number theory
greatest common divisor
modular arithmetic

[hide="solution"]
Using modular arithmetic it's easy to check that $ 15|f(n) \ \forall n\in \mathbb{N}$:

$ n \equiv 0,1,2 \pmod 3 \\ \implies 5n^3 \equiv 0,2,1 \pmod 3, \ 4n \equiv 0,1,2 \pmod 3 \\ \implies 9n^5\minus{}5n^3\minus{}4n \equiv 0 \pmod 3$

$ n\equiv 0,1,2,3,4 \pmod 5 \\ \implies 9n^5 \equiv 0,4,3,2,1 \pmod 5, 4n \equiv 0,4,3,2,1 \pmod 5 \\ \implies 9n^5\minus{}5n^3\minus{}4n \equiv 0 \pmod 5$

Factorizing $ f(n)$ gives us $ f(n)\equal{}n(n\minus{}1)(n\plus{}1)(9n^2\plus{}4)$. Now let's prove that $ 8|f(n)$:

If $ n$ is even $ 4|9n^2\plus{}4$ thus $ 8|n(9n^2\plus{}4)$.

If $ n$ is odd either $ 4|(n\minus{}1), \ 2|(n\plus{}1)$ or $ 4|(n\plus{}1), \ 2|(n\minus{}1)$ thus $ 8|(n\plus{}1)(n\minus{}1)$.

Now we have that $ 120|f(n)$.

Using factorization of $ f(n)$ we can quickly see that

$ f(17)\equal{}120 \cdot 2^2 \cdot 3 \cdot 17 \cdot 521$

$ f(18)\equal{}120\cdot 2 \cdot 3 \cdot 17 \cdot 19 \cdot 73$

$ f(19)\equal{}120\cdot 3\cdot 19\cdot 3253$

$ f(20)\equal{}120\cdot 2\cdot 7\cdot 17\cdot 19\cdot 53$

$ \gcd (f(17), f(18),f(19),f(20))\equal{}120 \implies \gcd (f(17), f(18),\ldots  f(2009))\equal{}120$[/hide]

$ f(n) \equal{} (n \minus{} 1)n(n \plus{} 1)(9n^2 \plus{} 4)$, considering all $ n\in [17,2009]$

Using mod arguments it is easy to show that
$ 8|f(n)$ but not necessarily $ 16|f(n)$
$ 3|f(n)$ but not necessarily $ 9|f(n)$
$ 5|f(n)$ but not neccessarily $ 25|f(n)$

Now we have $ 120|f(n)$. Assume there is a $ p > 5$ such that $ p|f(n)$ for all $ n\in [17,2009]$

Then $ (n \minus{} 1)n(n \plus{} 1)(9n^2 \plus{} 4)\equiv 0 \mod p$ for all $ n$

But if $ n\not\equiv 0, \pm 1 \mod p$ then $ 9n^2 \plus{} 4\equiv 0 \mod p$ for all $ n\in \{2,3,4,..,\frac {p \minus{} 1}{2}\}$

But if $ 9(2^2) \plus{}4 \equiv 9(3^2) \plus{}4 \mod p \Leftrightarrow 3^2\cdot 5\equiv 0 \mod p$ which is a contradiction since $ p$ is a prime and $ p > 5$

So the $ \gcd$ is $ \boxed{120}$

EDIT: fixed

Your factorization isn't correct, $ f(n) \neq n^5\minus{}n \ \forall n\in [17,2009]$.

oh thanks. the way to prove that f(n) mod 8 was zero was the key.


Well, since I was looking for a formula for the product of consecutive numbers, aslo known as factorial, I thought maybe there could be a forum with lots of useful formulas that will make long calculations a lot easier.  Hence, post and please explain whatever you know that will make calculations easier.  I will begin.

The formula for factorial is also known as Stirling's Approximation.  The formula is $n!=\sqrt{(2n+\frac{1}{3})\pi }*n^n e^{-n}$.

A formula that will simplify $\sum_{x=1}^{n}x=1+2+3+...+n=\frac{n(n+1)}{2}$.

Another useful formula that could be useful for younger students is $\sum_{x=1}^{n} x^2=1^2+2^2+3^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$.

Finally, a formula that will describe $\sum_{x=1}^{n}x^3=1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2$.

Masoud Zargar

I don't see the need for Stirling's Approximation. Firstly, it's an approximation. It's also unwieldy for mental estimations because of pi, square root and powers of e. It's easier to memorize the more common (smaller) factorials than the formula itself. In half the time it takes for you to enter it in a calculator, you could have used the factorial button or even multiplied it out. (Basically, read my sig.)
The other formulas could use some proofs. They're also featured in the proof without words in the left sidebar.

[quote="Bictor717"]
The other formulas could use some proofs. They're also featured in the proof without words in the left sidebar.[/quote]

I think they've been proved enough on the forums lately :P

Bictor717, I think you're not totally right (about stirling's formula being useless).
You should read that :) :
http://www.answers.com/topic/stirling-s-approximation
http://www.everything2.com/index.pl?node_id=565074

Any formula that has been discovered is very useful or it at least shows us how that has been discovered.  If so, then we may use that formula or method in doscovering other formulas that may or may not have practical uses.  They are all useful in some way.  Anyway, [b]does anyone else have other formulas that they would like to share?[/b]  Any formula...

Masoud Zargar

[quote="mathclass"]stirlings seems to be awfully complicated...

i don't remember any useful tricks right now but if i do i will post em![/quote]

Thanks.

Masoud Zargar


yes.

induction is the given proof in the textbook.


advanced fields
advanced fields unsolved

let $ \Gamma$ be a connected graph and $ P_0 \in vert~ \Gamma$. write $ l_\Gamma(P,Q)$ for the minimum length of paths joining $ P$ and $ Q$. show that there is a maximal tree $ \Delta$ of $ \Gamma$ such that $ l_\Gamma(P_0,P) \equal{} l_\Delta(P_0,P)$ for all $ P \in vert~ \Gamma$. (serre, trees, exiercise 2.3.1)

I've tried to use zorn's lemma (which also proves the existence of maximal trees in $ \Gamma$). but I don't know how to show that $ \Delta$ maximal under the condition that it's a tree in $ \Gamma$ containing $ P_0$ and $ l_\Gamma(P_0,P) \equal{} l_\Delta(P_0,P)$ is also a maximal tree ...

I'm a bit confused about the statement: if the connected tree $ \Delta$ contains all vertices (otherwise, how can we have $ \ell_\Delta(P,P_0)\equal{}\ell_\Gamma(P,P_0)$ for all $ P\in \operatorname{vert}\Gamma$, how can it be not maximal? I mean you cannot add edges to a connected tree containing all vertices and still have a tree. Am I missing something or it should be "for all $ P\in \operatorname{vert}\Delta$" instead :?

a maximal tree in a connected graph contains all vertices.

Here's one possible solution:

For a natural number $ n\ge 0$, let $ V_n$ be the collection of those vertices of $ \Gamma$ which have distance $ n$ from $ P_0$. In particular, $ V_0\equal{}\{P_0\}$, and $ V_1$ consists of the neighbors of $ P_0$. Let $ E_1$ be a collection of edges connecting $ V_1$ to $ V_0$ and maximal with respect to the property of not forming cycles. Then let $ E_2$ be a collection of edges connecting $ V_2$ to $ V_1$ and maximal with respect to the property of not forming cycles with the edges in $ E_1$. Now repeat the process inductively: for $ n\ge 3,\ E_n$ will be a collection of edges connecting $ V_n$ to $ V_{n\minus{}1}$ and maximal with respect to the property of not forming cycles with the edges in $ \bigcup_{i\equal{}1}^{n\minus{}1}E_i$.

Finding such maximal sets is possible by Zorn (or directly by Choice: in order to construct $ E_n$, simply choose for each vertex in $ V_n$ one edge connecting it to $ V_{n\minus{}1}$), and it should be easy enough to check that the tree formed by the edges in $ \bigcup E_n$ is what you're looking for.


function
inequalities
inequalities unsolved

Let be given $a,b,c,x,y$ are positive real numbers and $x>y$.Prove that
$\frac{a^x}{b^x+c^x}+\frac{b^x}{c^x+a^x}+\frac{c^x}{a^x+b^x}\geq \frac{a^y}{b^y+c^y}+\frac{b^y}{c^y+a^y}+\frac{c^y}{a^y+b^y}$

This is a wll-known one. It appeared even in G.M.B. Just the consider the function $f(x)=\frac{a^{x}}{b^{x}+c^{x}}+\frac{b^{x}}{c^{x}+a^{x}}+\frac{c^{x}}{a^{x}+b^{x}}$ and try to prove that $f$ is increasing.
 A direct consequence is Vasc's inequality:

    $\frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}}\geq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}.$ ;)

What's 'Vasc inequality'?:) Can anyone give me a link?:)

The last inequality from my post belongs to Vasc. :P

[quote="darij grinberg"]I remember "inventing" this once ( http://www.mathlinks.ro/Forum/viewtopic.php?t=17230 and http://www.mathlinks.ro/Forum/viewtopic.php?t=23416 ), and proving an even more general result ( http://www.mathlinks.ro/Forum/viewtopic.php?t=23752 post #2).

  darij[/quote]
Thank for all.

@Darij:This also my solution(23/10/2005).But myfriend talk:Use Bernuli's theorem.And I can't.  :(


By popular request, here's Macky's amazing act at the 2005 San Jose ARML site. Though the clip is only 20 seconds, it covers almost his entire performance.

See it to believe it!  :first:  :winner_first:  :clap:

http://www.artofproblemsolving.com/Images/rrusczyk/AVideo1.AVI

[i](Edited by rrusczyk to add link to uploaded AVI file.)[/i]

For those having trouble with the attachment, I have uploaded the full AVI file to our server and edited the post above to include a link to it.

That's like incredible. O.O Teach me! Teach me!

[quote="Kent Merryfield"]When Teddy said "it covers almost his entire performance." he was exaggerating. There's more, including the flourish at the finish. Did anyone else get video?[/quote]

A friend of mine got the whole thing. It's not great quality, though. Someone should've gotten it with good quality; that'd be something to watch.

[quote="LynnelleYe"]My comp still won't play it. It says that I need an update and that 'there are no available updates from RealPlayer that will play this movie'. :([/quote]

It's a Motion-Jpeg file, so you may not be able to view the video because you don't have a codec that can decode it.

Installing [url=http://www.pegasusimaging.com/picvideomjpeg.htm]this[/url] may fix the problem.

WOW this is cool...I cant even solve a regular rubiks cube.....

You SERIOUSLY have to be there to believe it....

During the last half I'm worried that Macky will get tired from juggling and drop it, tho...

[quote="Kent Merryfield"]There's more, including the flourish at the finish. ?[/quote]

The flourish at the end was spectacular.  For those who don't know, he threw one (all?) of the cubes into the air, spun around, and caught it.  The whole thing was quite amazing to watch :)

yeah, something like litoxe and a string of numbers behind it


limit
trigonometry
calculus
logarithms
irrational number
real analysis
real analysis solved

[i]Question:[/i] To find  $ \displaystyle \lim\limits_{n\to \infty} n\cdot\sin{(2\pi e n!)} $.

[b]Corollary:[/b]  Napier's constant $  e $ is an irrational number !
This nice  question  was proposed in Amer.Math.Monthly  many years ago   by D.J.Newman   in  Amer.Math.Monthly  E 924 [1950,416.]

This one's pretty simple. We use $e=\sum_{j=0}^{\infty}\frac1{j!}$, so
$2\pi e\cdot n!=2 \pi \sum_{j=0}^n \frac{n!}{j!} + 2 \pi \sum_{j=n+1}^{\infty} \frac{n!}{j!} \equiv  2 \pi \sum_{j=n+1}^{\infty} \frac{n!}{j!} (\mod 2\pi)$
Next, we use the estimate that for $j > n+1$, $0<\frac{n!}{j!}<(n+1)^{n-j}$. Summing these, $\frac1{n+1} < \sum_{j=n+1}^{\infty} \frac{n!}{j!} < \frac1n$
and $\sin (2\pi \cdot \frac1{n+1}) < \sin(2\pi \cdot \sum_{j=n+1}^{\infty} \frac{n!}{j!}) < \sin (2\pi\cdot \frac1n)$ for any $n\ge 4$ The limit of both sequences $n\sin\frac{2\pi }{n+1}$ and $n\sin\frac{ 2\pi}{n}$ is $2\pi\cdot \lim_{x\to 0} \frac{\sin x}{x}=2\pi$, so $\lim_{n\to \infty} n\cdot \sin(2\pi\cdot e\cdot n!)=2\pi$.

The corollary that $e$ is irrational is too easy to worry about; suppose $e=\frac{m}{n}$. Then $n!\cdot e = \sum_{j=0}^n \frac{n!}{j!} + \sum_{j=n+1}^{\infty} \frac{n!}{j!}$ is an integer, so $\sum_{j=n+1}^{\infty} \frac{n!}{j!}$ is an integer. As we proved above, $0<\frac{1}{n+1} < \sum_{j=n+1}^{\infty} \frac{n!}{j!} < \frac{1}{n} \le 1$, and we have a contradiction.

Indeed, this is the solution.

[b]A question[/b] : it's possible to use only elementary methods,e.g. without Taylor formula, or other differential calculus methods ? I have'nt any idea !

[b]Another question[/b]: If $ C:=\lim\limits_{n\to \infty}
(\sum\limits_{k=1}^n\frac{1}{k}- \ln{n} ) $, does  there exists a positive siquence  $ (x_n ) $ with $
\lim\limits_{n\to \infty}x_n =\infty , $  such that  $\displaystyle L:=\lim\limits_{n \to \infty} x_n\cdot \sin (2\pi\cdot C n! ) $ exists and $ L \in (0,\infty) $ ?

[i]It would be very nice !! [/i]
Regards, Alex

[quote="flip2004"]

 If $ C:=\lim\limits_{n\to \infty}
(\sum\limits_{k=1}^n\frac{1}{k}- \ln{n} ) $, does  there exists a positive sequence  $ (x_n ) $ with $
\lim\limits_{n\to \infty}x_n =\infty , $  such that  $\displaystyle L:=\lim\limits_{n \to \infty} x_n\cdot \sin (2\pi\cdot C n! ) $ exists and $ L \in (0,\infty) $ ?

Regards, Alex[/quote]

If such $x_n$ exist with the require condition then
$sin(2\pi.C.n!)$ would be equivalent to $\frac{L}{x_n}$ =>
$sin(2\pi.C.n!)$  has limit $0$ when n tends to +oo

Use $\sum_{k=1}^n\frac{1}{k} = ln(n) + C +1/(2n) - 1/(12n^2) +1/(120n^4) +
o(1/n^4)$ or maybe more term in this expansion, study the comportment of
$sin(2\pi.C.n!)$

If   such a sequence $ (x_n) $ exists, then as a corollary it follows that  C=irrational . After my informations, this is'nt yet known.
            Regards, Alex


logarithms
function
inequalities
number theory
relatively prime

Define $ p_n$ as nth prime.

Show that $ p_{n \plus{} 1}^2 < p_1p_2 \cdots p_n$.

Thanks.

Sorry... my mistake that I didn't type the square..

[quote="bjh7790"]Define $ p_n$ as nth prime.

Show that $ p_{n \plus{} 1} < p_1p_2 \cdots p_n$.

Thanks.[/quote]

We have $ p_1p_2 \cdots p_n\minus{}1$ not divisible by $ p_1,p_2,..., p_n$, so there must be a larger prime which is smaller than $ p_1p_2 \cdots p_n$ that divides it, done.

By Bertrand's Postulate $ p_{n\plus{}1}<2p_n<4p_{n\minus{}1}$. So $ p_{n\plus{}1}^2<8p_np_{n\minus{}1}$. If $ n\ge5$, then $ p_1p_2\ldots p_n\ge 2\cdot3\cdot5\cdots p_{n\minus{}1}p_n>8p_np_{n\minus{}1}>p_{n\plus{}1}^2$. We can check the cases $ n\equal{}1,2,3,4$ easily.

[quote="bjh7790"]Is there a sol not using bertrand's postulate?(Due to my stupidness)[/quote]

So you can see Bertrand's Postulate in my solution? :rotfl:

The new question is only valid for $ n \ge 4$.

Taking $ \ln$ of both sides gives the following formulation of the problem:
\[ 2\ln p_{n \plus{} 1} < \sum_{p \le p_{n \plus{} 1} \minus{} 1} \ln p \equal{} \theta(p_{n \plus{} 1} \minus{} 1)
\]
$ \theta(x) \equal{} \sum_{p \le x} \ln p$ is a known function which "behaves" like $ x$ (it is equivalent to the PNT, but weaker results that can be used in this problem, such as $ \theta(x) > cx$ for a constant $ c$, are elementary), so for $ n$ big enough, $ \theta(p_{n \plus{} 1} \minus{} 1)$ is much bigger that $ 2\ln p_{n \plus{} 1}$ ($ x$ is faster that $ 2\ln (x \plus{} 1)$)

You can probably just take the ideas of the proof of $ \theta(x) > cx$ to solve that problem.

Similarly, for any given $ k > 0$, $ p^k_{n \plus{} 1} < p_1p_2\cdots p_n$ holds for almost any $ n$, and $ p^n_{n \plus{} 1} > p_1p_2\cdots p_n$ also holds for almost any $ n$.

For completion's sake, here's sketch of proof to $ \theta(x) > cx$:
1. $ \binom{2n}{n} \ge 2^n$ because $ \binom{2n}{n}$ is the largest binomial coefficient in the expansion of $ (1 \plus{} 1)^{2n}$, so $ \binom{2n}{n} \ge \frac {4^n}{2n \plus{} 1} \ge 2^n$ for $ n \ge 3$, for $ n \equal{} 1,2$ it also holds.
2. $ p^{l_p} \parallel{} \binom{2n}{n} \implies p^{l_p} \le 2n$ (nice exercise!)
3. For $ 2n > p > \sqrt {2n}$, $ l_p \le 1$
4. Using previous lemmas,
\[ 2^n \le \binom{2n}{n} \implies n\ln 2 \le \sum_{p < 2n} l_p \ln p \equal{} \sum_{p < \sqrt {2n}} \ln p^{l_p } \plus{} \sum_{\sqrt {2n} \le p < 2n} l_p \ln p \le \sqrt {2n} \ln 2n \plus{} \theta(2n)
\]
Basically that's it.
(based on the proof in "Modern Introduction to Classical Number Theory", a book recommended by t0rajir0u a while ago...)

[quote="Johan Gunardi"][hide="A proof from Number Theory: Structures, Examples, and Problems"]Let us define $ A_k\equal{}p_1p_2\ldots p_k$ and $ a_k\equal{}kA_{n\minus{}1}\minus{}p_n$ for $ 1\le k\le p_n\minus{}1$. Observe that these numbers are relatively prime. Indeed, a prime common divisor of $ a_{k_1}$ and $ a_{k_2}$ would divide $ (k_1\minus{}k_2)A_{n\minus{}1}$ and since $ gcd(a_{k_1},p_n)\equal{}1$, this divisor would be $ p_1,\ldots,p_{n\minus{}1}$, which is clearly impossible. Of course, this implies that $ a_k\ge p_{n\plus{}k}$ (since $ a_k$ is relatively prime with $ p_1,\ldots,p_{n\minus{}1}$). Thus for $ k\equal{}p_n\minus{}1$ we have $ A_n\minus{}A_{n\minus{}1}\minus{}p_n>p_{p_n\plus{}n\minus{}1}$ and so $ p_1p_2\ldots p_n>p_{p_n\plus{}n\minus{}1}>p_{3n\minus{}1}$ for $ n\ge5$. From here we find that for $ n\ge 6$ we have $ p_1\ldots p_n>\left(p_1\ldots p_{\left[\frac{n}2\right]}\right)^2>p_{3\left[\frac{n}2\right]\minus{}1}^2>p_{n\plus{}1}^2$. For $ n\equal{}5$ one can easily check the inequality.[/hide][/quote]
can you give more details about this proof, i really don't understand why $ a_k\ge p_{n\plus{}k}$ and $ p_1\ldots p_n>\left(p_1\ldots p_{\left[\frac{n}2\right]}\right)^2>p_{3\left[\frac{n}2\right]\minus{}1}^2>p_{n\plus{}1}^2$ ? sorry for my silly question :D

[quote="Johan Gunardi"] Of course, this implies that $ a_k\ge p_{n\plus{}k}$ (since $ a_k$ is relatively prime with $ p_1,\ldots,p_{n\minus{}1}$).[/quote]
Why is this true?  I don't understand this part of the proof.


If the measure of angle $ C$ is double the measure of angle $ B$, what is the measure of angle $ A$ in triangle $ ABC$?
[metapost=554497509a333489b40b7b7318d8e511]save u, A,B,C; pair A,B,C;
 vardef mm_anglelabel(expr lbl, A, B, C, r) =
  save theta,tn;
  tn := turningnumber(A--B--C--cycle);
  theta := angle (tn*(unitvector(A-B)+unitvector(C-B)));
  label(lbl,B+r*dir(theta))
enddef;
 % Set the value of u to scale the entire image
 u:=1cm;

 B:=origin;
 A:=3u*left;
 C:=4.5u*dir(159);

 draw A--B--C--cycle;

 mm_anglelabel(btex $ 21^{\\circ}$ etex,A,B,C,40);

 label.llft(btex $ A$ etex,A);
 label.rt(btex $ B$ etex,B);
 label.urt(btex $ C$ etex,C);
[/metapost]

[hide="Solution based off of question in FTW"]
We have that angle $B$ is $21$ degrees and we are given the information that angle $C$ is double that so it is $42$ degrees. The sum of the angles in a triangle are 180 and we are looking for measure of angle A. Thus the answer is $180-(21+42)=\boxed{127}$[/hide]

[hide=According to FTW]Angle $B$ is $21$ degrees and $C$ is double, so that is 42 degrees. The sum of the angles in a triangle are 180.$180-21-42=\boxed{117}$[/hide]


Prove that if m>3, then the smallest index of a nonzero Fibonacci number divisible by m is at most m^2-4m+6 if m is odd, and m^2-4m+7 if m is even.

We can prove a very stronger result:

For every $ n \in \mathbb{N}$ there exist $ k \le 2n$ such that $ n \mid f_k$ (where $ \{f_i\}_{i \in \mathbb{N}}$ is the fibonacci sequence).
I have posted a proof [url=http://www.scienzematematiche.it/forum/viewtopic.php?f=12&t=898]here[/url]  using extension of finite rings (i'm jordan in that forum, sorry but it is in italian and it will be a bit long to traslate all).

However if we make exception on a few cases then the bounds of the previous statement holds also with $ \frac{3}{2}n$: it easily implies that the period of this sequence modulo $ n$ is at most $ 6n$ and seems that it cannot be improved..( some people said me that is a well-known result but i have never seen a proof on the web  :huh: )


I think I should have been more clear: k is an integer.

One thing to keep in mind: just because there are 21 points inside doesn't mean the area is 21!

[quote="baughmanhater"]ok, this is what i got and how i got it....feel free to correct me if it is wrong...
(i drew a picture of a triangle in standard position with the given vertices)

21=1/2 bh
42=bh , since you don't know b or h but they have to be the same b/c they are both k i substituted k in the equation for b and h.
42=k *k ... therefore 42=k 2
so k=  :sqrt:  42...

is this right or anywhere close to the right answer?[/quote]

Also, 21 is not the area. 21 is the number of dots totally (?) within the triangle.

A little further in. Is there any restrictions for the number of dots totally within a triange with vertices (0,0) (k,0) (k,k) ?

ok, thanx...i tried...hehe.....so, how would you go about solving it then? if you don't wanna tell everyone you can email it to me...i am most interested now...

Remember, we're not including the points that the edges go through.

meh! ok, well, i'll work on it some more...i don't have any more genius ideas right now....hopefully someone else will solve it and help me out...*wink wink*....

 :?  :maybe:  ;)

[hide]I got 7

A traingle with the vertices (0, 0) (k, k) and (k, 0) is a right triangle.  In a right triangle there is one point at the top, 2 points in the next row, and so on.  If there are 21 dots in the triangle, there are 6 dots at the bottom row.  You add  2 to 6 to get the number of integer x values the triangle crosses, and subtract 1 becuase of 0 to get 7.[/hide]

[hide] Just find the nth triangular number that is 21. n=6. Since it is the interior of the triangle, we must add 2, because of the top and bottom. Therefore, the answer is 6+2=8.[/hide]

[hide]k=8 because when it equals 1 or 2, the answer is 0.
When it equals 3, there's only one interior dot.  At four, there's three.  Formula is normally n(n+1)/2, but here we need to revise it to make it work for this case.  (n-2)(n-1)/2 is the formula we need and we get n or k in this case equaling 8 for the answer to be 21.[/hide]

k could be any counting number. The problem should be stated more clear by saying that 21 points are lattice points.

Similar to a question on warmup 6 or 7 for this year...

[hide][/hide]

Similar to a question on warmup 6 or 7 for this year...

They just had 15 instead of 21 and N instead of K.

[hide]If you notice, you are really only dealing with a trianglular number.   So, n(n+1)/2=21; n=6.  Then you need to go two more to expand 2 more colums to allow there to be 21 in the center.  6+2=8[/hide]

For those who don't understand, I have a PERFECT picture to show you.

[quote="rrusczyk"]What is k if there are exactly 21 points in the interior of the triangle with vertices (0,0); (k,k); and (k,0)?[/quote]
[hide="my work"]This has something to do with Pick's Formula.
So 21+B/2-1
20+B/2
No...Even Better!
It's just triangular numbers!
21 is the 6th triangular number.
But that's just the interionr, so it's 6+2=8
k=8
>>8<<[/hide]

well:
[hide]we see that in an isosceles right triangle the number of latice pts. inside is just:
(n-2)(n-1)/2 where n is the side lenght (look at the above diagram to see what I mean)
so:
(k-2)(k-1) = 42
k^2-3k -40 = 0
(k-8)(k+5)=0
k=8 or -5 (both are solutions since there was no specification about whether k had to be +ve)[/hide]

[hide="Click for solution"]These are triangular numbers because we have a 45-45-90 triangle. 21 is the sixth triangular number. 6+2 (for the two points on each side) is 8.[/hide]

As post #3 says, $k$ is an integer. Assuming that points means lattice points.

Case 1: $k$ is positive.
Note that the area is $\frac{k^2}{2}$.

The number of boundary points is $(k+1)+(k+1)+(k+1)-3=3k$.

So the area by Picks Theorem is also $20+\frac{3k}{2}$.

Thus, we get $\frac{k^2}{2}-\frac{3k}{2}-20=0\implies k^2-3k-40=0$.

So $(k-8)(k+5)=0\implies \boxed{k=8}$.

Case 2: $k$ is negative.
Then $-k$ must also be a solution. So $\boxed{k=-8}$.

HOLY COW the bump!!

anyways,
nice solution @bove! Didn't think that way before :)

[quote=gandalf61]An alternative approach - if you know Pick's theorem you can use it to show that k must satisfy

[hide]k^2 = 3k + 40[/hide][/quote]

wait what's pick's theorem again? i vaguely remember it but forgot how it works

[quote=Jwenslawski]Hey its richard rusczyk.[/quote]

thats why i bumped it


$ n!+1=(m!-1)^{2}$.  Find all pairs $ (m,n)$ that satisfy this.

Obviously $ m! \cdot \left(m!-2\right) = n!$ and $ n > m$. So write $ n = m+a$, then the equation reduces to \[ (m+a)\cdots(m+1) = m!-2\] Now look at the equation modulo $ 3$.

The right hand side will not be divisible by $ 3$ when $ m \geq 3$.

What information does this give you on the value of $ a$?


√3+(√3+(√3+.....=x

pls help me find the value of x. dun hav any idea.
thx in advance   :)

Do you mean ${ \sqrt{3+\sqrt{3+\sqrt{3}+\cdots}}}=x$?

[hide="Solution"]In that case, you square both sides of the equation to obtain ${ 3+ \sqrt{3+\sqrt{3+\sqrt{3}+\cdots}}}=x^2\implies{3+x=x^2}$(do you see why?). Solving the quadratic gives $ \boxed{\frac{1+\sqrt{13}}{2}}$ (we only take the positive solution.)[/hide]

Is this the only way to solve these type of questions? I have done this way many times but was wondering whether there are other ways to solve it.

-rts2007

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Ummmmmmm....???

This is has nothing to do with this forum. Please stop fooling around.

-rts2007

[quote="rd5493"]I don't quite understand where you got that answer from, could you please explain it again?[/quote]

$ \sqrt{3\plus{}\sqrt{3\plus{}...}}\equal{}x$

$ 3\plus{}\sqrt{3\plus{}...}\equal{}x^2$
$ x^2\minus{}3\equal{}\sqrt{3\plus{}...}$

But we defined $ \sqrt{3\plus{}...}\equal{}x$, so we substitute to get $ x^2\minus{}3\equal{}x$.

[quote="grn_trtle"][quote="rd5493"]I don't quite understand where you got that answer from, could you please explain it again?[/quote]

$ \sqrt {3 \plus{} \sqrt {3 \plus{} ...}} \equal{} x$

$ 3 \plus{} \sqrt {3 \plus{} ...} \equal{} x^2$
$ x^2 \minus{} 3 \equal{} \sqrt {3 \plus{} ...}$

But we defined $ \sqrt {3 \plus{} ...} \equal{} x$, so we substitute to get $ x^2 \minus{} 3 \equal{} x$.[/quote]

ok, thank you for the explanation, now i understand.

so if you had a number like 20 or 42 which is the product of 2 consecutive positive integers, could you result in an integer answer to this type of problem?

Let's say we have some integer n such that
sqrt(n+sqrt(n+...)) = x

Then...
n+x = x^2

so x = (-1 + sqrt(1 + 4n))/2
by the quadratic equation.

In order to make x integer, we must make the right side integer.
2x + 1 = sqrt(1+4n)
which are the odd numbers.
So x is integer if sqrt(1+4n) is odd...

There's a trick with these kind of problems
Sometimes they give you like sqrt(n+sqrt(n+...))=x
Try to find two consecutive numbers that multiply to n, and the largest number will be the answer
if it's sqrt(n-sqrt(n-...)=x then it would be the smaller consecutive number that multiply to n.

Obviously, it doesn't work with most numbers o.O


inequalities
function
integration
trigonometry
calculus
calculus computations

two different problemos:

1. Let $\phi(x)$ be a two times differentiable, non-zero function that is continuous in interval $\left[a,b\right]$ and there exists constant number $m>0$ such that for every $t\in\left[a,b\right]$ we have: $\phi^{'}(x)\geq m$. Prove that:

$\left|\int^{b}_{a}\sin \phi(t)dt\right|\leq \frac{4}{m}$

2. Prove that:

$\left|\int^{\sqrt{3}}_{1}\frac{e^{-x}\sin x}{x^{2}+1}\right|\leq\frac{\pi}{12e}$

Let $\psi$ be the inverse function for $\phi$, and make the change of variable $u=\phi(t)$:
$\int_{a}^{b}\sin\phi(t)dt=\int_{\phi(a)}^{\phi(b)}\sin u\, \psi'(u)du$
All we know about $\psi'$ is that $0<\psi'\le 1/m$.  What if $\psi'(u)$ is small when $\sin u<0$, and is $1/m$ otherwise? :huh:

Oh sorry, I had to say that $\phi^{''}(x)$ is continuous and non-zero in interval $[a,b]$.

$ 2)$ is really easy, we have $ 1\leq x\leq\sqrt{3}$ and $ \minus{}\sqrt{3}\leq\minus{}x\leq\minus{}1$ so $ e^{x}\leq e^{\minus{}1}$, now since $ \sin x\leq1$ :

$ \frac{e^{\minus{}x}\sin x}{x^{2}\plus{}1}\leq\frac{e^{\minus{}1}}{x^{2}\plus{}1}$

and

$ \left|\int^{\sqrt{3}}_{1}\frac{e^{\minus{}x}\sin x}{x^{2}\plus{}1}\right|\leq\int^{\sqrt{3}}_{1}\frac{e^{\minus{}1}}{x^{2}\plus{}1}\equal{} e^{\minus{}1}\left[\arctan x\right]^{\sqrt{3}}_{1}\equal{}\frac{\pi}{12e}$


Let abc=1,a,b,c >0 Prove that:
$ \sum {\frac{{a \plus{} b}}{{a^4  \plus{} b^4  \plus{} 2abc^2 }}}  \le \frac{3}{2}$ :)  :ninja:

[quote="TRAN THAI HUNG"]Let abc=1,a,b,c >0 Prove that:
$ \sum {\frac {{a \plus{} b}}{{a^4 \plus{} b^4 \plus{} 2abc^2 }}} \le \frac {3}{2}$ :)  :ninja:[/quote]
After $ a^4\plus{}b^4\geq(a^2\plus{}b^2)ab$ it's obvious.

[quote="arqady"][quote="TRAN THAI HUNG"]Let abc=1,a,b,c >0 Prove that:
$ \sum {\frac {{a \plus{} b}}{{a^4 \plus{} b^4 \plus{} 2abc^2 }}} \le \frac {3}{2}$ :)  :ninja:[/quote]
After $ a^4 \plus{} b^4\geq(a^2 \plus{} b^2)ab$ it's obvious.[/quote]
It's not really obvious.I cannot prove the inequality directly after using your method. :) Can you give more detail?

Arqady means that after using Am-Gm and abc=1, you'll have to prove:
$ \sum\ \frac{c(a\plus{}b)}{a^2\plus{}b^2\plus{}2c^2} \leq\ \frac{3}{2}$
Obviously trues by Am-Gm because:
$ \frac{c(a\plus{}b)}{a^2\plus{}b^2\plus{}2c^2} \leq\ \frac{1}{2}$

[quote="nguoivn"]Arqady means that after using Am-Gm and abc=1, you'll have to prove:
$ \sum\ \frac {c(a \plus{} b)}{a^2 \plus{} b^2 \plus{} 2c^2} \leq\ \frac {3}{2}$
Obviously trues by Am-Gm because:
$ \frac {c(a \plus{} b)}{a^2 \plus{} b^2 \plus{} 2c^2} \leq\ \frac {1}{2}$[/quote]
Yea, I see :lol:


Is there an algebraic/elegant method for number 1?

#1
What is the maximum number of acute interior angles a convex pentagon can have?

#2
Kendra starts counting with a and counts by d , where a and d are both positive
integers. For example, for a = 5 and d = 3 the sequence would be 5, 8, 11, …. The
sum of two terms in Kendra’s sequence is 10. How many different possible pairs
(a , d ) are there?

#1
The sum of the angles of a convex pentagon is $ 540\equal{}90\cdot 6$. Hence, it can have at most $ 3$ acute angles, since all acute angles are less than $ 90$ degrees, so the answer is $ \boxed{3}$.
#2
Suppose the two terms are $ a\plus{}bd$ and $ a\plus{}cd$. Then the sum of the two terms is $ 2a\plus{}(b\plus{}c)d\equal{}10$.
Case 1: $ (b\plus{}c)d\equal{}2$. There is $ 1$ way.
Case 2: $ (b\plus{}c)d\equal{}4$. There are $ 2$ ways.
Case 3: $ (b\plus{}c)d\equal{}6$. There are $ 3$ ways.
Case 4: $ (b\plus{}c)d\equal{}8$. There are $ 3$ ways.
In total, there are $ \boxed{9}$ ways.

[hide="another way for 1"]
Consider the exterior angles of the pentagon
[/hide]

[hide="2"]
Casework on $ a$:

Case 1: $ a\equal{}1$
Note that every sequence for which the sum of two terms is $ 10$ contains $ 9$.  Therefore the common difference $ d$ must divide $ 9\minus{}1\equal{}8$.  There are $ 4$ values of $ d$ that work here.

Case 2: $ a\equal{}2$
Similarly, every sequence we want contains $ 8$, so $ d$ divides $ 6$.  There are, again, $ 4$ values in this case.

Case 3: $ a\equal{}3$
There are $ 3$ divisors of $ 7\minus{}3$, so there are $ 3$ pairs $ (a,d)$ for this case.

Case 4: $ a\equal{}4$
Clearly there are only $ 2$ pairs for this sequence.

$ 4\plus{}4\plus{}3\plus{}2\equal{}\boxed{13}$

I'm pretty sure I only could have undercounted...
[/hide]

Just to make sure I'm understanding the problems, I will change the number on problem 2 from having a sum of 10, to having a sum of 14.

I did the casework and got 23 ways. Can someone check to see if thats right?

Also for #1 What if we changed the problem from a pentagon to a 37-gon?

[hide="using exterior angles"] The sum of the exterior angles of any polygon is always 360 degrees. For every acute interior angle we have, we have an obtuse exterior angle. There cannot be four or more, otherwise the sum of the exterior angles is more than 4(90)=360 degrees. Thus, we can have at most 3, for any polygon.[/hide]

[hide="a related question"] Find the maximum number of right angles a polygon can have[/hide].

[quote="xpmath"]Find the maximum number of right angles a polygon can have.[/quote]

Can someone not answer this question?  It's on my Chapter Sprint. :)

[quote="xpmath"]Bleh it was generalized off of 2008 state #3.[/quote]

Well... yeah, but I took it first. :P

[quote="mathemagician1729"]Either I'm being really stupid or to answer xp's question the maximum number of right angles is 3 regardless of the polygon. That seems a little weird though.[/quote]

Doesn't a rectangle and square have $ 4$ right angles though?

It is 3 for n-gons such that n>4. :wink:


2004 Warmup 12
#1 In a game with two standard six-sided dice, what is the probability that when both dice are rolled you get a six on at least one die or that the sum of both dice is 6?

Assuming you're asking for help :

[hide]probability of at least one 6 is 1/6 + 1/6 - 1/36 = 11/36  The probability thatthe sum is 6 is 5/36(this I just know since I know P(7) = 6/36 and it goes down on either side of 7)

So 11/36 + 5/36 = 16/36= 4/9
At least I hope so :) [/hide]

I guess you subract 1/36 because of 6,6.

[hide]When you have a probability problem that has an "or" situation, like the first die is a 6 "or" the second die is a 6, you add the two probabilities.  However if the two options could both happen at the same time such as both dice being 6s, you need to subtract the probability that both are 6s because you will otherwise double count that part.

the probability that both are 6 is 1/6 * 1/6 = 1/36[/hide]

[hide]THe answer is 4/9.
5/36 chance of getting a sum of six.
1/6 chance of getting one six on one die.
But we overcount 1/36 because of 6,6.
1/6+1/6+(5-1)/36=4/9.[/hide]
I did this problem earlier today.

Since you used the question mark i'm not going to hide the content:
There are 10 combinations of rolls with one five (6,1; 6.2; 6.3; 6.4; 6.5; 1;6; 2;6; 3;6; 4;6; 5,6), and there also is 6,6, which gives you 11 combinations with a six.
Then you have 5 combinations that add up to six (you must have done something like this at school, otherwise i just remember that there are six for seven, and just count down from there). 5+11/36 = 16/36 = 4/9


factorial
integration
logarithms
inequalities unsolved
inequalities

$\iff \Gamma (nx_{1}) \Gamma(nx_{2}) \cdots \Gamma(nx_{n}) \ge \Gamma(x_{1}+\cdots+x_{n})^{n}$.

When $\Gamma (x) = \int_{0}^{\infty}e^{-t}t^{x-1}dt$.

ETS : $(\log \Gamma (x))'' =\Psi_{1}(x) >0$ For all $x>0$.

But $\Psi_{1}(x) = \sum_{k=0}^{\infty}\frac{1}{(x+k)^{2}}$  (well-known) , it is obvious.


Update from Rudradev by email about India's [b]tentative performance[/b] at the IMO:

[b]Tentative scores: [/b]

1. Anirudh Gurjale : 24-28
2. Utkarsh Tripathi : 21-26
3. Prashant Sohani : 20-25
4. Rudradev Basak : 18-23
5. Ankit Baraskar : 16-21
6. Abhinav Garg : 16-21

If these do indeed turn out to be the scores, it'll be a great performance and a huge improvement from last year!

Ashwath

The results are out at : http://www.imo-official.org/year_individual_r.aspx?year=2008

We got 5 bronzes and one HM. Solid performance overall, a bit lower than the predicted total though. Abhinav missed the bronze by one point - hard luck..
Nice work guys!

Ashwath


FOr how many integers n does the cube root of n differ from the cube root of 1728 by less than 1?

[hide="Click to reveal hidden content"]There are $ 47$ integers between $ 122$ and $ 168$ inclusive.[/hide]

(Clicking on blue text does nothing on my computer so I hope this works.)

I think that is incorrect...
[hide="Solution"]
We can find that $ \sqrt [3]{1728} \equal{} 12$.  Thus, $ 12 \minus{} 1 \equal{} 11 < \sqrt [3]{n} < 13 \equal{} 12 \plus{} 1$.  Taking the cube of the inequality yields $ 1331 < n < 2197$.  Thus, $ n$ is an integer from $ 1332$ to $ 2196$, inclusive, which is $ \boxed{865}$ different integers.[/hide]


invariant
linear algebra
matrix
absolute value
algebra unsolved
algebra

We consider the matrices:
\[ A= \left( \begin{matrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 & 1 \\ 1 & -1 & -1 & -1 & 1 \\ 1 & 1 & -1 & 1 & -1 \\ \end{matrix} \right) \]

\[ B= \left( \begin{matrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 & -1 \\ 1 & 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 & 1 \\ 1 & -1 & 1 & -1 & 1 \\ \end{matrix} \right) \]

[b]We call elementary operations the following[/b]:

[i]a.)[/i] the change of sign for all elements of a line;
[i]b.) [/i] the change of sign for all elements of a column;
[i]c.)[/i] we can permute 2 lines;
[i]d.)[/i] we can permute 2 columns;

Prove that $A$ cannot be transformed into $B$ using a finite number of elementary operations.

[quote="harazi"]The absolute value of the determinant is conserved.  ;)[/quote]

Gabriel, what about a full solution?  :)

Orl, do you want me to compute the determinants? Or to prove that the absolute value of the determinant is conserved? Don't you think this is a little bit obvious?

well, you don't need to compute the determinants, but I think you do need to prove they're different? (I miss a trivial argument why this would be the case)

One determinant equals 16, the other equals 0.

As prof. Cuculescu wrote in his book "Olimpiadele Internationale de Matematica ale Elevilor" (Romanian :)), this problem was immediately discarded from the list because of this simple argument and the jury could not produce an example of two such matrices having the same absolute value of the determinants and such that one cannot transform one into the other using a finite number of elementary operations.

[quote="enescu"]the jury could not produce an example of two such matrices having the same absolute value of the determinants and such that one cannot transform one into the other using a finite number of elementary operations.[/quote]

Is that impossible to do?

Let \[ I(A)=\sum_{i,j} c(i,j)(i+j)(mod \ 5),  \] were c(i,j)=1 if a(i,j)=-1, else c(i,j)=0. elementary operators not change I(A). We have $I(A)\not =I(B)$.


Let $ \Omega_1$, $ \Omega_2$ be open convex sets in $ \mathbb{R}^n$, and let $ f_1 \in C^{\omega}(\Omega_1)$, $ f_2 \in C^{\omega}(\Omega_2)$.
Show that $ f_1(x)\equal{}f_2(x)$ for all $ x \in \Omega_1 \cap \Omega_2$, if $ D^{\alpha} f_1(y) \equal{} D^{\alpha} f_2(y)$ for all $ \alpha$ at a single point y in $ \Omega_1 \cap \Omega_2$.
Show by an example that this may not be true if the $ \Omega_k$ are just connected instead of convex.

[quote="lelouch"]Let $ \Omega_1$, $ \Omega_2$ be open convex sets in $ \mathbb{R}^n$, and let $ f_1 \in C^{\omega}(\Omega_1)$, $ f_2 \in C^{\omega}(\Omega_2)$.
Show that $ f_1(x) \equal{} f_2(x)$ for all $ x \in \Omega_1 \cap \Omega_2$, if $ D^{\alpha} f_1(y) \equal{} D^{\alpha} f_2(y)$ for all $ \alpha$ at a single point y in $ \Omega_1 \cap \Omega_2$.
Show by an example that this may not be true if the $ \Omega_k$ are just connected instead of convex.[/quote]

For the first part, $ \Omega_1 \cap \Omega_2$ is convex, hence connected, so it follows from the standard uniqueness theorem. For the second part, you want to think about the case where $ \Omega_1 \cap \Omega_2$ is not connected. Here you might try different versions of Arg z as you travel from 1 to - 1 clockwise versus counterclockwise.


$ |\log_x y \plus{} \log_y x|\geq 2$ is correct, because $ \log_x y$ is not always positive.

[quote="Adriana N."]If $ log_{x}{y} \plus{} log_{y}{x} \equal{} 2$  and  $ x^n \plus{} y \equal{} 12$ ($ n\in N$) ,then determin n such that there are $ (x,y) \in Z^2$,as a solution.

P.S. It's correct if I write $ log_{x}{y} \plus{} log_{y}{x}\ge2$,for every $ x,y\in R$?If not,please say me why.Thank you![/quote]

let $ log_{x}{y} \equal{}t$
Now $ t \plus{}\frac {1}{t}\equal{}2\implies t \equal{}1\implies x\equal{} y$
$ \implies x^n \plus{} x \equal{} 12 \implies x\ge 2$
for $ n\ge 4, x^n \plus{} x\ge 18\implies$ no solution in integers
Also for $ n\equal{}3$ no solution in integers
For $ n\equal{}2, x\equal{} 3,\minus{}4$
For $ n\equal{}1, x\equal{}6$

Possible $ n\equal{} 1,2$


In a convex hexagon $ABCDEF$, $AD=BC+EF$, $BE=AF+CD$, $CF=AB+DE$. Prove that $\frac{AB}{DE}=\frac{CD}{AF}=\frac{EF}{BC}$.

$|\vec AD|=|\vec BC|+|\vec EF| \geq |\vec BC+\vec FE|= |\vec BE+\vec FC| \Rightarrow (\vec AD)^{2}\geq (\vec EB+\vec CF)^{2}$. (*)

By emulating the process above,we can also have:

$(\vec EB)^{2}\geq (\vec AD+\vec CF)^{2}$ (**) and $(\vec CF)^{2}\geq (\vec EB+\vec AD)^{2}$(***).

Add the three inequations  (*),(**) and(***) up,and then we have:

$(\vec AD+\vec CF+\vec EB)^{2}\leq 0$,therefore  $\vec AD+\vec CF+\vec EB =0$, and so the three inequations must become equations,so$BC//AD//FE,AB//CF//DE,AF//BE//CD$!

Finally, we only need prove that $AD,BE,CF$ meet at the same point,and this very obvious!


What is the sum of all 4-digit numbers that are the same when their digits are reversed?

All $ 4$-digit numbers that are the same when the digits are reversed are palindromes. There are $ 9\cdot10$ $ 4$-digit palindromes. The average of these palindromes are $ 5\cdot10^3\plus{}4.5\cdot10^2\plus{}4.5\cdot10\plus{}5\equal{}5500$.
Since there are $ 90$ palindromes, the sum of them is $ 90\cdot5500\equal{}495000$. I hope.....  :maybe:


trigonometry
limit
calculus
integration
ratio
geometric series
calculus computations

Determine whether the series

$\sum^{\infty}_{k=1}k^{2}\sin^{2}(\frac{1}{k})$

converges.

The series diverges because $\lim_{x\to\infty}x^{2}\sin^{2}\left(\frac{1}{x}\right) =1$.

[quote="Kent Merryfield"]Now that you've seen Jimmy's (correct) explanation, here's a question for you, 236factorial:

Does $\sum_{k=1}^{\infty}k\sin^{2}\left(\frac1k\right)$ converge?[/quote]

[I hope I'm not being an idiot here...]

$\lim_{x\to\infty}x\sin^{2}\left(\frac{1}{x}\right) =0$, so this won't tell anything.

The only other test I know is the integral test, but I can not evaluate $\int^{+\infty}_{1}x\sin^{2}\left(\frac{1}{x}\right) dx$.

Use the Limit Comparison Test.
[hide]$\lim_{k\rightarrow\infty}\frac{k\sin^{2}\left(\frac1k\right)}{1/k}= \lim_{k\rightarrow\infty}k^{2}\sin^{2}\left(\frac1k\right) = 1$
Because $\sum1/k$ is divergent so is $\sum_{k=1}^{\infty}k\sin^{2}\left(\frac1k\right)$
[/hide]

It's not enough to say that $\sin x\to0$ as $x\to 0$. The question is, how fast does it tend to zero? And yes, the fact that $\lim_{x\to0}\frac{\sin x}x=1$ is very important.

You don't want to use that integral test, because $\sin^{2}\left(\frac1k\right)$ is a nasty thing to work with. But can we approximate it usefully?

As for knowing tests: this is a positive term series. A positive term series either converges or tends to infinity; we can tell which if we find an appropraite comparison.

The following are all comparison tests:

* Simple (one-sided) comparison, using inequalities.

* Limit comparison

* The integral test (compares to an integral)

* The ratio test (compares to a geometric series)

* The root test (also compares to a geometric series)

* The condensation test (not often taught in first calculus courses, but it has its uses)

For this problem: the ratio and root tests are out - nothing his shows any promise of reasonable comparison to the geometric. The integral test stalls out with an intractable integral.

The workhorse here - the "all-purpose de-clutterer" - is the limit comparison test. You've said that the terms go to zero; so tell me how fast they tend to zero.

[OK, kahsi beat me to the punch while I was composing this.]

could you please explain what the condensation test is, since I have never heard of it.

[quote="Kent Merryfield"]You don't want to use that integral test, because $\sin^{2}\left(\frac1k\right)$ is a nasty thing to work with. But can we approximate it usefully?
[/quote]

Do you mean approximating by Simpson's rule?

No, certainly not any method of numerical integration, and in fact, not the integral test at all. I really meant to approximate it the way kahsi did.


The arithmetic mean of four numbers is 12. When the largest of the numbers is removed, the mean of the remaining three numbers is 9.  What number was removed?

The sum of the $ 4$ numbers is $ 4\times12\equal{}48.$ The sum of the three numbers after the largest has been taken out, is $ 3\times9\equal{}27.$ Taking the difference gives the largest term, which is $ 48\minus{}27\equal{}\boxed{21}.$

Different way: After a number is taken out, the mean is 9. 9 is 3 less than 12 and since there are 3 numbers, $3\cdot3$ =9. In order to even the numbers out, we add 9 to 12 which gives us 21.


E spends his vacation in the Swiss Alps. Unfortunately, he ends up  spending the greater part of  his visit in the hospital, after having taken a nasty spill while skiing. On his return trip home to the United States the authorities do not ask him to show his passport.
Why not?

[hide="Because..."] He didn't survive.[/hide]

[hide]
He was already dead, and people were sending him back to the US to be buried.
[/hide]


inequalities
geometry
circumcircle
inequalities unsolved

[quote="Thaakisfox"]prove that for every triangle:
[b]2R(sa+sb+sc)>=a^2+b^2+c^2[/b]
where R is radius of outer circle sa, sb, sc are the medians, a,b,c are the sides of the triangle.[/quote]

In http://www.mathlinks.ro/Forum/viewtopic.php?t=16023 it was proven that $s_{a}\geq\frac{b^{2}+c^{2}}{4R}$, so that $2Rs_{a}\geq\frac{b^{2}+c^{2}}{2}$. Similarly, $2Rs_{b}\geq\frac{c^{2}+a^{2}}{2}$ and $2Rs_{c}\geq\frac{a^{2}+b^{2}}{2}$. Summing up these three inequalities, we get $2Rs_{a}+2Rs_{b}+2Rs_{c}\geq\frac{b^{2}+c^{2}}{2}+\frac{c^{2}+a^{2}}{2}+\frac{a^{2}+b^{2}}{2}$, what simplifies to $2R\left(s_{a}+s_{b}+s_{c}\right)\geq a^{2}+b^{2}+c^{2}$.

  darij

my solution:

let r the tangent to the circuncircle of ABC at A, and M the midpoint of BC.
let B' and C' and M' the feets of perpencicular lines to r trought B, C and M.
we have that $s_{a}= AM \leq MM' = \frac{(BB'+CC')}{2}$
but $BB' = c . sinC = \frac{c^{2}}{2R}$ and the same to CC'.


The formula $ F \equal{} \frac{9}{5} C \plus{} 32$ can be used to convert temperatures
between degrees Fahrenheit ($ F$) and degrees Celsius ($ C$). How many
degrees are in the Celsius equivalent of $ \minus{}22^\circ F$?

You simply have to plug in the numbers into the formula.

$ \minus{} 22 \equal{} \frac {9}{5} C \plus{} 32$

Then, isolate the C.

$ F \minus{} 32 \equal{} \frac {9}{5} C$

$ \frac {F \minus{} 32}{\frac {9}{5}} \equal{} C$

Invert and multiply.

$ C \equal{} \frac {5}{9} \cdot F \minus{} 32$

Then, plug in the numbers.

$ C \equal{} \frac {5}{9} \cdot \minus{} 22 \minus{} 32$

$ C \equal{} \frac {5}{9} \cdot \minus{} 54$

$ C \equal{} \minus{} 30$

Or you can just manipulate the equation to get $ C \equal{} \frac{5}{9}F \minus{} 32$, and plug in numbers to get $ \boxed{\minus{}30}$.

[quote="myyellowducky82"]no offense but i THINK thats wat ernie did.[/quote]

Yeah, I did exactly what izzy did, only I showed my work.  :D

[quote="isabella2296"]Erm...my reading skills have deteriorated under ernie's influence...bad boy, ernie![/quote]

Sorry, izzy.  :roll:

I think you mean $ C\equal{}\frac{5}{9}(F\minus{}32)$. Can't forget the parentheses.

Maybe I'm wrong, but I thought they weren't any parentheses?

[quote="ernie"]$ \frac {F \minus{} 32}{\frac {9}{5}} \equal{} C$

Invert and multiply.

$ C \equal{} \frac {5}{9} \cdot F \minus{} 32$[/quote]
There are parenthesis. This step is wrong. Both in case of ernie and izzy. It should be $ \frac{5}{9}\cdot (F\minus{}32)$

[quote="Akashnil"][quote="ernie"]$ \frac {F \minus{} 32}{\frac {9}{5}} \equal{} C$

Invert and multiply.

$ C \equal{} \frac {5}{9} \cdot F \minus{} 32$[/quote]
There are parenthesis. This step is wrong. Both in case of ernie and izzy. It should be $ \frac {5}{9}\cdot (F \minus{} 32)$[/quote]

Ernie may have written that step wrong, but he still got the correct answer.

[quote="isabella2296"]Hey, what'd I do?  :mad:[/quote]

lol

im jk-ing

[quote="isabella2296"][quote="ernie"]
im jk-ing[/quote]

You're just-kidding-ing?  :huh:[/quote]

SHouldn't it be I'm jk because people say I'm just kidding more than they say Im just kiddinging

Yes, ducky's right. Imagine this as LOCKED. Thank you.


AMC AIME USA(J)MO USAMO search Harvard college

It's a pretty well-established fact that the gender gap increases with level of ability/aptitude/test-taking strength - that is, the average (for example) SAT I math score only differs a bit, but if you look at the top 25% guys outnumber girls by a larger factor; if you look at the top 1%, by a much larger factor; and so on.

The AIME statistics, organized by gender, are pretty sad - not only does a smaller percentage of girls taking the AMC-10 or -12 qualify, but we don't do nearly as well on the AIME, especially if you look at the top scores.  Has a girl ever gotten a perfect score on the AIME?  For that matter, a 14?  (Either way, I'm going to try to remedy this personally next year  ;) but it seems quite odd that the opportunity might still be out there to be the first girl with a perfect score...)

Please don't get mad at me (or at least remember I'm a girl, and don't call me an arrogant male chauvinist), but it seems pretty reasonable that maybe guys are just innately better at math (or reasoning or spatial skills) as opposed to simply more likely to develop an interest in it, more likely to be encouraged in it, etc.  After all, there are all sorts of other pretty weird mental gender differences that we take for granted, without anyone getting as upset as when it seems the evidence points toward a fundamental difference in average reasoning ability.  Or perhaps at the level of ability we're looking at, the larger standard deviation in "intelligence" for men is playing a role (the whole XY v. XX chromosome thing... the "extra copy" we women have helps prevent some mutations from appearing).

Anyway, I'd be interested to know whether the girls who score well on the AIME tend to, like gifted middle school girls, lose interest in math; or whether at this stage most of the ones who'll get discouraged have already stopped caring about math.  Also, does anyone know whether a girl has in fact gotten a perfect AIME score before?

Melanie Wood was the first girl to compete for the US at the IMO I believe.  I do not know her AIME score however.  Also, last year I know that Alison Miller was on  the IMO team and got a gold.  There should be a few more that I do not know about but I will agree with you that there are few girls who make it to USAMO and above.  Looking at past stats the highest score for a woman is a 13.

There have been a few discussions recently about women in math.  You can use the search feature to find them.

LOL.  I don't think there's anything wrong with noting the differences in abilities in math between boys and girls.  I think that both hereditary and cultural traits play a part.  But anyhow, I'm glad to see that you're a girl who loves/and is good at math because math is awesome :lol: Anyhow, from the AIME website, no girl every got a 15 on the AIME, but girls have gotten 150s on the AMC 12.  Maybe you'll be the first.

[quote="tokenadult"]Just be glad you aren't the president of Harvard University or you'd be pilloried for saying what you just said.  :P[/quote]

Indeed.  I didn't see the actual remarks (was that a stipulation of his agreement to speak?  I think he may have prevented reporters from quoting him...) but I didn't understand what was so offensive about his suggestion that women tend not to [b]succeed in math careers[/b] because of gender - whether or not you're willing to consider that girls may not be as good at math, success in a career is dependent on far more than ability.  Namely, for many professorships/research positions, you need to be willing to put your career above a "balanced" life.  Women tend to be more social and more interested in starting a family (and obviously even if a woman isn't going to be involved much in raising the kids, she gets stuck with whole pregnancy thing ^^).

Perhaps the articles I've read have been biased, but the counterevidence provided by so-called feminists doesn't seem to explain the phenomenon described - namely, the large gap between male and female math ability [b]at several standard deviations from the norm[/b].  Sure, girls may do better on math tests without guys in the room; the *average* scores of girls in Iceland may be better; girls may get better grades and take more math classes than guys -- but none of this really addresses the, say, 50 or so mathematicians in a generation who will make significant strides in math.

That's not to say the others aren't important - having a lot of people well-educated in, interested in, and good at math - for instance, teachers, engineers, or statisticians, who USE but do not CHANGE mathematics - is crucial to our society.  But the insults thrown at Summers for acting condescending by suggesting, to a room full of women well-versed in the research done on the differences between *average* male and female math ability, that perhaps there *is* a difference once you look at significantly non-average populations, seem to me to be entirely undeserved.  Too, he wasn't insulting the women in the room personally; he was simply saying that there weren't enough of them (that is, a belief that women are inherently worse at math doesn't imply that a specific woman is unintelligent or limited in potential.)


geometry
calculus
integration
function
real analysis
calculus computations

[img]http://img232.imageshack.us/img232/4817/1copygh7.jpg[/img]

write a riemann sum and then a definite integral representing the area of the region, using the strip shown.

this is my main difficulties on how to use the strip in this question as a sum of riemann sum. I know how to find this using the [integral] from 0 to 1 of [sqrt]x - [integral] from 0 to 1 of x but not the riemann sum

dS = (y-y^2)dy, y∈[0,1]

Is that what you want?

I am not sure cause I don't know how this works. But my instructor give me this hint:

go to x = y, x = y^2 (inverse functions) to make the slices vertical in the choice of variable (that is y)

anyone??

You either need to go read your textbook or go talk to your professor (or TF or whatever) in order to figure out what's going on.  Go learn what a Riemann sum is and how they work and what their relationship is to integrals.

I know how the Riemann Sum Works and the relationship of it to integral. The difficulties that I am facing here is that I dont know how to find the area of the thin strips. The total area of this graph will then be the area of the strips * delta y. But how to find the area of the strips?


Use nine fours to make an equation that equals 3,748,106.

when i skimmed this i read four nines. so i did 9^9 to get a big number which turned out to be 387420489.  It looked like your number but I was stuck.

[quote="-Terminator-"][hide="Answer"]$44^4+4+4+(4/4)+(4/4)=3748106 $[/hide][/quote]
very nice, i didn't even get it. :10:

[quote="-Terminator-"][hide="Answer"]$44^4+4+4+(4/4)+(4/4)=3748106 $[/hide][/quote]
How did you get that answer?? Just guessing and trying??

i don't think there is a equation or a way to do it, besides guess and check

[quote="236factorial"]He must've gotten lucky, or otherwise he has done this problem before.[/quote]
not really  :?
looking at the size of the number, you just KNOW you'll have to work with something of the form $4^{4n}$ (or another expression instead of $4n$) or $44^4$...
and when you check $44^4$, you can see that you only need 10 more units...

Thats how i found it, trying different powers. I knew it was a power so i tried 4^4, 44^44, and 44^4 which led to the answer.

it was partly luck. :blush:


If $ a$ is a nonzero integer and $ b$ is a positive number such that $ ab^{2} \equal{} \log_{10}b,$ what is the median of the set $ \{0,1,a,b,1/b\}$?

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ a \qquad \textbf{(D)}\ b \qquad \textbf{(E)}\ \frac {1}{b}$

That should be $\log_{10}b$, not $\log_{1}0b$
[hide="Solution"]
Picking the trivial case really helps here.
Let's pick $a =-100; b = 0.1$
Clearly, this satisfies the equation.
$a =-100, b = 0.1, \frac{1}{b}= 10$
Arranging them, we get
$-100, 0, 0.1, 1, 10$
Thus the median is $\boxed{b}$, or Choice D
[/hide]

Revive, but more rigorous solution:
[hide]Clearly, $ b < 10^b$ for all $ b > 0$ (Sketch a graph to convince yourself if you have to), so we have $ \log_{10} b < b$. If $ b \geq 1$, then $ \log_{10} b < b \leq b^2$, or $ a \equal{} \frac {\log_{10}b}{b^2} < 1$. This value is also clearly positive, so we would have $ 0 < a < 1$, which is a contradiction, as $ a$ is an integer. Thus, $ b < 1$, which also implies $ a < 0$. Then we have $ a < 0 < b < 1 < \frac {1}{b}$, and the answer is $ \textbf{(D)}\ b$.[/hide]

[quote="basketball9"]Another trivial case is $a=0$ and $b=1$[/quote]
That's not a valid trivial case; please re-read the problem.

So I let $b=10$ on this and solved for $a$ to get $a=\frac{1}{100}$. But isn't the median of $0,\frac{1}{100},\frac{1}{10},1,10$ just $\frac{1}{10}$ which is $\frac{1}{b}$? Where did I mess up?

^above (I know this is really late but whatever): $a$ must be an [b]integer.[/b]


given any 11 distinct real numbers on[0,1000] prove that there are at least two of them say x and y such that
$0<x-y<1+3\sqrt[3]{xy}$

This is an interesting pidgeonhole problem, unless i am mistaken...
I can't come up with any progress. Only noticed so far that $\sqrt{x}+\sqrt{y}+\sqrt{xy}\geq 3\sqrt[3]{xy}$
we also have $(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})=x-y$

Do you come up with all these problems yourself, Ashrafmod?

[hide]

Let  and , so .  The left inequality easily falls from the solution of the right inequality. The right one is nonstrict, in fact. To show this, consider the 10 regions of equal size

We can show that (excluding a singular case) there always exist some  that are distance less than  apart.

Case 1 - If one of the numbers is , then either we have the set , and in this case we have equality for the right inequality or another number is in  and we can choose that number and  or there are ten numbers in the last nine intervals and we have, by pigeonhole, at least two in the same region and thus distance less than one apart.

Case 2 - No number is zero; then we have eleven numbers for ten regions, so at least two (by pigeonhole) are in the same region, and thus have distance less than 1.

So now, let's choose these two numbers of distance less than 1; let the larger one be  and the smaller be . We have

However, since  and  is positive (since each of  is postive), then we have  and clearly  equality only when each number is  for .

[/hide]


We have a sequence that is $1, 5, 13, 25,...$.  What is the 101th number iin this sequence.  By the way, this sequence does have a pattern.  Thanks to a friend. :)

Masoud Zargar

$1=1+0$ ; $5=4+1$ ; $13=9+4$ ; $a_n=n^2+(n-1)^2$ .
 (It is the same solution as previously posted, but i'm just guessing how it has been invented)

it is just when you find the difference between terms, you find the differences to be 4,8,12 (i.e. multiples of 4)

They are the same thing.  One term in the sequence might be this:
$n^2 + (n+1)^2$
The next term would be
$(n+1)^2 + (n+2)^2$
Note that $(n+1)^2 - n^2 = 2n+1$, so the difference between the terms is $(2n+1) + (2(n+1)+1) = 4(n+1)$ (you could just foil out both terms and subtract).  Therefore the difference between consecutive pairs will always in consecutive powers of 4.

[quote="Altheman"]there are a million patterns that this could follow

i suppose it could follow the recursion $a_{n+1}-a_n = 4n$ with $a_1=1$, and that is pretty easy
it would just be $a_{n+1}=1+2(n)(n+1)$ so $a_{100+1}=1+2(100)(101)=\boxed{20201}$[/quote]

I don't know much about recursions and sequences, but I understand most of it.
However, I have one question: How do you get from $a_{n+1}-a_n = 4n$ to $a_{n+1}=1+2(n)(n+1)$?

No, I understand that part...it was just I didn't see how he manipulated it from the 4n to 1+2(n)(n+1).


One morning each member of Angela's family drank an $ 8$-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

$ \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 7$

[hide]Let $m=$ the total amount of milk and $c=$ the total amount of coffee.  We are given that $\frac{m}{4}+\frac{c}{6}=8$ or $3m+2c=96$.  Also, we know that $m+c$ is a multiple of $8$.

Consider the ordered pair $(m,n)=(0,48)$.  This is not possible because every member of the family must have some milk and some coffee.  The next ordered pair of integers is $(2,45)$ (We don't need to consider non-integers because they will not have a sum that is a multiple of $8$).  Notice that $m$ increased by $2$ and $n$ decreased by $3$.  Also notice that the sum is one less than it was before.  For the sum to be a multiple of $8$, we must increase $m$ and decrease $n$ $8$ times each, which gives the ordered pair $(16,24)$.  If we decreased $n$ $8$ more times, we would have $n=0$, which violates the restrictions of the problem, so $(16,24)$ is the only possible ordered pair.

There is a total of $40$ ounces, so there are $\frac{40}{8}=\boxed{5}$ members of the family.[/hide]

[hide]If $m$ and $c$ represents the cups of milk and coffee respectively, and n represents the number of people, then we have the equation

$\frac{m}{4}+\frac{c}{6}=\frac{m+c}{n}$.

$3m(n-4)=2c(6-n)$.

Since both 3m and 2c are positive, we have that n-4>0 and 6-n>0.
Thus n has to be 5. [/hide]

[quote="Hriundel"]Shouldn't it be [b]6-n>0[/b]? :huh:[/quote]

Oh yes  :blush:  must've got carried away  :P

[quote="myc"]what level of math is this problem, because i dont get it.  Is it the AMC 12 level?

 :lol:[/quote]

Actually, it was on the AMC 10 (medium-hard on that test) [i]and[/i] on the AMC 12 (medium-easy on that test).

[quote="xfactor"]Since she drank one sixth of the coffee and more than one sixth of  the milk, she drank more than one sixth of the mixture, so there were fewer than 6.
Since she drank one fourth of the milk and less than one fourth of the coffee, she drank less than one fourth of the mixture, so there were more than 4.
Therefore the answer is 5.[/quote]

Oh!  That's a neat little solution.  Why didn't I think of that?

I have to say, xfactor's way is pretty much as easy as solutions get to this problem. It's quite neat.

What's cool about xfactor's answer is that it means that the answer is five regardless of the size of the cup. In fact, that 8 ounces is a distractor--probably most would come to xfactor's answer faster if the cup size were omitted.


Hello!
Also, I think there is a special usergroup for MC coaches.

[quote="FEROT"]I just would like to introduce myself. My name is Ferhat and I am mathcounts coach for Chesapeak Science Point . I would like to know if there is any invitational or mini mathcounts competition prior the official competitions. especially in MD, VA, DC, PA, NJ or in any close state in east coast.

thank u[/quote]
there is the school competitions...

I haven't heard of any, but you can always go on the AoPS bookstore, get some contests, and hold them at your school.
Hope that helps.

Never heard of any... But there's math club at school that you do mini mathcount competitions... doesn't really count though


Are you going to continue to add/change the problems?
I've encountered many of the problems 2 or 3 times already.

We will keep adding problems.  There are 1700 problems in the database now, and we should have closer to 3000 in the next couple weeks.

[quote="Klebian"]how about changing (variable) numbers? At least that stops memorization somewhat[/quote]

While this seems like a simple thing to do, unfortunately, this is a very complex task, especially when it comes to word problems, which is what a majority of the problems are.  By complex task, I mean it would take several years to do it effectively, not to mention the multi-thousands of dollars it would take to do.  I am looking into doing this for more rudimentary problems though.  However, we won't be looking at this until all the other bugs are fixed.


inequalities
linear algebra
matrix
vector
function
integration
calculus

It should be:

$1\le i, j \le n$ in the sum.

I have proved that inequality is true  when A is a positive integer.

You definitely need to include the diagonal terms- the inequality is trivially false for $n=2$ otherwise.

This amounts to showing that the matrix $M$ with entries $m_{ij}=(b_i+b_j)^{-A}$ is positive semidefinite. I will show that this matrix is positive definite by expressing it as a Gramian: $m_{ij}=\langle v_i,v_j\rangle$ for some (linearly independent) vectors $v_i$ in an inner product space.

Consider the space of bounded continous functions from $\mathbb{R}^+$ to $\mathbb{R}$ which decay at least exponentially as $x\to\infty$, with inner product $\langle f,g\rangle=\int_0^\infty\frac{x^{A-1}}{\Gamma(A)}f(x)g(x)\, dx.$
Now let $u_r(x)=e^{-rx}.$ We have
$\langle u_r,u_s\rangle=\int_0^\infty\frac{x^{A-1}}{\Gamma(A)}e^{-(r+s)x}\, dx=\frac1{\Gamma(A)}\int_0^\infty\left(\frac t{r+s}\right)^{A-1} e^{-t}\frac{dt}{r+s}$
$=\frac1{(r+s)^A}\cdot\frac1{\Gamma(A)}\cdot\int_0^\infty t^{A-1}e^{-t}\, dt=\frac1{(r+s)^A}.$

Now choose $v_i=u_{b_i}$; the matrix is a Gramian.

You thought it looks easy? It's not.

My solution uses integral in the case A = 2, A = n is similar. The idea is as  same as the previous solution, but it is more elementary.

Consider function:

$f(x,y) = \sum a_i x^{b_{i} - \frac{1}{2}} y^{b_{i} - \frac{1}{2}}$

$f(x,y)^2 \ge 0$

Taking the double integral

 $\int_{(0,1]^{2}} f(x,y)^{2} dxdy$

In my opinion, this question is very tough for students who do not know integral.

I was given this question in the case $b_i = i$ and $A = 1$, when I was a high school student. Some of my friend solved it by considering the inequality as a quadratic form of $a_n$, then used induction. The general form of this inequality comes from AMM (American Mathematical Monthly), as I know.

if A=1 .
 Put [tex] f(x)=\sum\limit_{1\le i;j\le n}\frac{a_ia_j}{b_i+b_j}x^{b_i+b_j}[/tex]
[tex] f'(x)=\sum\limit_{1\le i;j\le n}a_ia_jx^{b_i+b_j-1}[/tex][tex]=\frac{1}{x}(\sum\limit_{i=1}^na_ix^{b_i})^2[/tex]  .[tex]f'(x)\ge 0 \forall x>0 . => f(1)=\sum\limit_{1\le i;j\le n}\frac{a_ia_j}{b_i+b_j}\ge f(0)=0[/tex]

Here is a direct proof: $\int_{0}^{\inf}{(\sum_{i=1,n}{e^{-t^{\frac{1}{A}}x_i}})^2dt\geq 0}$. Develop and make a change of variable.

i don't understand your solution harazi;is it elementary?
The same question to tnk

it is not elementary, because the idea to prove is  at university's level.  It is probable that some of you dont understand well the notions of differentiality and integration. So it  may be hard to understand the proof.

Well!Now after becoming a student for 1 month,I think I have completely understood what tnk and harazi means.Maybe tnk made a typo in the last post.It should be:$ b_i^{\minus{}\frac12}$ in the both exponents of $ x$ and $ y$ in the initial two-variable function.
   But I beleive that I still misunderstood jmerry's solution.What is $ \Gamma(A)$ here?
   And what is Gramian matrix?


No solutions. I think it should be $\frac{1}{abc}$

[quote="jen7"]How many positive solutions (a, b, c, d) are there to the system of equations?
$\displaystyle a+b+c=d$
$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{d}$[/quote]

There is not a single solution. If all the numbers are positive, then $a<d$ because of the first equation, and then $\frac1d<\frac1a<\frac1a+\frac1b+\frac1c=\frac1d$ and that is, of course, a contradiction.


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
radical axis
power of a point

This problem was posted before, but I cannot find it.

Let $ \mathcal O, \mathcal O_1, \mathcal O_2$ be the 3 pairwise externally tangent circles with centers $ O, O_1, O_2.$ Let $ c$ be the common tangent of $ \mathcal O, \mathcal O_1$ at $ C$ and $ d$ the common tangent of $ \mathcal O, \mathcal O_2$ at $ D.$ Inversion with center $ C$ takes the circles $ \mathcal O, \mathcal O_1$ tangent at $ C$ into 2 parallel lines $ o' \parallel o_1',$ parallel to the their common tangent $ c.$ The line $ l_1$ tangent to the circles $ \mathcal O, \mathcal O_1$ goes to a circle $ \mathcal L_1'$ tangent to the parallel lines $ o' \parallel o_1'$ and passing through the inversion center $ C.$  The line $ l_2 \parallel l_1$ and tangent to the circle $ \mathcal O$ goes to a circle $ \mathcal L_2'$ tangent to the line $ o'$ and tangent to the circle $ \mathcal L_1'$ at the inversion center $ C.$ The circle $ \mathcal O_2$ tangent to the circles $ \mathcal O, \mathcal O_1$ and to the line $ l_2$ at $ B$ goes to a circle $ \mathcal O_2'$ tangent to the parallel lines $ o' \parallel o_1'$ and tangent to the circle $ \mathcal L_2'$ at $ B'.$ Let $ n$ be a line perpendicular to the parallels $ o' \parallel o_1' \parallel c$ through the center $ L_2'$ of the circle $ \mathcal L_2'.$ The circle $ \mathcal O_2'$ is a reflection of the circle $ \mathcal L_1'$ in $ n,$ therefore $ n$ is the perpendicular bisector of the segment $ CB'$ and the line $ CB'B$ is parallel to the lines $ o' \parallel o_1' \parallel c.$ Consequently, the parallels $ CB \parallel c$ through $ C$ are identical and $ B \in c.$ In exactly the same way, $ A \in d.$ Since the common tangents $ c \equiv CB, d \equiv DA$ are radical axes of the circle pairs $ \mathcal O, \mathcal O_1$ and $ \mathcal O, \mathcal O_2,$ their intersection $ Q$ is the radical center of the pairwise tangent circles $ \mathcal O, \mathcal O_1, \mathcal O_2$ and $ EQ$ is the radical axis and the common tangent at $ E$ of the circle pair $ \mathcal O_1, \mathcal O_2.$ It follows that $ QC \equal{} QD \equal{} QE,$ $ Q$ is the circumcenter of $ \triangle CDE.$

We know that line $ BC$ is the radical axis of the circles $ O$ and $ O_1$. Similarly we have line $ AD$ is the radical axis of the circles $ O$ and $ O_2$. We know that the radical axes of the above two pairs of circles intersect at $ Q$. This implies that the radical axis of the third pair of circles $ O_1$ and $ O_2$ passes through $ E$ and $ Q$. (Thus $ Q$ is the radical center for the three radical axes.) This implies that $ Q$ has equal power with regard to all three circles. Applying power of a point with regard to three circles gives $ CQ^2\equal{} DQ^2 \equal{} EQ^2$ which implies $ CQ\equal{}DQ\equal{}EQ$. So $ Q$ is the circumcenter of the triangle $ CDE$.

[quote="ishfaq420haque"]a circle $ O$ is tangent to two parallel lines $ l_1$ and $ l_2$.a second circle $ O_1$ is tangent to $ l_1$ at $ A$ and to $ O_1$ externally at $ C$.a third circle $ O_2$ is tangent to $ l_2$ at $ B$ and to $ O$ externally at $ D$ and to $ O_1$ externally at $ E$.let $ Q$ be the intersection of $ AD$ and $ BC$.prove $ Q$ is the ciercumcentre of $ CDE$[/quote]
[hide="Solution"]Let $ \mathcal O,\mathcal O_1,\mathcal O_2$ be the 3 pairwise externally tangent circles with centers $ O, O_1, O_2$. Let circle $ O$ is tangent to two parallel lines  $ l_1$ and $ l_2$ at points $ K,L$ respectively.
Let $ \angle AO_1C\equal{}2\phi$
$ \angle OO_1O_2\equal{}2\alpha$
$ \angle O_1O_2O\equal{}2\beta$
$ \angle O_1OO_2\equal{}2\gamma$
 At first points $ A,E,B$ - are collinear(becouse $ \angle O_1EA\equal{} \angle O_2EB$)
Then
$ \angle ECL\equal{}\angle ECD\plus{}\angle DCL\equal{} \alpha\plus{}\gamma\plus{}\frac{\angle DOL}{2}\equal{}\alpha\plus{}\gamma\plus{}\frac{180^o\minus{}\angle KOD}{2}\equal{}\alpha\plus{}\gamma\plus{}\frac{180^o\minus{}(180^o\minus{}2\phi\plus{}2\gamma)}{2}\equal{}\alpha\plus{}\phi$
$ \angle EBl\equal{}180^o\minus{}\angle BAK\equal{}180^o\minus{}(\alpha\plus{}\phi)\equal{}180^o\minus{}\angle ECL.$
So quadriletiral $ CEBL$ is inscribed. Therefore power of point $ A$ with respect to $ \mathcal O_2$ and $ \mathcal O$ is equal, so  line $ AD$ is radical axis of the circles $ \mathcal O$ and $ \mathcal O_2$.

With the same effort we will achieve that quadriletiral $ AEDK$($ \angle KAE\equal{}\alpha\plus{}\phi\equal{}180\minus{}\angle KDE$)  is inscribed. Therefore power of point $ B$ with respect to $ \mathcal O_1$ and $ \mathcal O$ is equal, so  line $ BC$ is radical axis of the circles $ \mathcal O$ and $ \mathcal O_1$.

   Therefore $ Q$ - is radical center of circles $ \mathcal O$, $ \mathcal O_1$, $ \mathcal O_2$
$ \equal{}>$ $ QE\equal{}QD\equal{}QC$ and point $ Q$ is the ciercumcentre of $ \triangle CDE$.[/hide]

[quote="ishfaq420haque"]we also have to prove that $ BC$ and $ AD$ are the radical axis of those circles(mentioned by you)and thats the main trap.i cant do it..have anyone any idea?[/quote]

The inversion circle $ \mathcal C$ with center $ C$ and arbitrary radius is in red. The inverted image (in blue) is symmetrical WRT the perpendiclar $ n \perp (o' \parallel o_1')$ through the center $ L_2'$ of the circle $ \mathcal L_2',$ which makes $ CB' \parallel (o' \parallel o_1').$ The lines $ o' \parallel o_1'$ are also parallel to the common tangent $ c$ of $ \mathcal O, \mathcal O_1$ at $ C$ and this puts $ B', B$ on $ c.$ I hope you see it now.

[quote="ishfaq420haque"]we also have to prove that $ BC$ and $ AD$ are the radical axis of those circles(mentioned by you)and thats the main trap.i cant do it..have anyone any idea?[/quote]

Sorry for the inadequacy of my proof. :oops:

thanks a lot to all of you for nice solutions.....it was from IMO shortlist

It's easy to prove that points $ B, E, A$ are collinear. Apply homotethy with center $ E$ which sends circle $ O_1$ to circle $ O_1'$ tangent to $ l_2$. It's cleat that $ O_1$ and $ O_1'$ are externally tangent and this homotethy sends point $ A$ to $ A'$ which is the tangency point of this circle with $ l_2$ hence $ O_2\equal{}O_1'$ and points $ A, B, E$ are collinear. Analogously we prove that $ B, D, K$ are collinear ($ K$ is tangency point of circle $ O$ with $ l_1$). Now by angle chasing $ \angle BAK\equal{}180^{\circ}\minus{}\angle ABM\equal{}\angle EDB$ where $ M$ is tangency point of circle $ O$ with $ l_2$ hence $ AEDK$ is inscribed quadrilateral. Using this fact we obtain that intersection of $ AE$ and $ DK$ lies on the radical axis of circles $ O$, $ O_1$ which means that $ BC$ is common tangent of these circles. Analogously we prove that $ AD$ i common tangent of $ O$ and $ O_2$. Since $ Q$ lies on the radical axis of $ O_1$ and $ O_2$ we see that $ EQ$ is tangent to $ O_1$ and $ O_2$ hence $ CQ\equal{}DQ\equal{}EQ$ QED


Find

$ \sum\limits_{k\equal{}2}^{\infty}\frac{(2k\plus{}1)H^{2}_{k}}{(k\minus{}1)k(k\plus{}1)(k\plus{}2)}$,

where $ H_{k}\equal{}\sum\limits_{j\equal{}1}^{k}\frac{1}{j}$, denotes the $ k^{th}$ harmonic number.


geometry
perimeter
algorithm
search
absolute value

[POL]
I think this is a bit easier if you want 50% marks because for half of the cases n<=100. But for tougher cases it get n=10^9.

I think we have to deal with grid distance. But how?

A O(n^2) solution is easy to make using "manhattan distance".

For the hard cases, you'll need math... :D

Well, by O(N^2) u mean that for each grid square u are going to see from which of the 3 points the distance is minimum right?

Well, have u found ur that second idea?

Well weeblie, using math I reduced O(N^2) to O(2N). I just scanned through all y values and thus solve the problem. But you know the limit is N<=10^9. Which is very large for O(2N) as well. Do you think this will get AC on judge computer? Or you have been able to get constant time solution?

O(2n) might be diffucult to make in time (depending on how "much" 2n the solution really is :D ).

Haven't thought so much of this task... Don't like math tasks. :P

But, if you look at the problem, you'll see that the areas is separated by a few lines. So, you can find the lines and calculate the solution using simple math. ;)

That line Weeblie mentioned is a line in a manhattan-plane;)

I didn't try much to solve it yet, but I think it won't be a big problem to calculate the POL problem in O(1) time. You might have to make your own large-number arithmetic, though.

I'm Jakub Zavodny, probably SVK2.

well actually the main obstacle to obtain O(1) soln is absolute value in inequality(In my method).

but still i can't understand what is the meaning of area and line of weeblie. i think area is those sqrs that is region for A, region for B and region for C. that means three areas. Right?

according to kubus, line is manhattan line. that means grid lines.

"areas seperated by few lines": hmm.. have to think more..

BTWnice code of DOM by Kubus. those worked really in a short time. and it was nothing actually just moderated form of Bipartite matching Hungarian Tree algortihm by DFS. Really nice! :)

Most probably mine will be BGD 1

[quote="mahbub"]Well what do u mean by line? Are u talking about the perimeter? Is it really possible to solve in this way?

BTW, what's ur and kubus's full name?

Mine is: Md. Mahbubul Hasan, full name will help us to meet there![/quote]

Yes, the perimeter of the "areas" is built up using a few lines (3 + border of flag for each area, I think).

I'm Chen Xing, probably SWE3 (we only have 3 persons in our team this year... the 4th couldn't come :( ).

[quote="kubus"]You mean, sweden could afford only 3 members for some reason or the person wasn't able to come? If this happened in Slovakia, I bet the fifth one in the qualification round would be happy to come...

(two years ago, the difference between the fourth and the fifth place was 0.13 points out of about 500;-)[/quote]

Dunno, he didn't say any reason. ;)

[quote="mahbub"]well, a region can have zig-zag border, then?[/quote]

Yes, it can... Actually, it doesn't matter...

The main thing here is that all rows have the form: aaaaaaaaaaaabbbbbbbbccccccccccccccc... :D

Btw, wrote a O(log(n)) solution today. :D

log(n).  :(  it is sounding a bit odd for this kinds of problem :?  yup i know that's aaa..bb..cccc. But what i am thinking is that without going to the level of each y/x OR not finding a complete formula how can it be lg(n).

 :blush: My head is going astray! Really! ;)

[quote="Emir Habul"]Hello, I am new to this forum
Country: Bosnia and Herzegovina (see you at IOI)

Weeblie, I used your method to "split" area in to pieces, thanks for that
but instead binary search I used math also to find shifts from Lucy to Bob etc.

Just for record O(1)[/quote]

That's of cource doable too. :D

friends, i am not sure whether i will be able to attend IOI 2005 or not. :(  Polish embessy at India was making so late that our representative had to come back without taking VISA. he returned on 12.. becoz 13, 14, 15 is holiday.. and 17 is our flight. so 16 would be late. that's why.. our leader has contacted with IOI officials and they are trying to contact with Polish minitry of foreighn for special permission (like VISA). Let me see what's happen..

well, can any one of you do favour for me?

as we are not going (yes.. it is finalled.. we are missing IOI this year.. and forever)

so, can u plz tell the guide of our country(Bangladesh) to contact us at shanto86@yahoo.com just tell it to ur guide.. he/she will tell it to our country guide.. it will be great help for me..  :?

Mahbub


Solve in $ R$ next equation:
$ 4^{x\plus{}\frac{1}{x}}\plus{}9^{x\plus{}\frac{1}{x}}\equal{}275$.

Let $ x \plus{} \frac {1}{x} \equal{} a$. By AM-GM, $ a\ge 2$. Then $ a\in R$ and $ 4^a \plus{} 9^a \equal{} 275$.
But if $ f(a) \equal{} 4^a \plus{} 9^a$, $ f'(a) \equal{} 4^a\ln(4) \plus{} 9^a\ln(9) > 0$ as $ \ln(4),\ln(9),4^a,9^a > 0$. So $ f(a)$ is increasing on $ (2,\infty)$ so at most one value of $ a$ exists such that $ 4^a \plus{} 9^a \equal{} 275$.
On the other hand, if $ a \equal{} \frac {5}{2}$, we get $ 4^a \plus{} 9^a \equal{} 275$. So the only value of $ a$ satisfying $ 4^a \plus{} 9^a \equal{} 275$ is $ \frac {5}{2}$.
So $ x \plus{} \frac {1}{x} \equal{} \frac {5}{2}$ which is equivalent to:
$ 2x^2 \minus{} 5x \plus{} 2 \equal{} 0$
$ (2x \minus{} 1)(x \minus{} 2) \equal{} 0$. So $ x \equal{} \frac {1}{2}$ or $ 2$.
[b]Answer[/b]: $ \frac {1}{2}, 2$.


How many solutions are there in NxN for the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{1995}$?

Please help and also tell some way to how to solve this type of problems.

Generally, you solve equations like this by getting rid of the fractions.

Edit:  Oops, thought it was 2005 instead of 1995  :blush:

[hide="Solution"]
$\frac{x+y}{xy} = \frac{1}{1995}$
$\frac{xy}{x+y} = 1995$
$xy = 1995x + 1995y$
$xy - 1995x - 1995y = 0$
$(x-1995)(y-1995) = 1995^2$ (Simon's Favorite Factoring Trick!)
$1995^2 = 3^2 \times 5^2 \times 7^2 \times 19^2$

The number of solutions (distinct ordered pairs of solutions) is equal to the number of positive integers that divide $1995^2$.  The way to count them is to note that any integer that divides a number can have prime factors up to as many as that number itself (for example, $3 \times 5^2$ will divide $1995^2$).  So, to count all the numbers that do that, we look at the exponent of each prime in the factorization - it can vary from 0 to the exponent it already is, for each prime, so the total number of factors of $1995^2$ is $(2+1) \times (2+1) \times (2+1) \times (2+1) = 81$, meaning there are $81$ distinct ordered pairs of solutions.[/hide]

[quote="t0rajir0u"]
...
$1995^2$ is $(2+1) \times (2+1) \times (2+1) \times (2+1) = 81$, meaning there are $81$ distinct ordered pairs of solutions.[/hide][/quote]

there is only one problem with this, it is looking for natural numbers (right?, that is what N means)

so you have to make sure that you don't count any negative solutions

I didn't.  The problem having 81 solutions means that there are 81 values of $k \in \mathbb{N}$ such that $k | 1995^2$.  Since each solution pair takes the form $(x, y) = (k + 2005, \frac{1995^2}{k} + 2005)$, they are both also guaranteed to be naturals.

sorry, i made a little confusion, i was thinking abut having negative divisors, but i just realized that cannot be since either of the expressions will be negative...you are right