trigonometry AMC AIME calculus geometry MATHCOUNTS quadratics \/closed

I'm not sure if this is the place to post this...

For the trigonometry class post-test, #1 is "What is the value of $\tan {75}$?" Also, there is a problem that says, "Simplify $\cos {12} \cdot \cos {24} \cdot \cos {48} \cdot \cos {96}$." Would you ever learn how to do this, if you didn't take the AoPS class? In other words, are these kinds of problems taught in pre-calc, trig, or calc?

Also, do these types of problems show up in AMC 12 or AIME?

Thanks in advance.

This should probably go in the online classes forum...

If you mean, will regular high school classes ever teach you how to do this type of material, the answer would probably be yes. The first problem can be easily done with regular pre-calculus trigonometry, as can the second problem. It seems like the online classes usually just try to review the basic tools you should have learned before the course, and then go more in depth with different strategies. So while the basic tools for the trig part of that class will most likely be taught in high school, creative problem solving approaches for contest math questions often will not.

Also, don't forget that the other half of the class is complex numbers. I don't recall having ever learned anything about those, from the very definition of $i$ to DeMoivre's theorem, in any of my courses so far.

The first one you'll see if you have a decent pre-calc class.  I did not have a decent pre-calc class, the only reason I know how to do it because I worked far ahead of my pre-calc class after finishing the boring classwork.  At our school, every math class, with a few exceptions, is a review of the previous one, adding perhaps one tiny thing each class (although we do have one good teacher, who got 2nd at mathcounts nationals.  He teaches good.)  At better schools, especially magnet schools, you'll learn how to do that.  I don't know how to do the second, however.  I don't know if better schools will teach you that in your normal curriculum.
Although that's weird.  We DID learn about $i$ in our algebra II class.  Not nearly as much as deMoirve (however you spell it), only enough to be able to solve quadratics in terms of $i$.  Although my algebra II teacher was the aforementioned good teacher.  I don't know if other teachers teach it.

You'll never see these problems on the AMCs.  Because you can use calculators on those, and anyone with a brain could solve those with a calculator (just plugging in the choices).  In AIME or other tests, it's possible that you'll see problems that have a shortcut requiring this knowledge, but I doubt you'll see a problem as "simple" as just simplifying a trig expression like that.  Maybe harder ones...

Now that I'm at it, would the MC problem series help with AMC 10?

[quote="towersfreak2006"]
Also, don't forget that the other half of the class is complex numbers. I don't recall having ever learned anything about those, from the very definition of $i$ to DeMoivre's theorem, in any of my courses so far.[/quote]Really?  I know at my school they teach complex numbers (including DeMoivre's i believe) in pre-cal, which i wil be taking next year.  I also own 3 pre-cal textbooks, all three of them have a chapter about polar coordinates/demoivre's.

Euler/DeMoivre's Theorem ($e^{i\theta} = \cos{\theta} + i\sin{\theta}$) is one of the greatest things you will learn in high school!! Such a beautiful and unexpected little formula!!

[quote="gauss202"]$e^{i\theta} = \cos{\theta} + i\sin{\theta}$[/quote]

...  :huh: what's the point of that?  :?

That really does depend on your school. Our pre-calculus curriculum barely goes beyond the scope of basic trigonometric identities. Complex numbers are not touched at all, and "Euler" is never mentioned until calculus.  :P

In fact, I've never seen it in any textbook at all (but I'm not very well read ^_^), so I'm really surprised that some high schools cover it. But in any case, I believe the instructors try to aim more at problem solving strategies than at textbook material.

[quote="DPatrick"][quote="chess64"]Thanks for the response. :) Would that class help with AMC 10's? I've never even heard of DeMoivre's Theorem except once or twice. Should I just take an Introduction to Geometry or something?[/quote]There is no trigonometry on the AMC 10.  This is one of the major differences between the AMC 10 and the AMC 12.[/quote]

And sometimes there is no trig on the AMC 12 either :P

[quote="chess64"][quote="gauss202"]
$e^{i\theta} = \cos{\theta} + i\sin{\theta}$
[/quote]

... :Huh?: what's the point of that? :Confused: [/quote]

$r(\cos \theta+i\sin \theta)$ is called the polar form of a complex number.  Instead of defining a complex number by the rectangular coordinate $(a,b)$ which represents $a+bi$ you are using $(r;\theta)$ which represents $r(\cos \theta+i\sin \theta)$.  What Euler's formula says is that $\cos \theta+i\sin \theta=e^{i\theta}.$  While you may not be able to immediately see why this could be beneficial, it is.  You can express everything in terms of exponents instead of $\sin$'s and $\cos$'s or $a+bi$'s.

It's hard to tell you the benefits without giving you a lecture which examples.  Until you actually see Euler's formula in action you won't appreciate it as much.


calculus
integration
geometry
trigonometry
limit
calculus computations

Hi.

$r=\frac{1}{\theta}$

$r=\frac{1}{Sin[\theta]}$

$0< \theta\leq \pi/2$

I need to find the area. I know that $\frac{1}{Sin[\theta]}$ is the bigger between the 2. But how do I find the limits of the integration?  :(

Thanks for any help!

Note that the region is unbounded and the integral to compute the area is an improper integral. It does turn out that the integral converges and the area is finite, but that may not be immediately evident from the setup.

It would be useful to draw a picture. $r=\frac1{\sin\theta}$ can be rewritten as $r\sin\theta=1.$ It is simply the horizontal line $y=1.$ The curve $r=\frac1{\theta}$ is a spiral. As $\theta\to0^{+},$ we have $r\to\infty$ as the spiral opens out without crossing the axis. In fact, our two curves are asymptotic, although we'd have to do some work to show it.

[hide="The problem worked."]We set it up as

$\int_{0}^{\frac{\pi}2}\int_\frac1{\theta}^{\frac1{\sin\theta}}r\,dr\,d\theta$

$=\frac12\int_{0}^{\frac{\pi}2}\frac1{\sin^{2}\theta}-\frac1{\theta^{2}}\,d\theta$

$=\frac12\int_{0}^{\frac{\pi}2}\csc^{2}\theta-\frac1{\theta^{2}}\,d\theta$

$=\frac12\left.\left[-\cot\theta+\frac1\theta\right]\right|_{0}^{\frac{\pi}2}$

$=\frac12\left[0+\frac2{\pi}\right]-\lim_{\theta\to0^{+}}\frac12\left[-\cot\theta+\frac1{\theta}\right]=\frac1{\pi}$

For you: can you show that

$\lim_{\theta\to0^{+}}\frac12\left[-\cot\theta+\frac1{\theta}\right]=0?$[/hide]


inequalities proposed
inequalities
3-variable inequality

Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ (a^2\plus{}b^2\plus{}c^2)^2 \geq 4(a\minus{}b)(b\minus{}c)(c\minus{}a)(a\plus{}b\plus{}c)\]

It's an old problem of Phan Thanh Nam. We can prove it easily by Am-Gm  :)

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 4(a \minus{} b)(b \minus{} c)(c \minus{} a)(a \plus{} b \plus{} c)\]
[/quote]

$ 0 \le a \le b \le c\equal{}1$

$ \Leftrightarrow (a^2 \plus{} b^2 \plus{} 1)^2 \geq 4(b \minus{} a)(1 \minus{} b)(1 \minus{} a)(a \plus{} b \plus{} 1)$

$ LHS$ - increase by "a"  and $ RHS$ - decrease by "a" , so we can assume $ a\equal{}0$

$ \Leftrightarrow ( b^2 \plus{}1)^2 \geq 4b(1 \minus{} b)( b \plus{} 1)\Leftrightarrow ( b^2 \plus{}2b\minus{}1)^2 \geq 0$

Equality holdes for $ a\equal{}0$ , $ b\equal{}\sqrt2\minus{}1$ , $ c\equal{}1$

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 4(a \minus{} b)(b \minus{} c)(c \minus{} a)(a \plus{} b \plus{} c)\]
[/quote]
$ f(a,b,c) \equal{} (a^2\plus{}b^2\plus{}c^2)^2$
$ g(a,b,c) \equal{} 4(a\minus{}b)(b\minus{}c)(c\minus{}a)(a\plus{}b\plus{}c)$
wlog assume $ g(a,b,c) \ge 0$.
$ k \ge 0$
$ f(a\plus{}k,b\plus{}k,c\plus{}k)\minus{}f(a,b,c) \equal{} (2(a^2\plus{}b^2\plus{}c^2)\plus{}2k(a\plus{}b\plus{}c)\plus{}3k^2)(2k(a\plus{}b\plus{}c)\plus{}3k^2) \ge 4k(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)$
$ g(a\plus{}k,b\plus{}k,c\plus{}k)\minus{}g(a,b,c) \equal{} 12k(a\minus{}b)(b\minus{}c)(c\minus{}a)$
Let $ h(a,b,c) \equal{} RHS \minus{} LHS \equal{} f(a,b,c) \minus{} g(a,b,c)$
$ 4k(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2) \ge 12k(a\minus{}b)(b\minus{}c)(c\minus{}a) \iff$
$ \sum_{cyc} a^3\plus{} a^2b \plus{} ab^2 \ge 3\sum_{cyc} 3ab^2 \minus{} 3a^2b \iff$
$ \sum_{cyc} a^3 \plus{} 4a^2b \minus{} 2ab^2$.
Since $ \sum_{cyc} a^3\plus{}a^2b \minus{} 2ab^2 \equal{} \sum_{cyc} b(a\minus{}b)^2 \ge 0$ we get: $ h(a\plus{}k,b\plus{}k,c\plus{}k)\minus{}h(a,b,c) \ge 0$. Wlog $ a \ge b \ge c \ge 0$. Then:
$ f(a,b,c) \equal{} f( (a\minus{}c)\plus{}c, (b\minus{}c)\plus{}c, (c\minus{}c)\plus{}c ) \ge f(a\minus{}c,b\minus{}c,c\minus{}c) \equal{} f(a\minus{}c,b\minus{}c,0)$.
Let $ x \equal{} a\minus{}c, y \equal{} b\minus{}c$. Then we have to prove:
$ (x^2\plus{}y^2)^2 \ge 4(y^2\minus{}x^2)xy \iff$
$ x^4 \plus{} y^4 \plus{} 2x^2y^2\minus{} 4xy^3 \plus{} 4x^3y \ge 0 \iff$
$ (x^2\plus{}2xy\minus{}y^2)^2 \ge 0$

Let $ f(a,b,c)\equal{}(a^2\plus{}b^2\plus{}c^2)^2\minus{}4(a\minus{}b)(b\minus{}c)(c\minus{}a)(a\plus{}b\plus{}c)$
W.L.O.G $ c\equal{}min(a,b,c)$
If $ a\ge b\ge c$, we have $ f(a,b,c)\ge 0$
If $ b>a\ge c$, we have $ f(a,b,c)\minus{}f(a,b,0)\equal{}(a^2\plus{}b^2\plus{}c^2)^2\minus{}(a^2\plus{}b^2)^2\plus{}4c(b\minus{}a)(a^2\plus{}b^2\plus{}ab\minus{}c^2)\ge 0$
And we have $ f(a,b,0)\equal{}(a^2\plus{}2ab\minus{}b^2)^2\ge 0$
$ \Rightarrow f(a,b,c)\ge 0$.QED

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3182964#p3182964


For positive integers $ x$ and $ y$, what is the least possible value of $ x$ if $ x^3\equal{} y^2\plus{} 2$?

Test out values for $ x$ (since it climbs quicker). $ x\equal{}1,2$ do not work. But $ x\equal{}3$ works (3^3=5^2+2). So the answer is $ \boxed{3}$.

We factor $x^3=y^2+2$ in $\mathbb{Z}[\sqrt{-2}]$. Then use algebra (someone else please explain) to show that 3, 5 is one of two possible solutions.


For each positive integers n find a set of positive integers whose sum is n and whose product is as large as possible.

A very interesting and very difficult problem.

Clearly no numer can be greater than two and none will equal one and all numbers are either 2 or 3, with at most two of them being 2. If one extends this to the continous case well find that the maximum product is given if we choose (e/n)^(1/n) if I recall correctly, henc the 2,3 combination.


[quote="kunny"]Let $ a,\ b\, c$ be positive real numbers. Find the minimum value of

$ \sqrt [3]{\frac {(a^{2008} \plus{} 2007b^{2007})(b^{2008} \plus{} 2007c^{2007})(c^{2008} \plus{} 2007a^{2007})}{a^{2008}b^{2008}c^{2008}}}$[/quote]
i don't know this problem .
  $ LHS  \geq 1 \plus{}  \frac{2007}{\sqrt[3]{abc}} \geq 1.$
 occur  equal  if only if :  $ a, b,c  \to \infty$

Sorry for that I made a typo. I have just edited.
Thank you for your indication.

kunny


inequalities
function
LaTeX
inequalities unsolved

My question is what are they?

help me with an another question:
i don't understand the following prove which i read in a book:

the question is in the picture
its solution is:As a function f of five variables, the left hand side is convex in each of a,b,c,d,e;hence,its maximum must occur when a,b,c,d,e∈{p,q}
Why could he say so??

one more:
how to show the picture without the form of attachment?

For the inequality $\left(a+b+c+d+e\right)\left(\frac 1 a+\frac 1 b+\frac 1 c+\frac 1 d+\frac 1 e\right)\leq 25+6 \left(\sqrt {\frac p q}-\sqrt {\frac q p}\right)^2$, see http://www.mathlinks.ro/Forum/viewtopic.php?t=49773 .

Schweitzer inequality: http://www.mathlinks.ro/Forum/viewtopic.php?t=36391 , also follows from a slightly stronger inequality given in http://www.mathlinks.ro/Forum/viewtopic.php?t=19244&postorder=asc&start=20 post #22 paragraph [b]4.[/b], and is a particular case of the Kantorovic inequality.

Kantorovic inequality: http://www.mathlinks.ro/Forum/viewtopic.php?t=44004 , http://www.mathlinks.ro/Forum/viewtopic.php?t=36732 , http://www.mathlinks.ro/Forum/viewtopic.php?t=4290 , http://www.mathlinks.ro/Forum/viewtopic.php?t=25557 .

  darij

[quote="Kondr"][quote="FishDragon"]
$(a+b+c+d+e)(\frac 1 a+\frac 1 b+\frac 1 c+\frac 1 d+\frac 1 e)\leq 25+6 (\sqrt {\frac p q}-\sqrt {\frac q p})^2$
[/quote]
First: try to learn Latex, it's easy.
Second: your ineq is useles until you define p and q.[/quote]
Ah,i know nothing about Latex.
could you recommand some book(on line) to me?
thanks

$(a-p)(a-q) \leq 0 \iff a^2 - (p+q)a +pq \leq 0 \iff a+ \dfrac{pq}{a} \leq (p+q)$
         likewise :
 $\sum a + pq\sum \dfrac{1}{a} \leq 5(p+q)$
  with eniq AM-GM , we have this proof

FishDragon, for the latex : http://www.mathlinks.ro/LaTeX/AoPS_L_About.php :)


Euler Gauss quadratics function geometry Support calculus

Based on sheer genius, who do you think is the greatest?

Think before you pick:!:  :!:  :!:

I would choose Srinivasa Ramanujan, but Sir Issac Newton was equally as great.

All the Indians will vote for Ramanujan.

Also, why do people always seem to think that modern mathematicians can't be amazing geniuses?? How about Terence Tao? Nicholas Katz? I was at a discussion today about careers in math, and the girl running it is a 4th year (or so) grad student here in NT/HFT. She said that her masters thesis at Cambridge was the first major result in characterizing in what fields 3 splits, or what primes split in fields of characteristic 3, or something like that, in the past 200 years. And she said that she needed to estimate some series, and she couldn't figure it out, and no one she could find could figure it out, and someone told her "You need to show this to Nick Katz." And she sent it to him, and a week or two later he had written a 27 page paper about it. And she said that it would have taken about 20 years for anyone else to develop the techniques that he used in solving the problem.

I know that historians don't fully judge a president and his presidency until about 25 years or so after he has left.  This is so they can view everything in the context of American history instead of just at a localized period of time.  I assume this is partly why the mathematicians and scientists of the day aren't compared to the agreed upon greats.  No doubt there are certain people who are recognized as outstanding figures in their fields (the first that comes to mind is Andrew Wiles for proving Fermat's Last).

Anyway, I'm not well enough versed in my mathematics history to make any kind of statement about who I think the greatest is.  Actually....looking at the choices I would have to say the Bernoulli family if we're talking about all of them combined :P

Despite the fact that I was miffed that there are no modern dudes (not even any 20th century, let alone currently living. What about Grothendieck? Hilbert?), I still think Euler wins. The only one even close on the list is Gauss (who, for example, figured out the Fast Fourier Transform on his own to use for computations 150 years before we rediscovered it and gave it a name).

[quote="Xevarion"]All the Indians will vote for Ramanujan.
[/quote]

I'm assuming that was intended as a joke, because otherwise it's pretty racist...

There are so many who aren't included here...are we forgetting early Greek and Egyptian mathematicians, who had little to build from?

[quote="K81o7"][quote="Xevarion"]All the Indians will vote for Ramanujan.
[/quote]

I'm assuming that was intended as a joke, because otherwise it's pretty racist...[/quote]
Yes. :P ;)

Now Gauss in winning! My Euler fanaticism drives me to a fiery rage over this! WTF!? So what if Gauss has quadratic reciprocity? [i]Euler[/i] totient function? [i]Euler[/i]-Mascheroni constant? And the base of natural log is E not G! :P :P :P :D

I think that Gauss was a bit arrogant (saying that he cold easily create conjectures that cannot be proven or disproven after failing proving FLT)
But I prefer Fermat (althought he wasn't officially a mathematician)

[quote="thatin"]But I prefer Fermat (althought he wasn't officially a mathematician)[/quote]
 :maybe:  :?:  :huh:
Why knot¿

Well, he had this habit of claiming to have proven something more than others (but to my knowledge this was quite common at thattime).

I think he was called Prince of Amateurs cause he was a judge

How come ancient mathematicians like Euclid aren't on there? Euclid basically laid the entire foundation for modern mathematics.

Erdos was pretty good too.

[quote="Xevarion"]Despite the fact that I was miffed that there are no modern dudes (not even any 20th century, let alone currently living. What about Grothendieck? Hilbert?)[/quote]
Okay apparently I'm an idiot. Not only did I not notice Poincare before, but I did notice Ramanujan and even talked about him. So oops. But Grothendieck was awesome. (I wonder if he's still alive...)
Still I don't regret my vote for Euler. :D :D Let me add some more Euler things: Euler line for geometry ppl. ;) Euler's formula $V-E+F=2$. ;) And the first argument (not proof :P) that $\zeta(2) = \frac{\pi^{2}}{6}$. :D

[quote="Albanian Eagle"]I voted Euler.
I don't understand what's P.Fermat doing in the list ??? :P
Gauss's good ,too :D[/quote]
cmon, Fermat rocks :D
well I just like Fermat because his stuff in Num theory and the Last theorem and stuff. I wouldve have voted Euler, but I was just tempted to vote for Fermat :D

The Goldbach conjecture was not 'formulated' by Euler, it was sent to Euler by Goldbach when a proof eluded him. Euler realised the magnitude of the conjecture and its popularity has much to owe him.
You guys can go on giving results by Euler, I mean he was one heck of a prolific mathematician. And he'd be my second choice.
However my vote goes to Bernhard Riemann, true his specialties were somewhat narrowed to analysis but he gave some brilliant results, even surprised Gauss(my 3rd choice, why isn't Cauchy or Dirichlet on here) with Riemann geometry which led the way to Einstein's general relativity.
And need I mention the Riemann Hypothesis?

Some known ones:
Fundamental theorem of Algebra (well, he was the first who did it formally enough).
Three-Squares-Theorem.
Some stuff on hyperbolic functions.

Conjectured: Dirichlet's theorem on primes.

[quote="bpms"]Ok, what did Gauss actually do? I'm not saying that sarcastically, but what proofs/conjectures did he make?[/quote]
I think ,the most famous discovery of C.F.Gauss is his distribution function in statistics.Widely known and used Gauss' distribution. :)
+ zillions of other remarkable things in science.
After his death ,it became clear he discovered noneuclidean geometries independently and even before Lobachevski.He never published that becouse of the fear he would be laughed at :(
What else to say about Great Genius Gauss?
Google  up and be stunned.

Although I'm helping to answer my own question, I found these:
Popularizer of complex numbers
Constructed regular 17-gon.

[quote="krustyteklown"]i'm pretty sure gauss invented congruences (modulo) and the QR theorem (at least thats what intro to the theory of numbers by niven is telling me)[/quote]

Yes, he did.

And he was the first to figure out
1+2+3+...+n probably

He was not the first, this was known looooooooooooooong ago.
Do not think that Gauß is _the_ mathematician and all others before were idiots.

I think all listed mathematicans were great,but ,anyway ,I think that the greatest is [b]Gauss[/b]!
Don't forget,he proved his theorem about constructing regular 17-gon in the age of 19,although it was unsolved problem for about 2000 years,since ancien Greeks.I have read that only after proving that fact he realized that he was born to be a mathematican,because beside mathematics he had a lot of other interests such as lingvistc,astronomy,physics.And untill his deth he was considering the result about 17-gon as his greatest result(although he had greater results than that)and ,as far as I know, now his statuy is standing on a 17-gon in Germany.

[quote="Ihatepie"]chose einstein (other)[/quote]
The question is not asking about the greatest physicist :P .

Whan gauss was in grade 1, he added up the numbers from 1-100 in less then 5 minutes! Grade 1!

This is my summary of Fermat.

Dude who came up with a bunch of BS randomly, and told everyone he was sure they were right and even said he proved some. 200 years later about half were identified as BS by other people, and it didn't even take half a brain to identify as BS, and the other half was proven.

Anyone talk about Arabian mathematician Al-Khwarizmi here? He discovered '0' digit, if i'm not wrong. Arab had many mathematician during its glory about 7-13 century ago.  :)

And also how about from China, Japan, and East Europe.

Not that countries i've said is very good, even  the dominators of  intenational science olympiad, IMO, IPhO, IChO. etc....

Anyway, I prefer Gauss as the greatest :maybe:


induction
number theory unsolved
number theory

n is a integer greater than 2

Prove that

lcm(1,2,...,n)>2^(n-1)

use this:
If $n=2k+1$ $\longrightarrow$ $[1,2,...,2k+1]\geq (2k+1).\binom{2k}{k}$,then use induction and prove$(2k+1).\binom{2k}{k}\geq 2^{2k+1}$for $k\geq 3$
If $n=2k$ and $k>1$ $\longrightarrow$ notice that $2[1,2,...,2k-1]\geq [1,2,...2k]$ :)


Here is a problem from Enumerative Combinatorics, from a chapter of the same name of Problem-Solving Strategies by Arthur Engel, a (former) German coach for the IMO team.

(a) The number d of diagonals  of a convex n-gon is equal to the number of all pairs of points minus the number of sides. So this is not hard.

(b) The number of intersection points of the diagonals, provided that no three lines are concurrent, is (n, 4) (an easier way to type 'n choose 2')

(c) We draw all diagonals of a convex n-gon. Suppose no three diagonals pass through a point. Into how many parts Tn is the n-gon divided?
This is not too difficult. Tn equals 1 + (n, 2) - n + (n, 4).

(d) The hardest (at least for me) is To count the number T of different triangles. Let T = T0 + T1 + T2 +T3, where Ti is the number of triangles with vertices with i vertices among the vertices of P (I don't understand this). This partition is decisive since each Ti can be easily evaluated. [There is a diagram in the book, which I can't draw here. So I think people who have the book can help to answer my questions.] We can assign some subsets of the vertices of P to the four types of triangles. The figures show that the assignments are 1:1, 1:5, 1:4, 1:1. Thus we have
T = (n, 6) + 5(n, 5) + 4(n, 4) + (n, 3)

I do not know how to type subscripts and superscripts. Can anyone help. I think the symbol ^, while quite universal, can be quite messy and looks sickening at times.

This website is good, and I hope it gets growing.

Hi Lek Huo Tang,

some similar problems can be found here:

http://www.mathlinks.ro/viewtopic.php?t=1207
http://www.mathlinks.ro/viewtopic.php?t=1028
http://www.mathlinks.ro/viewtopic.php?t=1206
http://www.mathlinks.ro/viewtopic.php?t=1205
http://www.mathlinks.ro/viewtopic.php?t=1203


geometry
trapezoid
perpendicular bisector
angle bisector
geometry unsolved

Give triangle $ABC$. Construct the point E in $\Delta ABC$ which has projections on $AB, BC, CA$ respectively are G, H, I and $AG=BH=CI.$

Let the perpendicular bisector $s_{b}$ of CA meet the bisector $l_{a}$ of the $\angle A$ at $P_{0}$ and let $G_{0}, I_{0}$ be the perpendicular projections of $P_{0}$ to AB, CA. $I_{0}$ is the midpoint of CA and obviously, $AG_{0}= AI_{0}= CI_{0}.$ Let $n_{a}\perp l_{a}$ be a normal to the angle bisector $l_{a}$ through $P_{0},$ let $P \in n_{a}$ be arbitrary point on this normal, and let G, I be the perpendicular projections of P to AB, CA.  $P_{0}PGG_{0}$ and $P_{0}PII_{0}$ are trapezoids with $PG \parallel P_{0}G_{0},\ PI \parallel P_{0}I_{0}$ and with equal angles, hence $GG_{0}= II_{0}$ and $AG = AG_{0}-GG_{0}= CI_{0}-II_{0}= CI.$ Thus the line $n_{a}$ is the locus of points P, such that AG = CI. Similar line $n_{b}$ can be constructed for the vertex B, the locus of points Q with the perpendicular projections H, G on BC, AB, such that BH = AG. The lines $n_{b},\ n_{a}$ meet at a point E with the prependicular projections G, H, I on AB, BC, CA, such that BH = AG = CI.


P is a point inside rectangle ABCD with PA=3, PD=4, and PC=5. Find PB.

[quote="i_like_pie"][hide="Solution"]Using the British Flag Teorem, $ PA^2 \plus{} PC^2 \equal{} PB^2 \plus{} PD^2$, so $ PB \equal{} \sqrt {3^2 \plus{} 5^2 \minus{} 4^2} \equal{} \sqrt {18} \equal{} \boxed{ 3\sqrt {2}}$.[/hide][/quote]
I think my proof is the proof of your proof  :D ...but the conclusion is the same:
Let $ P'$ be the symetrical point of $ P$ related cu $ BC$, $ A',D'$ the extension of the segments $ AB$ and $ DC$ such as $ AB\equal{}A'B$ and $ DC\equal{}CD'$.Therefore the rectangle $ BA'CD'$ is equal to the initial one.
$ P'B\equal{}3$
$ P'C\equal{}4$
The quadrilateral $ BPCP'$ has the diagonals perpendicular,therefore $ 9\plus{}25\equal{}16\plus{}PB^2$ $ \implies$ $ PB\equal{}3\sqrt{2}$

[quote="balan razvan"]I think my proof is the proof of your proof  :D[/quote]
The British Flag Theorem can be proven by dropping some altitudes and then applying the Pythagorean Theorem a bunch of times. :)

I didn't heard of it before....but it seems simple and nice....thanks :lol:


$ |x\minus{}3|\plus{}|2x\plus{}6|\le 8$

Solve over the set of reals.

|x-3|+|2x+6| <and=  8

if x >and= 3

so that x-3+2x+6 <and= 8
           x <and= 5/3  (no answer)

if -3 =and< x <3

           3-x+2x+6 <and= 8
           x <and=  -1   (ok)     -3 =and< x <and=-1

if x<-3

           3-x-2x-3 <and= 8
            x =and>-8/3   (no)

so       -3 =and< x <and=-1    [-3 , -1]

from   http://www.haiyanh.com


this is my 4th try at a game.(all the previous ones failed)

[u]rules[/u]
-a person posts a 5-10 letter word and everyone else has to find anagrams of it.
-when posting an anagram, it must come with the word used in a sentence
-you get points for the number of letters in the anagram and a bonus 3 points if you use all the letters in the original word
-if you cannot think of another anagram to post, post FORFIET. the last person to post an anagram gets the total number of anagrams listed for that word
-the angram must be a rearrangement of the letters(no roots). for example, from computer, you may not use compute.
-post all previously used anagrams
-names are not allowed
-abbreviations are not allowed

EX:
WORD: compact
a
am
map
cat
act
camp
tap
FORFEIT
whoever posted a gets 1 point
whoever posted am gets 2 points
whoever posted map, cat, act, or tap gets 3 points
whoever posted camp gets 4 points
whoever posted tap gets 7 points since there are 7 words listed.

yes i know these are not all the anagrams of compact
i left out the sentence part

post in the form
WORD:*word*
*1st word*
*2nd word*
..
..

lets begin

WORD: circular
[hide="scores"]
1. abacadaea-0[/hide]

The [b]air[/b] is um... outside?

+3 :lol:

It is [b]crucial[/b] to go to AOPS.

+7 yay

Can I post twice for one word?  :)

WORD: circular
air
crucial
He is a big fat [b]liar[/b] who lies all the time.

abacadaea: 0
leoxnlin: 7
james4l: 7

[b]rail[/b]

he grinded his skateboard on a rail
abacadaea: 0
leoxnlin: 7
james4l: 7
mz94: 4

WORD: circular
air
crucial
liar
rail
[b]curl[/b]

She had to curl her hair before going out to dinner.

abacadaea: 0
james4l: 7
leoxnlin: 7
lingomaniac88: 4

WORD: circular
air
crucial
liar
rail
curl
[b]lair[/b]

Do not enter the dragon's lair.
[hide="Scores:"]
abacadaea: 0
james4l: 7
leoxnlin: 7
lingomaniac88: 4
mathforlife: 4[/hide]

arc  :P

This voltaic arc is very strange...

not enough creative  :(

1: james4l: 7
1: leoxnlin: 7
3: lingomaniac88: 4
3: mathforlife: 4
5: pacman2812: 3
6: abacadaea: 0

WORD: circular
air
crucial
liar
rail
curl
lair
arc
[b]cardiology[/b]

I think he's going to study cardiology in college.

I'm not sure if I'm right with scores... but....

 mintybear: 10
 james4l: 7
 leoxnlin: 7
 lingomaniac88: 4
 mathforlife: 4
 pacman2812: 3
 abacadaea: 0

????

uh... cardiology isn't an anagram of circular...

That doesn't quite make sense. Neither, I think, does the title. Anagrams are only whole permutations. I think. But never mind.

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
[b]CAR [/b] :(
The car parked outside the house was cobalt.

Skipping the penultimate poster as of now, scores: (Jeez, how do I get this sound unoffensive?? No offense)

james4l: 7
leoxnlin: 7
lingomaniac88: 4
mathforlife: 4
pacman2812: 3
math_explorer: 3
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
[b]ail[/b]

Note: mz94 was accidentally deleted from the scores, but I put him back.

Scores:
james4l: 7
leonxlin: 7
lingomaniac88: 4
mathforlife: 4
mz94: 4
pacman2812: 3
math_explorer: 3
cf249: 3
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
[b]rural[/b]

Not liking the crowded feeling of the city, Alicia went to a university in a [u]rural[/u] area.

Scores:
lingomaniac88: 9
james4l: 7
leonxlin: 7
mathforlife: 4
mz94: 4
pacman2812: 3
math_explorer: 3
cf249: 3
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
[b]uracil[/b]

[u]Uracil[/u] is found in RNA but not in DNA.

Scores:
lingomaniac88: 9
james4l: 7
leonxlin: 7
krsattack: 6
mathforlife: 4
mz94: 4
pacman2812: 3
math_explorer: 3
cf249: 3
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa

Circa 1400 BC the Egyptian Middle Kingdom was formed.

Scores:
james4l: 12
lingomaniac88: 9
leonxlin: 7
krsattack: 6
mathforlife: 4
mz94: 4
pacman2812: 3
math_explorer: 3
cf249: 3
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
[b]a[/b]

I am too pathetic to think of a word other then a.

Scores:
james4l: 12
lingomaniac88: 9
leonxlin: 7
krsattack: 6
mathforlife: 4
mz94: 4
pacman2812: 3
math_explorer: 3
cf249: 3
michaelwong256: 1
abacadaea: 0

I won't object the previous word, but I must object the sentence. That's cheating.

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
[b]lira[/b]

The symbol for lira is L or ₤.

Scores:
james4l: 12
lingomaniac88: 9
leonxlin: 7
math_explorer: 7
krsattack: 6
mathforlife: 4
mz94: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I

I don't see anything wrong with michaelwong256's sentence, other than the fact that he used "then" instead of "than".
[quote]I am too pathetic to think of [b]a[/b] word other then a.
[/quote]
Scores:
james4l: 12
lingomaniac88: 9
leonxlin: 7
math_explorer: 7
krsattack: 6
mathforlife: 4
mz94: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial

In the Roman legislation, a curial assembly was used when one of the ten primitive subdivisions wanted to congregate.

Scores:
james4l: 18
lingomaniac88: 9
leonxlin: 7
math_explorer: 7
krsattack: 6
mathforlife: 4
mz94: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
[b]LAURIC[/b]

Lauric acid is found in coconut oil

james4l: 18
mz94: 10
lingomaniac88: 9
leonxlin: 7
math_explorer: 7
krsattack: 6
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

#21CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
lauric
[b]cur[/b]

A cur is a mean, cowardly person.

james4l: 18
lingomaniac88: 12
mz94: 10
leonxlin: 7
math_explorer: 7
krsattack: 6
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

#22CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
lauric
cur
curari

Curari is sometimes found in plants in South America, and the natives used this for poison darts.

james4l: 24
lingomaniac88: 12
mz94: 10
leonxlin: 7
math_explorer: 7
krsattack: 6
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
lauric
cur
curari
CIRCAR

Indians are ruled by a CIRCAR

james4l: 24
mz94: 16
lingomaniac88: 12
leonxlin: 7
math_explorer: 7
krsattack: 6
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

CIRCULAR
[hide="long list"]air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
lauric
cur
curari
circar
auric[/hide]

That gold sure has an [i]auric[/i] look, which would be expected since auric pertains to the color/metal of gold.

james4l: 24
mz94: 16
lingomaniac88: 12
leonxlin: 7
math_explorer: 7
krsattack: 6
Brut3Forc3: 5
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0[/quote]

CIRCULAR

[hide="List"]air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
lauric
cur
curari
circar
auric
IRREAL
[/hide]
The number 3+4i is, if you are an advocate of highly eccentric words, irreal.

james4l: 24
mz94: 16
math_explorer: 13
lingomaniac88: 12
leonxlin: 7
krsattack: 6
Brut3Forc3: 5
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

Also, "irreal" is not an anagram of "circular".

[quote="pacman2812"]Wait

I looked [url=http://www.scrabulous.com/twl_dictionary.php]this[/url] dictionary, and CIRCAR doesn't exist at it.

so mz94 is wrong :mad:[/quote]

oh well...
dont be so pissed... :mad:

james4l: 24
lingomaniac88: 12
mz94: 10
leonxlin: 7
math_explorer: 7
krsattack: 6
Brut3Forc3: 5
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

I know, that was pretty pathetic...  :oops: Thanks for changing the score for me.

Incidentally, CIRCAR [i]is[/i] in [url=http://www.scrabulous.com/sowpods_dictionary.php]this dictionary[/url], which is wider. I think the starter of this game should decide. Too bad (s)he's not posting. Is there a rule that the word-poster can't post anagrams?
[hide="Words"]
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
lauric
cur
curari
circar ????
auric
URIC
[/hide]
The formula for uric acid is $ C_{5}H_{4}N_{4}O_{3}.$
james4l: 24
lingomaniac88: 12
math_explorer: 11
mz94: 10
leonxlin: 7
krsattack: 6
Brut3Forc3: 5
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

[hide="ughhhhhhhhhhhhhhhhhhhhhhhhhh"]air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
lauric
cur
curari
circar ????
auric
CRURA[/hide]

The CRURA lie behind your femurs
james4l: 24
mz94: 15
lingomaniac88: 12
math_explorer: 11
leonxlin: 7
krsattack: 6
Brut3Forc3: 5
mathforlife: 4
pacman2812: 3
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0

CIRCULAR
air
crucial
liar
rail
curl
lair
arc
car
ail
rural
uracil
circa
a
lira
I
curial
lauric
cur
curari
circar ????
auric
crura
crural

He has a crural hernia...

james4l: 24
mz94: 15
lingomaniac88: 12
math_explorer: 11
leonxlin: 7
krsattack: 6
pacman2812: 6
Brut3Forc3: 5
mathforlife: 4
cf249: 3
michaelwong256: 1
33286:1
abacadaea: 0


The positive integers x and y satisfy
3x^2 - 8y^2 + 3(xy)^2 = 2008

then what is the value of xy?

please help.

Factorize, in order to get $ (3x^2 \minus{} 8)(y^2 \plus{} 1) \equal{} 2000 \equal{} 2^4 \cdot 5^3$, then find the solutions.


geometry
power of a point
radical axis
geometry proposed

Isosceles triangles $A_3A_1O_2$ and $A_1A_2O_3$ are constructed on the sides of
a triangle $A_1A_2A_3$ as the bases, outside the triangle. Let $O_1$ be a point
outside $\Delta A_1A_2A_3$ such that
$\angle O_1A_3A_2 =\frac 12\angle A_1O_3A_2$ and $\angle O_1A_2A_3 =\frac 12\angle A_1O_2A_3$.
Prove that $A_1O_1\perp O_2O_3$, and if $T$ is the projection of $O_1$ onto $A_2A_3$,
then $\frac{A_1O_1}{O_2O_3} = 2\frac{O_1T}{A_2A_3}$.

[quote="lasha"][hide]Denote $\angle{A_3O_2A_1}=x$ and $\angle{A_1O_3A_2}=y$. Take point $X$ different from $A_{3}$ on a line $O_1A_3$ so that $X$ belongs to the circle with center at $O_2$ and radii equal to $O_2A_3=O_2A_1$.(Call this circle $S_1$). Take point $Y$ on a line $A_2O_1$ in a similar way.(Call the second circle $S_2$). Then, $\angle{O_1XA_1}=x$ and $\angle{A_1YO_1}=y$, while $\angle{A_3O_1A_2}=180-x-y$ from triangle $A_3O_1A_2$. It means that points $X, A_1, Y$ are collinear. Assume the point $B$ belongs to $A_1O_1$ and circle $S_1$ as well(different from $A_1$). Than, $\angle{A_3BO_1}=\angle{A_3XA_1}=x=\angle{A_3A_2O_1}$. It means that $A_3BA_2O_1$ is cyclic and $\angle{O_1BA_2}=\angle{O_1A_3A_2}=y=\angle{A_1YA_2}$. The last equality means that $B$ belongs to the circle $S_2$ as well. So, line $A_1B$ is the common chord of circles $S_1$ and $S_2$. So, $O_2O_3$ is perpendicular to $A_1B$, the same as $A_1O_1$, q . e . d .
Denote by $M$ the projection of $O_1$ into the line $XY$. Easy to notice the similarity of triangles $O_1A_3A_2$ and $O_1YX$. As $O_1T$ and $O_1M$ are respective altitudes, $\frac{O_1T}{A_2A_3}=\frac{O_1M}{XY}$. It means we have to prove $\frac{A_1O_1}{O_1M}=\frac{2O_2O_3}{XY}$. $(*)$. Note the similiraty of triangles $O_2BO_3$ and $XBY$. (as $\angle{BXY}=\angle{BO_2O_3}$ and $\angle{BYX}=\angle{BO_3O_2}$. ) So, $\frac{2O_2O_3}{XY}=\frac{2O_2B}{XB}=\frac{1}{sin\angle{XA_1B}}$, while $\frac{A_1O_1}{O_1M}=\frac{1}{sin{\angle{XA_1B}}}$. Hence, we have proved $(*)$, which completes our proof.[/hide][/quote]

Very nice and elegant solution indeed,but with few errors.It will be $\angle{A_1O_3A_2}=2y$ and $\angle{A_1O_2A_3=2x}$.You can also prove that $A_1,B,O_1$ are collinear in a more simple way:$\angle{O_1A_2A_3}=\angle{O_1XY}=x \Rightarrow$ points $A_2,A_3,X,Y$ are concyclic $\Rightarrow O_1A_2*O_1Y=O_1A_3*O_1X \Rightarrow O_1 \in$ radical axis of the two circles.


induction
superior algebra
superior algebra unsolved

Let $A$ be a finite (corrected) ring with $1\ne 0$. Prove that the following are equivalent
i) A is not a field.
ii) for any $n\in\mathbb N^*$ the equation $x^n+y^n=z^n$ has solutions in $A^*$.

[i]Marian Andronache and Ion Savu[/i]

Are you sure about this¿
Setting $A= \mathbb{Z}$ would contradict Fermat's last theorem I think

I'm sorry.. A finit ring  :blush:
I will correct the pb

[b]a)[/b]
if $A$ is a field, there is (Langrange's theorem) an integer $n$ so that $x^n=1$ for all $x \in A'$.
Then you get $x^n+y^n=z^n \iff 1+1=1 \iff 1=0$ giving a contradiction.

[b]b)[/b]
if $A$ is not a field, there are $a,b \in A'$ so that $ab=0$
(otherwise you would have $xy=xz \iff y=z \implies xA'=A'$ for all $x \in A'$, so there is some $y \in A'$ fulfilling $xy=1$, but $A$ isn't a field, contradiction)

1. case: $a=-b \implies a^2=0$
for $n=1$ use $x^1+(y-x)^1=y^1$ for some $x,y \in A',x \neq y$ (if $|A'|=1$ it's trivial that $A$ is a field)
for $n \geq 2$ use $a^n+a^n=0+0=0=a^n$

2. case: $a \neq -b$
Then by induction:
$a^1+b^1=(a+b)^1$ and
$a^n+b^n=(a+b)(a^{n-1}+b^{n-1})=(a+b)^n$
so there is a solution for all $n$

q.e.d.

A is not commutative... from $ab=0\not\Rightarrow ba=0$

3. case: $ba \neq 0$:
Then $(ba)^n=0$ for all $n \geq 2$ and so you have $a^n+(ba)^n=a^n$ (case $n=1$ the same way as above)

Hope its correct this time  :blush:

I tought that you would consider an elemenet $x\in A$ s.t. $x^2=0$ and then $x^n+x^n=x^n$ for all $n\ge 2$ or if no such element exist you can apply what you wrote in your second post ;)


geometry
circumcircle
geometric transformation
reflection
parallelogram
geometry proposed

[color=darkred][b][size=117]Very nice problem ![/b][/size] Denote the distance $ \delta_{XY}(Z)$ of the point $ Z$ to the line $ XY$.

Prove that for any interior point $ P$ to the cyclic convex quadrilateral $ ABCD$ exists the relation

$ \boxed{\ AB\cdot\delta_{CD}(P)+CD\cdot\delta_{AB}(P) = BC\cdot\delta_{DA}(P)+DA\cdot\delta_{BC}(P)}$.

[b]Particular case ($ D\equiv A$).[/b] Denote the tangent $ AA$ in the point $ A$ to the circumcircle of $ \triangle ABC\ (\ b\ne c\ )$.

For a point $ M$ which belongs to the sideline $ BC$ choose a point $ P\in (AM)$. Then there is the relations :

$ \{\begin{array}{ccc}M\in (BC) &\implies & AB\cdot\delta_{AC}(P)\ +\ AC\cdot\delta_{AB}(P) = BC\cdot\delta_{AA}(P)\\\\ C\in (MB) &\implies &-\ AB\cdot\delta_{AC}(P)\ +\ AC\cdot\delta_{AB}(P) = BC\cdot\delta_{AA}(P)\\\\ B\in (MC) &\implies & AB\cdot\delta_{AC}(P)\ -\ AC\cdot\delta_{AB}(P) = BC\cdot\delta_{AA}(P)\end{array}$.

[hide="Hint for the particular case."]
Show easily that $ AL^{2}=\frac{abc}{|b^{2}-c^{2}|}$ and for any $ M\in BC$ exists the relation $ (b^{2}-c^{2})\cdot\overline{ML}=b^{2}\cdot\overline{MB}-c^{2}\cdot\overline{MC}$.[/hide][/color]

A start:

Let $ AB\cap CD \equal{} E$ and $ BC\cap DA \equal{} F$. Let the angle bisectors of $ \angle AEC$ and $ \angle AFC$ intersect at $ Z$. Reflect $ P$ over $ EZ$ and $ FZ$ to $ X$ and $ Y$ respectively. The midpoint of $ XY$ is $ Z$.

We have $ EZ\perp FZ$. Then we seek to show that $ [ABX]\plus{}[CDX] \equal{} [BCY]\plus{}[DAY]$.

I am not sure if this is easier, but I believe it is. It could be easily brute forced since we can take $ Z$ as the origin, $ E$ and $ F$ on the axes, then $ C$ and $ X$ as some points. We can easily generate the rest of the diagram relatively quickly from these variables.

Perhaps there is a more elegant way to finish the problem. This reminds me of many area locus problems. In particular, the locus of points, $ Q$, such that $ [ABQ]\plus{}[CDQ] \equal{} k$. In fact, if we translate segments $ AB$, $ BC$, $ CD$, $ DA$, to their respective places for that construction, we get a parallelogram!


geometry
3D geometry
algorithm
search
group theory

ummm well the subject basically says it all but HELP!!! :o

Wrong section. There is a lot of discussion in Strategy and Games forum.
First 2 layers is self-intuitive.
Last Layer you just memorize some algorithms
To get the Yellow Cross you do FRUR'U'R. Then you orient the last layer (repeat R U R' U R U2 R'). Then you switch the pieces that need to be swtiched using various algorithms. NOTE THAT ALGORITHMS IS NOT ALWAYS MATH. This is just an overview. It's not meant to instruct.

You will find that there are algorithms that allow you to solve it easily. It is a basic 7-step process, but involves more complicated algorithms if you want to be fast.

To avoid further discussion here (wrong forum), I suggest not talking about it, especially the algorithms.

There's

this        http://youtube.com/watch?v=HsQIoPyfQzM

and it's sequel         http://youtube.com/watch?v=IW_BBp3FPMQ&feature=related

There ya go!

Youtube is lame. It doesn't help you solve the first layer (pretty much) if you cannot already do it.

Or go to [url]http://www.rubiks.com[/url] and click Rubik's Academy. There are some nice solution links there.

Or go to Dan's Cube station cubestation.uk I think was the link.

yea that is the site where i learned to solve teh 3x3 and 4x4 cubes

buy a rubiks cube andtwist the sides til all of the sides have a equal amount of color

I like to use the following website:  http://lar5.com/cube/

Petrus method is pretty easy to pick up.  It improved my time from 120 seconds to 45 seconds.  Gotta practice more :/

Yeah, fridrich is faster if you don't mind memorizing a lot of algorithms.

These sites have great animations:

http://www.nascarjon.us/F2L.html
http://www.nascarjon.us/LLO.html
http://www.nascarjon.us/LLP.html

This one is my favorite for starting out. It introduces the basic notation used in rubik's cubes, and gives a fridrich method which doesn't have too many algorithms (the bare ones that you need to solve it). If you decide that you want to go for speed later, then it's easy to learn more.

[url]http://www.stanford.edu/~leyanlo/cube_solution.pdf[/url]

how do u do fridrich f2l intuitively without needing memorization?

theres a stupid intuitive f2l which is really dumb (keyhole/keyslot)

FRIDRICH F2L IS EASY TO FIGURE OUT.
They're all the same.

what's the keyhole/keyslot intuitive f2l?

me am very confused :o

i think that if you know advanced group theory and you are basically a human calculator (not really just exaggeration) you can devise solutions on the fly.  :o  :o  :o

Um no? I don't think math pros necessarily are just as pro at rubik's cube. It's the other way around too. If you're lame at math, you can still be pro at cube because it's memorization


function
logarithms
LaTeX
calculus
calculus computations

Find all $ continuous$ functions $ f: [1,\infty]\to R$ so that $ f(16^x)\plus{}2f(4^x)\plus{}f(2^x)\equal{}x,$for every$ x\in[0,\infty)$

Let $ g(t)=f(2^{2^t})-\frac19\cdot2^t$. The relation becomes $ g(t+2)+2g(t+1)+g(t)=0$, on all of $ \mathbb{R}$. Thinking of this as an integer recurrence, the solutions are of the form $ (-1)^n(a+bn)$.

If we just ask for continuity of $ g$, there are lots of solutions. In particular, any continuous periodic function with $ g(t+1)=-g(t)$ works. Requiring a limit at the lower end to exist, however, pins it down completely. The only solution to the integer recurrence with a limit at either $ \infty$ or $ -\infty$ is the constant zero. Looking at every possible value mod 1, $ g$ must be identically zero in this case. This makes $ f(x)=g(\log_2(\log_2 x))+\frac19\cdot \log_2 x=\frac19\log_2 x$.

By the way, don't put random words in $ \text{\LaTeX}$.


Forgive me if this can be considered a stupid question, but I am just wondering.

If the d/dx of x^2 is 2x, then what should be derive(is this the right term?) so that the derivative of shall we say z is equal to x. Im new to derivative.

Thanks

Please read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32837]this thread[/url]. This problem really should go in one of the Calculus forums.

However, as it's here and I can't move it, I'll answer it for you. If $f(z)=z$ and $f'(z)=x$, then we can say $\int x\, dx=\frac{x^2}{2}=z$. Since you're just getting started on derivatives, you may not understand integrals, but you can also think about it this way: if $f(x)=x^2 \Rightarrow f'(x)=2x$, then what you've done is moved the power down in front of the number and taken one off of the degree that $x$ was raised to. (For example, $f(x)=x^3 \Rightarrow f'(x)=3x^2$.) So, to have $f'(z)=x$, $z$ is simply equal to $\frac{x^2}{2}$, because if you say $f(x)=\frac{x^2}{2} \Rightarrow f'(x)=\frac{2x}{2}=x$. :)

And to be more technical, $z = \frac{x^2}{2} + C$ where C is a constant (oft called the constant of integration when used with indefinite integration).  Any constant will become 0 when you take the derivative of z.

whether this is stupid or easy or whatnot, this should not be HERE. calculus problems should not be here it should be in the Calculus forum even if it is VERY EASY or OBVIOUS or STUPID.

edit: i read someone's signature (forgot who) which said something like: there are no stupid questions but only stupid answers

You don't have to repeat what I already said. Either we can let this die down, since the question has been answered, or a mod can move it when one happens to see it.


Please read the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308193]stickies[/url].

If A, B, and C are the digits of the number, then the number would be
100A + 10B + C

so your equation would be

100A + 10B + C = 4ABC
100A + 10B = 4ABC - C
100A + 10B = C(4AB - 1)
C = (100A + 10B)/(4AB - 1)
= 10(10A + B)/(4AB - 1)

now a bit of logic.
you know the numerator ends in a zero, so the bottom has to be divisible by 10 or 5
something multiplied by 4 can only end in 4,8,2,6, or 0 so after subtracting the 1 the denominator can only end in
3,7,1,5, or 9 , never in 10

also neither A nor B can be zeros, since then the denominator would be -1

also A and B can only be between 1 and 9

so we are looking for A and B which makes 4AB - 1 end in a 5.

After a few 'guesses' I found A=3 and B=8 gives me C=4

so the number is 384

check 4(3x8x4) = 384

[quote="FullMetalNaruto"]What is the positive three-digit integer that is four times the product of its digits?

Once again, thank you!  :wow:[/quote]
It's a bit long and rough but:
[hide]Use expanding notation:
$ 100a \plus{} 10b \plus{} c \equal{} 4abc$
Since $ 4abc$ refers to an integer, that $ 3$ digit number must be divisible by $ 4$.
$ 100a \plus{} 10b \plus{} c \mod 4 \equiv 0(a) \plus{} 2b \plus{} c \mod 4 \equiv 2b \plus{} c \mod 4 \equiv 0$
Leave that for later, because that is a factor of a multiple of $ 4$ and $ 2b \plus{} c$.
Now, if it were to be divisible by 4, then $ c$ cannot be odd. Therefore, we have narrowed it to $ 2$, $ 4$, $ 6$, $ 8$, $ 0$. But, the chances for $ b$ slim signicantly. If $ c$ is $ 2$, then $ b$ is $ 1$, or $ 3$. But if that is true, then $ c$ must be $ > 4$. Thus, that isn't possible. For $ 4$, $ b$ can be $ 2$, $ 4$, or $ 8$. Given $ b$ is $ 2$, it isn't possible to get above $ 100$ with a single digit number, excluding $ 9$. So, we turn to $ 4$. $ 4 \times 4 \times 4 \equal{} 64$. This isn't possible, if you multiply by $ 2$, you'll see the problem. This isn't possible, so we try $ 8$. $ 8 \times 4 \times 4 \equal{} 128$. Therefore, the best suit is $ 3$, giving $ \boxed{384}$.[/hide]


This is a pretty classic problem so leave it to beginners. The key idea here is to substitute.

Find all solutions to the equation below.

$ \sqrt{x\plus{}7} \minus{} \frac{3}{\sqrt{x\plus{}7}} \equal{} 10$

$ \sqrt {x \plus{} 7} \minus{} \frac {3}{\sqrt {x \plus{} 7}} \equal{} 10$

let $ n \equal{} \sqrt {x \plus{} 7}$

$ \implies n \minus{} \frac {3}{n} \equal{} \frac {n^2 \minus{} 3}{n} \equal{} 10$

$ \implies n^2 \minus{} 10n \minus{} 3 \equal{} 0$

$ \implies n \equal{} \pm(2\sqrt {7} \plus{} 5)$

$ \implies \sqrt {x \plus{} 7} \equal{} \pm(2\sqrt {7} \plus{} 5)$

$ \implies \boxed{x \equal{} (2\sqrt {7} \plus{} 5)^2 \minus{} 7}$

edited.thankx hsiljak

Note there's no difference between your two solutions ;)

and of course, note that $ \sqrt {\alpha}\geq 0$, for $ \alpha\in \mathbb{R}^\plus{}$.

Hmm... I got $ n \equal{} 5 \pm 2 \sqrt{7}$ though.

I don't think you can say $ \pm (2 \sqrt{7} \plus{} 5)$. No, I don't think that's right. Your final answer, however, is right though for $ x$.

[hide]Let $ y\equal{}\sqrt{x\plus{}7}$. Then:
\[ y\minus{}\frac{3}{y}\equal{}10 \implies y^2\minus{}10y\minus{}3\equal{}0 \implies y\equal{}\frac{\minus{}b \pm \sqrt{b^2\minus{}4ac}}{2a} \\ \implies x\equal{}\frac{\left(\minus{}b\plus{}\sqrt{b^2\minus{}4ac}\right)^2}{4a^2}\minus{}7\]Plugging in, we find that:
\[ x\equal{}\frac{\left(10\plus{}\sqrt{100\plus{}12}\right)^2}{4}\minus{}7\equal{}\boxed{20\sqrt{7}\plus{}46}\][/hide]

123456789 - Remember that there's a negative value for the quadratic too!

Final answer would be $ x\equal{}46 \pm20\sqrt7$

((10+-sqrt112)-28)/4=-4.5+-sqrt7.or did I miscalculate it :?: ??[/quote]

@aguoman: Right, but you forgot as well what hsiljak pointed out and thus your answer is incorrect.

@123456789: Your solution would be complete if you put $ x>\minus{}7$.

@Cronus: Yes, you miscalculated. Read 123456789's solution.

@binomial_4eva: There are restrictions to $ n$ when you are working on the reals. After you apply the quadratic formula and simplify, you get $ n\equal{}5\pm2\sqrt7$, in which only $ n\equal{}5\plus{}2\sqrt7$ is acceptable.

Oh,Thans 10000thuser I didn't count the square on (10+-sqrt112). :blush:

maybe its 20sqrt7+-46
or           am i wrong :(

[quote="rt_08"]maybe its 20sqrt7+-46
or           am i wrong :([/quote]Yes it's wrong.

If you have $ \sqrt{x\plus{}7}$, $ \sqrt{x\plus{}7}\ge0$. Also, from the original equation, $ \sqrt{x\plus{}7}\ne0$, so $ \sqrt{x\plus{}7}>0$. You can use a calculator to find out that if you plug in$ x\equal{}46\minus{}20\sqrt7$ into the [u]original[/u] equation (and this is always important but often overlooked fact), you'll get -10 instead of 10 on the LHS; thus it is an [u]extraneous[/u] root.

thanx a lot ill always take care of that!

thanx again for correcting
 :)


This was forwarded to the coaches ARML list.  I personally think that Lehigh (and mathematics) definitely deserve more coverage :)

http://www.mcall.com/news/columnists/all-5stunningjun19,0,5872125.column?coll=all-news-col

[quote]Stunning triumph by math students gets short shrift

''Consider an analog clock with an hour and a minute hand but no numerals. At Time T a.m., a mirror image of the clock shows a time X a.m. that is 5 hours and 28 minutes later than T. Compute T in terms of hours and minutes.''

(The answer: 3:16 a.m.)

That was one of only a few questions I could even understand, much less solve, in an annual high school team competition held by the American Regions Mathematics League.......[/quote]

Agreed.  I think the Lehigh team actually got more coverage in the Morning Call last year when they didn't win.  Kind of sad, but it seems to be normal.  Here in Minnesota, the state Mathcounts competition generally gets a short blurb on page 2 of the local section of the Minneapolis paper.  If we're lucky, our state tournament for high schoolers makes page 1 of the local section.  This year was even worse: no newspaper article; the only news coverage we got was a rather disorganized 30-second segment at the end of the local news on one channel.

I will note that this is partially (or even primarily) the fault of those of us in the educational community.  We don't do nearly enough work seeking publicity, perhaps because the 'seek publicity' gene is so far removed from the 'do cool educational things' gene.

Not that we at AoPS are any better about this than anyone else - we have a notoriously hard time trying to focus on publicity tasks.

[quote="rrusczyk"]
Not that we at AoPS are any better about this than anyone else - we have a notoriously hard time trying to focus on publicity tasks.[/quote]

Just to add to what Richard said -- as we grow we plan to do more to help generate publicity for the awesome accomplishments of students in the mathematics and problem solving communities.  However, that's not an easy game to play as much of the news is entertainment-based.  Most people watch/read the news for the same reason they go to movies -- to stimulate the emotions that aren't being stimulated during the course of a regular day.

Fortunately there are some successful academic programs that generate publicity and perhaps we can learn a little about how they present themselves.  It shouldn't take top students until they're 40 and successful in their field to be recognized on the level of...every other social role in society.


geometry
trigonometry
trig identities
Law of Cosines
area of a triangle
Heron\u0027s formula
geometry unsolved

Given triangle ABC (AB = c,BC = a,AC = b).S is area of triangle ABC.$ m_a ,m_b ,m_c$ are lengths of median from vertex A,B,C to sides opposite respectively.
 Prove that:
a)If $ m_a^2  \equal{} m_b^2  \plus{} m_c^2$ then $ S \equal{} a^2 tanA$

b)If $ m_a^2  \plus{} m_b^2  \plus{} m_c^2\equal{} 3\sqrt 3 S$ then triangle ABC is equilateral triangle.

a) It can be derived from Stewart's theorem that $ m_a^2\equal{}\dfrac{2b^2\plus{}2c^2\minus{}a^2}{4}$, $ m_b^2\equal{}\dfrac{2a^2\plus{}2c^2\minus{}b^2}{4}$, and $ m_c^2\equal{}\dfrac{2a^2\plus{}2b^2\minus{}c^2}{4}$.
Substituting into the equation and multiplying by 4 we get
$ 2b^2\plus{}2c^2\minus{}a^2\equal{}(2a^2\plus{}2c^2\minus{}b^2)\plus{}(2a^2\plus{}2b^2\minus{}c^2)$ which is equivalent to
$ b^2\plus{}c^2\equal{}5a^2$.
By the Law of Cosines $ a^2\equal{}b^2\plus{}c^2\minus{}2bc\cos A$, so we can substitute $ 5a^2$ for $ b^2\plus{}c^2$ to get
$ a^2\equal{}5a^2\minus{}2bc\cos A$. Solving for $ \cos A$ we get
$ \cos A\equal{}\dfrac{2a^2}{bc}$.
A well known formula states that $ S\equal{}\dfrac{1}{2}bc\sin A$, so
$ S\equal{}\dfrac{1}{2}(2a^2)\dfrac{bc}{2a^2} \sin A\equal{}a^2\dfrac{\sin A}{\cos A}\equal{}a^2\tan A$ as desired.

b) Substituting into the equation and multiplying by 4 we get
$ (2b^2 \plus{} 2c^2 \minus{} a^2) \plus{} (2a^2 \plus{} 2c^2 \minus{} b^2) \plus{} (2a^2 \plus{} 2b^2 \minus{} c^2) \equal{} 12\sqrt {3}S$.
$ 3a^2 \plus{} 3b^2 \plus{} 3c^2 \equal{} 12\sqrt {3}S$, so $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 4\sqrt {3}S$.
By Heron's Formula $ S \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$ so
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 4\sqrt {3}\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)} \equal{} \sqrt {3\cdot 2s(2s \minus{} 2a)(2s \minus{} 2b)(2s \minus{} 2c)}$.
Since $ 2s \equal{} a \plus{} b \plus{} c$,
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} \sqrt {3(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c)}$.
By AM-GM $ (b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c) \le \left(\dfrac{a \plus{} b \plus{} c}{3}\right)^3$.
$ (a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c) \le 3 \left(\dfrac{a \plus{} b \plus{} c}{3}\right)^4 \le 3 \left(\dfrac{a^2 \plus{} b^2 \plus{} c^2}{3}\right)^2$ by AM-QM, so
$ 3(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c) \le (a^2 \plus{} b^2 \plus{} c^2)^2$.
Taking the square root of both sides, we get
$ \sqrt {3(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c)} \le a^2 \plus{} b^2 \plus{} c^2$.
Since we have equality $ a \equal{} b \equal{} c$, so we have an equilateral triangle.


limit
linear algebra
matrix
algebra
polynomial
function
calculus

Let $ \{c_n\}_{n\equal{}0}^{\plus{} \infty}$ be a sequence defined as: $ c_0,c_1 > 0$ and for every $ n\geq 1$ we have $ c_{n\plus{}1} \equal{} \frac{2}{c_n \plus{} c_{n\minus{}1}}$. Prove that there exists $ \lim_{n\to \plus{} \infty} c_n$.

[hide]My solution is based on the sub-sequence $ u_n : \equal{} \max\{c_n,c_{n\plus{}1},c_{n\plus{}2},c_{n\plus{}3}\}$, we will prove the existence of $ \lim u_n$, but the rest $ \varepsilon$-method is incredibe! Hope for a beautiful solution[/hide]

My incomplete thoughts:

Consider the mapping $ F$ from $ (0,\infty)\times (0,\infty)$ into itself given by $ F(x,y)\equal{}\left(y,\frac2{x\plus{}y}\right).$ The sequence in question is the iteration of this mapping.

We can show readily that $ F$ has only one fixed point, namely $ (1,1).$ Hence if the sequence $ c_n$ converges, its limit must be $ 1.$ Furthermore, the point $ (1,1)$ is stable for this mapping.

To be specific, the Jacobian of $ F$ is $ \begin{pmatrix}0&\minus{}\frac2{(x\plus{}y)^2}\\1&\minus{}\frac2{(x\plus{}y)^2}\end{pmatrix}.$

Evaluated at the fixed point, that Jacobian is $ \begin{pmatrix}0&\minus{}\frac12\\1&\minus{}\frac12\end{pmatrix}.$

The characteristic polynomial is $ \lambda^2\plus{}\frac12\lambda\plus{}\frac12,$ with eigenvalues $ \frac{\minus{}1\pm i\sqrt{7}}{4},$ which are less than $ 1$ in absolute value. That's how we know the point is stable. Once we get close enough to $ (1,1),$ our points will spiral inward. The exact values of those eigenvalues - in particular, their arguments - offer clues as the the somewhat alternating but not obviously periodic behavior of the signs of the differences between $ c_n$ and $ 1.$

To finish the argument globally, I think I'd like to find a function $ g(x,y)$ defined on the first quadrant with a global minimum at $ (1,1)$ such that $ g(F(x,y))<g(x,y)$ for all $ (x,y)\ne(1,1).$ I haven't found that function yet.

How about $ x\plus{}\frac 1x\plus{}2\left[y\plus{}\frac 1y\right]$   :P

Thanks, fedja. I'd looked at $ x \plus{} \frac1x \plus{} y \plus{} \frac1y,$ but somehow overlooked why the $ 2$ should be there. I see now.

Let me go ahead and put in the proof just so it's here in this thread.

Lemma 1: Let $ g(x,y) \equal{} x \plus{} \frac1x \plus{} 2\left(y \plus{} \frac1y\right),$ and let $ F(x,y) \equal{} \left(y,\frac2{x \plus{} y}\right)$, with both functions defined over $ (0,\infty)\times(0,\infty).$

Then $ g(x,y) \minus{} g(F(x,y))\ge 0,$ with equality only for $ x \equal{} y.$

Proof of lemma:

$ g(x,y) \minus{} g(F(x,y)) \equal{} x \plus{} \frac1x \plus{} 2\left(y \plus{} \frac1y\right) \minus{} \left[\left(y \plus{} \frac1y\right) \plus{} 2\left(\frac2{x \plus{} y} \plus{} \frac {x \plus{} y}2\right)\right]$

$ \equal{} x \plus{} \frac1x \plus{} y \plus{} \frac1y \minus{} \frac4{x \plus{} y} \minus{} (x \plus{} y)$

$ \equal{} \frac1x \plus{} \frac1y \minus{} \frac4{x \plus{} y} \equal{} \frac {(x \minus{} y)^2}{xy(x \plus{} y)}.$

The argument should be clear there.

Lemma 2: $ g(x,y) \minus{} g(F(F(x,y))\ge 0$ with equality only if $ (x,y) \equal{} (1,1).$

By Lemma 1, the only situation which needs to be addressed is that of $ x \equal{} y.$ But then $ F(x,x) \equal{} \left(x,\frac1x\right).$ We could only have $ g$ remain constant on this if $ x \equal{} \frac1x,$ and that only happens for $ x \equal{} 1.$

Lemma 3: On the open first quadrant, $ g(x,y)$ has only one local extremum and that is its global minimum at $ g(1,1) \equal{} 3.$ (Proof omitted - it's not difficult.) Furthermore, for all $ c\ge 3,$ the set $ \{(x,y): g(x,y)\le 3\}$ is a closed and bounded (hence compact) set.

Theorem:

Choose $ (x_0,y_0)$ arbitrarily in the open first quadrant and for $ n\ge 1$ define $ (x_n,y_n)$ by iteration: $ (x_n,y_n) \equal{} F(x_{n \minus{} 1},y_{n \minus{} 1})$ Then $ \lim_{n\to\infty}(x_n,y_n) \equal{} (1,1).$

(I think the proof of the theorem mostly writes itself now - the ingredients are here.)

Similar problem (the solution presented there works here) http://www.mathlinks.ro/Forum/viewtopic.php?t=296767. :)


This problem is adapted from the 2004 MMO.

Given the set, [tex]\displaystyle S=\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\}[/tex].

1.  Prove that the sum of the elements of S is infinite.

2.  Prove or disprove the following assertion:  Every positive rational number can be represented by the sum of the elements of a finite subset of S.

I remember doing #1 for fun a while ago... I posted my proof online:

[url]http://home.comcast.net/~carlosrd/series.pdf[/url]

[quote="belenos"]Can we take the same 1/n several times?[/quote]

No.  No element can be used more than once.

I agree belenos.  I think we're only allowed to sum each element of a subset once.  In which case fractions like 2/7 would be impossible to express as the sum of two distinct fractions simply because 7 is prime.  Therefore 2/7 can only be written as 1/7+1/7, which I don't believe is allowed.  Or did I misinterpret the problem?

Thanks Ravi B I thought I was overlooking something


just use the fact that: $ln(1+x)=x-x^2/2+x^3/3+...$
And you will obtain a double sum, absolutely convergent.


a,b,c are 3 real numbers such that $ 1 \leq a,b,c  \leq 2$.
Show that: $ (3a\plus{}2b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\leq \frac{45}{2}$.
Have fun! :P

[quote="nmtvn"]a,b,c are 3 real numbers such that $ 1 \leq a,b,c \leq 2$.
Show that: $ (3a \plus{} 2b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\leq \frac {45}{2}$.
Have fun! :P[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 3,,2008, and we can't discuss it now

you are right "chienthan" its solution must keep secret!!!
you are fair!!![/quote]

I don't understand, kihe_freety5 and chien than!

Ban nmtvn thường xuyên post bài đang thi trên THTT. Điều này không được phép. Không rõ mục đích của bạn là gì?


Ring Theory
algebra
polynomial
superior algebra
superior algebra unsolved

[b](1)[/b] Prove that $ ( xy \minus{} zw)$ is a prime ideal in $ \mathbb{Z}[x,y,z,w]$
[b](2)[/b] Prove that $ \mathbb{Z}[x,y,z,w] / (xy \minus{} zw)$ cannot be a UFD.

I appreciate any hint for it.  Danken!

The second part is fairly easy. Since $ xy\minus{}zw$ is homogeneous and the ideal it generates is prime, we can talk about degree on the quotient. Replacing instances of $ zw$ with $ xy$ doesn't change the degree of a monomial, and can only change the degree of a sum of monomials of the same degree by canceling them all out. Here we use the primeness: let $ f$ and $ g$ be two nonzero elements of the quotient, and let $ f'$ and $ g'$ be their highest-degree terms, reduced mod $ xy\minus{}zw$. If $ xy\minus{}zw$ divides $ f'g'$, it divides either $ f'$ or $ g'$, a contradiction. Thus the degree of $ f'g'$ mod $ (xy\minus{}zw)$ is the sum of the degrees of $ f'$ and $ g'$, and the same holds for $ f$ and $ g$.
Now, we can say that only constants are units in the quotient, and only $ \pm 1$ work because the ring is $ \mathbb{Z}$. This means that $ x,y,z,w$ are all distinct irreducible elements, and $ xy\equal{}zw$ is a failure of unique factorization.

That leaves the first part. Here, the polynomial ring is a UFD and $ (xy\minus{}zw)$ is irreducible; that's enough.

(Side note- I looked at the Wikipedia page while writing that last sentence. It contains [i]this very example[/i], with essentially the same argument)

See also my post below for much more on this ring: Krull, integrally closed, etc.
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=3157.0001.0001.0002


function
integration
algebra
polynomial
linear equation
calculus
calculus computations

${\chi+\beta f(x+\epsilon )+\alpha f'(x)=0}$

The solution of a linear recursive equation is a exponential function and the solution of a diferential linear equation is a exponential function too. So, we can try it with this differential recursive equation.

${\frac{\partial \left(\chi+\beta f(x+\epsilon )+\alpha f'(x)\right)}{\partial x}=0}$
${\beta f'(x+\epsilon )+\alpha f''(x)=0}$

To Simplify

${g(x)=f'(x)}$

Substituing

${\beta g(x+\epsilon )+\alpha g'(x)=0}$
${\alpha g'(x)=-\beta g(x+\epsilon )}$
${\frac{g'(x)}{g(x+\epsilon )}=-\frac{\beta }{\alpha }}$

Supossing it is exponential

${g(x)=a e^{x b}}$

Substituing

${\frac{g'(x)}{g(x+\epsilon )}=-\frac{\beta }{\alpha }}$
${b e^{-b \epsilon }=-\frac{\beta }{\alpha }}$

We need to find ${b}$. So we use the Lambert W function to solve for ${b}$.

${\kappa \in \mathbb{Z}}$
${b=-\frac{W_{\kappa }\left(\frac{\beta \epsilon }{\alpha }\right)}{\epsilon }}$

So ${g}$ is

${g(x)=a e^{-\frac{x W_{\kappa }\left(\frac{\beta \epsilon }{\alpha }\right)}{\epsilon }}}$

Since  ${f'(x)=g(x)}$

${f(x)=\int_{x_{0}}^{x}g(t) \, dt=\int_{x_{0}}^{x}a e^{-\frac{t W_{\kappa }\left(\frac{\beta \epsilon }{\alpha }\right)}{\epsilon }}\, dt}$
${f(x)=\frac{a \left(-e^{-\frac{x W_{\kappa }\left(\frac{\beta \epsilon }{\alpha }\right)}{\epsilon }}+e^{-\frac{W_{\kappa }\left(\frac{\beta \epsilon }{\alpha }\right) x_{0}}{\epsilon }}\right) \epsilon }{W_{\kappa }\left(\frac{\beta \epsilon }{\alpha }\right)}}$

That is the solution.
But... How to prove that this is the general solution!? And how to solve.

${f'(x)=\sum_{n=1}^{k}\alpha_{n}f\left(x+\omega_{n}\right)}$

For k=2,3 $\ldots$

[quote="Thales418"] And how to solve ${f'(x)=\sum_{n=1}^{k}\alpha_{n}f\left(x+\omega_{n}\right)}$ for k=2,3 $\ldots$[/quote]
There is a remarkable theorem by Schwartz that shows that the space of all smooth solutions is just the closure in $C^\infty$ of the linear span of exponential polynomials satisfying the equation. The proof is too long and complicated to post here. And in order to find all exponential polynomials satisfying the equation, one needs to find the complex roots of $z=\sum_{n=1}^{k}\alpha_{n}e^{\omega_{n}z}$. The latter form a fairly nice set on the complex plane but there is no explicit formula for them. It leaves open the question whether every solution can be represented as a convergent series in exponential polynomial solutions. That I don't know :( but the nice structure of zeroes allows to hope that the answer is "Yes".

As to the original problem you started with, all you need to do to find the general solution is to guess one paricular solution and  then add the general solution of the homogeneous equation.

i tryed to solve ${k=2}$ but it's too much difficult...


function
real analysis
real analysis unsolved

Let $ f : (0,\infty) \to \mathbb R$ be a continous function such that the sequences $ \{f(nx)\}_{n\geq 1}$ are nondecreasing for any real number $ x$. Prove that $ f$ is nondecreasing.

[hide="Quick and dirty"]It's trivial too see that if $ m>n$, then $ f\left( \frac mn x \right) \geq f(x)$, for all $ x$, so in particular $ f(p)\geq f(q)$ for all rational numbers $ p\geq q$. Now let $ y \in \mathbb R \minus{} \mathbb Q$, and let $ x\in \mathbb Q$, $ x<y$. If $ f(x)>f(y)$, let $ \delta \equal{} f(x)\minus{}f(y)$. Because $ f$ is continuous, there exists a rational number $ x_0$, $ x_0 \in (x,y)$ such that $ f(x_0)\equal{} f(y) \plus{} \frac {\delta}{2}$. But $ f(x)\leq f(x_0)$, so $ \delta \leq 0$ contradiction. This means that if $ x,z \in \mathbb Q$, and $ y\in \mathbb R\minus{}\mathbb Q$, such that $ x<y<z$ then $ f(x)\leq f(y)\leq f(z)$.

Now let $ x<y$ be any two real numbers. Then there exists $ z \in \mathbb Q$ such that $ x<z<y$, and using the above claim it follows that $ f(x)\leq f(y)$ which proves that $ f(x)$ is nondecreasing.[/hide]

It was given, I just made a typo in the statement.  In fact I think Darboux Property is the requirement (so it's valid for some non-continous functions - although they become continous afterwards).


[quote="math92"]I'm not sure if this goes here.

Express $\frac{\sqrt{-136}}{2}$ in terms of $i$
Express $\frac{\sqrt{-20}}{2}$ in terms of $i$[/quote]

[hide] $\frac{\sqrt{-136}}{2}=\frac{\sqrt{-1}\sqrt{4}\sqrt{34}}{2}=\frac{2i\sqrt{34}}{2}=\boxed{i\sqrt{34}}$ :D
$\frac{\sqrt{-20}}{2}=\frac{\sqrt{-1}\sqrt{4}\sqrt{5}}{2}=\frac{2i\sqrt{5}}{2}=\boxed{i\sqrt{5}}$ :D [/hide]


search
geometry
geometric transformation
topology

I have access to a vast number of excellent mathematical books in Russian. (An interesting fact is that over a quarter of the world's scientific literature is published in Russian.)
The problem is that my Russian is not that good and I was wondering is there a English-Russian and Russian-English dictionary of mathematical therms and other words commonly used in mathematical texts that can be downloaded?
Thanks in advance.

Hi! I often use [url=http://www.poiskknig.ru]this[/url] Russian site when I want to find some necessary books in Russian or English. Try to enter "словарь" in the search bar. I tried and found (possibly) relevant link, [i]http://lib.homelinux.org/_djvu/L_Languages/LEn_English/Neizv.Avtor.%20Matematicheskij%20russko-anglijskij%20slovar'%20(net%20s.288-289)(ru)(T)(907s)_LEn_.djvu[/i], (the book is called "Mathematical English-Russian dictionary"). I didn't download it and don't know its content, but its title seems to apt your request. In both fields you will see after clicking the link, you should enter "kolobok".

By the way, what is the address of the source with Russian books you refer to? Do you know some links to good maths books in English? I cannot found the book "Variational Methods, Application to Nonlinear Partial Differential Equations and Hamiltonian Systems" by M. Struwe.  :(

Thank you for the link. I will download the book and see if it solves my problem.
When I said I have access to a vast number of mathematical books in Russian I meant that I have a huge library of original Russian books available in my faculty and in Russian science and arts center. I do not have them in virtual form.

Here are some links I came by while searching for ebooks.

http://www.arxiv.org/
http://freescience.info/index.php
http://onlinebooks.library.upenn.edu/
http://www.comms.scitech.susx.ac.uk/fft/
http://www.poiskknig.ru/
http://lib.mexmat.ru/
http://www.eknigu.com/
http://www.ebookee.com/
http://www.ams.org/online_bks/onbk_list.html
http://ocw.mit.edu/OcwWeb/Mathematics/index.htm
http://avaxhome.org/ebooks/science_books/math
http://www.math.gatech.edu/~cain/textbooks/onlinebooks.html
http://onlinebooks.library.upenn.edu/webbin/book/subjectstart?QA
http://www.geocities.com/alex_stef/mylist.html

The dirtionary is very good. However, it is English-Russian only and I need Russian-English. Is there any Russian-English mahematical dictionary avaliable?

Try to use the electronic [url=http://www.lingvo.ru/lingvo/]Lingvo Dictionary[/url]. It has fairly big mathematical word stock. However, I have noted one flaw in translation: When I input the word "диффеоморфизм", the output was "homeomorphism" while the correct answer is "diffeomorphism". The "computer" counterpart of the disctionary does not have such a flaw, however.

Having received translation, you should pay particular attention on whether there is word [i][color=green]матем.[/color][/i] near the translation or not. If there is, you may be 98% sure that your math term or expression is translated correctly.

[b]P.S.[/b] Try to translate "локально тривиальное расслоение", "линейный оператор" etc.

Thank you for the link. It is very good.


Hey! I have a question which needs an answer with an explanation. If you can help, that would be great! Thanks.

[b]Question:[/b]

The sum of four numbers is 64. If 3 is added to the first number, 3 is subtracted from the second, the thrid is multiplied by 3, and the fourth is divided by 3, then all four results will be equal. The difference between the largest and the smallest of the original number is:

(A)128 over 3 "128/3"
(B)32
(C)27
(D)24
(E)11

I tried to figure it out, however, I got confused somewhere along the way. I would greatly appreciate an explanation to this question. Thanks. :)

[hide]If we let a,b,c,and d be the numbers, then
a+b+c+d=64 and
a+3=b-3=3c=d/3, so
b=a+6
c=(a+3)/3, and
d=3(a+3)
Substituting into the original equation:
a+(a+6)+((a+3)/3)+(3(a+3))=64
a+(a+6)+(a/3+1)+(3a+9)=64
5a+a/3+16=64
5a+a/3=48
16a=(48)(3)
a=(3)(3)=9
So b=15, c=4, and d=36
36-4=32[/hide]
or
[hide]
Let x be the results that are equal.Then
(x+3)+(x-3)+(3x)+(x/3)=64
So x is 12 and the four numbers are 6, 12, 4, and 36[/hide]


algebra
polynomial
function
algebra proposed

[b] If $x_1,x_2,x_3 \; , (x_1\in {\mathbb R}),$ are the roots of equation  $\; x^3-x^2+Bx -1 = 0$ where [/b]   $B \in {\mathbb R} \; ,\;$
 [color=green] [b]  to prove/ or disprove/   that  $| x_2-x_3 |$  is a [u]rational number[/u] .  [/b] [/color]

[size=75]
[color=blue][b]  No complete solution. [/b] [/color] [/size]

I do not see any reason why $|x_2-x_3|$ must be rational even in the simplest case of $B=0$.

Let $x_1,\ x_2,\ x_3$ are the roots of equation $z^3-z^2-1=0$ where $x_1$ is real and $x_2,\ x_3$ are conjugate complex numbers. Therefore, $x_2-x_3$ is purely imaginary and (up to a sign) $|x_2-x_3| = I(x_2-x_3)$ where $I$ is the imaginary unit.

Consider a polynomial
\[ \prod_{i,j=1}^3 (z-x_i+x_j)  \]
whose coefficients are symmetric functions of $x_1,\ x_2,\ x_3$ and can be expressed via coefficients of polynomial $z^3-z^2-1$. In particular, we have
\[ \prod_{i,j=1}^3 (z-x_i+x_j) = z^9 - 2 z^7 + z^5 + 31 z^3.  \]

Therefore, $|x_2-x_3|$ is a root of $z^6 + 2 z^4 + z^2 - 31 = 0$. But it is easy to see that this equation does not have rational roots.

P.S. With help of Maple $|x_2-x_3|$ can be computed explicitly:
\[ |x_2-x_3|=\sqrt{3}\left( \frac{1}{6}(116+12\sqrt{93})^{1/3} - \frac{2}{3 (116+12\sqrt{93})^{1/3}} \right).  \]

[quote="maxal"]I do not see any reason why $|x_2-x_3|$ must be rational even in the simplest case of $B=0$.[/quote]
Interesting ! My opinion is that for all $B$ ,  the polynomial has' nt all roots in ${\mathbb Q} .$ Thanks for interest.


find a real matrix B such that
$ B^4\equal{}\left[\begin{array}{ccc}2&0&0\\0&2&0\\0&\minus{}1&1\end{array}\right]$
 :lol:

First find $ P$ such that
$ PB^4P^{\minus{}1}\equal{}\text{diag}(2,2,1)$

which is equivalent to
$ (PBP^{\minus{}1})^4\equal{}\left(\text{diag}(\sqrt[4]{2},\sqrt[4]{2},1)\right)^4$ (*)

In order $ B$ satisfies (*) just take B so that
$ PBP^{\minus{}1}\equal{}\text{diag}(\sqrt[4]{2},\sqrt[4]{2},1)$, i.e.,

take $ B\equal{}P^{\minus{}1}\text{diag}(\sqrt[4]{2},\sqrt[4]{2},1)P$

The $ P$ in barra's post can be taken to be
\[ P\equal{}\begin{bmatrix}1&0&0\\0&1&0\\0&1&1\end{bmatrix},\quad P^{\minus{}1}\equal{}\begin{bmatrix}1&0&0\\0&1&0\\0&\minus{}1&1\end{bmatrix}\]

In which case $ B\equal{}\begin{bmatrix}2^{1/4}&0&0\\0&2^{1/4}&0\\0&\minus{}2^{1/4}\plus{}1&1\end{bmatrix}.$


Let $P(x)\in\mathbb{R}[x]$. Prove that there is a $Q(x)\in\mathbb{R}[x]$ such that $P(x)Q(x)$ is a polynomial in $x^{2001}$.

i am not sure how much i am allowed to "lemma bash"

[if p(x) is constant, take q(x)=1, we are done, else, we have a finite number of roots, hence the expression below represents all other polynomials]

if $p(x)=a\prod_{r| p(r)=0}(x-r)$, then let $p(x)q(x)=a\prod_{r|p(r)=0}\prod_{k=1,...,2001}(x*z_{k}-r)=\prod_{r|p(r)=0}(x^{2001}-r^{2007})$ where $z_{k}$ are the $2001$th roots of unity

we know that upon expansion, p(x)q(x) has coefficients that are symmetric sums of the roots... [a product of the roots to certain powers], however, we know that any such sum is representable as a polynomial of the elementary symmetric sums [lemma], these are the coefficients of p(x), which are real. since we can represent any coefficient of p(x)q(x) as a polynomial in the coefficients of p(x), the coefficients of p(x)q(x) are real

hence p(x)q(x) is real, note that $(x-r)|(x^{n}-r^{n})$, so $q(x)$ is a polynomial, by the division algorithm, $q(x)$ since $p(x)q(x)$ and $p(x)$ have real coefficients, so does $q(x)$


[quote="13375P34K43V312"]I think this is an important theorem to know the deriviation of.

Prove that in an arbitrary rectangle $ABCD$ with an arbitrary point $P$ chosen inside it, $(AP)^{2}+(CP)^{2}=(BP)^{2}+(DP)^{2}$.

[hide="Hint"] Drop perpindiculars to each side from $P$.[/hide][/quote]

Actually P doesn't have to be inside Rectangle $ABCD$ or even be on the same plane for that matter

[hide] Well, just draw the orthogonal projections of the point P to the sides of the rectangle: then, using repeatedly Pythagoras' Theorem, you obtain that those sums of squares are equal, just as claimed... [/hide]   :)


It seems lexington is one of the best towns.
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2008-12a/StateStats/MA.HTML

Indeed; 3 of the 4 state winners on the 12, and the state runner up on the 10 (to none other than Samlmaml). Also, state winners on every test except the 10B (again, Samlmaml).

how do you tell from those stats which school did better??

Yeah 'cause we're so well connected with other good math students.
-cough-
PERFECT HOW COULD YOU?
You ruined your 1000 posts.
:(

I prefer palindromes.

UH OH LOOK WHAT I DID

Eh, posts here don't count anyways.
Wait, I thought they did.
Hmm, maybe you got one of your other posts deleted.

Wow, Yongyi and I are the only 8th graders who took the AMC12B...
And then there was only one person who took the 12A...

Meh, my school didn't let us take the 12B.
Because you Acton kids came over and took the 5 free spots at our high school.
So basically it was either.
Get 10 people to do the 10B
Get 10 people to do the 12B
Since we had to pay for the test because there were no more free spots at the high school.
We chose the 10B.

Lawlz, you could have taken the 12B at Worcester, which is what I did...

Wow, who is the 7th grader that got 118.5 on 12A?!?

Mass Academy ranked 6th on the 12: 141, 130.5, 109.5

Satisfactory.  :wink:


[quote="buzzer11"][quote="xpmath"]Find a base 7 3-digit number which has its digits reversed when expressed in base 9.

Please help me, I'm not good with number bases.[/quote]

[hide]well:

$abc_{7}= cba_{9}$
$7^{2}(a)+7b+c = 9^{2}(c)+9b+a$
$48a-2b-80c = 0$
$24a-b-40c = 0$
$24a = b+40c$

some quick checking quickly reveals (2,8,1) as a solution.

$\boxed{a=2, b=8, c=1}$

Thus,

$\boxed{281_{7}= 182_{9}}$

That's one solution, at least.[/hide][/quote]
281 cannot be expressed in base 7 because "8" does not exist in base 7.  There is no such number that satisfies the problem.

[quote="buzzer11"]:wallbash_red:
I didn't see that.
haha, algebraically, it works. :)
Well, i guess there's no solution then?[/quote]

It's probably form a competition, xpmath is very big on competitions, and therefore there probably is a solution.

Two possibilities:
1) You made a mistake in your proof.
2) This can't be proven algebraically. I'll look over your proof to look for flaws...

EDIT: If we had a program that projects a 3-D graph of an equation, we could graph it and look at all the results.

EDIT 2: I see no flaws in your proof.

[quote="Teki-Teki"]here's how I did it - if you don't understand modular arithmetic, you won't get it, but mods make it far easier.

[hide]
Take the equation modulo 8.  $0 \equiv b$, hence $b = 0,8$.  But we already know that $b \neq 8$, hence $b = 0$.  Now we have $24a = 40c \leftrightarrow 3a = 5c$.  Since $a,c \le 6$, we have $(a,b,c) = (5,0,3)$ as the unique solution.
[/hide][/quote]

Oh, I get it! I don't see how the original conjecture doesn't work, though.

[quote="Teki-Teki"]here's how I did it - if you don't understand modular arithmetic, you won't get it, but mods make it far easier.

[hide]
Take the equation modulo 8.  $0 \equiv b$, hence $b = 0,8$.  But we already know that $b \neq 8$, hence $b = 0$.  Now we have $24a = 40c \leftrightarrow 3a = 5c$.  Since $a,c \le 6$, we have $(a,b,c) = (5,0,3)$ as the unique solution.
[/hide][/quote]
Haha nice! I didn't think of using mods.
And temperal: the method used earlier narrowed down the choices, but, this time, didn't reveal explicit solutions.
That is why many of us resorted to minimal trial and error.


inequalities
modular arithmetic
algebra unsolved
algebra

Given $ n \in N,n>1$,determine the number of ordered $ n$-tuples of integers $ (a_1,a_2,...,a_n)$ such that $ a_1\plus{}a_2\plus{}...\plus{}a_n$≥$ n^2$,and $ a_1^2\plus{}a_2^2\plus{}...a_n^2$≤$ n^3\plus{}1$

[quote="KDS"]Given $ n \in N,n > 1$,determine the number of ordered $ n$-tuples of integers $ (a_1,a_2,...,a_n)$ such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n$≥$ n^2\plus{}1$,and $ a_1^2 \plus{} a_2^2 \plus{} ...a_n^2$≤$ n^3 \plus{} 1$[/quote]

[b]Case 1:[/b] $ a_1\plus{}a_2\plus{}...\plus{}a_n \ge n\plus{}1$, then
$ n(n^3\plus{}1) \ge n(a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2) \ge (a_1\plus{}a_2\plus{}...\plus{}a_n)^2 \ge (n^2\plus{}1)^2$
$ \Rightarrow n \ge 2n^2\plus{}1$, contadiction.

[b]Case 2:[/b] $ a_1\plus{}a_2\plus{}...\plus{}a_n \ge n$, then

We have $ n^3\plus{}1 \ge a_1^2 \plus{} a_2^2 \plus{} ...a_n^2 \ge \frac{1}{n}(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^2\equal{}n^3$

Equality for right inequality holds $ \Leftrightarrow a_1\equal{}a_2\equal{}...\equal{}a_n\equal{}n$ as a solution.

Also we have $ a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2 \equiv a_1\plus{}a_2\plus{}...\plus{}a_n \equiv n \pmod{2}$ which is fixed.
So equality doesn't hold $ \Rightarrow n^3\plus{}1 \ge a_1^2 \plus{} a_2^2 \plus{} ...a_n^2 \ge n^3\plus{}2$, contradiction.

So there is only one tuple , $ (a_1,a_2,...,a_n)\equal{}(n,n,...,n)$.


Hi folks, I have a rather (maybe silly) query regarding the terms- "Hypothesis" and "Conjecture".

[b] Query [/b]

What is the difference (or distinction) between a 'hypothesis' and a 'conjecture'?
Can anyone elucidate the difference that exists between them with the help of examples? Can they be used interchangeably?

I do know that a hypothesis is a supposition, and a conjecture is an intuitive guess.

But I want to understand the subtle (or maybe major) difference that exists between them.

Can I say - [b] Riemann conjecture and Poincare Hypothesis? [/b] (I know I can't say that! :lol: ).

Do answer these queries of mine clearly.

Thanks!

"A conjecture" in mathematics is "a statement that has been proposed as true (or possible, or in need of determination) but not proven nor disproven."  "To conjecture" is "to propose as true (or possible, or in need of determination) without a proof."  "Hypothesis" is used relatively rarely in mathematics in the sense of "conjecture," with a few notable exceptions (e.g., the Riemann hypothesis, the continuum hypothesis).  See http://mathworld.wolfram.com/Hypothesis.html

Okay, Can the terms  'conjecture' and  'Hypothesis' being used interchangeably? Are the synonyms of each other?

What's the difference between the two?

As the mathworld article says, "hypothesis" is a rough synonym of "conjecture."  As I said, "hypothesis" is rarely used in mathematics in the role of "conjecture," which has the technical meaning I described.  "Hypothesis" is widely used (in a technical sense) in statistics and the sciences.  It is not always easy to articulate the precise distinction between two near-equivalent words.  In the case of mathematics, cultural norms are that we refer to conjectures as conjectures, not as hypotheses.

"Hypothesis" has more meanings, while the word "conjecture" is less ambiguous.  One meaning of "hypothesis," at least colloquially, is "assumption" or "premise," e.g. cryptography is done under the hypothesis that one-way functions exist (which is [b]also[/b] a conjecture).

Perhaps one reason the Riemann hypothesis is called a hypothesis is that much work is done under the assumption that it is true.  This is certainly not true of most conjectures.

[quote="t0rajir0u"]"Perhaps one reason the Riemann hypothesis is called a hypothesis is that much work is done under the assumption that it is true.  This is certainly not true of most conjectures.[/quote]
See also Schinzel's hypothesis.

Also, no, you can't say the "Reimann Conjecture" and "Poincare Hypothesis". Reimann Hypothesis and Poincare Conjecture are proper nouns; in particular, they are names.

Just like Fishman and Fisher probably mean the same thing, but if you wanted to talk about some, say, Andrew Fishman, you wouldn't say "Andrew Fisher".

interesting, proper nouns I see.

All I know is that the conjecture is 50 posts and the hypothesis is 100 posts.

what about yang mills theory.

Yang-Mills theory is a [i]scientific[/i] theory. Read about the associated [url=http://en.wikipedia.org/wiki/Millennium_Prize_Problems]Millenium Prize Problem[/url] at [url=http://en.wikipedia.org/wiki/Yang%E2%80%93Mills_existence_and_mass_gap]this Wikipedia page[/url].


geometry
circumcircle
incenter
geometric transformation
homothety
ratio
power of a point

Let $ABC$ be a triangle, $D$ be a point on side $BC$, and let $\mathcal{O}$ be the circumcircle of triangle $ABC$. Show that the circles tangent to $\mathcal{O},AD,BD$ and to $\mathcal{O},AD,DC$ are tangent to each other if and only if  $\angle BAD=\angle CAD$.

[i]Dan Branzei[/i]

[quote="hophinhan"][b](Romania, 1997) [/b]Let ABC be a triangle, let D be a point on side BC, and let (O) be the circumcircle of triangle ABC. Show that the circles tangent to (O),AD,BD and to (O),AD,DC are tangent to each other if and only if   \hat{BAD}  = \hat{CAD}.[/quote]
Let I be the incenter of triangle ABC.
+If AD is the bisector of $ \angle BAC$, let $ AD\cap (O) \equal{} \{E\}$
$ \Rightarrow E$ is the midpoint of arc BC
Then $ M,N,E$ are collinear. We have $ EM.EN \equal{} EC^2 \equal{} EI^2$
Construct a tangency EK of $ (O_2) (K\in AE)$ then $ EK^2 \equal{} EI^2$ so $ K\equiv I$
So $ (O_1)$ and $ (O_2)$ are tangent to each other at I.
+If $ (O_1)$ and $ (O_2)$ are tangent to each other. By Sawayama-Thébault's theorem and widen Lyness's theorem, it's easy to prove that they are tangent at I. Therefore $ \angle BAD \equal{} \angle CAD$

Dear Mathlinkers,
you can also see another way at
http://perso.orange.fr/jl.ayme   vol. 4  A new mixtilinear incircle adventure 3   p. 9-10
with a reference.
Sincerely
Jean-Louis

Please correct me if I say something wrong or stupid. But I think that the problem would be "better" if we are able to prove this problem:

[b]PROBLEM-[color=darkblue]Let $ ABC$ be a triangle. Let $ D$ be the point on $ BC$. Let $ (O)$ be the circumcircle of $ \triangle ABC$. Show that the $ (O_1)$, $ (O_2)$  tangent to each other at $ T$, where $ (O_1)$ is the circle that tangents to $ (O),AD,BD$, $ (O_2)$ is the circle that tangents to$ (O),AD,CD$, if and only if $ T$ is the incircle of $ \triangle ABC$[/color][/b].

Indeed, with problem that [b]hophinhanLHP[/b] has been proposed, we have a generalized problem for it, which is:

[b]LEMMA-[color=darkblue]Given $ (O)$ and $ A,B$ be $ 2$ points that lie on $ (O)$. Let $ (O_1),(O_2)$ be circles that lie on the same side wrt $ AB$ such that they both internallu tangent to $ (O)$ respectively at $ X_1,X_2$. Let $ \{C,D\}\equiv (O_1) \cap (O_2)$. Let $ K$ be the point that lie on the different side with $ (O_1),(O_2)$ wrt $ AB$. Prove that $ K$ is the midpoint of arc $ AB$ if and only if $ K\in CD$[/color][/b].

[i]Proof:[/i]
Let $ r,r_1,r_2$ be the radii of $ (O),(O_1),(O_2)$, respectively.$ K_1,K_2$ respectively be the tangency point of $ (O_1),(O_2)$ with $ AB$. Consider the homothety through point $ X_1$, ratio $ k_1 \equal{} \frac {r}{r_1}$. We have $ \mathcal {H}(X_1,k_1): (O_1)\mapsto (O)$, $ K_1\mapsto K'\Longrightarrow K'\in (O)$. Because $ O_1K_1\bot AB\Longrightarrow OK'\bot AB$, which implies that $ K'$ is the midpoint of arc $ AB$. With the same argument for $ (O_2)$, we conclude that $ X_1K_1,X_2K_2$ intersect at the point $ K\in (O)$ such that $ K$ is mipoint of arc $ AB$. We will prove that this will equivalent to $ K\in CD$. Indeed, when $ K$ is the midpoint of arc $ AB$, it is easy to prove that $ KA^2 \equal{} \overline {KK_1}\cdot \overline {KX_1}$, $ KB^2 \equal{} \overline {KK_2}\cdot \overline {KX_2}$. But $ KA \equal{} KB\Leftrightarrow \overline {KK_1}\cdot \overline {KX_1} \equal{} \overline {KK_2}\cdot \overline {KX_2}\Leftrightarrow \mathcal {P}_{K/(O_1)} \equal{} \mathcal {P}_{K/(O_2)}\Leftrightarrow K\in$ the radical axes of $ (O_1),(O_2)\Leftrightarrow K\in CD$. The result is lead as follow.

Back to the beginning problem.
Since $ A,D,T$ are collinear which is equivalent to $ \overline {ADT}$ is the radical axes of $ (O_1),(O_2)$. Let $ K\equiv AD\cap (O)$ ($ K$ is different from $ A$) $ \Leftrightarrow K$ is the midpoint of arc $ BC\Leftrightarrow AT$ is the bisector of $ \angle BAC$. It is easy to note that $ KB \equal{} KT \equal{} KC$. Therefore $ \angle KBT \equal{} \angle KTB$. But $ \angle KBD\equiv \angle KBC \equal{} \angle KAC \equal{} \angle BAD$, which implies that $ \angle TBA \equal{} \angle TBC\Leftrightarrow BT$ is the bisector of $ \angle ABC$. Thus $ T$ is the incircle of $ \triangle ABC$.

We'll use this little lemma :
[b]"If $(O;R);(O';R') [R>R']$ is internally tagent at point $K$ .Let $AB$ be a chord of $(O)$ and tagent to $(O')$ at $E$ then $KE$ passes through the midpoint of arc $AB$ "[/b](Easy to prove , right ?)
Let $J$ be midpoint of arc$BC$ not having $A$ .
Let $(O_1) ;(O_2)$ be the circles tagent to $(O);DA;DB$ and $(O);DA;DC$ respectively.
Call $E,F$ are tagency points of $(O_1);(O_2)$ with $(O)$ respectively .
$(O_1);(O_2) $ tagent to $BC$ at $P,Q$ respectively
According to the lemma , we have : $J,E,P$ are collinear ; $J,Q,F$ are collinear.
Furthermore , we have :$ \overline{JE}.\overline{JP}=JB^2 ; \overline{JQ}.\overline{JF}=JC^2$
$\Rightarrow \overline{JE}.\overline{JP}=\overline{JQ}.\overline{JF}$
$\Rightarrow  J$ lies on the racial axis of $(O_1);(O_2)$.
Due to that we get :
$ (O_1);(O_2)$ are tagent to each other $ \Leftrightarrow AD $ is their radical axis $\Leftrightarrow J \in AD$
$\Leftrightarrow \angle BAD =\angle CAD$
Done

Lemma-Let $ABC$ be a triangle.$I$ its incenter and $D$ a point on $BC$.consider the circle that is tangent to the circumcircle of $ABC$ but is also tangent to $DC$,$DA$ at $E,F$ respectively. Then $E,F$ and $I$ are collinear.

now in problem,if both circles are tangent together but $\angle BAD$ and $\angle CAD$ are different then clearly $I$ is not lies on $AD$.but the tangent points of the two circles collinear with $I$.which cant be possible(a tangent point is common from them).hence $BD$ is angle bisector.

Let $\omega_1$ be the circle which is tangent to $BC, AD$ and $\mathcal{O}$. Define $\omega_2$ similarly. Let $\omega_1$ touches $BC, AD$ at $P, Q$ and $\omega_2$  touches $BC, AD$ at $R, S$ respectively. Let $I$ be the incenter of $\triangle ABC$. By the Curvilinear Incircle lemma we know $P, I, Q$ are collinear. Similarly $R, S, I$ are collinear.

Suppose $D$ is the foot of $A-$angle bisector. Then $I=Q$ and similarly $I=S$. So $I$ will be the touch point of $\omega_1$ and $\omega_2$ on $AD$. Therefore $\omega_1, \omega_2$ will be tangent.

Now suppose $\omega_1, \omega_2$ are tangent on $AD$ which is $Q=S$, call it $I'$. Since by curvillinear we have $I$ lies on $PI'$ and $I$ lies on $RI'$ therefore $I=I'$ and the tangency point of these two circle is actually the incenter. Since the tangency point lies on $AD$ therefore $I$ lies on $AD$. As desired.


arithmetic sequence
algebra unsolved
algebra

I dont think this is an Algebra problem but if anyone can help me that would be awesome

On a twelve-digit watch, how many tiems do the individual digits, in order form an arithmetic sequence?

do you mean,say h:m:s,that how many times do h,m,s,  form an arithmetic sequence

Or do you mean, how many subsequences of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} can be found which form an arithmetic progression?
If that's so, the answer is clearly
$\displaystyle\binom{12}{2}=66$.

are you sure that the minute and second digits are from the set {1,2,..,12} saul,  i think they are from the set {0,1,2,...,59},if so ,answer is 354

to be honest guys i'm not sure...our wacky math teacher makes these up...i ended up getting 31


show that if p is the product of first prime numbers , the none of the two numbers $p-1$ and $p+1$ is a perfect square number.

You already posted it here : http://www.mathlinks.ro/Forum/viewtopic.php?t=66120 :)

$p$ is an even number, but not a multiple of $4$, so $n=4k+2$ for some $k\in\mathbb{N}$. Then $p+1=4k+3$, while perfect squares are always of the form $4k$ or $4k+1$.

$p$ is divisible by $3$, but not by $9$, so there is an $m\in\mathbb{N}$ such that $p=9m+3$ or $p=9m+6$. Then $p-1=9m+2$ or $p-1=9m+5$, while perfect squares are always of the form $9m$ or $9m+1$.


f(x)= 3x 2 + 7x + 5..  find all solutions of the congruence f(x)==0(modm)

m = 23
m=25
m=5

i have no idea how to start these problems.

$ax^2+bx+c=0$ (Mod n) with $n$ odd, and relative prime to a, is equivalent to: $(2ax+b)^2=b^2-4ac$ (mod n). Now you just find t such that $t^2=b^2-4ac$(mod n), and put $2ax=t-b$ or $2ax=-t-b$ (mod n) and solve this linear recurrences.


If f(x,y) --> (x-y, x+y) for every pair (x,y) in the plane, for what points (x,y) is it true that (x,y) --> (x,y) in my SParknotes SAT Math IIC prep guide (it has many erorrs so I'm wondering if its answer is just another error...).

The answer, according to the prep guide, is the set of points (x,y) such that x=y. The proof seems perfectly reasonable but when I tried out examples, it doesn't work. Frex...

(2,2) would be transformed into (0,4). So then it's wrong. But the proof seemed perfectly reasonable...

x = x - y
y = x + y

x + y = 2x
x = y

Anything wrong with it?

Thanks!

I would say that (0,0) is the only point that works.  We are looking for points where the transformation, (x+y,x-y),   is the same as the the original point, (x,y).  This can only occur if the the x-coordinates are equal and the y-coordinates are equal.  Or in math terms:

x+y=x   and x-y=y   (which is the same as what the book said)

Look at the first equation.  If you subtract x from both sides you get y=0.  Doing the same for the other gets x=0.

The book just didn't follow through with the problem.  It only found one fact about the ordered pairs(pair in this case), namely that x=y.

The only point that I can find that works with the description given is (0, 0).

Yes, I came to the same conclusion. Thanks! =)

Now one more bad review for Sparknotes's Math IIC guide! :p

There is only one point, (0,0).
The first two equations in the book are a good start, that
x = x - y
y = x + y

Then adding the 2 equations gives
 x + y = 2x
 x = y
(note that this just shows a property the point must have)

But also subtracting the 2 equations gives
x - y = -2y
x = -y

So the points must satisfy both those equations, and (0,0) is the only one that works.


linear algebra
matrix
counting
distinguishability
linear algebra unsolved

Covariance matrix,P, is P(1,1)=23, P(1,2)=P(2,1)=0.25 and P(2,2)=0.0027

The eigenvalues of P are  -1.7389e-005 and 23.0027

P is Hermitian symmetric and positive semidefinite - how can it have a negative eigenvalue?

You haven't given the entries to enough precision, of course. Those entries can't be exact (since $23\cdot 27<25^2$), and if we assume they're only accurate to two digits, the smaller eigenvalue is indistinguishable from zero.

Thank you kindly, you are absolutely spot on.  It turns out that the off diagonal terms must not exceed 0.249199... or the correlation coefficient would need to be greater than 1.  The problem looks so innocent but is an intersting demonstration of conditioning


[quote="GameBot"]Two dice are rolled. If both dice show a prime number, what is the probability that the total is odd?[/quote]

There are 9 total possibilites. There are 4 ways to show odd (2+3,2+5,5+2,3+2). Thus, $ \frac{4}{9}$


I'm currently a sophomore and planning on taking AMC12 on February 2009.
I'm completely new to AMC12, so I'm not sure what resources I should use to prepare for this exam.
Question #1: I just want to use one or two resources. Having said that, should I use Art of Problem Solving volume 2?
Question #2: I'm taking AP Calculus this year and got 34 on ACT math as a freshman, so I think I'm a little prepared. How exactly should I STUDY for AMC12 in 3 months(I can study about 2-3hours each day).

Welcome to AoPS. The question you asked is quite common, and if you post it in the search function, which is up in the blue bar, you'll get many threads that have answers.

The most recent one is [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227407[/url]. It's just a couple threads below this one.

In addition, there are many resources that can be found on AoPS. The contest section has LOTS of contests. If you don't understand something, you can look it up in the AoPS wiki (on the side). The wiki also has lots of contests in it. In addition, try to solve problems that users post in different parts of this forum. You can learn many things that way.


calculus
integration
derivative
function
logarithms
real analysis
real analysis unsolved

I'm stuck on something simple, I'd be grateful for any help.

My text ("Forgotten Calculus", Bleau) states the following integration formula:

integral [dx / (x ln x)] = ln |ln x| + C.

In the course of a problem where I've applied the formula, I realised I can't prove it.

Here's my attempt, for what it's worth.

(Re notation:
exponents are denoted with the "^" sign, e.g x squared = x^2;
first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))

f(x) = ln |ln x| + C

I want to differentiate this via the product rule, so I rewrite the function as

f(x) = (ln x^0) * (ln x) + C

Using the product rule,

f'(x) = [ (ln x^0) * (ln x)' ] + [(ln x^0)' * (ln x) ]
= [ (ln 1) * (1/x)] + [ (0/1) * (ln x)]
= [ ln / x ] + [0]
= ln / x.

I don't see how ln / x = ln |ln x|.

Any advice?

Thanks for your response, although I'm pretty new to calculus and so don't understand the implications of that. Which rule(s) would you use to take the derivative?

You'd use the chain rule:

$\ln()$ | $\ln(x)$
(take derivative)

$1/\ln(x)$ | $1/x$ -> multiply = $\frac{1}{x\ln(x)}$

In general, $\frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$.

Thank you for making that clear. For some reason or another, ln functions are the only thing so far to give me serious trouble as I teach myself calculus