search
real analysis
calculus
integration
derivative
function
abstract algebra

I'm planning to read the theory of fourier series and fourier transform on my own. Can anyone suggest a good book for the same? Most importantly since i'm just starting on it, i'd prefer something which is not too abstract but at the same time it should give me an idea of what is happening. So with this in mind please suggest some books. thank you

For the course I teach, I use my own notes and no textbook. But the textbook that does the most to inspire those notes was written by David Kammler - you can search for it, but I think it's out of print.

Before I had the notes I now use, I was using the book by Tolstov - since it's published by Dover, it should be available for a reasonable price. Tolstov is a very nice introduction to Fourier series; it doesn't particularly match the course I teach because (like Kammler), I want to deal with the Fourier transform on $\mathbb{T},\mathbb{R},\mathbb{Z},$ and $\mathbb{Z}_{N}$ all at the same time.

There are a number of choices to be made:

Lebesgue integral or not? I read the tone of your post as "not". (That's also the undergraduate-level course of mine that I'm talking about.) If you do want the Lebesgue integral and some level of abstraction, there's Katznelson (possibly also available through Dover.)

Real notation or complex notation? That is, $a_{n}\cos(nx)+b_{n}\sin(nx)$ or $c_{n}e^{inx}+c_{-n}e^{-inx}$ ? If "real" then your primary sources might be a number of different elementary partial differential equations textbooks, in which the Fourier series appear as needed to solve PDE's. Kammler's book, my notes, and Katznelson's book stay with the complex notation from the beginning. The biggest advantage of the complex notation is that it simplifies the expression of operators such as convolution and differentiation through the Fourier transform.

Just periodic functions (that is, Fourier series)? Or also functions on $\mathbb{R}$ and maybe even $\mathbb{R}^{n}$ (that is, the Fourier transform)? Or also include functions on $\mathbb{Z}$ and especially $\mathbb{Z}_{N}$ (the DFT, the FFT)? Or consider other locally compact abelian groups?

If just series, then try Tolstov. I'm not sure what the best book would be to answer the second questions. Kammler's book is positioned for the third question, as are my own notes. The fourth question takes you out of "elementary."

oh i didnt know there are so many choices to make. I am just interested in analysis and my university doesnt offer a course in fourier series this semester. So i thought of reading on my own.Dr Kent , if you have a pdf version of your notes, will you be so kind enough as to mail it to me?my email id is koushik.math@gmail.com

In case you dont have a pdf version, please tell me , among all the books you quoted which one would probably give an idea of most of the things you talked about? And yes you are right about lebesgue integral, i'd like to wait for some more time before that. thank you very much for any kind of help


You have already posted this problem here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124306[/url]

Do not double post.  This topic is locked.


[quote="Maks"]Prove that all positive integer numbers are interesting.
 :P[/quote]

Assume to the contrary that all positive integers are interesting. That would be impossible to prove, so all positive integers, [b]must[/b] be interesting.  :P

Sorry, just because it's unprovable doesn't mean it's false (points to Godel).

I don't think this is true. In fact, I can prove that every positive integer is boring.

[hide]Suppose, to the contrary, that not all positive integers are boring. Then there must be a smallest non-boring positive integer. But who cares?[/hide]

The number 0 is very interesting. Negative numbers are below 0, so they are not very interesting. However, positive numbers are even over 0, so they must be even more interesting than 0...  :D

assume that uninteresting integers exist, the smallest such number is interesting – a contradiction

:D found in the WOOT blog.


What does the word in the post before remind you of? Give the first thing that you think of!!!

Ok I'll start off:

mamma mia :D  :D  :D

Mud.  :D

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

Babies

i don't know... and don't ask why :)

donuts.
wow this is getting revived after a long time

u have a crush?
Anyway
that reminded me of love...
(i don't have a crush :o)
 :love:


geometry
circumcircle
inradius
trigonometry
incenter
trig identities
Law of Sines

Find the sum of the exradii of a triangle with side lengths $13, 14, 15$.

(Note:  The excircles of a triangle are found by drawing the largest circle externally tangent to each of the three sides and also tangent to extensions of the other two.)

[hide]I pulled this from Coxeter's [i]Introduction to Geometry[/i].

When $R$ is the circumradius of a triangle, $r$ is the inradius, and $r_{a},r_{b},r_{c}$ are the exradii, the equation $4R=r_{a}+r_{b}+r_{c}-r$ holds.  To answer the problem, all we have to do is compute the sum of $4R$ and $r$.  By the extended law of sines etc., $\displaystyle 4R=\frac{abc}{[ABC]}=\frac{13\cdot 14\cdot 15}{7\cdot3\cdot4}=\frac{65}{2}$, and $\displaystyle r=\frac{[ABC]}{s}=4$, so the answer should be $\displaystyle \begin{array}{|c|}\hline \frac{73}{2}\\\hline\end{array}$.

[hide="Derivation of Formula"] Let us consider triangle $ABC$, with semiperimeter $s$, incenter $I$ with inradius $r$, and excenters $I_{a},I_{b},I_{c}$, with excenters $r_{a},r_{b},r_{c}$, respectively.  Note that the area of the triangle is $rs=r_{a}(s-a)$ (consider $I_{a}AB+I_{a}AC-I_{a}BC$).  (Actually, this gives us another way of solving the problem, but I felt like learning this stuff.)  As we mentioned above, $4R[ABC]=abc$.  Now we take the extraordinarily rowdy expression

$s(s-b)(s-c)+s(s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c)$,

laboriously multiply it out (unless someone has a better way(?)), and note that it, too is equal to our tidy $abc$.  By Heron's, it is also equal to

$\displaystyle\frac{[ABC]^{2}}{(s-a)}+\frac{[ABC]^{2}}{(s-b)}+ \frac{[ABC]^{2}}{(s-c)}-\frac{[ABC]^{2}}{s}=[ABC](r_{a}+r_{b}+r_{c}-r)$

Comparing this to that with which we started, it is clear that $4R=r_{a}+r_{b}+r_{c}-r$.[/hide]
Q.E.D.[/hide]

Why are 13-14-15 triangles so popular, anyhow?

I just enjoy the fact that $\sin, \cos$ of all the angles are rational, it makes working with them more bearable :)

Wonderful problem.
[hide]
Let $\triangle ABC$ have sides $a,b, c$, where $a=15$,$b=14$, and $c=13$.
There are $3$ excircles, one each tangent to $a,b,c$.
Draw each of the excircles and draw a line tangent to each and parrelell to $a,b, and c$. Extend $AB$, $AC$, and $BC$ so that they meet these lines. Now the segments parrellell to $a,b,c$ will be called $a_1,b_1$, and $c_1$, respectively. Denote the points of intersection as $A_1, A_2$ for $a_1$, $B_1,B_2$ for $b_2$, and $C_1, C_2$ for $c_1$.

Notice we've formed similar triangles here.
$\triangle ABC \sim \triangle AA_1A_2$, $\triangle BAC \sim \triangle BB_1B_2$, $\triangle CAB \sim \triangle CC_1C_2$.

Given this, we can find the incircles of triangles $AA_1A_2, BB_1B_2, CC_1C_2$, which are the excircles of $\triangle ABC$.

To derive the proportions first we need to find a proportions inside $\triangle ABC$ and compare it with the ones in the larger circles.

Let the altitude of $\triangle ABC$ extending to side $a$ be $a_a$, and then the others being $b_a$ and $c_a$.

Using herons formula we have area of $\triangle ABC$:  $\sqrt{21(21-13)(21-14)(21-15)}=84$. So $a_a=56/8$, $b_a=12$ and $c_a=168/13$.

Let $r_e1, r_e2, r_e3$ be the exradii.

We have $\frac{a_a-2r}{a_a}=\frac{a_a}{2*r_e1}$, we get $r_e1=14$.
$\frac{b_a-2r}{b_a}=\frac{b_a}{2*r_e2}$, we get $r_e2=12$.
$\frac{c_a-2r}{c_a}=\frac{c_a}{2*r_e3}$, we get $r_e3=10.5$.

Therefore the sum of the exradii is $14+12+10.5=\boxed{73/2}$.[/hide]

13-14-15 triangles also have integer areas, and the hypotenuse to each side is rational.  They're just easier to work with overall than a lot of other triangles.

Wow, interesting problem, first time I did a problem with excircles...

[hide]We can draw a triangle using the radii of each of the excircles in which each of its sides touch one of the virteces of the triangle inside it. Now if we draw in the incircle of the bigger triangle, it's the circumcircle of the smaller triangle. So the circumradius of that triangle is:

$R=\frac{abc}{4A}$ and using Heron's, we have $A=\sqrt{21(21-14)(21-15)(21-13)}\Longrightarrow A=84$, therefore, $R=8.125$.  Now using $rs=A$ for the smaller triangle, we have $r \cdot 21=84 \Longrightarrow r=4$, since $4R+r$ is the sum of the exradii, the answer is $\boxed{36.5}$[/hide]

Interesting link that has to do with excircles: http://www.cut-the-knot.org/triangle/RelationsInTriangle.shtml#rrrrR


geometry
geometric transformation
reflection
symmetry
absolute value

Guys what does the graph of

$ (|x|\minus{}2)^2\plus{}(|y|\minus{}2)^2\equal{}5$

Look like? Any better way to find it other then plugging in values? Could anybody post a picture of it's graph?

[hide="Hint"]
Consider the cases $ x,y\ge0,\,\,x,\minus{}y\ge0,\ldots$.
[/hide]

[hide="Hint"]The set of points satisfying $ f(|x|,y)\equal{}0$ for any $ f$ is like saying, take the set of points satisfying $ f(x,y)\equal{}0$ for $ x\geq 0$ and reflect it over the y axis. We can say something similar for an absolute value of $ y$.

Therefore, what kind of symmetry does this graph have?

What is the graph of this in the first quadrant?
Therefore, what is the graph of this in the entire plane?[/hide]

It would be four circles. The following are the circles which would be included.

(x-2)^2 + (y-2)^2 = 5
(x+2)^2 + (y-2)^2 = 5
(x-2)^2 + (y+2)^2 = 5
(x+2)^2 + (y+2)^2 = 5

Pretty much imagine four circles positioned on the vertices of the square with points (2,2), (-2,2), (-2,-2) and (2,-2).

[hide="SnowEverywhere:"]The graph would not be 4 complete circles. It would be 4 "truncated" circles.
See TZF's hint for an explanation as to why.[/hide]

The graph will be like this:

[geogebra]dbe8ef03401e4221b6015ba2747ab7dea688ef9a[/geogebra]


I think that this forum was so active with "bus"'s posts i'm wondering what would be the reason that prevents him of participating i hope this situation won't last ,personally i'm gonna PM him .
To all the mathlinkers: Let this forum be more active plz :)

Wow, I sure didn't expect this... The truth is that I'm really busy right now. I haven't visited MathLinks / AoPS in quite a while mainly because of school and also because of my new work. I'd love to come back and take a look at the CS / Physics forums again - seems like so much has happend here since my last visit - but it's just not possible right now, I'm sorry. Perhaps in a couple of weeks...

C U all :)

OK, I just couldn't resist... but that's it for quite a while again. ;)

Thx for this tour :) and congratulations for the new job ;) keep in touch

Good to see u back BUS....

Really i was missin ur explanation style...

Have a GOOD DAY... :)


algorithm
real analysis
real analysis theorems

Hey,

I'm looking for a program that, given the appropriate initial conditions, will draw out a curve based on a differential equation [u]without [/u]necessarily solving for the function. Does anyone have any suggestions?

Thanks!

~ tZf

Anything that's capable of handling a numerical algorithm can do this- there are several efficient ways of calculating a solution to high precision at a sequence of sample points, and it's easy to graph that information once you have it.
There's no way to get a reasonable graph without at least an approximate solution.

[quote]Anything that's capable of handling a numerical algorithm ...[/quote]
I second the notion that this is a numerical analysis question. I'm not a user myself, but my first instinct concerning a place to look is MATLAB. Does MATLAB have built-in ODE solvers? I don't know, but I wouldn't bet against it.

Or there's always do-it-yourself work with a programming language. Look up the words "Runge-Kutta" or "Fourth Order Runge-Kutta."

Matlab, maple, and mathematica all have numerical ode solvers.

Thanks. I tried Mathematica, and it seems to be working pretty well. However, it is unable to graph delay differential equations; can matlab and/or maple do these?

I never worked with DDEs, but [url=http://www.radford.edu/~thompson/webddes/]Matlab can solve them numerically[/url].


Tonight is the first game of the world series, at Fenway. Let the probability that the Sox beat the rockies in a given game be $ 5/6$. What is the probability that they win the series at Fenway?

note the capitilizations.

When you say win the series at fenway, your implying the
[size=150]2-3-2[/size] format for home games right?
So basically find the chances the red soxs win game 6 or game 7 to win the series, whcih is winning 4 games..

so its just
game 6 = $ (\frac {5}{6})^{4}(\frac {1}{6})^{2}(\binom{5}{3})$
game 7 = $ (\frac {5}{6})^{4}(\frac {1}{6})^{3}(\binom{6}{3})$

add them up and its $ \frac {50000}{6^{7}} \equal{} \frac {3125}{17496}$, whcih is around $ 0.178$

A NOTE: the chances they win at Fenway (clinch the series) is small and thats becuase they said the red soxs had a 5/6 of winning a game, so most likely in that scenerio, they will win the series, and they will do it at coors field.

but just remember the rockies had abotu a 1/600000000000 chance of even making the playoffs and now they are here..(one strike away frmo the pads gettring in and down 2 in the 13th in an elim game)

and gess wat? the rockies are going to whoop! :rotfl:

im a yanks guy so i h8 bosox

[quote="DonkeyKong"]Which team winning the whole series?
You didn't state that well..even though it sounds like you are talking about Sox.

Well, I have to admit that I stink at this kind of problem...

But wouldn't it be just $ \frac56$??

That's most likely wrong...
Does anyone know how to solve this problem?[/quote]

To solve, you just use casework like firecricket did

Although the Sox did win in a harder league so comeback isn't as likely for the rockies.

Still, it's possible... (04)

Here's an extention to the problem.

The Red Sox are winning against the rockies in the series, 2-0. If the next three games are at coors field, what is the probability that the Red Sox win at Fenway, given that now, the probability that the Red Sox win a given game is 9/10?

[quote="firecricket91"]So basically find the chances the red soxs win game 6 or game 7 to win the series, whcih is winning 4 games..

so its just
game 6 = $ (\frac {5}{6})^{4}(\frac {1}{6})^{2}(\binom{5}{3})$
game 7 = $ (\frac {5}{6})^{4}(\frac {1}{6})^{3}(\binom{6}{3})$

add them up and its $ \frac {50000}{6^{7}} \equal{} \frac {3125}{17496}$, whcih is around $ 0.178$
[/quote]

I didn't quite get that..

Can someone explain little bit further..?
Sorry to bother you guys.. :blush:


function
calculus
derivative
real analysis

Let's f  be a real  function  on R .Prove if  f be differentiable in t with derivative A then :
For every e > 0 there be a d >0 such that for every open interval (x1,x2) containing t whose length |x2-x1|  is smaller than d ,| (f(x2)- f(x1))/(x2-x1) - A | < e

   Is its inverse true? Could you provide a contradictory example ? What conditions if  f had that would be true?

    This is useful to better insight to definition of derivatives of measures in topology.

This is not high school level.  You want one of the [url=http://www.artofproblemsolving.com/Forum/index.php?f=66]real analysis[/url] forums.


geometry
inequalities
ratio
analytic geometry
similar triangles

you have a triangle abd. you put a point c, between points a and b. you draw a line between the points c and d. how do you prve that the line dc is inside the traingle? (I think the definiyion of inside is that, if you choose a any point on cd, e, and draw a line from e to b, and from e to a, the line doesn't cut the two lines ad and bd.
I also have a problem with this http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI18.html why does he can assume that the new angle is part of the whole angle?
thanks in advance!

well... i am not sure about your definition of inside a triangle (or proving it), you could always just show that using analytic geometry, i.e. the set of points determined by the triangle satisfies a certain set of inequalities, and the points on that line satisfy those conditions

I think you are right. I tried, but I got complicated a bit, but i think it doesn't so difficult. but,does someone have a geometric proof?

well, here are a few ideas that i had in this notion....if you draw a parallel to AB, and have it intersect the 3 lines, you can easily show that the middle intersection is always inbetween the intersections with AD and BD using similar triangles on the ratios between $AB$, $AC$, and $BC$, and he says that the whole is greater than the part as his 5th common notion here:  [url]http://aleph0.clarku.edu/~djoyce/java/elements/bookI/cn.html[/url]


Assume that $ S \equal{} \{1,2,3,...,n\}$, and that $ A_{1}$, $ A_{2}$, ..., $ A_{k}$ are subsets of $ S$ such that for every $ 1\leq i_{1},i_{2},i_{3},i_{4}\leq k$, we have
\[ \left|A_{i_{1}}\cup A_{i_{2}}\cup A_{i_{3}}\cup A_{i_{4}}\right| \leq n \minus{} 2.
\]
Prove that $ k \leq 2^{n \minus{} 2}$.

[i]Edited by Myth[/i]

(I'm sorry if my English isn't so good, but it isn't my mother language...  :( )

#(A) = number of elements of A.
A={A1,...Ak}

[b]Lemma 1[/b]: Each set in A has at most (n-2) elements.
[u]Proof[/u]: obvious, from the intersection property with i1=i2=i3=i4.

[b]Lemma 2[/b]: If one set A* in A={A1, ... A*...Ak} has cardinality (n-2) then A* contains each set in A.
[u]Proof[/u]: suppose it is not, so exist an A' in A such that A' is not a subset of A*, so exist an a' \in A' : a' not  \in A ==> A'  \cup A is a proper superset of A so, #(A' \cup A)>#(A)=(n-2), which is an absurd according to lemma 1.

[b]Lemma 3[/b]: If one set A* in A={A1, ... A*...Ak} has cardinality (n-2)
then is the only set in A which contains (n-2) elements.
[u]Proof[/u]: it is an immediate conseguence of lemma 2; suppose 2 sets with (n-2) elements exist, say A' and A''. So According to lemma 2 A' conatins A'' and viceversa, so A'=A''.

[b]Hypotesis 1[/b]: No set has cardinality (n-2).
So, each set has at most (n-3) elements and the cardinality of the set: G={A subset of S : #(A) \leq n-3} is #G =  \sum C(n-3,i) , i \in {0,...,n-3} = 2^(n-3) \leq 2^(n-2) for all n \in[b]N[/b]

[b]Hypotesis 2[/b] (negation of Hypotesis 1): at least one set A* has cardinality (n-2).
By lemma 3 only one set of (n-2) elements exists, say A*. Every set in A is a subset of A*, the number of the possible subsets of A* is 2^#(A*)=2^(n-2).

So, because of Hp 1 and Hp 2 which cover every possibility, we can conclude that k = #(A) \leq 2^(n-2).  :D

[quote="Catraga"](I'm sorry if my English isn't so good, but it isn't my mother language...  :( )
[/quote] don't worry about it ... most of the people here don't have english as mother language ;)

[quote="iura"]I'm sorry but your solution is wrong.
Hypothesis 1 is wrong since the number of subsets of most n-3 elements is
2^n-(n 1)-(n 2)>2^(n-2) for large n. I saw the solution in 102 Combinatorial Problems.. It was the last problem( so almost the hardest).
the solution is not simple and very hard to find in contest time[/quote]
You can try to find your own simple solution ;)

[quote="Myth"][quote="iura"]I saw the solution in 102 Combinatorial Problems.. It was the last problem( so almost the hardest).
the solution is not simple and very hard to find in contest time[/quote]
You can try to find your own simple solution ;)[/quote]

I think it is originally a problem from the canadian bulletin of math.
I've seen only one kind of solution for this problem, and I agree it is a difficult problem.

Pierre.

[b]Solution. [/b] ([i]sketch[/i]) 1. Choose a set $A \subset S$ with a minimal possible number of elements such that $A \not\subseteq A_i \cup A_j$ for all $i,j$. By the minimality of $A$, there exist some $i_0,j_0$ such that the set $A \setminus A \cap \big( A_{i_0} \cup A_{j_0}  \big)$ contains a single element, and let $A \setminus A \cap \big( A_{i_0} \cup A_{j_0}  \big) = \{a\}$, i.e. $\big( A_{i_0} \cup A_{j_0} \big) \cap A = A \setminus \{a\}$.

2. Consider now the collections $\mathcal{S}_1 := \{A_1 \cap A, \ldots,A_k \cap A\}$ and $\mathcal{S}_2 := \{A_1 \cap (S \setminus A), \ldots,A_k \cap (S \setminus A) \}$. Show that $X \in \mathcal{S}_1$ implies $A\setminus X \notin \mathcal{S}_1$, and $Y \in \mathcal{S}_2$ implies $(S \setminus A) \setminus Y \notin \mathcal{S}_2$ (this is where we use that $A_{i_1} \cup A_{i_2} \cup A_{i_3} \cup A_{i_4}$ has at most $n-2$ elements, and in particular is not equal to neither $S$ nor $S \setminus \{a\}$).

3. Because $\mathcal{S}_1 \subset 2^A$ and $\mathcal{S}_2 \subset 2^{S \setminus A}$, conclude from 2. that $|\mathcal{S}_1| \leq 2^{|A| - 1}$ and $|\mathcal{S}_2| \leq 2^{|S \setminus A| - 1} = 2^{n - |A| - 1}$.

4. Finally, because every $A_j$, $1 \leq j \leq k$, can be recovered from $A_j \cap A \in \mathcal{S}_1$ and $A_j \cap (S \setminus A) \in \mathcal{S}_2$, it follows that
\[
k \leq |\mathcal{S}_1| \cdot |\mathcal{S}_2| \leq 2^{|A|-1} \cdot 2^{n-|A|-1} = 2^{n-2},
\]
as desired.

[b]Generalization. [/b] Let $n,\ell$ be fixed, and let $A_1,\ldots,A_k \subset [n]$ be such that, for every $i_1,\ldots,i_{2\ell}$, we have $|A_{i_1} \cup \cdots \cup A_{i_{2\ell}}| \leq n-\ell$. Show that $k \leq 2^{n-\ell}$.

[Follows easily by combining the above idea with an inductive argument on $\ell$.]

Pierre, do you know whether this generalization appears in the article in the CMB?

--Vesselin

[quote="vess"][b]Solution. [/b] ([i]sketch[/i]) 1. Choose a set $A \subset S$ with a minimal possible number of elements such that $A \subsetneq A_i \cup A_j$ for all $i,j$. By the minimality of $A$, there exist some $i_0,j_0$ such that the set $A \setminus A \cap \big( A_{i_0} \cup A_{j_0}  \big)$ contains a single element, and let $A \setminus A \cap \big( A_{i_0} \cup A_{j_0}  \big) = \{a\}$, i.e. $\big( A_{i_0} \cup A_{j_0} \big) \cap A = A \setminus \{a\}$.

i am not sure i understand-if we consider the obvious family of 2^(n-2) sets not containing n,n-1,then there is no set A which is contained properly in $A_i\cup A_j$ for all $i,j$.

or what did i misunderstand?

thanks!

[quote="Myth"]I thought, vess meant $A\not\subset A_i\cup A_j$.[/quote]

Oh, yes, of course, sorry for this; I've editted my most above.

[quote="vess"]

[b]Generalization. [/b] Let $n,\ell$ be fixed, and let $A_1,\ldots,A_k \subset [n]$ be such that, for every $i_1,\ldots,i_{2\ell}$, we have $|A_{i_1} \cup \cdots \cup A_{i_{2\ell}}| \leq n-\ell$. Show that $k \leq 2^{n-\ell}$.

[Follows easily by combining the above idea with an inductive argument on $\ell$.]

Pierre, do you know whether this generalization appears in the article in the CMB?
[/quote]

This is the proof that I known. No, they didn't mention this possible generalization. But it is said that the proposer (P. Frankl) thought that the same conclusion holds if the hypothesis involved only three subsets (instead of four) is an open problem.

Reference :
Canad. Math. Bull., vol.20 (2), 1977, p. 273-274.

Pierre.


If the perimeter of an isosceles triangle is 36cm and the altitude to its base is 12cm, what is the area, in square centimeters, of the triangle?

this was a countdown round, so do it fast

[hide]60 cm:^2: i did it in 10 sec. exactly[/hide]

im just curious, mystic

did u just try a 5-12-13 triangle, and it just happened to work, or did u actually ahve a method?

most likely 5-12-13, algebraically would prbly take more than 10 seconds.

in mc, [b]always[/b] try common right triangles. Hmmm...i think i estimated wrong, mebbe more like 5 sec.

notice that I say [i]try[/i]. Sometimes they will try to trick you up by making you think it's a common right triangle, but it isn't.

Yah. That's how I did it, and I think I saw this before.

Great first post! :) You even figured out how to use emoticons in your profile. (which took me like, half a month to figure out :) )

Btw, why are you so sad?

its funny cause he goes to my school. <refrains from insulting>. he has an awesome name though.

How do you all know whether or not one's in your school or not? I mean, practically everyone from Stuyvenescent...or somat :) recognizes each other and such.

i dont go to stuy. he pmed me and said who he be

Well then how come he knew you were in his school?

I think there's a paradox.

prbly chinaboy (who goes to my skoo also), told him to get on aops and told him that i was syntax error. or maybe he guessed from my signature (it has neal written in it)

Oh, I see. We need more people from the place where it's "great to be here!" :lol:
...a little local joke. Go to [url=http://wb56.trb.com/]WB56[/url] and find the commercial.

Addendum: I found the exact music.
[url=http://wb56.trb.com/extras/wlvi/gtbh1.mp3]Here's version 1 of the original[/url], [url=http://wb56.trb.com/extras/wlvi/gtbh2.mp3]version 2[/url], and [url=http://wb56.trb.com/extras/wlvi/gtbh3.mp3]the remix[/url].

Ever wonder if your city has a fan song? I mean, I think we're the only ones...our city sucks so much we need to brainwash people in order to make them like it :twisted:

Being rather tired, I solved it algebraically in about 35 seconds (But if I were awake I would have tried 9-12-15 and 5-12-13).

Hey! We did that problem in our countdown round in MC practice yesterday!!! I remembered this here and got it right (although I didn't remember how to do it, just the answer). :)  :D  :)

[quote="Syntax Error"]If the perimeter of an isosceles triangle is 36cm and the altitude to its base is 12cm, what is the area, in square centimeters, of the triangle?

[/quote]

thats just for reference.

mathfiend: lol, but ill go ahead and explain, cause whats the point of only knowing the answer.

so then, draw an isosceles triangle ABC. where A is on the vertex angle thingie, and B and C are on the base angles. draw the altitude to the base, and label that D. so AD is 12, AC=AB, and angle ADC=ADB=90 degrees. so then, just guess away at a 5-12-13 triangle. where CD and DB would be 5, and AC and AB would would 13. so adding the correct sides, you have a perimeter of 36. but it asks for the area, so the base is 10, and the altitude is 12, so 120/2=60cm:^2:

you could do that algebraically, but meh, its countdown, faster to make the pringles


The first four triangular numbers are pictured. The $ n$th triangular number is formed by drawing a row of $ n$ dots below the $ (n\minus{}1)$st triangular dot pattern. The $ k$th triangular number is represented by 120 dots. What is the value of $ k$?

[asy]pair A;
for(int i = 0; i < 4; ++i){
A = (i*(i+3)/2,0);
for(int x=0; x < i+1; ++x){
for(int y=0; y < x+1; ++y){
dot(A+x*dir(0)+y*dir(120));
}
}
}[/asy]

We use the formula $ \frac{n(n \plus{} 1)}{2}$.  In this case, $ \frac{n(n \plus{} 1)}{2} \equal{} 120$, so $ n(n \plus{} 1) \equal{} 240$.  Two such consecutive integers could be $ 15$ and $ 16$.  Thus, our answer is $ \boxed{15}$

[hide="answer"]we can use the equation n(n+1)/2 in this case the = will be 120 n(n+1)=240 solving for n we get the answer of 15 and 16 but the answer will need to be

15 or $\boxed{15}$[/hide]

[b][i][u]please rate my post thanks you ![/u][/i][/b]
please be fair though!! thanks [=

 :wow:   :wow:   :wow:   :wow:


There is a three volume set of books sitting on a shelf. The front and back covers of the books are each one eighth inch thick. The page section inside each book is exactly two inches thick. Now, if a cute little bookworm starts eating at page one of volume one and eats, in a straight line, through to the last page of volume three, how far will the cute little fellow travel?

[hide]There are 6 covers total. which is 3/4 of an inch. There are 3 page sections, which makes the whole thickness $3\frac{3}{4}$ inches. This is not the answer, however. The problem states that the bookworm starts at page one of volume one, not the front cover, so we subtract 1/8 of an inch from our previous answer. [b]This gives us $3\frac{5}{8}$ inches. THere's your answer.[/b][/hide]

[hide]
3 page sections and 4 covers are eaten so it goes:

$3(2)+4(\frac 18) = \boxed{6\frac 12 \text{ inches}}$

[/hide]


logarithms
limit
geometry
calculus
complex numbers
exterior angle
calculus computations

prove that:

sigma [ (-1)^(n+1) *sin(nx) / n ] = x/2 , where x is between -pi to pi (not including), and n=1,2,3,.....

thanks!

There are several ways to proceed here. One way would be to assume knowledge of Fourier inversion, and verify by calculation that $x/2$ on $(-\pi,\pi)$ has the appropriate Fourier coefficients.

I'll try a different method, using complex variables.

Consider the power series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}z^n}n.$

This has radius of convergence 1, and for $|z|<1,$ equals $\log(1+z),$ where this is a branch of the logarithm such that $\log 1=0.$

The series converges at every point on the circle $|z|=1$ other than $z=-1.$ (Proof postponed to another post.) Therefore, by Abel's Theorem,

$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}e^{in\theta}}n = \lim_{r\to1^-} \sum_{n=1}^{\infty}\frac{(-1)^{n+1}r^ne^{in\theta}}n$

$=\lim_{r\to1^-}\log(1+re^{i\theta}) =\log(1+e^{i\theta}).$

Now take the imaginary part of this:

$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin(n\theta)}n=\arg (1+e^{i\theta}).$

Now the argument becomes geometric. Draw the unit circle, and look at the triangle with vertices at $0,\,-1,$ and $e^{i\theta}.$ This is an isosceles triangle, with $\theta$ as an exterior angle at the origin. The interior angle at the $-1$ vertex is equal to $\arg(1+e^{i\theta})$ and, by the geometry of isosceles triangles, to $\frac{\theta}2.$

Thus, for $\theta\in(-\pi,\pi),\, \sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin(n\theta)}n= \frac{\theta}2.$

this answer is based on a lot of information yet not taught in the course of calculus I study...

we haven't studied fourier series also!

can somebody think of an answer, say, without using the previous methods?

Frankly, I don't see any way to approach this using methods from within a typical second semester calculus course. Even the convergence of the series: it's clearly not absolutely convergent, and does not meet the conditions of the alternating series test as usually taught in that course.

The proof of convergence must use "Abel's Criterion" or a similar summation by parts argument (which can be considered a generalization of the alternating series test.)

But actually computing the sum? That's going to take some serious machinery.

I have been taught abel's criterion ...

in other words, I look for a different solution where no fourier series and only real numbers are involved...

there must be one...

thanks...!


trigonometry
calculus
calculus computations

can you show :D

$ \sum_{k\equal{}1}^\infty\;\dfrac{1}{k\cdot \left(\dfrac{\pi}{2}\right)^{2k}\cdot\dbinom{2k}{k}}\;\equal{}\;\boxed{\dfrac{2\sin^{\minus{}1}\left(\dfrac{1}{\pi}\right)}{\sqrt{\; \pi^2 \minus{} 1\;}} }$

It is known that

$ \frac{2x\arcsin(x)}{\sqrt{1\minus{}x^{2}}}\equal{}\sum\limits_{n\equal{}1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}$,     $ |x|<1$,

If $ x\equal{}1/\pi$ the desired result follows.


function
set theory
advanced fields
advanced fields unsolved

well.. i'm not properly convinced that this is the proper section for set theory problems...
anyway, i've liked these ones...

[b]1.[/b] [i]without the use of axiom of choice[/i], find a function from $\mathcal{P}(\omega)$ [i]onto[/i] $\omega_{1}$. (*)
("onto" means it's surjective, right? that's what i actually want ;) )

[b]2.[/b] prove that $\prod_{n\in\omega}\aleph_{n}= \aleph_\omega^{\aleph_{0}}$.

[b]3.[/b] is it possible to find a continuous collection of subsets of $\mathbb{N}$ such that for any two of them, their intersection is finite?

(*) just let me recall, if needed, that $\mathcal{P}(A)$ denotes the set of all subsets of $A$, and $\omega_{1}$ is the least ordinal larger than $\omega$.

1. First, the set $\omega\times\omega$ is countable, therefore there is a bijection between $\cal P(\omega\times\omega)$ and $\cal P(\omega)$. Every element of $P(\omega\times\omega)$ defines a binary relation on $\omega$. Let $K$ be the subset of binary relations that define a well ordering on $\omega$ (not necessarily the standard one). Now map every element of $K$ to the (countable) ordinal it defines, map other elements to (say) $0$.

2. Hint: use Cantor-Bernstein theorem.

3. I do not understand what you mean by a continuous collection. But anyway I'll try to answer.

Consider subsets of $\mathbb Q$. Every real number $r$ defines a sequence of rational numbers that tend to $r$ (say, decimal approximations, if we do not want to use AC). Our collection consists of those sets - one set for every real number. Clearly, sets corresponding to distinct real numbers have a finite intersection. The cardinality of the family is continuum.

great :)

as a solution for $3$, a friend of mine thought of this: number the intervals you use to build up the cantor set $C$ (so that $I_{0}= [0;1], I_{1}=[0;\frac13], I_{2}=[\frac23,1]$ etc..), and to any point $x\in C$ assign the subset of $A_{x}\subset\mathbb{N}$ such that $n\in A_{x}\Leftrightarrow x\in I_{n}$...
of course, this is the same solution as yours...

anyway.. since problem two is still unsolved i'll give a little hint:
[hide]a countable union of countable sets is still countable...[/hide]
it sounds pretty obvious and unuseful, doesn't it? :ninja:


Google the title, you will find mountains of results...

今天是我的生日，女友早早的打来电话说晚上要到家里去为我祝贺生日，还要带给我惊喜！听了这个好消息！我今天工作起来是格外买力，一下跑了十几个客户！回到公司。都下午三点了，到食堂一看，只剩下可怜巴巴的一菜一汤了，肉炒三豆（肉炒黄豆、青豆、豌豆）和萝卜汤。没办法，跑了一上午客户，肚子早就咕咕的叫了，只好要了一大盘肉炒三豆和一大盆萝卜汤，吃了起来！没想到临下班了，我的肚子里就像一台越野吉普的发动机！--开始了剧烈的活塞运动！刹那间，一股股气体来势争先恐后的从我的体内冲了出来！我赶紧冲到没人的地方，肚子开始还是不好意思的轻声吟唱，但马上就变成了连珠炮似的噗噗做响！肚子好涨呀！而正在这时，女友却打来电话，说她已经到家了，叫我赶紧回家。唉！没办法只好回家去了，希望她不会看见我这幅狼狈样呀！……

　　　　在回家的路上我刻意努力的放了很多的屁。快到家了，肚子好受了很多，我觉得应该不会再出什么问题了。远远就看到了在门边等着我的女友，她看起来有点兴奋。她大叫着说，“亲爱的，今晚，我为你准备了一份非常奇妙的、一定会让你大吃一惊的礼物。”

　　　　还没进门，女友就用一块布把我的眼睛紧紧蒙了起来，说是要给我一个惊喜！还领着我坐到位于餐桌前头的椅子里，并且让我发誓不会偷看。突然，我感到又想放屁了。恰恰就在这时，女友的手机响了。这可救了我的命了！我找借口说嫌乱让她到另一间屋子里去接电话！她却非让我不能揭开蒙着眼睛的布，还让我发誓！之后才跑去另一间屋子里接电话。她一离开，我就抓紧时机，把全身的重量都移到一条腿上，把屁放了出来。这个屁放得不仅声音很大，而且气味就象是腐臭的鸡蛋散发出的臭味。我几乎不能呼吸，因此我摸到椅垫，使劲的向四周扇着，妄图扇掉这难闻的气味。

　　　　就在我刚感觉好一点的时候，另一个屁又来了。我又抬起腿开始放！它听起来就象是柴油发动机快速转动的声音，而且这一次气味更难闻了。为了不让自己窒息，我用胳膊挥舞着椅垫扇了起来，希望气味会尽快散掉。

　　　　又是在一切将要恢复正常的时候，另一个屁又迫不及待的冲来了。于是我站起来，弯下腰，把屁股向后上方撅了起来！把它放了出来。这个屁放得真正称得上是一流，连身后的报纸都被吹散到了地下..........

　　　　我侧耳倾听另一间屋子里女友交谈的声音，因为要遵守不偷看的诺言，我也不敢打开眼罩，只能在漆黑中不断放着屁，为了赶快把肚子中的气体全部排出，又不使屋里变得更臭！我解开了裤腰带，把内裤和长裤褪到了小腹以下，把屁股露了出来，并摸索着打开了身后阳台的门，几乎是将整个屁股都伸到阳台上，开始疯狂的放起屁来……，啊！好受多了！之后，我又手舞足蹈的用椅垫满屋乱扇，祈祷这股恶臭能赶快散去……，就这样，在接下来的十几分钟以内，我一边不断的放着屁，一边不断的扇着椅垫，终于，当我听到她在电话里说再见的时候，屋里的空气和我的肚子都已经好多了！我迅速的系上裤子，整理了一下头发，开始优雅地、微笑着等着我亲爱的她来带给我惊喜。

　　　　当她走近的时候，我脸上带着满足地微笑，一付温存的样子。女友首先为她打了这么长时间的电话向我道了歉，然后问我有没有偷偷掀开过布。在我向她保证没有偷看之后，女友移走了遮在我眼睛上的布，并对我说道，“意外吧！我的女友今天非让我带她们来看看你，她们说你在照片上很有风度，人长的很帅！喏！你看，坐在桌前的这五位都是我单位里的好姐妹，而站在阳台上的那六位是我上学时最要好的朋友！”

　　　　这时，我才极为震惊和恐惧的发现，有一大堆女孩正围坐在我对面的餐桌边，而身后的阳台上则站着另一堆，她们都是来参加这个令我感到非常意外的生日宴会的。现在，她们每个人脸上都带着一种无法言表的表情看着我，就像发现了火星人………………

[quote="asjsd"]:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
笑死我啦[/quote]
真的不禁笑了起來－－純粹是由於這些” :rotfl: ”


[quote="linker"]Prove that if $ a,b,c$ are positive numbers and $ a,b,c$ [b]could not be the lengths of the sizes of a triangle[/b], prove that:
\[ (a \plus{} b \plus{} c) \left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \geq 10.
\]
[/quote]
let a=max(a,b,c) so $ a\ge b\plus{}c$ let $ f(a)\equal{}(a \plus{} b \plus{} c) \left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right)$ so
$ f'(a)\equal{}\frac {1}{b} \plus{} \frac {1}{c}\minus{}\frac{1}{a^2}(b\plus{}c)\ge 0$ because $ a\ge b\plus{}c$ so $ f(a)\ge f(b\plus{}c)\equal{}2\plus{}2(b\plus{}c)(\frac {1}{b} \plus{} \frac {1}{c})\ge 2\plus{}8\equal{}10$ Is it true :maybe:


fie $f\in\mathbb{Z}[X]$. sa se arate ca exista $g\in\mathbb{Z}[X]$ de grad oricat de mare astfel incat polinomul $f(g(X))$ sa fie reductibil in $\mathbb{Z}[X]$.

Fie $\omega$ o radacina a lui $f$ si $m$ polinomul minimal al lui $\omega$. Luam un $h \in \mathbb Z[X]$, $\deg h \geq 1$, si consideram polinomul $g(X) = X+m(X) \cdot h(X)$. Avem $f \left( g \left( \omega \right) \right) = f \left( \omega \right) = 0$, deci $m \left| f \circ g \right.$, i.e. $f \circ g$ e reductibil.


geometry
3D geometry
conics
parabola
hyperbola
ellipse

A right circular cone is sliced by a plane perpendicular to its base. If the plane does not contain the vertex and the intersection of the cone, and the plane consists of more than one point, then the intersection is a portion of a:

(A). parabola     (B).ellipse     (C).hyperbola     (D).circle

C.  Hyperbola

Assuming that the conic section is not degenerate, a plane parallel to the base will form a circle, a plane parallel to the edge of the cone will form a parabola, a plane between the two will be an ellipse, and a plane at a steeper angle than the edge of the cone (including perpendicular to the base) will be a hyperbola.

[quote]Assuming that the conic section is not degenerate[/quote]Let me take this opportunity to ask, what is meant by a 'degenerate' conic section?


function
number theory proposed
number theory

Find all f:N $\longrightarrow$ N that:
[list][b]a)[/b] $f(m)=1 \Longleftrightarrow m=1 $
[b]b)[/b] $d=gcd(m,n) f(m\cdot n)= \frac{f(m)\cdot f(n)}{f(d)} $
[b]c)[/b] $ f^{2000}(m)=f(m) $[/list]

By $f^{2000}(m)$ do you mean $f(f(\cdots(m)\cdots))$ ?

It is so strange. It seems I know infinitely many such functions and they are not related to each other, i.e. there are no any regularity.

Actually, I don't see any solution at all.
Since you proposed this problem you know solution. Am I right?

Ok. I knew this answer, but I thought there are another solutions.

Omid Hatami! Did you mean such solution?

Well ,my solution is similiar,but I proved some different lemmas

$c)  f^{2000}(m)=m$
At book 2000-2001 This queston's (c) part is like this.

I only obtained that $ f(n) $ is the product of $ f(p) $ where $ p $ is a prime which divides $ n $.
how can I go on?


linear algebra
matrix
combinatorics unsolved
combinatorics

A set of poker has 52 cards with 13 cards in each suit (diamonds, clubs, hearts, spades). Each suit of cards are marked $2, 3, \cdots , 10, J, Q, K, A$. Calculate the number of subsets with 13 cards of the poker, where $2, 3, \cdots , 10, J, Q, K, A$ all appear and no two consecutive cards ($A$ and $2$, $2$ and $3$, …, $K$ and $A$) of the same suite both appear.

The most important part of this problem is to count the number of cases when the suit of $3$ is the same as $A$. This can be done by using a recurrence formula.

$4\times 3^{12}$?
Assuming that the ordering isn't important, there are four possible suits for the card $2$, and three possible suits for each of the following cards $3,4,\dots,A$ (so that we don't have two consecutive cards of the same suit).
Probably I'm missing something, because my answer is too evident. Where's the flaw? Thank you.

The catch is that the restrictions loop around.  The Ace not only needs to me differently suited than the King, but also from the Two, so the number of choices for the last card is either three or two depending on whether the King and Two are the same color.

Oh, I see. Sorry, I didn't see the loop. :blush:

I've been trying with graph theory. :idea:
If there are $4$ suits, the path matrix if you can't repeat the same suit is
$P=\left( \begin{array}{cccc}0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right)$
We want to compute the number $N$ of paths that go back to the same suit after $13$ steps, so our solution is $N=tr(P^{13})$.
[hide="A fast way to calculate that trace."]It's known that
$tr(P^{13})=\sum{{\lambda_{i}}^{13}}$,
where the $\lambda_{i}$ are the eigenvalues of $P$: $3,-1,-1,-1$.
Therefore our solution must be
$N=3^{13}+(-1)^{13}+(-1)^{13}+(-1)^{13}$.[/hide]$N=3^{13}-3$.

More generally, if we had a deck of cards with $m$ suits of $n$ cards each, the solution to our problem would be
$N=(m-1)^{n}+(-1)^{n}(m-1)$.

If you want to avoid advanced algebra, here's an entirely equivalent way to solve the problem.

Let's define:
$S_{n}$: the number of ways to go from one suit $i$ to the same suit $i$ after $n$ steps, without repeating suit at any two consecutive steps.
$O_{n}$: the number of ways to go from one suit $i$ to another suit $j\neq i$ after $n$ steps, without repeating suit at any two consecutive steps.

If we want the number $N$ of ways to go back to the same suit (for $i=1,2,3,4$) after $13$ steps, we want to compute $N=4\times S_{13}$.

Looking at all the possible outcomes after each step, there's the following recursion:
$S_{n+1}=3O_{n}$,
$O_{n+1}=S_{n}+2O_{n}$.
The initial values are
$S_{1}=0$,
$O_{0}=1$.

The way to solve the recursion is standard, and finally we have
$S_{n}=\frac{3^{n}+3(-1)^{n}}{4}$,
$O_{n}=\frac{3^{n}-(-1)^{n}}{4}$.

The solution to our problem is
$N=4\times S_{13}=3^{13}-3$.

[I edited an evident typo]

[quote="modeler"]Can anyone solve it without recurrence or matrix?[/quote]
Whenever I see a problem where the solution is a linear combination of powers, like in this case
$N=(m-1)^{n}+(-1)^{n}(m-1)$,
usually the easiest solution involves recursion/induction.


The game move right-move left works is a game involving digits and primes

1. Start with any prime number with 2 digits or more
2. If you choose to copy any digit or block of digits to the left substract (to the right add)
3. At each move you have to obtain a prime number.

Example

17 is prime

I copy the digit 7 to the left by substracting the digit 1 (7-1=6). 7 become 6 which I copy by concatenation at left

The new number will be 617

617 is prime

I move 6 to the right. I add 6 to 17 (17+6=23). 6 become 23 which I copy by concatenation at right

The new number will be 61723 which is prime

I move the digit 7 to the left. I substract 7 to 61 (61-7=54). 7 become 54 which I copy by concatenation at left.

The new number will be 5461723 which is prime..... and so on.

You can copy one digit or more and each you jump one or two or more digits you substract or add the block jumped.

or example

If you have as starting number 183317

You have a lot of possibilities the goal is to have new prime number.

If you copy the digit 8 at left you jump the digit 1 so the new number will be 7183317
If you copy the same digit at right you have 4 possibilities :
1- 18[b]3[/b][i]11[/i]317
2- 18[b]33[/b][i]41[/i]17
3- 18[b]331[/b][i]339[/i]17
4- 18[b]3317[/b][i]33125[/i]

Wich one is prime? You have to combine your digits at each step to find a new prime number.

My conjecture is that for at least some prime numbers you can do it indefinitly.
So my question is what is the shortest prime you can obtain in 5 steps?
So you will obtain 5 new prime numbers starting from the first one (2 digits) so the last one wil theorically have 7 digits.

8 digits!!! starting with 103

103
1033
13033
134033
1340033
13400033

All are prime!

103
move digit 0 to right
1033
move digit 3 (the third digit) to left
13033
move digit 1 to right
134033
move digit 3 (the last digit) to left
1340033
move the digit 0 (the 4th) to right
13400033


ratio
geometry
3D geometry
octahedron
tetrahedron
pyramid
Pythagorean Theorem

1.) Let a number $S_{n}$ have $n$ digits.

Find the value of: $\sqrt{111 \ldots 1_{2n} -  222 \ldots 2_{n}}$.

2.) Find the value of: $\sqrt{1988 * 1989 * 1990 * 1991 + 1} - 1989^2$.

3.) Simplify: $ \frac{\sqrt{6} + 4\sqrt{3} + 3\sqrt{2}}{(\sqrt{6} + \sqrt{3})(\sqrt{3} + \sqrt{2})}$.

4.) Which is greater: $\sqrt{4 + \sqrt{15}} + \sqrt{4 - \sqrt{15}}$ or $2\sqrt{2}$?

5.) Find the ratio of the volume of a regular octahedron with side length 1 to its surface area. Express as reduced common fraction.

1) [hide]Rewrite this as $\sqrt{111...1_{2n}-111...1_{n}-111...1_{n}}=\sqrt{111...1_n000...0_n-111...1_{n}}$. Now, our task is finding what the difference inside the square root is.

We see that starting from the right, the last digit of the difference would be 9. The second last digit 8, and the rest of them 8 up until the $n+1$ digit, which would be 0 (since the 1 is borrowed).

In the end, this is finally simplified to $\sqrt{111...1_{n-1}0888...8_{n-1}9}$. As customary of these problems, we look for a pattern. The last digit of the answer must be a 3 or a 7. We try 3 because $3^2=9$ and we see it can easily be extended over to around the 1s.

Since the digits are repeating under the square root, we try something like 33:

$33^2=1089$ (we're on to something)
$333^2=110889$

By now, we see that if $k$ is the amount of 3s there are, the final answer is $111...1_{k-1}0888...8_{k-1}9$. Therefore, $k=n$, so our answer is:

$333...3_n$
[/hide]

4) [hide]Conveniently, we can square both sides, and obtain:

$\left( \sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}} \right)^2=10$ and $(2\sqrt{2})^2=8$.

It's apparent that $\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}$ is larger.[/hide]

5) [hide]The surface area is the area of the eight equilateral triangles. Since the side length is 1, the area of each of the triangles is $\frac{\sqrt{3}}{4}$. All of 8 of them gives a final area of $2\sqrt{3}$.

To find the volume, we use Pythagorean Theorem to find that the height of the top tetrahedron of the octahedron is $\frac{\sqrt{6}}{3}$ so the volume of the octahedron is $\frac{2\sqrt{6}}{9}$. The ratio of volume to surface area is:

$\frac{\frac{2\sqrt{6}}{9}}{2\sqrt{3}}=\frac{\sqrt{2}}{9}$.

Hope that's the answer :D[/hide]

5
[hide]
The surface area is the easiest to find - I found mine the same way as Zyloch.  The formula for the area of an equilateral triangle is $\frac{s^2 \sqrt{3}}{4}$.  We need the area of eight, so the surface area becomes
$2s^2 \sqrt{3} = 2\sqrt{3}$

The volume can be defined as twice the volume of the two square pyramids.  The volume of a pyramid is $\frac{1}{3}Bh$.  We already know $B$ because the sides are all 1, $B = 1$.  We can find the height by a right triangle with legs $h$ and $\frac{1}{2}$ and the hypotenuse being the slant height of the pyramid, aka the height of one of the triangles.  The formula for the height of an equilateral triangle is $\frac{s\sqrt{3}}{2}$ which makes our hypotenuse $\frac{\sqrt{3}}{2}$.

Using the pythagorean theorem, $h = \sqrt{(\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2}$
This simplifies to $\frac{\sqrt{2}}{2}$.
Plugging this into the volume formula, the volume of the pyramid is $\frac{\sqrt{2}}{6}$ and thus the volume of the octahedron is $\frac{\sqrt{2}}{3}$

We need to find the ratio of the volume to surface area which is $\frac{\frac{\sqrt{2}}{3}}{2\sqrt{3}}$ which simplifies to $\frac{\sqrt{6}}{18}$.
[/hide]

[quote="Zyloch"]5) [hide]The surface area is the area of the eight equilateral triangles. Since the side length is 1, the area of each of the triangles is $\frac{\sqrt{3}}{4}$. All of 8 of them gives a final area of $2\sqrt{3}$.

To find the volume, we use Pythagorean Theorem to find that the height of the top tetrahedron of the octahedron is $\frac{\sqrt{6}}{3}$ so the volume of the octahedron is $\frac{2\sqrt{6}}{9}$. The ratio of volume to surface area is:

$\frac{\frac{2\sqrt{6}}{9}}{2\sqrt{3}}=\frac{\sqrt{2}}{9}$.

Hope that's the answer :D[/hide][/quote]

The height of the top tetrahedron is $\frac{\sqrt{6}}{3}$? That's the only mistake that I found.

[quote="white_horse_king88"][quote="deej21"]For 3
[hide]
Multiply by $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ and $\frac{\sqrt{6} - \sqrt{3}}{\sqrt{6} - \sqrt{3}}$
Simplify to $\sqrt{6} - 2\sqrt{2}$
[/hide]

I got something different for the last one, but I don't have time now.[/quote]

Hmm... check your arithmetic work. There's some mistake because the answer that I got is: [hide="Answer"]$\sqrt{6} - \sqrt{2}$[/hide][/quote]

Yeah, I don't doubt it... I did it really early this morning, and brute forced it.  I imagine I simplified some radical wrong somewhere in that mess.  I don't know if there's an easier way than to multiply everything out, but I did, and it was rough.  I guess at the least I could have checked my answer in the calculator before posting it...


geometry
trapezoid
circumcircle
geometry proposed

Let $ (O_1),(O_2)$ be $ 2$ circles intersect each other at $ P,Q$. Let $ AB$ be the external tangent of $ (O_1),(O_2)$, $ AB$ is nearer $ P$ than $ Q$ and $ A\in (O_2)$, $ B\in (O_1)$. The line through $ Q$ which is parrallel to $ AB$ intersects $ (O_1),(O_2)$, respectively, at $ C,D$. Let $ N$ be the second intersection of $ (ACQ)$ and $ (BDQ)$. Prove that: $ N,P,Q$ are collinear.

Let be the points $ A'\equiv (O'_{2})\cap AB$ and $ B'\equiv (O'_{1})\cap AB,$ where $ (O'_{1}),\ (O'_{2})$ are the circumcirles of the triangles $ \bigtriangleup AQC,\ \bigtriangleup BQD,$ respectively.

Because of the isosceles trapeziums $ AQCB',\ BQDA',$ we conclude that their circumcenters $ O'_{1},\ O'_{2},$ lie on the line segments $ O_{1}B,\ O_{2}A,$ as the midperpendiculars of the cords $ QC,\ QD,$ respectively.

Because of now, $ O'_{2}A\perp A'B$ and $ O'_{1}B\perp AB',$ we have that $ AA' \equal{} AB \equal{} BB'$ $ ,(1)$

Let be the point $ M\equiv AB\cap PQ$ and then we have $ MA \equal{} MB$ $ ,(2)$ as well $ ($ from $ (MA)^{2} \equal{} (MP)\cdot (MQ) \equal{} (MB)^{2}$ $ ).$

We denote the points $ N\equiv (O'_{1})\cap PQ$ and $ N'\equiv (O'_{2})\cap PQ.$

From the powers of $ M,$ with respect to the circles $ (O'_{1}),\ (O'_{2}),$

we have respectively that $ (MN)\cdot (MQ) \equal{} (MA)\cdot MB'$ $ ,(3)$ and $ (MN')\cdot (MQ) \equal{} (MB)\cdot (MA')$ $ ,(4)$

From $ (1),\ (2),\ (3),\ (4)$ $ \Longrightarrow$ $ (MN)\cdot MQ) \equal{} (MN')\cdot (MQ)$ $ \Longrightarrow$ $ MN' \equal{} MN$ $ \Longrightarrow$ $ N'\equiv N$ and the proof is completed.

Kostas Vittas.

[quote="vittasko"]Let be the points $ A'\equiv (O_{2})\cap AB$ and $ B'\equiv (O_{1})\cap AB.$ [/quote]
AB is the external tangent of $ (O_1) , (O_2)$ , so where is A' and B' ?

Last part of Vittako's proof can be seen with the "eye" of [b]radical axes[/b]. Indeed:
Because $ M$ is the midpoint of $ AB$ and $ BB' \equal{} AB \equal{} AA'$ hence $ MA' \equal{} MB'$. Therefore $ \overline {MA}.\overline {MB'} \equal{} \overline {MB}.\overline {MA'}$
$ \Longrightarrow \mathcal {P}_{M/(ACQ)} \equal{} \mathcal {P}_{M/(BDQ)}$, which implies that $ M$ belongs to the radical axes of $ (ACQ),(BDQ)\Longrightarrow M\in QN$. The result is lead as follow.

Inversion with center $ Q$ and arbitrary power takes $ (O_1)$ and $ (O_2)$ into the lines $ \ell_1 , \ell_2$ meeting at the inverse $ P'$  of $P$ and the common tangent $ \overline{AB}$ into a circle $\omega$ passing through $ Q$ and tangent to $ \ell_1 , \ell_2$ through $ B', A'.$ Circles $\odot (ACQ)$ and $\odot (BDQ)$ are taken into the lines $ A'C'$ and $ B'D',$ but $ \omega$ is the incircle of the triangle $ \triangle P'C'D'$ bounded by $ \ell_1,\ell_2,CD$ due to $ AB \parallel CD.$ Since $ C'A', D'B'$ and $ P'Q$ concur at the Gergonne point of $ \triangle P'C'D',$ then $ \odot(ACQ),$ $\odot (BDQ)$  and $ PQ$ meet again on $ PQ$ $ \Longrightarrow$ $N,P,Q$ are collinear.

[quote="long14893"]...AB is the external tangent of $ (O_1) , (O_2)$ , so where is A' and B' ?[/quote]
Thank you dear [b][size=100]long14893[/size][/b], for pointing out of my mistakes. I have corrected them (I hope).

Best regards, Kostas Vittas.


[quote="lccs"]Is this conclusion correct?
$a,b,c\in R^{+},Prove:$
$(2ab+2bc+2ca-a^{2}-b^{2}-c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\ge9$[/quote]

Generally no, since the left hand side can be negative. However, if a, b, c are the sidelengths of a triangle, then we can assert that

$\left(2bc+2ca+2ab-a^2-b^2-c^2\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\geq 9$.

In fact, since a, b, c are the sidelengths of a triangle, we can write a = y + z, b = z + x and c = x + y for positive reals x, y, z (if you don't know how to prove this, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=208379#p208379]http://www.mathlinks.ro/Forum/viewtopic.php?t=22198 post #3[/url]). Then, a calculation yields $2bc+2ca+2ab-a^2-b^2-c^2=4\left(yz+zx+xy\right)$, so that the inequality in question becomes

$4\left(yz+zx+xy\right)\left(\frac{1}{\left(y+z\right)^2}+\frac{1}{\left(z+x\right)^2}+\frac{1}{\left(x+y\right)^2}\right)\geq 9$,

or, equivalently,

$\frac{1}{\left(y+z\right)^2}+\frac{1}{\left(z+x\right)^2}+\frac{1}{\left(x+y\right)^2}\geq\frac{9}{4\left(yz+zx+xy\right)}$,

but this is the famous [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3547]Iran 1996 inequality[/url].

  Darij


calculus
integration
logarithms
calculus computations

Evaluate $ \int_1^e \{(1 \plus{} x)e^x \plus{} (1 \minus{} x)e^{ \minus{} x}\}\ln x\ dx$.

hello
[hide="solution"]
$ \frac {d(e^{x}(xlnx \minus{} 1)}{dx} \equal{} (x \plus{} 1)e^{x}lnx$
$ \frac {d(e^{ \minus{} x}(xlnx \plus{} 1)}{dx} \equal{} (1 \minus{} x)e^{ \minus{} x}lnx$
$ \int_1^e \{(1 \plus{} x)e^x \plus{} (1 \minus{} x)e^{ \minus{} x}\}\ln x\ dx \equal{} (e^{e \plus{} 1} \plus{} e^{1 \minus{} e}) \plus{} e^{ \minus{} e} \minus{} e^{e} \minus{} e^{ \minus{} 1} \plus{} e$
[/hide]
sorry i m writing the answer in a weird form.
thank u


[quote="Interval"]How many different types of geometry levels are there in the world of mathematics?

We know about Non-Euclidean Geometry and Euclidean Geometry, right?

Any others?  What are they?[/quote]
They're not really levels of geometry. They're different types of geometry. There is also spherical geometry. Those three are the only ones that I know of. (They may be the only kinds)

If you really want to know the answer: there are an infinite number of different types of geometries, each following different types of structures or transformations or topologies or whatever other kinds of meta-structures that mathematicians study.

Most popular ones are Riemannian geometry, projective, affine, hyperbolic, spherical, etc etc etc, the names describing certain types of properties pertaining to their respective geometrical worlds.

I understand very little geometry.

Perhaps, I'll take courses on other types to learn more.

Look at this :[url]http://en.wikipedia.org/wiki/Geometry[/url]
It's a wonderful world !! :wink:

[quote="Interval"]How many different types of geometry levels are there in the world of mathematics?

We know about Non-Euclidean Geometry and Euclidean Geometry, right?

Any others?  What are they?[/quote]
Correction: lots of non euclidean geometries, because of Euclid's parallel postulate which i am fairly certain is something worth wiki-ing if you really wanted to know

Euclid's parallel postulate is the defining characteristic of Euclidean geometry, which our world (mostly) obeys - hyperbolic/spherical geometries come into effect when spacetime is significantly warped (i.e. black hole, etc.)

it says that through any point not on a line there is exactly one line parallel to the first line.

but yeah, geometry is a really broad term, so basically any way you can figure out to measure properties of some type of space that's a geometry.

that's why you get weird stuff like differential geometry (aka Riemannian geometry?), etc etc etc.


I feel really stupid asking this, but I do not get how to solve equations with fractional exponents.

for example, $4^{x}=8$

How would you solve that?  :(  :(

when the bases have common factors, espicially like this one, just try some numbers for x and take roots when you've done that(or vice-versa). in this case, it's easy to first take the square root, then get 2. Then, clearly, raise to the third power. So, the answer is 3/2=x.

Also remember that $a^{p/q}=\sqrt[q]{a^{p}}$.

sometimes they don't have the same base
i suggest using that last method for exponents with the same base(it is disguised sometimes)

[hide="method i say dont do for this problem but if you forget everything but logarithms and your calc"]
$4^{x}=8$ (i have no clue with latex from here):

log4^x=log8
x(log4)=log8
with logarithm  table or calculator
x=(log8)/(log4)
x=1.5
[/hide]

[hide="suggested method"]
$4^{x}=8$
$2^{2x}=2^{3}$
we don't NEED logarithms this way if we take it for granted
$2x=3$
$x=1.5$
[/hide]

Eh funcia, instead of figuring out $\frac{\log 8}{\log 4}$ you can use $\log_{2}$ and have $\frac{\log_{2}8}{\log_{2}4}$, which is considerably easier to compute. (Basically, it's what you did in the second method.)

what can i say if i set the logarithms with same base i can do it with my calc but yeah second method is the only method you use in mathcounts...i think


[color=purple]Let ABC is a triangle.Prove that
    $\frac{ab+bc+ca}{4\sqrt{3}S}\ge \frac{1}{9}(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})$
  Have fun ! :D [/color]

[color=purple] No one can kill this problem ? [/color]

[quote="evarist"]Let ABC is a triangle.Prove that
    $\frac{ab+bc+ca}{4\sqrt{3}S}\ge \frac{1}{9}(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})$
  Have fun ! :D [/quote]
Are you sure ? :D try with triangle $(4,5,6)$ then
$\frac{ab+bc+ca}{4\sqrt{3}S}=1.08$ and $\frac{1}{9}(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})=1.11$
And moreover reverse is false too.

[color=purple] Sorry gemath i am wrong  :wink: [/color]

I don't know if anyone has attempted to prove this problem. I didn't. But clearly, one would be angry if he spent a lot of time trying to prove it.

So I would appreciate if in any Proposed and Owned Problems sections, one posts only the problems that are completely solved. If you propose it but haven't proved it, please either post it in the Unsolved Section or at least write something in the subject description or in your message so that everyone knows what to expect.

Thanks.

[color=purple]Thanks treegoner I'll do as you said.Here is the correct problem.[/color]
    $\frac{2(ab+bc+ca)}{ab+bc+ca+4\sqrt{3}S}\le \frac{1}{9}(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})$


inequalities
combinatorics unsolved
combinatorics

$ A_1,A_2,..,A_n$ are finite sets,such that $ |A_i\cap A_{i\plus{}1}|>\frac{n\minus{}2}{n\minus{}1}|A_{i\plus{}1}|$ ;for $ i\equal{}1,2..,n$.
Prove that:  $ \bigcap_{i\equal{}1}^{n}A_i\not\equal{} \oslash$

How do we interpret $ A_{n\plus{}1}$, ie: when $ i\plus{}1$ runs into $ n\plus{}1$?

Thanks,

[quote="math10"]$ A_1,A_2,..,A_n$ are finite sets,such that $ |A_i\cap A_{i \plus{} 1}| > \frac {n \minus{} 2}{n \minus{} 1}|A_{i \plus{} 1}|$ ;for $ i \equal{} 1,2..,n$.
Prove that:  $ \bigcap_{i \equal{} 1}^{n}A_i\not \equal{} \oslash$[/quote]

WLOG, we assume that $ A_n$ be the set with most elements, i.e. $ |A_n|\ge|A_i|$ holds for all $ i$. The condition $ |A_i\cap A_{i \plus{} 1}| > \frac {n \minus{} 2}{n \minus{} 1}|A_{i \plus{} 1}|$ implies that
$ |A_{k \plus{} 1} \minus{} A_{k}|\le \frac {1}{n \minus{} 1}|A_{k \plus{} 1}|$
And then we have the following inequalities with respect to $ k$:
\[ |\bigcap_{i \equal{} k}^{n}A_i| \equal{} |\bigcap_{i \equal{} k \plus{} 1}^{n}A_i \minus{} (\bigcap_{i \equal{} k}^{n}A_i \minus{} A_{k})|
\]

\[ |\bigcap_{i \equal{} k \plus{} 1}^{n}A_i| \minus{} |\bigcap_{i \equal{} k}^{n}A_i \minus{} A_{k}|\ge|\bigcap_{i \equal{} k \plus{} 1}^{n}A_i| \minus{} |A_{k \plus{} 1} \minus{} A_{k}|\ge |\bigcap_{i \equal{} k \plus{} 1}^{n}A_i| \minus{} \frac {1}{n \minus{} 1}|A_{k \plus{} 1}|\ge |\bigcap_{i \equal{} k \plus{} 1}^{n}A_i| \minus{} \frac {1}{n \minus{} 1}|A_{n}|
\]
Taking the summation of inequalities above for all $ k \equal{} 1, dots,n \minus{} 1$
we get that $ |\bigcap_{i \equal{} 1}^{n}A_i| > 0$


calculus
integration
trigonometry
calculus computations

Call the given integral $I$. Observe that $I = \int_{0}^{\pi}\frac{(\pi-x)\sin(2\pi-2x)\sin(\frac{\pi}{2}\cos(\pi-x))}{2(\pi-x)-\pi}dx=\int_{0}^{\pi}\frac{(\pi-x)\sin(2x)\sin(\frac{\pi}{2}\cos(x))}{\pi-2x}dx$. Therefore $2I = \int_{0}^{\pi}\sin(2x)\sin(\frac{\pi}{2}\cos(x))dx$. That is $I = \int_{0}^{\pi}\sin(x)\cos(x)\sin(\frac{\pi}{2}\cos(x))dx$. Set $u = \frac{\pi}{2}\cos(x)$. Thus $du =-\frac{\pi}{2}\sin(x)dx$ and subsituting in $I$ we get $I = \int_{-\pi/2}^{\pi/2}\frac{4}{\pi^{2}}u\sin(u)du = \frac{8}{\pi^{2}}\int_{0}^{\pi/2}u\sin(u)du = \frac{8}{\pi^{2}}$. Note $\int_{0}^{\pi/2}u\sin(u)du =-u\cos(u)|_{0}^{\pi/2}+\int_{0}^{\pi/2}\cos(u)du = 1$.

[quote="prithwijit"]$2I = \int_{0}^{\pi}\sin(2x)\sin(\frac{\pi}{2}\cos(x))dx$.[/quote]
why?I can't understand.

Sorry I jumped a step. Here it is:

$I = \int_{0}^{\pi}\frac{x\sin(2x)\sin(\frac{\pi}{2}\cos(x))}{2x-\pi}dx$ and

$I = \int_{0}^{\pi}\frac{(x-\pi)\sin(2x)\sin(\frac{\pi}{2}\cos(x))}{2x-\pi}dx$

Now add them to get the desired result.


geometry algebra polynomial search function

[quote="bleumoose"]i have a question about the problems that are posted mainly in the advanced section- has anyone noticed that the subject lines of these problems tell you absolutely [b]nothing[/b] about the problem itself? almost all of them are titled "nice" "very nice" "interesting" or "i like this problem"- could you please start putting something that is slightly descriptive on the subject line so i can find out if i want to look at it or not?
sorry if this is in the wrong part of the forum, by the way- im not sure where exactly it should go.[/quote] you can find many things about a problem, and not by looking at its title in the advanced section: for example depending on which sub-forum you have found it, it's either an unsolved algebra problem, or a solved geometry problem, etc.

I assure you that you cannot obtain the level of difficulty from one line in the subject field.

As if you are looking for say polynomial problems you can use the search function and narrow it down to say algebra - solved problems.

[quote="Valentin Vornicu"]you can find many things about a problem, and not by looking at its title in the advanced section: for example depending on which sub-forum you have found it, it's either an unsolved algebra problem, or a solved geometry problem, etc.[/quote]

I think the bleumoose's issue is when viewing problems in the AoPS "View posts since last visit".  The full title of the forum and subforum do not appear under the heading titled "Forum".  All we see is "Own Problems -- Nice".  I think bleumoose doesn't find this very helpful in deciding if he wants to click the link.

Yeah I agree. You can say that you find it interesting in the subject line without any difficulty, so a more descriptive title would be helpful.

[quote="tokenadult"]I further agree that subject lines should be more descriptive, in ALL forums, and I sometimes use my moderator powers to edit subject lines in Getting Started and in College, and I encourage other moderators to do the same.[/quote]

I have done this some time ago (you can see this from the []'s in the subject lines of some messages in the geometry forum), and I'll do this again when I have more time.

  darij

[quote="bleumoose"]exactly that, rcv![/quote]ups missunderstood you there :) I forgot about that problem. I will fix it as soon as possible.


http://img316.imageshack.us/my.php?image=hjul6db.png

The first picture is how it was before I started. The second picture is how it is now. I know that the two triangles are similar (I have the final answer of the task, but they don't explain how they got there), but I don't know why. Why is the small triangle ABC similar to the big DCA?
And why is DCA 90 degrees?

If $A$ is a point of tangency and $AB\perp AD$, then $AD$ has to be a diameter of the circle. Any triangle that has a diameter as one of its sides is a right triangle, and so $\angle DCA=90^\circ$.

Let $\angle BAC=\alpha$ and $\angle DAC=\beta$. We know that $\angle DAB=90^\circ$, so $\alpha+\beta=90^\circ$. Now look at $\triangle DCA$. Since $\angle DCA=90^\circ$, we have $\angle ADC+\angle CDA=90^\circ$. But $\angle DAC=\beta$, so $\angle ADC=90^\circ -\beta=\alpha$. Thus all three angles of $\triangle ABC$ and $\triangle DCA$ are the same, so they are similar.

This is my way of showing $\bigtriangleup ABC \sim \bigtriangleup DCA$

$\angle DCA = \angle ABC = 90^\circ$ (given)
$AC = AC$ (common side)
$\angle CAB = \angle ADC$ (AST)

$\bigtriangleup ABC \sim \bigtriangleup DCA$ by ASA.


For a,b,c \geq 0 such that a+b+c=1 prove that

2(a^2+b^2+c^2)+9xyz \geq 1

Classical (and xyz is abc): Schur. This can be immediately written 4(ab+bc+ca)<=9abc+1, which holds by Schur.

I know by Schur is simple but I have proved it without using Schur. If you are interested I can post my solution .

Of course I'm interested, since the beauty of any inequality consists especially in it's solutions. I'm waiting!

Suppose x, y, z are reals with x + y + z = 1.

(x - y + z)(x + y - z) \leq x
(y + x - z)(y + z - x) \leq y
(z + x - y)(z + y - x) \leq z

Taking square roots,

(x + y - z)(y + z - x)(z + y - x) \leq xyz

(This is well - known.)

Hence (1 - 2x)(1 - 2y)(1 - 2z) \leq xyz. Hence 4(xy + yz + zx) \leq 9xyz + 1.

I hope it is correct.

We can assume a \leq b \leq c .
As a+b+c=1 it must be  a \leq 1/3 and in particular 9a-4<0
Now we write

2a^2+2b^2+2c^2+9abc-1=2a^2 +2(b+c)^2 +(9a-4)bc-1 \geq 2a^2+

+2(b+c)^2+((9a-4)(b+c)^2)/4=2a^2+2(1-a)^2+((9a-4)(1-a)^2)/4-1=

=(a/4)(1-3a)^2 \geq 0

It is innatural and I have spent a lot of time to find it but it has the advantage not to use Schur.

It is a nice solution, but why did you say that "it has the advantage not to use Schur"?

I said that because when I solved it (about two years ago) I didn't know Schur.
I think that your solution by Shur is more elegant and neat than mine.


Determine all pairs (m,n) of positive integers m, n for which 2^m + 3^n is a perfect square.

^ - Raised to.

Can anybody tell me the whole solution?

[hide]Let $ 2^m\plus{}3^n\equal{}s^2$.  Taking it mod 3, we get $ 2^m\equiv 0\text{ or }1\pmod{3}$.  Thus, $ m\equal{}2r$.  If $ m\equal{}0$, we get the equation $ 3^n\equal{}s^2\minus{}1\equal{}(s\minus{}1)(s\plus{}1)\Rightarrow s\equal{}2$.  Otherwise, taking the equation mod 4 we get since $ m$ is even, $ 3^n\equiv 0\text{ or }1\pmod{4}$.  Thus, $ n\equal{}2t$.  So, we have that $ (2^r,3^t,s)$ is a pythagorean triple.  Thus, $ 2^r\equal{}2pq,3^t\equal{}P^2\minus{}q^2$ for some integer $ p,q$.  From $ 2^r\equal{}2pq$ we find that $ p\equal{}2^x,q\equal{}2^y$.  Thus, $ 3^t\equal{}p^2\minus{}q^2$ is odd when $ q\equal{}1$.  Thus, the equation is reduced to $ 3^t\equal{}4^x\minus{}1\equal{}(2^x\plus{}1)(2^x\minus{}1)$.  It is well known that the only solution to $ 3^k\equal{}2^x\minus{}1$ is $ k\equal{}2,x\equal{}3$ and for $ 3^k\equal{}2^x\plus{}1$ has solution $ x\equal{}1,k\equal{}1$ (for proof just go about it similar to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122734&sid=87f53a5926c048590815cf88b3229435).  Thus, we have we have solution $ (m,n)\equal{}(0,1)$.[/hide]


There is a bag weighing 11111 grams in the right pan of a balance. A person place weights of 1, 2, 4, 8,$\ldots$ grams in either of the pans; he first places the one weighing 1 gram, then the one weighing 2 grams, and so on. There is a moment when the pans equilibrate. Determine, with proof, the pan where the 16 grams weight is located.

I don't know if this problem is exactly belonging to combinatorics. I apologize if I've made a mistake placing the problem in this section. :)

We know that $2^{a_{1}}+2^{a_{2}}+...+2^{a_{k}}= 11111+2^{a_{k+1}}+...+2^{a_{n}}$ with $a_{i}\geq 0$. So it is sufficient to analize the equality modulo 4, 16, 32 and 64. From that 1, 2 and 16 are on the LHS, while 4 and 8 are on the RHS.

Does the problem imply that all weights must be used?  I find the problem's statement a little unclear.  You could write
\[\\ 11111_{10}+11111111111111_{2}=10101101100111_{2}+11111111111111_{2}\\ \implies 11111_{10}=10\underline{1}0\underline{1}00\underline{1}00\underline{1}\underline{1}00\underline{1}_{2}\]
Then you see that all the underlined weights go into the right pan, that is, the $\textbf{16}$ weight is placed into the right pan.

Suppose that  $2^{n}$ is the heaviest weight used in the equilibrium configuration. I think the problem requires that all the weights $2^{n-1},\ 2^{n-2},\ldots, 1$ must have been used too.

I'm glad to see this old topic has been answered after such a long time of being completely forgotten :D Thanks guys!


The number of real values of $x$ that satisfy the equation\[ (2^{6x+3})(4^{3x+6})=8^{4x+5} \]is:

$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{greater than 3}$

[hide]
(E)

Change the expression so the bases are all 2:

$(2^{6x+3})(2^{6x+12})=2^{12x+15}$

Then simplify by adding exponents.

$2^{12x+15}=2^{12x+15}$

Clearly all values for x work so the answer is E greater than 3.
[/hide]

[hide="solution"]Its obviously $\boxed{E}$ because for any value of $x$ $12x+15=12x+15$.

You basically have:

$(2^{6x+3})(2^{6x+12})=2^{12x+15}$ getting us
$12x+15=12x+15$.[/hide]

[hide]The solution is E since if we rearrange the question we get:
$2^{6x+3}\dot2^{6x+12} = 8^{4x+5}$
$2^{12x+15} = 8^{4x+5}$
$2^{12x+15} = 2^{12x+15}$
since 0 = 0, we know that there is an infinite number of solutions[/hide]

[quote="236factorial"][quote="Xantos C. Guin"][hide]
($(2^{6x+3})(2^{6x+6})=2^{12x+15}$

Then simplify by adding exponents.

$2^{12x+15}=2^{12x+15}$

[/hide][/quote]

6x+3+6x+6=12x+15?! (Did you mean 6x+12 in the first line of my quote?)[/quote]

Yes, thanks for pointing that out.


Two circles of radius $ 5$ and $ 8$ touch each other extenally. Calculate the length of the common tangent.

Could someone please show me their solution with work.

It quickly follows from the diagram that the length of the tangent is equal in length to the unknown leg of the $ 3\minus{}x\minus{}13$ right triangle, which has length

$ \sqrt{13^2\minus{}3^2}\equal{}\boxed{4\sqrt{10}}$.

[quote="ocha"][hide]
 [geogebra]72cfb5114bf231244935096a884cbd0dcf36a233[/geogebra]

See the traingle
The length between the two centers is $ 5 \plus{} 8 \equal{} 13$
The height of the triangle is $ 8 \minus{} 5 \equal{} 3$
So now its just pythag
$ 13^2 \equal{} 3^2 \plus{} x^2 \longrightarrow x \equal{} \sqrt {160} \equal{} 4\sqrt {10}$[/hide]

EDIT: Beaten  :([/quote]

Could you explain why you drew that diagram? Like, what made you draw it like that.

This is a pretty standard technique (i.e. drawing the extra line). The motivation is the fact that the tangent is perpendicular to the two parallel radii of the circles it is tangent to, and we think to apply the Pythagorean Theorem.


I would like to know how to write a document using $ \text{\LaTeX}$ in a language like Chinese or Japanese. I even can't typeset text in those languages in TeXnic Center... :maybe: What can I do??

Thanks for your help!! :wink:

LaTeX supports Asian languages but TexnicCenter doesn't as it has no unicode support. Perhaps you could try another editor like [url=http://www.xm1math.net/texmaker/index.html]Texmaker[/url] or [url=http://www.winshell.org/]WinShell[/url] which are free, runs on Windows and supports unicode.
See also [url=http://www.math.nus.edu.sg/aslaksen/cs/cjk.html]Chinese TeX Using the CJK LaTeX Package, Unicode TrueType Fonts and pdfTeX under Windows[/url]


algebra polynomial trigonometry quadratics

It seems that you can, without too much difficulty, reduce it to a 5th degree polynomial in $\sin^2 x$.  I don't know if that yields an obvious root, however.

[hide]Let cos(x)^2=n
You can use the trig substitution cos(x)^2=.5(1-cos(2x)), therefore cos(2x)=1-2cos(x)^2
sin(x)^2=1-cos(x)^2
So you have: (1-n)^5+n^5=(29/16)(1-2n)^2
You can put it into a calculator or something to get that
x=67.5 degrees or 3/8 pi[/hide]


limit
function
induction
calculus
calculus computations

Let $ (x_n)_{n \geq 0}$ be a sequence such that $ x_{n\plus{}2}\equal{}(a\plus{}b)x_{n\plus{}1}\minus{}ab \cdot x_n$ and $ x_0\equal{}1$, $ x_1\equal{}a\plus{}b$, $ 0<b<a$.

Find $ l\equal{} \lim_{n \to \infty} \frac{x_{n\plus{}1}}{x_n}\equal{}?$.

$ x_{n\plus{}2}\equal{}(a\plus{}b)x_{n\plus{}1}\minus{}ab \cdot x_n$

$ \frac{x_{n\plus{}2}}{x_n}\equal{}(a\plus{}b)\frac{x_{n\plus{}1}}{x_n}\minus{}ab$

$ \frac{x_{n\plus{}2}x_{n\plus{}1}}{x_{n\plus{}1}x_n}\plus{}ab\equal{}\frac{x_{n\plus{}1}}{x_n}(a\plus{}b)$

$ l^2\minus{}l(a\plus{}b)\plus{}ab\equal{}0\to l \equal{} \frac{(a\plus{}b) \pm (a\minus{}b)}{2}\equal{}a,b$

but thats assuming we know the limit exists

The reccurence can be written in two ways.

$ x_{n \plus{} 2} \minus{} bx_{n \plus{} 1} \equal{} a(x_{n \plus{} 1} \minus{} bx_n)$

$ x_{n \plus{} 2} \minus{} ax_{n \plus{} 1} \equal{} b(x_{n \plus{} 1} \minus{} ax_n)$

$ \therefore x_{n \plus{} 1} \minus{} bx_n \equal{} a^n(x_1 \minus{} bx_0) \equal{} a^{n \plus{} 1}$

$ \ \ \ \ \ \ \ x_{n \plus{} 1} \minus{} ax_n \equal{} b^n(x_1 \minus{} ax_0) \equal{} b^{n \plus{} 1}$

Subtracting both sides, we have $ (a \minus{} b)x_n \equal{} a^{n \plus{} 1} \minus{} b^{n \plus{} 1}$, yielding $ x_n \equal{} \frac {a^{n \plus{} 1} \minus{} b^{n \plus{} 1}}{a \minus{} b}\ \because 0 < b < a$.

$ \therefore \frac {x_{n \plus{} 1}}{x_n} \equal{} \frac {a^{n \plus{} 2} \minus{} b^{n \plus{} 2}}{a^{n \plus{} 1} \minus{} b^{n \plus{} 1}} \equal{} \frac {a \minus{} b\left(\frac {b}{a}\right)^{n \plus{} 1}}{1\minus{}\left(\frac {b}{a}\right)^{n \plus{} 1}}$.

Since $ 0 < b < a\Longrightarrow \lim_{n\to\infty} \frac {b}{a} \equal{} 0$, we have $ \lim_{n\to\infty} \frac {x_{n \plus{} 1}}{x_n} \equal{} a$.

Let $ y_n \equal{} \frac {x_{n \plus{} 1}}{x_n}.$ We can write the original two-step recursion for $ x_n$ as a one-step recursion for $ y_n:$

$ y_{n \plus{} 1} \equal{} a \plus{} b \minus{} \frac {ab}{y_n}.$

Now we can talk about the behavior of the sequence by talking about the function $ f(y) \equal{} a \plus{} b \minus{} \frac {ab}{y}.$

The only possible limits for $ y_n$ are the fixed points of $ f(y),$ meaning the solutions of $ f(y) \equal{} y.$ As kenn2000 indicated, there are two such points: $ a$ and $ b.$

Draw the graph of $ f$ on $ (0,\infty).$ We have $ f(y) < y$ for $ 0 < y < b,$ $ f(y) > y$ for $ b < y < a,$ and $ f(y) < y$ for $ y > a.$ This makes the fixed point at $ a$ stable and the fixed point at $ b$ unstable. Much depends on the initial conditions.

The given initial conditions give us that $ y_0 \equal{} a \plus{} b,$ with the key point being that $ y_0 > a.$ In this range, $ f(y) < y,$ but since $ f$ is an increasing function, we still have that $ f(y) > f(a) \equal{} a.$ Hence $ (y_n)$ is a decreasing sequence, bounded below by $ a.$ Thus it has a limit and that limit is at least $ a,$ leaving $ a$ as the only possible limit.

[color=green][Edited to fix the mistake that JBL noted.][/color]

[quote="Kent Merryfield"]This makes the fixed point at $ b$ stable and the fixed point at $ a$ unstable. Much depends on the initial conditions.[/quote]  That should be the other way around, of course.  Another way to get to the same place is to note that this recursion always has a solution of the form $ x_n \equal{} c a^n \plus{} d b^n$, and the $ a^n$ term will almost always dominate except in the rare cases when $ c \equal{} 0$.

hello, on can prove by induction the formula $ x_n=\frac{a^{n+1}-b^{n+1}}{a-b}$, so we have
${ \frac{x_{n+1}}{x_n}=\frac{a^{n+1}-b^{n+1}}{a-b}=\frac{a^{n+2}(1-(\frac{b}{a})^{n+2})}{a^{n+1}(1-(\frac{b}{a})^{n+1})}=a(\frac{1-(\frac{b}{a})^{n+2}}{1-(\frac{b}{a})^{n+1}}})$ and we get the limit as $ a$ for $ n$ tends to infinity.
Sonnhard.


email
geometry
geometric transformation
incenter
cyclic quadrilateral
national olympiad

China is one of the strongest team in IMO, so does anyone post the problems from Chinese Mathematical Olympiads

Well, the site http://www.kalva.demon.co.uk is vast collections of problems, also it included Chinese Mathematics Olympiad questions but I haven't found those in recent days. I think there were all the CMO problems and lots of solutions. Maybe it will be available in near future.

Can you  translate them into English?

I have them in Chinese too :D

zhaoli, would you send me all the chinese mathmatical olympiad problems and chinese IMO team selection test problems? I only have all CMO questions but not  team selection test problems. :(
Chinese is ok, in fact, I like reading chinese more than english. ;)

Email: samcrazywong@yahoo.com.hk

well, i think it is the best if you translate the team selection test into English. The Chinese MO we can get on internet, but the TST , it is really hard
Thx

mecrazywong:
I have e-mailed the problems of chineses TST problems to you.
Enjoy it!

zhaoli
 :D

I forget login, i am nntrkien and...Chinese TSTs for me too, thanks ntrkien88@yahoo.com

if you got english versions, please send them to me too :)

(peter.vdd@pandora.be)

I'm in too :) (please): ben_goss_ro@yahoo.com

Here is the translation of the first problem:

Let $I_A,I_B,I_C,I_D$ be the incenters of triangles BCD, ACD, ABD, ABC of a cyclic quadrilateral ABCD. Prove that $I_AI_BI_CI_D$ is a rectangle.

[Moderator edit: This problem is well-known; see
  http://www.mathlinks.ro/Forum/viewtopic.php?t=77259 ,
  [url=http://www.math.fau.edu/yiu/Geometry.html]Paul Yiu, [i]Notes on Euclidean Geometry[/i][/url], paragraph 10.2.6 (page 154);
  [url=http://groups.yahoo.com/group/Hyacinthos/message/8154]Barry Wolk, [i]Hyacinthos message #8154[/i][/url],
  [url=http://groups.yahoo.com/group/Hyacinthos/message/8178]Alexey A.Zaslavsky, [i]Hyacinthos message #8178[/i][/url]]

[quote="ThAzN1"]I found pdf files of 86-02 [url=http://olympiad.mathnet.or.kr/olympiad/list.php?mode=title&title=Chinese%20Mathematical%20Olympiads]here[/url], but only some of the files will open... oh well[/quote]
here are CMO 2004 problems (log in to download):

[quote="zhaoli"]I have all the chinese mathmatical olympiad problems and chinese IMO team selection test problems in hand, but in chinese. Does anyone like to see them?[/quote]

Can you send them to me plz? filletwho@yahoo.com Thx.

zhaoli, would you send me all the chinese mathmatical olympiad problems and chinese IMO team selection test problems?

my e-mail: edablo@hotmail.com

thanks  :)


[asy]defaultpen(linewidth(.8pt));

pair A = (0,11);
pair B = (2,0);
pair D = (4,0);
pair E = (7,0);
pair C = (13,0);

label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,S);
label("$E$",E,S);

label("$2$",midpoint(B--D),N);
label("$3$",midpoint(D--E),NW);
label("$6$",midpoint(E--C),NW);

draw(A--B--C--cycle);
draw(A--D);
draw(A--E);[/asy]

In triangle $ ABC$ in the adjoining figure, $ AD$ and $ AE$ trisect $ \angle BAC$. The lengths of $ BD$, $ DE$ and $ EC$ are $ 2$, $ 3$, and $ 6$, respectively. The length of the shortest side of $ \triangle ABC$ is

$ \textbf{(A)}\ 2\sqrt{10}\qquad
\textbf{(B)}\ 11\qquad
\textbf{(C)}\ 6\sqrt{6}\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ \text{not uniquely determined by the given information}$

A nice problem.
By the Angle Bisctor Theorem, let AB=2n AE=3n, AD=3m, and AC=6m. By Stewart's on $ \triangle ABE$, after simplification, we get $ 2n^2 \equal{} 3m^2 \plus{} 2$. Also, by Stewart's on $ \triangle ADC$, after simplification, we get $ 2m^2 \equal{} n^2 \plus{} 2$. Solving, $ m \equal{} \sqrt {6}$ and $ n \equal{} \sqrt {10}$. We find AB is the shortest side, with a length of $ 2 \sqrt {10}$.


Okay....I was going over homework last night and thought of this....

would all rectangles be similar? Since all rectangles have congruent angle measures, which are right angles

no. "same angles implies similar" only applies to triangles.

I think you need corresponding sides to be in proportion as well.

Two figures are said to be congruent if they have same size and shape.

If all the angles are equal, they only gives us same shape but not size.

Therefore 2 rectangles are not congruent if they have all the angles of equal measure.

[quote="rksmash"]

If all the angles are equal, they only gives us same shape but not size.

[/quote]

not true. a square and a non-square rectangle have the same angles, but aren't the same shape.

The "all angles means similar" that you are referring to is AAA similarity.  There are a bunch of theorems that can be used to prove triangles similar or congruent - SSS, SAS, AAS, AAA, HL, etc. etc.

However, these only work for TRIANGLES.

If you pick up a piece of paper and a pencil, it's easy to see why.  Draw a rectangle with sides of 10 cm and 12 cm, then draw a rectangle with sides of 35 cm and 1 cm.  (Or something similar.)  The angles are the same, because they are rectangles.  However, neither the shapes nor the sizes are the same.  We cannot say that all rectangles are similar.

Now, draw two triangles with the same angle measures but different side lengths.  It's clear that these triangles are similar.  The picture will show that the two obviously have the same shape.

Yes, I would go with Adithya's method. lol

.........Hey people! Do posts increase your rating on FTW?

You can ask these things in the FTW forum. :)

And the answer is no, applying both to number of posts and quality of posts.


Does anybody know the old AIME floor values for USAMO qualification?  The AMC sites only seems to list this years and last years floors.  I've been doing old tests off of Kalva and I'm wondering whether I would have qualified for any given test.  Thanks.

This was discussed a while back..

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27485

Of course, this year it was 9.

On that thread, this was said:

2004 - 7
2003 - 8
2002 - 6

And that's it. The idea of "floor" was invented in 2002, the third year of the existence of the AMC-10. Before that, qualification was based on index (AHSME + 10*AIME), with in some cases a different index needed for each grade level. As an example, I beleive that in 1998 (a relatively hard year), the minimum index for a 12th grader was 190 and for a 10th grader 180.

That was 200 and 190 in 1998. Since I was a 10th grader at exactly 200, I remember this.

Wow. seems very low compared to 226 in 2003, 210 in 2004, and 233 this yr.... AHSME must have been pretty hard compared to AMC. or the AIME, of course.

....the reason why I always train with hard year's AIME and AHSME!!!

One factor is that the 30-question AHSME was always (except 1994) harder to get very high scores on than the 25-question AMC-10 and AMC-12.

The larger factor is that the AIME was hard. The 200 qualifying index was low, but not out of line with the years around it- the usual standards were around 200-210.
Some numbers to compare the 1998 AIME with the 2005 AIME 1:
1998: 38 students got at least 10.
2005: 36 students got at least 13.
1998: 77 students got at least 9.
2005: 79 students got at least 12.
1998: 149 students got at least 8.
2005: 148 students got at least 11.
1998: 246 students got at least 7.
2005: 238 students got at least 10.
1998: 423 students got at least 6.
2005: 379 students got at least 9.

My 130 AHSME/7 AIME that year as a tenth grader would match pretty well with a 133.5 AMC-12/10 AIME 1 this year. If they had used an AIME floor system, 1998 and 1999 would probably each have had 6 as the floor score. I don't have enough information about other years to speculate; the above data was compiled by counting from the results handbook.

I can confirm, by having been a high school student in the early 90s, that the numbers were about the same as they were in 1998. It was generally true that an AHSME/AIME index of about 200 was enough to get invited to the USAMO; so an AIME of 6 or 7 was considered pretty good.

But I think that students studied for these exams MUCH less back then than they do now, and there weren't as many resources available (like this site and Kalva etc.). So that may be a reason the indexes were lower.

Perfect and near-perfect scores on the AHSME were also very rare (except for in 1994). I remember when Lenny Ng scored perfect for 4 years in a row (from 1990-1993) it was considered an amazing accomplishment. There usually weren't more than 3 or 4 perfect scorers in any given year, and only a handful of near perfect scores.


If $ a,b,c>0$, then

$ \prod[(a^2\plus{}2b\plus{}c^2)(2a^2\plus{}b^2\plus{}c^2)]\geq abc\prod(a\plus{}b)^3$

[quote="filomen"]If $ a,b,c > 0$, then

$ \prod[(a^2 \plus{} 2b \plus{} c^2)(2a^2 \plus{} b^2 \plus{} c^2)]\geq abc\prod(a \plus{} b)^3$[/quote]
The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ (2a^2\plus{}b^2\plus{}c^2)^2(a^2\plus{}2b^2\plus{}c^2)^2(a^2\plus{}b^2\plus{}2c^2)^2\geq8abc(a\plus{}b)^3(a\plus{}c)^3(b\plus{}c)^3\]

[quote="arqady"][quote="filomen"]If $ a,b,c > 0$, then

$ \prod[(a^2 \plus{} 2b \plus{} c^2)(2a^2 \plus{} b^2 \plus{} c^2)]\geq abc\prod(a \plus{} b)^3$[/quote]
The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ (2a^2 \plus{} b^2 \plus{} c^2)^2(a^2 \plus{} 2b^2 \plus{} c^2)^2(a^2 \plus{} b^2 \plus{} 2c^2)^2\geq8abc(a \plus{} b)^3(a \plus{} c)^3(b \plus{} c)^3\]
[/quote]

 $ \Longleftrightarrow \sum_{sym} (2a^{12} \plus{} 28a^{10}b^2 \plus{} 77a^8b^4 \plus{} 81a^8b^2c^2 \minus{} 8a^7b^4c \minus{} 24a^7b^3c^2 \plus{} 53a^6b^6 \minus{} 24a^6b^5c$

$ \plus{} 254a^6b^4c^2 \minus{} 72a^6b^3c^3 \minus{} 72a^5b^5c^2 \minus{} 312a^5b^4c^3 \plus{} 17a^4b^4c^4)\geq 0$

 true due to AM-GM Inequality and Muirhead theorem.

 :lol:

[quote="arqady"]hedeng123, $ (2a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} 2b^2 \plus{} c^2)\geq2c(a \plus{} b)^3.$ :wink:[/quote]

arqady  how to prove that the following inequality

$ (2a^2 \plus{} b^2 \plus{} c^2)^2(a^2 \plus{} 2b^2 \plus{} c^2)^2(a^2 \plus{} b^2 \plus{} 2c^2)^2\geq 8abc(a \plus{} b)^3(a \plus{} c)^3(b \plus{} c)^3\plus{}$ $ 2\sum {(a^3\minus{}b^3)^4}$.

 :lol:

[quote="arqady"][quote="hedeng123"][quote="arqady"]hedeng123, $ (2a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} 2b^2 \plus{} c^2)\geq2c(a \plus{} b)^3.$ :wink:[/quote]

arqady  how to prove that the following inequality

$ (2a^2 \plus{} b^2 \plus{} c^2)^2(a^2 \plus{} 2b^2 \plus{} c^2)^2(a^2 \plus{} b^2 \plus{} 2c^2)^2\geq 8abc(a \plus{} b)^3(a \plus{} c)^3(b \plus{} c)^3 \plus{}$ $ 2\sum {(a^3 \minus{} b^3)^4}$.

 :lol:[/quote]
After your expanding it's very easy. :)[/quote]

 :D


trigonometry function algebra polynomial Vieta inequalities Gauss

How do people use complex numbers to solve trigonometric identities easily?

For example, evaluate
\[ \left(\sin \frac {2\pi}{7} \plus{} \sin \frac{4\pi}{7} \minus{} \sin \frac {\pi}{7}\right)^2.
\]

The general idea is that writing trigonometric functions as sums of powers of roots of unity reduces everything to a calculation in one variable, and it is much easier to work with polynomials in roots of unity than with trigonometry.  Every trigonometric identity is, in principle, a consequence of a polynomial identity involving roots of unity.  If you're unaware of the actual technique, it is simply the pair of identities

$ \cos t \equal{} \frac {e^{it} \plus{} e^{ \minus{} it}}{2}$
$ \sin t \equal{} \frac {e^{it} \minus{} e^{ \minus{} it}}{2i}.$

I have seen two more specific techniques.  One involves using Vieta's formulas on [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]Chebyshev polynomials[/url], which can be derived and understood most easily with complex numbers, and the other is  some variant of a [url=http://en.wikipedia.org/wiki/Quadratic_Gauss_sum]Gauss sum[/url].

You wouldn't.  The algorithmic technique says just write $ \sin \frac{\pi}{7} \equal{} \frac{\zeta \minus{} \zeta^{13}}{2i}$ where $ \zeta \equal{} e^{ i \frac{\pi}{7} }$, etc., and the rest is algebra.

[hide="In fact..."] What we want to compute is just

$ \left( \frac{\zeta^2 \minus{} \zeta^{12} \plus{} \zeta^4 \minus{} \zeta^{10} \minus{} \zeta \plus{} \zeta^{13}}{2i} \right)^2$

and, using the fact that $ \zeta^7 \equal{} \minus{}1$, this is

$ \left( \frac{\zeta^2 \plus{} \zeta^5 \plus{} \zeta^4 \plus{} \zeta^3 \minus{} \zeta \minus{} \zeta^6}{2i} \right)^2$

and, using the fact that $ \zeta^6 \minus{} \zeta^5 \plus{} \zeta^4 \minus{} \zeta^3 \plus{} \zeta^2 \minus{} \zeta \plus{} 1 \equal{} 0$, this is

$ \left( \frac{2 \zeta^4 \plus{} 2 \zeta^2 \minus{} 2 \zeta \plus{} 1}{2i} \right)^2$.

I think you get the idea.  Squaring and applying the above two relations again, we find that the final answer is $ \boxed{ \frac{7}{4} }$. [/hide]

[quote="123456789"]How would I solve this problem specifically?

Would I take the imaginary part of $ \text{cis}^2 \frac {\pi}{7}$ and so on? How would I made the "imaginary" operator go away?[/quote]

Note that $ \text{Im}\,(\text{cis}^2\,\theta)\equal{}\sin 2\theta$ by De Moivre's theorem. It is not necessarily equivalent to $ \sin^2\theta$.

[quote="t0rajir0u"]using the fact that $ \zeta^6 \minus{} \zeta^5 \plus{} \zeta^4 \minus{} \zeta^3 \plus{} \zeta^2 \minus{} \zeta \plus{} 1 \equal{} 0$[/quote]

How do you know that?

EDIT: Wait, that follows from $ x^7\plus{}1\equal{}0$, right?

hello, i have computed that $ \left(\sin(\frac{2\pi}{7})\plus{}\sin(\frac{4\pi}{7})\minus{}\sin(\frac{\pi}{7})\right)^2\equal{}1.75$.
Sonnhard.

[quote="123456789"][quote="t0rajir0u"]using the fact that $ \zeta^6 \minus{} \zeta^5 \plus{} \zeta^4 \minus{} \zeta^3 \plus{} \zeta^2 \minus{} \zeta \plus{} 1 \equal{} 0$[/quote]

How do you know that?

EDIT: Wait, that follows from $ x^7 \plus{} 1 \equal{} 0$, right?[/quote]

Indeed it does.


Let a,b,c are positive numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that
$ \frac {a\plus{}b}{b\plus{}c}\plus{}\frac {b\plus{}c}{c\plus{}a}\plus{}\frac {c\plus{}a}{a\plus{}b}\le \frac {3}{abc}$ :D

[quote="thegod277"]Let a,b,c are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b}\le \frac {3}{abc}$ :D[/quote]
$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b}\le \frac {(a\plus{}b\plus{}c)^3}{9abc}$ :
$ \frac {(a \minus{}b)^2}{(a \plus{} c)(b\plus{}c)} \plus{} \frac {(b \plus{} c)(a\minus{}c)(b\minus{}c)}{(c \plus{} a)(a\plus{}b)(b\plus{}c)} \leq  \frac {(3c\plus{}b \plus{} a)(a\minus{}b)^2}{9abc }\plus{} \frac {(2a\plus{}2b\plus{}c)(a\minus{}c)(b\minus{}c)}{9abc}$ :

It is SOS-Shur tecnique you can find at http://reflections.awesomemath.org/2007_5.html

The solution is OK for $ c \equal{} min(a,b,c)$


A restaurant has a total of 30 tables which are of two types.  The first type seats two people at each table; the second type seats five people at each table.  A total of 84 people are seated when all seats are occupied.  How many tables for two are there?

[quote="math92"][hide="i have no idea how to do this"]
i set up a system of equations
2a+5b=84
a+b=30
solve those and you get...
2(30-b)+5b=84
60-2b+5b=84
3b=24
b=8
so..
a=22
[b]22[/b] two seater tables
but i don't know if this is right.[/hide][/quote]

You are exactly right. Sometimes you need to believe in yourself.  :)

ok. Start out assuming all of them are 5 seaters. A total of 30 tables means that 150 people are seated. Everytime you swap a table of 5 for a table of 2, 5 people are lost.
150-84=You need to lose 66 people
66/3=means 22 2 seaters.

I've been doing all of Crawford's systems of equations questions with matrices and have gotten a LOT better at them.  :D

I solved them simultaneously
[hide]
let there be x no of tables which can seat 2 people and
let there be y no of tables which can seat 5 people
x+y=30
2x+5y=84

therefore x=22 and y=8[/hide]


modular arithmetic
number theory unsolved
number theory

Determine all positive integers $m, n \ge 2$ such that \[ \frac{1 + m^{3^n} + (m^{3^n})^2}{n} \in \mathbb N  \]

the solutions are $(m, n) = (3k+1, 3)$, $k \geq 1$.

assume that  $n | 1 + m^{3^n} + (m^{3^n})^2$, with $m, n \geq 2$. then $m^{3^{n+1}} \equiv 1 \pmod n$. let $d$ be the order of $m \pmod n$. then $d|3^{n+1}$, so $d = 3^i$ for some $i$. moreover, $d|\phi(n) < n$, so $3^i < n$, which implies $i < n$. in particular, this means that $3^i | 3^n$. then $n | 1 + m^{3^n} + (m^{3^n})^2 \equiv 1+1+1 = 3 \pmod n$, which implies that $n = 3$. with $n = 3$, only $m \equiv 1 \pmod 3$ works.


calculus
geometry
probability
trigonometry
analytic geometry

Hello

I am have a dilemma. I keep obsessing about doing every single problem in a physics/math/chemistry textbook because only then do I feel that I have mastered the subject. But is it necessary to do every problem to know math/physics/chemistry inside and out? Would it be a waste of time, because you are spending a long time on one topic? Anybody else do the same thing?

Thanks

For a typical textbook, it's only necessary to do the hardest 10-25% of the problems. In general, you should try to work through any problem that you don't immediately know how to do. That should cut the amount of time you spend to something reasonable. If there's a problem that you can't solve for a really long time, it's fine to check the solution.

I'll paste in here a story about Donald Knuth. Knuth was interviewed in the book Mathematical People: Profiles and Interviews edited by Donald Albers and G. L. Alexanderson (Boston: Birkhäuser, 1985) and this is what he had to say about repetition of mathematics class exercises:

At Case, I spent hours and hours studying the mathematics book we used--Calculus and Analytic Geometry by Thomas--and I worked every supplementary problem in the book. We were assigned only the even-numbered problems, but I did every single one together with the extras in the back of the book because I felt so scared. I thought I should do all of them. I found at first that it was very slow going, and I worked late at night to do it. I think the only reason I did this was because I was worried about passing. But then I found that after a few months I could all of the problems in the same amount of time that it took the other kids to do just the odd-numbered [sic] ones. I had learned enough about problem-solving by that time that I could gain speed, so it turned out to be very lucky that I crashed into it real hard at the beginning.
(pages 185-186)

http://en.wikipedia.org/wiki/Donald_Knuth

http://www.softpanorama.org/People/Knuth/index.shtml

That said, I generally agree with Probability 1.01 that if you can find a good textbook, do all of the HARD problems in that textbook, and then find another textbook and do the hard problems in it.

Thanks for your help.

So basically it is best to do all the HARD problems then, to really understand a subject. What about a book like Spivak or Apostol or Kleppner/Kolenkow. Are all the problems in those books considered to be difficult?

Thanks

Spivak and Apostol are plenty difficult, and very good overviews of calculus.

yeah the book i'm working in is an honors book, so all the problems are difficult (very few seem to be routine). i've been trying to finish all most all of the theoretical problems, but it takes some time.

what would you suggest for a book like the one i'm working in?

what would someone think to be a good precalc worktext? an AoPS book is ok but i am just not sure...i have the same problem as the first poster being obsessed with each prob

[quote="tokenadult"]David Cohen's [url=http://www.amazon.com/Precalculus-Circle-Trigonometry-CD-ROM-Tutorial/dp/0534402305/]Precalculus with Unit Circle Trigonometry[/url] is quite good.[/quote]

Woo! Good precalc book. Really tough problems... especially in the random geometry with trig questions.

I think then that reading a textbook like a book would be a good way to learn the subject thoroughly?

Is that a question or a statement?

And it depends on certain factors, e.g. the textbook and the person reading it.


Set B = {55, 57, 58, 59, 61}.  Which number in set B
has the smallest prime factor?

The smallest prime factor is $ 2$, and $ \boxed{58}$ has it.


8. A deck of 52 cards is printed on a 4*13 rectangular grid. If stackign the pieces is allowed after each cut, but no folding is allowed, what is the elast number of cuts required to separate all the cards? (Cuts can be any length, but may go in only one directio per cut)

9. In triangle ABC, <A = 30 degress, and <C = 105 degrees. If AC = 8 units, what is the length of segment BC?

I got [hide]7[/hide] for the frist one, not sure if thats right, and I couldn't understand how to do 9 at all.

[hide="hint"]Does the angle 30 degrees ever ring a bell?

[/hide]

[hide="hint #2"]What about right triangles?[/hide]

[hide="hint #3"]Maybe... Extra lines?[/hide]

[hide="hint..."]9. In triangle ABC, <A = 30 degress, and <C = 105 degrees. If AC = 8 units, what is the length of segment BC?

split into 30-60-90 and 45-45-90...[/hide]

I got [hide="this"]6[/hide] for the first 1.

[hide]You use two cuts to get 4 1x13s stacked.
Then cut to get 4 1x6 and 4 1x7.
Cut to get 8 1x3 and 8 1x4.
Cut to get 16 1x2, and (8 1x1 and 4 1x2-The bottom half had a card sticking out because it was a stack of 6 on top of a stack of 7)
Since all of them are 1 or 2 thick, make one last cut.

6 cuts.[/hide]

For the second one...

[hide]
I oreinted the triangle so that AB was on the bottom, and drew an altitude to AB. Call the point the altitude ends on X, so I can refer to it. Then I had a 30-60-90 triangle, triangle AXC and I had AC = 8 so CX = 4. Then if you think about it, <XBC and <XCB are 45 degrees so that is a 45-45-90 triangle. Since we know CX = 4 then CB = 4*sqrt2
[/hide]
The hints were good :)

For the first one, I also got
[hide]6[/hide]but I am wondering whether there is a better way than drawing diagrams, and how do I know I have the least?

For the first 1 i got [hide]0
This is because it says you can stack the cards between cuts, then you don't need to cut the cards at all[/hide]

I think that means that if you cut it into a 2 by 13 and a 2 by 13 peice, for the next cut you can stack those two peices on top of each other.

I suppose 6 makes sense because it takes $2^2$ cuts to scale down 4 and $2^4$ cuts to scale down 13. Hence, 2+4=6

[quote="X_wing1redleader"]i dont quite understand the first at all  :( [/quote]

You have a deck of 52 cards arranged in a 4x13 grid. WHat's not to get?


Okay I'm stuck on the second last part (Already solved the last part) of a relatively long winded problem in Rudin (Ch. 9 Q#2). So far I have shown that $ f_n(x) \equal{} \frac {sin(x)sin(nx)}{x^2}$ has a fourier transform equal to a multiple of $ \chi_{[ \minus{} n,n]} \ast \chi_{[ \minus{} 1,1]}$, both of which are in $ L^1$ and I have shown that $ \|f_n\|_1 \rightarrow \infty$. The problem then states to conclude from this that the fourier transform mapping of $ L^1$ is therefore a proper subset of $ C_0$. Obviously I could just dig up an example but I can't understand how I could infer this from what I have just shown.

Any ideas would be much appreciated but please just a hint, as this is for an assignment. Thanks for the help!

The linear mapping $ L^1\ni f\mapsto \widehat f\in C_0$ between two Banach spaces is one-to-one and continuous. If it were onto, the inverse mapping would be (fill the blank).