linear algebra
matrix
function
algorithm

Hi,

[b]Question 1:[/b] (cutting all the crap)

There is a community hall where events are going on. People can come in and go out whenever they want. At one instant only one person may come in or go out. An observant fellow keeps track of times when ppl come in and go out in the form of integers (ie: 1 signifies, the guy has entered 1 second after the event as begin). At each instant so many people may be in the hall. Your aim is to find the maximum number of people present in the hall (this will be at one particular instant).

Sample Input: (the values at entry and exit are the no. of seconds after which they enter and leave)

Person            Entry Exit
Person1          1       7
Person2          2       4
Person3          6       9
Person4          3       8
Person5          5       10

Output is : 4

At the interval between second 6 and second 7, four ppl ( 1,3,4,5) are present in the hall, which is the maximum. So the answer is 4.

Given: No.of People <=10^5
Entry Time and Exit time are all distinct and t<=10^7

[b]Question 2::[/b]

There exists a 2 x N matrix (2 rows and n columns), which can be filled by dominoes of size 2x1, horizontally or vertically. Each cell in the matrix has a value. The Value the domino is said to be the difference between the bigger value and smaller value that it covers.

Ex:
8 6  2 3
9 7  1 1

Assume a domino covers like this [8 6]. Value of this domino is said to be 2 (8-6).
The task is find the maximum possible sum of values of the domino. (ie: You can cover it in so many ways, in which way will the possible sum of values of dominoes be max)

In above question:

Domino 1 - 8 9 (Value 1)
Domino 2 - 6 2 (value 4)
Domino 3 - 7 1 (value 5)
Domino 4 - 3 1 (value 2)

Sum is 12

Output: 12

Given: N<=10^5. Values of cells maybe from 0 --> 10^4

I've done well, what about you guys. Cutoff may be higher. [/img]

Got one sum fully right i think  . :D  :
 People say one fully right sum is enough to steer you through.    :maybe:
But i dont think so as almost everyone in my centre has got atleast one sum fully right
And i am hence not expecting anything.  :D  :
And i didnt even attempt the second so absolutely no chance of extra part marks
So if the cutoff is only one sum i may get through.
And it took me so much time bcos of one part in the first sum which worked in my house but which didnt work in the exam centre (and even Shreyas says he doesnt know wat is wrong in wat i did)

congrats shreyas and madness hope u both get through :)

I have done pretty decently and got both the problems.
The first one is fine for me and works for any permutation.
The second one works for low values in the grid but fails when the values are too big(dunno why). But I have no hope and will accept any result that comes my way. I am only happy that I atleast managed to have a crack at the paper.

from whatever i heard of i complete gets u thru

where i tends to 2 i suppose  :D  :D  :D

@rock: Is your first sum $ O(nlogn)$ or $ O(n^2)$?

nlogn logic isn't so straight forward, I dunno how rohit claims it is.

2nd was a std DP..

Over the previous years, getting 1 fully right and atleast 30% of 2nd sum has usually got ppl through. Might by 1 & 50% this year, so let's wait for 1 month.

[quote="chemrock"]Mine is O(n^2)
I didn't mind on the logic part because my primary aim is that the program should compile and give u the required result(Shre it worked for 10^7 I was surprised).[/quote]

Hey, you entered 10^7 values or are you talking about int??

Int is 4 bytes and accepts till $ 32767^2$ which is $ 10^9$

Incase of 2nd sum int is okay compared because values range from 0 --> 10^4.
The maximum sum can be 10^5 x 10 ^4 = 10^9. which falls within ints range.

@pardesi: how did your friend do info da?

About 20 students will qualify from INOI-2008 for the training camp in June to select the Indian team to IOI-2008. ====> This worries me a lot :(

[quote="pardesi"]shreyas he got the first one...not the second
a friend of his got the first one and wrote the program for second which didn't work
don't worry da u have done well ...u will make the cut[/quote]

There are lot of kattans in India. The guy who got bronze medal in IOI last year wrote INOI in our centre. And when the number is only 20. Only 24 students made the cut last year//http://www.iarcs.org.in/inoi/2007/inoi2007/results_inoi2007.php But lol the guy who had my roll number (90) last year went to the camp > :D:D

But ofcourse would be nice if I could go here (http://www.tisb.org/). .campus is nice. :D

[quote="madness"]well whatever logic i used is really simple accord to me (Well bcos i wrote it  :D ).
As u dont want me to discuss my logic wid u then how do u expect to find my logic simple  :P  (seeing u dont know wat it is)
And did u ask Venkatesh whether $ (1.6)\cdot 10^6$ will pass (im talking abt the memory)  :maybe:[/quote]

I have asked him. he hasn't replied..... And I still don't think yours might be nlogn... Perhaps I will hear your logic after results come out.

Hmm...I thought 2nd problem was much easier than first but ppl say otherwise. :((

It was to you (and could've been to me had i tried but i didnt  :D ).
And am sure that my logic is nlogn and am almost sure that it works completely.  :D

[quote="Cyber-Shreyas"]

And surely won't there be a memory limit within 40-64MB???? The region is pretty volatile. :) I've used 1.6 MB.

The inbuilt function in algorithm.h sort(a,a+n) where a is a pointer uses the most efficient sort. I am guessing quick sort but it might be merge sort as awell.

@sid: I don't think the cut-offs are class dependent. :( But not sure.  Who all have got both right??? Anyway many ppl can get both right but lot of other factors come into play. somebody in our centre i heard used an array of size 10^8, which obviously fails. :(

@Pradeep: Well, for an input where n=10^5, n2 = 10^10. 10^10 operations on the array might not pass.[/quote]

the memory limits at the ioi last year were in the range 16MB to 150MB . but since nothing was mentioned here i dont know if they will keep a limit

sort(a, a+n) is introsort, a mixture of quicksort and heapsort . for details , read this http://en.wikipedia.org/wiki/Sort_%28C%2B%2B%29

i dont have my code with me , but it will look somewhat like this ( I used printf and scanf but i 'm writing cin and cout here for better understanding

#include<iostream>

using namespace std;
int i,p,q,k=0,answer=0, a[10000002],n;
int main()
{
cin>>n;
for( i=0;i<n;i++)
{
cin>>p>>q;
a[p]++;
a[q]--;
}
for( i=0;i<10000002;i++)
{
k+=a[i];
if( k>answer) answer = k;
}
return 0;
}

i've simply written it now from my mind and haven't compiled it, so there might be some small errors.

[quote="bruteforce"]

forget about the first question  :D

the memory limits at the ioi last year were in the range 16MB to 150MB . but since nothing was mentioned here i dont know if they will keep a limit

sort(a, a+n) is introsort, a mixture of quicksort and heapsort . for details , read this http://en.wikipedia.org/wiki/Sort_%28C%2B%2B%29

i dont have my code with me , but it will look somewhat like this ( I used printf and scanf but i 'm writing cin and cout here for better understanding

#include<iostream>

using namespace std;
int i,p,q,k=0,answer=0, a[10000002],n;
int main()
{
cin>>n;
for( i=0;i<n;i++)
{
cin>>p>>q;
a[p]++;
a[q]--;
}
for( i=0;i<10000002;i++)
{
k+=a[i];
if( k>answer) answer = k;
}
return 0;
}

i've simply written it now from my mind and haven't compiled it, so there might be some small errors.[/quote]

Ahh..Privacy concerns. Okay, then. I've edited my post. :D

Your program is more or less same as vihangs. But I guess O(n) is better than O(n^2) (mine)...

[quote="Cyber-Shreyas"]
Your program is more or less same as vihangs. But I guess O(n) is better than O(n^2) (mine)...[/quote]

i've already said that the logic was same , but he wanted to see my code. I think the other logic mentioned here ( sorting the entry time and exit time individually ) is better than this. if a good sorting algorithm is used, that one would be O(nlogn) where n= 10^5. this method is O(n) where n is 10^7 . moreover , it uses a lot of memory.

[quote="bruteforce"][quote="Cyber-Shreyas"]
Your program is more or less same as vihangs. But I guess O(n) is better than O(n^2) (mine)...[/quote]

i've already said that the logic was same , but he wanted to see my code. I think the other logic mentioned here ( sorting the entry time and exit time individually ) is better than this. if a good algorithm is used, that one would be O(nlogn) where n= 10^5. this method is O(n) where n is 10^7 . moreover , it uses a lot of memory.[/quote]

Another classic case of Space was Memory. :D I guess nlogn for 10^5 and n for 10^7 is the same. Anyways lets wait for the results. we same to have saturated all discussions.

@bruteforce: Loads of people claim that they got both right. What about last time?? Dunno if cut-offs might be higher.

[quote="bruteforce"]Last time at the camp , there were  rumours like u need a positive score in both , u need atleast 1 perfect score , etc.

but our sir said that u can get  selected with just one bruteforce method. infact full marks in both questions are rare.

btw , almost half of the people in the camp were from chennai, i think there is something special about that place.[/quote]

how much did you get last time??

I hope this time too chennai rocks. Can I assume you too are from Chennai?? I can see some pride there in the post.

Lots of people claim that they got both right this time. :( Lets see.

I got both last time . i dont remember about the first . for the second one i got only partial mark because there was a solution with better complexity .

I've received my official certifcate. (letter) stating that I qualified in ZIO and have written INOI.  What about you??

hey u got through shreyas CONGRATS DA  :coool:  :coool:  :coool:  :coool:  :coolspeak:  :coolspeak:  :coolspeak:
treat le photo ve kude da  :D

[quote="pardesi"]hey u got through shreyas CONGRATS DA
treat le photo ve kude da  :D[/quote]

Avan photo va kudukamattan da  :D  He meant the official letter for qualifying to write the inoi  :D
I got it too  :P

INOI results have been declared. http://www.iarcs.org.in/inoi/2008/inoi2008/results_inoi2008.php

dei shreyas congrats da u shoudn't  have edited that
dei treata kude da :10:  :10:  :10:  :first:  :first:  :first:  :coolspeak:  :coolspeak:  :coolspeak:

Treat da shre
[color=blue][size=200][u][i][b]CONGRATS[/b][/i][/u][/size][/color]

i am locking this for obvious reasons :D


Let $ S(n)$ denote the sum of the digits of $ n$. Show that for positive integers $ a$, $ b$:

$ S(a\plus{}b)\le S(a)\plus{}S(b)$

with equality when there are no carry overs in the addition of $ a\plus{}b$.

[hide="Maybe not the best proof, but it is a start"]WLOG let $ a\ge b$.  Let $ A_{m}A_{m\minus{}1}A_{m\minus{}2}\ldots A_{1}$ represent the digits of $ a$ (where $ m$ is the number of digits in $ a$), and let $ B_{m}B_{m\minus{}1}B_{m\minus{}2}\ldots B_{1}$ the digits of $ b$.  If $ a$ and $ b$ have a different number of digits, then we let $ B_{m}\equal{}B_{m\minus{}1}\equal{}\underbrace{\ldots\ldots\ldots\ldots\ldots}_{\text{continue as necessary}}\equal{}0$.  For example, if $ a\equal{}1787569$ and $ b\equal{}12345$, then $ B_{7}\equal{}B_{6}\equal{}0$, while $ B_{5}\equal{}1$.  Clearly, $ S(a)\equal{}\sum_{i\equal{}1}^{m}A_{i}$ and $ S(b)\equal{}\sum_{i\equal{}1}^{m}B_{i}$

Define $ C_{n}\equal{}A_{n}\plus{}B_{n}$.  Therefore $ a\plus{}b\equal{}C_{m}C_{m\minus{}1}C_{m\minus{}2}\ldots C_{1}$.  However, if there exists a $ C_{n}\ge 10$, then we carry, which is actually subtracting 10 from $ C_{n}$ and adding 1 to $ C_{n\plus{}1}$.  Therefore we subtract 9 from $ S(a\plus{}b)$, hence $ S(a\plus{}b)\le S(a)\plus{}S(b)$, with equality holding when no carrying is necessary.

$ \mathbb{QED}$[/hide]

Let's see that for one digit number a,b.
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2/d/b/2db75d491ea39d87f646cf88d5517885269a3356.gif[/img]
It's true becuse each number 9<x<20   x=a+b is greater than sum of its digits which is always smaller than 11.
so if there is carry over kth+1 digit in a+b is always smaller then sum of digits below.


The sum of the last four digits of Marian’s phone number is 30
How many such four-digit sequences are there?

[quote="anirudh"]The sum of the last four digits of Marian’s phone number
How many such four-digit sequences are there?[/quote]

What? The problem isn't complete!

[hide="A very bad solution"]

  [b]Firstly, since this is a phone number I am assuming the order of the digits matters.[/b]

There are 4 slots we can put numbers into.  Suppose we fill the 4 slots with 9's.  We will get a sum of 36, six more than a desired number.  Thus the question can be re-worded into:  How many ways can we apply " six imaginary -1 tags" to our 4 nines.  This calls for some ordered partition work.

Partitions of 6:

6   :   4 ways

6,5:   4*3=12 ways

4,2:   4*3= 12 ways

4,1,1: 4* (3 choose 2)=12 ways

3,3:   (4 choose 2)= 6 ways

3,2,1: =24 ways

3,1,1,1:= 4 * (3 choose 3)=4 ways

2,2,2:=(4 choose 3)=4 ways

2,2,1,1 = (4 choose 2) * (2 choose 2) = 6 ways

2,1,1,1= 4 * (3 choose 3)=4 ways

  For a total of 88 different 4-digit sequences, when given regard to order.  That is assuming I didn't mess up my arithmetic. :D[/hide]

This was on the state team round. G-UNIT, you're close, but the answer isn't 88.
[hide]Let's start with 9.
9993
9984
9975
...
9939, 7 numbers.

9894
9885
9876
...
9849, 6 numbers.

9795
9786
9777
9768
9759, 5 numbers .

We see a pattern, the numbers go 7, 6, 5, ..., because you can also see the first numbers of the lists go 3, 4, 5, ...., and since we increase it by 1 each time, we must stop when the digit is 9.
7+6+5+4+3+2+1=28,

Let's try 8.
8994
8985
8976
8967
8958
8949, 6 numbers

8895
8886
8877
8868
8859, 5numbers,

Using the same principle, we can conclude that the rest will have 4, 3, 2, and 1, so 6+5+4+3+2+1=21.

Also, the first number of every list (9's, 8's) follow the pattern 3, 4, ... , like for 9, it was 999[b]3[/b], starting with 8, 899[b]4[/b] Let's see if starting with 7 we can start with a 5.

7995.

it works. The numbers starting with each digit go down like triangular numbers.
28+21+15+10+6+3+1=84. [/hide]

There are many solutions. I have like one or two more solutions.

G-UNIT has the correct solution, and it should be the correct answer

http://www.mathcounts.org/webarticles/articlefiles/510-05SolutionSetS.pdf and go to page 9.

YOu're on a roll of proving everyone wrong. But if he though 2+1+1+1 was 4, wouldn't that make him 4 less?

eDIT: incorrect interpretation

[quote="nat mc"][quote="G-UNIT"]
3,1,1,1:= 4 * (3 choose 3)=4 ways

...

2,1,1,1= 4 * (3 choose 3)=4 ways
[/quote]
Thats why you're four over. and anyways 2+1+1+1=5   :P[/quote]

lol, i copied down these stupid partitions from a stupid website, that was probably designed by a 5 year old, anyways thanks for catching that :)


probability
geometry
integration
calculus
real analysis
real analysis unsolved

The corporate offices for Big Q Business are trying to find a way to better control their office supplies expenses for tjeir branches. They have determined that the mean monthly office expense for a branch is 15,450 dollars with a standard deviation of 3,125 dollars.

A/ Whats is the probability that a randomly selected office has supply expenses between 12,000 and 18,000 dollars?

B/ Whats is the probability that an office has supply expenses over 27,000 dollars?

C/ If the company wants to investigate all offices which have supply expenses placing them in the top 5% of spenders what dollar value when exceeded will result in an investigation?

A/ z= (12000-15450) / 3125= - 1.10 and also z= (18000-15450) / 3125=0.82
P(-1.10<z<0.82)= 0.66

B/1-0.66=0.34

C/ 0.12= (x-15450)/3125 so x=15797.

Do u think questions B and C are right?

Thanks. Julien

The probability that the expenses are over 27,000 is the area under the normal distribution curve $f(x) = \frac{e^{-(x-\mu)^2/2\sigma^2}}{\sigma\sqrt{2\pi}}$ from 27,000 to infinity, and is thus
\[ \int_{27000}^\infty \frac{e^{-(x-15450)^2/2\cdot3125^2}}{3125\sqrt{2\pi}} \ dx \approx 1.095 \cdot 10^{-4}  \]

Another way you could do this is to standardize the value 27000, which is $\frac{27000-15450}{3125} = 3.696$, and then conult a chart or use the integral
\[ \frac{1}{\sqrt{2\pi}}\int_{3.696}^\infty e^{-x^2/2} \ dx \approx 1.095 \cdot 10^{-4}  \]

For part C, just look at a chart and see when the area is at least $.95$ for a given value $z$, then you have $z = \frac{\text{dollar amout} - 15450}{3125} \Rightarrow \text{dollar amount} = 3125z + 15450$.


function
integration
calculus
calculus computations

Could anyone show me how to prove the following question, thx alot.

Suppose $ (X, B, \mu)$ is a measure space, $ g$ is nonnegative measurable function and $ u$ is measure on $ B$.
Suppose $ f$ is a nonnegative measurable function on $ X$ then show that $ \int f du\equal{} \int fg d\mu$

As stated it doesn't really make any sense.  Do you want $ \nu$ to be the measure $ gd\mu$ and then show that
\[ \int_X fd\nu \equal{} \int_X fgd\mu
\]
?

I guess that I should have stated that $ vE\equal{} \int_E g d\mu$ and $ v$ is a measure on $ B$. With these conditions, how would you prove? Thanks.


The pages of a book with two parts are numbered 1, 2, 3, . . . ,N. The sum of the page numbers of the first part equals the sum of the page numbers of the second part. How many pages does the book have?

There are actually infinite possible values for $ N$:
We want a triangular number that is two times another triangular number so:
$ \frac{n(n+1)}{2}=2\frac{k(k+1)}{2}\Leftrightarrow 4n^{2}+4n+1-1=2(4k^{2}+4k+1)-2$
Factoring the squares: $ (2n+1)^{2}-2(2k+1)^{2}=-1$
Let $ x=2n+1$ and $ y=2k+1$ we have to find the integer odd solutions to:
$ x^{2}-2y^{2}=-1$ it's easy to see that if $ x,y$ are integers they must be odd (use modulo $ 4$)
So now we have a pell's equation that has infinitly many solutions, after finding the first one (e.g.: $ (1,1)$)
we have: $ (1+\sqrt{2})^{2j+1}=x_{j}+y_{j}\sqrt{2}$ and $ (x_{j},y_{j})$ is a solution to the equation.


inequalities
induction
number theory open
number theory

$s_n$ is sum the first n prime.Prove that for all $n$ then there exist $a$ is integer st $s_n\leq a^2\leq  s_{n+1}$

It is equavalent to $s_{n+1}=s_n+p_{n+1}>s_n+2\sqrt{s_n}+1$, and proved by chebeshev theorem.

The following inequality gives an elementary proof of this statement:

For all natural numbers n,

(1) $s_n < (p_{n+1}/2)^2$.

Proof: The inequality is obviously true for n=1 since $s_1 = p_1 = 2$ and $p_2 = 3$. Next assume (1) is true for $n=k-1$. Then

$s_k = s_{k-1} + p_k < (p_k/2)^2 + p_k < (p_k/2 + 1)^2 \leq (p_{k+1}/2)^2$.

Thus $s_k < (p_{k+1}/2)^2$, which completes the induction proof of the inequality (1).

Suppose this statement of "dieuhuynh" is false. Then  there must exits a natural number n such that the interval $[s_n,s_{n+1}]$ contains no integer squares. In other words, there is a natural number a such that

$(2) \;\; a^2 < s_n < s_{n+1} < (a + 1)^2$

which implies that

$2a + 1 = (a + 1)^2 - a^2 \geq (s_{n+1} + 1) - (s_n - 1) = s_{n+1} - s_n + 2 = p_{n+1} + 2$,

i.e. $p_{n+1} < 2a$. Now $a < \sqrt{s_n}$ by (2), hence $p_{n+1} < 2\sqrt{s_n}$. Therefore $s_n > (\frac{p_{n+1}}{2})^2$, contradicting the inequality (1).


Okay . . . Lets have some fun . . .  :lol:

Can you arrange 6   9s to make 100. You can use Multiplication, Division, Addition, Subtraction and decimal as well.

Good luck

[quote="nutz_for2.718281828"]No, that's 109. And I said 6 nines.[/quote]
Actually, you did say 9 nines.  And perfectnumber's solution can be corrected by saying $99+9\cdot\frac{999}{999}$
I'm still working on the 6 nines one

[quote="nutz_for2.718281828"]No, that's 109. And I said 6 nines.[/quote]
wow, my bad.  yeah that's supposed to be a multiplication there at the end.  and you said 9 9s.


For a positive integer k, let $f_{1}(k)$ be the square of the sum of the digits of k. (For example $f_{1}(123)=(1+2+3)^{2}=36$.) Let $f_{n+1}(k)=f_{1}(f_{n}(k))$. Determine the value of the $f_{2007}(2^{2006})$. Justify your claim.


Show that for infinitely many primes p
 $f(p)=p^{p}-(p-1)^{p-1}$ is not squarefree.

By computer can be find odd number k, suth that
(1) $a(k)=k^{k}-(k-1)^{k-1}$ is not freesquare.
Let q is prime and $q^{2}|a(k)$ (I show, that q>10). Then for all prime p, suth that $p=k(mod \ q(q-1))$ we have
(2) $p^{p}-(p-1)^{p-1}=k^{k}-(k-1)^{k-1}=0(mod \ q^{2})$.
Therefore if exist one k, satisfyed (1), then exist infinetely many primes p, satisfyed (2).

That's what I got also, but I want to see how you find it per computer.

$59^{2}|257^{257}-256^{256}$ :D
that together with the Dirichlet theorem on a.progressions solves the problem.


algebra
polynomial
limit
partial fractions
algebra unsolved

If $Q$ is a polynomial with roots $a_1,a_2,\ldots, a_n$, and if $P$ is a polynomial of degree $<n$, show that

$\frac {P(z)}{Q(z)}=\sum^n_{k=1} \frac {P(a_k)}{Q'(a_k)(z-a_k)}$

Multiply both sides by $Q(z)$. Then we have $f(z)=\sum\frac{Q(z)P(a_k)}{Q'(a_k)(z-a_k)}.$ (Take this as the definition of $f$)
The right-hand side is clearly a polynomial of degree less than $n,$ so $f$ is a polynomial of degree less than $n.$ Take the limit as $z\to a_k$. All terms except $\frac{Q(z)P(a_k)}{Q'(a_k)(z-a_k)}$ go to zero, and $Q'(a_k)=\lim_{z\to a_k}\frac{Q(z)}{z-a_k}$ by definition, so $\lim_{z\to a_k} f(z)=P(a_k)$ and $f(a_k)=P(a_k).$ Therefore $f$ is the unique interpolating polynomial of degree less than $n$ for these values, and $f(z)=P(z)$. Dividing by $Q$ gets us the original form back.

The general partial fractions expansion is also an interpolation formula.


Does anyone know where I can get some?
Thanks :]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189

Mods: Please lock.


vector
trigonometry
geometry
trapezoid
absolute value
trig identities
Law of Cosines

Circles $ C_1,C_2$ intersect at $ P$. A line $ \Delta$ is drawn arbitrarily from $ P$ and intersects with $ C_1,C_2$ at $ B,C$. What is locus of $ A$ such that the median of $ AM$ of triangle $ ABC$ has fixed length $ k$.

Of course, the locus of $ A$ is a circle with radius $ k$ centered at $ M$. Let's see what happens if we move $ \Delta$.

Let $ O_1$ and $ O_2$ be the centers of $ C_1$ and $ C_2$, respectively. Some investigation in the picture (I did it using Geogebra) leads to the conjecture that $ M$ lie on a circle with center $ O$, with $ O$ the midpoint of $ O_1O_2$, and radius $ OP$.

I tried to prove this with vectors, but couldn't finish the proof.

$ \vec{O}\equal{}\frac{\vec{O_1}\plus{}\vec{O_2}}{2}$
$ \vec{M}\equal{}\frac{\vec{B}\plus{}\vec{C}}{2}$
$ \vec{O}\minus{}\vec{M}\equal{}\frac{1}{2}(\vec{O_1}\minus{}\vec{B}\plus{}\vec{O_1}\minus{}\vec{C})$
$ \vec{MO}\equal{}\frac{1}{2}(\vec{BO_1}\plus{}\vec{CO_2})$

So we need to prove that the absolute value of this vector is a constant. Denote $ \theta_1$ and $ \theta_2$ the angles that $ BO_1$ and $ CO_2$ make with $ O_1O_2$, respectively. Let $ r_1$ and $ r_2$ be the radii of $ C_1$ and $ C_2$, respectively. By the Law of Cosines

$ |\vec{MO}|^2\equal{}r_1^2\plus{}r_2^2\minus{}2r_1r_2\cos(\theta_1\plus{}\theta_2)$

So we need to prove that $ \theta_1\plus{}\theta_2$ is constant, that is, the angle between the lines $ BO_1$ and $ CO_2$ is constant.

From here I couldn't finish. :(  Can someone fill in the rest of the proof? Can someone find a purely synthetic solution?

Yes, M moves on the circle with center O (midpoint of O1O2) and through P. And that's very easy.
However, when we move $ \Delta$ , finding the locus of A is really hard.

[quote="Sashsiam_2"]If we prove that M moves on (O,OP) circle, the problem is easy: the locus of A as we move M is a "ring" between two concentric circles, with center O and radii OP-k and OP+k. But we didn't prove that M actually moves on (O,OP).[/quote]
[hide]As [b]N.N.Trung[/b] said, it is easy to prove the result mentioned by [b]Sashiam_2[/b], because from $ O_{1}B\parallel O_{2}C$ and $ M,\ N$ as the midpoints of $ BC,\ O_{1}O_{2}$ respectively, by [b]Thales theorem[/b], we have that $ O_{1}B\parallel OM\parallel \ O_{2}C$ and then we have $ \frac {OM}{O_{2}C} \equal{} \frac {PO}{PO_{2}} \equal{} \frac {PO}{O_{2}C}$ $ \Longrightarrow$ $ OM \equal{} OP$ and from trapezium $ O_{1}BO_{2}C$ $ \Longrightarrow$ $ OM \equal{} \frac {O_{2}C \minus{} O_{1}B}{2}$[/hide]
Kostas Vittas.

[quote="Sashsiam_2"][geogebra]1226a16e0a6eb3eb96fe9bee452c3ca0b0d06100[/geogebra]

But $ O_1B$ is not parallel to $ O_2C$.[/quote]
Sorry dear all my friends because of I misunderstood the problem.

I thought that the circles $ (C_{1}),\ (C_{2}),$ tangent (externally, in my drawing) each other at point $ P,$ because of the problem states that they are intersected at one only point.

Kostas Vittas.

PS. Unfortunately, I can't view your attachment dear [b]Sashiam_2[/b] and I don't understand why.

Hello!

Problem:
$ C_{1}\left(O_{1}\right)$, $ C_{2}\left(O_{2}\right)$ are two circles and $ \left\{P,Q\right\} \equal{} C_{1}\cap C_{2}$. A line $ \Delta$ through the point $ P$
intersects the circles $ C_{1},C_{2}$ at the points $ B,C$ and $ M,O$ are the midpoints of $ BC$ and $ O_{1}O_{2}$.
Prove that $ OM \equal{} OP$.

Solution:
Construct the points $ B'\in C_{1}$ and $ C'\in C_{2}$ such that $ B'B\bot BC$ and $ C'C\bot BC$.
So $ O_{1}$, $ O_{2}$ are the midpoints of $ PB'$ and $ PC'$. It's easy to prove that the points $ B',Q,C'$ are collinear
and the quadrilateral $ BB'C'C$ is a right trapezoid ($ BB'\parallel{}CC'$ and $ \angle B'BC \equal{} \angle C'CB \equal{} 90^{\circ}$).
Let $ N$ be the midpoint of $ B'C'$. It's easy to prove that $ O$ is the midpoint of $ PN$ and $ NM\bot BC$
($ MN$ is the central median of the right trapezoid $ BB'C'C$). So $ OM \equal{} OP$.

P.S.
So, when the line $ \Delta$ is drawn arbitrarily, the point $ M$ moves on the circle $ \left(O,OP\right)$.

Best regards, Petrisor Neagoe


What makes something funny? Should it be considered childish to be amused by idiotic randomness? (Yes, you know what I mean. [url]http://www.albinoblacksheep.com/flash/demented[/url] At AwesomeMath I a group of people gathered in my dorm just to watch this.  :roll:  :| ) What makes puns punny or not punny? Are they just punny, and not funny? Parallelism also sometimes can be humorous (I can't think of any good examples at the moment).

I think in most situations you expect a certain story item to continue in a particular way based on your experience. That is you anticipate what will be likely to come next. And a joke lures you into that kind of thinking but that challenges your preconceptions by taking a sudden, unexpected turn of context.

I would crack up if I saw that. :rotfl: I would literally  :rotfl:  rofl  :rotfl:

[quote="leoxnlin"]...and that makes something humorous? A guy is walking down the street to the grocery store. Based on your experience, you might think he is probably going there to buy food. Instead, he approaches the store and turns straight back as soon as he touches the door, heading homeward.

Certainly unexpected.

Needs to be more out-of-the-ordinary?

He walks down the street toward the grocery store. He enters and heads straight for the aisle with beans. He grabs a bag and runs as fast as he can out of the store, tossing several hundred dollar bills behind him. He turns and says, "Thanks for the calculator!"

Certainly rather unexpected. Wasn't that hilarious?[/quote]
lol they're both funny!

I laughed at the first one, but the second one is slightly extraordinary.
Or maybe it's just me.

Two quotes from one of the greatest sci-fi novels of all time [i]Stranger in a Strange Land[/i] by Heinlein:

"I grok people. I am people… so now I can say it in people talk. I’ve found out why people laugh. They laugh because it hurts so much… because it’s the only thing that’ll make it stop hurting."

"I had thought — I had been told — that a ‘funny’ thing is a thing of a goodness. It isn’t. Not ever is it funny to the person it happens to. Like that sheriff without his pants. The goodness is in the laughing itself. I grok it is a bravery . . . and a sharing… against pain and sorrow and defeat."

One of the main points of the book is that laughter is the human way of escaping pain when we recognize our own mortality.

Does anyone ever crack up for no reason? I ALWAYS do that. It gets kinda annoying, I have to admit.

Apparently I have a very poor sense of humor. :dry:

*sigh*


Patrick and Ravi are playing the smiley game. The first person to post 4  :D s wins. Each person tries to be the first to post the  :D for a day, continuing until one person wins. Ravi is at his computer more often and has a 60% chance of winning the :D   for a certain day. What is the probability that Patrick wins the smiley game? Express your answer as a common fraction.

Er... lol...

[hide="Probability that Patrick wins 4-0"]$ \binom{4}{0} \times \left(\frac25\right)^4 \equal{} \frac{16}{625}$[/hide]

[hide="Probability that Patrick wins 4-1"]$ \binom{5}{1} \times \left(\frac25\right)^4 \times \frac35 \implies \frac{192}{3125}$[/hide]

[hide="Probability that Patrick wins 4-2"]$ \binom{6}{2} \times \left(\frac25\right)^4 \times \left(\frac35\right)^2 \implies 15 \times \frac{16}{625} \times \frac9{25} \implies \frac{432}{3125}$[/hide]

[hide="Probability that Patrick wins 4-3"]$ \binom{7}{3} \times \left(\frac25\right)^4 \times \left(\frac35\right)^3 \implies 35 \times \frac{432}{78125} \implies 7 \times \frac{432}{15625} \implies \frac{3024}{15625}$[/hide]

Summing yields $ \boxed{\frac{6544}{15625}}$.

Hope I didn't mess up. :)

u overcounted. for winning 4-1, u cant possibly have patrick, patrick, patrick, patrick, ravi cuz that don't make no sense cuz patrick would've already won before ravi posted the next smiley face.   :D


x,y,z are non-negative reals such that the condition $ x^{2}+4y^{2}+3z^{2}+2xyz=12$ holds.
Prove that:
$ x+y+z\leq\ 4 $


Suppose we have two polynomials in two variables $ f(x,y)$ and $ g(x,y).$

Then is it true that, the remainder on division of $ f(x,y)$ by $ g(x,y)$ will be the same, whether you take these polynomials as polynomials in the single variable $ x$ or in the single variable $ y$?

$ f \equal{} x$ and $ g \equal{} xy$ is a counterexample -- note that the degree of $ f$ is different depending on which of $ x$ and $ y$ we choose to be the variable and which we take to be a parameter.


geometry
geometric transformation
rotation
rectangle

Rectangle PQRS lies in a plane with PQ = RS = 2 and QR = SP = 6. The rectangle is rotated 90 degrees clockwise about R, then rotated 90 degrees clockwise about the point that S moved to after the first rotation. What is the length of the path traveled by P?

In the first rotation, P travels $ 90^{\circ}$ on a circle with radius $ \sqrt {2^2 \plus{} 6^2 \equal{} 2} \equal{} 2\sqrt {10}$. This is $ \frac {90}{360}(2)(2\sqrt {10})(\pi)$. The second rotation is $ 90^{\circ}$ on a path with radius $ 6$. This is $ \frac {90}{360}(2)(6)(\pi)$. Add them together to get $ \pi(\sqrt {10} \plus{} 3)$.

AMC 10A 2008 19


trigonometry
function
AMC
AIME
geometry
analytic geometry
3D geometry

A and C lie on a circle center O with radius $\sqrt{50}$. The point $B$ inside the circle is such that $\angle ABC=90^\circ, AB=6, BC=2$. Find $OB$.

[img]http://www.kalva.demon.co.uk/aime/soln/diag834.jpg[/img]

think if you knew $\cos{\angle OAB}$, you could find OB by the law of cosines, so extend AO to meet the circle at D, draw CD, you can find $\tan{\angle OAC}$, then you can also find $\tan{\angle CAB}$, and use the addition formula to find $\tan{\angle OAB}$, then convert that to cosines, and you are done

a little complicated due to my lack of knowledge of trig identities and formulas and stuff like that... :D

[hide]find the angle of BAC using right angle trig. now find AC using the pythagorean theorem. triangle AOC has an altitude from O to side AC. Draw it. we can then find the angle of OAC with more right angle trig and subtract angle BAC to get angle OAB, which is 45. Now use the law of cosines (for some reason, i knew it :) ) and u come up with the square root of 26. [/hide]

EDIT: oops, forgot about the calculator part... :mad:

[hide=" Another hint"]
Join  line AC and  call the midpoint of AC as M , Now consider OM as x-axis
[hide]Let M be mid point of AC, ($AC = \sqrt {36+4}$ so  $CM = \sqrt {10}$    and $OM = \sqrt {50-10} = \sqrt{40}$

Now draw a perpenicular from B to Line AC (Let it meet at point X on Line AC, Triangle BXC is symetric to ABC  so:
$BX = \frac 6 {\sqrt {10}}$ and $XC = \frac 2 { \sqrt {10}}$

You are done as  $OB^2 = (OM-BX)^2+(CM-CX)^2$
= $((\sqrt {40}- \frac 6 { \sqrt {10}})^2+(\sqrt{10} - \frac 2 {\sqrt {10}})^2 $
=$\frac {(20-6)^2+(10-2)^2} {10} $
Which does simplify to a nice round number equal to half the number of cards in a deck :) You may have to take a square root but that's all   [/hide]
[/hide]

plokoon51: you can find the trig functions of $A$ but not its value because AIME is a non-calculator test

Gyan: That's the right answer, but considering that it's #4, there should be a simpler solution... :P

[hide="basically what aitheman said"]
Extend $AO$ to $AD=2\sqrt{50}$. $AC=\sqrt{40} \implies CD=4\sqrt{10}$.

Now $\tan{OAC}=(4\sqrt{10})/(\sqrt{40})=2$ and $\tan{BAC}=1/3$. Now

$\tan{OAB}=\tan{OAC-BAC}=\frac{2-1/3}{1+2/3}=1\\ \implies \angle OAB = 45^\circ \\ \implies \cos {A} = \sqrt{2}/2$

Now by Law of Cosines, we have

$OB^2=50+36-2(6)(\sqrt{50})(\sqrt{2}/2)\\OB^2=86-60=26 \\ OB=\sqrt{26}$
[/hide]

Thanks everyone. :)

[quote="eryaman"]Gyan's solution is pretty simple...its basically just, draw some lines and use pythagorean theorem.  :P[/quote]

yeah, but it's hard to see it. i would never have thought of that...

[quote]yeah, but it's hard to see it. i would never have thought of that...[/quote]
Actually using co-ordinate geometry may   be   one of the  easy  and [b]straight forward  [/b]way to do these prolems.. Just choose a suitable origin, and see if you can find the coordinates easily.

I did  choose x-axis as  OM but you could have choosen any other, in fact, if you choose OA as the x axis, it would have been even easier..

[hide]If I use the radius of circle as $5$ (Instead of $5\sqrt 2$ and use a graph paper,  I  can put all points nicely  at:

$O \  (0,0)$
$A \   (5,0)$
$C \  (3 -4)$
$B \   (2,-3)$

(Check - out BC is equal to $\sqrt 2$  AB is equal to $3 \sqrt 2$ and  the angle is 90 degrees etc..)

$OB =  \sqrt {2^2+3^2} = \sqrt {13} $
(Of course you multiply everythig by $\sqrt 2$ for the actual problem, but that's trivial) :)
)[/hide]

Actually  perhaps easiest is to assume  B at $(0,0)$ , C at $(2,0)$ and A at $(6,0)$

[hide]Now if O is at $(x,y)$ you get
$50=(x-2)^2+y^2 = x^2+(y-6)^2$ which gives $x=-5,y=1$   [hide](If you are  lucky you may just notice that $50=5^2+5^2=7^2+1^2$ ) (Ignore  values of x which puts O on the othre side [/hide]
and $5^2+1^2=26$  :)
[/hide]

(So you see the  people who write these problems are actually very nice  :)

Added later:
PS - I left my previous solution here, even though the last one looks much simpler.. point is  that, these methods are fairly staraight forward, if you are lucky (in choosing the origin and axis etc )  you are done  in a short time, but even if you   unlucky and happen to  choose  a slightly different starting point, you will still get the answer soon enough..(certainly in less time that what is typically allocated for a AIME problem :)

Hope this helps.

Hey guys, I'm posting this again for the archive. :)

Chess64, do you mind if I edit your post to use it as a solution? For now, I'll just re-make one.

[hide="A Beautiful Solution"]With the current conditions, we can't do any effective angle chasing.

To fix this problem, we extend $AO$ to $D$ where $AD=2\sqrt{50}$ and is the diameter of the circle. Using the Pythagorean Theorem with triangle $ABC$, we find  that $AC=2\sqrt{10}$.

Because side $AD$ of triangle $ACD$ is the diameter of a circle and the triangle is inscribed in the circle, $ACD$ is a right triangle. Apply Pythagoras, $CD=4\sqrt{10}$.

Now, $\tan{OAC}=\frac{4\sqrt{10}}{2\sqrt{10}}=2$ and
$\tan{BAC}=1/3$.

We then apply the tangent subrtaction formula to obtain $\tan{OAB}=\tan{OAC-BAC}=\frac{2-\frac{1}{3}}{1+\frac{2}{3}}=1$, which implies that $\angle OAB = 45^\circ$

Now, $\cos {A} = \frac{1}{\sqrt{2}}$

We now use the Law of Cosines to find $OB$, which turns out to be $OB^2=50+36-2(6)(\sqrt{50})(\frac{1}{\sqrt{2}})$

The answer is thus $OB^2=86-60=26$

[/hide]

A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$cm, and that of $BC$ is $2$cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=398[/img]

Here is a fast solution from white_horse_king88:
[quote="white_horse_king88"]There is an interesting alternative that is extremely easy and elegant:

[hide]Extend perpendicular from O to AB, and label D.
Extend perpendicular from O to line containing BC, and label E.

Let $OE$ be $x$ and $OD$ be $y$.

We want $x^2+y^2$.

$OA^2 = OD^2 + AD^2$, and
$OC^2 = EC^2 + EO^2$.

So, $(\sqrt{50})^2 = y^2 + (6-x)^2$, and
$(\sqrt{50})^2 = x^2 + (y+2)^2$.

Solving the system, you get:

$x = 1$
$y = 5$

$1^2 + 5^2 = 26$.

TADA![/hide][/quote]And here is another solution:[quote="towersfreak2006"][hide]
Denote the center $O$ and draw radii to $A$ and to $C$ to form an isoceles triangle with $\overline{OA}=\sqrt{50}, \overline{OC}=\sqrt{50}, \overline{AC}=\sqrt{40}$ (radius, radius, hypotenuse of triangle with legs $2$ and $6$, respectively).
Draw the altitude down to $\overline{AC}$, call $Q$ the point of intersection with $\overline{AB}$ and call $P$ the point of intersection with $\overline{AC}$.
$\overline{OP}$, as the altitude of the isoceles triangle is $2\sqrt{10}$.

Note that $\bigtriangleup AQP\sim\bigtriangleup ACB$, so $\displaystyle\overline{QP}=\frac{(\overline{BC})(\overline{AP})}{\overline{AB}}=\frac{(2)(\sqrt{10})}{6}=\frac{\sqrt{10}}{3}$

Then $\displaystyle\overline{OQ}=\overline{OP}-\overline{OQ}=2\sqrt{10}-\frac{\sqrt{10}}{3}=\frac{5\sqrt{10}}{3}$

Similarly, $\displaystyle\overline{AQ}=\frac{(\overline{AC})(\overline{AP})}{\overline{AB}}=\frac{(2\sqrt{10})(\sqrt{10})}{6}=\frac{10}{3}$

Then $\displaystyle\overline{BQ}=\overline{AB}-\overline{AQ}=6-\frac{10}{3}=\frac{8}{3}$

Also, $\angle OQB\cong\angle AQP\cong\angle ACB$.

By Law of Cosines on $\bigtriangleup AQB$:
$\displaystyle(\overline{OB})^2=(\overline{OQ})^2+(\overline{BQ})^2-2(\overline{OQ})(\overline{BQ})(\cos{Q})$
$\displaystyle(\overline{OB})^2=\frac{250}{9}+\frac{64}{9}-2\left(\frac{5\sqrt{10}}{3}\right)\left(\frac{8}{3}\right)(\cos{C})$
$\displaystyle(\overline{OB})^2=\frac{314}{9}-\left(\frac{80\sqrt{10}}{9}\right)\left(\frac{2}{2\sqrt{10}}\right)$
$\displaystyle(\overline{OB})^2=\frac{234}{9}=26$
[/hide][/quote]For some reason, I couldn't solve this problem during a Mock AIME run two weeks ago. Yay for getting better over the course of a year... ;)

Yeah these people really are too nice

This was actually the first mock AIME I ever took, so I was pretty clueless on how to do things, but this was one of the ones I did get right.

[hide="15 second solution"]
[hide="not very mathematical..."]Draw a rough sketch. Keeping in mind that the square of OB is itself an integer, you notice that O, B, and C are nearly colinear, so OC = OB + BC;  sqrt(50) - 2 = OB; OB^2 = 50 + 4 - 2*2*sqrt(50) = 50 + 4 - 28 = 26[/hide][/hide]

I pecked at it later for a while after the mock test and did something more or less similar to what white-horse-king did

there are many similar solutions.
by the way, does anyone have any idea as to whether "guesstimating" the corect answer is a good strategy on the AIME (i.e. is it a good idea to just learn to be good at the quick guessing asnwers?)

Estimation is a very fast method on many AIME problems, and can raise your score by multiple points very quicly. Chances are though, if you can't estimate your answer in 3 or 4 minutes, you might as well forget about estimation. Also, if you want to solve a problem by estimation, the estimation should be about 75% of the solution or else you risk an uncomfortably high chance of inaccuracy.

Questions I can remember off the top of my head that I got through fast estimation:
- this one
- a recent AIME problem that gave four coordinate pairs and asked to find the side length/area (don't remember which) of the only square with one of those points on each side
- a problem involving connecting the center of each side of a regular tetrahedron to form a new one, and finding the ratio of volumes
- the circle problem on the recent AIME
- etc. etc.

[quote="The Zuton Force"]

Questions I can remember off the top of my head that I got through fast estimation:
- this one
- a recent AIME problem that gave four coordinate pairs and asked to find the side length/area (don't remember which) of the only square with one of those points on each side
- a problem involving connecting the center of each side of a regular tetrahedron to form a new one, and finding the ratio of volumes
- the circle problem on the recent AIME
- etc. etc.[/quote]

or in other words, only AIME geometry problems you can't do through simple calculations :rotfl:  :rotfl:


probability
algorithm
geometry
3D geometry

[quote="mathemagician1729"]A family has 3 children and at least one of them is a female. If the probability of having a male or female is equal, what is the probability that there is at least one male.

[hide] Is it 6/7?  [/hide]

If $ m^2 \plus{} p \equal{} 7308$ and $ m \plus{} p^2 \equal{} 6974$ where m and p are positive integers, what is $ m \plus{} p$ ?[/quote]

2.

the sqrt of 7308 is about 85. 85^2=7225, so p=83. Thus, 168.

To be completely rigorous you also have to show that $ m\equal{}84$ doesn't work (it will tip $ p$ way over).

Bleh.

$ p \equal{} 7408 \minus{} m^2$.

Assume $ m < 85$. Thus, $ p \geq 352$.

So $ p^2 \geq 123904$.

Ok obviously this doesn't work.

And please visit my topic on protesting everyone. It would mean a lot to me.

[quote="mathemagician1729"]oh does anyone know an algorithm like long division except for square and cube roots?[/quote]

well 85^2=7225<7308<86^2=7225+170+1, so yeah.

If a number ends in 25, look at the digits besides the 25. If this can be expressed as n(n+1), the number is [u]n5[/u], where [u]n5[/u] is a w/e digit number (not to be confused as 5*n).


How many centimeters are in the length of the longest side of a rectangle whose area is 108 square centimeters and whose perimeter is 42 centimeters?

One way: Brute Force.

Another way:

$ lw \equal{} 108, 2(l\plus{}w)\equal{}42 \implies l \equal{} 108 \div w \implies 2(\frac{108}{w} \plus{} w) \equal{} 42 \implies \frac{108}{w} \plus{} w \equal{} 21 \implies 108 \plus{} w^2 \equal{} 21w \implies w^2 \minus{} 21w \plus{} 108 \equal{} 0 \implies (w\minus{}12)(w\minus{}9) \equal{} 0 \implies w \equal{} \boxed{12}$

Please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=241933]here.[/url]

EDIT: Sorta beaten..

Sorry izzy, was gonna check the AoPS thread next. ;)


I am seeking NYC math penpals.  Is anyone interested?

interval

maybe i'm the one you're looking for.

i pm ed you yesterday

[quote="Interval"]It all  depends.  Are you male or female?

Tell me about you.  What is your connection to math?

Are you a math major?

Your age?

Race?

Tell me more about yourself.[/quote]
i don't think you should be asking kids these questions

[quote="Interval"](1) Did I know he or she is a kid?

(2) Obviously, I am seeking adult math pen pals NOT KIDS.

(3) What makes you think I am interested in contact with kids?[/quote]
well AoPS is primarily kids :roll:


Dear friends,

For a minimization problem which norm to choose out of  $ l_1$ or $ l_2$ or $ l_{\infty}$?

Thanks,

Bhupala.

Are you working in $ \mathbb{R}^{n}$?  Then choose the Euclidean norm $ \left\|\cdot\right\|_{2}$.

It depends on the problem. Any of them might be the right choice.


Dennis Phillips just got owned but hes still in 8th place! GO DENNIS!

Google him. Or look [url=http://www.worldseriesofpoker.com/]here[/url].

O.M.G. He had 70-30 odds of survival against Schwarz and got an Ace on the flop! GO DENNIS!

$ \text{AIME15} \neq \text{Dennis}$, however AIME15 pwnt dennis.  WTH?


If one decomposes water using electrolysis, the standard electrode potential for the acid half reaction is -1.229V whereas the one for the base half reaction is -.8227. How can I find the votalge required to make this non-spontaneous reaction spontaneous? I believe these are the right electrode potentials but I am not 100% sure.

[quote="Americandesi"]How can I find the votalge required to make this non-spontaneous reaction spontaneous?[/quote]

First a note: in an electrolysis one uses an applied voltage to drive a certain reaction in an non-spontaneous direction, so its perhaps more correct to say that the reaction is forced to go in the "unwanted" direction, instead of saying that it becomes spontaneous.

Now, the voltage required to drive a spontaneous reaction in the opposite way is theoretically the opposite of the potential that the spontaneous reaction would produce. In practice, the applied potential is usually greater.

[quote="Americandesi"]I believe these are the right electrode potentials but I am not 100% sure.[/quote]

That depends on the pH. For pH = 7, for example, we have $E^{0}(O_{2}/H_{2}O) =+0.81V$, $E^{0}(H_{2}O/H_{2})=-0.42V$, and so for the spontaneous reaction $E^{0}=+1.23V$. So, to drive this reaction in the opposite way at least a voltage of -1.23V must be supplied.


function
calculus
complex analysis
complex numbers
complex analysis solved

Find the radius of convergence of $\frac{x}{e^x-1}$ represented as an entire series.

According to a general theorem from complex analysis, this is just the distance from the point $x_0$ at which you want to write the Taylor series to the set $2\pi i (\mathbb Z\setminus \{0\})$ of the poles of the function $\frac z{e^z-1}$. If $x_0$ is real, the answer is $\sqrt{x_0^2+4\pi^2}$.

So in particular, if you want the series centered at zero (which is what I assume harazi meant), the radius of convergence is $2\pi.$ Note that the singularity at zero is removable.

"Entire" wasn't a good choice of word in the original statement. This function is meromorphic on $\mathbb{C}$ but not holomorphic.

Calculating the coefficients of this series also gives you the value of the zeta function at the negative integers and the positive even integers.

$\frac{t}{e^{t}-1}=\sum B_n\frac{t^{n}}{n!}$ with $0<|t|<2\pi$

$B_n$ Bernouilli numbers

[quote="Kent Merryfield"]Once you start talking about the radius of convergence of power series, the complex numbers appear whether you want them to or not. I like to tease calculus classes by asking them why the  series

\[\arctan x=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{2k+1}\]
only has radius of convergence $1$ when clearly nothing goes "wrong" with the arctangent function at $\pm1.$[/quote]

A little mean Kent! Do they know complex analysis?


probability
function
number theory
relatively prime

Hello everyone,

With regards to the two following math problems taken from contests, could someone please check the elegance of the solutions? In other words, would there be more intelligent methods to solve the problem?

For instance, in the second problem, a portion of my answer comes from educated guess and check.

Thank you very much!

---

[b] 1. A fraction $ \frac {p}{q}$ is in lowest termsif $ p$ and $ q$ have no common factor larger than 1. How many of the 71 fractions $ \frac {1}{72}, \frac {2}{72}, ... \frac {70}{72}, \frac {71}{72}$ are in lowest terms?

2. Suppose $ n$ and $ D$ are integers with $ n$ positive and $ 0 \leq D \leq 9$. Determine $ n$ if $ \frac {n}{810} \equal{} 0.\overline{9D5} \equal{} 0.9D59D59D5....$ [/b]

---

1. Since all nonprime, even numbers in the numerator can be reduced, verify all primes up to 71.

Primes up to 71: 2, 3, 5, 7, 11, 13, 15, 17, 19, 23, 29, 31, 37, 41, 43, 477, 53, 59, 61, 67, 71

Since $ \frac {2}{72}$ and $ \frac {3}{72}$ can be reduced, 18 of the 71 fractions are in lowest terms.

2. The equation can be written as:

$ \frac {n}{810} \equal{} \frac {9D5}{999}$

$ 810(9D5) \equal{} 999n$

$ 9D5 \equal{} \frac {37}{30}n$

Since $ 0 \leq D \leq 9$, try values of D:

If D = 0 $ \Rightarrow$ n = 733.78 $ \rightarrow$ not an integer

If D = 1 $ \Rightarrow$ n = 741.89 $ \rightarrow$ not an integer

If D = 2 $ \Rightarrow$ n = 750 $ \rightarrow$ solution since integer

[quote="catalan"]A fraction $ \frac {p}{q}$ is in lowest termsif $ p$ and $ q$ have no common factor larger than 1. How many of the 71 fractions $ \frac {1}{72}, \frac {2}{72}, ... \frac {70}{72}, \frac {71}{72}$ are in lowest terms?[/quote]Any numerator divisible bt 2 and/or 3 would give you a reducible fraction, so you need numerators that when divided by 6 give you a remainder of 1 or 5. (The other posters are going to be talking about mod 6).
Since 72/6=9, in the number from 1 to 71 there are 9 quotients (0, 1, ...5), each with both of those remainders, and 9X2=18 numbers that work.
EDIT:  :blush: I should not have been posting past my bedtime on a Friday. What I should have said is:
Since 72/6=12, among the integers from 1 to 71, as we divide them by 6, there are 12 quotients (0, 1, ...11), each with both of those remainders (1 & 5), and 12X2=24 numbers that work (all the numbers given by 6k+1 and 6k+5, with k belonging to {0, 1, 2, 3, 4, ...11}).

[quote="catalan"]Suppose $ n$ and $ D$ are integers with $ n$ positive and $ 0 \leq D \leq 9$. Determine $ n$ if $ \frac {n}{810} \equal{} 0.\overline{9D5} \equal{} 0.9D59D59D5....$ [/b]
.................................
The equation can be written as:

$ \frac {n}{810} \equal{} \frac {9D5}{999}$

$ 810(9D5) \equal{} 999n$

$ 9D5 \equal{} \frac {37}{30}n$[/quote]Since $ \frac {37}{30}n$ is an integer, $ n$ must be a multiple of $ 30$, but once divided by $ 30$ it is still a multiple of $ 5$, because $ 9D5$ is obviously a multiple of $ 5$. Then $ n$ must be a multiple of $ 150$. Since $ 600$ is obviously too low and $ 900$ too high, it must be $ n\equal{}750$.

[b]@ catalan:[/b] Regarding the first problem...

[hide="How to fix your solution to make it correct"]1. You missed an initial numerator of 1, which is not prime, but leads to the irreducible fraction 1/72

2. You missed numbers which are composite but do not share factors with 72. There are 5 of these.
[hide="Click to see what you missed"]5*5
5*7
5*11
5*13
7*7[/hide]

3. You accidentally included 15 in the list of primes, but you didn't count it. It doesn't make a difference, but I wanted to point it out.

The correct answer is 18+1+5 = 24[/hide]

[hide="A more effective approach"]$ 72 \equal{} 2^3 \cdot 3^2$. Therefore, a numerator will lead to an irreducible fraction if and only if it is divisible by neither 2 nor 3.

We have 72 integers (we're counting the number 72 for now -- it will be excluded anyway.)
We exclude the 72/2 = 36 integers divisible by 2
We exclude the 72/3 = 24 integers divisible by 3

Since we excluded the integers divisible by 6 two times, we need to add them back:
We add back the 72/6 = 12 integers we excluded twice

The answer is thus 72-36-24+12 = 24

This approach is called the [url=http://en.wikipedia.org/wiki/Inclusion-exclusion_principle]Principle of Inclusion and Exclusion (aka P.I.E.)[/url][/hide]

Alternatively, note that the probability that it is not divisible by $ 2$ is $ \frac12$ and the probability that it is not divisible by $ 3$ is $ \frac23$, so the probability that it is not divisible by either of them is $ \frac12\cdot\frac23\cdot72\equal{}24$.

Refer to [url=http://www.artofproblemsolving.com/Wiki/index.php/Totient_function]this.[/url]

Thank yuo very much to the responses!

They are very helpful.

[hide="first problem, higher tech"]The number of numbers less than or equal to $ n$ that are relatively prime to $ n$ are $ \phi(n)$, where that is the [[Phi Function]]. Thus $ \phi(72)\equal{}72*\frac{1}{2}*\frac{2}{3}\equal{}\boxed{24}$[/hide]

[hide="second problem, lower tech"]Let's call this number $ x$.

$ x\equal{}.9D59D59D5....$
$ 1000x\equal{}9D5.9D59D59D5....$
$ 999x\equal{}9D5$

$ x\equal{}\frac{9D5}{999}$

$ \frac{9D5}{999}\equal{}\frac{n}{810}$

$ \frac{9D5}{37}\equal{}\frac{n}{30}$

$ 9D5*30\equal{}37n$

The only $ D$ such that $ 9D5$ is a multiple of 37 is $ D\equal{}2$. Then we have $ n\equal{}25*30\equal{}750$.[/hide]


function
analytic geometry
inequalities
geometry
perimeter
combinatorics solved
combinatorics

find the minimum number of squares needed to draw a nxn board.

Sorry to ask, but please do not answer to this problem until 2 weeks, since it is used in some competition...

Thanks.

Pierre.

:D  Well...Valentin, this is a part of a solution...

Please, wait a little...
I'll give all informations you want and a solution too.

Pierre.

The solution's not too hard actually.

I take it that some of the people on this forum are participating in that competition?  If so, isn't your post going to be a "hint" in some way also?

sorry, i do not know that it is in any competition...  i thought this would be my own question  :D

although i want to know what competition this is for

I think we may be more interested in knowing what competition this question was used for than what the solution is ;)

sorry
i didn't understand the problem
n x n means that it has to have n squares on the line and n squares on the column???
and what means draw a bord... like puting squares, one beside other to form the board??

Hi Kuba!

Think of it as a chess board of n*n it's like that! :)

cheers! :)

The competition is a french training for the IMO pre-candidates. They have to solve a set of problems by month at home. The closing date is for January 2004....

Sorry for asking you to be patient...

Pierre.

Ok, the date line is behind us....thanks for not answering this problem before i tell you to do...

We use it in our correspondance training for the french team for the IMO.

Now solve it!!! :D ....or I will do it by myself!

Pierre.

Ok I am so waiting for this. You give me an N and I can solve it for you- Just give me enough CPU cycles that is. As for a function, I can't seem to find one.   Maybe I'm not understanding the question correctly.  I would be interested in seeing what you have for various N. In particular, prime N.

??? Well, there is no need of CPU here....

Let f(n) be the minimal number of squares required to draw (exactly) a nxn square.
it is easy to see that f(1) = 1 and f(2) = 3. Now, we assume that n \geq 3.
We will prove that f(n) = 2(n-1).

Consider that the nxn square S is made by n 2  unit squares whose corners are the points (i,j), where 0 \leq i,j  \leq n.

Now, consider the set E of the 4(n-1) points with coordinates (i,0) or (i,n) or (0,j) or (n,j) except the 4 corners of S.

Let's give a set A of squares from which we may form S.
Any square from A may only have two points from E as corners (at most two. The whole square has none). From the pigeon-hole principle, we deduce that A has at least 2(n-1) elements, that is f(n)  \geq 2(n-1).

To prove the reverse inequality, it suffices to give an exemple of such a drawning with only 2(n-1) squares, which will be boring with a keyboard...
let's M(0,0) and P(n,n) be two opposite vertices of the whole square. Let N(0,n),Q(n,0) be twe two other opposite vertices. Construct each of the squares with corner M (resp. P) and side 2,3,...,n-1. Then, we may finish the drawning of S with the square with corner N and the one with corner Q and both with sides n-1. Thus, a drawning does exist with only 2(n-1) squares.

Pierre.

Ok, I see my problem. I interpreted it as "find the minimum number of [b]tiles to cover [/b]an NXN checker board" - Very similar to the 'Mrs Perkins Quilt' problem. That's a considerably more difficult question. In that case we have
f(n) = 4 for n even and
n   f(n)
3    6
5    8
7    9
11  11
13  11
15  6
17  12
19  13

 And so forth.....

there doesn't seem to be any pattern in these numbers ..  :?

If there was a pattern, I would have already posted a solution. Now you see why I was waiting so eagerly  :D
 There is more to be said about my take on the problem. If you draw these out by hand you will notice a peculiar property at N=9 and 15 (and all odd composite numbers). I wont hijack this thread with the details. It's neat though.
Even still,  I havn't fully read the accepted solution yet but Im not sure how you can use 2(4-1)=6 squares to draw a 4X4 checkerboard. Did Pbornsztein mean instead that F(1)=1, F(2) = 2 and in general, F(n)=2*F(n-1) yielding F(n) = 2^(n-1)??? I guess I'm not sure on what drawing a NXN board really means. Is a checkerboard complete if each location either has a square or is bounded by 4 squares? The outside of the checkerboard could then also be considered a square so that the locations on the perimeter would need to either contain a square or be surrounded by 3 adjacent squares.
 I guess I need clarification from bugzpodder as to the details.


[quote="Ronald Widjojo"]Please explain what is CYH, PQR, SMV, EMV, SOS, and similar method that often used in this forum..

Thanks..  :D[/quote]
About $ EMV$ see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205183
About $ SMV$ see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113248
About $ SOS$ see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127


logarithms
function
real analysis
real analysis unsolved

n>0 integer
$u_n$  the smallest positive root of $e^x=x^n$

Give an asymptotic expansion of $u_n$ with two terms ?

Guess based on numerical twiddling (no proof):

$u_n\sim 1+\frac1n+\frac3{2n^2}$

(Note that $u_n$ is only defined for $n>e.$)

OK, amend that, I do have a derivation, given the assumption that an asymptotic expansion exists.

Write the equation to be solved as $x=n\ln x.$

Suppose $u_n=1+\frac an+\frac b{n^2}+O(n^{-3}).$

Then $\ln(u_n)=\frac an+\frac b{n^2}-\frac{a^2}{2n^2}+O(n^{-3})$

$1+\frac an+\frac b{n^2}+O(n^{-3})=a+\frac{2b-a^2}{2n}+O(n^{-2})$

Equating terms, $1=a$ and $a=\frac{2b-a^2}2,$ from which $b=\frac32.$

Let $z=1/n$. Then the equation to solve can be rewritten as $F(x,z)=x-e^{zx}=0$. Note that $F(x,z)$ is an analytic function of 2 complex variables and $\frac{\partial F}{\partial x}(1,0)=1\ne 0$. Thus, by the implicit function theorem, there exists an analitic function $g(z)$ in some neighborhood of the origin such that $g(0)=1$ and $F(g(z),z)=0$. This function gives our solution $u_n$ for small positive $z=\frac 1n$. But, as any analytic function,  it can be written as the sum of its Taylor series $g(z)=\sum_{n\ge 0}a_n z^n$. From here, it is clear that the desired asymptotic expansion of $u_n$ in inverse powers of $n$ exists and, moreover, the corresponding series converges to $u_n$.


1) Mixture of two liquids A and B is placed in a cylinder containing piston. Piston is pulled out isothermally so that volume of liquid decreases but that of vapour increases. When negligibly small amount of liquid was remaining, the mole fraction of A in vapour is $ 0.4$. Given $ P^0_A = 0.4atm$ and $ P^0_B = 1.2 atm$ at the experimental temperature. Calculate the total pressure at which the liquid has almost evaporated.(Assume ideal behaviour)

2) The latent heat of fusion of ice is $ 80cal/gm$ at $ 0^0C$. What is the freezing point of a solution of KCl in water containing $ 7.45$ gm of solute in $ 500gm$ of water, assuming that the salt is dissociated to the extent of $ 95%
$.

3)The molar volume of liquid benzene(density = 0.877 g/ml) increases by a factor of 2750 as it vaporizes at $ 20^0C$ and that of liquid toluene(density = 0.867 g/ml) increases by a factor of 7720 at $ 20^0C$, at a vapour pressure of 46 torr. Find the mole fraction of benzene in the vapour phase of the above solution.

4) An aqueous solution containing 288gm of a non volatile compound having stoichiometric composition $ C_xH_{2x}O_x$ in $ 90gm$ water boils at $ 101.24^0C$ at 1 atm pressure. What is the molecular formula?

Given : $ K_b(H_2O) = 0.512 K mol^{ - 1}kg$ and $ T_b(H_2O) = 100C$

[hide="Number 4"]From $ \Delta T \equal{} K_b \cdot \bar{m}$ we have $ 1.24 \,\,K \equal{} 0.512\,\,K\,kg\,mol^{\minus{}1} \cdot \frac{288}{M(0.288 \plus{} 0.090)} \equal{}> M \equal{} 314.6\,\,g\,mol^{\minus{}1}$, where M is the sought molar mass and $ \bar{m}$ is the solutions' molality. But $ M \equal{} 12.011x \plus{} 2 \cdot 1.0079x \plus{} 15.9994x \equal{} 30.0262x$, from which we get $ x \equal{} 10.5 \approx 11$.[/hide]

[quote]From $ \Delta T \equal{} K_b \cdot \bar{m}$ we have $ 1.24 \,\,K \equal{} 0.512\,\,K\,kg\,mol^{ \minus{} 1} \cdot \frac {288}{M(0.288 \plus{} 0.090)} \equal{} > M \equal{} 314.6\,\,g\,mol^{ \minus{} 1}$, where M is the sought molar mass and $ \bar{m}$ is the solutions' molality. [/quote]

How come this is the molality $ \frac {288}{M(0.288 \plus{} 0.090)}$ and not this $ \frac {288}{M.090}$.

Well actually the second gives the right answer x= 44 but I think since the solute is so much in excess , the situation might just be something else.

PLease try others also.

that's cent pc rite
most books use the conventional formula..without giving a look to solute amont..even NCERT :rotfl:  :P

But the question is from reliable source and we get answer using 288/M x .09 only.

Here are my answers. Please do tell if they are right. I shall post my method, if they are right.

[hide="1"]0.667 atm
[/hide]
[hide="2"]$ \approx \minus{}0.37^oC$[/hide]

[hide="3"]Either I have misinterpreted the question or the question has messed up with its data. Here is what I have gathered:
Pure liquid benzene vaporises and occupies a volume 2750 times its original value.
Pure liquid toluene vaporises and occupies a volume 7720 times its original value at a VP of 46 torr.
I checked the VP of toluene at 20C. It was 21.86 torr.
And all of a sudden, there is a mention of a solution...
 :o  :P [/hide]

[hide="4"]I agree with Carcul Sir. I did not do the calculations, but I agree with his method. Note that $ \Delta T\equal{}K_f\cdot m$ is an approximation. The correct formula is $ \Delta T\equal{}\frac {RT_m^2x_s}{\Delta H_f}$
where $ x_s$ is the mole fraction of solute and $ \Delta H_f$ is the enthalpy of fusion.[/hide]

A Useful link that include the derivation of the various colligative properties
http://www.lsbu.ac.uk/water/collig.html

1 is right.
2; the correct answer is -0.73
3; I dont understand the question as well.

Please go on explaining your answers.

1. $ P_s$ is VP of liquid mixture. X is mole fraction in liquid phase.
$ P_s\equal{}P_A^oX_A\plus{}P_B^oX_B$
Partial pressure of A =$ P_A^oX_A\equal{}0.4P_s$. Also $ X_A\plus{}X_B\equal{}1$.
This gives the values of $ X_A$ and $ X_B$. So, we know $ P_s$

2. Solution can be found in the link I gave.


inequalities
search
inequalities unsolved

while solving an inequality we came to have something like this

x/ :sqrt: (x 2 + :lambda yz) +y/ :sqrt: (y 2 + :lambda xz) + z/ :sqrt: (z 2 + :lambda xy) >= 3/ :sqrt: (1+ :lambda ).

Is this true for  :lambda  >= 8

You will definitely find many discussions on the forum about this inequality (from IMO 2001, I think). So, search the ssolved problems area.

Oh, my God!
Yet another one! A general method for such inequalities was also discussed.


limit
logarithms
integration
calculus
calculus computations

find this limit :D

$\lim_{n\to\infty}\;\left[ \;\left( 1+\frac{1}{n^{3}}\right) \left( 1+\frac{8}{n^{3}}\right) \left( 1+\frac{27}{n^{3}}\right) \cdots\;\;\;\cdots \left( 1+\frac{n^{3}}{n^{3}}\right) \;\right]^{\frac{1}{n}}$

[quote="yagaron"]${\lim_{n\to \infty }\, \log \left(\left(\prod_{i=1}^{n}\left(\frac{i^{3}}{n^{3}}+1\right)\right)^{1/n}\right)=\lim_{n\to \infty }\, \left(\sum_{i=1}^{n}\frac{\log \left(\frac{i^{3}}{n^{3}}+1\right)}{n}\right)=\lim_{n\to \infty }\, \int_{1}^{n}\frac{\log \left(\frac{i^{3}}{n^{3}}+1\right)}{n}\, di=\log (4)+\frac{\pi }{\sqrt{3}}-3}$

is this allowed?

i dont know, so i dont evaluate the integral, if someone say it is allowed i will evaluate[/quote]
 :lol:
or you have find this sum
${\sum_{k=1}^{+oo}\frac{(-1)^{k+1}}{k(3k+1)}=\log (4)+\frac{\pi }{\sqrt{3}}-3}$

the way you people write $\inf = oo$ is so funny.  :rotfl:


sorry but I can't post a new topic .it says; spam_title_all_caps. what does it mean?

It measn that all the letters in your title-text are uppercase. Make them lowercase and you will be fine.

Wait.  I thought that (when you are not a moderator) when you post a topic, no matter what, it all gets changed to lowercase except the first letter?  Has that been changed?


induction
number theory proposed
number theory


The binomial theorem

[img]http://img352.imageshack.us/img352/7360/mathwd7.png[/img]

If anyone can show the full step by step process to solving this it would be a great help!!! Please start from the beginning of the problem and work through it so I understand clearly.

To all those who do post a solutions...you guys are legends!![/img]

You're trying to find the wrong coefficient- specifically, that of $ c^{400}p^{\minus{}200}$. Beware of the pitfalls of exponent arithmetic.

That's what they told OP the first time this was posted:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193318
The second time this was posted, there was a different hint:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193470
The third time, no one has responded yet:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193472

Summary: DON'T DOUBLE POST.  It's exceedingly rude.


A circle with center $ C$ is tangent to the positive $ x$ and $ y$-axes and externally tangent to the circle centered at $ (3,0)$ with radius $ 1$. What is the sum of all possible radii of the circle with center $ C$?

$ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 8 \qquad
\textbf{(E)}\ 9$

[hide]If we draw the circle tangent to the positive axes with radius $ r$, then its center will be $ (r,r)$.

If this circle is tangent to the given circle, then the distance between its center and the center of the other circle will be $ r \plus{} 1$.

The distance between them, in terms of $ r$, is:

$ \sqrt {(r \minus{} 3)^2 \plus{} r^2} \equal{} r \plus{} 1$

$ (r \minus{} 3)^2 \plus{} r^2 \equal{} r^2 \plus{} 2r \plus{} 1$

This simplifies to $ r^2 \minus{} 8r \plus{} 8 \equal{} 0$. The sum of the roots of this will be $ 8 \rightarrow \boxed{\textbf{(D)}}$
[/hide]


[quote="Valentin Vornicu"]checking again the new version

\[ f(x) = \frac 34 { n\choose 2 } \cdot \sum^n_{k=1} x_k \]

or better yet

\[ f(x) - \frac {n(n+1)}2  = g(x) \]

or just plain old simple $ f(x) = \frac 34 $ ? :)[/quote]
and quoting works now as does use of double dollar sign
\[\frac 12\]

trying harder we obtain $ f(x) = g(x) $ ?

This is a very obscure bug - just fixing it now!

hmmm, let's test again with the fix up, $f(x)=g(x)$ ? let's also try the $ \$ ... $ \$ tag :)

it works, so now we can use the normal latex tags to do the work.

the dolar sign is to be made like this: [code] $\$ \ $\$ [/code].

hopefully you won't need it that much :)

[quote="Anonymous"]$\frac 12$ and \[\frac 12\] but
[code]$\frac 12$[/code] and [code]$$\frac 12$$[/code]
Yippee![/quote]
Forgot to log in so testing quote
Edited and still OK  :)

l maybe forgot some spaces or \... l dunno
what can you tell me about that ?

$\\28^x=\\19^y+\\87^z$

it's because you wrote \28^n = \19^n etc.
you must  use the backslash \  only for commands. \28 is not a command :D

[quote="arthur"]why some numbers vanished sometimes ?
l wrote 28^x=19^y+97^z and 8^x=9^y+7^z appeared
l try again :
$\28^x=\19^y+\87^z$

(you can solve it if you want)[/quote]
Commands in LaTeX always start with \ It tells the program that the next character is the start of a command and is to be treated in a special way. So when LaTeX sees \28 it thinks ... yes, 2 is the start of a command, but no command starts with that so it can be ignored. So never put in \ unless you want LaTeX to do something.
For example, \log 2 means that you want the word log to be the function , not the 3 variables l o and g.
Compare [code]$\log 2$[/code] which gives $\log 2$ with [code]$log 2$[/code] which gives $log 2$ and which means $l \times o \times g \times 2$

Does this makes sense?

[quote="stevem"][quote="arthur"]$\\28^x=\\19^y+\\87^z$[/quote]
Valentin beat me to it! In this case you had
[code]$\\28^x=\\19^y+\\87^z$[/code] and \\ means a new line so LaTeX puts a new line and is ready to deal with 28 as a number[/quote]

just a lucky shot :P

test :
$\sum^n_{p=1}a_p=2$ and $x_1=\frac{a}{\sqrt{px}}$


geometry
angle bisector
geometry proposed

A,B,C,D are 4 points on the line d. let a be a certain line on the plane. Show an approaching to construct the points M on that line, satisfies the following condition: angleAMB =angleCMD

We'll find all the points $M$ on the plane such that $\angle AMB=\angle CMD$

Suppose that the points $A,B,C,D$ lie on this order on the line $d$
If their order is like $A,C,B,D$, then just rename the points $B,C$ - it is equivalent
If their order is like $A,C,D,B$, then $\widehat{ AMB}> \widehat{CMD}$

Let $MX$ be the angle bisector of $\angle BMC \ (X \in d)$. Then of course $MX$ is also the angle bisector of $\angle AMD$

Let $Y\in d$ such that $MY \perp MX$ (the line $AY$ is external angle bisector of both $\angle BMC$ and $\angle AMD$)

The points $X,Y$ are harmonic conjugates of both $A,D$ and $B,C$. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16821]Here[/url] there is a way to construct the points $X,Y$

The locus of the point $M$ is the circle with diameter $XY$


limit
inequalities
real analysis
real analysis unsolved

In preparation for my multi-var. calc exam next week i came across this question:

Let $f(x,y) = \frac{|x|^{a}|y|^{b}}{|x|^{c}+|y|^{d}}$ where a, b, c, d are positive.
i) Prove that if $\frac{a}{c}+\frac{b}{d}> 1$ then $\lim_{(x,y)\to(0,0)}f(x,y)$ exists and equals zero

..im pretty much stuck...any help would b welcome

I think you mean:$\frac{a}{c}+\frac{b}{d}>1$
here is the proof(Using Inequality estimating):use the Young's Inequality:
(for $a+b=1,a>0,b>0$,then $x^{a}y^{b}\le ax+by$)
we have:
$\frac{ad}{bc}|x|^{c}+|y|^{d}\ge \left(\frac{ad}{bc}+1\right) |x|^\frac{acd}{ad+bc}|y|^\frac{bcd}{ad+bc}$
the rest will be easy. :)

[quote="mlok"]Consider $|x|^{c}$ and $|y|^{d}$ as new variables.[/quote]
?...sorry. i dont follow

Let $u=|x|^{c}$ and $v=|y|^{d}$. Note that $(u,v)\to (0,0)$ as $(x,y)\to (0,0)$. It remains to prove that \[\lim_{(u,v)\to (0,0)}\frac{u^{a/c}v^{b/d}}{u+v}=0\] where the limit is restricted to the first quadrant $u,v\ge 0$.

The degree of homogeneity of this fraction is $\frac{a}{c}+\frac{b}{d}-1>0$.

thats a nice logical way to look at it. Thank you!


induction
number theory unsolved
number theory

Let $x_0,x_1,x_2\ldots$ an infinite sequence of integer numbers.
$x_{n+2}=2x_n+x_{n-1}$
And
$x_0=1,x_1=0,x_2=2$
Prove that for all positive integer $m$ exists $m|x_k$ $m|x_{k+1}$ for some positive integer $k$

one easy induction prove that $x_n= F_{n-1} + (-1)^{n+1}$, for all $n \ge1$ ($F$ is the fibonacci sequence, whit $F_0 = 0$and $F_1 = 1$).

Now, we have to prove that for all positive integers $m$, exists $k$ such that $m| F_{2k} + 1$
and $m| F_{2k+1} - 1$.

(the same that prove that exists $k$ such that $m| F_{2k}$ and $F_{2k-1} = 1 (mod. m)$).

we know that for all positive integers $m$, exists $k$ such that $m| F_{2k}$, and the rest is easy.  :)


search
inequalities unsolved
inequalities

Please search here http://www.mathlinks.ro/resources.php?c=186&cid=41 .

[color=blue][b]Find the minimum value of $ f(x)\equal{}x^x$ for $ x>0$[/b][/color]

[quote="quocbao153"][color=blue][b]Let $ {a^2} \plus{} {b^2} \plus{} {c^2} \equal{} 3.$, Find the minimum value of the expression:
$ M \equal{} \frac {{{a^5}}}{{{b^3} \plus{} {c^2}}} \plus{} \frac {{{b^5}}}{{{c^3} \plus{} {a^2}}} \plus{} \frac {{{c^5}}}{{{a^3} \plus{} {b^2}}} \plus{} {a^4} \plus{} {b^4} \plus{} {c^4}$ [/b][/color][/quote]

It's a very weak ineq. It's obviously trues with the stronger condition: $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$
 :wink:


I am [i]really[/i] bad at problem solving and have absolutley no idea how to answer this question:

How many natural numbers less than 200 leave a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5?

My teacher skipped the sequences unit for some reason so I have no idea what they are or how to do them.  Also what is p?

P is an arbitrary natural number in his case I think.

[quote="Philmac"]How many natural numbers less than 200 leave a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5?[/quote]

Ok...help me out please...what did I do wrong?  :?

[hide]
If a number has a remainder of 4 when divided by 5, then it must have either a 4 or a 9 in it's ones' place.

Of the first ten multiples of 7 (7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ...) only 21 and 56 will have a unit's digit of 4 or 9 when 3 is added to it. So, for every 70 integers, 2 will leave a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5.

The smallest integer of $\frac{200/70}$ is 2, leaving a remainder of 60. Since 56 is less than 60, there must be a total of 6 numbers that will leave a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5.  :?
[/hide]

[quote="tarquin"][hide]Let the number you are looking for be $m$.

$m = 7x + 3$

$m = 5y + 4$

because it's four more than a multiple of 5 and three more than a multiple of 7.

$7x + 3 = 5y + 4$ so $y = \frac {7x - 1}5$

We must find all natural number values of $x$ and $y$.

Quickly we find that the values of $y$ are an arithmetic sequence in the form $7p - 3$. The maximum value of $p$ such that $7p - 3 < 200$ is $28$ so there must be $28$ values.[/hide][/quote]

Wait, what values are you looking for? x and y or m?

By the Chinese Remainder THeorem, there is at least one solution mod 35. If we examine all the integers less than 35, we see that 24 is the only such number. Therefore we can say that the only such m are: 24 59 94 129 164 199

[quote="mcalderbank"]Therefore we can say that the only such m are: 24 59 94 129 164 199[/quote]

Yay! Comforting! :P

Oh I messed up.

Was supposed to plug $7p-3$ for $y$ in $5y+4$

$35p-11 < 200$

$p<6.028$

So 6 values.

[quote="Philmac"]We haven't learned mod either, I only know what it is from computer programming  :(  Can you guys please come up with a dumber explanation that's at the eleventh grade level?[/quote]

Just scroll up and read my post. I don't think there are mods in there. (The idea behind it is still there, but it's disguised so you don't need knowledge of mods to understand my post...I think.) Let me know if you still have trouble.


function
real analysis
real analysis unsolved

Let $f: R\to R$ an additive function: $f(x+y)=f(x)+f(y) \forall x,y \in R$ which is bounded on $[0,1]$ and $f(1)=1$.
Prove that $f(x)=x \forall x \in R$

you know that $f(q)=q$ far every rational number $q \in \mathbb Q$. Now, suppose that there exists $a \in \mathbb R$ such that $f(a)=a+d$ where $d > 0$ for example. Then for every rational number $f(a-q)=(a-q)+d$, which shows that there exists arbitrary small real number $0<x<1$ such that $f(x)>d$. You can easily find a contradiction with $f$ bounded on $[0;1]$. The case $d<0$ is similar..


[quote="Bugi"]Find all two digit numbers $ \overline{AB}$ such that $ \overline{AB}$ divides $ \overline{A0B}$.[/quote]

We have $ 100x\plus{}y\equal{}k(10x\plus{}y)$ and so $ y\equal{}10\frac{10\minus{}k}{k\minus{}1}x$

We have obviously the solution $ k\equal{}10$ and $ \overline{A00}\equal{}10\cdot \overline{A0}$

If $ y>0$, $ 100x\plus{}y < 10(10x\plus{}y)$ and so $ k<10$. Then we need $ 10\minus{}k<k\minus{}1$ and so $ k>5$ So :

$ k\equal{}6$ $ \implies$ $ y\equal{}10\frac{10\minus{}6}{6\minus{}1}x\equal{}8x$ $ \implies$ $ \overline{108}\equal{}6\cdot \overline{18}$

$ k\equal{}7$ $ \implies$ $ y\equal{}10\frac{10\minus{}7}{7\minus{}1}x\equal{}5x$ $ \implies$ $ \overline{105}\equal{}7\cdot \overline{15}$

$ k\equal{}8$ $ \implies$ $ y\equal{}10\frac{10\minus{}8}{8\minus{}1}x\equal{}\frac{20}{7}$ $ \implies$ no solution.

$ k\equal{}9$ $ \implies$ $ y\equal{}10\frac{10\minus{}9}{9\minus{}1}x\equal{}\frac 54x$ $ \implies$ $ \overline{405}\equal{}9\cdot \overline{45}$

[quote="Bugi"]Find all two digit numbers $ \overline{AB}$ such that $ \overline{AB}$ divides $ \overline{A0B}$.[/quote]

[u]Alternate solution:[/u]

Let $\overline{AB}=x$

$10A\equiv -B\:mod\:x\Longrightarrow 100A \equiv -10B\:mod\:x$ but $100A\equiv -B\:mod\:x$ and so, $-B\equiv -10B\:mod\:x$

So, $x=\overline{AB}|9B$

$B=0\Longrightarrow A=1, 2, 3, \cdots 9$

$B=1$ has no solutions for $A$

$B=2$ has no solutions for $A$

$B=3$ has no solutions for $A$

$B=4$ has no solutions for $A$

$B=5\Longrightarrow A=1, 4$

$B=6$ has no solutions for $A$

$B=7$ has no solutions for $A$

$B=8\Longrightarrow A=1$

$B=9$ has no solutions for $A$

So, $\overline{AB}=\{10, 20, 30, 40, 50, 60, 70, 80, 90, 15, 45, 18\}$


pardon me if this problem has been discussed
A and B play a game on an $ n\times n board$. They take turn put either 0 or 1 into any of the blank squares. When all the squares have been filled, B wind if the sum of all numbers in each row is even. Otherwise A wins
Which player has a winning strategy?

hmm i'm not sure.... nevertheless

If i understand the problem correctly sum has to be even only in rows but not in colums. If that is the case then

If $ n$ is even and A starts the game B has a wining strategy. If A puts number $ i$ in some row then B only puts the same number $ i$ in the same row.

If $ n$ is odd and A starts than A has a winning strategy. Because he inserts the last number he can control the oddnes/eveness of the last row.

If $ n$ is even and B starts the game,A has a wining strategy. B begins by putting some number in some row. Now A puts a different number in the same row.Then everytime B puts a number in this row A just puts the same number in that row.

If $ n$ is odd and B starts than I don't know  :)
(in all cases the other player is eventualy forced to put the number in that row)


Ryt u guys! Wonder if u could give me a logical explanation to this.  It is really simple and some think quite stupid.....

3 brothers have saved up money towards buying a cd player for their bedroom.  They each save £10 and go to buy this £30 Cd player they found in the shops.  They take the player to the cash desk and the cashier runs it through and realises that there is a special £5 off bargain on the product.  The sneaky cashier puts £2 in his pocket and gives each of the brothers a £1 back.  So...........

Each brother had £10 but got a £1 back so each of the brothers have paid £9 each.

£9 X 3 = £27

and the cashier has £2 in his pocket.

£27 + £2 = £29... so where has the other £1 gone?????

Well that's just the bellhop problem in a different situation, and it's a very simple solution- $\$$27 and $\$$2 should not be added, although the problem is worded to make you think they should be.  Of the 30, each brother has 1 (3), the cashier has 2, and 25 went to buying the CD player.  3+2+25=30.

You should actually subtract the 2 (that the guy took) from the 27 (that they paid) to get the price of the CD player.


calculus
integration
trigonometry
calculus computations

evaluate

$\int\frac{1}{3\cos x+4\sin x+5}dx$

Any idea how to tackle it ?

I think it is becuase the lengths $3,4,5$ are a right triangle :wink:

ok i see. using $\tan(x)=t$ makes the integral kind of messy.


On a drive from your ranch to Austin, you wish to average 40mph.  The distance from your ranch to Austin is 70 miles.  However, at 35 miles (half way), you find you have averaged only 30 mph.  What average speed must you maintain in the remaining distance in order to have an overall average speed of 40mph? Answer in units of mph.

[hide]To average 40mph, you need to make the total trip in 70/40=7/4 hours. At the halfway point you have used 35/30=7/6 hours, so you must make the remaining journey in 7/4-7/6 = 7/12 hours. The speed you need is 35/(7/12)=60 mph[/hide]


let $ \lambda: [a,b]\rightarrow{\mathbb{R}^n}$ a closed path such that $ \lambda'(t)$ is continuous, ie, is $ C^1$. Show that there is some $ t\in(a,b)$ such that $ <\lambda(t),\lambda'(t)>\equal{}0$

Thanks

Hint: $ t \mapsto \parallel{}\lambda(t)\parallel{}^2$ is $ C^1$ and defined on a compact interval.

Hello dihuana, thanks for your help, I do not understande wery well, can you give some details?
Thanks a lot
Hugs

as dihuana stated, look at $ f(t)\equal{} \parallel \lambda(t)\parallel ^{2}\equal{} <\lambda(t),\lambda(t)>$
It is differentiable (as a sum of products of differentiable elements), and f(a)=f(b) since its a loop. now you can use a certain theorem from basic calculus.

thanks a lot guys, i understood it very well


Έστω $ p$ ένας πώτος αριθμός και $ m,n$ ακέραιοι μεγαλύτεροι του 1 . Υποθέτουμε ότι ο $ n$ διαιρεί τον

$ m^{p(n \minus{} 1)} \minus{} 1$ . Δείξτε ότι ο $ n$ και ο $ m^{n \minus{} 1} \minus{} 1$ έχουν μέγιστο  κοινό διαρέτη μεγαλύτερο της μονάδας .

Kat arxas euxaristw gia thn afierwsh!!(ektos k an uparxei allos hlias k den to 3erw! :P )
Profanws ta m,n den exoun koinous diairetes megaluterous ths monadas(alliws o diareths 8a diarouse to -1)
Estw q enas prwtos diaireths tou n.tote uparxei generator modq kai estw oti autos einai to g
estw epishs $ m\equiv g^x(modq)$
Tote exoume oti $ g^{xp(n\minus{}1)}\equiv 1(modq)\equal{}>q\minus{}1|xp(n\minus{}1)$ logw ths epiloghs tou g
alla p prwtos ara an $ q\neq2$ exoume oti
$ q\minus{}1|x(n\minus{}1)\equal{}>g^{x(n\minus{}1)}\equiv 1(modq) \equal{}>m^{n\minus{}1} \equiv 1(modq)$
dhladh $ q|(m^{n\minus{}1} \minus{}1)$ h alliws $ q|(n,m^{n\minus{}1} \minus{}1)$ dhladh $ (n,m^{n\minus{}1} \minus{}1)>1$
an twra $ q\equal{}2$ exoume oti $ 2|m^{p(n\minus{}1)}\minus{}1\equal{}>$ m perittos $ \equal{}>2|m^{n\minus{}1} \minus{}1$
An einai swsth h lush m tote thn afierwnw se esena!An oxi na 3ereis apla oti thn prospa8hsa eidika epeidh thn afierwses se emena!
Hlias

Σε σένα ήταν αφιερωμένη και η λύση σου είναι εξαιρετική  :)  Μπράβο .

Παιδιά, νομίζω ότι η (κατά τα άλλα πολύ ωραία :wink: ) απόδειξη του Ηλία έχει ένα μικρό κενό.

Συγκεκριμένα η σχέση $ q\minus{}1|xp(n\minus{}1)$ δεν συνεπάγεται απαραίτητα την $ q\minus{}1|x(n\minus{}1)$. Για να ισχύει αυτό πρέπει να δείξουμε επιπλέον ότι (με $ exp(t)$ συμβολίζω την μέγιστη δύναμη του $ p$ που διαιρεί τον $ t$) $ exp(q\minus{}1)<exp[xp(n\minus{}1)]$.

Αυτό το μικρό κενό μπορεί να καλυφθεί ως εξής:

Aρκεί να βρούμε έναν πρώτο διαιρέτη του $ n$ ώστε $ exp(q\minus{}1) \leq exp(n\minus{}1)$. Aν ο $ n$ έχει έναν πρώτο διαιρέτη $ q$ με $ (q\minus{}1,p)\equal{}1$, τότε επιλέγουμε αυτόν. Διαφορετικά $ n\equal{}\prod q_i^{a_i}$, όπου $ q_i \equiv 1 (mod p) \forall i$. Τότε επιλέγουμε τον $ q_j$ έτσι ώστε το $ exp(q_j^{a_j}\minus{}1)$ να είναι το ελάχιστο. Εύκολα μπορούμε να δείξουμε ότι $ exp(n\minus{}1) \geq exp(q_j^{a_j}\minus{}1) \geq exp(q\minus{}1)$, που είναι το ζητούμενο.

Διορθώστε με αν κάνω λάθος γιατί δεν το έχω δει με χαρτί και μολύβι  :wink: .


calculus function counting distinguishability

You know, I've seen a lot of things like this.

Try [url]http://www.openmathtext.org/downloads.html[/url].

I believe the one entitled "difference calculus generating functions" will be of use.  It explains how these things relate to generating functions and is in fact quite useful.

[quote="hyderman"]plerase explain this in details....

How many solutions are there to the equation
x1 + x2 + x3 = 12,
where x1 , x2 , x3 and x4 are integers satisfying
x1 ≥ 0,
x2 > 2 and x3 ≥ 1?[/quote]

$x_{1}\ge 0$
$x_{2}\ge 3$ (unless you meant "x2 ≥ 2" which would match the signs of the others.
$x_{3}\ge 1$

If we just let $x_{1}=0, x_{2}=3,$ and $x_{3}= 1$, then the sum would be $4$. So we need to add a total of $8$ units to the three variables.

consider each of these circles as a unit
00000000

We need to figure how to put these units into the variables. So we put in some seperators. For example,
000|00|00
would mean put 3 into $x_{1}$ ($x_{1}=0+3=4$), put 2 into $x_{2}$, ($x_{2}=3+2=5) and put 2 into$x_3$($x_3=1+2=3).

So there are 9 slots for the seperators, and two seperators, making $\binom{9}{2}+9=36+9=\boxed{45}$ (I added the 9 because the $\binom{9}{2}$ does not include the possibilities in which the seperators are in the same spot.

Using my own advice:

We are looking for the coefficient of $x^{12}$ in the expansion of $(1+x+x^{2}+\ldots)(x^{3}+x^{4}+x^{5}+\ldots)(x+x^{2}+x^{3}+\ldots)$.

Now notice that this can be rewritten as:

$\left(\frac{1}{1-x}\right)\left(\frac{x^{3}}{1-x}\right)\left(\frac{x}{1-x}\right)=\frac{x^{4}}{(1-x)^{3}}$

(Recall that $\frac{1}{1-x}= 1+x+x^{2}+x^{3}+\ldots$.  Multiply both sides by $(1-x)$ if this is not apparent.)

Now, since we are looking for the coefficient of $x^{12}$, it means that we need to find the coefficient of $x^{8}$ in the denominator (when written in power series form) so that the $x^{4}$ on top will multiply to an exponent of 12.  Now, we use the binomial formula:

$\binom{-3}{8}= \binom{8+3-1}{8}= \binom{10}{8}= 45$

This is handy for many problems like this, and the interesting thing is that you do not need to come up with a unique method for each type of problem, as this generalizes.

$y_{1}-1+y_{2}+2+y_{3}=12\implies y_{1}+y_{2}+y_{3}=11$, where $y_{i}\in\mathbb{N}$.  We have to choose 2 plus signs from the 10 spaces in 1_1_..._1, which can be done in $\binom{10}{2}=45$.

[quote="Altheman"]BALLS AND URNS

learn this...indistinguishable balls, and distinguishable urns for this problem[/quote]

How do you change formula to account for distinguishable urns (or distinguishable balls, for that matter)?

balls/urns

i/i no closed form

i/d bijection with walls and balls thing

d/i see stirling number of the second kind

d/d balls^urns

[quote="tjhance"]what are balls and urns?[/quote]balls are just balls, urns are just urns. Dictionary can help you if you need to look it up. If you want to know more about it mathematically, it refers the number of ways of placing $n$  ([size=150][b][color=red]d[/color][/b][/size]istinguishable/[size=150][b][color=red]i[/color][/b][/size]ndistinguishable)  balls into $m$ ([size=150][b][color=red]d[/color][/b][/size]istinguishable/[size=150][b][color=red]i[/color][/b][/size]ndistinguishable) urns. That's what Altheman is referring to. For more info, look up a good book with Combinatorics in it like Fred Roberts' Applied Combinatorics.


A standard deck of 52 cards is shuffled and the cards are dealt face up one at a time until an ace appears.  Show that the probability of getting the first ace on or before the ninth card is greater than 50%.

How would you solve this problem without having to write out all the probablities and then adding them all up?

Thanks!

[hide]The probability he doesn't get it in the first 9 turns is:

48/52*47/51*46/50...*41/45*40/44. We don't care what happens next. If you combine this, you notice that a lot of terms are repeated in the numerator and denominator, leaving 43*42*41*40/(52*51*50*49), which happens to be less than 50%. Therefore, the probability that you DO get an ace in the first 9 turns is greater than 50% (they have to add up to 100%)[/hide]

Would this be correct?

[hide]The deck is $ \frac{1}{13}$ aces so the probabilty that an ace would pop up in the first 9 cards would be $ \frac{9}{13}$ which is greater than $ \frac{1}{2}$[/hide]

SnipedYou, I don't think the probability an ace turns up is $ \frac {9}{13}$. Rather, it's easier to use complementary counting and find the probability that it [i]doesn't[/i] turn up:

$ \frac {48}{52} \times \frac {47}{51} \times \frac {46}{50} ... ... \times \frac{40}{44}$
$ \equal{} \frac {3526}{7735}$

So the probability it [i]does[/i] turn up is $ \frac {4209}{7735}$ (the complement) which is more than 50%.


Let a and b denote positive integers, such that $ a^n \plus{} n$ divides $ b^n \plus{} n$ for all positive integers n. Prove that a=b.

Let $ n(p)\equal{}1\plus{}(a\plus{}1)(p\minus{}1)$, then $ p|a^n\plus{}n\equal{}1\plus{}(a\plus{}1)(p\minus{}1)\plus{}a*a^{(a\plus{}1)(p\minus{}1)}$, therefore must be $ p|b^n\minus{}a^n\equal{}b\minus{}a\mod p$.
It mean, for any prime p $ p|b\minus{}a$ or $ b\equal{}a$.


Prove for $ a,b,c \in R_\plus{}$ that the if

$ ab\plus{}bc\plus{}ac \leq 3abc$ then

$ a^3\plus{}b^3\plus{}c^3 \geq a\plus{}b\plus{}c$

We have $ \sum a \le \frac {9a^2b^2c^2\left(\sum a \right)}{\left(\sum ab \right )^2} \le \sum a^3$, which is necessarily true by Muirhead.

[quote="NikolayKaz"]Prove for $ a,b,c \in R_ \plus{}$ that the if

$ ab \plus{} bc \plus{} ac \leq 3abc$ then

$ a^3 \plus{} b^3 \plus{} c^3 \geq a \plus{} b \plus{} c$[/quote]

We have:
$ ab\plus{}bc\plus{}ca\le 3abc\le \frac{1}{3}(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)$
so $ a\plus{}b\plus{}c\ge 3$
so that
$ a^3\plus{}b^3\plus{}c^3\ge \frac{(a\plus{}b\plus{}c)^3}{9}\ge \ a\plus{}b\plus{}c$ (by Holder)  :lol:


[quote="mdk"]For which real numbers b does the function f (x), defined by the conditions
f (0) = b and f' = 2f-x, satisfy f (x) > 0 for all x ≥ 0?[/quote]

First solve the differential equation,
$ f'-2f = x \implies f= Ce^{2x}-\frac{1}{2}x-\frac{1}{4}$

Now just find the conditions on the initial value problem.


can somone help?? I don't really know how yo count this

Question:
I need to get the number of sequences of length 20 consisting of 8 X chars, 7 Y chars and 5 Z chars in which no two Z chars appear adjacently, and X,Y do not appear adjacently.

Please Please Be specific
thanks
Alisha

Do not post the same problem more than once.  Continue your discussion here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193734[/url]


How do I prepare so I get at least a 50 on the local exam? :(

simple :)
do the past exams.
read books.

but where does one find these past tests?  and which books are highly recommended?

Sorry, cuz I'm sure these topics have been discussed before, but I went through a number of old topics and couldn't find much.

Try here:

http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education\student\co02.html

and for books?

(or more sites, but I'm lookin to get me a book soon)

The link above dosn't seem to work! Can you plz fix it!


Find the values of $ A,B$ such that:

$ \frac{A}{z\plus{}2}\plus{}\frac{B}{2z\minus{}3}\equal{}\frac{5z\minus{}11}{2z^2\plus{}z\minus{}6}$

Partial Fractions??
Multiply both sides by $ 2z^2 \plus{}z \minus{}6$ to get
$ A(2z\minus{}3) \plus{} B(z\plus{}2) \equal{} 5z \minus{} 11$
Let $ z \equal{} \minus{}2$ to get
$ \minus{}7A \equal{} \minus{}21$
so
$ A \equal{} 3$
Let $ z\equal{}\frac{3}{2}$ to get
$ \frac{7}{2}B \equal{} \minus{}\frac{7}{2}$
so
$ B\equal{} \minus{}1$


Given segments labeled 1, a, and b, construct a segment of length ab. Also, explain why a segment of length 1 is needed.

Given three segments with lengths $r,a,b$ their [i]fourth proportional[/i] is a segment $x$ such that $\frac{r}{a}= \frac{b}{x}\ \ (1)$

[hide="Construction 1"]Construct a segment $OA=a$.
Take a point $R\not\in OA$ such that $OR=r$
On the ray $(OR$ take a point $B$ such that $OB=b$
Through $B$ bring a parallel line to $AR$. This will intersect the ray $(OA$ at a point $X$. Let $x=OX$
Then from the similarity of the triangles $OAR,OBX$ we get $\frac{r}{a}= \frac{b}{x}$[/hide]
[hide="Construction 2"]I usually use a circle and power of point to construct the fourth proportional
Take $OA=a$ and on the ray $(OA$ choose a point $B$ such that $OB=b$
Take a point $R\not\in OA$ such that $OR=r$.
Construct the circumscribed circle of $\triangle ABR$.
Then the ray $OR$ intersects again the circle at a point $X$, where $OX=x$ is the fourth proportional of $r,a,b$, i.e. $ab=xr$[/hide]

From $(1)$ we get $x=\frac{ab}{r}$

But we want to construct the segment with length $x=ab$ . So it's necessary to choose $r=1$.

[color=brown]P.S.
In fact, independently of the length of [b]r[/b], the multiplication works if we set as unit of measure that length of [b]r[/b].

For example, suppose that [b]a=20cm[/b] and [b]b=30cm[/b]

If [b]r=1cm[/b] then we get [b]x=600cm[/b], and $20 \cdot 30 = 600$, ok!

If [b]r=10cm[/b], then [b]x=60cm[/b], and $20 \cdot 30 \not = 60$

But if we measure the lengths using as unit the length of [b]r=1dm[/b] then we can transform the lengths:
[b]a=20cm = 2 dm [/b]
[b]b=30cm = 3 dm[/b]
[b]x=60cm = 6 dm[/b]

Notice that $2\cdot 3 = 6$[/color]

we can easily employ power of a point to solve this problem


Here is a a problem:

If we have an n amount of points in a 2 dimensional plane, where n is bigger than 3, and these n points do not lie in a line, there exists a line which contains exactly two points.

Is theorm is Called Slyvester's Theorm.

Slyvesters Theorm: http://mathworld.wolfram.com/SylvestersLineProblem.html

Anybody know how to go abouts this?

[quote="E^(pi*i)=-1"][hide="A quick and probably incomplete proof"]
Assume for the sake of contradiction that the statement is false.  Consider the shortest possible distance between a point and a line. Say it is between point P and line m.

Line m must have at least 3 points on it, two of which must be on the same side of the perpendicular from P to m.  Call them Q and R, with R farther from the foot of the perpendicular than Q.  But clearly the distance from Q to line PR is shorter than the distance from P to line m, a contradiction.
[/hide][/quote]

I think it can be modified to work, but there's some ambiguity. Maybe you mean the shortest possible distance from any line containing at least 2 of the points and any point not on that line (the point exists because not all the points are on one line).

This is basically the same idea. Let $P$ and $m$ the be the point and line, respectively, with the shortest distance as defined above. Let $A$ be the foot of the perpendicular from $P$ to $m$. Then there are at least $3$ points on $m$, two on the same side of$A$, say $Q$ and $R$ with $QA < RA$. Let $B$ be the foot of the perpendicular from $Q$ to $PR$. We have $\triangle RBQ \sim \triangle RAP$ and $RQ < RA \Rightarrow \triangle RBQ$ is smaller, so $BQ < AP$, a contradiction.

[img]http://i71.photobucket.com/albums/i132/paladin8/1-27-07.jpg[/img]

paladin8, that is indeed what I intended - I didn't word it very well.

[quote="player"]Is theorm is Called Slyvester's Theorm.[/quote]
No, it is called Sylvesters theorem. You even gave a link with the correct spelling!
*posts this for the simple reason that a search for "sylvester" does not give this thread*

[quote="Altheman"]for the sake of contradiction, suppose that for every pair for 3 [or more] points, they lie on a line, anyway, take two points, A, B, then an unspecified point, X, then ABX is a line for all X, contradiction, since that makes all the points collinear[/quote]

Say what? If we take the easiest infinite counterexample, the lattice points, you can see that if we take two points, there are plenty of Xs that aren't on the line connecting those two points. I don't think your logic is complete.

Hopefully I didn't confuse anyone, but the infinite counterexample doesn't disprove the theorem because it wants a finite set of points.

[quote="t0rajir0u"]The infinite counterexample is irrelevant, but Altheman's logic is not sound.  The assumption by contradiction is $\exists X$, not $\forall X$.[/quote]

That was the point I was making by the infinite counterexample -_-


1.)  Prove by congruence that the integer $ 53^{103} \plus{} 103^{53}$ is divisible by $ 39$

2.) Prove that for any integer $ a$, $ a^3  \equiv 0, 1$ or $ 6 (mod 7)$.

[hide="1"]
Since $ \varphi(39) \equal{} 24$ (for Euler-Fermat),

$ 53^{103} \plus{} 103^{53} \equiv 14^{103} \plus{} 25^{53} \equiv (14)^7 \minus{} (14)^{5} \equiv 14^{5}(196\minus{}1) \equiv 0 \pmod{39}$
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[hide="2"]
You could just test every one of 7 possibilities for $ a \pmod{7}$.
[/hide]

1.
Since $ \varphi(39) \equal{} 24$ (for Euler-Fermat),

$ 53^{103} \plus{} 103^{53} \equiv 14^{103} \plus{} 25^{53} \equiv (14)^7 \minus{} (14)^{5} \equiv 14^{5}(196 \minus{} 1) \equiv 0 \pmod{39}$

2.
You could just test every one of 7 possibilities for $ a \pmod{7}$.

That was a direct quote removing the hidden information.  Please edit/delete your post.


conics
ellipse
geometry unsolved
geometry

The line V3V4 cut the tangent at P.The bissector of V3V4 is the major axis line support.V3V4 intersection wih bissector is point M.Draw the circle center M and diameter V3V4.Draw the tangent to this circle  through P.Through the tangency point T draw a paralel line o the major axis that cut the given tangent at pont J.This is the tangency point with the ellipse.Through J draw a perpendicular to he tangent that will cut the line V3V4 at point K.The circle with diameter PK will cut the majr axis at the points F1 and F2 that are the foci.