limit
calculus
derivative
Putnam
function
calculus computations

Let $ p,\ q,\ r$ be given constants. Find the values of $ a,\ b$ such that $ \lim_{x\rightarrow p} \frac {\sqrt {ax \plus{} b} \minus{} q}{x \minus{} p} \equal{} r$.

Note that we are not allowed to use l'Hopital's theorem here.

Is it

$ a \equal{} 2qr$ and $ b \equal{} p^2r^2 \plus{} q^2$ provided that $ p \neq 0$ and $ pr \plus{} q > 0$ ?

[quote="JBL"]$ (x \minus{} p)(x \plus{} p) \neq x^2 \plus{} p^2$ ;)  (Reality check: the denominator approached 0, then you multiplied it my something that approaches a constant; what must the result approach?)[/quote]

Yeah, I can't explain what I was thinking =p

That's incorrect. I can't understand what you did.
This problem seems to be difficult.

$ (x \minus{} p)(x \plus{} p) \neq x^2 \plus{} p^2$ ;)  (Reality check: the denominator approached 0, then you multiplied it my something that approaches a constant; what must the result approach?)

If you want to multiply by something to make this look nicer, you should be trying to deal with that square root up top -- how about rationalizing the numerator?

We cannot use L'Hopital's rule, but we can use the definition of the derivative.

Recall that

$ f'(p) \equal{} \lim_{x \to p} \frac{f(x) \minus{} f(p)}{x\minus{}p}$.

Thus, choose $ f(x) \equal{} \sqrt{ax\plus{}b}$, and we require that

$ f(p) \equal{} \sqrt{ap\plus{}b} \equal{} q$

$ f'(p) \equal{} \frac{a}{2f(p)} \equal{} r$.

This immediately gives us $ \frac{a}{2q} \equal{} r$, or $ a \equal{} 2qr$.  It then follows that

$ b \equal{} q^2 \minus{} ap \equal{} q^2 \minus{} 2pqr$.

Hum, your solution is almost using l'Hopital's rule, isn't it? :wink:

Anyway, why does it need to be $ \lim_{x\rightarrow p} (\sqrt{ax\plus{}b}\minus{}q)\equal{}0?$, which is the main key of the problem.

[quote="kunny"]Hum, your solution is almost using l'Hopital's rule, isn't it? :wink:

Anyway, why does it need to be $ \lim_{x\rightarrow p} (\sqrt {ax \plus{} b} \minus{} q) \equal{} 0?$, which is the main key of the problem.[/quote]

I don't think your remark is fair.  I did not require L'Hopital because your expression happened to be simple enough to work the way I solved it.  In other words, it was sufficiently weak and I solved it using a method any elementary calculus student who knows what a derivative is should be able to understand.

If you disagree, then I ask you, at what point in my solution did I claim that

$ \lim_{x \to p} \frac {f(x)}{g(x)} \equal{} \lim_{x \to p} \frac {f'(x)}{g'(x)}$?

If you want to make a problem that adequately addresses your question of why the numerator must also tend to zero if the denominator tends to zero, then construct one that is sufficiently abstract and then we can talk about epsilon-delta proofs, if that is how "elementary" you want to take things to.

[quote="kunny"]First of all, you should answer my question.[/quote]

I'm not a student.  I don't [b]have[/b] to answer anything unless I choose to.

Boring, that doesn't make sense.
Those who readily use l'Hopital's rule doesn't understand the principle.
My problem asks us more basic fact. :wink:

And I am going to say this only one more time.  If you are trying to make a point, construct a problem that can't be solved any other way than the way you want it to be solved.  Otherwise, don't complain when someone solves it using a method you did not anticipate.

Usually I enjoy solving your problems, but this is ridiculous.  From now on, I'm not going to play your games.  I'm done.

O.K.　You seem to be an expert of calculus, I dare to say, your work make 8/10 points, for not being logic mathematicaly.
I dare to make a opinion Ａnti-Ｔheze in a sense, whatever I am to blame.
Because I have been receiving sevaral private mails, "Why do you dislike using l'Hopital?''

Anyway, I do wait others opinion.

Kunny, your arrogance has lead you nowhere.

Guess I was among the first few who gave up answering your questions all together. Your problems are generally elementary but you make them sound like rocket science. Anyway, happy posting and try to offend fewer people. 'Pride goes before a fall'... not sure if you understand :rotfl:

[quote="IndoChina"]Kunny, your arrogance has lead you nowhere. [/quote]

Thank you for your advice, Indochina. Let keep me the argument a little longer.

Solution.

Let $ f(x)=\frac{\sqrt{ax+b}-q}{x-p}.$

Then $ \lim_{x \to p}{\sqrt{ax+b}=\lim_{x \to p}(f(x)(x-p)+q)}$

So $ \sqrt{ap+b}=q \Longrightarrow b=q^2-ap.$

So $ \lim_{x \to p} \frac{(\sqrt{ax+q^2-ap}-q)(\sqrt{ax+q^2-ap}+q)}{(x-p)(\sqrt{ax+q^2-ap}+q)}=\lim_{x\to p}\frac{a(x-p)}{(x-p)(\sqrt{ax+q^2-ap}+q)}=\frac{a}{2q}$

So $ a=2rq$ and $ b=q^2-2rpq$

[quote="Dimitris X"]Solution.

Let $ f(x) = \frac {\sqrt {ax + b} - q}{x - p}.$

Then $ \lim_{x \to p}{\sqrt {ax + b} = \lim_{x \to p}(f(x)(x - p) + q)}$

So $ \sqrt {ap + b} = q \Longrightarrow b = q^2 - ap.$

So $ \lim_{x \to p} \frac {(\sqrt {ax + q^2 - ap} - q)(\sqrt {ax + q^2 - ap} + q)}{(x - p)(\sqrt {ax + q^2 - ap} + q)} = \lim_{x\to p}\frac {a(x - p)}{(x - p)(\sqrt {ax + q^2 - ap} + q)} = \frac {a}{2q}$

So $ a = 2rq$ and $ b = q^2 - 2rpq$[/quote]

Yes, that's exactly what I wanted to say. The problem seems easy for experts, but few people can answer the basic question from beginner's question: Why does it need to be $ \lim_{x\rightarrow p} (\sqrt {ax + b} - q) = 0?$

[quote="atomicwedgie"][quote="kunny"]Hum, your solution is almost using l'Hopital's rule, isn't it? :wink:

Anyway, why does it need to be $ \lim_{x\rightarrow p} (\sqrt {ax + b} - q) = 0?$, which is the main key of the problem.[/quote][/quote]

The thing I mentioned before is the fact:
For some constants $ \alpha ,\ \beta$, $ \boxed{\boxed {\lim_{x\rightarrow a} f(x) = \alpha ,\ \lim_{x\rightarrow a} g(x) = \beta \Longrightarrow \lim_{x\rightarrow a} f(x)g(x) = \alpha \beta }}$. We Japanese high school students just learn the fact, but we aren't taught the proof.

After all, for my problem, $ \lim_{x\rightarrow p} \frac {\sqrt {ax + b} - q}{x - p}\cdot (x - p) = r\cdot 0 = 0$, yielding

$ \lim_{x\rightarrow p} (\sqrt {ax + b} - q) = 0$, which is the main key point.

Now the discussion is end up.

Finally, I appologize for my arrogance especially for atomicwedgie.

kunny

A post I made in a different thread, one that kunny responded to so there should be no excuse for him to claim that I don't know how to solve this question in the way that he expected:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1733931#1733931

Now combine that with my earlier post of the solution and the result is mathematically equivalent to the "ACCEPTED" solution.  It is the same because in both cases one rationalizes the numerator by multiplying numerator and denominator by $ \sqrt {ax \plus{} b} \plus{} q$.

I come here to comment on math problems and provide solutions that I think are clean, understandable, and illustrative.  I don't come here to be challenged by arrogant people.  I already have my degree and that was challenge enough.  I took the Putnam twice and scored in the top 200 without any preparation or training.  But I have not felt the need to say anything about my background because I feel that my posts speak for themselves.  Math is supposed to be enjoyable and the process of learning math should be better guided than this charade.

But thanks to you kunny I am no longer going to post answers to your questions.  You can take your presumptuous attitude and shove it.

This seems to have gotten a little out of hand. I think there's been overreaction and misinterpretation of intent on both side.

To atomicwedgie: Calm down. But find the threads and topics that you want to post on and post there.

To kunny: He does have this much of a point: far too often you put fences around your problems. "Solve by _ method," or "Solve without using _," or "Solve by methods that Japanese high school students should know." My reactions to that sometimes include, first, that I personally don't care what Japanese high school students should know, and second, I bet there are quite a few Japanese high school students who don't themselves care what they should know or not know.

I almost posted on this earlier, and this is approximately what I would have said:

I have a long record on this site as being one of the more anti-L'Hôpital posters around, and I'm the same in my own classes. I try very hard to wean my students away from using L'Hôpital as some kind of instant reflex. I try to get them to understand size - what grows or shrinks faster than what else. And I try to get them to use power series - to make power series something close to a native language when talking about limits. I dislike L'Hôpital mostly because it's a "black box" that doesn't enhance understanding, and because of that a user may be blind to its limitations.

But I don't say never. I use L'Hôpital myself when it's useful - and sometimes it is the most efficient way to do things. Do you want the residue of $ f(z)\equal{}\frac{1}{z^4\plus{}4}$ at $ z\equal{}1\plus{}i?$ L'Hôpital is as good a way as any to do that one.

That brings up this particular problem. It's one of the relative few that I might well use L'Hôpital on. Thinking about size, I see that it's a function with a simple zero (yes, we have to check that it's zero) divided by a function with a simple zero. L'Hôpital provides a simple-enough organization for that. Or maybe you do it algebraically, or there are other methods as well. It's just not a problem that I'd use in convincing my students not to use L'Hôpital.

As long as this thread isn't about math any more, I'd like to add that it seems that the biggest communication problem in this thread is not that kunny is arrogant (I've never detected arrogance in kunny's posts) but that kunny is [i]cryptic[/i] -- it's frequently unclear what kunny is actually saying.  I understand that English isn't kunny's first language, and that this has something to do with the lack of clarity, but it would certainly help to use additional words and/or sentences to make it clear what things mean.  For example, in kunny's first response to atomicwedgie, it's pretty clear that there was a quip followed by a question about the math, but it's not entirely clear what the mathematical question really says (and I don't think atomicwedgie noticed or understood the question); in follow-up comments, kunny never clarified.

Can anyone explain to me what the following sentence means?
[quote]I dare to make a opinion Ａnti-Ｔheze in a sense, whatever I am to blame. [/quote]

it doesn't have to start with $ \lim_{x\minus{}p} \sqrt{ax\plus{}b}\minus{}q\equal{}0$

you can simply starts with

$ r\equal{}\frac{(x\minus{}p) r}{x\minus{}p}$

then $ \sqrt{ax\plus{}b}\minus{}q\equal{}(x\minus{}p)r$

as $ x\to p, q^2\equal{}ap\plus{}b$

i think atomicwedgie solution is excellent

Sorry, could you make a more detail explanation?


calculus
analytic geometry
geometry solved
geometry

I've posted this in the calculus forum and haven't got replies, maybe here it will find it's answer.

You have n points in the plane (you know the coordinates of the points) and you have to find point A so that the sum of distances fron the n pointsto A is minimum. I know that for n = 3 point A is called Fermat's point. This is a computer programming task, and a friend of mine told me it is a classical problem in calculus and you can solve it using derivation, I couldn't solve it myself ... Any help would be appreciated. Thanks

This is the famous Fermat-Weber problem; for general values of n, the point A is not constructible with compass and ruler. There are lots of literature about this problem, and there are some approximative methods known. What do you mean by "solve it using derivation"?

  Darij

You can easily use gradient method for finding required point.
And simply derivation gives nothing, since you obtain nonlinear system of equaitions, which unsolvable in exact way.


quadratics
number theory unsolved
number theory

Let a be a non-perfect square number.Prove that ther are exist a prime number p such that p|x^2-a does not hold for all integer number x

I think you meant: there exists no $x$ such that $p|x^2-a$, the other one is trivial ;)

Or equivalently: such a number is not a quadratic residue $\mod p$.

Prove (but not elementary):
Let $a=2^s b = 2^s p_1^{v_1} p_2^{v_2} ... p_n^{v_n}$ be the prime factorisation and $b$ it's greatest odd factor. Since $a$ is not a square, either (WLOG) $v_1$ is odd or $s$ is odd.

1. case: all $v_i$ are even, so $s$ is odd.
Then simply taking any prime number $q$ not dividing $a$ and $\equiv 3\mod 8$ gives $\left( \frac{a}{q} \right) = \left( \frac{2^s}{q} \right) \left( \frac{b}{q} \right) = (-1)^s = -1$.

2. case: $v_1$ is odd.
Now let $c$ be a quadratic non-residue $\mod p_1$ and take a prime $q$ such that $q \equiv 1 \mod 8$, $q \equiv c \mod p_1$ and $q \equiv 1 \mod p_i$ for all other $p_i$.
Then we get:
$\left( \frac{a}{q} \right) = \left( \frac{2}{q} \right)^s \left( \frac{p_1}{q} \right)^{v_1} \left( \frac{p_2}{q} \right)^{v_2} ... \left( \frac{p_n}{q} \right)^{v_n} = \left( \frac{q}{p_1} \right)^{v_1} \left( \frac{q}{p_2} \right)^{v_2} ... \left( \frac{q}{p_n} \right)^{v_n}=(-1)^{v_1}=-1$
so $a$ is not a quadratic residue $\mod q$,.

Ok, I just see that it's really easy to make the proof above elementary:

Just forget that $q$ has to be prime in the 2. case (but still let it belong to that residue classes). Then we get $\left( \frac{a}{q}\right) =-1$, thus $a$ is not a quadratic residue $\mod$ at least one of $q$'s prime factors.

In case one wants to prove "if $a$ is quadratic residue $\mod$ all primes $p>N$, then $a$ is a square", just take $q$ such that it is also coprime to all primes $\leq N$.

[quote="ZetaX"]Ok, I just see that it's really easy to make the proof above elementary:

Just forget that $q$ has to be prime in the 2. case (but still let it belong to that residue classes). Then we get $\left( \frac{a}{q}\right) =-1$, thus $a$ is not a quadratic residue $\mod$ at least one of $q$'s prime factors.
.[/quote]

i think Legendre symbol is defined when q is prime


This is a reminder to all USA/Canada Mathcamp staff that I will attend mathcamp 2010.
I am chosen to be in there because I dreamed it.

okay.

The message is too small. Please make the message longer before submitting.


algebra
polynomial
logarithms
quadratics
algebra unsolved

[quote="james digol"]What complex numbers (if there exist) are the roots of some polynomial with positive coefficients?[/quote]

[i](nota : in the following, $ \mathbb R^\plus{}$ is the set of positive reals and $ \mathbb R_0^\plus{}\equal{}\mathbb R^\plus{}\cup\{0\}$[/i]

It's easy to check that $ z\equal{}\rho e^{i\theta}$ is a root of $ P(x)\equal{}0$ where $ P(x)\equal{}x^{2^n}\minus{}2\rho^{2^{n\minus{}1}}\cos(2^{n\minus{}1}\theta)x^{2^{n\minus{}1}}\plus{}\rho^{2^n}$

Then, for $ \theta\in(\minus{}\pi,0)\cup(0,\pi)$, consider $ n\equal{}1\plus{} \left[\log_2(\frac{\pi}{|\theta|})\right]$ such that $ \log_2(\frac{\pi}{|\theta|})\ge n\minus{}1 > \log_2(\frac{\pi}{|\theta|})\minus{}1$ and so $ \pi\ge 2^{n\minus{}1}|\theta| >\frac{\pi}2$.

We clearly have $ \cos(2^{n\minus{}1}\theta)<0$ and so $ P(x)\in\mathbb R_0^\plus{}[X]$

And for $ \theta\equal{}0$ or $ \rho\equal{}0$ , it obviously does not exist $ P\in\mathbb R^\plus{}[X]$ such that $ P(z)\equal{}0$

Then, multiplying $ x^{2^n}\minus{}2\rho^{2^{n\minus{}1}}\cos(2^{n\minus{}1}\theta)x^{2^{n\minus{}1}}\plus{}\rho^{2^n}$ (polynomial $ \in\mathbb R_0^\plus{}[X]$) by $ (1\plus{}x\plus{}x^2\plus{}...\plus{}x^{2^{n\minus{}1}})$, we clearly get the required polynomial $ \in\mathbb R^\plus{}[X]$

Hence rhe result : $ P(z)\equal{}0$ for some polynomial $ P\in\mathbb R^\plus{}[X]$ $ \iff$ $ z\in\mathbb C\backslash\mathbb R_0^\plus{}$

[b]Solution.[/b]

We show that every complex number that does not lie on the nonnegative real line is a root of such polynomial. (If we allow coefficients of terms of degree less than the degree of the polynomial to be $ 0$, then only positive real number are excluded.)

First, observe that if $ w$ is a nonegative real number and $ P(z)$ is a polynomial with positive coefficients, then clearly $ P(w)>0$. Thus no $ w\geq0$ is the root of a polynomial with positive coefficients.

Every $ w=a+bi \in \mathbb{C}$ is a root of the real polynomial \[ q(z)=(z-w)(z-\overline{w})=z^2-2az+a^2+b^2.\] If $ a<0$, then $ q(z)$ is a polynomial with positive coefficients. This shows that every complex number in the open left half-plane is the root of some such polynomial.

Now assume that $ a\geq0$ and $ b\neq0$. Thus $ w$ lies in the right half-plane, but not on the real axis, so $ 0<\left | \arg(w) \right |\leq \pi/2$. If $ n$ is the smallest positive integer for which $ \pi/2<\left | \arg(w^n) \right |$, then $ w^n$ lies in the open left half-plane. By our work above, there exists a quadratic polynomial $ q$ with positive coefficients such that $ q(w^n)=0$there exists, say $ q(z)=z^2+c_1z+c_0$ . We now let \[ P(z)=(z^{2n}+c_1z^n+c_0)(z^{n-1}+z^{n-2}+\cdots+z+1).\] It follows that $ P(z)$ has positive coefficients and $ P(w)=0$.


Hi,

I'd like to know which IMO shortlist problems you consider most beautiful?

For me its IMO 2001 N6 and 2004 G8


hi,
[u]problem:[/u]
$x > 0,x<0 \Rightarrow f(x)=\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$

$f(0)=1$

$g(x) = \frac{f(x)-f(0)}{x}$
compute $\lim_{x \to\ 0}g(x)$

thanks in advance.


binomial coefficients
algebra unsolved
algebra

Prove that there are no natural $n$ such that among numbers :
${{n} \choose 0},{{n} \choose 1},{{n} \choose 2}, ... ,{{n} \choose n}$
there are exactly half odd and half even numbers.

Sorry for my poor English

I think that
[hide]You should see the cases $n$ even and $n$ odd and use the fact that

${{n} \choose 0}={{n} \choose n}$ etcera
[/hide]

Hint: obviously, $n$ must be odd, say $n=2^mq-1$, with odd $q$. Observe that the binomial coefficients in each of the following groups:
\[ \{ \binom{n}{0},\binom{n}{1},\ldots ,\binom{n}{2^m-1}\},  \]
\[ \{ \binom{n}{2^m},\binom{n}{2^m+1},\ldots ,\binom{n}{2\cdot2^{m}-1}\},  \]
\[ \vdots \]
\[ \{ \binom{n}{(q-1)2^m},\binom{n}{(q-1)2^m+1},\ldots ,\binom{n}{q2^m-1}\},  \]
have the same parity, hence the conclusion.

A try with elementary arguments :

$\binom{n}{k} = \binom{n}{n-k}$, $\sum \binom{n}{k} = 2^n$ : so there is always an even number of odd binomial coefficients.

If n is even, there is n+1 (odd) binomial coefficients, so there is an odd number of even binomial coefficients.

If n is odd, $\binom{n}{k} = n/k \binom{n-1}{k-1}$ : this proves that in this case $\binom{n}{k}$ is odd for all even k, i.e for at least (n-1)/2 value. since $\binom{n}{1}$ and $\binom{n}{n}$ are even there are more odd binomial coefficients then even ones.

[quote="tµtµ"]

If n is odd, $\binom{n}{k} = n/k \binom{n-1}{k-1}$ : this proves that in this case $\binom{n}{k}$ is odd for all even k[/quote]
We have $\binom{11}{6}=462$, for instance.


inequalities
calculus
derivative
inequalities proposed

[quote="zhaobin"]If $ a,b,c,d\ge 0$,and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$,Prove that
\[ 3(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \plus{} \sqrt {d}) \ge 2(ab \plus{} bc \plus{} ca \plus{} ad \plus{} bd \plus{} cd)\]
[/quote]
Try $ a \equal{} b \equal{} c \equal{} \frac {4}{3}$ and $ d \equal{} 0.$

However, the following inequality holds and nice:
$ \frac{ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd}{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\plus{}\sqrt{d}} \le \frac{8}{3\sqrt{3}}.$

It is easy to transform this inequality into $ \sum_{cyc} f(a) \ge 16$, where $ f(x)\equal{}x^2\plus{}\frac{8}{3\sqrt{3}}\sqrt{x}$. This has an inflection point at $ x\equal{}\frac{1}{3}$, is concave on the left and convex on the right. Therefore, by Karamata on the first interval, then the second, it suffices to show the inequality when $ a,b,c,d \in \{0,m,n\}$, where $ m < \frac{1}{3}$, $ n\ge \frac{1}{3}$, and at most one of $ a,b,c,d$ is $ m$.

If now 3 or 4 of $ a,b,c,d$ are 0, the result is immediate by referring back to the original inequality. If none are $ m$, there are only 4 cases to consider, (0,0,0,4), (0,0,2,2), (0,4/3,4/3,4/3), and (1,1,1,1), which are essay to do through direct computation. Therefore, the only cases left to consider are $ (0,0,m,n)$, $ (0,m,n,n)$, and $ (m,n,n,n)$, up to permutation. But the first has $ n \ge \frac{10}{3}$, which has $ f(n) \ge f(\frac{10}{3}) \ge 16$, so the first case is done. For the second and the third, employ the substitutions $ m\equal{}4\minus{}2n$ and $ m\equal{}4\minus{}3n$ respectively, and we then have 2 one-variable inequalities which can be solved using the first derivative test, and some approximations. If anyone wants to actually write these out, be my guest :P

Cheers,

Rofler

P.S. what is your solution can_hang2007?

Thank you,I am wrong,sorry.
[quote="can_hang2007"][quote="zhaobin"]If $ a,b,c,d\ge 0$,and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$,Prove that
\[ 3(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \plus{} \sqrt {d}) \ge 2(ab \plus{} bc \plus{} ca \plus{} ad \plus{} bd \plus{} cd)\]
[/quote]
Try $ a \equal{} b \equal{} c \equal{} \frac {4}{3}$ and $ d \equal{} 0.$

However, the following inequality holds and nice:
$ \frac {ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd}{\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \plus{} \sqrt {d}} \le \frac {8}{3\sqrt {3}}.$[/quote]

[quote="Soarer"]One can of course use half convex half concave theorem to see that we only need to check a few cases, but is there a nice (and perhaps classical) solution?[/quote]
There is a nice solution. We can modify a little to get an equivalence inequality:

If $ a,$ $ b,$ $ c,$ $ d$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}3,$ then
$ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\plus{}\sqrt{d} \ge ab\plus{}ac\plus{}ad\plus{}bd\plus{}bd\plus{}cd.$

Now, take a look here to view my proof: http://can-hang2007.blogspot.com/2010/01/inequality-27-v-q-b-can.html


$ SOH,CAH,TOA$

look this up.
I doubt someone can teach you in one single post.

read about it around the internet if no one can help you personally.

I'm also learning trigonometry on my own.

Does any know of any great books I should read?
Any books that are designed for trig problem solving?

@tonypr: I found that learning just the trignometric functions was enough to figure out the rest.
              I was able to recreate much of the book using just these definitions on my own.

[quote="tonypr"]Is ther anyone here that can help me with trigonometry(the basics). I was supposed to have taken it in class last year but the teacher told me study for mathcounts instead.

All I need to know is how the Sin, Cos, And Tan work in a tringle, and what it is used for.[/quote]

The $ \sin$ of an angle is the opposite side length divided by the hypoteneuse side length of that angle placed in a triangle with a right angle.
The $ \cos$ of an angle is the adjacent side length over the hypoteneuse side length of that angle placed in a triangle with a right angle.
The $ \tan$ of an angle is the opposite side divided by the adjacent side of that angle placed in a triangle with a right angle.

A good way to remember other than $ SOHCAHTOA$ ($ SOHCAHTOA$ doesn't work that well for me) is to associate $ \sin$ with opposite, $ \cos$ with adjacent, and $ \tan$ with nonhypoteneuse (or hypoteneuse, whichever is easier to remember). Also remember that other than $ \tan$, they're divided by the length of the hypoteneuse.


Let $ f(n)$ denote the sum of the digits of n. Let $ N \equal{} 4444^{4444}$. Find $ f(f(f(N)))$.

[hide="Solution"]

First we see that $ 4444\equiv 7\pmod{9}$, and $ 4444^{3k\plus{}1}\equiv 7\pmod{9}$. This implies that the sum of the digits of $ 4444^{4444}$ will be congruent to $ 7\pmod{9}$.

Next we analize $ 4444^{4444}$, $ 4444^{4444}< (10^4)^{4444} \equal{} 10^{17776}$ so $ f(N)\le 17776 \cdot 9 \equal{} 159984 < 199999$, from that we get $ f(f(N))\le 46<49$ and $ \therefore f(f(f(N)))\le 13$. Since $ f(f(f(n)))\equiv 7\pmod{9}$ there is only one answer $ \boxed{7}$.

Any questions?  :ninja:
[/hide]

I got a similar solution, and I was wondering if there were any other solutions.


algebra
polynomial
number theory unsolved
number theory

The sequence $ (x_n)_{n \geq 1}$ is defined by:
$ x_1\equal{}3$
$ x_2\equal{}7$ and
$ x_{n\plus{}2}\equal{}3x_{n\plus{}1}\minus{}x_n$.
Prove that if $ x_n\plus{}(\minus{}1)^n$ is a prime number, then $ n\equal{}3^m$ $ (m \in N)$

I'm sorry if it had been posted already.I can not find this problem in the forum.Thanks a lot.

Trivially $ x_1\minus{}1\equal{}2 \in \mathbb{P}$ and in $ \mathbb{Z}/2\mathbb{Z}$ we obtain that if $ x_n \in \mathbb{P}$ for some $ n>1$ then $ 3 \mid n$. Working straighforward with charateristic polynomial $ \epsilon^2\minus{}3\epsilon\plus{}1$ (that incidentally have square of gold section as roots) we obtain $ y_n: \equal{}x_n\plus{}(\minus{}1)^n\equal{}z^n\plus{}z^{\minus{}n}\plus{}(\minus{}1)^n$, where $ z: \equal{}(3\plus{}\sqrt{5})2^{\minus{}1}\equal{}\left( (1\plus{}\sqrt{5})2^{\minus{}1}\right)^2$. Define $ q(n): \equal{}3^{\upsilon_3(n)}$ for all $ n \in \mathbb{N}_0$. We claim that if $ nq(n)^{\minus{}1}>1$ then $ y_{q(n)}$ divides (strictly) $ y_n$, that is $ z^{q(n)}\plus{}z^{\minus{}q(n)}\minus{}1 \mid z^{n}\plus{}z^{\minus{}n}\plus{}(\minus{}1)^n$. Now it is [i]equivalent[/i] to show that if a real $ t$ is fixed, and if $ p(x): \equal{}\phi_3(x)x^{\minus{}1}$ (where $ \phi_i(x)$ is the i-th cyclotomic polynomial) verify $ p(t^k) \in \mathbb{Z}$ for all $ k \in \mathbb{N}_0$ then $ p(t)$ divides $ p(t^q)$ for all $ q \in \mathbb{N} \setminus 3 \mathbb{N}$. We'll show it by PMI: if $ q\equal{}2$ then $ p(t^2)\equal{}p(t)(p(t)\minus{}2)$ and if $ q\equal{}4$ then $ p(t^4)\equal{}$ $ p(t)(p(t)\minus{}2)(p(t^2)\minus{}2)$. Suppose that the thesy is true up to $ q\equal{}3j\minus{}1$, we'll verify it for $ q\equal{}3j\plus{}1$ and $ q\equal{}3j\plus{}2$, infact $ p(t^{3j\plus{}1})\equal{}$ $ p(t)(p(t)\minus{}2)(p(t^{3j\minus{}1}\minus{}1)\minus{}p(t^{3j\minus{}1})\plus{}[p(t^{3j\minus{}3})\minus{}3]$ and $ p(t^{3j\plus{}2})\equal{}$ $ p(t)(p(t^{3j\plus{}1})\minus{}1)\minus{}p(t^{3j\plus{}1})\plus{}[p(t^{3j}\minus{}3]$. It mean that it is enough to show that $ p(t)$ divides $ p(t^{3v})\minus{}3$ for all $ v \in \mathbb{N}_0$, and we'll show it again by PMI: in fact if $ v\equal{}1$ then $ p(t^3)\minus{}3\equal{}p(t)(p^2(t)\minus{}3p(t))$, if $ v\equal{}2$ then $ p(t^6)\minus{}3\equal{}p(t)(p^2(t)\minus{}3p(t))(p(t^3)\minus{}2)$ and if the thesy is shown up to $ v\minus{}1$ then $ p(t^{3v})\minus{}3\equal{}p(t)(p^2(t)\minus{}3p(t))(p(t^{3v\minus{}3})\minus{}1)\plus{}2(p(t^{3v\minus{}3})\minus{}3)\minus{}(p(t^{3v\minus{}6})\minus{}3)$.


Assuming you need a $ 120$ to qualify for AIME, and your goal is qualifying for AIME, you must answer at least $ 19$ to qualify. I would answer at least $ 22$ questions, because stupid mistakes crop up inevitably.

I probably should have mentioned that I want to qualify for the AIME. Thanks.

If I answered 22, the maximum number of answers I could get wrong and still qualify would be 2. The same is true if I answered 21 questions. My next question is - should I answer 21 or 22? Would the extra time in solving one less problem improve my chances?

Don't try to game the test. Answer as many as you can solve - this will usually give you the best chance of making any target score.

[quote="nuclearbob"]With the new scoring system of 1.5 points for every blank answer, how many questions should I answer on the AMC 10?[/quote]

If it's possible, answer every single one. :lol:

However, in the case where you are unable to, guess on the problems that you have narrowed down to like 2 answers, and just skip the rest.

AMC 10 = 6 points per correct answer, 1.5 per blank answer, 0 points incorrect answer, qualifying score 120 and/or top 1%
AMC 12 = same as AMC 10 except qualifying score is 100 and/or top 5%

[quote="Truffles"]Wait, what it's only 1.5 for every blank answer now?  Is this the same for the amc 12 too?[/quote]

Yes, that is correct.

I would answer all of them, if possible.


KS does not have a team, so you have to join MO (there were 6 Kansans on MO ARML last year). The site seems to be down. Please contact the head coach "RArmstrong@stlcc.edu" to join the mailing list. An application is sent out in the spring, and there's a fee, like 200-300 dollars. MO sends 2 teams and a few alternates, so a little more than 30 people, which in the last two years, is everyone that applied. MO does have practices about once a month in Springfield and St. Louis, though they are not required.

i am trying to revive the KS team for this year (i started the team in 98 but it kind of died when i moved to CA).  i'll be recruiting from AMC and GPML results, but if you're interested send an email to mathleague@mathleague.org and i'll put you on a mailing list for more information when it becomes available..

Whats GPML? Also, what are the typical scores to get in?

Great Plains Math League: http://mathleague.org/

is a competition with about 6 qualifying meets per year, then a state meet, and then a championship meet. The meets are hosted throughout MO, IA, KS, and CA.

It's kinda like ARML and/or Mathcounts, with a sprint, target, team, and relay round. (I'm not sure how they do power round).


function
limit
modular arithmetic
algebra
polynomial
real analysis
real analysis unsolved

Let $ f: [0,1]\to\mathbb{R}$ a continuous function .Prove that
\[ \lim_{n\to\infty}\frac{1}{2^n}\sum _{k\equal{}1}^n (\minus{}1)^kC_n^k\cdot f\left( \frac{k}{n}\right)\equal{}0\]

[hide="hint"]
since $ f$ is continious there exists a maxima $ M$ and a minima $ m$
we have then $ m\sum_{k \equal{} 1,k \equiv 0 \pmod 2}^{n}\binom{n}{k} \minus{} M \sum_{k \equal{} 1,k \equiv 1 \pmod 2}^{n} \binom{n}{k} \leq \sum_{k \equal{} 1}^{n} ( \minus{} 1)^{k} \binom{n}{k} f(\frac {k}{n}) \leq M\sum_{k \equal{} 1,k \equiv 0 \pmod 2}^{n}\binom{n}{k} \minus{} m \sum_{k \equal{} 1,k \equiv 1 \pmod 2}^{n} \binom{n}{k}$
also we have if
$ \lim_{n\to\infty}\frac {1}{2^n}\sum _{k \equal{} 1}^n ( \minus{} 1)^kC_n^k\cdot f\left( \frac {k}{n}\right) \equal{} L$
then
$ \frac {m\sum_{k \equal{} 1,k \equiv 0 \pmod 2}^{n}\binom{n}{k} \minus{} M \sum_{k \equal{} 1,k \equiv 1 \pmod 2}^{n} \binom{n}{k}}{2^{n}} \leq L \leq \frac {M\sum_{k \equal{} 1,k \equiv 0 \pmod 2}^{n}\binom{n}{k} \minus{} m \sum_{k \equal{} 1,k \equiv 1 \pmod 2}^{n} \binom{n}{k}}{2^{n}}$
thsi along with the know binomial sum and that as $ n \to \infty$  m=M can be proved...
i am leaving will return hope i am right sorry :( [/hide]

Actually, pardesi's hint is not very useful :(. What one should notice instead is that the statement is true for every polynomial (the expression gets exactly $ 0$ as soon as $ n$ exceeds the degree of the polynomial) and that polynomials are dense in $ C[0,1]$. This is not the only possible approach, of course.

sorry...i wiil try if possible to continue :rotfl:  :)


[quote="Ulanbek_Kyzylorda KTL"]Find maximum value of x>0 such that for every a,b,c with conditions 0<a,b,c=<1 satisfies to this inequality: 1/a+b+c>=1/3+x(1-a)(1-b)(1-c)[/quote]
If $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ we getting $ x\leq\frac{9}{4}.$
Id est, we need to prove that $ \frac{1}{a\plus{}b\plus{}c}\geq\frac{1}{3}\plus{}\frac{9}{4}(1\minus{}a)(1\minus{}b)(1\minus{}c).$
Let $ a\plus{}b\plus{}c\equal{}t.$
Hence, $ (1\minus{}a)(1\minus{}b)(1\minus{}c)\leq\left(\frac{3\minus{}a\minus{}b\minus{}c}{3}\right)^3\equal{}\left(\frac{3\minus{}t}{3}\right)^3.$
Thus, it remains to prove that $ \frac{1}{t}\geq\frac{1}{3}\plus{}\frac{9}{4}\left(\frac{3\minus{}t}{3}\right)^3.$
But  $ \frac{1}{t}\geq\frac{1}{3}\plus{}\frac{9}{4}\left(\frac{3\minus{}t}{3}\right)^3\Leftrightarrow4(3\minus{}t)\geq t(3\minus{}t)^3$ and $ t(3\minus{}t)^2\equal{}4t\cdot\frac{3\minus{}t}{2}\cdot\frac{3\minus{}t}{2}\leq4\left(\frac{t\plus{}\frac{3\minus{}t}{2}\plus{}\frac{3\minus{}t}{2}}{3}\right)^3\equal{}4.$ Done!.


calculus
integration
number theory
prime numbers

[color=darkblue][size=75]<split from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39242]this thread[/url] by moderator rcv.>[/size][/color]

We can use $n! + 1$ to prove that there are infinitely many primes (a nicer proof than the usual one, in my opinion).

As the first post noted, every prime factor of $n! + 1$ must be greater than $n$.  In particular, there is a prime greater than $n$.  That means there are arbitrarily large primes.  That means there are infinitely many primes.  QED.

Hmm... ~QED.

First we have to prove that there are infinitely many n such that n!+1 is composite. Besides this, the proof is rather unsatsifying; it seems to have a lot of holes. I'll finish it up, while answering the first question simultaneously:

Suppose n is one less than a prime (p). Then n!=pm-1 (for some integral m), a famous theorem. Add one to get pm. Therefore, not all n!+1 are prime.

Now take n!+1=mp (where n=p-1 and p is the largest prime known). Since n!+1=mp is not divisible by anything less than p, there exist other prime factors greater than p. Given such a prime factor, repeat to find something larger. QED

[quote="jb05"]First we have to prove that there are infinitely many n such that n!+1 is composite. Besides this, the proof is rather unsatsifying; it seems to have a lot of holes.[/quote]
Why do we have to prove that $n! + 1$ is composite for infinitely many $n$?  If $n! + 1$ is prime, then it is already a prime greater than $n$.

Can you point out one hole in the proof?

My math book gave a very simple explanation, starting with the fact that there are infinitely many integers.

[quote="Hamster1800"]bah. didn't euclid prove this one way back?

assume $p_{1},p_{2},p_{3},...,p_{n}$ is all the prime numbers (assume a finite number)

$p_{1}p_{2}p_{3}...p_{n} + 1$ is not divisible by any $p_{x}$, therefore it is prime. This contradicts the fact that there is a finite number of primes.[/quote]

yep, i believe so  :D

[quote="Hamster1800"]$p_{1}p_{2}p_{3}...p_{n} + 1$ is not divisible by any $p_{x}$, therefore it is prime.[/quote]
This step is false.  For example, 2 * 3 * 5 * 7 * 11 * 13 + 1 is not prime.  (It factors into 59 * 509.)

but 59 and 509 are not prime because only 2, 3, 5, 7, 11, and 13 are primes so therefore 59 and 509 are not prime factors of that number making it prime (not divisible by any primes)

I think there is a little confusion here, or would be to some reading this.  What is being said is that you can prove by contradiction.  Assume there is a finite number of primes, say $p_i, 1 \le i \le k$.  Then if we consider $\prod_{i=1}^k p_i + 1$, not one of the existing primes is a factor.  This is a contradiction, since all numbers have at least one prime factor.

Look inside this book:

http://www.amazon.com/exec/obidos/tg/detail/-/3540404600/qid=1117417130/sr=8-1/ref=pd_csp_1/103-4841712-2342228?v=glance&s=books&n=507846

5 of those proofs are amazing.

http://www.amazon.com/gp/reader/3540404600/ref=sib_dp_pt/104-5819939-1591118#reader-page

Cool  :)


Hey, guys do you know precamp material is in the mail?
I am really excieted!!

hmm perhaps they will send them after may 19, since that's when they send out emails for the C registration?

anybody from C registration get the materials yet?

Hmm...Last year I took test C and got my acceptance like the end of May. Looking back at my email, I think that precamp materials were mostly sent out the first week of July. I'm not sure about that. For the time being, you have most of what you need. Take a look at the curriculum pdf on the site. The only other things that the sent out were like rules, maybe some forms to fill out, packing advice, and how to meet everybody at the airport.

Can we drive there if we live close-ish-far-ish (but mostly close-ish)?

I couldn't see the directors saying that must fly there... As long as your parents drive you, I don't see a problem. Last year everybody sorta went to this one building to say hello and all. Your parents could easily drop you off there. I think there was some opening ceremony after everybody arrived that parents could attend.

On another note, the Awesomemath website says "AMSP Pre-camp materials are in the mail." so thats the answer I guess.


What's the most efficient rigorous way to calculate limits like
$\lim_{x\to \infty}[\sqrt[2007]{x^{2007}+x^{2006}}-\sqrt[2007]{x^{2007}-x^{2006}}]$
without using L'Hopital's Rule?

Well, I'm always a big fan of Newton's binomial series:

$(1+u)^{\alpha}=1+\alpha u+\frac{\alpha(\alpha-1)}2u^{2}+O(u^{3})$ as $u\to0.$

The first rule of thumb in using this: first factor out the thing that is big. So, as $x\to\infty,$

$(x^{2007}+x^{2006})^{1/2007}-(x^{2007}-x^{2006})^{1/2007}$

$=x\left(1+\frac1x\right)^{\frac1{2007}}-x\left(1-\frac1x\right)^{\frac1{2007}}$

$=x\left[\left(1+\frac{1}{2007x}+O(x^{-2})\right)-\left(1-\frac{1}{2007x}+O(x^{-2})\right)\right]$

$=\frac2{2007}+O(x^{-1}),$ which tends to $\frac{2}{2007}$ as $x\to\infty.$

So basically, degree 2005 and lower is insignificant under the 2007th root and you can approximate $x^{2007}+x^{2006}\approx (x+1/2007)^{2007}$?  :maybe:

oh, allways $\sqrt[n]{ax^{n}+bx^{n-1}+...}\sim \sqrt[n]{a}\mid x+\frac{b}{na}\mid$.


Prove that:
$ sin^{2n}(\theta)cos^{2n}(\theta)\le \frac {sin^{4n}(\theta)}{log_{2}(2 \plus{} 2^{tan^{2n}(\theta)})} \plus{} \frac {[log_{2}(2 \plus{} 2^{tan^{2n}(\theta)}]\cos^{4n}(\theta)}{4}$
is true for all $ n\in\mathbb{N^{*}}$ and $ (\theta)\neq \frac {\pi}{2} \plus{} k\pi, k \in \mathbb{Z}$
When equality?

[hide="Solution?"]Simplify the logarithm first. \begin{align*}\log_2(2+2^{\tan^{2n}\theta})&=\log_2[2(1+2^{\tan^{2n}\theta-1})]\\&=1+\log_2(1+2^{\tan^{2n}\theta-1})\qquad(1)\end{align*}
To make the inequality look "nicer" and a lot easier to work with (and type), let
\[ a = \sin^n\theta\newline b = \cos^n\theta\newline x = \log_2(2 + 2^{\tan^{2n}\theta}).\]
Then the previously complicated-looking inequality becomes the simple task of proving
\[ a^2b^2\le\frac {a^4}{x} + \frac {b^4x}{4}.\]
This can easily be shown using AM-GM since all the terms are nonnegative. This is easy to see: by the Trivial Inequality, $ a^2,b^2\ge0$ and from $ (1),$ the quantity is $ 1+k$ with $ k$ being the logarithm of a quantity greater than $ \frac{1}{2}$, meaning that the logarithm must be greater than $ -1$ (even though the exact value is unknown), so that is also nonnegative.
\[ \frac {\frac {a^4}{x} + \frac {b^4x}{4}}{2}\ge\sqrt {\left(\frac {a^4}{x}\right)\left(\frac {b^4x}{4}\right)}\newline\frac {a^4}{x} + \frac {b^4x}{4}\ge2\sqrt {\frac {a^4b^4}{4}}\newline\frac {a^4}{x} + \frac {b^4x}{4}\ge2\left(\frac {a^2b^2}{2}\right)\]
$ \text{Q.E.D}$
[/hide]


let (C1),(C2),...,(Cn) be n circles .

find the locus of point M,in wich : the power of M to (C1)+the power of M to (C2)+...+the power of M to (Cn)=stable

I think that is that is quite hard .
where did you find it ashegh??
 :o

No Kheirkhah come on  it's so easy try to solve it.I will send my solution if U couldn't solve it.

the problem isn't hard. plz use this fact.

do u know what will be the locus of points like M in wich:

the power of point M to (C)+the power of point M to (C')=stable

i talked about it recently in fisaghoures(pytagoras)theorem.

the locus will be a circle wich its center is the mid point of OO'. (wich OO' is the centers of the circles (C) and (C').

we can also solve it by using gravity  :!:

I'll hide my solution, because there are people who want to post their own

[hide]I'll start inversely

Let $O_1,O_2,...,O_n$ the points of the circles and $G$ their center of gravity $\Rightarrow \sum_{i=1,..,n}\overline{GO_i}=0$

The power of $M$ to the circle $C_i$ is $MO_i^2-r_i^2$

Let $S$ the sum of all powers

$S=\sum{(MO_i^2-r_i^2)}=constant \Leftrightarrow$

$\sum{MO_i^2}=constant$

$\overline{MO_i}=\overline{MG}+\overline{GO_i}$

$|\overline{MO_i}|^2=\overline{MO_i}^2=(\overline{MG}+\overline{GO_i})^2$

$|\overline{MO_i}|^2=|\overline{MG}|^2+|\overline{GO_i}|^2+2\overline{MG} \cdot \overline{GO_i}$

$\sum{|\overline{MO_i}|^2}= \sum{|\overline{MG}|^2}+\sum{|\overline{GO_i}|^2}+2\sum{\overline{MG} \cdot \overline{GO_i}}$

$\sum{|\overline{MO_i}|^2}= n \cdot |\overline{MG}|^2+\sum{|\overline{GO_i}|^2}+2\overline{MG} \cdot\sum{ \overline{GO_i}}$

$\sum{|\overline{MO_i}|^2}= n \cdot |\overline{MG}|^2+\sum{|\overline{GO_i}|^2}+2\overline{MG} \cdot\sum{ \overline{0}}$

$\sum{|\overline{MO_i}|^2}= n \cdot |\overline{MG}|^2+\sum{|\overline{GO_i}|^2}$

In the last equation, $\sum{|\overline{GO_i}|^2}$ is constant and $n \cdot |\overline{MG}|^2$ depends only on $R=|\overline{MG}|$

Finally we find that

$S=n \cdot R^2 +c$, where $c$ is a constant ($c=\sum{|\overline{GO_i}|^2}-\sum{r_i^2}$)

Each of the circles $(G,R)$ is a locus of points, which gives its own sum $S=n \cdot R^2 +c$

Given the value of $S$, we can find the value of $R$, iff $S\geq c$

If $S=c=\sum{|\overline{GO_i}|^2}-\sum{r_i^2}$ then the locus is the point $G$ itself[/hide]

Actually, this problem was really nice, but not so easy, until ashegh gave the hint.

I'd like to see other solutions,  particularly with pure geometry!

excelent... :)

u wrote the thing that i said...


trigonometry
inequalities unsolved
inequalities

Find the max of $sinA_{1}sinA_{2}\cdots sinA_{n}$ such that $tanA_{1}tanA_{2}\cdots tanA_{n}= 1$.


Hello,

I think this is as off topic as is possible in a forum about Mathematical Problem Solving, but hey, that's what The Round Table is for, right?

I'm working on a robot design with two sets of wheels - one needs to be lowered using a motor while the robot is running. I plan to do this by building a frame to hold the second set of wheels and gluing the end of a bolt to the center of said frame. This bolt will run through a nut glued to the body of the vehicle. Now when the bolt turns the frame will be lowered or raised, thus accomplishing the goal of switching wheels. To make things easy, I'm looking for a bolt that had a head with teeth so that I can turn it with a geared motor.

So my question is, where could I buy a bolt with a toothed head? Preferred length is about an inch, but I could always cut off excess.

I'm also open to any better ideas for changing wheels. Note that I'm trying to limit the size of the body to about 9 by 3 by 3 inches, with a micro-controller board coming vertically off of the top.

Thanks!

Probably the part of the forum that deals with engineering would be better ;)  I've moved it there.


linear algebra
matrix
complex numbers
superior algebra
superior algebra solved

Let a,b,c,d,e,f,g,h,i reals or complex numbers
$A=\left (\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right )$

$B=\left (\begin{array}{ccc}a&-c&b\\-g&i&-h\\d&-f&e\end{array}\right )$

Prove that A and B are similar.

What is the definition of two similar matrices, A and B?

There is an invertible matrix $X$ s.t. $A=XBX^{-1}$.

[tex]X=I_{2,3}I_{3}(-1)=\left(\begin{array}{ccc}
1&0&0\\0&0&-1\\0&1&0\end{array}\right)
[/tex]

And [tex]XAX^{-1}=B[/tex] , so A and B are similar.


[color=blue][size=150] Rusia 1997[/size][/color]

Sea $\Gamma$ un circulo con centro $O$ e inscrito al triangulo $ABC$,
al cual toca en sus lados $AC, AB, BC$ en los puntos $K,M,N$ respectivamente. La mediana $BB_1$ del triangulo $ABC$
corta al segmento $MN$ en $D$. Mostrar que los puntos $O,D,K$ son colineales.

(Sugerencia)
[hide="Hint"] Consideremos  $L = KO \cap MN$, y probemos que $L = D.$ Luego juega con algo de semejanza. [/hide]

Carlos Bravo :)
Lima -PERU

he  encontrado  una solución  bastante fea para el  problema.  utilizando  ley de senos,  teorema  de la  bisectriz  generalizada , identidades trigonométricas probé  que la  mediana  y  la linea OK   dividen  al segmento MN  en  razón AB : BC  Y por lo  tanto  MN,OK Y BB_1 concurren .  pero  me  gustaría  ver  una solución  mas bonita (la de semejanza)
puedes  colgarla carlos?? :)   gracias :P  :lol:

Bueno el Hint realmente contiene el key ..
aqui envio la solucion referida en la sugerencia ...

 [color=blue][size=150]Solucion:[/size][/color]

 Sea $L = KO \cap MN$, y sea $A_1, C_1$ los puntos de interseccion de $AB$ y $BC$ respectivamente con la paralela trazada a $AC$ que pase por $L$, portanto tenemos $A_1L =LC_1$, lo cual implica que $L$ es el punto medio de $BB_1$ y entonces $L= D.$

Notar que: $\angle A_1MO$ y $\angle A_1LO$ son rectos, entonces el cuadrilatero $MA_1OL$ es ciclico, entonces $\angle MLA_1 = \angle MOA_1$ como tambien $\angle C_1LN = \angle C_1ON$

( de manera analoga, notemos que  $\angle MLA_1$ y $\angle C_1LN$ son rectos, entonces  $\angle MOA_1 = \angle C_1ON$). De aqui tenemos que  los triangulos $OMA_1$ y $ONC_1$ son congruentes, entonces $OA_1 = OC_1$ con lo cual concluimos que $A_1L = LC_1.$

Carlos Bravo :)
Lima -PERU

Supongamos que $ KO$ corta a $ MN$ en $ D.$ Tenemos en el triangulo isósceles $ \triangle BMN$

$ \frac { \sin \angle DBC}{\sin \angle DBA} \equal{} \frac {DN}{DM} \equal{} \frac {KN}{KM} \cdot \frac {\sin \angle DKN}{\sin \angle DKM} \equal{} \frac { \cos (\frac {C}{2}) \sin (\frac {C}{2})}{ \cos (\frac {A}{2}) \sin (\frac {A}{2})} \equal{} \frac { \sin C}{ \sin A} \equal{} \frac {c}{a}$

$ \Longrightarrow D$ yace en la mediana $ BB_1.$

Denotemos $ T_{\infty}$ el punto del infinito de la recta $ AC,$ el cual es polo de $ KD$ respecto a $ (O).$ $ B$ es el polo de $ MN$ respecto a $ (O)$ y el polo de $ BB_1$ es el punto $ P$ en el rayo $ MN$ tal que $ (M,N,D,P) \equal{} \minus{} 1,$ pero como $ (A,C,B_1,T_{\infty}) \equal{} \minus{} 1$ $ \Longrightarrow$ $ PB \parallel AC.$ Como los polos $ B,P,T_{\infty}$ estan alineados, entonces las polares $ MN,BB_1,DK$ concurren $ \Longrightarrow$ los puntos $ O,D,K$ estan alineados.


Let $n\neq 2^{k}$ be a positive integer such that $\phi(n)\mid n$. Show that $\frac{n}{\phi(n)}=3$. (here $\phi$ is the Euler function.)

when we simplify that product, we get like

a product with terms $\frac{p}{p-1}$, etc

the maximal term on the bottom must be prime...

but the only adjacent primes are $2,3$, so ta da

It is not so hard to show that $\phi(n) | n$ if and only if $n=2^{n}3^{m}$ where $n,m$ are non-negative. Since $n\not = 2^{k}$ it means $m\not =0$ so $\phi (n) = 2^{n}3^{m-1}$ when we divide  $n=2^{n}3^{m}$ we are left with $3$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ K$ be the splitting field of $ x^3\minus{}2$ over $ \mathbb{Q}$
(i) Describe $ K$ explicitly by giving generators for $ K/\mathbb{Q}$.

(ii) Prove that $ Gal(K/\mathbb{Q})$ is isomorphic to $ S_3$, the symmetric group on three letters.

(iii) For every subgroup $ H$ of $ S_3$, find the fixed field $ K^H$ and check that $ Gal(K/K^H)$ is equal to $ H$.

(iv) There is a unique subfield $ F$ of $ K$ s.t. $ F$ has degree $ 2$ over $ \mathbb{Q}$.
Thus, $ F \equal{} Q(\sqrt{D})$ for some squarefree integer $ D$. What is the value of $ D$?

For (i), first find the roots of x^3 - 2 and see what needs to be added.

For (ii), what is the degree if you  add the (two) things from (i)?  is there an
obvious map into S_n?

Do you really expect people to do all of your homework here?  parts 3 and 4, I leave to
you.


У меня есть 4 вопроса для вас
Объясните мне 4 идиома или фразологии:
1. терять голову
2. Влюбиться с первого взгляда
3 жить за каменной стеной
4.как две капли воды
5 Весь в отца .
Я понял но не могу объяснить по-русски.

Спасибо ранее!!

[quote="Erken"][quote="Erken"]Насколько я понимаю, ты хочешь увидеть перевод на английский язык.
But at first several suggestions about russian language.
Instead of "фразологии" you should say "фразеологии",and also "спасибо заранее" sounds better and it is correct, unlike this "спасибо ранее"
1.It means to lose control.Usually used to express strong love or just being crazy on something.
2.Love at first sight.
3.To be extremely good protected,or it means to have a men who can protect you whenever it is needed.
4.It is often used to emphasize similarity.This phrase can also be used towards two things or two men.
5.To have the same nature,character as father.Sometimes,but slightly rare,to emphasize having similar face.[/quote]
Aha...Only now I have finally realized what you've been asking.
1.Потерять контроль, часто используется чтобы выразить сильную любовь по отношению к противоположенному полу или выразить страсть к какому-нибудь виду занятий,в частности к азартным играм.
2.There is nothing to explain in russian...It is not a phraseologism,I think,though I doubt it is possible,to be in love with someone after first sight.
3.Быть обеспеченным материально,быть морально и физически защищенным.Иметь защитника.
4.Очень схожие по физическому строению.Обычно используется по отношению к людям,чтобы указать на сходства их лиц,немного реже,по отношению к неодушевленным предметам.
5.Быть похожим на отца,чаще используется чтобы указать на общие черты характера или род деятельности,реже на внешние сходства.[/quote]
 Спасибо за все.

Каждому второму – язык бесплатно!
   Беспрецендентную акцию начинает в рамках подготовки к новому учебному году международный информационно - образовательный и культурный Центр “Interllingva”  (Хайфа, Израиль)
   Каждый второй сотрудник любого предприятия в мире сможет пройти базовый разговорный курс одного из основных иностранных языков, преподаваемых в проекте «Лингва-Класс» – совершенно бесплатно. Напомним, что базовый курс – это 60 астрономических часов учебных занятий, которые проходят в режиме он-лайн, в прямом контакте с преподавателем и другими учениками группы, два раза в неделю по полтора часа.
  В международные группы принимаются учащиеся различного возраста (начиная с 12 лет), для обучения необходимо иметь вход в интернет, наушники (или колонки) и микрофон. При наличии колонок, в уроке могут принимать участие все члены семьи.
Акция пройдёт в период с 1  по 31 августа 2009 года.
Подробную информацию можно найти на сайте [url]http://www.lingvaclass.com[/url]


1. Find a 5-digit number for which all of the following are true:
a) The number's square root is a prime palindrome;
b) The number's first two digits express a perfect square;
c) The number's middle digit is a perfect square;
d) The number's last two digits express a perfect square;
(Nats 92 #8)
[hide="Answer"]36481[/hide]
2. A science test has 10 questions worth 5 points each. 7 questions worth 6 points each, and 4 questions worth 2 points each. How many scores between 0 and 100 are impossible to score?(Nats 98 #7)
[hide="Answer"]4 scores[/hide]
3. A sign in New York City indicates the amount of money Americans had saved using one telephone company instead of another. At 3:50 PM on January 1, 1999, the sign read $ \$2,519,206,712.$ The sign adds $ \$1$ each 0.1 seconds. What month and year will the sign read $ \$9,999,999,999?$(Nats 99 #2)
[hide="Answer"]September 2022[/hide]
4.On Tuesday, Megan biked to school but walked home. The total travel time Tuesday was 35 minutes. If the round-trip travel time by bicycle is 30 minutes less than the round-trip tavel time walking, how many minutes does it take Megan to walk one way?
[hide="Answer"]25 minutes[/hide]
5. Two marbles are randomly selected without replacement from a bag containing blue and green marbles. The probability they are both blue is $ \frac16$ If three marbles are randomly selected without replacement, the probability that all three are blue is $ \frac1{21}$ wHt is the fewest number of marbles that must have been in the bag before any were drawn?
[hide="Answer"]9 marbles[/hide]

[hide="1."]

Well, the first $ 2$ digits must be $ 16, 25, 36, 49, 64, or 81$.

Same with the last $ 2$ digits.

We look to the 3-digit square root.

The first digit will be the same as the last digit.

It must be $ 1$ or $ 3$, if it will be prime.

If it is $ 3$ our option for the second digit is $ 0$ or $ 1$, which give

$ 91809$ and $ 97969$, obht of which don't work.

So it must be $ 1$, and we just use trial and error, to get $ 191^2\equal{}36481$.[/hide]

[hide="5"]
Equation time... ($ b$ represents blue marbles and $ x$ represents the total amount of marbles)

\begin{align*} \frac bx \cdot \frac {b - 1}{x - 1} & = \frac 16 \\
\frac bx \cdot \frac {b - 1}{x - 1} \cdot \frac {b - 2}{x - 2} & = \frac 1{21} \\
\frac {b - 2}{x - 2} & = \frac 6{21} \end{align*}

Cross multiply...
$ 7b = 2x + 10$

At this point, we know that $ b$ has to be even. However, $ b$ cannot be $ 2$, since that would force $ x$ to be 2, therefore making the probability that two blue marbles are drawn $ 1$. The next largest value of $ b$ is $ 4$, which makes $ x = 9$. We can plug those values in to make sure they work.

$ \frac 49 \cdot \frac 38 = \frac 16$ Check
$ \frac 49 \cdot \frac 38 \cdot \frac 27 = \frac 1{21}$ Check

Therefore, the answer is $ \boxed{9}$.
[/hide]

2.
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1792095425&t=186413

[hide="3"]

so first find the number of dollars left to get, it is 7,480,793,281.

Since 1 dollar is added .1 seconds, 10 is added every second. 600 a minute
36000 an hour, 864000 a day. times 365.25 to get 315,576,000 a year.

You get approximately 23.70520344 years when you divide 7,480,793,281 and 315,576,000. So 23 years after 1999 is 2022. .70520344/1=x/365

so approximately 257.3992559th day, which falls in september.

So you get September 2022.
[/hide]

[hide="4"]
so x is the time it takes to walk one way on foot while y is the biking time.
x+y=35
2y+30=2x (2x-2y=30) (x-y=15)
add to together.
2x=50
x=25.

[/hide]

[hide="5-easier way"]

The quotient of 1/21 divided by 1/6 is 2/7, so there is a 2/7 chance of choosing a blue on the 3rd draw if the first two were blue. So Just add the first two marbles to the denominator to get 7+2=9.
[/hide]


A Wii game disc and a Gamecube game disc are carefully balanced on the ground a couple of feet apart, standing straight up.  The Wii disc is rotated about its diameter perpendicular to the ground until the 2 discs lie in the same plane (the discs are 2D for our purposes). The Wii disc is then rolled towards the Gamecube disc until they touch.
Let $w$ and $g$ be the respective radii of the Wii game disc and the Gamecube game disc.  Assume the sun is directly overhead and the sun's rays are all perpendicular to the ground.  Find the length of the shadow created by the formation of game discs, in terms of $w$ and $g$.

There would not be a shadow if the sun is perpendicular to the ground and the discs are two dimensional.  :maybe:

[hide]Maybe he means $2w+2g$... if we pretend the discs are not there...[/hide]

Well, the shadow would be a line.  Lines only have length, not width.
And buzzer11, keep in mind the discs are of different sizes.  $2w+2g$ is not correct.

ok, so would it be
[hide]
g+2sqrt(wg)+w...since they have different sizes,when they are tangent to each other, the line through both of the circles becomes diagonal, and you get a right triangle with hypotenuse (w+g), and a leg of (g-w) or (w-g), depending on which one's larger, so the second leg = 2sqrt(wg), and then you have to add the two radii, g and w. [/hide]

[quote="erdogankerem123"]my solution[/quote]
Your solution is correct.  And just for general knowledge, Wii game discs are larger.


Suppose that $a$ and $b$ are natural numbers such that:

$p=\frac{4}{b}\sqrt \frac{2a-b}{2a+b}$

is a prime number. What is the maximum value of $p$?

It's Iran 1998. If I'm remember correctly the maximum value of $p$ was 5.

It is not correct. When a and b natural 2a-b<2a+b and it give p is not integer.

oh, it doesn't matter ,you should find the maximum positive integer number. :wink:

Excuse me this is not Iran ... :blush:
But the solution
Obviously $p\leq 4$. If $p=3$ then $b\leq\frac43$. So $b=1$. So $(2a+1)^{2}p=4(2a-1)^{2}$. Left side is odd and right side is even. Contradiction.
For $p=2$ then $b$ must be even, easily you see that there does not such $a,b$.
So the equation doesn't have any solution.


trigonometry
real analysis
real analysis unsolved

Does the sequence $ n(1\plus{}\sin(n))$ tend to $ \plus{}\infty$ ?

take $ n\equal{}\frac{\pi}{2}\plus{}2k\pi$

then $ n(1\plus{}\sin{n})$ $ \rightarrow \infty$

take $ n\equal{}\frac{3\pi}{2}\plus{}2k\pi$

then $ n(1\plus{}\sin{n})\equal{}0$

But, the unstated assumption, when he says "sequence":  $ n$ is an integer.  Xingyan used non-integer values for $ n$.

So this problem is then something about the rational approximation of some irrational number.

An answer, more general than asked for here:
The sequence $ n(1 \plus{} \sin(n\theta))$ goes to $ \infty$ if and only if $ \theta \equal{} \frac {a}{b}\pi$ for some integers $ a,b$ with $ b$ odd. In fact, we could say the same about $ n^2(1\plus{}\sin(n\theta))$.


Wow.

That really came in handy. THanks.

@all NM mc competitors: GJGJGJGJ

Ummmm, not for me.

EPIC PHAIL :mad:
13 points of reading/copying/arithmetic errors
and a bad thinking day
worse than last year too
unfortunate that my mc career had toi end in heartbreak

....and our team should have gotten top 10...... oh well
team round was the only thing that wasn't bad........

oh well, disney, meeting ppl, mastres, and cd were fun though

everyone has their bad days. all new mexicans had bad days. andy only got 8 on target, robert had 13 points of errors, drew had like 10 pnts in errors, and EPICLY PHAILED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I'm not to say thati didn't do well... i did ok... but i should've gotten a perfect target, and 5 more sprint if there were no stupid mistakes...

but anyway, nm: Good Job!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Also, we got 21st at nats!!!!!!!!!! is that exciting or what!!!!!!!!!!!!!!!!!! even though our cutoff was only 28 to get into nats, (sorry nm, i phailed at states. our cutoff should've been like 33) [excessive screaming] WE ARE THE BEST NM TEAM!!!!!!!!!!!!!!!!!!!!!!!!! Exciting? YES!!!!!!!!!!!!!!!!!!!! WE ARE THE MOST IMPROVED STATE IN ALL OF THE US!!!!!!!!!!!!!! WE ARE THE 21ST BEST STATE/TERRITORY IN THE US!!!!!!!!!!!!!!!!!!!!!!!!! WE SHOULD BE VERY HAPPY!!!!!!!!!!!!!!!!! AND WE ARE!!!!!!!!!!!!!!! [/excessive sreaming] sorry about that...

anyway, who is mathlover?? he says he competes in nm.

[quote="pinkmuskrat"]No, the best team ever was 1999, I think. They were high teens in place.
BTW I'll hopefully volunteer next year at chapter and state. GL repeating at going to nats, connaissance and prezcoin. Also, I've made my top 10 at state predictions already for next year, though subject to change based upon 1. unknown 6th graders 2. some kids making or not making our school team. PM if you want the list.[/quote]

I hope that NM will do well again. We could have easily done better this year, though...

oh, and I'm sorry for the hypocrisy.

I'm thinking 2011 could be very strong...

hmmmmm... u seem to be over thinking this too much...


a,Find the maximum value of $ k(k>0)$ such that
$ a\plus{}b\plus{}c\equal{}1(a,b,c>0)$
and:
$ \frac{a^2\plus{}kb}{b\plus{}c}\plus{}\frac{b^2\plus{}kc}{c\plus{}a}\plus{}\frac{c^2\plus{}ka}{a\plus{}c} \ge \frac{3k\plus{}1}{2}$
b,try with $ k\equal{}7$

Is the denominator in the third fraction supposed to be $ a\plus{}b$?


Ahaha... Epic fail at square roots. I forget, when you have something like say $ \sqrt5 \plus{} 4$ Would I add the inside first THEN find the square root, or the square root of each number and then find then add? And how about multiplication? (E.g $ \sqrt(4)(4)$) Oh by the way, the parentheses mean multiplication if you didn't know, AND ALSO; I couldn't seem to stretch the square root sign over the whole entire expression; could someone tell me how to do that?
Also, I forget what you do when you divide an expression by a number for example... $ \frac{4\minus{}4x}{4}$. Would it be $ 1\minus{}x$ or what? Do you divide everything by it or...
Thanks; and if you think these are homework problems; they aren't I'm just studying for finals... Epic fail... I have a bad memory!  :D

hello, here is what i would do
$ \sqrt{5}\plus{}4$ at first you must calculate the square root and then you have to add 4, but it is sensless, because you can only calculate an approximate value of this term.
$ \sqrt{4}\cdot 4\equal{}2\cdot 4\equal{}8$
$ \frac{4\minus{}4x}{4}\equal{}\frac{4}{4}\minus{}\frac{4x}{4}\equal{}1\minus{}x$.
Sonnhard.

I didn't know how to stretch the square root sign over the whole thing... it wouldn't be the same thing would it?

If you have the square root stretched over the whole thing, it's like $ \sqrt(5\plus{}4)$, so you solve the thing under the square root and THEN find the square root.

Say something is a variable and you don't know what "$ x$" is equal to? Say something like; $ \sqrt(54 \plus{} 3x)$?
I'm sorry; my mind seems to have imploded this information...

Here is how to stretch the square root sign:

[code]
$\sqrt{(4)(4)}$
[/code]

You have to put curly brackets around the place where you want it to go over.

[hide="EXPONENT LAWS"]

$ a^n \times b^n=(a\times b)^n$ and vice versa.

$ n^a\times n^b=n^{(a+b)}$ and vice versa.

$ \frac{a^n}{b^n}=(\frac{a}{b})^n$ and vice versa.

$ \frac{n^a}{n^b}=n^{a-b}$ and vice versa.

$ (n^a)^b=n^{a\times b}$ and vice versa.

There is no laws concerning adding or subtraction.
[/hide]

[hide="YOUR QUESTIONS"]

You can rewrite $ \sqrt{n}$ as $ n^{\frac{1}{2}}$.

Example: You asked what is $ \sqrt{4\times4}$. You can do this two ways:

Way 1: $ \sqrt{4\times4}=\sqrt{16}=\boxed{4}$

Way 2: $ \sqrt{4\times4}=\sqrt{4]\times\sqrt{4}=2\times2=\boxed{4}}$

For your last question ($ \sqrt{54+3x}$), you can't simplify it further. If it is part of an equation, you have to get it onto its own side and then square both sides.
[/hide]

Hope that helps!

Ah... That makes much more sense. Thanks! ^________________^
So what was the exponent laws about?
Also, considering the fact that I have spontaneously become idiotic and stupid; would I do the same thing for the division thing?

Exponent laws was explained because essentially, square roots ARE exponents because $ \sqrt{n}$ can be written as $ n^{\frac{1}{2}}$.

And yes, it's the same for division because division is merely multiplication of the reciprocal. So $ \sqrt{\frac{n}{m}} \equal{} \sqrt{n * \frac{1}{m}}$

The exponent laws is for anything else you have to ask. This is because square roots ARE EXPONENTS. Use the exponent laws for any more questions, such as the division thing. If you don't understand anything, just ask.

EDIT: I guess two posts were posted at the same time.

[quote="joyofpi"] I forget, when you have something like say $ \sqrt5 + 4$ Would I add the inside first THEN find the square root, or the square root of each number and then find then add? [/quote]

The $ \sqrt$ symbol counts as a grouping symbol, so follow GEMDAS (grouping symbols first) and add $ 4+5$ first, then take the Square root.


Putnam
induction
inequalities
percent
college contests
Putnam easy

Yes.  At some point for say, N-1 throws S(N-1)<4/5*(N-1) and the next throw (being necessarily succesful) gives us S(N)>=4/5*N.   Since the Nth throw was succesful S(N-1)=S(N)-1.

4/5*(N-1)>S(N)-1>=4/5*N-1

=>    4N-4>5S(N)-5>=4N-5

subtract 4N and add 5 to everything
=>  1>5S(N)-4N>=0

Since 5S(N)-4N is an integer the only possibility left is that

 5S(N)-4N = 0  so therefore  S(N)=4/5*N=.8N

If someone wants to TeX  this answer go ahead.

[u][b]The author of this posting is : blahblahblah[/b][/u]
____________________________________________________________________

[quote="Kent Merryfield"]Basketball star Shanille O'Keal's team statistician keeps track of the number, $S(N),$ of successful free throws she has made in her first $N$ attempts of the season. Early in the season, $S(N)$ was less than 80% of $N,$ but by the end of the season, $S(N)$ was more than 80% of $N.$ Was there necessarily a moment in between when $S(N)$ was exactly 80% of $N$?[/quote]

Yes.

Let $P\left(n\right) = \frac{S\left(n\right)}{n}$.

Say that over a certain interval, we have $P\left(n\right) <0.8$. If there is a multiple of $5$ in this interval, we must have $P\left(5i\right) \leq 0.8 - \frac{1}{5i}$. Thus we must have $P\left(5i+j\right) <0.8$ for $1\leq i\leq 4$.
By induction, we can establish a similar result for $P\left(5i+5\right)$ and conclude that we can only have $P\left(n\right) \geq 0.8$ if $n=5j$, and the best we can do is equality, so the answer is yes.

[u][b]The author of this posting is : Jrthedawg[/b][/u]
____________________________________________________________________

Another solution:

Sorry, I don't know how to use that fancy font. [color=red][Moderator edit: This "fancy font" is probably LaTeX...  :D ][/color]

We prove by contradiciton: Assume there exists some N that satisfies

S(N)/N<4/5<(S(N)+1)/(N+1).

This statement basically says that there exists some N for which O'Keal's percentage is below 80%, and with one more make, her percentage jumps above 80%.

This leads to the inequality 4N-1<5S(N)<4N.

We know 5S(N) is an integer, and no integer exists between 4N-1 and 4N, thus, a contradiction, and our original assumption is false. If no such N satisfies our original inequality, then there is no way O'Keal could have progressed past 80 percent without being at exactly 80 percent.

[u][b]The author of this posting is : Ravi B[/b][/u]
____________________________________________________________________

Another way to say it. Consider the quantity 5S(N) - 4N. It is an integer. Early on it is negative and later it is positive. After each throw, the quantity changes by either +1 or -4. Thus it is clear the quantity must hit 0 at some point in between. To be more precise, consider the first time (in between) that the quantity is nonnegative. The moment before it must have been negative, that is at most -1. Because jumps are at most +1, at the current time it must be at most -1 + 1 = 0. The only nonnegative number at most 0 is 0 itself, so at the current time the quantity is 0.

someone said something about this working for all percentages of the form 1-1/n, $n\geq 2$

i proved it for .9, i think (i can't remember but i'm pretty sure i scrawled it in my notebook).

it would be cool to see a general proof, maybe i will do it tommorrow as it is 2 am here.

For luther_driggers:

Suppose the target percentage is $p=1-\frac1k,k\ge2.$ (In the problem as stated, $k=5.$) $S(N)<pN$ for small value of $N$ but eventually $S(N)>pN$.

Consider the quantity $Q(N)=kS(N)-(k-1)N.$ The quantity $Q(N)$ is an integer. With each made free throw, $Q(N)$ increases by 1; with each missed free throw, $Q(N)$ decreases by $k-1.$ Since $Q(N)$ integer valued and it starts out negative and becomes positive, and since it increases by 1 each time it increases, then it must equal 0 at some point.

Note that this is exactly Ravi B's proof.

It doesn't work if the target percentage has a different form. Suppose we ask if the percentage must ever be 60%. It doesn't. Suppose she misses the first one, then makes the next two. The percentages go 0$, 50%, 67% and never equal 60%.

sorry kerry i couldn't see your code, u must have an xtra dollar sign or whatever   :)

i was just gonna prove it the way i did the first one.

[hide="I like hidden proofs"]

$p=\frac{k-1}{k},k\geq 2$.

$S(N-1)<  \frac{k-1}{k} (N-1)$

$S(N)-1=S(N-1)$

$S(N)\geq \frac{k-1}{k}N$

combining and multiplying by k

$(k-1)(N-1)>kS(N)-k \geq (k-1)N -k$

subtract $(k-1)N$ and add $k$

$1>kS(N)-(k-1)N \geq 0$

which is an integer so

$kS(N)-(k-1)N=0$

and

$S(N)= \frac{k-1}{k}N$

[/hide]

it would be cool to c the fixed version of yours thou.

Wow, that was easy.

The only Putnam question I'll probably ever get.

[quote=CrazyBeast2.0]Hey I am no expert in college math but isn't the graph of S(N) continuous. If it is, then definitely when it goes from less than 80% to more than 80% at one point it must be 80%.[/quote]

Not really, since $S$ is defined over $\mathbb{N}$, not over $\mathbb{R}$. In fact, if we change $80\%$ to $60\%$ then the answer is negative (misses the fiirst one, scores the second and third, and never shoots again).

Assume the contrary. It follows that there's a $n$ such that $S(n)<80\%$ and $S(n+1)<80\%.$ So $$S(n)/n<80\%<S(n+1)/n.$$ But it's obvious that $S(n+1)=S(n)+1.$ So it follows that $$5S(n)<4n<5S(n)+1,$$ a contradiction since an integer cannot be in between two consecutive integers. $\square$

[hide=solution]We claim that the answer is yes.

Let the first time $S(N)$ becomes at least $\frac{4}{5}N$ (we know this exists by assumption) be when Shanille O'Keal makes $M$ of his first $N$ shots. Then, since this is the first time $S(N)$ becomes at least $\frac{4}{5}N,$ we know that $\frac{M-1}{N-1}<\frac{4}{5},$ which gives $5M<4N-1$ after rearranging. However, we also know that $\frac{M}{N}\ge \frac{4}{5},$ so $5M\ge 4N.$

The only way for both $5M<4N-1$ and $5M\ge 4N$ to hold for positive integers $M,N$ is if $5M=4N.$ This implies that there was indeed a moment during the season where $S(N)$ was precisely $\frac{4}{5}N.$[/hide]

Incomplete.
[rule]
The answer is [b]yes[/b].

Suppose otherwise. Then, there would be a point at which $\tfrac{S(N)}{N}<80\%$ and $\tfrac{S(N+1)}{N+1}>80\%$. Suppose that this occurs at $N=n$. Then, $$\frac{S(n)}{n}<\frac{4}{5} \implies 5S(n)<4n, \quad \frac{S(n+1)}{n+1}>\frac{4}{5} \implies 5S(n+1)>4n+4.$$Since $\tfrac{S(n+1)}{n+1}>\tfrac{S(n)}{n}$, it follows that $S(n+1)>S(n)$, which is only true if the $(n+1)$th free throw was successful. Thus, $S(n+1)=S(n)+1$.

Then, $5(S(n)+1)>4n+4 \implies 5S(n)+1>4n$. We also have $5S(n)<4n$, so $$5S(n)<4n<5S(n)+1.$$ However, $5S(n)$ and $5S(n+1)$ are consecutive integers, and no integers can exist strictly between two consecutive integers, a contradiction.

To-do: Add in the vice versa case.

As mentioned by the folks above, the easiest way to go about it is contradiction. Suppose right before going from below an $80$ win streak to a above an $80$ win streak, we have $a$ wins and $b$ losses. So clearly $a < 4b$. Now after this game, we will have $a+1$ wins and $b$ losses with us purportedly obeying $a+1 > 4b \implies a > 4b-1$. So $4b-1 < a < 4b$. But $a$ is an integer, contradiction.

This is true, so let's give a proof my contradiction. We can test cases to see that we can't have $S(N)>0.8N$ for $N\leq 5$, meaning at some $N$, we have $S(5N)=4N-\epsilon_1$ and $S(5(N+1))=4(N+1)+\epsilon_2$, with $\epsilon_1, \epsilon_2 > 0$, giving $4+\epsilon_1+\epsilon_2\geq 6$. However, we arrive at
\begin{align*}
    &4 +\epsilon_1 + \epsilon_2\\
    &= 4(N+1)+\epsilon_2 - (4N-\epsilon_1)\\
    &= S(5(N+1))-S(5N)\\
    &\leq 5(N+1)-5N = 5,
\end{align*}
contradiction.

The answer is yes.

	Notice that at $S(N)$ can only increase when Yang Liu scores, and so there has to be a point $N'$ before $S(N)\ge 80\%$ from which on Yang Liu keeps on scoring consecutively until $S(N)>80\%$. Say that $S(N')=\frac{a}{b}$, and let $\frac{4}{5}=\frac{4(b-a)}{5(b-a)}$, where we note that $5(b-a)>b$ as $\frac{4}{5}>\frac ab\iff 4b>5a$. We note that $S(N'+m)=\frac{a+m}{b+m}$ now, and we see that the difference between denominator and numerator stays constant. Now clearly, $S(N'+m)$ runs through all fractions with difference between denominator and numerator $b-a$ such that the denominator is bigger than $b$. Eventually, $b+m'=5(b-a)$. But now $b+m'-a+m'=b-a$ and $5(b-a)-4(b-a)=b-a$, so $S(N'+m)=\frac{4(b-a)}{5(b-a)}=\frac45$, as desired.

[hide]
The answer is $\boxed{\text{Yes}}$; otherwise, there are $a,b\in\mathbb{Z}$ so that $\tfrac{a}{b}<\tfrac{4}{5}$ and $\tfrac{a+1}{b+1}>\tfrac{4}{5}$, which by cross-multiplying we find is absurd. \qed

[quote=OTIS Version]
Basketball star Yang Liu’s team statistician keeps track of the number, $S(N)$, of successful free throws he has made in his first $N$ attempts of the season. Early in the season, $S(N)$ was less than $80\%$ of $S(N)$, but by the end of the season, $S(N)$ was more than $80\%$ of $N$. Was there necessarily a moment in between when $S(N)$ was exactly $80\%$ of $N$?
[/quote]

The answer is $\boxed{\text{yes}}$. For the sake of contradiction, assume that there was never necessarily a moment when $S(N)$ was $80\%$ of $N$.

Suppose that in the moment right as the percent of successful free throws jumps from below $80\%$ to above $80\%$, Yang Liu had made $a$ shots and missed $b$ shots. It is clear that $\frac{a}{b}<4 \implies a<4b$. After making a shot, we must have $\frac{a+1}{b}>4 \implies a>4b-1$. Thus,

\[4b-1<a<4b,\]

a contradiction as $a$ is an integer.


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

In a cyclic quatrilateral ABCD with AB=AD point M , N lie on the sides BC and CD respectively so that MN=BM+DN. lines AM and AN meet the circumcircle of ABCD agian at point P and Q respectively. Prove that the orthocenter of the trianglr APQ lies on the segment MN.

[hide="Hint only"]Take point $ H$ on $ MN$ such that $ BM\equal{}MH$ and $ HN\equal{}DN$. Point $ H$ is the orthocenter of triangle $ APQ$.[/hide]

[color=darkblue]As [b]Johan Gunardi[/b] said, consider point $ H$ on $ MN$ such that $ DN \equal{} NH$ and $ MH \equal{} BM$. Consider the perpendicular bisectors of the segments $ DH$ and $ BH$ (they are also external bisectors of the angles $ CNM$ and $ CMN$), they intersect at $ A'$, which is $ C$-excenter in the triangle $ CMN$. We get that points $ A$, $ A'$ and $ C$ lie on ther bisector of $ \angle NCM$ and $ A'D \equal{} A'H \equal{} A'B$ (draw the exradius and get some congruent triangles).

So we have two possibilities: $ CD \equal{} CB$ (this case is easy) and, by a known lema, $ A' \equal{} A$.
If $ A' \equal{} A$ then reflections of $ H$ wrt sides $ AP$ and $ AQ$ lie on the circle... You can easy prove that the unique point with this property is orthocenter...[/color]

P.S. could you post all problems form the Zhautykov olympiad???


calculus integration geometry function derivative trigonometry conics

My first post here, please bear with my newbishness.

I'm currently taking Calculus in High School, and i've been able to grasp pretty much everything about Graph analysis using Differentiation.  Now my class is into Integration and area under a curve.  One thing popped up in my mind and my teacher can't give me an answer.  I try to work it out but my Integration skill is not strong enough  :(

I clearly understand the relationship between the graph of a function and the graph of its derivative when it comes to finding its rel max and min, its inflection point, and all. But when it comes to the use of the antiderivative of a function in THE FIRST FUNDAMENTAL THEOREM OF CALCULUS to find an area under a curve, i suspect there has to be a link between the graph of f and F (similar doing the opposite in  using the derivative of a function, anyone can see that when the derivative is set to 0, you find the rel max n min. And when you set the second derivative to 0, you find the possible inflection point)

My way of trying to find any relationship was to use the graph of a semicircle and the graph of its antiderivative but i can't integrate x 2 +y 2 =r 2  :(

[b]So can anyone please show me if there is a link between the graph of a function and the graph of its antiderivative concerning the area underneath of a curve or something else?[/b]

maybe I don't understand, but the relationship between int (f) and f is the same as the relationship between f and f'.

I understand that clearly but what i'm talking about is: when finding the area of under a curve in  [a, b], you find the antiderivative of the function and substitute a and b in F(b)-F(a), though i accept the theorem, looking at the the graphs( f and F) it makes no sense to me why i substitute a and b in the antiderivative :( .

[b]But can anyone do a graphical analysis for me to show why the antiderivative is used to find the area.  Maybe i'll find my way out of what seems to be a labyrinth.
[/b]
PS:I'm not good at asking question so if it doesn't make sense tell me and i'll try to rephrase what i'm trying to ask

For a semicircle with radius $r$, the equation is $y=\sqrt{r^2-x^2}$.
$\int \sqrt{r^2-x^2} dx = \frac{\sin^{-1}{\left(\frac{x}{r}\right)}r^2}{2} + \frac{x\sqrt{r^2-x^2}}{2}+C$.
Graph both of these and you will see the relationship.

For a simpler example look at the graph od $y=x$, which is the derivitive of $y=\frac{1}{2}x^2$, so the integral of a parabola is a line.

The reason we use the integral to find the area under a curve is because each strip we take has a with $dx$, a height $f(x)$, and an area $dA$, and the integral undoes the derivative, $dA = f(x)dx \Rightarrow A=\int_a^{b} f(x) dx$.  This is equivilent to taking infinit infinitely thin strips from $a$ to $b$ and adding their areas.  This concept of infinitely thin is similar to the definition of a derivative $f'(x) = \lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}}$.

Thanx Jimmy, i was working with it using y=x and i came up to that same conclusion. I was stupid enough  :lol: trying to use a semicircle to analyze it.  I completely overlooked the simple function f(x)=x thinking that it would be easier for me using a semicircle since i already know how to find its area using simple geometry.

[quote="Peter VDD"]also PLEASE DON'T WRITE YOUR TITLES IN CAPITAL LETTERS ;) thank you[/quote]

no problem


At a certain intersection, $ 30\%$ of the cars turn left, $ 40\%$ turn right, and $ 30\%$ go straight. What is the probability that the next car turns left and the following one turns right? Express your answer as a common fraction.

uh to clarify nike's answer...

$ 30\%\equal{}\frac{3}{10}$ so the probability the first car turns left is $ \frac{3}{10}$.
$ 40\%\equal{}\frac{2}{5}$ so the probability the second car turns right is $ \frac{2}{5}$.

Since these are independent events, meaning one choice doesn't affect the other, the probabilities are multiplied.

$ \frac{3}{10}\times\frac{2}{5}\equal{}\boxed{\frac{3}{25}}$ which is the same answer nike got :)


algebra
polynomial
limit
real analysis
real analysis unsolved

Prove that we can't find two polynomials $f,g \in \mathbb C[X]$ such that $\frac{f(n)}{g(n)}= 1+\frac1{1!}+\frac1{2!}+\ldots+\frac1{n!}$ for all integers $n \geq 1$.

Assume that such polynomials can be found. Then  we have that $degree (f)=degree (g)=k$.

Now $\frac{f(n+1)}{g(n+1)}-\frac{f(n)}{g(n)}=\frac{1}{(n+1)!}$.
Hence there is $c_{n}\in (n, n+1)$ such that

$\frac{f'g-fg'}{g^{2}}(c_{n})=\frac{1}{(n+1)!}$.

Now let $degree (f'g-fg')=a\leq k-1$ and let $b=k^{2}-a$. Then observe that  we have

${(c^{b}_{n})\frac{f'g-fg'}{g^{2}}(c_{n})\frac{(n+1)!}{c^{b}_{n}}}=1$.
Now observe that

$\lim (c_{n})^{b}\frac{f'g-fg'}{g^{2}}(c_{n})$ is finite and nonzero. The limit is zero provided that $f'g=fg'$ but this is excluded.

Hence we get that $\lim \frac{(n+1)!}{(c_{n})^{b}}$ is finite which is definitely false since $c_{n}\approx n$.

$lim\frac{f(n)}{g(n)}=e$ hence $f(x)=ex^{k}+\dots$ and $g(x)=x^{k}+\dots$. Moreover $e-\frac{f(n)}{g(n)}\leq\frac{1}{n!}$; it's easy to show that $e-\frac{f(n)}{g(n)}\sim\frac{a}{n^{r}}$, a contradiction.


Give  x, y, z >0 and xyz=1. Prove
Sorry ... wait ...

[hide="the sign is flipped"]
The correct form of it is:
$ x^{2}y+y^{2}z+z^{2}x\leq x^{2}y^{4}+y^{2}z^{4}+z^{2}x^{4}$ $ \leftrightarrow$ $ xyz(x^{2}y+y^{2}z+z^{2}x)\leq x^{2}y^{4}+y^{2}z^{4}+z^{2}x^{4}$
If I denote $ xy^{2}=a$, $ yz^{2}=b$ and $ zx^{2}=c$ the inequality is equivalent to: $ ab+bc+ca\leq a^{2}+b^{2}+c^{2}$
[/hide]


videos
counting
distinguishability
number theory

[quote]Just wondering... that's how its supposed to be, right?[/quote]

it was and it still is the same with me ... but i don't have any ilusion that i am able to solve every problem (if i havn't heard for anything simular before)... that's basicly why i like solving problems ... because i always learn something new...(and that's why i dislike homework because it doesn't teach you anything new)

:rotfl: Welcome to my universe.
If anyone ever tells you that they know how to solve every problem, they're either highly delusional, egotistical/kallikakian/narcissistic, or they've only been exposed to problems that are far too easy for them. Something of the sort.

This 'phase' of learning how to problem solve will never end. You will always be frustrated. You will always be triumphant. You will always be amused and amazed. You will always be bewildered and lost. You will always be found and rediscovered.
That's part of the great beauty of it. :) It never ends.

Just learn your stuff. Learn the concepts, derive the formulas, grasp the concepts, and smile while you do it. Be tortured and enjoy it.

...However, if you need a self-esteem boost, you could always do easy problems. :P

Haha... nice question.

I won't say that I'm an expert in problem solving (because I am absolutely not) but I see where you come from. When I try problems I spend long fruitless hours on them, racking my brain, and I HATE looking at the solutions before I'm finished. It just goes against everything I am and do to look at the answers, and I sympathize with you. But I keep coming back again and again, just because it's so fun to finally say, "Yes! I did it!" and have that feeling of accomplishment.

It's good that you're still smashing your head against problems like that because it shows your dedication and how much you're trying. I sometimes wish I had that kind of dedication and spirit... so many times I look back and see how much time I have wasted that could have been used to do math... aieeeee....

Sometimes I wish I could be nearly as good as all of you posters seem. Because I'm really really intimidated with the level of math that you are at, and I wish I could be at least near that good. One of my friends is Kevin Li (budi713), and looking at his level of accomplishments makes me look up to him in awe. I just wish I could figure out what makes him so good.

--

And then, what comes after problem solving? It's great to see all these "fun" problems with clever answers, but unfortunately real life is a different story. Equations don't always solve with integers in real life, and the world of applied mathematics is unfortunately a lot tougher.

So what do we do? We don't weep, we don't cry, we keep plugging away and try to learn the talents used later in life. We try, fail, try and try again, and get to do things with the tools we use. It's all great.

I'm just glad to see so many people that feel the way I do.

Many people (including me, and i'm not from USA) have learnt how to solve problems by solving competition problems...AoPS is another way ... i found it in december last year and i thought 'oh another  site that has high level books for IMO' but when i've found the experts form Intro books i found out that i basicly know almost everything form intro books...so i realy liked that this was 'all levels' site. and then i found this forum and I love it ... probably because there are so many people here that know so much so you can talk about interesting problems. and i think the most inportant thing is to try ... cids from my school will probably never know this things because whenever they see a more difficult problem they look at it for 5 min find it boring and just to difficult and move on...

I like to take a short break when I don't understand something. This could mean playing some video games, taking a nap, or going on AoPS. It usually relaxes the mind from it's state of failure.

Also, reviewing past material might help. I was looking at the Number Theory portion of AoPS vol 2 and I a  bit unsure of the material. I looked back in AoPS vol 1 and reviewed; it helped me immensely in understanding that chapter (as well as further on in the Diophantine Equations).

Moved to Other Problem Solving Topics.

[quote="t0rajir0u"]There are, in fact, books that discuss problem-solving technique.  A short list:

[url=http://www.amazon.com/How-Solve-Aspect-Mathematical-Method/dp/0691023565]Polya[/url] (Problem-solving in general:  probably what you're looking for)
[url=http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191]Engel[/url] (Olympiad training)
[url=http://www.amazon.com/Art-Craft-Problem-Solving/dp/0471789011/ref=pd_sim_b_title_5]Zeitz[/url] (Contest training)

One of the most useful things is to do lots of problems, but that's just so you can become familiar with common strategies.  What you're probably looking for is ways to figure those strategies out if you didn't know them already, which is much more difficult to teach.  I don't actually know if the creative aspect of problem-solving can be taught.

Part of your problem, specifically, might just be that you're letting yourself get frustrated too easily.  Don't just use one method and be disappointed when it fails; try as many plausible avenues as you can, and develop the ones that look promising.  Like the previous poster said, take short breaks, or even long breaks.  Your brain works on problems on a subconscious level; returning to a problem after a break might give you the bright idea you need.[/quote]

To follow up on that advice, Zeitz is the best book for Problem Solvers on the AoPS level (and it follows a similar regime to what t0rajir0u was talking about).

With all due respect, y'all are kinda math maniacs who would spend all day on math problems if your schedule allows it.

I probably shouldn't be speaking since I more or less live in the shadow of people like y'all, yet I must give credit to user aufha for bringing up this very interesting topic.

Thanks for the awesome book recommendations from the others as well, which were both practical and extremely helpful.

However, I would strongly advise against extremist comments such as the one posted by user zhh:

[quote]When I try problems I spend long fruitless hours on them, racking my brain, and I HATE looking at the solutions before I'm finished. It just goes against everything I am and do to look at the answers, and I sympathize with you. But I keep coming back again and again, just because it's so fun to finally say, "Yes! I did it!" and have that feeling of accomplishment.

It's good that you're still smashing your head against problems like that because it shows your dedication and how much you're trying. I sometimes wish I had that kind of dedication and spirit... so many times I look back and see how much time I have wasted that could have been used to do math... aieeeee....
[/quote]

A crazy 14-year-old who applauds you for smashing your head in frustration, such as user zhh, poses a threat not only to himself but others as well, whether you know him on a personal level or across your computer screen......lol

I personally get a lot of my solutions to math problems when I'm half-heartedly fingering through a Chopin or Richard Clayderman piece on my piano (or singing indisinguishable tunes to myself in the showers :p).

I would say instead of hurting yourself the way user zhh does, if you find that a math solution comes to you whenever Randy Moss catches a Tom Brady TD, watch all the Patriots games you can find!

[quote="aufha"]How do you learn to problem solve?

When I practice math contest problems, there is a lot of running into dead ends, hair pulling, head banging, punching my chair in frustration, walking around in circles telling myself "I don't know how to do this," getting angry when the problem required a trick that would have been very difficult for me to derive myself, ripping my paper up then reprinting it to try the problems again, and laying my head down on the desk in submission during hard times.

Just wondering... that's how its supposed to be, right? That is, is that the basic method in learning how to problem solve in general? It isn't linear, like... section 6, now we know how to factor integers, section 7, now we know how to do long division... right?[/quote]

i have the same problem

I [i]would[/i] spend close to all day on math every now and then if i had the time. Math is fun, however lately i've been kinda out of it- not sure why. Dedication is important to developing problem solving skills though there is a fine line between dedication and obsession.

well i am sorry for replying to this topic after such a long time.
but now only i came to know about this. This is an interesting question.Though i am not good in maths myself i would like to give some advice.Well the previous posters speak about problem solving and especially some of them speak about learning concepts from books.
i am not telling that is wrong but i would like  to do books which dont spoon feed and state some conjectures which are left as an exercise.This is because more than the theorem or concept the idea or the base on which the theorem was constructed is very  important.This might help you when to use them and apply them correctly.When a constructive problem is given ,the student should not just solve the problem ,but should also look for spaces where the  dynamics and boundary of the problem can be extended.By doing this the student himself might get some new concept.Ultimately given a problem it is for the student to explore the theory part of it by analyzing and working on it.It is not about doing x books and y problems ,even if  one problem is done the student should analyze its beauty and should get benefited from it and you will never forget the theorem. This is what a great maths thinker told me.I would like to implement it more vigorously and i feel a student should do books which provide the above quailities


conics
ellipse
analytic geometry
graphing lines
slope
function
geometry unsolved

[quote="shyong"]Let the equation of an ellipse be $\frac{x^{2}}{4}+\frac{y^{2}}{12}=1$ .
$A(1,3),B,C$ are three points located on the ellipse such that $ABC$ is isosceles triangle with $AB=AC$ . Find the slope of $BC$

p/s : I have posted this question in pre olympiad months ago but remain unsolved,so I repost again here .[/quote]

The conditions specified do not make $\triangle ABC$ unique.  This is easy to see because for a circle of radius $0 < r < 6.557570355\ldots$ centered at $(1,3)$, there are exactly 2 points of intersection with the given ellipse.  These points of intersection $B$ and $C$, along with $A$, form the vertices of an isosceles triangle with $AB = AC$ by construction.  However, it should be obvious that the slope of $BC$ is not independent of $r$, and hence there is no unique answer.

Following from the previous post, we can see that different values of r will result in a set of parallel chords BC.
Theorem.The midpoints of a system of parallel chords of gradient m of an ellipse ${\frac{x^{2}}_{a^{2}}}$+${\frac{y^{2}}_{b^{2}}}$ =1 lie on the line y=$\frac{-b^{2}x}_{a^{2}m}$(such a line is called a diameter).
In this particular problem the line joining A to the origin is a diameter for the system of chords BC.

Then prove that the locus of segment BC as a function of the congruent legs of triangle ABC are parallel lines.  You can't, because they're NOT, as I pointed out.  When r = AB = AC is small, BC approximates the tangent at A.  This is obviously not the case as r increases!

If you're not convinced, solve the system

$\frac{x^{2}}{4}+\frac{y^{2}}{12}= 1$
$(x-1)^{2}+(y-3)^{2}= r^{2}$.

Then compute the slope of the intersection points as a function of r, and you will immediately find that it is a variable function of r, and not at all a simple one at that.


function
algebra
polynomial
ceiling function

There exists a function so that f(0)=12, and f(x)=x for x=1-11 (integers).
Find the simplest function possible, even though there are polynomials, I'd like a function using the floor/ceiling) function (perhaps dividing by 12 and taking the ceiling?).

Interesting question... but this is probably not what you're looking for:
[hide="an answer"] $ \left\lceil\frac{x}{12}\right\rceil (x\minus{}12)\plus{}12$[/hide]

Also...
[hide]
$ 12+(x-12)\lceil\frac{x}{1+x}\lceil$
[/hide]

That would be better, as it doesn't crash after 12.

@archimedes: you might want to put the denominator as 1+x.

Although putting $ x\plus{}k$ in the denominator woud make the function not "crash" after $ 12$, putting $ x\plus{}k$ in the denominator would also make $ x\equal{}\minus{}k$ undefined....
How should we define "simplest function possible"?

You can also use fractional part function: $ f(x) \equal{} 12\left\{{x\minus{}1\over 12}\right\}\plus{}1$ (which is periodic)

Or you can use absolute values: $ f(x)\equal{}6|x\minus{}1|\minus{}5x\plus{}6$ (which extends as lines over both sides)

[hide="The simplest function possible"] $ \begin{equation*}f(x) =\begin{cases}12 &\text{if }x=0,\\ x &\text{if }x\in\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}.\\ \end{cases}\end{equation*}$ [/hide]
;)

Perhaps you should be more specific in what you mean by "function" even before you specify what you mean by "simplest."  The above is a function, after all.

[quote="t0rajir0u"][hide="The simplest function possible"] $ \begin{equation*}f(x) =\begin{cases}12 &\text{if }x = 0,\\ x &\text{if }x\in\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}.\\ \end{cases}\end{equation*}$ [/hide]
;)

Perhaps you should be more specific in what you mean by "function" even before you specify what you mean by "simplest."  The above is a function, after all.[/quote]

"Simplest" in the sense "which requires the least amount of time to think of it" ;)

I guess he meant "simplest formula", since he mentions polynomials, floor, ceiling etc.

[quote="t0rajir0u"][hide="The simplest function possible"] $ \begin{equation*}f(x) =\begin{cases}12 &\text{if }x = 0,\\ x &\text{if }x\in\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}.\\ \end{cases}\end{equation*}$ [/hide]
;)

Perhaps you should be more specific in what you mean by "function" even before you specify what you mean by "simplest."  The above is a function, after all.[/quote]
I'm trying to avoid a piecewise function.
Archimedes' works for what I need, thanks.

What is it that you need, anyway?  If you'd been more specific from the start, we wouldn't need to guess what function you would consider "simplest" :|