Find all functions  $ f\colon \mathbb{N}\to\mathbb{N}$     such that    $ f(f(n))\plus{}(f(n))^2 \equal{}n^2 \plus{}3n\plus{}3$

__________________________________
$ \textbf{Azerbaijan Land of Fire }$ :huh:  :ninja:

Simple... Let $ n \equal{} 1$ and $ f(1) \equal{} a$. Then we get:

$ f(a) \plus{} a^2 \equal{} 7$

This means that $ a \equal{} 2$ and $ f(2) \equal{} 3$. Now using math induction. Suppose $ f(k) \equal{} k \plus{} 1$ for some $ k$. Then put $ n \equal{} k$: $ f(k \plus{} 1) \plus{} (k \plus{} 1)^2 \equal{} k^2 \plus{} 3k \plus{} 3$. You get $ f(k \plus{} 1) \equal{} k \plus{} 2$. ;)

Nice. You get that a = 2 because since the function's codomain is the natural numbers one can only consider \[f(1) + 1 = 7\]  and \[f(2) + 4 = 7\]

Since  \[f(1) = a\] we would get \[ 2= 7\] which is clearly false. Thus \[a =2\].

Concider two cases $f(x)\leq x$ and $f(x)\geq x+2$ and get cntradiction;thus $f(x)=x+1$.

Sorry to revive , how can we get the contradiction in the case f(x)<= x .

[quote=i_am_not_the_one_who]If deg(f(n))>=2, deg(LHS)>=4, which shouldn't be the case. So, degree of f is 0 or 1. Since RHS is increasing, degree of f is 1. Let f(n)=an+b. Substituting back in the given equation gives us a=1 and b=1. Or, by keen observation, one can rewrite RHS as (n+1)^2 + {(n+1)+1}. So, f(n)=n+1.[/quote]

Wrong.
You can not use "degree" since the problem did not say that $f(x)$ is a polynomial.

Put x=0 yeilds
f(f(1))+(f(1))^2=7
So f(1)can be 1 or 2
Case1 f(1)=1
 Which yeilds f(1)=6 a contradiction
 So f(1) =2 and rest is done by yuriy's solution

Let $P(x)$ be the assertion.
Notice that $f(f(n)) + (f(n))^2=n^2 + 3n +3 = (n+1)^2 + (n+1)+1.$

Also, $P(n-1) : $ $f(f(n-1)) + (f(n-1))^2 = n^2 + n+ 1.$

Therefore, we can guess that $f(n)=n+1$ may be a possible solution.

$\textbf{Claim:}$ $f(n)=n+1$.
     Proved easily by induction.

We show that there are no more solutions.

First, assume that there exists some solution such that $f(n)<n+1$.
    $\Leftrightarrow$ $f(f(n))<f(n+1)<n+2.$ and $(f(n))^2<n^2 + n +1$
  $\implies$  $f(f(n)) + (f(n))^2 < n^2 + 3n +3.$, and thus we have got a contradiction.
Similarly, we can obtain a contradiction for $f(n)>n+1.$

Therefore, the only solution is $f(n)=n+1.$

Can we use one-to-one function to prove that f(x)<=x is wrong?

[quote=Fermat_Theorem]Let $P(n)$ be the assertion.
Notice that $f(f(n)) + (f(n))^2=n^2 + 3n +3 = (n+1)^2 + (n+1)+1.$

Also, $P(n-1) : $ $f(f(n-1)) + (f(n-1))^2 = n^2 + n+ 1.$

Therefore, we can guess that $f(n)=n+1$ may be a possible solution.

$\textbf{Claim:}$ $f(n)=n+1$.
     Proved easily by induction.

We show that there are no more solutions.

First, assume that there exists some solution such that $f(n)<n+1$.
    $\Leftrightarrow$ $f(f(n))<f(n+1)<n+2.$ and $(f(n))^2<n^2 + 2n +1$
  $\implies$  $f(f(n)) + (f(n))^2 < n^2 + 3n +3.$, and thus we have got a contradiction.
Similarly, we can obtain a contradiction for $f(n)>n+1.$

Therefore, the only solution is $f(n)=n+1.$[/quote]

I think there are some small mistakes here.
It would be $P(n)$ in the first line and $(f(n))^2<n^2 + 2n +1$, not $n^2 + n +1$ as otherwise we would not get the contradiction.

We can easily use induction on $n$

$n=1:f(f(1))+f(1)^2=7$
It's obvious that $f(1)=2$ and so $f(2)=3$
Similarly $f(f(2))+f(2)^2=13\rightarrow{f(3)=4}$

So we claim intuitively $f(k)=k+1$ ,for all $k\in{\mathbb{N}}$

$(1)$ Assume it holds for $n-1$ that $f(n-1)=n$

$(2)$ it's known that $f(f(n-1))+f(n-1)^2=(n-1)^2+3(n-1)+3=n^2+n+1$
$\Rightarrow{n^2+n+1=n^2+f(n)\rightarrow{\boxed{f(n)=n+1}}}$

I claim the answer is $f(n) = n+1$, which clearly works. I prove this by induction.

Base Case:
When $n=1$, we get
$$f(f(1)) + f(1)^2 = 3$$
Clearly, $f(1) \neq 1$ or else the LHS would be 2. Also, $f(1) < 3$ because if it weren't, then the RHS would be at least 9, due to the $f(1)^2$ term.

Inductive step:
Say our claim works for $n = k$. We want to prove that our claim works for $n = k+1$. From n=k,
$$f(f(k)) + f(k)^2 = k^2 + 3k + 3$$
$$f(k+1) + (k+1)^2 = k^2 + 3k + 3$$
$$f(k+1) = k + 2$$
so our induction is complete, and our claim is proven.

[hide = Solution]

The answer is $f(x) = x+1.$

First plug in $n=1.$ We obtain $f(f(1)) + f(1)^2 = 7.$ Note that $f(1) \neq 1$ as then $f(f(1)) + f(1)^2 = 2. $ Additionally, we have $f(1) < 3$ since otherwise we would have $f(3) = -2$ contradicting the codomain of the function. Hence, we know $f(1) \neq 1$ and $f(1) <3$ and $f(1) \in \mathbb{N}$ so we know $f(1) = 2.$ Now we proceed by induction to prove that $f(n) = n+1$ for all positive integers $n.$ The base case $n=1$ has already been established, so assume that $f(k) = k+1$ for some integer $k.$ Note that $f(f(k)) + f(k)^2 =k^2+3k + 2 \implies f(k+1) + (k+1)^2 = k^2+3k+2 \implies f(k+1) = k+2$ which completes the inductive step. Hence, we know $f(n) = n+1$ for all $n \in \mathbb{N}$ and we are done.

Assume that if
$$f(n)<n+1 \Leftrightarrow f(f(n)) + (f(n))^2<n+3+n^2+2n+1=n^2+3n+3$$
we have a contradiction.
similarly for $f(n)>n+1$
Thus: $\boxed{f(n)=n+1}$

None of the non-inductive solutions given above are correct. Here is my non-inductive solution to the problem.
$f(n)^2 < f(f(n)) + f(n)^2 = n^2 + 3n + 3 < (n + 2)^2 \Rightarrow f(n) < n + 2 \Rightarrow f(n) \leq n + 1 \Rightarrow f(f(n)) \leq f(n) + 1$.
$n^2 + 3n + 3 = f(f(n)) + f(n)^2 \leq f(n) + 1 + (n + 1)^2 \Rightarrow n + 1 \leq f(n) \Rightarrow f(n) = n + 1 \;\forall n \in \mathbb{N}$.

[quote=IMOStarter][quote=nicetry007]None of the non-inductive solutions given above are correct. Here is my non-inductive solution to the problem.
$ f(f(n)) \leq f(n) + 1$.
.[/quote]

Why do you have this?  :huh:[/quote]

It comes from the statement "$f(n) \leq n + 1 \Rightarrow f(f(n)) \leq f(n) + 1$", which I think is correct. This is because $f(n) \in \mathbb{N}$.

Therefore, substituting $n = f(n)$, the relation is derived.

e.g. $f(n) = 3$. then $f(3) \le 3 + 1$


What defines an extra curricular? Can you give me some examples?  What are some good ones that colleges like to see?

An extracurricular is more or less anything you don't do for a grade (that is, as part of your school's curriculum).  Examples include debate team/club, math team/club, a sport team, an instrument (outside of school), volunteering...

Don't focus on what colleges want.  Focus on what's interesting to you.  If you present yourself, your interests, and your passions accurately in your extracurriculars, the college that fits you best will be impressed anyway.

So any sports, clubs, and instruments are ECs?

Yep.

And don't forget jobs, internships, hobbies, research, or any other enriching experiences that show interest/motivation/commitment/skills.


In the formula for the area of a triangle, $ A\equal{} \sqrt{s (s\minus{}a) (s\minus{}b) (s\minus{}c)}$, $ a$, $ b$, and $ c$ are the lengths of the sides of the triangle and $ s\equal{}\frac{1}{2}$ of the length of the perimeter. Find the number of square units in the area of $ \triangle ABC$, expressed as a decimal to the nearest hundredth.
[asy]size(150);
draw((0,0)--(6,0)--(2,3)--(0,0));
dot((0,0));
dot((6,0));
dot((2,3));
label("16.9",(3,0),S);
label("15.8",(1,1.5),W);
label("23.4",(4,1.5),E);[/asy]


function
limit
topology
real analysis
real analysis solved

Let $f:[0,\infty)\to\mathbb R$ be a continuous function s.t. $\lim_{x\to\infty}\frac {f(x)}x=0$. Let $(x_n)_n$ be a sequence of positive real numbers s.t. $\left(\frac{x_n}n\right)_n$ is bounded. Prove that $\lim_{n\to\infty}\frac{f(x_n)}n=0$.

[i]Dorin Andrica, Eugen Paltanea[/i]

Please, add links for these topics to http://www.mathlinks.ro/Forum/viewtopic.php?t=30108
Thank you!

[quote="Kuba"]Let $f:[0,\infty)\to\mathbb R$ be a continuous function s.t. $\lim_{x\to\infty}\frac {f(x)}x=0$. Let $(x_n)_n$ be a sequence of positive real numbers s.t. $\left(\frac{x_n}n\right)_n$ is bounded. Prove that $\lim_{n\to\infty}\frac{f(x_n)}n=0$.[/quote]
Hmm... What am I missing?
Suppose it is untrue, then there is $\{n_k\}$ s.t. $\frac{f(x_{n_k})}{n_k}\geq \epsilon$. It means that $x_{n_k}$ is unbounded. We will assume wlog that $x_{n_k}\to \infty$. Then we have $\frac{f(x_{n_k})}{n_k}=\frac{f(x_{n_k})}{x_{n_k}}\cdot \frac{x_{n_k}}{n_k}\to 0$.  :?

Yeap, myth. I took on every interval $[0,Mn]$, with $M$ to be that bound, the sup of the function and constructed a sequence of the maximums on each interval $[0,Mn]$ for which the upper sup is the value of the function in that point. And for this sequence $\frac {f(a_n)}{n}$ is bigger than $\frac {f(x_n)}{n}$ and it tends to 0 by an immediate statement.
Well, for the fact that I didn't prove that a continuos function reaches its maximum on a set that contains it's sup on a closed set I got my score cut by 3.5  :(
Some absolute values should be taken there, but it's all the same.


integration
calculus
real analysis
real analysis unsolved

Let $f: [0,1]\rightarrow(0,+\infty)$ be a continuous function.

a) Show that for any integer $n\geq 1$, there is a unique division $0=a_{0}<a_{1}<\ldots<a_{n}=1$ such that $\int_{a_{k}}^{a_{k+1}}f(x)\, dx=\frac{1}{n}\int_{0}^{1}f(x)\, dx$ holds for all $k=0,1,\ldots,n-1$.

b) For each $n$, consider the $a_{i}$ above (that depend on $n$) and define $b_{n}=\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}$. Show that the sequence $(b_{n})$ is convergent and compute it's limit.

I think part a) is actually incorrect: the division is not unique. Here is a counterexample.
Let $f(x) = \cos(3\pi x)$, $n=2$. One division is $a_{0}=0$, $a_{1}= 1/3$, $a_{2}= 1$. The other is $a_{0}=0$, $a_{1}=2/3$, $a_{2}=1$.

Or a simpler case being f(x) = 0, which can be partitioned infinitely different ways to satisfy the equality.k

Are you sure a is correct?

If f(x)=Sin[2Pi*x] on [0,1] then for n>2 you can't partition [0,1] in any way so that the integral vanishes on every subinterval. It just can't be done.  :maybe:

Am I right?

ups...sorry  :blush: i have edited the post now.

[hide]
1. Let $\Phi(t) = \int_{0}^{t}f(x)dx$. Since $f(x)$ is positive, $\Phi(x)$ is increasing; since $f(t)$ is continuous, $\Phi'(x) = f(x)$. Note that $\Phi(x)$ is a bijection between $[0,1]$ and $[0, I]$, where $I = \int_{0}^{1}f(x) dx$. Therefore, numbers
\[a_{i}= \Phi^{-1}\left(\frac{i I}{n}\right)\] are defined uniquely, and $a_{i}< a_{i+1}$.

2. We have \[b_{n}= \frac{1}{n}\sum_{i=0}^{n}a_{i}= \frac{1}{n}\sum_{i=0}^{n}\Phi^{-1}\left(\frac{i I}{n}\right) \to \frac{1}{I}\int_{0}^{I}\Phi^{-1}(t) dt = \frac{1}{I}\int_{0}^{1}x d\Phi(x) = \frac{1}{I}\int_{0}^{1}x f(x) dx.\]
Answer: \[\frac{\int_{0}^{1}x f(x) dx}{\int_{0}^{1}f(x) dx}.\]
[/hide]


Countdown:

What is the remainder of 8^6+7^7+6^8 is divided by 5?

no calculator of course, paper isn't needed either, but sure.

You find the sum of the units digit of each term, which doesn't require multiplication(look for the pattern). Add the units digits and find the remainder(shouldn't require division) ;  if it ends in 3, the remainder is 3, and if it is 8 the remainder is (8-5) also 3.

[quote]3^6=243[/quote]
Check your calculation again.

my bad, its actually 729, which makes it 4+3+1=8=3 (mod 5)

However, if the number is like 6^13 or 8^24 the modular representation can be situated, but it takes longer to find the answer.

hmm, 6^13=1(mod 5) and 8^24=3^24 and that is easy patten since 3^4=1 we get 1

Yeah. My mistake. I knew that 6 = 1(mod5), and the whole pattern, so that was a bad example.


linear algebra
matrix
linear algebra unsolved

Let A be a (real) orthogonal matrix.

(a) let e, f in Rn are eigenvectors of A belonging to distinct eigenvalues, prove that e.f=0

(b) if c in Cn is a complex eigenvector with complex eigenvalue not in R. prove that z.z=0 (here x.y is the usual inner product).

Here $ e.f\equal{}\sum{x_{i}y_{i}}$ and $ e: f\equal{}\sum{\bar{x_{i}}y_{i}}$
For a):
if $ Ae\equal{}\lambda{e}$ with $ e\in\mathbb{C}^{n}\minus{}\{0\}$ and $ Af\equal{}\mu{f}$ with $ f\in\mathbb{C}^{n}\minus{}\{0\}$ and $ \lambda\not\equal{}\mu$ then
$ Ae: f\equal{}\bar{\lambda}({e}: f)\equal{}e: A^{\minus{}1}f\equal{}(1/\mu)(e: f)\equal{}\bar{\mu}(e: f)$ and $ e: f\equal{}0.$
For b):
if $ Ae\equal{}\lambda{e}$ with $ e\in\mathbb{C}^{n}\minus{}\{0\}$ and $ \lambda\notin\mathbb{R}$ then $ A\bar{e}\equal{}\bar{\lambda}\bar{e}$; a) implies that $ e:\bar{e}\equal{}0$ or $ e.e\equal{}0$


It says:
The ordered pair of four-digit numbers (2025; 3136) has the
property that each number in the pair is a perfect square and each digit
of the second number is 1 more than the corresponding digit of the
first number. Find, with proof, all ordered pairs of five-digit numbers
and ordered pairs of six-digit numbers with the same property: each
number in the pair is a perfect square and each digit of the second
number is 1 more than the corresponding digit of the first number.

Ok.....[i]question deleted by admin; see reply below[/i]

thanks!

We cannot clarify the wording of the questions while the round is in progress.  Please also note the one of the rules (#7 on [url=http://www.usamts.org/Rules/U_RulesComplete.php]this page[/url]) specific prohibits asking for help on an online forum.

If you are having trouble interpreting the question try:

1. If you think that both interpretations are valid, choose one and solve the problem based on this. If there are 2 interpretations, you won't be penelized

2. If you think the interpretation is reasonably clear, but there is an alternate interpretation that seems unlikely, solve both interpretations.

Solutions:

(in case 1)

I interpreted this problem as follows:

blah blah blabbity blah blah.

The solution is:

blah blah blabbity blah blah.

(in case 2)

I believe the problem should be be interpreted like "blah blah blabbity blah blah"

<solution>

However, if the porlbme should be interpreted like "blah blah blabbity blah blah", the solution is:

<solution>

Case 2 is the better choice assuming you have enough time even if you think both interpretations are vlaid, but w/e.

Hope this helped.

This is somewhat dangerous advice.  All USAMTS problems have only 1 correct interpretation.  It would be a substantial failure of our internal vetting process if we had a problem on the contest with more than one valid interpretation.  The problems are read by several experienced people--both USAMTS staff members and people outside the contest who volunteer to read the problems--before a round is made public to try to eliminate any ambiguities in wording.

Saying "you won't be penalized" for solving the problem with an incorrect "interpretation" is not accurate.  We also do not want to see solutions of the form "If the problem means (A), then my solution is (X), but if the problem means (B), then my solution is (Y)."  Part of the problem is figuring out what the problem means.

Read the problems carefully.  Think about them carefully.  There's no rush to solve them: the Round 1 deadline is not until mid-October.  We believe that the problem statements are as unambiguous as possible, and we are definitely not trying to trap you with tricky wording.


November 16,2004 Contest 1 Division E

A piece of paper is cut out as is shown.
It is folded along the inside lines.
The box is placed so that the top is open. What is the letter on the bottom of the box?
Excuse me for the bad drawing. :oops:

[hide] fold it to get a $W$ at bottom .  :D  [/hide]

w. i remember answering this question. this one was ez

i hate these types of folding questions -- this one was simple (w), but some are as nasty as %@#$. They are hard to post because they require diagrams.

I remember this 1 also. And how do u post those pics. (Not scanned images but like drawing on the computer)

You add it as an attachment which is in the part that says options under Attachment Posting Control Panel


calculus
integration
trigonometry
calculus computations

For $ a\in\mathbb{R}$ evaluate $ \int_{0}^{2\pi}\cos(a\sin(\theta))e^{a\cos(\theta)}d\theta$.

That smells the flavour of Bessel function.

This can be computed by complex integral, just consider $ \oint_C \frac {e^z}{z}\,dz$, where $ C = \{z: |z| = R \text{ and } R > 0\}$.

for details on this:
[hide="click it"]
Since $ f(z) = e^z/z$ is not analytic everywhere inside $ C$, it doesn't vanish, the integral can be evaluated by the substitution $ z = r\exp(i\theta)$. Moreover, by deformation theorem, we are free to set $ r\to 0$, that yields $ \oint_C \frac {e^z}{z}\,dz = 2\pi i$.

On the other hand, we let $ z = a\exp (i\theta)$,  since $ \frac {dz}{z} = id\theta$ (this is also a clue to how the integrand comes), we have
\begin{align*}\oint_C \frac {e^z}{z}\,dz & = i\int_0^{2\pi}\exp (a(cos\theta + i \sin \theta))\,d\theta \\
& = i\int_0^{2\pi}e^{a\cos \theta}\cos (a\sin \theta)\,d\theta - \int_0^{2\pi}e^{a\cos \theta}\sin (a\cos\theta)\,d\theta \end{align*}
this shows that desired answer is $ 2\pi$.
[/hide]

[quote="Sugaku Koshiki II (Mathematical Formulas II) (Iwanami Shoten, Tokyo, 1987) [in Japanese]"]
$ \sum_{n \equal{} 1}^\infty \frac {a^n}{n!}\sin nx \equal{} e^{a \cos x} \sin (a \sin x)$

$ \sum_{n \equal{} 1}^\infty \frac {a^n}{n!}\cos nx \equal{} e^{a \cos x} \cos (a \sin x)$
[/quote]

Though it was I had looked somewhere problem, I have forgotten a link.


Hi,

How do you calculate the summation of primes up to a certain number? E.g. Summation of the first 9297 primes.

Could someone enlighten me?

Thank you.

PS: Found this challenge on http://www.rankk.org

With a... program.  There's not really a better way.

Yeesh, t0rajir0u is right; the primes follow no easily summable pattern (for lack of a better word).

Rankk is a nice riddle sometimes, and there are some challenges .

But I'm curious: xlogin, did you have another intention when posting this¿  :D  :rotfl:
(Those who did it may know why  :wink: )


[quote="Leopoldo Taravilse"]Does anybody knows when will the APMO 2006 results will be available on the APMO's page or maybe here in Mathlinks?[/quote]

Leopoldo: On May the second the results will be sent to all countries that participate.....


AIME 1994
Hello, I looked at the AoPSWiki solution to problem 13 of the AIME 1994, but I was not able to understand how $ t\overline{t} \equal{} 170 \minus{} 2\cos\left(\frac{(2k \plus{} 1)\pi}{10}\right)$ after expanding, because I am getting $ t\overline{t} \equal{} 170 \minus{} 26\cos\left(\frac{(2k \plus{} 1)\pi}{10}\right)$. (Basically, when I expand, I am getting $ 13(e^{\theta i} \plus{} e^{\minus{}\theta i}) \equal{} 13(2\cos\theta)$ for the middle term where $ \theta \equal{} \frac{(2k \plus{} 1){\pi}}{10}$.)
Can someone tell me where I am going wrong?
Thanks!

It was probably a typo. Notice that, in the end, the cosines cancel out anyway, so it's very possible that the author made a typo, got the right answer anyway, and didn't realize s/he made a mistake.

Since it's a Wiki, why don't you go in and fix it yourself? :)

Oh, I see. Yeah, sure I'll do that. Thanks!


How do you find the answer to 1/ x^2 - x if x= 1 − i(sqrt 3)/ 2

Um...two questions:

1) $ x\equal{}\frac{1\minus{}i\sqrt3}2$ or $ x\equal{}1\minus{}\frac{i\sqrt3}2$?
2) $ \frac1{x^2}\minus{}x$ or $ \frac1{x^2\minus{}x}$?

The first choice in #1 and second choice in #2

OK.

$ x(x \minus{} 1) \equal{} \left(\frac {1 \minus{} i\sqrt3}2\right)\left(\frac { \minus{} 1 \minus{} i\sqrt3}2\right) \equal{} \minus{} \left(\frac {1^2 \minus{} (i\sqrt3)^2}4\right) \equal{} \minus{} \frac44 \equal{} \minus{} 1$, so

$ \frac1{x^2 \minus{} x} \equal{} \frac1{x(x \minus{} 1)} \equal{} \frac1{ \minus{} 1} \equal{} \boxed{ \minus{} 1}$.

Alternatively, note that $ x \equal{} \minus{} \left( \minus{} \frac12 \plus{} i\frac {\sqrt3}2\right) \equal{} \minus{} \left(\cos\left(\frac {2\pi}3\right) \plus{} i\sin\left(\frac {2\pi}3\right)\right)$, so $ \minus{} x$ is a third root of unity (not $ 1$). All third roots of unity $ r$ other than $ 1$ satisfy $ r^2 \plus{} r \plus{} 1 \equal{} 0$, and since one root is $ r \equal{} \minus{} x$, we find that $ ( \minus{} x)^2 \plus{} ( \minus{} x) \plus{} 1 \equal{} 0\implies x^2 \minus{} x \equal{} \boxed{ \minus{} 1}$.


What is the least positive integer that has each of the first eight positive integers as factors?

We want to find the greatest common factor of 1, 2, 3, 4, 5, 6, 7, 8. First, prime factorize each number.

$ 1 \equal{} 1$

$ 2 \equal{} 2$

$ 3 \equal{} 3$

$ 4 \equal{} 2^2$

$ 5 \equal{} 5$

$ 6 \equal{} 2 \cdot 3$

$ 7 \equal{} 7$

$ 8 \equal{} 2^3$

Thus, our answer is simply $ 2^3 \cdot 3 \cdot 5 \cdot 7 \equal{} \boxed{840}$.

What I did was find the least common multiple one step at a time. 1,2,3,4,5, and 6 all have a least common multiple of 60, which I thought of right off the top of my head(you'll know this too if you practice Mathcounts a lot). Now let's deal with 7 and 8. We'll do 8 first since it's easier. 8=2^3, and 60 only has 2 2's, so we multiply it by another 2 to get 120, the LCM of 1,2,3,4,5,6, and 8. 120 has no common factors with 7, so we multiply them to get the LCM. Our answer would be $\boxed{840}$.


geometry
3D geometry
sphere
analytic geometry
function
real analysis
real analysis unsolved

For each point $ x \ne 0$, we denote by $ x^\star$ the inversion of $ x$ with respect to the unit sphere which is defined by $ x^\star \equal{} \frac{x}{|x|^2}$. The Kelvin transform $ f^\star$ of $ f$ with respect to the unit sphere is
\[ f^\star(x^\star) \equal{}  \frac{1}{|x^\star|^{n\minus{}2}}f(x).
\]Therefore,
\[ {f^ \star }(x) \equal{} \frac{1}
{{|x{|^{n \minus{} 2}}}}f({x^ \star }) \equal{} \frac{1}
{{|x{|^{n \minus{} 2}}}}f\left( {\frac{x}
{{\left| x \right|^2}}} \right).\]
We have the rule
\[ \Delta {u^ \star }({x^ \star }) \equal{} \frac{1}
{{|{x^ \star }{|^{n \plus{} 2}}}}{\left( {\Delta u} \right)^ \star }\left( {{x^ \star }} \right).\]

My question is how to show that
\[ \Delta {u^ \star }\left( x \right) \equal{} \frac{1}
{{{{\left| x \right|}^{n \plus{} 2}}}}\Delta u\left( {{x^ \star }} \right).\]

Could you tell me in details, please? Thank you.

Denote $ x = (x_1,...,x_n)$ and $ \xi = {x^\sharp } = \left( {\frac {{{x_1}}} {{{{\left| x \right|}^2}}},...,\frac {{{x_n}}} {{{{\left| x \right|}^2}}}} \right)$. We have the following identities
\[ \frac {{\partial {\xi _j}}} {{\partial {x_k}}} = \frac {1} {{{{\left| x \right|}^2}}}\left( {{\delta _{jk}} - 2\frac {{{x_j}{x_k}}} {{{{\left| x \right|}^2}}}} \right)
\]
and
\[ \sum\limits_{l = 1}^n {\frac {{\partial {\xi _l}}} {{\partial {x_j}}}\frac {{\partial {\xi _l}}} {{\partial {x_k}}}} = \frac {1} {{{{\left| x \right|}^4}}}{\delta _{jk}}.
\]
Thus,
\[ \sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}.
\]
Next, think of $ \xi$ as a system of orthogonal curvilinear coordinates for $ x$, we deduce that the metric tensor of the Euclidean space in curvilinear coordinates
\[ {g_{j,k}}\left( {{\xi _j},{\xi _k}} \right) = \sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}.
\]
This implies that the so-called Lam\'{e} coefficients is
\[ {h_j} = \sqrt {{g_{j,j}}\left( {{\xi _j},{\xi _j}} \right)} = \frac {1} {{{{\left| \xi \right|}^2}}}.
\]
Now Laplacian of a function can be computed as following
\[ \Delta u = \frac {1} {{\prod\limits_{k = 1}^n {{h_k}} }}\sum\limits_{j = 1}^n {\frac {\partial } {{\partial {x_j}}}\left( {\frac {{\prod\limits_{k = 1}^n {{h_k}} }} {{h_j^2}}\frac {{\partial u}} {{\partial {\xi _j}}}} \right)}
\]
which implies that
\[ \Delta u\left( x \right) = {\left| \xi \right|^{2n}}\sum\limits_{j = 1}^n {\frac {\partial } {{\partial {\xi _j}}}} \left( {\frac {1} {{{{\left| \xi \right|}^{2n - 4}}}}\frac {{\partial u}} {{\partial {\xi _j}}}} \right).
\]
Next using the rule
\[ \Delta \left( {uv} \right) = v\Delta u + 2\nabla u \cdot \nabla v + u\Delta v
\]
we get
\begin{eqnarray*}
  \frac{1}{{{{\left| \xi  \right|}^{2 - n}}}}\frac{\partial }{{\partial {\xi _j}}}\left( {\frac{1}{{{{\left| \xi  \right|}^{2n - 4}}}}\frac{{\partial u}}{{\partial {\xi _j}}}} \right) &=& 2\frac{{\partial u}}{{\partial {\xi _j}}}\frac{\partial }{{\partial {\xi _j}}}\left( {\frac{1}{{{{\left| \xi  \right|}^{n - 2}}}}} \right) + \frac{1}
{{{{\left| \xi  \right|}^{n - 2}}}}\frac{{{\partial ^2}u}}{{\partial \xi _j^2}}\\
   &=& \frac{{{\partial ^2}}}{{\partial \xi _j^2}}\left( {\frac{1}{{{{\left| \xi  \right|}^{n - 2}}}}u} \right) - u\frac{{{\partial ^2}}}{{\partial \xi _j^2}}\left( {\frac{1}{{{{\left| \xi  \right|}^{n - 2}}}}} \right)
\end{eqnarray*}
and finally conclude with the help of
\[ \Delta \left( {\frac {1} {{{{\left| \xi \right|}^{n - 2}}}}} \right) = \sum\limits_{j = 1}^n {\frac {{{\partial ^2}}} {{\partial \xi _j^2}}\left( {\frac {1} {{{{\left| \xi \right|}^{n - 2}}}}} \right)} = 0
\]
that
\begin{eqnarray*}
\Delta u\left( x \right) &=& {\left| \xi  \right|^{n + 2}}\sum\limits_{j = 1}^n {\frac{{{\partial ^2}}}
{{\partial \xi _j^2}}} \left( {\frac{1}
{{{{\left| \xi  \right|}^{n - 2}}}}u} \right) = \frac{1}
{{{{\left| x \right|}^{n + 2}}}}\sum\limits_{j = 1}^n {\frac{{{\partial ^2}{u^\sharp }}}
{{\partial \xi _j^2}}}  \\
&=& \frac{1}
{{{{\left| x \right|}^{n + 2}}}}\Delta {u^\sharp }\left( \xi  \right) = \frac{1}
{{{{\left| x \right|}^{n + 2}}}}\Delta {u^\sharp }\left( {{x^\sharp }} \right).
\end{eqnarray*}
Therefore,
\[ \Delta u\left( {{x^\sharp }} \right) = \frac {1} {{{{\left| {{x^\sharp }} \right|}^{n + 2}}}}\Delta {u^\sharp }\left( x \right).
\]
Finally,
\[ \Delta \left( {\frac {1} {{{{\left| x \right|}^{n - 2}}}}u\left( {\frac {x} {{{{\left| x \right|}^2}}}} \right)} \right){\text{ }} = \Delta {u^\sharp }(x) = {| {{x^\sharp }} |^{n + 2}}\Delta u\left( {{x^\sharp }} \right) = \frac {1} {{{{\left| x \right|}^{n + 2}}}}\Delta u\left( {{x^\sharp }} \right) = \frac {1} {{{{\left| x \right|}^{n + 2}}}}\Delta u\left( {\frac {x} {{{{\left| x \right|}^2}}}} \right)
\]
which completes the present proof.


What did you get on your SAT?

I got:
700 in math
620 in writing (9 out of 12 on my essay)
530 in critical reading

What did you get on your SAT?

Please include what I included. (if possible)

which one? If you mean January, there is already a thread on it on the Colleges Forum


induction
number theory proposed
number theory

[quote="N.T.TUAN"]I think, we can find $n_{0}$ such that $6^{n}>3^{n}(7n-12)+2^{n}(2n-14)+24n\;\forall n\geq n_{0}$.  :wink:[/quote]

and for $n\leq n_{0}$?

Yes, $n_{0}$ is not large. It is easy to see by induction that

$\frac{6^{n}}{2}>3^{n}(7n-12)$ for $n\ge 2$

$\frac{6^{n}}{3}>2^{n}(2n-14)$ for $n\ge 7$

$\frac{6^{n}}{6}>24n$ for $n\ge 4$

So $6^{n}> 3^{n}(7n-12)+2^{n}(2n-14)+24n$ for $n\ge 7$

We do not need to check all $n=0,1,2, ... , 6$ because considering  modulo 2 and 3 gives $n\equiv 4 (\mod 6)$. It remains to check n=4 only which is successful.


There are four positive numbers $ a,b,c,d$ such that $ cd\equal{}1$. Prove that $ ab \leq n^2 \leq (a\plus{}c)(b\plus{}d)$ where $ n$ is an integer.

[hide="this is similar to what i wrote during the competition :D "]Assume the contrary that there is no such $ n$. Hence $ ab$ and $ (a\plus{}c)(b\plus{}d)$ lie between two consecutive perfect squares. So we must have some $ k$ such that $ k^2<ab<(a\plus{}c)(b\plus{}d)<(k\plus{}1)^2$. So $ (a\plus{}c)(b\plus{}d)\minus{}ab<(k\plus{}1)^2\minus{}k^2$ or $ ad\plus{}bc\plus{}cd\le2k\plus{}1$ or $ ad\plus{}bc\le2k$. So $ 2k\ge ad\plus{}bc\ge2\sqrt{abcd}$ or $ k^2\ge abcd$. But $ abcd\equal{}ab>k^2$. Contradiction. Q.E.D.[/hide]


function
induction
algebra
functional equation
algebra proposed

Let $n\geq 3$ be an integer. Find the number of functions $f:\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$ such that
\[ f(f(k)) = f^3(k) - 6f^2(k) + 12f(k) - 6 , \ \textrm{ for all } k \geq 1 . \]

First rewrite the functional equation as $f(f(k)) - 2 = (f(k)-2)^3.$ (1)

Let $b$ such that $f(a)=b.$
Then, from (1), we have : $f(b)-2 = (b-2)^3.$
A straightforward induction leads to $f^{(k)} (b) - 2 = (b-2)^{3^k}$, where $f^{(k)}$ denotes the $k^{th}$ iterate of $f$.
If $b \geq 4$ we deduce that $n-2 \geq f^{(k)}(b) - 2 \geq 2^k$ which is absurd for $k$ sufficientely large.
Thus $b=1,2,3.$ In each case we have $b-2 = (b-2)^3$ so that $f(b)=b.$
It is then easy to verify that if $f(k)=b$ where $b \in \{1,2,3 \}$ and $f(b)=b$ then the initial relation is satisfied.

Thus, thereare three cases :
- The function $f$ takes only one value : There are three choices namely $1,2,3$
- The function $f$ takes exactly two distinct values (three choices). Say $1$ and $2$. Then $f(1)=1$ and $f(2)=2$. For each $k = 3, \cdots, n$ the number $f(k)$ is $1$ or $2$.
Thus there are $3 \cdot 2^{n-2}$ functions in this case.
- The function $f$ takes exactly three distinct values, which are $1,2,3.$ Then $f(1)=1,f(2)=2,f(3)=3.$ And there are three choices for each other $f(k).$
Thus there are $3^{n-3}$ functions in that case.

Then the desired number is $3+3\cdot 2^{n-2} + 3^{n-3}.$

Pierre.

-must. learn. to. read. question. not. $\mathbb{N} \rightarrow \mathbb{N}$. bleh.-


At the moment, when I look at the Round Table forum, the top thread is "Personality Types" with 199 replies and the most recent post is Today, at 4:53 pm by cf249.  When I click on the "View latest post" button, it takes me to the top of the last page (usually it sends me to the bottom), with url http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1305348#1305348 , and the last posts are number 199 (yesterday at 3:25 pm, by Quickster94) and 100 (yesterday at 5:04 pm, by leoxnlin.  Reloading doesn't change anything.  I'm using Firefox 1.0.7 on Windows XP.

Screenshot?

As a side note, why are you using Firefox 1.0.7 which is an antique version of the browser?

Sorry, it's now fixed so I can't produce a screenshot -- basically, the last post just wasn't appearing for me (but now it is).

This computer is an antique and will be replaced in the near future, so not much value-added in upgrading the browser :)

Edit: Wait, I take it back, it's not really fixed, it's just that only the one post by cf249 is broken.  If I put in the URL of that single post (http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1305348#1305348), it sends me to exactly the same place as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88374&start=180 , i.e., the top of the 10th page of responses.  Otherwise, everything looks normal.  Also, there is currently a link to the 11th page of responses, but it wasn't there when I made the original post.


limit
function
calculus
calculus computations

Suppose $ y'(x)$ satisfies the differential equation $ y' \equal{} y^2 \minus{} x^2$, and $ y(0) \equal{} 0$. Prove that $ \lim_{x \rightarrow \infty} y'(x)$ exists, and determine its value.

hello, i think we have $\lim_{x \to \infty}y'(x)=0$.
Sonnhard.

Actually, Basic numerical methods show that as $x \to \infty$, $y(x) \to -\sqrt{x^2-1}$, and $y'(x) \to -1$. Proving this is much harder. This differential equation is known as a [url=http://en.wikipedia.org/wiki/Riccati_equation]Riccati Differential Equation[/url].

hello, yes, you have right, the solution of this equation contains the Bessel-function and we get
$\lim_{x \to \infty}y'(x)=-1$.
Sonnhard.


calculus
superior algebra
superior algebra solved

Hi.  This is my first post here   :D

This is part of a larger word problem.  I've worked it this far, but don't know how to isolate d.

[tex] x = \frac{M}{d^2} - \frac{M}{(d+2)^2}[/tex]

d is the variable for distance, 2 means 2 meters, d >> 2  (millions of times greater)

x = 147328230865.604
M = 1.98911e30

and d = 3781979, but I got this answer using another non-algebric method.

But how can I do this algebraically?  Re-write it so it says "d= blah blah blah..."

Not easily

Re-write as a common fraction but then you end up with a quartic for D

I figure it out  :) .  Fully expanding it gave me d^4 and a bunch of other terms that were insignificant compared to d^4, so I just got rid of them.

[quote="jmerry"]Which terms did you get rid of? Since we know the value of $d$ will be large, we have $\frac xM=\frac{4d+O(1)}{d^4+O(d^3)}=\frac4{d^3}+O(d^{-4})$ and $d\approx\sqrt[3]{\frac{4M}x}$

This is essentially a calculus idea.[/quote]

Thanks, I got the same thing you did, except I replaced x with
[tex] \frac{Mass_{Earth}}{Radius_{Earth}^2}[/tex]
and "2+" with "h+"

So my final formula is

[tex]
\sqrt[3]{\frac{2*hMr_{Earth}^2}{M_{Earth}}}[/tex]

This approximated answer and the real answer agree within 1/2 kilometer.  Not bad considering the real answer is about 3700 km.

The original question is:

[i]You(a 2 meter-tall person) decide that to keep a safe distance you might want to stay far enough away from the black hole (1 solar mass, or 1.9891e30kg) that the tidal force would be comparable to the force of gravity on the surface of the Earth. What would this distance be?[/i]


quadratics
algebra
polynomial
sum of roots

[quote="warut_suk"]One root of $mx^2-10x+3$ is two-thirds of the other root. What is the sum of the roots?[/quote]

[hide]
Let $\alpha$ and $\frac{2}{3}\alpha$ be the roots . Then

$\frac{5}{3}\alpha$ = $\frac{10}{m}$ ; $\frac{2}{3}\alpha^2$ = $\frac{3}{m}$

Solving this gives us $\alpha$ = $\frac 34$ since $\alpha>0$ so sum of roots = $\frac 54$[/hide]


Let $ABC$ be a triangle. Prove that :
$sin^6(A/2)+sin^6(B/2)+sin^6(C/2)\geq\frac{3}{64}$.


Prove that $ (\forall)x>0$ we have :

$ \frac{x^6\plus{}5}{x^5}\ge \frac{6}{\sqrt[3]{25}}$


Roll 3 dice, and multiply the three numbers showing. What is the probability that the product is even?

[hide]In order for the product to be even, you have to have at least one even # in the three rolls. Therefore, do some complementary counting (in other words, figure out the probablity the product is odd, and then subtract it from 1):

Probability it's odd (all 3 rolls are odd): (1/2)*(1/2)*(1/2)=1/8
The answer is: 1-(1/8)=7/8[/hide]

Reminder: Hide your answers!

[hide]Correct![/hide]

[hide]Well, the only way that it's not even is when it's all odd. The chances of that is 33/63, which is 1/8, therefore the probability of it being even is 1-1/8=[b]7/8[/b].[/hide]


In a tournament each player played exactly one game against each of the other players.  In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned 1/2 point if the game was a tie.  After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points.  (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten).  What was the total number of players in the tournament?

for your very first post you sure know a lot of LaTeX...

Sorry to revive an old topic, but I'm doing the 1985 test and checking my answers right now, and how can there be 2 separate answers?

There aren't.  I don't quite have the patience to re-digest this problem, but there was another topic that wrangled with this problem for awhile and I'm not sure their final answer coincided with rjt's.

http://www.kalva.demon.co.uk/aime/soln/answ85.html
25 it is

[quote="rjt"][hide]
Let $ x$ equal the number of players. The total number of points won by all players in the tournament is $ \binom{x}{2}$, that is, $ \frac {x(x \minus{} 1)}{2}$. The number of points won againt the lowest ten players is half this, or $ \frac {x(x \minus{} 1)}{4}$. The number of matches played by the lowest ten is $ (x \minus{} 1) \plus{} (x \minus{} 2) \plus{} ... \plus{} (x \minus{} 10) \equal{} 10x \minus{} 55$. Letting $ x \equal{} 10$, we find that the lowest 10 players score $ 45$ points against themselves. We know that they score the same amount of points against themselves as they do against the other players, so the 10 lowest score $ 45$ points against the other players. The number of matches played by all of the players, except the 10 lowest, against the 10 lowest is $ 10(x \minus{} 10)$.

But the number of points won against the 10 lowest = (number of points the lowest 10 won against themselves) + (number of points the others won against the 10 lowest). Substituting in the values we found above:
$ \frac {x(x \minus{} 1)}{4} \equal{} 45 \plus{} 10(x \minus{} 10) \minus{} 45$
$ \Rightarrow x^2 \minus{} 41x \plus{} 400 \equal{} 0$
$ \Rightarrow x \equal{} 16,25$
[/hide][/quote]

$ \frac {x(x \minus{} 1)}{4} \equal{} 45 \plus{} 10(x \minus{} 10) \minus{} 45$
Why did you subtract 45 from above equation?

Could someone please explain why one answer is right?  Thanks.

I can't explain it, but the correct answer is $ 25$
You can find all the answers to old AIMEs and most AMCs at AoPS Wiki
Here's the link for this particular one: http://www.artofproblemsolving.com/Wiki/index.php/1985_AIME_Answer_Key

Basically only 25 works because if there were 16 players, then at least one of the "top 6" would've scored below at least one of the "bottom 10."

[quote="rjt"][hide]
Let $x$ equal the number of players. The total number of points won by all players in the tournament is $\binom{x}{2}$, that is, $\frac{x(x-1)}{2}$. The number of points won againt the lowest ten players is half this, or $\frac{x(x-1)}{4}$. The number of matches played by the lowest ten is $(x-1)+(x-2)+...+(x-10) = 10x-55$. Letting $x=10$, we find that the lowest 10 players score $45$ points against themselves. We know that they score the same amount of points against themselves as they do against the other players, so the 10 lowest score $45$ points against the other players. The number of matches played by all of the players, except the 10 lowest, against the 10 lowest is $10(x-10)$.

But the number of points won against the 10 lowest = (number of points the lowest 10 won against themselves) + (number of points the others won against the 10 lowest). Substituting in the values we found above:
$\frac{x(x-1)}{4} = 45+10(x-10)-45$
$\Rightarrow x^2-41x+400 = 0$
$\Rightarrow x=16,25$
[/hide][/quote]

I'm kind of confused.. I think I'm missing something.  Why isn't it just $\frac{x(x-1)}{4} = 10x - 55$?  Because after each match one point gets awarded to someone regardless of who wins.  So why isn't 10x - 55 equal to the number of points won against the 10 lowest scorers?

$45$ needs to be subtracted because the $10(x-10)$ points didn't all go to the people not in the lowest ten. Since we know $45$ points went to the lowest ten, $10(x-10) - 45$ points went to the people not in the lowest ten.

If we didn't subtract, we would be including an extra "number of points won by lowest ten against those not in the lowest ten" term on the right.

[hide="Alternate Method"]

Let $x$ be the number of players not in the bottom $10$. Note that of the $10x$ games played between the top $x$ players and the $10$ bottom players, $\binom{10}{2}=45$ must have been won by thte $10$ bottom players.

Hence, against the $10$ bottom players, the top $x$ players won $10x-45$ points. This is half of their total points, so it must be equal to the total number of points they won amongst themselves, which is simply $\binom{x}{2}$.

Hence $x=6, 15$, so our total, $x+10=25$. (As before, note that $16$ does not work)

[/hide]

We can see that 25 is the correct answer. For 16 players, everyone scored half their points on the 10 people, who by hypothesis are the bottom 10, and scored the other half on the six highest. This means that more points were scored on the top six than the bottom 10, which means the average points scored in top six is strictly less than that scored in top 10.

[quote=Skupp3][quote=matt276eagles]By the way, if you type this in $\LaTeX$,

\binom{n}{2}

...it looks like this:

$\binom{n}{2}$. ;)

Also, welcome to AoPS!  :D[/quote]

Lmao. He said welcome to a site admin[/quote]

He was referring to rjt............................

Let there be $x+10$ members in the tournament.
The bottom $10$ scored $\binom{10}{2}=45$ points against each other.
When the top $x$ competed with each other, they scored $\binom{x}{2}$ points.
When the bottom $10$ competed with the top $x$, they scored $45$ points because half of the points the bottom $10$ were scored against the top $x$.
When the top $x$ competed with the bottom $10$, they must have scored $\frac{\binom{x+10}{2}}{2}-45$.
In all, since $\binom{x+10}{2}$ points were scored, $\frac{\binom{x+10}{2}}{2}-45+90+\binom{x}{2}=\binom{x+10}{2} \implies$ $\frac{\binom{x+10}{2}}{2} = 45 + \binom{x}{2} \implies$ $\frac{(x+9)(x+10)}{4}=45+\frac{x(x-1)}{2} \implies$ $x^2+19x+90=180+2x^2-2x \implies$ $x^2-21x+90=0 \implies$ $(x-15)(x-6)=0$. Hence, $x$ is either $15$ or $6$. If $x$ is $6$, then the top $x$ must have scored $\binom{6}{2} + \frac{\binom{16}{2}}{2}-45 = 30$ (average of $5$ points) and the bottom $10$ would have scored $90$ points (average of $9$ points), which is a contradiction because then, the top $x$ would perform worse than the bottom $10$. Hence, our answer is $15+10 = \boxed{25}$.

[b]Solution.[/b]
There are $\binom{n}{2}$ points in total. Note that the condition tells us that half of the points of the total, $\frac{1}{2}\cdot \binom{n}{2}$ is from the games between the $10$ and the $n-10$ players, so
$$\frac{1}{2}\cdot \binom{n}{2} = (n-10)10$$
$$n = 16 \text{ or } n = 25$$
If $n$ is $16$, it's easy to calculate that the average of the $6$ biggest players is $\frac{30}{6}=5$, while the average of the $10$ smallest is $\frac{120-30}{10} = 9$, a contradiction. So $n=\boxed{25}$.

This problem has messed with my brain for so long  :wallbash_red:
The key is to [hide]use a global approach and consider the sum of points in certain cases.[/hide]
[hide=Solution]
Let there be $n$ players. Since $\binom{n}{2}$ games are played, there are $\binom{n}{2}$ points total.
Within the $10$ worst players, $\binom{10}{2}$ games are played, so the sum of their points is $2\binom{10}{2}$.
Within the other $n-10$ players, each player scored half of their points with the other $n-11$. The sum of the points in this group of players is thus $2\binom{n-10}{2}$.
Therefore, we have
$$2\binom{10}{2}+2\binom{n-10}{2}=\binom{n}{2}$$
$$180+2(n-10)(n-11)=n(n-1)$$
$$2n^2-42n+400=n^2-n$$
$$n^2-41n+400=0$$
$$n=16,25$$
However, $16$ is not possible because it would cause the mean score of the $10$ worst to be $9$, and the mean of the $6$ best to be $5$. Hence, $\boxed{25}$ is the answer.

We say that a team is a loser if it is in the bottom 10 teams, and we say that a team is a winner otherwise. Let $n$ be the number of winners.\\

We break this up into three seperate tournaments, which we denote as $T_1,T_2,T_3.$ In $T_1$, each possible pairing of two losers is played. In $T_2$, each possible pairing of a loser and a winner is played, and in $T_3$, each possible pairing of two winners is played. The total number of games (and thus total points scored) is 45 in $T_1$, $10n$ in $T_2$, and $n\choose 2$ in $T_3.$\\

Note that by summing the condition over all loser teams, we see that the number of points scored by losers against other losers (which is 45 from $T_1$) is the same as the number of points scored by losers against winners, so the losers score a total of 45 points in $T_2$. Similarly, the winners score a total of $n\choose 2$ points in $T_2$ against losers since they also scored that many points against other winners. This gives us the identity $$45+{n\choose 2}=10n.$$ Expanding out the quadratic, the solutions are $n=6$ and $n=15$. However, $n=6$ is not possible since that would imply that each loser score more points on average than each winner on average, which is not possible since the losers are the bottom 10, so $n=15$ and our answer is $\textbf{25}.$

Let there be $n+10$ players in the tournament. Split these players into two groups, one group of the bottom $10$ players, called team $B$, and another group of the remaining top $n$ players, called team $A$. Notice that we always have a constant $\binom{n+10}{2}$ points circulating in the system. Between $A$ group, there are $\binom{n}{2}$ points, and between $B$ group, there are $\binom{10}{2}$ points. Since each group earned half of its points from people in their group and the other half from the other group, we also have that $2\left[\binom{n}{2}+\binom{10}{2}\right]=\binom{n+10}{2}$.

Expanding the quadratic equation and solving, we have that either $n=15$ or $n=6$, but $n=6$ yields a contradiction to the fact that the $6$ players must be the top $6$ players[color=#9a00ff]*[/color], meaning that we must have $25$ players in this tournament, and we are done.

[color=#9a00ff]*If we have $16$ total players, we have that the group of top $k$ players score a total of $30$ points while the bottom $10$ players score a total of $90$ points. However, by average properties, we have that out of the top $6$ players, the maximum of the minimum of points scored out of the top $6$ players is $\frac{30}{6}=5$, meaning that the $10$ other players could have scored a maximum of $10*5=50$ points, a contradiction. Therefore $k=6$ is extraneous, giving us that $k=15$, for a total of $25$ players in the tournament. [/color]

This took me 45 minutes, to find the elegant way. We can let there be $x+10$, players in the tournament. We then make $2$ teams, one with the bottom $10$ which is team $A$, and the other team with the remaining players, called team $B$, there is always $\binom{x+10}{2}$, points. We have $2\binom{x}{2}+2\binom{10}{2}=\binom{x+10}{2}$, solving we get $x=6$, or $x=15$, if $x=6$, leads to a contradiction. So $x+10=\boxed{25}.$

We can see that the ten lowest players score $\binom{10}{2} = 45$ points among themselves. This means that they score $45$ points against all of the other players. Let there be $x$ players in the tournament discluding the $10$ lowest scoring players. This means that those players score $10x-45$ point against the $10$ lowest scoring players. Also they score $\binom{x}{2}$ points among themselves. We can easily see that $$\binom{x}{2} = 10x - 45$$

We can easily extract with quadratic formula that $x = 15 \Rightarrow \boxed{025}$. $\blacksquare$

[hide=Solution]Say that there are $x$ teams not in the bottom $10$. The total number of points scored in games with the bottom $10$ is $45+10n$
We obtain from the problem statement $45 + \binom{n}{2}=10n$
So $n=15$. So there are $25$ players.

Let there be $k$ other people, not in the bottom ten. In the bottom ten, there are $45$ points in total, so the bottom ten must have scored $45$ points against the $k$ other people. This means the $k$ other people scored $10k-45$ which is equal to the amount of points they scored among themselves, or $\binom{k}{2}$. Thus,

\[\binom{k}{2}=10k-45\]
\[\implies k=6,15\]

Then, note that $k=6$ does not work, so our answer is $10+15=\boxed{025}$.

Call the ten lowest scoring players "bad players" and call everyone else "good players".

Note that the bad players collectively got half of their points from the good players, and the good players collectively got half of their points from the bad players. If the bad players collectively scored $x$ points and the good players collectively scored $y$ points and there are $n$ good players, we have
\[\frac x2+\frac y2=10n\]
\[x+y=\binom{10+n}{2}\]
therefore
\[40n=(n+10)(n+9)=n^2+19n+90\Longrightarrow n^2-21n+90=0.\]
So $n=6$ or $n=15$. From here it is intuitive that $n=6$ doesn't make sense(since then the players all did better against the good players than the bad players), so the answer is $10+15=\boxed{25}$.


algebra
polynomial
sum of roots
algebra proposed

Let $n$  be a positive integer and $(a)_0 : =1 , \; (a)_k=a(a+1)...(a+k-1)$  for $k=1,2,...$ .
[b]Question:[/b] [color=green] [b] Find the sum of roots of polynomial [/b] [/color]
\[ P_n(x)=\sum\limits_{k=0}^{n} (-1)^{n-k}{n\choose k}x^{n-k}(x)_k \; . \]

Indeed, $P_n(x) $ is a linear combination of polynomials of degree n. This means that [b]degree[/b]$(P_n(x) \le n .$ In case of this problem, the degree seems to be $= \left[\frac{n}{2}\right] $ . Perhaps here is a little difficult question, to prove this assertion. Moreover , regarding your tentative to solve the problem (see your posted  message) let us remind that
\[ \sum\limits_{k=0}^{n}(-1)^{n-k}{n \choose k}k^p = 0   \; , \; \;  n\ge 1  \; ,\;  p=0,1,...,n-1. \]


A triangle has sides measuring $5$, $6$, and $7$ cm. What is the length, in centimeters, of the median to the $6$ cm side?

[quote="paladin8"][quote="rnwang2"]A direct application of Stewart's Theorem yields

 [hide="Answer"]
$2\sqrt{7}$ [/hide][/quote]

Pulling out the big guns, eh? Didn't think of that during the competition. There is a slightly more tedius elementary solution, though.[/quote]

you can be totally cool and rotate 180 about the midpoint, and you get a parallelogram

at this point, you can be like OMG i read aops and use the fact that the sum of the squares of the sides is equal to the sum of the squares of the diagonals, or a parallelogram,

[quote="Altheman"]you can be totally cool and rotate 180 about the midpoint, and you get a parallelogram

at this point, you can be like [b]OMG i read aops[/b] and use the fact that the sum of the squares of the sides is equal to the sum of the squares of the diagonals, or a parallelogram,[/quote]
I remember the eyeball that was next to that exercise.  :rotfl:

Altheman's method is probably the best way to solve this problem. Of course, instead of being totally cool and rotating 180 degrees about the midpoint, you could just extend the median its own length to get a parallelogram.

[quote="mathgeniuse^ln(x)"][quote="bpms"][hide="elementriest solution"]Okay, the altitude to the 6 side has length $2\sqrt{6}$. The distance from where the altitude hits to the nearest vertex is 1. The rest of the segment has length 5. Subtract half of the segment cut by the median, to get a right triangle with legs 2 and $2\sqrt{6}$ with the median being the hypotenuse. So, the length of the median is $2\sqrt{7}$[/hide]
Is the solution you were referring to? Or is there a simpler one (this one only requires pythagorean)?[/quote]

Sure...but I believe it would take some work to find the altitude, if it was much more screwed up, the dimensions of the triangle that is.[/quote]
What?
Heron's is the quick way, but set the altitude to be h, and have it breat the segment into lengths of x and 6-x and apply pythag twice and solve. It isn't that bad at all.

[quote="Altheman"][quote="adalton"][quote="alex__ib"]I highly sugest you to do this:

[hide]$m^{2}=\frac{2(5^{2}+7^{2})-6^{2}}{4}\\ m^{2}=28\\ m=2\sqrt{7}$[/hide]

I think it is called median theorem...i don't know.[/quote]

That's actually Stuart's Theorem where $m=n$[/quote]

no....there are other ways to derive that theorem, which are outlined above[/quote]
Indeed. I've heard it called appolonius's theorem by the way, never heard of median theorem, though that name would make sense.

My mistake, I was recently looking through some of Mildorf's lectures @ TJ and I saw the two combined and he mentioned that if you set m=n in Stuarts theorem you obtain that result.


In the recent posts bar, I would find it helpful if it was stated in which forum the post is. A lot of inequalities and number theory problems are posted without suggestive titles, and I don't care for reading those

At some point in the future I play to let people choose which forums should display recent posts from. You would then be able to take inequalities out. As for displaying the forum name, that would just take out too much space.

yes that's a nice idea  :)  and is there any specific reasons why posts posted in setions other than HSB and related,Olympiad,College playground don't show up?

Also how about directinga  new user directly to his national forum by identifying his place of posting thru his ip address.. :) ?

[quote="Cyber-Shreyas"][quote="pardesi"]

Also how about directinga  new user directly to his national forum by identifying his place of posting thru his ip address.. :) ?[/quote]

Well is that feasible and possible? And some users are not interested in posting in their national forums. They are more interested in using the other forums.[/quote]
It's feasible and possible, but error prone. Also, many countries/states do not have forums, i.e. I would have nowhere to start because there's no Nebraska forum. Also, I really wouldn't want to start in a Nebraska forum.

That is something that I thought about but it will have to wait a little while. All this other stuff which is more important right now has priority.


Let $x,y,z$ be real numbers satisfying
 $x^{2}+y^{2}+z^{2}+\frac{16}{25}xy=a^{2}$($a$ is positive real number we have known)
Find $Max (xy+yz+xz)$


[quote="SpaceWeather.com"]
After a 13.6 year absence, Comet 8P/Tuttle is once again traveling through the inner solar system. On Jan. 1 and 2, 2008, it makes its closest approach to Earth--only 24 million miles away. The emerald-colored comet will brighten to a predicted magnitude of 5.8, visible to the unaided eye from dark-sky sites and a fine target for backyard telescopes anywhere.

Two nights before closest approach, on Dec. 30th and 31st, something extraordinary will happen: Comet 8P/Tuttle has a beautiful close encounter with spiral galaxy M33. The comet and the galaxy may even overlap!  This is a can't-miss opportunity for astrophotographers around the world.
[/quote]
That should be very interesting.  :)

Wow. That's exciting. If anyone could get any pictures, that would be excellent!

Yes, pictures, especially when it overlaps M33.  :)


calculus
derivative
function
limit
calculus computations

I am looking for the derivation of the following formula using limits

1.) $[f(x)+g(x)]^{'}= [f(x)^{'}+g(x)]^{'}$
2.) $[f(x)g(x)]^{'}= f(x)^{'}g(x)+f(x)g(x)^{'}$
3.) $[\frac{f(x)}{g(x)]}^{'}= \frac{g(x)f(x)^{'}-f(x)g(x)^{'}}{g(x)^{2}}$

You can answer it or you can also post a link just in case there is. Thanks.

[quote="benorin"]1 is straight forward, just write it out using the limit definition of the derivative and then use the limit property that the limit of a sum is the sum of the limits.[/quote]

I'll do it and post it later, maybe you can correct me.

[quote="guile"]hey, i can't get it. i'm so new to this. can somebody help me please. :blush:[/quote]
Okay, just to get things started, let $w(x) = f(x)+g(x).$
We have to prove that $w'(x) = f'(x)+g'(x).$

Using the definition of the derivative of a function, we have
$w'(x) = \lim_{h\rightarrow 0}\frac{w(x+h)-w(x)}{h}$

So, $w'(x) = \lim_{h\rightarrow 0}\frac{(f(x+h)+g(x+h))-( f(x)+g(x))}{h}$

Now, what could you do to get the desired result?

part 3 follows from part two if you write the fraction as multiplying two function with possibly negative exponents, then doing chain rule possibly...

Here you go:

http://tutorial.math.lamar.edu/AllBrowsers/2413/ProductRuleProof.asp

http://tutorial.math.lamar.edu/AllBrowsers/2413/QuotientRuleProof.asp

once you have proven #2 basically just let the exponent on $g(x)$ be $-1$

[quote="me@home"]I don't think our calculus teacher ever showed us how to derive the product rule, yet it is so painless!![/quote]

I think derivation is seldom shown in the undergrad studies

Today we had a calculus review test of last year (i.e. first two semesters calculus) and we were asked for the derivative of $f(x)g(x)$. I derived it on the test :lol:


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

[hide="Solution"]
We are asked to find how many such four-digit sequences there are, such that the sum of the four digits is $ 30$.  First, let's just list the sets of $ 4$ digits:

If the largest number in the sequence is $ 9$, we have $ 9993, 9984, 9975, 9966, 9885, 9876$, and $ 9777$.

If the largest number in the sequence is $ 8$, we have $ 8886$ and $ 8877$.

The largest number in the sequence cannot be less than $ 8$, because $ 7 \times 4\equal{}28$.

If the sequence has $ 3$ of the same digit and the next different there are $ \frac{4!}{3!}\equal{}4$ arrangements.  These sequences are $ 9993,9777,$ and $ 8886$.

If the sequence has $ 2$ digits the same and the other two are different, there are $ \frac{4!}{2!}\equal{}12$ arrangements.  These sequences are $ 9984, 9975,$ and $ 9885$.

If the sequence has $ 2$ sets of $ 2$ digits the same there are $ \frac{4!}{2!2!}\equal{}6$ arrangements.  These sequences are $ 9966$ and $ 8877$.

If the sequence has $ 4$ different digits, there are $ 4!\equal{}24$ arrangements.  This sequence is $ 9876$.

Finally, we add up to get $ (4 \times 3)\plus{}(12 \times 3)\plus{}(6 \times 2)\plus{}(24 \times 1)\equal{}12\plus{}36\plus{}12\plus{}24\equal{}\boxed{84}$.
[/hide]

This is the method I used.  Is there an easier way that is more general?

Not that I know of. Actually, even that method is somewhat elegant to me: if I was taking a test with that problem on it, I would have been drooling and probably writing out all the numbers only to give up 2 seconds later.

If the last four digits are $ \overline{abcd}$, then we must have $ 0\leq a,b,c,d\leq9$ and

$ a \plus{} b \plus{} c \plus{} d \equal{} 30\implies(9 \minus{} a) \plus{} (9 \minus{} b) \plus{} (9 \minus{} c) \plus{} (9 \minus{} d) \equal{} 6$ if

$ 0\leq9 \minus{} a,9 \minus{} b,9 \minus{} c,9 \minus{} d\leq9$, so by Balls and Urns, there are

$ \dbinom{6 \plus{} 4 \minus{} 1}{4 \minus{} 1}\equal{}\boxed{84}$ sequences.

That made absolutely no sense to me. What is Balls and Urns? Math154 you seem to very knowledgeable: you answer basically every question. However, even though I'm not a mod, that question seems to be too difficult to be in classroom math...

I agree--the MATHCOUNTS forum is more appropriate for this.

As for my solution, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1323456#1323456]this explains Balls and Urns (case 4)[/url].


Can anyone give me tips on what is important when preparing for MathCounts?

[hide="1"][url=http://www.artofproblemsolving.com/Wiki/index.php/How_should_I_prepare%3F]Search[/url][/hide]
[hide="2"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1315377612&t=258459]Search more.[/url][/hide]
[hide="3"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1366783574&t=196992]Search even more.[/url][/hide]
[hide="4"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1773083379&t=188105]Search a lot more.[/url][/hide]
[hide="Finally"]You can find a lot more if you [url=http://www.artofproblemsolving.com/Forum/search.php]search.[/url][/hide]

Akhil's right; he's not trying to be funny. There's a lot of information if you use the search function.

Did you actually click on the links? You find 200 pages of tips; [b] look before accusing [/b].If you actually felt that, you would've posted some tips instead of accusing me for trying to be funny; spamming in an indirect way.

isabella 2296 said:
Before posting, think hard whether your post is necessary or not. Remember, you do not have to press the "Submit" button.
Was that post really necessary?

[i]Now[/i] tell me who's at fault.

[quote="JITTT"]I bet you feel like that was funny akhil, but it wasn't.[/quote]
Akhil gave good tips. Also, you could find some extra help and questions on the following sites:
[url=http://www.mathcountscentral.com]MathCountsCentral[/url]
[url=http://mathcounts.saab.org/mc.cgi]Saab's MathCounts Drills[/url]
[url=http://mathcountsmn.org/tests.php]Minnesota MathCounts[/url]

Thanks for pointing it out to me.
I guess your right- I should have thought of searching for tips and I'm sorry I didn't think of that before posting.

well... Why don't you try FTW?

eh... I didn't even make state MC... MY state/region is very difficult! Now that I learned about AoPS, I think I have a shot!


I feel really stupid, but the answer to this question is 24mph:

A messenger pigeon can fly at a constant speed of 25 mph in a wind free environment.  What is its average speed on a round trip where it faces a 5 mph headwind on the way out and a 5mph tailwind on the return leg?

I don't understand the problem... I said 25mph but I still can't understand why it would be 24mph... any comments?

Oh wow thanks... I see - I was just guessing.  I wish we had something like Mathcounts back when I was still in middle/elementary school...


induction
algebra
polynomial
algebra proposed

We have $P(x)\equiv\frac{1}{2009}\sum_{k=1}^{2009}(-1)^{\delta_{k,2008}}\binom{2009}{k}B_{2009-k}x^{k}$, where $\delta_{i,j}$ is Kronecker delta and $B_{i}$ is Bernoulli number. Now, we need some property of Bernoulli number.

From http://mathworld.wolfram.com/BernoulliNumber.html and http://mathworld.wolfram.com/PowerSum.html

Wow, It is simply and no new, but I can not  found it  :(  olorin is good :P
[quote="olorin"]Nice problem!
$P(n+1)=P(n)+(n+1)^{2008}$ for all $n\in\mathbb{N}$, so $P(x)-P(x-1)-x^{2008}\in\mathbb{R}[x]$ has an infinite number of roots.
This gives $P(x-1)=P(x)-x^{2008}$.
Then $P(1)=1\Rightarrow P(0)=0\Rightarrow P(-1)=0$, and then by induction $P(-n)=-P(n-1)$ for all $n\in\mathbb{N}$.
So $P(-x)+P(x-1)\in\mathbb{R}[x]$ has an infinite number of roots,
which gives $P(-x)+P(x-1)\equiv 0$ and $x={1\over 2}$ gives $P(-{1\over 2})=0$.[/quote]

[quote="olorin"]Nice problem!
$P(n+1)=P(n)+(n+1)^{2008}$ for all $n\in\mathbb{N}$, so $P(x)-P(x-1)-x^{2008}\in\mathbb{R}[x]$ has an infinite number of roots.
This gives $P(x-1)=P(x)-x^{2008}$.
Then $P(1)=1\Rightarrow P(0)=0\Rightarrow P(-1)=0$, and then by induction $P(-n)=-P(n-1)$ for all $n\in\mathbb{N}$.
So $P(-x)+P(x-1)\in\mathbb{R}[x]$ has an infinite number of roots,
which gives $P(-x)+P(x-1)\equiv 0$ and $x={1\over 2}$ gives $P(-{1\over 2})=0$.[/quote]
It is the simplest solution!

Also N.T.TUAN had forgot that half of that Bernoulis numbers must be zero.And that formula is more simple.We can use induction for polinomial $P_{2k}(n)=1^{2k}+2^{2k}+3^{2k}+...+n^{2k}$
to prove that $P_{2k}(-1/2)=0$
$\sum_{i=1}^{n}(i+1)^{k}= \sum_{i=1}^{n}(i)^{k}+\sum_{i=1}^{k-1}((C_{k}^{i})P_{i}(n))$
$(n+1)^{k}-1=\sum_{i=1}^{k-1}((C_{k}^{i})P_{i}(n))$(*)
$\sum_{i=1}^{n}(i-1)^{k}= \sum_{i=1}^{n}(i)^{k}+\sum_{i=1}^{k-1}((-1)^{i}(C_{k}^{i})P_{i}(n))$
$(n-1)^{k}-1=\sum_{i=1}^{k-1}((-1)^{i}(C_{k}^{i})P_{i}(n))$(**)
from (*) and (**) we can get equation in which there is no $P_{2i+1}$.We will use that in induction.

No, I said ''Can you continuous on what I HAVE ABOVE''?  :oops:

[quote="olorin"]Nice problem!
$P(n+1)=P(n)+(n+1)^{2008}$ for all $n\in\mathbb{N}$, so $P(x)-P(x-1)-x^{2008}\in\mathbb{R}[x]$ has an infinite number of roots.
This gives $P(x-1)=P(x)-x^{2008}$.
Then $P(1)=1\Rightarrow P(0)=0\Rightarrow P(-1)=0$, and then by induction $P(-n)=-P(n-1)$ for all $n\in\mathbb{N}$.
So $P(-x)+P(x-1)\in\mathbb{R}[x]$ has an infinite number of roots,
which gives $P(-x)+P(x-1)\equiv 0$ and $x={1\over 2}$ gives $P(-{1\over 2})=0$.[/quote]

why P(X-1)=P(X) - X^2008?

[quote="spix"]Because the polinomial $Q(x)=P(x-1)-P(x)+x^{2008}$ has as roots all the natural numbers, that makes it the null polinomial ( a polinomial has no more than its degree number of roots).[/quote]

I know this...but wasnt about that P(X-1) - P(X) - X^2008 has an infinite nomber of roots instead of P(X-1) - P(X) + X^2008

[quote="spix"]You are misplacing $x$ and $x-1$.[/quote]

no`you didn`t understand me. I mean that there should be + instead of -

[quote="N.T.TUAN"][quote="Shishkin"]
Also N.T.TUAN had forgot that half of that Bernoulis numbers must be zero.And that formula is more simple[/quote]
Can you continuous on what I have above?  :oops:[/quote]
I think I undestood your joke! :lol:

[quote]
Then $P(n)-P(n+1)+(n+1)^{2008}=0$ and $x=n+1$ gives $P(x-1)-P(x)+x^{2008}=0$ for an infinite number of $x$[/quote]

wrong! i think you are making a stupid mistake. there is P(n)-P(n+1)-(n+1)^2008=0 , NOT P(n)-P(n+1)+(n+1)^2008=0

[quote="Zamfirmihai"][quote]
Then $P(n)-P(n+1)+(n+1)^{2008}=0$ and $x=n+1$ gives $P(x-1)-P(x)+x^{2008}=0$ for an infinite number of $x$[/quote]

wrong! i think you are making a stupid mistake. there is $P(n)-P(n+1)-(n+1)^{2008}=0$ , NOT $P(n)-P(n+1)+(n+1)^{2008}=0$[/quote]
Unbelievable! :o
We have $P(n)+(n+1)^{2008}=P(n+1)$. Clear so far?
And this gives $0=(P(n)+(n+1)^{2008})-P(n+1)=P(n)-P(n+1)+(n+1)^{2008}$.

I think you have edited your post or I am just really confused. Forget it...


quadratics
geometry
geometric transformation
rotation
conics
ellipse
calculus

Classify and plot the quadratic curve given by $ 4x^2 \minus{} 3xy \plus{} y^2 \plus{} 2x \plus{} y \minus{} 1 \equal{} 0$.
~~~

I have used the Principal Axes Theorem, $ \lambda_1 (x')^2 \plus{} \lambda_2 (y')^2 \plus{} \left[\begin{array}{ccc} d & e\end{array}\right]PX' \plus{}f \equal{} 0$, to perform a rotation of axes.

My result is $ \frac{5\plus{}3\sqrt{2}}{2} \cdot (x')^2 \plus{} \frac{6\minus{}3\sqrt{2}}{\sqrt{26\minus{}12\sqrt{2}}} \cdot x' \plus{} \frac{5\minus{}3\sqrt{2}}{2} \cdot (y')^2 \plus{} \frac{6\plus{}3\sqrt{2}}{\sqrt{26\plus{}12\sqrt{2}}} \cdot y' \minus{} 1 \equal{} 0$.

I know this is an ellipse, but I do not know how to simply this such that it can be easily plotted.

You need to complete the squares and find the center.

Or you could have centered it first, then rotated it.

Here's one way to figure out what to do: consider

$ f(x,y)\equal{}4x^2\minus{}3xy\plus{}y^2\plus{}2x\plus{}y\minus{}1$ and find its critical point. Setting partial derivatives equal to zero:

$ 8x\minus{}3y\equal{}\minus{}2$
$ \minus{}2x\plus{}2y\equal{}\minus{}1$

(Notice the appearance of essentially the same matrix that we're going to diagonalize later.) Solve this to get $ x\equal{}\minus{}1,y\equal{}\minus{}2.$ Now let $ u\equal{}x\plus{}1,v\equal{}y\plus{}2,$ or more usefully, $ x\equal{}u\minus{}1,v\equal{}y\minus{}1.$

The original equation becomes

$ 4u^2\minus{}3uv\plus{}v^2\equal{}2.$

Now rotate that, as you have above.


[hide="Solution"] $ \sum_{k\equal{}2}^{\infty} \sum_{j\equal{}2}^{2008} \frac{1}{j^k}$ is a double summation; exchanging its order, we find that

$ \sum_{j\equal{}2}^{2008} \sum_{k\equal{}2}^{\infty} \frac{1}{j^k} \equal{} \sum_{j\equal{}2}^{2008} \frac{ \frac{1}{j^2} }{ 1 \minus{} \frac{1}{j} } \equal{} \sum_{j\equal{}2}^{2008} \frac{1}{j^2 \minus{} j} \equal{} \sum_{j\equal{}2}^{2008} \frac{1}{j(j\minus{}1)}$

which is a telescoping series with sum $ \boxed{ 1 \minus{} \frac{1}{2008} }$. [/hide]

[quote="t0rajir0u"][hide="Solution"] $ \sum_{k \equal{} 2}^{\infty} \sum_{j \equal{} 2}^{2008} \frac {1}{j^k}$ is a double summation; exchanging its order, we find that

$ \sum_{j \equal{} 2}^{2008} \sum_{k \equal{} 2}^{\infty} \frac {1}{j^k} \equal{} \sum_{j \equal{} 2}^{2008} \frac {\frac {1}{j^2} }{ 1 \minus{} \frac {1}{j} } \equal{} \sum_{j \equal{} 2}^{2008} \frac {1}{j^2 \minus{} j} \equal{} \sum_{j \equal{} 2}^{2008} \frac {1}{j(j \minus{} 1)}$

which is a telescoping series with sum $ \boxed{ 1 \minus{} \frac {1}{2008} }$. [/hide][/quote]
That is *really* nice.

[hide="Solution"]
$ \sum_{r \equal{} 2}^{\infty} \sum_{j \equal{} 2}^{2008} \frac {1}{j^r} \equal{} \sum_{j \equal{} 2}^{2008} \sum_{r \equal{} 2}^{\infty} \frac {1}{j^r}$

The inner summation simplifies as a geometric series: $ \sum_{r \equal{} 2}^{\infty} \frac {1}{j^r} \equal{} \frac {1/j^2}{1 \minus{} \frac {1}{j}} \equal{} \frac {1}{j(j \minus{} 1)} \equal{} \frac {1}{j \minus{} 1} \minus{} \frac {1}{j}$

The entire summation is then $ \sum_{j \equal{} 2}^{2008} \sum_{r \equal{} 2}^{\infty} \frac {1}{j^r} \equal{} \sum_{j \equal{} 2}^{2008} \left( \frac {1}{j \minus{} 1} \minus{} \frac {1}{j} \right)$, which telescopes to $ \frac {1}{2 \minus{} 1} \minus{} \frac {1}{2008} \equal{} \boxed{\frac {2007}{2008}}$
[/hide]

[b]Edit:[/b] [i]Beaten by t0rajir0u! We have the same solution[/i]


This is from Kalva... I can't connect the answer to the question in any way...

5. Find the volume of the set of points that are inside or within one unit of a rectangular 3 x 4 x 5 box.
[hide]
$154+\frac{40\pi}{3}$
The box is $60$. There are 4 quarter-cylinders length $3$ and similarly for the other edges, total $3\pi + 4\pi + 5\pi = 12\pi$. There are 8 one-eighth spheres at the vertices, total $\frac{4\pi}{3}$, and there are boxes height $1$ on each face, $12+12+15+15+20+20 = 94$. Total $154 + \frac{40\pi}{3}$

Wtf?
[/hide]

so you have a box of length 5, width 4, and depth 3.  take the set of all points inside the box.

now consider the set of all points within 1 unit of the box.  that is, all points P such that the distance between P and the nearest point in\on the box is less than or equal to 1.

find the volume of set of all such P

diagram:

edit: i probably should mention the diagram is not entirely complete.

Diagram isn't exact.
Don't forget the 1.4th cylinders and 1/8th spheres lying in between.

edit: is the hidden your answer or kalva's? cuz it looks right to me...

[quote="Kalva"]I have changed the wording, but not the substance, of the problems.[/quote]Hmm... :P

The hidden answer is Kalva's, but Nosepuppy explained it well enough, so off I go to do an AIME :D

Be very aware that nosepuppy's diagram is by no means correct....

[quote="white_horse_king88"]Well, it's hard to see the original 3 by 4 by 5 if you add the quarter cylinders in no?  :P[/quote]

But it's hard to be correct without the quarter cylinders.

Well its basically 3x4x1, 4x5x1, 3x5x1 boxes and sphere of 1 radius plus the 3x4x5 box, the diagram is missing the eighth-spheres at its vertices.


I have an idea:

to avoid some people getting confused we should have a traditional Chinese community and a modern one...

haha sounds like a good idea. I know a few chinese people who use this forum

[quote="tim26270746"]Yeah, maybe we should make a few new communities like "Taiwan", "Hong Kong", "Macau" and "China"[/quote]
If so, I suggest making "Taiwan", "Hong Kong", "Macau" subforums of the "China" community.


maybe for confidence issues? I don't know how to elaborate.

or easier goals

You could apply that same argument to chapters and you'll end up with chapter competitons only slightly easier than nationals.

[quote="13375P34K43V312"][quote="bpms"]Okay, I have a question. The whole point of states is to choose a state team, right? So why doesn't mathcounts make states harder so that it is only slightly easier than nationals?

[size=75][color=darkred]Split into a new topic.
-nebula42[/color][/size][/quote]

Yeah most likely I wouldn't have made nationals if this was the case, considering how much I bombed nationals by, at least this year.

Anyway, I think it's because of the number of people at state. The point of mathcounts is to provide a reasonable challenge for the participants. If it is national level, the scores will obviously be lower, and people would get discouraged. Nationals is harder because the level of the competitors is higher.

Also, the point of state is not neccessarily only to choose a state team. The point is to enrich the mathematical study of the average middle school student.[/quote]
You and splash D make good points. I guess it is majorly because they don't want to discourage the students at states but they want to challenge them.

In fact, if you're assuming the only point of having a state and chapter competition is to choose the next level of competitors, then why not just have a nationals?

It's too inconvient and expensive.

But in that case, why don't we just make 3 national level tests, they don't even have to be easier than each other, the only point of creating multiple ones is so it's easier to administer.

And pretty soon, you have discouraged kids.

I think that if the state chose the level of their competition, we might end up with state competitions being closer to national level.  This is probably why some coaches don't use the school round to decide their team.  But since mathcounts makes the competitions, there is no reason why they would want the absolute best scoring people at national competitions from a state.


$ x \geq y \geq z \geq 0$
Proof
$ \frac{x^2y}{z}\plus{}\frac{y^2z}{x}\plus{}\frac{z^2x}{y} \geq x^2\plus{}y^2\plus{}z^2$

[quote="colorfuldreams"]$ x \geq y \geq z \geq 0$
Proof
$ \frac {x^2y}{z} \plus{} \frac {y^2z}{x} \plus{} \frac {z^2x}{y} \geq x^2 \plus{} y^2 \plus{} z^2$[/quote]

Hence

$ (x\minus{}y)(x\minus{}z)(y\minus{}z)(yz\plus{}zx\plus{}xy) \ge 0$.

[quote="frankvista"][quote="colorfuldreams"]$ x \geq y \geq z \geq 0$
Proof
$ \frac {x^2y}{z} \plus{} \frac {y^2z}{x} \plus{} \frac {z^2x}{y} \geq x^2 \plus{} y^2 \plus{} z^2$[/quote]

Hence

$ (x \minus{} y)(x \minus{} z)(y \minus{} z)(yz \plus{} zx \plus{} xy) \ge 0$.[/quote]

terrible :surrender:


Let be $ m,n\in \mathbb{N}$ such that $ 5\ |\ 2^n\plus{}3^m$ . Prove that $ 5\ |\ 2^m\plus{}3^n$ .

Exponents of 2 end with :2,4,8,6
Exponents of 3 end with :3,9,7,1
Possible combinations: 2+3,4+1,8+7,6+9
Then $ 2^m\plus{}3^n$ end with 5 always also (look mod 4).

[quote="alex2008"]Let be $ m,n\in \mathbb{N}$ such that $ 5\ |\ 2^n \plus{} 3^m$ . Prove that $ 5\ |\ 2^m \plus{} 3^n$ .[/quote]
$ 2^m \plus{} 3^n \equal{} 2^{m \minus{} n}.(2^n \plus{} 3^m) \minus{} 3^n.(6^{m \minus{} n} \minus{} 1) \equal{} 2^{n \minus{} m}.(2^m \plus{} 3^n) \minus{} 3^m.(6^{n \minus{} m} \minus{} 1)$ which is divisible by $ 5$ since $ 5\ |\ 2^n \plus{} 3^m$ and $ 5$ is divisible by either $ 6^{m\minus{}n}\minus{}1$ or $ 6^{n\minus{}m}\minus{}1.$

[quote="alex2008"][quote="Bugi"]Exponents of 2 end with :2,4,8,6
Exponents of 3 end with :3,9,7,1
Possible combinations: 2+3,4+1,8+7,6+9
Then $ 2^m \plus{} 3^n$ end with 5 always also (look mod 4).[/quote]

You forgot 1+9 when n=0[/quote]

Yes, but it's still correct :)


Consider a matrix C which is the product of two other matrices, A and B. That is C = AB. If C is invertible, show that both A and B are invertible.

Invertibility implies that $\det(C)\not=0$.  so $\det(C)=\det(A)\det(B)\not=0$, and so $A,B$ have nonzero determinants and hence are invertible.

You could do this without using the determinant, as long as you know that a matrix has a left inverse if and only if it has a right inverse, and the left and right inverses are the same.

$I=CC^{-1}=(AB)C^{-1}=A(BC^{-1})$ so $BC^{-1}$ is the right inverse, hence inverse, of $A$.

Similarly, $C^{-1}A$ is the left inverse of $B.$


Suppose that $T$ is an invertible linear transformation, $U(T(x))=x$ for all $x$ for all $x$ in $R^{n}$, and $S(T(x))=x$ for all $x$ in $R^{n}$. Prove that $U(x)=S(x)$ for all $x$ in $R^{n}$.

Linearity is irrelevant: this is a fact which applies to any (two-sided) invertible map between sets $X$ and $Y$. What happens when you compose $S\circ T$ and $U\circ T$ with $T^{-1}$?

I know that linearity is irrelevant. However, I chose to keep the word in there since the source I took it from decided to include it in the problem (because of the context of the problem).
That's one way. Do you know of another way to prove it?

When you have an elementary one-line proof, why look for anything else? There is no value in a convoluted solution which avoids the obvious.

you can look for alternative proofs because it forces you to understand the subject better.


function algebra domain calculus derivative

Prove that there is no function $f: \mathbb{Z^+}\to\mathbb{Z^+}$ satisfying the condition
$f(f(x))=x+2007$

[quote="Bictor717"]Doesn't $f(x)=x+1003$ work?[/quote]

I would think so because $(x+1003)+1003 = x+2006$... unless he meant that the domain of the function is $\mathbb{Z^+}$, in which case it would need like an even root or a log.... im not sure.

I suspect there is a typo -- perhaps where $x$ is replaced with $2x$.  The likely solution could not possibly use $f'$, as a derivative is not defined for functions on the positive integers (and even if the function were defined on the reals, you would still need to prove that a derivative existed before you could use it!).

[quote="me@home"]I think the solution he was shooting for would manipulate $f(x)$ and $f'(x)$, with some formulas. I thought a little about it...[/quote]
Oops! sorry  :blush:
I didn't mean to use the derivative, I meant to write the inverse function $f^{-1}(x)$. I didn't mean to make it worse! Wow, two typos in the same problem, you should be confused..

Sorry, man. I was tired when I posted, so I make a typo. Edited. :blush:

By the given condition:
$f(f(f(x)))=f(x)+2007$ ( Replace x by f(x))
Therefore, $f(x+2007)=f(x)+2007$ ( Replace f(f(x)) by x + 2007).
Inductively, we can show that $f(x+2007n)=f(x)+2007n$..........


[quote="Facci"]Prove that:
$(xy-z)(z-1)+(xz-y)(y-1)+(yz-x)(x-1)\geq 0$[/quote]
Try $y=z\rightarrow0^{+}$ and $x=2.$


The function $ f$ satisfies $ f(x) \plus{} 2f \left ( \frac{1}{x} \right ) \equal{} 3x$ for any $ x \neq 0$. Find all numbers $ b$ for which $ f(b) \equal{} f(\minus{}b)$.

Let $ x \equal{} \frac 1y$. Then $ f(\frac 1y) \plus{} 2f(y) \equal{} \frac 3y$.
But then again, $ f(y) \plus{} 2f(1/y) \equal{} 3y$, so adding:
$ 3f(y) \plus{} 3f(1/y) \equal{} 3y \plus{} 3/y \implies f(y) \plus{} f(1/y) \equal{} y \plus{} 1/y$

Have to go now... :( Will continue later.

[color=darkred]Edit: Grrrr... 100000th user will shame me for the rest of my life :P[/color]

From above, we have

$ f(x) \plus{} 2f\left(\frac1x\right) \equal{} 3x\text{    }(1)$

$ f\left(\frac1x\right) \plus{} 2f(x) \equal{} \frac3x\text{    }(2)$

$ 3f(x) \plus{} 3f\left(\frac1x\right) \equal{} 3x \plus{} \frac3x\implies f(x) \plus{} f\left(\frac1x\right) \equal{} x \plus{} \frac1x\text{    }(1) \plus{} (2)$

$ f(x) \minus{} f\left(\frac1x\right) \equal{} \frac3x \minus{} 3x\text{    }(2) \minus{} (1)$

$ 2f(x) \equal{} \frac4x \minus{} 2x\implies f(x) \equal{} \frac2x \minus{} x\text{    }((1) \plus{} (2)) \plus{} ((2) \minus{} (1))$

In order for $ f(b) \equal{} f( \minus{} b)$, we must have

$ b \minus{} \frac2b \equal{} \frac2b \minus{} b\implies2b \equal{} \frac4b\implies b$

$ \frac2b\implies b^2 \equal{} 2\implies b \equal{} \boxed{\pm\sqrt2}$.


$ m,n \in \mathbb Z$
$ (m\minus{}n)^2\equal{}\frac{4mn}{m\plus{}n\minus{}1}$

$ a)$ Show that $ m\plus{}n$ is perfect square
$ b)$ Find all pairs $ (m,n)$ satisfy the equation

[hide]My approach:
First we see that (-n,n) is a solution of this equation.
So, we examine the equation in the case m#-n
a) We have: (m+n)(m-n)^2=(m+n)^2(some algebraic manipulation)
Since (m+n)#0, canceling both sides we get: (m+n)=(m-n)^2 (1), so m+n is a perfect square(since m,n are integers)
b) W.L.O.G, assume that m>n(consider that m=n=1 or m=n=0 are solutions of this equation when m=n).We easily infer that m and n>0. In addition, from (1) we infer that m+n=k^2 with k>=2 and k belongs to N, so we have the system of equation: m-n=k
                                        m+n=k^2
From these equations we get: m=k(k+1)/2 and n=k(k-1)/2 and m,n are positive integers(since k(k+1) is divisible by 2 and k is a positive integers).
In conclusion, the equation above has infinite integers solutions:(m,n)=(k(k+1)/2, k(k-1)/2) with k>=2 and k belongs to N(Q.E.D)
@sinankaral53: By the way, what is "TR"?[/hide]


ratio
limit
geometric sequence
algebra unsolved
algebra

Is that "$x_{n+1}^2$" in your definition supposed to be $u_{n+1}^2$?

Let's assume that's what you meant. From the quick numerical observation that the limit appears to be $\frac{\pi}2$ (spreadsheets can be useful), we rewrite it so that $u_n=\frac{\pi}2-x_n$. Putting this into the sequence and rearranging, we get the following recursive equation for $x_n$:

$x_{n+2} = \frac{2\pi}5\left(1-\cos(x_n)\right)+\frac25\cdot x_{n+1}-\frac4{25\pi^2}\cdot x_{n+1}^2$

If we now assume that both $x_n$ and $x_{n+1}$ are small, this shows that
$x_{n+2}=\frac25\cdot x_{n+1}+O(x_n^2)+O(x_{n+1}^2)$.

Thus, once we get sufficiently close, $x_n$ goes to zero at a rate approaching that of a geometric sequence with ratio 2/5. Observation of the first few terms shows that we do get sufficiently close. Thus $\lim u_n = \frac{\pi}2$.


inequalities
trigonometry
limit
real analysis
real analysis unsolved

Let's call a set A of real numbers good if there exists sequences $a_n\geq 0$ and $b_n\in R$ such that $\sum_{n\geq 1}{ a_n|cos(nx-b_n)|}$ converges for all $x\in A$ but the series $\sum_{n\geq1}{a_n}$ diverges. Prove that A is good if and only if there exists a sequence $c_n$ of nonnegative real numbers with the sum of all terms divergent and such that for all $x\in A$ the series $\sum_{n\geq 1}{c_n|sin(nx)|}$ converges.

[quote="harazi"]Let's call a set A of real numbers good if there exists sequences $a_n\geq 0$ and $b_n\in R$ such that $\sum_{n\geq 1}{ a_n|cos(nx-b_n)|}$ converges for all $x$[/quote]
Should it be "for all $x\in A$" both times?

Yes, fedja, sorry. I've edited the post.

It is a difficult one (at least, I couldn't find a really short and nice solution). So, if you like the problem but get stuck, I put a small hint [hide="here"] Show first that you can  make all $b_n$ in the definition of a good set the same number $b$. Then use the inequality $|\sin(\alpha-\beta)|\leq |\cos\alpha|+|\cos\beta|$. [/hide] I wonder whether the official solution follows the same lines as mine  :) .

Suppose that there exists a sequence of real numbers $ c_{n} \geqslant 0 $ such that
$ \sum_{n=1}^{\infty} c_{n} = + \infty $ and $ \sum_{n=1}^{\infty} c_{n}|sin(nx)|
 < + \infty $ for all $ x \in A $. If we take $ a_{n} = c_{n} $ and $ b_{n} = \pi/2 $
then we will obtain one direction.

Now suppose there exist $ a_{n} \geqslant 0 $, $ b_{n} \in \mathbb{R} $ such that
$ \sum_{n=1}^{\infty} a_{n} = + \infty $ and $ \sum_{n=1}^{\infty} a_{n}|cos(nx-b_{n})|
 < + \infty $ for all $ x \in A $. Define $ d_{n} = b_{n} + \pi/2 $ , then we obtain that
$ \sum_{n=1}^{\infty} a_{n}|sin(nx-d_{n})| < + \infty $ for all $ x \in A $.
Substituting $ x = 0 $ we get that $ \sum_{n=1}^{\infty} a_{n}|sin(d_{n})| < + \infty $
for all $ x \in A $. Now , $ sin(nx-d_{n}) = sin(nx)cos(d_{n}) - cos(nx)sin(d_{n}) $,
but $ |cos(nx)sin(d_{n})| \leqslant |sin(d_{n})| $ , so because of
$ \sum_{n=1}^{\infty} a_{n}|sin(d_{n})| < + \infty $ and
$ \sum_{n=1}^{\infty} a_{n}|sin(nx-d_{n})| < + \infty $ we get that
$ \sum_{n=1}^{\infty} a_{n}|cos(d_{n})||sin(nx)| < + \infty $
for every $ x \in A $. Now I claim that we can take $ c_{n} = a_{n}|cos(d_{n})| $.
Indeed, if no, then $ \sum_{n=1}^{\infty} a_{n}|cos(d_{n})| < + \infty $
but we also have $ \sum_{n=1}^{\infty} a_{n}|sin(d_{n})| < + \infty $ and because of
$ |sin(d_{n})| + |cos(d_{n})| \geqslant sin(d_{n})^{2} + cos(d_{n})^{2} = 1 $
we get that $ \sum_{n=1}^{\infty} a_{n} < + \infty $, which is not true.

I hope this is correct  :)  .

[quote="Leva1980"]Substituting $x = 0$  we get that $\sum_{n=1}^{\infty} a_{n}|sin(d_{n})| < + \infty$  [/quote]   What if $0\notin A$?

A month passed since Harazi posted the problem and nobody solved it, so let me post a full solution (I'll prove the hard implication only). The solution I know consists of two parts, the first one being rather boring and dealing with various standard properties of divergent  series with non-negative terms. The second part is a nice trick that finishes the problem.

[b]Part 1: [/b] We shall show that, without loss of generality, we may assume that all $b_n$ in the definition of a good set are the same number $b$.
Lemma 1: Suppose that $a_n\geqslant 0$ and $\sum_n a_n=+\infty$. Let $b_n$ be any bounded sequence. Then there exists an increasing sequence $n_k$ such that $\sum_k a_{n_k}=+\infty$ and $b_{n_k}$ has a limit.
[hide="Proof of Lemma 1"]
Let $I$ be a bounded interval containing all $b_n$. Let $I_1$ and $I_2$ be two halves of $I$. Consider the sets $\mathcal N_j=\{n>0 \,: \, b_n\in I_j\}$. Note that $\sum_n a_n=\sum_{n\in\mathcal N_1}a_n+\sum_{n\in\mathcal N_2}a_n$, so at least one of the series on the right diverges. Let $j_1$ be the corresponding index ($1$ or $2$). Choose $n_1<n_2<\dots<n_{K_1}\in \mathcal N_{j_1}$ such that $\sum_{k=1}^{K_1} a_{n_k}>1$. Then split the interval $I_{j_1}$ into two halves $I_{j_11}$ and $I_{j_12}$. Again, define $\mathcal N_{j_1j} =\{n> n_{K_1}\,: \, b_n\in I_{j_1j}\}\subset N_{j_1}$ and choose $j_2$ so that $\sum_{n\in\mathcal N_{j_1j_2}}a_n=+\infty$. Then choose $n_{K_1+1}<\dots<n_{K_2}\in \mathcal N_{j_1j_2}$ such that $\sum_{k=K_1+1}^{K_2} a_{n_k}>1$. Continue this process. The $b_{n_k}$ will be restricted to shorter and shorter subintervals of $I$ and, during each step, the sum of $a_{n_k}$ will increase by at least $1$.
[/hide]
Lemma 1 allows us to say only that, WLOG, we may assume that $b_n$ tend to some limit $b$ (just replace all $a_n$ outside our subsequence by $0$(or some convergent series) and replace the corresponding $b_n$ by $b$). To make all $b_n$ exactly the same, we need
Lemma 2: Suppose that $a_n\geqslant 0$ and $\sum_n a_n=+\infty$. Let $b_n\to b$ be any convergent sequence. Then there exist non-negative $a_n'\leqslant a_n$ such that $\sum_n a_n'=+\infty$ but $\sum a_n'|b_n-b|<+\infty$.
[hide="Proof of Lemma 2"]
WLOG, we may assume that all $a_n\leqslant 1$ (just replace too big $a_n$ by $1$, the series will still diverge). Now choose $n_1>0$ such that $|b_n-b|<2^{-1}$ for all $n\geqslant n_1$. Replace $a_1,\dots a_{n_1-1}$ by $0$ and choose the least $N_1\geqslant n_1$ such that $\sum_{n=n_1}^{N_1}a_n\geqslant 1$. Since each $a_n\leqslant 1$, this sum is automatically less than $2$. Now choose $n_2>N_1$ such that $|b_n-b|<2^{-2}$ for all $n\geqslant n_1$. Replace $a_{N_1+1},\dots a_{n_2-1}$ by $0$ and choose the least $N_2\geqslant n_2$ such that $\sum_{n=n_2}^{N_2}a_n\geqslant 1$. Again, this sum is automatically less than $2$. Proceeding in this way, we shall get $\sum_n a_n'=\sum_k \sum_{n=n_k}^{N_k} a_n\geqslant \sum_k 1=+\infty$ while  $\sum_n a_n'|b_n-b|=\sum_k \sum_{n=n_k}^{N_k} a_n|b_n-b|\leqslant \sum_k 2\cdot 2^{-k}<+\infty$.
[/hide]
Lemma 2 allows to replace $a_n$ by $a_n'$ and then $b_n$ to $b$.
That is essentially the end of Part 1. The last thing that we'll need to have Part 2 formally correct is
Lemma 3:  Suppose that $a_n\geqslant 0$ and $\sum_n a_n=+\infty$. Then there exist non-negative $a_n'\leqslant a_n$ such that $\sum_n a_n'=+\infty$  and $\lim_{n\to\infty}a_n'=0$.
I leave the proof to the reader: it is much simpler than Lemma 1 or 2.

[b]Part 2: [/b] From Part 1, we know that there exist $a_n\geqslant 0$ and $b\in\mathbb R$ such that $\sum_n a_n=+\infty$ but the series $\sum_n a_n|\cos(nx-b)|$ converges for all $x\in A$. Let $a_{n_1}\geqslant a_{n_2}\geqslant\dots$ be the decreasing rearrangement of the sequence $a_n$. Then $\sum_k a_{n_k}=\sum_n a_n=+\infty$ and, since $a_{n_k}$ is non-increasing, we may also conclude that $\sum_k a_{n_{2k}}=+\infty$. Now observe that $a_{n_{2k}}|\sin(|n_{2k}-n_{2k-1}|x)|\leqslant a_{n_{2k-1}}|\cos(n_{2k-1}x-b)|+a_{n_{2k}}|\cos(n_{2k}x-b)|$. Therefore, $\sum_k a_{n_{2k}}|\sin(|n_{2k}-n_{2k-1}|x)|<+\infty$ for all $x\in A$. We can rewrite the last series as $\sum_n \alpha_n |\sin nx|$ where $\alpha_n=\sum_{k: |n_{2k}-n_{2k-1}|=n}a_{n_{2k}}$, so $\sum_n \alpha_n=\sum_k a_{n_{2k}}=+\infty$. This is the end if all $\alpha_n$ are finite. But if $\alpha_n=+\infty$, we conclude that $\sin nx=0$ for all $x\in A$ and we can use the series $|sin nx|+|\sin 2nx|+\dots$.

I still wonder if Harazi has a simpler solution.  :)

Sorry, mine is also long and difficult. :(


1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

[hide]

1)The nth term of the 2nd sequence minus the nth term of the 1st sequence can be expressed algebraicly:

[tex]n(n+1) - n^2[/tex]

[tex]n^2 + n - n^2=n[/tex]

Thus, if you subtract the first k terms of the first sequence from the first k terms of the second sequence, you essentially get the sum of the first k natural numbers, which can be expressed by

[tex]\frac{k(k+1)}{2}[/tex]

[tex]\frac{k(k+1)}{2}=210[/tex]

[tex]k^2+k=420[/tex]

[tex]k^2+k-420=0[/tex]

[tex](k+21)(k-20)=0[/tex]

Thus, the answer is 20.[/hide]


analytic geometry
geometry
3D geometry
function
calculus
derivative
calculus computations

Hey, got a problem with my AP Calc homework actually two questions.

1.) Assuming ax^3 + bx^2 + cx + d with a, b, c, d > 0 has maxima/minima and points of inflection, find both the x and y coordinates of the maxima/minima and points of inflection.

That one I'm completely stuck on, this one I'm kind of getting

2.) The strength of a rectangular beam is proportional to its width times the square of its depth. The stiffness of a rectangular beam is proportional to its width times the cube of its depth. If I have a log with a diameter of 25 inches, then find the dimensions of:
a: the board with maximum strength
b: the board with maximum stiffness

So far all I've got is
w = depth d = diameter
f(strength) = wd^2
f(stiffness) = wd^3

I differentiate with respect to d since the width would be constant (right?).
After differentiating I should be able to find the value where its strength or stiffness if max but nothing seems to work out.

Any help would be appreciated.

1) The graph of such a function goes up, then down, then up again. There is one point of max, one point of min, and one inflection point. Equating the first derivative to zero, you will get two values of $x$: the smaller one is the maximum, the greater one is the minimum. Equating the second derivative to zero, you will find the $x$-coordinate of the point of inflection.


I have a few questions.
1.  Whats a bot?  Are they suspended from the site?
2. How do you become a moderator?
3.  How do you use that cool italicy thing?

1. A bot is an internet bot. It indexes pages, such as GoogleBot.
2. You become a moderator by working hard, solving problems and stuff like that.
3. Do you mean $ \text{\LaTeX}$? Then just put dollar signs around what you want to latex-i-cize.

A bot is a program Google uses to get the information so you can search it.

Moderators make sure [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19760]you don't do bad things[/url]. You do it by being a good representative of AoPS and catching an admin's eye for being good.

Do you mean $ \text{\LaTeX}$? Go check out [[LaTeX]] in the [[AoPSWiki]].

Not all bots are from Google. In fact most search bots are from Yahoo and MSN.

Also, not all bots are indexing bots. There are spam bots (which basically just post random things and/or ads on the boards, thankfully we don't have any of these, at least not that I've noticed), archive bots, malicious bots that try to create Denial of Service crashes, and others. Bots are recognized by IP, so [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=20369]some humans can be mistaken for bots.[/url]


geometry
MATHCOUNTS
LaTeX
analytic geometry
rectangle
rhombus
incenter

this is like the mathcounts marathon, except its only geometry problems.
rules:
1. give the solution, not just the answer.
2. post a problem with your solution.
3. if someone answered your problem wrong, tell them! the next problem will be postponed until the last one is solved.
4. please LaTex your answers, if not asymptote.
5. use $ \pi$ for $ \pi$. that's it.

first question.
a circle is inscribed inside a square.
the square has a diagonal $ d$.
solve for the circle's area in terms of $ d$.

circle in square, not square in circle.
and you didn't post a new problem. :noo:

Bah when will I learn to read!

Hm if a diagonal is d then side is $ \frac{d}{\sqrt2}$ so radius is hlaf that or $ \frac{d}{2\sqrt2}$ so $ \frac{\pi d^2}{8}$

Find the area of a triangle that has vertices (0,0), (1,10) and (1337,1)

hmmm...

the length of the sides are:
$ \sqrt{10^2\plus{}1^2}$
$ \sqrt{1336^2\plus{}9^2}$
$ \sqrt{1337^2\plus{}1^2}$
simplified, that is:
$ \sqrt{101}$
$ \sqrt{1784977}$
$ \sqrt{1787570}$
then hero's formula:
$ \sqrt{\frac{\sqrt{101}\plus{}\sqrt{1784977}\plus{}\sqrt{1787570}}{2}(\frac{\sqrt{101}\plus{}\sqrt{1784977}\plus{}\sqrt{1787570}}{2}\minus{}\sqrt{101})(\frac{\sqrt{101}\plus{}\sqrt{1784977}\plus{}\sqrt{1787570}}{2}\minus{}\sqrt{1784977})}$...gah! its too long! i get an error if i try to fit it all
don't feel like simplifying.
maybe later.

The more effecient way.  Use shoelace.

$ \frac{1}{2} |0 \plus{} 1 \plus{} 0 \minus{} 0 \minus{} 13370 \minus{} 0|$

$ \frac{1}{2} |\minus{}13369|$

$ \boxed{\frac{13369}{2}}$

Find the area of a pentagon with coordinates (0,0), (2,4), (4,1), (-1,3), and (-3,7).  Express your answer as a common fraction if necessary.

Why not just use the MathCounts Marathon? :wink:

isabella2296:
it goes way too far off mathcounts level.
And I need practice specifically in geometry.
anyway... answer... guess i should try the shoelace method.

$ \frac{1}{2}|0\minus{}9\minus{}4\plus{}2\plus{}0\minus{}0\plus{}7\minus{}6\minus{}16\minus{}0|$

$ \frac{1}{2}|\minus{}26|$

$ \frac{26}{2}$

so... i guess the answer is 13.
thanks for using the shoelace, ernie. i would have never known about it.
new problem:
A side of square ABCD is equal to $ \frac{1}{2}$ the diameter of a circle with a center B and a square EFGH is inscribed within that circle. What is the area of everything within circle B [b]AND[/b] square ABCD (meaning it has to be in both, not one of them) but is not in square EFGH if line AB=2 in.

[geogebra]56db8b401a49b244e41c814f69b0cd3e768de214[/geogebra]

Intersection of squares: 2 sq in
Intersection of ABCD and circle: pi
So, pi-2

New question:

Find the area of a cyclic quadrilateral with sides of 1,2,3,4.

We use Brahmagupta's Formula, or $ A\equal{}\sqrt{(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)}$. Plugging in the numbers, we get $ A\equal{}\sqrt{4*3*2*1}\equal{}\sqrt{24}\equal{}2\sqrt{6}$. I don't think that was MC level.

A point P is within a rectangle ABCD. AP=6, BP=7, and CP=8. Find DP.

Can you tell a bit more on Brahmagupta's Formula? to lazy to google.

For a cyclic quadrilateral, A=sqrt((s-a)(s-b)(s-c)(s-d))

where s is semiperimeter

It's Heron with a 4th side extension.

This is not Mathcounts level...in general, cyclic quadrilaterals are not a MC topic.

A point P is within a rectangle ABCD. AP=6, BP=7, and CP=8. Find DP.

 [geogebra]a9f5967b7c51dba3c41f0d0e271bac62a10410ef[/geogebra]

Drop perpendiculars from P to sides, denote resulting segment division as a,b,c,d

$ 6^2\plus{}8^2\equal{}a^2\plus{}b^c\plus{}c^2\plus{}d^2\equal{}7^2\plus{}x^2$
Or
100=49+x^2
x^2=51
x=sqrt(51)

You have rhombus ABCD. Draw its incircle (incenter O). If AC is 16 and AO is $ 8\sqrt{2}$, what is the area of ABCD?

what happened here? why does the applet not open my computer?

[quote="james4l"]You have rhombus ABCD. Draw its incircle (incenter O). If AC is 16 and AO is $ 8\sqrt {2}$, what is the area of ABCD?[/quote]

Hm? This situation is impossible. All rhombi have 4 equal sides.


modular arithmetic
number theory
number theory unsolved

Use Fermat's Little Theorem to find the smallest odd prime number which divides $ {1155}^{55}\minus{}1$.

I know $ 1155\equal{}3\cdot5\cdot7\cdot11$, and for any prime $ p$ that does not divide $ 1155$, $ {1155}^{p\minus{}1}\equiv 1 (mod \ p)$, but I don't know how to put it together. Thanks.

Let $ p$ is smallest odd prime divisor of $ 1155^{55}\minus{}1$

$ (p,3)\equal{}(p,5)\equal{}(p,7)\equal{}(p,11)\equal{}1$

$ 1155^{p\minus{}1}\minus{}1\equiv 1\pmod{p}$  $ \implies$  $ 1155^{(p\minus{}1,55)}\equiv 1\pmod{p}$

$ q$ is prime $ (p\minus{}1,55)\equal{}d$,$ q|d$  $ \implies$  $ q\equal{}5$  or $ q\equal{}11$

if $ q\equal{}5$  $ \implies$  $ p\equiv 1\pmod{5}$

$ \implies$  $ p\equal{}31$ satisfies this condition;  $ (30,55)\equal{}5$ ,$ 1155^{5}\equiv 1\pmod{31}$

$ (1155\equiv 8\pmod{31},2^{15}\equiv 1\pmod{31})$

if $ q\equal{}11$  $ \implies$  $ p\equiv 1\pmod{11}$  $ \implies$ $ p\equal{}23$ isn't satisfies because;

$ (22,55)\equal{}11$ ,$ 1155\equiv 5\pmod{23}$  ,$ 5^{11}\equiv 32.5\equiv \minus{}1\pmod{23}$

$ \implies$  $ p\equal{}31$

Many thanks cnyd. I understand the method much better now.