probability
expected value
probability and stats

There are $ 15$ white and $ 10$ black balls in the basket. We randomize one ball without returning and finish when we pull out all black balls. Let $ X$ - number of white balls that remain in the basket. Find $ EX$.
How should I find $ EX$ using conditional expectation and the fact that $ EX \equal{} E(E(X|Y))$?  :maybe:

Here's a solution with recursion:  let $ f(a, b)$ be the expected number of white balls left after all black balls get chosen, starting with $ a$ white and $ b$ black balls.  Then $ f(0, b) \equal{} 0$, $ f(a, 0) \equal{} a$ and $ f(a, b) \equal{} \frac {a f(a \minus{} 1, b) \plus{} b f(a, b \minus{} 1)}{a \plus{} b}$.  This lets us compute as many values as we like, in particular $ f(15, 10)$.  If we look at a few computed values, it also lets us conjecture that $ f(a, b) \equal{} \frac {a}{b \plus{} 1}$, which then follows immediately from the recursion by induction.  Once we know what the answer is, it shouldn't be too hard to prove it more directly with your desired method (and in fact there are lots of reasonably nice ways to solve this problem).


geometry
rotation
calculus
integration
function
trigonometry
geometry unsolved

It's a quite unusuall problem ;).
There are $n$ segments on the plane, sum of their lengths is $1$.
Prove that their exists a line on the plane for which the sum of the lengths of pependicular projections of those segments on this line is less than $\frac{2}{\pi}$.

I think it's a classic :).

Take a line and rotate it. The integral from $0$ to $2\pi$ of the length of the projection (regarded as a function of the angle our variable line makes with a fixed direction) is $\int^{2\pi}_0|\cos \theta|\mbox{d}\theta\ (*)$ (as can easily be seen), and this is equal to $4$. There must be a time when the projection is at least as large as the average of $(*)$, and this is $\frac 4{2\pi}=\frac 2\pi$.

Pick some horizontal axis and define the inclination $\{\theta_i\}_{i=1}^n$ of each segment as its angle with respect to this axis as well as its length $\{\ell_i\}_{i=1}^n$. For some line with inclincation $\phi$ with respect to the horizontal axis, the sum of the lengths of the projections will be $f(\phi) = \sum_{i=1}^n \ell_i \cos (\theta_i - \phi)$. Suppose, for sake of contradiction, that $\forall \phi \in [-\frac{\pi}{2}, \frac{\pi}{2}], f(\phi) \ge \frac{2}{\pi}$. Then $2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2}{\pi}\, d\phi \le \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(\phi)\, d\phi$ $= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sum_{i=1}^n \ell_i \cos (\theta_i - \phi)\, d\phi$ $=  \sum_{i=1}^n \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ell_i \cos (\theta_i - \phi)\, d\phi =$ $\sum_{i=1}^n \ell_i [\sin(\theta_i - \phi)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \le \sum_{i=1}^n 2\ell_i = 2$, so we must have equality everywhere, so $\forall i, \sin (\theta_i - \frac{\pi}{2}) = 1 \Rightarrow \forall i, \theta_i = \pi$, but then simply take $\phi = \frac{\pi}{2}$.

Nice one :),  I've already seen the "integral" solution and I'm interested whether there is a more elementary solution (that's why it's in the olimpyc section).

[Edit] there are millions of math problems and I know you know most of them grobber :), but this problem indeed seems quite unusuall :).


number theory
least common multiple
MATHCOUNTS
function

I think you're wrong for b...

Find possible ways with 1 quarter, then 2 quarters, then three quarters

For one quarter: find ways for one dime, two dimes, 3 dmes etc.

And so on.

[quote]But isn't that exactly the same thing as brute forcing it?  :)

Please correct me if I'm terribly wrong like I usually am.
Since 25 cents is their LCM, there has to be a way to make 25 cents (there's no way you can "miss" it somehow)... So we can take it 25 cents at a time to make it more convenient.[/quote]

Nevermind, I'm wrong again... Sigh. One way is 8 dimes, which obviously doesn't fit into the 25c pieces.

For number 1, I think it might be 14.  I just made it a coin problem ( coins worth 1, 2, 3, how many possibilities to get 10).  I made a table and counted 14 combinations.  Though something tells me I'm wrong...

It's not 14, because order matters. (3-3-2) is different from (3-2-3).

And what I'm really asking is, is there a way without writing everything else, like a really concise pattern?

I don't know a generic formula for this but, systematic counting shouldn't take too long.

a) Use a sequence.

$ a_1 \equal{} 1$
$ a_2 \equal{} 2$
$ a_3 \equal{} 4$
$ a_x \equal{} a_{x \minus{} 1} \plus{} a_{x \minus{} 2} \plus{} a_{x \minus{} 3}$
Then
$ a_4 \equal{} 7$
$ a_5 \equal{} 13$
$ a_6 \equal{} 24$
$ a_7 \equal{} 44$
$ a_8 \equal{} 81$
$ a_9 \equal{} 149$
$ a_{10} \equal{} 274$
So our answer is 274.

You can look at [url]http://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers#Tribonacci_numbers[/url]
for the tribonacci numbers.

[quote="Textangle"]How do you tackle the following kinds of problems, in general?

a) How many different ways can you add 1, 2, and 3 to get 10 if order matters?

b) I sorta have an idea for this one:
How many different ways can you make 80 cents with pennies, nickels, dimes, and quarters?
[hide="Can someone please tell me if this method is right"]The LCM of their values is 25 cents, and there are 13 ways to "compose" each 25 cents using pennies, nickels, dimes, and quarters... So taking it every 25 cents at a time, there are 5 cents left over, which can be composed in 2 ways. So is the answer $ 13\times 13\times13\times2 \equal{} 4394$?
[/hide]

c) What is the largest whole number that cannot be made by adding 4, 5, and 13 together?[/quote]

For $ c$, if there's two numbers, the formula is $ ab\minus{}a\minus{}b$.  I'm guessing at $ abc\minus{}(a\plus{}b\plus{}c)$, so it would be $ 238$.

[quote="Textangle"]
b) I sorta have an idea for this one:
How many different ways can you make 80 cents with pennies, nickels, dimes, and quarters?
[hide="Can someone please tell me if this method is right"]The LCM of their values is 25 cents, and there are 13 ways to "compose" each 25 cents using pennies, nickels, dimes, and quarters... So taking it every 25 cents at a time, there are 5 cents left over, which can be composed in 2 ways. So is the answer $ 13\times 13\times13\times2 \equal{} 4394$?
[/hide]
[/quote]

[hide="b"]
The "Mathcounts level" way is to systematically list the possibilities.
Basically, make a table, and start with the most number of quarters possible, then the most number of dimes, etc.
Your table would look something like this--remember, you can ignore the pennies, as they are determined by the Q,D,N.
Q D N
3 0 (1,0)--2
2 3 (0)--1
2 2 (2,1,0)--3
2 1 (4-0)--5
2 0 (6-0)--7
1 5 (1,0)--2
1 4 (3-0)--4
1 3 (5-0)--6
1 2 (7-0)--8
1 1 (9-0)--10
1 0 (11-0)--12
0 8 (0)--1
0 7 (2-0)--3
0 6 (4-0)--5
0 5 (6-0)--7
0 4 (8-0)--9
0 3 (10-0)--11
0 2 (12-0)--13
0 1 (14-0)--15
0 0 (16-0)--17
Adding up, we get $ 141$. I may have made a mistake.
[hide="Not Mathcounts level"]
You can also use generating functions.
The generating function of the pennies is $ 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots$.
For the nickels, it's $ 1\plus{}x^5\plus{}x^{10}\plus{}\cdots$.
For the dimes, it's $ 1\plus{}x^{10}\plus{}x^{20}\plus{}\cdots$.
For the quarters, it's $ 1\plus{}x^{25}\plus{}x^{50}\plus{}x^{75}\plus{}\cdots$.
See a pattern?
Now, we can multiply these, and the coefficient of $ x^{80}$ is our answer. Do you see how this intuitively makes sense?
Of course, for our purposes, the systematic approach is probably faster.
[/hide]
[/hide]

a and b are most simply solved by listing all the possibilities. Use tables and charts to organize the possibilities.

As for c, I think the answer is 11. Plugging 4 and 5 into the ab - a - b yields 11. Anything above that can be made with 4's and 5's (The presence of the 13 in the problem does not change the answer. In fact, if the 13 was replaced with any number greater than 11, the answer would not change.).

Oooh, markkram, I like what you did for c...disregard the 13.

After some thinking, I have devised a method for solving problems like c.

Firstly, know that ab - a - b doesn't work if a and b have any common factors other than 1.

If two numbers are given:

1. Perform ab - a - b if it can be done. The result is the answer. If ab - a - b cannot be done, then the problem has no answer.

Example:

"What is the largest whole number that cannot be made by adding 4 and 5 together?"

Since there are only two numbers, perform ab - a - b. The result is 11, which is the answer.

If more than two numbers are given:

1. Find the smallest pair of numbers in the given set on which ab - a - b can be performed. If there is no such pair, then the problem has no answer

2. Backcount from the result of ab - a - b to find the answer (if necessary).

Example1:

"What is the largest whole number that cannot be made by adding 4, 5, and 13 together?"

4 and 5 is the smallest pair of numbers in the given set on which ab - a - b can be performed. Performing ab - a - b yields 11. Since 11 < 13, there is no need to backcount. 11 is the answer.

Example2:

"What is the largest whole number that cannot be made by adding 4, 5, and 11 together?"

4 and 5 is the smallest pair of numbers in the given set on which ab - a - b can be performed. Performing ab - a - b yields 11. We now know that everything above 11 can be made. From here, we can backcount to the answer. We know that 11 can be made using one 11. We also know that 10 can be made with two 5s, 9 can be made with one 4 and one 5, and that 8 can be made with two 4s. However, 7 cannot be made. Thus, 7 is the answer.

Tell me if you find any problems with my method so I can revise it. I am also working on a method for problems like b. I will hopefully post it sometime soon if I have time to complete it.

http://en.wikipedia.org/wiki/Chicken_McNugget_Theorem

This is also the Chicken McNugget / Froebius Coin Problem.

Yes, that is where the ab - a - b comes from. Remember that a and b must be relatively prime in order for it to work.

The answer to b is indeed 141. To learn how to solve problems like b, see my post in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236178. I devised this method myself in chemistry class today.

[quote="james4l"][quote="Textangle"]Thank you James :) .

But why does it work?[/quote]

It is the definition of the problem.

f(10)=f(9)+f(8)+f(7)
because
we can attach a 1 from a 9-sequence, a 2 from a 8-sequence, or a 3 from a 7-sequence to make a 10-sequence.[/quote]

I'm just not getting this...  :huh:  Can you please explain again? (I'm sorry I'm so slow...)

@Mark... Nice method... that sure makes things simpler.

I solved problem a using pure listing and got a different answer from james. I'll edit the work into this post tomorrow afterschool.

Actually, james4l is correct.

$ a_n \equal{} a_{n \minus{} 1} \plus{} a_{n \minus{} 2} \plus{} a_{n \minus{} 3}$ because this includes all possible cases:
1. The highest you get to is $ n \minus{} 1$ which has $ a_{n \minus{} 1}$, and then you take one to get to $ n$.
2. The highest you get to is $ n \minus{} 2$ which has $ a_{n \minus{} 2}$, and then you take two to get to $ n$.
3. The highest you get to is $ n \minus{} 3$ which has $ a_{n \minus{} 3}$, and then you take three to get to $ n$.
Notice how these cases are all-encompassing and mutually exclusive.

However, your method was also correct. You just made a little error.

 3 | 2 | 1 |
| 3 | 0 | 1 | -> 4!/3! = 4
| 2 | 2 | 0 | -> 4!/2!2! = 6
| 2 | 1 | 2 | -> 5!/2!2! = 30
| 2 | 0 | 4 | -> 6!/2!4! = 15
| 1 | 3 | 1 | -> 5!/3! = 20
| 1 | 2 | 3 | -> 6!/2!3! = 60
| 1 | 1 | 5 | -> 7!/5! = 42
| 1 | 0 | 7 | -> 8!/7! = 8
| 0 | 5 | 0 | -> 5!/5! = 1
| 0 | 4 | 2 | -> 6!/2!4! = 15
[b]| 0 | 3 | 4 | -> 7!/3!4! = 70[/b]
| 0 | 2 | 6 | -> 8!/2!6! = 28
| 0 | 1 | 8 | -> 9!/8! = 9
| 0 | 0 |10 | -> 10!/10! = 1

Ya sure? :wink:


[quote="lim phalkun"]find f(x) if f(x)+xf(-x)=x+1[/quote]

$ f(x) \equal{} 1$??

 :lol:

[quote="lim phalkun"]find f(x) if f(x)+xf(-x)=x+1[/quote]

Changing $ x\to \minus{}x$ gives $ f(\minus{}x)\minus{}xf(x)\equal{}\minus{}x\plus{}1$ and multiplying then by $ \minus{}x$ gives $ \minus{}xf(\minus{}x)\plus{}x^2f(x)\equal{}x^2\minus{}x$
The original equation is $ f(x)\plus{}xf(\minus{}x)\equal{}x\plus{}1$

Adding these two lines gives $ (1\plus{}x^2)f(x)\equal{}1\plus{}x^2$ and so $ f(x)\equal{}1$ $ \forall x$ which indeed is a solution.

Hence the unique solution : $ \boxed{f(x)\equal{}1}$ $ \forall x$


vector
calculus
derivative
parameterization
linear algebra
matrix
advanced fields

Stuck with problem:

There is a local surface $\alpha(u) = (f_1(u), f_2(u), f_3(u), f_4(u)) \in R^4$. I need to find basis vectors of tangent space on that surface in some point p. It is not difficult to calculate tangent space for that surface, we should choose some curve on the surface and then it's derivative, but how to find 3 vectors that will be a basis for tangent space? Is it Frenet trihedron?

And some extra data:
so:
$u=(u_1, u_2, u_3)$
and specifically
$f_1(u)=sinu_1 sinu_2 sinu_3$
$f_2(u)=cosu_1 sinu_2 sinu_3$
$f_3(u)=cosu_2 sinu_3$
$f_4(u)=cosu_3$

Can anyone give me a hint? Should be easy.

You want three different curves, one for each parameter coordinate.
At $\alpha(u)$, the curves will be $x_1(t)=\alpha(u+(t,0,0))$, $x_2(t)=\alpha(u+(0,t,0))$, $x_3(t)=\alpha(u+(0,0,t))$.

The tangent space will be the span of the tangent vectors to these curves. If that span is 2-dimensional, $\alpha(u)$ is a singular point.

Are you sure? I have found 3 linearly independent vectors $\gamma_1, \gamma_2, \gamma_3$ and they are came from Jacobian matrix, other words $\gamma_1, = d\gamma/du_1$, $\gamma_2, = d\gamma/du_2$ and $\gamma_3, = d\gamma/du_3$. And these vectors are summation of each row for 3 columns in Jacobian matrix (number of columns - number of variables u1,u2,u3, number of rows - number of components)... or am I wrong?


z^6+z^4+z^3+z^2+1=0.  The product of the roots with positive imaginary part can be expressed as r(cos  :theta: + i*sin :theta: ).  Find :theta:.

This is off of kalva.  I feel so cheated by their solution (in spoiler).

[hide]11.   We need the factorisation: z6 + z4 + z3 + z2 + 1 = (z4 + z3 + z2 + z + 1)(z2 - z + 1). So the roots are ei72o, ei144o, ei216o, ei288o, ei60o, ei300o. The roots with positive imaginary part are ei72o, ei144o, ei60o, with product ei276o.

Note from Sam:all of those o's wish that they were  :^o:'s. [/hide]

This question has been batted about here before -- do a search.


group theory
abstract algebra
invariant
algebra
polynomial
superior algebra
superior algebra unsolved

Consider the subgroup of $ \text{Aut}_{K}K(x)$ consisting of the three automorphisms induced by $ x\mapsto x$, $ x\mapsto\frac{1}{1\minus{}x}$, and $ x\mapsto\frac{x\minus{}1}{x}$, ie $ \frac{f(x)}{g(x)}$ is sent to $ \frac{f(\frac{x\minus{}1}{x})}{g(\frac{x\minus{}1}{x})}$ by the third automorphism.  What is the fixed field of this subgroup?

Note: It's easy to show that the above subset is actually a subgroup, and you can show that those are all automorphisms since they're fractional linear transformations.

Here's one solution, but if anyone can give a method that is more easily generalizable, it'd be appreciated.

First, create an invariant polynomial fraction via symmetric functions.  One example is $ x\plus{}\frac{1}{1\minus{}x}\plus{}\frac{x\minus{}1}{x}\equal{}\frac{x^{3}\minus{}3x\plus{}1}{x(x\minus{}1}$.  If $ F$ is the fixed field, then the claim is that $ F \equal{} K\left (\frac{x^{3}\minus{}3x\plus{}1}{x(x\minus{}1}\right )$.  One inclusion is obvious.  For the other, note that $ \left [ K(x) : K\left (\frac{x^{3}\minus{}3x\plus{}1}{x(x\minus{}1}\right )\right ] \equal{} 3$, and so either $ [K(x) : F] \equal{} 1$ or $ 3$.  If it's three, we're done.  It can't be 1 though since the subgroup acts nontrivially, it doesn't fix everthing.

Any suggestions for getting the second inclusion direction when the index isn't prime?


Hello,

I was just wondering if there is a reason for not allowing html or txt files as attachments( for instance in pm's).
I was sending a very simple example of a html webpage(just a few lines of text) to a member of Mathlinks, but both the txt and the html version were denied.

Is there a reason for this (is it a "dangerous" extension?)?  If not, is it technically possible to arrange that it is allowed?

Thank you!

.txt extension is allowed (starting now).

Thanks.

Can html (simple text, no images along with it) be allowed too, or is there a difference between html and txt that I not know of?  Anyway, txt is a decent replacement.

Oh and by the way : welcome back Valentin  :welcomeani:
(It was weird without you, some got worried like me, but everyone's at ease now  :) )


ye masale hendeseye kheili asoon:
dar mosalase motasaviosagheine ABC makane hendesie noghati ra peida konid ke fasele anha az ghaede barabare ba vasete hendesie faselehaye an az do sagh bashad

mishe ,kamani az dayere ke bar saghhaye mosalas dar do noghteye entehaye ghaede momas ast. :D

in soal mano yade esbat e vojood e noghteye brocard mindazeh.


[quote="white_horse_king88"]When are round 2 problems coming?[/quote]
I think they said mid-October.

Hopefully next week, but the week after at the latest.

Hopefully they come out soon.
DPatrick, if you were to guess at the date that they would be put out, what would you say?

[quote="rrusczyk"]The next round will be up next week. We will send emails out for each.[/quote]

So... I don't think it's really set. Probably just whenever they manage to get them all together and up.


function
analytic geometry
topology
real analysis
real analysis unsolved

Take some $ x \in l_p$.

Define $ Q$ to be the set of all rational sequences whose elements are $ 0$ from some point on. Then $ Q$ is readily seen to be countable. Let $ \epsilon > 0$ be given. We show we can find an element in $ q$ which is closer than $ \epsilon$ to $ x$.

We know that because $ x$ is in $ l_p$ that there is some $ n \in \mathbb{N}$ such that $ \sum\limits_{k \equal{} n}^\infty |x_k|^p < \frac{\epsilon^p}2$.

For the coordinates $ x_i$, $ i \equal{} 0,\dots , n\minus{}1$, we can find rational numbers $ q_i$ such that for each $ i$ we have $ |x_i \minus{} q_i| < \frac{\epsilon^p}{2 n}$. This can be done because it is known that the rationals are dense in the reals.

Now, define $ q$ by $ q \equal{} (q_0, \dots, q_{n\minus{}1}, 0, \dots)$. Then we have $ q \in Q$. Furthermore, we have:

$ \parallel{}x \minus{} q\parallel{}_p^p \equal{} \sum\limits_{k \equal{} 0}^\infty |x_i \minus{} q_i|^p \equal{} \sum\limits_{k\equal{}0}^{n\minus{}1} |x_i \minus{} q_i|^p \plus{} \sum\limits_{k\equal{}n}^\infty |x_i \minus{} q_i|^p$
$ < \sum\limits_{k\equal{}0}^{n\minus{}1} \frac{\epsilon^p}{2 n} \plus{} \sum\limits_{k \equal{} n}^\infty |x_i|^p < \frac{\epsilon^p}2 \plus{} \frac{\epsilon^p}2 \equal{} \epsilon^p$

This implies that $ Q$ is dense in $ l_p$, which means $ l_p$ is separable.

[quote="marko avila"]A set B is said to be "at most numerable" if there exists and injective function from B to N.
 A metric space X is said to be "separable" if there exists a subspace C of X that is at most numerable and dense in X.

Prove that $ l_p$ is separable for p greater or equal to 1 but less than infinity.[/quote]

Let $ F_n$ be the sequences $ a_1, a_2, ...$ in $ l^p$ that are 0 after the $ nth$ term. Because $ F_n \approx R^n$, you know that  each $ F_n$ is separable. Then show the union of all $ F_n$ is dense in $ l^p$ and you're done.

Here's a proof that $ \ell^{\infty}$ is NOT separable: consider the set of all sequences that whose entries are made up of zeroes and ones.  Obviously this is a subset of $ \ell^{\infty}$.  As we saw in chapter 3, each sequence is the binary expansion of some number in $ (0,1)$, and every number in $ (0,1)$ has a binary expansion, so we get a bijection between $ (0,1)$ and our subset.  Hence our subset is uncountable.  Now because of our choice of metric on $ \ell^{\infty}$ (the sup-norm), any two (distinct) elements in the set are distance one apart.  So if we place an open ball of radius $ r<1/2$ around each point, then none of these balls will intersect.  But since any dense subset of $ \ell^{\infty}$ must have an element in each ball, any dense subset of $ \ell^{\infty}$ must be uncountable, so $ \ell^{\infty}$ is not separable.


Let $ x,y$ such that $ 2x^2\plus{}y^2\minus{}2xy\equal{}1$. Find min,max of
$ A\equal{} 2x^4\plus{}y^4\minus{}2x^2y^2$

The minimum and maximum are roots of irreducible cubic polynomials...


inequalities
analytic geometry
geometry
rectangle
probability
3D geometry
ratio

Sorry about the title messup, everybody, these are problems from last year's nationals sprint round. My bad.  [color=darkblue][size=75][Moderator rcv's note:  Title has been updated.][/size][/color]

Okay, I worked through last year's Nationals sprint round recently in preparation for nationals this year and here are some toughies I could use a little help on. My coach is going to e-mail me with some help, but the more varied solutions I get, the more likely I find the easiest one for me, right? Oh, and one more thing: I don't need the answer. That's right, I don't need the answer; I have an answer key. What I need are solutions. So if you post just an answer, I will hunt you down, barge through your front door, and...um...put a cold compress on your head. So here they are:

16. Two years ago there were 20 trailer homes on Elm Street with an average age of 18 years. At that time, a group of brand new trailer homes was then added to Elm Street. Today, the average age of all the trailer homes on Elm Street is 14 years. How many new trailer homes were added two years ago?

22. The intersection of the graphic solutions of all three inequalities listed below form an enclosed region in the coordinate plane. What is the area of this region? Express your answer in terms of (pi).
                     (x - 4)^2 + y^2 (< or =) 16
                     y (> or =) x - 4
                     y (> or =) –(1 / 3)x

23. Three positive integers a, b, and c satisfy abc=8! and a < b < c. What is the smallest possible value of c – a?

25. If a + b = 7 and a^3 + b^3 = 42, what is the value of the sum (1/a) + (1/b)? Express your answer as a common fraction.

26. The product 2^2 * 3^3 * 5^7 can be rewritten in the form a^2 * b^3 * c with positive integers a, b, and c. What is the greatest possible value of a * b – c?

29. From a circular piece of paper with radius BC, Jeff removes the unshaded sector shown. Using the larger shaded sector, he joins edge BC to edge BA (without overlap) to form a code of radius 12 centimeters and of volume 432pi cubic centimeters. What is the number of degrees in the measure of angle ABC of the sector that is not used?
(The given diagram shows a shaded circle B with one unshaded sector bordered by radii BC and BA. This sector is missing from the circle; the rest of the circle is filled in with a dotted line.)

30. A 2 by 2003 rectangle consists of unit squares as shown below. The middle unit square of each row is shaded. If a rectangle from the figure is chosen at random, what is the probability that the rectangle does not include a shaded square? Express your answer as a common fraction.
(The diagram shows a 2 by 2003 rectangle that is broken twice by dotted lines, so that three segments are shown: two end segments and one that is supposedly in the exact middle as the middle squares (one for each row) of the middle segment are shaded.)

I know, I didn't do too well. There are very few of the problems I don't understand or can't find some method to; most of these problems are just ones I wouldn't be able to complete fast enough for the sprint round. So please, [i]please[/i], if you have a heart, help!

I'm going to post the answers now so you can check your work before you post.

[hide="answer to 16"]10 trailer homes[/hide]
[hide="answer to 22"]6 :pi: +2  or equivalent[/hide]
[hide="answer to 23"]4[/hide]
[hide="answer to 25"]21/43 - I need help on this one badly![/hide]
[hide="answer to 26"]749[/hide]
[hide="answer to 29"]72 degrees[/hide]
[hide="answer to 30"]1001/2003[/hide]

well, for 16, you just set x as number of new trailers, so u set an equation:
(400+2x)/(20+x)=14
(400 because the first twenty trailers are 20years old now and 2x because the new ones are 2 years old now)
solving for x u get x=10 so 10 new trailers

second one dont u just graph it?

for 25:
expanding a^3+b^3=42 we have (a+b)(a^2-ab+b^2)=42
since we know a+b=7 we now know (a^2-ab+b^2)=6
if u square (a+b) u have a^2+2ab+b^2=49
subract a^2-ab+b^2 from it to get 3ab=43, thus ab=43/3
1/a+1/b=a+b/ab or 7/(43/3) or [b]21/43[/b]

Wow, I could never have gotten 25. I'll have to look into that.

Setting an equation for the first one was my first method, but for some reason I never got an integer number...I'll look into my work. I must have set it up wrong or made a miscalculation.

Abour graphing the second one, how would you graph that first inequality? I could easily do the other 2. Is it a circle with center (4,0) and radius 4? If that's it, I'm gonna kill myself.

The formula for graphing a circle is $(x-h)^2 + (y-k)^2 = r^2$ where $(h, k)$ is the center of the circle and $r$ is the radius of the circle, so yes, $(x-4)^2 + y^2 = 16$ is a circle centered at $(4,0)$ with radius 4.

I think I got 23.

[hide]For 23, you first prime factorize 8!, so you get 2^7x3^2x5x7.
From there you get the 3 closest multiples which would be 35, 32, and 36. So the answer would be 4.[/hide]

I think I got 29 too.

[hide]First you figure out what the height of the cones is.
The height is 9 cm. Then you use the pythagorean theorem to figure out the radius of the original circle, because the radius of the original circle is the slant height of the cone. You get 15. You know that the ratio of the radius of the cone to the radius of the original circle is equal to the number of degrees of the partial circle to 360. Therefore 12/15 = x/360 x= 288 It asks for the number of degrees not used so 360 - 288 is [b]72[/b].[/hide]

I think I know 30 too but it takes too long to explain.

Shoot, now I know why I wasn't getting 29! I thought the height was 3 because I was under the impression that the formula for the volume of a cone was pi*r 2 *height. I forgot to factor in the 1/3. That explains it.
Thanks for all your help so far, guys.

30.
[hide]To simplify the problem a little bit, we can get the same answer if the rectangle was 1x2003, because all the rectangles are in the same places and the probabilities would be equivelent. Now, for every two vertical lines that are chosen, we can make a rectangle. Therefore, the total number of rectangles is $nCr(2004,2)$. Then, the total number of rectangles that don't contain the shaded square is, since the left line that forms the middle square is the 1002nd line, $nCr(1002,2)$ and that multiplied by two because we have the other side. Putting everything together, $\frac {\frac {1002!*2}{2!1000!}}{\frac {2004!}{2!2002!}} = \frac {1002*1001}{1002*2003} = \frac {1001}{2003}$.[/hide]

22
[hide]The circle has radius 4 centered at $(0,4)$, and we are considering only everything contained within the circle. The 2nd equation divides the circle in two, and the x-axis and that 2nd equation create a 45-degree arc, or if you look at it like a piece of pie, $\frac 18$ of the circle. The 2nd equation cuts away most of the lower piece of pie leaving triangle with base 4 and height 1 jutting out the other $\frac 38$ of the semicircle. Then, adding all the pieces we have $\frac 38 * 16\pi + 2 = 6\pi + 2$.[/hide]

Forgive my ignorance - what does nCr mean? I understand your logic up to that point. Thanks for the help with 22 - remembering the formula for a circle, I figured it out this morning :)

$nCr$ is basically just the first number choose the second number -- combinations.

26.
[hide]You want the smallest value of c, so you make it 1. Then, you have 57, which is 53 *522, so you put put one 5 in b, and 52 in a, then you have 33, which you also put in b, so b is 3*5=15, then you put the 22 in a, making a 52*2=50, so you have 50*15-1=749. Sorry if this is kind of unclear.[/hide]

It's perfectly clear, thanks everybody for your help. I think I was slightly drugged doing some of those problems  :blush:

And thanks to rcv for the title change, I wasn't sure how to do that.

30.

[hide]First, there are three cases, rectangles of width one in the top row, rectangles of width two on the bottom row, and rectangles of width two on both rows. Each have equal probability, so just do the top row.

To find the number of rectangles, find a pattern. Length 1, there is 2003. Length 2, there is 2002 because the 2002nd ends in the last square.  This continues until 2003, where there is 1.  The pattern is for length l, there are 2004-l rectangles.  So there are 2003x 2004/2 or 2003x1002.

It is easier to find the rectangles that do have the square in it. Again find a pattern. Length 1, there is 1. Length 2, there is 2. This continues until 1002. On 1002, since the shaded square is the 1002nd, all rectangles have the square. This continues to the end.  For length 1003, there are 2004-1003=1001 squares till length 2003 with 1. So there are 1+2+3...+1001+1002+1001+...+3+2+1= 1002x1002

So the probability it does have the square is 1002x1002/2003x1002=1002/2003 Therefore the probabailtiy it does not have the square is 1001/2003[/hide]

16.[hide="the non-algebraic way"]2 years ago, the average after adding the trailers would've had to have been 12, so you are trying to figure out how many to add to get the same number before taking the average as before.

18 x 20 = 360

360/12 = 30

30 - 20 = 10[/hide]

Another way for #25:[hide]
(a + b)^3 = 7^3 = 343 = a :^3:  + 3a 2 b + 3ab 2  + b :^3:
a :^3: + b :^3: = 42.
Subtract this from the first equation to get 3a 2 b + 3ab 2  = 301. Simplify to get 3(a + b)(ab) = 301. Plugging in 7 for (a + b), you get 21ab = 301, or ab = 43/3.
You want to find (1/a) + (1/b), which equals (a + b)/(ab). You now know the values of (a + b) and (ab), so substituting  . . . (1/a) + (1/b) = 7/(43/3) = [b]21/43[/b].[/hide]

A pretty way for number 30:

[hide]Because the two rows have equal probability, deal with one row. Because any two vertical lines you pick will make a rectangle, pick any of the 2004 vertical lines. Then of the remaining 2003 lines, the 1001 on the side of the unit square you picked WILL work and the other 1002 will not. Thus you have 1001/1002[/hide]

Correction on solution for 30:

[hide]Sorry, me in all my lack of brilliance said the final answer was 1001/1002, when it was 1001/2003![/hide]


I have seen a recent outburst in tourneys over the past week (including mine, but this revived my old tourney).  In my opinion, I believe we should  stop this nonsense. I believe the tourneys are for fun.  But too man ytourneys can ruin that.  And I also think that all these tourneys defeat the actual purpose of a [i]tourney[/i].

We need a tourney that finishes.

Maybe is there was some prize for winning.

a prize would be hard to issue
how is someone gonna receive the prize?
i'm not gonna give my address out to someone on the internet....

[quote="ernie"]You can always use the classic 3 doors idea.  (This was posted in the Butterscotch Tourney thread, it works!)

[hide="Door 1"]:first:[/hide]

[hide="Door 2"]:winner_first:[/hide]

[hide="Door 3"]HA HA! YOU'RE A WEINER! :P[/hide][/quote]

lol :rotfl:
i like #3!! XD

I believe this has become excessive. The whole point of a tournament is to decide whos [b]best[/b] in one tournament. Ever wonder why Wimbledon and the Australian OPen don't occur [b]at the same time[/b]? I think we should let all the tournaments finish or die down before starting any new ones. I for one will not join any tournaments made after this point.

[hide="A proposed solution"]I say we should have a group of people who run the tournaments, and we should have a monthly tournament. In this monthly tournament we should have a bunch of divisions, like the sub-1300 division, the sub-1400, etcetera. We should also have team tournaments. If we have one group organizing tournaments, no chaos will ensue. Maybe the people of AOPS could vote who will run tournaments? Please comment on my idea.[/hide]

Good idea, mewto.

As for my tourneys:

PPT: Status Pending
CD Cup: Cancelled
Ernie's Tourney: Cancelled
_____________________________________________________________________

I would like to add that any of these tourneys I signed up on, I will be dropping out.

i think that's a good idea, but i'm gonna continue my tourny
hopefully, it'll finish
(once it starts, that is)

i hated the ads that ppl copied and pasted into ftw
but the tourneys are good... i mean you don't HAVE to join
i just hate how people advertise

There is no use practically making 10 tourneys.  Especially if more than half of them are the same format...

well, you might be able to play more people if there is a tourny
people that might not normally play would play each other and they might get better
anyway, if your eliminated from a tourny, it's always nice to know that there's another one you can play in  :D
omg!!! i found more emoticons!!!
i no, that's sad...
o well
i like this one  :starwars:

But if there are so many tournaments, everyone gets caught up in the new ones and no one finishes them.

Why do you want it locked, ernie?

I really do hate advertising on For The Win!, if people want to join your tourney they will, and you keep pushing and pushing tourneys and never finish them, so people lose trust in you. You are [i][b]very[/b][/i] annoying when you post "JOIN MY TOURNEY!" in multiple games multiple times. The people are there to countdown against each other, not to receive advertisement to boiling point. I only joined a few of your tourneys because I wanted you to be quiet and stop your immature shenanigans. I really sought another solution, because I knew that joining your tourney would only provoke you further.

I am very sorry.  :(

And apparently no one understands my ideas...

May as well give up.

:(

idk what ideas ur giving up, but that's a good idea anyway

Give up my idea on the limit of tourneys in a month.  It was a good idea, and the tourneys could actually finish.  Oh well.  :(  :(  :(

hmm... here's my idea:

[hide]There should be a limit to how many tourneys can be created for a month. We'll make a topic about it in FTW Other Stuff and the first n people to sign up can make a tourney. We can decide about n later. And because the advertisements are sooooooo annoying we can just ban them or only allow them in the lobby (if thats ok with everyone) But I don't agree with Mewto on the committee or something. It kinda excludes everyone else.[/hide]

If you took the time to read my idea... thanks :D ... if you think I'm too stupid to propose a decent solution... well you may be right...  :(

It's just that someone has to be in charge to [i]organize[/i] all the votes and the system and so on,... and would we have to vote for that person? OK, then someone'll have to be in charge of that vote...

Angie, I don't think you completely understand my idea(or maybe I failed to articulate what I meant). In any case, I believe this commitee should really have a few appointed leaders, but everyone should be a member, and instead of making the tournaments, this group [b]delegates[/b] the making of tournaments, so if everyone wnnats to make a tournament, we'll decide who makes it. The point is to have everyone involved in some large group to make tournaments,so if too many tournaments occur, the people in the group, who will hopefully be a majority of FTW players, won't join and the unauthorized tournament can't start.

I think that all the tourneys are all the same now, and there is nothing different about them other than the names. We need something different in these tourneys, or else might as well not have tourneys.


function
integration
trigonometry
calculus
limit
calculus computations

For any given (suitable) function $ f$, the Laplace trasform of $ f$ is the function $ F$ defined by

$ F(s)\equal{} \int_{0}^{\infty}\, e^{\minus{}st}f(t)\,dt$ $ (s>0)$

(i) Show that the Laplace transform of $ e^{\minus{}bt}f(t)$, where  $ b>0$, is $ F(s\plus{}b)$,

(ii) Show that the Laplace transform of $ f(at)$, where $ a>0$, is $ a^{\minus{}1}F(\frac{s}{a})$,

(iii) Show that the Laplace transform of $ f'(t)$ is $ sF(s) \minus{} f(0)$.

(iv) In the case $ f(t) \equal{} \sin t$, show that $ F(s) \equal{} \frac{1}{s^{2}\plus{}1}$.

Using only these four results, find the Laplace transform of $ e^{\minus{}pt} \cos qt$, where $ p>0$ and $ q>0$.

[b]a)[/b]:
$ F(s) \equal{} F[e^{ \minus{} bt}f(t)])(s) \equal{} \int\limits_0^\infty e^{ \minus{} st} e^{ \minus{} bt}f(t) \, dt \equal{} \int\limits_0^\infty e^{ \minus{} (s \plus{} b)t} f(t) \, dt \equal{} F(s \plus{} b)$.  Note that $ b > 0$ to ensure existence.

[b]b)[/b]:
$ \int\limits_0^\infty e^{ \minus{} s t} f(t) \, dt \equal{} \int\limits_0^\infty \frac1a e^{ \minus{} \frac {s}a t a} f(t a) \, a dt \equal{} \int\limits_0^\infty \frac1a e^{ \minus{} \frac {s}a u} f(u) \, du \equal{} \frac1a F(\frac {s}a)$ by substitution $ u \equal{} a t$ with $ du \equal{} a dt$.

[b]c)[/b]:
$ \int\limits_0^\infty e^{ \minus{} s t} f'(t) \, dt \equal{} \left[ f(t) e^{ \minus{} st} \right |_0^\infty \minus{} \int\limits_0^\infty ( \minus{} s e^{ \minus{} s t}) f (t) \, dt \equal{} \minus{} f(0) \plus{} s F(s)$, by partial integration (note we assume $ \lim\limits_{x \to \infty} f(x) e^{ \minus{} s x} \equal{} 0$ etc).

[b]d)[/b]:
$ F(s) \equal{} \int\limits_0^\infty e^{\minus{} s t} \sin(t) \, dt \equal{} \int\limits_0^\infty e^{\minus{} st} \text{Im}(e^{i t}) \, dt \equal{} \text{Im} \left(\int\limits_0^\infty e^{(i \minus{}s)t}\, dt\right)$
$ \equal{} \text{Im} \left( \left[\frac1{i\minus{}s} e^{(i\minus{}s)t}\right|_0^\infty \right) \equal{}  \text{Im} \left( \frac1{s\minus{}i} \right) \equal{} \text{Im} \left( \frac1{s\minus{}i}\frac{s\plus{}i}{s\plus{}i} \right) \equal{} \text{Im} \left( \frac{s\plus{}i}{s^2 \minus{} i^2} \right)$
$ \equal{} \frac1{s^2 \plus{} 1}$.

[b]e)[/b]:
Define $ g(t) : \equal{} e^{\minus{}p t} \cos(q t)$,  $ f(t) : \equal{} \cos(q t)$. By a), the Laplace transform of $ g$ is $ F(s \plus{} p)$, where $ F$ is the Laplace transform of $ f$, as we have $ g(t) \equal{} e^{\minus{} p t} f(t)$.

Note that $ f(t) \equal{} \frac{d}{dt} \frac1q \sin(q t)$. The Laplace transform is a linear operator, so we can suffice by finding the Laplace transform of $ \frac{d}{dt} \sin(q t)$. Using c), this is equal to $ s H( s) \minus{} \sin(q 0)$ where $ H$ is the Laplace transform of $ \sin(q t)$. By b), this Laplace transform is equal to $ \frac1q K(\frac{s}q)$, where $ K$ is the Laplace transform of $ \sin(t)$, which by d) is equal to $ \frac{1}{s^2\plus{}1}$. Putting this together, we get:

Laplace transform $ \equal{} F(s \plus{} p) \equal{} \frac1q \left[ (s \plus{} p) H(s \plus{} p) \minus{} 0\right] \equal{} \frac{s\plus{}p}q \left[\frac1q K(\frac{s\plus{}p}q)\right]$
$ \equal{} \frac{s\plus{}p}{q^2} \frac1{1\plus{}\left(\frac{s\plus{}p}q\right)^2} \equal{} \frac{s \plus{} p}{q^2 \plus{} (s \plus{} p)^2}$.

If I haven't made any mistakes, that is.


Isn't USAMTS round 3 over? When are the solutions going to be posted? Can we discuss the questions?

The solutions will be posted tomorrow (Wednesday).  Please do not discuss the problems or solutions until they are posted.


calculus
integration
logarithms
calculus computations

find
$\int_{e^{\sqrt2}}^{e^{2}}\left\{ \ln(\ln x)+\frac{1}{\ln(\ln x)}\right\}dx$

sorry,editted :(

[quote="kunny"]$\int_{e^{\sqrt2}}^{e^{2}}\left\{ \ln(\ln x)+\frac{1}{\ln(\ln x)}\right\}dx=[x\ln (ln x)]_{e^{\sqrt{2}}}^{e^{2}}=e^{2}\ln 2-e^{\sqrt{2}}\ln \sqrt{2}.$[/quote]

$\frac{d}{dx}\left(x\ln (\ln x)\right) = \ln(\ln x)+\frac{1}{\ln x}$.

Nice thought, though.  Mathematica doesn't think this integral has a nice form, so perhaps you gave a solution to the intended problem.


limit
trigonometry
function
algebra
functional equation
real analysis
real analysis unsolved

find:

$\lim_{n\rightarrow \infty}\sin \frac 1{2}+\sin\frac 1{2^{2}}+\dots+\sin\frac 1{2^{n}}$

$\sum_{n=1}^\infty \sin \frac{x}{2^{n}}= \sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!(2^{2n+1}-1)}$, by expanding the sine function into a power series. Note that the resulting function $f$ is entire and solves the functional equation $f(x) = \sin \frac{x}{2}+f(\frac{x}{2})$.  I'm not sure how much this helps.


linear algebra
matrix
probability
modular arithmetic
limit
vector
function

What is the probability that the determinant of a random nxn integer matrix is even?

One difficulty with this question is that since the set of all integer matrices is infinite, it is not clear how we should regard it as a probability space. Here's a question - possibly equivalent, depending on your defnition of what the first question means - in which the probability space is a finite set in which the points can be taken as equally likely.

Choose $A$ at random from the $n\times n$ matrices with entries in $\mathbb{Z}/2\mathbb{Z}.$ (The field of integers mod 2.)

What is the probability that $\det(A)=0$?

(All arithmetic is being done in $\mathbb{Z}/2\mathbb{Z}.$ If it makes things any clearer, call that $\det(A)\equiv0\pmod{2}.$)

[Added in edit.]

This is the same as the following: for each entry of the $n\times n$ matrix, flip a coin. If it is heads, assign some arbitrary choice of even number; if tails, assign some arbitrary choice of odd number. It turns out that the question of whether the determinant is even or odd does not depend on your arbitrary choices; just on the coin flips. Now what is the probability that the determinant is even?

Does this capture what you mean, jrthedawg?

Speculation time. Let $P_n$ be the solution to the problem I just stated. It is true that $\lim_{n\to\infty}P_n$ exists. What do you think its value is?

$0\,?$
Between $0$ and $\frac12\,?$
$\frac12\,?$
 Between $\frac12$ and $1\,?$
$1\,?$

It is a bit easier to find the probability $Q_n=1-P_n$ of the event that the determinant is odd, or, which is the same, of the event that $n$ vectors in $V=(\mathbb Z/2\mathbb Z)^n$ are linearly independent over the field $\mathbb Z/2\mathbb Z$. The idea is that the probability of the event that a random vector $v$ does not belongs to a $k$-dimensional subspace $W\subset V$ does not depend on $W$ and equals $1-2^{k-n}$. From here we easily conclude that $Q_n=(1-2^{-n})(1-2^{1-n})\dots (1-2^{-1})$ (just multiply the probabilities that each next vector does not belong to the linear span of the previous vectors). It follows that the limit of $Q_n$ exists and is strictly between $0$ and $\frac 12$ (so the answer to Kent's question is "between $\frac12$ and $1$"). I do not think there is any closed form expression for the infinite product $\prod_{k\geqslant 1}(1-2^{-k})$ (though there may be some special standard notation for this number).

But easy enough to approximate with a calculator or spreadsheet.

$\prod_{k\geqslant 1}(1-2^{-k})\approx .28879$

So, 29% chance of invertible, 71% chance of singular.

There are some interesting generating functions here; $\prod_{k\ge1}\frac1{1-x^k}=\sum p_nx^n$ where $p_n$ is the number of (unlabeled) partititions of $n$. That doesn't come any closer to a closed form evaluation at any of the key points.

Professor Merryfield, that is exactly what I mean, although I am unclear why the way I stated it is ambiguous.  Could you explain more clearly?

Maybe I can answer that by posing a different question with the same structure:

What is the probability that a the quadratic equation $x^2+bx+c=0$ for randomly chosen integers $a$ and $b$ has real roots?

Can you see how that is an ambiguous question? If you think it's clear, how would you answer it?

I see your point, but as you demonstrated, the only property of the integers in the nxn matrix that is significant to this problem is whether or not they are odd or even.  Also, is not the probability that a random integer is even 1/2?  So it seems to me that your reduction of the problem to essentially a matrix of 1's and 0's, each having a fifty percent chance of being selected for a certain position in the matrix, is an equivalent problem to my original question.

[quote="Jrthedawg"]Also, is not the probability that a random integer is even 1/2?[/quote]It depends: How do you define a "random integer"?  What probability distribution are you using?

I don't understand.  Please be more clear.  I am unsure as to what probability distribution means.

For your case (when the probability space is the discrete set of integers), a probability distribution is an assignment of a probability to each integer so that the sum of all the probabilities is equal to 1.

I recommend that you get a basic book on probability.  The concept of probability distribution is arguably the most foundational concept in probability.  You can't really ask a question about probability without a probability distribution in mind.

Oh, you can certainly do what Ravi B asked for. There are some other properties you can't have when you're doing that. Notably, you can't have your measure be translation invariant. More simply, you can't give each point the same measure.

Just as an example in $\mathbb{Z}^2$ (and I'm sure Ravi B was thinking in terms of something like this):

Choose $0<r<1.$ Let $P\{(m,n)\}=\frac{(1-r)^2}{(1+r)^2}\,r^{-|m|-|n|}.$

[quote="gongli"][i]For your case (when the probability space is the discrete set of integers), a probability distribution is an assignment of a probability to each integer so that the sum of all the probabilities is equal to 1. [/i]
----------------------------------------------------
I don't believe that can be done.  Maybe it can be done for a finite set, but for an infinite set such as the integers  you have to define a probability MEASURE which is defined on suitable subsets of the sample space. It  is not enough to define a probability for every integer.[/quote]The set of integers is countable.  Once you have assigned a probability to each integer, it automatically lifts to a probability measure by countable union.  And vice versa -- given a probability measure, you can just restrict to the singleton sets.

Here's the problem with trying to assign prob to singleton integers and extending that prob to all subsets:
You can't assign the same probability to each singleton integer.
If you assign 0 to each integer then the sum of probs for all the integers wont' add up to 1. If you assign a finite but not zero value prob to each integer then then sum won't converge to 1.

So you have to assign different probs to each integer. But then the probability of
a subset of integers will depend on which integers are in the subset, even though
we intuitively want the probability of same sized subsets to be the same.

I.e. P({1,2}) = P(1)+P(2) will probably not be equal to P({3,4}) = P({3}) + P({4}).

Can such a prob on singletons be defined so that you do get P(S) = P(T)
for all finite subsets S and T.
And we would like P(odd integers)  = P(even integers) = 1/2.

You would also have to have that P({100,102,104,106,...}) = 1/2
(shouldn't the prob be translation invariant.

So all in all it's not clear you can get  a prob measure that would correspond to our
intuitive ideas by assigning probs to singleton integers and extending.

For an example of a way to assign probs to all subsets of {1,2,3...}
see Billinglsey, Prob and Measure, problem 2.15 in chapter 1. Note it does not assign probs to singleton subsets and then extend!

I agree that a probability distribution on the integers can't assign the same probability to each integer.  But that is a flaw with the notion of random integer -- it has nothing to do with my definition or another definition.

I don't have the Billingsley book.  Does his measure give the same probability to every finite subset of a particular size?  Is his measure translation-invariant?  Is his measure countably additive?  I don't see how all these are possible.  Could you supply some details of his measure?

Billingsley  gives the following sequence of probability measures on
{1,2,3,4...}.

For a given n , define P_n(A) = 1/n * number of elements of A intersect {1,2,...n}

P_n is a probability measure for all n defined for all subsets A of {1,2,3...}

If you define D(A) = lim P_n(A) as n goes to infinity for all A for which the limit exists,
then D has some of the properties
we want. D(even integers) = 1/2, and I think it's translation invariant.
But D isn't a probability measure because it's not countably additive on the the set of all subsets for which the limit exists.

So I don't know if there is probablity measure on the integers that has the properties one would want.

D(A) is usually called the "density" of the set A.  That's a familiar concept, and it is true that the density of odd integers is 1/2.  It's vaguely like a probability measure, but not exactly, because as you said, it is not countably additive.

There is no (countably additive) probability measure that gives each integer the same probability.  Every (countably additive) probability measure can be specified in the manner I originally said: Assign a probability to each integer, and lift from there.  That's because every set of integers is a countable union of singleton sets.

So I think we have reached agreement, yes?


inequalities
geometry
area of a triangle
Heron\u0027s formula
geometry unsolved

Given equilateral triangle $ABC$, let $M$ be an arbitrary point in space.
a)Prove that one can construct a triangle from the segments $MA,MB,MC$.
b)Suppose that $P,Q$ are two points symmetric with respect to the center $O$ of $ABC$. Prove that the two triangles constructed from the segments $PA,PB,PC$ and $QA,QB,QC$ are of equal area.

a) Using Ptolémé Inequalities, we have $MA.BC \leq\ MB.CA+MC.AB \Rightarrow MA \leq\ MB+MC$.
b) Construc the equilateral triangle $MCM'$ we have $\Delta BCM = \Delta ACM'$ so we constructed a triangle from three segments $MA;MB;MC$; it's $\Delta AMM'$. We have:
$S_{AMM'}= \frac{1}{2}MA.MM'.sin (\hat{AMM'}) = \frac{1}{2}MA.MB.sin (\hat{AMB}-60^{o})\\=\frac{1}{4}MA.MB(sin (\hat{AMB})-\sqrt{3}cos (\hat{AMB}))$
Now we prove that:
$PA.PB.(sin (\hat{APB})-\sqrt{3}cos (\hat{APB}))=QA.QB.(sin (\hat{AQB})-\sqrt{3}cos (\hat{AQB})) \\ \Leftrightarrow 4 (S_{OPA}+S_{OPB}) = \sqrt{3}(\vec{PA}.\vec{PB}-\vec{QA}.\vec{QB})= 2\sqrt{3}(\vec{OA}+\vec{OB})\vec{PO}\\ \Leftrightarrow 2(S_{OPA}+S_{OPB}) = \sqrt{3}(\vec{OA}+\vec{OB}).\vec{PO}\\ \Leftrightarrow sin (\hat{AOQ})+sin(\hat{BOQ}) = \sqrt{3}(cos(\hat{AOQ})+cos(\hat{AOQ})) \\ \Leftrightarrow 2sin(\frac{x+y}{2})cos(\frac{x-y}{2}) =2\sqrt{3}cos(\frac{x+y}{2})cos(\frac{x-y}{2}) \ \ (\hat{AOQ}=x , \hat{BOQ}=y )\\ \Leftrightarrow tan (\frac{x+y}{2})=\sqrt{3}$
It's true because $x+y= \frac{2 \pi}{3}$  :wink:
[hide][img]http://i108.photobucket.com/albums/n9/PDatK40SP/4324235.jpg[/img][/hide]

i have two solutions! the first is similary to PDatK40SP solution.

the other is with complex numbers. let O = 0, A = 1, B = w and C = w², P = p and Q = -p. (where w is the 3-root of unity)
now, use heron's formula and many computations!  :lol:


$ n>2$ is a given integer.Consider finite sequence $ a_1,a_2,..,a_n$ which satisfies next three conditions:

1)$ a_1\equal{}1,a_2\equal{}2$
2)$ a_1<a_2<\cdots <a_n$
3)For any integer $ k$ $ (1<k<n)$,there exists a pair $ (i,j)$ $ (1\le i< j\le n)$ such that $ a_k\equal{}\frac{a_i\plus{}a_j}{2}$.

Find the maximum value of $ a_1\plus{}a_2\plus{}\cdot \plus{}a_n$ in terms of $ n$.

[hide="very easy problem"]

Consider $ a_k$. If $ a_i$ is minimized (namely, equal to $ a_1\equal{}1$), then $ a_j$ is maximized (namely, thus, equal to $ 2a_k\minus{}1$). Since $ a_i$ is an increasing sequence, the series is maximized if $ j\equal{}k\plus{}1$.

So the sequence is defined by $ a_{k\plus{}1}\equal{}2a_k\minus{}1$ for $ k>2$, and the sum is $ 2^{n\minus{}1}\plus{}n\minus{}1$[/hide]


[quote="Beginner"]The numbers a,b are integers, and the roots of $x^2+ax+b+1=0$  are natural number. prove that $a^2+b^2$ is not a prime.[/quote]

[hide]
From Viete's Formulas, we have
$a = -r_1-r_2$
$b = r_1r_2-1$

So $a^2+b^2 = (-r_1-r_2)^2+(r_1r_2-1)^2 = r_1^2r_2^2+r_1^2+r_2^2+1 = (r_1^2+1)(r_2^2+1)$ but since $r_1$ and $r_2$ are natural numbers, it can't be prime.[/hide]


vector
function
integration
real analysis
real analysis unsolved

Let $ X$ be the vector space of all measurable real-valued functions on $ [0,1]$ with addition and scalar multiplication defined in the usual way. Define $ \sigma(x)\equal{}\int_{0}^{1} \frac{|x(t)|}{1\plus{}|x(t)|} \; dt$.
Then $ \rho(x,y)\equal{}\sigma(x\minus{}y)$ defines a metric on $ X$.

Show that in this metric, $ x_{n} \rightarrow x$ if and only if $ x_{n} \rightarrow x$ in measure.

[I have only managed to show the "if" part.]

For the "only if" part, this is what I have:
Suppose $ (x_{n})$ does not converge to $ x$ in measure. There exists $ \varepsilon>0$ such that for all $ N$ there is $ n \geq N$ with $ m\{ t: |x_{n}(t)\minus{}x(t)| \geq \varepsilon \} \geq \varepsilon$. This produces a subsequence $ (x_{n_{k}})$ where $ m\{ t: |x_{n_{k}}(t)\minus{}x(t)| \geq \varepsilon \} \geq \varepsilon$ for all $ k$. Now $ \rho(x_{n_{k}},x) \geq \int_{\{ t: |x_{n_{k}}(t)\minus{}x(t)| \geq \varepsilon \}} \frac{|x_{n_{k}}(t)\minus{}x(t)|}{1\plus{}|x_{n_{k}}(t)\minus{}x(t)|} \geq \int_{\{ t: |x_{n_{k}}(t)\minus{}x(t)| \geq \varepsilon \}} \frac{\varepsilon}{1\plus{}\varepsilon} \geq \frac{\varepsilon^{2}}{1\plus{}\varepsilon}$ for all $ k$ so this subsequence does not converge to $ x$ in the metric, which means the sequence $ (x_{n})$ does not either.

you can refer to " Measure theory" by G de Barra. You can get the complete solution.

Thanks for the reference. Unfortunately my college library doesn't have a copy of the book.


function
topology
real analysis
real analysis unsolved

maybe it seems idiot :blush:  do we have retract of $\mathbb{R}^2$ to $\mathbb{S}^1$ (i mean that can give me continuous function $r:\mathbb{R}^2 \rightarrow \mathbb{S}^1 $such that restriction of r to $\mathbb{S}^1$ is identity)

No, there is no such retract. This is a lemma used in the proof of Browuer's Fixed Point Theorem, for example, but the only proof I know relies on the notion of fundamental group:

Suppose there is such a retract $r$, and let $r^*$ be the homomorphism it induces on the fundamental groups, $r^*:\pi_1(\mathbb R^2,1)\to\pi_1(\mathbb S^1,1)$ ($1$ is the point $(1,0)$, which I identified with the complex number $1$). In this case (whenever $r$ is a retract between two spaces), $r^*$ is surjective (this is pretty obvious), but in our case we have a contradiction, because we have a surjective homomorphism from the null group (the fundamental group of $\mathbb R^2$) to $\mathbb Z$ (the fundamental group of $\mathbb S^1$).

Look above. I've edited my previous post. :)


Let $A=\sum x_i$ and $B=\sum \frac{1}{x_i}$. Rewrite the inequality as: $AB+3n \ge nA+3A$ or it is equivalent to the following two inequalities: 1. $(A-n)(B-3) +n(B-A) \ge 0$ or 2. $(A-3)(B-n)+3(A-B) \ge 0$.
Thus we have to prove that $max(A,B) \ge 3$. Assume it is not correct. Then $6 \ge A+B$ wich deffinitely is not corect for $n \ge 2$.

And what?  :?
Am I missing something: suppose $A>B$, $A>3$, then we know nothing about $(A-n)(B-3) +n(B-A)$ and we know nothing about $(A-3)(B-n)+3(A-B)$...

We have $A,B \ge n$ from the condition of the problem. So if $A>B$ and $A>3$ we have $(B-n)(A-3) \ge 0$ and $(A-B) \ge 0$ thus their sum is positive. Hope there is no mechanical mistake i haven't seen.

[quote="Andrei"].................................Then $6 \ge A+B$ wich deffinitely is not corect for $n \ge 2$.[/quote]

$n \geq 4$ ;)

I think I didn't clrealy clarify what I meant to say. Maybe the solution will be of some help.
Now I guess it is enough to prove the second step from the end to be clear with what I wrote.
Lets mention for the time beeing  that $x_ix_{i+1}...x_n \leq 1$ for all $i \leq n-1$.
Now :
$f(x_1,x_2,...,x_{n-2}x_{n-1}x_n,1,1)-f(x_1,x_2,.....,x_{n-1}x_n,1)$ is :
$\frac{(x_{n-2}-1)(1-x_{n-1}x_n)}{x_{n-2}x_{n-1}x_n}+3n\frac{(1-x_{n-2})(x_{n-1}x_n)}{(x_1+x_2+...+x_{n-1}x_n+1)(x_1+x_2+...+x_{n-2}x_{n-1}x_n+1+1)}$ $(*)$
Now if $x_{n-2} \geq 1$ the whole expression is positive and we are done. Now let's see that if there is an another number from the sequence $>1$ than we can use the same method. So simply we can say that while there is a number strictly greater than $1$ than my method works. After some of this steps we will be left only with numbers less or equal to $1$ and with their product $1$ so thwy are all one :?

[quote="Andrei"] Rewrite the inequality as: $AB+3n \ge nA+3A$ or it is equivalent to the following two inequalities: 1. $(A-n)(B-3) +n(B-A) \ge 0$ or 2. $(A-3)(B-n)+3(A-B) \ge 0$.
[/quote]
2. is wrong.

Yes, Vasc :( Thanx for pointing out my mistake. It should be $(A-3)(B-n) +3(B-A) \ge 0$ and we need only to solve in the case: $A \ge B \ge n \ge 3$

Let $f(x_1,x_2,...,x_n)=\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}+\frac{3n}{x_1+x_2+...+x_n}$. Now, let $t=\sqrt{x_1x_2}$
\[
\begin{gathered}
  f\left( {x_1 ,x_2 ,...,x_n } \right) - f\left( {t,t,...,x_n } \right) \hfill \\
   = \frac{1}
{{x_1 }} + \frac{1}
{{x_2 }} - \frac{2}
{t} + 3n\left( {\frac{{2t - x_1  - x_2 }}
{{(x_1  + ... + x_n )(2t + x_3  + ... + x_n )}}} \right) \hfill \\
   = \left( {x_1  + x_2  - 2t} \right)\left( {\frac{1}
{{x_1 x_2 }} - \frac{{3n}}
{{(x_1  + ... + x_n )(2t + x_3  + ... + x_n )}}} \right) \hfill \\
   \geqslant \left( {x_1  + x_2  - 2t} \right)\left( {\frac{1}
{{x_1 x_2 }} - \frac{{3n}}
{{n^2 }}} \right) \hfill \\
   = \left( {x_1  + x_2  - 2t} \right)\left( {\frac{1}
{{x_1 x_2 }} - \frac{3}
{n}} \right) \hfill \\
\end{gathered}
\]
If $x_1x_2 \le 1$ we have $f\left( {x_1 ,x_2 ,...,x_n } \right) - f\left( {t,t,...,x_n } \right) \ge 0$.
Now WLOG assume $x_1 \le x_2 \le ... \le x_n$, then
\[
\begin{gathered}
  f\left( {x_1 ,x_2 ,...,x_n } \right) \hfill \\
   \geqslant f\left( {t,t,...,x_n } \right) \hfill \\
   \geqslant f\left( {\sqrt {tx_3 } ,t,\sqrt {tx_3 } ,...,x_n } \right) \hfill \\
   \geqslant ... \hfill \\
   \geqslant f\left( {y,y,...,y,x_n } \right) \hfill \\
\end{gathered}
\]
where $y_n=(x_1x_2...x_{n-1})^{\frac{1}{n-1}}$
(This is a step i'm unsure of, but I think it's true although I don't have a rigorous proof of it)
Then the given inequality becomes
\[
\frac{{n - 1}}
{y} + \frac{{3n}}
{{(n - 1)y + x_n }} \geqslant n + 3
\]
with $y^{n-1}x_n=1$ and $y \le x_n$, after substitution,
\[
\left( {\frac{{n - 1}}
{y} + y^{n - 1} } \right)\left( {(n - 1)y + \frac{1}
{{y^{n - 1} }}} \right) + 3n \geqslant \left( {n + 3} \right)\left( {(n - 1)y + \frac{1}
{{y^{n - 1} }}} \right)
\]
\[
 \Leftrightarrow \left( {n^2  + n + 2} \right) + \left( {n - 1} \right)y^n  + \frac{{n - 1}}
{{y^n }} \geqslant \left( {n + 3} \right)\left( {n - 1} \right)y + \frac{{n + 3}}
{{y^{n - 1} }}
\]
Now,
\[
\begin{gathered}
   = \left( {n + 2} \right)\left( {n - 1} \right) + \left( {n - 1} \right)y^n  + 4 + \frac{{n - 1}}
{{y^n }} \hfill \\
   \geqslant \left( {n + 3} \right)\left( {n - 1} \right)y^{\frac{{n(n - 1)}}
{{\left( {n + 3} \right)\left( {n - 1} \right)}}}  + \left( {n + 3} \right)y^{\frac{{n + 3}}
{{n(n - 1)}}}  \hfill \\
\end{gathered}
\]
I bet the remaining can be done by grouping and the fact $y \le 1$ as I have done for n=4 case...

It forces me to remember about more than one layers of notion "nice problem".

[quote="siuhochung"]Let $f(x_1,x_2,...,x_n)=\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}+\frac{3n}{x_1+x_2+...+x_n}$. Now, let $t=\sqrt{x_1x_2}$
\[
\begin{gathered}
  f\left( {x_1 ,x_2 ,...,x_n } \right) - f\left( {t,t,...,x_n } \right) \hfill \\
   = \frac{1}
{{x_1 }} + \frac{1}
{{x_2 }} - \frac{2}
{t} + 3n\left( {\frac{{2t - x_1  - x_2 }}
{{(x_1  + ... + x_n )(2t + x_3  + ... + x_n )}}} \right) \hfill \\
   = \left( {x_1  + x_2  - 2t} \right)\left( {\frac{1}
{{x_1 x_2 }} - \frac{{3n}}
{{(x_1  + ... + x_n )(2t + x_3  + ... + x_n )}}} \right) \hfill \\
   \geqslant \left( {x_1  + x_2  - 2t} \right)\left( {\frac{1}
{{x_1 x_2 }} - \frac{{3n}}
{{n^2 }}} \right) \hfill \\
   = \left( {x_1  + x_2  - 2t} \right)\left( {\frac{1}
{{x_1 x_2 }} - \frac{3}
{n}} \right) \hfill \\
\end{gathered}
\]
If $x_1x_2 \le 1$ we have $f\left( {x_1 ,x_2 ,...,x_n } \right) - f\left( {t,t,...,x_n } \right) \ge 0$.
Now WLOG assume $x_1 \le x_2 \le ... \le x_n$, then
\[
\begin{gathered}
  f\left( {x_1 ,x_2 ,...,x_n } \right) \hfill \\
   \geqslant f\left( {t,t,...,x_n } \right) \hfill \\
   \geqslant f\left( {\sqrt {tx_3 } ,t,\sqrt {tx_3 } ,...,x_n } \right) \hfill \\
   \geqslant ... \hfill \\
   \geqslant f\left( {y,y,...,y,x_n } \right) \hfill \\
\end{gathered}
\]
where $y_n=(x_1x_2...x_{n-1})^{\frac{1}{n-1}}$
(This is a step i'm unsure of, but I think it's true although I don't have a rigorous proof of it)
[/quote]
I don't sure about it.The condition $x_1x_2 \le 1$ isn't enough to prove $n-1$ number $x_i$ equal.

I found a nice solution for the weaker inequality

$\displaystyle \frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}+\frac{4n}{n+x_1+x_2+...+x_n} \ge n+2$.

Is your solution use lemma $n-1EV$ equal,Vasc?
I ask that only because I don't like inequality problem which can be solved by one-way :( .

Without using (n-1)EV Theorem, of course.

Yeah.I've that realize that in my inequality with $2n$ number and:

$x_{n+1}=x_{n+2}=...=x_{2n}=1$

(The stronger inequality still true:

$\frac{1}{x_1}+\frac{1}{x_2}+..+\frac{1}{x_n}+\frac{6n}{x_1+x_2+...+x_n+n} \ge n+3$  )

But,Indeed,This inequality is nice,too.And maybe can you post your solution?

[quote="Vasc"]Use Mac Laurian ineq.

$(\frac{t_1+t_2+...+t_n}{n})^{n-1} \ge t_1t_2...t_n(1/t_1+1/t_2+...+1/t_n)/n$.[/quote]

But how?

I have put $x_i=\frac {1}{t_i}$ to obtain

$\sum t_i + \frac{4n}{n +\sum \frac {1}{t_i}} \geq \sum t_i + \frac{t_1....t_n}{(\frac{\sum t_i}{n})^{n-1} +t_1...t_n} \geq  \sum t_i + \frac{t_1..t_n}{(\frac{\sum t_i}{n})^n+(\frac{\sum t_i}{n})^{n-1}} \geq ......?????$

Am I wrong? Can you please give me another hint, Vasc?

Thank you very much.

Manlio,
Denote $x=\frac{t_1+...+t_n}{n}$. It suffices to show that for $x \ge 1$  the ineq. is valid

$nx+\frac{4n}{n+nx^{n-1}} \ge n+2$.

Notice that for $f(x)=nx+\frac{4}{1+x^{n-1}}$, we get

$f'(x)> n-1-\frac{4(n-1)x^{n-1}}{(1+x^{n-1})^2}$.  :P

Thank you very much, Vasc.

Very nice proof !!


absolute value
number theory
superior algebra
superior algebra unsolved

Prove that every Archimedean Field is order isomorphic to a subfield of R.

In this context Archimedean Field is defined as follows (I'm not sure if other books do it the same way).  We've proven that every ordered field has a subset order isomorphic to the integers:  call this set Z.  A field F is said to be archimedean then if it is ordered and for every f in F, there exists an n in Z, such that n>f.

Here's another problem I've proven, but am looking for a better solution to.
Prove that any two ordered fields with the lub property are order isomorphic.

Any ideas?

For the first problem, here's a general outline of a proposed approach: start by showing that $ F$, the Archimedean field, contains a copy of $ \mathbb{Q}$ with the usual ordering (you basically did this already). So your desired order isomorphism $ \varphi: F \to \mathbb{R}$ should be defined in the obvious way on this subfield. Now given an arbitrary element $ x \in F$, you need to determine how to define $ \varphi(x)$. The trick is that $ \varphi$ being order-preserving forces it to be continuous, in the sense that if $ q_n$ are rational approximations of $ x$ then $ \varphi(q_n)$ must be rational approximations of $ \varphi(x)$.

To make this precise, show how to sandwich $ x$ between arbitrarily close rationals using the Archimedean property applied to integer multiples of $ x$. Use this to prove that $ \varphi$ exists and is well-defined, and must preserve the field operations since it preserves the rationals and the field operations are continuous.

I hope this is enough to get you started.

Take the completion in respect to the absolute value defined by this ordering. You get a complete ordered archimedian field, thus by Ostrowski's theorem just $ \mathbb R$.

Heh. I thought the whole point was to prove this from scratch... this is pretty much the way you prove Ostrowski. If you take it as a given, I agree that there's not a whole lot left to do :)

Well, the proof I know for Ostrowski (every archimedean complete field $ K$ is $ \mathbb R$ or $ \mathbb C$) doesn't use this intermediate step (see e.g. Neukirch, Algebraic Number Theory, chapter 2): You show that $ \mathbb R \subset K$, then you show that every element of $ K$ is algebraic over $ \mathbb R$.

Fair enough. I guess that to take a completion you need to work with Cauchy sequences which is similar to the approach I proposed, bu I see your point.

How do you like Neukirch's book? It always intrigued me.


Nobody ?! Means that this problem is no "easy (simple)" or "nice" !

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle for which $ BC > AB$ . Consider the point $ \{\begin{array}{c} D\in (BC) \\
\ \widehat {ADB}\equiv\widehat {BAC}\end{array}$ .

The bisector of the angle $ \widehat {ABC}$ cut the lines $ AD$ , $ AC$ in the points $ E$ , $ F$ respectively.

Prove that $ DC = 2\cdot EF$ $ \Longleftrightarrow$ $ 2\cdot BF = AB + BC$ .[/color][/quote]

1)  $ \widehat {ADB}\equiv\widehat {BAC}$ -> $ %Error. "bigtriangleright" is a bad command.
ABC\sim%Error. "bigtriangleght" is a bad command.
DBA$ ->

$ \frac{AB}{BD}=\frac{BC}{AC}$  ->$ AB^{2}=BC*BD$ -> $ BD=\frac{AB^{2}}{BC}$
 $ AB=c,BC=a,AC=b,EF=x,DC=2x$ ->  $ BD=\frac{c^{2}}{a}=a-2x$ -> $ c^{2}=a(a-2x)$  -> $ x=\frac{a^{2}-c^{2}}{2a}$
2)  $ %Error. "bigtriangleright" is a bad command.
ABF\sim%Error. "bigtriangleght" is a bad command.
DBE$ -> $ \frac{AB}{BD}=\frac{BF}{BE}$ --> $ \frac{c}{\frac{c^{2}}{a}}=\frac{BF}{BF-x}$ -->
$ BF=\frac{ax}{a-c}=\frac{a}{a-c}*\frac{a^{2}-c^{2}}{2a}=\frac{a+c}{2}$  ->

$ 2BF=a+c$

[color=darkred][b][u]PP[/u].[/b] Let $\triangle ABC$ with $ a > c$ . Consider $ \left|\begin{array}{c} D\in (BC)\\\\
\widehat {ADB}\equiv\widehat {BAC}\end{array}\right|$ . The bisector of $ \widehat {ABC}$

cut $ AD$ , $ AC$ in $ E$ , $ F$ respectively. Prove that $ DC \equal{} 2\cdot EF$ $ \Longleftrightarrow$ $ 2\cdot BF=AB+BC$ .[/color]

[b][u]Proof[/u].[/b] $\triangle DBA\sim\triangle ABC\iff$ $\frac {DB}{AB}=\frac {BA}{BC}=\frac {AD}{CA}\iff$ $\frac {DB}{c}=\frac ca=\frac {l_a}{b}\iff$ $\left|\begin{array}{c}
BD=\frac {c^2}{a}\\\\
l_a=\frac {bc}{a}\end{array}\right|$ .

Apply the [u][b]Menelaus[/b]' theorem[/u] to the transversal $\overline{AFC}$ and $\triangle BDE\ :\ \frac {AE}{AD}\cdot\frac {CD}{CB}\cdot\frac {FB}{FE}=1\iff$

$\frac {a}{a+c}\cdot\frac {CD}{a}\cdot\frac {l_b}{FE}=1\iff$ $\boxed{\frac {DC}{EF}=\frac {a+c}{l_b}}$ . In conclusion, $DC=2\cdot EF\iff$ $2l_b=a+c$ .

[b][u]Remark (Sunken Rock)[/u].[/b] Sorry, this problem is without consistence. Indeed, $l_b=\frac {a+c}{2}\iff$ $\frac {2ac\cdot\cos\frac B2}{a+c}=\frac {a+c}{2}\iff$

$\cos \frac B2=\frac {(a+c)^2}{4ac}$ . But $\cos \frac B2\le 1\le \frac {(a+c)^2}{4ac}$ . In concluzion, $\cos \frac B2=1=\frac {(a+c)^2}{4ac}\iff$

$a=c$ and $B=0$ ,  what is an absurdity. In general, the relation  $2l_b<a+c$ is truly.

[u]Thank you[/u], [b]Sunken rock[/b] for your remark. I"ll modify my proposed problem so :

[color=darkred][b][u]PP[/u].[/b] Let $\triangle ABC$ with $ a > c$ and let $k>2$ be a real constant. Consider $ \left|\begin{array}{c} D\in (BC)\\\\
\widehat {ADB}\equiv\widehat {BAC}\end{array}\right|$ . The bisector

of $ \widehat {ABC}$ cut $ AD$ , $ AC$ in $ E$ , $ F$ respectively. Prove that $ DC \equal{} k\cdot EF$ $ \Longleftrightarrow$ $ k\cdot BF=AB+BC$ .[/color]


On her birthday, Jennifer always has a birthday cake decorated with the number of candles that corresponds to her age. So far, she has used a total of 253 candles on her cakes. How many years old is Jennifer?

$ 253$ is the $ 22$nd triangular number. Thus, Jennifer is $ \boxed{22}$ years old.

please dont consider as spam

but can just someone tell me what is the triangular number ?

thank you so much !

you can PM me thank you !!!!!!!!!!

@markoboy: Putting "please dont consider as spam" doesn't help your point; people will decide for themselves what they think of it.

The $n^{th}$ triangular number if $1+2+3+\ldots+n=\frac{n(n+1)}{2}$.

The nth triangle number is going to be
$1+2+3+4+5+6+7+8+9+10+\cdots (n-1)+n$

The formula for finding a triangular number is going to be $\frac{n(n+1)}{2}$
\[ \begin{array}{c|c}\text{N}&\text{Nth triangular number}&\\ \hline 1&1 &2&3\\ 3&6 & 4&10 \\ 5&15\\ 6&21\\ 7&28\\ 8& 36\\ 9&45\\ 10&55\\ 11&66\\ 12&78\\ 13& 91 \\ 14& 105\\ 15&120\\ 16&136\\ 17&153\\ 18& 171\\ 19&190\\ 20& 210\end{array} \]

[hide="Solution to the problem"]
When she turned one, she got one candle, 2 she got two candles, and so forth until we have her age.  Let's call her age n.
$1+2+3+4+5+6+7+8+9+\cdots n=253$
$\frac{n(n+1)}{2}=253$
Multiply by 2 to get:
$n(n+1)=506$
$n^2+n=506$
$n^2+n-506=0$
Using the quadratic formula, we get:
$n=\frac{-1 \pm \sqrt{1^2-4*506*1}}{2*1}$

$n=\frac{-1 \pm \sqrt{1+2024}}{2}$

$n=\frac{-1 \pm \sqrt{2025}}{2}$

$n=\frac{-1 \pm 45}{2}$

$n=\frac{-1+45}{2}\implies \frac{44}{2}=22$

$n=\frac{-1-45}{2}\implies \frac{-46}{2}=-23$

Since we can't have a negative value for n, we only have one possible value which is $\boxed{22}$ [/hide]


modular arithmetic
quadratics
number theory
number theory unsolved

If $ x, y$ are intergers, find all solutions of $ x^{5}\plus{}31 \equal{} y^{2}$.

[quote="SAPOSTO"]The equation is equvalent to $ x^5 \plus{} 32 \equal{} y^2 \plus{} 1$. Use this and use the lemma: if p is prime of the form $ 4k \plus{} 3$ and p divide $ a^2 \plus{} b^2$ then p|a and p|b.
Maybe you may also use arithmetic in $ \mathbb{Z}[\sqrt {31}]$[/quote]

Hmm...if you are right, this equation has no solutions.

But please, would you mind explain the lemma?

[quote="worthawholebean"][hide="Solution"]Consider the equation $ \pmod{11}$. $ x^{10}\equiv 1\pmod{11}$, so $ x^5$ is $ 1$ or $ \minus{} 1\pmod{11}$. The LHS is thus either $ 8$ or $ 10\pmod{11}$. However, the quadratic residues $ \pmod{11}$ are $ \{0,1,4,9,5,3\}$; thus, the LHS cannot be a perfect square. There are no solutions.[/hide][/quote]

Thanks.

The idea of mod 11 is good when you have 5 as power. However you miss the case $ x\equiv0\left(\mod 11\right)$
Let $ x\equiv0\left(\mod 2\right)$. Then we get contradiction by modulo 4.
The equation is equvalent to $ \left(x\plus{}2\right)\left(x^4\minus{}2x^3\plus{}4x^2\minus{}8x\plus{}16\right)\equal{}y^2\plus{}1$
If $ x\equiv1\left(\mod 4\right)$ then since $ x\plus{}2\equiv3\left(\mod 4\right)$ LHS has a prime divisor of the form $ 4k\plus{}3$  so it divides $ y^2\plus{}1$ and from here it divides 1. Contradiction!
If $ x\equiv3\left(\mod 4\right)$ then $ x^4\minus{}2x^3\plus{}4x^2\minus{}8x\plus{}16\equiv3\left(\mod 4\right)$ and continue in the same way!

And about the lemma.
Assume $ p \equal{} 4k \plus{} 3$ is prime which doesn't divide neither a nor b but it divides $ a^2 \plus{} b^2$. We have $ a^2\equiv \minus{} b^2\left(\mod 4k \plus{} 3\right)$ $ \Leftrightarrow$ $ a^{4k \plus{} 2}\equiv \minus{} b^{4k \plus{} 2}\left(\mod 4k \plus{} 3\right)$. From the Fermat's Little Theorem $ 1\equiv \minus{} 1\left(\mod 4k\plus{} 3\right)$ which is obviously contradiction!

[quote="SAPOSTO"]The equation is equvalent to $ x^5 \plus{} 32 \equal{} y^2 \plus{} 1$. Use this and use the lemma: if p is prime of the form $ 4k \plus{} 3$ and p divide $ a^2 \plus{} b^2$ then p|a and p|b.
Maybe you may also use arithmetic in $ \mathbb{Z}[\sqrt {31}]$[/quote]

I saw a question that was asking prove that $ p^2$ divides $ a^2 \plus{} b^2$ with conditions as in the lemma and that can be proved directly from the lemma.


[quote="N.T.TUAN"][quote="gunes"]Prove that $\dbinom{2n}{n}=\sum_{i=1}^{\infty}{(\dbinom{n}{k})^{2}}$[/quote]
Please edit it! If true, this is a corollary of Vandermoon's theorem![/quote]
I don't know what it is, it was our final exam question.

[quote="N.T.TUAN"][quote="gunes"]Prove that $\dbinom{2n}{n}=\sum_{i=1}^{\infty}{(\dbinom{n}{k})^{2}}$[/quote]
What is $i$?  :huh:[/quote]
it's $k$, also the upper bound is not infinity, it's $n$, sorry, check out the question again

[quote="spix"]This is just classic. Consider two sets of $n$ balls, one white and one black. In how many ways can we choose $n$ balls? Use the fact that $\dbinom{n}{k}^{2}=\dbinom{n}{k}\cdot\dbinom{n}{n-k}$[/quote]
or $(x+1)^{2n}=(x+1)^{n}(x+1)^{n}$, now see at $x^{n}$ at both sides!

[quote="spix"]This is just classic. Consider two sets of $n$ balls, one white and one black. In how many ways can we choose $n$ balls? [/quote]
Thank you :)
apparently, I haven't got that much info about combinatorics

[quote="gunes"]also the upper bound is not infinity, it's $n$, sorry, check out the question again[/quote]

The problem is not the upper bound because it is ordinary to state that $\binom{n}{p}= 0$ when $n<p$.
The problem is the lower bound because the given answers assume that it is $0$ instead of $1$.

Moreover, it is Vandermonde's indentity (not Vandermoon  :wink:  ).

Pierre.


Prove that (p-q)!(q-1)!≡(-1)^q 9mod p) where p is an odd prime and q is a positive integer less than or equal to p.

perhaps someone has a better solution, but here goes mine

[hide]

(p-1)_C_(q-1)= (p-1)!/((p-q)!(q-1)!)

since (p-1)!=-1 mod p we have to prove that (p-1)_C_(q-1)= (-1)^(q-1) (mod p)

now

 :Sigma: (p-1)_C_(q-1) = 2^(p-1) (taken from q=1 to q=p)

 :Sigma: (p-1)_C_(q-1) = 1 (mod p)

since (p-1)_C_(n-1) + (p-1)_C_(n) = p_C_n = 0 mod p (for all n does not equal 0 or p)

so now we knew that

(p-1)_C_(q-1) = k(-1)^q (mod p) or k(-1)^(q-1) (mod p)("adjacent" terms in the sumation sum to 0 (mod p)

since there are p terms in the sumation, and they sum to 1, the first and last are 1. (the sum is equal to the last term since there is an odd number of terms)

and thus (p-1)_C_(q-1)= (-1)^(q-1) (mod p) since 1-1 and p-1 are even.

[/hide]

I just d/led and installed the LaTeX stuff so I'm going to practice by TeXing my solution

$\displaystyle{p-1\choose q-1}= \frac {(p-1)!}{(p-q)!(q-1)!} $
since $(p-1)!\equiv -1 \mod {p}$

we have to prove that  $\displaystyle{p-1\choose q-1} \equiv (-1)^{q-1} \mod {p} $

$ \displaystyle\sum_{q=1}^{p}{p-1\choose q-1}=2^{p-1} \equiv 1 \mod {p}$

since $\displaystyle {p-1 \choose n-1} + {p-1 \choose n} = {p \choose n} \equiv 0 \mod {p}$ (for all $n \ne p,0$)

So now we know that
$\displaystyle {p-1 \choose q-1} \equiv k(-1)^q \mod {p}$ or $k(-1)^{q-1} \mod {p}$ ("adjacent" terms in the sumation sum to $0 \mod {p}$)

since there are an odd number of terms, and they sum to 1, the value of k is 1 and,
$\displaystyle {p-1 \choose q-1} \equiv (-1)^{q-1} \mod {p}$


Let think about:

If $x, y, z > 0, x^3 + y^3 + z^3 = 3$ then $x^4y^4 + y^4z^4 + z^4x^4 \le 3$

(If I'm not wrong, this problem was proposed by Harazi and solved with nice trick:

use $3xy \le x^3 + y^3 + 1$)

I think last prb have straight connection with arqady's one.

Thank you very much, Namdung!
It is wonderful!:lol:

i was following your directions but i can't prove

$\sqrt[4]{x^3} + \sqrt[4]{y^3} + \sqrt[4]{z^3} \geq x^3 + y^3 + z^3$

so can you please post whole solution...

[quote="Albanian Eagle"][quote="arqady"]Let $x,y,z$ are non-negative real numbers and $xy+xz+yz=3.$
Prove that $\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}\geq3,$[/quote]
this is equivalent to:
$xy+yz+zx=1$and $\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}\geq3^{\frac{5}{8}}$
but we can substitute $x=ctg A\2$ etc and use the fact that (ctg x)^{\frac{3}{4}}
is convex

PS if it is not convex use IPT :D[/quote]

IPT= Inflection point theorem. Is it correct?
  I heard that it is a secrect weapon of Myth. Can anyone give me a link to that method, please? That would be very nice! :lol:

[quote="mathmanman"][b]Gibbenergy[/b], it seems that you're right!  :lol:

http://www.mathlinks.ro/Forum/viewtopic.php?t=19812
http://www.mathlinks.ro/Forum/viewtopic.php?t=20422
http://www.mathlinks.ro/Forum/viewtopic.php?t=32332
http://www.mathlinks.ro/Forum/viewtopic.php?t=36027
http://www.mathlinks.ro/Forum/viewtopic.php?t=39925[/quote]

Many thank for your meaningful post. I appreciate that! :)


function
search
calculus
calculus computations

Hi

Does there exist an injective function from an infinite sequence of reals to R? I believe it is true for finite sequences.

Thanks

Yes it is true.You can always find an injection from an infinite sequence of reals to $ \mathbb{R}$ since the set $ \{x_n| n \in \mathbb{N}\}$ is countable , but $ \mathbb{R}$ is uncountable.From this you can notice that conversely is not true.

[quote="amitie"]could you please explain some more?
thanks :)[/quote]

You can search for more information about countable and uncountable sets on the internet (try wiki :) ).I can give you an injective function just to make it clear for you: Let $ f: \{x_n | n\in \mathbb{N}\} \to \mathbb{R}$ , with $ f(x_0)\equal{}0$ , $ f(x_1)\equal{}0,1$ and so on $ f(x_n)\equal{}0,\underbrace{11..1}_{n \text{ times}}$.It is clear that $ f$ is injective.


Find the smallest positive integer n>3 such that there exists a convex polygon which can be dissected into finitely many non-convex n-gons.

Enjoy! :D

I think the answer is $n=5$. It's easy to find an example to prove that this is possible, but my 'proof' that it can't be done for $n=4$ is shaky.  Basically the idea is that a non-convex quadrilateral has to be of a particular form (like a triangle with another triangle removed).  I think we can show that if the union of some number of these is convex, then there must be infinitely many of them.

your answer concurs with mine, actually i have a nice proof for the impossiblility of n=4, but i didn't find an example for n=5, perhaps we can share?

Okay, here is my proof for that no convex polygon can be dissected into finitely many non-convex 4-gons. Suppose there is such polygon P that can be dissected into m non-convex 4-gons. Then the sum of the internal angles of all these  4-gons is 360*m. Every non-convex 4-gon has an internal angle greater than 180 at the point of concavity. Now the m points of concavity of these m concave 4-gons are all pairwise distinct, and are all inside P. Hence the sum of internal angles of all m 4-gons is greater than 360*m. Contradiction. :D


Support
abstract algebra
topology
Ring Theory
superior algebra
superior algebra unsolved

Let $ A$ be a ring and $ M$ a Noetherian $ A$-module. Prove that $ Supp(M)\subset Spec(A)$ is a Noetherian topological space.

We can show the stronger result: $ A/Ann_A(M)$ is a Noetherian ring (*).

$ Ann_A(M)$ denotes the annihilator ideal of $ M$ consisting of such $ a\in A$ with $ aM\equal{}0$.

From this it follows that $ Supp(M)$ is a Noetherian topological space: Since $ M$ is finitely generated, $ Supp(M)\equal{}V(Ann_A(M))$ and the latter is homeomorphic to $ Spec(A/Ann_A(M))$ which is Noetherian as prime spectrum of a Noetherian ring.

[hide="Proof of (*):"]Let $ M$ be generated by $ m_1,\dots,m_n$. Defining the ideals $ I_i\equal{}Ann_A(Am_i)$, $ A/I_i\cong Am_i$ is a Noetherian $ A$-module, hence $ A/I_i$ is a Noetherian ring. Since $ Ann_A(M)\equal{}\cap_i I_i$, we are done with the following lemma:

If $ I$ and $ J$ are ideals in a ring $ A$ such that $ A/I$ and $ A/J$ are Noetherian rings, then so is $ A/(I\cap J)$.

[hide="Proof of lemma"]Let $ K$ be an ideal in $ A$ containing $ I\cap J$. Then $ I\cap J\subset I\cap K\subset K$ and it is enough to show that the subsequent quotients are finitely generated $ A$-modules. The first stems from $ K/(I\cap K)\cong (K\plus{}I)/I$ which is an ideal in $ A/I$, the second quotient $ (K\cap I)/(I\cap J)$ is a sub-module of $ I/(I\cap J)\cong (I\plus{}J)/J$ which is an ideal in $ A/J$.[/hide][/hide]


function
algebra
functional equation
algebra proposed

Find all function from $R$ to $]0,\infty)$wich satisfyse
$f(x^{2}+y^{2})=f(x^{2}-y^{2})+f(2xy)$.

Interchanging $x$ and $y$ we get $f(x)=f(-x)$ for any $x$. We work only with positive numbers.
Let $a=2xy,b=x^{2}-y^{2}$. So $x^{2}+y^{2}=\sqrt{a^{2}+b^{2}}$ and $f(\sqrt{a^{2}+b^{2}})=f(a)+f(b)$.
The function $g(x)=f(\sqrt{x})$ verifies $g(a^{2}+b^{2})=g(a^{2})+f(b^{2})$ for all $a,b$ and $g(x)>0$ thus from the Cauchy equation $g(x)=kx$ and $f(x)=kx^{2}$  for arbitrary real k.

I assume you meant $f: \mathbb{R}\rightarrow [0,\infty)$.
$y=\pm\frac{1}{2}$ gives $f(-x)=f(x^{2}+\frac{1}{4})-f(x^{2}-\frac{1}{4})=f(x)$ for all $x\in\mathbb{R}$.
For $a,b\geq 0$ take $x=\sqrt{\frac{\sqrt{a+b}+\sqrt{a}}{2}}$ and $y=\sqrt{\frac{\sqrt{a+b}-\sqrt{a}}{2}}$.
Then $x^{2}+y^{2}=\sqrt{a+b}$,$x^{2}-y^{2}=\sqrt{a}$ and $2xy=\sqrt{b}$.
This gives $f(\sqrt{a+b})=f(\sqrt{a})+f(\sqrt{b})$ for all $a,b\geq 0$.
So $g(x): =f(\sqrt{x}): [0,\infty)\rightarrow [0,\infty)$ satisfies Cauchy's functional equation, and
$g(b)=g(a)+g(b-a)\geq g(a)$ for all $0\leq a\leq b$.
So for $c: =g(1)=f(1)\geq 0$ we get $g(x)=cx$ for all $x\in [0,\infty)$ and
$f(x)=f(|x|)=g(|x|^{2})=c|x|^{2}=cx^{2}$ for all $x\in\mathbb{R}$.


geometry
circumcircle
incenter
perimeter
geometry unsolved

In triangle $ ABC$, we have $ \angle A\equal{}60^{\circ}$. Let $ O,H,I,I'$ be the circumcenter, orthocenter, incenter, and excenter opposite $ A$, respectively, of $ ABC$. Let $ B'$ and $ C'$ be points on the segments $ AC$ and $ AB$ such that $ AB\equal{}AB'$ and $ AC\equal{}AC'$. Prove that:
$ (a)$ The eight points $ B,C,H,O,I,I',B',C'$ are concyclic.
$ (b)$ If $ OH$ intersects $ AB$ and $ AC$ at $ E$ and $ F$, respectively, the perimeter of triangle $ AEF$ equals $ AB\plus{}AC$.
$ (c)$ $ OH\equal{}|AB\minus{}AC|.$

it is easy to get (1) notice H,I,O,B,C are concyclic and CC'//BB'
(2)EH=BE  OF=FB'  OH =B'C
(3)OH =B'C=AC-AB


[hide=" Solution"]

I think the answer is $ 5$:

$ 26$ is the first solution, then since $ 5, 6$ and $ 7$ are coprime, the next ones are got by successively adding $ 5\times 6\times 7 \equal{} 210$, giving $ 236, 446, 656$ and $ 866$, that is $ 5$ in all.

[/hide]

[hide="hint"]
Apply Chinese Remainder Theorem:

Doing so will give your answers to be in the form $ 210a\plus{}26$ where $ a$ is an integer.
[/hide]

[quote="Truffles"]How many of the first 1000 integers have a remainder of 1 when divided by 5, 2 divided by 6, and 5 when divided by 7.[/quote]
[hide="Using CRT"]We want to solve the simultaneous congruences:
$ x\equiv 1\bmod{5}$
$ x\equiv 2\bmod{6}$
$ x\equiv 5\bmod{7}$.

Due to the CRT, our answer should be in the form $ x\equiv a\bmod{210}$.
Taking the first congruence, it gives us that $ x\equal{}1\plus{}5m$, where m is an integer. Plugging that into the 2nd congruence we have $ m\equiv 5\bmod{6}$, or $ m\equal{}5\plus{}6l$, where l is an integer. Plugging this into the 3rd congruence gives us $ l\equiv 0\bmod{7}$, or $ l\equal{}7n$, where n is an integer. Plugging this back we have that $ m\equal{}5\plus{}42n$, and then $ x\equal{}26\plus{}210n$. Then we simply have the inequality to solve that $ 210n\plus{}26<1000$. This solution gives us that $ n<4.638...$. Since n is an integer, n can take on 5 values, 0-4.[/hide]


analytic geometry
geometry
floor function
ceiling function
modular arithmetic
number theory
divisibility tests

1. Find the number of point with integer coordinates which are at a distance of less than 5 units from point(1,2) on the X-Y Plane.
a.	61
b.	69
c.	81
d.	105

Whats the simplest wasy of doing this in under 1 to 2 mins... i know i can list out all the points...and so on....but that is going to be time consuming.

2. Let S be a set of positive integers such that every element n of S satisfies the conditions: a) 10000 <= n <= 20000, b) every digit in n is oddm then how many elements of S are divisible by 11?
a.	30
b.	20
c.	10
d.	15

How to solve this fast?

For Question 1, the area of a circle with radius 5 should give a good approximation, enough to eliminate the other 3 choices.

Edit: Actually, I think the approximation won't be so good.  This method works better when the distance "5" is larger.

Instead, maybe a quick enumeration might do instead.  Here is a list of integer pairs such that $x^{2}+y^{2}< 25$:

$x = 0, \pm 1, \pm 2$: $y = 0, \pm 1, \pm 2, \pm 3, \pm 4$
$x = \pm 3$: $y = 0, \pm 1, \pm 2, \pm 3$
$x = \pm 4$: $y = 0, \pm 1, \pm 2$

We count 69 pairs.

[quote="amcavoy"][hide]Even multiples of $11$ will produce even numbers, so since every element in $\mathcal{S}$ is odd, you have \[\frac{\left\lfloor\frac{20000}{11}\right\rfloor-\left\lceil\frac{10000}{11}\right\rceil }{2}=\boxed{454}.\] [/hide][/quote]

odd multiples of 11 will also produce numbers with even digits, take 12221 for example

Oh, every [b]digit[/b] in $n$.  I just did the number of odd values of $n$.  In that case, might the divisibility by $11$ rule be useful?

For the second one, we're looking for $|\mathcal{S}|$ if $\mathcal{S}= \left\{n\in\mathbb{N}\ |\ \exists a,b,c,d\in\{1,3,5,7,9\}: n=\overline{1abcd}\ \wedge\ 11|n\right\}$. But $11|n$ is equivalent to $0\equiv n \equiv 10^{4}+10^{3}a+10^{2}b+10c+d \equiv 1-a+b-c+d\pmod{11}$, i.e. $b+d\equiv a+c+1\pmod{11}$. Now remark that $b+d\leq 18$ and $a+c+1\leq 19$ and $b+d$ is even and $a+c+1$ is odd. Therefore, $|(a+c+1)-(b+d)| = 11$. Now let $b+d = 2,4,6,\cdots,18$ and find a total of $4+6+6+4+0+0+3+4+3 = 30$ solutions. (End is boring, anyone a better idea?)

Kurt,

I'm lost.... This is exactly the same answer in the answer key...[quote]Now remark that  and  and  is even and  is odd. Therefore, . Now let  and find a total of  solutions. [/quote] Could you clarify this better??? i dont understand it.... sorry i'm no maths wiz....any other simpler method?

No problem, fizsle!  :)

[quote="Kurt Gödel"]Now remark that $b+d\leq 18$ and $a+c+1\leq 19$ and $b+d$ is even and $a+c+1$ is odd. Therefore, $|(a+c+1)-(b+d)| = 11$. Now let $b+d = 2,4,6,\cdots,18$ and find a total of $4+6+6+4+0+0+3+4+3 = 30$ solutions. (End is boring, anyone a better idea?)[/quote]

(I'll elaborate a bit.) We have $b+d\equiv a+c+1\pmod{11}$. So $b+d = a+c+1+11n$ for some integer $n$. We know that $11n \leq a+c+1+11n \leq 19+11n$ and $2 \leq b+d \leq 18$. Now suppose $n\geq 2$, we have $b+d \leq 18 < 22 \leq 11n \leq a+c+1+11n = b+d$, which is impossible. Suppose $n\leq-2$, then $b+d = a+c+1+11n \leq 19+11n \leq 19-22 =-3 < 2 \leq b+d$, which is impossible. So $n\in\{-1,0,1\}$. But if $n=0$, then $b+d = a+c+1$, which is impossible because $b+d$ is even and $a+c+1$ is odd. Therefore, $n\in\{-1,1\}$. If $n=-1$ we have $a+c+1-11 = b+d$, so $(a+c+1)-(b+d) = 11$, so $|(a+c+1)-(b+d)|=11$. If $n=1$, we have $a+c+1+11 = b+d$, so $(a+c+1)-(b+d) =-11$, so $|(a+c+1)-(b+d)| = 11$.

For the second part, I'll do some examples:
If $b+d = 2$, then $(b,d) = (1,1)$ and $|a+c+1-2| = |a+c-1| = 11$, so $a+c = 12$ (because $a+c-1 =-11$ is impossible), i.e. $(a,c) = (3,9),(5,7),(7,5),(9,3)$, so $1\cdot 4 = 4$ solutions.
If $b+d = 4$, then $(b,d) = (1,3),(3,1)$ and $|a+c-3| = 11$, so $a+c = 14$ (because $a+c-3 =-11$ is impossible), ie. $(a,c) = (5,9),(7,7),(9,5)$, so $2\cdot 3 = 6$ solutions.
If $b+d = 12$, then $(b,d) = (3,9),(5,7),(7,5),(9,3)$ and $|a+c-11| = 11$, so $a+c \in\{0,22\}$, which is impossible.
Etcetera.

[quote]Edit: Actually, I think the approximation won't be so good. This method works better when the distance "5" is larger. [/quote]
FWIW   one can get upper and lower bound with a simple argument
$\pi (r-\sqrt 2)^{2}< N(r) < \pi (r+\sqrt 2)^{2}$
Assumng you are also counting the points on the circle


algebra
polynomial
number theory unsolved
number theory

Prove or disprove the following problem:
Let $R(x) = x(x-1)(x+1)P(x) + 1$ be a polynomial, where $P(x)$ is an integer polynomial.
Prove there are infinitely many values $t$ with $R(t!)$ being composite (ie. not prime).

It is true for all R(x). Let m=R(x),|m|>1,p|m, p - prime. Then p|R(x+kp) for all k. Only finite of them equal 0 or p or -p.

i think you made a mistake.
note the problem says infinitely many $t$ s.t. $R(t!)$ is compositie, not $R(t)$.


function
analytic geometry
graphing lines
slope
inequalities

[url]http://www.artofproblemsolving.com/Wiki/index.php/1995_AHSME_Problems/Problem_12[/url]

This is a 1995 AHSME #12.

Isn't a linear function just any graph that is made up of lines that follows the definition of a function?

So why is the solution correct? Is my definition wrong or has the definition of a linear function changed since 1995?

Thanks

ac-king

A linear function is f(x)=ax+b for some real a,b.

A linear function does not have a graph made up of lines, but is simply the graph of a straight line. It is a function of form $ ax\plus{}b$, as mentioned above, where $ a$ is the slope and $ y\equal{}b$ is where the function intersects the y-axis.

[hide="Better solution"]Let $ f(x)\equal{}ax\plus{}b$.
$ f(1)\leq f(2)\implies a\plus{}b\leq 2a\plus{}b\implies a\leq 2a$
$ f(3)\geq f(4)\implies 3a\plus{}b\geq 4a\plus{}b\implies 3a\geq 4a$.
Subtract $ a$ from both sides of the first inequality to get $ 0\leq a$.
Subtract $ 3a$ from both sides of the second inequality to get $ 0\geq a$.
The only value of $ a$ that satisfies both of these is $ a\equal{}0\implies f(x)\equal{}b$.
Then $ f(5)\equal{}5\implies b\equal{}5\implies f(0)\equal{}5$.[/hide]

This is pretty much the solution given in the Wiki, but it helps to put the function in terms of constants and variables, rather than cite various properties.

Just think of a line.  It either goes up or it goes down.  In other words, as x gets bigger,
either y gets smaller or y gets bigger or y stays the same.  That's all they're saying.  Because the inequalities
they give tell you that f(1) is either less than or equal to f(2) BUT f(3) is greater than
or equal to f(4), you KNOW that they have to be equal because otherwise it wouldn't be
a linear function.  That's all.  THen you know that all of the y's are equal, so since
f(5)=5, you know that f(0)=5.  Of course, they could've said f(12137891723)=5. Same answer.


function
geometry
inradius
3D geometry
inequalities
national olympiad

[quote="fourierseries"]
I have materials from old China National Olympiads (you can get a small pamphlet of those here for about 33 cents).  I also have their camp materials, national olympiads, girl's olympiad, etc. for years 2004 and 2006.
[/quote]

Could you publish the problems of the camp materials possibly here? Thanks.

Well, there are quite a few problems. I'll start with the 2004 IMO Mock Tests (there are 39 problems which start easy and become increasingly difficult).  These  are given to the selected IMO team as practice:

1. If $a,b,c$ are positive reals, prove:

$\frac{a+b+c}{3}\ge^{3}\sqrt{\frac{(a+b)(b+c)(c+a)}{8}}\ge \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{3}$.

2. Let a and b be distinct positive rationals such that there exist infinitely many n for which $a^{n}-b^{n}$ is a positive integer.  Prove that a and b are positive integers.

3. Find all functions f,g: (0, +$\infty$) to (0 +$\infty$) such that for all x in positive reals, we have:

$f(g(x))=\frac{x}{xf(x)-2}$ and $g(f(x))=\frac{x}{xg(x)-2}$

4. Let ABC be a triangle with semiperimeter s and inradius r.  Use AB, BC, CA as diameters to draw three circles.  Let the circle which is externally tangent to all three constructed circles have radius t. Prove:

$\frac{s}{2}<t< \frac{s}{2}+(1-\frac{\sqrt{3}}{2})r$

I'll post more when I have the time! Enjoy!

5. In isoceles ABC, BC=a and AB=AC=b.  Let M and N be points on AC and AB respectively such that:

$a^{2}AM AN = b^{2}BN CM$.

If the BM and CN intersect at P, find the locus of P.

Thanks fourierseries! And keep on posting more problems!  :)

6. Let a, b, c be positive reals.  Prove:

$\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^{2}\ge 4 \sqrt{3abc(a+b+c)}$

7. Find all functions $f: (0,+\infty) \to  (0,+\infty)$ such that the function is monotonically increasing on $[1,+\infty)$, and for all positive reals a,b,c, the folliowing is satisfied:

$f(ab)f(bc)f(ca)=f(a^{2}b^{2}c^{2})+f(a^{2})+f(b^{2})+f(c^{2})$

8. Let n be a positive integer and $a_{1},...,a_{n}, b_{1},...,b_{n}, c_{2},...,c_{2n}$ be $4n-1$ positive reals such that $c_{i+j}\ge a_{i}b_{j}, 1 \le i,j, \le n$.  If we let $m=max_{2 \le i \le 2n}c_{i}$, prove:

$(\frac{m+c_{2}+...+c_{2n}}{2n})^{2}\ge (\frac{a_{1}+...+a_{n}}{n})(\frac{b_{1}+...+b_{n}}{n})$

Note: This is a very well-known, famous problem. However, it was not that widespread in 2004 and I'm sure the team leaders thought that it was beautiful.

9. Let {$f_{n}$} be the Fibonacci sequence with the initial conditions $f_{0}=1 f_{1}=1$. Solve the equation:

$n f_{n}f_{n+1}=(f_{n+2}-1)^{2}$.

In response to me@home, I will begin sorting problems by subject and posting them in their respective olympiad forums, so that people can post solutions.

If anyone has other suggestions, please make them.

[quote="orl"][quote="Hawk Tiger"][quote="harazi"]Interesting enough this set of problems, as I can see these are almost exactly the problems also given in Mathlinks Contest (I can even recognize two of my problems). Isn't this a bit strange, to give in IMO test problems that appear on some forums?[/quote]
No ,it is not test problem,but just training problem.
And the 2004 is the only year when the online problem is used.
In 2006 ,Only 20 problems has been published.which are  six problem of American TST,and five geometry problem of Mr.Ye, and some other totally old problem.[/quote]

Fourierseries keep on posting the problems for training 2004 please. Hawk Tiger could you please post the problems of 2004 in this thread. Maybe annotating them with 'old', 'Armerican TST' etc. that we can either create a new posting for them or link them to previously created problems on the forum. And Hawk Tiger which other years do you have? Thanks.[/quote]
I have the problem of 2003-2006,and I will have the problem of 2007 :)
But I am a little bit busy these days ,because I have to finish the  homework (for national team) ,and about a week later ,I will go to Beijing for the training of IMO .So I only can post some nice problem for you before IMO.  When I am back ,I will post all.Sorry.And My English is too bad ,I only can transelate some problems with easy word :blush:
Problem 14 (2006):
Prove that every postive rational numbers can be expressed as a sum of three cube of rational numbers.
And find all the postive rational numbers can be expressed as a sum of two cube of rational numbers.(Fermat Therem for the $n=3$ case)
Problem 11. the very famous inequality of Iran 1996(But there is an another solution without SOS)
Problem 10. for any rational number $r$,ther exist integers $a,b,c,d$ ,such that $r=\frac{a^{3}+b^{3}}{c^{3}+d^{3}}$
Problem 13.a>=b>=c>=d>0,
Find the miminum of $(1+\frac{c}{a+b})(1+\frac{d}{b+c})(\frac{a}{c+d}+1)(\frac{b}{d+a}+1)$

I really hope fourierseries can post them ,becuase it is quite hard to translate some problems.......

[quote="kdano"][quote="Hawk Tiger"]
Problem 14 (2006):
Prove that every postive rational numbers can be expressed as a sum of three cube of rational numbers.
And find all the postive [b]rational[/b] numbers can be expressed as a sum of three cube of [b]rational[/b] numbers.(Fermat Therem for the $n=3$ case)[/quote]

These are supposed to be integer, aren't they? Otherwise it makes no sense after the first task..[/quote]
Sorry ,It is a typo.Not that hard, just Fermat Theorem( I have edited it)


I wonder how much people practice (or how lazy I am)

Does team practice count?
without team I am probably 5-6 hours, and team 4 hours a week.

If you still have team practice after state, I guess it does

Um... I practice 3-5 hours a day by myself.

I wont disclose the amount of Michigan practice time though.

wow!  if I did that, this would be my schedule

1. school
2. track
3. hw, Viola, eat
4. Mathcounts
5. any other extra cirricular activity (or more mathcounts if none)
6. more Mathcounts
7. sleep

that would drive me completely mad

Our team practices for 90 minutes a week.

should there be a separate poll for team practices instead?


group theory
abstract algebra
superior algebra
superior algebra unsolved

If a group fails to have a normal subgroup of order $ p$, then it has an element of order $ p^2$, where $ p$ is a prime number. Thanks in advance.

$ A_4$ has no normal subgroup of order $ 2$ and no element of order $ 4$.

For any $ p$, $ S_{n}$ is a counterexample whenever $ p\le n<p^2$.

I'm sure everyone is thinking it, but the group $ \{1\}$ is a counterexample as well.  This problem needs more conditions. (though of course jmerrry covers the case where $ n$ is divisible by $ p^2$ at least)

It's true when $ |G|\equal{}p^k$, but every such $ G$ has a normal subgroup of order $ p$ so the conclusion is vacuous.


combinatorics solved
combinatorics
Combinatorial sum

Prove that: \[ \left(\sum_{k=0}^{n} 2^k \binom{2n}{2k}\right)^2 - 2 \left(\sum_{k=0}^{n-1} 2^k \binom{2n}{2k+1}\right)^2=1 \]

$\binom{a}{b}$ denotes the binomial coefficient, $\frac{a!}{b!(a-b)!}$


function
floor function
algebra open
algebra

I want to find a function that takes a number which is not divisible by 4 (even or odd) and then produces a number which is divisible by 4.. can someone help me?

sorry let me rephrase the question then say I have a function y:

y = 2x - 8

and x is a negative odd or even number which is not divisible by 4..

I know some results of y still is not divisible by 4, but what if I those that are not divisible by 4, I subtract 2 from it.. then will that new number will it always be divisible by 4?

[quote="t0rajir0u"]I don't follow.  Please be more specific.  Are you looking for a function with the following properties:

- if $ x$ is divisible by $ 4$, then $ f(x)$ is not divisible by $ 4$.
- if $ x$ is not divisible by $ 4$, then $ f(x)$ is divisible by $ 4$.

If so, define the appropriate piecewise function with period $ 4$.[/quote]

no that's not what

Say I have a function y, where y = 2x-8 and therefore the result of y varies, some can be divisible by 4 and some can also not be divisible by 4.
My concern is for those y values that is not divisible by 4,  I want to find the next closest value greater than y that is divisible by 4, how can I do this?

and the constriction once again is that x is a negative number which is not divisible by 4

The closest integer to a given real number $ y$ is $ \lfloor y \plus{} \frac{1}{2} \rfloor$.  The closest integer to a given real number $ y$ divisible by $ 4$ is $ 4 \lfloor \frac{y}{4} \plus{} \frac{1}{2} \rfloor$ (why?).


[quote="trannguyenquyND_VN"]let $ a.b.c.d,p,q$ such :
$ p^2\plus{}q^2\minus{}a^2\minus{}b^2\minus{}c^2\minus{}d^2\ge 0$

$ (p^2\minus{}a^2\minus{}b^2)(q^2\minus{}c^2\minus{}d^2)\le (pq\minus{}ac\minus{}bd)^2$[/quote]
It's easy, but very nice! :lol:

This follows straight from ACZEL inequality.
Because we have;

$ p^2 \minus{} a^2 \minus{} b^2\geq0$ and $ q^2 \minus{} c^2 \minus{} d^2\geq0$

It follows that;
$ (p^2\minus{}a^2\minus{}b^2)(q^2\minus{}c^2\minus{}d^2)\le (pq\minus{}ac\minus{}bd)^2$

[quote="trannguyenquyND_VN"]let a.b.c.d,p,q such :
p^2+q^2-a^2-b^2-c^2-d^2$ \ge$ 0

(p^2-a^2-b^2)(q^2-c^2-d^2)$ \le$ (pq-ac-bd)^2[/quote]
see it here;
http://www.mathlinks.ro/viewtopic.php?t=261027 :)  :wink:

[quote="Papalakis"]This follows straight from ACZEL inequality.[/quote]
What is that?