modular arithmetic
function
number theory
least common multiple
probability
MATHCOUNTS

I always thought 0 was a multiple of anything because 0 * x = 0, but a CML I took a couple months ago said otherwise. Is 0 a multiple of 3, or any number for that matter?

i thought it was because -1*3=-3, 0*3=0, 1*3=3.  if the other two are, why can't zero times three be a multiple?

[quote="ragnarok23"]It is disputed over a lot, but I am pretty sure it is not considered a multiple. There was a question on MOEMS in which the answer was affected if 0 was considered a multiple or not. MOEMS considered it a multiple, however it was challenged and the challenge succeeded.[/quote]

I remember that question. I spent like 10 minutes debating with myself over that question...

I think N is a multiple of m, if there is some integer k such that N=km, so yes (k=0).

Yeah, I think 0 is a multiple of everything.

Reason 1:
$N$ is a multiple of $N$ (duh)
To create another multiple, you must add or subtract $N$, and $N-N=0$, so 0 is a multiple.

Reason 2 (similar to pianoforte's):
To be a multiple of a number $x$, you have $N\equiv0\pmod{x}$.  This can be expressed as $N=0+Ax$, for some integer $A$.  When $A=0$, $N=0$, so 0 is a multiple of $x$

Reason 3:
Why shouldn't it be a multiple of all numbers?

The divisor function (number of divisors) considers 0 to have $\infty$ factors. ie. It's a multiple of everything.

Well, as common sense, if somebody tells you, "Start counting multiples of 3 from -6."

You would do, or at least I would do, $-6,-3,\boxed{0},3,6, \text{etc.}$.  It would be weird if you just skipped zero...

I think 0 is the multiple of any number except.

Math is my life :!:

Hmmm...so 0 is like the lord of all composite numbers :idea:

Actually it depends on your definition of a multiple.

If it means that for $a$ to be a multiple of $b$, then for some INTEGER $k$, $ak=b$, then  0 is a multiple of all integers $b$.

If you exchange the integers with natural number, then 0 is NOT a multiple. Dependes on your definition.

For this level, I agree with biffanddoc that 0 is a multiple. However, if 0 were a multiple, wouldn't the LCM of any group of numbers be 0? Weird...

0 is a multiple of everything. The first multiple of any number is the number. 0 would be the zeroth multiple. Thus, the least common multiple is positive because we don't consider the zeroth multiple.

From previous posts, it looks like negatives are also multiples, so that would complicate things earlier.

Correct me if I'm wrong, but generally on math competitions 0 is not a multiple and LCM is always positive.

1.  Well-editted contests would avoid this issue by using 'positive multiples' or 'natural numbers' as descriptors.  If not so limited on a contest, 0 should count as a multiple.  But....
2.  Editors are human. Humans make errors.  'Little' things are easy to miss. :roll:

Well, sometimes it's indirect:

E.G. 2004 Nats #28(?), there's a number line and there's some probability you go there, here, etc. The probability of landing on a multiple of three would differ if 0 was counted as a multiple of 3.

I think I counted 0 as a multiple of 3 and got it wrong

Well, I'd consider $0$ as a multiple of $3$

Kind of late, but no I don't believe 0 is ever a multiple of anything in math competitions. Generally in math competitions, correct me if I'm wrong, it is implied to be the positive multiples unless otherwise stated.

Hey someone else actually does CML? Amazing! What did you get? Does anyone know if you get anything for a perfect score?

[quote="Fanatic"]E.G. 2004 Nats #28(?), there's a number line and there's some probability you go there, here, etc. The probability of landing on a multiple of three would differ if 0 was counted as a multiple of 3.[/quote]Right, and to get the correct answer of $\frac{31}{90}$, you had to assume that 0 was a multiple of 3.  If you assumed it wasn't, then you would get the wrong answer.

For folks who don't have the problem in front of them, here is a paraphrase.  You pick a random integer from 1 to 10.  Then, with probability $\frac{2}{3}$, you move one up; otherwise you move one down.  For a second time, with probability $\frac{2}{3}$, you move one up; otherwise you move one down.  What is the probability that you end up on a multiple of 3?

[quote="BFG"]0 is not a multiple of 3.  In Number Theory, if a is divisible by b, then b is a divisor of a, and a is a multiple of b.  Note that a and b are all natural numbers (excluding "0").[/quote]

0 can be a multiple by this if you mess with the numbers for a while.
put a as 0, and b as 3. 0 is divisible by 3 (0/3=0) 3 is a divisor of 0 (3 goes into 0 zero times)  and 3*0=0.  Why does it have to be natural numbers?

i like to break mathematical laws... :lol:

[quote="lotrgreengrapes7926"][quote="life=tennis+math"]Hey someone else actually does CML? Amazing! What did you get? Does anyone know if you get anything for a perfect score?[/quote]
I think I got 1 wrong or something. I don't know what you get, but it's probably not an all-expense paid trip to Washington DC. :D[/quote]

what is CML?

who makes the mathcounts tests??? i want to know their opinion on this so i dont mess up on possible question like this next year.

[hide]0 is a multiple of any number.  At least that is what I think.

do you think they might have said positive somewhere in the problem?[/hide]

No, it does not say positive in the problem, I checked

Right.  That means the MathCounts writers agree that 0 is a multiple of everything.  Otherwise, they would not have written that $\frac{30}{91}$ is the answer to the probability question.

There is no doubt about it. 0 is a multiple of any number. It said it in AoPS Volume 1. You can always trust it. :D

A much better question would be what is a multiple of 0?

CML = continental math league, I got a perfect score =o

CML for me:

6-6-6-6-6-6-5-4... (perfect score is 6/6)

Our school gives a (small) award to the grade level winners... I thought my 4 would make me not win it but I still had by far the highest score =O

I missed like 1 or 2 ( my teacher never told us what we got on the last one...)
I know I ended up with second and what does the first place person get? name on the morning announcements (which second and 3rd people do too) and its almost as bad of a thing as a good thing....(because then your ELA teacher who just likes you too much comes up to talk to you while you are talking to your friends)

For the 0 question...Im confused cuz I thought I got it wrong since I put 0 as a not multiple and later my friend was like you did put 1 right and I was like um whoops.


trigonometry
limit
calculus
calculus computations

How do you prove the following formula?: \[\frac{2}{\pi}=\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots\]


Can anybody recommend an effective book to study from in order to learn HMMT-level calculus?

hello, could it be that he meant this
http://web.mit.edu/hmmt/www/
Sonnhard.

I guess you're right, Sonnhard.

Having had a look at the website, I can't think of any book which would be appropriate or useful.

Sorry.


[removed] upon the person below's request

You should remove this problem for now bugzpodder. I'll explain why later.

.......  okay i removed it (i will do it right now actually)... why??

It was similar to a problem on the AMC-12B. Wait til tomorrow then we can discuss it. It's a nice problem.


calculus integration function derivative geometry rectangle trapezoid

Can someone explain to me the Fundamental Theorem of Calculus?

I can do the exercises in my textbook using it, but I have no intuitive or geometric understanding of it. In particular, I am confused about the change of variables. Why is there both a "t" variable and an "x" variable? Is the integral a function of x or t? Also, why can you arbitrarily choose a constant on the lower end of the integral and still not affect the answer to a problem?

Okay, this is going to be hard to explain without pictures to illustrate - but I'll try my best...

First of all, lets discuss why the Fundamental Theorem of Calculus is so important. Before The Fundamental Theorem of Calculus you have two completely different kinds of integrals. You have indefinite integrals (or anti-derivatives) and definite integrals. They are two very different things. The indefinite integral of a function is a function, whereas the definite integral of a function is a number. And they are defined in two very different ways. The fundamental theorem of calculus ties these two things together - making it possible to actually compute definite integrals without resorting to actually taking the limit of Riemann Sums.

Basically, the FTC gives you a fancy way of adding things up - without actually having to add them up!!

Okay, now that the motivation is out of the way, lets talk about what a definite integral is...

Intuitively you know that the definite integral of a positive function f(t) from t = a to b, tells you the "area" under that function f(t) between a and b. For a specific function f(t), and a and b, it gives you a number. Well first of all, there's nothing special about that variable t. It's just a "dummy variable". The function would still be the same if we replaced that t by an x or a y or anything else. It's just something used to signify that this is a function of this one thing, t, and wherever else I use it I mean the same thing.

Now the a and the b are specific set numbers. But if we replace that fixed number b with something that can move, say x. We don't get a number from the definite integral, we get something that changes with x, i.e. a function of x!!

Lets call this function G(x). So G(x) = the integral from a to x of f(t) dt. So what if we wanted to find the derivative of G(x)? Well we apply our old definition of derivative:

G'(x) = lim h->0 [G(x+h) - G(x)]/h

But what is G(x+h) - G(x) graphically? It is the area under f(t) between x and x+h! And as h gets smaller and smaller, this gets closer and closer to being like one of our rectangles from our Riemann Sum - with a height of f(x), and a width of h!

So G'(x) = f(x). This is the fundamental theorem of calculus. Lets take a second to think about what it means. It tells us that this thing, G(x), defined by a definite integral of a function, is equal to the anti-derivative (i.e. indefinite integral) of the same function! This is the first thing that you have to says that definite integrals have anything to do with anti-derivatives; and it is very good news! Because taking Riemann Sums is very hard (think Trapeziod Rule and Midpoint Approximation and all of that), but finding anti-derivatives is (well, relatively) easy.

The distinction between that x and the t though, is that G is a function of the x, which is in the endpoint of the integral, and t is just a dummy variable. It can be replaced by anything. It's almost like a place holder. And that number that's on the bottom, the a - that can be replaced by anything because all it is going to do is change G by a constant. It's basically the "constant of integration" that you get when you take the indefinite integral of f. But since the FTC says that G'(x) = f(x), you see that changing G by a constant doesn't affect the statement of the theorem. So in a sense, the lower endpoint, a, of the integral is sort of like a place holder as well. The x is the thing that "does all the work" and ties G and the f together. All the "action" happens at x!!

That was a really good explanation, Gauss.

The weird thing is TI-89s are sort of stupid. If you ever try int(x^2,x,0,x) or anything that has x in both the interval and the name. it will give you an error, although it is legal and should give you a 1/3*x^3.

No, the TI is right. While the variable of integration (the thing you integrate with respect to) is just a "dummy variable", you should never use that same variable in the endpoints of your integral. For one thing, when you use a variable in a problem it's understood that it should have the same context throughout the problem. You shouldn't use the same variable for two different things.

Second of all, from the standpoint of how the calculator computes integrals, you're telling it to use the same memory slot (x) for two different things.

You [i]can[/i] use a variable both within the integral and in the endpoints of integration. You just can't use it as the variable of integration. Once it's chosen, that variable has a specific meaning.

Thanks! That must mean that Maple is lying to me.  :cry:

I think I understand it now; the thing that I had the most difficulty understanding was how we went from the expression for G'(x) to just f(x). So-- let me restate it and correct me if I am wrong-- basically, as you said, G(x + h) - G(x) is the area from some number a to x + h minus the area from a to x, which gives us a rectangle with area f(x) * h, and lim (h->0) G(x + h) - G(x) / h = lim (h->0) f(x) * h / h = f(x)? Does this sound right?

Exactly! We're leaving out some of the pedantic details about how f(t) needs to be continuous and blah blah blah.. but basically that's it.


LaTeX
trigonometry
real analysis
real analysis unsolved

how would i cancel out (2n)! with (2(n+1))!; how would i manipulate the second term to make it easier

its a part a question which leads to

{ [(-1)^(n+1) x^(2(n+1))] / [(2(n+1))!] } { [(2n)!] / [(-1)^n x ^(2n) ]

also, how do you get the symbols like some people have in their post.  when i hover over these symbols it says stuff like what i have above so i'm guessing it's some sort of script they type.  it looks much neater that way and i havent been able to figure out how to do it

[quote="aitch"]

also, how do you get the symbols like some people have in their post.  when i hover over these symbols it says stuff like what i have above so i'm guessing it's some sort of script they type.  it looks much neater that way and i havent been able to figure out how to do it[/quote]

That's latex.  In the end everyone here ends up learning it.  There is a button if you scroll up : Latex help and it takes you to a wonderful part of this site explaining it [url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]

the point basically is that you put dollar signs before and after your math text
$\sqrt{x}$ is an example of what happens when you put \sqrt{x} between dollar signs

now about your question

$\frac{\frac{ (-1)^{n+1} x^{2(n+1)}} { (2(n+1))! } } { \frac{ (-1)^n x ^{2n} }{(2n)!}}$
was this what you meant

you probably are proving that the taylor sequence for cosine converges absolutely?

well $(2(n+1))!=(2 n) ! (2 n+1)(2n +2)$


superior algebra
superior algebra unsolved

Prove that the absolute Galois group of a number field has trivial center.

And something more difficult: Show that every automorphism of the absolute Galois group of $\mathbb{Q}$ is inner.


Which section of the aops forum would it be appropriate to ask where to find other other cool forums related to academics?

Hm.. Okay! :)

Aops has always been the main center for learning new math and such online, and I'm curious whether there are other cool academic websites that are focal points for learning.

What other academic related forums do you like?

[url=http://www.biology-online.org/biology-forum/]Biology-Online[/url] is a nice forum.

Of course [url=http://www.physicsforums.com/]Physics Forum[/url] is good for Physics.


Consider the expressions of the form  $ x\plus{}yt\plus{}zt^2$ with $ x,y,z$ rational numbers and $ t^3\equal{}2$.

Prove that:
If $ x\plus{}yt\plus{}zt^2\neq0$ then there $ u,v,w$ (rational numbers) such that $ (x\plus{}yt\plus{}zt^2)(u\plus{}vt\plus{}wt^2)\equal{}1$

Thanks

This follows from the fact that $ \{x + yt + zt^2: x,y,z\in\mathbb Q\}$ is the subfield $ \mathbb Q(t)$ of the reals and so every nonzero element has a multiplicative inverse. If you want to prove it directly, I suppose you could solve the following simultaneous equations for $ u,v,w$ in terms of $ x,y,z.$
\begin{align*} xu + 2zv + 2yw & = 1 \\
yu + xv + 2zw & = 0 \\
zu + yv + xw & = 0 \end{align*}


Prove that the product of $r$ consecutive integers is divisble by $r!$.

hmmm..

for r consecutive integers, at least one is divisible by r because from n to n+r (mod r), the range is from 0 to r-1 no matter what. The same can be used with integers from 1 to r, making the product of r consecutive integers divisible by r!

thats one way of looking at it, but more generally the formula for the the integer C( n, r ) shows that the product of $r$ consecutive integers is divisible by $r!$.


[quote="kimigtsc"]For how many positive integers a does there exist a real number x so that \sqrt{a-x}+\sqrt{a+x}= a?

A. 0
B. 1
C. 2
D. 3
E. 4[/quote]

For $\LaTeX$, embed your code with dollar signs, as follows:
[code]$\sqrt{a-x}+\sqrt{a+x}= a$[/code]

[hide="Anyways,"]$\sqrt{a-x}+\sqrt{a+x}= a$

Squaring both sides,
$a-x+2\sqrt{a^{2}-x^{2}}+a+x = a^{2}$
$a^{2}-a = 2\sqrt{a^{2}-x^{2}}$

Squaring both sides again,
$a^{4}-2a^{3}+a^{2}= 4(a^{2}-x^{2})$
$-a^{4}+2a^{3}+3a^{2}= 4x^{2}$

Notice that if the LHS is bigger than or equal to $0$, then we are done. Thus,
$-a^{4}+2a^{3}+3a^{2}\geq 0$
$a^{4}\leq 2a^{3}+3a^{2}$

Since $a\neq 0$ (it is positive), we may divide by $a^{2}$:
$a^{2}\leq 2a+3$
$a^{2}-2a-3\leq 0$
$(a-3)(a+1)\leq 0$
$-1\leq a\leq 3$

So, the values are:
$1, 2, 3$

So $3$, $(d)$[/hide]

I got the same answer, but a different set for $a$

[hide]
$\sqrt{a-x}+\sqrt{a+x}\leq 2\sqrt{a}\implies a \leq 2\sqrt{a}\implies a \leq 4$ (x=0 case)
$\sqrt{a-x}+\sqrt{a+x}\geq \sqrt{2a}\implies a \geq \sqrt{2a}\implies a \geq 2$ (x=a case)

$a \in \{2,3,4 \}$

$\boxed{D. \; 3}$

[/hide]

[hide]Square, $\implies 2a+2\sqrt{a^{2}-x^{2}}=a^{2}$.
Rearrange and square, $\implies 4a^{2}-4x^{2}=a^{4}-4a^{3}+4a^{2}$.
Cancel, rearrange, $a^{3}(4-a)=4x^{2}$.
Since the right side has to be non-negative, $a\in\{1,2,3,4\}$. $E.4$[/hide]

$\sqrt{1+x}+\sqrt{1-x}=1 \implies 2+2\sqrt{1-x^{2}}=1 \implies \sqrt{1-x^{2}}=-\frac{1}{2}$

Hence $a=1$ does not produce a real solution in $x$.

Refer to my previous post.


Prove or disprove:

For all primes p>=3, there exists a prime between (p-1)/2 and p.

The Goldbach conjecture is not allowed.

I thought of this problem while trying to solve another problem. I don't know the solution or whether there is some well known result that trivializes this conjecture.

Assume there does not exist a prime in that range. Take the largest prime less than or equal to $ \frac{p\minus{}1}{2}$, say prime $ m$. By [b]Bertrand's Postulate[/b] there exists a prime $ l$ between $ m$ and $ 2m$. Therefore, there exists a prime $ l$ greater than $ \frac{p\minus{}1}{2}$ (by construction of $ m$) but less than $ p$ as $ m\leq \frac{p\minus{}1}{2}$. Then $ l$ lies in that range from $ \frac{p\minus{}1}{2}$ to $ p$ and we get a contradiction. Therefore, there always exists a prime in that range.

Thanks a lot.

I need to learn more number theory.
 :)

Do you have a proof for Bertrand's Postulate , it is a famous method  :)

Here is a proof of Bertrand's Postulate: http://www.math.niu.edu/~rusin/known-math/97/bertrand. This is a very famous theorem, so searching the proof in google will give you what you want.

Oh yes ,rem . Some one can retype it on the different version (pdf,djvu), in this link, the proof is difficult to pronounce  :)

You can read on the book:Elementary theory of number' of Author Sierpinsky.

Thanks TTsphn. Can you upload this book to the forum ?  :D . I don't have much books .  :(


vector
parameterization
calculus
calculus computations

Say I have a plane:

$ 3x \minus{} 3y \plus{} 8z \equal{} 21$

and it's normal is therefore $ 3\vec{i} \minus{} 3\vec{j} \plus{} 8\vec{k}$

so if I want a vector parallel with the plane I should do a dot product of that normal vector with this unknown vector and it should be 0. However I am quite confused on how to do it....

Can anyone help?

So you want a vector $ \left< a, b, c \right>$ such that $ 3a \minus{} 3b \plus{} 8c \equal{} 0$.  It should not be hard to come up with small integers $ a, b, c$ such that this is true; speaking more systematically, you've got an overdetermined system and the set of all solutions has two free parameters, so you're free to set, for example, $ a \equal{} b \equal{} 1$ and solve for $ c$.

[b]1).[/b]$ 3x \minus{} 3y \plus{} 8z \minus{} 21 \equal{} 0\Leftrightarrow 3.(x \minus{} 2) \minus{} 3(y \minus{} 3) \plus{} 8.(z \minus{} 3) \equal{} 0.$
[b]2).[/b]$ \mbox{Let }P_0(2;3;3), P(x;y;z) \mbox{ and } \overrightarrow{v} \equal{} 3.\overrightarrow{i} \minus{} 3.\overrightarrow{j} \plus{} 8.\overrightarrow{k},\mbox{ then: }$
$ \overrightarrow{P_0P} \equal{} \overrightarrow{OP} \minus{} \overrightarrow{OP_0} \equal{} (x.\overrightarrow{i} \plus{} y.\overrightarrow{j} \plus{} z.\overrightarrow{k}) \minus{} (2.\overrightarrow{i} \plus{} 3.\overrightarrow{j} \plus{} 3.\overrightarrow{k}) \equal{} \\
\equal{} (x \minus{} 2).\overrightarrow{i} \plus{} (y \minus{} 3).\overrightarrow{j} \plus{} (z \minus{} 3).\overrightarrow{k}.$
[b]3).[/b]$ \overrightarrow{P_0P}\perp\overrightarrow{v}\Leftrightarrow\overrightarrow{P_0P}.\overrightarrow{v} \equal{} 0\Leftrightarrow 3.(x \minus{} 2) \minus{} 3(y\minus{} 3) \plus{} 8.(z \minus{} 3) \equal{} 0.$

how did you get the point 2,3,3 from? is it just a random guess??

I am still confused, if $ 3(x\minus{}2)\minus{} 3(x\minus{}3) \plus{} 8(x\minus{}3)$ is the answer, that is an equation of a plane and it's not a vector... Is it the same?

One thing is that, if you look at no.1, it doesnt have an y and z, because everything is there

$ 3(x \minus{} 2) \minus{} 3(y \minus{} 3) \plus{} 8(z \minus{} 3) \equal{} 0$ is not 'the answer', it is the equation of the original plane you gave.

I think you should give some more thought to t0rajir0u's post.  The vectors parallel to the plane are those $ \left < a,b,c\right >$ satisfying $ 3a \minus{} 3b \plus{} 8c \equal{} 0$.  Two such vectors are, for example, $ v_1 \equal{} \left < 1,1,0\right >$ and $ v_2 \equal{} \left < 0,8,3\right >$.  All linear combinations of these two vectors (i.e. $ c_1v_1 \plus{} c_2v_2$ where $ c_1,c_2$ are real numbers not both zero) are also parallel to the original plane (try to think about why this is true from both a geometric and algebraic point of view).

The equation that the components of the vector satisfy looks very similar to the original equation of the plane, and this is not by coincidence -- the set of vectors parallel to a plane themselves form a plane passing through the origin and parallel to the original plane.  Trying to visualize this might yield some insight that will make the algebra more transparent.

$ \mbox{If the vectors } \\
\overrightarrow{v_1} \equal{} \overrightarrow{i} \plus{} \overrightarrow{j} \mbox{ and } \overrightarrow{v_2} \equal{} 8.\overrightarrow{j} \plus{} 3.\overrightarrow{k} \\
\mbox{parallels to the plane } \alpha, \mbox{ then the vector: } \\
\overrightarrow{v} \equal{} \overrightarrow{v_1}\times\overrightarrow{v_2} \equal{} \left|\begin{array}{ccc}\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\
1 & 1 & 0 \\
0 & 8 & 3 \end{array}\right| \equal{} 3.\overrightarrow{i} \minus{} 3.\overrightarrow{j} \plus{} 8.\overrightarrow{k} \mbox{ is normal to the plane }\alpha.$


A polynomial $ f(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}c$ is such that $ b<0$ and $ ab\equal{}9c$. Prove that the polynomial $ f$ has three different real roots.

$ f(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}\frac{ab}{9}$

Let $ b\equal{}\minus{}d$ with $ d>0$

$ f(x)\equal{}x^3\plus{}ax^2\minus{}dx\minus{}\frac{ad}{9}$
If $ a\equal{}0$ it is very easy the proof.

Case 1: If $ a>0$
$ f(\minus{}a)\equal{}8ad$
$ f(0)\equal{}\minus{}ad$
$ f(\sqrt{d})\equal{}8ad$
Then $ f(x)$ has at least two different real roots, then it has three real roots.

Case2: If $ a>0$
$ f(\minus{}a)\equal{}8ad$
$ f(0)\equal{}\minus{}ad$
$ f(\minus{}\sqrt{d})\equal{}8ad$
Then $ f(x)$ has at least two different real roots, then it has three real roots.

We only have two prove that if it has two different real roots, then it has three different real roots.
By contradiction, suppose that it has two equal roots. Then:

$ f(x)\equal{}(x\minus{}r)^2(x\minus{}t)$
By Viette`s formulas:
$ a\equal{}\minus{}(2r\plus{}t)$
$ b\equal{}r^2\plus{}2tr$
$ ab\equal{}\minus{}9r^2c$

$ \implies \minus{}(2r\plus{}t)r(r\plus{}2t)\equal{}\minus{}9r^2t$
$ \implies (r\minus{}t)^2\equal{}0$
$ \implies r\equal{}t$
Which is an contradiction because $ f$ has at least two different roots.


inequalities trigonometry calculus derivative

$ x$ is real number such that $ 0 \le x \le \frac{\pi }{2}.$

Prove this inequality: $ \frac{{\sin x \plus{} 2\cos \frac{x}{2}}}{{\cos x \plus{} 2\sin \frac{x}{2}}} \ge 1 \plus{} \frac{{\sqrt 2 }}{2}$

It's very nice!!!  :D

Clue: You can use derivation to solve thí problem!!! :wink:

[hide="Along the lines of..."]We have equality for $ x=\frac{\pi}{2}$ We take the derivative of the expression, and we get ${ \frac{-2\cos x (\sin x+2\cos \frac{x}{2})}{(\sin x + 2\cos \frac{x}{2})^2}}$, which is obviously nonpositive for $ 0\le x \le \frac{\pi}{2}$. Thus the expression is stricly decreasing for $ 0 \le x \le \frac{\pi}{2}$, and the inequality holds.[/hide]

Although this problem is very easy, but can you give me a detail solution? Thank you very much!  :oops:

yep.

since the ineqaulity says $ \frac{{\sin x \plus{} 2\cos \frac{x}{2}}}{{\cos x \plus{} 2\sin \frac{x}{2}}}$ is greater or equal to $ 1\plus{}\frac{\sqrt{2}}{2}$, we must consider the lower bound.

a fraction gets small when the numerator is small and the denominator is large.

smallest when $ x\equal{}\frac{\pi}{2}$.

then the inequality becomes $ \frac{1\plus{}2*\frac{\sqrt{2}}{2}}{0\plus{}2*\frac{\sqrt{2}}{2}}\equal{}\frac{1\plus{}\sqrt{2}}{\sqrt{2}}\equal{}\frac{\sqrt{2}\plus{}2}{2}\equal{}1\plus{}\frac{\sqrt{2}}{2}$

Q.E.D.

easy for a pre-olympiad problem

That result was not very rigorous.  At least the calculus solution clearly demonstrates that it decreases.

With $ f(x) \equal{} \frac{{\sin x \plus{} 2\cos \frac{x}{2}}}{{\cos x \plus{} 2\sin \frac{x}{2}}}$, I solve to $ f'(x) \equal{} \frac{{\sin \frac{{3x}}{2} \minus{} 1}}{{\left( {\cos x \plus{} 2\sin \frac{x}{2}} \right)^2 }}$  :wink:


A school is creating a four-digit student code system. For security reasons, the code cannot start with an even number. How many codes are possible?

There are $ 5$ possibilities for the first digit and $ 10$ possibilities for each of the other digits. $ 5\times{10^{3}}\equal{}\boxed{5000}$.

:)

Exactly, gaussintraining

5 for first because 1,3,5,7,9 are odd, so 5*10*10*10...
Very weird "security", don't you think?


I think Intro to NT has a typo but i'm not sure.
It states that the binomial expression when $ \ 10_(b \plus{} 1)$
When b is a positive integer,
$ \ (b \plus{} 1)^{2} \equal{} b^{2} \plus{} b \plus{} 1$ I thought it should be $ \ b^{2}\plus{}2b\plus{}1.$
Am i right?

[quote="cgyao15"]
It states that the binomial expression when $ \ 10_(b \plus{} 1)$
[/quote]

what does that mean?

He means to say $ 10_{b \plus{} 1}$ which is equivalent to $ 1\cdot (b \plus{} 1) \plus{} 0 \equal{} b \plus{} 1$. From there, I assume that you are considering the square of $ 10_{b \plus{} 1}$ in general which is $ (b \plus{} 1)^2 \equal{} b^2 \plus{} 2b \plus{} 1.$ This is most likely a typo so I recommend that notify AoPS staff.

What page is this on? I will check to see if I also have that typo. If not, it means that you have an older edition.

According to [url=http://www.artofproblemsolving.com/BookLinks/IntroNumberTheory/links.php#errata]this,[/url] it seems like they haven't caught onto it yet.

@b555 err yes i put parenthesis instead of curvy brackets.
Thx ne ways guys.


I wish the board wouldn't automatically close my scripts.  That is, when I post italics, which begin with "[i ]," I have a habit of typing in the [/i].  It's really annoying when that also appears at the end of my post, because the board auto-closed the script.

Would it be able to make that feature an on-off feature?

I have never enoutered this [i]problem[/i] :) Maybe it's you doing it wrong?  :?

Yeah, it happens all the time to me also. And I've seen it many times on other peoples post - a [ /spoiler ] or something like that which just appears. Surely it wouldn't be too hard to either stop it doing that altogether or make it actually check they aren't closed in the first place.

edit - note that it only happens when you click the [b]B[/b] button or any of the buttons above the message but close it by hand yourself. Not if you type both or if you click and unclick both.

[quote="TripleM"]edit - note that it only happens when you click the [b]B[/b] button or any of the buttons above the message but close it by hand yourself. Not if you type both or if you click and unclick both.[/quote]Ah wait, that's a totally different story. If you use the button once, you make sure you use it again.

[code]After all it's harder to type [/b] than [b] isn't it?! ...[/code]


geometry
trigonometry
incenter
parallelogram
inradius
circumcircle
angle bisector

It's simple to prove that $ a \sin\frac{\gamma}2 \cos\frac{\beta}2 \equal{} (p\minus{}b)\cos\frac{\alpha}2$ cause it's equal to $ a \sqrt{\frac{(p\minus{}b)(p\minus{}a)}{ab}} \sqrt{\frac{p(p\minus{}b)}{ac}} \equal{} (p\minus{}b) \sqrt{\frac{p(p\minus{}a)}{bc}}$.

Then $ CQ \equal{} \frac{\overline{CC'} \sin\angle CC'Q}{\sin \angle CQC'} \equal{} \frac{(p\minus{}b)\cos{\frac{\alpha}2}}{\cos\frac{\beta}2}\equal{}a \sin\frac{\gamma}{2} \equal{} \overline{BC} \cos \angle BCQ$ and so $ BQ \perp CQ$ and $ BQ \parallel CI$ and equally we can say that $ CP \parallel BI$ and so that the triangle $ \triangle BIC$ and $ \triangle BI_aC$ are equal.

[quote="Amir.S"]Let $ I_{a}$ be excenter of triangle $ ABC$ wrt $ A$.excircle of triangle $ ABC$ wrt $ A$ is tangent to $ AB$ and $ AC$ at points $ B'$ and $ C'$.Let $ P$ and $ Q$ be intersecttions of lines $ BI_{a}$ and $ CI_{a}$ with $ B'C'$ respectively and let $ M$ be intersection of $ PC$ and $ BQ$. Prove that if $ L$ is the foot of the prependicular drawn from $ M$ to $ BC$ then $ ML \equal{} r$($ r$=inradi)[/quote]
[hide="Solution"]
Let $ I$ be the incenter of $ \triangle ABC$, let the incircle be tangent to $ BC$ at $ A'$, and let the midpoint of $ BC$ be $ X$.  Also, let $ \angle BAC \equal{} 2\alpha$ and $ \angle ABC \equal{} 2\alpha$, so $ \angle ACB \equal{} 180 \minus{} 2\alpha$.  Also, $ I$ and $ I_a$ lie on the angle bisector of $ \angle BAC$, so $ A$, $ I$, and $ I_a$ lie on a line.  Now, notice that $ \angle IBA \equal{} \angle IBC \equal{} \beta$, $ \angle IAB \equal{} \angle IAC \equal{} \alpha$, $ \angle ICA \equal{} \angle ICB \equal{} 90 \minus{} \alpha \minus{} \beta$, $ \angle B'BC \equal{} 180 \minus{} 2\beta$, and $ \angle BCC' \equal{} 2\alpha \plus{} 2\beta$.  It follows that $ \angle I_aBB' \equal{} \angle I_aBC \equal{} 90 \minus{} \beta$ and $ \angle I_aCB \equal{} \angle I_aCC' \equal{} \alpha \plus{} \beta$.  Thus, $ \angle IBI_a \equal{} 90 \minus{} \beta \plus{} \beta \equal{} 90$ and similalry, $ \angle ICI_a \equal{} 90$.  Thus, $ ICI_aB$ is cyclic.  It follows that $ \angle BI_aI \equal{} \angle ICB \equal{} 90 \minus{} \alpha \minus{} \beta$ and $ \angle II_aC \equal{} \angle IBC \equal{} \beta$.  Since $ AC'$ and $ AB'$ are tangents to the $ A$-excircle, we see that $ AC' \equal{} AB'\implies \angle AB'C' \equal{} \angle AC'B' \equal{} 90 \minus{} \frac {\angle BAC}{2} \equal{} 90 \minus{} \alpha$.  It follows that
\[ \angle BB'Q \equal{} 90 \minus{} \alpha \equal{} 180 \minus{} (90 \plus{} \alpha) \equal{} 180 \minus{} (\frac {\angle BAC}{2} \plus{} 90) \equal{} 180 \minus{} \angle BIC \equal{} \angle BI_aC
\]so $ BRI_aB'$ is cyclic.  Thus, $ \angle BRI_a \equal{} 180 \minus{} \angle I_aB'B \equal{} 90\implies BR\perp I_aC$ and similarly, $ CQ\perp I_aB$.  This means that $ M$ is the orthocenter of $ \triangle I_aBC$, so $ I_aM\perp BC$.  It follows that $ L$ is the tangency point between the $ A$ excircle and $ BC$.  Also, $ BR\perp I_aC$ and $ IC\perp I_aC$, so $ MB\parallel IC$ and similarly, $ IB\parallel MC$, so $ ICMB$ is a parallelogram, from which it follows that $ IM$ passes through the midpoint of $ BC$, which is $ X$.  Since $ L$ is the $ A$-excircle tangency point with $ BC$, we see that $ A'X \equal{} XL$, $ \angle MLX \equal{} 90 \equal{} \angle IA'X$, and $ \angle IXA' \equal{} \angle MXL$, so $ \triangle MXL\cong \triangle IXA'\implies r \equal{} IA' \equal{} LM$, as desired. [/hide]

[quote="Amir.S"] [color=darkred]The $ A$ - exincircle $ w_a \equal{} C(I_a,r_a)$ of $ \triangle ABC$ is tangent to $ AB$ , $ AC$ at the points $ B'$ , $ C'$ . Denote the incircle

$ w \equal{} C(I,r)$ of $ \triangle ABC$ and $ \begin{array}{ccc} P\in BI_{a}\cap B'C' & ; & Q\in CI_{a}\cap B'C' \\
 \\
M\in PC\cap BQ & ; & L\in BC\ ,\ ML\perp BC\end{array}$ . Prove that $ ML \equal{} r$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b]

$ \blacktriangleright$ The point $ I$ is the orthocenter of the triangle $ I_aI_bI_c$ (evidently). Prove that the quadrilateral $ BPQC$ is cyclically.[/color]

[hide="Proof."]
[color=darkblue]$ \left\|\begin{array}{c} m(\widehat {BB'P}) \equal{} m(\widehat {BB'C'}) \equal{} 90^{\circ} \minus{} \frac A2 \\
 \\
m(\widehat {B'BP}) \equal{} m(\widehat {B'BI_a}) \equal{} 90^{\circ} \minus{} \frac B2\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c} m(\widehat {QPI_a}) \equal{} m(\widehat {B'BP}) \equal{} 90^{\circ} \minus{} \frac C2 \\
 \\
m(\widehat {BCQ}) \equal{} m(\widehat {BCI_a}) \equal{} 90^{\circ} \minus{} \frac C2\end{array}\right\|$ $ \implies$

$ \widehat {QPI_a}\equiv\widehat {BCQ}$ , i.e. $ BPQC$ is cyclically.[/color][/hide]
[color=darkblue]$ \blacktriangleright$ Denote $ T\in I_aM\cap BC$ . Show that $ \frac {TB}{TC} \equal{} \frac {p \minus{} c}{p \minus{} b}$ , i.e. $ \overline {I_aMT}\perp BC$ (the point $ T\equiv L$ and belongs to the circle $ w_a$ ).[/color]

[hide="Proof."]
[color=darkblue]$ \left\|\begin{array}{c} \frac {PB}{PI_a} \equal{} \frac {B'B}{B'I_a}\cdot\frac {\sin\widehat {PB'B}}{\sin\widehat {PB'I_a}} \equal{} \frac {p \minus{} c}{r_a}\cdot\frac {\sin\left(90^{\circ} \minus{} \frac A2\right)}{\sin\frac A2} \equal{} \frac {p \minus{} c}{r_a}\cdot\cot\frac A2 \\
 \\
\frac {QC}{QI_a} \equal{} \frac {C'C}{C'I_a}\cdot\frac {\sin\widehat {QC'C}}{\sin\widehat {QC'I_a}} \equal{} \frac {p \minus{} b}{r_a}\cdot\frac {\sin\left(90^{\circ} \minus{} \frac A2\right)}{\sin \frac A2} \equal{} \frac {p \minus{} b}{r_a}\cdot\cot \frac A2\end{array}\right\|$ $ \implies$ $ \frac {PB}{PI_a}\cdot\frac {QI_a}{QC} \equal{} \frac {p \minus{} c}{p \minus{} b}$ .

Apply [b]Ceva's theorem[/b] to $ M$ in $ \triangle BCI_a$ : $ \frac {PB}{PI_a}\cdot\frac {QI_a}{QC}\cdot\frac {TC}{TB} \equal{} 1$ $ \implies$ $ \frac {p \minus{} c}{p \minus{} b}\cdot\frac {TC}{TB} \equal{} 1$ $ \implies$ $ \frac {TB}{TC} \equal{} \frac {p \minus{} c}{p \minus{} b}$ .[/color][/hide]
[color=darkblue]$ \blacktriangleright$ Apply[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=202247][color=red][b]lemma I[/b][/color][/url] [color=darkblue]to the triangle $ BCI_a$ , where $ I_aL\perp BC$ and for the point $ M\in I_aL$ the quadrilateral $ BPQC$

is cyclically. In conclusion, the point $ M$ is the orthocenter of $ \triangle BCI_a$ , i.e. $ CM\perp BI_a$ and $ BM\perp CI_a$ .[/color]

$ \blacktriangleright$ [color=darkblue]Apply[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=202252][color=red][b]lemma II[/b][/color][/url] [color=darkblue] to  $ \triangle I_aI_bI_c$ with the orthocenter $ I$ , where $ M$ is the orthocenter of $ \triangle BCI_a$ .

In conclusion, $ ML$ is equally to the inradius of the orthic $ \triangle ABC$ w.r.t. the triangle $ I_aI_bI_c$ , i.e. $ ML \equal{} r$ .[/color]

We denote as $ P,\ Q,$ the orthogonal projections of $ C,\ B$ respectively and it is easy to show that the points $ B',\ P,\ Q,\ C',$ are collinear.

$ ($ as [b][size=100]SpongeBob[/size][/b] wrote in the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199330]Excircle & collinear[/url] lately, we have $ \angle C'QC \equal{} \angle LQC \equal{} \angle LBI_{a} \equal{} \angle B'BI_{a} \equal{} \angle B'QI_{a}$ $ \Longrightarrow$ $ B',\ Q,\ C',$ are collinear and similarly the points $ B',\ P,\ C',$ are collinear $ ).$

So, the point $ M\equiv BQ\cap CP,$ is the orthocenter of the triangle $ \bigtriangleup I_{a}BC$ and then, the point $ L,$ as the orthogonal projection of $ M,$ on $ BC,$ is the tangency point of the $ A$-excircle to $ BC.$

It is well known that $ KL \equal{} KL'$ $ ,(1)$ where $ K$ is the midpoint of the side-segment $ BC$ and $ L'$ is the tangency point of the incircle $ (I),$ to $ BC.$

Because of $ IB\perp BI_{a}$ and $ IC\perp CI_{a},$ we have that the quadrilateral $ IBI_{a}C$ is cyclic with circumcircle $ (A'),$ taken as diameter the segment $ II_{a}.$

We denote the point $ M'\equiv (A')\cap I_{a}M$ and so, we have $ ML \equal{} LM'$ $ ,(2)$

Because of $ (1)$ $ \Longrightarrow$ $ LM' \equal{} IL' \equal{} r$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ ML \equal{} r$ and the proof is completed.

Kostas Vittas.


Given $ a, b, c \geq\ 0$ satisfy $ a^2\plus{}b^2\plus{}c^2\equal{}2(ab\plus{}bc\plus{}ca)$.
Prove that: $ \sqrt{1\plus{}\frac{a}{b\plus{}c}}\plus{}\sqrt{1\plus{}\frac{b}{c\plus{}a}}\plus{}\sqrt{1\plus{}\frac{c}{a\plus{}b}} \geq\ 1\plus{}2\sqrt{2}$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} bc \plus{} ca)$.
Prove that: $ \sqrt {1 \plus{} \frac {a}{b \plus{} c}} \plus{} \sqrt {1 \plus{} \frac {b}{c \plus{} a}} \plus{} \sqrt {1 \plus{} \frac {c}{a \plus{} b}} \geq\ 1 \plus{} 2\sqrt {2}$
 :)[/quote]
It's stronger than the well-known of Le Trung Kien:
Given $ a, b, c \geq\ 0$ and $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that:
$ \frac{1}{\sqrt{a\plus{}b}}\plus{}\frac{1}{\sqrt{b\plus{}c}}\plus{}\frac{1}{\sqrt{c\plus{}a}} \geq\ 2\plus{} \frac{1}{\sqrt{2}}$
 :)


Hi all,

I am a college student with a very quantitative background. In the past couple years, I have been interested in going down the career path of a quant. However, recently, after doing more detailed investigation into the different possible careers in finance, I have heard that a lot of quants who have been in the industry for a few years decide that they want to move into the role of a trader. I have also heard that quants often play second fiddle to traders (the work that quants do is supposed to support traders after all), and thus end up being less respected and receiving less compensation than traders.

Therefore, I would like to ask those of you in the know about the comparison between quants and traders (well, the more quantitative traders, that is). What are the pros and cons of each type of job? Is it easy to switch between the two after working in the industry for a couple of years? How is each role generally compensated (of course, presumably we can only compare quants and traders working at the same firm)?

Any thoughts/input/advice you may have would be appreciated. Thanks!

This is going to vary significantly depending on the firm you work for. There's not going to be a straight and simple answer.

Well, I don't really have a specific firm in mind to be honest. I guess two different classes of firms I am wondering about would be investment banks (Goldman, Morgan Stanley, etc...) and the larger, more well-known quant shops/hedge funds (Shaw, Jane Street, Citadel, Two Sigma, etc...). However, I do not know whether there is an answer that would apply to all/most of the firms in either of these two categories.

Hey guys, I'm double majoring in business(emphasis finance) and mathematics. I'm in my second year and I still don't know what I want to do. Can you tell the difference between a quant and a trader. Also does a trader need to have a mathematical baackgroud? And what is the level of math needed for a quant, and for a trader?

thank you...


You can do it almost the same way as you normally would [b][i]except[/i][/b] put in a couple of carriage returns after the starting dollar sign. Thus
[code]$\textdollar$

\begin{align*}
\left(x^n\right)' &= nx^{n-1} & \left(e^x\right)' &= e^x & (\sin x)' &= \cos x \\
\left(\frac{1}{x^n}\right)' &= -\frac{n}{x^{n+1}} & \left(a^x\right)' &= a^x\ln a & (\cos x)' &= -\sin x
\end{align*}$\textdollar$[/code] produces
\begin{align*}
\left(x^n\right)' &= nx^{n-1} & \left(e^x\right)' &= e^x & (\sin x)' &= \cos x \\
\left(\frac{1}{x^n}\right)' &= -\frac{n}{x^{n+1}} & \left(a^x\right)' &= a^x\ln a & (\cos x)' &= -\sin x
\end{align*}

(Example taken from Kopka & Daly's A Guide to LaTeX)

[quote="Negative_3"]Can you align equations in this forum?

In LaTeX you use the &.[/quote]
..
use equnarry, or you use it in texnincenter and upload the pdf file here.[hide][/hide]


LaTeX
videos
algebra
binomial theorem
Pascal\u0027s Triangle

Using the Binomial Theorem Part 4

We use the Binomial Theorem to prove a pattern in Pascal's Triangle

For the first pattern ($ 2^n$), I found another explanation. Let's look at the transition between the 3rd and 4th rows:

[color=orange]1[/color] [color=violet]3[/color] [color=blue]3[/color] [color=brown]1[/color]
([color=orange]1[/color]) ([color=orange]1[/color]+[color=violet]3[/color]) ([color=violet]3[/color]+[color=blue]3[/color]) ([color=blue]3[/color]+[color=brown]1[/color]) ([color=brown]1[/color])

When we add the numbers in the 4th row, we see that it is twice the 3rd row, because each number in the 3rd row is found twice in the 4th row.

For the second pattern (the alternating subtraction), we can simplify each line to (x-y)^n. For example. (x-y)^3 is
(3 c 0)x^3 - (3 c 1)(x^2)y + (3 c 2) (x)(y^2) - (3 c 3)y^3

Let x = y = 1, and so we have (0)^3 = 1-3+3-1 = 0

This will work for all rows of the triangle, because (1-1) ^n  = 0

Please forgive the lack of Latex

[quote="trivea"]For the second pattern (the alternating subtraction), we can simplify each line to $ (x - y)^n$. For example. $ (x - y)^3$ is
$ \binom{3}{0}x^3 - \binom{3}{1}x^2y + \binom{3}{2}xy^2 - \binom{3}{3}y^3$

Let $ x = y = 1$, and so we have $ (0)^3 = 1 - 3 + 3 - 1 = 0$

This will work for all rows of the triangle, because $ (1 - 1) ^n = 0$

Please forgive the lack of Latex[/quote]
$ \text{\LaTeX}$'d  :)

Mr. Rusczyk, please make more videos!!!!!

Another way we could think about the 2^n power pattern would be to use the binomial theorem verbatim, i.e. with the (x+y) instead of the (x-y) format, but substitute y=-1.

P.S. the videos were great, it's a shame Mr. Rusczyk won't be producing any more :(

[quote="rashi101"]Yes, that was easy. The harder question is: is there a counting problem that proves it? I can't think of any... :maybe:  :wink:[/quote]

The entries in the nth row of Pascal's triangle are simply nC0, nC1 ... nCn. Therefore the sum of all of them is nC0 + nC1 + ... + nCn, or the numbers of ways to choose a subset from a set of n elements as the sum represents the number of ways to choose a set of 0 elements plus for 1 element, 2 elements and so on up to n. To choose a set there are two options for each element (to include it or not), therefore there are 2^n ways to choose a subset. Thus the sum in a row will simply be 2^n.

There's the counting argument for it :)


A number $ x$ is twice its reciprocal. What is $ x^6$ ?

We know that:

$ x \equal{} \frac {2}{x}$

Multiply both sides by $ x$.

$ x^2 \equal{} 2$

$ x \equal{} \sqrt {2}$

$ \sqrt {2}^6 \equal{} \boxed{8}$

EDIT: Oops. As Dr Sonnhard Graubner said, the other solution is $ \minus{}\sqrt{2}$, which, when raised to the 6th power, also yields 8.

hello isabella, $ x\equal{}\minus{}\sqrt {2}$ is also a solution of your problem,
$ \minus{}\sqrt{2}\equal{}\frac{2}{\minus{}\sqrt{2}}\equal{}\frac{2\sqrt{2}}{\minus{}2}\equal{}\minus{}\sqrt{2}$ and
$ \left(\minus{}\sqrt{2}\right)^6\equal{}8$.
Sonnhard.

Oh... I see. $x^6$ can be converted to that.


[quote="stvs_f"]There are a finite number of red and blue lines in the plane, no two parallel. There is always a third line of the opposite color through the point of intersection of two lines of the same color. Show that all the lines have a common point.[/quote]
From condition ,for each color there exist at least a line that colored in this color . A hint to consider two lines that have different color and angle between them is smallest . All line must be pass through the intersect of two lines .


logarithms
limit
calculus
derivative
calculus computations

Does there exist a positive real number k such that for all natural n, log(n) is bigger than $ n^k$?

no. by L'Hopital's Rule, $ \lim_{x\to \infty} \frac{\log x}{x^k} \equal{} \lim_{x\to \infty} \frac{1}{kx^k}\equal{} 0$.

Let $ \sim$ mean comparison:
$ \log n \sim n^k$
$ k \log n \sim k n^k$
$ \log n^k \sim k n^k$
$ \log x \sim k x$

Note that the right side is always increasing at rate $ k$ w.r.t. $ x$ while the left side's derivative decreases to $ 0$, so RHS eventually surpasses LHS.


hey hey hey ,Who had chosen OMID HAMATI as the moderator of IRAN community ,

 :P  :P  Just kidding ,Congratulations  OMID for being a moderator ,but who on the earth had chosen you ????? :!:  :?:

ANSWER to my question:

One of the ADMINS: we chosed him  :mad:

My answer:
I was just kidding ,but you should have taked votes for chosing the moderator, :D  :D

PS.in general just a joke

Haha, cool, but the admins ALWAYS decide who will become moderator, you see :)

[They are ALMIGHTY :D]

Arne, I was wondering if i can become a moderator. ;)

the moderators are cool in math but i think if you use the others too,it will be better and more useful,why don't you test it? :D

I don't understand what you are saying...

lomos_lupin I have choosen Omid Hatami. I hope you have no problem with that. I highly respect his professionalism.

PS In choosing the moderators there is no democracy on this board. It's pure evil :diablo:  :evilgrin: :P

[quote="cezar lupu"]Valentin, I'm interrested in becomming a moderator. Do you need one? :?[/quote]I doubt the Iranians would like you, Cezar :)

I was wondering if you need moderators for Romanian Section, or inequalities or algebra?

That is not a discussion relevant for this topic.

And since there isn't any thing left to say about this topic "who had chosen this moderator?" (the answer is above :D), I declare topic closed ;)


limit
trigonometry
real analysis
real analysis solved

If $a_{n}={ \prod_{k=1}^{n}} cos(2^{k-1}*x)$,  $x{\neq}k{\pi}$ then what is the limit of $ \lim_{n {\rightarrow} {\infty}} a_{n} ?
$

does anyone have a clue how it can be solved ?

Evaluate the finite products using the identity $\cos x=\frac12\cdot\frac{\sin 2x}{\sin x}$.

$a_n=\prod_{k=1}^n \cos (2^{k-1} x)=\left( \frac{1}{2} \right)^n \frac{\sin 2^n x}{\sin x} $.
$ \sin x \ne 0 \, \mbox{is}\,\mbox{fixed} \, \mbox{here}$.
$a_n=\left( \frac{1}{sin x} \right)\,\left( \frac{1}{2} \right)^n\,\sin 2^n x$.
$\mbox{Since} \lim_{n \to \infty}\left( \frac{1}{2} \right)^n=0$
$\mbox{and}\left| \sin 2^n x \right| \le1$.
$\mbox{then} \lim_{n \to \infty}a_n = \left( \frac{1}{sin x} \right)\,*\mbox{0}=0$.

And if $x=k\pi,$ we have

$\cos(k\pi)\cos(2k\pi)\cos(4k\pi)\cdots=(-1)^k\cdot1\cdot1\cdots=(-1)^k.$

If $x=2^{-j}k\pi$ for some odd $k$ and some $j>0$, then franck's work is still valid. In that case, the sequence $a_n$ becomes zero at some point and remains zero afterwards.


I am contemplating on whether to take the fermat or the cayley.

For cayley, I usually finish the whole contest in less than a half an hour, but I might have trouble if there is a problem I can’t solve or if its a tough contest.
For fermat, I usually get perfect on the practice tests, but it will definitely be harder.

Should I take the cayley and try to get perfect or take the fermat for a challenge?

Fermat. If you come top in your province you get 200 dollars and an invitation to the Waterloo Seminar.

Depends on what your goal is.  I believe that in previous years, cayley winners have also been invited to Waterloo seminars.  Most students in your position would go with Fermat... it also depends on your geographic location.  If you live in Saskatchewan or something... go for it.  If you live in Toronto, you'll have a tougher time coming home with cash.

lol im not in it for the money (though it would be nice). maybe i should take advantage of the fact that all the brilliant mathematicians are now in gr 12. i live in toronto, so a perfect score is probably needed to become the provincial champion. Do you get an invitation only if your the provincial champion or can ppl who get 144 still get it?

They invite on cayley too you know.  Everyone seems to think its Fermat only.  Its pascal they don't invite on at all.  Hehehe its your choice.  Good luck on whichever one you decide to take!

OMG!  Sarah got a new avatar!  It's so... infectious!

And in other news... apparently she agrees with me that they also invite on Cayley... so write whatever you feel like.

[quote="Sunny"]They invite on cayley too you know.[/quote]Apparently not. Last year I was of your opinion so I asked my math teacher to send an email.[quote]Steven’s score of 150 on the Cayley contest was outstanding. However, we invited very few Cayley contestants to the seminar, and those invitations were based on a number of criteria.

Once we have processed the results of the Fryer/Galois/Hypatia/Euclid contests, we will be issuing some more invitations. It is possible, though not guaranteed, that Steven could be invited based on those results.

Many of the seminar invitees were Fermat contestants. I hope that Steven will continue with his excellent participation in the contests and write the Fermat contest next year.

Sincerely,
Judith Koeller[/quote]

I was invited to Seminar twice based on Cayley (and the Grade 10 invitational when they still had it).  Its true that they don't invite as many as Fermat, but more people probably take Fermat.  You know...just cuz you get a certain score doesn't mean you'll be invited for sure.  No one really knows for sure how people are invited.  Honestly, every year there seems to be a few people who really shouldn't have been there, and lots of people who should have.  Then again, the organizer changed last year, so maybe that's why priority is given to Fermat.  Hehehe, though the Seminar wasn't any different, we still had Ian around to bug.

But if they are inviting based on the awards they give out, then much more people should have been invited based on Euclid, especially those in grade 12, since the Euclid is the only contest they can write.  Ah well.  We'll see what happens this year.

Good luck on the Fermat tomorrow everybody! :nhl:

If theres no chance at getting an invitation for writing the Cayley then i should probably write the fermat.
Last year about how many people were invited to the seminar and how many were in gr 10?

BTW, Good luck on the contests 2morw!

is there a way to get invited to the Waterloo Seminars with 144 in Fermat, ie. Open score, Hypatia score, etc.

ps have they posted solutions to this year's Fermat yet

It's best not to discuss the contest on the forum immediately after it's written, for some people may not have taken it yet (for whatever reason).

Hey, could someone message me how they did #25?

[quote="Sunny"][quote="chilly_snow"]Approximately how many people are invited to the Waterloo Seminar each year? Do they invite from COMC?[/quote]

60 ish people.  No, they don't invite based on COMC.[/quote]
oh ya.. what about CMO? i forgot to ask that... :P
(I didn't do so well on the Fermat contest.. made two really STUPID mistakes...  :blush: )

[quote="chilly_snow"]
oh ya.. what about CMO? i forgot to ask that... :P
(I didn't do so well on the Fermat contest.. made two really STUPID mistakes...  :blush: )[/quote]

Not really.  If you happen to be there, then they'll give you the awards for CMO.  They might give invites to 1st, 2nd and 3rd, but chances are they were invited anyways based on other contests.

Write Fermat. If you don't win, you still gain experience to allow you to win next year :D

i would write cayley.
but for some ambitious people they might choose fermat...
so it's up to you
and remember if u have the ability to get perfect, the grade difference won't trouble u


How and when will my math teacher, who runs the AMC/AIME at my school, be notified that I qualified for the AIME through the USAMTS?

If you have completed the CEEB field and the teacher field in your Profile on the USAMTS page, they will be contacted by the AMC some time this week (if you have not filled in those fields, please let me know - I have already passed the info to the AMC, and if you did not have these fields completed, you were not designated as an AIME qualifier).


number theory
greatest common divisor
modular arithmetic
quadratics

:lol:   Suppose x,y,z are three integers which are in arithmetic progression. If x is of the form 8n + 4 where n is an integer and each of y,z is expressible as a sum of squares of two integers, show that [b]gcd[/b] (x,y,z) cannot be odd.

Since x is even according to the given condition,hence it does not have any odd factor hence ....... G.C.D cannot be odd.

The fact that x is even doesn't mean that it won't have any odd factors.

[hide]Clearly $x$ is even, and if $y$ is even, then $z$ will also have to because the common difference would be even; this would make $\gcd (x,y,z)$ even.

The squares mod 8 are 0, 1, and 4, so $y$ and $z$ must each be 0, 1, 2, 4, or 5 mod 8.

Suppose $y$ were odd. Then $y$ would have to be 1 or 5 mod 8. But $y\equiv 1\Rightarrow z\equiv 6$ and $y\equiv 1\Rightarrow z\equiv 6\pmod{8}$ so $z$ is not a sum of two squares. Contradiction.

Therefore $y$ is even, so so will $z$, making $\gcd(x,y,z)$ not odd.[/hide]

Let's play with mod :

$x = 8n+4$ (*)
$y = 8n+4 + r = a^2 + b^2$ (**)
$z = 8n+4 + 2r = c^2 + d^2$ (***)

Now gcd is odd iff $r$ is odd

$c^2+d^2$ is even so $mod \ 8$ it must be $0$ or $2$ (looking quadratic residues $mod\ 8$). Only possibilty with (***) is $r = 1 \ mod \ 8$

Now an odd sum of two square (**) must be $1$ or $5$ mod 8.

$8n+4 = c^2+d^2 - 2 (a^2+b^2)$ would give $4 = 0\ mod \ 8$ contradiction


Find the smallest value of x satisfying the condition x^3+2x^3=a, where x is an odd integer, and a is the square of an integer.

[hide]$ 3x^{3}=b^{2}\Rightarrow (x)^{2}(3x)=b^{2}$, so $ 3x$ must be a square, clearly, $ x=3$.[/hide]


Find, with proof, all real numbers $x \in \lbrack 0, \frac{\pi}{2} \rbrack$, such that $(2 - \sin 2x)\sin (x + \frac{\pi}{4}) = 1$.

If $y=\sin (x+\frac{\pi }{4})$ then $\sin{2x}=2y^2-1$.
It give equation $2y^3-3y+1=0$ with solution $y=1,y=\frac{-1\pm \sqrt 3 }{2}.$


function
real analysis
real analysis unsolved

Let  $f: [a;b] \rightarrow [a;b]$ to be a continuous function.Prove that f has a fixed point.

Study the function $g(x)=f(x)-x$
 :cool:

To make it complete: $g(a)\geq 0$ and $g(b)\leq 0$. Therefore, because g is continuous (as a difference of 2 continuous functions), we can find a point c in [a,b] such that g(c)=0, i.e. f(c)=c.


[hide="hint!"]
enough to prove that:

$ \left[\frac{n}{2}\right]+\left[\frac{n}{2^{2}}\right]+\cdots+\left[\frac{n}{2^{e_{1}}}\right]=n-r$
[/hide]

Another way

[hide] Let $ D(n)=\alpha$ such that $ \alpha$ is the greatest non-negative integer such that $ 2^{\alpha}$ divides $ n$. Prove by induction that $ D(n!) = n-r$ where $ r$ is the number of $ 1$'s in the binary representation of $ n$.
 [/hide]


[quote="friendlist"]Find all functions: $ f: R \to R$ such that $ f(x \plus{} y) \equal{} x^2.f(\frac {1}{x}) \plus{} y^2f(\frac {1}{y})$for all $ x;y \in R^*$[/quote]
Please lock this topic. This is problem in this issue in Mathematics and Young Magazine.

My solution is very long, if I have time i will post my solution.

My result: $ f(x) \equal{} {\rm{ax}}$ with $ a \in R,x \in R*$


logarithms
function
induction
calculus
derivative
calculus computations

Study the convergence of $(x_{n})_{n\geq0}$ defined by: $x_{0}=\frac{1}{2}$ and $5^{x_{n}}=2x_{n-1}+3$ for $n\in [1,\infty)$.

We have that $x_n = \frac{\ln(2x_{n-1} + 3)}{\ln 5}$. Or we could say $x_n = f(x_{n-1})$ where $f(x) = \frac{\ln(2x + 3)}{\ln 5}$. f is obviously a continous strictly increasing function for all $x > -\frac{3}{2}$. A obvious proof by induction shows for example that $x_n > 0$, so we have no problems with coming into undefined numbers.

$(x_n)$ is a increasing sequence:
$n=1$: $x_0 = \frac12$, $x_1 = \frac{\ln 4}{\ln 5} > \frac12 = x_0$.
Assume that $x_{p+1} > x_{p}$ for some $p$. Then $x_{p+2} = f(x_{p+1}) > f(x_p) = x_{p+1}$ since $f$ is strictly increasing. The principle of induction gives us our conclusion.

All possible values that $x_n$ might converge to are roots of the equation $A = f(A), A > \frac12$. Set $g(A) = A - f(A)$, so that $g'(A) = {\frac{2\,\ln \left( 5 \right) A+3\,\ln \left( 5 \right) -2}{ \left( 2\,A+3 \right) \ln \left( 5 \right) }}$ Knowing that $\ln5 > \frac12$ we see that $g'(A) > 0$ for $A > \frac12$. Therefore, by the intermediate value theorem there is precisely one root to the equation $A = f(A), A > \frac{1}{2}$. By a lucky guess we see that this root is $A = 1$.

We now show that that $x_n < 1$ for all $n$:
$n=1$: $x_0 = \frac12 < 1$.
Assume that $x_p < 1$ for some $p$. Then $x_{p+1} = f(x_p) < f(1) = 1$ since f is a strictly increasing function.

Since $(x_n)$ is also increasing the axiom of upper bound (might be called something else in english) shows that $(x_n)$ is convergent, and one of the middle paragraphs shows that is convergent with the value 1.

Nice. I don't undersand some parts because there are some things that i haven't leaned yet. Is there a solutino that uses only basic theory?...  :?

Hmm. You probably have. It's just that I express myself too short sometimes. Here is the idea:

1) We note that the function by which the sequence is defined is strictly increasing.

2) Since $x_1 > x_0$ we can make a very simple induction proof that that $x_n$ is an increasing series using
 1). If we had found that $x_1 < x_0$ we could have used the exact same proof to show the opposite. In such sequences where the function $f$ is strictly increasing, the first two values always determine whether the sequence is increasing or decreasing.

3) For any given recursive sequence of the form $x_{n+1} = f(x_n)$ (where $f$ is continous. disregard if you don't know what this is) we have that if the sequence has a limit A, then it must be the case that $A = f(A)$. This is due to the fact that $x_{n+1} \to A$ and $f(x_n) \to A$ (since f is continous). Such a point where $x = f(x)$ is called a fixed point.

4) We use 3) to determine any such possible limits of the sequence. However, since the equation cannot actually be solved because of the nature of the function $f$, I instead create the difference of $A$ and $f(A)$ and show  that the derivative of this difference is always positive. Thus, the difference function is strictly increasing. Therefore there can be at most one solution to the equation, and we can "guess" this as $A = 1$.

5) Now, whenever we have such a fixed point it is easy to show that that the sequence is bounded upwards by it. I think this induction proof is easy to understand.

6) There is a theorem (actually an axiom), that you hopefully know, that says that if a sequence is increasing (decreasing) it has a limit if and only if it is bounded upwards (downwards). So due to this, we have shown that the sequence converges. Since we have also found that there is only possible limit, i.e. 1, it must converge to 1.

Feel free to ask about anything. Have you not learned these methods?


abstract algebra
superior algebra
superior algebra unsolved

Sorry for the many question. But I finally decided to return to all the things I did not understand previously from about 500 pages commutative algebra ;)

[b]Firstly:[/b]
[quote="Some Theorem"] Let $ A$ be a Noetherian ring, $ M \neq 0$ a finite $ A$-module. Then there exists a chain $ 0 \equal{} M_0 \subset M_1 \subset ... \subset M$ of submodules of $ M$ such that for each $ i$ we have $ M_i/M_{i \minus{} 1} \cong A/P_i$ with $ P_i \in Spec(A)$.
Proof: Choose any $ P_1 \in Ass(M)$, then there exists a submodule $ M_1$ of $ M$ with $ M_1 \cong A/P_1$. If $ M_1 \neq M$ choose $ P_2 \in Ass(M/M_1)$, then there exists $ M_2 \subset M$ such that $ M_2/M_1 \cong A/P_2$. Continuing this way and using the a.c.c. we arrive at $ M_n \equal{} M$.
[/quote]
-I don't see why we need $ A$ noetherian. $ M$ a finite $ A$-module and $ A$ being noetherian implies that $ M$ is noetherian. But why don't we just start the theorem with $ M$ noetherian, and $ A$ any ring or where else do we use that $ A$ is noetherian?
-If we take another $ P_1' \neq P_1 \in Ass(M)$ will we find a chain of the same length?

[b]Secondly:[/b]
[quote="Some corollary"]
If $ P \subset Q$ are prime ideals of $ A$, then $ (A_Q)_{P A_Q} \equal{} A_P$.
[/quote]
-I don't understand the transitive localization rules very well (also seen in my other post). Does someone want to elaborate?

1. existence of associated primes? I don't remember exactly, but I think A noetherian is important.
2. use the universal property.

1) Yes indeed. The noetherian property is important:

Let $ \mathcal{F} \equal{} \{ ann(x) \ | \ x \in M \}$, then every maximal element in $ \mathcal{F}$ is prime. Let $ ann(x)$ be a maximal element. Let $ ab \in ann(x)$, $ b \notin ann(x)$, then by maximality $ ann(bx) \equal{} ann(x)$, so $ ax \equal{} 0$, and $ a \in ann(x)$.

2) I'll review that part :)


geometry
perimeter
cyclic quadrilateral
number theory unsolved
number theory

1.5

Cyclic quadrilateral $ABCD$ has integer side lengths. It is known that $AD =2005$, $\angle ABC=\angle ADC=90^\circ$, and $\max\{AB, BC, CD\}<2005$.

Determine that minimum and maximum possible values for the perimeter of $ABCD$.

Very Nice problem.. my solution use Ptolomeu´s Theorem, divisibility and Pitagoras Theorem  :)


Question 1
3x+6+2x=4x-3

I got x= -9

Question 2
-11a-(5-6a)=-(6-3a)+1

I got a= 0

Question 3
-[p-(4p+2)]=3+(4p+7)

I got p=1

Question 4
M+M= -9
~   ~
6   4

Thats M over 6 + M over 4

I got M=-4

Question 5
P-2+p=1
~    ~   ~
3     4    2

I got P=1

.06x=.14(x+500)=130

I got x=300

.35(140) +.15W=.05(w+1100)

I didnt find an awnser to this one

2[3-4(5-R)]=2(3r-11)
I didnt find an awnser to this one

2(2y-5)-3(4-y)=7y-22
I didnt find an awnser to this one

6(3-4x+10=-15x+3(2-3x)
I didnt find an awnser to this one

[i][size=134]New here? WELCOME TO AOPS!!![/size][/i]
3x+6+2x=4x-3

I got x= -9

[b]right.[/b]

Question 2
-11a-(5-6a)=-(6-3a)+1

I got a= 0

[b]right.[/b]

Question 3
-[p-(4p+2)]=3+(4p+7)

I got p=1

[b]I got a different answer, check your work again... if you are really sure... then I would check mine.

Here's what I got[/b][hide]-p=8
p=-8[/hide]

Question 4
M+M= -9
~   ~
6   4

Thats M over 6 + M over 4

I got M=-4

[b]Check your work.
My answer[/b]
[hide]M=21.6[/hide]

Question 5
P-2+p=1
~    ~   ~
3     4    2

I got P=1

p-2 all over 3 right? if so then...

[b]check your work.
My answer[/b]
[hide]7p=14
p=2[/hide]

.06x=.14(x+500)=130

I got x=300

[b]are you sure there's two equal sign here?[/b]

.35(140) +.15W=.05(w+1100)

I didnt find an awnser to this one

[b]where did you got stuck?

answer:[/b]
[hide]0.1w=6
w=6/0.1=60[/hide]

2[3-4(5-R)]=2(3r-11)
I didnt find an awnser to this one

[b]um... show me where you got stuck[/b]

otherwise
[hide]2r=-12
r=-6[/hide]

2(2y-5)-3(4-y)=7y-22
I didnt find an awnser to this one

[b]Is there an answer AT ALL? or no solution?[/b]

6(3-4x+10=-15x+3(2-3x)
I didnt find an awnser to this one
[b]I think you missed one paranthesis[/b]

WARNING: You have to really know how to get to the correct answer. In other words, I don't guarantee you that mine are correct  :D . But do check over the ones I marked.

[b]For future posts, try to include what you have done for the problem so far. [/b]
sorry, I don't have time to type out full solutions. :(

If you have any more questions and are in short of time, AIM me.
my screenname is probsolving.

6(3-4x+10)=-15x+3(2-3x) woops

Yea I had so much trouble with this one, I got to the point after I distributed in the () and added like terms.

2[3-4(5-R)]=2(3r-11)

2[3-20-4R=

6-20+4r=2(3r-11)

there is were i got stuck

 :?:  2(2y-5)-3(4-y)=7y-22  :?:

After working on the others for so long, My brain stopped.

This was a pop quiz that i took and didnt count "thank god" But My teacher sugested that I go online and get some help. Fridays are test days for all my classes. Math was last period, so I hope you understand.

Thank you for your help BTW   :)

6(3-4x+10)=-15x+3(2-3x) woops
18 - 24x + 60 = -15x + 6 - 9x and move the x to one side, the others to the other
-24x + 15x + 9x = 6 - 60 - 18
0y = -72

SO

[b]NO SOLUTIONS[/b]

Yea I had so much trouble with this one, I got to the point after I distributed in the () and added like terms.

2[3-4(5-R)]=2(3r-11)

2[3-20-4R= [b]Remeber to multiply 2 by 20 and by 4r! (DISTRIBUTION)[/b]
6-20+4r=2(3r-11)

[b]it is actually 6 - 40 + 8r = 6r - 22
which is equal 6 + 22 - 40 = -2r
so -12 = -2r
so r = 6[/b]

Woops, I made a mistake on that earlier :blush: .

there is were i got stuck

 :?:  2(2y-5)-3(4-y)=7y-22  :?:
using distribution property:
4y - 10 - 12 + 3y = 7y - 22
4y +3y - 7 y = numbers...
0 = numbers...

so [b]NO SOLUTION[/b]

After working on the others for so long, My brain stopped.

This was a pop quiz that i took and didnt count "thank god" But My teacher sugested that I go online and get some help. Fridays are test days for all my classes. Math was last period, so I hope you understand.

Remember... some problems in algebra have no solution. The thing is... you must REALIZE when there is no solution by going through all the normal steps and remember to CHECK YOUR WORK. Math is all about accurary. So make sure ALL of your steps are accurate in order to arrive at a correct response.

Thank you for your help BTW   :)

No problem! Happy to help!  :)


[quote="hmoonster"]it says "the output PDF file is not finished" and won't let me see it in Adobe Acrobat.
What does this mean, and what can I do?[/quote]

OK, when and where does it gives you that error?

If I assume that: You are taking about a PDF generated with $\LaTeX$, using TeXNicCenter, and you are trying to open it with the output option of the TeXNic, well, that has happened to me a few times too, and I don't know exactly why that happens. I just either open the file directly with Acrobat Reader (usually I don't have problems with that).

It also can be that you tried to compile a file whose PDF output was open, or that you try to open a pdf file that hasn't finished the compilation yet.

If you use linux, I strongly recomend to compile for the regular DVI file, and after that use the dvipdf command. Usually the output is a lot smaller, and there is no lose of quality of the digital display.


geometry
incenter
circumcircle
geometry unsolved

Dear Mathlinkers,
let ABCD a square, M a point on segment CD, I, X, Y the incenters wrt MAB, MAD, MBC,
Px the parallel to AD passing through X, Py the parallel to AD passing through Y,  Pm the parallel to AD passing through M.
Prove that the isogonal lines of Px, Py, Pm wrt triangle MXY are concurrent at I.
Sincerely
Jean-Louis

Let $ M_1, M_2$ be arbitrary points following on the square side $ DC$ in this order, let $ AM_1$ cut $ BM_2$ at $ M_3,$ and let $ M$ be incenter of $ \triangle M_1M_2M_3.$ Let $ (X), (Y), (I)$ be incircles of $ \triangle M_1AD, \triangle M_2BC, \triangle M_3AB.$ The common external tangent of $ (X), (Y)$ other than $ DC$ cuts $ AM_3, BM_3$ at $ E, F.$ The square $ ABCD$ has an incircle $ (J)$, i.e., the common external tangents $ \overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{CD}, \overrightarrow{DA}$ of the points $ A, B$ and the circles $ (X), (Y)$ touch a single circle $ (J).$ Then their other common external tangents $ \overrightarrow{BA}, \overrightarrow{AE}, \overrightarrow{EF}, \overrightarrow{FB}$ also touch a single circle, the incircle $ (I)$ of the $ \triangle ABM_3.$ But then $ X, I, Y, M$ are intersections of the external angle bisectors of the general quadrilateral $ M_1EFM_2,$ which means that $ XIYM$ is cyclic with a circumcircle $ (O).$ By the continuity principle, this holds even in the limiting case, when the points $ M_1, M_2$ coincide and the $ \triangle M_1M_2M_3,$ together with its incenter $ M,$ degenerate into a single point $ M \in DC.$ See [url]http://www.mathlinks.ro/viewtopic.php?t=247411[/url] for the general case.

Assume $ M$ is a point on $ DC.$ Let $ (I_m)$ be excircle of the $ \triangle ABM$ against $ M$ and let $ K, L$ be feet of the perpendiculars from $ I$ to $ MX, MY,$ so that $ KL$ is Simson line of $ \triangle MXY$ with the pole $ I \in (O).$ The quadrilaterals $ IKML \sim I_mAIB$ are centrally similar, having parallel sides and corresponding diagonals $ IM, I_mI,$ therefore their other corresponding diagonals $ KL \parallel AB$ are also parallel and $ KL \perp AD.$ But this means that $ I$ is the isogonal conjugate WRT the $ \triangle MXY$ of the infinite point of any line parallel to $ AD.$ See [url]http://www.mathlinks.ro/viewtopic.php?t=129699[/url], click on "Show details" and scroll down to the 3rd paragraph "Let D be pole of the Simson line $ d \perp k$...

Dear Mathlinkers,
for a proof see:
http://perso.orange.fr/jl.ayme  vol.5,  le résultat de Larrosa Canestro, p.37
Sincerely
Jean-Louis


symmetry
geometry
parallelogram
angle bisector
geometry unsolved

(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A'$, $ B'$; $ C_1$ is the meet of lines $ AA'$ and $ BB'$. Prove that $ \angle C_1CA \equal{} \angle C_0CB$.

see http://www.mathlinks.ro/Forum/viewtopic.php?t=208506
in my solution ,I obtain ∠PAD=∠QAF ∠PAF=∠QAD (different letters)
thus our problem is solved

It’s not a big deal to show that C1 lies on (ABO), ÐAC1B=2ÐA, then C1C is its angle bisector: the triangles A’CA and B’BC are isosceles, then ÐAC1B=ÐA+ÐA’AC+ÐB’BC=2ÐA. Further, draw perpendiculars from A and B on CC0, call M and N their feet on it (obviously AM=BN, C0 being the middle of AB) and the perpendiculars from C on AA’ and BB’, call P and Q their feet onto these lines respectively. By symmetry, the triangles ACM and ACP are congruent, therefore AM=CP, similarly BN=CQ  ( Here, the sign Ð means "angle").

Best regards,
sunken rock

In may solution i made a parallelogram ACBC2 (C2 is a point) and i have  a point S so <CSA=<AC2C so quadritetral AC2SC is cyclic,therofore by angle chassing we have <CBS=<ACB and we have <CSA=<CSC1=<CBC1 ,therofore C1BSC is cyclic,therofore <CBS=<CC1S = <ACB, therofore we have <AC1C= 180-<ACB and therofore we have <ACC1=<BCB1, as desired.

Dear Mathlinkers,
just to remind that CC1 is the C-symmedian of ABC.
Perhaps another approach without angle is possible.
Sincerely
Jean-Louis


calculus
derivative
number theory unsolved
number theory

Waks. you say it's easy but I solved it using derivative, monotonic blah-blah.
I'm too shy to post a solution. Can you give a sketch of your solution please?

[quote="X-man"]Solve in positive integers \[ x^y + y = y^x + x  \][/quote]

Obviously $x=y$ works... anybody with $x\neq y$ ???

WLOG consider $x\geq y$ then we make the substitution $x=y+k, k\geq 0$

So we have:

$(y+k)^y+y=y^y^+^k+(y+k)$

Dividing by $y^y$ we get $(1+k/y)^y=y^k+k/(y^y)$

The left side is less than $e^k$ so we have $e^k>y^k+k/(y^y)$

If $k=0$ then $y=x$.

If $k\geq 1$ then $y<e$ which implies that $y=1, 2$

If $y=1$ then $x=1$.

If $y=2$ we have $2^x+x=x^2+2$ which gives us $x=2, 3$


probability
function
geometry
3D geometry
integration
logarithms
probability and stats

* I'm new here and I was hoping that you guys could give me some guidance with my probability homework!  Thanks!  I hope I posted this in the right section.

1)

Suppose you choose a real number X from the interval [2, 10] with a density
function of the form
f(x) = Cx ,
where C is a constant.
(a) Find C.
(b) Find P(E), where E = [a, b] is a subinterval of [2, 10].
(c) Find P(X > 5), P(X < 7), and P(X2 − 12X + 35 > 0).

* With this question, I'm not sure how to approach it to answer the first part.  Should C = f(x)/x?  That seems not very useful.  I'm just starting probability and this notation is kind of confusing to me.

2)  Assume that a new light bulb will burn out after t hours, where t is chosen
from [0,infinity) with an exponential density
f(t) = xe^−(xt) .

(a) Assume that  X= 0.01, and find the probability that the bulb will not
burn out before T hours. This probability is often called the reliability
of the bulb.
(b) For what T is the reliability of the bulb = 1/2?

* With this question, I'm uncertain of what f(t) is actually measuring and how probability plays a part here.

3)  Take a stick of unit length and break it into three pieces, choosing the break
points at random. (The break points are assumed to be chosen simultaneously.)
What is the probability that the three pieces can be used to form a
triangle? Hint: The sum of the lengths of any two pieces must exceed the
length of the third, so each piece must have length < 1/2.

* With this question, I'm not sure how to set this up.   I believe I should use a cube of [0,1]^3 and perhaps use the idea that A+B+C = 1 and A, B & C are all less than 1/2.  But I don't really fully understand the concept.   Or perhaps I just graph those two equations and get the area that overlaps?

Thanks!

1) $ f(x) \equal{}Cx$ on [2,10]

$ 1\equal{}\int_2^{10} f(x)dx \equal{} 48C \to C\equal{}\frac{1}{48}$

$ P([a,b]) \equal{} \int_a^b f(x) dx \equal{} \frac{b^2\minus{}a^2}{96}$

$ P(X>5) \equal{} P([5,10]) \equal{} \frac{75}{96}, P(X<7) \equal{} P([2,7]) \equal{} \frac{45}{96}$

$ X^2\minus{}12X\plus{}35 \equal{} (X\minus{}5)(X\minus{}7) > 0 \to X \in [2,5) \cup (7,10]$

$ P([2,5) \cup (7,10]) \equal{} P(X<5)\plus{}P(X>7) \equal{}$
$ (1\minus{} P(X>5))\plus{} (1\minus{}P(X<7)) \equal{} \frac{72}{96}$

2) i cant read some of the characters, sorry

3) think of the line [0,1] with two break points $ x,y \in (0,1)$

WLOG 0<x<y<1

the three segments would then be of lengths: x, y-x, 1-y

for now, fix x and lets focus on how likely we are to have a successful y:

$ 2x \ge 1 \to$ no y>x works, so assume $ x<.5$

then we have a triangle only when $ y\in (.5,x\plus{}.5)$

so for fixed x, $ P( y good) \equal{} \frac{x}{1\minus{}x} \equal{} \frac{1}{1\minus{}x} \minus{}1$

so$ P( all good) \equal{} \int_0^\frac{1}{2} \frac{1}{1\minus{}x}\minus{}1 dx \equal{} \ln 2 \minus{}\frac{1}{2}$


let the group  A={1,2,3,.........,n} such that n is odd.

X1,X2,X3,X4,...........,Xn are different elements two by two from A.

prove that it is k elements from A such that Xk-k is even.

remark: Xk isn't X x k.

Please learn to use $ \text{\LaTeX}$ , greatestmaths  :mad:

Let the group $ \text{A}=\{1,2,3,..,n\}$ such that n is odd.

$ X_{1},X_{2},X_{3},X_{4},..,X_{n}$ are different elements two by two from $ \text{A}$.

prove that it is $ k$ elements from $ \text{A}$. such that $ X_{k}-k$ is even.

[code]Let the group $\text {A}=\{1,2,3,..,n\}$ such that n is odd.

$X_1,X_2,X_3,X_4,..,X_n$ are different elements two by two from $\text{A}$.

prove that it is $k$ elements from $\text{A}$. such that $X_k-k$ is even.[/code]

Do you mean exist $ k$ ,greatestmaths ? . If i am not wrong , this problem very easy. Suppose not exist $ k$ so that $ S_{k}\minus{}k\vdots 2$ then $ \sum_{k \equal{} 1}^{n}(S_{k}\minus{}k)\equiv n(\text{mod 2})\Leftrightarrow n\equiv 0(\text{mod 2})$contracdiction because $ n$ is an odd number  :)

[quote="greatestmaths"][code]Let the group $\text {A}=\{1,2,3,..,n\}$ such that n is odd.

$X_1,X_2,X_3,X_4,..,X_n$ are different elements two by two from $\text{A}$.

prove that it is $k$ elements from $\text{A}$. such that $X_k-k$ is even.

[/code][/quote]

What do you want to say , greatestmaths  :(

Let the group $ \text{A}\equal{}\{1,2,3,..,n\}$ such that n is odd.

$ X_{1},X_{2},X_{3},X_{4},..,X_{n}$ are different elements two by two from $ \text{A}$.

prove that it is $ k$ elements from $ \text{A}$. such that $ X_{k}\minus{}k$ is even.


$ \binom{2n}{n\plus{}1} \plus{} \binom{2n}{n} \equal{} \dfrac{1}{2} \binom{2n \plus{} 2}{n\plus{}1}$

$ \dfrac{(2n)!}{(n\plus{}1)!\times(n\minus{}1)!} \plus{} \dfrac{(2n)!}{n!\times n!} \equal{}
\dfrac{(2n \plus{} 2)!}{2(n\plus{}1)!\times(n\plus{}1)!}$

$ \dfrac{n! \times n! \times (2n)! \plus{} (n\plus{}1)! \times (n\minus{}1)! \times (2n)!}{n! \times n! \times (n\plus{}1)! \times (n\minus{}1)!} \equal{}
\dfrac{(2n \plus{} 2)!}{2(n\plus{}1)!\times(n\plus{}1)!}$

$ \dfrac{ (2n)!  \times (n! \times n! \plus{} (n\plus{}1)! \times (n\minus{}1)!)}{n! \times n! \times (n\plus{}1)! \times (n\minus{}1)!} \equal{}
\dfrac{(2n \plus{} 2)!}{2(n\plus{}1)!\times(n\plus{}1)!}$

$ \dfrac{ (2n)!  \times ((n\minus{}1)!)^2 \times (2n^2 \plus{} n)}{n! \times n!\times (n\minus{}1)!} \equal{}
\dfrac{(2n \plus{} 2)!}{2(n\plus{}1)!}$

$ \dfrac{ (2n)!  \times ((n\minus{}1)!)^2 \times (2n^2 \plus{} n)}{(n!)^2} \equal{}
\dfrac{(2n \plus{} 2)!}{2(n^2 \plus{} n)}$

$ \dfrac{ (2n)!   \times (2n^2 \plus{} n)}{n^2} \equal{}
\dfrac{(2n \plus{} 2)!}{2(n^2 \plus{} n)}$

$ \dfrac{(2n^2 \plus{} n)}{n^2} \equal{}
\dfrac{(2n \plus{} 1)(2n \plus{} 2)}{2(n^2 \plus{} n)}$

$ 2(n^2 \plus{} n)(2n^2 \plus{} n) \equal{} n^2(2n \plus{} 1)(2n \plus{} 2)$

$ 2n^2(n \plus{} 1)(2n \plus{} 1) \equal{} 2n^2(n \plus{} 1)(2n \plus{} 1)$

tada! I don't know if that's exactly what you wanted, but it works!

Well that's not what I was looking for exactly.  The part of the question before it asks me to prove it by substitution, which is what you did.  Or at least I assumed that's what was meant by "prove by substitution."  Although I am glad that we got the same answer.  At least I did one thing right.

Anyway, I think by combinatorial proof is asking me to prove it by using C(n,k) type identities and properties.

Two applications of Pascal's Identity can do it for you:

[b]Pascal's Identity: [/b] $ \binom{n}{k} = \binom{n - 1}{k} + \binom{n - 1}{k - 1}$

Applying this to the RHS,
\begin{eqnarray*} \frac {1}{2}\binom{2n + 2}{n + 1} & = & \frac {1}{2}\left[\binom{2n + 1}{n + 1} + \binom{2n + 1}{n}\right] \\
& = & \frac {1}{2}\left[\binom{2n}{n + 1} + \binom{2n}{n} + \binom{2n}{n} + \binom{2n}{n - 1}\right] \\
& = & \binom{2n}{n + 1} + \binom{2n}{n} \\
\end{eqnarray*}
where the last step follows from the fact that $ \binom{2n}{n - 1} = \binom{2n}{n + 1}$.

Wow.  That was extremely simple.  I'm almost mad at myself for not seeing that on my own.

Thank you very much.

[hide="Another way"]Consider a set of $ 2n\plus{}2$ elements, $ 2$ of which are "special". (Like the hint says! :P ) Call the rest of the elements "regular".
We will now consider the amount of ways to choose $ n\plus{}1$ elements of this set.

Method 1: Simply choose $ n\plus{}1$ of the the $ 2n\plus{}2$ elements. There are $ \binom{2n\plus{}2}{n\plus{}1}$ ways to do this.
Method 2: Choose from the special elements, then choose from the regular elements. If we choose $ 0$ special elements, then there are $ \binom{2n}{n\plus{}1}$ ways to choose from the regular elements. If we choose $ 1$ special element in one of $ 2$ ways, then there are $ \binom{2n}{n}$ ways to choose from the regular elements, giving $ 2\cdot\binom{2n}{n}$ ways total. If we choose $ 2$ special elements, there are $ \binom{2n}{n\minus{}1}$ ways to choose from the regular elements. In total, there are $ \binom{2n}{n\plus{}1}\plus{}2\cdot\binom{2n}{n}\plus{}\binom{2n}{n\minus{}1}$ ways to do this.

These two methods yield the same amount of combinations, so we may equate them.
We have \[ \binom{2n\plus{}2}{n\plus{}1}\equal{}\binom{2n}{n\plus{}1}\plus{}2\cdot\binom{2n}{n}\plus{}\binom{2n}{n\minus{}1}\]
Since $ \binom{2n}{n\minus{}1}\equal{}\binom{2n}{n\plus{}1}$, this is equivalent to \[ \binom{2n\plus{}2}{n\plus{}1}\equal{}2\cdot\binom{2n}{n\plus{}1}\plus{}2\cdot\binom{2n}{n}\]
Dividing by $ 2$ gives \[ \frac{1}{2}\binom{2n\plus{}2}{n\plus{}1}\equal{}\binom{2n}{n\plus{}1}\plus{}\binom{2n}{n}\]
which was what we wanted. :D[/hide]
Note: I believe that the way I proved it (counting the same thing in two different ways) is the kind of proof that the problem asked for. Sorry I wrote it poorly. :P

[quote="Brut3Forc3"]
Note: I believe that the way I proved it (counting the same thing in two different ways) is the kind of proof that the problem asked for. [/quote]

That's essentially the same idea behind Pascal's Identity.

Also, I think that a simple block walking argument can be used to prove the identity.


geometry
rectangle
calculus
integration
trigonometry
calculus computations

Hi everyone,

I've been having trouble with this problem for a few days. I managed to get an expression using a bit of geometry but it doesn't work perfectly in all situations.

Given a disk centered on $ (x_{c},y_{c})$ with radius $ r$ and a rectangle with $ x_{r} \in [a,b]$ and $ y_{r} \in  [c,d]$, how would you express the area of the intersection of the circle with the rectangle ?
Simplification: $ x_{c} \in [a,b] \vee y_{c} \in [c,d]$

I suppose a double integral would do the trick but I'm not sure how to formulate the problem  :(

S.

It's going to split into an annoyingly large number of cases.

I would look for a procedure, rather than trying for any sort of piecewise formula.

1. Find the intersections between the circle and the right side $ x\equal{}b$ of the rectangle. If there are no intersections, take the point $ (x_c\plus{}r,y_c)$ (which is inside the rectangle) twice instead. If a corner $ (b,c)$ or $ (b,d)$ is inside the circle, take that corner instead of the lower/upper intersection.
2. Repeat this process for the other three sides in counterclockwise order.
3. The eight points $ (x_i,y_i)$ generate eight wedges (some likely degenerate). The four coming from pairs of points on the same side are triangles with area $ \frac12(x_iy_{i\plus{}1}\minus{}x_{i_1}y_i)$*, and the four coming from pairs on different sides are circular wedges with area $ r^2\cos^{\minus{}1}\left(\frac{x_ix_{i\plus{}1}\plus{}y_iy_{i\plus{}1}}{r^2}\right)$*. Add them up.

* Area formulas stated as if $ x_c\equal{}y_c\equal{}0$. Replace $ x_j$ with $ x_j\minus{}x_c$ and $ y_j$ with $ y_j\minus{}y_c$ for the generic version. Also, indices are interpreted mod 8, so that $ i\plus{}1$ is the next point even if it means going back to the first one.

Thanks for the answer.

My goal is to be able to express the distribution of the area for a fixed $ r$ and a center $ (x,y)$ wandering about a surface where the rectangle lies, so a formula would be more useful if I don't want to spend more time waiting for my computer to compute that numerically.

How large is "annoyingly large" ?  :D

S.


Let $ f: \mathbb R\to\mathbb R$ be continuous such that $ f(x)\ne0,\,\forall\,x\in\mathbb R$ and $ f(0)\equal{}\sqrt2.$ Show that $ f$ is positive.

Look at the Bolzano-Cauchy intermediate value theorem

How'd you do it? Because I don't know how to apply such theorem.

Roughly, in order to take a negative value $ f$ must cross (hence intersect) the $ x$-axis, and we are given that this does not occur.  Specifically, if $ f(0) \equal{} \sqrt{2}$ and $ f(a) < 0$ for some $ a$ then on $ (0, a)$ we know that $ f$ takes every value between $ \sqrt{2}$ and $ 0$ (this is the content of the Intermediate Value Theorem); in particular, $ \exists c \in (0, a) : f(c) \equal{} 0$, contradiction.


The original problem was :

If a>1, find a bounded sequence of real numbers x_i with :
|x_i-x-j|*|i-j|^a>=1 for all distinct i,
j.

My question is : what if a=1 ?

I thought that such a sequence was always unbounded, but the author of the problem told that one could construct such a bounded sequence (and that candidats succeded in doing it during the 1991 olympiads !).
Could someone explain me how to construct it ?

Also, contrast problem A2 from ISL 2002: [url]http://www.kalva.demon.co.uk/short/soln/sh02a2.html[/url].

Sketch of a proof for this problem:

I think something that would work would be to construct it by thinking of it as placing $a_1, a_2, \dots$ on the real line one by one in sequential order in a manner as below: (the integer n in the diagram represents the relative position of $a_n$)

1
1         2
1    3    2      4
1 5  3 6  2   7  4  8
... and so on (distances not to scale).  When $a_n$ is inserted, say it is directly between $a_i$ and $a_j$, it suffices to make sure the condition holds for the pairs $i, n$ and $j, n$, for all others will be greater than pairs checked before.

Also, one should be able to get an upper bound on the the size interval needed to contain the first $2^t$ terms for any $t$:  one can then start placing the integers in that interval one by one so as to always assure there is enough space for the integers following.

There is a similar problem in ASU 1978.

http://olympiads.win.tue.nl/imo/soviet/RusMath.html
(problem 257)

Prove that there exists such an infinite sequence $\{x_i\}$, that for all $m$ and all $k$ ($m\neq k$) holds the inequality $|x_m-x_k|>\frac{1}{|m-k|}$.

I think the sequence $ x_n=7\cdot f(n\cdot\sqrt{2}) $ is good for $ a=1$, where f is the function fractional part . It is also bounded!

If 0<a<1, you can use the following theorem by Erdos :

From every real sequence of length n^2+1 you can extract a monotonic subsequence of length n+1.

So take the first a_1, ..., a_{n^2+1} terms, extract a monotonic subsequence a_{i_1}<...<a_{i_{n+1}} and then you can prove that

a_{i_{n+1}}-a_{i_1} > n^{1-a}

So the sequence canot be bounded


By a given point $ P$ inside angle $ BAC$ construct the straight line wich cuts $ AB$ and $ AC$ at $ M$ and $ N$ respectively, and wich minimize the sum $ \frac{1}{MP}\plus{}\frac{1}{NP}$

 :)

There is no local minimum; do you mean maximum?

Is it asking for an answer, or is it a construction problem?  And unless I'm interpreting it incorrectly, there isn't a minimum.  You could keep moving the line away from P.  There is a maximum when you draw the line through P to make isosceles $ \triangle MNA$.

Perhaps ElChapin intended for P to be outside triangle AMN.

I'm sorry it was supposed to say maximize.

 :)

[quote="ZzZzZzZzZzZz"]In which case you'd just make isosceles $ \triangle MNA$ where $ P$ is on $ \overline{MN}$.[/quote]

No, that is clearly incorrect.  What makes you think that the sum of the reciprocals of the distances will be maximized for the isosceles case?  The correct answer is that the sum of reciprocal distances is maximized when the line through P is drawn such that PA is the altitude of triangle MNA.

Let $ \angle BAP\equal{}y$ and $ \angle CAP\equal{}z$. Then $ \frac{1}{MP}\plus{}\frac{1}{NP}\equal{}AP\cdot (\cot y\plus{}\cot z)\cdot \sin \angle MPA$. The max occurs when $ \angle MPA\equal{}90^\circ$. We construct $ M$ and $ N$ by drawing the line perpendicular to $ AP$ through $ P$.


When a certain number is increased by 4, the result is equal to that number divided by 7 What is the number?

I think that it's 14/3.  I'm not sure if it should be -14/3.

My calcs may not make sense. I don't know what I did.

x + 4 = y   =   y/7

(2/3) + 4 = y

14/3 + 4 = 4 2/3 + 4 = 8 2/3

(26/3) * (1/7)

You only need one variable for this problem.

Let the number be $ x$.

$ x\plus{}4\equal{}\frac{x}{7}$.

$ 7x\plus{}28\equal{}x$

$ 28\equal{}\minus{}6x$

$ \frac{\minus{}14}{3}\equal{}x$

[quote="cf249"]You only need one variable for this problem.

Let the number be $ x$.

$ x \plus{} 4 \equal{} \frac {x}{7}$.

$ 7x \plus{} 28 \equal{} x$

$ 28 \equal{} \minus{} 6x$

$ \frac { \minus{} 14}{3} \equal{} x$[/quote]

Thank you.

Oh and the for the first post, y/7 does not equal y.

quoting: "oh and for the first one y does not equal y/7."
couldn't y = y/7 if y was zero? or is that undefined?

$ \frac{y}{7}\equal{}y$
$ y\equal{}7y$
Only works when $ y\equal{}0$

But the point was that y/7 does not ALWAYS equal y. That would be like making a formula that works only for one number, or only one set of points.

so how can it SOMETIMES equal y/7? though i think i get what ur saying.

Point is that it doesn't always work.

$ X\equal{}2x$ when $ x\equal{}0$. IT DOES NOT HOLD WHEN $ X\equal{}anything else$


logarithms
calculus
calculus computations

Hello everyone... I'm really stuck on some problems testing the convergence of these sequences:

1) (n>=2)∑ [1 / root(n^(2)-1)]

2) (n>=1)∑ [((-1)^(n)*ln(n)) / (1+ln(n)]

3) (n>=1)∑ [(cos(n(pi)/2) / (n^2)]

Help would really be appreciated!

Think about size, and "de-clutter."

For large $n,$, how big is $\frac1{\sqrt{n^2-1}}?$ It looks a lot like $\frac1n,$ doesn't it?

For large $n,$ how big is $\frac{\ln n}{1+\ln n}?$ A lot like $\frac{\ln n}{\ln n},$ and that doesn't go to zero, does it?

In the third case, you have a fancy variant on a $\pm$ sign, multiplied by $\frac1{n^2}.$ What do you know about $\sum\frac1{n^2}?$

For the second one, you said that ln(n)/ln(n) approaches 0, but does it not approach 1 for all values of n? Also, since it is an alternating series, would it converge to a specific value?

Read that sentence again: "and that [b]doesn't[/b] go to 0..."

Since the values alternate between about 1 and about -1, the partial sums jump around; in this case, the sequence of partial sums should have two limit points.

Oops, my mistake... well wouldn't it be divergent if it went to two different limit points?

Yes, it would be divergent. jmerry provided more information than you really need - the two subsequential limts is a nice touch, but it's more than you were originally asked.

Theorem: if a series converges, then its terms go to zero.

Contrapositive: if the terms of a series do not go to zero, then the series diverges.


Hi everyone,
As the Canadian representative at AoPS, I feel like it is partially my duty to keep this board alive, even though I think it's doing pretty well on its own.  So, I'll try to post some more from now on.

At AoPS, one thing we'd like to have is more exposure in Canada.  I've written to the CEMC and the CMS, so I believe they're aware of us, but I'm wondering if there's anyone else I should be writing to.  If you have any ideas, I would love to hear them.

[quote="Sunny"][quote="lightrhee"]:huh: There is a Canadian admin?!?! I didn't know that.[/quote]

If I'm not mistaken, he was at the National Camp I went to.  At least in my year, Naoki?  Hehehe, good times those were.[/quote]Not only that he is Canadian, but he was your deputy at some point, and won a couple of silvers for you guys ;)

[quote="lightrhee"]:huh: There is a Canadian admin?!?! I didn't know that.[/quote]
Yeah, born and raised in Toronto, very proud.

[quote="Sunny"]If I'm not mistaken, he was at the National Camp I went to.  At least in my year, Naoki?  Hehehe, good times those were.[/quote]
I remember you Sarah, good to see you on the board, all the way from beautiful downtown Okotoks.  That camp was a lot of fun, I kinda wish I could do more of those.  Felicitations on your recent engagement to Singular/Alex, I'm sure you'll make a happy couple.

[quote="Valentin Vornicu"]Not only that he is Canadian, but he was your deputy at some point, and won a couple of silvers for you guys ;)[/quote]
Well, only one silver actually, but who's counting? :P  I prefer to think I've helped other students win medals, but who knows for sure.

[quote="Singular"]hear that sparky?  Naoki Sato says we make a good couple!   :lol:  :lol:[/quote]

...oh boy....

[quote]hear that sparky? Naoki Sato says we make a good couple![/quote]

Congratulation Alex... you seem to have found a soulmate there..........
And also, Sarah? why is he calling you sparky??

[quote]I've written to the CEMC and the CMS, so I believe they're aware of us[/quote]

I'm very sure they're aware of us, considering the last seminar..  :lol:  :blush:

[quote="jpark"][quote]hear that sparky? Naoki Sato says we make a good couple![/quote]

Congratulation Alex... you seem to have found a soulmate there..........
And also, Sarah? why is he calling you sparky??[/quote]

Apparently because I'm bright and slightly evil.

[quote][quote]I've written to the CEMC and the CMS, so I believe they're aware of us[/quote]

I'm very sure they're aware of us, considering the last seminar..  :lol:  :blush:[/quote]

lol...wow that was really scary....they knew everything....

[quote][quote]I've written to the CEMC and the CMS, so I believe they're aware of us[/quote]

I'm very sure they're aware of us, considering the last seminar..  :lol:  :blush:[/quote]

Why, what happened?

... no comment on that please.. :blush:  :blush:

[quote="Elyot"]LMFAO!!!  Yea seminar was great.  Good times all around.  This whole forum is getting quite funny.[/quote]

I can't believe they found the forum t.t


geometry
geometric transformation
reflection
geometry proposed

Given a triangle $ ABC.$ Let $ A_{1},B_{1},C_{1}$ be the feet of the altitudes of triangle $ ABC$ from the vertices $ A,B,C,$ respectively. Denote by $ Y \equal{} CA\cap C_{1}A_{1}, Z \equal{} AB\cap A_{1}B_{1}.$ Point $ P$ lies in the plane of triangle $ ABC$. Let $ P_{1},P_{2},P_{3}$ be the reflections of $ P$ in sidelines $ BC,CA,AB,$ respectively. Let $ l_{1},l_{2}$ be lines through $ Y,Z,$ respectively, parallel to $ P_{3}P_{1},P_{1}P_{2}.$ Prove that if $ AP_{1},BP_{2},$ and $ CP_{3}$ are concurrent, then so are $ l_{1},l_{2},$ and $ AP_{1}.$

[quote]Let $ P_{1},P_{2},P_{3}$ be the reflections of $ P$ in sidelines $ BC,CA,AB,$ respectively [/quote]
[quote]$ AP_{1},BP_{2},$ and $ CP_{3}$ are concurrent $\Longleftrightarrow$ $ \ell_{1},\ell_{2},$ and $ AP_{1}$ are concurrent[/quote]
[hide="Hint"]Locus of P is the Darboux cubic of ABC.[/hide]

Very nice problem and ... very hard!

A synthetic proof for the Sondat theorem is here:
[url]http://pagesperso-orange.fr/jl.ayme/Docs/Le%20theoreme%20de%20Sondat.pdf[/url]

The triangles $ ABC$ and $ A_{1}B_{1}C_{1}$ are orthologic and the points $ O$ and $ H$ are the orthologic centers of these
triangles. The lines $ AA_{1},BB_{1},CC_{1}$ are concurrent at the point $ H$. The Sondat theorem implies that $ OH\bot YZ$.   (1)

Let $ P^{*}$ be the isogonal conjugate of the point $ P$ wrt $ ABC$, $ \left\{Y'\right\}\equal{}P_{1}P_{3}\cap CA$,  $ \left\{Z'\right\}\equal{}P_{1}P_{2}\cap AB$.
The triangles $ ABC$ and $ P_{1}P_{2}P_{3}$ are orthologic and the points $ P$ and $ P^{*}$ are the orthologic centers
of these triangles. The lines $ AP_{1},BP_{2},CP_{3}$ are concurrent at a point $ Q$.
The Sondat theorem implies that the points $ P,P^{*},Q$ are collinear and $ PP^{*}\bot Y'Z'$.   (2)

Now, $ P_{1},P_{2},P_{3}$ are the reflection points of $ P$ in $ BC,CA,AB$, respectively and the lines $ AP_{1},BP_{2},CP_{3}$
are concurrent. From here [url]http://pagesperso-orange.fr/bernard.gibert/Exemples/k001.html[/url]
we have that the point $ P$ lies on the Neuberg cubic, so $ PP^{*}\parallel{}OH$.   (3)
I don't have a synthetic solution to prove that $ PP^{*}\parallel{}OH$ or $ PQ\parallel{}OH$. :wink:

(1),(2),(3) $ \Rightarrow YZ\parallel{}Y'Z'$.

$ \left\{M\right\}\equal{}P_{1}P_{3}\cap YZ$, $ \left\{N\right\}\equal{}P_{1}P_{2}\cap YZ$.
$ MN\parallel{}Y'Z'\Rightarrow \frac{P_{1}M}{P_{1}Y'}\equal{}\frac{P_{1}N}{P_{1}Z'}$   (4)
Let $ V'$ be the intersection point of $ AP_{1}$ with the parallel through $ M$ to $ AC$,
and $ V"$ be the intersection point of $ AP_{1}$ with the parallel through $ N$ to $ AB$.
$ MV'\parallel{}Y'A\Rightarrow \frac{P_{1}M}{P_{1}Y'}\equal{}\frac{P_{1}V'}{P_{1}A}$   (5)   and   $ NV"\parallel{}Z'A\Rightarrow \frac{P_{1}N}{P_{1}Z'}\equal{}\frac{P_{1}V"}{P_{1}A}$   (6)
(4),(5),(6) $ \Rightarrow\frac{P_{1}V'}{P_{1}A}\equal{}\frac{P_{1}V"}{P_{1}A}\Rightarrow V'\equal{}V"\equal{}V$.

$ \left\{L\right\}\equal{}AP_{1}\cap YZ$
$ YA\parallel{}MV\Rightarrow \frac{LY}{LM}\equal{}\frac{LA}{LV}$   (7)   and   $ ZA\parallel{}NV\Rightarrow \frac{LZ}{LN}\equal{}\frac{LA}{LV}$   (8)
(7),(8) $ \Rightarrow\frac{LY}{LM}\equal{}\frac{LZ}{LN}$   (9)

$ \left\{W'\right\}\equal{}l_{1}\cap AP_{1}$ and $ \left\{W"\right\}\equal{}l_{2}\cap AP_{1}$.
$ YW'\parallel{}MP_{1}\Rightarrow\frac{LY}{LM}\equal{}\frac{LW'}{LP_{1}}$   (10)   and   $ ZW"\parallel{}NP_{1}\Rightarrow\frac{LZ}{LN}\equal{}\frac{LW"}{LP_{1}}$   (11)
(9),(10),(11) $ \Rightarrow\frac{LW'}{LP_{1}}\equal{}\frac{LW"}{LP_{1}}\Rightarrow W'\equal{}W"\equal{}W$.

So, the lines $ l_{1},l_{2},AP_{1}$ are concurrent at $ W$.

Best regards, Petrisor Neagoe :)

Dear luis,

The points $ P_{1},P_{2},P_{3}$ are the REFLECTION of $ P$ in sidelines $ BC,CA,AB$.
The triangle $ P_{1}P_{2}P_{3}$ IS NOT THE PEDAL TRIANGLE of the point $ P$.

Best regards


vector
trigonometry
calculus
derivative
calculus computations

The position vector of a point on a circular helix is given by

$ \mathbf{r}\equal{} a[(\cos{t})\mathbf{i}\plus{}(\sin{t})\mathbf{j}\plus{}(t\cot{\alpha})\mathbf{k}]$

Find the unit tangent vector to the helix at any point.
Show that the curvature at any point is

$ \kappa \equal{}\frac{\sin^{2}{\alpha}}{a}$

Find the unit normal, and show that it always passes through the $ z$-axis.

(I got stuck at finding the unit normal. You may think I am a bit silly but shouldn't there be a whole plane of normals at any one point?)

Back from a long vacation, let me see if I still remember any math...  :lol:

The unit tangent vector is given by:

$ T(t)\equal{}|r'(t)|\equal{}\frac{ (\minus{}a\sin t, a\cos t, a\cot\alpha)}{a\csc\alpha}$

The curvature is given by:

$ \kappa(t) \equal{}\frac{ |T'(t)| }{|r'(t)|}$

Now $ T'(t)\equal{}\frac{ (\minus{}a\cos t,\minus{}a\sin t, 0) }{a\csc\alpha}$, so we get:

$ \kappa(t)\equal{}\frac{\frac{a}{a\csc\alpha}}{a\csc\alpha}\equal{}\frac{\sin^{2}\alpha}{a}$

The unit normal vector is given by:

$ N(t)\equal{}\frac{T'(t)}{|T'(t)|}\equal{}\frac{\sin^{2}\alpha}{a}\cdot (\minus{}a\cos t,\minus{}a\sin t, 0)$.

Since the $ z$-component is always zero, it must always pass through the $ z$-axis.

[quote="sludgethrower"](I got stuck at finding the unit normal. You may think I am a bit silly but shouldn't there be a whole plane of normals at any one point?)[/quote]

Yes, there is.  However, we often (as edoo did) take the (normalized) derivative of the unit tangent and refer to it as [i]the[/i] unit normal.


Prove that there are infinitely many n such that a^n+1 is divisible by n (with a is an integer, a>1)
It seem old but i don't know the solution.

note that $ a \plus{} 1 > 2$ which means that this number has a prime divisor,let's call it $ q$;
then obviously $ q^b |a^{q^b} \plus{} 1$ where $ b$ is a poistive integer.so there exists infinitely many such natural numbers.

shoki: if $ a \equal{} 3$, then $ q \equal{} 2$, but $ 2^2\not | 3^{2^2} \plus{} 1$

sorry u are right i didn't consider the case $ a\equal{}2^k\minus{}1$.anyway ,in this case just put $ n\equal{}2$ then we must have an odd prime dividing  it let it  be $ q$ then we have $ 2q^b | (2^k\minus{}1)^{2q^b}\plus{}1$