probability
limit
function
topology
geometry
3D geometry
sphere

Assuming a fair die, one expects any given face to appear exactly once in $ N$  rolls, where $ N$ is the number of faces that is $ \sum_{j\equal{}1}^{N} a_i \equal{} Na_i \equal{} 1\Rightarrow \forall i\,a_i \equal{}\frac{1}{N}$ .

The expected face-value of an $ N$  sided die is then $ \sum_{i\equal{}1}^{N}a_ii\equal{}\frac{1}{N}\sum_{i\equal{}1}^{N} i \equal{}\frac{N\left(N\plus{}1\right)}{N2}\equal{}\frac{N \plus{} 1}{2}$

The question now is “With what probability is $ \frac{n \plus{} 1}{2} \equal{} n$  ?”

$ \Rightarrow \frac{n}{2}\equal{}\frac{1}{2}$

$ \Rightarrow n \equal{} 1$

$ \therefore \lim_{n\to\infty} p_n \equal{} 0$

1) It wasn't a fair die. A fair die with infinitely many faces just does not exist as a mathematical object

2) The expected value for the fair die with $ N$ faces is, indeed, $ \frac{N\plus{}1}2$.

3) The definition of $ p_n$ has nothing whatsoever to do with the expected value. You may want to reread alekk's original post to see what $ p_n$ is.

[quote="fedja"]1) It wasn't a fair die. A fair die with infinitely many faces just does not exist as a mathematical object[/quote]

And a die with infinitely many sides, fair or not, does?

In any case, having re-read the problem, I believe it is much simpler than any of us had thought.

Rather than casting the $ \left\{a_i\right\}$  as probabilities in themselves let $ a_i$  represent the event that the [i]i-th[/i] face is showing at the end of a given throw.

Now let $ I\left(a_i\right)$  be an indicator function, a Bernoulli random variable, for $ a_i$  .

That is,

	 \[ I\left(a_i\right) \equal{} \begin{cases}1 & \text{if }  a_i\text{ occurs }\\0  &\mbox{if }  a_i\text{ does not occur }\end{cases}
\]

	$ E\left[I\left(a_i\right)\right]\equal{}P\left(a_i\right)\cdot 1 \plus{} P\left(\neg a_i\right)\cdot 0 \\ \\ \,\therefore E\left[I\left(a_i\right)\right]\equal{}P\left(a_i\right)$

From the weak law of large numbers,

	\[ \lim_{n\to \infty} P\left(\left|\overline{I}_n\left(a_i\right)\minus{}E\left[I\left(a_i\right)\right]\right|> \delta\right)\equal{}0
\]

	\[ \overline{I}_n\left(a_i\right) \equal{}\frac{\sum_{j \equal{} 1}^n I_j\left(a_i\right)}{n}
\]

	\[ \Rightarrow \lim_{n\to \infty} P\left(\left|\frac{\sum_{j \equal{} 1}^n I_j\left(a_i\right)}{n}\minus{}P\left(a_i\right)\right|> \delta\right)\equal{}0
\]

Therefore, $ P\left(a_i\right)\equal{} \lim_{n\to \infty} \frac{\sum_{j \equal{} 1}^n I_j\left(a_i\right)}{n}$

I believe this is what alekk had in mind by $ p_n$  , i.e. $ \lim_{n\to \infty}p_n \equal{} P\left(a_i\right)$ .

[quote="Dr. No"]And a die with infinitely many sides, fair or not, does?[/quote]
In a theoretical point of view: yes.

And $ \lim_{n\to \infty}p_n \equal{} P\left(a_i\right)$ is likely not to be what he had in mind. I think he stated very clearly that $ p_n$ is the probability that at some number of throws, the sum of all thrown numbers is $ n$.

[quote="ZetaX"][quote="Dr. No"]And a die with infinitely many sides, fair or not, does?[/quote]
In a theoretical point of view: yes.[/quote]

Then could you please explain why an object with infinitely many faces can exist in theory but one is disallowed from further theorizing that said "theoretical object" is fair?

[quote="ZetaX"]And $ \lim_{n\to \infty}p_n \equal{} P\left(a_i\right)$ is likely not to be what he had in mind. I think he stated very clearly that $ p_n$ is the probability that at some number of throws, the sum of all thrown numbers is $ n$.[/quote]

That's what I had thought originally -- so I guess we're back to the same intractably messy problem, which you real analysis wizards, have already given an approach to if not a full blown solution.

The fair dice doesn't exist because of additivity:

Assume that it has faces numbered $ 1,2,3,...$ and let the probability that some fixed face $ k$ occures be $ a_k$. We call it fair if all these $ a_k$ are equal, let's write $ a$ instead (as it does not depend on $ k$). We know that if throwing a dice, there always occurs exactly one number, so by the law of additivity, the probability that any number occurs is $ \sum_{k\equal{}1}^\infty a_k \equal{} \sum_{k\equal{}1}^\infty a$. But also, this probability is $ 1$ (one number will occure, we just don't know which one). Thus we have $ \sum_{k\equal{}1}^\infty a\equal{}1$ which cannot be true (the sum is $ 0$ iff $ a\equal{}0$ and is infinite otherwise). So the assumption at the beginning was wrong, no such dice exists (even theoretically).

If we don't need it to be fair, we can easily construct nonzero $ a_k$ with sum $ 1$, so a non-fair dice exists.

[quote="JBL"]A die exists as the mathematical object known as a probability distribution, and a fair die exists as the mathematical object known as a uniform distribution.  One of the properties of a probability distribution is countable additivity, and then ZetaX's comments are a proof that there is no uniform distribution on a countable set, i.e. no fair die with countably infinitely many sides.[/quote]

Perhaps, but it seems to me that such a distinction is ambiguous because intuitively (meaning that I have no idea how to prove it) there appears to be an homeomorphism between an N-hedron and the sphere, as $ N\to \infty$. The uniform distribution can easily be mapped onto a sphere and thus "a fair die" with infinitely many faces can indeed exist in theory just as easily as can any other 'infinitely' sided theoretical object.

[quote="JBL"]A sphere consists of more than countably many points.  There is a uniform distribution on a sphere.[/quote]

Ah, that whole $ \aleph_0$ vs. the continuum thing. So what you're saying is that there are more states available on the sphere than on the proverbial die by definition. Why couldn't one then simply partition the interval into a countably infinite number of bins? That way the i-th face would correspond to the i-th bin (sub-interval).

One certainly can. What one cannot do is to make these bins of equal length.


Hello!
I've been accepted to study mathematics and computer science at Oxford University. I was wondering, are there any fellow AoPSers who were accepted and consider going? For graduates of this university, how strong do you think their program was, compared to US universities? Any feedback would be greatly appreciated.

Thanks!

I've been accepted to Oxford for engineering (I'm from America), PMing you.


does anyone know how you could plot a single point on the graphing calculator?  pls && thx :!:

ii have to make a snowman on my calculator for extra credit....if ur wondering

If your calculator can plot parametric equations, and you want to plot a point, for example (3,4),  just put the calculator into parametrics mode, let $x = 3$ and $y=4$ (or $x =$ whatever x-value you want, and $y =$ whatever y-value you want.)  The calculator will graph that point.

What kind of graphing calculator do you have? On a TI-83/84 I know you can draw on the blank screen using the pen tool (GRAPH, 2nd, PRGM, up, ENTER) by using ENTER to toggle write/move. You can also use the Circle tool in programs or on the home screen (2nd, PRGM, 9) and doing Circle(x-coordinate, y-coordinate, radius).

Or you can graph a really small semicircle.

Assuming yours is like mine, go to mode parameter then just plug in x and y values of the points. Then,  you will have to know how to make the circles in parameter form though, its y = r sin(t)+k and x = cos(t)+h if I remember correctly

You can plot a set of points by going to STAT => 1: Edit... Then plug in the x- and y-values. :D

draw a circle with radius 0
if the calculator comes up with an error, then make the radius something like 0.000000000000000001
 :D

Try going to mde and changing the third setting to parametric
then just imput the coordinate of the point you want to graph

[2nd] [PRGM] [RightArrow] and select Pt-On(entercoordinateshere)

Tedious, but...


A point $ (x,y)$ is randomly and uniformly chosen inside the square
with vertices (0,0), (0,2), (2,2), and (2,0).  What is the probability
that $ x\plus{}y < 3$?

[hide]Because of the common difference $ 2$, we notice that the square has side length $ 2$. Then, it's area (our denominator) is $ 4$. Now, we draw the portion of the line $ x\plus{}y\equal{}3$ inside the square. We see two regions. But which region is $ x\plus{}y<3$? We notice that the origin satisfies this inequality. Thus, it is the below region's area we are looking for. \[ \dfrac{4\minus{}\frac{1\cdot1}{2}}{4}\equal{}\boxed{\frac{7}{8}}\] $ \Box$[/hide]

[hide]Graphing, the satisfying area is $3.5$ units squared and the total area is $4$ units squared. The answer is $\frac{3.5}{4}=\boxed{\frac78}$ :pilot:[/hide]


Let $ S_1,S_2,\ldots,S_{10}$ be ten subsets of $ S(n)\equal{}\{1,\ldots,n\}.$ Determine the least $ n$ such that it is possible for $ S(n)$ to be the union of five or more of the subsets $ S_i$ but not the union of four or fewer of the subsets $ S_i.$

Please help. :maybe:

I think $ n\equal{}210.$

Note that in order for any union of five subsets to be $ S(n),$ each number $ 1,\ldots,n$ must belong to at least six of the subsets $ S_i.$ If not, say $ k\in S(n)$ belongs to at most five of the $ S_i,$ then it would be possible to find five subsets $ S_{i_1},\ldots,S_{i_5}$ such that $ k$ is not in any of them – from which it follows that $ \bigcup_{r\, \equal{} \,1}^5S_{i_r}\ne S(n).$ Thus given each $ k\in S(n)$ there are at most $ 4$ subsets to which $ k$ doesn’t belong. In the extreme case, each $ k$ does not belong to exactly $ 4$ subsets. Since there are only $ \rm^{10}C_4$ distinct selections of four subsets, it follows that $ n\leqslant{\rm^{10}C_4} \equal{} 210.$

So far we have not made used of the condition that no union of four of the subsets $ S_i$ is equal to $ S(n),$ i.e. that given any four members of $ \mathcal S\equal{}\{S_1,\ldots,S_{10}\},$ there is a $ k\in S(n)$ which does not belong to any of the four members of $ \mathcal S.$ Let $ \{S_{i_1},S_{i_2},S_{i_3},S{i_4}\}$ and $ \{S_{i_5},S_{i_6},S_{i_7},S{i_8}\}$ be two distinct selections of four subsets of $ \mathcal S$ and suppose there is a $ k\in S(n)$ such that $ k\notin S_{i_1},\ldots,S_{i_4}$ and $ k\notin S_{i_5},\ldots,S_{i_8}.$ Since the two selections are different, there is some $ S_j\in\{S_{i_1},S_{i_2},S_{i_3},S{i_4}\}$ such that $ S_j\notin\{S_{i_5},S_{i_6},S_{i_7},S{i_8}\}.$ But then $ k\notin S_{i_5},\ldots,S_{i_8}$ implies that $ k$ is in all of the other six members of $ \mathcal S$ [i]and[/i] $ S_j$ is one of the other six members of $ \mathcal S$ – contradiction! Hence, given distinct selections of four of the subsets $ S_i,$ there must be distinct $ k\in S(n)$ not belonging to any of the four subsets. Since there are $ \rm^{10}C_4$ distinct selections of four subsets, it must be that $ n\geqslant{\rm^{10}C_4} \equal{} 210.$

This proves that $ n\equal{}210.$ :jump:


calculus
integration
function
ratio
trigonometry
logarithms
calculus computations

[img]http://img57.imageshack.us/img57/5647/integraljr9.jpg[/img]

Can anybody please teach/explain how to do this step-by-step by using the techniques of integration? The "cot" with the even exponential power part confuses me (I understand how to do the normal ones like "sin, cos, and tan" but the other three functions confuses me.). I tried switching the "cot" to "cos/sin" (ratio identity) and then used the half-angle/double identities, but the expression turned out quite "ugly" and now I'm stuck.

Thank you. :)

P.S. When an integral only has one trig. function alone with an even exponential (e.g. Tan^4[x]), do you like split it in separate parts or do we solve it like the other types with two trig. functions (e.g. Tan^3[x]Sec^4[x])? This is another thing that confuses me. Please explain.

Make the substitution $x=\tan^{-1}u\Rightarrow dx = \frac{du}{1+u^{2}}$, then the integral becomes:

$\int\frac{du}{(1+u^{2})u^{4}}= \int \left(\frac{1}{u^{2}+1}+\frac{1}{u^{4}}-\frac{1}{u^{2}}\right) \ du$

Do you see where to go from here?

[color=darkblue]For $n\in N$ denote $I_{n}=\int \cot^{n}x\ dx\ .$ I recall the formula $(\cot x)'=-1-\cot^{2}x\ .$ Therefore,
$I_{n+2}=\int \cot^{n}x(\cot^{2}x+1-1)\ dx=$ $-I_{n}-\int \cot^{n}x(\cot x)'\ dx$ $\Longrightarrow$ $\boxed{\ I_{0}=x+C\ ;\ I_{1}=\ln\left|\sin x\right|+C\ ;\ I_{n+2}=-I_{n}-\frac{\cot^{n+1}x}{n+1}\ ,\ n\in N\ }\ .$

In the our case, $I_{4}=-I_{2}-\frac{1}{3}\cot^{3}x=-\left(-I_{0}-\cot x\right)-\frac{1}{3}\cot^{3}x\Longrightarrow \boxed{\ I_{4}=x+\cot x-\frac{1}{3}\cot x\ }\ .$[/color]


geometry
3D geometry
complex numbers
imaginary numbers

Hope it's ok to write more than one problem in this post  :P. I have some difficulties with the following ecuations radicals.

$\#1)$ Which are the values for $a$ such that the ecuation $a^2x^2+ax\sqrt{ax}=1+ax$ has one unique root.

$\#2)$ Find $a$ such that the ecuation $\sqrt{x^2-1}+a=x$ has no real roots.

$\#3)$ Solve the ecuation $\sqrt[3]{2x+1}-\sqrt[3]{2x+3}=1$

and the last one...

$\#4)$ Find the solution of the folloing ecuation $\sqrt[3]{x+1}+\sqrt[3]{4x+7}=\sqrt[3]{x+3}+\sqrt[3]{4x+5}$

Have fun :D

First of all, having noticed this in both of your posts, I would like to point out that the term is "equation", not "ecuation". In fact, it is usually true of English that we use "qu" instead of "cu" for the kw sound, and "cu" for the kyu sound.

[hide="1"]Make $f(x)=\sqrt{ax}$. Our equation is $f(x)^4+f(x)^3-f(x)^2-1=0$. 1 is clearly a root, so $x=\frac 1a$ is a root. In the original equation, a cannot be 0, so this is the one root we are looking for. We can now simplify the quartic to a cubic: $f(x)^3+2f(x)^2+f(x)+1$, which clearly does not have positive roots. Since a negative root is impossible (f(x) is strictly positive), the next best thing would be an imaginary number (not complex, just imaginary). Again, there are no such imaginary numbers, so all real $a\neq 0$ work.[/hide]

[hide="2"]Since $\sqrt{x^2-1}=x-a$, $x^2-1=x^2-2ax+a^2$ and $a^2+1=2ax$.
We now have $x=\frac{a^2+1}{2a}$. The only way to make that nonreal is to set a=0.[/hide]

Thanks for your observation. I will know that from now on :). I thoght that [i]ecuation[/i] can be abreviate as "ec" and [i]equality[/i] to "eq". I know that are many mistsakes in my written english and I don\t mind if you correct me ;)

jb05, in question 2 you have committed the terrible sin ( ;) )of squaring a radical and then not back-substituting to make sure your solution is valid.

[hide="3"]
Not entirely confident of my solution.  We have
$\sqrt[3]{2x+1}=1+\sqrt[3]{2x+3}$    (*)
Cube boths sides
$2x+1=1+3\sqrt[3]{2x+3}+3(\sqrt[3]{2x+3})^2+(2x+3)$
Rearranging and factoring:
$-3=3\sqrt[3]{2x+3}(1+\sqrt[3]{2x+3})$
Dividing by 3, and substituting in from (*),
$-1=\sqrt[3]{2x+3}(\sqrt[3]{2x+1})$
Cubing again
$-1=(2x+3)(2x+1)$
$4x^2+8x+4=0$
($2x+2)^2=0$
$\Rightarrow x=-1$
But plugging this back into the original equation, we find $-2=1$, which is a false statement, so there are [b]no solutions[/b].

[/hide]

[quote="shobber"][quote="Psycho"]Question four
[hide]
let $a=x+1$ and $b=4x+5$
then the equation becomes $\sqrt[3]{a}+\sqrt[3]{b+2}=\sqrt[3]{a+2}+\sqrt[3]{b}$
which clearly has a solution when $a=b$
$\Rightarrow x+1=4x+5$
$\Rightarrow x=-\frac{4}{3}$

don't know how to show this is the only solution though  :|  maybe i overlooked something obvious  :oops: [/hide][/quote]
Can you please show your steps on how you get $a=b$?[/quote]
 :huh:
Psycho is not saying that $a=b$ is the only answer, he is saying for $a=b$ the equality holds true,(i think it's the only answer, though)...


conics
parabola
algebra
function
domain
calculus
calculus computations

You are given a parabola (1 - x^2). You have to find points (P and Q) on the parabola so that the triangle ABC formed by the x-axis and the tangent lines at P and Q is an equilateral triangle.

So we have for the parabola: $y=1-x^{2}$ and we want to find points $P$ and $Q$ such that the tangents at these points form an equilateral triangle.

[hide]This means that points $P$ and $Q$ must be between the x-intercepts of the parabola (in order for the x-axis to be the base), so: $0=1-x^{2}\iff x= \pm 1$.  So our domain for $P$ and $Q$ is $(-1,1)$ where $x \not = 0$ (because then it wouldn't be a triangle).

To be an equilateral triangle, the distance from the y-intercept to the x-intercept must be equal to the distance between the two x-intercepts.  To simplify things we'll just say that the distance of an x-intercept must be half of the distance between the x and y-intercepts.

At any point $(h,k)$ the equation of the tangent line will be:

$y-k=-2h(x-h)$; and we know $y=1-x^{2}\implies k=1-h^{2}$
$y-(1-h^{2})=-2h(x-h)$

Now let's find the y-intercept of this tangent line by letting $x=0$:
$y_{int}-(1-h^{2})=-2h(0-h)$
$y_{int}=h^{2}+1$

Next, the x-intercept by letting $y=0$:
$0-(1-h^{2})=-2h(x_{int}-h)$
$2hx_{int}=1+h^{2}$
$x_{int}=\frac{h^{2}+1}{2h}$

So like I said, the x-intercept must be half of the distance between the x and y intercept.

$\frac{h^{2}+1}{2h}=\frac{1}{2}\sqrt{ \left( \frac{h^{2}+1}{2h}\right)^{2}+(h^{2}+1)^{2}}$
$h=\frac{\sqrt{3}}{2}\iff P,Q: ~( \pm h,k) \implies \boxed{ \left( \pm \frac{\sqrt{3}}{2}, \frac{1}{4}\right) }$[/hide]


geometry
AMC
AMC 8
MATHCOUNTS
HMMT
AIME
blogs

I heard some high schools in the east have math teams, but my school doesn't have one, which is very disappointing. :(  So if any of you guys are actually in a math team, what do you guys do?  Are there any math competitions for teams?

Well, I'm on the math team at my school, but its not all that great.  We have "practices" once a week, during which we superficially cover topics like geometric progressions, algebraic manipulations, logs, etc.  However, I imagine this is because of the students at my school (highest AMC 10 score last year was a 60!).  We do go to a few competitions, which are fun, and significantly more challenging then "practices".  So, you (like me) would probably be bored out of your mind on my math team, but I'm sure other people have much better programs.

Well the thing is there usually aren't regular events for a math team to go to, so most schools wouldn't have an official team, although areas with a healthy number of math competitors would certainly have a group of people who regularly go to contests.

I have a middle school math club, organized by a high school student, at my school, but the level is quite low and we only have AMC8, MathCounts, and a couple local competitions we go to.

our high school has an organized math team
we meet once a week for about 90 minutes

we participate in a competition in which, once a month, we go to different schools and compete
there might be a total of 6 schools there
there are 6 rounds, 10 people on a scored team, each person takes 3 rounds
at the end, a team round is done in which the whole team does 6 (harder) problems in 12 minutes
the scores are added up and a winning school is decided

we also participate in the HMMT (i live on the east coast)

we also have NEML contests each month

there are also the AMC tests, with invitations to AIME and beyond if you do well

that's basically the contests we participate in

for our practices, we mostly do practice problems of the competitions and go over some more difficult concepts

Our high school has an organized math team.
Since our school is fairly large, we meet twice a week, once for the JV division and once for the Varsity division.
In our meetings, we usually go over tough/reappearing concepts and prepare for upcoming competitions.

I live in a fairly rich area (close to Atlanta, GA), so there is a lot going on. We participate in math competitions about once or twice a month. In most local competitions, we compete with about 10 other teams (except for the big ones, where you might find around 20-30 teams). We are even organizing our own math competition as I type (you can read about last year's on my blog on AoPS).

We also participate in the AMC tests, and some of us have gone on to AIME. Two people in our school have gone on to the next round, the USA Math Olympiad.

Now, all of this might dazzle you, but I go to a fairly rich school.

Thank you guys for the information.
And for Benjamin Hu and vallon 22, how many students are there in your schools?


MIT
college
CMU
Stanford
Carnegie Mellon

This is my first post in a while...

I figured that many of you are doing math majors...and Im planning on it so this seemed like the perfect place to post.  I wanna know about all of the stuff in a college thats not on paper.

First off....I would like to attend a laid back and liberal institution...thats already a high request

Second... Id like to either double major or major-minor in math and computer science ... and Id like a nice applied math program...

Third... no uptight schools or schools with incredulous demands....I just want a peaceful relaxing learning environment...

I have been considering Carnegie Mellon and UMICH...anyone have any other suggestions?

I am going to Carnegie Mellon in  :approx: 2 months for college.

[quote]First off....I would like to attend a laid back and liberal institution...thats already a high request [/quote]

Laid back?  I didn't get the impression that it was really hardcore where everyone was stressed out all the time (like MIT was when my Dad went there) but I would not describe it as laid back.

I don't know about being liberal-minded, but I don't think that there are many Republicans there or anything like that.

[quote]Second... Id like to either double major or major-minor in math and computer science ... and Id like a nice applied math program... [/quote]

Very good choice to study these things.

[quote]I wanna know about all of the stuff in a college thats not on paper. [/quote]

I will be the perfect person to ask this to...in 2 months.  :wink:

[quote]no uptight schools or schools with incredulous demands....[/quote]

I don't think it is uptight.  I don't think it has "incredulous demands" in terms of getting in there either.  What you want to do is apply to both MCS for math and SCS for CS.  SCS is very difficult to get into, but MCS is a little easier.  Once you get into CMU, you can probably work on getting into both if you don't get into SCS originally.  ED helps a lot.

You should read Yooper's post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=94366&highlight=#94366]here[/url]. He has some good advice related to your questions.

I would suggest Stanford, even though that's really hard to get into, but I think it fits your description exactly.  Also, maybe Brown University (Providence, RI) except for it does not have a really good math or CS dept. but it does have a good applied math dept.


[quote="anorithdialga"]find all reals x,y,z such that
\[ (1 \minus{} x)^2 \plus{} (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} z^2 \equal{} 1/4
\]
[/quote]

[hide="Solution"]

The eq is equivalent to

$ \frac{4}{3}(x\minus{}\frac{3}{4})^2\plus{}\frac{3}{2}(y\minus{}\frac{2}{3}x)^2\plus{}2(z\minus{}\frac{1}{2}y)^2\equal{}0$.

Since $ x,y,z$ are reals we see $ x\minus{}\frac{3}{4}\equal{}y\minus{}\frac{2}{3}x\equal{}z\minus{}\frac{1}{2}y\equal{}0$,then the solution is only

$ \boxed{x\equal{}\frac{3}{4},y\equal{}\frac{1}{2},z\equal{}\frac{1}{4}}$.

[/hide]


limit
function
abstract algebra
calculus
calculus computations

[quote="Salvador"]Show that for any positive integer m

$\mathop{\sum}_{n=1}^{\infty}\frac{m}{m^{2}+n^{2}}> \frac{1}{2}$[/quote]

We can write:
$\lim_{n\to \infty}\sum_{k=1}^{n}\frac{m}{m^{2}+k^{2}}$
Be $f(x)=arctg\frac{x}{m}$ and aplying Lagrange's theorem for the intervals $[k,k+1]$ we got $\exists c_{k}\in (k,k+1),f'(c_{k})=f(k+1)-f(k)$. But the function $f'(x)=\frac{m}{m^{2}+x^{2}}$ decreasing so $f'(k+1)<f'(c_{k})=f(k+1)-f(k)<f'(k)$, therfore
$\sum_{k=1}^{n}f'(k)>f(n+1)-f(1)$ so  $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{m}{m^{2}+k^{2}}\ge \frac{\pi}{2}-arctg\frac{1}{m}$


Find $ x \in \mathbb{C}$ if : $ (x \plus{} i)^{10} \equal{} i(x \minus{} i)^{10}$ , where $ i^2 \equal{} \minus{} 1$ .

$ \frac{x\minus{}i}{x\plus{}i}\equal{}\cos \frac{4n\plus{}1}{20}\pi \plus{}i\sin \frac{4n\plus{}1}{20}\pi \ (n\equal{}0,\ 1,\ 2,\ \cdots 9)$.


geometry
circumcircle
induction
trigonometry
inequalities
geometry unsolved

Prove that for every convex polygon there exist three consecutive vertixes $A,B$ and $C$, such that there is no point of that polygon outside a circumcircle of triangle $ABC$.

May be use the induction w.r.t. $n\in N,\ n\ge 4$, ( the number of the sides) !

Let $A_{1}A_{2}\ldots A_{n}$ be the convex polygon and consider the set $T$ of all triangles $A_{i}A_{i+1}A_{k}$ such that $\angle A_{i}A_{k}A_{i+1}<90^{\circ}$ (of course, $A_{n+1}=A_{1}$). The set $T$ is nonempty, since, for instance, one of the angles $\angle A_{2}A_{1}A_{3},\angle A_{2}A_{3}A_{1}$ must be less than $90^{\circ}.$ Take from $T$ the triangle whose circumradius is maximal. Assume that this is $A_{1}A_{2}A_{k}$ and let $R$ be its circumradius.

We claim that no vertex lies outside the circumcircle of $A_{1}A_{2}A_{k}.$ Indeed, suppose by way of contradiction that $A_{m}$ lies outside that circle. Then
\[
\angle A_{1}A_{m}A_{2}<\angle A_{1}A_{k}A_{2}<90^{\circ},
\]
so $A_{1}A_{m}A_{2}\in T$ and the sine law shows that its circumradius is greater than $R,$ a contradiction.

We claim that $A_{k}$ coincides with either $A_{3}$ or $A_{n}.$ Suppose, for instance, that $\angle A_{2}A_{1}A_{k}\leq\angle A_{2}A_{1}A_{k}.$ We prove that $A_{k}=A_{3}.$ Indeed, if $A_{k}\neq A_{3},$ since $A_{3}$ lies inside (or on) the circumcircle of $A_{1}A_{2}A_{k},$ we have $\angle A_{2}A_{3} A_{k}\geq180^{\circ}-\angle A_{2}A_{1}A_{k}>90^{\circ},$ thus $\angle A_{2}A_{k}A_{3}<90^{\circ},$ therefore $A_{2}A_{3}A_{k}\in T.$ The sine law (using the above inequalities) again yields that the circumradius of $A_{2}A_{3}A_{k}$ is greater than $R,$ a contradiction.

Thanks for the solution... i am going to read it later but i am sure that it is correct

So Enescu get's a CANDY again   :D  :D  :D

All I can see that this problem is not so easy, not for me, only tricky part is trigonometry because i don't know that stuf yet...

But thanks


In the attached pic, how do we get from our equations to the last line $ [H_3A]_{total} \equal{} 8,29*10^{ \minus{} 5} mol/dm^3$?

I did not cut out any steps in the solution; I just took screen shots of the original question and its solution. Source: 2008 icho preparatory problems


Let be $ (a_n)_{n\in \mathbb{N}^*}$ a sequence of real numbers such that $ a_1>1$ and $ a_{n\plus{}1}\minus{}a_n\ge 2\ ,\ (\forall)n\in \mathbb{N}^*$ . $ (\forall)n\ge 3$ let be $ S_n\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n$ and $ S_{n\plus{}1}\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n\plus{}a_{n\plus{}1}$ . Prove that in the interval $ [S_n;S_{n\plus{}1}]$ exists at least one perfect square .

[hide="Solution"]Suppose $ S_{n}>k^{2}$ for some positive integer $ k$.  Assume for contradiction that $ a_{n}<2k\minus{}1$.  But then $ S_{n}<(2k\minus{}1)\plus{}(2k\minus{}3)\plus{}\cdots \plus{}1\equal{}k^{2}$, a contradiction.  Therefore, $ a_{n}\ge 2k\minus{}1$, so $ a_{n\plus{}1}\ge 2k\plus{}1$, and $ S_{n\plus{}1}>k^{2}\plus{}2k\plus{}1\equal{}(k\plus{}1)^{2}$

We have $ a^{2}<S_{n}\le (a\plus{}1)^{2}$ for some positive integer $ a$.  By the above we have $ S_{n\plus{}1}>(a\plus{}1)^{2}$, so $ (a\plus{}1)^{2}$ is in $ [S_{n},S_{n\plus{}1}]$.[/hide]

[quote="alex2008"]Let be $ (a_n)_{n\in \mathbb{N}^*}$ a sequence of real numbers such that $ a_1 > 1$ and $ a_{n \plus{} 1} \minus{} a_n\ge 2\ ,\ (\forall)n\in \mathbb{N}^*$ . $ (\forall)n\ge 3$ let be $ S_n \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n$ and $ S_{n \plus{} 1} \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n \plus{} a_{n \plus{} 1}$ . Prove that in the interval $ [S_n;S_{n \plus{} 1}]$ exists at least one perfect square .[/quote]

This is reminiscent of USAMO 1994 #1 right?  The only thing is that it was open parentheses after S_(n+1)

Nice solution matt276eagles.


calculus
integration
function
calculus computations

hello. can someone please verify my answer :

question :

If  $ f$  is a differentiable function such that  $ f(x)$  is never  $ 0$  and
$ \int_{0}^{x} f(t) dt \equal{} [f(x)]^2$ for all  $ x$, find  $ f$.

I got  $ f(x)\equal{}\frac{x}{2}$

Up to a constant, that is correct.  Although, the problem says that $ f(x)$ is never 0 and $ f$ is 0 at $ x\equal{}0$.  Is it never zero on $ x>0$, or something like that?

oh yes, i suddenly noticed that  $ f(0) \equal{} 0$ in my solution.  so does that mean
$ f(x) \equal{} \frac{x}{2}$ is wrong ?  and how about $ \frac{x}{2}\plus{}C$ ?  I think we will be forced to have
$ C \equal{} 0$ if the conditions of the problem will be satisfied, ain't it ?   :?:

nope, no typo, that's what the book said.  i have attached here the scanned page.
it's number 3 in the problem list.

maths486, the problem assumes two conditions:
1. $ f$ is never zero
2. $ f$ is differentiable and satisfies $ \int_0^xf(t)\,dt \equal{} [f(x)]^2$ for all $ x$.

As stated, these two clearly contradict each other at $ x \equal{} 0$ since $ \int_0^0 f(t)\,dt \equal{} [f(0)]^2 \equal{} 0 \implies f(0) \equal{} 0$.
Your answer was not wrong, the problem itself was inconsistent with regard to $ x \equal{} 0$.

--

But if one tossed 1. and instead asked to find all differentiable $ f$ such that 2. is satisfied, we get that both $ f(x) \equal{} \frac {x}{2}$ for all $ x$ and $ f(x) \equal{} 0$ for all $ x$ are the possible solutions, since 2. is equivalent to solving the following differential equation with the given initial condition (which I believe is what you did to get $ \frac {x}{2}$):
\[ 2ff' \equal{} f, f(0) \equal{} 0\]


Prove that for any integer $ n>3$, $ \gcd(n,6)\equal{}1$ there exist three odd, distinct positive integers $ a,b,c$ so that $ \displaystyle\frac3n\equal{}\frac1a\plus{}\frac1b\plus{}\frac1c$.
[i]AMM[/i]

If anyone have the official solution of AMM for this problem, I would like to see it. Thanks...
Here is my solution :

We have $n\equiv 1,5 \mod{6}$, so we consider the two following cases.
Case 1: If $n\equiv 1 \mod{6}$, then there exist a positive integer $k$ such that $n=6k+1$. Now, let $a=2k+1$ et $b=4k+1$.Hence we have

\[ \frac{3}{n}=\frac{3}{6k+1}=\frac{1}{a}+\frac{1}{ab}+\frac{1}{bn} \]

Case 2 : If $n\equiv 5 \mod{6}$, then we have $n\equiv 2 \mod{3}$. Let $n=\sum_{i=0}^{\infty}a_i*3^{i}$ the ternary representation of n. Let $s \ge 1$ be the smallest integer such that $a_i=2$ for all $i < s$ and $a_s=t= 0$ or $1$. We have

\[ n\equiv 2+ 2.3 +.....+ 2.3^{s-1}+ t.3^{s}\mod{3^{s+1}} \]
Assume  $t=0$, then $ n\equiv  2.3^{s}-1 \mod{3^{s+1}}$ which imply  $ (3^{s}+1) n\equiv -1 \mod{3^{s+1}}$. Hence there exist an odd positive integer $a$ such that  $ (3^{s}+1)n +1=3^{s+1}a$.
Now, if we write  $b= 3^{s}a$, then
\[ \frac{3}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{bn} \]
Since $a$, $b$ and $nb$ are distinct odd, positive integers, then we have  the desired decomposition.
Now if  $t=1$ then $ n\equiv  -3^{s}-1 \mod{3^{s+1}}$. There exist an odd  positive integer  $a$ such that $n+1+3^{s}= 3^{s+1}a$. Now, if we write  $b= 3^{s}a$, then
\[ \frac{3}{n}=\frac{1}{an}+\frac{1}{b}+\frac{1}{bn} \]
Since $an$, $b$ and $nb$ are distinct odd, positive integers, then we have  obtained the desired decomposition.


Consider a grid of red and green points with 20 rows and 20 columns.
Two adjacent points (on the same line or column) are connected with a segment.
If 2 adjacent points have the same colour, the segment connecting them is also of that colour. If 2 adjacent points have different colours the segment connecting them is black.
Now, there are 219 red points in the grid, 39 of which are placed on the borders.
No red point is placed in one of the corners. Also there are 237 black segments.
Determine the number of green segments.

I think I have a proof, but I'm not sure abt it:

There are totally 760 segments. The red points on the edges have 3 segments going out from them and the other ones have 4 segms each. If we add these we get 3*39+4*180 and this is 2*r+b (r is the number of red segms and b the number of black segms), because the red segments are counted twice (having 2 red extremities) and the black ones counted once (they have a red extremity and a green one). This means that 2r+b=837, so 2r=600, so r=300, so r+b+g=760=300+237+g, so g=223.

(as you already imagined, g is the number of green segments).

Is it correct?

I think so, it's the same as my own solution.
When you see the solution, it looks very very easy.
But it took me a reasonable amount of time to get the point.
Nice one, isn't it ?


generaly $ det((cos^h(a_i \minus{} b_i))_{1\le i,j\le n}) \equal{} 0$ for $ h > n \plus{} 1$
here $ det(A) \equal{} 0$ if $ n > 5$

$ \left[\begin {array} {c} cos^{h}(\alpha_{1} \minus{} \beta_{j}) \\
cos^{h}(\alpha_{2} \minus{} \beta_{j}) \\
... \\
... \\
cos^{h}(\alpha_{n} \minus{} \beta_{j}) \end {array} \right ] \equal{} \sum_{k \equal{} 0}^{h}C_{h}^{k}cos^{k}(\beta_{j})sin^{h \minus{} k}(\beta_{j}) \left[\begin {array} {c} cos^{k}(\alpha_{1})sin^{h \minus{} k}(\alpha_1) \\
cos^{k}(\alpha_{2})sin^{h \minus{} k}(\alpha_2) \\
... \\
... \\
cos^{k}(\alpha_{n})sin^{h \minus{} k}(\alpha_n) \end {array} \right ]$
then $ dim(vect(\{C_j: \ j\in[|1,n|]\}))\le h \plus{} 1.$ where $ C_j \equal{} \left[\begin {array} {c} cos^{h}(\alpha_{1} \minus{} \beta_{j}) \\
cos^{h}(\alpha_{2} \minus{} \beta_{j}) \\
... \\
... \\
cos^{h}(\alpha_{n} \minus{} \beta_{j}) \end {array} \right ]$
that gives $ det(A) \equal{} 0$ for $ n > h \plus{} 1$

[b] Solution of yours is Very nice !!! [/b]  :roll:
Thanks! Thanks! Thanks! Thanks!

$ C_{k}^{h} or C_{h}^{k}$ ?
Edit to perfect !!!! :roll:


A quadrilateral inscribed in a circle has side lengths whose squares are 17, 99, 19, and 97. Find the area of the circle.

[hide="Hint 1"] $17+99 = 19+97$ [/hide]
[hide="Hint 2"] Hint 1 implies that one look for right triangles. [/hide]
[b]Follow-up:[/b]  Find the area of the quadrilateral.

well the area of the quad is just brahmagupta's formula

so you can form two right triangles and join the two equal hypotenuses, which forms the diameter of the circle. ok you constructed an appropriate cyclic quad but it is not unique (is it?) so there may be others that... etc

[quote="iamagenius"]ok you constructed an appropriate cyclic quad but it is not unique (is it?)[/quote]

Call the quadrilateral $ABCD$ and the center of the circle $O$.  The four triangles $ABO, BCO, CDO, DAO$ have invariant area dependent only on the relevant angles $\angle AOB, \angle BOC, \angle COD, \angle DOA$ (which in turn depend on the side lengths $AB, BC, CD, DA$) and the circumradius, so the area of the quadrilateral is invariant.


inequalities
trigonometry
function
geometry proposed
geometry

Let ABC is a acute triangle. Prove that:

$ \frac{tanA\plus{}tanB\plus{}tanC}{tan^2{\frac{A}{2}}\plus{}tan^2{\frac{B}{2}}\plus{}tan^2{\frac{C}{2}}} \ge \frac{1}{tan{\frac{A}{2}}\cdot tan{\frac{B}{2}}\cdot tan{\frac{C}{2}}}$

Let $ x\equal{}\tan\left(\frac{A}{2}\right)$, $ y\equal{}\tan\left(\frac{B}{2}\right)$, $ z\equal{}\tan\left(\frac{C}{2}\right)$, where $ 0<x,y,z<1$. By a well known identity we have $ xy\plus{}yz\plus{}zx\equal{}1$. After substituting $ \tan\left(A\right)\equal{} \frac{2x}{1\minus{}x^{2}}$ etc., the inequality becomes
$ \left(x\plus{}y\plus{}z\right)\left(\frac{2x}{1\minus{}x^{2}}\plus{} \frac{2y}{1\minus{}y^{2}}\plus{} \frac{2z}{1\minus{}z^{2}}\right)\geqslant x^{2}\plus{}y^{2}\plus{}z^{2}$.
It is easy to prove that the function $ f\left(t\right)\equal{} \frac{2t}{1\minus{}t^{2}}$ is convex for $ t\in\left(0;1\right)$, so by the Jensen inequality
$ \frac{2x}{1\minus{}x^{2}}\plus{} \frac{2y}{1\minus{}y^{2}}\plus{} \frac{2z}{1\minus{}z^{2}}\geqslant 3\cdot \frac{2\cdot \frac{x\plus{}y\plus{}z}{3}}{1\minus{}\left(\frac{x\plus{}y\plus{}z}{3}\right)^{2}}$.
Let $ s\equal{}\left(x\plus{}y\plus{}z\right)^{2}$. By the condition $ xy\plus{}yz\plus{}zx\equal{}1$ we have $ s\geqslant 3$. It is enough to prove
$ \left(x\plus{}y\plus{}z\right)\left(3\cdot \frac{2\cdot \frac{x\plus{}y\plus{}z}{3}}{1\minus{}\left(\frac{x\plus{}y\plus{}z}{3}\right)^{2}}\right)\geqslant x^{2}\plus{}y^{2}\plus{}z^{2}$
or
$ \frac{2s}{1\minus{}\frac{s}{9}}\geqslant s\minus{}2$
which is equivalent to
$ 2s\minus{}\left(1\minus{}\frac{s}{9}\right)\left(s\minus{}2\right)\geqslant 0$
or
$ s^{2}\plus{}7s\plus{}18\geqslant 0$
which is true for any $ s\in\mathbb{R}$.


If it is now 6 AM Tuesday, what day of the week will it be in
50,000 minutes?

$ 1 \ \text{day} \equal{} 24 \ \text{hours} \equal{} 24 \times 60 \ \text{minutes} \equal{} 1440 \ \text{minutes}$

So $ \left\lfloor \frac {5000}{1440} \right\rfloor \equal{} 3 \ [\text{days}]$
$ 5000 \minus{} 1440 \times 3 \equal{} 680 \ [\text{minutes}]$
$ \left\lfloor \frac {680}{60} \right\rfloor \equal{} 11 \ [\text{hours}]$
$ 680 \minus{} 60 \times 11 \equal{} 20 \ [\text{minutes}]$

Hence $ 5000 \ \text{minutes} \equal{} 3 \ \text{days} \ 11 \ \text{hours} \ 20 \ \text{minutes}$.

Thus, $ \text{AM} 6: 00 \ \text{Tuesday} \plus{} 5000 \ \text{minutes} \equal{} \text{PM} 5: 20 \ \boxed{\text{Friday}}$.

Just use the 50,000 minutes from the problem.  The solution technique above will yield Monday (very nearly Tuesday).

Using a calculator ...
50,000 minutes * (1 hour / 60 minutes) * (1 day / 24 hours) = 34.72222 days
To start from midnight, at the start of Tuesday, instead of 6am, add .25 days for the 6 hours, giving 34.972222 days.
35 days, or 5 weeks would bring us back to Tuesday at midnight.  50,000 minutes is just a bit short of 5*7*24*60.

If one memorized that a week is 10,080 minutes (= 7*1440) it would be easier to see that the 6 hours plus 50,000 minutes falls 40 minutes short of the 50,400 needed for 5 weeks.


Hi!  I happen to know that the number of ways to make change for  a dollar using any coins is 293.  However, the only solution I have heard of is listing them all out.  it seems like something that could be done with some slick generating functions or something... :maybe:
anybody have a good solution?

yeah, thats kinda what I thought it would be like.  is there any easy way to find the coefficient other than casework?

[url]http://www.cut-the-knot.org/ctk/GeneratingFunctions.shtml[/url]

Scroll down some and you'll see the dollar problem.


inequalities
blogs
inequalities unsolved

Given $ a,b,c\ge0.$Prove that:
$ a^2\plus{}b^2\plus{}c^2\plus{}\frac{\sqrt{3}.\sqrt[3]{abc}(ab\plus{}bc\plus{}ca)}{\sqrt{a^2\plus{}b^2\plus{}c^2}}\ge 2(ab\plus{}bc\plus{}ca)$

[quote="can_hang2007"][quote="leedt26"]Given $ a,b,c\ge0.$Prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {\sqrt {3}.\sqrt [3]{abc}(ab \plus{} bc \plus{} ca)}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}}\ge 2(ab \plus{} bc \plus{} ca)$[/quote]
See here: http://canhang2007.wordpress.com/2009/11/05/inequality-49-m-rozenberg/[/quote]
Very nice , can_hang. Thank you !

The first proof of can_hang is nicer than my proof (the second)  :lol:

I think it's quite easy.We can suppoes $ ab\plus{}bc\plus{}ca\equal{}1$,then it becomes
\[ a^2\plus{}b^2\plus{}c^2\plus{}\frac{\sqrt{3}\cdot\sqrt[3]{abc}}{\sqrt{a^2\plus{}b^2\plus{}c^2}}\geq 2\]
 (i) if $ a^2\plus{}b^2\plus{}c^2\geq 2$,it is obvious.
 (ii) if $ 1\leq a^2\plus{}b^2\plus{}c^2\leq 2$,then $ \sqrt{3}\leq a\plus{}b\plus{}c\leq 2$.
by Schur we have $ r\geq\frac{4p\minus{}p^3}{9}$,then it's enough to prove that
\[ p^2\minus{}2\plus{}\frac{\sqrt{3}\cdot\sqrt[3]{\frac{4p\minus{}p^3}{9}}}{\sqrt{p^2\minus{}2}}\geq 2\]
\[ \Longleftrightarrow (4\minus{}p^2)(p^2\minus{}2)^{\frac{3}{2}}[(4\minus{}p^2)^2(p^2\minus{}2)^{\frac{3}{2}}\minus{}\frac{\sqrt{3}}{3}p]\leq0\]
\[ \Longleftrightarrow (4\minus{}p^2)^4(p^2\minus{}2)^3\leq \frac{1}{3}p^2\]
By AM-GM we have
\[ (4\minus{}p^2)^4(p^2\minus{}2)^3\leq (4\minus{}p^2)\cdot\left(\frac{3(4\minus{}p^2)\plus{}3(p^2\minus{}2)}{6}\right)^6\equal{}4\minus{}p^2\leq \frac{1}{3}p^2\]
which is true because $ 3\leq p^2\leq 4$,where $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$.

 :( I always can't open the links given by Can.....

Hey tourish,since you can't open the link i just decided to copy it here so you can also see the solution :).By the AM-GM Inequality and the Cauchy-Schwarz Inequality, we have

$ \begin{aligned} \displaystyle \frac{\sqrt{3}\sqrt[3]{abc}(ab\plus{}bc\plus{}ca)}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}} &\geq \frac{3\sqrt{3}abc}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}\equal{}\frac{3abc\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}}{a^{2}\plus{}b^{2}\plus{}c^{2}} \\ &\geq \frac{3abc(a\plus{}b\plus{}c)}{a^{2}\plus{}b^{2}\plus{}c^{2}}. \end{aligned}$

Therefore, it suffices to prove that

$ \displaystyle a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\frac{3abc(a\plus{}b\plus{}c)}{a^{2}\plus{}b^{2}\plus{}c^{2}}\geq 2(ab\plus{}bc\plus{}ca).$

Using some simple computations, we can write this inequality as

$ \displaystyle \sum a^{4}\plus{}abc\sum a\geq \sum ab(a^{2}\plus{}b^{2})$,

which is true because it is the fourth degree Schur’s Inequality. Equality holds if and only if $ a\equal{}b\equal{}c, or a\equal{}b and c\equal{}0$, or any cyclic permutation.

[quote="Tourish"] :( I always can't open the links given by Can.....[/quote]
I am really sorry for this inconvenience, Tourish. Actually, I want to create my own blog about inequalities. This will be more convenient for me to edit posts and proofs because I know in Mathlinks, we can't edit our post after 2-3 days. So if you can't see my links, just tell me if you are interested in seeing my proofs. Otherwise, I think you can use the following website to open my blog: http://zend2.com

Regards


inequalities
trigonometry
inequalities unsolved

$ a\plus{}b\plus{}c\equal{}1,a,b,c>0$    prove that      $ a^2\plus{}b^2\plus{}c^2\plus{}2\sqrt{3abc}\le1$

[quote="sxgfly"]$ a \plus{} b \plus{} c \equal{} 1,a,b,c > 0$    prove that      $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2\sqrt {3abc}\le1$[/quote]

We can easily prove that

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\ge a^2bc \plus{} b^2ca \plus{} c^2ab \equal{} abc(a \plus{} b \plus{} c) \equal{} abc$.

So, $ 2abc(a \plus{} b \plus{} c) \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\ge2abc \plus{} abc \equal{} 3abc$.

So, $ (ab \plus{} bc \plus{} ca)^2\ge3abc$.

So, $ ab \plus{} bc \plus{} ca\ge\sqrt {3abc}$.

So, $ 2(ab \plus{} bc \plus{} ca)\ge2\sqrt {3abc}$.

If we add $ a^2 \plus{} b^2 \plus{} c^2$ on both sides, we are done because $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca) \equal{} (a \plus{} b \plus{} c)^2 \equal{} 1$.

Substitute $ x\equal{}\sqrt{\frac{ab}{c}}\ ,\ ...\Rightarrow b\equal{}xy\ ,\ ...$ and the inequality is equivalent with :

$ \sum_{cyc}x^2y^2\plus{}2\sqrt{3}xyz\le 1$ , with $ xy\plus{}yz\plus{}zx\equal{}1$ .

$ \Leftrightarrow (xy\plus{}yz\plus{}zx)^2\plus{}2\sqrt{3}xyz\le 1\plus{}2xyz(x\plus{}y\plus{}z)\Leftrightarrow x\plus{}y\plus{}z\ge \sqrt{3}$

And now using trigonometric substitution the inequality becomes $ \sum_{cyc}\tan \ \frac{\alpha}{2} \ge \sqrt{3}$ , which is well-known .

[quote="sxgfly"]$ a \plus{} b \plus{} c \equal{} 1,a,b,c > 0$    prove that      $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2\sqrt {3abc}\le1$[/quote]
$ \iff a^2 \plus{} b^2 \plus{} c^2 \plus{} 2\sqrt{3abc(a \plus{} b \plus{} c)}\le (a \plus{} b \plus{} c)^2$
$ 3abc(a \plus{} b \plus{} c)\leq (ab \plus{} bc \plus{} ca)^2$
that is obvious


Gauss
linear algebra
matrix
vector
calculus
induction
algebra

Rigorously prove that in order to solve a system with n distinct variables, you need n unique equations.
[hide="thoughts"]I don't know if this is particularly difficult or extremely trivial, but a friend of mine brought this point up and we both agreed that we had never seen a proof of this. Is this one of those fundamental properties or can it actually be proven with rigor?
Some of my thoughts as to how to solve it: trying to solve an arbitrary n-system with matrices (using Gauss-Jordan method), you see that you can't get an identity matrix on the left side. This may or may not work. It's not particularly rigourous, but maybe someone can build off of this. For some reason, I think that the Fundamental Theorem of Algebra may be related, but it's probably a stupid unrelated hunch. [/hide]

This is not true with real solutions, as posted before. Take $ (x \minus{} 1)^2 \plus{} (y \minus{} 2)^2 \equal{} 0$, the only real solution is (1,2) since all squares are at least 0.

By equations I assume you mean linear equations, and by unique I assume you mean linearly independent.

[hide]Basic idea without very much linear algebra: You know that n linearly independent equations have a solution.  So add in linearly independent equations until you have n of them which defines a solution, now change the last equation (just the value that they add too, not the coefficients), and you have a different solution.  So two solutions means that the equations you started off with didn't define a unique solution.[/hide]

Essentialy, if we have linear a system of equations:

$ ax\plus{}by\equal{}c$ and $ dx\plus{}ey\equal{}f$,

and the corresponding determinant of the coefficient matrix:

$ \begin{pmatrix}
a & b\\
d & e\\
\end{pmatrix}\ne0$

, then the system is linearly independent and a unique solution exists.

[quote="The Zuton Force"]I assume you're talking about linear systems, based on the talk of Gauss-Jordan matrices.

You want to explore the [url=http://en.wikipedia.org/wiki/Determinant]matrix determinant[/url]. Your condition, having n linearly independent, consistent equations, is equivalent to having a non-zero determinant. Now, consider [[Cramer's Rule]] -- what happens when the determinant of the coefficient matrix is zero?[/quote]

Thank you all for your meaningful posts.
Excuse my non-mathematical terms.
Based on some of the above posts, my question about the need for n-linear independent equations to solve for n-variables is the equivalent of proving that a system with n-equations is linearly independent and has a unique solution?
I know how to solve determinants and I have used Cramer's Rule, but I thought the determinant of a system with n variables and n-1 (for example) equations does not exist. I'm somewhat confused about the Zuton Force's last question. The determinant of the coefficient matrix is 0 when you have a non-linearly independent system, right? How does that prove my query?
*edit-I actually think I see where you're going with this. Correct me if I'm wrong--having a system with n-1 equations for n variables is the equivalent of having a system with n-equations for n variables but having the system not necessarily independent. For example, assume a system of three equations
$ 3x\plus{}2y\plus{}z\equal{}a$, $ 6x\plus{}4y\plus{}2x\equal{}b$, and $ x\plus{}y\plus{}z\equal{}c$, disregarding the constants. If you find the determinant of the coefficient matrix, you get 0, which is because the first two equations are equivalent. Thus if you were to solve for this system using Cramer's Rule, you would have an undefined solution, due to division by 0. Is this sufficient to prove that it is necessary to have an nxn system?
I think I'm understanding this. Thanks!

A linear system with more variables than (linearly independent) equations is called an [b]underdetermined[/b] system.  What you're really asking is to prove that an undetermined system cannot have a unique solution.

[b]Basic idea:[/b]  If you use Gaussian elimination, you'll either run into a contradiction or you'll be able to express all the other variables in terms of one (or more).  That means you can set them to any value you want.

[b]Demonstration (with linear algebra):[/b]  An undetermined system has an $ m \times n$ matrix (call it $ \mathbf{A}$) where $ m < n$.  Its [url=http://en.wikipedia.org/wiki/Rank_%28matrix_theory%29]rank[/url] is at most $ m$, so by the [url=http://en.wikipedia.org/wiki/Rank-nullity_theorem]rank-nullity theorem[/url] its nullity (the dimension of its [url=http://en.wikipedia.org/wiki/Kernel_%28matrix%29]nullspace[/url], the set of solutions to $ \mathbf{Ax} \equal{} 0$) is positive.

What this means is that if $ \mathbf{Ax} \equal{} \mathbf{b}$ has a nontrivial solution $ \mathbf{x}_0$, then any vector $ \mathbf{v}$ in the nullspace of $ \mathbf{A}$ generates nontrivial solutions $ \mathbf{x}_0 \plus{} \mathbf{v}$ to the original equation.  So either

- there aren't any solutions at all (if the equations contradict), or
- there are an infinite number of solutions.

Another way of saying this is the row reduced echelon form of the coefficient matrix cannot have all leading ones since there are more variables than equations. So there are free variables. Once you have one solution, you have an infinity of solutions.

[quote="Altheman"]All of these questions can be answered by cracking open a linear algebra book.[/quote]

I'm not that far ahead in math. I am planning to start calculus in June. :D
I didn't know this question required math that advanced (that is for the simplest, most clear solution).

I think that you can learn a fair amount of linear algebra before calculus. Granted, my linear algebra class is not really college level, but pretty much of the stuff is elementary. I suppose that you get into a lot of proofs though...

[quote="t0rajir0u"][quote="mihail911"]I didn't know this question required math that advanced (that is for the simplest, most clear solution).[/quote]

The basic idea is not that "advanced"; we're reasoning about the possible outcomes of Gaussian elimination.  The very simplest (and possibly the clearest) proof I can think of is by induction.

[b]Inductive hypothesis:[/b]  A system of $ n$ (linearly independent) equations in $ n \plus{} 1$ variables has either no solution or infinite solutions.

[b]Base case:[/b]  $ n \equal{} 1$.  A single equation in two variables defines a line, which is clearly an infinite number of solutions.

[b]Inductive step:[/b]  Suppose every system of $ k$ linearly independent equations in $ k \plus{} 1$ variables has either no solution or infinite solutions.  Suppose by contradiction that there exists a system of $ k \plus{} 1$ linearly independent equations in $ k \plus{} 2$ variables with a unique solution.

Take any of those equations, isolate some variable $ x$, and plug it into the other equations.  Now we have a system of $ k$ linearly independent equations in $ k \plus{} 1$ variables, which either has no solution or infinite solutions.  In the former case there are no solutions to our original system; in the latter there are an infinite number of solutions.  Contradiction.

Here we're assuming that a system of linear equations has no solution, one solution, or an infinite number of solutions.  Do you accept this or should this be proven too?

Edit:  The reasoning you wrote about a system of $ n$ not necessarily independent equations in $ n$ variables is correct and leads to another proof.  In other words, given less than $ n$ equations, we can add extra copies of any of the equations to give another system with more equations equivalent to the first.  The reason I gave the demonstration using more "advanced" concepts from linear algebra is that it lets us quantify the geometric nature of our infinite solutions.[/quote]

thanks for the help torajiru!
in your proof, was the induction really necessary? the majority of your proof utilized contradiction. The only reason i see you did this is by repetition of substituting and isolating and you would normally solve a system, you will eventually end up with a single equation in two variables which as you mentioned in your [b]inductive[/b] step, results in a line, with an infinite number of solutions.

"Repetition of substitution and isolation" is easiest to make rigorous inductively (it's clearly not rigorous as is).  At any step, we might run out of variables and instead get an equation like $ 13 \equal{} 21$ (which corresponds to the "no solutions" case).

An equivalent formulation would be something like "let $ k$ be the minimum positive integer such that a system of $ k$ equations in $ k \plus{} 1$ variables exists with a unique solution...."  This is well-known; it is referred to as the equivalence of induction and [url=http://en.wikipedia.org/wiki/Well-ordering_principle]the well-ordering principle[/url].

[quote="mihail911"]I'm not that far ahead in math. I am planning to start calculus in June. :D
I didn't know this question required math that advanced (that is for the simplest, most clear solution).[/quote]
Linear algebra, as mentioned, doesn't really require calculus to learn, nor is it terribly advanced. After precalculus or so you have the tools to tackle many undergraduate level disciplines such as linear algebra. Just because you're "not that far ahead in math" doesn't mean you can't figure it out.

If you're seriously interested in finding out all of the answers to this stuff (and much more) you should find a book or some other resource on the subject.


I am so excited for arml :clap: :icecream: :w00t:. Get to meet new people :)

Hmm, I have not prepared at all though....well I did a few practice problems in school, then lost the packet of arml problems, so I can't check answers. Darn.

Weird, since I happen to have this huge arml book with past problems and solutions (yea I am not sure I think many people had to practice for arml with knowing the right answers like me last year), for some years.

don't you guys leave tommorow?
either way GOOD LUCK!!

hmmmmm, i thought alabama arml was canceled...

Thanks chestnut. Hopefully I shall get more than half the questions right on individual round, unlike last year  :lol:

Well, yea, it was canceled, but then Grissom decided to make the trip, and I managed to get onto their team.

Also, hopefully my relay team can get at least one question right. (to not regress from last year's performance)

Yay, I got a 7. I could have done better, and could have done much worse. On round 3, I kept messing up the easy log problem, so that left me no time to do number 6, so I missed both problems. However, I probably wouldn't have gotten 6 right even if I had more time (I think I could get it right only if I had to write up a proof for it, or otherwise I would drown in a sea of holes in my rigor).

argh, I can't believe I actually tried triple angle formula for tangent for 10, I should have known to try complex numbers or something, since that is quite common.

So, considering time constraints and my slowness on certain problems, I could get 8 or 9 on a better day? But I also could have done a lot worse I guess on a worse day.

So yea, I am pretty content with my 7. I am also pretty content that my relay team got at least one question right, since there were only 2 problems (and I had thought there were more). I am also pretty happy that I was productive in team round and solved more than 0 problems (3 problems in fact). I get the feeling that this will be my best year at arml.

Dang I can't believe a 6th grader won arml..... Then again, I think Alex Song has made usamo for the last year or two.

We also got first place division B at the UGA site :)

Most importantly, I met my goal "meeting ppl", like Nick, Ash, Sid, the other Grissom ppl, and some other aopsers. :)

Okay, let me just say right now that Xinkes signature goal of being nicer to people is totally ridiculous because he was about the nicest person I've ever known this weekend. he was probably either 1st or 2nd when it came to team productiveness, too.

Disclaimer: The dorms were crap. One sheet, a cup (?) and a towel. Oh, and a mattress. We had a sink, no running water though.
I got a 4 and I was beating everyone (yes, even Xinke!) after the first two pairs of individual problems, and i think the high schoolers were surprised that I was not an idiot. Then I made stupid mistakes. Should have gotten a 6 or 7, but thats life.
I was an alternate even though I got 5th or 6th of the 15 Grissom people. Fail.
The sign on the dorms said "AMLR" instead of "ARML". Fail.
Got third in speed Scrabble. Fail.
Yeah ARML was really fun. I especially liked walking around downtown eating a ton of food and chilling with the team, and late night poker and speed scrabble. I hope Alabama sends lots of teams next year cuz then I can meet even more people and get connections. Mwahahah.

your signature turned out to be more ridiculous. My signature was related to some incidents last year...but I'll change it

[hide="dynamo dont read this"]Actually I was tied with you after 2 rounds, in fact, 3 rounds[/hide]

Yea the senority thing is kind of ridiculous. Last year, the 3 freshman not on the A team (Eugene, Owen, Kyle) took this test on the Friday before ARML to see who was going to be the one alternate out of the 31 people. I was thinking "What? They are clearly probably better than some of the sophomores." And guess what, the freshman average was higher than the A team average... (I got 4, Kyle got 5, Owen got 3, Eugene got 3).

Speed scrabble was fun.

The dorms were better than last year, believe it or not. Last year, it was the same, except we didn't even have a sink, nor 2 study tables with lights built into the wall, nor soap, nor cup, and the showers were way better this year. Also, believe it or not, the room we practiced in on Friday was better than last year. Dynamo, did you finish all of your fries on Friday night?

On power round, 4 of the team members just gave up after 10 minutes and started chatting and drawing random cartoons for our answer on 9 and 10. So yea, dynamo should have been on the team.

who takes and drives you guys?

hmm... i wanna do it next year! although i'll probably fail the test...

Next year, mostly likely (well hopefully), we will have a state team and send 2 or 3 teams, so we will take a (hopefully charter) bus like last year. This year, Grissom parents drove, and my parents drove me

Last year there were um 2 or 3 or 4 parent chaperones. A total of 5 (I think) Grissom parents drove the students this year. (yes the grissom students carpooled since there's no point in everyone driving).

OH MAN we beat number 2 division B at our site by 1 point.

Oh and yea ash and sid were pretty dif

For the purposes of my post #5 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280090]here[/url], I treated you guys as the continuation of the Alabama team, and if the state team is restored next year, I'll let "Grissom" be part of the state team's history.

Out of curiosity, why did the Alabama team not compete this year?

Well, we had someone who thought that they would be the state coach and kinda organize it, but in the end they weren't able to do it, and asked if anyone else wanted to step up and be the state coach, but no one did. Grissom decided to send a team as a school, but one of our competitors (hurdler) doesnt go to Grissom.

[quote="dynamo729"]Yeah ARML was really fun. I especially liked walking around downtown eating a ton of food and chilling with the team, and late night poker and speed scrabble. I hope Alabama sends lots of teams next year cuz then I can meet even more people and get connections. Mwahahah.[/quote]

Yeah, ARML is always the most chill thing we did in math team. Ridiculous that Vestavia didn't send anyone.

whats up alabama youngsters. I don't think I'm a mod here anymore...

[quote="yaofan"]do your parents [b]HAVE[/b] to come?[/quote]

No, your parents don't [b]HAVE[/b] to come?


geometry
geometric transformation
rotation
vector

If an object rotates at a rate of  $ \theta_x$ per second about the $ x$ axis, $ \theta_y$ per second  about the $ y$ axis, and $ \theta_z$  per second about the $ z$ axis, is it possible to 'add' all of these rotations together and find an overall axis and rate of rotation about that axis? How would i calculate it?

Thankyou

It is possible, but not pretty without doing some tricky stuff. The best way (as far as I know), is to represent the rotations as quaternions, then multiply together the quaternions. There might be a better way though.

I just had an idea. If we represented a rotation with a vector pointing in the axis of rotation, and its magnitude is the rate of rotation, to to sum multiple rotations we just add the vectors yeah?

Rotation $ \theta_x$ along $ x$ axis  -> $ \theta_x \left( \begin{array}{c} 1 \\0\\0 \end{array} \right) \equal{}  \left( \begin{array}{c} \theta_x \\0\\0 \end{array} \right)$

The same applies to the $ x$ and $ y$ axis.

Adding them we get $ \left( \begin{array}{c} \theta_x \\0\\0 \end{array} \right) \plus{} \left( \begin{array}{c} 0 \\\theta_y\\0 \end{array} \right) \plus{} \left( \begin{array}{c} 0 \\0\\\theta_z \end{array} \right) \equal{} \left( \begin{array}{c} \theta_x \\\theta_y\\\theta_z\end{array} \right) \equal{} \omega$

The rate of rotation = $ |\omega| \equal{} \sqrt{\theta_x^2 \plus{} \theta_y^2 \plus{} \theta_z^2}$.
And the unit vector in the direction of the axis = $ \frac{1}{|\omega|}\left( \begin{array}{c} \theta_x \\\theta_y\\\theta_z\end{array} \right)$.

I think this is correct. Anyone see anything wrong?

Thankyou

Sounds reasonable.  I think it's a concept from physics, where you denote the rotation of a spinning electron by a vector in the same way as you said.


calculus
integration
logarithms
real analysis
real analysis unsolved


prove that x(x+1)(x+2)(x+3)+1 is a square with the most "elegant" way :P

Personnaly, i've found a method that is quite nice :
[hide]
We know that x(x+2) +1 is a square.(1)
We have x(x+1)(x+2)(x+3)+1=(x²+3x)(x²+3x+2)+1(2)
From (1) and (2) we have the proof
[/hide]

$ A\equal{}x(x\plus{}1)(x\plus{}2)(x\plus{}3)\plus{}1\equal{}(x^2\plus{}3x)(x^2\plus{}3x\plus{}2)\plus{}1$

Let $ y\equal{}x^2\plus{}3x$

thus $ A\equal{}y(y\plus{}2)\plus{}1\equal{}y^2\plus{}2y\plus{}1\equal{}(y\plus{}1)^2\equal{}(x^2\plus{}3x\plus{}1)^2$


If f(x) is a fifth degree polynomial and f(0)=3, f(1)=5, f(2)=6, f(3)=10, f(4)=11, and f(5)=12, find f(6).

I'm looking for an efficient way to do this (I mean, besides plugging in all the values and getting a big system of equations).

Write $ f(x)$ in terms of [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]Newton polynomials[/url].

It is pretty easy to write out a difference table.

That approach is equivalent to writing $ f(x)$ in terms of Newton polynomials, but the end result when the problem is explicitly cast in terms of Newton polynomials is the exact formula of $ f(x)$.

[b]N.B.[/b]  The approach by Newton polynomials generalizes to the arbitrary interpolation problem; that is, if you are given $ n\plus{}1$ arbitrary values of an $ n$-degree polynomial, write $ f(x) \equal{} a_0 \plus{} a_1 (x \minus{} x_1) \plus{} a_2 (x \minus{} x_1)(x \minus{} x_2) \plus{} a_3 (x \minus{} x_1)(x \minus{} x_2)(x \minus{} x_3) \plus{} ...$ where the given values are $ (x_1, y_1), (x_2, y_2), ... (x_{n\plus{}1}, y_{n\plus{}1})$.


symmetry
algebra
polynomial
quadratics
algebra proposed

Let $ a>b>c$ be $ 3$ roots of the equation $ x^3\minus{}3x\plus{}1\equal{}0$.Compute $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$.

[quote="KDS"]Let $ a > b > c$ be $ 3$ roots of the equation $ x^3 \minus{} 3x \plus{} 1 \equal{} 0$.Compute $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$.[/quote]

$ a\plus{}b\plus{}c\equal{}0$
$ ab\plus{}ac\plus{}bc\equal{}\minus{}3$
$ abc\equal{}\minus{}1$

http://www.mathlinks.ro/viewtopic.php?t=309978 -->

$ b^{2}\minus{}c\equal{}a^{2}\minus{}b\equal{}c^{2}\minus{}a\equal{}2$

$ a^{2}\equal{}b\plus{}2$
$ b^{2}\equal{}c\plus{}2$
$ c^{2}\equal{}a\plus{}2$

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\equal{}\frac{a^{2}c\plus{}b^{2}a\plus{}c^{2}b}{abc}\equal{}\frac{(b\plus{}2)c\plus{}(c\plus{}2)a\plus{}(a\plus{}2)b}{abc}\equal{}\frac{ab\plus{}ac\plus{}bc\plus{}2(a\plus{}b\plus{}c)}{abc}\equal{}\frac{\minus{}3\plus{}2*0}{\minus{}1}\equal{}3$

With this cyclic problem, we can "[i]symmetricalize[/i]" it by putting $ A: \equal{}\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$ and $ B: \equal{}\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c}$, then $ A\plus{}B$ and $ AB$ becomes symmetry with respect to $ a,b,c$, now Viete's symmetry polynomials work. Moreover $ a>b>c$ so $ A>B$, and the greater solution of the quadratic equation $ t^2 \minus{} (A\plus{}B)t \plus{} AB \equal{}0$ is equal to $ A$.

[quote="Dimitris X"][quote="Mashimaru"] Moreover $ a > b > c$ so $ A > B$[/quote]
How???
Can you prove it please???[/quote]

I am sorry, this is a typo :P  It should be:

$ A \minus{} B \equal{} \frac {(a \minus{} b)(b \minus{} c)(c \minus{} a)}{abc} < 0 \Rightarrow A < B$.

Nope i think $ A>B$.....i misread it before.....

Let me now post the full solution by the brilliant mashimaru's method.

First of all we can easily see that:

$ a\plus{}b\plus{}c\equal{}0$
$ ab\plus{}bc\plus{}ca\equal{}\minus{}3$
$ abc\equal{}\minus{}1$.

Let $ A\equal{}\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$ and $ B\equal{}\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c}$.

$ A\plus{}B\equal{}\sum\frac{a\plus{}b}{c}\equal{}\frac{\sum_{sym}a^2b}{abc}\equal{}\minus{}[(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}3abc]\equal{}\minus{}3$.

So $ A\plus{}B\equal{}\minus{}3$ $ \boxed{1}$.

$ AB\equal{}3\plus{}\sum \frac{a^2}{bc}\plus{}\sum \frac{ab}{c^2}\equal{}3\plus{}\frac{abc(a^3\plus{}b^3\plus{}c^3)\plus{}a^3b^3\plus{}b^3c^3\plus{}c^3a^3}{a^2b^2c^2}\equal{}3\minus{}[(a\plus{}b\plus{}c)^3\minus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)]\plus{}(ab\plus{}bc\plus{}ca)^3\minus{}3(ab\plus{}bc)(bc\plus{}ca)(ca\plus{}ab)\equal{}3\plus{}6(a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}27\equal{}3\plus{}6[(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}abc]\minus{}27\equal{}3\plus{}6\minus{}27\equal{}\minus{}18$

So $ AB\equal{}\minus{}18$.

$ A,B$ are the solutions of the quadric formula:

$ x^2\minus{}(A\plus{}B)x\plus{}AB\equal{}0 \Longleftrightarrow x^2\plus{}3x\minus{}18\equal{}0 \Longleftrightarrow (x\plus{}6)(x\minus{}3)\equal{}0$.

So $ (A,B)\equal{}(3,\minus{}6)$ or $ (A,B)\equal{}(\minus{}6,3)$.

BUT $ A\minus{}B\equal{}(a\minus{}b)(a\minus{}c)(b\minus{}c)>0$, so $ A\equal{}3$.....

Nice "symmetrization".  :)
Also note that $ (a,b,c)\equal{}(2\cos20,2\cos100,2\cos140)$. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=445583]here[/url] post #44-46.  :P


[quote="brady00n"]Factor this in integers as much as possible.

$ a^{3}-2a-1=0$

This is part of an AIME problem (problem #9, 1997).  Seems like the factorization's hard to see...are there general methods for factoring stuff like this?[/quote]

Well, there is a general formula, but it is so complicated that few people bother to do it.

[hide="You can rather do this..."]Guess a root. Actually, you can use the Rational Roots Theorem to find that the only possible [b]rational[/b] roots (not all roots) are $ \pm 1$. Plugging in $ -1$, we see that $ a=-1$ is a solution. Thus, by the Factor Theorem, we see that $ a+1$ must be a factor. So, we can divide through using synthetic division:
$ a^{3}-2a-1 = (a+1)(a^{2}-a-1)$

Since you wanted to factor in integers, this is the farthest we can go.[/hide]

[quote="brady00n"]Factor this in integers as much as possible.

$ a^{3}-2a-1=0$

This is part of an AIME problem (problem #9, 1997).  Seems like the factorization's hard to see...are there general methods for factoring stuff like this?[/quote]

If $ a=-1$ then $ (-1)^{3}-2(-1)-1 =-1+2-1 =0$ so $ a+1$ is a factor.


$ H$ is the orthocentric of triangle $ ABC$.the incircle with center $ I$ is tangent to $ AB,AC$ at $ E,F$.if $ H,F,E$ are lies on a line prove that $ HI$ pass fromthe midpoint of $ BC$

its realy easy.is there anybody that have a strange or very nice solution


vector
quadratics
function
analytic geometry
advanced fields
advanced fields open

Hey!

I'm currently working on a small vector graphics rasterizer to try out some different techniques. Basically, what I do is to take a spline outline consiting of straight lines and quadratic bezier curves, and fill that outline. I'm not satisfied with the performance and accuracy of my current scanline rasterizer...

The new approach is more similar to how triangles in "regular 3d graphics" are rasterized. I evaluate if pixels lie within a triangle by using the edge functions of the three edges (implicit line ax + by + c) and checking the sign.

But there are alot of bezier curves in the outline... It takes some time to convert the bezier curves into line segments (and then to proceed as described above).

So my question is, is there any way to rewrite a bezier curve into something that resembles the edge function? I can't seem to figure it out quite...

I've tried to rewrite a curve on the form ax + by + cx^2 + dx^2 + e (+ 0*xy) = 0, but it gets very messy, with roots and stuff which is not really cpu-friendly. I also tried converting the pixel coordinates to be evaluated into barycentric coordinates, but this also gets a bit messy, and I can't get rid of the divisions (two, one per coordinate)...

I'm very thankful for any help, or even just a hint in any direction, because I fell a bit stuck :/

I think I need a break and some coffee :)

Thanks in advance!

/Magnus
[color=olive][reformatted by a moderator][/color]

Ok, perhaps I expressed myself in a too complicated way :)

What I basically want to do is:
For any given point, determine on which side of a quadratic bezier curve that point is.

Secondarily I would like to know if it anyone knows if you can write a (all) quadratic bezier curve on the form, (and how to get it):

Ax + By + Cx^2 + Dy^2 + E = 0

I havn't found any info on the net regarding this :/

Thanks!
/Magnus


Apologies if i'm posting this in the wrong section. I have the following question which i was hoping to get some help with. The question is .

"Let A be a set. Prove that there is no onto function f:A -> P(A)"

Thanks in advance.

We can use a diagonal argument here, can't we?

That is, given a mapping $A \to P(A)$ we can construct an element of $P(A)$ to which no $A$ maps.

Consider $A_{F}=\left\{x\in A\ | \ x\not\in f\left(x\right)\right\}\subset P\left(A\right)$.  Assume that $f: A\rightarrow P\left(A\right)$ is onto.  This means that there exists $y\in A$ such that $f\left(y\right)=A_{F}$.  If $y\in A_{F}$, then $y\not\in f\left(y\right)=A_{F}$, so $y\not\in A_{F}$. But by the definition of $A_{F}$, $y\not\in A_{F}\implies y\in f\left(y\right)=A_{F}$, which is a contradiction.  Therefore there doesn't exist such a function.

Also, it should be clear that an onto function doesn't exist when $A$ is finite, as $\left|A\right|=n$ and $\left|P\left(A\right)\right|=2^{n}$.  The proof above is needed when dealing with $\left|A\right|=\infty$.  Since we have shown that an infinite set has a smaller size than its power set, we have shown that there are different sizes of infinity.


i started to read [url=http://www.maa.org/pubs/Calc_articles/ma063.pdf]this article[/url] explaining through elementary methods
how one derives Green's formula but i got stuck along the way on understanding it
at page 6 where Mean Value Theorem is said to be used.
i don't understand how exactly it is beeing used.
if someone is kind enough to explain i will remain forever indebted.
thank you

We have that $y \left( t_{i}\right)-y \left( t_{i-1}\right) = \left( t_{i}-t_{i-1}\right) \cdot y^\prime \left( v_{i}\right)$ for some $v_{i}\in \left[ t_{i-1}, t_{i}\right]$. [url=http://en.wikipedia.org/wiki/Mean_value_theorem]What isn't clear?[/url]

that's the way i knew mean value theorem also lol,
it  seems i didn't pay enogh attention as i thought mean value theorem
was used exactly after the figure,but as you denote it seems to be used
in the 2nd formula after the figure.
:)
now i'm happy
thank you
next time i'll try to be more carefull and focus better


abstract algebra
number theory
superior algebra
superior algebra unsolved

[quote="yannick"]What are all groups with 3 conjugacy classes? So far, I only managed to figure out that such a group needs to have $6$ elements. So it is probably $\mathcal{S}_{3}$???
However, I did not manage to prove this. Can someone do this for me or correct me if such a group doesn't necessarily has $6$ elements.

thanks in advance[/quote]

I figured out that $Cl(e)=\{e\}$ so you have to divide $(n-1)$ into two classes, one with $k>0$ elements one with $(n-1-k)$ elements. Then I said that for one particular $k$ we need to have that $k|n$ and $(n-k-1)|n$ because $|G|=|Cl(x)\parallel C(x)|$ for all $x \in G$.

But that seemed to be a number theory problem which I could not solve. If you take $n=6$ and $k=2$, it appaers to be true.

That number theory is really pretty easy. We have $k|n,m|n,k+m=n-1$. WLOG, $k\ge m$. Since $k<n$, $k\le \frac{n}{2}$. Then $m\ge n-1-\frac{n}{2}=\frac{n}{2}-1$ and $m\le\frac{k+m}2=\frac{n-1}2$.
If $n$ is even, we must have $m=\frac{n}{2}-1$ and $\frac{n}{2}-1|n$. If $n$ is odd, we must have $m=\frac{n-1}2$ and $\frac{n-1}2|n$.
In the first case, $\frac{n}{2}-1|n-2(\frac{n}{2}-1)$, so $\frac{n}{2}-1|2$ and $n=4$ or $n=6$. In the second case, $\frac{n-1}2|n-2\frac{n-1}2=1$ and $\frac{n-1}2=1$, so $n=3$.

Checking all five groups of order 3,4, and 6, the solutions are $S_{3}$ and the cylic group of order 3. The abelian groups of order 4  and 6 have 4 and 6 conjugacy classes respectively.


geometry
AMC
USA(J)MO
USAMO
AIME
summer program
MathPath

I'm not yet sure whether I have enough stuffs in my brain to do this but I think many people in here should be advance to intermediate level and leave the getting started forum.  So, even I'm not moderator or administrator, I'll soon post the problems that deals in variety area of math that I know that can help people to decide on THEIR OWN for advancing to next level or not.

Questions will be posted soon so please wait.

Thank You

Silverfalcon

That's all fine but how does someone really advance to the next level in terms of the forums?  I mean they still have access to this one.  Also the problems in each forum are posted by the users so the difficulty of a forum depends on what ppl post.  The only thing that you can really do is tell them to stop answering challenge questions.  I also don't think that you should label a level to someone because they might be in between.  I understand the forum levels to be tentative and open to range of ppl.  Yet this doesn't mean that someone at USAMO level should be answering challenge questions from the beginner forum.  I understand where you are coming from but I am just pointing out the difficulty in doing all of that.

tokenadult wrote:Here in the United States, the television talk show host and humorist David Letterman has used a top-ten list format for years to make jokes about various subjects. With his example in mind, I'd like to share a top-ten list about how you can know you're not a getting started level mathlete anymore.

Top Ten Signs You're Not Just Getting Started in Math Problem-Solving Anymore

10. You've qualified for AIME twice.

9. You've aced the AMC 8.

8. You've been to MathPath (or MathCamp).

7. You've had two commended problems in USAMTS.

6. You've gone on to nationals in Mathcounts.

5. You've qualified for USAMO (or the equivalent, in your country).

4. You've been to MOP (or the equivalent, in your country).

3. You think the problems on AoPS Intermediate Forum are awfully easy.

2. You've been a member of your country's IMO team.

And the number-one sign you're not just getting started in math problem-solving anymore is

(in spoiler, to build suspense)

[hide]
1. You're a co-author of the book series The Art of Problem Solving. [/hide]

I qualified to be at Getting Started Level.

Only qualified once for AIME, I'm still stuck here :-(

I personally think beta is smart enough to go to further level as well as joml88 but I didn't been in Intermediate level neither but here are the questions I made people to see, they aren't very hard but hope to do some good (please! Don't post answers in here but PM to me, okay?)

1. [5^(4+m)/5^(3+m) + ( :sqrt: 900/  :sqrt: 400) * 2] !

Give solution if m is only positive integer.

2. 3 log x 6 = 4

3. a  n = 6n; n = 25

Define nth partial sum, Sn, of each arithmetic sequences.

4.  What is the first larger number that comes after 111, 111, 111 and is divisible by 2, 3, 6, and 9?

5. Compute: (n/2   +   1)^7

6.  Suppose two 30-60-90 degree triangles are added and formed a equailateral triangle.  Height is 71  :sqrt:   3.  Find the area of new equilateral triangle.

7. Two chords with 40 cm each are drawn and put into circle, facing each other.  If another chord is drawn from each one end point from first two chords diagonally with middle point as center, and is 41 cm long, find the distance from center to either of first two chords.

8. (1+2+3+......n) :^3: /(n-3-2+10) :^3: = 16 :^2:  - 40

Those questions may not be challenging at all because I myself is not a such high mathematician yet.  But it is worthwhile to do those problems so enjoy!

Edited: It should been 16^2 - 40 at first time but I accidentally put -50. Sorry.

Question:  If I should go to intermediate, which consequently means that everyone as good as me or better should go to intermediate, who would be left in the beginner forum?  I mean there aren't many ppl in this forum who know less than me or whatever.  Also, I periodically look in the intermediate and it is way to hard for me.  If you start "sending" ppl up to the intermediate it will make that forum too beginner-like.  I really suggest that you look at som e of the topics in that forum and maybe you will understand why I don't think that ppl should be sent to there as fast as you want them to be.

You misunderstood concept of what I said.

I said let THEY decide, not me.

[quote="beta"][quote="tokenadult"]Top Ten Signs You're Not Just Getting Started in Math Problem-Solving Anymore [details omitted] [/quote]
I qualified to be at Getting Started Level.[/quote]

Thanks for bringing the Top Ten List to people's attention. As is usual with the top-ten list format, all of those examples are a little bit over the top. In other words, to my mind even the easiest of those accomplishments (and there was some debate about whether they are correctly ordered in my list) is a sign someone is CLEARLY not just getting started anymore. How long a person stays at the Getting Started level depends, of course, on how much opportunity that person has to practice math competition problems and to accelerate in curricular math and to get mathematical enrichment from outside activities.

I made a joking reference to JBL and to TripleM once when I mentioned people who shouldn't be answering challenge problems on this forum. They are, in fact, very helpful in answering informational questions and we're glad to have them here. They are just some of the DOZENS of people here who can eat me for breakfast when it comes to mathematical problem-solving. Chinaboy and Rep123Max in their moderation of this forum, and I in mine since I became a co-moderator, have been pretty tolerant of people answering the challenge problems they feel like answering, as long as they attempt to put the answers in spoiler text (Question: does latex stay in spoiler, or does it "break through" and become visible?) and show solutions as well as numerical answers. Anyone is always welcome to answer a challenge question that has already been answered--I encourage that, and hope a lot of you will post YOUR solutions to challenge problems (or questions about other people's solutions) even after someone else has posted two or three other solutions. We all need to post solutions (me too) to practice writing solutions clearly.

Carry on as before.  :D  I will try in the next year or so to move up into the Intermediate level on a few challenge problems, although I am not gaining ability as fast as most of the rest of you are. Try what you can try, but try the hardest stuff you can, in general, and save the really, really easy stuff for the youngest beginners.

Thanks to everyone for discussing the issue in a public thread.

[quote="tokenadult"](Question: does latex stay in spoiler, or does it "break through" and become visible?)[/quote]

It breaks through, unfortunately. I fathom it'd be possible to program something that sets the opacity of LaTeX text (png images) to 0% and toggle it back to 100% when called for, though...

[quote="JBL"]I have not seen many [i]individuals[/i] posting here and solving questions who I thought should be in Intermediate.  I have seen a fair number of math problems that shouldn't have been in this forum.  When you get too good to solve problems here, someone will complain about it, don't worry.[/quote]

I personally don't sure who should go but I already told it's THEMSELVES that decide on advancing.  Furthermore, like AoPS books, these are just approaches to them, not exactly test you take to go like AMC10 to AIME.

[quote="yif man12"]One problem is that many of the problems in the intermediate forum involve trig. some people don't know trig at all. Also, if you had to place them in a math contest, what would you put the intermediate problems in?[/quote]

Hmm... Yes, that's quite true that many higher level math involves trigonometric skills such that one of stuffs you do in calculus is trigonometric function.  I don't think any of my questions I posted here involves trig but logarithrms...

No one pm me with their answers yet.  No one wants to do them for just fun or at least practice maybe?

I don't know, I'm having some trouble deciding which forum to be in. Technically speaking, I should be in intermediate, based on what math courses I've taken and I qualified for AIME this year. But there are still getting started problems that I don't understand, and I'm bouncing between the two right now. Or maybe I'm just stupid.
But anyhow, I don't think there's a clear-cut divide between Getting Started and Intermediate.  I mean, some stuff isn't taught as part of school curriculum so I've never seen it before, while other stuff is.  I'm sure people will move up by themselves naturally, once problems start getting too easy. After all, you wouldn't be on this site unless you liked math, and if you liked math you would seek out more challenging problems once you mastered one set.
[quote]Try what you can try, but try the hardest stuff you can, in general, and save the really, really easy stuff for the youngest beginners. [/quote]
Yeah, there isn't really a point to doing stuff that's not hard for you, unless other people don't get it and need help.
Ah well.  I'm off to do math!

I am probably Getting Started since I didn't qualify for AIME. However I found a lot of Getting Started problems very easy. Most of my posts are in Intermediate Forum and I barely post in Getting Forum.

Rather than that, I would say "Know what the partial sum is first, then find the formula, and submit it to find the answer."  Hint:  This is actually arithmetic question.

Also for Number 5, do you want us to just expand it?

[quote="Osiris"]So you mean that ak is an arithmetic sequence, whose first 25 terms have a sum of 150? What are you asked to find? In any case this doesn't feel like enough information to solve the problem.

(I'm not solving these questions, by the way. Just helping those who might want to...)[/quote]

I think you misunderstood. What I meant is that 25th term's value is 150, not the sum of first 25 terms.  Actually, the sum of first 25 terms are partial sums, I think.

P.S. This is actually from algebra II but I hardly doubt whether it is kind of thing that's widely covered in school (at least my school didn't).

I'm not really sure what you were asking in the problem but it seems to be an arithmetic sequence.  At my school those are covered in algebra 2 and in precal(I think). As are geometric sequences.

I think it's supposed to be given in general that the nth term is equal to 6n.  I don't understand what you are then supposed to find.

It's hard for me to explain.

So far, I received two people's answer and the second person got that arithmetic sequence question right.  It's basically formulas and understanding what Partial Sum is.  I can't really explain because it's not something I covered with a lot experience.  I just got this question and studied them for couple of days ago.. So...

Hey Silverfalcon. When are you going to send out the results?

I already posted it.

Here, go to:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12344

[quote="Osiris"]Still doesn't feel like enough information.

So you have the twenty-fifth term of an arithmetic sequence equalling 150. I don't know either the first term or the common difference; I can only derive a relationship between them and no more.[/quote]

Here is nice expression that summarize the question:

[tex]\displaystyle \sum_{n=1}^{25} a_n = 6n [/tex]

Is that the whole question[tex] \displaystyle \sum_{n=1}^{25} a_n=6n[/tex]?  I still don't quite understand that?  I intrerpret that to mean [tex]a_1+a_2+...+a_{25}=6n[/tex]   ...  maybe you can clarify.  I know that you have tried explaining this one before but I still don't understand.

hmm... your probably right.  That would make a whole lot more sense.

[quote="Ravi B"]Maybe he means that if an = 6n, compute
[tex]\displaystyle \sum_{n=1}^{25} a_n .[/tex][/quote]

Yep. You're right... But joml88, it's definitely not like what u said..

[tex]a_1 + a_2 + a_3 +........ a_{25} = 6n[/tex]


Two sides of an isosceles triangle measure 3 and 7.  What is the measure of the third side?

[hide]
its either $3$ or $7$

if its $3$ then it fails the triangle inequality $3+3<7$
Therefore it is $7$ as:

$3+7>7$
$7+7>3$
$7+3>7$
[/hide]

What do you mean by rules?

It basically states that the sum of the lengths of two sides of a triangle must be greater than the length of the third side.

a + b > c

[quote="Interval"]Good reply but tell me, what are the rules used in triangle inequality math?[/quote]
Triangle inequality states that any two sides added up has to be bigger than the third side. So $a+b>c,b+c>a,a+c>b$.

Exactly what I was looking for.

Thanks!


Let $ ABC$ be an equilateral triangle. In each vertex we put a test particle of mass $ m$ and charge $ q$. We join each particle with a massless spring with constant $ k$ that is initialy in rest.

the initial velocity of each particle is zero and they are in the free space. Don't have any gravitational field near it.

Let $ A_{max}$ be the maximum area for the triangle and $ A_{min}$ the minimum one.

Find $ \frac {A_{max}}{A_{min}}$

are the particles attached at the either end of the spring?

Just like that.

The solution is:

[hide]
Are you sure that you want to know?
[hide]
Had you already solved that?
[hide]
Thats is

$ \frac{1}{4} \left(\sqrt{\frac{8 K q^2}{k l^3}\plus{}1}\plus{}1\right)^2$

[/hide][/hide][/hide]


I don't have enough options for team-by-team, so which division do you think reigns supreme? Please explain who you think will win in your post.

Um, how do we know what two vote. There are two of each region. Which one is the East and whick one is the West? :what?:  :wacko:  :huh:

First three are in the east, last three are in the south.

I don't really care about division, but GO LAKERS even though I don't live in LA.

I think the Piston's have a chance to get to the finals, but winning it may be shaky.

GO SPURS!

haha i just think they willwin

[quote="kstan013"]Well you all know who I am backing...
But it is legitimate? Maybe, maybe not.[/quote]

I believe they may have a chance, the league gets pretty scrambled every year.

[quote="neelnal"]The big three will lead hte NETS!!![/quote]

I also believe that the NETS will win but they need a solid front line. The big three are simply awesome.

[quote="Iversonfan2007"]isn't it obvious that the Sixers will win....UNDEFEATED BABY![/quote]

The 76ers are pretty good, especially Iverson. I'm pulling for the Houston Rockets and Charlotte Bobcats.

Bulls!

But seriously, for this poll the good choice is Central, hands down.

They have 5 legitimate title contenders:
Cavaliers
Pistons
Pacers
Bulls
Bucks

In fact, all 5 teams made the playoffs last season.  :o

[quote="jackdillon"]you guys are nutz!!  My team Dallas are gona cream ya'll!![/quote]

in your dreams buddy.

[quote="goldendomer"]The 76ers are pretty good, especially Iverson. I'm pulling for the Houston Rockets and Charlotte Bobcats.[/quote]

Whoa me too

[quote="Iversonfan2007"]3-0 BABY! oh snap, sixers are up by 15 against the heat...uh oh[/quote]

They won anyways. (probably because Mr. Brick throwing free throw shooter/mad dunker sat out :rotfl: )


calculus
integration
logarithms
trigonometry
calculus computations

Let $x=\tan t$ and the integral becomes \[ \int_0^{\tfrac{\pi}{2}} \ln(\tan t+\cot t)\dfrac{\sec^2 tdt}{\sec^2 t} = \int_0^{\tfrac{\pi}{2}} \ln \dfrac{2}{\sin t\cos t} dt = \dfrac{\pi}{2}\ln 2 -\int_0^{\tfrac{\pi}{2}} \ln \sin 2t dt = \dfrac{\pi}{2}\ln 2 + \dfrac{\pi}{2}\ln 2 = \pi\ln 2 \]

Thanks !!! I forgot the substitution! The rest fo :mad:  :censored:  :furious:  :cursing: llows logically!!! Damn!!!


ratio
geometry
perimeter
3D geometry
pyramid
rotation
floor function

These are the problems I got wrong on my first practice test. Help me if you can. Sorry there are so many.

1. How many ways can you rearrange the letters in "ALGEBRA" ?

2. A password is created by randomly selecting 3 digits to form a 3-digit number. How many passwords are possible?

3. The ratio of the perimeters of two similar polygons is 2:3. What is the ratio of their aeas.

4. A boy ran up a hill at 3/2 mph and came down it at 9/2 mph. The trip took him 6 hours. How many miles is it to the top of the hill?

5. A ream has 2000 sheets of paper. A box has nine reams. How many 200 page books can be printed with one box?

6. As Sarah continues preparing for her backpack trip she found that lard has 120 calories/13 grams. She wants to know how many calories are in each ounce of lard. Use the following information she found on the package to help her out. Each package is one pound. Each package has 35 servings. Each serving is 13 grams.

7. What is the volume pf a pyramid with base area 18 and height 27?

8. What is the distance from the point (33,17) to the point (33, -51)

9. 500 cars in a lot using a character followed by a single digit. How many cars will have the same catalog code as another car?

10. What is the area of a triangle with vertices at points (0,0), (5,0), and (2,4)?

11. A hubcap with radius 24 inches falls off the Math Team's land rover and rolls down the street. How many times will it rotate if the street is 200 feet long?

1. Algebra has 7 letters. Thus there are 7x6x5x4x3x2x1 (denoted as 7!) ways to arrange it. However, Algebra is the same as algebrA (i made caps so you could see). Thus we overcounted by 2! (since there are 2 A's), so we do 7!/2!=5040/2=2520 ways.

2. If 0 is allowed for first one, we have 10 ways to choose first number (0-9), 10 for 2nd, and 10 for 3rd. Thus 10x10x10=1000

If we must have 1-9 for first number, then its 9x10x10=900.

3. Pretend they are both squares. Let the perimeters be 8 and 12, making the sides 2 and 3. The area of one is 4 and the area of the other is 9. Thus, 4/9.

[hide="4"]
Since $ t \equal{} \frac ds$, we have $ 6 \equal{} \frac x{\frac32} \plus{} \frac x{\frac92} \equal{} x\left(\frac69 \plus{} \frac29\right) \equal{} \frac89x\implies x \equal{} \frac98\cdot6 \equal{} \boxed{\frac {27}4}$.
[/hide]
[hide="5"]
One box has $ 9\cdot2000$ pages, so we can have $ \frac {9\cdot2000}{200} \equal{} 9\cdot10 \equal{} \boxed{90}$ two hundred paged books.
[/hide]
[hide="6"]
Based off of our information and the fact that $ 1\,\text{pound} \equal{} 16\,\text{ounces}$,

$ \frac {120\,\text{calories}}{13\,\text{grams}} \equal{} \frac {120\,\text{calories}}{1\,\text{serving}} \equal{} \frac {120\,\text{calories}}{\frac1{35}\,\text{packages}} \equal{} \frac {120\,\text{calories}}{\frac1{35}\,\text{pounds}}$

$ \equal{} \frac {120\,\text{calories}}{\frac1{35}\cdot16\,\text{ounces}} \equal{} \boxed{\frac {525\,\text{calories}}{2\,\text{ounces}}}$.
[/hide]
[hide="7"]
The volume of a pyramid is $ \frac13Bh$, so our answer is $ \frac13\cdot18\cdot27\equal{}\boxed{162}$.
[/hide]
[hide="8"]
Draw this--it is easy to see that the answer is $ 17\minus{}(\minus{}51)\equal{}\boxed{68}$.
[/hide]
[hide="9"]
I don't understand the question--there are many possibilities. For example, everybody could have $ A0$, so the answer would be $ 500$, or everybody except for one person could have $ A0$, making the answer $ 499$.
[/hide]
[hide="10"]
This is a triangle with base $ 5$ and height $ 4$, so the area is $ \frac12\cdot5\cdot4\equal{}\boxed{10}$.
[/hide]
[hide="11"]
The circumference of the hubcap is $ 2\cdot24\pi\equal{}48\pi$ inches, which is equal to $ 4\pi$ feet.

Thus, it will rotate $ \frac{200}{4\pi}\equal{}\boxed{\frac{50}\pi}$ times, and $ \left\lfloor\frac{50}\pi\right\rfloor\equal{}\boxed{15}$ complete times.
[/hide]


conics
ellipse
Putnam
geometry unsolved
geometry

How many ways do you have to solve this problem ? Do you have ideas to generalize it ?

Here's a generalization: the problem is perfectly fine if you consider the whole thing on a conic insttead of a circle. I think I posted this a long time ago. The case when the conic is an ellipse is an old Putnam problem.

I have a generalization : Let a circle (O) and a line d. OI is perpendicular to d, I on d. 4 points A, A', B, B' on the circle. AI, A'I cut the circle again at A1 and A1' , BA1, B'A1' cut d at P and P'. Prove that : IP=IQ iff AA', BB', d are concurrent .

what is Butter Fly problem, and why does it have anything to do with a circle?


Let
$ f(x)\equal{}\sum_{k\equal{}1}^n a_k x^k$ and $ g(x)\equal{}\sum_{k\equal{}1}^n \frac{a_k x^k}{2^k \minus{}1}$ be two polynomials with real coefficients.
Let g(x) have $ 0,2^{n\plus{}1}$ as two of its roots. Prove That $ f(x)$ has a positive root less than $ 2^n$.

[quote="ith_power"]Let
$ f(x) \equal{} \sum_{k \equal{} 1}^n a_k x^k$ and $ g(x) \equal{} \sum_{k \equal{} 1}^n \frac {a_k x^k}{2^k \minus{} 1}$ be two polynomials with real coefficients.
Let g(x) have $ 0,2^{n \plus{} 1}$ as two of its roots. Prove That $ f(x)$ has a positive root less than $ 2^n$.[/quote]

[hide="My solution"]

It's immediate to see that $ g(x)\equal{}\sum_{i\equal{}1}^{\plus{}\infty}f(\frac{x}{2^i})$ (which is a convergent serie).

So $ g(2^{n\plus{}1})\equal{}0\equal{}\sum_{i\equal{}1}^{\plus{}\infty}f(2^{n\plus{}1\minus{}i})$ and so the terms of this serie cant be all $ >0$ or all $ <0$ and some sign changes exist.

So, with continuity, some zero of $ f(x)$ exists in $ (0,2^n]$. If $ f(2^n)\equal{}0$, then $ \sum_{i\equal{}2}^{\plus{}\infty}f(2^{n\plus{}1\minus{}i})\equal{}0$ and, for the same reason, some zero of $ f(x)$ exists in $ (0,2^{n\minus{}1}]$

So, as required, some zero of $ f(x)$ exists in $ (0,2^n)$
[/hide]

But this problem is from USA TST 2005. Is there an official solution?

Here is what our members talked about.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=661463&sid=593911e24e913256be05c282b9e71099#p661463[/url]


exterior angle
algebra
system of equations

[quote="humblemastermind"]If you have a 5-pointed star (not necessarily regular) what is the positive difference between the sum of the measures of all of the exterior angles and the sum of the measures of all of the interior angles?[/quote]

What do you mean by "interior angles"?

[quote="pn12345"][quote="humblemastermind"]If you have a 5-pointed star (not necessarily regular) what is the positive difference between the sum of the measures of all of the exterior angles and the sum of the measures of all of the interior angles?[/quote][/quote]

OK... so:

Sum of interior angles: 180 deg.
Sum of exterior angles: 720deg.
Difference: 540

Right?

[quote="humblemastermind"]Hmm. Explain your reasoning (and hide your solutions)  :wink:[/quote]

OK here (example for regular pentagon):

FORM THE STAR BY CONNECTING THE VERTEXES OF A PENTAGON

The pentagon inside the star has each interior angle of 108 deg.
Then calculate that the vertex of the triangle on top of the pentagon has a measure of 36 deg.
36 x 5 = 180 deg. for interior angles.

Each outer angle = 180 - (interior angle)

So:
outer angle = 144 deg

144 x 5 = 720 deg.

[quote="i_like_pie"][quote="pn12345"]The pentagon inside the star has each interior angle of 108 deg.[/quote]
[quote="humblemastermind"]If you have a 5-pointed star [color=red][b](not necessarily regular)[/b][/color] what is the positive difference between the sum of the measures of all of the exterior angles and the sum of the measures of all of the interior angles?[/quote]
[/quote]

That does not matter,  I just used the average degree measure.

[quote="pn12345"]
Each outer angle = 180 - (interior angle)
[/quote]
Not quite...
[hide="hint"] See that the exterior angles and the angles of the pentagon are vertical angles....[/hide]

And pn12345 is right - it's always going to be the same whether or not it's regular.

[hide="good extension-how about an n pointed star"]First, you can make any 5 pointed star, and connect all vertices other than the star's points. Thus, the sum of the exterior angle is 540, as they're vert. angles with the angles of this newly formed pentagon. there can also be 5 triangles formed, such that 2 vertices are the star's points, and the third is one of the above mentioned pentagon's, each of the star's vertices are used twice, and each of the pentagon's once. The sum of the measures of the angles of all these triangles is 5*180=900. Thus, twice the sum of the star's points is 900-540=360, and the sum is 180. Thus, the desired value is 540-180=[b]360[/b][/hide]

perfect is right.

[hide="extension solution"]
I like setting it up a different way - make an n-gon using each point as a vertex. Assume they're regular. Each angle will be equal to $\frac{180(n-2)}{n}$. Now each angle is composed of 3 smaller angles: $x$ - the interior angle - and 2 $\alpha$s. There are also triangles with 2 angles of $\alpha$ and one angle of $y$ - the exterior angle. Thus, you can set up a system of equations like this:
$2\alpha+x=\frac{180(n-2)}{n}$
$180-2\alpha=y$
Then, substituting the value of $2\alpha$ into the first equation:
$180-y+x=\frac{180(n-2)}{n}$
$180-\frac{180(n-2)}{n}= y-x$
$y-x$ is the difference, which is almost what we wanted. As a final step, you need to multiply everything by $n$, so you get
$180n-n \left(\frac{180(n-2)}{n}\right) = n(y-x)$.
Now, when you go to clean the equation up: SURPRISE!! Everything cancels, and you find that it is, in fact, an identity: every n-pointed star has 360 as the difference between the interior and exterior angles. [/hide]


trigonometry
limit
induction
calculus
calculus computations

Induction on $n$ we have $0\leq a_{n}\leq 1\forall n\geq 0$(*).

By use $|\sin{x}|\leq |x|\forall x\in\mathbb{R}$ and (*) and hypothesis we have $a_{n+1}-a_{n}\geq 0\forall n$ therefore exists $\lim_{n\to\infty}a_{n}$ and $\lim_{n\to\infty}a_{n}=1$.

Final, by Cesaro's theorem $\lim_{n\to\infty}\frac{a_{1}+a_{2}+...+a_{n}}{n}=1$.


The local theater has one ticket window. In how many ways can six people line up to buy a ticket?

[quote="236factorial"][quote="mathmanman"]6!  :?:[/quote]

Yes, which is 6*5*4*3*2*1=[/quote]

Yes, which is : 720.  :P

[hide]6 factorial ways or 720 ways. :) [/hide]