calculus
derivative
vector
function
limit
logarithms
calculus computations

In an economics problem i face the following difficulty:

$ F \equal{} f_{1}(q_{1}) \plus{} ... \plus{} f_{n}(q_{n})$
$ q \equal{} q_{1} \plus{} ... \plus{} q_{n}$

I have to compute this derivative:
$ \frac {\partial F}{\partial q} \equal{} \frac {\partial (f_{1}(q_{1}) \plus{} ... \plus{} f_{n}(q_{n}))}{\partial (q_{1} \plus{} ... \plus{} q_{n})}$

Could someone help me with it please? (I have knowledge of multiple variable calculus, but i can't find the right way to compute this).

Thanks very much in advance

Now, i thought i could write:

$ \frac {\partial F}{\partial q} \equal{} \frac {\partial f_{1}}{\partial q_{1}} \frac {\partial q_{1}}{\partial q} \plus{} ... \plus{} \frac {\partial f_{n}}{\partial q_{n}}\frac {\partial q_{n}}{\partial q_{n}}$

and q1 = q - q2 - ... - qn
      ...
      qn = q - q1 - ... - qn-1

so i can compute the $ \frac {\partial q_{1}}{\partial q}$ and so on...

Are the previous correct?
And what if for example: q1 = q1(q2,...qn)   / q2 = q2(q1,q3,...,qn) and so on...
Is there any change?

Thanks!!

Since $ F$ doesn't depend on $ q$ alone, its derivative with respect to $ q$ is somewhat ambiguous. The most natural interpretation is probably a directional derivative in the $ \frac1n(1,1,\cdots,1)$ direction.

Here's what that looks like:
Let $ x$ be the vector $ (x_1,x_2,\cdots,x_n)^T$, and consider a differentiable function $ F(x)$. If $ v$ is a vector $ (v_1,v_2,\cdots,v_n)^T$, the directional derivative $ \frac {\partial F}{\partial v}$ at $ x$ is $ \lim_{t\to 0}\frac {F(x \plus{} tv) \minus{} F(x)}{t}$ by definition, and $ F'(x)\cdot v$ if you want to practically compute it.
That derivative $ F'$ is the gradient, of course, in its natural form as the total derivative. Here, that becomes $ \frac1n\left(\frac {\partial F}{\partial x_1},\frac {\partial F}{\partial x_2},\cdots,\frac {\partial F}{\partial x_n}\right)\cdot (1,1,\cdots,1)^T \equal{} \frac1n\left(\frac {\partial F}{\partial x_1} \plus{} \frac {\partial F}{\partial x_2} \plus{} \cdots \plus{} \frac {\partial F}{\partial x_n}\right)$

There's a further ambiguity regarding the norm in the underlying version of $ \mathbb{R}^n.$ jmerry seems to have implicitly used the 1-norm on that space rather than the 2-norm (Euclidean norm), which is what you see most often in vector calculus classes. But the Euclidean norm implies some kind of geometric sense, and that geometric sense is likely to be absent in an economics problem. It's not clear what the proper norm should be.

To further explain what I'm talking about: a directional derivative uses a unit vector. $ \frac1n(1,1,\dots,1)$ is a unit vector with respect to the 1-norm but not with respect to the Euclidean norm; for the Euclidean norm, you'd say $ \frac1{\sqrt{n}}(1,1,\dots,1).$

Actually, it's all about the 2-norm; $ \frac1n(1,1,\cdots,1)$ is the unique point on the hyperplane $ x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n \equal{} 1$ of smallest 2-norm. It comes from the correspondence between non-origin points and hyperplanes not through the origin, which involves the standard inner product but not unit vectors of any sort.

By this convention, for any linear map $ a(x)\equal{}a\cdot x$, I would write $ \frac{\partial F}{\partial a(x)}\equal{}\frac{1}{\|a\|^2}F'\cdot a.$

I'm actually somewhat confused.

If for example, i have 3 cost functions:
$ f_{1}(q_{1}) \equal{} 10 \plus{} 3q_{1}$
$ f_{2}(q_{2}) \equal{} 5 \plus{} 10\ln q_{2}$
$ f_{3}(q_{3}) \equal{} 5 \plus{} 8\sqrt {q_{3}}$

so the total cost:

$ F \equal{} F(f_{1}(q_{1}),f_{2}(q_{2}),f_{3}(q_{3})) \equal{} 10 \plus{} 3q_{1} \plus{} 5 \plus{} 10\ln q_{2} \plus{} 5 \plus{} 8\sqrt {q_{3}}$

how do i compute the partial derivative, $ \frac {\partial F}{\partial q}$
where $ q \equal{} q_{1} \plus{} q_{2} \plus{} q_{3}$
???

Thanks very much for your time. It is important for me to figure it out.

That's what I tried to answer in my first post. It would help if you understood what your intended question was.

Yes, but i don't have a vector $ v\equal{}[v_{1},...v_{n}]$

I have an $ \Re ^{n}: \Re$ function $ q\equal{}q(q_{1},...,q_{n})\equal{}q_{1}\plus{}q_{2}\plus{}q_{3}$

Then the question doesn't make sense at all. With only one new variable, a directional derivative is the only way I could make sense of it at all. If we take a new system of variables $ (q,y_2,y_3)$ and consider $ F$ with respect to them, the $ q$ partial derivative can be anything.

By the way, it's \mathbb{R}: $ \mathbb{R}$.

Well, the problem is exactly the following:

$ P(q)\plus{}q\frac{\partial P(q)}{\partial q}\minus{}\frac{\partial \sum_{i\equal{}1}^{n}F_{i}(q_{i})}{\partial q}\equal{}0$

where, $ q\equal{}\sum_{i\equal{}1}^{n}q_{i}$

It is a known cartel (enterprise) problem but i can't compute thw partial derivative in the last term.

Although i get your clue, i believe it does have a way of computing it without ambiguity.


Hi!
I am currently working on a program to calculate pi using the BBP formula (http://mathworld.wolfram.com/BBPFormula.html). I have been taking a look at it, however I do not fully understand it. I have been looking at some c++ versions of this, however this just myths me more. Both of the programs I have looked at have a limit of 2^24, however a project like pihex (http://www.cecm.sfu.ca/projects/pihex/index.html) was able to use this kind of formula to work out the quadrillionth bit of pi.

Now, as far as I am aware this formula works in base 16 to return a hexa-decimal value of pi, however when it is run through a base converter 16->10 it does not give the decimal version of pi. So I would like to know if there is a way to compute the n'th digit of pi in decimal/ convert the hex value to the deciaml one.
Thanks for all of your help :d (sorry if this is in the wrong section :blush: ).

I'm not too sure I understand your question, I ran the sumation from $0$ to $10$ and got $3.14159265359$, it had nothing to do with hexadecimal though.  I have a different question though, how do you get more decimal places when writing a prgram in C++?


find all primes such that $ p^{3}\minus{}q^{7}\equal{}p\minus{}q$.

[hide="Solution"]
Rewrite the equation to $ p(p^2 \minus{} 1) \equal{} (q^2 \plus{} q \plus{} 1)(q^2 \minus{} q \plus{} 1)(q \minus{} 1)(q \plus{} 1)q$.
Obviously $ p > q$ so we have three cases:
1.
$ p \mid (q \plus{} 1)$. Then $ p \equal{} q \plus{} 1$ so $ p \equal{} 3,q \equal{} 2$, does not satisfy the equation.
2.
$ p \mid (q^2 \minus{} q \plus{} 1)$. But $ p^2 \minus{} 1 < p^2 \leq (q^2 \minus{} q \plus{} 1)^2 < \frac {q^7 \minus{} q}{q^2 \minus{} q \plus{} 1} \leq p^2 \minus{} 1$ if $ q \geq 2$ so this yields no solutions, too.
3.
$ p \mid (q^2 \plus{} q \plus{} 1)$. But $ p^2 \minus{} 1 < p^2 \leq (q^2 \plus{} q \plus{} 1)^2 < \frac {q^7 \minus{} q}{q^2 \plus{} q \plus{} 1} \leq p^2 \minus{} 1$ if $ q \geq 4$ so $ q < 4$.
If $ q \equal{} 2$ then $ p|7$ so $ p \equal{} 7$, no solution.
If $ q \equal{} 3$ then $ p|13$ so $ p \equal{} 13$, which is a solution.

So the unique solution is $ (p,q) \equal{} (13,3)$.
[/hide]

[hide]Note that $ p^3 \minus{} p \equal{} q^7 \minus{} q > q^3 \minus{} q$, so $ p > q$. We have $ p^3 \minus{} q^7 \equal{} p \minus{} q > 0$, so $ p > q^{7/3}$. Also note that $ p$ divides $ q^7 \minus{} q \equal{} q(q^2 \minus{} 1)(q^2 \minus{} q \plus{} 1)(q^2 \plus{} q \plus{} 1)$. So one of $ q,q^2 \minus{} 1,q^2 \minus{} q \plus{} 1,q^2 \plus{} q \plus{} 1$ is divisible by $ p$, which gives $ q^2 \plus{} q \plus{} 1\ge p$. Therefore $ q^2 \plus{} q \plus{} 1 > q^{7/3}$. Note that $ q^2 > q \plus{} 1$, so $ 2q^2 > q^{7/3}$, which gives $ q < 8$. $ q \equal{} 2$, $ q \equal{} 5$, or $ q \equal{} 7$ gives no solution, and $ q \equal{} 3$ gives $ p \equal{} 13$. So the only solution is $ (13,3)$.[/hide]


Does anyone know what the cutoffs typically are to qualify for the USA Physics Olympiad? I  was looking at some of the sample problems and wondering if I'd actually have a chance of making it in.

Ahhh, so you're John Schulman.  Okay, this is a little off topic, but you do NCIML correct?  I live in Nassau too and I'm a tenth grader and currently in the lower part of the top 50.  The thing is though... most of the questions are SO easy and I can figure out how to do them but during the tests, I dont know, maybe I freeze or more likely the time factor, but I just can't do as well as I really could.  Perhaps, you've had the same or similar problems, how would you suggest dealing with this?

Oh, and I play jazz piano too, its fun.

Good to see another Nassau County guy on the forums. You're the second one I've met this way. :)
I've never really been good at NCIML/NYSML/ARML format with two questions at a time, probably for the same reasons as you, so I don't know what advice to give you. It's very hard to simulate the conditions of that sort of contest.
It will probably comfort you to know that if you stay in the top 50, you'll almost certainly be invited to be on the county all-star team that goes to NYSML and ARML.

Yea, I heard that.  So I'll try to stay in it.  Who else on the forums does NCIML?

[quote="schulmannerism"]Last year I emailed the director about that question. The cutoff for the semifinal exam varies each year, but is in the ballpark of 75 out of 140, where 140 = 2 * (multiple choice score out of 20) + (open response score out of 100.)

I don't have a similar figure regarding semi-final scores that will qualify you for the actual Team (the training camp.) Being younger than senior year gives you a huge boost. I qualified last year as a junior, and I estimate that I got about 2/3 of total credit.[/quote]

Aren't there 30 multiple choice questions, though?

yea there are 30
I don't know why he said 2*20 questions

your score= # right on multiple choice (max 30?)+ score on 4 free response(max 100)

And about 70 should qualify you.


What is the sum of the reciprocals of the equation

(2003x/2004) + 1 + (1/x) = 0?

Here's what I got so far:

[hide]We can multiply both sides by 2004 and then by x to get 2003x 2  + 2004x + 2004 = 0.  This simplifies to 2003x 2 + 2004(x+1) = 0.  I'm not sure where to go from here.[/hide]

Hints would be greatly appreciated

What do you mean by sum of the reciprocals?

If, by the sum of the reciprocals, you mean the sum of the reciprocals of the roots, this is how you do it from here (in spoiler):

[hide]If you have 2003x^2+2004x+2004=0, we first divide both sides by 2003 to get x^2+(2004/2003)x+(2004/2003)=0.  If the solutions of this are p and q, the above factors to (x-p)(x-q)=0, so we know (if we expand this) that -(p+q)=(2004/2003) which means p+q=-(2004/2003), and pq=2004/2003.[/hide]

This is the clever part (again, in spoiler):

[hide]We wish to find 1/p+1/q.  Adding the fractions, we get (p+q)/(pq), which, from the above, is equal to (-2004/2003)/(2004/2003), so the answer is -1.[/hide]

No, it wasn't out of cleverness that I got the above.  I have seen similar problems before, so I knew how to do this one.  One of the best ways to learn new methods is to look at problems like this and remember how to solve them.

I'm sorry, the question should have been "What is the sum of the reciprocals of the roots of the equation..."
I forgot to put roots.


How come sometimes when I hit "enter" in the classroom, it sends my message, but sometimes it just goes down another line? Thanks!

What do you mean by "goes down another line"?

He means that instead of entering the message, the cursor goes down a line.

This is because you are holding the shift key down (I'm pretty sure about this).

[quote="$ LaTeX$"]He means that instead of entering the message, the cursor goes down a line.

This is because you are holding the enter key down (I'm pretty sure about this).[/quote]
Ahh, now I understand what he is saying, I think.

Make sure you aren't accidentally pressing shift at the same time as enter.

Ah! That's what I was doing! Thanks everyone! :)


[hide] The problem reduces to $x<y$ and $x+y\leq 10$.

The equation $x+y=n$ has $\left \lfloor \frac{n-1}{2} \right \rfloor$ solutions: namely $(1,n-1)$, $(2,n-2)$, and so on. Once $x$ is greater than $\left \lfloor \frac{n-1}{2} \right \rfloor$, then $x\geq y$ so there are no more solutions.

The answer is then $\sum_{n=3}^{10} \left \lfloor \frac{n-1}{2} \right \rfloor=2(1+2+3+4)=\boxed{20}$.[/hide]

There's more than nine solutions to $x+y\le 10$

How many $(x,y,z)\in N$ is available such that $x+y=z>10$ and $x<y< 10$?


Let $ n,k$ be positive integers.Determine the possible values of $ k$ satisfying the next condition.

Condition:There exists $ n$ such that $ n$ has just $ k$ distinct prime factors and $ \frac {S(n)}{n} \equal{} k$.

,where $ S(n)$ is the sum of the positive divisors of $ n$ including $ n$ itself.

Trivially we cannot have a solution for $ k \equal{} 1$. For $ k \equal{} 2$, we find $ 2 \cdot 3 \equal{} 6$ works, for $ k \equal{} 3$, we find $ 2^3 \times 3 \times 5 \equal{} 120$ works, and for $ k \equal{} 4$, we find $ 2^5 \times 3^3 \times 5 \times 7$ works.

Let $ n \equal{} p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$. Note that $ k \equal{} \frac {S(n)}{n}$ $ \equal{} \left(\prod_{i \equal{} 1}^k 1 \plus{} p_i \plus{} p_i^2 \plus{} \cdots \plus{} p_i^{e_i}\right) \left/ \left(\prod_{i \equal{} 1}^k p_i^{e_i} \right)\right.$ $ \equal{} \prod_{i \equal{} 1}^k \frac {p_i^{e_i \plus{} 1} \minus{} 1}{(p_i \minus{} 1)p_i^{e_i}} < \prod_{i \equal{} 1}^k \frac {p_i}{p_i \minus{} 1}$. To maximize this last product, we minimize the $ p_i$. Thus $ k < \underbrace{\frac {2}{1} \cdot \frac {3}{2} \cdot \frac {5}{4} \cdot \frac {7}{6} \cdots \frac {P_k}{P_k \minus{} 1}}_{k\ \text{times}}$, where $ P_n$ is the $ n$th prime. For $ k \equal{} 5$, we have $ 5 \le 2 \cdot \frac 32 \cdots \frac {P_n}{P_n \minus{} 1} \cdots \frac {11}{10} \equal{} 4.8125 < 5$, contradiction, and by induction on $ \frac {P_n}{P_n \minus{} 1} < \frac {n}{n \minus{} 1}$, we see that $ k > 5$ also does not work.


function
induction
algebra unsolved
algebra

Let $ f: N\minus{}\minus{}\minus{}>N$ such that
(i): $ f(n\plus{}f(n))\equal{}f(n) ( \forall n\in N)$
(ii):for some $ n_0\in N, f(n_0)\equal{}1$
Show that:$ f(n)\equal{}1$

1. Prove $ f(n)\equal{}1,\forall n\geq n_0$
2. Prove $ f(k)\equal{}1,\forall k\equal{}1,..,n_0\minus{}1$

$ f(n_0 \plus{} 1) \equal{} f(n_0) \equal{} 1$ so using induction,its easy to see that $ \forall n\geq n_0: f(n) \equal{} 1$

in the other hand:

$ (n_0 \minus{} 1) \plus{} f(n_0 \minus{} 1)\geq n_0\Rightarrow 1 \equal{} f(n_0 \minus{} 1 \plus{} f(n_0 \minus{} 1)) \equal{} f(n_0 \minus{} 1)$

so again by induction we can show that $ \forall n < n_0: f(n) \equal{} 1$

so $ \forall n\in\mathbb{N}: f(n) \equal{} 1\Rightarrow f\equiv 1$


For all positive reals x, y determine the smallest possible value of

max(x, 1/y, 1/x + y).

Well, it may look horrible but actually there's a two - line solution. :)

I have a solution, but it definitely doesn't qualify as 2-line.

Because in the intermediate forum, I feel like I should give other people a longer chance than I would in advanced, where I usually will post as soon as I can answer something.

Well ? No solution ? :? Should I post it ?

Well if nobody else is going to..

[hide]If "x" is the maximum, then "1/x + y" is also a maximum and vice versa (otherwise changing x until they are equal will decrease the max of these) and if "1/y" is the maximum so is "1/x + y" and vice versa. Basically that shows that all three must be equal with a little thought.. so xy=1 and x:^2:=1+xy = 2, so the answer is :rt2:.[/hide]


Are all Borel sets $ F_\delta$ or $ G_\sigma$ sets ?

[quote="mlok"]Using Baire category, it's easy to show that $ \mathbb Q$ is not $ G_{\delta}$ in $ \mathbb R$. Therefore, $ ([0,1]\cap Q)\cup ([2,3]\setminus\mathbb Q)$ is neither $ F_{\sigma}$ nor $ G_{\delta}$.[/quote]

Sorry for possibly stupid question, it should probably be easy.

I agree that $ Q$ isn't $ G_\sigma$.

I don't see how you prove that this union is neither $ F_\delta$ nor $ G_\sigma$. As far as i understand the first element of the union $ [0,1]\cap Q$ is $ F_\delta$, but i have i'm not sure i can say something definite on the second one element of the union. It doesn't seem to be either closed or open.

I would appreciate if you could give a precise idea or proof how to do that.

[quote="mlok"]The complements of $ G_{\delta}$ sets are $ F_{\sigma}$ and vice versa. Since $ \mathbb Q$ is not $ G_{\delta}$, its complement in $ [2,3]$ is not $ F_{\sigma}$.[/quote]

Ok, and why cannot the union of $ F_{\sigma}$ and something, which is not $ F_{\sigma}$ be  $ F_{\sigma}$ nor $ G_{\delta}$ ?

Thanks.

My example is the union of two "separated" sets. If $ A\equal{}([0,1]\cap \mathbb Q)\cup ([2,3]\setminus\mathbb Q)$ is $ G_{\delta}$, then so is $ A\cap (\minus{}\infty,3/2)\equal{}[0,1]\cap \mathbb Q$, a contradiction. If $ A$ is $ F_{\sigma}$, then so is $ A\cap [2,3]\equal{}[2,3]\setminus\mathbb Q$, a contradiction.

Ok. Now i've got it. Thank you and sorry for indeed stupid questions.


calculus
derivative
function
trigonometry
calculus computations

Given $ f(x)$ is a function such that
$ f(x) = \{\begin{array}{ccc} x^{\alpha}\sin{\frac {1}{x}} & , & x > 0 \\
 \\
0 & , & x = 0 \end{array}$
 where $ \alpha >0$ is a constant
Find $ \alpha$ satisfy $ f(x)$ has derivative $ \forall x \ge 0$

When $ x\neq 0$,
$ f'(x)\equal{}\alpha x^{\alpha\minus{}1}\sin{\frac{1}{x}}\minus{}x^{\alpha\minus{}2}\cos{\frac{1}{x}}$


calculus
integration
number theory unsolved
number theory

Find all positive integral triples (a,b,c) such that:
i)$b|a^2-1;c|b^2-1; a|c^2-1$
ii)$a+b+c>2003$

I have checked out here http://www.pdmi.ras.ru/~olymp/ [Russian] and:
At the olympiad [Saint Petersburg, 2003] the problem was given under different conditions. It was sufficient to prove that there exist a triple $(a,b,c)$ satisfying those conditions. It was proposed to 7th grade.  :)

Well, a solution is $(a, 2a-1, 2a-2)$, where $a$ is an odd integer. To have $a+b+c>2003$, it is sufficient to have $a\geq402$.


not sure if this is the correct section to post but seemed most suitable to me...please move if needed.

here's the problem integral problem
[img]http://xs107.xs.to/xs107/06413/Clipboard01.jpg[/img]

the way i've been doing it, and the way i want to is to find the antiterivative and then sub in the limits, however, have some trouble doing so.
i'l use '~' as the integral symbol
and 'sqrt' as square root symbol

first i let u = sinx   where du=cosxdx
therefore

=-sinx 0~1/2 sqrt(u)*du       if x=0, u=0      if x=1/2, u=0.009

= -sinx 0~0.009 u^1/2*du

=-sinx [2/3U^2/3]

=[-sinx(2/3sinx^3/2)]

and the answers are completely wrong. i have check it on my calculator and it should be 0.031 or thereabouts.

secondly, solving the systems using the Gaussian elimination method, i haven't been able to come across any results that satisfy and of the equations.

here is the equations giving me trouble
[img]http://xs107.xs.to/xs107/06413/Clipboard02.jpg[/img]

and here is what i worked out

  [img]http://xs207.xs.to/xs207/06413/Clipboard03.jpg[/img]

if someone could smash up what they thinks right, thatd be awesome as i'm just blind to what i've done wrong (most people are to their own mistakes!), or how i should go about it. these are the methods i need to use to solve them to so please dont smash up easier ways, i'd be using them if i could!
thanks,
dave

[quote="1600dave"]=-sinx 0~1/2 sqrt(u)*du[/quote]

Not sure what that extra term is doing out there.  It should just be

$\int_{0}^{ \frac{1}{2}}\sqrt{u}\, du$

still a bit lost, could you work through it a bit more?

t0rajir0u, you're wrong. It should be $\int_{0}^{\sin{1\over 2}}\sqrt{u}du$ (take care of the boundaries!)

thanks, figured out the integral! the answer is 0.221 :)

And I had an error in my previous post, which is now edited.


algebra
polynomial
abstract algebra
superior algebra
superior algebra unsolved

show that $ \frac {Z_2 [x]}{(x^3 \plus{} x \plus{} \hat 1)} \simeq \frac {Z_2 [x]}{(x^3 \plus{} x^2 \plus{} \hat 1)}$ show that they are $ Z_2$ isomorph

[quote="Ottem"]Try $ x\mapsto (x \plus{} 1)$.[/quote]

$ X^3\plus{}X\plus{}\hat{1}\equal{}X^3\plus{}X^2\plus{}\hat{1}$ in $ Z_2[X]$
so it's basically $ Z_2[X]$ factored over the same polynomial no ? that should be isomorphic
do we need to apply some isomorphism theorem maybe ?

$ X^3 \plus{} X \plus{} 1$ and $ X^3 \plus{} X^2 \plus{} 1$ are distinct in $ \mathbb{Z}_2[X]$. The point is that
$ (X\plus{}1)^3 \plus{} (X\plus{}1) \plus{} 1 \equal{} (X^3 \plus{} X^2 \plus{} X \plus{} 1) \plus{} (X\plus{}1) \plus{} 1 \equal{} X^3 \plus{} X^2 \plus{} 1.$ So $ x \mapsto x\plus{}1$
is the map from left to right, and $ x \mapsto x\minus{}1$ inverts this as a map from right to left.

[quote="spx2"] $ X^3 \plus{} X \plus{} \hat{1} \equal{} X^3 \plus{} X^2 \plus{} \hat{1}$ in $ Z_2[X]$ so it's basically $ Z_2[X]$ factored over the same polynomial no ? ?[/quote]

No, you're confusing (formal)  polynomials vs. polynomial functions. Over $ \mathbb{Z}_2$ the equality $ X^2 \equal{} X$ is true functionally but not formally.
Said equivalently, it is an identity of $ \mathbb{Z}_2$ but it is not an identity of $ \mathbb{Z}_2$-algebras (rings containing a central image of $ \mathbb{Z}_2$).


Company WizRuz is selling magic cards through the Internet. In 1995, they sold 100 cards. In 2005, they sold 400. Assuming that the number of cards sold increases linearly, how many cards will they sell in 2011?

[hide]they should sell 580 cards in 2011. they sell 300 additional cards every 10 years, so that's 180 additional cards every 6 years. so, 400+180=580.[/hide]

[hide="is it"]180?[/hide]
whoops wrong answer. hehe
[hide="real answer"]580[/hide]

Correct.

Now try to solve the problem using a linear equation.

[hide]2005 is the 11th term from 1995.

$a_n=a_1+(d)(n-1)$, where $a_n$ is the nth term, d, is the common difference, and $a_1$ is the first term. (Learn this formula, it may be useful).

$400=100+d(11-1)$
$d=30$

Now, 2011 is the 17th term from 1995, so
$a_n=100+30(17-1)$
$a_n=580$.

It is sure refreshing to write such a long solution in the basics forum  :D [/hide]


Find all functions $ f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfy \[ f(f(x)\plus{}y) \equal{} f(x^2 \minus{} y) \plus{} 4f(x)y\] for all $ x,y \in \mathbb{R}$.

[quote="rajelaz"]1. Find all function $ f(x) \mathbb{R} \rightarrow \mathbb{R}$ that satisfy  $ f(f(x) \plus{} y) \equal{} f(x^2 \minus{} y) \plus{} 4f(x)y$  for $ \forall x,y \in \mathbb{R}$  [/quote]
Set $ y\equal{}\frac{x^2\minus{}f(x)}2$ and we get $ f(x)(f(x)\minus{}x^2)\equal{}0$ and so : $ \forall x\in\mathbb R$, either $ f(x)\equal{}0$, either $ f(x)\equal{}x^2$

Suppose now $ \exists u\neq 0$ such that $ f(u)\equal{}0$ and $ v\neq 0$ such that $ f(v)\equal{}v^2$
Using $ x\equal{}u$ and $ y\equal{}v$, we get $ v^2\equal{}f(u^2\minus{}v)$ and so $ f(u^2\minus{}v)\equal{}(u^2\minus{}v)^2$ and so $ v\equal{}\frac{u^2}2$ and so contradiction : just choose another $ x\notin\{0,\minus{}u,u,v,\frac {u^2}2\}$ and either $ f(x)\equal{}0$ and we get $ v\equal{}\frac{x^2}2\equal{}\frac {u^2}2$, either $ f(x)\neq 0$ and we get $ x\equal{}\frac{u^2}2$

So two possibilities :
Either $ f(x)\equal{}0$ $ \forall x$
Either $ f(x)\equal{}x^2$ $ \forall x$

And it is easy to check that there two functions indeed are solutions.


My teacher said that we will get our amc 8a scores right before winter break (which is what happend last year). So... I was just wondering if we can email the amc director and get our scores from them because I've heard that other people did that. Because honestly, I can't wait another month until I get my scores! If this is possible, who do we email?

Also, what is the minimum score you can get on the amc 8 and still make it to the amc 10?

Thanks. This is my first time taking the Amc8 and i am utterly confused.

you don't qualify for AMC 10, you sign up. it is exactly like AMC 8, there are no requirements to take AMC 10.

No, do not email the AMC office or me for scores, we do not have your score in any accessible database.   We are still receiving packages of answer sheets from schools, we are still logging them in, we are still scanning the answer sheets, we are still compiling and sorting the data.  It takes time, about 2 to 3 weeks, to receive and compile all the data.   Therefore, we do not have the capability to find your score and respond to your individual request.   As soon as your score does become accessible, it is emailed to your contest manager.  Therefore, there is no advantage to trying to contact the AMC, your teacher will have your scores as soon as we do.

When your teacher receives the score reports, there will be information about registering for the AMC 10.  Talk to your teacher about the AMC 10.

Steve Dunbar
MAA Director, American Mathematics Competitions


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

PS: My proof is only 2 lines  :oops:

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {1}{2a^2 \plus{} bc} \plus{} \frac {1}{2b^2 \plus{} ca} \plus{} \frac {1}{2c^2 \plus{} ab} \geq\ \frac {1}{ab \plus{} bc \plus{} ca} \plus{} \frac {2}{a^2 \plus{} b^2 \plus{} c^2}$

PS: My proof is only 2 lines  :oops:[/quote]
We need to check only $ b\equal{}c\equal{}1$ and $ c\equal{}0.$ :wink:

hello, what do you mean with $ b\equal{}c\equal{}1$ and $ c\equal{}0$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, what do you mean with $ b \equal{} c \equal{} 1$ and $ c \equal{} 0$?
Sonnhard.[/quote]
I mean that for a proof of this inequality enough to check only two cases:
1) b=c;
2) c=0.

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {1}{2a^2 \plus{} bc} \plus{} \frac {1}{2b^2 \plus{} ca} \plus{} \frac {1}{2c^2 \plus{} ab} \geq\ \frac {1}{ab \plus{} bc \plus{} ca} \plus{} \frac {2}{a^2 \plus{} b^2 \plus{} c^2}$

PS: My proof is only 2 lines  :oops:[/quote]
$ \frac{1}{2a^2\plus{}bc}\plus{} \frac{1}{2b^2\plus{}ac}\plus{} \frac{1}{2c^2\plus{}ab} \ge \frac{4(a\plus{}b\plus{}c)^2}{6(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\plus{}ab(a^2\plus{}b^2)\plus{}bc(b^2\plus{}c^2)\plus{}ca(c^2\plus{}a^2)\plus{}4abc(a\plus{}b\plus{}c)} \ge \frac{(a\plus{}b\plus{}c)^2}{ab(a^2\plus{}b^2)\plus{}bc(b^2\plus{}c^2)\plus{}ca(c^2\plus{}a^2)\plus{}abc(a\plus{}b\plus{}c)}\equal{} \frac{2}{a^2\plus{}b^2\plus{}c^2}\plus{} \frac{1}{ab\plus{}bc\plus{}ca}$

Nice proof, secrets  :)
I also use CS but with a different way:
By CS, we have: $ \sum\ \frac{bc}{2a^2\plus{}bc} \geq\ 1$
and $ \sum\ \frac{a(b\plus{}c)}{2a^2\plus{}bc} \geq\ \frac{2(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2}$
Add 2 ineqs above, we obtain q.e.d  :)


trigonometry
ratio
geometry
trig identities
Law of Sines

Pop Eye takes his family sailing. They leave dock A and sail 1.5 miles on a course of 30 degrees to buoy B. They turn and travel 1.75 miles on a bearing of 110 degrees to buoy C. How far is it from buoy C to dock A?

I found a solution using a Law of Sines, setting the ratio of 1.75 to the sine of 30 degrees equal to the ratio of $ x$ to the sine of 110 degrees.

But something tells me there's a much easier way. Anybody know?

I interpreted the problem differently, supposing that the angles are given in respect to a fixed axis (relative to the boat). Not very pretty answer, though.

[hide]Then I get a triangle with a central angle of $ (180\minus{}110)\plus{}30\equal{}100$ degrees bordered by two sides of length 1.5 and 1.75. That gives that $ c$, the desired distance, is $ c \equal{} \sqrt{1.5^2\plus{}1.75^2\minus{}2(1.5)(1.75)\cos{100}}\approx2.49483$[/hide]


Κάθε σημεί του επιπέδου χρωματίζεται είτε κόκκινο είτε πράσινο.Έστω $ ABC$ ένα σταθερό τρίγωνο. Να αποδείξετε ότι υπάρχει ένα τρίγωνο $ DEZ$ στο επίπεδο έτσι ώστε το τρίγωνο $ DEZ$ να είναι όμοιο με το τρίγωνο $ ABC$ και οι κορυφές του τριγώνου $ DEZ$ να έχουν όλες το ίδιο χρώμα.

[quote="kritikos"]Θα έλεγα οτι το τριγωνο DEZ είναι αυτό που σχηματίζουν τα συμμετρικά του ορθοκέντρου του τριγώνου ABC ως προς της πλευρές του.[/quote]

Περίεργο μου φαίνεται αυτό :huh: πώς γίνεται αυτό το συγκεκριμένο τρίγωνο που λες να είναι πάντα μονοχρωματικό? Θα μπορούσε κάλλιστα να μην είναι...


How many unlabeled families of subsets of a $ 3$-set are there?

So we know that there are $ 2^{2^3} \equal{} 256$ labeled families. How do we get the number of unlabeled?

You've already posted this once before.  Please don't double-post.  Also, "Calculus computations and tutorials" is a completely ridiculous place to put this problem.


Find the smallest whole number n such that (5^n)+n is divisible by 31.
A. 23
B. 30
C. 51
D. 68
E. 88

5^1=5mod31
5^2=25mod31
5^3=1mod31
5^4=5mod31
Now we have a cycle.
so we want to attain 31 mod 31.
5^23+23 does not equal 0 mod 31
5^30+30 equals 0 mod 31.
because 5^30=1mod31 +30mod31=0mod31.
So we are done.

For $ n\geq 3$, we have that $ 5^n \plus{} n \equiv 5^{n \mod 3} \plus{} n \equiv 0 \mod 31$, since $ 5^3 \equiv 1 \mod 31$.

So for $ 5^n$, it cycles between the numbers $ 1, 5,$ and $ 25$, for $ n \equiv 0, 1,2 \mod 3$.

So we have 3 cases:

Case 1: $ n \equiv 0 \mod 3$

$ 1 \plus{} n \equiv 0 \mod 31$

$ n \equiv 30 \mod 31$

Smallest integer satisfying this is $ n \equal{} 30$.

Case 2: $ n\equiv 1 \mod 3$

$ 5 \plus{} n \equiv 0 \mod 31$

$ n \equiv 26 \mod 31$

Smallest number is $ n \equal{} 88$.

Case 3: $ n \equiv 2 \mod 3$

$ 25 \plus{} n \equiv 0 \mod 31$

$ n \equiv 6 \mod 31$

Smallest number is $ n \equal{} 68$.

$ \therefore n \equal{} 30 \rightarrow \boxed{B ) 30}$

$ 31$ is prime
$ a^{30}\equiv1\mod31$ by fermat's theorem where $ GCD(a,31)\equal{}1$
thus the answer is $ 30$

modularmarc101 gave the best solution since it doesn't depend on given possible answers but directly proves that $ n\equal{}30$ is the desired minimal value.
krishkoushik and implex, you didn't prove that $ 23$ isn't desired answer.

$ {n^5\plus{}n}\equal{}n(n\minus{}1)(n\plus{}1)(n^2\plus{}1)$.
Minimality is $ {n\minus{}1}$ i.e $ {31\minus{}1}\equal{}30$

this is really easy if you do modular arithmatic

5^3=remainder is 1
5^4=remainder is 5
5^5=remainder is 25

using this patter you began experimenting with the choices

let us try A

5^23 + 23
well if we use our pattern we have to find the remainder of 5 ^23 when divided by 31
if we subtract 23 -2=21 and 3-2=1
we the remainder is 25

then we add 25 to 23which is 48 and 48/31 leaves a remainder of 17 so that is not right

now let us try B

5^30+30 we do the same thing use our pattern to find out what 5^30 divided by 31 is

so we subtract 30 -2=28 and 3-2=1
using the pattern the remainder is 1

so we add 1+30= 31

it evenly divides!
so the answer is B 30

Yeah that is a better solution for when it's a multiple choice test.

Looking for patterns. I started by just listing out the first couple of $ 5^n$. I noticed that $ 125 \equiv 1 \mod 31$. So it becomes pretty clear that for all $ n \geq 3$ you can write $ 5^n$ as $ 125 \cdot 5^{n \minus{} 3} \equiv 5^{n \minus{} 3} \equiv 5^{n \minus{} 6} \equiv 5^{n \minus{} 9}...$

if it is not aq multiple choice test all you have to do is take the remainders and see what power adds to 31

but it wont work till thirty

because you have to try out the power

or you can use crazy number theory skills if you are really good at number theory


A function is so defined that for all x, $ [f(x)]^n\equal{}f(nx)$. Prove that $ f(x).f'(nx)\equal{}f'(x).f(nx)$, where f'(x) is ofc the derivative of  f(x) wrt x.

Assuming [] doesn't stand for GIF, :D

[hide="solution"]
divide by $ f(nx)$ on both sides and differentiate wrt $ x$
ofcourse this is assuming f(x) not = 0
and i'm too lazy to take the case of $ f(x) \equal{} 0$

[/hide]

[quote="hell_ever"]A function is so defined that for all x, $ [f(x)]^n \equal{} f(nx)$. Prove that $ f(x).f'(nx) \equal{} f'(x).f(nx)$, where f'(x) is ofc the derivative of  f(x) wrt x.[/quote]

Dunno where you got this, but this was a test question in some career point test.

[hide]
$ [f(x)]^n \equal{} f(nx)$
Diff both sides w.r.t x

$ nf(x)^{n \minus{} 1} . f'(x) \equal{} nf'(nx)$
Multiply both sides by f(x)

$ [f(x)]^ n . f'(x) \equal{} f'(nx) . f(x)$

But $ [f(x)] ^n \equal{} f(nx)$

Hence proved :D

[/hide]


Let $ a,b,c$ be nonnegative real numbers. For any $ 1\le k \le2$ or $ k \ge 3$,

$ a^k \plus{} b^k\le (a \plus{} b)^k \minus{} (2^k \minus{} 2)ab(\frac {a \plus{} b}{2})^{k \minus{} 2}$.

For $ 0\le k \le 1$ or $ 2\le k\le 3$, the reverse inequality holds.

Let $ f(x)\equal{}x^k\plus{}(1\minus{}x)^k\plus{}\frac{2^k\minus{}2}{2^{k\minus{}2}}x(x\minus{}2)$ for $ 0\leq x\leq 1$,

$ f'(x)\equal{}k\{x^{k\minus{}1}\minus{}(1\minus{}x)^{k\minus{}1}\}\plus{}\frac{2^k\minus{}2}{2^{k\minus{}2}}(1\minus{}2x)$, $ f(0)\equal{}f(1)\equal{}1,\ f\left(\frac{1}{2}\right)\equal{}\left(\frac{1}{2}\right)^{k\minus{}1}$, yielding

$ f(x)\leq 1.$ Plugg $ x\equal{}\frac{a}{a\plus{}b}$ into the inequality, we can obtain our inequality.


Prove that the three circles whose diameters are the diagonals of a complete quadrilateral are coaxal.

(That is, if A, B, C, D are four points, and AB meet CD at E, AD meet BC at F, then the circles with diameters AC,BD and EF have a common radical axis.)

Oh well, this is really well-known. In an equivalent version, it's Theorem 2.43 in Coxeter & Greitzer, [i]Geometry Revisited[/i], Toronto - New York 1967.

  Darij

Moreover, the axis is the line of the four orthocenters of the triangles formed by the lines and it's perpendicular to the Gauss line of the quadrilateral (well, that's obvious, of course :)).


group theory
abstract algebra
superior algebra
superior algebra unsolved

I'm new in group theory and I want a little help.

1)Prove that the number of solutions of $ x^3\equal{}e$ are odds.

2)Let $ G$ be a group and $ a,b \in G$ , $ n \in \mathbb{N}$ and $ xy^n\equal{}y^{n\plus{}1}x , yx^n\equal{}x^{n\plus{}1}y$. Show that $ x\equal{}y\equal{}e$.

3)Let $ G$ be a group and $ A,B$ be two subgroups of $ G$ and $ |A|\plus{}|B|>|G|$ .Prove that $ AB\equal{}G$.

4)Prove that in any group $ G$ the number of $ a \in G$ which $ a^2 \not\equal{} e$ are evens and show that if $ G$ be a group ($ |G|\equal{}2k$) there exists an element $ a \not\equal{}e$ s.t. $ a^2\equal{}e$.

1. Other than $ e$, for each $ x$ such that $ x^3 \equal{} e$, $ x^{\minus{}1}$ is another solution different from $ x$. Group $ x$ with $ x^{\minus{}1}$, then you can see that the number of solutions is odd.

3. I suppose your G is finite. In this case, it means that if $ |A| \ge |B|$, then $ 2|A| > |G|$, thus $ |A| \equal{} |G|$, and actually $ A \equal{} G$. Of course $ AB \equal{} GB \equal{} G$.

4. Similar to 1. If $ a^2 \neq e$, then $ a \neq a^{\minus{}1}$. Group $ a, a^{\minus{}1}$.
If $ |G|$ is even. Then there are even number of elements $ a$ with $ a^2 \equal{} e$ (it's just the complement of the set defined just now). But $ e$ is in this set, being even it has at least one other element, nontrivial, in this set. We are done.

Is it possible to use that solution for other values of $ n$?

[quote="vict85"]2.
$ xy \equal{} y(yx) \equal{} yx^2y \equal{} x^3y^2 \equal{} x^2(xy^2) \equal{} x^5y \equal{} x^4y^2x \equal{} x^3xy^2x \equal{} x^6yx \equal{} x^8y$
$ xy \equal{} x^5y \equal{} x^8y\ \ \to \ \ x \equal{} x^5 \equal{} x^8$
$ x^5x^{ \minus{} 1} \equal{} e \ \ \to \ \ x^4 \equal{} e$
$ x^8x^{ \minus{} 5} \equal{} e \ \ \to \ \ x^3 \equal{} e$
$ x^4 \equal{} xx^3 \ \ \to \ \ x \equal{} e$
$ y^2e \equal{} ey \ \ \to \ \ y^2y^{ \minus{} 1} \equal{} e \ \ \to \ \ y \equal{} e$[/quote]

thanx, but for $ n\equal{}1$ it's trivial ,can you solve it for $ n\equal{}2$ ?

[quote="ehsan2004"]3)Let $ G$ be a group and $ A,B$ be two subgroups of $ G$ and $ |A| \plus{} |B| > |G|$ .Prove that $ AB \equal{} G$.[/quote]

Well, probably it should be:

Let $ G$ be a group and $ A,B$ be two sub[b]sets[/b] of $ G$ and $ |A| \plus{} |B| > |G|$. Prove that $ AB \equal{} G$.

It wasn't the case n=1...
Anyway...

$ xy^{n} \equal{} y^{n \plus{} 1}x$
$ xy^{n}x^{ \minus{} 1} \equal{} y^{n \plus{} 1}$
$ y^{n \plus{} 1} \in Orb(y^n)$

But $ y^{n \plus{} 1} \in G_{y^n}$

$ G_{y^n} \cap Orb(y^n) \equal{} \{ y^n \}$

So $ y^{n \plus{} 1} \equal{} y^n\ \ \to \ \ y \equal{} e$

$ x^{n \plus{} 1}e \equal{} ex \ \ \to \ \ x^{n \plus{} 1}x^{ \minus{} n} \equal{} e \ \ \to \ \ x \equal{} e$

[quote="Soarer"][quote="LydianRain"]What is $ G_{y^n}$?[/quote]

I suppose he means the stabilizer of the conjugacy action. But I have no idea why $ G_{y^n} \cap Orb(y^n) \equal{} \{ y^n \}$ is true.[/quote]

 :blush: in fact it's not true... I consider in the wrong way the relation $ |Orb(x)|\equal{}[G : G_x]$

Counterexample:
$ (456)(123)(465) \equal{} (123)$


induction
number theory proposed
number theory

Prove , without induction  :D (sorry Ulanbek_Kyzylorda KTL) , that :

$ 147\ |\ 5^{4n \plus{} 1} \minus{} 3\cdot 10^{2n} \minus{} 2^{4n \plus{} 1}\ ,\ (\forall)n\in \mathbb{N}$ .


Use the half-angle formulas to simplify the expression.
$ -\sqrt{\frac{1-\cos 8x}{1+\cos 8x}}$.

cos 8x= 2 (cos 4x)^2 -1
cos 8x= 1-2 (sin 4x)^2
The expression can be simplified :
-tan 4x

You've already posted this here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161460[/url]

Do not double post.


geometry
circumcircle
inradius
trigonometry
geometry unsolved

Hi!

I'd like to show that a triangle $ ABC$ is right-angled iff $ r\plus{}R\equal{}\frac{a\plus{}b}{2}$, where $ r$ and $ R$ are the inradius and the circumradius.

I can prove that $ 2R\plus{}r\equal{}s$ is a sufficient condition, but I'm stuck with the other one.

Maybe someone can help?

Thanks
/Kon

We has some result:
$ 4\prod {\sin{\frac {A}{2}}\sin{\frac {B}{2}} + 1 = \frac {r + R}{R}}$
$ \frac {a}{2R} = \sin{A}$
Apply we has
$ \Leftrightarrow \cos{A} + \cos{B} + \cos{C} = \sin{A} + \sin{B}$
$ \Leftrightarrow (\cos{A} + 1)(\cos{B} + 1) = (\sin{A} + 1)(\sin{B} + 1)$
$ 2\cos{\frac {A}{2}}.cos{\frac {B}{2}} = (sin{\frac {A}{2} + \cos{\frac {A}{2}})(\sin{\frac {B}{2}} + \cos{\frac {B}{2}}}$
So $ cos\frac {A}{2}\cos{\frac {B}{2}} = \sin{\frac {2A + \pi}{4}}\sin{\frac {2B + \pi}{4}}$
So $ \cos{\frac {A + B}{2}} = - cos{\frac {A + b + \pi}{2}}$
so $ C = \frac{\pi}{2}$

Hi and thanks for replying.

Here is what I [i]do[/i] get:
$ r\plus{}R\equal{}\frac{a\plus{}b}{2}\Leftrightarrow \left(\cos A \plus{} 1\right)\left(\cos B \plus{} 1\right)\equal{}\left(\sin A \plus{} 1\right)\left(\sin B \plus{} 1\right)$

and

$ 2\cos\frac{A}{2}\cos\frac{B}{2}\equal{}\left( \sin\frac{A}{2}\plus{}\cos\frac{A}{2} \right)\left( \sin\frac{B}{2}\plus{}\cos\frac{B}{2} \right)\Leftrightarrow C\equal{}\frac{\pi}{2}$.

What I [i]don't[/i] get, however, is how you get

$ \left(\cos A \plus{} 1\right)\left(\cos B \plus{} 1\right)\equal{}\left(\sin A \plus{} 1\right)\left(\sin B \plus{} 1\right)\Leftrightarrow 2\cos\frac{A}{2}\cos\frac{B}{2}\equal{}\left( \sin\frac{A}{2}\plus{}\cos\frac{A}{2} \right)\left( \sin\frac{B}{2}\plus{}\cos\frac{B}{2} \right)$.

I ought to be pretty straightforward, but I can't seem to get it right. Could you show me what you did?

Thanks,

/Kon

[quote="kontinuert"]I'd like to show that a triangle $ ABC$ is right-angled iff $ r \plus{} R \equal{} \frac {a \plus{} b}{2}$, where $ r$ and $ R$ are the inradius and the circumradius.[/quote]

Your proposition is false, $ r \plus{} R \equal{} \frac {a \plus{} b}{2}$ does not imply that the $ \triangle ABC$ is right angled. $ \angle C \equal{} 90^\circ$ is possible, but not necessary.
See the problem [url]http://www.mathlinks.ro/viewtopic.php?&t=128227[/url].

First of all, thanks for the feedback!

If I modify the proposition to be the following then it holds true:

>>>>>
Let $ \Delta ABC$ be a triangle with $ C>A,B$, then we have:

$ r\plus{}R\equal{}\frac{a\plus{}b}{2} \Leftrightarrow C\equal{}90^{\circ}$.
<<<<<

[b]Brief proof:[/b]

$ r\plus{}R\equal{}\frac{a\plus{}b}{2} \Leftrightarrow$

$ \cos A \plus{} \cos B \plus{} \cos C \equal{} \sin A \plus{} \sin B \Leftrightarrow$

$ \cos C \equal{} \sin A \plus{} \sin B \minus{} \cos A \minus{} \cos B$

and

$ \cos^2 C \equal{} \left(\sin A \plus{} \sin B \minus{} \cos A \minus{} \cos B\right)^2 \Leftrightarrow$

$ \left( 1\minus{}\sin C \right)\left( 1\plus{}\sin C \right)\equal{}2\left( 1\minus{}\sin C \right)\left( 1\plus{}\cos \left( A\minus{}B \right) \right) \Leftrightarrow$

$ C\equal{}90^{\circ}\, \vee \, \sin C \equal{} 1\plus{}2\cos\left( A\minus{}B \right)$.

-----

If $ C\equal{}90^{\circ}$ then $ \cos C\equal{}0$ and $ \cos A \equal{} \sin B$ and $ \cos B \equal{} \sin A$, so $ \cos A \plus{} \cos B \plus{} \cos C \equal{} \sin A \plus{} \sin B$ is trivially true.

If $ r\plus{}R\equal{}\frac{a\plus{}b}{2}$ and $ C \neq 90^{\circ}$ then  $ \sin C \equal{} 1\plus{}2\cos\left( A\minus{}B \right)$ and consequently $ \cos\left( A\minus{}B \right)<0$ and thus $ A>A\minus{}B>90^{\circ}$ or $ B>B\minus{}A>90^{\circ}$ contradicting $ C>A,B$.

-----
/Kon


geometry
incenter
circumcircle
geometry proposed

A line passing through the incenter $I$ of the triangle $ABC$ intersect its incircle at $D$ and $E$ and its circumcircle at $F$ and $G$, in such a way that the point $D$ lies between $I$ and $F$. Prove that:
$DF \cdot EG \geq r^{2}$.

$DF.EG = (IF-r)(IG-r) = IF.IG-r(IF+IG)+r^{2}= (R^{2}-OI^{2})-r.FG+r^{2}= 2Rr-r.FG+r^{2}= r(2R-FG)+r^{2}$

but FG is one chord of the circuncircle, so, $2R \geq FG$, and the problem is solved!

$ CI$ meets the circumcircle of triangle $ ABC$ at $ C'$.

$ |CI|\equal{}\frac {r}{sin{\frac {C}{2}}}$ and $ |C'I|\equal{}2Rsin{\frac{C}{2}}$.

$ |GI|.|FI|\equal{}|CI|.|C'I|\equal{}2Rr   \Longrightarrow$

$ |DF|.|EG|\plus{}r^2\plus{}r(|GF|\minus{}2r)\equal{}2Rr \Longrightarrow$

$ |DF|.|EG|\equal{}2Rr\minus{}r^2\plus{}2r^2\minus{}r.|GF| \geq 2Rr\plus{}r^2\minus{}r.2R\equal{}r^2$ and solution is done.

$ Remark$: Equality holds if and only if $ GF$ is passing through $ O$.

$IF \cdot IG\equal{}(r\plus{}FD) (r\plus{}EG)\equal{} r^2\plus{}r(FG\minus{}2r)\plus{}DF \cdot EG$

Substituting $ IF \cdot IG\equal{}2Rr$ (power of I with respect to the circumcircle) gives:

$ DF \cdot EG\equal{} 2Rr \plus{} r^2\minus{}r \cdot FG.$

$DF \cdot EG$ attains its minimum when $ FG$ is maximum $\Longleftrightarrow$ $FG \equal{}2R$.

$\Longrightarrow \ DF \cdot EG\ge r^2.$


Is $ \mathbb{R}$ equipotent to $ \mathbb{R}\times\mathbb{R}$?

Does there exist such bijiective mapping?

I wonder if the way to treat $ \mathbb{N}\times\mathbb{N}$ can be applied?!

every infinite set $ X$ is equipotent to $ X\times X$ (a proof can be found in a.e. introduction in set theory). the proof requires the axiom of choice. but in the case of $ X \equal{}\mathbb{R}$, there are two more concrete possitibilites:

a) first, $ x\mapsto x/(1\plus{}|x|)$ yields $ \mathbb{R}\cong ]\minus{}1,1[$, and of course $ ]\minus{}1,1[\cong ]0,1[$. now use hilbert's hotel to show $ ]0,1[\cong ]0,1]$. now, binary representation shows that elements $ ]0,1]$ can be identified with infinite subsets of $ \mathbb{N}$ (without $ 0$). the subset consists of the indices where the $ 0$ occurs in the binary representation. infinite subsets of $ \mathbb{N}$ can be identified with strictly increasing sequences and these - via the differences of consecutive elements - with arbitrary sequences in $ \mathbb{N}$. thus we have shown $ \mathbb{R}\cong\mathbb{N}^{\mathbb{N}}$. finally, $ \mathbb{N}\equal{} 2\mathbb{N}\cup (2\mathbb{N}\minus{}1)$ is equipotent to the disjoint union $ \mathbb{N}\dot{\cup}\mathbb{N}$, so that $ \mathbb{N}^{\mathbb{N}}\times\mathbb{N}^{\mathbb{N}}\cong\mathbb{N}^{\mathbb{N}\dot{\cup}\mathbb{N}}\cong\mathbb{N}^{\mathbb{N}}$. it follows $ \mathbb{R}\times\mathbb{R}\cong\mathbb{R}$.

the resulting bijection $ ]0,1]\to ]0,1]\times ]0,1]$ is very clear: for example, if $ 0.01110001100...$ is the binary representation of a number in $ ]0,1]$, the according sequence is $ 1,4,1,1,3,1,...$.  the new two sequences are $ 1,1,3,..$ and $ 4,1,1,...$, i.e the corresponding element in $ ]0,1]\times ]0,1]$ is $ (0.00110...,0.11100...)$.

b) the mapping $ \mathbb{R}\to\wp(\mathbb{Q}), x\mapsto\mathbb{Q}_{ > x}$ is injective (take infima). thus there is an injection $ \mathbb{R}\to\wp(\mathbb{N})$. on the other hand, the map $ \wp(\mathbb{N})\to\mathbb{R}, T\mapsto\sum_{i\in T}2^{\minus{}i}$ is injective. the theorem of schröder-bernstein yields $ \mathbb{R}\cong\wp(\mathbb{N})$, thus $ \mathbb{R}\times\mathbb{R}\cong\wp(\mathbb{N}\dot{\cup}\mathbb{N})\cong\wp(\mathbb{N})\cong\mathbb{R}$.


Let
$ \frac{a}{x}\equal{}p,$

$ \frac{b}{y}\equal{}q,$

$ \frac{c}{z}\equal{}r.$

The given says that $ p\plus{}q\plus{}r\equal{}1$ and $ \frac{1}{p}\plus{}\frac{1}{q}\plus{}\frac{1}{r}\equal{}0$.  This last equation we can multiply by $ pqr$ and get

$ pq\plus{}qr\plus{}rp\equal{}0$.

If we multiply this by 2 and subtract it from the square of the other piece of information we are given, we get

$ (p\plus{}q\plus{}r)^2\minus{}2pq\minus{}2qr\minus{}2rq\equal{}1$

or

$ p^2\plus{}q^2\plus{}r^2\equal{}1$ which is what we wanted to prove.

[quote="tonypr"]I was at a math competition today and there was this problem that I had struggled with, unfortunately they didn't give solutions so, I am asking for any hints or solutions to this problem:

If $ a,b,c,x,y,z$ are Real numbers, and:
$ \dfrac{a}{x} \plus{} \dfrac{b}{y} \plus{} \dfrac{c}{z} \equal{} 1$

$ \dfrac{x}{a} \plus{} \dfrac{y}{b} \plus{} \dfrac{z}{c} \equal{} 0$

Prove that:

$ \left(\dfrac{a}{x}\right)^2 \plus{} \left(\dfrac{b}{y}\right)^2 \plus{} \left(\dfrac{c}{z}\right)^2 \equal{} 1$[/quote]
$ \dfrac{a}{x} \plus{} \dfrac{b}{y} \plus{} \dfrac{c}{z} \equal{} 1$
If we square it we got
$ \sum (\frac {a}{x})^{2} \plus{} 2(\sum\frac {ab}{xy})\equal{}1$
So we finaly have to prove that:
$ \sum\frac {ab}{xy} \equal{} 0\rightarrow \sum abz \equal{} 0$ So if we divide by $ abc$ we got
$ \frac {z}{c} \plus{} \frac {x}{a} \plus{} \frac {b}{y} \equal{} 0$ ,so we prove it....


Let S(n) be the sum of the reciprocals of the first n primes. How is it proved that S(n) is asymptotically equal to $ log(log(p_n))$ where $ p_n$ denotes the n-th prime.

hello, i think this will help you
http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges
Sonnhard.


An open box is constructed by starting with a rectangular sheet of metal $ 10$ in. by $ 14$ in. and cutting a square of side $ x$ inches from each corner. The resulting projections are folded up and the seams welded. The volume of the resulting box is:

$ \textbf{(A)}\ 140x \minus{} 48x^2 \plus{} 4x^3 \qquad \textbf{(B)}\ 140x \plus{} 48x^2 \plus{} 4x^3\qquad \\\textbf{(C)}\ 140x \plus{} 24x^2 \plus{} x^3\qquad \textbf{(D)}\ 140x \minus{} 24x^2 \plus{} x^3\qquad \textbf{(E)}\ \text{none of these}$

[hide="Solution"]
When a square of side length $ x$ is taken out, the side becomes $ \text{The original side length}\minus{}2x.$ That is one dimension of the base.
$ (10\minus{}2x)$ and $ (14\minus{}2x)$ are the dimensions of the base. Their product is the area of the base. $ (\minus{}2x\plus{}10)(\minus{}2x\plus{}14)\equal{}4x^2\minus{}48x\plus{}140$ is the area of the base.
The height of the prism is $ x,$ so the volume is $ x(4x^2\minus{}48x\plus{}140)\equal{}4x^3\minus{}48x^2\plus{}140x\equal{}\boxed{\textbf{(A)}}.$[/hide]


geometry
3D geometry
sphere
calculus
calculus computations

[quote="asdfG"]1. A plane intersects a sphere h units from the center of the sphere. The sphere as an area of 9pi square units and h increases at a rate of 1/5 unit per second. Circle Q is the circle of intersection for the plane and the sphere. Find the rate at which the area of circle Q changes when h is 1/2 unit. Give your answer in square units per second.
[/quote]

Finally I'm back in the game! It's been a while since I've answered some questions!

Let's say the area of circle $Q$ is $\pi r^{2}$ and we will arbitrarily decide to situate the plane parallel to the $xy$-plane where $Q$ is the intersection of this plane and a sphere of radius $\frac{3}{2}$ centered arbitrarily at the origin (we get the radius of the sphere by solving $4\pi r^{2}= 9\pi$.

If $h$ is the distance from $Q$ to the center of the sphere and $r$ is the radius of circle $Q$, then the radius of the sphere can be represented as $r^{2}+h^{2}=\frac{9}{4}$.

Substituting into the formula for area of circle $Q$  in terms of $h$ units from the center of the sphere we get $\text{Area of Q}=\pi \left(\frac{9}{4} - h^{2}\right)$.

Taking instantaneous rate of change with respect to time, $\frac{dQ}{dt}=-2\pi h \frac{dh}{dt}$ Substituting given values we get $\frac{dQ}{dt}=-2\pi \frac{1}{2}\cdot \frac{1}{5} = \boxed{\frac{-\pi}{5}\frac{\text{units}^{2}}{\text{sec.}}}$.

=)   i wonder why the surface area is given


Let $ a,b,c$ be real positive numbers and $ k \ge max(a^2,b^2,c^2)$.Prove that
  $ f(a,b,c) \ge f(a^2\plus{}k,b^2\plus{}k,c^2\plus{}k)$

where $ f(a,b,c)\equal{}\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\minus{}3\ge 0$

[quote="Hamddi"]Let $ a,b,c$ be real positive numbers and $ k \ge max(a^2,b^2,c^2)$.Prove that
  $ f(a,b,c) \ge f(a^2 \plus{} k,b^2 \plus{} k,c^2 \plus{} k)$
where $ f(a,b,c) \equal{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\ge 0$[/quote]
Assume without loss of generality c = min(a, b, c). We rewrite G(a, b, c)
in the following way
$ G(a,b,c)\equal{}(\frac{a}{b}\plus{}\frac{b}{a}\minus{}2)\plus{}(\frac{b}{c}\plus{}\frac{c}{a}\minus{}\frac{b}{a}\minus{}1)$
$ \equal{}\frac{(a\minus{}b)^2}{ab}\plus{}\frac{(a\minus{}c)(b\minus{}c)}{ac}$
It is enough to prove that
$ \frac{(a\minus{}b)^2}{ab}\plus{}\frac{(a\minus{}c)(b\minus{}c)}{ac} \geq \frac{(a\minus{}b)^2}{\frac{(a^2\plus{}k)(b^2\plus{}k)}{(a\plus{}b)^2}}\plus{}\frac{(a\minus{}c)(b\minus{}c)}{\frac{(a^2\plus{}k)(c^2\plus{}k)}{(a\plus{}c)(b\plus{}c)}}$
But this is true, because $ k \geq max(a^2,b^2,c^2)$ and $ (a\minus{}c)(b\minus{}c)\geq 0$. The proof is completed and
the equality holds when a = b = c.

totaly same with the official solution   :)


1. Let A1 A2 A3...........An be the vertices of an N sided regular polygon such that.
    1/(A1*A2)= 1/(A1*A3) + 1/(A1*A4). Find the value of n.

2. Find the Least value of the expression:

    cot2x-tan2x/ (1+ sin (5pi/2-8x)

    for 0<x<pi/8.

3. Prove that if Angles A and B satisfy the relation , sinB/sin(2A+B)= n/m where , |m|>|n|,

   then : (1+ tanB/tanA)/m+n = (1-tanAtanB)/m-n

4.  Show that the expression (ax-b)(dx-c)/(bx-a)(cx-d) will assume all real values when x is real ,
if (a^2-b^2) and (c^2-d^2) have the same sign.

Some more inequalities + quad. and log problems will be added to this thread later when the above are solved and answers are confirmed.
:)

No one as yet?? Or is this not exactly pre-olympiad?

So for the first problem, what do you mean by the product of two points? Do you mean the distance between them? Or are you referring to the products of two side-lengths, where $ A_i$ are actually the lengths of the sides counterclockwise of their respective vertices? Or something else?

[quote="thealchemist"]I think he means the distance between the two points... then its a standard problem. If I remember right, the number of sides is 7.[/quote]
Oh , bad of me. Well yes , distance between the vertices . Sorry. And 7 is correct.


inequalities
logarithms
inequalities proposed

[hide="Hint for 1st  ineq"]For 1st inequality: Put $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x$ then $ a,b,c$ are sides of a triangle.[/hide]

For the first one, if we apply Popoviciu inequality to $ f(x) \equal{}\ln x$ we are done, because

$ f(x)\plus{}f(y)\plus{}f(z)\plus{}3f\left(\frac{x\plus{}y\plus{}z}{3}\right)\leq 2\left[ f\left(\frac{x\plus{}y}{2}\right)\plus{}f\left(\frac{y\plus{}z}{2}\right)\plus{}f\left(\frac{z\plus{}x}{2}\right)\right]$,

and then

$ xyz\left(\frac{x\plus{}y\plus{}z}{3}\right)^{3}\leq\left(\frac{x\plus{}y}{2}\right)^{2}\left(\frac{y\plus{}z}{2}\right)^{2}\left(\frac{z\plus{}x}{2}\right)^{2}$.

[quote="gemath"]Given $ x,y,z > 0$ prove that
\[ \frac{(x\plus{}y\plus{}z)^{3}}{27xyz}\ge\frac{(x\plus{}y)^{2}(y\plus{}z)^{2}(z\plus{}x)^{2}}{64x^{2}y^{2}z^{2}}\]

\[ \frac{3\sqrt{3xyz}(x\plus{}y\plus{}z)}{2(x\plus{}y)(y\plus{}z)(z\plus{}x)}\plus{}\frac{8xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\ge 5\]
An question what is the best const $ k$ for
\[ \frac{3\sqrt{3xyz}(x\plus{}y\plus{}z)}{2(x\plus{}y)(y\plus{}z)(z\plus{}x)}\plus{}k\frac{8xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\ge 4\plus{}k\]
[/quote]
It's wrong, gemath. Try $ x\rightarrow0^{\plus{}}.$ :wink:

Yes, inequailties are on the reverse side, in fact, the second inequality is a consequence of the first one, and the third one holds for all $ k\geq 0$.

May be he means the following :
Let a,b,c >0. Prove that :
$ \frac{3\sqrt{3}(x\plus{}y)(y\plus{}z)(z\plus{}x)}{2\sqrt{xyz(x\plus{}y\plus{}z)^{3}}}\plus{}\frac{8xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\ge 5$
And also , find the best constant of the inequality :
$ \frac{3\sqrt{3}(x\plus{}y)(y\plus{}z)(z\plus{}x)}{2\sqrt{xyz(x\plus{}y\plus{}z)^{3}}}\plus{}\frac{kxyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\ge 4\plus{}\frac{k}{8}$
The best constant is : $ 9.3062334328912$

Ok sorry you are right, I mean
Given $ x,y,z > 0$ prove that
\[ \frac{(x\plus{}y)^{2}(y\plus{}z)^{2}(z\plus{}x)^{2}}{64x^{2}y^{2}z^{2}}\ge\frac{(x\plus{}y\plus{}z)^{3}}{27xyz}\]

\[ \frac{3\sqrt{3}(x\plus{}y)(y\plus{}z)(z\plus{}x)}{2\sqrt{xyz(x\plus{}y\plus{}z)^{3}}}\plus{}\frac{8xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\ge 5\]
An question what is the best const $ k$ for
\[ \frac{3\sqrt{3}(x\plus{}y)(y\plus{}z)(z\plus{}x)}{2\sqrt{xyz(x\plus{}y\plus{}z)^{3}}}\plus{}k\frac{8xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\ge 4\plus{}k\]
But about best const, can you prove it,or you shown only because we can show the const by the way let $ x \equal{} y$ and $ z\to 0$ however is it the best const in all cases?

Notice that :
$ \large sin (\frac{A}{2})sin (\frac{B}{2})sin (\frac{C}{2}) \equal{}\frac{r}{4R}$
$ \large cos(\frac{A}{2})cos(\frac{B}{2})cos(\frac{C}{2}) \equal{}\frac{p}{4R}$
So we need to prove :
$ \frac{6\sqrt{3}R}{p}\plus{}\frac{2kr}{R}\geq\ 4\plus{}\frac{k}{8}$
Use the following link  :lol:  :

http://www.mathlinks.ro/Forum/viewtopic.php?t=125623

Done ! That is the best constant of all cases !

I don't understand , can you prove that inequality with the const you find out? :D

[quote="gemath"]Ok sorry you are right, I mean
Given $ x,y,z > 0$ prove that
\[ \frac {(x \plus{} y)^{2}(y \plus{} z)^{2}(z \plus{} x)^{2}}{64x^{2}y^{2}z^{2}}\ge\frac {(x \plus{} y \plus{} z)^{3}}{27xyz}
\]

[/quote]
We can rewrite this inequality:
$ (x\minus{}y)^2[\frac{2z[(x\plus{}y)(y\plus{}z)(x\plus{}z)\plus{}8xyz]}{64x^2y^2z^2}\minus{}\frac{x\plus{}y\plus{}2z}{27xyz}]\plus{}(y\minus{}z)(x\minus{}z)[\frac{(x\plus{}y)[(x\plus{}y)(y\plus{}z)(x\plus{}z)\plus{}8xyz]}{64x^2y^2z^2}\minus{}\frac{2x\plus{}2y\plus{}z}{27xyz}] \geq 0$
 :blush:


What did you guys think about that #6 geo. problem on the last GML? I finally found the solution with about 5 minutes left but made a mistake in a 3rd case (which was actually equivalent to one of my previous cases) and got 18 as an answer, in addition to the correct answers of 2 and 14.


linear algebra
matrix
probability
real analysis
real analysis unsolved

Hello friends...the following problem is breaking my head..please help.

We all know that if $ a_n$ and $ b_n$ are two sequences of real nos. such that $ a_n\to 0$ and $ b_n\to 0$ as $ n\to \infty$, then $ a_n+b_n\to 0$ as $ n\to \infty$

Can we extend this when we have countably infinite no. of sequences?
That is, my question is the following.

Suppose that for each fixed integer $ j\ge 1$,         $ a{}_n{}_j$ is a sequence of positive real numbers such that  for fixed $ n$, $ \sum_j a{}_n{}_j<\infty$.

Under what condition(s), $ a{}_n{}_j\to 0$ as $ n\to \infty$ for each $ j$  iff  the sequence $ \{a{}_n{}_1+a{}_n{}_2+a{}_n{}_3+...\}\to 0$ as $ n\to \infty?$

Thanks in advance.

No.
Counterexample #1: $ a(i,j)=\begin{cases}1&i=j\\0&\text{otherwise}\end{cases}$
Couterexample #2: $ a(i,j)=2^{j-i}$.

Under what conditions that statement could be true?

[quote="jmerry"]No.
Counterexample #1: $ a(i,j) = \begin{cases}1 & i = j \\
0 & \text{otherwise}\end{cases}$
Couterexample #2: $ a(i,j) = 2^{j - i}$.[/quote]

Jmerry... your 1st example doesn't satisfy my hypothesis that all $ a{}_i{}_j$'s are positive real nos.
You 2nd example doesn't satisfy $ \sum{}_ja{}_i{}_j<\infty$ for every fixed $ i$

[quote="jmerry"]Do you really care about the difference between "positive" and "nonnegative"? For the other, I was just using a different labeling. Reverse the indices if you feel like it.[/quote]

Ya,  I really care about positivity, because this question has got some connection with a probability question.
Moreover, if you look at my first post, I have the double sequence such that for each  $ i$, $ \sum{}_ja{}_i{}_j<\infty$ and for each $ i$, $ a{}_i{}_j\to 0$ as $ n\to \infty$. I think your 2nd example doesn't meet these requirements altogether(If I'm not mistaken!)

Probability works just fine with zeros mixed in. I don't know any applications which actually care about being strictly positive everywhere.

The answer is still no, no matter how much you complain about any particular example. If you want convergence in a different direction than you started with, you basically need to guarantee it directly.

Sorry...in my just previous post it should be: For each  $ i$, $ \sum{}_ja{}_i{}_j < \infty$ and for each $ j$, $ a{}_i{}_j\to 0$ as $ i\to \infty$. There is some typo.
Actually the $ a{}_i{}_j$'s are strictly positive because my probability measure assigns positive probability at each of the countably infinite points.

Thanks for your valuable replies Jmerry.


analytic geometry
induction
modular arithmetic
geometry
perimeter
calculus
integration

A polygon on a coordinate grid is built of $ 2005$ dominoes $ 1 \times 2$. What is the smallest number of sides of an even length such a polygon can have?

We claim the best we can do is two, achievable by putting them all end-to-end in either of the two obvious ways. We prove by induction on the number of dominoes that there will always be at least two sides of even length. For $ n\equal{}1$, the result is obvious. Now assume it for $ n\equal{}k\minus{}1$. Consider any polygon created by $ k$ of these dominoes; we claim it has at least two even-length sides. This is because if we remove a domino, the resulting figure will have at least two, and to "get rid" of an even-length side, we would have to position the domino either so that it extended the side or interrupted the middle of the side. In either case, the "exposed" 2-unit side would give another even face - unless one of the existing dominoes covers or adjoins part of it. However, then we would in turn need another domino to cover that domino's exposed side, and in turn another, etc. Hence either the dominoes form a loop, contradicting the definition of a polygon, or there are infinitely many dominoes, which is of course impossible. We therefore have a contradiction, and the minimum is 2, as desired.

This is probably clearer: Every angle of the polygon is $ 90$ or $ 270$ and the sum of its angles is $ 0 \pmod{180}$. This means the polygon has an even number of sides. It also must have even perimeter; suppose we start at an arbitrary vertex and travel along the whole perimeter. If we travelled $ u$ units up, $ d$ units down, $ l$ units left and $ r$ units right, then $ u \equal{} d$, $ l \equal{} r$, and the perimeter is $ u \plus{} d \plus{} l \plus{} r \equal{} 2u \plus{} 2l$ which is even. Then the polygon having one side of even length means it has an odd number of sides of odd length and an odd perimeter, which is a contradiction. So at least two sides of even length are needed.

MellowMelon, that argument doesn't rule out the possibility that all sides are odd, does it?

You're right; somehow I thought that was the easier case. First solution still seems a bit imprecise to me though.

I think you can show a polygon made of unit squares on a checkboard, all side lengths odd, has a different number of black and white squares. Haven't managed to prove it though.

Hmm, let's clear this up a bit. The only hole I could see here is if the loop's interior was actually stacked with dominoes, but by the inductive step, I don't actually see this is a problem, because if this were the case, then the side of even length we're considering wouldn't have even length in the first place.

As for your colouring approach - which is pretty ingenious - I couldn't find a proof in the five minutes I thought about it, but I just realized something: does the problem stipulate that the vertices of the dominoes must fall on lattice points? Even without that restriction, I don't believe it's possible to get less than two.

fwolth: the vertices must be on lattice points or else we wouldn't get integral side lengths.

MellowMelon, I agree (from empirical evidence) that if a simply-connected polygonal region made of dominoes has all odd sides then it has different numbers of black and white squares, but it's not true for non-simply-connected regions (look at a $ 3 \times 3$ square with its center square removed, for example).  On the other hand, I've only been able to find such regions using an even number of dominoes (actually, a number divisible by 4, I think), and I imagine that non-simply-connected regions probably do not qualify as "polygons," but it does mean that any proof has to exclude this case in some fairly obvious way.  fwolth's proof attempted to do so, but I admit I don't find it incredibly convincing as written.  (Basically, the argument that there are only two cases, the one in which induction works and the one in which it's not simply connected, seems awfully much like intuition-driven handwaving.)

Yes, there would be nonintegral sidelengths - however, there would also still have to exist integral ones; and in particular, at least two even-sided ones.

JBL: Agreed that regions with stuff missing in the middle causes bad things to happen.

Here's the idea I have been trying to use to tackle this: consider the smallest rectangle of unit squares enclosing the entire polygon. You want to use an algorithm to modify the polygon and "fill" this rectangle, all while keeping careful count of the number of black squares and white squares added. The main thing you can do in this algorithm is something to this effect:
[code]XXXX
Xabc
Xdef
Xghi[/code]

Given two odd sides $ a \times b$ joined by a 270 degree angle, you can extend them and fill the rectangle they leave out, so you would add all of squares abcdefghi in. This would do two things though: create even sidelengths and change the difference of the black and white squares. One option to fix both is to leave out square i, but I think it's too hard to get the algorithm to terminate, and to fill the rectangle you have to make even sides eventually.

I posted [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=292336]a conjecture[/url] (mostly empirical) in my blog which shows that the number of 270 degree angles (aka the number of sidelengths) is critical, so having the algorithm target those would seem to be on the right track.

The main reason (among a few) I can't carry this through to completion is because I cannot figure out how to specify what to do in situations like this.

[code]XXXXX.X
XX....X
XX....X
XX,XXXX
XX,XXXX
XX,XXXX
XXXXXXX[/code]
Most of the rectangles that we might "fill in" as described above are blocked somehow. I don't know how to identify the spaces labeled with commas in the general case, as filling these is the obvious first step here.

[quote="MellowMelon"]I think you can show a polygon made of unit squares on a checkboard, all side lengths odd, has a different number of black and white squares. Haven't managed to prove it though.[/quote]Just for the sake of completeness, since I believe the blog post is no longer there...

Suppose the polygon $P$ has all odd sides and let $V$ be the set of its vertices. Color the unit lattice in a chessboard fashion, and for every point $p\in\mathbb{Z}^2$ define $f(p)$ to be the difference between the number of black and white squares (in the chessboard coloring) incident to $p$. Then $f(p)=0$ for all $p\in\mathbb{Z}^2\setminus V$, whence
\[4B-4W=\sum_{p\in\mathbb{Z}^2}f(p)=\sum_{p\in V}f(p).\]But $P$ has all odd sides, so considering 90 and 270 degree turns separately, we see that $f(p)\in\{-1,1\}$ is constant for all $p\in V$. In particular, $B\ne W$.

Alternatively, we can use [url=http://en.wikipedia.org/wiki/Green's_theorem]Green's theorem[/url] to get something equivalent---in the notation of the article, $L=-e^{i\pi(x+y)}$ and $M=e^{i\pi(x+y)}$ works. Intuitively, $e^{i\pi t}$ "extends" $(-1)^t$ to non-integer $t$. See pythag011's Mock USAMO Problem 1 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=458669]here[/url] for another application of the above ideas.

To prove the weaker statement that no domino tiling exists, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2554428#p2554428]height[/url] [url=http://math.stackexchange.com/questions/36099/transformations-of-domino-tilings/36109#36109]functions[/url] suffice. (For any cycle enclosing a region tiled by dominoes, the sum of the height function over the points/vertices of the cycle must vanish, but if we take the cycle to be the perimeter of $P$, the odd-side condition makes this impossible. [url=http://turgor.allzd.ru/lktg/2009/4/articles/conway.pdf]Conway and Lagarias[/url] give a reasonably detailed proof of this in Section 2, Theorem 2.1.)


geometry
3D geometry
sphere
calculus
integration

This was from a "math team tryout" that i just took today, not quite sure how to do it:

A sphere of clay has radius 13. A circle with radius 5 cuts through the center of the sphere. Squish the "bead" so it makes a new sphere. What is the radius of this new sphere?

I think what it meant was basically to cut out a cylinder with radius 5, but also with some "caps" on the ends. I'm having trouble finding the volume of these extra regions.

you need to use calculus to find the volume and then solve for the radius, it is pretty easy calculus problem
think of the cross section formula (a cross section would be done so the hole would make a circle on the cross section, and interms of the height x, the size of the cross section would be $\pi(13^2-x^2-5^2)$ (the first part for the outer circle, and the second part to remove the middle, and then you would integrate $[-12,12]$, but since it is symmetric, we can do $[0,12]$ and multiply by 2 (it is only to 12 because the cut out in the middle makes it so it only goes to a height of 12
$2\pi*\int_{0}^{12}144-x^2 = V$
$2\pi (144x-\frac{x^3}{3}) \, ^{12}_0] = 2304 \pi = V$
$\frac{4}{3}\pi r^3 = 2304 \pi$
$\boxed{r = 12}$
i think that i did it right, i am amazed if that is the answer, it worked out so well

that's what my guess got me, lol

i did it without the two hubcaps, and then rounded down to the nearest perfect cube...

lol...they should just have given a problem that does not have a nice answer so you could not do that, but test taking skills like that are useful

This is an old "classic" problem  - Generally asked as:  find the remainig volume  - if a  24 unit long hole is present  in the sphere of radius R.  If a "hole" is  $2\cdot 12$ " long" ,  (as it is here), it does not matter what the  radius of the original sphere (as long a it is large enough)   is , answer is alway the same!

(Limiting case: radius 12 units and the hole is  "zero"  unit thin and thus 24units long) :)


calculus
quadratics
number theory
number theory unsolved

Let $p$ be a prime number such that $ p \mid a^2 + 2$ for a natural number $a$. Prove that $p$ or $2p$ can be written on the form $x^2 + 2y^2$, with $x,y \in \mathbb{N}$.

Note that $2p=x^2+2y^2\Rightarrow x$ is even, so $x=2x_1\Rightarrow 2p=4x_1^2+2y^1\Rightarrow p=2x_1^2+y^2$, so the two conclusions of the problem are, in fact, one and the same :).

Assume $p$ is a prime in $A=\mathbb Z[i\sqrt 2]=\mathbb Z[u]$. Since $p|(a+u)(a-u)$, $p$ must divide one of $a+u$ and $a-u$, so there must be some $c,d\in\mathbb Z$ s.t. $a+u=p(c+du)$ or $a-u=p(c-du)$, which is absurd. This means that $p$ isn't prime in $A$, so we can find $a,b,c,d\in\mathbb Z$ s.t. $p=(a+bu)(c+du)$ s.t. $a+bu,c+du\notin\{\pm 1\}$. Now apply the norm: $p^2=(a+2b^2)(c+2d^2)$, and since $a^2+2b^2,c^2+2d^2\notin\{\pm 1\}$, we get $a^2+2b^2=p$, Q.E.D.

Very nice solution,grobber. All number theory problems should be killed with Calculus or Algebra :D .

Misha

Good solution Grobber.
I think it can be generalized that any number written
in the form x^2 + ty^2 where (x,y) =1  t  Z  t ≠ r^2.
has prime divizors of the form (u^2 + tv^2).
Could you write where can be found theory about
polinomials Zp, Z[i√2] and others.

The solution only works because $\mathbb Z[i\sqrt 2]$ is Euclidean, and thus an element is prime iff it's indecomposable. That solution doesn't work for general $t$, only for a few $t$'s.

Regarding info on quadratic integers rings, just input stuff like "algebraic number theory" in Google.


[quote="jasonwingofcarmelin"]When is an even root of a real number not a real number?

and

If rons speed is 13 km/hr faster then mike and in the time kike takes to bike 52 km, ron can bike 78 km.  what is rons speed in km/hr?

Whats 4096-2048+1024..if its geometric and has 16 terms?[/quote]
well for number 1, I just put if that orgininal real number is less than 0

number 2 is 39?  I think..thats what I got

Oh well, its kinda sad i wont get a perfect on the core 40 :rotfl: because I misread a question...and I put some guys speed when you were suppose to find his dad's speed..

1. not gonna bother cause i'm sleepy as hell

EDIT: yeah. what Art of Owna said.. there were too many "real"'s and "number"'s in there <_<

[hide="2"]
$\frac{78}{52}= \frac{3}{2}$ so apparently ron is 150% as fast as mike.

so that extra 50% corresponds to the extra 13km/h,

ron's speed is three times that much to get 150%, or $\boxed{39\text{km/h}}$
[/hide]

[hide="3 (copied from other post)"]
for geometric series,

$S_{n}= a_{1}\cdot\frac{1-r^{n}}{1-r}$

in this case, $a_{1}= 4096$, $r =-\frac{1}{2}$, and $n = 16$.

so $S_{n}= 4096\cdot\frac{1-(\frac{1}{2})^{16}}{1+\frac{1}{2}}= \boxed{\frac{21845}{8}}= 2730.625$

or you can just type in $\sum_{k=0}^{15}{(4096\cdot(-\frac{1}{2})^{k})}$ into a calculator :D
[/hide]

Number 1 was an open ended one..but I think theya re looking for less than 0, I hope :maybe:

undefined btw what did you get on usamo?

omg i misunderstotd the qustion..Is this really going on transcrpit?  Or is it useless like detentions on transcrpts>

wow transcripts?  rofl :rotfl:  :rotfl:  :rotfl:

no but your gpa is going to go on the transcript...they will like it when they see a 4.0..................................isnt that yours?


One or two times in a normal round, a problem like this came up:

"What is the remainder when 2^2005 is divided by 7?"
(it might not be 2005 but it was definitely something 1 mod 3)

The answer is definitely 2, but on the scoreboard it says the answer's 20!
20 cant even be an answer for the remainder mod 7!!  :noo:

Yeah, I meant to report error, on that, but I don't think I did.

I saw that when playing you and a bunch of other people, xpmath.  I reported it.  It was probably some kinda typo when entering the problem into the database, if that's even what happens.


calculus
integration
calculus computations

Suppose $f(x,y)$ is smooth enough and it satisfies that $\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2} = e^{-(x^2+y^2)}$. Prove that \[\iint_{x^2+y^2<1} \left( x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}\right)dxdy = \frac{\pi}{2e}.\]

I've managed to do it.
Consider the line integral
$\int_{x^2+y^2=1}(x^2+y^2)(-\frac{\partial f}{\partial y}dx+\frac{\partial f}{\partial x}dy)$

Apply Green's Theorem, we see that it is
$\iint_{x^2+y^2\leq 1} (2x\frac{\partial f}{\partial x}+(x^2+y^2)\frac{\partial^2 f}{\partial x^2}+2y\frac{\partial f}{\partial y}+(x^2+y^2)\frac{\partial^2 f}{\partial y^2})dxdy$
$=2\iint_{x^2+y^2\leq 1} (x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y})dxdy + \iint_{x^2+y^2\leq 1} (x^2+y^2)(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2})dxdy$
$=2\iint_{x^2+y^2\leq 1} (x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y})dxdy + 2\pi \int_0^1 r^3 e^{-r^2} dr$

On the other hand, the linear integral is exactly
$\int_{x^2+y^2=1} -\frac{\partial f}{\partial y}dx+\frac{\partial f}{\partial x}dy$
and from Green's Theorem it is
$\int_{x^2+y^2=1} (\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}) = 2\pi \int_0^1 e^{-r^2}rdr$

Then we can work out the answer easily.


It seems to me that the AMC tests are gradually getting easier.  What do you think?

Please post your thoughts.  What effect does a gradual shift to easier or harder questions have on the scores and/or qualification for AIME or USAMO?  If the tests are getting easier, does this make it easier or harder to qualify for USAMO because while your score may go up, the cutoff also goes up?

is it necessarily harder for a 10th grader or below though. because all they really have to do is get the floor aime, and if the aime is easier that year, then its not that hard to match the others...

[quote="Fierytycoon"]AMC people want more people to participate => AMC12's get much easier => Much higher cutoff scores for USAMO => Less room for careless mistakes => Fierytycoon freaks out about contests[/quote]

You forgot a couple:

=> Fierytycoon spends more time studying for AMC and AIME => Fierytycoon doesn't have time to be a chickmagnet.   :lol:

[quote]You forgot a couple:

=> Fierytycoon spends more time studying for AMC and AIME => Fierytycoon doesn't have time to be a chickmagnet.[/quote]

Remember that goal about getting a perfect score on the AMC12 so that I could get my picture up on the perfect scorers page so rippledance could see and be in awe of my handsomeness? :D :D :D

Why don't you just post/send a picture of yourself?  That's how must people let other people see their smiling mugs.  (Only a rare few show off their picture in the AMC perfect scores page.)

Oh, I see, that makes it more special, right?

[quote]Oh, I see, that makes it more special, right?[/quote]

It wouldn't be as slick any other way now, would it? :)

if the aime floor is at 8, ill be a happy neal


Putnam
algebra
polynomial
function
college contests

Let $ f$ be a polynomial with positive integer coefficients. Prove that if $ n$ is a positive integer, then $ f(n)$ divides $ f(f(n)\plus{}1)$ if and only if $ n\equal{}1.$

The statement is in error. It should have said "Let $ f$ be a [b]nonconstant[/b] polynomial with positive integer coefficients." Constants will break the conclusion.

Yeah. I just wrote the counterexample of a constant function for the only if part. Do you think that I will get points off for that?

Obviously $ f(f(n) \plus{} 1) \equiv f(1) \bmod f(n)$.  Moreover, since $ f(x)$ has positive integer coefficients, $ \forall n > 1 : f(n) > f(1)$.  The conclusion follows.

[quote="FibonacciFan"][quote="t0rajir0u"]Obviously $ f(f(n) \plus{} 1) \equiv f(1) \bmod f(n)$. [/quote]

Why is that obvious?[/quote]
Write the polynomial as $ \sum_k a_kx^k$ and plug in.

[b]Lemma:[/b]  Given a polynomial $ f(x) \in \mathbb{Z}[x]$ and integers $ a, b \in \mathbb{Z}$,

$ a \minus{} b | f(a) \minus{} f(b)$

Proof:  Write $ f(x) \equal{} \sum c_k x^k$.  Then $ f(a) \minus{} f(b) \equal{} \sum c_k (a^k \minus{} b^k)$, and $ a \minus{} b$ divides each term.

Then $ (f(n) \plus{} 1) \minus{} 1 | f(f(n) \plus{} 1) \minus{} f(1)$.

The more useful form of the lemma is that if $ a \equiv b \bmod m$ then $ f(a) \equiv f(b) \bmod m$.

This is what I wrote down, although I'm not sure if it's rigorous enough:

Let $ f\left(n\right)=a_{0}+a_{1}n+a_{2}n^{2}+\cdots +a_{m}n^{m}$ for $ a_{k},n\in\mathbb{N}$.  Then we have

\begin{align*}f\left(f\left(n\right)+1\right) &= a_{0}+a_{1}\left(f\left(n\right)+1\right)+\cdots +a_{m}\left(f\left(n\right)+1\right)^{m} \\ &= a_{0}+a_{1}\left(f\left(n\right)+1\right)+\cdots +a_{m}\sum_{r=0}^{m}\binom{m}{r}f\left(n\right)^{r} \\ &= f\left(1\right)+a_{1}f\left(n\right)+\cdots +a_{m}\sum_{r=1}^{m}\binom{m}{r}f\left(n\right)^{r} \\ &\equiv f\left(1\right)\bmod f\left(n\right).\end{align*}
But since the coefficients and $ n$ are natural numbers, $ f\left(n\right)<f\left(n+1\right)$ for all $ n\in\mathbb{N}$.  Therefore if $ f\left(n\right)\ |\ f\left(f\left(n\right)+1\right)$ then $ f\left(1\right)\equiv 0\bmod f\left(n\right)$ (see above) and we must have $ n=1$ since for larger $ n$ we would have $ f\left(n\right)>f\left(1\right)$.

Then, if we let $ n=1$, it is clear that $ f\left(1\right)\ |\ f\left(f\left(1\right)+1\right)$ by the third line in the above equations, which completes the proof.

[quote="amcavoy"] since for larger $ n$ we would have $ f\left(n\right) > f\left(1\right)$.[/quote]
This, of course, is only true for a nonconstant $ f.$ It's why we need nonconstant as a hypothesis.

[quote="amcavoy"]I just assumed that $ f$ was nonconstant on the test because of the wording of the problem.  Do you think the graders will take off a lot of points for not mentioning this?[/quote]

Yeah, same. I hope they don't because it really was the problem's fault... not to mention it confused me for a while since I kept checking to make sure I read the problem right.

I have no idea what they do in a case like this, where the confusion comes from the question-writer's mistake.

The way I would handle it:
Note the error in the statement. Restate the problem with the error fixed (adding "nonconstant" in this case), and solve the restated problem.

The same issue came up once for me, on problem B3 in 2000.

[quote="t0rajir0u"]The more useful form of the lemma is that if $ a \equiv b \bmod m$ then $ f(a) \equiv f(b) \bmod m$.[/quote]

Did this require proof on the exam? I directly stated that $ f(f(n) \plus{} 1) \equiv f(1) \bmod f(n)$, as you did; it seemed evident enough.

As for the nonconstant condition, I don't think it's quite fair to expect solvers to point out an oversight in the problem statement. I'm certainly biased because I didn't notice it, but an 8 or 9-point deduction for it seems harsh beyond the point of believability, even for the Putnam.

I would think showing $ f(n) \equal{} \sum_{k \equal{} 0}^{m}a_kn^k$ for some natural number $ m$ and
\[ f(f(n) \plus{} 1) \equal{} \underbrace{\sum_{k \equal{} 0}^{m \minus{} 1}f(n)^{m \minus{} k}\sum_{i \equal{} 0}^ka_{m \minus{} i}\left(\begin{array}{c}m \minus{} i \\
k \minus{} i\end{array}\right)}_{\mathrm{divisible\ by\ }f(n)\ \mathrm{because\ }n \minus{} k,a_{m \minus{} i},\ \mathrm{and\ combinations\ are\ all\ integers}} \plus{} f(1)
\]
is a more direct way to show that $ f(f(n) \plus{} 1)\equiv f(1)\mod f(n)$ than to state its "obvious", but I guess if you can bypass showing that then its a huge time saver.

Edit: I fixed the limit so now the sum is correct.  Sorry!  :D

That sum isn't quite right- you need to separate out the $ f(n)^0$ term, which isn't divisible by $ f(n)$.

I definitely wouldn't assume you can just state that fact on a B1- you need to justify it.
The proof I like: factor $ f(x)\minus{}f(y)\equal{}\sum_{k\equal{}0}^na_k(x^k\minus{}y^k)\equal{}\sum_{k\equal{}1}^n a_k(x\minus{}y)\sum_{j\equal{}0}^{k\minus{}1}x^jy^{k\minus{}j\minus{}1}$. All terms are integers, and there's a common factor of $ x\minus{}y$.

Do you suppose I'll lose points for not pointing out that f can't be constant?  The problem should have said so, so it seems like they shouldn't be docking points for that.

I just claimed at one point that since f has positive integer coefficients and n is positive, f is increasing.

[quote="t0rajir0u"]I thought $ a \minus{} b | f(a) \minus{} f(b)$ was a well-known lemma.  Does it really need justification as long as it's cited?[/quote]

I certainly hope not since i just quoted it. :)
the problem to me seemed trivial given this though...

Hmm, I'm not sure if they will take off marks for not pointing it out.

I think they will take off marks because $ f(1)<f(2)<..$ does not hold if $ f$ is constant and I guess you have to point that out for it to be a complete proof. Even if the question said "let $ f$ be a non-constant polynomial..." you would still have to state it to justify your reasoning for the only if part.

I'm not sure what they'll do but I'm already assuming that I'm getting 1-2 on that question for not pointing that out  :blush:.

Similar to others.

[hide=Short]
Because the coefficients of $f$ are integers, the lemma $a-b | f(a)-f(b)$ for $a, b \in \mathbb{Z}$ holds.
Using this lemma, we find that $f(n) | f(f(n)+1)-f(1)$. Hence, $f(n) | f(f(n)+1)$ is equivalent to $f(n) | f(1)$.
Because the coefficients are also positive and $f$ is nonconstant, we get $f(n) \ge f(1)$ for all $n \in \mathbb{N}$. Hence, $f(n) | f(1)$ is equivalent to $n=1$.
[/hide]

[hide=Solution]Notice that $f(n)+1-1\mid f(f(n)+1)-f(1),$ so $f(n)\mid f(f(n)+1)$ iff $f(n)\mid f(1).$ Since $f(n)>f(1)$ if $n>1,$ $n=1$ is both necessary and sufficient.$\square$[/hide]

[hide=Note]As others have noticed, $f$ must be a nonconstant polynomial for this to be true otherwise not.[/hide]
[hide=Proof.]Let $$f(x)=\sum_{i=0}^m a_ix^i.$$
We get \begin{align*} f(f(n)+1)=\sum_{i=0}^m a_i(f(n)+1)^i \\ \equiv \sum_{i=0}^m a_i (\mod f(n)) \\ \implies f(f(n)+1) \equiv f(1) (\mod f(n)). \end{align*} If $n=1,$ $$f(f(1)+1)\equiv 0 (\mod f(1)) \implies f(1)| f(f(1)+1).$$
If $n>1$, we have $f(n)|f(1).$ Since $f$ is a nonconstant polynomial with positive integer coefficients, we have $$f(n)>f(1)>0, \text{ a contradiction.}$$[/hide]

As $$f(n) \mid f(f(n) + 1) - f(1),$$ $f(n)$ also divides $f(1)$. But $f(n) \geq f(1)$ as $f$ is increasing, so $f(n) = f(1)$ and in particular $n=1$.

[hide=Note]This solution uses the fact that $a-b \mid f(a)-f(b)$ for a polynomial $f$ with integer coefficients. This can be proved easily by expansion.[/hide]

Notice that $$f(1) \mid f(f((n)+1) \implies f(n)+1-1 \mid f(f(n)+1)-f(1) \implies f(n) \mid f(1).$$ But since $f$ is increasing, it is clearly necessary that $n=1$, as desired. $\blacksquare$

Effectively, $f(1) \equiv f(m+1) \bmod m$. Then for $n \geq 2$, $f(1) < f(n)$ and $f(1) \equiv f(f(n)+1) \not\equiv 0 \bmod f(n)$ which completes the proof.

[Note: I am working under the assumption that $f$ is nonconstant, as written in the book Number Theory Concepts and Problems]
We state the following well known lemma:
If $f$ is a polynomial with integer coefficients, then $a - b \mid f(a) - f(b)$ for integers $a, b$ with $a \ne b$.
Now let us work on the proof.
Observe that from the given lemma, $$f(n) = (f(n) + 1) - 1 \mid f(f(n) + 1) - f(1)$$
so we have
$$f(n) \mid f(f(n) + 1)$$$$ \iff f(n) \mid f(1) $$$$ \iff n = 1,$$ where the last part holds because otherwise we clearly have $f(n) > f(1), f(1) \ne 0$ as $f$ has positive coefficients. $\square$


Prove that $ \frac{1}{2!}\plus{}\frac{2}{3!}\plus{}\ldots\plus{}\frac{n\minus{}1}{n!}$ isn't integer for all n values :!:  :?:

[hide="induction pwnage"] taking a base case of 3, we get 5/6 which is obviously not an integer so we're good. Now for the inductive hypothesis, we assume that this is true for n and prove it for n+1. That means the equation would become:

$ \frac{1}{2!}\plus{}\frac{2}{3!}\plus{}...\plus{}\frac{n\minus{}1}{n!}\plus{}\frac{n}{(n\plus{}1)!}$

the part all the way up to n-1/n! is obviously less than 1 from our inductive hypothesis. Therefore, we just have to prove that n/(n+1)! is not an integer and less than 1. We can simplify n/(n+1)! to:

$ \frac{n\minus{}1}{n!}\plus{}\frac{1}{n\plus{}1}$

obviously both those fractions are less than 1 and are not equal to each other, so they equal a non-integer. adding this with our other non-integer number we see that the sum must be a non integer.(since the fractions cannot equal each other)
[/hide]

not exactly a full proof but I don't have much time right now

Induction is not the best way.

[hide="Solution"]
$ \sum_{k \equal{} 2}^{n}\frac {k \minus{} 1}{k!} \equal{} \sum_{k \equal{} 2}^{n}\frac {1}{(k \minus{} 1)!} \minus{} \frac {1}{k!} \equal{} 1 \minus{} \frac {1}{n!}$

Therefore, $ 0 < \sum_{k \equal{} 2}^{n}\frac {k \minus{} 1}{k!} < 1$.
[/hide]

Edit: Also, adding two non-integers doesn't necessarily give you a non-integer, so you're going to have to rethink that proof quite a bit.

Edit 2: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=317069
:mad: Don't double post

[quote="grn_trtle"]
[hide="Solution"]
$ \sum_{k \equal{} 2}^{n}\frac {k \minus{} 1}{k!} \equal{} \sum_{k \equal{} 2}^{n}\frac {1}{(k \minus{} 1)!} \minus{} \frac {1}{k!} \equal{} 1 \minus{} \frac {1}{n!}$

Therefore, $ 0 < \sum_{k \equal{} 2}^{n}\frac {k \minus{} 1}{k!} < 1$.
[/hide]
[/quote]

Yes,the best way(also my way).


geometry
geometric transformation
rotation
3D geometry

We have a normal red 6-sided die and a normal green 6-sided die.  They are identical, besides color.  (On each die, the opposite faces sum to 7.)  The dies are glued to one another, so that a red face is attached to a green face.  For how many different glued figures will the sum total of the visible green numbers be at least 3 greater than the sum total of the visible red numbers?  (A unique figure is not merely defined by which 2 faces are eliminated.  Rotations of any one dice while the other stays in place will create different figures.  Also, distinguish between the red and green dice.)

[hide="hint"]Consider what the numbers will be on the showing faces and work from there.[/hide]

[hide]There are $ 6$ possible combinations of the numbers that will be glued together:

Green, Red

1, 4
1, 5
1, 6
2, 5
2, 6
3, 6

Also, one of the cubes in each of these combinations can be rotated $ 4$ ways, creating a total of $ 24$ different glued figures.[/hide]


Replace $ x$ by $ 1\minus{}x$ and then add due to which the exponential  cancels out.Such questions have only this approach


given a line and 2 points on one side of the line, how do you construct a circle that is tangent to the line and passes through the 2 points (i mean like using a ruler and compass construction)

Let $A,B$ the two points and $d$ the line

Let $(O,r)$ is the circle and $K$ its tangent point with the line $d$.

First of all we may notice that $OA=OB$ ($O$ belongs on the perpendicular bisector of $AB$) and $OK \perp d$.
So, if we find the point $K$, the problem is solved.

If $AB \parallel d$ then the problem is easy

Bring the perpendicular bisector $f$ of $AB$, which intersects line $d$ at the point $K$.
Then $OA=OB=OK=r$

($OK$ is perpendicular to line $d$, and $K$ is a point of $(O,r)$, so $d$ is tangent to the circle)

We bring the perpendicular bisector $g$ of $KB$ and the intersection point of $f$ and $g$ is the point $O$.

Now, assume that the line $BA$ intersects the line $d$ at the point $P$ (such that $A$ is between $P,B$).

$PK$ is tangent to the circle $(O,r)$ and $PB$ intersects the circle at the points $A$ and $B$.

Hence $PK^2=PA \cdot PB$

Hence $PK$ is the geometric mean of $PA,PB$.

The most popular way to find the length of $PK$ is to construct any circle passing through $A,B$ and the tangent segment from $P$ to this circle has the length of $PK$.

But I prefer this way:

Construct a circle $w$ with diameter $PB$.
$AH \perp AB, H \in w$

Since $PB$ is diameter, $\angle PHB=90^0 $

The right triangles $PAH,PHB$ have a common angle in $P$, so they are similar.

$PA/PH = PH/PB \Rightarrow  PH^2 = PA \cdot PB$

So the length $PH$ is what we are looking for.

We can see that, there are two points $K,L$ on $d$ such that their distance from $P$ is equal to $PH$.

We bring a perpendicular line $h$ from $K$ to $d$, and the perpendicular bisector $f$ of $AB$. Where they intersect, there is the point $O$.

We do the same from point $L$, so we get another point $O'$ , such that $(O',O'A)$ pass through $A,B$ and is tangent to line $d$.

So, when $AB$ is not parallel to the line $d$, we get two different circles.

wow, thanks for the response, i could not solve this


function
inequalities
inequalities proposed

If  $ a,b,c$ are [b]real[/b] numbers such that $ a\plus{}b\plus{}c\equal{}3$, then

$ 7\left(a^2\plus{}b^2\plus{}c^2\right) \plus{} 8\left(b^2c^2\plus{}c^2a^2\plus{}a^2b^2\right) \plus{} 4a^2b^2c^2\geq49,$

with equality if $ a \equal{} 2,b\equal{}  c \equal{} \frac{1}{2}.$

[quote="Ji Chen"]If  $ a,b,c$ are [b]real[/b] numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ 7\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} 8\left(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2\right) \plus{} 4a^2b^2c^2\geq49,$

with equality if $ a \equal{} 2,b \equal{} c \equal{} \frac {1}{2}.$[/quote]

We will prove with $ a,b,c\ge 0$
Let $ f(a,b,c) \equal{} LHS \minus{} RHS$
Assume $ c \equal{} max a,b,c$ so $ f(a,b,c) \minus{} f(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2},c) \equal{} (a \minus{} b)^2(\frac {7}{2} \plus{} 4c^2 \minus{} \frac{(c^2 \plus{} 2)(6ab \plus{} a^2 \plus{} b^2)}{4}\ge 0$
Because $ c\ge 1,a \plus{} b\le 1$
The last we only need check a function with one-variable, I think it is not hard
	I hope I am right  :maybe: :blush:

The hint: Using Am-Gm, we only need to prove:
$ 7(a^2\plus{}b^2\plus{}c^2)\plus{}8(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\plus{}4abc \geq\ 50$
And can_hang's lenma works here  :)

[quote="Allnames"][quote="arqady"]Allnames, nguoivn $ a,$ $ b$ and $ c$ are real numbers. :wink:[/quote]
Yes,arqady,with other cases(one number $ \le 0$,two numbers$ \le 0$,) are quite easy[/quote]
I don't see it and I don't think that it's easy. Explain please. Thank you!
[quote="nguoivn"][quote="arqady"]Allnames, nguoivn $ a,$ $ b$ and $ c$ are real numbers. :wink:[/quote]
But canhang's lenma is sum of square. It's trues with a, b, c are the real numbers.  :wink:[/quote]
About which lemma you speak? How it helps? Thank you!

[quote="Ji Chen"]If  $ a,b,c$ are [b]real[/b] numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ 7\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} 8\left(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2\right) \plus{} 4a^2b^2c^2\geq49,$

with equality if $ a \equal{} 2,b \equal{} c \equal{} \frac {1}{2}.$[/quote]
The pqr technique works very well here :)

[quote="arqady"]
About which lemma you speak? How it helps? Thank you![/quote]
Sorry, I can't find the links of it.
But I think U can recollect. It's a lenma about the inequalities have degree 4.
All of the ineq have degree with the equality occurs at 1 point or 2 points can be solved by it.
$ 7(a^2\plus{}b^2\plus{}c^2)\plus{}8(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\plus{}4abc \geq\ 50$
After using Am-GM, we have a ineq degree 4 and after expand, we can use his lenma  :)

[quote="nguoivn"][quote="arqady"]
About which lemma you speak? How it helps? Thank you![/quote]
Sorry, I can't find the links of it.
But I think U can recollect. It's a lenma about the inequalities have degree 4.
All of the ineq have degree with the equality occurs at 1 point or 2 points can be solved by it.
$ 7(a^2 \plus{} b^2 \plus{} c^2) \plus{} 8(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \plus{} 4abc \geq\ 50$
After using Am-GM, we have a ineq degree 4 and after expand, we can use his lenma  :)[/quote]
I think we can't use the AM-GM to deduce the inequality into fourth degree, nguoivn. :(
Taken notice at the equality cases of the original are $ a\equal{}b\equal{}c\equal{}1$ and $ a \equal{} 2,b \equal{} c \equal{} \frac {1}{2}$ ;)

Sr, I'm so silly  :huh: . I thought that equality only occurs then $ a\equal{}4b\equal{}4c$
If two cases, maybe expand and SOS works here  :wink:

[quote="can_hang2007"][quote="Allnames"][quote="arqady"]Allnames, nguoivn $ a,$ $ b$ and $ c$ are real numbers. :wink:[/quote]
Yes,arqady,with other cases(one number $ \le 0$,two numbers$ \le 0$,) are quite easy[/quote]
Are you sure with it ;)[/quote]
Yes,sorry,It is not easy like I thought
The case with two numbers $ \le 0$,is really easy
Suppose $ a,b\le 0$ then $ c\ge 3$ hence $ 7c^2\ge 63$
Else,I cant finish with $ a,b\ge0,c\le0$now,let me think more.Sorry :blush:


What is the slope of the line passing through $ (\minus{}3,5)$ and $ (2,\minus{}5)$?

The equation for slope is $ \frac{y_2\minus{}y_1}{x_2\minus{}x_1}$. Plugging the numbers in we get $ \minus{}2$.


calculus
AMC
USA(J)MO
USAMO
geometry
inequalities
number theory

I have recently gained a huge interest in learning more about math.  I think it's the fact that I have always taken my math education for granted, especially in algebra.  I memorized the formulas, I did well on the tests, but I never had a deep understanding of how math really works.  Before I take calculus, I want to gain a deep and solid comprehension on all topics relevant to precalculus.

I have also made a goal of qualifying for the USAMO by my junior year (I'm a sophomore right now).  Right now, I am learning Algebra II ( :blush: ), but I plan to self-study.  I also recently bought a book called "Mathematical Olympiad Challenges," and that should be coming in a couple days.

So my question is, what should I do, and where should I start?  Is it a reasonable goal to shoot for qualifying for the USAMO (and doing reasonably well in it?).  I plan to study slowly and develop my knowledge over this year, and study all through summer and maybe attend a math camp.  Which branch of math should I start on, and what's more important, the problem solving techniques learned by practicing AIME/AMC problems, or by actually learning the material, and should they go together, or one before the other?

Thanks a lot for all of your advice.  :)

well, just make sure that you can qualify (by taking practice tests)

also, one topic is not really more important than the other, you need to know pretty much everything, but the biggest thing is that you have to develop you math creativity, and that happens from doing (difficult) math all the time

[quote="brady00n"]
So my question is, what should I do, and where should I start?  Is it a reasonable goal to shoot for qualifying for the USAMO (and doing reasonably well in it?).  I plan to study slowly and develop my knowledge over this year, and study all through summer and maybe attend a math camp.  Which branch of math should I start on, and what's more important, the problem solving techniques learned by practicing AIME/AMC problems, or by actually learning the material, and should they go together, or one before the other?
[/quote]

These are reasonable goals if you have the requisite desire to work.  I would suggest working some old tests first and finding out what areas you need work on.  Don't get discouraged if you have a lot of trouble at first - it's very different from the math you've been taught in school.  Stick with it and you'll pick it up.

One warning - that Math Olympiad Challenges book will likely be *very* hard for you right now.  Don't get discouraged by that; it's a very tough book, and you'll be ready for it eventually if you keep working.

Hang out on the message board, do some of the problems you find here, and speak up when you have questions about something.  Ask lots of questions!

If you're in Algebra II and set this goal, you must expect that your goal of USAMO won't be easy. Yet it can be possible.

There aren't any special things you need to learn. You just have to learn almost EVERYTHING used in high school math competitions ---> Number theory, algebra, geometry, inequality, trig/complex number/log, etc..

And if you don't have the AoPS books, you should get them; they really help, especially for someone like you who doesn't have much problem solving background.

Welcome by the way.  :) :welcome:


calculus
derivative
calculus computations

I have a problem here... what is needed is the length of the arc of the curve to be solved by using integrals... the problem goes like this:

Find the length of arc of the curve 6y² = x(x-2)² from (2,0) to (8,4²√3)

i've tried the problem myself but get stuck up during the differentiation process

thanks :D

well
if u took the square root and then diffrentiated it then don't do it just simply diiferentiate both sides
$ \frac{dy^{2}}{dx}=2y\frac{dy}{dx}$

since us said  u had roblem diffrentitaing it i sid do this way
$ 6y^{2}=x(x-2)^{2}$

thus $ 12y\frac{dy}{dx}=(x-2)^{2}+2x(x-2)$
i think u get that


function
real analysis
real analysis unsolved

A function $f$ is holder continuous of order $\alpha$ if there is a constant $C$ such that $|f(x)-f(y)| \leq C|x-y|^\alpha$. Define $||f||_\alpha = \max |f(x)| + \sup |f(x)-f(y)|/|x-y|^\alpha$ .

for $0 < \alpha \leq 1$, I want to show that the set of functions $f \in C[0,1]$ such that $||f||_\alpha \leq 1$ is compact.

do I need to prove this set is equicontinuous (and use Ascoli) or is there a better way? overall this is just giving me a headache.

thank you

[quote="brooklynmth"]
do I need to prove this set is equicontinuous (and use Ascoli)
[/quote]
Yes, you can use the Ascoli theorem here. Equicontinuity immediately follows from the Holder condition and uniform boundedness is clear.