geometry algebra polynomial search calculus integration rectangle

[quote="binomial_4eva"]is it possible to find the nth term or formula for a sequence by finding its area? any relation or connection between them ?

 1,4, 9, 16, 25, 36, 49, 64,........
for the above sequence the formula is Un=N^2

is there any other way to find the formula?
is there other ways to find formula for sequences ? other than how we normally find.

how about area? how about limits?  if no why ? any other way ? how ?[/quote]
The answer to your question, whether it is possible to find the nth term for a sequence, depends on the context. What you are asking is like 'can we fulfill our hunger without digesting food?'. There is no 'general way' of doing it. There are many ways to find formulas for sequences. At a higher point of view, there are an infinite number of different formulas that can describe a certain sequence. For example,

1,4, 9, 16, 25, 36, 49, 64,........

You don't know what the next term is. I can find a 7th degree polynomial using Lagrange's interpolation to describe the above terms and yet end up with a different 8th term. Now if you are telling me that 'but the next term is 81', you already know that the sequence is $U_{n}=n^{2}$. I can say that the next term is 101 or 1001.

Second thing is I don't know what you are talking about 'area' of a sequence. If you are asking how people derive area formulas like the area of a circle or the area of a triangle, there are posts in this website and you just need to search them or google the internet, there are way too many sites explaining how to derive such formulas.

What about limits? You just calculate them...

Asking 'any other way? how?' means you need to reread your book.

And lastly, stop double posting, it's SPAM

:spam:

As stated, u_n=n^2 whis is the area of a square with side n. I think this is what he/she meant by area.

[quote="binomial_4eva"]u didnt quite get what i mean,
i know how to find area of shapes, i am not a school boy, what i mean is something comletly different,
MDK got what i mean.
if we can have square number, 1, 4, 9, 16, 25, 36,49, 64, 81...
with this formula Un=N^2 which is the formula for finding area of a square with side n. then its possible that any formula for any sequence will have something to do with the area of a shape.for example trianguler numbers and formula for the shape triangle.and so+...
but my question was that, if we have area of a shape or area under a graph ( *integrals) can we find the formula for the sequence of that shape.in other word is it possible to find the next point on a graph by looking at its area ?
I am sure there is always a strong relation between this two, but bcoz its something new i cant put it in word, i wanna know if anyone agrees with me, i know its something that no one agrees at first coz its new and seems nonescence.
apologies for postin it twice at first, i am new in this.[/quote]

So you want a geometric approach to the problem?

Well, if a regular shape has n sides and a radius of length r, the area of it is $r^{2}ncos(\frac{180^{o}}{n})sin(\frac{180^{o}}{n})$  or $r^{2}ncos(\frac{\pi}{n})sin(\frac{\pi}{n})$ if you prefer radians.

[quote="binomial_4eva"]if we can have square number, 1, 4, 9, 16, 25, 36,49, 64, 81...
with this formula Un=N^2 which is the formula for finding area of a square with side n. then its possible that any formula for any sequence will have something to do with the area of a shape.for example trianguler numbers and formula for the shape triangle.and so+...[/quote]

Well, think about this... the area of a rectangle is base x height and it so happens that the square has base = height so that the area is the square of the side. I understand that you are trying to find a 'pattern' to come up with area formulas starting out from a sequence. A sequence is a just a bunch of numbers following a certain numerical pattern. Your example of triangular numbers can't deduce the area of an equilateral, not even an isosceles right triangle. In fact, they are called 'triangular numbers' because they are based on the pattern like

*
**
***
****

giving a resemblance to a triangle. It is hardly a geometrical fact.

[quote="binomial_4eva"]but my question was that, if we have area of a shape or area under a graph ( *integrals) can we find the formula for the sequence of that shape.in other word is it possible to find the next point on a graph by looking at its area ?
I am sure there is always a strong relation between this two, but bcoz its something new i cant put it in word, i wanna know if anyone agrees with me, i know its something that no one agrees at first coz its new and seems nonescence.
apologies for postin it twice at first, i am new in this.[/quote]

That's what integrals are for. It may not be a sequence that you get, but you will get an 'area function'. For example, integrating $\sqrt{a^{2}-x^{2}}$ from $-a$ to a certain value $x_{0}$ gives you the 'area function' of a shape bounded by the semicircle and the line $x=x_{0}$. It is not possible to find the next point on a graph because we can have different shapes with the same area; in other words, you can't predict the next point unless you have a pre-determined shape assigned, which happens the same with sequences, as I explained before.

The strong relation you are looking at is the numerical fact that links them. Same with the sequence 1, 8, 27, 64, 125, ... The first thing that comes to your mind is the volume of a cube with side n, isn't it? That's because it is a numerical fact.

Sorry if I sounded mean or disappointing or anything, but I hope you can learn something here and therefore continue studying the wonderful world of math :).

for different shapes different formulas, like the area under any shape  adds up to 1 in probability.
why 1 ?
i think  area of graphs always adds up to a particuler number.
if not to that number then to another number if the shape is different.it might have different formulas all together at the same time.

if i am saying nonescence, sorry but , because i always think something is there hidden from us.why should probabilities not add up to another number but 1 always? we might be able to make something else add up to 2.

[quote="binomial_4eva"]for different shapes different formulas, like the area under any shape  adds up to 1 in probability.
why 1 ?
i think  area of graphs always adds up to a particular number.
if not to that number then to another number if the shape is different.it might have different formulas all together at the same time.

if i am saying nonsense, sorry but , because i always think something is there hidden from us.why should probabilities not add up to another number but 1 always? we might be able to make something else add up to 2.[/quote]
The total probability of any space adds up to 1 because that's how it is defined. Why choose 1? It has some nice properties and easy to work with; it applies to real life problems reasonably in a simple way. You can choose 2 if you want, but you will have to develop a whole new theory of probability, and I'm sure it will be way more complicated than necessary, let alone that you have to check if your theory is consistent or not. Sometimes when you ask 'why' too much, too deep down, the questions become more philosophical than mathematical. And when that happens, it is not a part of this subforum's discussion. Why is $\pi=3.14159...$? Why not $\pi=3$? Why is a circle 'circular'? It all depends on your perception. A good thing learned from Geometry is that there are 'undefined terms' and 'axioms' or 'postulates'.

PS EDIT: I don't know if you are one of those savants, but certain people represent numbers with colors and shapes as if they were 'alive'. So yea...

"You can choose 2 if you want"

its impossible that the total probability of any space add up to 2.even if u try to build new prbblty theory.
so if impossible then why 1.
its not philosophical, its more mathematical,mathematics was invented with reasoning.not solving easy equations.


function
trigonometry
advanced fields
advanced fields unsolved

Let $A$ be the open connected subsets in $R^2$ with natural Topology.Is the closure of $A$ path-connected?

Saman Gharib

No, not necessarily. Take $A$ to be the region of the plane situated strictly between the graphs of the functions $f(x)=\sin\frac 1x,\ \forall x>0$ and $g(x)=\sin\frac 1x-x,\ \forall x>0$.


linear algebra
matrix
trigonometry
geometry
geometric transformation
rotation
vector

Let $ S$ be the set of all $ diagonal$ matrices belonging to $ M_{2X2}(C)$ (all $ 2X2$ complex matrices)

Let $ T$ be the set defined as:

$ A\in T  if  \exists \theta and B\in S; A\equal{}\begin{pmatrix}cos\theta &\minus{}sin\theta\\sin\theta &cos\theta\end{pmatrix}B\begin{pmatrix}cos\theta &sin\theta\\\minus{}sin\theta &cos\theta\end{pmatrix}$

let $ U$ be the set of all matrices expressable as products of finitely many elements belonging to $ T$.

the question is : Is $ U\equal{}M_{2X2}(C)$? if not, what are the conditions to be satisfied by an arbitrary matrix, so that it belongs to $ U$?

The set $ U$ actually represents all $ polarizers$ which can be manufactured using '$ waveplates$'.

$ A \in T$ means: there is an orthonormal basis of $ C^2$ made up of eigenvectors of $ A$.
We say $ A$ is a "normal" matrix.
Your trigonometric factors are rotations.  The two rotations are inverse of each other.  The columns of one of them are the vectors of the orthonormal basis I mentioned above.
Yes, any matrix is a product of normal matrices.  Probably you need only two factors.

Hi transseries,
$ A\in{T}$ implies $ AA^*\equal{}A^*A$ and $ A$ is normal.
The converse is false; let $ A$ be a normal matrix. Then there exists an unitary matrix $ U$ s.t. $ A\equal{}UBU^{*}$ where $ B$ is a diagonal matrix. $ U$ is not necessarily a real matrix.
More precisely if $ U$ is real then there exists a real orthonormal basis of $ A$-eigenvectors; thus you can't reach all the normal matrices.


ratio
arithmetic sequence
geometric sequence

Let $a_{n}$ be the $n$ th term of the arithmetic sequence with $a_{1}=7,$ the common difference $2,$ and $b_{n}$ be

 the $n$ th term of the geometric sequence with $b_{1}=\frac{1}{3},$ the common ratio $\frac{1}{3}.$
For the sequence $\{c_{n}\},$ If $\sum_{k=1}^{n}a_{k}b_{k}c_{k}=\frac{1}{3}(n+1)(n+2)(n+3)$ holds, then find $c_{n}$ and evaluate $\sum_{n=1}^{\infty}\frac{1}{c_{n}}.$

Regrettably your answer is incorrect. :(

Oh yeah, [hide]$c_{1}=\frac{24}{7}$, forgot that...[/hide]

There was something else I messed up in the original post, so I fixed that too...


I'm surprised that there isn't any mention of RTS.  They're not much worse than chess.

Turn based strategies maybe, but there's no way real time strategies can come close to chess. I can't do 100 keystrokes per minute!

I own in Rtss and FPSs but they dont have anything to do with strategy... other then this game I play that combines both.
 :lol:


Do you think Democracy is a good form of government? Before answering, really think about it. Here are some opening ideas:

Pros:
-the contentment of the people is a high priority in maintaining internal stability.
-people must have direct power over the government to prevent otherwise-inevitable corruption

Con:
-most people do not know enough to vote intelligently; even most "informed" people, do not have the Ph.D. training to know, for instance, what the economic policy of a government should be for a country as big as, say, the USA.
-people can be manipulated and thus cannot really prevent corruption

Ideas?

[quote="bubka"]Democracy is an okay form of government. But the associated ideas of liberty and freedom are just some poor jokes.[/quote]

If you don't think that democracy is that great, what do you think is better?

Bubka believes that communism is best I think.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100949[/url]

EDIT: I Do not belive that communism is best!

I might have been misunderstood

[quote="kheg_k"]Bubka believes that communism is best I think.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100949[/url][/quote]

I definately DON't agree with that! Why would you like communism?

First, about that communism thing : if people are so happy about communism, why can't they just vote for it.  Elect a communist president, maybe re-elect and then stop being communist for a couple of years again?
It seems like communism can't be pronounced without oppression :mad:  :o

If you have democracy, at least dictators can be ousted, people like Niyazov or Stalin would stand no chance.
Crucial is of course free participation in elections and freedom of speech, if the government decides what their people know, it's easy to make them choose you.

Although the actions of certain countries, especially like USA, sometimes make me wonder whether or not people who can't find their country on a map should be allowed to vote. It is true that if the people are ignorant the system fails.  Oh well, we don't even know for sure Bush was elected in 2000.

I do not believe that communism is the best form of government.

After reading the thread, I'm pretty certain that bubka isn't being serious.

Let us put aside the question of communism for the time being.
What do you people think of dictatorship as the alternative of democracy?

Dictatorship how? I think if the leader(s) of any group of people have power but are not accountable for their actions, things will not work well. Look at the Roman empire -- a few of the emperors were good leaders, but most of them just abused their power and left us with a bunch of horrific stories with which to entertain our friends.

in theory, the best government form would probably be a good monarch or dictator who looks out for the people but can also quickly and efficiently take action when needed.  in theory, communism is a great form of government: no one is ever hungry or without a home or unemployed, and people do what they can to help.  however, in practice, in long-term monarchy, eventually there would be a corrupt monarch, while communism is too easily manipulated by people taking advantage of the system.  a bad dictatorship is one of the worst government forms.

in practice, democracy is one of the best long-term government forms.  keep in mind a country that elects a leader with communist tendencies is still a democracy.  although the general masses are often unfit to rule themselves, although they are easily influenced by pre-existing prejudices, the founders in the beginning made certain aspects of the current u.s. system that some things cannot be changed.  if a majority wanted to, for example, banish non-christians, the constitution wouldn't allow it even if it was overwhelming majority vote.  however, the general masses also prevent certain people or parties from becoming self-sufficiently authoritative.  so as long as people have indirect, not excessive power in the government, while retaining some influence over the politicians, democracy will work well and last long.  i believe in a government where that balance is found, democracy, although flawed, is the least dysfunctional government system.

I like that characterization... the "least disfunctional" in practice.

i think Democracy has different meaning in different region.
in some parts of the world Dictatorship is the best choice form of Gov.

[quote="bubka"]Good. Now let us come to anarchy. What do you think of it?[/quote]
The man with the biggest gun wins.

Anarchy is a catch-22. You wouldn't be able to maintain anarchy without some kind of structure...

any government that gives its citizens the fundamental human rights is fine.  Personally i favor jehr314piocracy

[quote="PenguinIntegral"][quote="Ignite168"]I favor socialism. But I would never want to run a big country in the first place. As the bigger a country is, the harder it is for socailism to work. I would try and bring socailism to a small country if I could.[/quote]
With what restrictions on individual freedom? Just Economically? You'd be hard pressed to make it non-corrupt.[/quote]

I'd have some radical differences from the traditional thought of Socialism. First, a unit of currency is important for trade to the outside world. Second, citizens should have freedoms, they should be allowed to voice their mind about ideas that will bring the group collectively to greater success.

I sure wish the US was a democracy...nationstates.net reminds you of your facist roots...I think I'd like socialism if I were in the OTHER country.


geometry
parallelogram
geometry proposed

this is another one :
$ABCD$ is a parallelogram and $DE=BF$. prove that $HC$ is the bisector of $\angle DHB$.
i dont know why,but i like this problem very much! :!:

That's interesting, since comparable with http://www.mathlinks.ro/Forum/viewtopic.php?t=5873 ...

  darij

thank u darij
here is my solution :
$\triangle DEC$ and $\triangle FBC$ have the same area (half of pallerogram) but $DE=BF$ so the distance of
$C$ of $BF$ and $ED$ are equal that means $HC$ is bisector of $\angle DHB$.
 :P
Payman


First, I didn't quite catch the problem's meaning.
We take the first example. $ 4x+2y-5>0$.
For $ P0(x0,y0)$, we can change r to a bigger number, surely we can find $ r0,x1,y1$, when $ r>r0$, $ (x1-x0)^{2}+(y1-y0)^{2}<r^{2}$, but $ 4x1+2y1-5\leq 0$ and $ (x1,y1)$ is not in $ M$. Simplely, we draw the line $ 4x+2y-5=0$, and a circle $ (x-x0)^{2}+(y-y0)^{2}=r^{2}$, and if the line goes across the circle, the points in the left part of the circle aren't in $ M$. Why we say it is a good set?
Maybe I didn't catch it...


[quote="Peter"]Let $\{u_{n}\}_{n \ge 0}$ be a sequence of positive integers defined by \[u_{0}= 1, \;u_{n+1}= au_{n}+b,\] where $a, b \in \mathbb{N}$. Prove that for any choice of $a$ and $b$, the sequence $\{u_{n}\}_{n \ge 0}$ contains infinitely many composite numbers.[/quote]

We can easily find that
\[u_{n}=a^{n}+b\frac{a^{n}-1}{a-1}\]

for all $n\in{N_{0}}$.

Also it is obvious that in the case when $gcd(a;b)>1$ then all the members of the sequence would be composite numbers.

Let's discusse the case when $gcd(a;b)=1$.
\[u_{n}=a^{n}+b\frac{a^{n}-1}{a-1}=(a^{n}-a)+b\frac{a^{n}-a}{a-1}+(a+b)\]

Let's denote by $p$ any prime divisor of $a+b$.
We have that $(a;p)=1$=>
\[p^{m}|a^{\varphi(p^{m})+1}-a\]

Therefore there is a $m_{0}$ number ($m_{0}=\parallel a-1\parallel _{p}$) that for any $m>m_{0}$
\[p^{m}|u_{\varphi(p^{m})+1}\]

and of course $p^{m}<u_{\varphi(p^{m})+1}$. So the infinite set of numbers $u_{\varphi(p^{m})+1}$ ,where $m>m_{0}$, includes only composed numbers.
 :wink:


If so, this problem said that $\mathbb{Q}[\sqrt[3]{2}]$ is a field.  :wink:

[quote="spix"][quote="ElChapin"]Consider expresions of the form $x+yt+zt^{2}$ with x, y, z rational numbers and $t^{3}=2$.
Prove that:

if $x+yt+zt^{2}<>0$, then exists u, v, w rationals such that:

     $(x+yt+zt^{2})(u+vt+wt^{2})=1$.

 :)  :lol:[/quote]

I think that $x+yt+zt^{2}$ is always different from zero, therefore that "if" does nothing :)[/quote]
x=y=z=0 :wink:


algebra
polynomial
inequalities
geometric sequence

Okay, this one was number 13 on one of the AIME's and I had no clue how to work it, and I didn't want to TeX it all up so I copied word for word from Kalva:

The roots of the polynomial (1 + x + x2 + ... + x17)2 - x17 are rk ei2piak, for k = 1, 2, ... , 34 where 0 < a1 < a2 < ... < a34 < 1 and rk are positive. Find a1 + a2 + a3 + a4 + a5.

And another one:

If x,y, and z are positive real numbers such that $x+y+z=1$, prove $(x+\frac{1}{x})^2 +(y+\frac{1}{y})^2 +(z+\frac{1}{z})^2 > 33.$

[quote="eryaman"]For the second one is the minumum $33\frac{1}{3}$.?[/quote]
I think so. When $x=y=z=\frac{1}{3}$

Cool.  I don't know anything about inequalities, but for some reason I just intuitively knew the minimum would be reached when they were all $\frac{1}{3}$.

Edit: this is my $9^3$ post!.  :D

the first one is not so bad.

write the thing as a geometric sequence, square it, then subtract $x^17$, then you'll get $\frac{x^{36}-x^{19}-x^{17}+1}{(x-1)^2}$, but the numerator factors as $(1-x^{19})(1-x^{17})$, so you're looking at the 17th and 19th roots of unity except for 1.  so the a's are 1/17, 2/17, 1/19, 2/19, 3/19.

[quote="mna851"]Shobber, your solution is correct but I believe the solution with Cauchy still works even if the coefficeints of the second group of terms is say 1.  Try it, it worked for me I got the same answer.[/quote]
I can't see it...
I have just started doing inequalities and i know Cauchy from the sticky of this forum...So can you please show your work?

wow... you all made those problems look ridiculously easy.  :blush: thanks a bunch lol.

and btw, mna851, could you show your process? i always like to see multiple solutions to problems i couldn't understand

and you got that the sum of the reciprocals was at least 9 by what? AM>HM?

[quote="matt276eagles"]What is $AM>HM$?[/quote]

Arithmetic Mean $\ge$ Harmonic Mean...

$\frac{\sum a_i}{n} \ge \frac{n}{\sum \frac{1}{a_i}}$

so for the case $n=3$ we have

$\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

for all positive reals $a,b,c$.


Could someone tell me how Aitken's method works (involving Iterative methods for a nonlinear equation) in order to accelerate the convergence to a solution??

And then, applied this method in the example below (it's only a sequence):

\[ x_{n\plus{}1}\equal{}2e^{\minus{}x_n}\, ,x_o\equal{}0.8\]

From now, thanks.

Yeah. I have seen that, but I can't get it anyways. I need an example, like the one I posted before.


Prove or disprove: $ (0,0,0)$ is the only solution of
$ \{ (a,b,c) \in \mathbb{N}^3,$ such that $ (a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(a \plus{} c)$ is the fourth power of an integer $ \}$

[hide="Is it..."]
Infinite Descent? If $ (a,b,c)$ satisfies this condition, so does $ \left(\frac{a}{2},\frac{b}{2},\frac{c}{2}\right)$
 :maybe:
[/hide]

Don't forget that it asks for $ (a,b,c) \in \mathbb{N}^3$. So, you must prove that $ \dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2} \in \mathbb{N}$.

Sorry but whats $ \mathbb{N}^{3}$, i also want to know what $ \mathbb{R}^{2}$ is. Thanks  :)

$ \mathbb{N}$ is the set of natural numbers.

$ \mathbb{N} \times \mathbb{N} \equal{} \mathbb{N}^2$ is the set of all ordered pairs $ (a,b)$ such that $ a,b \in \mathbb{N}$.

Similarly, $ \mathbb{R} \times \mathbb{R} \equal{} \mathbb{R}^2$ is the set of all ordered pairs $ (a,b)$ such that $ a,b \in \mathbb{R}$.

$ \mathbb{N} \times \mathbb{N} \times \mathbb{N} \equal{} \mathbb{N}^3$ is the set of all ordered pairs $ (a,b,c)$ such that $ a,b,c \in \mathbb{N}$.

Of course, $ (0,0,0)\notin \mathbb N^3$ in the first place... :roll:

To be unambiguous about whether zero is included or not, sometimes an index "0" is added in the former case, so I meant $ \mathbb{N}\equal{} \{0, 1, 2, 3, \ldots \}$, but if you prefer I can write $ \mathbb{N}_0$ :)


Here's a cute, albeit easy problem :):

Let $X=\{1,2,\ldots,n\}$, and let $P_1,\ldots,P_n$ be $n$ distinct pairs of elements of $X$. We know that $P_i,P_j$ have an element in common iff $\{i,j\}$ is one of the pairs. Prove that every element of $X$ belongs to exactly two pairs.

it suffices to note that no vertex(element of $X$) can be in $3$ or more of the $P_i$'s; the rest follows easily since, if we think of the corresponding graph with labelled vertices and edges, then in its connected components,of sizes $k_1,k_2,\ldots,k_r$ with number of edges $m_1,m_2,\ldots,m_r$ we have $2m_i\leq 2k_i$; also $\sum_i k_i=\sum_i m_i=n$, so it follows necessarily that $k_i=m_i$ ,i.e each connected component has same number of vertices as edges and hence necessarily a cycle. But then, unless the induced graph on that component is a cycle by itself there will be a vertex of degree $3$ or more.
To illustrate this we do the following:
suppose say the vertex $1$ has degree atleast $3$,i.e. $\{1,i\},\{1,j\},\{1,k\}$ are among the $P$'s($i\neq j\neq k\neq i$). Since $\{1,i\}\cap\{1,j\}\neq\emptyset$ these two sets are $P_1=\{1,i\}$ and $P_j=\{1,j\}$(say). Then, similarly one gets that $P_k=\{1,k\}$. But then $P_j\cap P_k\neq\emptyset$ and neither of these is $\{j,k\}$ .

[quote="grobber"]Here's a cute, albeit easy problem :):

Let $X=\{1,2,\ldots,n\}$, and let $P_1,\ldots,P_n$ be $n$ distinct pairs of elements of $X$. We know that $P_i,P_j$ have an element in common [b]iff[/b] $\{i,j\}$ is one of the pairs. Prove that every element of $X$ belongs to exactly two pairs.[/quote]
note that you can forget that 'iff' and set in an 'if' instead.

seshadri: I don't understand the last part of your solution. can you please explain¿

i guess the part that you don't understand is where i show that the vertices have degree less than $3$.
 suppose $1$ has degree atleast $3$. then there exist three distinct $i,j,k$ all different from $1$ and such that $\{1,i\},\{1,j\},\{1,k\}$ all belong to the given collection of $P's$. since $\{1,i\},\{1,j\}$ have nontrivial intersection, it follows from the hypothesis that these two sets are (wlog-otherwise we can relabel) $P_1,P_j$. hence there are two possibilities: $P_1=\{1,i\},P_j=\{1,j\}$ or $P_1=\{1,j\},P_j=\{1,i\}$. in either case a similar argument can be given: if $P_1=\{1,i\}$ and $P_j=\{1,j\}$ then consider the pairs $\{1,i\},\{1,k\}$. if $P_{\alpha}=\{1,k\}$ then we know that $\{1,\alpha\}$ is one of these two sets by the hypothesis so that $\alpha\in\{i,k\}$. if $\alpha=i$ then consider the sets $P_j,P_i$(these have just been labelled now). they have non empty intersection and neither of the sets is $\{i,j\}$ and that is a contradiction. if $\alpha=k$, again the sets $P_j,P_k$ have non empty intersection and neither of these sets is $\{j,k\}$.
you can argue out each case that way. and this is what i meant.


function
linear algebra
matrix
quadratics
complex numbers
linear algebra unsolved

Let $ K$ be a subfield of complex numbers and $ n$ be a postive integer. Let $ j_{1},j_{2},\cdots,j_{n}$ and $ k_{1},k_{2},\cdots,k_{n}$ be positive inetegrs $ \leq n$. For each $ n\times n$ matrix $ A$ on the subfield $ K$ we define:

$ D(A)\equal{}A(j_{1},k_{1})A(j_{2},k_{2})\cdots A(j_{n},k_{n})$

Prove that $ D$ is a $ n$-linear function if and only if $ j_{1},j_{2},\cdots,j_{n}$ are distinct.

Note: Let $ K$ be a commutative ring with indentity element and $ n$ be a positive integer, Let $ D$ be a function that corresponds scalar $ D(A)$ from $ K$ to each $ n\times n$ matrix like $ A$. Now $ D$ is a $ \textbf{n\minus{}linear}$ function if for all $ 1\leq i\leq n$, $ D$ is a linear function from the $ i$'th row which keeps the other $ (n\minus{}1)$ rows fixed.

If $ j_{1},...j_{n}$are distinct, checking that D is n-linear is straightforward (if you are using Hoffman and Kunze, the proof is given). Just a matter of manual verification.

Now if $ j_{i}\ne j_{l}$consider D as a function of the ith row so that we can write $ D(a_{i}) \equal{} A_{j_{i},k_{i}}A_{j_{i},k_{l}}b$, where b is a fixed element in K. If $ a_{i}'$ is the ith row of the matrix A' whose other rows are those of A, then

$ D(ca_{i}\plus{}a_{i}') \equal{} c^{2}D(a_{i})\plus{}D(a_{i}')\plus{}cb[A_{j_{i},k_{i}}A'_{j_{i},k_{l}}\plus{}A_{j_{i},k_{l}}A'_{j_{i},ki}]$

This is a quadratic (in c) and so D cannot be linear.


Εστω $ f: R\to R$ μια πολυωνυμικη συναρτηση βαθμου n με συντελεστες πραγματικους αριθμους. Υποθετουμε οτι ολες οι ριζες της f ειναι πραγματικοι αριθμοι. Ν.δ.ο

$ (n - 1)\left(f'(x)\right)^2 - nf(x)f''(x)\geq0$

[hide="Λυση"]Υποθετουμε οτι $ r_1,r_2,...,r_n$ ειναι ριζες της $ f(x)$

$ \frac {f'(x)}{f(x)} = \frac {1}{x - r_1} + \frac {1}{x - r_2} + ... + \frac {1}{x - r_n}$

Παραγωγιζοντας κατα μελη εχουμε

$ \frac {f''(x)f(x) - (f'(x))^2}{(f(x))^2} = - \frac {1}{(x - r_1)^2} - \frac {1}{(x - r_2)^2} - ... - \frac {1}{(x - r_n)^2}$

$ \frac {(f'(x))^2 - f''(x)f(x)}{(f(x))^2} = \frac {1}{(x - r_1)^2} + \frac {1}{(x - r_2)^2} + ... + \frac {1}{(x - r_n)^2}$

$ \forall x\in R-\lbrace r_1,r_2,...,r_n\rbrace$

Απο την BCS εχουμε οτι

$ n\cdot\left[\frac {(f'(x))^2 - f''(x)f(x)}{(f(x))^2}\right] = n\cdot\left[\frac {1}{(x - r_1)^2} + \frac {1}{(x - r_2)^2} + ... + \frac {1}{(x - r_n)^2}\right]\geq \left(\frac {1}{x - r_1} + \frac {1}{x - r_2} + ... + \frac {1}{x - r_n}\right)^2 = \left(\frac {f'(x)}{f(x)\right)^2}$

Αρα $ (n - 1)\left(f'(x)\right)^2 - nf(x)f''(x)\geq0$  :)[/hide]


queria preguntarles sobre el cuarto problema por que no he encontrado como atacarlo, no se si hay algo que aun no conozco con lo que pueda resolverse.de antemano les agradezco cualquier ayuda.

Hint: Esas matrices forman un grupo isomorfo al grupo simetrico de permutaciones. ¿Que significa transoponer en terminos de ciclos y permutaciones.

Sea $A^t$ la transpuesta de $A$. Entonces $A\cdot A^t = I$, pues el producto de dos líneas distintas es cero y el producto de dos líneas iguales es la suma de los cuadrados de sus entradas, que es $1$.

Por lo tanto basta probar que existe $m > 0$ tal que $A^m = I$ y tomar $n = m-1$ (si $m = 1$ $A$ es la identidad, y el resultado es inmediato). Pero $A$ es inversible (su inversa es $A^t$) y el producto de dos matrices en las condiciones del enunciado también tiene la misma propiedad (de hecho, la entrada $a_{ij}$ del producto es igual a $1$ si y sólo si la linea $i$ e la columna $j$ contienen $1$ en una misma posición $k$; además, fijado $i$ existe exactamente un valor de $j$ tal que esto ocurre; en los otros casos, $a_{ij} = 0$). Por lo tanto, como la cantidad de tales matrices es finita, por el princípio de los palomares existen $p > q$ tales que $A^p = A^q \iff A^{p-q} = I$. Entonces basta tomar $m = p - q$.


He he.  :rotfl:  Try the last one too. I think it looks nice too. But it is hard (I believed):
If $a,b,c$ are the side of triangle. Prove:
$3[bc(b+c)+ab(a+b)+ac(a+c)] \geq a^3+b^3+c^3+15 abc$
 :D

As I said in topic Gibb 1 and some other topics , The ineq is homogeneous , so we can change the ineq into $\sum A(b-c)^2 \geq 0$ , then this problem can be solved easily

$\sum (5c-a-b)(a-b)^2 \ge 0$
Now WLOG assume $a \ge b \ge c$, then we obviously have $5a-b-c \ge 0$ and $5b-a-c \ge 0$.
If $5c-a-b \ge 0$, we are done.
Else,
$(5b-a-c)(a-c)^2+(5c-a-b)(a-b)^2 \ge (5b-a-c+5c-a-b)(a-b)^2 = (4b+4c-2a)(a-b)^2 \ge 0$. Done.

Oh? Why don't we take $a=y+z,b=z+x,c=x+y$.
Then abit expending gives us an easy problem.
It's all and I shouldn't post it here.

[quote="ehku"]Easy. The following inequality is stronger :
$2[ab(a+b)+bc(b+c)+ca(c+a)] \ge a^3+b^3+c^3+9abc$[/quote]

 Yeah, it is very nice and shape inequa. I can prove it by using Schur . But I want to see how others kill this one.
   :lol:

Nice sols Soar!
My sols is different from yours.
Put $x=a+b-c$ $y=b+c-a$ $z=c+a-b$
We have $x^3+y^3+z^3 +3xyz \geq xy(x+y)+yz(y+z)+zx(z+x)$
Expand it and we get the inequa.

[quote="Soarer"]$\sum (4c-a-b)(a-b)^2 \ge 0$, then it 's easy to proceed.[/quote]
  I think it should be : $\sum(3c-a-b)(a-b)^2 \geq 0$


Find $ a_3,a_4,...,a{}_2{}_0{}_0{}_8$, such that $ a_i =\pm1$ for $ i=3,...,2008$ and

$ \sum\limits_{i=3}^{2008} a_i2^i = 2008$

and show that the numbers $ a_3,a_4,...,a_{2008}$ are uniquely determined by these conditions.

we suppose that $ \sum_{i\equal{}3}^{2008}a_i2^i\equal{}\sum_{i\equal{}3}^{2008}b_i2^i$ where $ a_i,b_i\equal{}\pm 1$
$ \sum_{i\equal{}3}^{2008}(a_i\minus{}b_i)2^i\equal{}0$ where $ a_i\minus{}b_i\in\{\minus{}2,0,2\}$
let $ A_(h)\equal{}\{j: \ a_i\minus{}b_j\equal{}h\}$
if $ A_{(2)},A_{(\minus{}2)}\neq \emptyset$ we will have $ (*): \ \sum_{i\in A_(2)}2^{i\plus{}1}\equal{} \sum_{i\in A_{(\minus{}2)}}2^{i\plus{}1}$
then $ A_{(2)}\equal{}A_{(\minus{}2)}$ if we see $ (*)$ in base $ 2$.so neccesary $ A_{(2)}\equal{}A_{(\minus{}2)}\equal{}\emptyset,A_{(0)}\equal{}[|3,2008|]$
then $ a_i\minus{}b_i\equal{}0$ for $ i\equal{}3,4,...,2008$

then there is uniquely solution $ (a_3,...,a_{2008})\in\{\minus{}1;1\}^{2006}$ for the equation $ \sum_{i\equal{}3}^{2008}a_i2^i\equal{}2008$

$ \left(x \plus{} x^{\minus{}1} \right) \left( x^2 \plus{} x^{\minus{}2} \right) \left( x^4 \plus{} x^{\minus{}4} \right) ... \equal{} ... \plus{} x^{\minus{}4} \plus{} x^{\minus{}2} \plus{} 1 \plus{} x^2 \plus{} x^4 \plus{} ...$.

See [url=http://en.wikipedia.org/wiki/Balanced_ternary]balanced ternary notation[/url].


algebra
polynomial
AMC
USA(J)MO
USAMO
inequalities
algebra unsolved

Let $ P$ be a polynomial with integer coefficients, such that  $ P(a)\equal{}b, P(b)\equal{}c, P(c)\equal{}a$ for  some integers $ a,b,c$.
Prove that there exists an integer solution of the equation $ P(x)\equal{}x$.

It give $ a\equal{}b\equal{}c$, see Russian forum.

Very similar to USAMO 1974, problem no. 1 : [url]http://www.kalva.demon.co.uk/usa/usa74.html[/url]
If we show that a=b or b=c or c=a, we are done (one of a,b,c must be a fixed point). Let's assume that a,b,c are distinct.
P has integer coefficients, therefore, for every integers x,y: $ x\minus{}y | P(x)\minus{}P(y)$, hence $ a\minus{}b|P(a)\minus{}P(b)\equal{}b\minus{}c, b\minus{}c | c\minus{}a, c\minus{}a | a\minus{}b$.
We get - $ |a\minus{}b| \le |b\minus{}c|, |b\minus{}c| \le |c\minus{}a|, |c\minus{}a| \le |a\minus{}b|$. If we multiply these inequalities, we get $ |a\minus{}b\parallel{}b\minus{}c\parallel{}c\minus{}a| \le |a\minus{}b\parallel{}b\minus{}c\parallel{}c\minus{}a|$. But we can get equality iff $ |b\minus{}c|\equal{}|a\minus{}c|\equal{}|a\minus{}b|$. a,b,c are distinct, therefore b-c=c-a, a-c=b-a, a-b=b-c, which gives a=c $ \implies$ a contradiction.


Find positive integer solution:
$ x^2\plus{}(x\plus{}y)^2\equal{}(x\plus{}9)^2$

First, expand to get \[ x^2 \plus{} x^2 \plus{} 2xy \plus{} y^2 \equal{} x^2 \plus{} 18x \plus{} 81\] Now move around the terms to get \[ x^2 \plus{} 2xy \plus{} y^2 \equal{} 81\] and then simplify as \[ (x \plus{} y)^2 \equal{} 81\] Therefore, \[ x \plus{} y \equal{} 9\] and \[ x \plus{} y \equal{} \minus{}9\] The positive integer solutions are then $ (x,y) \equal{} (1,8), (2,7), (3,6), (6,3), (7,2), (8,1)$.

@isabella2296: You forgot the $ 18x$ term.

[hide="Solution"]$ x^2 \plus{} (x\plus{}y)^2 \equal{} (x\plus{}9)^2$.

$ 2x^2 \plus{} 2xy \plus{} y^2 \equal{} x^2 \plus{} 18x \plus{} 81$

$ x^2 \plus{} 2xy \plus{} y^2 \equal{} 18x \plus{} 81$

$ (x\plus{}y)^2 \equal{} 18x \plus{} 81$

$ x\plus{}y \equal{} \pm \sqrt{18x\plus{}81}$

$ y \equal{} \minus{}x \pm \sqrt{18x\plus{}81}$.

We only want positive solutions, so $ y \equal{} \minus{}x \plus{} \sqrt{18x\plus{}81}$. We get an upper bound on $ x$ when $ x \equal{} \sqrt{18x\plus{}81}$, or when $ x^2 \equal{} 18x \plus{} 81$. This occurs when $ x \equal{} 9 \plus{} 9\sqrt{2}$, so $ x < 22$.

We brute-force solutions to get integers. Two solutions arise: $ \boxed{(8,7)}$ and $ \boxed{(20, 1)}$.[/hide]

[quote="darkdieuguerre"]@isabella2296: You forgot the $ 18x$ term.

[hide="Solution"]$ x^2 \plus{} (x \plus{} y)^2 \equal{} (x \plus{} 9)^2$.

$ 2x^2 \plus{} 2xy \plus{} y^2 \equal{} x^2 \plus{} 18x \plus{} 81$

$ x^2 \plus{} 2xy \plus{} y^2 \equal{} 18x \plus{} 81$

$ (x \plus{} y)^2 \equal{} 18x \plus{} 81$

$ x \plus{} y \equal{} \pm \sqrt {18x \plus{} 81}$

$ y \equal{} \minus{} x \pm \sqrt {18x \plus{} 81}$.

We only want positive solutions, so $ y \equal{} \minus{} x \plus{} \sqrt {18x \plus{} 81}$. We get an upper bound on $ x$ when $ x \equal{} \sqrt {18x \plus{} 81}$, or when $ x^2 \equal{} 18x \plus{} 81$. This occurs when $ x \equal{} 9 \plus{} 9\sqrt {2}$, so $ x < 22$.

We brute-force solutions to get integers. Two solutions arise: $ \boxed{(8,7)}$ and $ \boxed{(20, 1)}$.[/hide][/quote]

Solution (the point of view from x solved in terms of y)
[hide]
$ x^2 \plus{} 2xy \plus{} y^2 \equal{} 18x \plus{} 81$

$ x^2 \plus{} 2yx \minus{} 18x \plus{} y^2 \minus{} 81 \equal{} 0$

$ x^2 \plus{} (2y \minus{} 18)x \plus{} y^2 \minus{} 81 \equal{} 0$

Use the quadratic formula and simplify:

A = $ 1$
B = $ 2y \minus{} 18$
C = $ y^2 \minus{} 81$

$ x \equal{} 9 \minus{} y \pm 3\sqrt{18 \minus{} 2y}$

Use the "positive" branch so $ x$ will not be negative:

$ x \equal{} 9 \minus{} y \plus{} 3\sqrt{18 \minus{} 2y}$

So it must be the case that:

$ 18 \minus{} 2y \ge 0$

$ 9 \minus{} y \ge 0$

$ y \le 9$

$ y$ must be a positive integer, so (only) check $ y \equal{}$ integers $ 1$ through $ 9$, inclusive.

$ y \equal{} 1$ and $ y \equal{} 7$ are the only positive values that make the radicand a perfect square.

For $ y \equal{} 1, x \equal{} 20$
For $ y \equal{} 7, x \equal{} 8$

Answer:  $ (8, 7)$ and $ (20, 1)$

[/hide]

$ m > n > 0$ are integers.
Put $ x \equal{} m^2\minus{}n^2,\ x\plus{}y \equal{} 2mn,\ x\plus{}9 \equal{} m^2\plus{}n^2.$
Subtracting 1st equation from the 3rd one $ \Longrightarrow$ $ 2n^2\equal{} 9,$ impossible.
Put  $ x \equal{} 2mn,\ x\plus{}y \equal{} m^2\minus{}n^2,\ x\plus{}9 \equal{} m^2\plus{}n^2.$
Subtracting 1st equation from the 3rd one $ \Longrightarrow$ $ (m\minus{}n)^2 \equal{} 9,\ m \equal{} n\plus{}3.$
Subtracting 1st equation from the 2nd one $ \Longrightarrow$ $ y \equal{} (m\minus{}n)(m\plus{}n) \minus{} 2mn \equal{} 9 \minus{} 2n^2 > 0$
$ \Longrightarrow$ $ n \equal{} 1, 2$ $ \Longrightarrow$ $ \{m,n\} \equal{} \{4,1\},\ \{5,2\}$  $ \Longrightarrow$ $ \{x,y\} \equal{} \{8,7\},\ \{20,1\}.$


Let $P$ be an arbitrary point inside of a triangle $\triangle ABC$. Let the cevians $AP, BP$ meet the sides $BC, CA$ at points $D, E$, respectively. If the point $P$ is selected inside of the triangle $\triangle ABC$ with a constant probabilty density, what is the probability that the area of the triangle $\triangle APE$ is greater than $\frac 1 2$ of the area of the triangle $\triangle ABC$ ?

notice that the locus of $P$ for which $\triangle APE$ has area exactly $\frac 12$ the area of $\triangle ABC$ is a certain curve connecting $B$ to the midpoint $M$ of $BC$ .  the set of $P$ for which $\triangle APE$ has area *greater* than $\frac 12$ the area of $\triangle ABC$ is then the region bounded by this curve and $BM$.

for $P$ on the curve, and $E$ the intersection of $BP$ and $AC$, we must have $\frac{AE}{AC} \cdot \frac{EP}{EB} = \frac 12$.  this is then a problem for parametrization.  i haven't worked out the details yet...

Using a parallel projection, we can transform an arbitrary triangle $\triangle ABC$ into any triangle we desire. The ratio of areas are preserved in a parallel projection. The 1st thing to do is to project the triangle $\triangle ABC$ into an isosceles right angle triangle $\triangle A'B'C'$ with the angle $\angle A' = 90^o$, which substantially simplifies the equation of the curve you are looking for.


Contest tomorrow! Get ready for the race of speed and good luck to everyone!

PS: As seen from the topic http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1401197#1401197, please restrain discussions on the AMC problems until further notice.

That link doesn't seem to lead me to an existing page.

Can we discuss now? It's been a day.

let's just wait for a few more hours (just in case)
the math jam will be today, i think..

yeah it's OK to discuss it now. I just checked.

Alright.

I did AMC12, and don't know anything about AMC10.

I got 22 questions, one of which was wrong, and left 3 blank, so it adds up to 130.5, if I'm lucky. :)

And I feel bad, since I now know how to solve all but #23, and they're so easy. :(

If anyone has the answer to #23, please give some biref solution. That was too complicated to look at, so I never even tried it.

well, the answers are already posted : http://www.mathlinks.ro/resources.php?c=182&cid=44&year=2009
I did 21, left 4 blank, and got one wrong...so..126..

i took amc 12 and got 21 questions right, 3 questions wrong, and 1 left blank- so 127.5

Hmm..It seems that you guys all did (much) better than I did last year.

well i dropped  by 3 from last year's score :(

I took AMC10, and got 124.5. 2 wrong, 3 blank. I could have done better though, because the 5 questions I messed up on were pretty easy and I found all the answers shortly after the test.

i got 127.5... :(

Made REALLY REALLY STUPID mistakes... lol
I was supposed to get the number of non-real zeroes..
I found all of it, but instead, I putin the number of REAL zeroes :wallbash_red:
andI also failed to notice that 5^k + 1 is an even number  :cursing:

I, too, failed to realize that 5^k + 1 was even. Must be a common mistake.

Best of luck to everyone writing AMC 10/12 B tmr
As it was for AMC 10/12 A, please don't discuss it until further notice.

Epic failure! I believe that I can write AIME off (unless, by a stroke of luck, the cutoff would be 96 or lower)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=260664

cutoff for AMC 10/12 A was 120/97.5 respectively..


Let $ a,b,c$ be real numbers.Prove that

[b]Problem 1:[/b] $ 5(a^4 \plus{} b^4 \plus{} c^4) \minus{} 4(ab^3 \plus{} bc^3 \plus{} ca^3)\ge 3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} 2(a^3b \plus{} b^3c \plus{} c^3a)$

[b]Problem 2:[/b] $ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2(a^3b \plus{} b^3c \plus{} c^3a) \plus{} 3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\ge 4(ab^3 \plus{} bc^3 \plus{} ca^3) \plus{} 2abc(a \plus{} b \plus{} c)$

We might try to write it in the form $ \sum_{cyc} (ax^2\plus{}by^2\plus{}cz^2\plus{}dxy\plus{}eyz\plus{}fzx)^2$ where a,b,c,d,e,f are undetermined constants and $ x,y,z$ are variables.

[quote="Altheman"][quote="zaizai-hoang"][quote="Altheman"]We might try to write it in the form $ \sum_{cyc} (ax^2 \plus{} by^2 \plus{} cz^2 \plus{} dxy \plus{} eyz \plus{} fzx)^2$ where a,b,c,d,e,f are undetermined constants and $ x,y,z$ are variables.[/quote]
Of course, this is the best way to solve them. In Vasc's book, it was solved with a fouth-degree's inequality. Some people in Vietnam also found out a very useful technique to factor expression to sum of squares as above  :wink:[/quote]

Well if you are desperate, you can expand and equate coefficients.

These cyclic systems are interesting to solve. (that is when each factor is zero). For example, you can find a,b,c,d,e,f so you get solutions x=y=z, and (t-x)(t-y)(t-z)=t^3-pt^2+qt-r[/quote]

Yes, we can do like that but I am not desperate  :D
I think you want to only some indentity ... For example:
\[ \sum a^4\minus{}\sum a^2b^2\equal{}\frac{1}{2}\sum (a^2\minus{}b^2)^2\]
\[ 3(\sum a^3b\minus{}\sum a^2bc)\equal{}\sum (a^2\minus{}b^2)(ab\plus{}ac\minus{}2bc)\]
\[ 3(\sum a^2b^2\minus{}\sum a^2bc)\equal{}\frac{1}{2}\sum(ab\plus{}ac\minus{}2bc)^2\]

Of course, your way is more complicated  :) Above way is also the way in Vasc's book  :wink:
Let's try the following inequality by this way:
[Pham Van Thuan - pvthuan]
Let $ a,b,c$ be real numbers. Prove that:
\[ a(a\plus{}b)^3\plus{}b(b\plus{}c)^3\plus{}c(c\plus{}a)^3\ge \frac{8}{27}(a\plus{}b\plus{}c)^4\]

[quote="hungkhtn"]Problem 2 is in my book, "Secrets In Inequalities", problem 89, chapter 2, part (b), page 226.

Problem 1 is pretty easy. We have
\[ 5(a^4 \plus{} b^4 \plus{} c^4) \minus{} 4(ab^3 \plus{} bc^3 \plus{} ca^3) \minus{} 3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \plus{} 2(a^3b \plus{} b^3c \plus{} c^3a)
\]

\[ \equal{} 2(a^4 \plus{} b^4 \plus{} c^4 \plus{} a^3b \plus{} b^3c \plus{} c^3a \minus{} 2ab^3 \minus{} 2bc^3 \minus{} 2ca^3) \plus{} 3(\sum a^4 \minus{} \sum a^2b^2)\ge 0.
\]
[/quote]

Really? I found out two problems few days ago .Here my solution:
ineq(1)<=> $ \sum (ab\plus{}ac\minus{}2bc)^2\ge 0$

and ineq(2)<=> $ \sum (a^2 \plus{} 2ab \minus{} 2ac \minus{} b^2)^2\ge 0$

How do you show problem 2,hungkhtn?

@Hung: Cuon "Secrets In Inequalities" co the mua o dau? Tet nay Hung co ve VN ko?

[quote="dduclam"][quote="hungkhtn"]Problem 2 is in my book, "Secrets In Inequalities", problem 89, chapter 2, part (b), page 226.

Problem 1 is pretty easy. We have
\[ 5(a^4 \plus{} b^4 \plus{} c^4) \minus{} 4(ab^3 \plus{} bc^3 \plus{} ca^3) \minus{} 3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \plus{} 2(a^3b \plus{} b^3c \plus{} c^3a)
\]

\[ \equal{} 2(a^4 \plus{} b^4 \plus{} c^4 \plus{} a^3b \plus{} b^3c \plus{} c^3a \minus{} 2ab^3 \minus{} 2bc^3 \minus{} 2ca^3) \plus{} 3(\sum a^4 \minus{} \sum a^2b^2)\ge 0.
\]
[/quote]

Really? I found out two problems few days ago .Here my solution:
(1)<=> \sum (ab+ac-2bc)^2\ge 0

and (2)<=> $ \sum (a^2 \plus{} 2ab \minus{} 2ac \minus{} b^2)^2\ge 0$

How do you show problem 2,hungkhtn?

@Hung: Cuon "Secrets In Inequalities" co the mua o dau? Tet nay Hung co ve VN ko?[/quote]

Problem 2 is also in Sang Tao BDT, chapter 2, somewhere from problem 2.90 to 2.100. It is expressed as

If $ (a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2 \equal{} 12$ then
\[ a^3b \plus{} b^3c \plus{} c^3a \plus{} abc(a \plus{} b \plus{} c)\le 6.
\]

@ Lam. I did not go back to Vietnam this holiday. Thanks for concern.

[quote="hungkhtn"]Problem 2 is also in Sang Tao BDT, chapter 2, somewhere from problem 2.90 to 2.100. It is expressed as

If $ (a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2 \equal{} 12$ then
\[ a^3b \plus{} b^3c \plus{} c^3a \plus{} abc(a \plus{} b \plus{} c)\le 6.
\]
.[/quote]
Thanks,Hung. I saw it. A idea concur  :wink:

[quote="hungkhtn"]@ Lam. I did not go back to Vietnam this holiday. Thanks for concern.[/quote]
No problem  :)


Two circles of radii $1$ and $4$ are tangent externally. What is the length of the part of the common tangent line between the points of tangency?

Since both radii make a right angle with the tangent line, we can extend a line parallel to the tangent line going thru the radius of the smaller circle, creating a rectangle with the 4th point being on the radius of the larger circle.  Then we have a right triangle with hypotenuse $5$ (the sum of the radii), and one leg $3$, because the rectangle has width $1$, and thus the leg of the triangle is 1 less than the radius.  So the other leg of the triangle, and the length of the rectangle and segment of the tangent, is $4$.

Although that seems wrong...

Yes, that's exactly what I got, but what exactly did you mean the rectangle? What I did was I constructed a trapezoid with the bases as the radii of the circles, and the slanted leg, as the distance between the centers $5$ and then the height of the trapezoid, which was clearly $4$ was the tangent line, was the rectangle the part of the trapezoid?

[quote="236factorial"]this looks really bad, but you have to assume what is shown is true  :D

We see a right triangle with sides 3 and 4+1=5. The other side must be 4, the desired value.[/quote]

Ah, so that was the rectangle you were talking about, I simply drew the whole trapezoid out, very well, fairly simple problem.

I didn't talk abaout the rectangle  :mad:


I used ICT independent study AP CS (http://ict.org) and then I read a review book. I felt it was very good, although most of the ICT stuff was too simple since I had already had comp sci experience.  I guess it worked pretty well because I got a 5 on the exam.

APCS AP Exam is (unfortunately) in JAVA. It would probably be good to know that.

My main suggestion is just, know how to code, know data structures, know big O notation, know recursion, and you're fine.

[quote="quantum leap"]My school doesn't offer AP computer science AB, but i'm interested in taking it as independent study.  Does anyone know a good text to use for this?  Also, basic LaTeX is all my programming experience :oops: Is there an especially good book to learn C++ from?[/quote]

What are "AB" & "AP"? => "AP computer science AB"

I suggest 2 books that were my first step to become profession in C++,

Practical C++ Programming [Steve Oualline, Oreilly]
Essential C++ [Stan Lippman]

I would suggest that the first language you learn be Java rather than C. However, I can't recommend a good book because I haven't really used any programming textbook. My dad teaches programming, so he just taught me lots of stuff. But it's not that hard to learn, so you could probably look for an internet tutorial.

AP is advanced placement (you get college credit and it's *supposedly* harder) and AB means first semester as opposed to BC. At least, that's what it is for calculus here. I think AP is called IB or IP or something like that in some places.

IB is "international baccalaureate" which i think is to the british what ap is to america.  Some schools in america do have an IB as well as AP program.

For Computer Science there's an "A" and an "AB" test (as apposed to the "AB" and "BC" names given for calculus.)

Edit: I'll be taking the A test this May.

Read R.Sedgewick-"Algorithms" for algorithms and
        Bruce Eckel  -"Thinking in C++" for c++

For review book, I'll recommend "Be Prepared for the AP Computer Science Exam" by Dave Wittry (my comp sci teacher last year :lol:), Sally Bellacqua (a teacher at TJST), Roger Frank, and Maria Litvin.  That's what the students at my school use for preparing the AP AB test (although half the class doesn't prepare and still get 5's...)


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

Hello, let $ f \in \mathbb F[x]$ an irreducible poly and $ \mathbb E$ a splitting field for f over F. Also suppose that $ Gal(\mathbb E | \mathbb F)$ is Abelian. If $ \alpha_1,...,\alpha_n \in \mathbb E$ are the roots of f, prove that $ \mathbb E \equal{} \mathbb F[\alpha_i]\ \forall i\equal{}1,...,n$

It doesn't say whather the roots are distinct or not and I don't know how to use the information about the Galois group.. I don't even know whether $ \mathbb E | \mathbb F$ is Galois (don't know further information about $ \mathbb E$ and $ \mathbb F$). What can I do?

Let $ E_1 \equal{} F(a_1)$ be a proper subfield of $ E$. Since $ Gal(E/F)$ is abelian, all subgroups are normal and hence, $ E_1$ is a normal extension of $ F$. Since $ a_1\in E_1$, all the other roots of the irr. polynomial are in $ E_1$.

Of course the Abelian Galois group was given to make us say every subgroup of his is normal, but can we say this corresponds to normal extensions of F if we don't know whether E|F is Galois or not?

the claim does not hold in general, we have to assume E/F to be galois.

Yes, I think would suffice to say that the roots of the poly are distinct, wouldn't it?


In a Czech physics correspondence seminar FyKoS, there was an interesting problem a couple years ago. The name of the problem was "basic instinct". (I provide an almost exact translation)

"An ice pick - the famous killing tool from this movie (for those who haven't seen it, it looks like a screwdriver with a sharp pointy end), was set on its tip by the murderer. Using the uncertainity relations, estimate the maximal time this corpus delicti can stay in this unstable position. (neglect all the perturbing influences)"

You don't have to give the result with five significant digits, just give an estimate supported by some figures. Is it a few seconds? minutes? hours? days? or even years?

These Eastern European problems (like those in 200 Puzzling Physics Problems) are all non- routine which want the solver to analyse out of the box!!!

Give some hints plz!!!


Let $ ABCDE$ be a convex pentagon with:

$ ABC \equal{} ADE$ and $ AEC\equal{}ADB$ (angles)

Prove that:

$ BAC\equal{}DAE$

Split to http://www.mathlinks.ro/Forum/viewtopic.php?t=17016 .

This was very easy to find: just search for "ABC ADE AEC ADB BAC DAE".

  darij


Ν.δ.ο. για οποιαδηποτε θετικο ακεραιο $ n$ και οποιαδηποτε διανυσματα που σχηματιζουν μεταξυ τους γωνια $ 60$ μοιρων ισχυει
$ |\vec{a}\plus{}\vec{b}|<2\sqrt{n}|\vec{a}\minus{}\vec{b}|\prod_{k\equal{}2}^n\frac{2k\minus{}2}{2k\minus{}1}$

Καταρχήν για $ n \equal{} 2$ μπορούμε εύκολα να επαληθεύσουμε ότι το ζητούμενο ισχύει (μία ύψωση στο τετράγωνο και το ότι $ \vec{a}\cdot \vec{b} \equal{} \frac {1}{2}|\vec{a}\parallel{}\vec{b}|$ αρκούν ).

Για $ n\geq 3$ ισχύει (απόδειξη πολύ απλή με επαγωγή) ότι $ \prod_{k \equal{} 2}^n\frac {2k \minus{} 2}{2k \minus{} 1} > \frac {\sqrt {3}}{\sqrt {4n \minus{} 1}}$

Άρα λοιπόν αρκεί να αποδείξουμε ότι

$ |\vec{a} \plus{} \vec{b}| < 2\sqrt {n}|\vec{a} \minus{} \vec{b}| \frac {\sqrt {3}}{\sqrt {4n \minus{} 1}}$

Υψώνοντας και τα δύο μέλη στο τετράγωνο (και κάνοντας χρήση του ότι $ \vec{a}\cdot \vec{b} \equal{} \frac {1}{2}|\vec{a}\parallel{}\vec{b}|$ ) παίρνουμε τελικά $ (8n \plus{} 1)|\vec{a}|^2 \minus{} (16n \minus{} 1)|\vec{a}|\vec{b}| \plus{} (8n \plus{} 1)|\vec{b}|^2 > 0$ το οποίο ισχύει αφού έχει αρνητική διακρίνουσα (ως προς το $ |\vec{a}|$).

Παρατήρηση: Το κύριο πρόβλημα και η δυσκολία της άσκησης (κατά την δική μου άποψη) νομίζω ότι έγκειται στην εύρεση του κατάλληλου φράγματος για το γινόμενο $ \prod_{k \equal{} 2}^n\frac {2k \minus{} 2}{2k \minus{} 1}$ (Του οποίου όμως η επιλογή δεν έγινε τυχαία αλλά μετά από κατάλληλες παρατηρήσεις).

Χρήστο, μήπως έχεις κάποιο διαφορετικό τρόπο προσέγγισης?

Αλέξανδρος

Aλεξανδρε ωραια η λυση σου .
Αυτη ειναι η ιδεα της κατασκευης και λυσης .
Απλα εχω λιγο μικροτερο φραγμα και ......βρισκω το μεγιστο της $ |\frac{\vec{a}\plus{}\vec{b}}{\vec{a}\minus{}\vec{b}}|$ με κλασικη ανισωση


Given a triangle ${ABC}$ and points ${X,Y,Z}$ such that $\angle{ABZ} = \angle{XBC}, \angle{BCX} = \angle{YCA}, \angle{CAY} = \angle{ZAB}$, prove that ${AX,BY,CZ}$ are concurrent.

Hint : use trigonometric version of Ceva theorem...with it everything is easy :)

This is the Jacobi theorem, proved in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=186725#p186725]http://www.mathlinks.ro/Forum/viewtopic.php?t=30114 post #2[/url].

  Darij


probability logarithms calculus geometry integration

Billy Bob wants to buy a calculator from Calculators R Us in the next 10 days. At Calculators R Us, the price of a calculator changes every day and is completely unpredictable. How can Billy Bob maximize his chances of buying a calculator on the day when the price is the lowest? For example, if he is impatient and buys a calculator on the first day, the price may go down the next day. However, if he waits until the 10th day, the price may have been lower on a previous day, but it is now too late.

Sorry if the wording is really bad.

He should buy after about 3.67 days...optimally the evening of the third day or the morning of the fourth...

[quote="eryaman"]Uhh, why is that?  If it's completely random...and getting the calculator earlier doesn't have any preference over getting it later...[/quote]

This objection is absolutely correct.  But I suspect Elemenop really meant to say something different than he did say, because it is in fact possible to game this system.  While "about 3.68" is the key number, it is of course possible to decide whether "3" or "4" is really better.  If you don't already know the solution, my previous sentence shouldn't make any sense to you.

Hmmm, How would you prove this? Wouldn't the probability of any particular day been the lowest price be 1/10? so why is it that it is best to buy on the 3.67th day?

[quote="JBL"]If you don't already know the solution,[/quote]

I'm trying to figure out the solution.

i've heard a problem like this where you use information about the amounts for the first few days to decide when to buy one.  and if you do that then you won't be always buying it on the 3.67th day.  but i don't know the solution to that one either.

wouldn't the probability for each day be the same?  what's "special" about the 3.67th day?

OK, so can anyone provide a solution for this please? I'm dying to know why. I was just think about it, can I thought you could like use n to be the day you should buy it and then find the chances that the prices will be lower on the days before and the days after and see which n maximizes the probability, but thne the probability that it is higher or lower than a certain day is the same for each day so I got confused. Does this have anything to do with if something already happened, the probability of that is counted out of the total? Like the problem with the prizes in three doors, you choose on, and someone opens a door and the prize is not in there and you are asked if you want to switch thing?

Here's a [non-optimal, but pretty good] strategy you could use, and a very rough calculation of its probability of working:
[hide]replace the "10" in the question with any even number $2n$, where $n$ can be as big as you like.  You already know what $n$ is, of course.  Let the first $n$ days go by, without buying anything.  All you have to do is to record the prices on the various days.  Suppose $p$ is the lowest price you see on the first $n$ days.  Then, if at any point you see a price less than $p$, you buy immediately, or if you never see such a price, buy on the last day.

So, how good is this strategy?  Well, it works very nicely if the cheapest day will be in the second half and the second-cheapest day will be in the first half, because then you have a $100\%$ probability of picking the lowest price.  This event happens exactly $25\%$ of the time.  So even this incredibly weak calculation [there are certainly lots of other circumstances in which you could win under this strategy] gives you a lower bound of $25\%$, for any $n$.  That's pretty impressive.[/hide]

When the number is as small as 10, this strategy has only 10 ways it can work [one of which is equivalent to "buy on the first day" and one of which is equivalent to "buy on the last day"], and you could do [by hand or by computer program] the probabilities of winning in each of those cases.

One important thing to note about this question is that it is [i]not[/i] a question of expected value.  Because we know nothing about the 10 numbers that will be chosen, there's no way we can really minimize our expected cost, which would be a more usual thing for a shopper in this situation to try and do.

[quote="mathnerd314"][quote="SplashD"]Is 3.67 like close to 11/3 days or something?[/quote]

It [b]is[/b], but that's just a coincidence. The real reason that the answer is 3.67 is because [hide]3.67 is approximately $\frac{10}{e}$[/hide]

Actually, I just did it on my calculator, and if you round instead of truncating, the answer is 3.6[b]8[/b]. Minor detail...[/quote]

why, more quietly:
[hide]
why $e$? Does the natural log have some tie into probabilitiy that I'm not familire with?

(it obviously does, i always not familiar with things!  :blush:  :D )[/hide]

[quote="rzsolt"]why $e$?[/quote]

Yes, I'm wondering that too. All I know is the answer--I don't know how it's derived.

The problem is classic , Was very popular among Indian graduate students  in  those days :) ...

Basically it is "How to optimize your  selection process of (arranged) marriage?"

(Sorry for being a little 'politically incorrect' - but here  a person wants to optimize and  wants to select the best possible match for a spouse  from  $n$  prospective candidates.)

Problem went something like:

Your "Auntie" (traditional match maker) has lined up  $n$ prospects , you interview them (one by one) and try  to  select the best prospect. Of course, if you  once select someone the process ends.  (You can say yes/no after each interview but once you said no (or yes) you can not change your mind)

The solution (which is well known) is to try to check out the market with  first ($n/e$)  samples,  and then  first one you meet which is better than the best from  the market sample - is to be chosen.  :)

(When $n$ is large, the probablity that you will select the best candidate is about $1/e$ )

How large does $n$ have to be for the probability to be $\frac{1}{e}$

"$n$ is large" just means that the little deviations that depend on $n$ become negligible.  Exact values of "large" aren't really consistent with the definition of "negligible"  :P

so, back to the original problem, you would buy the first time the price is lower than what it was on the first $\frac{10}{e}$ days.

how do you prove this is the best strategy?

[quote="Gyan"]Here is quick reasoning to get the answer.

[hide]We assume in this "optimizing  spousal candidate" :)  problem:
a)	–  There is one (and only ONE) BEST candidate.
b)	-   When it comes to success or failure the result is Binary (That is either you select the BEST or  its a  failure.   (second best or the last ugly one :)  are  counted as same degree of failure)

Now obviously the probability that you would select the  best one in the kth selection is:
(1/n)

Now suppose you let go first (j-1)  and then   use the strategy  (First one I see better than what I have seen) then the probability you would select the best at  $k$th  $(j \le k \le n)$ is:

$(\frac{1}{n}) ( \frac{j-1}{k-1})$

[hide="because "]Becasue,  of the probablity that first runner up (up till  now) has occured in fist  (j-1)  and not later)
[/hide]
So probability of success =

$(1/n) \sum_{j}^{n}\frac{j-1}{k-1}= (\frac{j-1}n ) \sum_{j}^{n}\frac 1{k-1}$
When n is large:
This is  =  $\frac{j-1}n \ln (\frac n{j-2})$

You differentiate and equate it to zero  you get something like:

$\frac{j-1}{j-2}= \ln \frac{n}{j-2}$

LHS is nearly  1 so  $n/(j-2)$  is nearly $e$  etc…

Rest is simple ..
[/hide]

Also as JBL said, Using (1/2) as sample your chances are about (better than) 25%, more optimized sample this increases to about (1/e ) or a little better than (1/3) when n is large.[/quote]

Nice proof, but how did you get to the part after n is large ($\frac{j-1}n \ln (\frac n{j-2})$)? And why do you add up the probabilities?

[quote="computer_nuke"]how did you get to the part after n is large ($\frac{j-1}n \ln (\frac n{j-2})$)? [/quote]

[b]Theorem:[/b]  $\ln n < 1+\frac{1}{2}+...+\frac{1}{n}< \ln (n)+1$

This requires some calculus to prove, but approximately:  $\sum_{k=1}^{n}\frac{1}{n}$ is an approximation of the area under the curve $f(x) = \frac{1}{x}$ (edit: excuse my lazy typing  :blush: )

[quote="t0rajir0u"]This requires some calculus to prove, but approximately:  $\sum_{k=1}^{n}\frac{1}{n}$ is an approximation of the area under the curve $f(x) = \ln x$.[/quote]

Nice try -- $\sum_{k=1}^{n}\frac{1}{n}$ is a very good approximation of the area under the curve $f(x) = \frac{1}{x}$ between $x = 1$ and $x = n$ (or $n+1$?  Whatever, it's basically just epsilons :)).  It also happens that this area is given (exactly) by the definite integral of $f$, which happens to be $\ln n$.


How can the value under the first radical be positive?

[quote="archimedes1"][quote="mewto55555"]It can be 0...[/quote]
Good point... I guess that leads to the solution  :)[/quote]

[hide="Beginning of solution"]The part under the first radical factors to become $ \frac {\cos^2{x}}{16}\times{(\cos^2{x} \minus{} 8)}$ For that to equal 0, $ \cos^2{x} \equal{} 0,8$ If $ \cos^2{x} \equal{} 0$ there is no solution, but if $ \cos^2{x} \equal{} 8$, the second radical ends up not equaling $ \frac {1}{4}$ I think I did something wrong, or there is no solution.[/hide]

[hide="Solution"]The value under the first radical must be nonnegative.  Thus,
$ \frac {1}{16}\cos^4x \minus{} \frac {1}{2}\cos^2x \equal{} \frac {1}{16} \cos^2 x (\cos^2 x \minus{} 8) \ge 0$.

Since $ \cos x \le 1$, we have $ \cos^2 x \minus{} 8 < 0$.  Thus, we must have $ \cos^2 x \equal{} 0$.
But plugging it in, we see that this is not a solution.  Hence, there are no solutions to this equation.[/hide]

Oh lol I forgot that $ \cos^2{x}$ can't $ \equal{} 8$ (how do you do not equal sign)??  :blush:


The mean of three numbers is 10 more than the least of the numbers and 15 less than the greatest. The median of the three numbers is 5. What is their sum?

$ \textbf{(A)} \ 5 \qquad \textbf{(B)} \ 20 \qquad \textbf{(C)} \ 25 \qquad \textbf{(D)} \ 30 \qquad \textbf{(E)} \ 36$

[hide]Let $x$ be the least of the 3 numbers and $y$ be the greatest. So $\displaystyle\frac{x+5+y}{3}=x+10=y-15$. We find that $x=y-25$. Plug this into $\displaystyle\frac{x+5+y}{3}=y-15$ to get $y=25$. With this, we can find that $x=0$. So the 3 numbers are $0,5,25$, and the sum is $30$. $\boxed{D}$[/hide]

[hide="Answer"]The middle number is $5$, and we have $\frac{5+(m+15)+(m-10)}{3}=m\Rightarrow 2m+10=3m\Rightarrow m=10$, so $3m=30\Rightarrow \boxed{D}$.[/hide]

I believe @2above's answer is correct, but the approach is wrong, I think this is the correct approach:

[hide=solution]Let $x$ be the least of the 3 numbers and $y$ be the greatest. So $\dfrac{x+5+y}{3}=x+10=y-15$. We find that $x=2y-25$. Plug this into $\dfrac{x+5+y}{3}=y-15$ to get $y=25$. With this, we can find that $x=0$. So the 3 numbers are $0,5,25$, and the sum is $30,$ which is $\boxed{D}$
[/hide]


Is $C_{60}$ more electrically conductive  than graphite? If so, why ?

i believe Graphite has a better electron conductivity,
becasue if you look at the structure of graphite and $C_{60}$ you'll see that in graphite every Carbon has a hybridization of $SP3$, this is the same for $C_{60}$ but the structure of $C_{60}$ is different form graphited. $C_{60}$ contains $60$ carbon atoms in a spherical structure consisting of $20$ hexagons and $12$ pentagons arranged on the surface. this structure is very similar to a soccer ball (btw, the more scientific name for this form of carbon is buckminsterfullerene (also called "fullerene")), so in $C_{60}$ the free electrons of carbon can move harder than than in graphite.(because in graphite there is no special form of carbons, all there is, is layers of carbon atoms)
btw i believe $C_{60}$ is a semicondutor. but graphite is a conductor.

i think also b/c in graphite, the $\pi$ molecular orbitals are closer together, with each having delocalized electrons, making it a better electrical conductor

We cannot say for sure. We know that graphite only conducts electricity due to its pi bond (delocalised electrons in graphit's case) since every carbon atom is covalently bonded to 3 other carbon atoms. In Buckminsterfullerence, the molecule is shaped like a soccer ball.

Modern chemists had a hypothesis that proposed the idea that maybe Buckminsterfullerence is actually a yellow superconductor after analyzing the data the obtained from various calculations. However, modern techniques could only produce this allotrope of carbon in minute amounts and so we cannot say for sure whether this hypothesis is true or false.

So I believe that it is fair to say that Buckminsterfullerence could possibly be of a higher rank in terms of electroconductivity but in the years to come, it could be proved false.


abstract algebra
algebra
polynomial
superior algebra
superior algebra solved

can you example a submodule of a finetely generated that is not noetherian?

What is "noetherian"? (I am not English, please in mahematical formulas).

sepidetaghizade: This should go to the "College Playground - Superior Algebra Unsolved Problems" Forum :)

I don't think I understand the question

let the module be a ring, as module over itself

take as ring the polynomials over any field with infinitely many variables : $X_{1},X_{2},\ddots$

any ring is generated by the identity element as module over itself

however this ring, is not noetherian, consider the ideal generated by all the variables


ratio
quadratics
algebra
polynomial
continued fraction
number theory
relatively prime

[quote="drunner2007"]Prove that if $n$ is an irrational number that it cannot be written as a finite, proper continued fraction.[/quote]

[hide]If a continued fraction is finite then it can be expressed as a ratio of two relatively prime integers which you can't do with an irrational number...[/hide]

[hide="Formally"] We will prove the contrapositive:  If a real number can be written as a finite, proper continued fraction, then it is not irrational (i.e. it is rational).

We need to remember a few simple facts.

- The rationals are closed under addition.
- The rationals are closed under multiplication.
- The nonzero rationals are closed under multiplicative inverse.  (This implies that the nonzero rationals are closed under division.)

Let $r$ be a real number with a finite continued fraction.  The
$r = a_{0}+\frac{1}{a_{1}+\frac{1}{...+\frac{1}{a_{n}}}}$

Define the sequence

$s_{0}= a_{n}$
$s_{k}= a_{n-k}+\frac{1}{s_{k-1}}$

Of course, $s_{n}= r$.  Now, since $a_{i}$ are all rational, it follows that $s_{0}$ is rational, and therefore that $s_{k}$ is rational for all $k$ by induction.

Hence, $s_{n}= r$ is rational.  QED. [/hide]

[b]Follow-up:[/b]  Show that the continued fraction of any real root of an integer quadratic polynomial is either terminating or periodic.

[quote="t0rajir0u"]

[b]Follow-up:[/b]  Show that the continued fraction of any real root of an integer quadratic polynomial is either terminating or periodic.[/quote]

I'm not used to continued fractions; do you think you could give us a hint on this one  :P

[hide="Hint"] $\phi = 1+\frac{1}{1+\frac{1}{1+... }}$ is the solution to the quadratic equation

$\phi = 1+\frac{1}{\phi}$ [/hide]


Find the removable points as well as the horizontal and vertical asymptotes for
\[ f(x) \equal{} \frac {2x \minus{} 1}{x^2 \minus{} x \minus{} 2}
\]

[hide="Solution"]$ \frac{2x\minus{}1}{(x\minus{}2)(x\plus{}1)}$

Vertical asymptotes: $ x\equal{}2,\minus{}1$.

$ \lim_{x\to\infty}\frac{2x\minus{}1}{x^2\minus{}x\minus{}2}\equal{}0$, so horizontal asymptote at $ y\equal{}0$.[/hide]

It helps to write it in the form:

$ f(x) \equal{} \frac {1}{x \minus{} 2} \plus{} \frac {1}{x \plus{} 1}$


Let $d$ be a real number. For each integer $m\ge0$, define a sequence ${a_{m}(j)} , j= 0,1,2, \dots$ by the condition

$a_{m}(0)= d/2^m$ and $a_{m}(j+1)= (a_{m}(j))^2+ 2a_{m}(j)$for$j\ge0$
Evaluate $lim_{n\rightarrow\infty}a_{n}(n)$

[hide]
So i started out by testing a few terms and came out with the general term
 $a_{n}(x)=2^x*a_{n}(0)$
however im slightly confused with how to evaluate this limit, i thinking that i can somehow change it into the form (1+1/x) but i cant directly see how to raise it to a power.

Finally for an official write up should i prove this via induction?[/hide]

Let $b_j=a_m(j).$ Rewrite the given reccurence as $b_{j+1}=b_j^2+2b_j\Longleftrightarrow b_{j+1}+1=(b_j+1)^2.$ Thus $b_j+1=(b_{j-1}+1)^2=(b_{j-2}+1)^{2^2}=\cdots =(b_0+1)^{2^n},\ b_0=\frac{d}{2^m},$ yielding $a_m(j)=\left(1+\frac{d}{2^n}\right)^{2^m}-1.$ Therefore $a_n(n)=\left(1+\frac{d}{2^n}\right)^{2^n}-1.$ Use the limit $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e,$ the answer is $\lim_{n\to\infty} a_n(n)=e^d-1.$

alot of putnam questions are starting to seem quite accesible

This is like the easiest year of the Putnam in (semi)-recent memory (I don't think I can say recent memory when I was born in 1986!), so don't expect most year's problems to be that simple.

Still, that's not to take anything away - if they're starting to seem accessible, that's a good thing!


How would you put a curly brace underneath something with like a 3 underneath it? I don't know any examples to post.

It's like:

a
{
( the bracket is flipped so it looks like upside down mustache)
1

Use \underbrace as in $ \underbrace{xx\dots x}_{n}$


calculus
geometry
vector
real analysis
real analysis theorems

I want a good Calculus textbook that will help me patch holes in my basic calculus knowledge and review and drill other areas and serve as a good reference. My goal is to brush up my basic calc knowledge first before I tackle some other advanced math topics. I completed a course in Calc 3 plus a course in differential equations many years ago and do not plan on retaking any calculus classes.

I recently learned non-trivial statistics by buying the textbook recommended by the local university. I thought the textbook was a fantastic way to learn the material (much faster than taking a night class for a full semester or two) and am hoping to find something similar for Calculus.

I bought some Schaum's Outline books and I don't like them. The explanations are too short and hard to follow. They lack the writing, clarity, and graphics and diagrams of a good college textbook.

My local university uses the James Stewart Calculus textbook; normally I just buy what they use. However, after reading many online reviews of this book, I suspect this isn't the book that I want. I'd rather not sink $120+ and time into it if the reviews are any indication.

Does anyone have a good recommendation?

Look around used textbook websites a little more thoroughly. You can get much better prices on costly advanced textbooks.

If you can get a hold on it, Calculus and Analytical Geometry (4th edition) by George B. Thomas. It'll be really old, but you know what? It's miles better than my school calc. book.

Again, that's 4th edition. 3rd edition is pretty much the same, but the newer (esp 7th+ edition) ones are far inferior.

[quote="SridharaQ"]I bought some Schaum's Outline books and I don't like them. The explanations are too short and hard to follow. They lack the writing, clarity, and graphics and diagrams of a good college textbook.[/quote]

I have taken a large number of calculus and analysis courses, and I found that the Schaum's outline on 'Advanced Calculus', by Spiegel, remains one of my more useful references.  I have an older version (it's my dad's) but I imagine the newer ones haven't changed much.

Marsden and Tromba's book is pretty nice for multivariable and vector calculus, and for a more rigorous introduction, 'everyone' seems to like Spivak's Calculus (and those who don't like Apostol), though I confess I have never opened either.


I think there's a typo
it should be
$ \frac{2^{\minus{}1}\plus{}3^{\minus{}1}}{6^{\minus{}1}}$
so instead the answer is
$ \frac{\frac56}{\frac16}\equal{}5$

[quote="AceOfDiamonds"]I think there's a typo[/quote]
Really?
[quote="GameBot"]$ \frac {2^{ - 1} + 3^{ - 1}}{6^\color{red}1\color{black}}$.[/quote]
[quote="AceOfDiamonds"]$ \frac {2^{ - 1} + 3^{ - 1}}{6^{\color{red}-1\color{black}}}$[/quote]

yes. the answer in ftw is $ 5$ not $ \frac{5}{36}$

[quote="BOGTRO"]Yes, there is a typo. It should be $ 6^{\minus{}1}$.[/quote]
Yeah, it's a typo.


An analog clock is 15 minutes late every day.  The clock shows the exact time on January 1st, at noon.  Which day will the clock be exact at noon for the first time?

[hide]The clock has to slow by 24 hours to restore back to noon. Since every day it slows 1/4 of a hour, it will take 96 days. Assuming it's not a leap year, it should come back on April 7th [/hide]

[hide]
it has to be 24 hours to make it noon again.  It becomes slower by 15 minutes everday.  So it would take 24*4 days.  So it would take 96 days.  So the date would be April 6th.  I think...[/hide]

i don't know if it involes leap years...i think i counted the days wrong in my answer...

I used my friendly calculator  :D , and found that it's April 7th if it's a normal year, April 6th if it's a leap year. So we're both right  :lol:


function
inequalities proposed
inequalities

[quote="XAKKER"]x,y,z>0 xyz=1 prove that: (xy+yz+zx)(x+y+z)+6>=5(x+y+z)[/quote]
The Mixing Variables method kills it easy enough.
We need to prove only that $ 2t^4 \plus{} 4t^3 \minus{} 4t^2 \minus{} t \plus{} 2\geq0$ for all $ 0 < t\leq1.$

[quote="mathwizarddude"][quote="bunhiacovski"][quote="ABRORBEK"]What is  The Mixing Variables method ?[/quote]

The Mixing Variables are deficit variable prothesis proof easy :D

problem that prove by Mixing Variable :

we need prove $ f(a,b,c) \ge f(a,\sqrt {bc},\sqrt {bc})$ and $ f(a,\sqrt {bc},\sqrt {bc})\ge 6$

with $ f(a,b,c) \equal{} (xy \plus{} yz \plus{} zx)(x \plus{} y \plus{} z) \minus{} 5(x \plus{} y \plus{} z)$[/quote]

I don't think I understand your $ f(a,b,c)$.  $ f$ is clearly a function of $ a, b$, and $ c$, but then how come only $ x, y$, and $ z$ appear on the RHS?   :huh:[/quote]
sorry. :wink:
$ f(x,y,z)$ not $ f(a,b,c)$


[b]Edit[/b]: I solved it! After a drink [tea, people!], I saw the pattern. Which is really obvious NOW. So if anyone wants the solution...you know, reply or something.

Welcome to art of problem solving!

Id guess what you do would go along these lines:

We know $f(1)+f(2)+\cdots+f(n)=n^{2}f(n)$
We also know $f(1)+\cdots+f(n)+f(n+1)=(n+1)^{2}f(n+1)$
Just by plugging in $n+1$ instead of $n$.
Now pluggin the first into the second for the first $n$ terms gives:

$n^{2}f(n)+f(n+1)=(n+1)^{2}f(n+1)$

After some re-arranging:

$f(n+1)=\frac{nf(n)}{n+2}$

Can you see how this will now telescope down nicely?

If you want to see the answer, click below! But dont spoil it!

[hide]$f(n)=\frac{4}{n(n+1)}$[/hide]

Thank you for the warm welcome. I can't believe I didn't discover this site until now. But, I shall make up for lost time!

Yes, about 30 mins after posting this, I solved it. With a slightly different approach. I devised another function for the denominators of f(1) and f(2) until f(4) when I realised that it was a quadratic function---if you rewrite equivalent fractions with 2 as the numerator.

Then, at the end, I wound up with the same simplified fraction as yours so I guess it all works out! =)

Thanks for your help!

Thats how i was trying it at the start, but i felt lazy and didnt bother going through all that! Where did the problem come from?

Ah, my teacher posts a problem of the day every night on her website. It's an extension for those who are frustrated by the simplicity that is our math curriculum.


What's the formula to calculate mean(expected value) for a geometric distribution?

On the princeton review book it's (1-p)/p but on my text book it's 1/p.

It depends on whether 0 is an allowed value for the random variable.


In the adjoining figure, $ CDE$ is an equilateral triangle and $ ABCD$ and $ DEFG$ are squares.  The measure of angle $ GDA$ is

$ \textbf{(A)}\ 90^{\circ} \qquad \textbf{(B)}\ 105^{\circ} \qquad \textbf{(C)}\ 120^{\circ} \qquad \textbf{(D)}\ 135^{\circ} \qquad \textbf{(E)}\ 150^{\circ}$

[geogebra]1f261c66b2dbf010cbbe994c5925227dca12d674[/geogebra]

The desired angle is just $ 360\minus{}(90\plus{}90\plus{}60)\equal{}\boxed{120^{\circ}}$


[quote="t0rajir0u"]Calculator gives $x \approx 18.245203...^{\circ}$, so I doubt that this question has an elementary solution.  Are you sure you wrote the question correctly?[/quote]

I think it should be \[ \tan^23x=3 \]which gives x=20° AND x=80°

By power reducing formula $\sin ^2\alpha = \frac{1- \cos 2\alpha}{2}$ and $\cos ^2\alpha = \frac{1+ \cos 2\alpha}{2}$ so $\frac{\sin ^2\alpha}{\cos ^2\alpha} = \frac{\frac{1- \cos 2\alpha}{2}}{\frac{1+ \cos 2\alpha}{2}} = \frac{1- \cos 2\alpha}{1+ \cos 2\alpha}$ Because in this problem $\alpha = 3x$, we get $\tan ^2 3x = \frac{1- \cos 6x}{1+ \cos 6x} = 2$ and by multiplying over the denominator (assuming $x \neq 0, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6},$ or  $\pi$, we get $1- \cos 6x = 2+2 \cos 6x \Rightarrow 3 \cos 6x = -1 \Rightarrow \cos 6x = \frac{-1}{3}$ at this point there's nothing you can do other than splitting $\cos 6x$ by double angle formula, but I don't think you can do anything but use your calculator as it isn't $\frac{\sqrt2}{2}, 1, \frac{1}{2}, \frac{\sqrt3}{2}$ etc.
Hope this was helpful.

Hokkage, it is much easier to simplify by $\tan^2 x = \sec^2 x - 1$, which gives the equation $\cos 3a = \pm \frac{1}{\sqrt{3}}$ for the original problem, which has no solution that is a rational multiple of $\pi$.

For the modified problem, it gives $\cos 3a = \pm \frac{1}{2} \implies 3a = \frac{\pi}{3} + k\pi, \frac{2\pi}{3} + k\pi \implies a = \frac{\pi}{9} + \frac{k\pi}{3}, \frac{2\pi}{9} + \frac{k\pi}{3}$.

I definitely like that solution a lot more. Forgot the $\tan ^2x= \sec ^2x - 1$.   :oops:


Find all integer roots of following equation :
36x 2 + 18y 2 +6z 2 =5t  2

This is an old problem from the Asian Pacific Mathematical Olympiad (APMO).


Lets a,b,c are in $ [1;3]$ such that $ a\plus{}b\plus{}c\equal{}5$. Prove the inequality
  $ 9abc\plus{}100 \geq 17(ab\plus{}bc\plus{}ca)$

[quote="AngelHolder"]Lets a,b,c are in $ [1;3]$ such that $ a \plus{} b \plus{} c \equal{} 5$. Prove the inequality
  $ 9abc \plus{} 100 \geq 17(ab \plus{} bc \plus{} ca)$[/quote]

\[ \Leftrightarrow \sum_{cyc}(a\minus{}b)^2(a\plus{}b\minus{}c\minus{}1)\ge 0\]

which is obvious.


Hey, we're doing a chem project, and our group doesn't really have a good idea.  It's on a specific element, and ours is hydrogen.  For example, the mercury group did a crime scene investigation of a poisoned person.  There has to be one main focus of the topic (such as hydrogen car fuel cells, H-bombs, etc), and it should be funny and informative at the same time.

I'd really appreciate any ideas.

Thanks!

Try looking in the chemistry section of the forum, you might have better luck there.


Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that
\[ ({a^2} \plus{} 2bc)({b^2} \plus{} 2ac)({c^2} \plus{} 2ab) \ge \sqrt {1 \plus{} 36{a^2}{b^2}{c^2}{{\left[ {(a \minus{} b)(b \minus{} c)(c \minus{} a)} \right]}^2}}\]
PS:We can make this ineq more stronger. But with $ k\equal{}36$, it exists a nice solution by Am-Gm  :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that
\[ ({a^2} \plus{} 2bc)({b^2} \plus{} 2ac)({c^2} \plus{} 2ab) \ge \sqrt {1 \plus{} 36{a^2}{b^2}{c^2}{{\left[ {(a \minus{} b)(b \minus{} c)(c \minus{} a)} \right]}^2}}\]
PS:We can make this ineq more stronger. But with $ k \equal{} 36$, it exists a nice solution by Am-Gm  :)[/quote]
The gereral problem:
Find the best $ k$ for this ineq still trues (with the same condition):
\[ ({a^2} \plus{} 2bc)({b^2} \plus{} 2ac)({c^2} \plus{} 2ab) \ge \sqrt {1 \plus{} k{a^2}{b^2}{c^2}{{\left[ {(a \minus{} b)(b \minus{} c)(c \minus{} a)} \right]}^2}}\]
 :)


Here's the expression that I'm having trouble with:
[quote](2^-1x^-2y^-1)^-2(2x^-4y^3)^-2(16x^-3y^3)^0 / (2x^-3y^-5)^2[/quote]
The Answer is:
[quote]x^18y^6 / 4[/quote]

I've worked this problem over and over as many different ways as I could, but I've never got the correct answer.  I would appreciate very much it if someone could post the steps to solving this problem.

Thanks for your time!

$ \frac{(2^{\minus{}1}x^{\minus{}2}y^{\minus{}1})^{\minus{}2}\cdot(2x^\minus{}4y^3)^{\minus{}2}\cdot (16x^{\minus{}3}y^3)^0}{ (2x^{\minus{}3}y^{\minus{}5})^2}$

Is this what you mean? Parentheses on the negative exponents would have helped.

This is not an olympiad-level problem, and should be moved.

Yes that's what I meant!  I'm sorry that I've posted under the wrong forum.  Initially I wasn't paying attention to the forum path.

[hide="complete solution"]
$ \frac {(2^{ \minus{} 1}x^{ \minus{} 2}y^{ \minus{} 1})^{ \minus{} 2}\cdot(2x^ \minus{} 4y^3)^{ \minus{} 2}\cdot (16x^{ \minus{} 3}y^3)^0}{ (2x^{ \minus{} 3}y^{ \minus{} 5})^2} \equal{}$

$ \frac {(2^2x^4y^2)(2^{ \minus{} 6}x^2y^{ \minus{} 6})}{2^2x^{ \minus{} 6}y^{ \minus{} 10}}$

$ \frac {2^{ \minus{} 4}x^6y^{ \minus{} 4}}{2^2x^{ \minus{} 6}y^{ \minus{} 10}}$

$ 2^{ \minus{} 6}x^12y^6 \equal{} \frac {x^{12}y^6}{2^6}$
[/hide]
Is it me or the answer is incorrect??

I would suggest you in first place to work as ordered and clean as you can.

Then you may want to clear the exponents on the agrupations signs.

You could simplify fisrt the numerator and denominator and then, the whole fraction.

At least this works for me jiji.

 :)

Actually, you fell victim to something I missed- the second term is $ (2x^{\minus{}4}y^3)^{\minus{}2}$


Source: USAMO shortlist
Let $a_{1}=a_{2}=97$ and
$a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_{n}^{2}-1)(a_{n-1}^{2}-1)}, n>1$.
Prove that
a) $2+2a_{n}$ is a perfect square.
b) $2+\sqrt{2+2a_{n}}$ is a perfect square.


I dont understand where the 0's acme from...

[hide]1995[/hide]

[quote="jimli"][hide]1995[/hide]
i'm on a roll.[/quote]
yea, your on a roll

[hide]I got 1995. I was just basic math really. :D [/hide]

1995
easier than  :starwars:  :play_ball:

Reviving old posts=Spam
This was first posted in May, 2005!

Yes, this is very hard : :starwars:

[quote="cgyao15"]no need to revive a post that old  :wink:[/quote]

*cough*COUGH* HYPOCRITE

Yeah, haha, by saying to not revive old threads, you revive the old thread....
 :rotfl:

Please do not post asking someone to not revive if they are saying something useful. The person was reminding the other poster to not revive very old topics. This is useful, and is thus not reviving. And in the normal definition of reviving, the poster actually wasn't reviving by telling the other poster to not revive, since the thread was no longer old  :wink:

But seriously, do lock.

[quote="cgyao15"]1995
easier than  :starwars:  :play_ball:[/quote]

This is what I meant when I called him a hypocrite...

[quote="cgyao15"]no need to revive a post that old  :wink:[/quote]

It's kinda weird how he revives a 4 year old post only to do the same thing to someone after reviving a 4 month old post...

also, did anyone notice cygao15 and cyao15 with 3 minute apart posts?
are they the same person!?!  :maybe:

ok, mods, lets get this thing locked up tight now... :P

[quote]nick42 no, i think they're actually brothers...
 [/quote]
we're brother an sister, i'm the boy happy?


I hope someone's help.

If someone has some idea of following thing, please write your idea here.

$Problem$: There is a spring(spring constant: $k$) fixed on the floor with massless board fixed upside of it.

Peter put a box of $m$$(kg)$ on the board and make spring in a state of equilibrium.

Peter want to make the box off from the board.

What is the least value of $x$, the distance Peter should press the spring, to make box off from the board?

I would suppose that the spring should accelerate down at g at the point of greatest elongation at the top... however, since everything's massless... =/

the motion is simple harmonic, which is defined by the equation relating the acceleration of the mass on the board a and the displacement of that mass from an equilibrium position, x, by:

a = -kx

The mass will lose contact with the board at the top of the path when the acceleration a needed to maintain simple harmonic motion exceeds that which can be provided by the gravitational field, g.

so the limiting case is when g = -kx, and I dont think there is enough information to go any further from here

but if you had the spring constant for example (this is not k in the equation above, confusingly.. k here means the square of the angular frequency w of the oscillations) then you'd be able to determine the frequency of oscillations using the two equations:

T = 2*pi*sqrt(m/k)
f = 2*pi / T

And then from this determine k = w^2 = (2*pi*f)^2, and then you'd know the actual displacement.

I hope a proof using forces, such as gravity, spring force, and $N$.

or using Energy, such as $mgh$, $\frac{1}{2} kx^2$, etc..

Can anyone help me?

[quote="Jimmy"]Well the force of gravity on the box is $mg$, so the restoring force of the spring $kx$ must be greater than $mg$, so $x>\frac{mg}{k}$.  But that seems too simple.... I must have missed something.[/quote]

I think you need more explanation of your solution.  :D

So you have a massless spring with massless board and a weight... OK. First of all, massless things are tricky. They have no momentum nor kinetic energy so they can move however they like. Fortunately, there is a force acting on the spring whenever it's not in the equilibrium position and this makes the spring's movement somewhat reasonable. When you put the box on top of the spring, the equilibrium position will be in distance $x_0$ from the original equilibrium where:

$kx_0=mg$

$x_0=\frac{mg}{k}$.

This is actually quite irrelevant but I put it here anyway. Now suppose that you push the spring even further to distance $x>x_0$ from the original equilibrium, what happens? The box will either start oscillating while touching one end of the spring at all times or it will accelerate upwards and in some instant leave the spring behind - it will jump off the board (and then fall back on it and repeat the whole thing). This jumping off the board can't happen before reaching the zero position because while the spring is still compressed, it is forced to accelerate as fast as possible upwards (and because it's massless it can accelerate arbitrarily fast) so it has to keep up with the box. On the other hand, as soon as the spring reaches the zero position it can't go any further because the spring force would immediately stop it. Therefore, if the box reaches the equilibrium position of the spring with nonzero velocity, it will jump off the board. We can calculate the necessary $x$ using energy:

$\frac{1}{2}kx^2=mgx+\frac{1}{2}mv^2$

$kx^2-2mgx=mv^2$

$\frac{x(kx-2mg)}{m}=v^2$

If we get negative $v^2$, box will never reach the zero position. What we need is $v^2>0$:

$\frac{x(kx-2mg)}{m}>0$

$kx>2mg$

$x>\frac{2mg}{k}$

Thank you for nice solution bus  :D

In your solution, you implied that the least value of $x$ to press the spring from the equillibrium is $\frac{mg}{k}$

Hm... I'm afraid I don't quite understand :(. But anyway, my solution is $\frac{2mg}{k}$ while yours is only $\frac{mg}{k}$ - or did I get you wrong?


Prove that if $ n \plus{} 1$ is a perfect square, then $ n \plus{} 1 | 2(n \minus{} 1)!$.

[hide]If $ k^2\equal{}n\plus{}1$, then we have to prove $ k^2|2(k^2\minus{}2)!$

For $ k\equal{}2$ we have $ 4|2\cdot 2!$, which is true.

For $ k\geqslant 3$ we have $ (k\minus{}1)^2\geqslant 4\implies k^2\minus{}2k\minus{}2\geqslant 1>0$, hence $ k^2\minus{}2>2k$, therefore both $ k$ and $ 2k$ are the factors of $ (k^2\minus{}2)!$ product and the conclusion follows.[/hide]


Can anyone help me with this?

In the disk of radius 16 there are given 650 points. Show that there exists an annulus of an outer radius 3 and an inner radius 2 containing at least 10 of these points.

See page 5-6 (problem 7) [url=http://www.usamts.org/Forms/ExpectedValue.pdf]here[/url].


http://www.geocities.com/scmathteam/

check it out...actually...wait a little bit till it's completely finished which will hopefully be in a few days :lol:

hahahaha good one..."a few days..." lol.  It's mostly resources & bios right now.

well, if it's not gonna be done in a few days, then we agreed to just making this forum the official sc website.

that way, members of the team will have to sign up to the forum

[quote="Iversonfan2005"]
that way, members of the team will have to sign up to the forum[/quote]

 :yup:

Btw, ads are annoying in case you were wondering ;)

[quote="joml88"]

Btw, ads are annoying in case you were wondering ;)[/quote]

no joke :mad:

i think you should talk to mr. r into getting the members to join aops, besides, it'll be really cool if we have a lot of people in the carolina's forum (perhaps even South Carolina forum :D ) and discuss problems here. Especially if Mr. R or Mr. Neil are free, they can post lessons and problems for us.  ;)

[quote="furious"]1) my picture still isn't showing up
2) joe only listed about half of his accomplishments
3)that's still the ghetto site. i thought you were gonna buy one sean.
4)joe's aops username is wrong

being as it is, the carolinas sections would be better to use as the "official" site.[/quote]

lo (freaking) l

why don't we take off accomplishments :blush: it takes too much space and time anyway :lol: , we don't wanna brag now do we?

:mad:  :mad:, hey why can you excede the smiley limit....those tyrannical powers

[quote="Iversonfan2005"]:mad: , hey why can you excede the smiley limit....those tyrannical powers[/quote]

what are you talking about  :huh:

[quote="joml88"][quote="Iversonfan2005"]we "normal people" can only have two smilies per post[/quote]

I only two smilies per post...so I don't know what you're talking about.  :ninja:[/quote]

nice try...hehe ;)  ;)   ;)

omg shobhit...you just posted three smileys...wtf?

hacker...

(# of smilies allowed) if a function of (# of non-smilie words in post)


geometry
calculus
integration
AMC
AIME
rhombus
Pythagorean Theorem

An integer quadrilateral is a quadrilateral in which all side lengths are of integer length, the area is an integer, and the pieces that the diagonals divide each other into are also of integral length.  Compute the area of an integer quadrilateral with at least three sides of length 13.

What number do you think this would be on the AIME?

[hide="Quick"] A rhombus with side length $13$ and diagonals $10, 24$.  It has area $120$ and its diagonals bisect each other into lengths $5, 5, 12, 12$.

It's a little too contrived for an AIME problem. [/hide]

Give that this is inspiration for a future fake AIME problem, you'd need to make sure that's the only possible answer.

I did make sure.

The non-guess&check way is with stewart's theorem.

What if exactly 3 sides are 13? Is there such an integral quadrilateral? Something more involved to ponder :)

I don't think there is.  I tried every possibility before making up the problem.  I might have missed a possibility though... :maybe:

Well, let $AB=BC=CD=13, \; AD=k, \; AC=r, \; BD = s$.
Let the intersection of $AC$ and $BD$ be $E$

You know that $k$ and $a$ are integral, so $BE$ and $ED$, by the Pythagorean Theorem, are expressible in the form $\sqrt{n}$.

What do you know about two square-roots that add to an integer (namely $b$)?


[quote="moogra"]Two noncongruent triangles have the same area but different perimeters. Triangle A has sides lengths of 8, 8 and 6 cm. Triangle B has two sides with length of 8 cm. Find the length of the third side length of Triangle B. Round your answer to the nearest hundredth of a cm.[/quote]

[hide]$2\sqrt{8^{2}-(\frac{6}{2})^{2}}\approx \boxed{14.83}$[/hide]


[quote="Michael Niland"]IMO 2000 question 2 :
If a,b,c are positive real numbers such that abc=1 Prove that

P = (a-1+$\frac{1}{b}$).(b-1+$\frac{1}{c}$).(c-1+$\frac{1}{a}$)
 then P $\leq$ 1
This is considered to be an easy question.
Can you also show that P$\geq$ -1 ?[/quote]
No, since it is wrong. Try $a=4,$ $b=c=\frac{1}{2}.$ :wink:


Ok i need help with this one.

Square SQUA with midpoints M and N of sides SQ and SA respectively, has an area of 400 square units. What is the number of square units rounded to the nearest integer, in the area of the largest circle that can be inscribed in the pentagon MNAUQ ? Take Pi=3.14 .

[quote="rcv"][quote="GoBraves"]Ok i need help with this one.[/quote]
By symmetry, we can argue that the inscribed circle will be tangent to segment MN at the midpoint of MN.  The circle will also be tangent to segments QU and UA at unknown points.

Now, if you draw the diagonal, SU, of the square, you can subdivide that diagonal into three sections.  The section from point S to the midpoint of MN.  The section from the midpoint of MN to the center of the circle.  And the section from the center of the circle to point U.

The total length of the diagonal is known (or computable), given the area of SQUA.  The first of the subdivided sections has a known (or computable) length.  The second two sections can be written in terms of the circle's radius, r.

Is that enough to get you started?[/quote]
Thanks

I got 353 square units. Is that right?

[quote="GoBraves"]Thanks
I got 353 square units. Is that right?[/quote]
A sanity-check says that can't be correct.  A circle inscribed in the 400 square-unit square would have an area of 314 square units.  But the actual inscribed circle is smaller.  So its area cannot exceed 314.

Sanity checks tell me the radius must be:  $5\sqrt{2}<r<10.$

[quote="GoBraves"]Sorry, A slight mistake caused that. Ok I got 155. Is this correct? If im wrong this time, please tell me the answer.

Thanks[/quote]
I don't recall seeing this problem, so I don't have an answer key.  My setup follows.  BTW, I will gladly give guidance, but I don't usually just give answers, especially without knowing the source of this problem.
\begin{eqnarray*}
d&=&s_1+s_2+s_3 \\
20\sqrt{2}&=&5\sqrt{2}+r+r\sqrt{2} \\
&\cdots&
\end{eqnarray*}