for her. She has completed an algebra course, so I'm guessing she's around my level, but I don't know exactly what she should take.

She'll have to wait until the fall, not the spring.  Also, the Intro Counting or Intro Number Theory classes might be more appropriate for her.

So are you allowed to ever sign up for a class that has already started?

If it's within 10 days of the class start date, yes (unless the class is full).


function
topology
calculus
calculus computations

Suppose $ (f_n)$ is a sequence of continuous real-valued functions on a metric space $ X$ and $ (f_n)$ converges uniformly to a function $ f$. Also all the terms of $ (f_n)$ are bounded. Prove that $ f$ is bounded.

So for every $ \epsilon > 0$, there exists $ N \in \mathbb{N}$ such that if $ n > N$ and $ x \in X$ then $ |f(x) \minus{} f_{n}(x)| < \epsilon$. Also $ |f_{n}(x)| \leq M_{n}$ for all $ n$ in $ \mathbb{N}$ and $ x \in X$. Then $ |f(x) \minus{} f_{n}(x)| < \epsilon$ for $ n > N$. So $ |f(x)| \leq M_n \plus{} \epsilon$. Let $ M \equal{} \min \{M_{n}: n \in \mathbb{N}, n > N \}$ by well ordering principle. Then $ |f(x)| \leq M \plus{} \epsilon$. Hence $ f$ is bounded.

Is this correct?

Choose $ M$ to be the maximum of the $ M_{n}$.  The proof seems fine.

$ |f(x)| \le M_n \plus{} \epsilon$ already shows that $ f$ is bounded, and no, you don't need to take either a minimum or a maximum, although you can show that $ \sup M_n$ exists using the Cauchy criterion for uniform convergence.

Could you do this: Let $ S \equal{} \{M_{n}: n \in \mathbb{N} \}$. Can we show that $ S$ is compact by considering an arbitrary open cover $ \{U_{\alpha}: \alpha \in A \}$. Show that it contains a finite subcover. And since $ S$ compact define a function $ f: S \to \mathbb{R}$ by $ f(M_{n}) \equal{} M_{n}$. And since this is continuous $ S$ is bounded. So $ \max S$ exists? And then we can show uniform boundedness.

Or Is $ S$ not necessarily compact? Can we add any conditions to ensure that $ S$ is always compact?

$ S$ isn't necessarily compact, and that's not relevant to the problem.  For example, if $ M_n \equal{} 2^{\minus{}n}$ then $ S$ doesn't contain its limit point $ 0$.  All you need is that $ S$ is bounded, which follows by the Cauchy criterion for uniform convergence.


[hide]
$\fbox{19}$
because
you can have sums of
$1$
to
$18$
and you just need $1$ more student
in order for the teacher to be right[/hide]

But doesnt the question say three students have the same sum, not 2.  I get 35 because the sums 2-17 can be achievable in 2 ways and 1 and 18 in 1 way, so 16*2+2=34, you only need one more sum and there must be one with three ways so 35.

[quote="Aznwithabrain"]oh forgot about the three person thing

[hide]
$18*2+1=\fbox{37}$
[/hide][/quote]

The numbers each student writes on the board must be distinct, so your answer is slightly off.

Like Art of Owna said, theres theres 1 number if sum is 1 or 18, 2 numbers if sum is 2 or 17, and for the rest theres $\ge$ 3 ways to do it. So $1+1+2+2 = 6$ students can write down the numbers w/ sum 1, 18, 2, 17, then there is $14 \cdot 2+1 = 29$ more, so answer = 35.

Or, take your number and subtract two to make up for that only 1 number can be done for sum = 1, 18.

It is impossible to get a sum of 18 as the question states that the digits are different.

[quote="zhou_qiang"]Each of the students in a class writes a different 2-digit number on the whiteboard. The teacher claims that no matter what the students write, there will be at least three numbers on the whiteboard whose digits have the same sum. What is the smallest number of students in the class for the teacher to be correct?[/quote]

Maybe the question in the source you have (if you didn't write this problem) says it has different digits, but there is nothing here that requires that the digits of each number be distinct.


inequalities
real analysis
real analysis unsolved

can infinity be a solution of $x \geq 2x$ ?

confused because if $x$ is infinity, you seems not be able to substract both sides by $x$ to have $0 \geq x$.

on the other hand $\infty \geq \infty$ holds.

I felt something is wrong here, but where?

Let's try to extend the set of real numbers by adding $\infty$ and $-\infty$ to it. The ordering of reals extends to the new elements in an obvious way: $-\infty<x<\infty$ for any $x\in\mathbb R$. But if you also want to do arithmetics with $\infty$ and $-\infty$, you'll have to come up with definitions for $x+\infty$, $\infty-\infty$, $2\cdot \infty$, $0\cdot \infty$, $\frac{\infty}{\infty}$... And if you want to use associativity, commutativity, or distributivity, it's your responsibility to make sure that these "laws" actually hold for your arithmetical operations. You can't take them for granted.

The answer is up to the choice of number system. For example, the real numbers has no infinity by archimedean property, while the extended real numbers has $\infty$ and $\infty \geq \infty$ holds. But be careful! As mlok pointed out, usual arithmetic manipulation may fail to give a well-defined quantity or equality, inequality, etc... (ex: you cannot have $0 = \infty$ by subtracting $\infty$ from both side of $\infty = 2\infty$.)


rotation calculus geometry rectangle integration algebra polynomial

Find the volume of the solid of revolution generated when the region bounded by y=sqrt(x-1), y=1, x=5 is revolved about the y-axis.

I am wondering if it's revolved about the y-axis... do I have to convert the y= equations to the x= form (e.g. y=sqrt(x-1) convert to x=y^2+1)? And if the problem was revolved around the x-axis instead... do I convert the x= equations to y= form?

My prof worked out some examples and she did it on some and some she didn't so I'm confused about this. If possible, can somebody please show me the step-by-steps to work this example out? Thanks.

Would it start out to be something similar to this?:
[img]http://img208.imageshack.us/img208/4281/eqld9.jpg[/img]

What you wrote there is not the volume rotated around the $y$-axis. Rather, it is the volume you get rotating the figure about the horizontal line $y=1.$

Now, back to rotating around the $y$-axis: you ask about working with $y$ in terms of $x$ and integrating $dx$ verus solving for $x$ in terms of $y$ and integrating $dy.$ They both work. One may be more convenient than the other, but not always the same one - that's a problem by problem judgment. Calculus textbooks tend to call one the "shell method" and the other the "washer method." You're going to ask me which is which and I'll say: don't make rules, draw pictures.

The first picture to draw is a flat one; a picture of the two-dimensional object before it's rotated. We see that this has three boundaries. It's bounded below by the horizontal line $y=1.$ It's bounded on the right by the vertical line $x=5.$ And it's bounded both above and to the left by the curve $y=\sqrt{x-1}$ which we could just as easily rewrite as $x=y^{2}+1.$ The figure has corners (this may take a little algebra) at $(2,1),\ (5,1)$ and $(5,2).$

Now we have to slice this flat object into thin strips and see what happens when we rotate those strips around the $y$-axis. You can either slice it vertically or slice it horizontally.

(To be continued.)

Let's try slicing into vertical strips. Each such strip is located at a horizontal location which we will call $x.$ Its width is $dx.$ It extends from $y=1$ to $y=\sqrt{x-1}$ so its length is $\sqrt{x-1}-1.$ Now rotate that thin strip about the $y$-axis. What you get is a thin cylindrical shell, something like the paper label on a metal can. The radius of this cylinder is the distance from the axis, namely $x.$ Slice this open and lay it out flat as a rectangle. The width of this rectangle is the circumference of the cylinder, namely $2\pi x.$ The height is the height of the strip: $\sqrt{x-1}-1.$ And the thickness is $dx.$ Hence our volume element is

${dV=2\pi x(\sqrt{x-1}-1})\,dx.$

Hence to compute the volume, you have to integrate this over all values of $x$ that matter - and that's why we needed the good two-dimensional picture. From that we see that $x$ ranges from 2 to 5. Hence

${V=\int_{2}^{5}2\pi x(\sqrt{x-1}-1})\,dx.$

Or, go back to the original flat picture and slice the other way. The slices are horizontal. Each slice has a vertical location $y$ and width $dy.$ The left edge of each slice is at $x=y^{2}+1$ and the right edge is at $x=5.$

Now rotate that slice about the $y$-axis. (You have to make the effort to visualize this.) If the slice had extended to the $y$-axis, you would get a thin disk. But it doesn't go all the way, so you get a "washer" a thin disk with a smaller disc removed from it. Looked at from above, it would be an annulus. The outer radius of this annulus is $5$ and the inner radius is $y^{2}+1.$ The area of an annulus is simply the area of the bigger circle minus the area of the smaller circle: $\pi5^{2}-\pi(y^{2}+1)^{2}.$ Now multiply this by the thickness $dy$ to get the volume element:

$dV=\pi(25-(y^{2}+1)^{2})\,dy.$

To get the whole volume, we must integrate this over all $y$ that matter. Consulting the flat diagram again, that would be $1\le y\le 2.$ So

$V=\pi\int_{1}^{2}(25-(y^{2}+1)^{2})\,dy.$

Since that's a polynomial integral, this second method is a little easier for this particular problem. That comment is made in hindsight, and either integral can be computed by ordinary means.

My #1 rule here: draw pictures. Draw good, careful pictures. If you need more than one picture draw it in more than one way. And visualize what you're doing.

Wow, I really appreciate the detailed explaination! It helped alot. However, I am still unsure about one thing.

For the dx function: [img]http://img291.imageshack.us/img291/149/xhk6.gif[/img] is that "x" always written as "x" alone for any dx function of the shell method or could it also be something like "4-x", "2-x", etc.? The lectures given by my prof says "x=distance from y-axis to the axis of revolution" so I am wondering if this "x" in this problem should be "x-1" for the dx function of shell method?
Sorry, if this doesn't make sense. :huh:

As I said, don't make rules - visualize.

Where is this thin vertical slice located? At $x.$ What's it being rotated around? The $y$-axis, otherwise known as the line $x=0.$ How far is the slice from the axis of rotation? $x,$ and that will be the radius of the cylindrical shell (meaning the circumference, which is what you'll actually use, is $2\pi x.$)

You'd have $2\pi(x-1)$ if you were rotating this figure about the line $x=1,$ and you'd have $2\pi(10-x)$ if you were rotating it about the line $x=10.$

Always try to make a picture when you do this rotation stuff.
Remember label on your picture what is R and r (outer and inner radius) or what is radius and length. Then figure it out using equations.


integration
function
inequalities
real analysis
triangle inequality
real analysis unsolved

Prove that $ \left(L^1[0,1],\|\cdot\|_1\right)$ is a normed space , where :

$ L^1 \cong$  Collection of equivalence classes according to the relation R , we say $ f \ R \ g$ iff $ f = g , a.e$ .

and $ \parallel{} f\parallel{}_1 = \int\limits_{0}^1 |f(t)| dt$ . (Leb Int )

Note that : $ f = [f]$ , the class of functions which they equal a.e

Just I need how to prove :

$ \parallel{}f\parallel{} = 0 \ \Longrightarrow [f] = [0]$

 :)

The key here is Chebyshev's (Markov's) Inequality: for any $ \alpha>0,$ $ m(\{|f|>\alpha\})\le\frac{\int|f|}{\alpha}.$

Hence if $ \|f\|_1\equal{}0,$ then for every $ \alpha>0,$ the set $ \{|f|>\alpha\}$ has measure zero. Use a sequence of $ \alpha_k$ that tend to zero and we have that $ \{|f|>0\}$ is a countable union of sets of measure zero, hence of measure zero.

This is what you wanted - the proof that $ |f|\equal{}0$ almost everywhere.

I'll assume that you already have the other norm properties proved: the scaling property and the triangle inequality.

It's the standard argument, and it applies equally well to $ L^p$ for all $ 1\le p<\infty.$ (You have to work harder to get the triangle inequality for $ p>1.$) It's also clear that we didn't need to specify $ [0,1].$ Any measure space would have worked.

The arguments have to be a little different for $ L^{\infty}$ but are generally a little easier. (As long as we don't go messing around with measure spaces that aren't $ sigma$-finite.)

Does that mean u can solve the 1st the previous condition without using Markov's inequality ?

I know how to prove the other conditions of normed for any $ p\geq 1$ .

Thx


Hello,

I don't know if the problem was done like that on purpose, or is a bug, but when I press one botton to include some comment (as hide or quote, or an emoticon) it is not included at the end of the text and not where the cursor is in that moment. It is a little uncomfortable because sometime I edit what I write (not all the time) and include such commands afterwards, and suddently the cursor goes to another place to introduce the command.

Best regards,

This is not new; it's always been like that. You could just learn the codes for the ones you use more often and type them in.

As jmerry said it's not a bug, rather an inconvenient way to insert smillaes. Will be changed at some point in the future :)


solve equation
$ 3x^2 \plus{} 11x \minus{} 1 \equal{} 13 \sqrt { 2x^3 \plus{} 2x^2 \plus{} x \minus{} 1} \equal{} 0$

are you sure about your problem  i think there is some mistakes


function
algebra
functional equation
algebra unsolved

Find all the continuous functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$ f(x)f(y)\equal{}f(\sqrt{x^2\plus{}y^2})$, for all $ x,y \in \mathbb{R}$

[hide="Solution"]Consider $ g(x)\equal{}f(\sqrt{x}) \Leftrightarrow f(x)\equal{}g(x^2)$. Making the substitution, we have $ g(x^2)g(y^2)\equal{}g(x^2\plus{}y^2)$. For $ g(k): k \geq 0$ (though this condition on $ x$ is irrelevant in the end), this is the Cauchy exponential equation with continuous solutions given by $ g(x)\equal{}b^x$ for $ x \geq 0$ and constant $ b \geq 0$ ($ b\equal{}0$ being the generate case, and $ b<0$ implying that the range of $ g(x)$ is not $ \Re$). (Making the substitutions $ a\equal{}x^2$ and $ b\equal{}y^2$ yields the standard form of the equation, $ g(a)g(b)\equal{}g(a\plus{}b)$, holding for $ a,b \geq 0$.)

We conclude that the solution to the functional equation is $ f(x)\equal{}g(x^2)\equal{}b^{x^2}$ for non-negative $ b$ (we can do this as $ x^2 \geq 0$). In particular, note that either $ \forall x, f(x)\equal{}0$ or $ \forall f(x), f(0)\equal{}1$.[/hide]

Thank you for the solution !!!

[url=http://www.mathlinks.ro/viewtopic.php?t=262799]Functional equation 1[/url] uses some of the ideas above.


algebra
polynomial
quadratics
Irrational numbers

Let $p(x)=x^{3}+a_{1}x^{2}+a_{2}x+a_{3}$ have rational coefficients and have roots $r_{1}$, $r_{2}$, and $r_{3}$. If $r_{1}-r_{2}$ is rational, must $r_{1}$, $r_{2}$, and $r_{3}$ be rational?

I assume it is given that there are real roots, else the answer is trivially no.

If I don't have calculation mistakes, this should do it. Denote $ r_{1}\equal{}a,r_{2}\equal{}b,r_{3}\equal{}c$ for simplicity, then we have $ a\minus{}b,a\plus{}b\plus{}c,ab\plus{}bc\plus{}ca,abc\in\mathbb{Q}$ by de Viete.

So, $ c^{2}\plus{}2ab\equal{}a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}(a\minus{}b)^{2}\equal{}(a\plus{}b\plus{}c)^{2}\minus{}2(ab\plus{}bc\plus{}ca)\minus{}(a\minus{}b)^{2}\in\mathbb{Q}$, and $ c^{2}\minus{}2ac\minus{}2bc\in\mathbb{Q}\Rightarrow c(c\minus{}2a\minus{}2b)\in\mathbb{Q}$. Now, $ c(a_{1}\minus{}3(a\plus{}b))\in\mathbb{Q}\Rightarrow c(a_{1}\plus{}3c)\equal{}3c^{2}\plus{}ca_{1}\in\mathbb{Q}\Rightarrow c^{2}\plus{}\frac{ca_{1}}{3}\in\mathbb{Q}$.

Since we had $ c^{2}\plus{}2ab\in\mathbb{Q}$ this gives $ 6ab\minus{}ca_{1}\in\mathbb{Q}\Rightarrow6ab\minus{}c(a\plus{}b\plus{}c)\in\mathbb{Q}$, adding the rationals $ c^{2}\plus{}2ab$ and $ ab\plus{}bc\plus{}ca$ this yields $ 9ab\in\mathbb{Q}$, so $ ab\in\mathbb{Q}$.

And then $ c^{2}\in\mathbb{Q}$, so $ a_{1}^{2}\plus{}ab\minus{}a_{1}(a\plus{}b)\equal{}(a_{1}\minus{}b)(a_{1}\minus{}a)\equal{}(c\plus{}a)(c\plus{}b)\equal{}c^{2}\plus{}ab\plus{}bc\plus{}ca\in\mathbb{Q}$, meaning $ a\plus{}b\in\mathbb{Q}$. Together with $ a\minus{}b\in\mathbb{Q}$, this solves the problem.

You mean $ a_{1}\equal{}a\plus{}b\plus{}c$?

Sorry, I thought $ p(x)\equal{}x^{3}\minus{}a_{1}x^{2}\plus{}a_{2}x\minus{}a_{3}$. But substituting $ a_{1}\mapsto\minus{}a_{1}$ should repair that.

Suppose that $ r_2\not\in\mathbb{Q}.$
Let $ r\equal{}r_{1}\minus{}r_{2}\in\mathbb{Q}.$
[b]Case: [/b]$ r\neq0$
$ \Rightarrow r_{2}$ is a root of the polynomial:
$ Q(x)\equal{}p(x\plus{}r)\minus{}p(x)\equal{}3rx^{2}\plus{}\left(3r^{2}\plus{}2a_{1}r\right)x\plus{}\left(r^{3}\plus{}a_{1}r^{2}\plus{}a_{2}r\right).$
Note that $ Q(x)$ is a quadratic polynomial with rational coefficients,
hence $ r_{2}\equal{}t\plus{}v$ for some $ t\in\mathbb{Q}$ and $ v$ such that $ v^{2}\in\mathbb{Q}.$
From $ r_{2}\not\in\mathbb{Q},$ we have $ v\not\in\mathbb{Q},$
therefore $ t\minus{}v$ is a root of $ p(x).$
$ \Rightarrow\{r_{1},r_{2},r_{3}\}\equal{}\{t\plus{}v,t\minus{}v,\minus{}a_{1}\minus{}(t\plus{}v)\minus{}(t\minus{}v)\}\equal{}\{t\plus{}v,t\minus{}v,\minus{}a_{1}\minus{}2t\}$
$ \Rightarrow r_{1}\minus{}r_{2}\not\in\mathbb{Q},$ contradiction.

[b]Case: [/b]$ r\equal{}0$
$ \Rightarrow p(x)$ has multiple root $ r_{2}\equal{}r_{1}$
$ \Rightarrow r_{2}$ is a root of the polynomial: $ R(x)\equal{}\frac{d\left(p(x)\right)}{dx},$
which $ R(x)$ is a quadratic polynomial with rational coefficients,
similarly to the first case, we get contradiction.
Therefore $ r_{2}\in\mathbb{Q},$
and $ r_{1}\equal{}r_{2}\plus{}r\in\mathbb{Q},$ $ r_{3}\equal{}\minus{}a_{1}\minus{}r_{1}\minus{}r_{2}\in\mathbb{Q}.$

[quote="wya"]
therefore $ t\minus{}v$ is a root of $ p(x).$[/quote]
Can somebody tell me why we have the above remark? :)


An eight-digit number contains two 5's, two 6's, two 7's and two 8's. The 5's are separated by one digits, the 6's are separated by two digits, the 7's and separated by three digits, and the 8's are separated by 4 digits. What is the greatest possible value for the number?

Since, even with these restrictions, we still have freedom, we place the largest digit, 8, in the first space, and another in the 6th, meeting the requirement.

8 _ _ _ _ 8 _ _

We would like to put a 7 in the second spot, but then the other 7 would have to be in the 6th spot, which is occupied. So, we place a 6 instead.

8 6 _ _ 6 8 _ _   However, we have no places to put the 7 and 5 now.

We must place a 5 in the second space.

8 5 _ 5 _ 8 _ _

We finish with the requirements for 7 and 6

8 5 7 5 6 8 7 6


I saw this problem recently somewhere, I had noted it down but have forgotten the source. I think it is some contest problem.

In triangle ABC, M is a point on AC such that the inradii of ABM and CBM are equal. Show that BM^2 ( read as BM squared) = cot(B/2) area ABC.

Sathej

See http://www.mathlinks.ro/Forum/viewtopic.php?t=60412 .

  darij


Let me get this straight.

Introductory Algebra consists of Algebrae 1 and 2.
Intermediate Algebra is Algebra 3.


algebra
polynomial
induction
algebra proposed

Is it possible to find cubic polynomial $ P$ such that, for any positive integer $ n$, the polynomial $ P(P(..(P(x))...)$ ($ n$ times $ P$) has exactly $ 3^{n}$ distinct real roots?

$ P(x): \equal{} x^3 \minus{} 3x$
has roots $ 0,\pm \sqrt {3}$,
and $ P(1) \equal{} \minus{} 2$ and $ P( \minus{} 1) \equal{} 2$ and $ P(2) \equal{} 2$ and $ P( \minus{} 2) \equal{} \minus{} 2$
and $ P(x)$ is strictly increasing on $ [1,\infty)$, as $ P'(x)>0$ for $ x>1$.

$ P(x)$ is odd, and so is $ P^n(x): \equal{} \underbrace {P\circ P\circ\cdots \circ P}_{n\text{ times}}(x)$ for all $ n\in\mathbb{N}$, and therefore has root $ 0$.
Take $ \alpha_n\ge 0$ the largest real root of $ P^n(x)$ for all $ n\in\mathbb{N}$.
Let's show $ \alpha_n < 2$ for all $ n\in\mathbb{N}$ by induction.

This is clear for $ n \equal{} 1$, as $ \alpha_1 \equal{} \sqrt {3} < 2$.
Assume $ 0\le\alpha_n < 2$, then $ P(1) \minus{} \alpha_n \equal{} \minus{} 2 \minus{} \alpha_n < 0$ and $ P(2) \minus{} \alpha_n \equal{} 2 \minus{} \alpha_n > 0$,
and using the intermediate value theorem we find $ \lambda\in(1,2)$ with $ P(\lambda) \equal{} \alpha_n$.
Then $ P^{n \plus{} 1}(\lambda) \equal{} P^n(P(\lambda)) \equal{} P^n(\alpha_n) \equal{} 0$, so $ \lambda$ is a real root of $ P^{n \plus{} 1}(x)$ and $ \alpha_{n \plus{} 1}\ge \lambda$ by choice of $ \alpha_{n \plus{} 1}$.
But if $ \alpha_{n \plus{} 1} > \lambda > 1$, then $ P(\alpha_{n \plus{} 1}) > P(\lambda) \equal{} \alpha_n$ and $ P(\alpha_{n \plus{} 1})$ is a real root of $ P^n(x)$.
This contradicts the choice of $ \alpha_n$, which gives $ \alpha_{n \plus{} 1} \equal{} \lambda < 2$.

And now by induction $ P^n(x)$ has $ 3^n$ different real roots for all $ n\in\mathbb{N}$.

Again this is clear for $ n \equal{} 1$, and if we assume $ P^n(x)$ has $ 3^n$ different real roots,
then for every such root $ \lambda\in\mathbb{R}$ we need to find $ 3$ different real roots of $ P(x) \minus{} \lambda\in\mathbb{R}[x]$.

But $ P^n(x)$ is odd, so $ \pm \lambda$ are both real roots of $ P^n(x)$ and $ \pm \lambda\le \alpha_n$ by choice of $ \alpha_n$.
This gives $ \minus{} 2 < \minus{} \alpha_n\le\lambda\le \alpha_n < 2$
and $ P( \minus{} 2) \minus{} \lambda \equal{} \minus{} 2 \minus{} \lambda < 0$
and $ P( \minus{} 1) \minus{} \lambda \equal{} 2 \minus{} \lambda > 0$
and $ P(1) \minus{} \lambda \equal{} \minus{} 2 \minus{} \lambda < 0$
and $ P(2) \minus{} \lambda \equal{} 2 \minus{} \lambda > 0$.
The intermediate value theorem completes the proof.


If the angles $ a,b,c,d,$ and $ e$ are known, what does the angle $ u$ equal?

[img]http://i250.photobucket.com/albums/gg265/geometry101/GeometryImage1.jpg[/img]

A. $ 108^\circ$
B. $ 2a$
C. $ 3a$
D. $ c\plus{}d$
E. $ a\plus{}c\plus{}d$

[quote="Silverfalcon"]If the angles $ a,b,c,d,$ and $ e$ are known, what does the angle $ u$ equal?

[img]http://i250.photobucket.com/albums/gg265/geometry101/GeometryImage1.jpg[/img]

A. $ 108^\circ$
B. $ 2a$
C. $ 3a$
D. $ c \plus{} d$
E. $ a \plus{} c \plus{} d$[/quote]

[hide="Solution"]
Let the star be $ ABCDE$, with $ \angle DAC \equal{} a$, $ \angle CEB \equal{} e$, $ \angle ADB \equal{} d$, $ \angle ACE \equal{} c$, and $ \angle DBE \equal{} b$.  Allow $ DB$ and $ EC$ to meet at $ F$ and let $ AC$ meet $ EB$ and $ DB$ at $ X$ and $ Y$ respectively.  Therefore, from $ \triangle FEB$, we have that $ u \plus{} e \plus{} b \equal{} 180\implies u \equal{} 180 \minus{} (e \plus{} b)$.  Also, $ a \plus{} d \equal{} \angle XYB$ and $ e \plus{} c \equal{} \angle YXB$.  Thus, $ 180\equal{}\angle B \plus{} \angle BXY \plus{} \angle BYX \equal{} b \plus{} (e \plus{} c) \plus{} (a \plus{} d)\implies (a \plus{} c \plus{} d) \equal{} 180 \minus{} (e \plus{} b) \equal{} u$.  Therefore, the answer is $ \boxed{E}$. [/hide]


How much days you don't go to school when you have math comp.(municipal, regional, republic )?
Does your teacher's give you extra A marks, at the end of year(from non-natural subjects)?

If you want you can start a poll about that.

I only had Flemish competition (and IMO in summer vacation) : I was dismissed for a couple of hours two times in my two years of participation, and one time I could get out half an hour earlier to eat a bit.

It never affected any results what so ever.

well, ussualy I don't go to school for one week when i have republick competitions. For bigger cometitions i don't go to school at all.

Sounds fun, reminds me of how the government cut all fundings for IMO participation, including plain tickets for the partipicants and team leaders, after inquiring what the Biology olympiad in Belgium in 2001 was about.  If IMO 2002 wasn't in Glasgow, I probably would not have participated.


Everyone should go here and take the Mock AMC:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1634175#1634175


inequalities
trigonometry
geometry
trig identities
Law of Sines
inequalities solved

Let x,y,z and a \leq b \leq c be the sides and the angles of a triangle. Prove that

(x^2+y^2+z^2)(1/x^2+1/y^2+1/z^2) \leq 9(1+(sena)^2)/4(sena)^2

we have sina \leq sinb \leq sic \leq 1
From the Law of Sines we have that  \sum x 2  *  \sum 1/x 2 =  \sum (sina) 2  *  \sum 1/(sina) 2
From P. Schweitzer's  inequality we have:
 (p+q+r)(1/p+1/q+1/r) \leq  9(m+M) 2  /4mM
if we take p=(sina) 2  q=(sinb) 2  r=(sinc) 2  m=(sina) 2   M=1
So we have:
 \sum (sina) 2 * \sum 1/(sina) 2   \leq  9(a+(sina) 2 ) 2 /4(sina) 2

Have you a proof of Schweitzer's inequality?

I am very interested in it.

Thank you very much.

Sorry, Schweitzer's inequality is very simple to prove by AM-GM.


[quote="shortlist"]Prove that exist infinite $ n \in N$ such that : $ 3^{n \minus{} 1} \minus{} 2^{n \minus{} 1}$ is composite[/quote]

Choose any prime $ n>3$ and apply Fermat little theorem.

[quote="shortlist"]Prove that exist infinite $ n \in N$ such that : $ 3^{n \minus{} 1} \minus{} 2^{n \minus{} 1}$ is composite[/quote]

you mean $ n$ is not prime ... right ??

[b]Hint[/b]: Choose $ n\equal{}3^{2^k}\minus{}2^{2^k}$  $ (k \in Z  \ ; \ k \ge 1)$

[quote="hophinhan"][quote="shortlist"]Prove that exist infinite $ n \in N$ such that : $ 3^{n \minus{} 1} \minus{} 2^{n \minus{} 1}$ is composite[/quote]

you mean $ n$ is not prime ... right ??

[b]Hint[/b]: Choose $ n \equal{} 3^{2^k} \minus{} 2^{2^k}$  $ (k \in Z \ ; \ k \ge 1)$[/quote]

Do you mean that $ 3^{3^{2^k} \minus{} 2^{2^k}\minus{}1}\minus{}2^{3^{2^k} \minus{} 2^{2^k}\minus{}1}$ is composite for any integer $ k\ge 1$ ?

[quote="pco"][quote="hophinhan"][quote="shortlist"]Prove that exist infinite $ n \in N$ such that : $ 3^{n \minus{} 1} \minus{} 2^{n \minus{} 1}$ is composite[/quote]

you mean $ n$ is not prime ... right ??

[b]Hint[/b]: Choose $ n \equal{} 3^{2^k} \minus{} 2^{2^k}$  $ (k \in Z \ ; \ k \ge 1)$[/quote]

Do you mean that $ 3^{3^{2^k} \minus{} 2^{2^k} \minus{} 1} \minus{} 2^{3^{2^k} \minus{} 2^{2^k} \minus{} 1}$ is composite for any integer $ k\ge 1$ ?[/quote]

any mistake, dear pco ?

$ (3^m \minus{} 2^n) \mid (3^{mk} \minus{} 2^{nk})$ and $ 2^k \mid (3^{2^k} \minus{} 2^{2^k} \minus{} 1 )$ $ \Longrightarrow (3^{3^{2^k} \minus{} 2^{2^k} \minus{} 1} \minus{} 2^{3^{2^k} \minus{} 2^{2^k} \minus{} 1}) \ \vdots  \ (3^{2^k} \minus{} 2^{2^k})$

[quote="shortlist"]Prove that exist infinite $ n \in N$ such that : $ 7^{n \minus{} 1} \minus{} 3^{n \minus{} 1}$ is composite[/quote]

Choose any prime $ n>7$ and apply Fermat little theorem.

For the second one. Choose $ n$ such that the expression is at least $ 5$ and $ n\minus{}1$ composite, factor as in elementary algebra.

:rotfl: what a funny !! @shortlist : if $ n$ is a prime $ (n > 7)$ : ${ 2^{n - 1} \equiv 3^{n - 1} \equiv 7^{n - 1} \equiv 1 \ (\ \text{mod} \ n})$

check your problem carefully   :rotfl: and read all posts above

for second one, similarly choose $ n=7^{2^k}-3^{2^k}$  $ (k\in N)$


The desired triangle $\triangle XYZ$ is centrally similar to the outer Napoleon triangle of $\triangle ABC.$ This is, construct outwardly on the sides of $\triangle XYZ$ the equilateral triangles $\triangle BCP,$ $\triangle CAQ$ and $\triangle ABR.$ Centers $O_1,O_2,O_3$ of these triangles form the outer Napoleon triangle. Through $A,B,C$ draw parallels to $O_2O_3,$ $O_3O_1,$ $O_1O_2,$ respectively. These three lines bound $\triangle XYZ.$

[b]Proof: [/b]Let $\triangle X_0Y_0Z_0$ be any equilateral triangle circumscribed to $\triangle ABC$ $\Longrightarrow$ $X_0 \in (O_1),$ $Y_0 \in (O_2)$ and $Z_0 \in (O_3).$ If $M,N$ are the midpoints of $AY_0,AZ_0,$ then from the right trapezoid $O_2O_3NM,$ we have $O_2O_3 \ge MN=\tfrac{1}{2}Y_0Z_0$ $\Longrightarrow$ $2 \cdot O_2O_3 =YZ \ge Y_0Z_0$ $\Longrightarrow$ $[XYZ] \ge [X_0Y_0Z_0].$


inequalities
trigonometry
inequalities proposed

Let x, y, z be positive reals. Prove that

$(xy+yz+zx)^3 \ge$

$xyz(\sqrt{(x+y)(x+z)}+x)(\sqrt{(y+z)(y+x)}+y)(\sqrt{(z+y)(z+x)}+z)$.

Substitute a = yz, b = zx, c = xy. Then $x = \sqrt{\frac{bc}{a}}$, $y = \sqrt{\frac{ca}{b}}$, $z = \sqrt{\frac{ab}{c}}$. And your inequality is transformed into

$\left( a+b+c\right) ^3 \geq$
$\sqrt{abc} \cdot \left( \sqrt{\frac{bc}{a} +a+b+c} + \sqrt{\frac{bc}{a}} \right) \cdot \left( \sqrt{\frac{ca}{b} +a+b+c} + \sqrt{\frac{ca}{b}\right) \cdot \left( \sqrt{\frac{ab}{c} +a+b+c} + \sqrt{\frac{ab}{c} \right)}}$.

But this follows from

$\left( a+b+c\right) ^3$
$= \left( \frac{\left( a+b\right) + \left( a+c\right)}{2} + \frac{b+c}{2} \right) \cdot \left( \frac{\left( b+c\right) + \left( b+a\right)}{2} + \frac{c+a}{2} \right) \cdot \left( \frac{\left( c+a\right) + \left( c+b\right)}{2} + \frac{a+b}{2} \right)$
$\geq \left( \sqrt{\left( a+b\right) \left( a+c\right)} + \sqrt{bc} \right) \cdot \left( \sqrt{\left( b+c\right) \left( b+a\right)} + \sqrt{ca} \right) \cdot \left( \sqrt{\left( c+a\right) \left( c+b\right)} + \sqrt{ab} \right)$ (after AM-GM, applied 6 times!)
$= \left( \sqrt{bc + a\left( a+b+c\right)} + \sqrt{bc} \right) \cdot \left( \sqrt{ca + b\left( a+b+c\right)} + \sqrt{ca} \right) \cdot \left( \sqrt{ab + c\left( a+b+c\right)} + \sqrt{ab} \right)$
$= \sqrt{a} \cdot \left( \sqrt{\frac{bc}{a} +a+b+c} + \sqrt{\frac{bc}{a}} \right) \cdot \sqrt{b} \cdot \left( \sqrt{\frac{ca}{b} +a+b+c} + \sqrt{\frac{ca}{b}\right) \cdot \sqrt{c}}$
$\cdot \left( \sqrt{\frac{ab}{c} +a+b+c} + \sqrt{\frac{ab}{c} \right)}$
$= \sqrt{abc} \cdot \left( \sqrt{\frac{bc}{a} +a+b+c} + \sqrt{\frac{bc}{a}} \right) \cdot \left( \sqrt{\frac{ca}{b} +a+b+c} + \sqrt{\frac{ca}{b}\right) \cdot \left( \sqrt{\frac{ab}{c} +a+b+c} + \sqrt{\frac{ab}{c} \right)}}$.

  Darij

Nice proof! I didn't thought about using AM-GM only could crack the problem.

My proof, by Cauchy-Schwarz,

$\sqrt{yz}(\sqrt{(x+y)(x+z)}+x) = \sqrt{z(x+y)}\cdot\sqrt{y(x+z)} + \sqrt{xy}\cdot\sqrt{zx}$

$ \leq \sqrt{(z(x+y) + xy) (y(x+z) + zx)} = xy+yz+zx$

Similarly, we have

$\sqrt{zx}(\sqrt{(y+z)(y+x)}+y) \leq xy+yz+zx$   and

$\sqrt{xy}(\sqrt{(z+y)(z+x)}+z)\leq xy+yz+zx$.

Multiplying all three together yields the desired result.

If we multiply both sides by $(\sqrt{(x+y)(x+z)}-x)(\sqrt{(y+z)(y+x)}-y)(\sqrt{(z+y)(z+x)}-z)$ and canceling $(xy+yz+zx)^3$ from both sides, the problem is transformed into:

[quote]Let $x,y,z>0$. Prove that

$(\sqrt{(x+y)(x+z)}-x)(\sqrt{(y+z)(y+x)}-y)(\sqrt{(z+y)(z+x)}-z) \geq xyz$.
[/quote]

Cauchy-Schwarz (or some other way) would easily give us $\sqrt{(x+y)(x+z)} \geq x + \sqrt{yz}$, therefore the result.

Finally, darij , congratulations! You just proved a nice trig inequality proposed by yourself!

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4972

Cool proof, too, but you actually didn't really use Cauchy-Schwarz. Instead you used the inequality

$\sqrt{a} \cdot \sqrt{b} + \sqrt{c} \cdot \sqrt{d} \leq \sqrt{\left( a+c\right) \cdot \left( b+d\right) }$

for any four real numbers a, b, c, d. And this is a very easy corollary from AM-GM: We rewrite the inequality as

$\sqrt{ab} + \sqrt{cd} \leq \sqrt{\left( a+c\right) \cdot \left( b+d\right)}$.

Now, squaring this yields

$ab + cd + 2 \sqrt{abcd} \leq \left( a+c\right) \cdot \left( b+d\right)$,

or, equivalently

$ab + cd + 2 \sqrt{abcd} \leq ab + ad + cb + cd$,

what simplfies to $2 \sqrt{abcd} \leq ad + cb$, and this directly follows from AM-GM.

So you see, our proofs are not so different...

  Darij

[quote="darij grinberg"]Cool proof, too, but you actually didn't really use Cauchy-Schwarz. Instead you used the inequality

$\sqrt{a} \cdot \sqrt{b} + \sqrt{c} \cdot \sqrt{d} \leq \sqrt{\left( a+c\right) \cdot \left( b+d\right) }$
[/quote]

Hmm, in my view, it is Cauchy-Schwarz.

Cauchy-Schwarz takes the general form in the inner product space, $\left|({\bf v}, {\bf w})\right| \leq \sqrt{({\bf v}, {\bf v})({\bf w}, {\bf w})}$.

In $\mathbb{R}^2$, ${\bf v} = (\sqrt{a}, \sqrt{c})$ and ${\bf w}= (\sqrt{b}, \sqrt{d})$ gives us $\sqrt{a} \cdot \sqrt{b} + \sqrt{c} \cdot \sqrt{d} \leq \sqrt{\left( a+c\right) \cdot \left( b+d\right) }$.

I guess in this case, AM-GM and Cauchy-Schwarz are equivalent, they all come down to $x^2\geq0$.  :lol:


AMC
AIME
algebra
polynomial
LaTeX
geometric series

P_1, P_2, and P_3 are polynomials defined by:

    P_1(x) = 1+x+x^3+x^4+\cdots+x^{96}+x^{97}+x^{99}+x^{100}
    P_2(x) = 1-x+x^2-\cdots-x^{99}+x^{100}
    P_3(x) = 1+x+x^2+\cdots+x^{66}+x^{67}

Find the number of distinct complex roots of P_1 \cdot P_2 \cdot P_3.

To use $ \text{\LaTeX}$ you must enclose your code in dollar signs:

Original Problem:

$ P_1$, $ P_2$, and $ P_3$ are polynomials defined by:

$ P_1(x) = 1+x+x^3+x^4+\cdots+x^{96}+x^{97}+x^{99}+x^{100}$
$ P_2(x) = 1-x+x^2-\cdots-x^{99}+x^{100}$
$ P_3(x) = 1+x+x^2+\cdots+x^{66}+x^{67}$

Find the number of distinct complex roots of $ P_1 \cdot P_2 \cdot P_3$.

[quote="007math"]To use $ \text{\LaTeX}$ you must enclose your code in dollar signs:

Original Problem:

$ P_1$, $ P_2$, and $ P_3$ are polynomials defined by:

$ P_1(x) = 1 + x + x^3 + x^4 + \cdots + x^{96} + x^{97} + x^{99} + x^{100}$
$ P_2(x) = 1 - x + x^2 - \cdots - x^{99} + x^{100}$
$ P_3(x) = 1 + x + x^2 + \cdots + x^{66} + x^{67}$

Find the number of distinct complex roots of $ P_1 \cdot P_2 \cdot P_3$.[/quote]
[hide]Factoring each one individually results in:
$ P_1(x)=\frac{x^{101}-1}{x-1}$
$ P_2(x)=1-x+x^2-\cdots-x^{99}+x^{100}-x^{101}+x^{101}$
$ P_2(x)=(1-x)(x^2+x^4+\cdots+x^{50})+x^{101}$
$ P_2(x)=x^{2}(1-x)\frac{x^{51}-1}{x-1}+x^{101}$
$ P_2(x)=x^{2}(1-x^{51})+x^{101}=x^{2}$
$ P_3(x)=\frac{x^{68}-1}{x-1}$
So $ P_1(x)\cdot P_2(x)\cdot P_3(x)=\frac{x^{101}-1}{x-1}\cdot\frac{x^{68}-1}{x-1}\cdot x^{2}$
These are the 101st roots of unity and the 68th roots of unity. Subtract one root since we can't have $ 1$ as a root for 100 roots, 67 roots, and of course, $ 0$ as a root for a total of $ 168$ roots. [/hide]

I got a different answer.

[hide="My Solution"]
Using the geometric series formula, we get:

$ P_1(x) \equal{} \frac {x^{101} \minus{} 1}{x \minus{} 1} \\P_2(x) \equal{} \frac {x^{101} \plus{} 1}{x \plus{} 1} \\P_3(x) \equal{} \frac {x^{68} \minus{} 1}{x \minus{} 1}.$

Because $ \gcd(68,101) \equal{} 1$, we know that $ P_1(x)$ and $ P_3(x)$ don't share common roots. The roots of all the polynomials are distinct, so there are $ 100 \plus{} 100 \plus{} 67 \equal{} \boxed{267}$ distinct roots.
[/hide]

Oh I know what I did wrong...heh yeah I think your answer's right  :lol:


inequalities
trigonometry
inequalities unsolved
algebra

$b=\frac{a+c}{1-ac}$,and then suppose $tanA=a,tanB=c$,then $tan(A+B)=b$.And then use the formulas of cosine and sine.

Uhm, $ tan A \equal{} a , tan B \equal{} b$
~~> $ tan A \plus{} tan B \equal{} b (1 \minus{} tan A tan B)$

~~> P = $ 2 cos^2 A \minus{} 2 cos^2(A\plus{}B) \plus{} 3 cos^2 B$ = $ 2 sin B sin(2A\plus{}B) \plus{} 3(1 \minus{} sin^2 B)$     ( use $ cos^2 u \minus{} cos^2 v \equal{} sin(v\minus{}u) sin(v\plus{}u)$ )

~~> P $ <\equal{} 2 sin B \plus{} 3\minus{}3sin^2B  \equal{} \minus{}3 (sin B \minus{} 1/3)^2\plus{} 10/3 <\equal{} 10/3$
So, max = $ 10/3$ ^^
P/s : this is my first post , so if I have a mistake, please forgive me  :blush:

Given are three positive real numbers $ a,b,c$ satisfying $ abc \plus{} a \plus{} c \equal{} b$. Find the max value of $\frac{2}{a^{2}+1}-\frac{2}{b^{2}+1}-\frac{4c}{\sqrt{c^{2}+1}}+\frac{3c}{(c^{2}+1)\sqrt{c^{2}+1}}$
http://www.artofproblemsolving.com/community/c6h1209582p5988768

Given are three positive real numbers $ a,b,c$ satisfying $\frac {2}{a^2 \plus{} 1} \minus{} \frac {2}{b^2 \plus{} 1} \plus{} \frac {3}{c^2 \plus{} 1}=\frac{10}{3}$. Prove that$$abc+a+c \leq b$$

[quote=James Potter]Given are three positive real numbers $ a,b,c$ satisfying $ abc \plus{} a \plus{} c \equal{} b$. Find the max value of the expression:
\[ P \equal{} \frac {2}{a^2 \plus{} 1} \minus{} \frac {2}{b^2 \plus{} 1} \plus{} \frac {3}{c^2 \plus{} 1}.\][/quote]
Replace $b = \frac{c+a}{1-ca}$ then
\[ P = \frac {2}{a^2  +1} - \frac {2(1-ca)^2}{(c^2+1)(a^2+1)} +\frac {3}{c^2 + 1} = \frac{3a^2+4ac+2c^2+3-2a^2c^2}{(c^2+1)(a^2+1}.\]
Note
\[P = \frac{10}{3}\left[1-\frac{(16c^2a+a-6c)^2+(8c^2-1)^2}{(c^2+1)(a^2+1)(16c^2+1)}\right] \leqslant \frac{10}{3}.\]
Equality occur when $a=\frac{1}{\sqrt2},\,b=\sqrt2,\,c=\frac{1}{2\sqrt2}.$ Therefore $P_{\max} = \frac{10}{3}.$

[quote=James Potter]Given are three positive real numbers $ a,b,c$ satisfying $ abc \plus{} a \plus{} c \equal{} b$. Find the max value of the expression:
\[ P \equal{} \frac {2}{a^2 \plus{} 1} \minus{} \frac {2}{b^2 \plus{} 1} \plus{} \frac {3}{c^2 \plus{} 1}.\][/quote]
\[ P = \frac {2}{a^2  +1} - \frac {2(1-ca)^2}{(c^2+1)(a^2+1)} +\frac {3}{c^2 + 1} =  \frac{10}{3}-\frac {(a-2c)^2+(1-4ac)^2}{(c^2+1)(a^2+1)} \leq \frac{10}{3}.\]Equality holds  when $a=\frac{1}{\sqrt2},\,b=\sqrt2,\,c=\frac{1}{2\sqrt2}.$ Therefore $P_{\max} = \frac{10}{3}.$

[b][size=150]or[/size][/b]

$$P=2cos^2\frac{A}{2}-2sin^2\frac{B}{2}+3cos^2\frac{C}{2} \le\frac{10}{3}.$$
[hide=here]$$2cos^2\frac{A}{2}+2cos^2\frac{B}{2}+3cos^2\frac{C}{2} \le\frac{16}{3}.$$
https://artofproblemsolving.com/community/c6h495081p2779244
[/hide]


How do I prove the following? (using epsilon and delta definitions)

   1.) lim x^2*sin(2/x)=0
        x->0

   2.) lim (2x-5)/(4x-x^2)=1/3
       x->3

   Thanks a lot.

Please , for ε>0, take δ = $\sqrt{\epsilon}$. Then for |x| <δ we' ll have:

  |......| <= x^2 <=ε, since |sin(2/x)| <= 1

 For the other(2) you need some more techniq. Assume |x-3|<1. Then |x|<4 ect.Try to find a solved problem in a book. The way is similar. I have not time this moment to write. I must go in the classroom.Sorry.

Babis


algebra
polynomial
advanced fields
advanced fields unsolved

Hi i dont know if I am in the right topic... Can someone please tell me how to solve this system?

x^14 + y^6=1
x^6   + y^17=1

I want to find the roots in R.
Thank you very much! (sorry for my poor english)

Start a new topic for new problems, this obviously is completely unrelated!

Sorry, I thought it was in the right topic because I assume that its solution based in polynomial system

f1(x1,x2,x3...,xn)=0
f2(x1,x2,x3....,xn)=0 , where f1 and f2 polynomials with n variables....


function
calculus
derivative
real analysis
real analysis unsolved

I know this must exist, but it is putting up a good fight for actually constructing it.

What is a function $ f: \mathbb{R}\to\mathbb{R}$ that is both open and closed (as a mapping) but is not continuous?

One idea is to construct a function for which the image of every interval is $ \mathbb{R}$.  I'm sure there are easier ways to accomplish this than what I describe in the following, but this just happened to occur to me because I'm currently reading up on generalized derivatives.

Let $ f: \mathbb{R}\to \mathbb{R}$ be a continuous nowhere differentiable function (e.g. [url=http://en.wikipedia.org/wiki/Nowhere_differentiable]this one[/url]).  Denote by $ f_ \plus{} '$ the upper [url=http://en.wikipedia.org/wiki/Dini_derivative]Dini derivative[/url] of $ f$.  Define $ g: \mathbb{R}\to \mathbb{R}$ by
\[ g(x) \equal{} \begin{cases} f_ \plus{} '(x) & f_ \plus{} '(x) \text{ exists} \\
0 & \text{ otherwise} \end{cases}
\]
I [i]believe[/i] it can be shown that $ g(I) \equal{} \mathbb{R}$ for each interval $ I$, so $ g$ is a discontinuous function with the properties we want.

Edit: See sections 1-3 from Chapter 4 of [url=http://books.google.com/books?id=fXfEG-F2zJUC&dq=0821869906][i]Differentiation of Real Functions[/i][/url] by A. M. Bruckner (pages 39-44).  In particular, note Theorem 3.8:[quote]If $ F$ is a continuous nowhere differentiable function on $ I$, then $ D^ \plus{} F$ takes on every real value on every interval.  In fact, the set $ \{x: D^ \plus{} F(x) \equal{} \alpha\}$ has cardinality $ c$ for each $ \alpha \in \mathbb{R}$.[/quote]

You give an example of an open, non-continuous, map.
Are the images of closed sets (with empty interior) closed?

You're right, for uncountable, closed, and totally disconnected sets there's no reason to think the images would be closed.  It seems I was somehow thinking that closed intervals form a basis for the closed sets under unions. :oops:


induction
number theory unsolved
number theory

Consider the sequence of integer such that:
$ a_1 \equal{} 2$
$ a_2 \equal{} 5$
$ a_{n \plus{} 1} \equal{} (2 \minus{} n^2)a_n \plus{} (2 \plus{} n^2)a_{n \minus{} 1}, \forall n\ge 2$

Find all triplies $ (x,y,z) \in \mathbb{N}^3$ such that $ a_xa_y \equal{} a_z$.

Ps. The real [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=242972]final round[/url] is still open..

[quote="xxp2000"]It is easy to show by induction that $ a_n\equiv 2\mod3$. Hence, there is no such $ (x,y,z)$.[/quote]

Eventually  :D  :D


Let ABCD be a square, and O its center. A line through A intersects BC and CD at M and N respectively.
Prove the following sentences:
a) If MN does not pass through the interior of the square, then the angle MON equals the sum of the angles DMC and BNC
b) if the line passes through the interior of the square and intersects (CD) at N and BC at M, then the angle MON equals the difference of the angles DMC and BNC.

Best regards,
Sunken Rock

In fact I wrongly posted it here, at the time of posting I already had a solution for it, rather difficult, I look for other ideas/solutions.

Best regards,
sunken rock

An easy problem?
[quote="mr.danh"]Let ABCD be a square, and O its center. A line through A intersects BC and CD at M and N respectively. Prove that if MN does not pass through the interior of the square, then $ \angle  MON\plus{}\angle DMC\plus{}\angle BNC\equal{}180^o$ (similar to the second case) [/quote]
$ \triangle ABM\sim\triangle NDA\Rightarrow BM.DN\equal{}AD.AB\equal{}BD.OD$
$ \angle ODM\equal{}\angle DBM\equal{}135^o\Rightarrow \triangle MBD\sim\triangle ODN\Rightarrow DMC\equal{}\angle DMB\equal{}\angle DON$
Analogously, $ \angle BNC\equal{}\angle BOM$. Hence result.

Dear Mathlinkers,
an article concerning "Geometric miniatures on a square" with original proofs has been put on my website
http://perso.orange.fr/jl.ayme    vol. 7 Miniatures géométriques sur un carré  p. 49
Sincerely
Jean-Louis


I think this is a fun rate-distance-time problem which I saw in my book.

[b]Question:[/b] Let two cities be located at point A and B. Galois is located at A and Riemann at B. Exactly at sunrise Galois and Riemann travel in opposite directions (A goes to B and B goes to A). Galois and Riemann meet eachother at noon. Galois reaches B at 5:00 P.M. and Riemann reaches A at 11:15 P.M.. At what time was sunrise? (They are traveling at the same speed throughout).

I get the following answer....
[hide]
4:30 A.M.
[/hide]

[quote="sylow_theory"]I get the following answer....
[hide]
4:30 A.M.
[/hide][/quote]
[hide="I agree"]

Let $N$ be the point were they meet

[code]
G ->

|----------  x ----------|----  5  ----|

A                        N             B

|--------  11.25 --------|----  x  ----|

                                    <- R

[/code]

Galois walks $x$ hours from $A$ to $N$ and $5$ hours from $N$ to $B$

Riemann walks $x$ hours from $B$ to $N$ and $11.25$ hours from $N$ to $A$

$\frac{x}{11.25}=\frac{5}{x}\Rightarrow$

$x^{2}= 5\cdot 11.25=56.25\Rightarrow$

$\boxed{x=7.5}$

and $12-7.5=4.5$[/hide]

I join.  :lol:
The sunrise is at $t=12-\sqrt{(a-12)(b-12)}$, where $a$ and $b$ are reaching times in 24-hours time-format.
Here $a=17, \ b=23.25$ gives $t=4.5$

I solve it in the strangest way  :P
Want to see it? Okay (see picture below).

The line down the middle is where they meet.
$L_{1}$ is the distance until to point the meet.
$L_{2}$ is the distance from the other side.
$S_{1}$ be the speed of Galois.
$S_{2}$ be the speed of Riemann.
$T$ be the time at T a.m. in the morning what we want to find.

Now we can start.

We know that the amount of time it takes Galois to reach the line and Riemann to reach the line is the same. So:
$\frac{L_{1}}{S_{1}}= 12-T\Rightarrow L_{1}= (12-T)S_{1}$
$\frac{L_{2}}{S_{2}}= 12-T\Rightarrow L_{2}= (12-T)S_{2}$

We know Galois travels $L_{2}$ in 5 hours.
We know Riemann travels $L_{1}$ in 11.25 hours.

So,
$\frac{L_{2}}{S_{1}}= 5\Rightarrow L_{2}= 5S_{1}$
$\frac{L_{1}}{S_{2}}= 11.25 \Rightarrow L_{1}= 11.25 S_{2}$

We end up with the equations,
$\{ \begin{array}{cc}(12-T)S_{1}-11.25 S_{2}= 0 \\ 5S_{1}-(12-T)S_{2}=0\end{array}$

Now this is a homogenous linear system of equations. So it [b]always[/b] has a solution no matter what $T$, i.e. the trivial, that is, $S_{1}=S_{2}=0$.
But this solutions doesn't make sense physically.
So we want [b]non-trivial[/b] solution.
That happens when,
$| \begin{array}{cc}12-T &-11.25 \\ 5 &-(12-T) \end{array}| = 0$
So,
$(12-T)^{2}= 56.25$
And the rest is easy.

[quote="kunny"]You can solve in using similar triangles, that is to say, geometrically.[/quote]

Yes, the answer I gave was immediate from the string:

$\frac{12-t}{a-12}=\frac{b-12}{12-t}$

which is obvious because each of Galois and Riemannne has passed up to noon the same distance what another has passed after noon - the same thing I see in the post of [b]pontios[/b].
Of course I've noticed that it is eqvivalent to finding perpendicular $12-t$ from a right angle on a  hypotenuse if it divides the hypotenuse into parts $a-12$ and $b-12$.
But I do not know stright gemetrical applying.

.


What is the remainder of $\frac{x^{203}-1}{x^4-1}$?

Since the denominator is of the fourth degree, the remainder will be of the third degree, hence

$x^{203}-1=(x^4-1)Q(x)+ax^3+bx^2+cx+d$

for some polynomial $Q(x)$

Now substitute the roots of $x^4-1$, which are $\pm 1,\pm i$:

$a+b+c+d=0$
$-a+b-c+d=-2$
$-ia-b+ic+d=-i-1$
$ia-b-ic+d=i-1$

From the last two equations $-a+c=-1\implies c=-1+a$ and $-b+d=-1\implies d=-1+b.$ Substituting that into the first two equations we get

$a+b-1+a-1+b=0\implies a+b=1$
$-a+b+1-a-1+b=-2\implies -a+b=-1$

From these two we get $a=1, b=0$, which gives $c=0, d=-1$

Hence the remainder is $x^3-1$


limit
real analysis
real analysis unsolved

Suppose that $ 0 < x_1 < 1$ and $ x_{n \plus{} 1} \equal{} 1 \minus{} \sqrt{1\minus{}x_n}$ for n $ \in$ N. Prove that $ x_n\downarrow\: 0$ as n$ \rightarrow\infty$ and $ x_{n\plus{}1}$/$ x_n\rightarrow$1/2, as n$ \rightarrow\infty$.

$ z_n \equal{} \sqrt {1 \minus{} x_n}$
$ z_{n \plus{} 1}^2 \equal{} z_n$ so
$ z_{n \plus{} 1} \equal{} \sqrt {z_{n}}$
Easy to check $ z_n$ is monotone and bound.so it has limit
Call it is $ l$
then
$ l^2 \equal{} l$ so $ l \equal{} 1$ so $ \lim_{n\to\infty}x_n \equal{} 0$
$ \frac {x_{n \plus{} 1}}{x_n} \equal{} \frac {1 \minus{} \sqrt {1 \minus{} x_n}}{x_n}\to \frac {1}{2}$ when $ x_n\to 0$

since
\[ \begin{gathered}
  f\left( x \right) \equal{} 1 \minus{} \sqrt {1 \minus{} x}  \hfill \\
  f'\left( x \right) \equal{} \frac{1}
{{2\sqrt {1 \minus{} x} }} > 0 \hfill \\
\end{gathered}
\]
and the fact that $ x_2  \equal{} 1 \minus{} \sqrt {1 \minus{} x_1 }  < x_1$ we deduce that $ \left\{ {x_n } \right\}$ is decreasing. Letting $ n \to \infty$ we get $ \mathop {\lim }\limits_{n \to \infty } x_n  \equal{} 0$.


Can someone tell me how to install Asymptote on a Mac?  I've downloaded Asymptote 1.37, but it just comes as a ton of files ending in .asy or some other file type, and there's no application.  Is there supposed to be an installer or an application to download like there is for Latex, or am I supposed to do something else to get Asymptote?  I have  OS 10.3.9, and my Mac meets all requirements.  Thanks if anyone can help :)

There's a universal binary at http://mac.softpedia.com/get/Developer-Tools/Asymptote.shtml
Does that have an installer?

That's the one I downloaded.  Is there even supposed to be an installer?  Because I can't find one in the file I downloaded.  I'm not too familiar with the term "universal binary."  Does it mean that there's a different way to install it than normal?

A universal binary is a program that runs both on the old Macs with the old chips and the new one with Intel chips. Binary means it is a program that you run so it should be the installer.