algebra
polynomial
induction
AMC
USA(J)MO
USAMO
modular arithmetic

No 3? (which is what I hope I did right)

[quote="noneoftheabove"]2 full 2 partial are we allowed to discuss the questions?[/quote]
Think so. Mr. Patrick said we could discuss it.

2 full, which I hope I got right
and random stuff for the other 4

2 full 2 partial. I'm sure of 1 and 4, and I partly solved 2 and I sent something in for 6 but it was way off the mark. They might give me a 1 on 6 if I'm lucky :D What'd everyone get for 4? I got 8025.

Yay I put 8025! I was so sure it was 4013 until last hour.

Solved 1, 3, 4 completely, partial on 2, some observations on 5, nothin on 6.  Expected score: around 23, 24.

Yeah, I got the same for 4, 8025.  They decided to call it a dinosaur, so I decieded to give its body parts names like "torso" and "legs".

Nah, I didn't: I called them "connecting part" and "appendages", but I wish i had said the above...

[quote="kstan013"]with the t shape right?[/quote]
Yeah with 2006 in each "leg" and one in the cneter

4 was definitely 8025, but proving it was less trivial than finding it.

I solved 1,2,4,6. My solution for 6 dawned on me in the last 10 minutes so I was super rushed in writing up my solution, but hopefully I'll get a 6 on it. I know #1 was watertight, and #2 may or may not be quite rigorous enough, but I'm pretty sure that it deserves a 6 at least. #3 was very hard, I actually spent about 2 hours on it and didn't get anywhere. Still sent in something to try to get 1 point. #5 I reduced it to showing that $7^{6\cdot7^{n}}-7^{5\cdot7^{n}}+7^{4\cdot7^{n}}-7^{3\cdot7^{n}}+7^{2\cdot7^{n}}-7^{7^{n}}+1$ was composite, but I had no idea how to do it. I wrote down the roots of unity factorization and said it was obvious (which it's not).

I called the one in the center the "hub" :rotfl:

what are you talking about? I don't even know what Bono's inequality is and I never mentioned triangles once in my proof! and stop paraphrasing my proof. It has nothing to do with Bozo's inequality.

i predict $8\le x\le14$ where $x$ is my score (probably toward the lower end)

[quote="Phelpedo"]No need to start slinging mud. Now, kstan, I must admit that the proof you have posted is quite unrigorous. Things like "we want the maximum number of cells, so we want the maximum number of legs" and "twists and turns were not useful" are not rigorous. Also, we are trying to prove the conclusion for all situations, not just the simplest ones. For a [i]rigorous[/i] proof, look at the one on the AMC site.[/quote]

The thing to understand is that, unlike is probably the case in your school classes, you can't get away with stating your intuition. Your proof must be traceable back to the accepted axioms of mathematics, though generally your arguments are not going to make this noticeable for anything but the most simple concepts. The point, however, is that there can be no doubt that what you say is true...no one should be able to say "but what if <blah>?" after having thoroughly read your proof. Another thing to watch out for is to make sure you have defined all of your terms precisely. If someone can legitimately ask "what do you mean by <blah>?" then again you are lacking rigor. This seems to be a major problem with your proof -- what do you mean by "twist around"? How can you be sure that it is impossible, or suboptimal?

[quote="Teki-Teki"]go look at the history of Ono's inequality on MathWorld.  (http://mathworld.wolfram.com)

To paraphrase your proof using an extended metaphor around Ono's inequality:

In this inequality, the left-hand side must be greater than or equal to the right-hand side.  To find the boundary conditions, we want the two to be almost equal.  Hence we want the left-hand side to be just a little bit bigger than the right-hand side.  Consider the following triangle.  The inequality holds.  How can this be?  If you notice, there are a few things that make this triangle special.  All three sides are of the same length.  We will call such a triangle 'equilateral'.  Since we want the left-hand side to be bigger than the right-hand side, their difference should be positive.  For this to happen, their difference cannot be negative, or zero.  Thus the minimum difference is epsilonically greater than zero.  Notice how, in this triangle, the difference is very small.  Thus increasing the lengths of the sides does not matter.

Therefore, the simplest solution will consist of equal sides.  As stated in the beginning, we want the  left-hand side to be just a little bit bigger than the right-hand side.  The smallest difference possible is nonnegative.  Thus the left-hand side is greater than the right-hand side.

BUT ONO'S INEQUALITY IS FALSE FOR OBTUSE TRIANGLES! WHERE DOES YOUR 'PROOF' ADDRESS THIS ISSUE?  YOU MUST CONSIDER ALL CASES, NOT JUST ONE SPECIAL CASE THAT MAKES YOUR PROOF REALLY SHORT AND NICE AND GIVES YOU THE DESIRED ANSWER! YOUR ARGUMENTS NEED TO BE COMPLETELY RIGOROUS! RIGHT NOW YOU ARE HAND-WAVING!

sorry.  i'm not saying my proof was good, and this is a little mean, but hopefully it will help you understand why your proof merits probably at most one point.  also, although arguing along boundaries is good (the extremal principle is powerful), you need to argue along all points of those boundaries.  your proof only addresses one special case.

that is what jb05 meant by his comment.[/quote]

That was really funny... but kstan's proof wasn't that redundant.

@kstan: Chill! People are trying to help you and you're flaming them... and don't forget that most first year USAMOers (like me) overestimate their scores.

Right now I'm thinking 3/0/0/1/0/0 for msyelf because they should at least give 1 point to me for getting the right answer on #4 and although when I looked at it my proof was very similar to the AoPSWiki proof #1 I probably screwed up my English or something...

Hmm.
I don't see a flaw in number 1, but just in case I'm thinking a minimum of
3/0/0/0/1/0=4
and a maximum of
7/0/0/0/1/1=9
Probably 7ish though.

You probably got more for your work on #6...

Did you at least get the answer for #4? They'll probably award a point for that.

Nah he got 3n-2 instead of 4n-3 I believe

@kstan: umm...yeah.  sorry about the mock proof.  mine was not so great either, maybe i'll scrape like 2-3 points out of it somehow.  i think JSteinhardt's explanation as to why your proof is not rigorous makes more sense.

but yeah, i mean, using extremal principle is good, and you got the right answer.  so it's not like you completely failed, you just failed to tighten up your proof.  a common problem.

but i guess i was trying to show how you were handwaving and only considering a special case or two.  consider what would have happened to your proof if someone says no, you're wrong: there really is a ridiculously complicated primitive dinosaur with like 10000 cells.  then all your arguments would go up in flames and you wouldn't really be able to support your argument.  whereas proof's like Altheman's, maybe JSteinhardt's (i dunno, haven't really looked at his) would give you enough basis to say: NO, you're wrong, there can be no primitive dinosaur with more than 8025 cells.

EDIT: same thing happened on my first USAMO last year.

I'm kind of worried about my solution for #4. I think I posted it somewhere. Basically, I made an algorithm that finds a partition of a dinosaur of size 8026 or greater into two dinosaurs. The algorithm does work, but I was not completely rigorous in proving that my algorithm works during the contest. In other words, I solved the problem, and proved the answer is correct, but I didn't prove my proof was correct. How bad is this?

post your proof under the USAMO #4 thread and maybe it will get some responses: i'm not going to find your proof for you. (nobody else is, either)

Well the proof was in this thread, so I'm not going to be redundant and post it again here, but I'll post it in the USAMO #4 thread.

I wonder if my proof that 4n-3 was the maximum was rigorous enough. I just said that anything that's greater than 4n-2 can be created by inserting tiles into 4n-2 and various partitions of 4n-2 which only make them bigger therefore it is obvious that it is only necessary to show 4n-2 is not possible. Then I drew a diagram with a partition line and a dotted partition line 1 cell width down. I said to get it so there are at least dinosaurs created as possible, put n-1 in any arrangement touching the line in at least one area at the top of the solid partition line. Next, since the partition line can be changed to the dotted one, put n-1 under the dotted line. This leaves 2n in the middle which can be divided in half which means that 4n-2 is obviously not possible. Decreasing the middle means you have to put the cell at the top or bottom which creates a dinosaur there.

However, I made a misinterpretation of the problem and assumed partitions could only be straight lines so both my partitioning lines were straight and didn't account for non straight ones, I wonder how much that'll hurt me :(

Doesn't that interpretation essentially trivialize the problem? If so, I think, unfortunately, that you would probably only get one point for the correct answer, if they decide to award a point for that.


Let $a_1;a_2;...;a_n \in N$ are not square numbers. Prove that: $\displaystyle \sum_{i=1}^n \sqrt{a_i} \notin Q$.

You can find a link [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6174&highlight=]here[/url].

Hxtung, could please look at problems posted by others before posting yours.
Moreover, could you be more explicit in the references, because all of them are 'unknown', which is quite boring after a while, and you have found them somewhere yet.

Pierre.

I agree with Pierre. Why do you "always" fake the sources and write unknown. It is stupid to write book, web, internet etc. I also could write bus, car, holiday etc. Books have titles, authors, ISBNs etc. Web sites have URLs, journals have names, years etc. too. Please consider my order before posting.

probably he doesn't want us to look up the solution in the book :maybe:


the amount of water that had evaporate after 30 days = 30*0.5=15

200-15=185.

Answer : 185

[quote="GameBot"]A child's wading pool contains 200 gallons of water. If water
evaporates at the rate of 0.5 gallons per day and no other water
is added or removed, how many gallons of water will be in the pool
after 30 days?[/quote]

Well, first of all: "A child's wading pool contains 200 gallons of water. If water
evaporates at the rate of 0.5 gallons per day"

That gives us a function:

$f(x)=200 - \frac{1}{2}x$

where $x=$ number of days.

Then we get to the part: "how many gallons of water will be in the pool after [b]30[/b] days?"

So we plug in the $x-\text{value}$ as $30,$ therefore getting:

$f(x)=200 - \frac{1}{2}(30)=200-15=\boxed{185}.$


Anybody knows where the excursions for the IMO will be? Or in what city the jury will be meeting before the competition? Or anything else interesting that hasn't been posted on the official website?

I'll try to guess (my guess is based on the list of sponsors of IMO 2008).
The jury meeting will take place in Segovia
http://en.wikipedia.org/wiki/Segovia

and the excursions will be to:
1) Toledo: http://en.wikipedia.org/wiki/Toledo,_Spain
2) Aranjuez: http://en.wikipedia.org/wiki/Aranjuez
3) Teatro Circo Price: http://www.esmadrid.com/circoprice/portal.do

Damn. I wish I could see Toledo, but I'll probably be coordinating at that time.


[quote="4everwise"]Have another simple question.  :blush:

For the page numbers on the bottom of each page, do we write it as something out of the pages for each problem or as something out of all the pages for the problems in round 1?

For example, page 2 out of 3 (for a problem)

or page 2 out of 13 (all five problems)[/quote]

We graders look at the solutions one problem at a time. So it is easier for us if you label the pages by problem.

Occasionally a page of a solution is lost. Faxed solutions are particularly vulnerable. If we graders notice, we write a comment asking for the missing page. If we graders don't notice, we grade a solution that has an enormous hole in it and our comments don't mention a missing page, so the grade must be confusing to the person who submitted the solution. Page numbers help prevent that problem.

Erin Schram
USAMTS grader

[quote="4everwise"]Have another simple question.  :blush:

For the page numbers on the bottom of each page, do we write it as something out of the pages for each problem or as something out of all the pages for the problems in round 1?

For example, page 2 out of 3 (for a problem)

or page 2 out of 13 (all five problems)[/quote]
Hey, that's crossed my mind, too.

Luckly, I guessed to label them by problem.  :D


Let $n_1 < n_2 <\dots$ be a sequence of natural numbers such that for $i < j$ the decimal representation of $n_i$ does not occur as the leftmost digits of the decimal representation of$n_j$ . (For example, $137$and $13729$ cannot both occur in the sequence.) Prove that:
$\sum_i\frac{1}{n_i}\leq 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac 18+\frac{1}{9}$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=3692 .

[hide="Guess what keyword I searched for..."]13729  :D[/hide]

  darij


The ﬁrst four terms in an arithmetic sequence are $ x \plus{} y$, $ x \minus{} y$, $ xy$, and $ x/y$, in that order. What is the ﬁfth term?

$ \textbf{(A)}\ \minus{}\frac{15}{8} \qquad
\textbf{(B)}\ \minus{}\frac{6}{5} \qquad
\textbf{(C)}\ 0 \qquad
\textbf{(D)}\ \frac{27}{20} \qquad
\textbf{(E)}\ \frac{123}{40}$

We have the equations
\begin{align*}
x+y+d&=x-y \\
x-y+d&=xy \\
xy + d &= \frac{x}{y}
\end{align*}
Simplifying and rearranging, we can keep substituting and find that $ \displaystyle (d, x, y)=\left(\frac{6}{5}, -\frac{9}{8}, -\frac{3}{5}\right)$.

Then
\[ \frac{x}{y}+d=\frac{9 \cdot 5}{8 \cdot 3}+\frac{6}{5}=\frac{123}{40} \rightarrow \boxed{\textbf{E}}.\]

hello, from the problem we get the following equation system
$ x\minus{}y\equal{}x\plus{}y\plus{}d$
$ xy\equal{}x\plus{}y\plus{}2d$
$ \frac{x}{y}\equal{}x\plus{}y\plus{}3d$
Solving this we get $ x\equal{}\minus{}\frac{9}{8}$, $ y\equal{}\minus{}\frac{3}{5}$ and $ d\equal{}\frac{6}{5}$, so our fifth term is
$ \minus{}\frac{9}{8}\minus{}\frac{3}{5}\plus{}4\cdot\frac{6}{5}\equal{}\frac{123}{40}$, so the right answer is $ \text{e}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, from the problem we get the following equation system
$ x \minus{} y \equal{} x \plus{} y \plus{} d$
$ xy \equal{} x \plus{} y \plus{} 2d$
$ \frac {x}{y} \equal{} x \plus{} y \plus{} 3d$
Solving this we get $ x \equal{} \minus{} \frac {9}{8}$, $ y \equal{} \minus{} \frac {3}{5}$ and $ d \equal{} \frac {6}{5}$, so our fifth term is
$ \minus{} \frac {9}{8} \minus{} \frac {3}{5} \plus{} 4\cdot\frac {6}{5} \equal{} \frac {123}{40}$, so the right answer is $ \text{e}$.
Sonnhard.[/quote]
how did u solve the equations please show it

hello, from the first equation we get $ d\equal{}\minus{}2y$, inserting this in the second one we have $ y\equal{}\frac{x}{x\plus{}3}$,with this equation and $ d\equal{}\minus{}2y$ we get in the last one $ \frac{x}{\frac{x}{x\plus{}3}}\equal{}x\minus{}\frac{5x}{x\plus{}3}$, this is equivalent with $ x\equal{}\minus{}\frac{9}{8}$.
Sonnhard.

[hide = Solution]The difference between consecutive terms is $(x-y)-(x+y)=-2y.$ Therefore, we can also express the third and fourth terms as $x-3y$ and $x-5y.$ Then we can set them equal to $xy$ and $\frac{x}{y}$ because they are the same thing.

\begin{align*}
xy&=x-3y\\
xy-x&=-3y\\
x(y-1)&=-3y\\
x&=\frac{-3y}{y-1}
\end{align*}
Substitute into our other equation.

$$
\frac{x}{y}=x-5y$$$$\frac{-3}{y-1}=\frac{-3y}{y-1}-5y$$$$-3=-3y-5y(y-1)$$$$0=5y^2-2y-3$$$$0=(5y+3)(y-1)$$$$y=-\frac35, 1$$
But $y$ cannot be $1$ because then the first term would be $x+1$ and the second term $x-1$ while the last two terms would be equal to $x.$ Therefore $y=-\frac35.$ Substituting the value for $y$ into any of the equations, we get $x=-\frac98.$ Finally,

$$ \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\frac{123}{40}}$$[/hide]

[quote=Piano_Man123][quote=wizkid28]what problem on what contest is this?[/quote]

Dude, that's not cool. Why did you bump this thread?

(on a sidenote, this is on AMC 10 problem Series hw but I finished so ye)[/quote]

He clearly bumped the thread to ask "what problem on what contest is this?". Honestly, *your* bump was more unnecessary.

AHHHHHH

Edit: farpecan reported it for no reason...

[hide=Solution]Notice that $$x-y-(x+y)=xy-x+y=\frac{x}{y}-xy$$
Thus,
$$-2y=xy-x+y=\frac{x-xy^2}{y}$$
Therefore, $$-2y^2=x-xy^2 \implies \frac{2y^2}{y^2-1}=x$$
Thus,
$$\frac{2y^2}{y^2-1}=3y+\frac{2y^3}{y^2-1}$$
Getting rid of fractions,
$$2y^2=3y^3-3y+2y^3 \implies 5y^3-2y-3y=0 \implies y=0,-\frac{3}{5},1 \implies x=-\frac{9}{8}$$
Thus, the difference is
$$\frac{x}{y}-xy=\frac{6}{5} \implies \frac{15}{8}+\frac{6}{5}=\boxed{\frac{123}{40}}$$


try integration by parts...thats the only way i know to do this
remember, pick the u and dv wisely...

[quote="stephenryan"]Right - silly question probably but I cant find the answer anywhere....

How do you integrate arcsin(x) with respect to x?[/quote]
Let  dv = dx, u = arcsin x

Then int (udv ) = xarcsinx - int ( x/sqrt(1-x^2)

Let u = x^2, so that we have du/2sqrt(1-u) --> -sqrt(1-u) -->

int ( arcsin x ) = xarcsinx + sqrt(1-x^2) + C

You know, I saw this thread's title, and the first thing that came to mind was integration by parts.  I must be psychic or something.

given an arcsinx graph, draw a vertical line that passes through x, which is the upper limit, as 0 being the lower limit.  now, do the arcsin of x, and you'll be given the place where the vertical line intercepts the arcsin graph.
Now, if you flip the graph to resemble that of a sinx graph then you can find the area between what was your y value on the arcsin graph and 0 on a sinx graph. so finding the area by doing the integral of sinx will give you the area of the INSIDE of part of the arcsinx enclosed by the y axis.  we are trying to find the area outside of this region, but we know that it has same y (arcsinx) and x (x) values of the area of the inside region.  so it makes a rectangle.  So [(arcsinx)(x)-  (Integral{limits-0, upper-arcsinx} of sinx)] should give you the integral with the lower limit being 0 and the upper, not exceeding one.


calculus
integration
algebra
polynomial
number theory unsolved
number theory

If f(m) = s^2 and f(m) is monic and irreducible and m is known

how do you find

f(n) = t^2, in integers? (I don't think there is an anwer to this.)

Is the question:

Given a irreducible monic integral polynomial $f(x)$ such that $f(m)$ is a square for some integer. Find another integer such that it is a square¿

Yes, that's what I'm asking. m and s^2 are known. Is there any way to find n and t^2?

btw, f(m) is a quadratic.


Ok, here they go:
[code]UKR1 (Mishchenko)   7 7 7  7 7 0 (35)
UKR2 (Semikina)     7 2 0  7 0 0 (16)
UKR3 (Senin)        7 5 0  6 2 0 (20)
UKR4 (Serdyuk)      7 5 0  1 7 7 (27)
UKR5 (Shyshatskyy)  7 5 7  6 7 0 (32)
UKR6 (Sobolyev)     7 7 2  4 3 0 (23)[/code]

Medal cutouts will be known only tomorrow.

[quote="torbich"]Ok, here they go:
UKR1 (Mishchenko)  7 7 7  7 7 0 (35)
UKR2 (Semikina)     7 2 0 7 0 0 (16)
UKR3 (Senin)          7 5 0 6 2 0 (20)
UKR4 (Serdyuk)      7 5 0  1 7 7 (27)
UKR5 (Shyshatskyy) 7 5 7 6 7 0 (32)
UKR6 (Sobolyev)     7 7 2 4 3 0 (23)

Medal cutouts will be known only tomorrow.[/quote]

UKR1  :first:

I feel sorry for UKR 4 :/ it looks like he doesn't like algebra...

UKR4 is very good mathematic, I don't know why has he written so bad (for him). I was sure that he would win.

I wouldn't say that UKR4 is not good at Algebra. You never know why somebody doesn't solve a problem. Last year ARG5's score was 2 7 0 7 7 0. If you see that score maybe you should think that he couldn't solve problem 1 but he had 4:30 hours and he knew that he could probably solve geometry problems so he spent almost all the time in problem 2 and when he started thinking problem 1 he just had a few minutes. And cases like that I think there may be a lot, so I think that having 0, 1 or 2 in a problem doesn't mean that you couldn't solve that problem, maybe you just didn't think the problem for enough time.

i haven't said he isn't good at it, i just think it isn't his favourite section of math, and anyone who solves P6, especially this one, is a really damn good mathematician... makes me said to see someone that brilliant doesn't win gold...

i am also brilliant but i don't have gold in my pocket :(  :D

I am surprised of UKR4's result, because he, being in grade 10, take part in olympiads for grade 11. And he was absolutely first at the Kiev city olympiad, Ukrainian olympiad and training olympiads before UMO and IMO. And many people waited a very good result from him.

He solved 6-th problem ;) I think it that's enougth. But it's strange about his result in 4-th problem. Maybe he spent a lot of time on 6. Who is Yuliya Semikina?

[quote="Yuriy Solovyov"] Who is Yuliya Semikina?[/quote] I have already answered you. She finished the lyceum "Leader".

And who are you? I don't know any ukrainians from this site.

I'm Maxim Chornyy, I have finished the 9th form in "Leader". I had the first prizes in Ukrainian olympiads
2007 and 2008. And who are you?:)

I studied in the liceum of foreign languages in Lugansk. Took part in math national olympiad in 2001 in 10-th grade, in 11 - third prize in Ukraine national physics olympiad in 2002. And 3-rd in Ukraininan students math olympiad in 2006. About city and regional olympiads I'll not write. :))

2 Solovyov&BlackMax
 And you both decided to discuss it here, in English...


Prove that if the matrix $ A$ has a right inverse, then for each $ b$
the equation $ Ax \equal{} b$ has at least one solution. If $ A$ has a left inverse,
then that equation has at most one solution.

If $ A$ has a right inverse, then it has a full row rank and hence your system is consistent. If $ A$ has a left inverse, then you can multiply both sides of your system by $ A$'s left inverse to find $ x$, if it exists of course!


Hi all,

Have been struggling with some issues in math and would like to find out what you guys think of them.

I went through a period of time where I struggled heavily with mathematics. I don't mean performing calculations, but I mean having difficulty with the meaning of mathematics. For example, I drove myself crazy over the meaning of '1 + 1 = 2', or 'x^2 - 2x + 1 = (x-1)^2'. I mean, yes, if we choose to accept the system of mathematics that has been taught to us so far, then these things are 'intuitively' obvious, but if we don't, then there is no reason for '1 + 1 = 2', or for any algebraic expansion, or anything else based on such a system to work.

Likewise, if we choose not to accept the logical system of mathematics, where something cannot be both true and false at the same time, then automatically the proving techniques of direct and indirect proof fail, because they rely on the assumption that something cannot be both true and false at the same time.

Eventually I figured that it would be better to stop trying to find so much meaning in math and treat it as a game. If we[i] choose[/i] to accept certain logical rules, and if we[i] choose[/i] to accept certain things as givens, then we can draw 'logical' conclusions, 'logical' meaning that they are consistent with the rules of the game.

That was a few years ago, but I'm getting interested in college mathematics and problem solving, and it has begun to tickle my overanalytical mind and unearth all these questions.

So what do you guys think? Am I confusing philosophy and math? Should I just learn to play the game, and accept that mathematics makes no inherent sense (in the way I just described)?

Thanks for your thoughts
Joshua

P.S. On a side note, I find that being too analytical makes you unable to do math, because you start asking silly questions like what is the meaning of addition, and then you can't do grade school problems  :)

Study the basic theory of integers (usually found in the beginning of a a good college calculus book). It shows you these definitions and how we rigorously build up mathematics.

Something cannot be true and false based on the definitions of true and false, and similarly 1+1=2 because that is how we define 2.

Mathematicians as a whole went through a similar crisis in the 1800s when they realized that, roughly, they didn't know what they were talking about.  This period in mathematical history is rather fascinating, and you would do well to read about it, as it might help answer a lot of your questions.  Unfortunately, I can't recommend a good resource for you, but some keywords:  Russell and Whitehead, Godel...

Edit:  You're making the implicit assumption that there is only one logical system of mathematics.  Some more keywords:  the axiom of choice, the continuum hypothesis, the Paris-Harrington theorem...

The things that are true in mathematics are true for 3 reasons (or combinations of these three reasons):

1. Definitions
2. Axioms
3. Logical deductions from definitions and axioms

There are commonly accepted definitions and axioms from which we derive what is known as standard "mathematics". There are also some controvertial axioms which are accepted by some and rejected by others and which either extend or restrict the standard body of mathematics.

Some of your examples, like why does "1 + 1 = 2?", essentially boil down to what are your definitions and axioms. So what are the definitions of 1 and 2, and what are the axioms/meaning of addition and equality? Once those questions are answered, the reason why 1 + 1 = 2 will follow. [If you come up with your own definitions and axioms I be they'll be very close to the "official" ones. But the official ones are dressed up in much more formal terms to make them as free from languange ambiguity as possible. If you want to see the "standard" definitions for integers and addition see [url=http://en.wikipedia.org/wiki/Peano_Axioms]Peano Axioms[/url]].

As for your second question, why does $ x^2 \plus{} x \plus{} 1 \equal{} (x \plus{} 1)^2$, it follows from the distributive property (axiom) and commutative property (axiom) of real numbers (assuming x represents a real number).

By the way, if you like these kind of questions, you might make a good logician or algebraist! A good book to read on these kind of philosophical qustions is [url=http://en.wikipedia.org/wiki/The_Mathematical_Experience]The Mathematical Experience[/url] by Philip J. Davis and Reuben Hersh. A more difficult book would be [url=http://en.wikipedia.org/wiki/G%C3%B6del,_Escher,_Bach]Gödel, Escher, Bach[/url] by Douglas Hofstadter.

[quote="gauss202"]As for your second question, why does $ x^2 \plus{} x \plus{} 1 \equal{} (x \plus{} 1)^2$, it follows from the distributive property (axiom) and commutative property (axiom) of real numbers (assuming x represents a real number).[/quote]

we must be using different axioms!

Hi, I do still (I am in seventh grade now) have questions like joshuatan. For instance, I often find myself asking questions about the fundamentals about mathematics. And, following you guys' advise, i have started reading The Mathematical Experience, and it is great at answering those questions.

I, however, have this other problem that has developed only recently. It is that although I agree with the logic of a problem, I often find myself "overthinking" and denying the concept of the problem, which in most cases leads to nowhere. For example, if there were 25 people in my gym, and we stand in a line and are assigned the numbers 1 through 5. So, persons $ A, B, C, D, E$ are given 1 through 5, respectively. So in order to be in the same team as my friend, I need to be 5 away for him. Simple concepts like those are driving me nuts. Although I do partly agree with what I am doing, I don't know why I am having such ridiculous and "silly" questions, as joshuatan put it.

I think this may be due to some lack of canfidence maybe. What are your ideas? And any remedies?

Thank you  :blush:
limac

P.S. I know this is a really stupid problem, but...it is really bugging me.

[quote="t0rajir0u"]1800s[/quote]
1900s  :wink:

[quote="limac"]
I think this may be due to some lack of canfidence maybe. What are your ideas? And any remedies?
[/quote] Maybe you could start by not reviving year-old topics.

"Do not revive an old topic even if this means opening an already existing one again"¿ I see in general no reason not to revive topics. Possibly the new post is offtopic (not sure about that), then there is some reason.


inequalities
algebra
three variable inequality

a,b,c>0 abc=1

1/a^3(b+c) + 1/b^3(c+a)+ 1/c^3(a+b)  \geq  3/2

2 or more solutions please! Cause I think it's best to find as much solutions as we can.

Well, here's one soln:

We can rewrite our left-side expression as E= \sum (bc)^2/(ab+ac). Now we use that famous lemma: if A, B, C, x, y, z positive then  \sum a^2/x >= (  \sum a)^2/( \sum x) and we apply it for A=bc, B=ac, C=ab, x=ab+ac etc. We get E>=(ab+bc+ca)^2/2(ab+bc+ca)=(ab+bc+ca)/2 >= 3/2 by AM-GM.

Here's another soln:

We make the substitutions a=yz/x^2, b=xz/y^2, c=xy/z^2.
Then the ineq becomes S = sum (1/a^3(b+c)) = sum (x^5/(y^4z+yz^4)) (check it out, I might have made a mistake) >= sum (x^5/(y^5+z^5)) (because y^4z+yz^4 <= y^5+z^5 by rearrangement ineq). Now let u=x^5, v=y^5, t=z^5. We should prove sum (u/(v+t)) >= 3/2, which is well known.

I hope I can come up with more ways of solving this..

okay we have to prove that:

sum (1/(a^3(b+c))>=3/2 with abc=1

this ineq is equivalent to: sum (x^2/(y+z))>=3/2 with xyz=1 (x=1/a etc)

now we write the ineq like so:

[(y+z)+(z+x)+(x+y)][sum(x^2/(y+z))]>=(x+y+z)^2>=3(x+y+z)

the last ineq is true as we have x+y+z>=3*(xyz)^(1/3)=3. We have equality when x=y=z=1 and we have that a=b=c=1.

cheers! :D :D :D :D

Just an observation: to get [(y+z)+(z+x)+(x+y)][sum(x^2/(y+z))]>=(x+y+z)^2 we can apply that lemma I applied in the first proof. So you can see that this comes in really handy. Many ineqs are easier if we use that. There are also some interesting generalisations. Here's one:

a_i and x_i are positive, then  \sum((a_i)^m/x_i) >=                           (\sum a_i)^m/n^(m-2)*( \sum x_i) where i goes from 1 to n. It's proved using generalised AM-GM.

Well I kind of found the same solutions:
1) \sum a 2 b 2 c 2 /a^3(b+c) \geq 3/2
 \sum b 2 c 2 /a(b+c) \geq  (ab+bc+ca) 2 /2(ab+bc+ca)= (ab+cb+ca)/2  \geq  3*(abc)^2/3 /2=3/2
2)We have:
 \sum 1/a^3(b+c) \geq 3/2(abc)^4/3
a=1/x......

 \sum x 2 /y+z  \geq 3/2  *(xyz)1/3
(x+y+z) 2 /2(x+y+z)  \geq  3/2  * (xyz)^1/3
x+y+z \geq 3*(yxz)^1/3
3)Suppose a \geq b \geq c
1/c \geq 1/b \geq 1/a
 1/(ac+bc) \geq 1/(ab+bc) \geq 1/(ab+ac)
1/c(ac+bc) \geq 1/b(ab+bc)+1/a(ab+ac)
ab \geq ac \geq bc
 ab/c(ac+bc)+ac/b(ab+bc)+bc/a(ab+ac) \geq  ac/c(ac+bc) + bc/b(ab+bc) + ba/a(ab+ca)
ab/c(ac+bc) + ac/b(ab+bc)+bc/a(ab+ac)  \geq  bc/c(ac+bc) + ab/b(ab+bc)+ac/a(ab+ac)
Adding we get:
2(ab/c(ac+bc) + ac/b(ab+bc)+ bc/a(ab+ca) ) \geq  1/a+1/b+1/c
but 1/a+1/b+1/c \geq 3 So we finally get:
 \sum 1/a^3(b+c) = ab/c(ac+bc) + ac/b(ab+bc)+ bc/a(ab+ac) \geq 3/2

Let A be the given expression.
Let B = a(b + c) + b(c + a) + c(a + b). Let C = 1/a + 1/b + 1/c.
Cauchy gives AB \geq C. But B = 2(1/a + 1/b + 1/c).
So A \geq C/B = C/2 \geq 3(abc)^(1/3)/2 = 3/2 by AM GM.

I don't know if I  made a mistake cause the post was prety long so if I have I'm sorry.

A brute force solution (using Muirhead) exists and is straightforward.
But I won't type it here :?

[quote="Maverick"]$ a,b,c > 0, abc \equal{} 1\Longrightarrow$

$ \frac {1}{a^3(b \plus{} c)} \plus{} \frac {1}{b^3(c \plus{} a)} \plus{} \frac {1}{c^3(a \plus{} b)} \geq \frac {3}{2}.$[/quote]$ \sum{\frac {1}{a^3(b \plus{} c)} } \minus{} \frac {3}{2} \equal{} \sum{\frac {b^3c^3}{b \plus{} c} } \minus{} \frac {3(abc)^{\frac {5}{3}}}{2}$

$ \equal{} \frac {2\sum{b^4c^4} \plus{} 2\sum{a^4(b^3c \plus{} bc^3)} \minus{} 3\sum{a^{\frac {11}{3}}(b^{\frac {8}{3}}c^{\frac {5}{3}} \plus{} b^{\frac {5}{3}}c^{\frac {8}{3}})} \plus{} 2\sum{a^2b^3c^3} \minus{} 6a^{\frac {8}{3}}b^{\frac {8}{3}}c^{\frac {8}{3}}}{2(b \plus{} c)(c \plus{} a)(a \plus{} b)},$

which is clearly non-negative by [url=http://en.wikipedia.org/wiki/Muirhead's_inequality]Muirhead theorem[/url].

This problem is introduced as Example 2.2.2 on page 39 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://book.beifabook.com/Product/BookDetail.aspx?Plucode=754284848]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

See also : http://www.mathlinks.ro/viewtopic.php?t=7854

By Cauchy - Schwarz we have
\[ \frac{1}{a^{3}(b\plus{}c)}\plus{}\frac{1}{b^{3}(c\plus{}a)}\plus{}\frac{1}{c^{3}(a\plus{}B)}\ge\frac{(ab\plus{}bc\plus{}ca)^{2}}{2(ab\plus{}bc\plus{}ca)}\equal{}\frac{ab\plus{}bc\plus{}ca}{3}\ge\frac{3(abc)^{\frac{2}{3}}}{2}\equal{}\frac{3}{2}\].

I`m sorry, my LaTeX is not good.
$ \frac{1}{a^{3}(b\plus{}c)}\plus{}\frac{1}{b^{3}(c\plus{}a)}\plus{}\frac{1}{c^{3}(a\plus{}b)}\ge\frac{(ab\plus{}bc\plus{}ca)^{2}}{2(ab\plus{}bc\plus{}ca)}\equal{}\frac{ab\plus{}bc\plus{}ca}{2}\ge\frac{3}{2}$


[quote="ice_age"]Let P1, P2, ...; Pn (n>= 2) be any rearrangement of 1, 2,... n.
Show that
  1/(P1+P2)+1/(P2+P3)+...+1/(Pn-1+Pn)>(n-1)/(n+2)[/quote]
 we have :
  $ A \equal{} 1/(P1 \plus{} P2) \plus{} 1/(P2 \plus{} P3) \plus{} ... \plus{} 1/(Pn \minus{} 1 \plus{} Pn) \geq \frac { (n \minus{} 1)^2 } { n(n \plus{} 1) \minus{} P_1 \minus{} p_n} \geq \frac {(n \minus{} 1)^2 }{n(n \plus{} 1) \minus{} 3} \geq \frac {n \minus{} 1}{n \plus{} 2} .$
 obvious  right   therefor : $ \minus{} 2 \geq \minus{} 3$


Jacob bought a CD for $ \$$15 and sold it for $ \$$20. He then bought
it back for $ \$$25 and sold it again for $ \$$28. How many dollars
profit did he make?

Jacob:

Bought for 15:  -15
Sold for 20:  -15+20=5
Bought for 25:   5-25=-20
Sold for 28:  -20+28=8

TOTAL PROFIT= 8 DOLLARS


rotation
geometry
geometric transformation
reflection
3D geometry
group theory

I've been searching the internet but I've been having a difficult time trying to look for a definition of group orbits that I can understand.

An easy to understand explanation of what group orbits are or a link to a good website on intro to group theory would be greatly appreciated.

thx

This isn't a high school topic; you'll want to post a question like this in [url=http://www.artofproblemsolving.com/Forum/index.php?f=65]here[/url].

Orbits are talked about in the context of [url=http://en.wikipedia.org/wiki/Group_action]group actions[/url]. Groups were originally developed to study the symmetries of objects. A group action of $ G$ on $ X$ is basically turning each element of a group $ G$ into a transformation on a set (or other structure) $ X$ that preserves $ X$ setwise but may move elements around.

Examples:
1. $ X$ is a square, and you want to study all the ways it can be rotated and reflected. The dihedral group $ D_4$ describes these rotations and reflections, so we can have $ D_4$ act on the square to produce the set of transformations we want to look at. If you exclude we get the cyclic group with 4 elements instead.
2. $ X$ is a Rubiks cube, with each element of $ X$ being a square face of one of the cubes (so $ X$ has 54 elements). You want to study all the positions of the Rubiks cube. Writing $ G$ is a little trickier in this case; usually you write each possible move on a Rubiks cube as an element of $ S_{54}$ (the [url=http://en.wikipedia.org/wiki/Symmetric_group]symmetric group[/url] of 54 elements), and then have these be the generators of $ G$. When $ G$ acts on $ X$, you get the ways in which the Rubiks cube can be moved.

The orbit of a given element $ x$ in $ X$ is all the places that $ x$ can go to under the transformations in $ G$. In example 1, you can move any vertex to any other with one of the transformations in $ D_4$, so the orbit of $ x$ is $ X$ for every $ x$. In example 2, the places to which a given face of the Rubiks cube can move to is substantially more limited. For instance, the center square on each face can only move to the center of another face, so if $ \{x_1, x_2, x_3, x_4, x_5, x_6\} \in X$ are those 6 centers, the orbit of $ x_1$ is $ \{x_1, x_2, x_3, x_4, x_5, x_6\}$ (and the orbit of $ x_2, x_3, \ldots x_6$ is the same thing).

Orbits are very tied to stabilizers; both are discussed in one section in the wikipedia article on group actions I linked to above.


function probability calculus integration Support geometry rectangle

The probability density function for a pair of continuous random variables X and Y is:

f(x,y)= kx^2y for 0≤y≤x≤2

to determine the value of k I:

did the double integral of x^2y dydx with the limits of the dx integral from 0 to 2 and the dy integral from 0 to x. After all of the calculations I received a value of k=5/16.

to determine the marginal probability density functions for X and Y I:

took the integral from y to 2 of (5/16)x^2ydx which gave me (40y-5y^4)/48 for 0≤y≤2

took the integral from 0 to x of (5/16)x^2ydy which gave me (5x^4)/32 for 0≤x≤2

They are thus not independent as the product of these two m.d.f. do not give 5/16 x^2y.

I have done the work I just do not trust that I am getting the right answers... if you see any mistakes please let me know.

Those look right. In particular, they make sense when integrated.

Your answers are correct.

Note that from the beginning, there was never a chance that $X$ and $Y$ could be independent. Why not? Because for independent random variables, the "support" of the joint density function must be a "retangular set" (that is, a cartesion product of some set of $x$'s times some set of $y$'s.) (The support of a function is, more or less, the set on which it is anything other than zero.)

The support of this joint density function is a triangle, not a rectangle - and that dooms any chance of independence right there. To see that, just pick a point $(x,y)\in [0,2]\times[0,2]$ such that $y>x$ - that is, inside the square but outside the triangle. Without even computing the integrals, you know that the marginal density for that $x$ is positive and the marginal density for that $y$ is positive, yet the joint density at that point is zero.

--

I happen to be teaching this course right now. We were doing this material about a week and a half ago. (Textbook: Wackerly, Mendenhall, Schaefer.) If this were my homework problem, I'd probably have some followup questions.  Here are some possible additional questions:

1. Compute $E(X),E(Y),\text{Var}(X),$ and $\text{Var}(Y).$

2. Compute $\text{Cov}(X,Y)$ and the correlation coefficient of $X$ and $Y.$ (Hint: just based on which way that triangle is slanting, would you expect positive correlation or negative correlation?)

3. Compute the conditional expectations: $E(X|Y=y)$ and $E(Y|X=x).$

Thank you both for the input. I appreciate the follow up questions and I will work on them. Because right now I am taking this course and the prof that I have isn't giving me context or explanation for what we are learning, so the fact that I'm able to get these questions right leaves me in disbelief.
Thanks again.


Let $ p \equal{} (a_1, a_2, \dots, a_{n\minus{}1})$ be a permutation of the elements of $ \mathbb{Z}_n \backslash \{0\}$
so that no consecutive subsequence $ (a_i, a_{i\plus{}1}, \dots, a_j)$ $ (1 \le i \le j \le n\minus{}1)$ of $ p$ has

$ a_i \plus{} a_{i\plus{}1} \plus{} \dots \plus{} a_j \equal{} 0$.

How many valid permutations are there for $ Z_n \backslash \{0\}$?

Notes:

- if $ n$ is odd then $ 1 \plus{} 2 \plus{} \dots \plus{} n\minus{}1 \equal{} 0$, which is a contradiction: therefore $ n$ must be even
- it's easy to see that $ (a_1, a_1 \plus{} a_2, \dots, a_1 \plus{} a_2 \plus{} \dots \plus{} a_{n\minus{}1})$ must be a permutation of $ \mathbb{Z}_n \backslash \{0\}$
- multiplying the elements of a permutation by a number $ q$ coprime to $ n$ gives us another valid permutation

The existence problem was posted already here (and probably elsewhere?):

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202477

(It's from the Russian Math Olympiad (ARMO) 1995.)  The thread I linked to has only a partial solution -- the other part is to come up with a clever construction in the even case.  There has also been discussion of the enumerative problem (which is the one you actually asked).  See my various posts at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199166 for some results -- exact enumeration seems hard, but (much) better asymptotic bounds seem like a realistic possibility.

Thanks for answering! I had missed those threads in my search for the problem.
I browsed through them and they were very interesting reading: at least I can
be more confident about the hardness of the problem now. Deriving tighter bounds
for the number of permutations would be interesting, though. :)


Two travelers are walking on the same path. The first is 8 km ahead of the second one and is traveling at the speed of 4 km/h, the second is walking 6 km/h. One of the travelers has a dog which exactly at the same time as the travelers started walking ran from the owner to the other traveler at the speed of 15 km/h and then came back to the owner and ran back to the other traveler and kept running back and forth until the 2 travelers met. How many km's did the dog run? Very simple!

Im not sure but maybe this was sepose to be posted in the begginer section?

I think intermediate is fine.

[hide="Here is my solution"]
It will take $\frac{8}{6-4}=4$ hours for them to meet, during which the dogs travels constantly at 15 km/h, so the dog travelled $15 \cdot 4=\boxed{60}$ km. This is a bit tricky because some people would try to separate the dog's path and come up with an infinite series.

[/hide]

Well if it's infinite, then won't they keep meeting?  But the first ship that's sent will meet with exactly 7 ships.  The second sent from the same city will meet 8, the third 9, and so on, until you get to the 7th, which will meet 13.

[quote="Vash854"]Can you solve another simple one then? Here goes, A company sends a ship from city A to city B [b]Every day[/b] and the same company sends one from city B to city A, [b]Every day[/b], it takes 7 days to get from one city to another. How many times will 2 ships of the same company meet?[/quote]
you have to give a time interval...
also you should state, that the ships leave right after they get to the city...right?

Let me restate the question. How many times will 2 ships of the company meet until all the ships from city A to B reach city B and all the ships from city B to city A reach city A, if they send 1 ship every day from each city and it takes 7 days for a ship to get to the other city. [b]We will assume the company has 7 ships in each city.[/b]

Do you guys need it more simple or is it understandable?

Well in that case, each ship from city A will meet each one of the ships from city B exactly once.  And there are 7 ships from each city.  So it must be 49.

o.0 I think I translated the question wrong   :?


calculus
integration
calculus computations

Determine
\[\int_{C}\frac{yzdx+xzdy+xydz}{1+x^{2}y^{2}z^{2}}\]
where $C$ is the curve $r(t)=(t^{22},t^{3},t^{2007})$
between $t=0$ and $t=1$.
I dont understand how the answer becomes $\pi/4$


geometry
complex analysis
advanced fields
advanced fields theorems

I am learning algebraic geometry by myself,But I find it is very hard.I read the Hartshorne's book,but I find I can't understand even Chapter 1(Varieties).Does anyone can give some suggestion to me foe learning algebraic geometry?

By the way,does anyone have solutions to the Hartshorne's Book(Just Chapter 1).(I find solutions here :http://math.berkeley.edu/~reb/courses/alggeom/,but it is not completed)

You could try to read Liu's book http://www.math.u-bordeaux.fr/~liu/Book/index.html, which is more self-contained (less knowledge in commutative algebra required) and doesn't rely as much on the exercises as Hartshorne. On the other hand, it gives less geometric intuition and avoids homological algebra which renders the cohomology chapter quite unusable (here, Hartshorne is better, but avoids spectral sequences, which make live easier). Hartshorne is more geometric, Liu more arithmetic, see the later chapters. You can read both books and as an introduction Mumford's red book and The Geometry of Schemes (Eisenbud, Harris).

The answer to your question depends on what your background is.  Hartshorne is the "standard" introductory algebraic geometry textbook, but it is famously difficult, and should not be someone's first experience with the subject (unless the student is experienced in closely related topics at the graduate level).

Algebraic geometry can be taught in either a "classical" way (focusing on varieties first) or a "modern" way (focusing on schemes right away, as in Hartshorne).  In *most* cases, a student should learn about varieties first.  This is not only for the reason that varieties are easier, but also for the reason that one has no context for appreciating the purpose of the scheme-y machinery without first understanding the classical problems.

So, in short, look for a book on varieties first.  I'm far from knowledgeable about algebraic geometry texts, but I learned about varieties from Shafarevich, which was decent.  (Never had reason to tackle Hartshorne.)

One elementary book to try is [i]Undergraduate Algebraic Geometry[/i] by Miles Reid. This is a fairly elementary introduction to varieties and some classical applications.

If you're really interested in algebraic geometry, and you also like complex analysis, I highly recommend the book [i]Algebraic Curves and Riemann Surfaces[/i] by Rick Miranda. This book looks at algebraic curves from the complex analytic point of view, but eventually develops sheaves and cohomology. Many who read this book have suggested that it provides great preparation for the modern language of algebraic geometry, particularly for schemes. The book may not be as difficult as some other books, but I feel that is irrelevant as it has a lot of interesting material, and it does a fantastic job preparing the reader for more advanced texts.

As a final note, I've heard many stories about people who went into algebraic geometry too quickly (such as starting with schemes too quickly) and eventually were turned away from it. I think if one takes time to build the intuition necessary for schemes, algebraic geometry can be conquered, not shuddered at  :wink:


At a meeting with of $ 25$ businessmen, twelve businessmen drank coffee, and ten businessmen drank tea.  Five businessmen drank both coffee and tea.  How many businessmen drank neither coffee nor tea?

[hide]$ 12\minus{}5 \equal{} 7$ drank coffee

$ 10 \minus{} 5 \equal{} 5$ drank tea

$ 25 \minus{} 7 \minus{} 5 \minus{} 5 \equal{} \boxed{8}$ drank neither.[/hide]

[hide]Using the Principle of Inclusion Exclusion, we have the number of people who drank coffee or tea.

$ 12 \plus{} 10 \minus{} 5 \equal{} 17$ people

To find the number of businessmen who didn't drink either, we subtract our number of people who did from the total 25 people, thru complementary counting.

$ 25 \minus{} 17 \equal{} \boxed{8}$

EDIT Posted just as an addition to a clarification.

In modularmarc's solution, when he says "drank coffee" and "drank tea", it means ""just drank coffee"" and ""just drank tea"". =)[/hide]

I have a suggestion if you are having a hard time with this: read intro to C&P

[hide=my way]$$25-12-10+5=\boxed{8}$$Simple PIE.[/hide]

EDIT: also @2above there's no need to be THAT dramatic... just sayin'.

lol

[hide=sol]

by PIE, we have 12+10-5=17

we get 25-17, hence the answer 8


quadratics
geometry
calculus
ratio
derivative
trigonometry
trig identities

Find the minimum value of $ x^2 \plus{} y^2$ for the equation  $ (x\plus{}2)^2  \plus{} (y\plus{}3)^2 \equal{}25$

Please explain it easy and clear =D

[hide="Hints"]Think geometrically.

[hide="Hints for one way"]If $ M$ is the minimum value, then the circles $ x^2 \plus{} y^2 \equal{} M$ and $ (x \plus{} 2)^2 \plus{} (y \plus{} 3)^2 \equal{} 25$ are tangent.

[hide="Hint"]The perpendicular lines through tangents to circles pass through the center.[/hide][/hide]

[hide="Hints for another way"]If you don't want to think about it this way, do the substitution $ y \equal{} \sqrt {M \minus{} x^2}$ (why positive?) and minimum $ M$.

[b]Note:[/b] You can also just start out trying both $ y \equal{} \pm\sqrt {M \minus{} x^2}$ separately and see which one gets a smaller value (they are both extrema; one is the maximum and the other is the minimum).

[hide="Hint"]After simplifying, treat it as a quadratic in $ x$ and note that because $ x$ is real, the discriminant is nonnegative...[/hide][/hide][/hide]

[b]Edit:[/b] I can't read... blah.

[hide="Happy geometry (essentially what math154 said)"]
Find the minimum distance from that circle to the origin.[/hide]

I'm doing it this way. But i can't get to a result  :wink:

First i made the issolation of $ y$ in terms of $ x$, $ y \equal{} \sqrt {(x \plus{} 7)( \minus{} x \plus{} 3)} \minus{} 3$

then the value of $ x^2 \plus{} y^2$ is     $ x^2 \plus{} (\sqrt {(x \plus{} 7)( \minus{} x \plus{} 3)} \minus{} 3)^2$  the problem reduces to find the minimum value of this function....but how to find it without using differential calculus!!??

PD: Mewto...the distance from the origin to any point of the circle is always the same!

@That last part: No, its different.

I.e, you can have the point -7,-3 and 3,-3, the distance is clearly different.

ooops :blush: ......You are right....i was thinking in the distance between ratio and any point.

How to solve it using the procedure i am doing?

It wouldn't simplify very nicely so, yes, the derivative would be your best friend here.

@Mewto55555: Happy 4000th post!!

Yes, the derivative would be very useful, but i am trying to solve it without differential calculus.

[b]Mewto55555[/b] meant to minimize the distance from the circle $ (x\plus{}2)^2\plus{}(y\plus{}3)^2\equal{}25$ to the origin. You can do this if you think about my hints.

But you can also do this without calculus... expand and simplify the thing you got to get $ 30\minus{}2(2x\plus{}3\sqrt{(x\plus{}7)(3\minus{}x)})$, so we want to find the maximum value of $ 2x\plus{}3\sqrt{(x\plus{}7)(3\minus{}x)}$ to get the minimum of the desired expression. Now set $ 2x\plus{}3\sqrt{(x\plus{}7)(3\minus{}x)}\le k\implies3\sqrt{(x\plus{}7)(3\minus{}x)}\le k\minus{}2x$ and square both sides.

I believe you get a nonnegative quadratic, which means that the discriminant is nonpositive.

Now you can solve the equation from the discriminant to get the maximum value of $ k$ and thus solve the equation.

Or you can look at my hints for a nicer way.

Your way seems easy....but i don't understand why  if $ M$ is the minimum value, then $ x^2 \plus{} y^2\equal{}M$ and $ (x \plus{} 2)^{2} \plus{} (y \plus{} 3)^{2} \equal{} 25$ are tangent, neither why the minimum distance from the origin to the first circle is the minimum value. :(

$ x^2 \plus{} y^2$ is the square-distance from $ (x,y)$ to the origin.
$ (x \plus{} 2)^2 \plus{} (y \plus{} 3)^2 \equal{} 25$ means that $ (x,y)$ represents a point on the circle centered at $ ( \minus{} 2, \minus{} 3)$ with radius $ 5$.

Thus, we want to minimize the square-distance from $ (x,y)$ to the origin, given that $ (x,y)$ is on that particular circle.
In order to do so, we select the point on the circle closest to the origin as the optimal $ (x,y)$.

Now, consider the circle $ x^2 \plus{} y^2 \equal{} M$, where $ M$ is the minimum square-distance. There is only one point on the given circle that is also on this circle, since there is only one point on the given circle that has the minimum square distance. Thus, the two circles just touch.

[b]Alternate Solution:[/b] [i]This is another geometrically inspired solution, but does not directly use circles.[/i]

[hide="Solution"]Consider the triangle formed by vertices $ A(0,0)$, $ B( \minus{} 2, \minus{} 3)$, and $ X(x,y)$.

We know $ AB \equal{} \sqrt {13}$ by the distance formula, and $ BX \equal{} 5$, which is the squareroot of the given condition on $ x$ and $ y$. Let the third side be $ AX \equal{} \sqrt {M}$. Let the angle $ \angle ABX$ have measure $ \theta$. By the law of cosines:

$ 13 \plus{} 25 \minus{} 2\cdot 5 \cdot \sqrt {13} \cos \theta \equal{} M$
$ \implies M \equal{} 38 \minus{} 10 \sqrt {13} \cos \theta$

We want to minimize $ M$, so we want to maximize $ \cos \theta$, so we have $ \theta \equal{} 0 \implies M \equal{} 38 \minus{} 10\sqrt {13}$[/hide]

[b]Chalao:[/b] Sorry for my hasty explanations; I think [b]TZF[/b] explained pretty well why they are tangent.

[hide="Solution using my first approach"]Now, remember that the perpendicular line to a tangent of a circle passes through the center of that circle. Because the two circles are tangent to each other at a point, they share a tangent line, and thus the perpendicular line to that tangent passes through both circles. This means that the point of tangency lies on the line through the two centers $ (0,0)$ and $ ( \minus{} 2, \minus{} 3)$, which is easily found to be $ y \equal{} \frac32x$.

Now just plug this in $ (x \plus{} 2)^2 \plus{} (y \plus{} 3)^2 \equal{} 25$ (because the point of tangency is on this circle) to get $ (x^2 \plus{} 4x \plus{} 4) \plus{} \left(\frac94x^2 \plus{} 9x \plus{} 9\right) \equal{} 25\implies$, which has solutions $ x \equal{} \minus{} 2\pm\frac {10}{\sqrt {13}}$.

Finally, $ x^2 \plus{} y^2 \equal{} x^2 \plus{} \left(\frac32x\right)^2 \equal{} \frac {13}4x^2 \equal{} \frac {13}4\left(4\pm\frac {40}{\sqrt {13}} \plus{} \frac {100}{13}\right)$

$ \equal{} 13 \plus{} \frac {100}4\pm10\sqrt {13} \equal{} 38\pm10\sqrt {13}$, and because we want the minimum value, the answer is $ \boxed{38 \minus{} 10\sqrt {13}}$.[/hide]

[hide="Solution using my second approach"]
$ x^2\plus{}y^2\equal{}M\implies y\equal{}\pm\sqrt{M\minus{}x^2}$

$ \implies25\equal{}(x\plus{}2)^2\plus{}(y\plus{}3)^2\equal{}(x^2\plus{}4x\plus{}4)\plus{}(M\minus{}x^2\pm6\sqrt{M\minus{}x^2}\plus{}9)$

$ \equal{}M\plus{}4x\plus{}13\pm6\sqrt{M\minus{}x^2}\implies M\plus{}4x\minus{}12\equal{}\pm6\sqrt{M\minus{}x^2}$

$ \implies M^2\plus{}16x^2\plus{}144\plus{}8Mx\minus{}96x\minus{}24M\equal{}36M\minus{}36x^2$

$ \implies52x^2\plus{}8(M\minus{}12)x\plus{}(M^2\minus{}60M\plus{}144)\equal{}0$ (1).

$ x$ is always real, so the discriminant must be nonnegative:

$ 64(M\minus{}12)^2\minus{}4\cdot52(M^2\minus{}60M\plus{}144)\ge0$

$ \implies\minus{}144(M^2\minus{}76M\plus{}144)\ge0\implies M^2\minus{}76M\plus{}144\le0$

$ \implies(M\minus{}(38\minus{}10\sqrt{13}))(M\minus{}(38\plus{}10\sqrt{13}))\le0$, so

$ M\ge\boxed{38\minus{}10\sqrt{13}}$.

Equality can be found by plugging $ M$ in (1).[/hide]

[hide="Trigonometric approach"]With reference to the pythagoras theorem,
$ (x\plus{}3)$=base
$ (y\plus{}3)$=perpendicular
$ 25$=hypotenuse

then we are left to find out the minimum value of $ (25\sin\theta\minus{}3)^{2}\plus{}(25\cos\theta\minus{}3)^{2}$ which is easy.[/hide]


[quote="Genecks"]I'm starting to do some equations with rational exponents. I'm able to figure out the even types of rational exponents, but I don't know how to calculate the odd types of rational exponents, such as (-3125)^1/5

I mean, that's a pretty big number to start using process of elmination on. And I really don't think I'll have the time to do such a thing on a test.

What's the best way to figure out problems with odd rational exponents?[/quote]
Odd rational exponents? What do you do for even? I do the same method for both ways. In this as soon as I saw the 5 in the units digit I thought it must be $5^{5}$. So that would be $-5$.

the $\frac{1}{5}$ exponent means take the fifth root then take the power of one.
in all fractional exponets in the form of x/y, you take the xth power of the number, then the yth root, or visa versa, both give you the same result.

How do I take the fifth root? I can't exactly do that on a plain calculator. I've been using the root function for even numbers, so that has simplified some things.

Is there some kind of inductive/deductive logic or method to it all?

[quote="Genecks"]So there is really no way? It's all a bunch of critical, quick thinking?[/quote]
Or memorizing some. :P

[quote="Genecks"]Is there some kind of inductive/deductive logic or method to it all?[/quote]

Well, if you suspect that the answer is an integer, you can take fifth powers of numbers until you get what you want.

$1^{5}= 1$
$2^{5}= 32$
$3^{5}= 243$
$4^{5}= 1024$
$5^{5}= 3125$

And there you have it.  (Many problem-solvers have some common powers of small integers memorized, such as $3125$.  This is especially true of powers of $2$.)

[quote="Genecks"]So there is really no way? It's all a bunch of critical, quick thinking?[/quote]

I would call critical, quick thinking the exact opposite of "no way."   :)

If you mean that there's no quick way to calculate roots by hand, then you would be correct, and that is why you take advantage of the following law:

[b]t0rajir0u's law of problem-writing:[/b]  If a problem looks hard, assume that the answer is simple because that is often how a problem is written.  (For example, assuming that the answer was an integer worked well here.)

In this case since the units digit is 5, you can assume that the answer ends in a 5.

[quote="cincodemayo5590"][quote="davidlizeng"]what i mean is if you get some weird radical that you can simplify by hand like $\sqrt[3]{2}$[/quote]

Unless you've got basic radicals memorized (they do come in handy  in competitions, but I personally don't know that many off the top of my head), I'm pretty sure that you can't calculate this directly.

What you can do is guess-and-check your way through the decimal; i.e:
$1.2^{3}<2<1.3^{3}$; guessing: $1.24^{3}<1.25^{3}<2<1.26^{3}$; and so on, but this becomes time consuming to do by hand after the first few places.

There are few that I'm sure more than a couple people know, just because they come up so often:
$\sqrt 2 \approx 1.414, \ \sqrt{3}\approx 1.73, \ \sqrt 5 = 2 \phi-1 \approx 2.236$ from the golden ratio, etc.[/quote]

wtf who knows that last one???

also contests don't really ask you to do square roots, etc

[quote="Altheman"][quote="cincodemayo5590"][quote="davidlizeng"]what i mean is if you get some weird radical that you can simplify by hand like $\sqrt[3]{2}$[/quote]

Unless you've got basic radicals memorized (they do come in handy  in competitions, but I personally don't know that many off the top of my head), I'm pretty sure that you can't calculate this directly.

What you can do is guess-and-check your way through the decimal; i.e:
$1.2^{3}<2<1.3^{3}$; guessing: $1.24^{3}<1.25^{3}<2<1.26^{3}$; and so on, but this becomes time consuming to do by hand after the first few places.

There are few that I'm sure more than a couple people know, just because they come up so often:
$\sqrt 2 \approx 1.414, \ \sqrt{3}\approx 1.73, \ \sqrt 5 = 2 \phi-1 \approx 2.236$ from the golden ratio, etc.[/quote]

wtf who knows that last one???[/quote]
Well, $\phi=\frac{1+\sqrt5}2$, so $\sqrt5=2\phi-1$.

[quote="davidlizeng"]you should have a button that looks like $\sqrt[y]{x}$
then just type in the values :)
if you dont, i suggest you get a calculator that does.
its useful[/quote]

or you can just type  in (-3125)^(1/5)

If you have a graphing calculator of the family of TI-83 and TI-84, you would have to go to the $MATH$ key and scroll down until you see a sign that looks like this: $\sqrt[x]{}$ (or you could just press math and then 5).

For any exponent in the form $y^\frac{1}{x}$, it is the same as $\sqrt[x]{y}$ but I don't know how you would do this problem by hand...

How do calculators do it then? They must have some formula they use.

Probably logarithms.

And if you're wondering how a calculator calculates logarithms, it would be with power series, which is too complicated a subject to get into here.


probability
linear algebra
matrix
algebra
system of equations
probability and stats

A coin is tossed repeatedly, heads appearing with probability $ p$ on each toss.
Let $ X$ be the number of tosses until the first occasion by which three heads have appeared successively.
Find $ E(X)$.

Let $ T_k$ be the expected remaining time in the process, conditioned on the last $ k$ flips having been heads. Note that $ T_3\equal{}0.$

Considering what happens on the next flip gives us the following system of equations:

$ T_0\equal{}1\plus{}\frac12T_0\plus{}\frac12T_1$
$ T_1\equal{}1\plus{}\frac12T_0\plus{}\frac12T_2$
$ T_2\equal{}1\plus{}\frac12T_0$

This $ 3\times 3$ linear system can be solved readily. We get that $ T_2\equal{}8,T_1\equal{}12,$ and $ T_0\equal{}14.$

$ T_0$ is the quantity asked for. $ E(X)\equal{}14.$

Conditional expectation is a powerful tool.

Let $ X_m$ be the number of tosses until the the first time $ m$ successive heads appear.

$ E(X_1)\equal{}2,\ E(X_2)\equal{}6,$ and from above, $ E(X_3)\equal{}14.$

Claim: $ E(X_m)\equal{}2^{m\plus{}1}\minus{}2.$

Let $ Y_m$ be the number of tosses preceeding the first sequence of $ m$ successive heads. Looking at when the first tail appears, we get
$ EY_m\equal{}\frac 12 (1\plus{}EY_m)\plus{}\frac 14(2\plus{}EY_m)\plus{}\frac 18(3\plus{}EY_m)\plus{}\dots\plus{}\frac 1{2^{m}}(m\plus{}EY_m)$
(the $ k$-th term corresponds to the situation when you start with $ k\minus{}1$ heads followed by a tail). Solving this equation, we get $ EY_m\equal{}2^{m\plus{}1}\minus{}2\minus{}m$, which is equivalent to Kent's claim ($ X_m\equal{}m\plus{}Y_m$).

My own solution is to set up the system of equations for $ T_0,T_1,\dots,T_{m\minus{}1}$ exactly as above. Moving the variables to the left-hand side and multiplying by $ 2$ to clear the fractions, we get the following augmented matrix:

$ \begin{bmatrix}1&\minus{}1&0&0&\cdots&0&0&|&2\\
\minus{}1&2&\minus{}1&0&\cdots&0&0&|&2\\
\minus{}1&0&2&\minus{}1&\cdots&0&0&|&2\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&|&\vdots\\
\minus{}1&0&0&0&\cdots&2&\minus{}1&|&2\\
\minus{}1&0&0&0&\cdots&0&2&|&2\end{bmatrix}$

This matrix is easy to row-reduce to the following echelon form:

$ \begin{bmatrix}1&\minus{}1&0&0&\cdots&0&0&|&2\\
0&1&\minus{}1&0&\cdots&0&0&|&4\\
0&0&1&\minus{}1&\cdots&0&0&|&8\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&|&\vdots\\
0&0&0&0&\cdots&1&\minus{}1&|&2^{m\minus{}1}\\
0&0&0&0&\cdots&0&1&|&2^m\end{bmatrix}$

And from that we get $ T_{m\minus{}1}\equal{}2^m, T_{m\minus{}2}\equal{}2^m\plus{}2^{m\minus{}1}$ and so on until we get $ T_0\equal{}2^m\plus{}2^{m\minus{}1}\plus{}\cdots\plus{}4\plus{}2\equal{}2^{m\plus{}1}\minus{}2.$


logarithms algorithm algebra polynomial inequalities invariant

Is there somebody that has read and has understood the paper http://www.math.princeton.edu/~annals/issues/2004/Sept2004/Agrawal.pdf ?

I have 2 question :
1) Why $\sqrt{r}\log n < r$  (page 788) ? We know that $r < \max\{3, \log^{5}n\}$ and that $n > r$ (from line 4 of algorithm -page 784).

2) In the original article, is explained that $\frac{x^{r}-1}{x-1}$ factorises into irruducible polynomials each of degree $o_{r}(p)$ with $r$ and $p$ primes in hypothesis. But in the new article [b]r isn't said prime[/b].
Is there a proof of theorem for r composite or  for all r ( r-th polynomial cyclotomic)?

Thanks and excuse me for my english.

I did not read the article in details, so I may make a mistake. But note that the choice of $r$ is such that $o_{r}(n)>\log(n)^{2}$ and since $o_{r}(n)$ is a divisor of $\phi(r)$, thus smaller than $r$, you have the first inequality that interests you.
   Secondly, it is well-known that the $n$-th cyclotomic polynomial decomposes over $F_{p}$ as a product of $\phi(n)/k$ polynomials of degree $k$, where $k$ is the order of $p$ in $U(Z/nZ)$ (if $p$ is a prime and does not divide $n$). This answers the second question.

I have understood the first answer :-) !  THANKS

But in the second answer, you write : "Se $p$ is a prime and does not divide $n$". But $p$ is prime and $p$ divides $n$.

Where can I find the proof of this theorem ?

Oh, come on! I said that the $n$-th cyclotomic polynomial decomposes like that if $n$ and $p$ are relatively prime. Here you have to replace $n$ by $r$ and it's done. The $n$ in my statement is not the $n$ in the article, it's actually $r$.

AHHH OK!!! :P I have understood, I'm stupid!

Where can i find the proof that "$n$-th cyclotomic polynomial decomposes...." ?

Can you give me a title? or a internet address?

I really hope this is the last question. Ok, better than a title, a proof. So, take $n$ and $p$ relatively prime and take $x$ a primitive root of $1$ of order $n$, where $x$ lies in $F_{p^{k}}$, the field with $p^{k}$ elements. Then the minimal polynomial of $x$ also has roots $x^{p}, x^{p^{2}},...,x^{p^{k-1}}$, where $k$ is the order of $p$ mod $n$ (I used here the well-known fact-and don't ask where it can be found!- that the Galois group of $F_{p^{k}}$ over $F_{p}$ is generated by the Frobenius automorphism and has order $k$). Thus the minimal polynomial of $x$ is a multiple of the product of $X-x^{p^{j}}$ for $0\leq j<k$. Or, this last polynomial has coefficients in $F_{p}$, since it is invariant by Frobenius's automorphism. Thus, it is actually the minimal polynomial of $x$ and it has degree $k$. This shows that the $\varphi(n)$ primitive roots of the unity of order $n$ can be partitioned into classes of $k$ elements, each class forming an irreducible factor of $\phi_{n}(X)$. This finishes the proof.


I will check your solution, but I think it exists a simpler solution that I cannot find. :oops:

[quote="Little Gauss"]I think that inequality is so strong.
This is my ugly solution.
$LHS\le \sqrt{2(a+b+c)}\sqrt{\sum_{cyc}\frac{1}{(b+c)(a^{2}+bc)}}$ (Jensen)
ETS : $(a+b+c)\sum_{cyc}\frac{1}{(b+c)(a^{2}+bc)}\le (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})^{2}$.
$\iff [8,1,1]+4[7,2,1]+2[6,4,2]+2[6,3,1]+4[6,2,2]+3[5,5,0]+4[5,4,1] \ge 6[5,3,2]+6[4,4,2]+8[4,3,3]$
Majorize![/quote]
I think my solution is much simpler, using Cauchy Schwarz inequality, we have
\[\left(\sum\frac{1}{\sqrt{a^{2}+bc}}\right)^{2}\le \left(\sum\frac{1}{(a+b)(a+c)}\right)\left(\sum\frac{(a+b)(a+c)}{a^{2}+bc}\right)=\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}\left(\sum\frac{a(b+c)}{a^{2}+bc}+3\right) \]
Hence, it suffices to prove that
\[\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}\left(\sum\frac{a(b+c)}{a^{2}+bc}+3\right) \le \left(\sum\frac{1}{b+c}\right)^{2}\]
or
\[\sum\frac{a(b+c)}{a^{2}+bc}+3 \le \frac{(a^{2}+b^{2}+c^{2}+3ab+3bc+3ca)^{2}}{(a+b)(b+c)(c+a)(a+b+c)}\]
or
\[\sum (a-b)(a-c)\left(\frac{1}{a^{2}+bc}+\frac{1}{(b+c)(a+b+c)}\right) \ge 0 \]
Now, WLOG, suppose that $a \ge b \ge c$, then we have
\[LHS \ge \frac{(a-b)(b-c)}{b}\left(a\left(\frac{1}{a^{2}+bc}+\frac{1}{(b+c)(a+b+c)}\right)-b\left(\frac{1}{b^{2}+ca}+\frac{1}{(a+c)(a+b+c)}\right) \right) =\frac{c(a-b)^{2}(a+b)(b-c)(a^{2}+b^{2}-ab+ac+bc)}{b(a+c)(b+c)(a^{2}+bc)(b^{2}+ca)}\ge 0 \]
We are done. :) Equality holds if and only if $a=b=c$.

[quote=manlio]For any positive real numbers $a,b$ and $c$,
\[\underset{cyc}{\sum}\frac{1}{\sqrt{a^{2}+bc}}\le \underset{cyc}{\sum}\frac{\sqrt{2}}{a+b}\\ \][/quote]

[img]http://s6.sinaimg.cn/middle/0018zOAxgy70grQRwCF15&690[/img]


geometry
3D geometry
number theory open
number theory

It is well known that all numbers are expressible as the sum of four squares.
Is there any upper bound on n where all numbers being expressible as the sum of n cubes? Has it been proven that there exists no n?

See the following links:

[code]http://en.wikipedia.org/wiki/Waring's_problem
http://mathworld.wolfram.com/WaringsProblem.html[/code]

The problem is interesting but how can we prove it for n=3 or n=4 , ...

For $ n\equal{}3$ it's wrong, for $ n\equal{}4$ it's open. But note that the full Waring theorem states that "enough" _nonnegative_ cubes suffice.

you can see this in solved questions (in Number theory)


Basic formula applicable here: $ t_n = a + (n - 1)d$
$ S_n = \frac {n}{2} [2a = (n - 1)d]$ -or- $ \frac {n}{2} [first term - last term]$
Where $ t$=term, $ a$=first term, $ d$=common difference, $ n$=number of terms, $ S$=sum

Here are few Arithmatical Progression problems which I am unable to solve:

1. [hide]In a children potato race $ n$ potatoes are placed 1 metre apart in a straight line. A competitor starts from a point in the line which is 5m from the nearest potato. Find an expression for the total distance run in collectin the potatoes one, at a time and bringing them back one at a time to the starting point. Calculate the value of $ n$ if the total distance run is 162 metres.
[/hide]

2. [hide]If $ a, b, c$ are in A.P. then show that $ b + c, c + a, a + b$ are also in A.P.[/hide]

3. [hide]If $ (b - c)^2, (c - a)^2, (a - b)^2$ are in A.P. , then prove that $ \frac {1}{b - c}, \frac {1}{c - a}, \frac {1}{a - b}$ , are also in A.P.[/hide]

4.[hide] If $ a_1, a_2, ....... a_n$ are in A.P. , then show that $ \frac {1}{a{}_1a{}_2} + \frac {1}{a{}_2a{}_3} + \frac {1}{a{}_3a{}_4} + ........\frac {1}{a_n_ - {}_1a{}_n} = \frac {n - 1}{a{}_1a{}_n}$[/hide]

5. [hide]If the $ p{}^t{}^h, q{}^t{}^h and r{}^t{}^h$ terms are $ P, Q, R$ respecitvely, prove that $ p(Q - R) + q(R - P) + r(P - Q) = 0$[/hide]

[quote="CyanSnow"]Basic formula applicable here: $ t_n = a + (n - 1)d$
$ S_n = \frac {n}{2} [2a = (n - 1)d]$ -or- $ \frac {n}{2} [first term = last term]$
Where $ t$=term, $ a$=first term, $ d$=common difference, $ n$=number of terms, $ S$=sum

Here are few Arithmatical Progression problems which I am unable to solve:

1. [hide]In a children potato race $ n$ potatoes are placed 1 metre apart in a straight line. A competitor starts from a point in the line which is 5m from the nearest potato. Find an expression for the total distance run in collectin the potatoes one, at a time and bringing them back one at a time to the starting point. Calculate the value of $ n$ if the total distance run is 162 metres.
[/hide]

2. [hide]If $ a, b, c$ are in A.P. then show that $ b + c, c + a, a + b$ are also in A.P.[/hide]

3. [hide]If $ (b - c)^2, (c - a)^2, (a - b)^2$ are in A.P. , then prove that $ \frac {1}{b - c}, \frac {1}{c - a}, \frac {1}{a - b}$ , are also in A.P.[/hide]

4.[hide] If $ a_1, a_2, ....... a_n$ are in A.P. , then show that $ \frac {1}{a{}_1a{}_2} + \frac {1}{a{}_2a{}_3} + \frac {1}{a{}_3a{}_4} + ........\frac {1}{a_n_ - {}_1a{}_n} = \frac {n - 1}{a{}_1a{}_n}$[/hide]

5. If the $ p{}^t{}^h, q{}^t{}^h and r{}^t{}^h$ terms are $ P, Q, R$ respecitvely, prove that $ p(Q - R) + q(R - P) + r(P - Q) = 0$[/quote]

For the first two,

[hide="Question One"]
Try writing down the first terms ran, so

$ 2.5, 2(5 + 1),..$

Note the pattern,

Then try it with $ a = 0$ in your normal equation.

Then look at the sum. Set it to the answer and solve for an unknown
[/hide]

[hide="Question Two"]
Look at the A.P with difference 0!
[/hide]

Question 1:
I'm sorry to say I don't understand your explanation. I don't understand the whole question, if i did then I'd be able to solve it by myself. Please clarify the question by step by step explanation.

Question 3:
common difference: $ \frac {1}{c \minus{} a} \minus{} \frac {1}{b \minus{} c} \equal{} \frac {b \minus{} 2c \plus{} a}{cb \minus{} c^2 \minus{} ab \plus{} ac}$ is not equal to zero  :huh:

OK

Question 1

Say there was one potato.

They would run 5 metres to it, and 5 metres back. Thats $ 2.5$. Say there were two potatoes, he would run $ 2.5$ for the first one, and then $ 2.(5 \plus{} 1)$ for the second one, so in total $ 2.5 \plus{} 2(5 \plus{} 1)$, continue this for $ n$ potatoes. So now we see an A.P. with $ a \equal{} 10, d \equal{} 2$

Once you set that up in the sum equation, set it to 162 (or w/e it was) and solve for n.

Question 2

Say a=1,b=1

Then the A.P = 1,1,1,1,1,1,etc

Hence a+b isn't in the AP

[quote="SimonM"]Question 2

Say a=1,b=1

Then the A.P = 1,1,1,1,1,1,etc

Hence a+b isn't in the AP[/quote]

Different A.P.'s...

If $ a,b,c$ are in A.P., then $ b\minus{}a\equal{}c\minus{}b$. Now look at $ b\plus{}c,c\plus{}a,a\plus{}b$. This forms another A.P.

[hide="Question 3"]$ (c\minus{}a)^2\minus{}(b\minus{}c)^2\equal{}(a\minus{}b)^2\minus{}(c\minus{}a)^2$

$ (b\minus{}a)(2c\minus{}a\minus{}b)\equal{}(2a\minus{}b\minus{}c)(c\minus{}b)$

$ \frac{(b\minus{}a)(2c\minus{}a\minus{}b)}{(b\minus{}c)(c\minus{}a)(a\minus{}b)}\equal{}\frac{(2a\minus{}b\minus{}c)(c\minus{}b)}{(b\minus{}c)(c\minus{}a)(a\minus{}b)}$

$ \frac{2c\minus{}a\minus{}b}{(b\minus{}c)(c\minus{}a)}\equal{}\frac{2a\minus{}b\minus{}c}{(c\minus{}a)(a\minus{}b)}$

$ \frac1{b\minus{}c}\minus{}\frac1{c\minus{}a}\equal{}\frac1{c\minus{}a}\minus{}\frac1{a\minus{}b}$[/hide]

For Question 4, assume all terms are non-zero or the argument breaks down

[hide="5"]
$ \indent\sum_{p,q,r}p[(a_1 \plus{} (q \minus{} 1)d) \minus{} (a_1 \plus{} (r \minus{} 1)d)]$

$ \Rightarrow \sum_{p,q,r} p[d(q \minus{} r)]$

$ \Rightarrow \sum_{p,q,r} pdq \minus{} pdr$

$ \Rightarrow pdq \minus{} pdr \plus{} qdp \minus{} qdr \plus{} rdp \minus{} rdq \equal{} 0$

$ \boxed{\text{QED}}$[/hide]


Source: Massachusetts Mathematics League March 2003

A telephone operator makes 755 connections in a day with a probability of 0.001 of connecting to the wrong number each time a connection is made. To the nearest thousandth, what is the probability that in one day at least one connection is wrong.

[hide="My solution"]
The probability that one particular connection isn't wrong is $0.999$. So the probability that all 755 connections aren't wrong is $0.999^{755}$. So the probability that NOT all 755 connections were correct—that is, the probability that at least one connection was wrong—is $1-0.999^{755} \approx 0.530$.
[/hide]

[hide="My solution"]

P(at least 1 wrong) = 1 - P(all correct)
                                = 1 - (.999)^755[/hide]

Don't have my calculator here, so I can't give you the exact answer...  :oops:


geometry
function
inequalities
advanced fields
advanced fields theorems

Does anyone knows a reference where I can found a proof of Weil's theorem. I'm interested to see what kind of ideas lies in the proof? Of course the proof is in original article (foundation of algebraic geometry) but often there are books in which theorem are explained with more details. I tried once read Stepanov's proof in his original paper but I got confused by those algebraic manipulations.

I think I found a proof from the book Fried, Jarden: "Field arithmetic". However, I have some difficulties to understand the proof. Namely, they let $ K$ to be a finite field of characteristic $ p$ with $ q$ elements. Let $ F$ be a function field of one variable over $ K$ and let $ g$ be the genus of $ F$. Let $ r$ be a fixed positive integer and denote the unique extension of $ K$ of degree $ r$ by $ K_r$ (similarly $ F_r$). Then they prove the following:

TH1: The Riemann hypothesis holds for the function field $ F/K$ if and only if it holds for the function field $ F_r/K_r$.

The prove that theorem by proving the following:

TH2: Let $ F$ be a function field of one variable over a finite field $ K$ and let $ \sigma$ be an automorphism of $ F$ over $ K$ of finite order. Then $ K$ has a finite extension $ K'$ with $ q'$ elements and there exists a positive constant $ c$ such that for every positive integer $ r$ we have
\[ |N^{(\sigma )}(F_r^\prime )\minus{}((q^\prime )^r\minus{}1)|\leq c(q^\prime )^{r/2},\]
where $ K_r^\prime$ is the unique extension of $ K^\prime$ of degree $ r$, the field $ F_r^\prime$ is equal to $ K_r^\prime F^\prime$ and $ \sigma$ extends to an automorphism, also denoted by $ \sigma$, of $ F_r^\prime$ over $ K_r^\prime$.

Two questions: They don't define $ q'$. Is $ q'\equal{}q^r$ because the extension is of degree $ r$?

They say that TH1 follows by choosing $ \sigma\equal{}1$. But why this is enough to show TH1 because in TH_2 if $ \sigma\equal{}1$ is an automorphism then
\[ |N^{(\sigma )}(F_r^\prime )\minus{}((q^\prime )^r\minus{}1)|\equal{}|N(F_r^\prime )\minus{}((q^\prime )^r\minus{}1)|.\]
But in Weil's theorem the inequality is of the form $ |N\minus{}(q\plus{}1)|\leq 2q\sqrt{q}$, where $ N$ is the number of prime divisors of $ F/K$, not $ |N\minus{}(q\minus{}1)|\leq 2g\sqrt{q}$


I'm new to LaTeX, so I copy-pasted the Hello World file in the LaTeX section and tried to compile. This is my exact code:

\documentclass[11pt]{article}
\begin{document}
Hello, world!
\end{document}

When I try to compile, it gives the source path of the latex.exe file in MiKTeX\bin\ and then says

Cannot execute command.Error:%1 is not a valid Win32 appli...

Can someone please help me solve this problem?

Also, when I installed TeXnicCenter, it never asked me for a DVI viewer. Is this supposed to happen?

As for the DVI viewer, you may already have a system default viewer on your computer, so it didn't need to ask.  As for the other, how did you install MiKTeX?  Also, what version of Windows are you using?

I installed MiKTeX from the link in the LaTeX section of this website. My windows version is XP.

It sounds like at least one of the installations went awry - did TeXnicCenter ask for a MiKTeX to be used as the TeX compiler?

I might try reinstalling both.

Yes, you should reinstall both, and make sure you got the latest version, and the version for your operating system :)

Seems just not the normal installation error :?

I reinstalled everything and I got the same error. I downloaded MiKTeX from http://miktex.org/setup.html, and TeXnicCenter Setup, Version 1 Beta 6.21 ('Fawkes') from http://toolscenter.texniccenter.org/front_content.php?idcat=50. I did change the path of installation. I don't know if that would affect it.
Thanks for your help! I hope I can get this working because it sounds like really nice software.

Try installing without changing the path.

Also, are you installing the 'small' package or the whole deal?  (You should be fine with the 'small')


geometry
3D geometry
percent
Stanford
college
percent increase
AMC 8

A cube has edge length 2. Suppose that we glue a cube of edge length 1 on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area from the original cube to the new solid formed is closest to:

A) 10  B) 15 C) 17 D) 21 E) 25

How would you do this problem?

Let's see. I am taking the AMC 8 on tuesday as well. first time for me

cube has edge length 2 so its surface area is 24. with 6 faces of 4.

then since a cube with edge length is completely ontop of the cube, you subtract 1x1 equals 1 length from 24 to get 23.

Then, with 5 faces showing of 1 area each, you add 5 to 23 to get 28 which is the answer. so 24 to 28 is like a 1/6  increase which is 16 2/3 percent so it is answer C. 17 percent

now a ranger, kind of a non-sequitur, but are you taking the AMC 8 at stanford university?
My bro is... He might see you there.

Nah. I'm from Alabama.    :(

I am taking it in class since it's only like 40 minutes.


Find $ x,y,z \in Z$ such that :$ x!y! \equal{} z!$

It is obvious that $ (x,y,z)\equal{}(0,1,0),(1,0,0),(1,1,1),(1,2,2),(2,1,1)$ are all solutions, but I do not know how to get here :(

[quote="Potla"]It is obvious that $ (x,y,z) \equal{} (0,1,0),(1,0,0),(1,1,1),(1,2,2),(2,1,1)$ are all solutions, but I do not know how to get here :([/quote]

Not all solutions.  :wink:
Note that $ \boxed{(0,a,a),(1,a,a),(a,0,a),(a,1,a),(a,a,0),(a,a,1)}$ are solutions also :).

And $ (n!,(n! \minus{} 1)!,(n!)!)$ and it's permutation, too  :wink:
But $ (a,a,0)$ or $ (2,1,1)$ (and some others) aren't a solution  :wink:

Don't forget $ 6! \times 7! \equal{} 10!$.

In general, we need to find $ n$ consecutive integers $ x_1,x_2,...x_n$ such that $ \prod x_i\equal{}k!$ for some $ k$.

I just ran some code, which showed there are no sequences of 2-21 consecutive integers on the interval [3,30000] which multiply to a factorial  other than $ {7,8,9,10}$ and $ {8,9,10}$.

I'm working on proving those are the only ones. Currently trying to approach it based on the relative factorizations of $ {a,a\plus{}1,a\plus{}2,...}$ for a given $ a$.


Hi and welcome to the CTC Winter Break Tourney 2008, hosted by isabella2296 and ernie! Our players are as shown in the first post of the Sign-Ups thread. Here's the seeding and matches, as determined by ernie using random.org:

Seeding:

1 CA Math
2 gogators
3 GoldenArcher
4 AIME15
5 Dojo
6 FantasyLover
7 Drig0nfive52
8 BOGTRO
9 mathblitz
10 jjx1
11 AIME_is_hard
12 lotsofmath
13 Discrete_Math
14 Jongao
15 mewto55555
16 someperson01

Games:

game A- 1 CA Math vs. 16 someperson01
[color=red]game B- 2 gogators vs. 15 mewto55555(mewto55555 leading 1-0)[/color]
game C- 3 GoldenArcher vs. 14 Jongao
game D- 4 AIME15 vs. 13 Discrete_Math
game E- 5 Dojo vs. 12 lotsofmath
game F- 6 FantasyLover vs. 11 AIME_is_hard
[color=darkred]game G- 7 Drig0nfive52 vs. 10 jjx1(Drig0nfive52 wins!)[/color]
game H- 8 BOGTRO vs. 9 mathblitz
game I- 1/16 vs. 8/9
game J- 2/15 vs. Drig0nfive52
game K- 3/14 vs. 6/11
game L- 4/13 vs. 5/12
game M- I vs. L
game N- J vs. K
game O- winner(M) vs. winner(N), 1st place match
game P- loser(M) vs. loser(N), 3rd place match

The games are due 12/24 at midnight but finish as fast as possible! Right now we're doing Games A-H. I am sending out reminder pms after typing this.

Rules:

- You must play exactly 5 Countdowns. Whoever wins the majority is the winner of the match.

- You may play as a guest but this is not mandatory.

- Have fun!

Mewto55555 won Game 1 of 5 against gogators, 4-0.

Drig0nfive52 won 3 cds.
they were 4-1, 4-2, 4-0.

Jongao leads GoldenArcher 2-0 (4-3,4-3).
I'll update this later once we finish.

FantasyLover vs. AIME_is_hard

Game# - Score - Winner

1 - 1-4 - AIME_is_hard
2 - 4-0 - FantasyLover
3 - 3-4 - AIME_is_hard
4 - 4-1 - FantasyLover
5 - 4-2 - FantasyLover

hm i think fantasylover wins

Games:

[color=green]game A- 1 CA Math vs. 16 someperson01 (CA Math wins!)[/color]
[color=orange]game B- 2 gogators vs. 15 mewto55555 (mewto55555 leads 1-0)[/color]
[color=green]game C- 3 GoldenArcher vs. 14 Jongao (Jongao wins!)[/color]
[color=red]game D- 4 AIME15 vs. 13 Discrete_Math[/color]
[color=red]game E- 5 Dojo vs. 12 lotsofmath[/color]
[color=green]game F- 6 FantasyLover vs. 11 AIME_is_hard (FantasyLover wins!)[/color]
[color=green]game G- 7 Drig0nfive52 vs. 10 jjx1 (Drig0nfive52 wins!)[/color]
[color=red]game H- 8 BOGTRO vs. 9 mathblitz[/color]
[color=indigo]game I- 1 CA Math vs. 8/9[/color]
[color=indigo]game J- 2/15 vs. 7 Drig0nfive52[/color]
[color=darkblue]game K- 14 Jongao vs. 6 FantasyLover[/color]
[color=violet]game L- 4/13 vs. 5/12[/color]
[color=violet]game M- I vs. L[/color]
[color=violet]game N- J vs. 14/6[/color]
[color=violet]game O- winner(M) vs. winner(N), 1st place match[/color]
[color=violet]game P- loser(M) vs. loser(N), 3rd place match[/color]

The games are due 12/24 at 11:59 pm but finish as fast as possible!

[i]Since mewto55555 is being cranky on color, I'll have to change it up.[/i]

Colors:
Violet- not yet ready to be played
Indigo- almost ready to be played
Dark Blue- ready to be played, but not this round
Green- completed
Orange- in progress
Red- ready to be played, due at above date

Why does that look totally familiar? Orange for games partway done, green and red for games done and not started (in a different order). The seeding at the top? All you are missing is a hide tags to have a total copy of my CTC tournament. I guess CTC is standardized now?

anyway I sent a pm on the first day to gogators. She has picked this up but has not replied. The PM was asking a time that was good to play.

[quote="mewto55555"]Why does that look totally familiar? Orange for games partway done, green and red for games done and not started (in a different order). The seeding at the top? All you are missing is a hide tags to have a total copy of my CTC tournament. I guess CTC is standardized now? [/quote]

Not rly.  I don't see you as president.

Sent out another round of private messages. Ernie, want to extend the deadline another day?

Games:

[color=green]game A- 1 CA Math vs. 16 someperson01 (CA Math wins!)[/color]
[color=green]game B- 2 gogators vs. 15 mewto55555 (mewto55555 wins!)[/color]
[color=green]game C- 3 GoldenArcher vs. 14 Jongao (Jongao wins!)[/color]
[color=red]game D- 4 AIME15 vs. 13 Discrete_Math[/color]
[color=red]game E- 5 Dojo vs. 12 lotsofmath[/color]
[color=green]game F- 6 FantasyLover vs. 11 AIME_is_hard (FantasyLover wins!)[/color]
[color=green]game G- 7 Drig0nfive52 vs. 10 jjx1 (Drig0nfive52 wins!)[/color]
[color=red]game H- 8 BOGTRO vs. 9 mathblitz[/color]
[color=indigo]game I- 1 CA Math vs. 8/9[/color]
[color=indigo]game J- 2/15 vs. 7 Drig0nfive52[/color]
[color=darkblue]game K- 14 Jongao vs. 6 FantasyLover[/color]
[color=violet]game L- 4/13 vs. 5/12[/color]
[color=violet]game M- I vs. L[/color]
[color=violet]game N- J vs. 14/6[/color]
[color=violet]game O- winner(M) vs. winner(N), 1st place match[/color]
[color=violet]game P- loser(M) vs. loser(N), 3rd place match[/color]

The games are due 12/24 at 11:59 pm but finish as fast as possible!

Colors:
Violet- not yet ready to be played
Indigo- almost ready to be played
Dark Blue- ready to be played, but not this round
Green- completed
Orange- in progress
Red- ready to be played, due at above date

* Mew now leads 2-0.
* Mew wins!

Games:

[color=green]game A- 1 CA Math vs. 16 someperson01 (CA Math wins!)[/color]
[color=green]game B- 2 gogators vs. 15 mewto55555 (mewto55555 wins!)[/color]
[color=green]game C- 3 GoldenArcher vs. 14 Jongao (Jongao wins!)[/color]
[color=red]game D- 4 AIME15 vs. 13 Discrete_Math (I haven't seen Discrete_Math.)[/color]
[color=green]game E- 5 Dojo vs. 12 lotsofmath (lotsofmath wins!)[/color]
[color=green]game F- 6 FantasyLover vs. 11 AIME_is_hard (FantasyLover wins!)[/color]
[color=green]game G- 7 Drig0nfive52 vs. 10 jjx1 (Drig0nfive52 wins!)[/color]
[color=red]game H- 8 BOGTRO vs. 9 mathblitz (I haven't seen mathblitz.)[/color]
[color=indigo]game I- 1 CA Math vs. 8/9[/color]
[color=green]game J- 15 mewto55555 vs. 7 Drig0nfive52 (mewto55555 wins!)[/color]
[color=darkblue]game K- 14 Jongao vs. 6 FantasyLover[/color]
[color=indigo]game L- 4/13 vs. 12 lotsofmath[/color]
[color=violet]game M- I vs. L[/color]
[color=indigo]game N- 15 mewto55555 vs. 14/6[/color]
[color=violet]game O- winner(M) vs. winner(N), 1st place match[/color]
[color=violet]game P- loser(M) vs. loser(N), 3rd place match[/color]

The games are due 12/25 at 11:59 am EST but finish as fast as possible! (ok, extended deadline by 12 hours)

Colors:
Violet- not yet ready to be played
Indigo- almost ready to be played
Dark Blue- ready to be played, but not this round
Green- completed
Orange- in progress
Red- ready to be played, due at above date

here but i havent seen bogtro on lately....

PMed Discrete_Math, he said he could be on from 1:00-5:00 central, I PMed him back saying OK, but he never came on today.

Hey, what do you mean both inactive? he doesn't respond. I'm winning 2-1.

Mewto55555 defeats BOGTRO 3-2,3-1, or 3-0, depending on how much of a liar i am.

Wait, why did we change the bracket?  Why do I have to play mew now?

I've beat him the last 5 times I played him, with him getting no more than 2 points each time...

since nobody is posting

FantasyLover vs. AIME15

4-1 (FantasyLover)
3-4 (AIME15)
4-1 (FantasyLover)
4-0 (FantasyLover)

heh fantasylover wins

quick question
why did AIME play mew for 2nd place?
because of previous matches, AIME already got a higher standing than mew so.....
i'm not sure why the 2nd place match was put in there

That's what I told ernie...

Apparently he "thought of it this morning".

Easy. Due to the random seeding, this can only truly determine who is 1st. The only reason I came in 3rd originally is because of the bracket, I played the winner of the tourney a round earlier than AIME did. Thus the random seeding put my at a disadvantage. To truly determine 2nd place, ernie had me play AIME. If AIME deserved 2nd, he would have won  :rotfl:

Not in this tourney...
Mewto is right, even better would be league system (everyone vs. everyone) at least when come to the Final Four
But ernie should have told that earlier.

[quote="Mewto55555"]He's the moderator, he can do whatever the heck he wants.[/quote]

Of course, if it was the other way around, your opinion would be different. :wink:

I am sorry for the late call, but I didn't think all the games would be finished in a day... Anyways, mewto55555 is right (his first post).  Because FantasyLover beat both of you, we can't determine who's better 2nd or 3rd, AIME15 or Mewto55555.  If AIME15 had beat FantasyLover, there would have been no 2nd place match.

In the MC Nats Countdown round, Bobby Shen didn't have to cd the person Darryl beat in the semifinals...

CONGRATULATIONS!!!
a tourney has finally ended!!!!
great job everyone!

[hide="in third place"]AIME15  :winner_third: [/hide]

[hide="our runner up"]Mewto55555  :winner_second: [/hide]

[hide="our champion!!"]FantasyLover!!!  :first: [/hide]

now, to prevent uncontrollable spamming in this thread, this topic is LOCKED


geometry
cyclic quadrilateral
geometry proposed

1. Points A,B,C,D,E and F lie on the circumference of a circle. Prove that angle ABC + Angle CDE + Angle EFA = 2pi.

2. In a triangle ABC, angle BAC= 80 degrees, and Q is a point on BC such that AQ bisects the angle BAC. There is also a point P on AB such that QPC = 40 degrees and P is distinct from A. Prove that PQ = QC.

[quote="Gib Z"]1. Points A,B,C,D,E and F lie on the circumference of a circle. Prove that angle ABC + Angle CDE + Angle EFA = 2pi.
.[/quote]
This one is really easy. Just use the arc theorem, the angle of a inscribed angle is half the arc. Now looks at the arcs of those angles and note that they are $ 4\pi$ because you count each peice twice of the full circle. So (1/2) of that are the actual angles.

[quote="Gib Z"]2. In a triangle ABC, angle BAC= 80 degrees, and Q is a point on BC such that AQ bisects the angle BAC. There is also a point P on AB such that QPC = 40 degrees and P is distinct from A. Prove that PQ = QC.[/quote]

since $ \angle QAP=\angle QAC=\angle QPC=40^{\circ}$, we conclude that the quadrilateral $ APQC$ is cyclic and thus $ \angle QAP=\angle QCP$. but $ \angle QAP=40^{\circ}$ so $ \angle QCP=40^{\circ}=\angle QPC\iff PQ=QC$.

I'm really sorry :( But i'm still having serious problems..Please bear with me. How do we conclude the quadrilateral is cyclic from what you said >.<? Also, even if they are cyclic, why do those results follow? After proving angle QAP and QCP are equal, I understood rest, which i know isnt much :(

And for the first one, sylow_theory...Are you sure that theorem is correct?Can you prove it please...I can't see why it is correct. Also, If the inscribed angles are half the arcs, and the 3 inscribed angles' arcs make a circle, then the arc is 2pi, so the angle would be pi by the theorem >.<...:(

I solved the first one :D Thanks sylow_theory :) When I tried to prove your theorem on my own, I managed to find a route to the proof, heres how it goes:

Label some point O as the centre of the circle. Construct lines AO and CO. We use the following theorem (this one I know how to prove =] ) : The angle at the centre is twice the angle at the cirfumferece if the angles are standing on the same arc. For points of simplicity, let the angle ABC be equal to x. So then the reflex angle at the centre  AOC is equal to 2x. So the obtuse interior angle AOC is (2pi-2x).

Now we use the relation $ l = r\theta$ (which I also know how to prove, thank god =] ).

So lets denote the arc length of the angle ABC as l_x. So we know $ l_{x}= (2\pi-2x)r$. If we were to repeat the process for angles CDE and EFA, letting them equal y and z, we have the relations $ l_{y}= (2\pi-2y)r, l_{z}= (2\pi-2z)r$.

Now we add all of them: $ l_{x}+l_{y}+l_{z}= 2r (\pi-x)(\pi-y)(\pi-z)$. Since we know the LHS represents the entire circles circumference, which is 2(pi)r, we can simplify to $ \pi = (\pi-x)(\pi-y)(\pi-z), x+y+z = 2\pi$, which concludes the proof :)

But still troubled with the second one...

[quote="Gib Z"]I'm really sorry :( But i'm still having serious problems..Please bear with me. How do we conclude the quadrilateral is cyclic from what you said >.<? Also, even if they are cyclic, why do those results follow? After proving angle QAP and QCP are equal, I understood rest, which i know isnt much :( [/quote]

try drawing a diagram, it'll help. :)

there is an elementary theorem: if there exists four points $ A,B,C,D$ in a plane such that $ \angle ABC=\angle ADC$ and both points $ B$ and $ D$ lie on the same side of $ AC$, then the points $ A,B,C,D$ are concyclic. it immediately follows from the reverse: if $ ABCD$ is a cyclic quadrilateral, then $ \angle ABC=\angle ADC$. this is the only theorem i used in my proof. :wink:

ask me if you need more help.


[color=darkblue]Hi evry One! Now I want find some Geometry problems about apply the [u][b]Simson line[/b][/u] to done and some Geometry problems about apply the [u][b]Stainer line[/b][/u] to done. So I want every body give me some problem about this or give me a ebook same. Thanks you very much.[/color]

[quote="jayme"]Dear Thanhnam,
See for example R. Johnson, Advanced Euclidean Geometry
Coxeter, Geometry Revisited
Sincerely
Jean-Louis[/quote]

Thank jayme very much. But I have not this book.

Who can help me? I want give me the links about these books?


[quote="GameBot"]What is the probability, expressed as a decimal, of drawing one
marble which is either red or blue from a bag containing 3 red,
2 blue, and 5 yellow marbles?[/quote]

Required Probability $ \ \equal{} \frac {3 \plus{} 2}{3 \plus{} 2 \plus{} 5} \equal{} \frac 12$

[color=red]
Required Probability $ \ \equal{}0.5$[/color]

Thanks alex song i edited my post

It is actually 0.5  :)

Note the "expressed as a decimal"

---
Thanks for reading,
Alex Song