Coloring
Extremal combinatorics
combinatorics
polygon
diagonals
IMO Shortlist

Suppose we have a $n$-gon. Some $n-3$ diagonals are coloured black and some other $n-3$ diagonals are coloured red (a side is not a diagonal), so that no two diagonals of the same colour can intersect strictly inside the polygon, although they can share a vertex. Find the maximum number of intersection points between diagonals coloured differently strictly inside the polygon, in terms of $n$.

[i]Proposed by Alexander Ivanov, Bulgaria[/i]

What restrictions are there to subtract from $(n-3)^2$?

Well no two diagonals of the same colour are allowed to intersect inside the polygon.

Btw, I'm assuming epitomy meant "$n-3$ are colored black and $n-3$ are colored red"...

[quote="vulgarfraction"]But... if, for example, your n-3 diagonals all come from the same vertex for each colour. Then they won't intersect again. Then you have $(n-3)^2$ intersections.[/quote]

Not necessarily, actually.  So let's say the you have all possible black diagonals coming from one vertex.  Consider one of those diagonals.  Let's say it stems from vertex $A$ and goes to vertex $B$.  If some red diagonal stems from vertex $C$ and goes to vertex $B$, then the two do not intersect.
Nevertheless, I must say a clever idea...

exactly, they have to meet strictly inside the polygon, so meeting at a vertex is not good enough.
note the scenario where you have 2 vertices spanning all the $n-3$ edges is not optimal for most cases .. eg try $n=6,7$.
and yes i did have a small typo which i have now corrected.

any ideas people on this difficult problem?
i hear its from the shortlist 2005 also ...

My solution:

the intersection points are at most

$\frac{3n^2 - 18 n + 27 + p}{4}$, where $p$ depends on the parity of $n$ and is $0$, if $n$ is odd, $1$ otherwise.

I have not yet written it down properly, but the main idea is to prove the following:

[size=75][b]Claim:[/b] [i]The number of blue diagonals with at least $\scriptstyle n-3$ intersection points is at most $\scriptstyle 1$;
the number of blue diagonals with at least $\scriptstyle n-4$ intersection points is at most $\scriptstyle 3$;
the number of blue diagonals with at least $\scriptstyle n-5$ intersection points is at most $\scriptstyle 5$;
...
the number of blue diagonals with at least $\scriptstyle n-3-k$ intersection points is at most $\scriptstyle 1+2k$.[/i][/size]

This gives you an upper bound that turns out to be optimal: is it possible infact to draw diagonals in such a way that the maximum is attained.

The construction is not so obvious, so have fun in finding it! The rest is just counting how many intersection points are there and writing the ugly polynomials of the solution.

M.

Answer is

$\frac{3}{4}*(n-3)^2+p$, where $p=1/4$ if $n$ is even, and $p=0$ if $n$ is odd.

The example is easy.

The proof that it can't be more is much harder, i don't have time to write it now.

Consider a chord $XY$, and suppose $a$ points are on one side of it, and $b$ points are on the other side (so in particular $a+b=n-2$). Define the [i]length[/i] $\ell(XY)$ to be $\min\{a,b\}$. We have the following main lemma.

[b]Lemma:[/b]  Suppose we have two non-intersecting chords $AB$ and $CD$ of lengths $x$ and $y$. For any triangulation of the $n$-gon, the number of interior intersections with these two chords is bounded above by $x+y+n-4$.

[i]Proof:[/i]  Starting form $A$, let the points be \[A,P_1,\ldots,P_x,B,U_1,\ldots,U_t,C,Q_1,\ldots,Q_y,D,V_1,\ldots,V_s,\] where $x+y+t+s=n-4$. Call a point $P_i$ or $Q_i$ $U$-good if the triangulation has some edge from that point to some $U_j$, and define $V$-good similarly.

Let's first count the number of intersections of edges that go from a $P_i$ or $Q_i$ to a $U_j$ with $AB\cup CD$. Let $P_{a_1},\ldots,P_{a_g}$ be the $U$-good $P_i$s where $a_1<\cdots<a_g$, and let $Q_{b_1}<\cdots<Q_{b_h}$ be the $U$-good $Q_i$s where $b_1<\cdots<b_h$.

Let the set of $j$s such that $U_j$ is connected to $P_i$ be called $S_{P_i}$, same with $Q_i$. It is clear that all the intervals $[\min S_{P_i},\max S_{P_i}]$ and $[\min S_{Q_i},\max S_{Q_i}]$ can not intersect in their interiors (but can intersect on their boundaries). It is easy to see that this implies that there are at most \[t + (g+h-1)\] such intersections.

This means that the number of intersections of edges that go from $P_i$ or $Q_i$ to a $U_j$ with $AB\cup CD$ is bounded above by $t-1$ plus the number of $U$-good points. Similarly, the number of intersections of edges that go from $P_i$ or $Q_i$ to a $V_j$ with $AB\cup CD$ is bounded above by $s-1$ plus the number of $V$-good points.

Now we have the count the number of edges between $P_i$s and $Q_j$s (each of these edges count towards to intersections with $AB\cup CD$). Suppose there are $g_u$ $U$-good $P_i$s and $g_v$ $V$-good $P_i$s. It is easy to see then that there are at most $\max\{\min\{x+2-g_u-g_v,x\},0\}$ possible $P_i$s that can even connect to a $Q_i$. We assume that $g_u,g_v\ge 1$ (in they case where they are not we can have a better bound for $P_i$ to $U_j$s or $V_j$s which compensates). If $x+2-g_u-g_v<0$, then we're fine as there are then $0$ possible such edges anyway. Thus, we have that the number of possible edges is at most  \[(x+2-g_u-g_v)+(y+2-g_u'-g_v')-1\] edges intersecting $AB\cup CD$ that go from some $P_i$ to some $Q_j$ (where $g_u'$ is the number of $U$-good $Q_i$s and $g_v'$ is similar). Therefore, the total number of intersections is at most \[(t-1)+g_u+g_u'+(s-1)+g_v+g_v'+2[(x+2-g_u-g_v)+(y+2-g_u'-g_v')-1].\] As mentioned above, we are assuming $g_u,g_v,g_u',g_v'\ge 1$, so this is at most \[2(x+y)+t+s+4-(g_u+g_v+g_u'+g_v')\le 2(x+y)+t+s = n-4+x+y,\] as desired. $\blacksquare$

The final bound now follows by summing bounds given by the above lemma.

Suppose first that $n$ is odd. Let $d_1\le d_2\le\cdots\le d_{n-3}$ be the lengths of the chords of the red triangulation. Applying the lemma above bounds above the total red-black intersections by \[(d_1+\cdots+d_{n-3})+\frac{n-3}{2}(n-4).\] We claim that $d_1,d_2\le 1$, $d_3,d_4\le 2$, so on till $d_{n-4},d_{n-3}\le \frac{n-3}{2}$. This is clear by induction, since we can start by removing two chords of length $1$ to get an $(n-2)$-gon, and now chords of length $1$ with respect to that $(n-2)$-gon have actual length at most $2$, and so on. This shows that the total number of intersections is at most \[\frac{n-3}{2}\cdot\frac{n-1}{2}+(n-4)\frac{n-3}{2} = \frac{3(n-3)^2}{4}.\] Suppose now that $n$ is even. Note that the number of red-black intersections with any given red chord is at most $n-3$. Thus, a similar bound as above bounds above the number of red black intersections by \[\left(1+1+2+2+\cdots+\frac{n-4}{2}+\frac{n-4}{2}\right)+\frac{n-4}{2}(n-4)+n-3=\frac{3(n-3)^2+1}{4}.\] Therefore, there are always at most $\boxed{\lceil\tfrac{3}{4}(n-3)^2\rceil}$ intersections.

We now provide a construction for each $n$. We label the points $1,2,\ldots,n$ in order. If $n$ is even, let the black triangulation have edges \[\{1,3\},\{1,4\},\ldots,\{1,\tfrac{n+1}{2}\},\{\tfrac{n+1}{2}+2,\tfrac{n+1}{2}\},\{\tfrac{n+1}{2}+3,\tfrac{n+1}{2}\},\ldots,\{n-1,\tfrac{n+1}{2}\}\] and the red triangulation have edges given by $\{x+1,y+1\}\pmod{n}$ where $\{x,y\}$ are the black edges. The construction for odd $n$ is similar.

Sorry for the bump, but it's important to mention that the $n$-gon is convex (which the official shortlist statement does).


[quote="Svejk"][b]2[/b]
Is $ f(n)=$ the power of $ 2$ from the factorization of $ n!$ ?[/quote]
Sorry for my poor english.Your f(n) is correct.
How to find $ f(1)+f(2)...+f(2003)=?$

For the second problem, calculate $ \sum_{n=1}^{2003}\sum_{j=1}^{\infty}\left\lfloor\frac{n}{2^{j}}\right\rfloor$.


If $ f: \mathbb{R\rightarrow\mathbb{R}}$ is a function defined by the expression $ f(x\minus{}2)\equal{}x^{3}$, then the value of $ f(3)$ is equal to:

a) 1
b) 27
c) 8
d) 125
e) 0

Let $ 3 \equal{} x\minus{}2$, then $ x \equal{} 5$ and $ f(3) \equal{} 5^3 \equal{} \boxed{D) \ 125}$


search
geometry
geometric transformation

(a) Is it possible to find a sequence of two or more consecutive whole numbers which add up to $ 1000$? If not, prove that it is impossible; if so, find all possible sequences.

(b) Find a number greater than $ 1000$ but less than $ 2000$ which is impossible to write as the sum of two or more consecutive whole numbers.

b: I am thinkin 1002 is one. haha idk, just a guess because 1001 and 1003 are wrong

(a) Yes, it's possible!

Let's imagine a sequence of $ n$ integers $ k,k+1,k+2,k+3,...,k+n-1$ $ (k\in\mathbb{Z})$.
We can write it as: $ nk+n(n-1)/2 = n(\frac{2k+n-1}{2})$.

So, $ n(\frac{2k+n-1}{2})=1000$ which means that $ n|1000$. Let's set $ A = n$, $ B=\frac{2k+n-1}{2}=k-\frac{A-1}{2}$.
B must be integer, so A must be odd. There are only 3 odds divisors of 1000: 5, 25, 125.

So, the sequences are:
198 to 202  (found using $ A=n=5$)
28 to 52 (using $ A=25$)
-45 to 70 (using $ A=125$)

b) The number would have to be prime.  So a number in this set would be 1009.

For part (b) we have to find an integer $ j$ between 1000 and 2000 with no odds divisors. So, the number is written as $ 2^k$
It is 1024.

Yeah, thanks Gabuntu, I missed that one.

Yes.

By the way, should your last sequence for (a) be $ 55$ to $ 70$, that is, $ 16$ consecutive integers which sum to $ 1000$?

Oh sorry...
I meant -54 to 70, not -45 to 70 as I wrote.

Of course we can write -54 to 70 as
$ 1\plus{}2\plus{}...\plus{}70\minus{}(54\plus{}53\plus{}52\plus{}...\plus{}1)\equal{}55\plus{}56\plus{}...\plus{}70$  :)

There was a topic (I'm too lazy to search for it), essentially stating:

[b]Theorem.[/b][color=blue] All positive integers can be written as a sum of two or more consecutive positive integers, except the powers  of $ 2$.[/color]

Try to prove it, it's not difficult at all.

[quote="AndrewTom"](a) Is it possible to find a sequence of two or more consecutive whole numbers which add up to $ 1000$?[/quote]

Whole numbers is the set that often is defined as {0, 1, 2,, 3, ...} or
{1, 2, 3, ...} (depending on which source you read).

Regardless, they don't include the negative integers.

Gabuntu94,
if (zero with or without) positive integers are meant, then please don't
use the phrase "whole numbers."

As it is, the part of your solution (-54) + (-53) + ... + 69 + 70 doesn't
count in its form because it includes negative numbers.

(I know you did a positive integer rewrite of it.)

Thanks Arrange your tan!

I'm really glad someone's finally taught me that!

I had never understood the meaning of "whole numbers" in English (wikipedia says:

[quote]The term whole number does not have a consistent definition.  Various authors use it in one of the following senses:

    * the nonnegative integers (0, 1, 2, 3, ...)
    * the positive integers (1, 2, 3, ...)
    * all integers (..., -3, -2, -1, 0, 1, 2, 3, ...).
[/quote]

wich is not clear at all!)

I hope I won't get wrong again :)

What shall I say instead of "whole numbers"? Positive integers is right?

Thanks for your patience  :oops:

Gabriele  :)

The Oxford English Dictionary defines a whole number as an integer, that is a whole number which can be negative, zero or positive. This is what I think it is.

I should have specified "positive" in the question.

So... what's wrong? what's right?  :huh:

I seem to have been wrong in accepting the definition in the OED.

Garrett Birkhoff and Saunders MacLane in "A Survey of Modern Algebra" (page 1) write "... all the positive integers (whole numbers). A related but somewhat larger system is the collection $ \mathbb Z$ of all integers $ 0, \pm1, \pm2, \pm3, ...$."

John William Archbold in "Algebra" (page 1) writes "The first number system encountered is that of the [i]natural numbers [/i](also referred to as [i]whole numbers [/i]or [i]cardinal integers[/i]) $ 1, 2, 3, ...$"

I hope this helps (and doesn't add to the confusion).

So... it could mean positive and negative integers as well as just positive ones...
But the most widely used meaning is "natural numbers"...

Well, now it's clear  :roll:

You know, every language have different features that make it special.
For example, in my language (Italian), the literal translation of "whole number" is "numero intero", which means integers in general, both positive and negative...

It's been a pleasure to learn that with you(plural)!
[You see, this lack of a plural objective pronoun is another typical feature that makes the language so different from mine... I always doubt if people who will read a "you" will understand that I meant it was plural. How do you understand it?  :P
Besides, just for curiosity: how old are you, where are you from? I think other cultures are extremely interesting and help us growing with the feeling of what is the world...
If you want, and have some time, please answer these two questions in a PM :) I'll be glad to read your answers.]

I hope I haven't bored you!
Again, thanks for your patience!   :oops:

Gabriele  :lol:

[quote="Gabuntu94"]But the most widely used meaning is "natural numbers"...

[/quote]

No, I would not get that conclusion from these dictionary example references.

As taught in American schools and universities as I have attended, in their
math courses in general and in Mathematics for Elementary Teachers
courses, the natural numbers are {1, 2, 3, ...} and the whole numbers are
{0, 1, 2, 3, ...}.

---------------------------------------------------------------------------------------------------------

So, now I won't even use the phrase "whole numbers" because
it is not universal and often causes confusion.

Birkhoff and MacLane were American; Archbold was Scottish. (I seem tor emember reading somewhere that $ 0$ is considered a natural number in Australia.) There must be a definitive answer.

I was taught that whole numbers included zero and the natural numbers didn't. I guess it depends on the region.


Three primes $p,q,$ and $r$ satisfy $p+q = r$ and $1 < p < q$. Then $p$ equals

$ \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 17 $

[hide="Answer"]All primes other than $2$ are odd, so they are separated by an even distance. If $p<q$, then $q>2$ for any value of $p$, and the difference between $q$ and $r$ must be even, so $p=2\Rightarrow \boxed{A}$.[/hide]

[hide]Since primes numbers are all odd except for one, we must have two numbers p and q in which one is even and one is odd. Thus p or q must be 2. Since 2 is the smallest prime, and q>p, p=2. [/hide]


Does this really belong in this forum or should it go in the Puzzles and Brainteasers forum?

[quote="MathFiend"]What five letter word completes this equation?

99+9=9+_____[/quote]
[hide]
i think it is not dozen
[/hide]


More than 6 months ago, I wrote a 'mathematical fiction' story, about 6 pages long. First, it was something I wrote just for my pleasure, but recently, I came to the idea of publishing it. Being sick and tired of publishing stories on blogs etc. (I've had experiences of finding my stories on other people's blogs, chainmails and fora [unsigned] after they were published on my blog-it's a compliment for a writer, but also frustrating), I looked for a journal to publish it.

Going through 'standard' science fiction journals and magazines, I've noticed that mathematical fiction isn't so popular in these circles, which is understandable (it's easier to read/write about aliens than it is to read/write about Riemann Hypothesis :lol: ).

I wonder: is there a magazine/webzine/etc. specialized in publishing this kind of works? My search for such journals gave no results, so it gave me an idea to try to find people with interest in this kind of literature to make a webzine in which 'mathematical art and humor' could be published, in a level higher than blogging...

Any comments?

You could start your own webzine...

Would you consider making your own website and putting your own stuff on there... on "a higher level then blogging"?

I can't imagine that you've missed [url]http://kasmana.people.cofc.edu/MATHFICT/default.html[/url] (or [url]http://kasmana.people.cofc.edu/MATHFICT/search.php?go=yes&medium=ss&orderby=title[/url]), it being the first result on a Google search for "mathematical fiction" (perhaps you were not interested in it for some reason?). Searching around seems to suggest that there is not really "a webzine in which 'mathematical art and humor' could be published" as you are looking for, and that most mathematical fiction is published like other unconventional or uncommon genres:

-in compilations
-somehow inserted into big journals that are somewhat related
-included in small journals or magazines that tend to include such content (random example: [url]http://www.ms.unimelb.edu.au/~paradox/archive/[/url])
-famous authors can basically publish whatever they want... so if you're a famous author...

[hide="_"]Sorry for the rather unhelpful post.[/hide]

[quote="leoxnlin"]

Would you consider making your own website and putting your own stuff on there... on "a higher level then blogging"?
[/quote]

Well, my hope would be that it would gradually evolve into something bigger, given the fact that other people would publish too, and the works would go through review... My idea was to make it bit more serious, but there is agiant question mark above it-how can you make something which in its basis is not serious- serious.

[quote]I can't imagine that you've missed [url]http://kasmana.people.cofc.edu/MATHFICT/default.html[/url] (or [url]http://kasmana.people.cofc.edu/MATHFICT/search.php?go=yes&medium=ss&orderby=title[/url]), it being the first result on a Google search for "mathematical fiction" (perhaps you were not interested in it for some reason?).[/quote]

I haven't missed it. It's just the fact that it's a collection of hardprint published publications in mathematical fiction, which couldn't answer my question where I can publish ['officially'].

[quote]most mathematical fiction is published like other unconventional or uncommon genres:

-in compilations
-somehow inserted into big journals that are somewhat related
-included in small journals or magazines that tend to include such content (random example: [url]http://www.ms.unimelb.edu.au/~paradox/archive/[/url])
-famous authors can basically publish whatever they want... so if you're a famous author...[/quote]

second option would need me to be a four (famous author  :lol: ), so it leaves me with one and three. For now, I'm pursuing the third option (still waiting for a response from one of the magazine editors, and considering sending the story to some other too).

[hide="_"]Thank you for this rather helpful post!  :lol: [/hide]


Let $ {X_n}$ and $ {Y_n}$ denote two sequences of integers defined as follows:
$ X_0\equal{}1$, $ X_1\equal{}1$, $ X_{n\plus{}1}\equal{}X_n\plus{}2X_{n\minus{}1}$ $ (n\equal{}1,2,3, \ldots)$
$ Y_0\equal{}1$, $ Y_1\equal{}7$, $ Y_{n\plus{}1}\equal{}2Y_n\plus{}3Y_{n\minus{}1}$ $ (n\equal{}1,2,3, \ldots)$
Thus, the first few terms of the two sequences are;
$ X: 1,1,3,5,11,21, \ldots$
$ Y: 1,7,17,55,161,487, \ldots$
Prove that, except for "1", there is no term which occurs in both sequences.


induction
inequalities
real analysis
real analysis unsolved

Let $ 1 \leqslant p < q \leqslant \infty$. Prove that $ l_p \subset l_q$ and $ \parallel{}x\parallel{}_{l_q} \leqslant \parallel{}x\parallel{}_{l_p} \forall x \in l_p$.
I tried to prove by induction on $ p$ - okay, no problems with $ l_1 \subset l_2$, but how to prove it for $ p \plus{} 1$ if its correct for $ p$ - tried to use Hölder's inequality, but seems it can't help here. Any ideas? Thanks.

How do you expect induction to be useful when $ p$ and $ q$ don't even have to be natural numbers?

Oops, seems realy stupid, thanks  :blush:
So can anyone please help with an idea about this?

The second result can't be true of, say, both the sequences $ x \equal{} 2, 0, 0, ...$ and $ x \equal{} \frac {1}{2}, 0, 0, ...$.  Are we talking about integer sequences?

As for the first result, how did you prove $ l_1 \subset l_2$?  The first proof that comes to mind generalizes with no trouble to the general case.

I am sorry, misstyped in the first post - the secound result is $ \parallel{}x\parallel{}_{l_q} \leqslant \parallel{}x\parallel{}_{l_p} \forall x \in l_p$.
And it seems I mistaken when proving $ l_1 \subset l_2$ - gonna find another way now.

Here's the start of the argument:

Suppose $ p<q$ and $ x\in\ell^p.$

Start with the assumption that $ \|x\|_p\equal{}1.$ If you start there, note that we have $ |x_j|\le 1$ for all $ j.$ And if $ |x_j|\le 1,$ what can you say about the relationship between $ |x_j|^p$ and $ |x_j|^q\ ?$

Now, back to the top: is there any way to make that assumption that $ \|x\|_p\equal{}1$ into a WLOG?

Okay, thanks, sorry for late reply - had Christmas holidays. As I understood, if we have $ \parallel{}x\parallel{}_p \equal{} 1, p<q$, then $ |x_j|\le 1$ for all $ j$, and $ 0 < |x_j|^q < |x_j|^p$, then if $ \sum\limits_{n\equal{}0}^{\infty} {|x_n|^p}$ converges, then so does $ \sum\limits_{n\equal{}0}^{\infty} {|x_n|^q}$, so $ l_p \subset l_q$. Now about inequality $ \parallel{}x\parallel{}_{l_q} \le \parallel{}x\parallel{}_{l_p}$ - can we again consider the case $ \parallel{}x\parallel{}_p \equal{} 1$, so then $ |x_j|\le 1$ for all $ j$, and we have $ S_q \le S_p$, where $ S_q \equal{} \left(\sum\limits_{n\equal{}0}^{\infty} {|x_n|^q}\right)^{1/q}, S_p \equal{} \left(\sum\limits_{n\equal{}0}^{\infty} {|x_n|^p}\right)^{1/p}$. But about the way to make that assumption that $ \parallel{}x\parallel{}_p \equal{}1$ into a WLOG - still can't deal with it; as far as I know, we can use something like that when proving Helder's inequality for sums $ \sum\limits_{k\equal{}1}^{n} {|x_k b_k|} \le \left(\sum\limits_{k\equal{}1}^{n} {|a_k|^p}\right)^{1/p} \left(\sum\limits_{k\equal{}1}^{n} {|b_k|^q}\right)^{1/q}$ (I mean its homogeneous, i.e., if it holds for two points $ (a_1,...,a_n)$ and $ (b_1,...,b_n)$, the it holds for any two points $ (\lambda a_1,...,\lambda a_n)$ and $ (\mu b_1,...,\mu b_n)$, where $ \lambda$ and $ \mu$ are arbiraty real real numbers, so therefore we need only to prove it for the case $ \sum\limits_{k\equal{}1}^{n} {|a_k|^p} \equal{} \sum\limits_{k\equal{}1}^{n} {|b_k|^q} \equal{} 1$). So its something like that here, right? Can't formulate it correctly :blush: . Thanks.

I am sorry, but about this inequality you wrote, feniks, - how can I use it here? And can we consider $ \frac x {\parallel{}x\parallel{}_p}$ if we have just normal space? Sorry, not clear enought  :maybe:

You have proved it for $ \lVert x \rVert_p \equal{} 1$, so if $ \lVert x \rVert_p \neq 1$ then just put $ y \equal{} \frac{x}{\lVert x \rVert_p}$.


Let $S$ be a set of $\displaystyle { 2n \choose n } + 1$ real numbers, where $n$ is an positive integer. Prove that there exists a monotone sequence $\{a_i\}_{1\leq i \leq n+2} \subset S$ such that
\[ |x_{i+1} - x_1 | \geq 2 | x_i - x_1 | , \]
for all $i=2,3,\ldots, n+1$.

I think that the question involves an application of the pigeon-hole principle, but as

far as a solution goes, im drawing a complete blank.

Does anybody know the solution to this?

Valentin, you went to the competition. Do you know the solution?

by some chance i know the solution. all in all two of our teammates solved the problem correctly.the trick here is to generalise the induction.
 let a_m_n be the minimal number of points such that we have an m-term ascending or n-term descending sequence of points with the given property. clearly a_m_n=a_n_m and a_m_n<=a_m-1_n   +   a_m_n-1   - 1 since we can glue two sets of points with this lengts and what comes out must contai either m-term descending or n-term descending. now all is clear . using this inequality we obtain a_m_n<= combinations of n elements out of m+n       -1 (dont care the technical details

sorry for bumping this topic, but
[quote=Valentin Vornicu] Prove that there exists a monotone sequence $\{a_i\}_{1\leq i \leq n+2} \subset S$ such that
\[ |x_{i+1} - x_1 | \geq 2 | x_i - x_1 | , \]
for all $i=2,3,\ldots, n+1$.[/quote]
what is the connection between $a_i$ and $x_i$? I suppose that it is a typo and should be $a_i$ for all terms not $x_i$.


Prove ${ {7}^n -1}$ is divisible by $ 6$

Here is what I have so far:

Let ${ P(n)={7}^n -1}$
$ P(1)={7}^1 -1=6$ thus true for $ n=1$
Assume result is true for $ n=k$:
${ P(k)={7}^k -1}$

Let $ n=k+1$

${ P(k+1)={7}^{k+1} -1}={(7)}({7}^k)-1$

but can't see the end.

$ 7\equal{}1\pmod 6$
$ 7^n\equal{}1^n\pmod 6 \equal{} 1 \pmod 6$.

Enough said.

Use the factorization $ a^k\minus{}b^k\equal{}(a\minus{}b)(a^{k\minus{}1}\plus{}\dots\text{junk}\dots)$ if mods aren't your 'thing'

Thanks but could someone just write out the end of the induction proof for me please?

$ 7 \cdot 7^k \minus{} 1 \equal{} (6 \plus{} 1) \cdot 7^k \minus{} 1 \equal{} 6 \cdot 7^k \plus{} (7^k \minus{} 1)$, which is the sum of two things you know are divisible by $ 6$.

JasonSullivan,

I'm sorry that you were given the wrong method (and attitude)
by user Yongyi781.

For the induction proof you should be clearly indicating:

1) This is the base step, and it is...
2) This is what you are assuming for the nth (or kth step)...
3) This is the work to show the (n+1)st step (or (k+1)st step).
4) For the conclusion, write/type a blurb that states that by
   the Principle of Mathematical Induction you have shown
   the original statement to be true.

Arrange your tan, please do not condescend and assume no one knows what induction is. JasonSullivan is not asking what is induction, he's asking how to show the (n+1)st step.

(1) You do not flame someone for not knowing something or having a bad attitude.

(2)
[quote="JasonSullivan"]Thank you. Much appreciated.
[/quote]
This problem has already been solved.

Read all the posts in this topic before posting.

Arrange your tan, thank you for being so helpful. I do know what induction is but as I'm new to it any reiteration is great.

Yongyi781, you expect me to read all the posts on this board in the hope it is there?   :huh:

Doesn't matter anyway. Thanks for the help everyone this board is great!

Just to clear a few things up: "Arrange your tan" (henceforth ayt) is a troll, and is usually best ignored.  JasonSullivan, you mis-parsed Yongyi's post: he was using your quote to address ayt -- the lines that followed it were addressed to ayt, not to you.  Also, Yongyi was refering to this thread, not to the entire forum.  Finally, a general suggestion: you post an awful lot of questions that (from a math competition perspective, which many of the members of this forum have) are relatively basic.  I suggest that when you post more of these problems, you include in your post (e.g., in hidden text) a description of how you think you should proceed and what you've tried already -- it's much better to learn by doing than to learn by having other people tell you what to do.

Hi JBL,

I don't really want to waste my time going in to it or name calling. I don't know anyone on here personally so I will take it nothing was meant by anybody. I do see what you mean about Yongi referring to the thread however.

Also, in regards to my posts, I do try the questions before I put them up; often several times over. I don't really know how to use latex so have to keep looking symbols and things up so I usually write as little as possible. It works extremely well for me to work backwards through a solution once I have tried it many times myself.

LaTeX issue: You can go to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=224]test forum[/url] and look at other people's tests(quote the post to see the code) and learn how to use LaTeX that way.


calculus
induction
trigonometry
geometric sequence
geometric series

can somebody please explain the formula for the sum of a geometric sequence?? I've read explanations but i haven't been able to understand them.

For example:

What is the formula for questions like find $\sum_{k = 1}^{n}{ar^{k}}$ and why?

If r[u]>[/u]1, then the sum of the sequence as n goes to infinity appraoches infinity. That seems self-explanatory because, for instance, (1.00....001)^(infinity) = infinity, then adding it to the sequence will bring the sum to infinity.

If 0[u]<[/u]r<1, then we know that the sum reaches a finite limit.

Perhaps the most common example of this is (1/2)^k

1/2 + 1/4 + ... 1/2n = 1 when n approaches infinity.

In this case, there is a formula you can use to find the sum. Just look it up in any calculus book.

If -1<r<0, then you have a  series that is alternating.

The formula is $a(\frac{r^{n}-1}{r-1})$, proven by induction.

Well, here is my ghetto non-induction deriviation with factoring. It's easy for me to understand so I don't need to memorize the formula. I will derive it for a=1 after which you can multiply the formula by a to get the general version.

$1+r+r^{2}+r^{3}+...+r^{n}=x$
$1+r(1+r+r^{2}+...+r^{n-1})=x$
$1+r(x-r^{n})=x$
$1+rx-r^{n+1}=x$
$1-r^{n+1}=x(1-r)$
$x=\frac{r^{n+1}-1}{r-1}$
I came up with this one day when I was thinking about how easy it should be to derive the formula, though someone else probably derived it this way before me...

Oops, typo in my other post, too lazy to fix it, w/e

Extension:

Prove, IN AS MANY WAYS POSSIBLE, that sum of an infinite geometric series with $-1<r<1$, is $\frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio.

[hide="It is a geometric series after all"] Consider $\triangle ABC$ with $\angle ACB = 90^{\circ}$ and $AB = 1$.

Let $\angle ABC = \theta$.  Then $AC = \sin \theta, BC = \cos \theta$.

Now, drop a perpendicular from $C$ to $AB$ and call the intersection point $C'$.  Then $\triangle ABC \sim \triangle BCC'$.  Then drop a perpendicular from $C'$ to $BC$ to create a point $C''$ and a new triangle.  Repeat this procedure infinitely.

There are several equivalent ways to handle the proof from here, but let's calculate the length of $AB$ in terms of our infinite triangles.  It would be clearer with a diagram, but this length is

$1 = \sin^{2}\theta+\sin^{2}\theta \cos^{2}\theta+\sin^{2}\theta \cos^{4}\theta+...$

And therefore we know that

$\frac{1}{1-\cos^{2}\theta}= 1+\cos^{2}\theta+\cos^{4}\theta+...$

Letting $r = \cos^{2}\theta$ (of course, this proof is only valid for positive $r$), we have our desired result. [/hide]


Let $ a,b,c \in \mathbb{R^+}$ Prove that:
\[ {\sum_{cyc}\sqrt{(a^2+bc)(b^2+ca)}\ge\sqrt{3(a^2b^2+b^2c^2+c^2a^2)}+\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})}\]

[quote="apollo"]Let $ a,b,c \in \mathbb{R^ + }$ Prove that:
\[ {\sum_{cyc}\sqrt {(a^2 + bc)(b^2 + ca)}\ge\sqrt {3(a^2b^2 + b^2c^2 + c^2a^2)} + \sqrt {abc}(\sqrt {a} + \sqrt {b} + \sqrt {c})
}\]
[/quote]
By Cauchy we obtain:
$ \sum_{cyc}\sqrt {(a^2 + bc)(b^2 + ca)}\geq\sum_{cyc}\left(\sqrt {a^3b} + \sqrt {a^3c}\right).$
Let $ a = x^2,$ $ b = y^2$ and $ c = z^2,$ where $ x,$ $ y$ and $ z$ are non-negative numbers.
Then it remains to prove that $ \sum_{cyc}(x^3y + x^3z - x^2yz)\geq\sqrt {3(x^4y^4 + x^4z^4 + y^4z^4)}.$
After squaring we obtain that the last inequality is equivalent to:
$ \sum_{cyc}(x^6y^2 + x^6z^2 - x^4y^4 - 2x^5y^2z - 2x^5z^2y + 2x^6yz - 3x^4y^2z^2 + 4x^3y^3z^3)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x - y)^2((x + y)^2z^4 - 2xy(x^2 + 3xy + y^2)z^2 +$
$ + 2xy(x^2 + y^2)(x + y)z + 2x^2y^2(x + y)^2)\geq0.$
But $ (x + y)^2z^4 - 2xy(x^2 + 3xy + y^2)z^2 +$
$ + 2xy(x^2 + y^2)(x + y)z + 2x^2y^2(x + y)^2\geq4xyz^4 - \frac {5}{2}xy(x + y)^2z^2 + xy(x + y)^3z =$
$ = \frac {xyz(x + y)^3}{2}\left(8\left(\frac {z}{x + y}\right)^3 - 5\left(\frac {z}{x + y}\right) + 2\right) =$
$ = \frac {xyz(x + y)^3}{2}\left(8\left(\frac {z}{x + y}\right)^3 + 1 + 1 - 5\left(\frac {z}{x + y}\right)\right)\geq$
$ \geq\frac {xyz(x + y)^3}{2}\left(6\left(\frac {z}{x + y}\right) - 5\left(\frac {z}{x + y}\right)\right) = \frac {xyz^2(x + y)^2}{2}\geq0.$


[quote="imber.matutinis"]A circle is inscribed in a right triangle with sides 6, 8, 10. (GBML problem -- local competition)

How do you find the radius? Is there some sort of formula? How do you get the formula?[/quote]

[hide]Hm. If i remember correctly, inradius = area $\div$ semi-perimeter, making the "answer" $\frac{24}{12}$, or ..$2$[/hide]

The inradius formula is $A=rs$, where $A$ is the area of the triangle, $r$ is the inradius, and $s$ is the semiperimeter.

it's (area)=(1/2)(inradius)(perimeter), which can be shown by splitting triangle $ABC$ with incenter $I$ into triangles $AIB$, $BIC$, and $CIA$.

or since it is a right triangle, you can do a walkaround using the congruent tangent segments to get it is $\frac{l_1+l_2-h}{2}=\frac{6+8-10}{2}=2$


Εστω τετραγωνο με κορυφες Α.Β,Γ,Δ .Στο εσωτερικο του θεωρουμε σημειο Μ ωστε ΜΓ=ΜΔ και οι γωνιες ΜΓΔ=ΜΔΓ=15.
ΝΔΟ=>το τριγωνο ΑΒΜ ισοπλευρο.

Ωραίο προβληματάκι!

Λύνεται πολύ έυκολα αν ληφθεί υπόψη ότι το Μ είναι μοναδικό. Συγκεκριμένα τότε μπορούμε να δουλέψουμε αντίστροφα και να θεωρήσουμε ένα $ M'$ τέτοιο ώστε $ ABM'$ ισόπλευρο. Τότε $ AM' \equal{} AD$, $ BM' \equal{} BC$. Τότε δουλεύοντας στα τρίγωνα $ ABD, BCD$ επειδή $ \angle{M'BC} \equal{} \angle{M'AD} \equal{} 30$ ki epeidi ta trigona einai isoskeli $ \angle{M'DA} \equal{} M'CB \equal{} 75\iff \angle{M'DC} \equal{} \angle{M'CD}\equal{}15$

Αρα το $ M'$ ταυτίζεται με το $ M$ και τέλος :)

Υπάρχουν πολλές προσεγγίσεις. :)  Μία είναι να πάρεις ένα ισόπλευρο έξω από το τετράγωνο. :o

Είχα βρει παλιά μια πιο τριγωνομετρική.

Αν η $ CM$ τέμνει την $ AD$ στο $ K$ και η κάθετη από το $ M$ στην $ AD$ την τέμνει στο $ L$ είναι $ KMD$ ισοσκελές (λόγω γωνιών) άρα $ KL\equal{}LD$.

Είναι $ tan15\equal{}2\minus{}sqrt3$ άρα $ DL\equal{}(2\minus{}sqrt3)a/2$ άρα $ AL\equal{}a\minus{}(2\minus{}sqrt3)a/2$
Όμως $ ML\equal{}a/2$ οπότε εύκολα βρίσκουμε την εφαπτομένη της $ <DAM$... :wink:

Τυχαια την θυμηθηκα...Θυμαμαι στην γ λυκειου που την προσπαθουσα και ειχα βρει 3 λυσεις....Αυτην την ασκηση δεν την βαριεσαι να ψαχνεις για οσο το δυνατον περισσοτερες αποδειξεις... :)Απο τις λιγες γεωμετρικες που θυμαμαι... :)

Αλλα πολυ γρηγορα νομιζω οτι λυθηκε η ασκηση....
ΑΣΚΗΣΗ 2
Θεωρουμε τετραγωνο ΑΒΓΔ.Να βρειτε τον γεωμετρικο τοπο των σημειων Μ ωστε $ (MM_1)^3 \plus{} (MM_2)^3 \plus{} (MM_3)^3 \plus{} (MM_4)^3 \equal{} K^3$ οπου M_1,M_2,M_3,M_4 οι προβολες του Μ πανω στις πλευρες του τετραγωνου και Κ δοσμενο ευθ.τμημα.

[quote="Αρχιμήδης 6"]ΑΣΚΗΣΗ 3
Θεωρουμε την  περιφερεια (Κ,R), τα σταθερα σημεια Α,Β και την μεταβλητη περιφερεια (C) η οποια τεμνει την (Ο,R) στα Γ,Δ και διερχεται απο τα σταθερα σημεια Α,Β.
ΝΔΟ=>Ηκοινη χορδη ΓΔ δερχεται απο σταθερο σημειο.[/quote]

Φίλε Αρχιμήδη6 (btw πώς σε λένε αν επιτρέπεται; ) μπορείς να εξηγήσεις λίγο το σημείο που λες ότι τέμνει την (Ο,R) . Βασικά λέω μήπως ήθελες να γράψεις (Κ,R) ή απλώς πρόκειται γαι διαφορετικό κύκλο? Αν θες να γράψεις (Κ, R) είναι βέβαια περίεργο αφού απ' τα σημεία A,Β,Γ,Δ μόνο ένας κύκλος μπορεί να περνάει. Αν απ' την άλλη πρόκειται για δαιφορετικό κύκλο (τυχαίο) τότε μου φαίνεται λίγο περίεργη η άσκηση. Σε κάθε περίπτωση θα ήθελα να το εξηγήσεις λίγο αν μπορείς.

Θεωρώ δύο τυχαίους κύκλους που περνούν από τα Α ,Β και τέμνουν την περιφέρεια Κ(Ο,R) στα σημεία Γ,Δ (ο ένας)

και Γ' ,Δ΄ . Τότε οι ευθείες ΑΒ ,ΓΔ,Γ'Δ' συντρέχουν στο ριζικό κέντρο των τριων αυτών κύκλων .Άρα όλες οι ευθείες $ C_iD_i$ τέμνονται πάνω στην ΑΒ  :)

Σωστα Σιλουανε... :)Παρακαλω για την [color=red]ΑΣΚΗΣΗ 2[/color] καμια ιδεα?

[quote="Αρχιμήδης 6"]ΑΣΚΗΣΗ 3
Θεωρουμε την  περιφερεια (Κ,R), τα σταθερα σημεια Α,Β και την μεταβλητη περιφερεια (C) η οποια τεμνει την (Ο,R) στα Γ,Δ και διερχεται απο τα σταθερα σημεια Α,Β.
ΝΔΟ=>Ηκοινη χορδη ΓΔ δερχεται απο σταθερο σημειο.[/quote]

sorry tora thn eida...

Koita...eykola boreis na deikseis oti an $ AX,BY$ oi apostaseis ton A,B apo th CD tote $ \frac{AX}{BY}\equal{}\frac{\Delta^{A}_K}{\Delta^{B}_K}\equal{}ct$ ara h CD dierxetai apo stathero shmeio....qed.

[quote="Nick Rapanos]
Koita...eykola boreis na deikseis oti an $ AX,BY$ oi apostaseis ton A,B apo th CD tote $ \frac {AX}{BY} \equal{} \frac {\Delta^{A}_K}{\Delta^{B}_K} \equal{} ct$ ara h CD dierxetai apo stathero shmeio....qed.[/quote]

Διάβασε δύο ποστ παραπάνω την λύση μου  :P

ox..sorry silouan..eida to post tou arximhdh gia thn askhsh 2 kai opos anevaina pros ta pano berdeuthka kai koitaksa thn askhsh 3 apo oti fainetai.
tespa den peirazei etsi ki allios h lysh pou edosa einai diaforetiki alla de xreiazetai na epektathoume.

otan vro xrono tha koitakso kai th 2.


function
algebra
domain
inequalities proposed
inequalities

Let x,y,z>0 satisfied that:$xy+yz+zx$>=$max(x^2,y^2,z^2)$.Prove that:
(xy+yz+zx)(1/(x+y)(y+z)+1/(x+z)(z+y)+1/(y+x)(x+z))<= 5/4+2/3$\sum(\frac{x} {\sqrt(x+y)(x+z)})$
 I think my ineq is not really strong! ;) Can this ineq be stronger?

Are you sure it's not $\leq \frac 94$? If you take $z=0$ (you can take it sufficiently close to $0$ because the numbers have to be $>0$) then you get $2\ge \frac 94$, which is obviously wrong.

[quote="dreammath"]Latex'd out: Let $x,y,z>0$.Prove that:
$(xy+yz+zx)\cdot \left( \frac{1}{(x+y)(y+z)}+\frac{1}{(x+z)(z+y)}+\frac{1}{(y+x)(x+z)} \right) \geq \frac94$[/quote]

rewritten: $(xy+yz+zx)\cdot (2x+2y+2z) \geq \frac94\cdot (x+y)\cdot (y+z)\cdot (z+x)$

So $8\cdot (x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+3xyz)$ $\geq$ $9\cdot (x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+2xyz)$

Which means $x^2y+y^2x+x^2z+z^2x+y^2z+z^2y \leq 6xyz$

Which is clearly not true. You need to reverse the sign: it's $\cdots\leq\frac94$
(and 1,1,1 gives equality so it cannot be done stronger).

Sorry for my mistake...I am careless when I don't remember it,but I hope it is correct for this edition.
I have no time to check the comparision between it and the original problem.So sorry everybody. :blush:

It seems to me this isn't true!

$LHS = (xy+yz+zx)(1/(x+y)(y+z)+1/(x+z)(z+y)+1/(y+x)(x+z)) = [2(xy+yz+zx)(x+y+z)] / [(x+y)(y+z)(z+x)] \geq 2$

If we let $x=1$ and $y=z$ are very close to 0, we can make the RHS as close to $5/4+2/3 = 23/12$ as we like.

This disproves the inequality.

After edit this problem again,perhaps I will show my proof ;if I have any mistake which make the ineq is false,moderators can delete it. :(
We can easily find that any x,y,z>0 there exits an acuted triangle ABC satisfied:
x/ :sqrt: (x+y)(x+z)=cosA
y/ :sqrt: (x+y)(x+z)=cosB
z/ :sqrt: (x+y)(x+z)=cosC
(Because (cosA)^2+(cosB)^2+(cosC)^2+2cosAcosBcosC=1)
$LHS=(sinA)^2+(sinB)^2+(sinC)^2$ (*)
Due to xy+yz+zx>= max(x^2,y^2,z^2) so cosA,cosB,cosC lie in the interval I=(0,1/ :sqrt: 2] or pi/2>A,B,C>=pi/4
Lemma: If -f is a convex function and a,b,c lie in its domain then
$f(a)+f(b)+f(c)+f(\frac{a+b+c}{3}) <= 4/3[f(\frac{a+b}{2})+f(\frac{a+b}{2})+f(\frac{a+b}{2})]$ (Ref:"A less than B",via Titu Andreescu)
Use the lemma :with f(x)=(sinx)^2 , x lies in [pi/4,pi/2) then f"=2cos2x<=0 and we have:
(sinA)^2+(sinB)^2+(sinC)^2+3/4<=4/3((cosA/2)^2+(cosB/2)^2+(cosC/2)^2)(*')
From (*) (*') we have:
LHS<=-3/4+2/3(cosA+cosB+cosC)+2=RHS

[quote="dreammath"]Let $ x,y,z > 0.$ Prove that

$ (yz \plus{} zx \plus{} xy)\left[ \frac {1}{(z \plus{} x)(x \plus{} y)} \plus{} \frac {1}{(x \plus{} y)(y \plus{} z)} \plus{} \frac {1}{(y \plus{} z)(z \plus{} x)} \right] \leq \frac {9}{4}.$

I think my ineq is not really strong! ;) Can this ineq be stronger?[/quote]$ \sum{\frac {yz \plus{} zx \plus{} xy}{(z \plus{} x)(x \plus{} y)}}\leq\frac {2}{3}\sum{\frac {(x \plus{} y \plus{} z)^2}{(x \plus{} y \plus{} z)^2 \minus{} yz}} \leq \frac {9}{4}.$

The second one is [b]Bài toán 4.18[/b] at page 130 of the following book:

Phạm Văn Thuận, Lê Vĩ, [url=http://www.mathlinks.ro/viewtopic.php?t=156796]Bất đẳng thức, Suy luận & Khám phá[/url], NHÀ XUẤT BẢN ĐẠI HỌC QUỐC GIA HÀ NỘI, 2007.


let $ a,b,c,d$ such as $ a^2\plus{}b^2\plus{}1\equal{}2(a\plus{}b)$ and $ c^2\plus{}d^2\plus{}36\equal{}12(c\plus{}d)$. Prove that
$ 5\sqrt{2}\minus{}7\leq\sqrt{(a\minus{}c)^2\plus{}(b\minus{}d)^2}\leq 5\sqrt{2}\plus{}7$

the two equations you have represent circles and the min and max of the expression $ \sqrt{(a\minus{}c)^2\plus{}(b\minus{}d)^2}$ are simply the smallest and the largest distances between the two circles...


geometry
circumcircle
power of a point
radical axis

Let $ ABCD$ be a quadrilateral inscribed in a circle of center $ O$. Let M and N be the midpoints of diagonals $ AC$ and $ BD$, respectively and let $ P$ be the intersection point of the diagonals $ AC$ and $ BD$ of the given quadrilateral .It is known that the points $ O,M,Np$ are distinct. Prove that the points $ O,N,A,C$ are concyclic if and only if the points $ O,M,B,D$ are concyclic.

[i]Proposer[/i]: [b]Dorian Croitoru[/b]

Let $\tau_a,\tau_b,\tau_c,\tau_d$ be the tangents of $(O,R)$ through $A,B,C,D,$ respectively. $X \equiv \tau_b \cap \tau_d,$  $Y \equiv \tau_a \cap \tau_c.$ Inversion with respect to  $(O)$ transforms $A,B,C,D$ into themselves and $M \mapsto Y,$ $N \mapsto X,$  because of $R^2 \equal{} OM \cdot OY \equal{} ON \cdot OX.$  If $O,N,A,C$ are concyclic, then $X,A,C$ are collinear, i.e. $X \equiv \tau_b \cap \tau_d \cap AC$ $\Longleftrightarrow$ $ABCD$ is harmonic $\Longleftrightarrow$ $Y \equiv \tau_a \cap \tau_c \cap BD,$ i.e $Y,B,D$ are collinear $\Longleftrightarrow$ $O,M,B,D$ are concyclic.

Suppose that $OMBD$ is cyclic. Let $\{P\}=OM\cap BD$ we have $PM\cdot OP = PB\cdot PD = OP^2 - R^2$ or
$R^2 = OM\cdot OP$. Let $\{Q\}=ON\cap AC$,it remains to show that $ON\cdot OQ = R^2$.
We have also that $OQ\perp BD$ and $OP\perp AC$ $\Longrightarrow$ $PMNQ$ is cyclic,
so $ON\cdot OQ = OM\cdot OP= R^2$.
Similarly we prove the reciprocal.

[hide="A little different"]

Invert about pole $O$ with power $r^2$. $A,B,C,D$ go to themselves. Call the intersection of $ON$ and $AC$ $N'$. Call the intersection of $OM$ and $BD$ $M'$. Clearly, $NN'MM'$ is cyclic since $N,M$ are midpoints of chords. Therefore, $\triangle ONM' \sim \triangle OMN'$, yielding $ON \cdot ON' = OM \cdot OM'$. Now, if $ANOC$ is cyclic $I(N)$ must lie on $AC$, and is therefore $N'$. From there we obtain $I(M)=M'$ easily, which means $M$ is on the circumcircle of $\triangle OBD$. It works identically for the other direction.  [/hide]

Let $ ON \cap AC=T $ and $ OM \cap BD=S $.We  know that the points $ M $ and $ N $ lies on circle with dimametr $ OP $.Let $ Q $ is midpoint of $ OP $,then we know that $ TS \perp QO $.Also know that  the points $ B,M,O,D $ are concyclic if and only if the radical axis of $ (O),(Q),(BMOD) $ are concurrent at $ T $,and the radical axis of $ (O) $ and $ (Q) $ is perpendicular to $ (OQ) $, so is $ TS $, similarly for $ ANOC $.Hence we have $ ONAC $ is cyclic if and only if $ BMOD $ is cyclic if and only if $ TS $ is radical axis of $ (O) $ and $ (Q) $, so we are done !

Suppose, w.l.o.g. $AB>AD$.
If $BOMD$ was cyclic, $\angle OBD+\angle OMD=180^\circ\ (\ *\ )$, but $\angle OBD=90^\circ-\angle BAD$ and, for $(*)$ to be true, since $\angle OMC=90^\circ$, we need $\angle CMD=\angle BAD$, or $\angle ADM=\angle BAC=\angle BDC$, i.e. $BD$ is symmedian of $\triangle ADC$ and $ABCD$ is a harmonic quadrilateral, thus we are done.

Best regards,
sunken rock

[quote=patrickhompe]

Invert about pole $O$ with power $r^2$. $A,B,C,D$ go to themselves. Call the intersection of $ON$ and $AC$ $N'$. Call the intersection of $OM$ and $BD$ $M'$. Clearly, $NN'MM'$ is cyclic since $N,M$ are midpoints of chords. Therefore, $\triangle ONM' \sim \triangle OMN'$, yielding $ON \cdot ON' = OM \cdot OM'$. Now, if $ANOC$ is cyclic $I(N)$ must lie on $AC$, and is therefore $N'$. From there we obtain $I(M)=M'$ easily, which means $M$ is on the circumcircle of $\triangle OBD$. It works identically for the other direction.  [/quote]

Can someone explain why the following holds:

Now, if $ANOC$ is cyclic $I(N)$ must lie on $AC$, and is therefore $N'$.

[img]https://i.imgur.com/7erk2Th.png[/img]

We start off with a little lemma:
[b][u][color=#f00]Lemma:[/color][/u][/b] The quad $ONPM$ is cyclic
[b][u][color=#00f]Proof:[/color][/u][/b] $ \angle OMP = \angle ONP = 90 \implies \angle OMP + \angle ONP = 180$

So now we assume that $ONAC$ is cyclic.
We start off by defining a inversion $\psi(O,OC)$
Then by that inversion the circle around $ONAC$ is sent to $\overline{AC}$, and the circle around the quad $ONPM$ is sent to a straight line.
The inverses of points we denote by adding a $'$ to it.
So now we have since $N$ belongs to two circles we have that $N'$ is the intersection of two lines.
Now we get that the points $C,M,P,A,N'$ are all colinear.
So now we take a look at $M'$, we want to show that $\angle ONM' = 90$, if we show that we are done, since then we would get that the points $M',D,P,N,B$ are all colinear and they invert into a circle, hence proving that the quad $OMBD$ is cyclic.

To show this we just take a look at the angles $\angle ONM' = \angle OMN' = 90$, the last relation is true because of colinearity.
Hence we have shown that the quad $OMBD$ is cyclic . . . :D


ok, it sure was quite fun  :D

I, however, did horribly, ending up with a 64 :(

bleh, stupidity still reigns :|

YES IT WAS FUN!!!

TIME WAS KILLLLERRR!!!

8th grade- im slow, i probably got a 40 or something

You think only you got pwned...

I GOT PWNED TOO!

CARELESSNESS,

and a few were way above my head with limited time grrr..!!

oh well... bigger math tournamnet tomorrow

I did horribly and got a [hide="bad"] 34 [/hide] score

Oh well. It's just a mail in tournament, and I have 4 more years to go since I am currently in 8th grade.

I thought the test was a lot harder than last year.  I got an 85 (I think) last year, and althought I don't know what I got this year yet, I'm sure its much worse.

was it really that much worse? I have no prior experience in this contest, and I got a 34.

I don't know how I did compared to other contestants.

I think it was hard because I didn't know how to work 2 out of every three problems.

My scores on each round were 7,3,3,3,3,3,3,3,3,3

it wasn't hard... just hard to do it in the time limit

When the scores come up, I will have completely forgotten about it. Just like the Fall Startup Event.

It will be a few days from Grissom's tournament when it comes out ( January 8 ).  :oops:

Do you think that the scores will be down from last year?
Like if someone of the same ability got a 40 last year, would he got a 30 or something this year? Or not.

[quote="Elemennop"][quote="drunner2007"][quote="Elemennop"]I didn't think it was that bad..[/quote]

yeah i agree. they were 9 points for the first minute right?

i got an 81. i missed the dice problem because i only looked at the primes up to 6  :mad:[/quote]

I don't think there were divisions in time for the questions..[/quote]

oh ok. someone trying to tell me the point scheme without really knowing the rules.

Every round has three problems. If you get two problems right, you get 7 points. 1 problem right, 3 points. All three problems right, 10 points.  :)

75  :D

When are the results coming up on the website?

You can't see them yet, but if your local proctor graded them you should know.

(Ask Mr. Holbrook on math team this saturday)

oh yea
i remember he said that my score was around mid-50s, but i want the real score  :(

http://www.natassessment.com/2008_CTT/2008_CTT_Results/2008_CTT_Results_FrameDef.htm

FIRST PLACE OMG YESZ

Why were all the middle schoolers listed as 8th graders in the results? :maybe:

Exactly. I'm in 6th grade, not 8th. And I got 3rd. GO AAST!


[hide="A"]1-digit numbers arranged in alphabetical order?[/hide]
[hide="C"]Look and say: 1 -> one one -> 11 -> two ones -> 21 -> one two, one one -> 1211 -> etc.[/hide]

i_like_pie

did not identify which letters his answers go to.
I am taking his first line to go with (a).

Then his last two lines would go with (c).

-  - - -  -

For part (b), my best explanation is:

Multiply the 1st and 2nd terms and get the 3rd. 2*3 = 6

Multiply the 2nd and 3rd terms and get 18, which can be separated into
two terms, the 4th and 5th terms.  3*6 = 18

Multiply the 3rd and 4th terms and get 6, the 6th term (which is right after
the "18")  6*1 = 6

Multiply the 4th and 5th terms and get 8, the 7th term (which is right after
the "6")  1*8 = 8

Multiply the 5th and 6th terms and get 48, which can be separated into
two terms, the 8th and 9th terms (which is right after the 7th term,
the "8")   8*6 = 48

Continue this method to come up with other subsequent digits.

[quote="box of rocks"]i_like_pie

did not identify which letters his answers go to.
I am taking his first line to go with (a).

Then his last two lines would go with (c).[/quote]
Notice the "names" of the hide tags: A and C. :)

I agree with your answer for B.

i_like_pie,

in my corrupted version on AoPS, I can't see your hide tags, BUT
I can when I go on someone else's computer.

Thank you for correcting me.


Hello!
Don't know for sure wether this subject fits here, sorry if it doesn't.
I have to makea project involving the computation of the confluent hypergeometric function 1F1 with very high precision. I am interested if anybody can find, or knows a site where to be mentioned several practical domains where such high precision computation of this function is usefull. I need to write at the end of the project practical applications of the function's computation with high precision, and i know there is a site with such thing, but i cannot find it. Maybe you can help!
Thanks!!!

well i do not know what is hyper geometric function. Can u plz tell me?

(I am really  :blush: for not knowing this :mad: )

i will gladly answer your question but i have no idea how to use Latex, in order to write the formula for it.
anyway i have another program for writing formulas at home and i will write it when i get there, that is this evening. :)
and don't be  :blush: because it's not quite the most common thing.

well, i hope u will not mind to meet up my curiosity. Actually i want to know about urself. r u going to IMO or IOI? do u take part in online programming contest? do u practise in some sites?

Hehe.. so many Questions. don't hesitate to ask me any question! :P

no problem answering your questions.
but first i wanna know why you ask them. :)
after that i will answer you

no any particulur aim. just for meeting up curiosity. if u don't want to do that here then u may simply PM me. btw what about that function?


Άιντε Σιλουανέ, για να μη μας παραπονιέσαι ( :P ). Σίγουρα θα τη λύσεις σε κλάσματα του δευτερολέπτου (πολύ πιθανόν να την έχεις λύσει κιόλας, είναι κλασικό ασκησάκι), αλλά ήταν η πρώτη πρόχειρη που βρήκα.

Δίνεται $ \triangle ABC$ και ο εγγεγραμμένος σε αυτό κύκλος $ (O)$. Οι κάθετες στο $ O$ στις $ OA,OB,OC$, τέμνουν τυχαία εφαπτομένη του $ (O)$ στα σημεία $ M,N,P$, αντίστοιχα. Να αποδειχθεί ότι οι $ AM,BN,CP$, περνούν απ' το ίδιο σημείο.

Είναι γεγονός πως έχω παρατήσει το forum λόγω υποχρεώσεων ( :( ), αλλά I'll be back σε λίγο καιρό!

Στέλιος

Εχω την εντυπωση οτι βγαινει διανυσματικα... αλλα θελει παρα πολυ γραψημο και γινεται χαμος απο πραξεις, σιγουρα υπαρχει ευκολοτερος τροπος

Η άσκηση είναι εξαιρετική  :)  και ευχαριστώ για την αφιέρωση (αλλά μου πήρε κάτι παραπάνω από κλάσμα τα του δευτ να τη λύσω Ένα 15-λεπτο το έφαγε :P )
Έστω $ D,E,F$ τα σημεία επαφής του εγκεγραμμενου κύκλου με τις πλευρές $ BC,AC,AB$ αντίστοιχα .
Έστω $ S$ το σημείο του κύκλου από άγεται η τυχαία εφαπτομένη . Είναι εύκολο να δείξει κάποιος ότι ο πόλος της $ AM$ είναι το ίχνος της καθέτου από το $ S$ στη
$ EF$ .Όμοια και για τους πόλους των $ BN,CP$ . Τώρα όμως οι πόλοι αυτοί είναι συνευθειακά σημεία από το θεώρημα Simpson άρα και οι πολικές τους διέρχονται από το ίδιο σημείο  :)  QED

(με διανύσματα ή αναλυτική θέλει κανά δυο μέρες πράξεις  :P )

Πολύ ωραία Sil!  :)

Η λύση σου είναι ίδια με τη δική μου.

Η άσκηση είναι καθαρή εφαρμογή στις πολικές!

Στέλιος


Alyssa hammered some nails into her garage wall to hang her spare motorcycle tires on. If the tires have a radius of 10 inches, how far apart would the nails have to be so that the tires would not touch each other when hung next to each other?

[hide]they have to be separated by greater than 2 times the radius, or greater than 20 inches.[hide][/hide][/hide]

[hide]to seperate the tires they have to be twice the radius apart or 20 inches[/hide]


This question is for 8th graders and below only. Just like to see how middle schoolers (MAINLY, MATHCOUNTS STUDENTS) are doing at the AMC10.

What did you get on the AMC 10?

[quote="BFG"]This question is for 8th graders and below only. Just like to see how middle schoolers are doing at the AMC10.

What did you get on the AMC 10?[/quote]

^^^^^^^^^^^^^^^^^^^^^^^^^^^^

shouldn't this be in another forum?

I got 150. I found it much easier than last year.

i think this was harder than the past few years, except 2003.

what does AMC stand for?
how do we take it?

American Math Competition

ask your math teacher

-jorian

This was probably the hardest year according to me. i got a 145.5 on last years when practicing but got a way lower score this time...

it was my first time this year. I didnt prepare for it, so i think i got a 123.

dang! i got a 117...darn it.

@bpms: yes, SHE meant a 117.

[quote="now a ranger"]how can u get a 147?[/quote]
I'm pretty sure he meant 117.
I personally got a 132 grrrrrrrr...

you can't

the top scores can be:

150
25 correct
145.5
24 correct, 1 omit
144
24 correct, 1 wrong
141
23 correct, 2 omit
139.5
23 correct, 1 omit, 1 wrong
138
23 correct, 2 wrong

[quote="jhredsox"]what happened to you being really horendous under pressure?

jw,
jorian[/quote]

Huh? I got a 120 on the real thing  :(

But I got a 145.5 on a practice 2003

i don't even want to say what i got.........
i made soooooo many stupid mistakes, its not even funny........

I really suck, i screwed up the most stupid problems... I missed the cutoff by a #$\$$%@! 1.5 points!!!

I mean... how do you miss number 14?!

 I suck at geometry... I skipped 18 and 19

have you read the topic created by DPatrick "complaints about high scores"? well, this would qualify as an insulting complaint.  What if someone missed 14  and thought it was hard?

he said it to raise his own self-esteem.

[quote="ehehheehee"][quote="mathgeniuse^ln(x)"][quote="ehehheehee"]You got 14th at nationals.  If you don't make a mistake, you will get a perfect score.

How else can you prepare for the AMC 10, but do past competitions of it?  Probably AMC 12's are easier to find, though.

I don't agree with your signature.[/quote]

What do I have in my signature?

I have a quote of Altito
formula for pi^4/90
14th place at nats.[/quote]

Hmm, I don't see that...I see this:
$4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...)=\pi$
I such at Math[/quote]
That's his old sig.

[quote="ehehheehee"]Does anybody but me see his old sig?  Is it just my computer?[/quote]
No, I mean what he said is his old sig.
What you see is what I see.

an example of "brute forcing" a solution would be when a question asks how many factors does 720 have and you actually list them all out unstead of using a counting technique for finding how many there are without actually listing any of them.

[quote="life1"]hi - noticed several people on this forum including prolific H34N1 saying they brute forced their way through a problem - exactly what do you mean by that???[/quote]
I think you misused the word prolific.

Brute forcing is when you use The shoestring formula for example to find the area of a square with vertices (0,0), (0,1), (1,1), (1,0).
Basically, it's like overkill.

[quote="now a ranger"]a good example is on number 25 of AMC 10A 2007. I actually tried to find the years instead of using mods.[/quote]
Actually you're supposed to find the years.
Mods just make brute forcing easier.

really?

lol

okay then I'm wrong  :lol:

I was stupid and I wasted 15 minutes listing out every single year from 1960 to 2007.

thanks for the brute force explanation - I get it now

also, note to bpms - "Prolific": "turning out many products of the mind; abounding" -- I saw that other person post quite a bit; that's why I used the word - was meant to be sort of complimenatary/factual -- not to be taken SO exactly literally...


[quote="Bob"]Besides the AoPS books (I already have vol. 2) and Geometry Revisited (which I hear is too complex), are there any other books that will prepare you for typical AMC/AIME geometry problems?[/quote]

Richard is currently working on a Geometry book aimed at the levels from MATHCOUNTS through AMC and would inherently help with many AIME level geometry problems.  The books will include an enormous number of problems and exercises subject by subject and the subject coverage will be far broader than in AoPS Volume 1.

[quote="joml88"][i]Challenging Problems in Geometry[/i][/quote]

Who's the author Joe? I might check that out :lol:


The roots of x^4 - x^3 - x^2 - 1 = 0 are a, b, c, d. Find p(a) + p(b) + p(c) + p(d), where p(x) = x^6 - x^5 - x^3 - x^2 - x.

Anyone?

Hmm, I managed to solve this using the Newton-Girard identities (Newton's Sums). I let the desired sum be (a^6 +b^6 + c^6 + d^6) - (a^5 + b^5 + c^5 + d^5) - (a^3 + b^3 + c^3 + d^3) - ...

However, does anyone know of a different method to solve this problem, since using Newton-Girard identities gets tedious with high exponents...

If you want a hint.. (haven't gone all the way through but it should simplify it):

[hide]

you can simplify things by noting that substituting a,b,c,d into the initial quartic gives you things equal to 0. Take the 'a' case, maybe multiply by a:^2: and subtract from p(a), do that again and I think you may get everything down to quadratics..

[/hide]

Ah, I think I see. So, after tinkering around with the polynomials a bit, I get P(x) = (x^2 + 1)Q(x) + x^2 - x + 1. After applying Newton's Sums to a much smaller set of numbers, we have the sum S = 3 - 1 + 4 = 6. Thanks for the insight.


If $ N$ stands for the set of natural numbers, then of the following, the unbounded set is -

[b](A)[/b] $ X\equal{}\{x |x \equal{} \frac{1}{n} , n\in N.$

[b](B)[/b] $ X\equal{}\{x |x \equal{} 2^n , n\in N.$

[b](C)[/b] $ X\equal{}\{x |x\equal{}\frac{1}{2^n},n \in N\}$

[b](D)[/b] $ W\equal{}\{x|x \in N, x < 4532\}$

Please give answer with answer. Thanks a lot.

I think this does not belong here, but rather in a HS Basics/Interm Forum.
I don't know this though.

You seem to have omitted \} on the right of (A) and (B).

Also, that should really be n\in \mathbb{N} (that is, $ n\in\mathbb{N}$) instead of n\in N.

And come to think of it, for (A), I might consider \left\{ and \right\} instead of \{ and \}:

$ X\equal{}\left\{x\mid x\equal{}\frac1n,\ n\in\mathbb{N}\right\}.$


symmetry
number theory
relatively prime
probability
expected value

For each permutation $ a_1, a_2, a_3, \ldots,a_{10}$ of the integers $ 1,2,3,\ldots,10,$ form the sum
\[ |a_1 \minus{} a_2| \plus{} |a_3 \minus{} a_4| \plus{} |a_5 \minus{} a_6| \plus{} |a_7 \minus{} a_8| \plus{} |a_9 \minus{} a_{10}|.\]
The average value of all such sums can be written in the form $ p/q,$ where $ p$ and $ q$ are relatively prime positive integers.  Find $ p \plus{} q.$

Consider the sum of the described sums.  The term |a-b| (a and b in that order) will occur 8! times.  The value 9 (can be a result of |10-1| or |1-10|) occurs 2(8!) times.  The value 8 (can be a result of |10-2|, |2-10|, |9-1|, or |1-9|) occurs 4(8!) times.  By similar reasoning, 7 occurs 6(8!) times, 6 occurs 8(8!), and 5 occurs 10(8!) times.

So, the average is \[ \frac{9(2)(8!)+8(4)(8!)+7(6)(8!)+6(8)(8!)+5(10)(8!)}{10!} = \frac{18+32+42+48+50}{(9)(10)} = \frac{190}{90} = \frac{19}{9}  \]

So, \[ p+q = 19+9 =28  \]

wat about the differences such as 1, i got a different answer.  The average is basically the average of the possible values times 5.  so basically, we just sum up all possible values which comes up to 165, divide by the number of them, and multiply by five giving 55/3.  I checked it up on Kalva.

Also, how can it possibly be 19/9.  The total sum must be at least 5.

[hide]
$\boxed{\frac{55}{3}}$ - I checked it with the official solutions too -

Because of symmetry, we may find all the possible values for $|a_n - a_{n-1}|$ and multiply by the number of times it appears $5 * 8!$ and take that over $10!$ since that's the number of total permutations.

To find all possible values for $|a_n - a_{n-1}|$ we have $(1 - 10) + (1 - 9) + \ldots + (1-2) + (2-1) + (2-3) + \ldots + (2 - 10) + \ldots (10 -9)$

This is equivalent to \[ 2\sum\limits_{k=1}^{k=9}\sum\limits_{j=1}^{k}j = 330 \] Now we multiply it by $5 * 8!$ because if you fix $a_n$ and $a_{n+1}$ there are still 8! spots for the others and you do this 5 times because there are 5 places $a_n$ and $a_{n+1}$ can be.

Therefore, the answer is $\frac{330 * 8! * 5}{10!} = \frac{55}{3}$[/hide]

[hide="Quick Solution"]

Let $X_1$ be the expected value of $|a_1-a_2|$.  Then by expected linearity, the expected value of the other sums is the same, so the answer is hence $5$ times $E[X_1]$.  Now, notice that

\begin{eqnarray*} E[X_1]&=&\frac{9\times 1+8\times 2+7\times 3+6\times 4+5\times 5+4\times 6+3\times 7+2\times 8+1\times 9}{\binom{10}{2}} \\ &=& \frac{11}{3} \end{eqnarray*}
So the answer is therefore $5\times \frac{11}{3}=\frac{55}{3}$. [/hide]

[quote=akubisa]Please help me[/quote]

Can you translate to english  :)


algebra
polynomial
group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Peter Scholze told me the following nice problem:

Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$.
Let $\alpha \in \overline{\mathbb{Q}}\backslash \mathbb{Q}$ and let $K \subset \overline{\mathbb{Q}}$ be a maximal extension of $\mathbb{Q}$ in respect to not containing $\alpha$ (so $\alpha \not\in K$, but $\alpha$ in every nontrivial extension of $K$).
Let $G$ be the Galois group of $\overline{\mathbb{Q}}$ over $K$.

Show that either $G = \mathbb{Z}/ 2 \mathbb{Z}$ or $G = \mathbb{Z}_{p}$ ($p$-adic integers) for some prime $p$.

(Looking back at my proof, I'm not sure if I really used that we work over $\mathbb{Q}$ only)

Nice indeed :).

Let's first take care of the case $\left[\overline{\mathbb Q}: K\right]<\infty$. In this situation, $K$ is a field which is not algebraically closed, but whose algebraic closure is a finite extension. By the Artin-Schreier Theorem, the degree of this finite extension must be $2$, so the desired result follows in this case.

From this point onwards we assume that $\overline{\mathbb Q}$ is an infinite extension of $K$, and the aim will be to prove that the Galois group is some $\mathbb Z_{p}$. If we manage to show that the finite extensions of $K$ in $\overline{\mathbb Q}$ form a tower of cyclic extensions of degrees $p^{n},\ n\ge 0$ (one for each $n$) for some prime $p$, then we're done: it will follow from this that the Galois group of $\overline{\mathbb Q}/K$ is the projective limit of the groups $\mathbb Z/p^{n}\mathbb Z$, which is precisely $\mathbb Z_{p}$.

First we prove that $K(\alpha)/K$ is cyclic of degree $p$ for some prime $p$.

This extension is Galois: let $\alpha=\alpha_{1},\ldots,\alpha_{n}$ be the roots of the minimal polynomial $f$ of $\alpha$ over $K$. Each $K(\alpha_{i})$ contains $K$ strictly, so, by the maximality assumption on $K$, it must contain $K(\alpha)$ as well. Since all $K(\alpha_{i})/K$ have degree $n$, they all coincide, which means that $K(\alpha)$ is the splitting field of $f$, and hence a normal extension of $K$. It's also finite and separable, of course, so the extension is indeed Galois. If its Galois group were not cyclic of prime order, it would have a non-trivial proper subgroup. The fixed field of this subgroup would then be an extension of $K$, lying strictly between $K$ and $K(\alpha)$. Again, this contradicts the maximality assumption on $K$, and finishes the proof of the fact that $K(\alpha)/K$ is cyclic of degree $p$.

Now let's show that all finite extensions of $K$ in $\overline{\mathbb Q}$ have order $p^{n}$ for some $n\ge 0$.

Assume the contrary. Then we can find a Galois extension $F/K$ whose degree is not a power of $p$. The fixed field of a Sylow $p$-subgroup of $\mbox{Gal}(F/K)$ has degree greater than $1$ and coprime to $p$ over $K$. Hence it is a field strictly larger than $K$ which cannot contain $\alpha$; contradiction.

The final step of the proof consists in showing that for each $n$ there is precisely one extension of $K$ in $\overline{\mathbb Q}$ of degree $p^{n}$, which is cyclic.

By our assumption that $\overline{\mathbb Q}/K$ is infinite, it follows that there are extensions of $K$ of degree $p^{n}$ for any $n$. Choose an extension $F/K$ of degree $p^{n}$, which we may assume Galois (otherwise replace it with a Galois extension of even larger degree). $G=\mbox{Gal}(F/K)$ is a group of order $p^{n}$, which by elementary group theory has subgroups of all possible orders (i.e. all divisors of $p^{n}$). Moreover, a subgroup of $G$ of order $p^{m}$, say, is part of a chain of subgroups of orders $1,p,\ldots,p^{m-1},p^{m},\ldots,p^{n}$. The fixed field of a subgroup of $G$ of order $p^{n-1}$ has degree $p$ over $K$. Since there is only one such field, $K(\alpha),\ G$ has only one subgroup $H$ of order $p^{n-1}$. All subgroups of $G$ of order $\le p^{n-2}$ are thus subgroups of $H$, by the preceding discussion, so we can repeat the procedure with $H$ instead of $G$ and conclude that $G$ has precisely one subgroup of order $p^{m}$ for every $m\le n$, which, in turn, implies that $G$ is cyclic. The existence and uniqueness of cyclic extensions of $K$ of degrees $p^{n},\ n\ge 0$ follow from this.

Grobber's proof immediately generalizes to arbritary PERFECT fields $F$ in place of $\mathbb{Q}$; indeed, grobber uses separability at one point. If we took some element $\alpha\in \overline{F}\backslash F^{\text{sep}}$ with $F$ nonperfect, then there was some field $K$ satisfying the hypotheses of the problem s.t. $F^{\text{sep}}\subseteq K$; therefore $\text{Gal}(\overline{F}/K)\subseteq \text{Gal}(\overline{F}/F^{\text{sep}})=0$. Of course, this gives just one more possibility in the general case.


why is the field just outside a conductor supposedly perpendicular to it;s surface

Is it? I haven't seen that claim anywhere...

yes it is ...let me see if i can get u some site

i just started trying wll get you sorry forgot busy with... :(

see  [url=http://physics.bu.edu/~duffy/PY106/Electricfield.html]here[/url] in "The field around a charged conductor"  :wink:

the book had the same explaination as thsi guy has written but that doesn't seem that logical because they are calculating the field outside the conductor so hoew does t ever matter if it has a tangential component outside... :maybe:

Ah, well, that's another proof of the type "if it's like that, then it isn't electrostatics" ;) It's pretty logical to me...

[quote="Djole"]if it's like that, then it isn't electrostatics"[/quote]
 :rotfl:
but that's what most laws in electrostaics are good physical approximations :maybe:

Well, looked at like that, yes, they are approximations, just as the entire field of electromagnetism is an approximation of quantum electrodynamics and so on...


calculus integration function derivative vector geometry rectangle

You would learn it in a multivariable calculus course.  I have no idea how to explain it without calculus, though.  I don't have a good understanding of what it really means in a geometric sense.

The more general theorem is Stokes's Theorem, which, when written in the language of calculus, says

[tex]\displaystyle \int\limits_{\partial\Omega} \omega = \int\limits_\Omega d\omega.[/tex]

What this means is that when we integrate a differential form (think of this as a function, more or less) on the boundary of a region, we get the same value as we do when integrating the derivative of the form throughout the entire region. If you think of integrals as ``areas'' or ``volumes,'' you can sort of understand what it says. To me, it's a completely nonobvious result, but maybe it's obvious to other people.

Green's Theorem is a special case of Stoke's Theorem. And both theorem's are really just the Fundamental Theorem of Calculus in 2 and 3-Dimensions respectively.

A geometric interpretation of Green's Theorem is that the total amount of microscopic "curl" of a vector field inside of a closed curve equals the macroscopic "curl" of the vector field around the closed curve.

It would be easier if I drew a picture, but I'm too lazy right now. But think of it like this. If you draw a grid of rectangles in the plane, and imagine going around each one in a counterclockwise direction - you will go along most of the edges twice (once for each rectangle that shares that edge), but you will go along the edges in the opposite direction for each rectangle. So each edge that you go along twice you can think of as canceling each other out.

But there are some edges that you won't go along twice (can you see which ones?). They are the ones along the outside of your grid of rectangles because they're not mathched by the edge from another rectangle. So these edges won't get canceled out - and in total they all add up to go counterclockwise around the outside of the grid of rectangles.

That.. in a nutshell is Green's Theorem. You will learn about this in a class on multivariable calculus. Green's Theorem and Stoke's Theorem are two of the major topics of any such course. Multi is usually considered a sophomore college math course, and it's usually required for anyone wanting to major in engineering or a physical science.

[quote="JS1527"]I asked my dad how much math he took in college and he said that "he learned a lot past Green's theorem."  Can anyone explain Green's theorem to a person who does not know calculus? If not, can you tell me what math class you would learn it in?  Thanks![/quote]

Basically, after reading everyone else's explanation, and thinking about it for a while myself... I don't think it's possible to explain Green's or even worse Stoke's Theorem without any reference to calculus ideas. There is a reason that the topics are covered in a very advanced (to most of the population) calculus course.

There are many things that a lot of people aren't taught until they get quite far in math not because they won't understand the statement but because much more mathematics must be developed to prove the result. For example, consider the Brouwer Fixed Point Theorem:

Consider a continuous bijection function from the closed unit disc to itself. There is at least one point that remains unchanged under this function.

In simpler language and more dimensions, suppose you have a cup of coffee. Now stir the coffee. It is reasonable that stirring is performed in a continuous fashion. There is at least one point (ok you have to suppose you can break it up beyond the molecules and consider it as a math problem) that ended up where it started -- although it may have moved in the process.

But how would you prove this result with only basic calculus? I would be stunned if someone could come up with a proof that relied only on basic calculus or something like that. In general one would use rather deep topological ideas such as the fundamental group to prove this theorem.


Given the nine-sided regular polygon $ A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$, how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $ \{A_1,A_2,...A_9\}$?

$ \textbf{(A)} \ 30 \qquad \textbf{(B)} \ 36 \qquad \textbf{(C)} \ 63 \qquad \textbf{(D)} \ 66 \qquad \textbf{(E)} \ 72$

I don't really know how to attack this problem or generally, this "type" of problem. Can anyone give me a full good solution?

By the way, I put in Intermediate because there is a similar problem in AIME 2002.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=378347#p378347

For anyone who wondered.  :D

Thank you!!

[hide]for each pair of verticies, there are 2 equilateral triangles associated with it (the other 2 verticies lie on opposite sides of the line connecting the initial 2 points), so we have

$2*\binom{9}{2}$

but we have over counted the triangles that have all 3 verticies on the nonagon (we have counted them 3 times, one time for each pair of verticies of that triangle)

so we subtract twice the number of equilateral triangles that have all 3 verticies on the nonagon

and it is easy to see that there are only 3 of these, so

$2*\binom{9}{2}-2*3=66 \implies \boxed{D}$[/hide]

Okay.. I understand the first part (especially about multiplying by 2) but I don't understand how did you know that we overcounted 3 of them. Sorry if this sounds stupid but I'm not sure how you got it.  :blush:

[hide]consider the triangle $\triangle A_1A_4A_7$, when we do the count, we choose every pair of 2 verticies triangles, so we count

$A_1A_4$
$A_1A_7$
$A_4A_7$

(the other vertex is implied to be there, it is just not one of the 2 verticies that we chose initially)

so that is counting them each 3 times, so we need to subtract 2 of those over counts[/hide]

Thanks. That clears things up a lot.

Just wondering, is there anyway to solve general cases for this type of problem though? Say for the $n$-gon and you're picking at least 2 vertices to form a equilateral triangle (assume that the $n$-gon is regular). What will be this amount in terms of $n$?

[quote="Altheman"][hide]well, if the $n$-gon is not divisible by 3, it is just

$2*\binom{n}{2}$, but if it is divisibly by 3, then it is

$2*\binom{n}{2}-2*\frac{n}{3}=n^2-\frac{5n}{3}$[/hide][/quote]

Hmm.. I'm not sure how you derive these though.. So, if 3 doesn't divide $n$, there is no "repeated" count? And is this just working for triangles or can be applies to other polygons (regular)?

I appreciate this a lot. I always need long time to comprehend combinatorics..  :oops:

[quote="JesusFreak197"]If $3$ doesn't divide $n$, then you can't possibly have equilateral triangles with all three vertices on the $n$-gon, because you can't pick three vertices with equal distances between each pair. I'm not sure how else to explain it. :?[/quote]

For a n-gon , you have n vertices . To form equilateral triangle , you have to divide to $n$ vertices into three equal parts . And that implies $n$ must be a multiple of $3$  ;)


How many pairs of integers can be picked from the set $\ {1000, 1001, 1002, . . . , 2000}$ can be added without carrying?

[hide]

For any of the numbers from $1000-1999$:

For the units, tens, and hundreds digits of the two numbers, we have $25$ different possible pairs: $0$ can be paired with $9$ numbers greater than or equal to it; $1$ with $7$, $2$ with $5$, ..., $4$ with $1$.

For the thousands digits, we only have one choice: $1$.

The total number of possible pairs is then $25^3\times2^3 = 125000$ (since order matters).

For the number $2000$, it can be paired with any other number in the set, so $1000$ pairs there.

Their sum equals $\boxed{126000}$.

[/hide]


inequalities
geometry
parallelogram
geometry proposed

If point $M$ is in  triangle $ABC$ then $MA+MB+MC\geq 2{at_{a}+bt_{b}+ct_{c}\over a+b+c}$, where $a,b,c$ are lenghts of sides triangle $ABC$ and $t_{a},t_{b},t_{c}$ are lenghts of medians triangle $ABC$.

[hide="Hint"]
   This problem is an application of Ptolemei s inequality:

   [i]Ptolemei's inequality[/i] Let $A, B, C, D$ be four arbitrary points. Then $AB.CD + AD.BC \geq AC.BD$

   If we call $A'$ the fourth vertex of the parallelogram $ABA'C$, then by applying Ptolemei's theorem to four points $B, M, C, A'$, we have $BM. A'C + CM. A'B \geq MA'.BC \geq (2t_a - MA).BC$. This yields that $c.BM + b.CM +a.Ma \geq 2at_a$

   By doing it similarly to the other vertices and adding all up, we obtain the inequality.
[/hide]


given the endless arithmetic progression with the positive integer members. one of those is an exact square---prove it.


How is coming in Serbia this year?? Say the names of your national teams.

Andi Reçi , Ornela Xhelilaj , Erixhen Sula, Ridgers Mema, Arbana ....., Niko .....

Greece Team: Giehaskiel Ilias, Papelis Constantine, Papadimitriou Dimitris,  Pagonakis Dimitris, Logothetis Fotis and Taratoris Euaggelos. Unfortunately I am first of the "deputy team" . If I do well on the BMO I will replace someone of the above in the IMO team.


Construct a triangle $ABC$ knowing: $\measuredangle A$, $AH_a=h_a$ (altitude from $A$) and $BM_b=m_b$ (the median from $B$).

I would also like to know if there is a trigonometric proof that this is possible.

Thanks

Let $AD$ the altitude and $BM$ the median

The perpendicuar from $M$ to the side $BC$ intersects $BC$ at $F$.

From the triangle $ADC$ we know that $AD=2MF$

So the length $MF$ is known ,so we can construct the right triangle $BMF$

[size=150][color=blue]Construction[/color][/size]

We construct the right triangle $BMF$ with hypotenuse $BM$ and the known leg $MF$
We find the symmetric point $F'$ of $F$ with respect to $M$, and we bring a parallel line $\ell$ from $F'$ to $BF$.
It's obvious that $A$ is somewhere on $\ell$ (because $AD=2MF$)

Now, we draw an arc, such that $\angle BKM = \angle A$ for every point $K$ on the arc.

Let $A$ be some of the intersections (at most two) of this arc with line $\ell$.

$AD = d(A,BF)=2 \cdot d(M,BF)$
So the ray $AM$ intersects the line $BF$, say at $C$.
$M$ is the midpoint of $AC$ (in the triangle $ADC$ the segment $MF$ is parallel to the base $AD$ and equal to its half)
So,in the triangle $ABC$ we have:
$AD$ is the altitude from $A$ , $BM$ is the median from $B$, and $\angle BAC$ is the given angle

Of course, we will not always have a solution.
The first condition is that $BM>\frac{AD}{2}$ so that the right triangle $BMF$ can be constructed.

The second condition is that the line $\ell$ intersects the arc.


What's the fastest way or how to check inqualities that true or not true?

If we knew that, then inequalities wouldn't appear on Olympiads anymore.

I want to know that when I use AM-GM or others theorem in an inequality.How to know that I'm in a right way?I'm not expert in inequalities. :blush:


vector
combinatorics unsolved
combinatorics

Let $ v_1, v_2, \ldots, v_{1989}$ be a set of coplanar vectors with $ |v_r| \leq 1$ for $ 1 \leq r \leq 1989.$ Show that it is possible to find $ \epsilon_r$, $1 \leq r \leq 1989,$ each equal to $ \pm 1,$ such that \[ \left | \sum^{1989}_{r\equal{}1} \epsilon_r v_r \right | \leq \sqrt{3}.\]

This is from IMO1989 LongList; obviously a very weak result, since $\sqrt{3}$ can be replaced with the lower value $\sqrt{2}$. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=386586&hilit=vectors[/url].


least common multiple
superior algebra
superior algebra unsolved

Prove that if G and H are cyclic groups of orders m and n, with (m,n) = 1. then G x H is cyclic. How many different generators does it have?

Thanks in advance!

[quote="ZetaX"]Hint: what order can the element $(a,b)$ have when $a,b$ are generators of $G,H$ respectively¿[/quote]

i am confused... the order would be m and n respectively... and we know m and n are relatively prime... right??

Yes, their orders would be $m,n$, but now we are looking for the smalles $k$ with $(a,b)^k = (a^k,b^k) = (e,e)$, so ...

so... i have no clue where to go from there

Well, you are looking for some $k$ such that $a^k=e$ and $b^k=e$.

Now a claim: every $k$ with $a^k=e$ is a multiple of $m$.
To prove this, write $k=qm+r$ with $0 \leq r < m$ (division with remainder) and look at what is given.

Similar, when $b^k=e$, then $k$ is a multiple of $n$.

Both conditions together tell us that $k$ is multiple of $m$ and $n$, and by the $\gcd(m,n)=1$ condition, we have that their least common multiple is $mn$ itself.

from that you get that $G X H$ is cyclic now how do you find generators....

We already have one: $(a,b)$, and $a,b$ where any generators of $G,H$, so all such pairs work.
Now look at $(c,d)$ where at least one of them (WLOG $c$) is not a generator, then do the same again as before to find the order of $(c,d)$, it will be to low.


How many solutions does the equation $ |x\plus{}2| \equal{} 2x$ have?

[quote="nikeballa96"]either x+2=2x, or -x-2=2x. solvint these, we get x=2 and x=-2/3, for $ \boxed{2}$ solutions.[/quote]

I thought this too, but the answer says $ \boxed {1}$.  Here's why.  When you do $ \mid x\plus{}2\mid$ for $ \frac{\minus{}2}{3}$, you get $ 1$ $ \frac {1}{3}$.  But when you do it for $ 2x$, you get $ \minus{}1$ $ \frac{1}{3}$.  So only $ x\equal{}2$ works.  Our answer is actually $ \boxed {1}$


Find all functions $f$ which maps integer to integer such that $2000f(f(x))-3999f(x)+1999x=0$ for all integer $x$

It is $2000( f(f(x) - f(x) ) = 1999( f(x) - x)$

Fix x.

So the RHS has a factor 2000.  Thus, $f(x) = 2000k_x + x$ for some $k_x$.

Then, $f(f(x)) - f(x) = 2000(k_{2000k_x + x} - k_x)$.

So the RHS has a factor $2000^2$.  Thus, $f(x) - x = 2000^2k_x + x$, for some $k_x$,

implying the LHS has a factor $2000^3$.. etc.

Contradiction.  So $f(x) - x = 0$.


modular arithmetic
number theory proposed
number theory

Find all $a,c,d,e,f,k \in Z^{+}$ such that:
          $2^{e}(2^{c}(2^{a}-3^{d+f+k-2})-3^{f+k-1})=3^{k}+1$
I have the solution for this problem.But it is very long (close 5 pages) and so that I want the other solution :)  :)

Looks like something related to the 3n+1 conjecture....(?)

We will prove that the only solutions are $ (a,c,d,e,f,k)\in\{(2,2,1,2,1,1),(5,1,2,2,2,1),(4,1,1,1,1,2)\}$.

Here is a labelling for a few equations we will refer to.
$ A: 2^e(2^c(2^a\minus{}3^{d\plus{}f\plus{}k\minus{}2})\minus{}3^{f\plus{}k\minus{}1})\equal{}3^k\plus{}1$
$ B: 4(2^c(2^a\minus{}3^{d\plus{}f\plus{}k\minus{}2})\minus{}3^{f\plus{}k\minus{}1})\equal{}3^k\plus{}1$
$ C: 2^{a\plus{}c\plus{}1}\minus{}2^{c\plus{}1}3^{d\plus{}f\plus{}k\minus{}2}\minus{}2\cdot 3^{f\plus{}k\minus{}1}\equal{}3^k\plus{}1$
$ D: 2^c(2^a\minus{}3^{d\plus{}f\minus{}1})\equal{}3^f\plus{}1$
$ E: 4(2^a\minus{}3^{d\plus{}f\minus{}1})\equal{}3^f\plus{}1$
$ F: 2(2^a\minus{}3^{d\plus{}f\minus{}1})\equal{}3^f\plus{}1$
$ G: 2^a\equal{}3^d\plus{}1$
$ H: 2^a\minus{}5\equal{}3^{d\plus{}1}$
$ I: 2^{a\plus{}c}\minus{}2^c3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$
$ J: 2^{a\plus{}2}\minus{}4\cdot 3^{d\plus{}f\plus{}k\minus{}2}\minus{}2\cdot 3^{f\plus{}k\minus{}1}\equal{}3^k\plus{}1$
$ K: 2^{a\plus{}3}\minus{}8\cdot 3^{d\plus{}f\plus{}k\minus{}2}\minus{}2\cdot 3^{f\plus{}k\minus{}1}\equal{}3^k\plus{}1$
$ L: 2^{a\plus{}1}\minus{}2\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$
$ M: 2^{a\plus{}2}\minus{}4\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$
$ N: 2^{a\plus{}3}\minus{}8\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$

During the course of proof, we shall follow this scheme:

1) $ A$ is reduced to $ B$ (with $ e\equal{}2$) or $ C$ (with $ e\equal{}1$).
2) $ B$ is reduced to $ D$ (with $ k\equal{}1$).
3) $ D$ is reduced to $ E$ (with $ c\equal{}2$) or $ F$ (with $ c\equal{}1$).
4) $ E$ is reduced to $ G$ (with $ f\equal{}1$).
5) $ F$ is reduced to $ H$ (with $ f\equal{}2$).
6) $ G$ is solved.
7) $ H$ is solved.
8) $ C$ is reduced to $ I$ (with $ k\equal{}2$), $ J$ (with $ c\equal{}1$) or $ K$ (with $ c\equal{}2$).
9) $ I$ is reduced to $ L$ (with $ c\equal{}1$), $ M$ (with $ c\equal{}2$) or $ N$ (with $ c\equal{}3$).
10) $ L$ is solved.
11) $ M$ is solved.
12) $ N$ is solved.
13) $ J$ is reduced to a special case of $ I$.
14) $ K$ is reduced to a special case of $ I$.

[b]Accomplishing 1):[/b]

When divided by $ 8$, for $ k\equal{}1,2,3,4,\dots$, RHS of $ A$ gives the following remainders: $ 4,2,4,2,\dots$. On the other hand, if $ e\geqslant 3$, the LHS is divisble by $ 8$. Therefore, the only possibilities are $ e\equal{}2$ and $ e\equal{}1$, what reduces $ A$ to $ B$, respectively $ C$ (because $ 2(2^c(2^a\minus{}3^{d\plus{}f\plus{}k\minus{}2})\minus{}3^{f\plus{}k\minus{}1})\equal{}3^k\plus{}1\iff C$.

[b]Accomplishing 2):[/b]

Modulo $ 179,\!361$, for $ i\equal{}0,1,2,\dots$ remainders of numbers of the form $ 2^i$ are:
$ [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 82783, 165566, 151771,$
$ 124181, 69001, 138002, 96643, 13925, 27850, 55700, 111400, 43439, 86878, 173756, 168151, 156941, 134521, 89681]$
(square brackets denote the period which is further repeated), and of numbers of the form $ 3^i$:
$ 1,3,9,[27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 172719, 159435, 119583]$.
Among remainders from periods there is no solution of $ 4(2^c(2^a\minus{}3^{d\plus{}f\plus{}k\minus{}2})\minus{}3^{f\plus{}k\minus{}1})\equiv 3^k\plus{}1\pmod{179,\!361}$ (because we are asking that the RHS equals $ 28, 82, 244, 730, 2188, 6562, 19684, 59050, 177148, 172720, 159436$ or $ 119584$, while it is not possible to get any of these values on the LHS if we are allowing only remainders from periods—a computer help is handy at this moment). Therefore, at least one power of $ 3$ from $ B$ equals $ 9$ or $ 3$ (because the exponents are from $ \mathbb{Z}^\plus{}$). The following cases arise:
• $ 3^k\equal{}9$: Therefore, $ k\equal{}2$. This reduces $ B$ to $ 4(2^c(2^a\minus{}3^{d\plus{}f})\minus{}3^{f\plus{}1})\equal{}10$, what is immediately impossible.
• $ 3^k\equal{}3$: Therefore, $ k\equal{}1$. This reduces $ B$ to  $ 4(2^c(2^a\minus{}3^{d\plus{}f\minus{}1})\minus{}3^f)\equal{}4\iff 2^c(2^a\minus{}3^{d\plus{}f\minus{}1})\minus{}3^f\equal{}1$, and $ B$ is reduced to $ D$.
• $ 3^{f\plus{}k\minus{}1}\equal{}9$: Therefore, $ f\plus{}k\minus{}1\equal{}2$, what is possible only if $ k\equal{}2$ (and $ f\equal{}1$) or $ k\equal{}1$ (and $ f\equal{}2$), both of which cases were already handled.
• $ 3^{f\plus{}k\minus{}1}\equal{}3$: Therefore, $ f\plus{}k\minus{}1\equal{}1$, what is possible only if $ k\equal{}1$, what is already handled.
• $ 3^{d\plus{}f\plus{}k\minus{}2}\equal{}9$: Therefore, $ d\plus{}f\plus{}k\minus{}2\equal{}2$, what again implies $ k\equal{}2$ or $ k\equal{}1$.
• $ 3^{d\plus{}f\plus{}k\minus{}2}\equal{}3$: Therefore, $ d\plus{}f\plus{}k\minus{}2\equal{}1$, and again $ k\equal{}1$.

[b]Accomplishing 3):[/b]

This is done absolutely analogously to 1).

[b]Accomplishing 4):[/b]

Modulo $ 1,\!971$, for $ i\equal{}0,1,2,\dots$ remainders of numbers of the form $ 2^i$ are:
$ [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 77, 154, 308, 616, 1232, 493, 986]$, and of numbers of the form $ 3^i$: $ 1, 3, 9, [27, 81, 243, 729, 216, 648, 1944, 1890, 1728, 1242, 1755, 1323]$. Among remainders from periods there is no solution of $ 4(2^a\minus{}3^{d\plus{}f\minus{}1})\equiv 3^f\plus{}1\pmod{1,\!971}$. Therefore, at least one power of $ 3$ from $ E$ equals $ 9$ or $ 3$. The following cases arise:
• $ 3^f\equal{}9$: Therefore, $ f\equal{}2$. This reduces $ E$ to $ 4(2^a\minus{}3^{d\plus{}1})\equal{}10$, what is immediately impossible.
• $ 3^f\equal{}3$: Therefore, $ f\equal{}1$. This reduces $ E$ to  $ 4(2^a\minus{}3^d)\equal{}4\iff 2^a\minus{}3^d\equal{}1$, and $ E$ is reduced to $ G$.
• $ 3^{d\plus{}f\minus{}1}\equal{}9$: Reduces to $ f\equal{}2$ or $ f\equal{}1$.
• $ 3^{d\plus{}f\minus{}1}\equal{}3$: Reduces to $ f\equal{}1$.

[b]Accomplishing 5):[/b]

Absolutely analogously to the previous case (the same modulo works), we find out that at least one power of $ 3$ from $ F$ equals $ 9$ or $ 3$. The following cases arise:
• $ 3^f\equal{}9$: Therefore, $ f\equal{}2$. This reduces $ E$ to  $ 2(2^a\minus{}3^{d\plus{}1})\equal{}10\iff 2^a\minus{}3^{d\plus{}1}\equal{}5$, and $ F$ is reduced to $ H$.
• $ 3^f\equal{}3$: Therefore, $ f\equal{}1$. This reduces $ F$ to $ 2(2^a\minus{}3^{d\plus{}1})\equal{}4\iff 2^a\minus{}3^{d\plus{}1}\equal{}2$, what is immediately impossible.
• $ 3^{d\plus{}f\minus{}1}\equal{}9$: Reduces to $ f\equal{}2$ or $ f\equal{}1$.
• $ 3^{d\plus{}f\minus{}1}\equal{}3$: Reduces to $ f\equal{}1$.

[b]Accomplishing 6):[/b]

When divided by $ 8$, for $ d\equal{}1,2,3,4,\dots$, RHS of $ G$ gives the following remainders: $ 4,2,4,2,\dots$. On the other hand, if $ a\geqslant 3$, the LHS is divisble by $ 8$. Therefore, the only possibilities are $ a\equal{}2$ and $ a\equal{}1$. In the first case, $ G$ reduces to $ 4\equal{}3^d\plus{}1\iff 3^d\equal{}3\iff d\equal{}1$; this gives solution: $ (a,c,d,e,f,k)\equal{}(2,2,1,2,1,1)$. In the second case, $ G$ reduces to $ 2\equal{}3^d\plus{}1$, what is impossible (because $ d\in\mathbb{Z}^\plus{}$).

[b]Accomplishing 7):[/b]

Modulo $ 1,\!088$, for $ i\equal{}0,1,2,\dots$ remainders of numbers of the form $ 2^i$ are: $ 1, 2, 4, 8, 16, 32, [64, 128, 256, 512, 1024, 960, 832, 576]$, and of numbers of the form $ 3^i$: $ [1, 3, 9, 27, 81, 243, 729, 11, 33, 99, 297, 891, 497, 403, 121, 363]$. Among remainders from periods there is no solution of $ 2^a\minus{}5\equiv 3^{d\plus{}1}\pmod{1,\!088}$. Therefore the only possibilities are $ a\in\{1,2,3,4,5\}$, which give the following equations: $ \minus{}3\equal{}3^{d\plus{}1}$, $ \minus{}1\equal{}3^{d\plus{}1}$, $ 3\equal{}3^{d\plus{}1}$, $ 11\equal{}3^{d\plus{}1}$ and $ 27\equal{}3^{d\plus{}1}$. Only the last one of these has a solution over $ \mathbb{Z}^\plus{}$, namely $ d\equal{}2$, what gives solution: $ (a,c,d,e,f,k)\equal{}(5,1,2,2,2,1)$.

[b]Accomplishing 8):[/b]

Modulo $ 3,\!734,\!640$, for $ i\equal{}0,1,2,\dots$ remainders of numbers of the form $ 2^i$ are:
$ 1, 2, 4, 8, [16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576,$
$ 2097152, 459664, 919328, 1838656, 3677312, 3619984, 3505328, 3276016, 2817392, 1900144, 65648, 131296,$
$ 262592, 525184, 1050368, 2100736, 466832, 933664, 1867328]$,
and of numbers of the form $ 3^i$:
$ 1, 3, 9, [27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 1048329, 3144987, 1965681,$
$ 2162403, 2752569, 788427, 2365281, 3361203, 2614329, 373707, 1121121, 3363363, 2620809, 393147, 1179441,$
$ 3538323, 3145689, 1967787, 2168721, 2771523, 845289, 2535867, 138321, 414963, 1244889]$.
Among remainders from periods there is no solution of $ 2^{a\plus{}c\plus{}1}\minus{}2^{c\plus{}1}3^{d\plus{}f\plus{}k\minus{}2}\minus{}2\cdot 3^{f\plus{}k\minus{}1}\equiv 3^k\plus{}1\pmod{3,\!734,\!640}$. Therefore, either at least one power of $ 3$ from $ C$ equals $ 3$ or $ 9$, or at least one power of $ 2$ equals $ 4$ or $ 8$ (since both are $ \geqslant 4$). The following cases arise:
• $ 3^k\equal{}3$: Therefore, $ k\equal{}1$. This reduces $ C$ to $ 2^{a\plus{}c\plus{}1}\minus{}2^{c\plus{}1}3^{d\plus{}f\minus{}1}\minus{}2\cdot 3^f\equal{}4\iff 2^{a\plus{}c}\minus{}2^c3^{d\plus{}f\minus{}1}\minus{}3^f\equal{}2$, what is immediately impossible.
• $ 3^k\equal{}9$: Therefore, $ k\equal{}2$. This reduces $ C$ to $ 2^{a\plus{}c\plus{}1}\minus{}2^{c\plus{}1}3^{d\plus{}f}\minus{}2\cdot 3^{f\plus{}1}\equal{}10\iff 2^{a\plus{}c}\minus{}2^c3^{d\plus{}f}\minus{}3^{f\plus{}1}\equal{}5$, and $ C$ is reduced to $ I$.
• $ 3^{f\plus{}k\minus{}1}\equal{}3$: Reduces to $ k\equal{}1$.
• $ 3^{f\plus{}k\minus{}1}\equal{}9$: Reduces to $ k\equal{}1$ or $ k\equal{}2$.
• $ 3^{d\plus{}f\plus{}k\minus{}2}\equal{}3$: Reduces to $ k\equal{}1$.
• $ 3^{d\plus{}f\plus{}k\minus{}2}\equal{}9$: Reduces to $ k\equal{}1$ or $ k\equal{}2$.
• $ 2^{c\plus{}1}\equal{}4$: Therefore, $ c\equal{}1$, what reduces $ C$ to $ J$.
• $ 2^{c\plus{}1}\equal{}8$: Therefore, $ c\equal{}2$, what reduces $ C$ to $ K$.
• $ 2^{a\plus{}c\plus{}1}\equal{}8$: Reduces to $ c\equal{}1$.

[b]Accomplishing 9):[/b]

Modulo $ 65,\!520$, for $ i\equal{}0,1,2,\dots$ remainders of numbers of the form $ 2^i$ are: $ 1, 2, 4, 8, [16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768]$, and of numbers of the form $ 3^i$: $ 1, 3, [9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 46107, 7281, 21843]$. Among remainders from periods there is no solution of $ 2^{a\plus{}c}\minus{}2^c3^{d\plus{}f}\equiv 3^{f\plus{}1}\plus{}5\pmod{65,\!520}$. Therefore, at least one power of $ 2$ from $ I$ equals $ 2$, $ 4$ or $ 8$ (because both powers of $ 3$ are at least $ 9$, what is in the period). The following cases arise:
• $ 2^c\equal{}2$: Therefore, $ c\equal{}1$, what reduces $ I$ to $ L$.
• $ 2^c\equal{}4$: Therefore, $ c\equal{}2$, what reduces $ I$ to $ M$.
• $ 2^c\equal{}8$: Therefore, $ c\equal{}3$, what reduces $ I$ to $ N$.
• $ 2^{a\plus{}c}\equal{}4$: Reduces to $ c\equal{}1$.
• $ 2^{a\plus{}c}\equal{}8$: Reduces to $ c\equal{}1$ or $ c\equal{}2$.

[b]Accomplishing 10):[/b]

Modulo $ 5,\!440$, for $ i\equal{}0,1,2,\dots$ remainders of numbers of the form $ 2^i$ are: $ 1, 2, 4, 8, 16, 32, [64, 128, 256, 512, 1024, 2048, 4096, 2752]$, and of numbers of the form $ 3^i$: $ [1, 3, 9, 27, 81, 243, 729, 2187, 1121, 3363, 4649, 3067, 3761, 403, 1209, 3627]$. Among remainders from periods there is no solution of $ 2^{a\plus{}1}\minus{}2\cdot 3^{d\plus{}f}\equiv 3^{f\plus{}1}\plus{}5\pmod{5,\!440}$. Therefore the only possibilities are $ a\in\{1,2,3,4\}$, which give the following equations: $ 4\minus{}2\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$, $ 8\minus{}2\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$, $ 16\minus{}2\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$ and $ 32\minus{}2\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$. RHS of all of them is $ \equiv 5\pmod 9$, while LHS of all but the last one is not. We are, therefore, solving $ 32\minus{}2\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$. If it were $ f\plus{}1\geqslant 4$, then $ d\plus{}f\geqslant f\plus{}1\geqslant 4$ and the RHS would be $ \equiv 5\pmod{81}$, while LHS would not. Therefore, the only possibilities are $ f\in\{1,2\}$, which give the following equations: $ 32\minus{}2\cdot 3^{d\plus{}1}\equal{}14\iff 3^{d\plus{}1}\equal{}9\iff d\equal{}1$, what gives solution: $ (a,c,d,e,f,k)\equal{}(4,1,1,1,1,2)$, and $ 32\minus{}2\cdot 3^{d\plus{}2}\equal{}32\iff 3^{d\plus{}2}\equal{}0$, what is impossible.

[b]Accomplishing 11):[/b]

When divided by $ 8$, for $ f\equal{}1,2,3,4,\dots$, RHS of $ M$ gives the following remainders: $ 6,0,6,0,\dots$. On the other hand, the LHS is divisble by $ 4$ and not divisible by $ 8$. Therefore, $ M$ does not have solution.

[b]Accomplishing 12):[/b]

Modulo $ 16,\!320$, for $ i\equal{}0,1,2,\dots$ remainders of numbers of the form $ 2^i$ are: $ 1, 2, 4, 8, 16, 32, [64, 128, 256, 512, 1024, 2048, 4096, 8192]$, and of numbers of the form $ 3^i$: $ 1, [3, 9, 27, 81, 243, 729, 2187, 6561, 3363, 10089, 13947, 9201, 11283, 1209, 3627, 10881]$. Among remainders from periods there is no solution of $ 2^{a\plus{}3}\minus{}8\cdot 3^{d\plus{}f}\equiv 3^{f\plus{}1}\plus{}5\pmod{16,\!320}$. Therefore the only possibilities are $ a\in\{1,2\}$, which give the following equations: $ 16\minus{}8\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$ and $ 32\minus{}8\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$. RHS of both of them is $ \equiv 5\pmod 9$, while LHS of the first one is not. We are, therefore, solving $ 32\minus{}8\cdot 3^{d\plus{}f}\equal{}3^{f\plus{}1}\plus{}5$. If it were $ f\plus{}1\geqslant 4$, then $ d\plus{}f\geqslant f\plus{}1\geqslant 4$ and the RHS would be $ \equiv 5\pmod{81}$, while the LHS would not. Therefore, the only possibilities are $ f\in\{1,2\}$, which give the following equations: $ 32\minus{}8\cdot 3^{d\plus{}1}\equal{}14\iff 16\minus{}4\cdot 3^{d\plus{}1}\equal{}7$ and $ 32\minus{}8\cdot 3^{d\plus{}2}\equal{}32\iff 3^{d\plus{}2}\equal{}0$, both of which are clearly impossible. Therefore, $ N$ does not have solution.

[b]Accomplishing 13):[/b]

Analogously to steps 4) and 5) (the same modulo works), we find out that at least one power of $ 3$ from $ J$ equals $ 9$ or $ 3$. However, this has already been dealt with in step 8) (as $ J$ is a special case of $ C$).

[b]Accomplishing 14):[/b]

Modulo $ 78,\!840$, for $ i\equal{}0,1,2,\dots$ remainders of numbers of the form $ 2^i$ are:
$ 1, 2, 4, [8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 52232, 25624, 51248,$
$ 23656, 47312, 15784, 31568, 63136, 47432, 16024, 32048, 64096, 49352, 19864, 39728, 616, 1232, 2464,$
$ 4928, 9856, 19712, 39424]$,
and of numbers of the form $ 3^i$:
$ 1, 3, 9, [27, 81, 243, 729, 2187, 6561, 3363, 10089, 13947, 9201, 11283, 1209, 3627, 10881]$.
Among remainders from periods there is no solution of $ 2^{a\plus{}3}\minus{}8\cdot 3^{d\plus{}f\plus{}k\minus{}2}\minus{}2\cdot 3^{f\plus{}k\minus{}1}\equiv 3^k\plus{}1\pmod{78,\!840}$. Therefore, at least one power of $ 3$ from $ K$ equals $ 9$ or $ 3$ (because $ 2^{a\plus{}3}\geqslant 16$). However, this has already been dealt with in step 8).

The problem is finally solved. I think that this does not fill "close to five pages", more likely about three. :)

@Bojan: Wow! That is magic!! How on earth did you guess to look $ \bmod 179,\!361$, and $ \bmod 3,\!734,\!640$?

Thank you for your comment.

I was not really "guessing", it was a real pain in the a** to find those moduli. :) You have a few hints in this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=48431, but they are not as helpful as they seem at the first glance, because they tend to generate much larger moduli then needed (i. e., instead of $ 3,\!734,\!640$, we would get $ 272,\!628,\!720$), and it is of crucial importance here to use as small moduli as one can find, because the number of cases grows too fast. At some points, however, I had to bite the bullet and go on with these methods, but after a modulus had been found, I looked for a way to decrease it (looking at its divisors and trying them as possible moduli often turned out as a good idea). Most of the others were found by something what could be called "smart bruteforcing" (essentially bruteforcing, with a few minor shortcuts).

I cannot guarantee that these are the smallest possible moduli, though I am pretty confident that they are. All in all, they produce a reasonable number of cases.

If you have any more questions, feel free to ask, I will be glad to help if I can.


ABCD is a parallelogram and X and Y are the midpoints of BC and CD respectively. Prove that the area of $\bigtriangleup$ AXY is equal to $\frac{3}{8}$ the area of the parallelogram

[XCY]=[BCD]/4=[ABCD]/8 since 2CX=BC & 2CY=CD.
[ABX]=[ABC]/2=[ABCD]/4.
[ADY]=[ACD]/2=[ABCD]/4.
Therefore [AXY]=3[ABCD]/8.


Four numbers $ a, b, c$, and $ d$ satisfy $ 2^a2^b3^c3^d \equal{} 16,384 \times 59,049$. What is the average value of the four numbers?


[quote="DrKevin"]$ 1 \plus{} 2^3 \minus{} 4^5 \plus{} 6\times 7\times 8\times 9 \equal{} 2009$
$ 12 \minus{} 3 \minus{} 4^5 \plus{} 6\times 7\times 8\times 9 \equal{} 2009$
$ 9 \plus{} 8\times 7 \plus{} 6\times 54\times 3\times 2\times 1 \equal{} 2009$
$ 9 \plus{} 8\times 7 \plus{} 6\times 54\times 3\times 2^1 \equal{} 2009$
$ 9 \plus{} 87 \plus{} 65 \plus{} 43^2 \minus{} 1 \equal{} 2009$
$ 9\times 8 \minus{} 7 \plus{} 6\times 54\times 3\times 2\times 1 \equal{} 2009$
$ 9\times 8 \minus{} 7 \plus{} 6\times 54\times 3\times 2^1 \equal{} 2009$
$ 98 \minus{} 7 \plus{} 6\times 5\times 4^3 \minus{} 2\times 1 \equal{} 2009$
$ 98 \minus{} 7 \plus{} 6\times 5\times 4^3 \minus{} 2^1 \equal{} 2009$[/quote]

only on aops...
:roftl:

[quote="pinkmuskrat"]WOW, amazing!

Where did you get those?[/quote]
It's not hard, really  :D