Since |z| is real, then z must have imaginary part 8.

It's simple after that.

[quote="Morgoth"]$z+|z| = 2+8i,$
where $z$ is a complex number. Find $|z|^{2}$.[/quote]

This was on the IntermTrig finals.

[hide]
$z+|z|=2+8i$

$a+bi+\sqrt{a^{2}+b^{2}}=2+8i$

Since $b$ is real, $b=8$.

$a+8i+\sqrt{a^{2}+64}=2+8i$

$a+\sqrt{a^{2}+64}=2$

Let $m=a^{2}+64$

$a+\sqrt{m}=2$

$\sqrt{m}=2-a$

$a^{2}+64=2^{2}-4a+a^{2}$, after substituting $m$ back in

$64=4-4a$

$60=-4a$

$a=-15$

$|z|^{2}=a^{2}+b^{2}=225+64\implies\boxed{289}$

In my opionion, Complex Numbers belong in Intermediate.
[/hide]


Prove that there is a positive integer $ k$ such that $ k.2^{n}\plus{}1$ is a composite for every positive integer value of $ n$.


A sequence $a_1, s_2, a_3, \ldots$ is defined as follows: $a_1 = 1$ and $a_{2^k+j} =  -a_j$ for $j = 1,2,3, \ldots, 2^k.$ Prove that the sequence has no period.

Looks like $a_k=(-1)^{n_k+1}$, where $n_k$ is the number of 1 in the binary representation of $k$. Hence the result.


Anyone can join and leave the classroom whenever they wish (that is, if they paid for it). You can quickly read the transcript when you get back and then join the class (or at least figure out what problem we're on).

[quote="Textangle"]I live in the middle of the Pacific Ocean :laugh: , and I'm taking Intermediate Algebra course, which, just so happens to start at 2:30 here once DST ends (hopefully it will, I'm not good with time). My school ends at 2:45. Is it possible to still enter the classroom 15 minutes late, after class starts???[/quote]
It's ok. It doesn't matter how late you are...when I was late to a class I just quickly read the transcript and asked any clarifying questions I needed.


[i]Show that every positive integer can be written as a sum of distinct Fibonacci numbers.[/i]

Masoud Zargar

I try to write an idea...
[hide]
consider a number $n$. Clearly, for some $F_{i}$, where $F_{n}$ is the nth Fibonacci number, we will have $F_{i}\le n < F_{i+1}$. Now we want that $n-F_{i}$ is a Fibonacci number. Consider $F_{i+1}-F_{i}=F_{i-1}$ This number can be written as $F_{i-1}=F_{i-2}+F_{i-3}$. If one beween $F_{i-2}$ and $F_{i-3}$ is equal to $n-F_{i}$, we have done. If it isn't, we have 2 possible cases:$F_{i}<n<F_{i}+F_{i-2}$ or $F_{i}+F_{i-2}<n<F_{n+1}$ If $n$ is in the first case, consider that $F_{i-2}=F_{i-3}+F_{i-4}$, if $n$ is in the second, consider that $F_{i-3}=F_{i-4}+F_{i-5}$ and repeat the reasoning. Being our succession of $F_{n}$ monotonic  strict decreasing, it will have a lower bound in a finite number of passages. Therefore $n$ can be written as a finite sum of distinct $F_{n}$. [/hide]

EDITED


analytic geometry
geometry
calculus
derivative
vector
function
trigonometry

What do you think are the most brilliant ideas in math?

There was just a big thread about everyone's favorite formulas.  But how about this--what are your favorite mathematical *ideas*?  For example....Descartes' idea of introducing a coordinate system when studying geometry?  The idea of a Dedekind cut?  The idea of using complex numbers to figure out useful things about real numbers?

What's your favorite idea in math?

Thanks!

I'm curious to hear what people say.

Interesting question--- I personally tend to marvel more at the elegance of the finsished product rather than the insight that produced it, but here are a few ideas that I find quite impressive:

-forcing
-goedel numbering
-delta-epsilon as a basis for doing calculus

Hmm...what's forcing?

I do love the epsilon-delta definition of a limit.  It's just so clear.

gauss's sum inverse thing, the story's so genius

1+1=2.
The foundation of all mathematics.

EDIT:
Probably the idea of a derivative==>spawns calculus, and a quantum LEAP in math.

just cuz its used a lot as a "stupid" problem doesn't mean its the foundation of math : |

My favorite idea is countability (though I wouldn't rank it as the "most brilliant").

Good one, JBL.  That's a great example.

Dog of Justice, I love the idea of countability.  Personally I was shocked to find out that there is a bijection between the rationals and the integers.  It just seemed to me that there were so many more rational numbers than integers.

Some people, when they are introduced to the subject, feel that there should be a bijection between any two infinite sets.  I never felt that way.

The idea of "generalization" should not be neglected!

[quote="JBL"]The idea that it makes sense to attack problems of number theory with the tools of analysis. (This, for instance, is where the infamous Riemann Hypothesis comes from.)[/quote]
The same holds for attacking problems of number theory with algebra! (not the boring Algebra, but the Superior one, with rings etc.)

The coordinate system is a brilliant idea, and really useful.

how about the idea/discovery of $0$, i mean isn't that where mathematics stems from?

To be honest, ZetaX, it doesn't seem brilliant at all to me to try to attack number theory with algebra.  I mean, what else is algebra but number theory stripped of many of its properties?  (Admittedly, I haven't taken a class on algebraic number theory, so I don't know what the subject looks like in practice -- it just strikes me as more of an "obvious," and thus less brilliant, idea.)

I think I would generalize the IversonFan's suggestion: the idea that numbers have independent existence (that is, that I don't need to say "2 of something," but can just say "2" and have it make sense) is quite deep and took people a very long time to figure out.  Indeed, even when they had some of the basics (like 0 or fractions), further steps of acceptance (negative numbers and irrationals, and then of course complex numbers) were often still difficult to make.

Hilbert's program of trying to build formal axiomatic systems.

[quote="JBL"]To be honest, ZetaX, it doesn't seem brilliant at all to me to try to attack number theory with algebra.  I mean, what else is algebra but number theory stripped of many of its properties?  (Admittedly, I haven't taken a class on algebraic number theory, so I don't know what the subject looks like in practice -- it just strikes me as more of an "obvious," and thus less brilliant, idea.)[/quote]

Sure, but why would you say that something like the structure of $\mathbb{Z}[\sqrt{-3}]$ should, offhand, have any deep implications for number theory?

[quote="easyas3.14159..."][quote="IntrepidMath"]1+1=2.
The foundation of all mathematics.[/quote]

Not in a base besides 10.[/quote]

haha actually, except base 1 or  2, it's right for all bases, not only base 10.  :D

[quote="easyas3.14159..."][quote="IntrepidMath"]1+1=2.
The foundation of all mathematics.[/quote]

Not in a base besides 10.[/quote]

Doesn't 1+1=2 work for all bases but base 2? (of course there is no base 1)

I personally find Green's Theorem quite nice. :)

How about the analogy between vectors and functions?

For example, we know the spectral theorem for finite-dimensional vector spaces...Can we do a similar thing for some kind of infinite-dimensional vector space of functions?  This question leads to the study of Sturm-Liouville problems, eigenvalues of the Laplacian...lots of math that is useful for quantum mechanics and differential equations.

[quote="Halluci"]inductive reasoning  :)[/quote]

Yes!!!! That's perfect!!!!!!!!!!!.

I don't think 1+1=2 is the base of mathematics.
It's just a definition!!!!!!!!!
How would you define 2?Exactly as 1+1!
3=1+1+1,etc.

Something which is more closer to the base would be,for example,the equality relation.

My vote:  http://mathworld.wolfram.com/LanglandsProgram.html

And more generally, the idea that a problem in one particular field of mathematics can be tackled more easily by translating it to another field - for example, using trigonometry to solve a sequence problem, or relating an algebraic system to intersection of geometrical objects.  I find this concept very powerful.

[quote="leg"]For me the most genius thing was the discovery of the natural
log base "e" ~ 2.71828 due to its enhancement of finance. Of course to others it would be П.[/quote]

I would hardly call the enhancement of finance the most important impact of the number $e$.  It only has an impact at all because the economy often concerns itself with quantities $Q$ describable by $dQ = k Q$, and that's where the real power of $e$ comes from, namely, $\int_{1}^{x} \frac{dt}{t} = \log_{e} x$.

[quote="hockeyadi"](of course there is no base 1)[/quote]

Untrue.  It is entirely possible to construct a base $1$ consisting only of the numeral $1$ repeated over and over again, which would have value $1^k$ in each place.  :)  (No zero is required, you see, because of the non-need for placeholding.)  For example:

$1 = 1_{1}$
$2 = 11_{1}$
$3 = 111_{1}$

what about the pythagorean theorem or eulers invention of the varaible i

@all those that give special theorems: a theorem often is not a 'idea' but more a result

PS: Euler I 'invented' variables¿ I doubt...

Definitely Godel's Incompleteness Theorem.

[quote="shadysaysurspammed"]How do you write the inverted "?" :D[/quote]
Hold "Alt", press"1", "6", "8" on the number pad, release "alt"

nice question :

Well the Greeks were among the first to think along the lines of theorem->proof
At first I didn't realise that that is a major step

The theory of derivatives and integrals is not that old but extremely useful, and applied in many really practical problems and not that hard to master.

Galois theory : connecting field properties to group theory is quite brilliant

The ideas of abstract algebra are powerful. To study the properties and symmetries of systems becomes very powerful in modern mathematics.

[quote="Dave Vather"]The idea... ¡Of making math olympiads! ¿Isn't one of the greatest ideas?[/quote]

Yes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[quote="shadysaysurspammed"]How do you write the inverted "?" :D[/quote]

Don't you have the "¿" button??????  Maybe because in english you don't put it to start a question. In spanish we do...

really important: to combine all the different aspects of some special property (like e.g. holomorphic=analytic=...) or even all the different types of math (Number Theory, Algebra, Geometry, ...).

PS: Well, I don't have an inverted ¿ on my board, but use Alt-1-6-8 ;)


superior algebra
superior algebra solved

I have a trouble when I try to show that $\text{Rad }I$ is homogeneous whenever $I$ is.

Anybody can help me?

Here $I$ is an ideal in a $\mathbb{Z}$-graded ring $R$

Let $f=g_{-n}+\ldots+g_{n}$ be an element of $\sqrt I$ with $g_{i}$ homogeneous of degree $i$ for all $i$. Such a writing always exists for some $n$ with part of the $g_{i}$s probably $0$, but we chose it such that $g_{-n}$ or $g_{n}$ is non-zero.
There exists $m\in\mathbb N$ such that $f^{m}\in I$. It is clear that the homogeneous part of degree $\pm mn$ in $f^{m}$ is $g_{\pm n}^{m}$. Since $I$ is homogeneous we get $g_{\pm n}^{m}\in I$, so $g_{\pm n}\in \sqrt I$.
Consider $f'=f-g_{-n}-g{n}$ which is then also an element of $\sqrt I$ and continue the procedure described above.
In the end we prove that all the homogeneous elements of $f$ belong to $\sqrt I$ which is what we wanted to prove.

This proof works for any ring graded by a totally ordered group.
A proof can also be given for a ring of prime characteristic $p$ graded by a group with no $p$-torsion.


For some reason I just love the font that LaTeX uses (I think it's Computer Modern Font or something). I think it's much better than Times New Roman, especially for certain types of documents. Does anybody know if it's possible to get this font installed on Microsoft Word?

Thanks.

well, in my personnal opinion, the best way to have the $\LaTeX$ default font is....to use $\LaTeX$  :D  :P

but if you really need to use word, I think there's a file for every font.  You would need to copy this file in the right directory in word.  This is just an idea, I never tried it.  You can also try to look for help on word.

And for the font itself, I think you have the right name.

ledurt

I just use Latex and copy it into word as a picture but you could use bold and Euclid together with italics for the words.


$ 0\leq x,y,z\leq \frac{\pi}{2}$ and $ x\plus{}y\plus{}z\equal{}\pi$  prove:

$ 2cotx\plus{}3coty\plus{}4cotz\geq\sqrt{23}$

[quote="zaya_yc"]$ 0\leq x,y,z\leq \frac {\pi}{2}$ and $ x \plus{} y \plus{} z \equal{} \pi$  prove:

$ 2 \cot x \plus{} 3\cot y \plus{} 4\cot z\geq\sqrt {23}$[/quote]
The problem is equivalent to
$ a,b,c \ge 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that
\[ 2a \plus{} 3b \plus{} 4c \ge \sqrt {23}
\]

\[ \Leftrightarrow (2a \plus{} 3b \plus{} 4c)^2 \ge 23(ab \plus{} bc \plus{} ca)
\]

\[ \Leftrightarrow 4a^2 \minus{} a(11b \plus{} 7c) \plus{} 9b^2 \plus{} 16c^2 \minus{} bc \ge 0
\]

\[ \Leftrightarrow 64a^2 \minus{} 16a(11b \plus{} 7c) \plus{} 144b^2 \plus{} 256c^2 \plus{} 16bc \ge 0
\]

\[ \Leftrightarrow (8a \minus{} 11b \minus{} 7c)^2 \plus{} 144b^2 \plus{} 256c^2 \plus{} 16bc \minus{} (11b \plus{} 7c)^2 \ge 0
\]

\[ \Leftrightarrow (8a \minus{} 11b \minus{} 7c)^2 \plus{} 23(b \minus{} 3c)^2 \ge 0
\]
which is true. :)
We have done. :)


I dont get this...

suppose you tie a stone to one end of a string and from the other end you rotate it above your head $horizontally$, while its revolving fast, the strone doesnt fall down , moment yu stop revolving it , it falls down. WHY? gravity was acting all the time!? how come it dropped only when we stoped revolving it???

The funny thing is that it's actually your muscles that hold the stone in the air even though you might not realize it. The only difference is that when you swing the rope you transfer your muscle power to the stone in a different way and you can hold the stone higher.

just to add an interesting conceptual question, do you do any work in keeping the stone in the air by spinning it?

No, because there is no force component acting parallel to the direction of motion.
Work could perhaps be done if there were a force component parallel to the motion in which case the motion would non-uniform circular.
e.g. when you are speeding up the string

[quote="degreestudent"]first there is a slight mistake in what you said - you will never be able to swing it horizontally, the stone will always be below the point where you hold the rope. when you do swing it, the rope trail will make the shape of a cone, and the rope is diagonal when you look at it at any one point (and will become closer to horizontal the faster you spin it). the tension in the rope which is being felt by the stone is therefore diagonal as well, and so it has both horizontal (keeping it in the path of a horizontal circle) and, essentially, vertical (keeping it from falling by cancelling the effect of gravity out) components.

when you are not swinging the rope, there is no tension and nothing to keep the stone from falling, until the rope becomes vertical and the stone will just hang.

this is similar to why a clothes line with washing on it can never be horizontal, no matter how taught the line is, because if it were horizontal, there would be no vertical component of tension to act on the clothes and they would fall under gravity, which would then cause the line to bend.[/quote]

so basically the vertical component of the tension in the string = the gravitational pull, correct?

that is correct
otherwise, well.... that stone would be falling, wouldn't it?

heheh , thats what my question was.... before degreestudent told the answer. i thought like " shouldnt the damn stone fall down??"


In a certain city the rate of taxation is the following: $ x\%$ tax is collected for an income of $ x$ thousand dollars. What income, in dollars, will yield the greatest take home pay? (\textit{Take-home pay }  is the income minus the tax on that income.)

You get $ \frac{100\minus{}x}{100}\cdot{1000x}\equal{}(100\minus{}x)(10x)\equal{}1000x\minus{}10x^2\equal{}\minus{}10(x^2\minus{}100x)\equal{}\minus{}10((x\minus{}50)^2\minus{}2500)$ which maximizes when x=50, so our answer is 50000


Sa inteleg ca $a,b,m,n\in \mathbb R$ ? :maybe:

Daca $a,b\in (0,1)$ sau $a,b\in (1,\infty)$ si $m,n\geq 0$ avem:
Din Cauchy$\rightarrow$ $\sum\frac{log_{a}b^{m}c^{n}}{b+c}\geq \frac{(\sum\sqrt{log_{a}b^{m}c^{n}})^{2}}{2(a+b+c)},(*)$
Fie $x,y,z>0$ a.i. $x=log_{a}b$ $y=log_{b}c$ si $z=log_{c}a$ $\rightarrow$ $xyz=1$
Folosind aceste notatii si avand in vedere relatia $(*)$ este suficient sa aratam ca:
\[\sqrt{mx+\frac{n}{z}}+\sqrt{my+\frac{n}{x}}+\sqrt{mz+\frac{n}{y}}\geq 3\sqrt{m+n}\]
Dar membrul stang al acestei inegalitati $\geq 3\sqrt[6]{(mx+\frac{n}{z})(my+\frac{n}{x})(mz+\frac{n}{y})}\geq 3\sqrt{m+n}$
Suficient sa aratam ca:
\[(mx+\frac{n}{z})(my+\frac{n}{x})(mz+\frac{n}{y})\geq (m+n)^{3}\]
Efectuan calculele si tinand seama de faptul ca $xyz=1$ aceasta ultima inegalitate este echivalenta cu:
\[m^{3}+(x+y+z)m^{2}n+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})mn^{2}+n^{3}\geq (m+n)^{3}\] $\leftrightarrow$
\[m^{2}n(x+y+z-3)+mn^{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3)\geq 0\]
dar $x+y+z\geq 3\sqrt[3]{xyz}=3$ si $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \sqrt[3]{\frac{1}{xyz}}=3$


Sea $c\in \mathbb{R}^{+}$ . Demuestre que:

$\frac{c^{4}+1}{c^{2}+c+1}+\frac{c^{3}}{c^{3}+c+1}+\frac{c^{2}}{c^{3}+c^{2}+1}\ge \frac{4c}{3}$

La cree pensando en una solucion y la pongo aqui para ver si tendra otra solución trivial o si la mia es la mas corta.

Suerte  :D

Holas... acaso nadie puede sacarlo... yo creo que si... esta muy dificil? o es demasiado facil??  :maybe:

Hola Zaratustra,
para ser sincero, no parece muy directa la solucion,

podria ayudar si nos aclaras cual fue el caso "particular" q resolviste ....
de todos modos estoy intentando algunas ideas ..  :lol:

Carlos Bravo :)
Lima - PERU

Ya bueno, pero igual si les doy el caso no se si sea tan particular, loa verdad que es mas bien una variació; se hace mucho mas facil el problema, sobre todo por que es uno de los que estan resueltos en el tema de INECUACIONES. Si lo quieren ver aun; aqui esta:

[hide]$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$[/hide]

Saludos  :)

He estado ocupado estudiando para postular a la UNI pero este problema llamo mi atencion asi q me puse a intentarlo.

Primero dividamos la expresion entre c para obtener:

$\frac{c^{3}}{c^{2}+c+1}+\frac{c^{2}}{c^{3}+c+1}+\frac{c}{c^{3}+c^{2}+1}+\frac{1}{c^{3}+c^{2}+c}\geq \frac{4}{3}$

Con esto el problema esta casi resuelto :D

Luego reemplazamos: $c^{3}=x$, $c^{2}=y$, $c=x$, $1=t$

Ahora tenemos que demostrar: $\boxed{ \frac{x}{y+z+t}+\frac{y}{x+z+t}+\frac{z}{x+y+z}+\frac{t}{x+y+z}\geq \frac{4}{3}}$

Bueno podemos usar $MA-MH$ o $Cauchy$, yo escogi la segunda:

 $\frac{x}{y+z+t}+\frac{y}{x+z+t}+\frac{z}{x+y+z}+\frac{t}{x+y+z}=$
$(x+y+z+t)(\frac{1}{y+z+t}+\frac{1}{x+z+t}+\frac{1}{x+y+z}+\frac{1}{x+y+z})-4=$
$\frac{1}{3}[(y+z+t)+(x+z+t)+(x+y+t)+(x+y+z)])(\frac{1}{y+z+t}+\frac{1}{x+z+t}+\frac{1}{x+y+z}+\frac{1}{x+y+z})-4 \geq \frac{1}{3}(1+1+1+1)^{2}-4 =\boxed{\frac{4}{3}}$


algebra
polynomial
Vieta
number theory unsolved
number theory

Let $ a,b,\frac{a^2\plus{}b^2}{ab\minus{}1}$ are positive integers.
Prove that \[ \frac{a^2\plus{}b^2}{ab\minus{}1}\equal{}5\]

http://www.mathlinks.ro/Forum/viewtopic.php?t=22810
Vieta Jumping works well in this type of problems


$ a,b,c>0$ and
$ a\plus{}b\plus{}c\geq abc$
$ a^{2}\plus{}b^{2}\plus{}c^{2}\geq kabc$ . Find $ max(k)\equal{}?$

$ k\equal{}\sqrt{3}$

$ (a^2 \plus{} b^2 \plus{} c^2 )^2 \geq (ab\plus{}bc\plus{}ca)^2 \geq 3abc(a\plus{}b\plus{}c) \geq 3(abc)^2$, therefore the inequality holds for $ k\equal{}\sqrt{3}$.

Moreover, substituting $ a\equal{}b\equal{}c\equal{}\sqrt{3}$, we have that $ k \leq \sqrt{3}$.
This completes the proof.


Two cirlces (w1),(w2) cuts at A,B. MN is external common tangent (M on (w1), N on (w2)). Let P on (w1), Q on (w2) such that PQ||MN. Lines PM,QN cut at K. Prove that the line BK is bisecs PBQ.


Whatt's your favorite band?

I go for the Beatles!

try this: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33802

this should get moved to Games and Fun Factory, according to what Valentin said on the link above.

yep. now it's in fun and games

favorite band is rage against the machine (notice avatar)
also like zed leppelin and pink floyd for softer stuff.

there's plenty of other topics here about music. please go there rather than starting a new topic. thank you.

[b]LOCKED[/b]


probability
number theory
advanced fields
advanced fields theorems

Does anybody here know where I can find articles about [b]probabilistic methods in number theory? [/b]
Any author other than Mihail Kolountzakis would be great, as I've already got his papers.
Kind regards

Andrews' book (Dover) is cheap and goes through a lot of the principle concepts in number theory (divisibility, additive, multiplicative, partitions) using combinatorial techniques. Even only learning a couple new techniques from this book is great since it is about \$10.


is n fixed? (it probably is but i ask anyway  :P because the text isn't  clear on this matter):D

well anyway here is the solution for n fixed (if n isn't fixed the solution is even easier)
take $x_2=x_3=... x_n=0$ so $f(0)+(n-1)f(x_1)\geq nf(x_1)$ so $f(0)\geq f(x)\forall x\in R$.
then lets take $x_1=-2x_2$ and  $x_3=x_4=... x_n=0$ .We obtain $f(-2x_2) +f(x_2) +(n-2)f(-x_2)\geq nf(0)$ but
$f(0)\geq f(x)$  $\forall x\in R$ so $f(0)\geq f(-2x_2) ,f(0)\geq f(x_2),f(0)\geq f(-x_2) $   so we actually have equality   and   $f(x)=f(0)$  $\forall x \in R$ and $f(0)$ can take any real value.
P.S. this solution is for n>1.For n=1 the text becomes $f(x_1)\geq f(x_1)$ so any function will do ,not to mention n=0 :lol:


LaTeX geometry rectangle articles Asymptote vector AoPSwiki

[code]% Fig POLYLINE object %\linethickness = 1pt\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 5.425 24.202 4.238 25.063 / }% % % Fig POLYLINE object %\linethickness = 1pt\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 4.238 25.063 2.701 24.644 / }% % % Fig POLYLINE object %\linethickness = 1pt\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 4.123 24.862 2.902 24.530 / }% % % Fig POLYLINE object %\linethickness = 1pt\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 4.238 21.874 2.747 22.293 / }% % % Fig POLYLINE object %\linethickness = 1pt\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 2.747 22.293 2.165 23.480 / }% % % Fig POLYLINE object %\linethickness = 1pt\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 2.385 23.406 2.822 22.513 / }% % % Fig POLYLINE object %\linethickness = 1pt\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 2.165 23.480 2.701 24.644 / }%\linethickness = 0pt\putrectangle corners at 2.119 25.110 and 8.755 21.827\endpicture}[/code]
thsi is a latex code for a picture can this be opened/executed here

Not sure why you are asking the question since either it works or it doesn't and you can try that for yourself.
The code you've given has every line commented out so produces nothing and is difficult to read as it's laid out poorly, so no idea if it could be adapted.

i mean to say whether i need to add any dollar before the lines or something like that.Also does commented line create any kind of problem in execution ?

Yes, you would need to use dollar signs before and after. You've commented out all the lines so nothing will happen.
Here's an example of a picture environment on the forum
[code]\setlength{\unitlength}{1cm}
\begin{picture}(3,1.6)
\put(1,1){\circle*{0.2}}
\put(1,1){\circle{1.2}}
\put(1,1){\vector(0,1){0.6}}
\put(2.5,1){\circle*{0.5}}
\end{picture}[/code] which gives
$ \setlength{\unitlength}{1cm}\begin{picture}(3,1.6)\put(1,1){\circle*{0.2}}\put(1,1){\circle{1.2}}\put(1,1){\vector(0,1){0.6}}\put(2.5,1){\circle*{0.5}}\end{picture}$
It does appear though that the forum doesn't like any comments , even small ones.

can u please point out what lines to be removed it just doesn't seem to work

I have worked on this and discovered that you haven't given all the information.
1. it is an ancient piece of TeX ([i]not[/i] LaTeX) which has been obsolete for at least 15 years. You do NOT want to start learning TeX;
2. it uses the pictex package which clearly would not be supported on the forum;
3. you've missed code out at the beginning, which is obvious from the fact that the final } after endpicture is not balanced by a { at the beginning.
The following does work (I have changed the last linethickness to 0.5 pt otherwise there is no rectangle)
[code]\documentclass[12pt]{article}
\usepackage{pictex}
\usepackage{color}
\begin{document}
{\beginpicture
\setcoordinatesystem units <1.04987cm,1.04987cm>
\unitlength=1.04987cm
% Fig POLYLINE object %
\linethickness = 1pt
\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 5.425 24.202 4.238 25.063 / }% % % Fig POLYLINE object %
\linethickness = 1pt
\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 4.238 25.063 2.701 24.644 / }% % % Fig POLYLINE object %
\linethickness = 1pt
\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 4.123 24.862 2.902 24.530 / }% % % Fig POLYLINE object %
\linethickness = 1pt
\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 4.238 21.874 2.747 22.293 / }% % % Fig POLYLINE object %
\linethickness = 1pt
\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 2.747 22.293 2.165 23.480 / }% % % Fig POLYLINE object %
\linethickness = 1pt
\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 2.385 23.406 2.822 22.513 / }% % % Fig POLYLINE object %
\linethickness = 1pt
\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}){\color[rgb]{0,0,0}\plot 2.165 23.480 2.701 24.644 / }%
\linethickness = 0.5pt
\putrectangle corners at 2.119 25.110 and 8.755 21.827
\endpicture}
\end{document}[/code]which gives this
[img]http://sixthform.info/gblog/pictures/c7cc3ab422f0ae792b5f79cb06428b16.gif[/img]

Now forget it - do not use such ancient code.

To use modern latex code to draw benzene rings someone gave an example on comp.text.tex of a box he defined which he then used for a stilbene molecule $ C_{6}H_{5}\minus{}HC \equal{} CH\minus{}C_{6}H_{5}$

[code]\setlength{\unitlength}{1mm}
\newsavebox{\benzolring}
\savebox{\benzolring}(10, 10.59){
   \put(5, 5.3){\circle{5.5}}
   \put(5, 0){\line(-2, 1){5}}
   \put(5, 0){\line(2, 1){5}}
   \put(0, 2.5){\line(0, 1){5.59}}
   \put(10, 2.5){\line(0, 1){5.59}}
   \put(5, 10.59){\line(-2, -1){5}}
   \put(5, 10.59){\line(2, -1){5}}
   }
\begin{picture}(50, 29)
 \put(5, 24){\makebox(5, 5){\textbf{\textsf{H}}}}
 \put(37.5, 24){\makebox(5, 5){\textbf{\textsf{H}}}}
 \put(15, 14){\makebox(5, 5){\textbf{\textsf{C}}}}
 \put(27.5, 14){\makebox(5, 5){\textbf{\textsf{C}}}}
 \put(10, 24){\line(1, -1){5}}
 \put(37.5, 24){\line(-1, -1){5}}
 \put(20, 15.5){\line(1, 0){7.5}}
 \put(20, 17.5){\line(1, 0){7.5}}
 \put(10, 9){\line(1, 1){5}}
 \put(37.5, 9){\line(-1, 1){5}}
 \put(0, 0.91){\usebox{\benzolring}}
 \put(37.5, 0.91){\usebox{\benzolring}}
\end{picture}[/code]The code before the picture is the box which is then used twice to make the molecule. You should be able to use the same procedure for the molecules that you want and, best of all, it will show on the forum like this:
$ \setlength{\unitlength}{1mm}\newsavebox{\benzolring}\savebox{\benzolring}(10, 10.59){\put(5, 5.3){\circle{5.5}}\put(5, 0){\line(\minus{}2, 1){5}}\put(5, 0){\line(2, 1){5}}\put(0, 2.5){\line(0, 1){5.59}}\put(10, 2.5){\line(0, 1){5.59}}\put(5, 10.59){\line(\minus{}2,\minus{}1){5}}\put(5, 10.59){\line(2,\minus{}1){5}}}\begin{picture}(50, 29)\put(5, 24){\makebox(5, 5){\textbf{\textsf{H}}}}\put(37.5, 24){\makebox(5, 5){\textbf{\textsf{H}}}}\put(15, 14){\makebox(5, 5){\textbf{\textsf{C}}}}\put(27.5, 14){\makebox(5, 5){\textbf{\textsf{C}}}}\put(10, 24){\line(1,\minus{}1){5}}\put(37.5, 24){\line(\minus{}1,\minus{}1){5}}\put(20, 15.5){\line(1, 0){7.5}}\put(20, 17.5){\line(1, 0){7.5}}\put(10, 9){\line(1, 1){5}}\put(37.5, 9){\line(\minus{}1, 1){5}}\put(0, 0.91){\usebox{\benzolring}}\put(37.5, 0.91){\usebox{\benzolring}}\end{picture}$

simply brilliant hope there were more than excellent in those ratings .  sheer patience and brilliance.
where do i get to know all these commands.any link .i really want to be doing this in the forum :)  :omighty:

See [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url] Section 5.2 The picture environment.
Of course [[Asymptote (Vector_Graphics_Language)]] is also fully supported by the forum, but I haven't seen any chemical examples using it.

is it  possible to make a minimal finite collection of codes such that a permutation of them would yield almost all the chemical structures nad then put it as standards in the chemistry forum.that is save them
say the

[code]\benzene[/code]
yields the structure u have done

You can't do much in the forum. The best you can do is make a post listing the various codes, which are easier to use in a savebox as in one of the above posts. Or you could put them in the AoPSWiki and refer people to that to copy and paste. The problem is that whatever you do, each post would need to have all the code in it.

In a document it's much easier as you can make a new command called \benzene using
[code]\newcommand{\benzene}{
\setlength{\unitlength}{1mm}
\begin{picture}(50, 29)
 \put(5, 0){\line(-2, 1){5}}
 \put(5, 1){\line(-2, 1){4}}
 \put(5, 0){\line(2, 1){5}}
 \put(0, 2.5){\line(0, 1){5.59}}
 \put(10, 2.5){\line(0, 1){5.59}}
 \put(9.2, 2.9){\line(0, 1){4.5}}
 \put(5, 10.59){\line(-2, -1){5}}
 \put(5.1, 9.5){\line(-2, -1){4}}
 \put(5, 10.59){\line(2, -1){5}}
\end{picture}}[/code]


geometry
3D geometry
rotation
probability

In how many different positions could a 3X3X3 cube be? What about a 4X4X4 and a 5X5X5?

[hide="generalization"]an n*n*n cube can have 720^n arrangements.

Because there are n^2 each of colors. The first color can have 6 for the first sticker. 6 for the next. 6 so on. Then for the next color, since there is an average of one square filled up, 5*5*...*5 and so on until yo get (6!)^n[/hide]

So a 1X1X1 cube has 720 arrangements? interesting....
Now honestly, I don't know what you're doing, it's not that simple. It's asking how many positions we can get it to be by turning the sides(I think. if not, I don't know why he called it a Rubik's cube, since he could've called it just s 3x3x3 cube thingy).
For example, we can't get the cube so that one edge piece is just flipped, while every other pieces are in their correct positions.

A 3x3x3 cube has 45 quintillion combos, definitely not $720^{3}$.
I saw this in intro to c&p

Let's see: There are 8 corners and 12 edges. The corners can be arranged in 8! ways, and the edges can be arranged in 12! ways. Multiply by 2^12 and 3^8 for orientations of edges and corners. Then divide by 2 because there is an edge orientation parity, and divide by 3 for corner orientation parity. Then divide by 2 for permutation parities.

Ubemaya's solution is convincing. But is it right?

[quote="Ubemaya"]Then divide by 2 because there is an edge orientation parity, and divide by 3 for corner orientation parity. Then divide by 2 for permutation parities.[/quote]

By "edge orientation" and "corner orientation", do you mean that a corner or edge is placed in a different manner?  And what do you mean by "permutation parities"?

Orientation is when you rotate something; you can rotate an edge in 2 ways and a corner in 3 ways. When you rotate one edge, you can't solve the cube (so it's impossible to reach that position from the starting position); if two different edges are rotated, you can rotate both of them back. Similarly, one out of 3 total corner orientations work. By permutation, I mean how you can move cubies around, and 1/2 of them work.

Interesting fact using this: If you take the pieces out and randomly put them back, there's a 1/2 * 1/3 * 1/2 = 1/12 chance that it's still solvable.

Are there any other interesting combinatorics/ probability questions that have something to do with rubik's cubes?


find the common tangents to the parabola (y)^2=4ax and (x)^2=4by....
(both calculus and intuitive soluytions are welcome) :)

1. $ x\equal{}{y^2\over 4a}\implies x'\equal{}{y\over 2a}$, hence the equation of the tangent in the point $ \left({y_0^2\over 4a},y_0\right)$ would be

$ x\minus{}{y_0^2\over 4a}\equal{}{y_0\over 2a}(y\minus{}y_0)\iff\dots\iff y\equal{}{2a\over y_0}x\plus{}{y_0\over 2}$

2. $ y\equal{}{x^2\over 4b}\implies y'\equal{}{x\over 2b}$, hence the equation of the tangent in the point $ \left(x_0,{x_0^2\over 4b}\right)$ would be

$ y\minus{}{x_0^2\over 4b}\equal{}{x_0\over 2b}(x\minus{}x_0)\iff\dots\iff y\equal{}{x_0\over 2b}x\minus{}{x_0^2\over 4b}$

Comparing the two equations, we get

$ {2a\over y_0}\equal{}{x_0\over 2b}\land{y_0\over 2}\equal{}\minus{}{x_0^2\over 4b}$

$ y_0\equal{}{4ab\over x_0}\implies {2ab\over x_0}\equal{}\minus{}{x_0^2\over 4b}\implies x_0\equal{}\minus{}2\sqrt[3]{ab^2}\implies y_0\equal{}\minus{}2\sqrt[3]{a^2b}$

That yields the equation $ y\equal{}\minus{}\sqrt[3]{a\over b}x\minus{}\sqrt[3]{a^2b}$


In art class, we've started a unit where we make our own books from scratch. I couldn't decide of a subject for my book, so I finally picked out xkcd. I plan to use about 20-24 of xkcd's funniest comics ever. Anyone have any good ones?

Please note that the comic [i]must[/i] be school appropriate, so that means any comic with language or sexual contents cannot be selected (although that cuts off over half of the funny ones  :| ).

yeah. the ones i really like are the inappropriate ones.. i like the one about the hamster ball and how they are giving it a soul. :)

(that one's pretty recent)

Here are some recent ones that I particularly liked:

[url]http://xkcd.com/416/[/url]
[url]http://xkcd.com/413/[/url]
[url]http://xkcd.com/410/[/url]
[url]http://xkcd.com/406/[/url]
[url]http://xkcd.com/405/[/url]
[url]http://xkcd.com/399/[/url]
[url]http://xkcd.com/396/[/url]
[url]http://xkcd.com/394/[/url] (very good - if you know about the intel FPU bug)
[url]http://xkcd.com/393/[/url]
[url]http://xkcd.com/384/[/url]
[url]http://xkcd.com/382/[/url]
[url]http://xkcd.com/380/[/url]
[url]http://xkcd.com/379/[/url]
[url]http://xkcd.com/378/[/url]
[url]http://xkcd.com/376/[/url]
[url]http://xkcd.com/375/[/url]

And this is also one of my favorites of all time
[url]http://xkcd.com/138/[/url]

This one's also great:
[url]http://xkcd.com/225/[/url]

Oh and, for any who didn't know
[url]http://xkcd.com/404/[/url]

http://xkcd.com/132/
http://xkcd.com/135/
http://xkcd.com/156/
http://xkcd.com/178/
http://xkcd.com/242/
http://xkcd.com/418/

May have been mentioned before... I didn't bother to check

[url]http://www.xkcd.com/86/[/url]
[url]http://www.xkcd.com/156/[/url]
[url]http://www.xkcd.com/208/[/url]
[url]http://www.xkcd.com/228/[/url]
[url]http://www.xkcd.com/249/[/url]
[url]http://www.xkcd.com/284/[/url]
[url]http://www.xkcd.com/303/[/url]
[url]http://www.xkcd.com/327/[/url]
[url]http://www.xkcd.com/386/[/url]

Oh!  Here's another one:

http://xkcd.com/81/

Ignore the title?
http://xkcd.com/290/
:|
It's the only one I've read lol, from your blog thingy...

[quote="lingomaniac88"]You could use this as an intro:
http://xkcd.com/207/[/quote]

Yes I love the part about the floor tiles. That is totally me.

Also this one is random but funny

http://www.xkcd.com/337/
http://www.xkcd.com/235/

[quote="tjhance"][url]http://www.xkcd.com/327/[/url][/quote]
I showed my computer science teacher this one.  He thought it was funny.

[quote="lingomaniac88"]You could use this as an intro:
http://xkcd.com/207/[/quote]

Yeah, I tried re-cropping the boxes but I didn't think it would work. I'll try again later.

Thanks for all the replies so far!

Also, my cover needs work; it seems a bit bland simply saying "XKCD." Any ideas on what to put on it (not just pictures, but items or fabric can be used, etc.)?

EDIT: My book size is approximately 7.5 by 5 inches (length-wise). So this also unfortunately reduces my comic choices. Although I do like the "Death Speaking" one.  :lol:

To those that read MegaTokyo, comic strip number 157 is hilarious  :rotfl:

EDIT: [url]http://xkcd.com/184/[/url] is awesome.


[quote="aspiringchemlete"]Which combination of solutions of HCl and NaOH would produce the largest delta T?

A) 50 mL of 1 M HCl with 50 mL of 1 M NaOH

B) 50 mL of 2 M HCl with 50 mL of 2 M NaOH

C) 100 mL of 1 M HCl with 50 mL of 2 M NaOH

D) 100 mL of 1 M HCl with 100 mL of 1 M NaOH[/quote]

1M 1L HCl and 1M 1L NaOH combine to give out 57.2KJ/mol.  So calculate

Ans should be (b), (c) (d) cause maximum no of moles of water are formed so greatest temperature change.in each case 100 millimoles of water are formed. :)

But I think that alone is wrong because that is only for the heat. We also have to consider the mass of water in soln for Delta T.

So, the ans is B. Am I right?


Sorry about the double post, but here is the answer.
[hide="ANSWER"]$\frac{2^{-1}+4^{-2}+8^{-3}}{8^{-1}+4^{-2}+2^{-3}}$  is the equation.  Combine terms on the bottom and terms on the top until you are only multiplying.  Then you can cancel terms out easier.  First, write both in terms of powers of 2.  Then change the top to $2^{-36}$.  Finally, simplify the fraction.  Start with

\[ \frac{2^{-1}+2^{-4}+2^{-9}}{2^{-2}+2^{-4}}\\=\frac{(2^{-1}\cdot2^{36})(2^{-4}\cdot2^{36})(2^{-9}\cdot2^{-26})}{5\cdot2^{-4}}\\=\frac{(2^{35}+2^{32}+2^{27})2^{-36}}{5\cdot2^{-4}}\\=\frac{(2^{35}+2^{32}+2^{27})2^{-32}}{5}\\=\frac{2^3+2^0+2^{-5}}{5}\\=\frac{9+2^{-5}}{5}  \]

I think I simplified it all the way.[/hide]


6. An isoscleles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the radius, in inches, of the circle? Express your answer as a mixed number.

I think you either divide the triangle into right triangle or use the median ratio...but I don't know..

[hide="Answer"]$ 3\frac18$[/hide]

[hide]draw the median line of the base (=6) which is also altitute, and it passes through center of circle

connect center to one of the end point on base

look at two right triagles

[5,3,r+x=4] and [[r,x,3]

r^2=(4-r)^2+9  [/hide] :P

[hide="Solution"]The circumcircle is found by finding the intersection of all perpendicular bisectors.  There intersection is the center of the circle.
Or... using extended laws of sines (which I shouldn't be using) $ 2R \equal{} \frac {a}{\sin\alpha} \equal{} \frac {b}{\sin\beta} \equal{} \frac {c}{\sin\gamma} \equal{} \frac {abc}{2S}$, where $ \alpha,\beta,\gamma$ denote the angles and $ S$ is the area of the triangle.  obviously, the easiest way is to use $ 2R \equal{} \frac {abc}{2S}$.

Draw the altitude to get two $ 3,4,5$ right triangle.  $ S \equal{} 12$.

So. $ 2R \equal{} \frac {6*5*5}{2*12}\rightarrow R \equal{} \frac {25}{8} \equal{} 3\frac {1}{8}$[/hide]
Didn't feel like using geometry like policecap, who seems to be getting to all the posts before me by like 1 second -_-.


Great work dear Fang Jianhua  :first: Thank you very much!


The lines $ x \equal{} \frac 14y \plus{} a$ and $ y \equal{} \frac 14x \plus{} b$ intersect at the point $ (1,2)$.  What is $ a \plus{} b$?

$ \textbf{(A) } 0 \qquad \textbf{(B) } \frac 34 \qquad \textbf{(C) } 1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } \frac 94$


trigonometry geometry circumcircle LaTeX

given $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi,$ then prove that $ \sin \alpha \plus{}\sin \beta \plus{}\sin \gamma \equal{}4\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\cos \frac{\gamma }{2}.$

This is a little bit messy, but I think it works:

[hide="Solution"]

Let $ A = \frac{\alpha}{2}$, $ B = \frac{\beta}{2}$, and $ C = \frac{\gamma}{2}$.
Since $ \alpha+\beta+\gamma=\pi,$, we know that $ A + B + C = \frac{\pi}{2}$, so $ C = \frac{\pi}{2} - (A + B)$.
Since $ \sin (\frac{\pi}{2} - x ) = \cos x$ and $ \cos (\frac{\pi}{2} - x ) = \sin x$
$ \sin C = \sin (\frac{\pi}{2} - (A + B) ) = \cos (A + B)$
and
$ \cos C = %Error. "sincos" is a bad command.
(\frac{\pi}{2} - (A + B) ) = \sin (A + B)$

Then we have:
$ \sin\alpha+\sin\beta+\sin\gamma$

$ \sin 2A + \sin 2B + \sin 2C$

$ 2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C$

$ 2\sin A \cos A + 2\sin B \cos B + 2\cos (A + B) \sin (A + B)$

$ 2\sin A \cos A + 2\sin B \cos B + 2( \cos A \cos B - \sin A \sin B )(\sin A \cos B + \cos A \sin B )$

$ 2\sin A \cos A + 2\sin B \cos B + 2\sin A \cos A \cos ^ 2 B + 2\sin B \cos ^ 2 A \cos B - 2\sin ^ 2 A \sin B \cos B - 2\sin A \sin ^ 2 B \cos A$

$ 2\sin A \cos A ( 1 - \sin ^ 2 B + \cos ^ 2 B ) + 2\sin B \cos B ( 1 - \sin ^ 2 A + \cos ^ 2 A)$

$ 2\sin A \cos A ( 2\cos ^ 2 B ) + 2\sin B \cos B ( 2\cos ^ 2 A )$

$ 4\sin A \cos A \cos ^ 2 B + 4\sin B \cos ^ 2 A \cos B$

$ 4\cos A \cos B ( \sin A \cos B + \cos A \sin B )$

$ 4\cos A \cos B \sin (A + B)$

$ 4\cos A \cos B \cos C$

$ 4\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}$

So, we have that $ \sin\alpha+\sin\beta+\sin\gamma = 4\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}$.

[/hide]

[hide="Solution"]Let $ A \equal{} \alpha, B \equal{} \beta, C \equal{} \gamma$. Clearly, these are the angles of a triangle. Without loss of generality, let the radius of its circumcircle be 1. Then $ a \equal{} 2\sin A$, $ b \equal{} 2\sin B$, and $ c \equal{} 2\sin C$.

We proceed as follows. Note that $ 4\cos(A/2)\cos(B/2)\cos(C/2) \equal{} 4s \sqrt{\frac{s(s \minus{} a)(s \minus{} b)(s \minus{} c)}{(abc)^2}} \equal{} \frac{2(a \plus{} b \plus{} c) [ABC]}{abc}$ (follows $ \cos(A/2) \equal{} \sqrt{\frac{s(s \minus{} a)}{bc}}$.) Since [ABC] = $ \frac{abc}{4}$, mass simplification yields that the LHS is equal to $ \sin A \plus{} \sin B \plus{} \sin C$, as desired.
[/hide]

[quote="Zhero"][hide="Solution"]Let $ A \equal{} \alpha, B \equal{} \beta, C \equal{} \gamma$. Clearly, these are the angles of a triangle. Without loss of generality, let the radius of its circumcircle be 1. Then $ a \equal{} 2\sin A$, $ b \equal{} 2\sin B$, and $ c \equal{} 2\sin C$.

We proceed as follows. Note that $ 4\cos(A/2)\cos(B/2)\cos(C/2) \equal{} 4s \sqrt {\frac {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}{(abc)^2}} \equal{} \frac {2(a \plus{} b \plus{} c) [ABC]}{abc}$ (follows $ \cos(A/2) \equal{} \sqrt {\frac {s(s \minus{} a)}{bc}}$.) Since [ABC] = $ \frac {abc}{4}$, mass simplification yields that the LHS is equal to $ \sin A \plus{} \sin B \plus{} \sin C$, as desired.
[/hide][/quote]

It's nowhere said these are the angles of a triangle. Any one of them can be negative.

[quote="Zhero"]Ah snap. My bad. I was working under the assumption that that was implicitly assumed, since that would have resulted in a fairly nice problem. :/

In that case, I can't think of anything better than a brute force (like MathWise did.) Any ideas?[/quote]

Basically, what you need to add to your solution is the proof that $ \alpha,\beta,\gamma$, no matter if either of them is negative or greater than $ 2\pi$, can always be REDUCED to the angles of a triangle without harming the validity of the given expression. Once you do that, everything will stand.

I've considered that, but I haven't made any progress. The /2's in the cosines are really killing me.

[hide="Another brute-force approach"]Start from

$ {1\over 2}(\sin\alpha \plus{} \sin\beta) \equal{} \sin{\alpha \plus{} \beta\over 2}\cos{\alpha \minus{} \beta\over 2} \equal{} \cos{\gamma\over 2}\left(\cos{\alpha\over 2}\cos{\beta\over 2} \plus{} \sin{\alpha\over 2}\sin{\beta\over 2}\right)$

and write similar identities for $ \alpha,\gamma$ and for $ \beta, \gamma$. Sum them all up, plug into the given expression and simplify.

Then it remains to prove:

$ \cos{\alpha\over 2}\sin{\beta\over 2}\sin{\gamma\over 2} \plus{} \sin{\alpha\over 2}\cos{\beta\over 2}\sin{\gamma\over 2} \plus{} \sin{\alpha\over 2}\sin{\beta\over 2}\cos{\gamma\over 2} \equal{} \cos{\alpha\over 2}\cos{\beta\over 2}\cos{\gamma\over 2}$

Put the first term to the other side and simplify:

$ \sin{\alpha\over 2}\sin{\beta \plus{} \gamma\over 2} \equal{} \cos{\alpha\over 2}\cos{\beta \plus{} \gamma\over 2}$

which obviously can be simplified further into

$ \cos{\alpha \plus{} \beta \plus{} \gamma\over 2} \equal{} 0$

which holds by the initial condition.[/hide]

[hide="hint"]
You can solve almost any problem of this kind by substituting $ \gamma$. In this example $ \gamma\equal{}\pi\minus{}(\alpha \plus{} \beta)$.[/hide]
[hide="solution"]
Now we have:

$ \sin\alpha \plus{} \sin\beta \plus{} \sin(\pi\minus{}(\alpha\plus{}\beta))\equal{}4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos(\frac{\pi}{2}\minus{}\frac{\alpha\plus{}\beta}{2})$

$ \sin\alpha \plus{} \sin\beta\plus{}\sin(\alpha\plus{}\beta)\equal{}4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\sin\frac{\alpha\plus{}\beta}{2}$

$ \sin(\alpha\plus{}\beta)\equal{}4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\sin\frac{\alpha\plus{}\beta}{2}\minus{}2\sin\frac{\alpha\plus{}\beta}{2}\cos\frac{\alpha\minus{}\beta}{2}$

$ \sin(\alpha\plus{}\beta)\equal{}2\sin\frac{\alpha\plus{}\beta}{2}(2\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\minus{}\cos\frac{\alpha\minus{}\beta}{2})$

$ \sin(\alpha\plus{}\beta)\equal{}2\sin\frac{\alpha\plus{}\beta}{2}(\cos\frac{\alpha\minus{}\beta}{2}\plus{}\cos\frac{\alpha\plus{}\beta}{2}\minus{}\cos\frac{\alpha\minus{}\beta}{2})$

$ \sin(\alpha\plus{}\beta)\equal{}2\sin\frac{\alpha\plus{}\beta}{2}\cos\frac{\alpha\plus{}\beta}{2}$

$ \sin(\alpha\plus{}\beta)\equal{}\sin(\alpha\plus{}\beta)$

$ QED$[/hide]

sinA+sinB+sinC
=2sin(A+B)/2.cos(A-B)/2 + sinC
= 2cosC/2. cos(A-B)/2 + 2sinC/2 cosC/2
= 2cosC/2 {cos(A-B)/2 + sinC/2}
=2cosC/2{cos(A/2-B/2)+ cos(A/2+B/2)
=2cosC/2.2cosA/2.cosB/2
=4cosA/2.cosB/2.cosC/2 proved
But I have a better problem
Prove that cosA/2-cosB/2+cosC/2 = 4 .cos(p+A)/4.cos(p-B)/4.cos(p+C)/4
P is pi radian or 180 degrees. A,B,C angles of triangle

@communist Really, you should learn basics of Latex, it's annoying to read such posts and when you do, you are not even sure what it says.

[hide="Some examples:"]
[code]$1$[/code] gives $ 1$

[code]$\sin x$[/code]
gives $ \sin x$

[code]$\frac{a}{b}$[/code]
gives $ \frac{a}{b}$

[code]$2 \cdot \cos (\pi - \frac{A}{2})$[/code]
gives $ 2 \cdot \cos (\pi \minus{} \frac{A}{2})$

[code]$\left(\pi - \frac{A}{2}\right)$[/code]
gives $ \left(\pi \minus{} \frac{A}{2}\right)$

Full list can be found [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]here[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands]here[/url].[/hide]


For each positive integer $ c$, the sequence $ u_{n}$ of integers is defined by

$ u_{1} \equal{} 1, u_{2}\equal{}c, u_{n}\equal{} (2n\plus{}1)u_{n\minus{}1} \minus{} (n^{2}\minus{}1)u_{n\minus{}2}, (n \geq 3)$.

For which values of $ c$ does this sequence have the property that $ u_{i}$ divides $ u_{j}$ whenever $ i \leq j$?

[$ 4$ divides $ \minus{}12$.]

[quote="AndrewTom"]For each positive integer $ c$, the sequence $ u_{n}$ of integers is defined by

$ u_{1} \equal{} 1, u_{2} \equal{} c, u_{n} \equal{} (2n \plus{} 1)u_{n \minus{} 1} \minus{} (n^{2} \minus{} 1)u_{n \minus{} 2}, (n \geq 3)$.

For which values of $ c$ does this sequence have the property that $ u_{i}$ divides $ u_{j}$ whenever $ i \leq j$?

[$ 4$ divides $ \minus{} 12$.][/quote]

$ u_3 \equal{} 7c \minus{} 8$ and so $ u_2|u_3$ $ \iff$ $ c|8$ $ \iff$ $ c\in\{1,2,4,8\}$

For $ c\equal{}1$ : $ u_5\equal{}\minus{}240$ and $ u_6\equal{}\minus{}2280$ and so $ u_5\not | u_6$

For $ c\equal{}2$ : It's easy to show with induction that $ u_n \equal{} n!$ which indeed is a solution

For $ c\equal{}4$ : It's easy to show with induction that $ u_n \equal{} \frac {(n \plus{} 2)!}6$  which indeed is a solution

For $ c\equal{}8$ : $ u_3\equal{}48$ and $ u_4\equal{}312$ and so $ u_3\not | u_4$

Hence the answer : $ \boxed{c\in\{2,4\}}$


superior algebra
superior algebra unsolved

Let $G$ be a finite group such that the set of orders of the elements of $G$ is precisely $\{1,2,\ldots,n\}$. Prove that $n \leq 8$.

Bertrand's Postulate: there is always a prime between $n$ and $2n$.

You of course noticed that it is not necessary to have $|G|=n$.

Could someone give a complete solution? Group theory is still "new" to me :D

It's a hard problem- I don't think anyone here has completely solved it yet.

Some examples for possible values of $n$:
$n=1$: Trivial group.
$n=2$: $\mathbb{Z}_2^k$
$n=3$: $S_3$ or $A_4$
$n=4$: $S_4$
$n=5$ $A_6$
$n=6$: $S_5$ or $S_6$
$n=7$: $A_7$

I don't know how to do $n=8$, or if it is even possible.


[hide="Click to reveal AWESOME content"]
Set the bases equal to each other.
$2^{2xy+8}=2^{3x^2-3y-3}\\ 2xy+8=3x^2-3y-3$
Then, all you need to do is substitute $x+1$ in for $y$, solve, and solve.
$2x(x+1)+8=3x^2-3(x+1)-3\\ 2x^2+2x+8=3x^2-3x-6\\ x^2-5x-14=0\\ x=(x-7)(x+2)\\ x=7,-2\\ (7,8),(-2,-1)$
[/hide]

-BLt

[hide]Substituting $y=x+1$ and rewriting it with a base of 2 gives us

$2x^2+2x+8=3x^2-3x-6$.

$x^2-5x+14=0$.

$x=7, -2$.

Solutions: $(7,8), (-2,-1)$. [/hide]


In the equation: $ tan(\alpha\plus{}\beta) \equal{}\frac{tan\alpha\plus{}tan\beta}{1\minus{}tan\alpha\cdot tan\beta}$ Where $ tan(\alpha\plus{}\beta)$ can be rewritten as: $ \frac{sin(\alpha\plus{}\beta)}{cos(\alpha\plus{}\beta)}$, Prove that:

A. $ sin(\alpha\plus{}\beta)\equal{}tan\alpha\plus{}tan\beta$
B. $ cos(\alpha\plus{}\beta)\equal{}1\minus{}tan\alpha\cdot tan\beta$

I think this should fall under easy.  :)

i am sorry but....are you declaring the validity of the two statements above for any $ \alpha$ and $ \beta$ . If so i believe you are wrong.
I think it is true only for all $ x ,y$ such that -
$ (x \minus{} 1)(y \minus{} 1) \equal{} \minus{} 1$
where $ x \equal{} tan^2(\alpha) ,y \equal{} tan^2(\beta)$

For part a, suppose that it were true. Then, by the addition formula for $ \sin$

$ \sin\alpha\cos\beta\plus{}\sin\beta\cos\alpha\equal{}\sin(\alpha\plus{}\beta)$

(We will assume that $ \cos\alpha\neq0$, $ \cos\beta\neq0$).

We next may write

$ \sin\alpha\cos\beta(\frac{\cos\alpha}{\cos\alpha})\plus{}\sin\beta\cos\alpha(\frac{\cos\beta}{\cos\beta})\equal{}\sin(\alpha\plus{}\beta)$.

$ \implies \cos\alpha\cos\beta\tan\alpha\plus{}\cos\alpha\cos\beta\tan\beta\equal{}\tan\alpha\plus{}\tan\beta$.

$ \implies \cos\alpha\cos\beta(\tan\alpha\plus{}\tan\beta)\equal{}\tan\alpha\plus{}\tan\beta$

Hence, in order for this to be true, we must either have

$ \tan\alpha\plus{}\tan\beta\equal{}0$ or $ \cos\alpha\equal{}cos\beta\equal{}\pm1$.

Clearly, this is not true for [i]all[/i] $ \alpha$ and $ \beta$.

$ \frac12\equal{}\frac24$ implies $ 1\equal{}2$ and $ 2\equal{}4$ ;)


Well this post is not a Questions, Suggestions, & Announcements , it is just a general information.

Have you guys seen the words that are writen under your ids in your posts like
[b]P versus NP[/b](as mr.vornicu ordered , no words in latex mode ,we should jus use [b] commands :D  ) , i just wondered what are they and i find this :
[url]http://en.wikipedia.org/wiki/Millennium_Prize_Problems[/url]

Does any one have more informations?

PS.it seems that administators are really interested in open problems . :D

I don't quite understand what do you want from this topic?  :?

I jus want some more informations about those open problems Mr.vornicu and i dont find any better place to post this message. Sorry if this is nonsense :blush:

And i Heard that the The Poincaré conjecture is almost solved , is it true?

lol half of those I can't even understand the statement of the problem let alone the solution.

Go to [url=http://www.claymath.org/millennium/]this site[/url] for more information.

There are books devoted to the Riemann Hypothesis.  I'm pretty sure there is a book just on the Millenium problems.  There is also plenty of information available online (use Google).

I have that book, it is pretty interesting, but I can't understand half of it...


say the number of unit servings of a pie is n.

how many different ways are there to cut the pie, if for each piece size, only multiples of a unit are allowed
(not necessarily the same multiple, ie pieces of different sizes are allowed for a given instance of a way to cut the pie)

?

If I understood you correctly, you want to represent n as a sum of positive integers (corresponding to the sizes of pieces). Such representations are called either compositions (if you care about their order) or partitions (if you don't care about their order) of n.
See http://mathworld.wolfram.com/Composition.html and/or http://mathworld.wolfram.com/PartitionFunctionP.html correspondingly.
Compositions are easier to count: the total number of compositions of n is simply $2^{n-1}.$


combinatorics unsolved
combinatorics
USAMO

[hide]Well, there are 20 players, and 28 playing positions.  Therefore, there are at most 8 games in which a player is playing who has already played.  That means there are 6 games in which neither player has played.  This set of six games is the desired set.  whatev.  [/hide]

we use graphs.[color=#008000]we take a maximal matching[/color].if possible suppose we have p[<=5]edges.it is easy to see that there are 20-2p who are not in this matching. all vertices have degree >=1.the vertices who are not in the matching do not span edges. so, [color=#FF0000]we need at least (20-2p)+p=20-p >=15 edges [/color], which is a contradiction as only 14 games are there. :lol:

Consider an arbitrary graph $G$ with $20$ vertices and $14$ edges. Separate it into $k$ connected components of $v_1,v_2,\dots,v_k$ vertices and $e_1,e_2,\dots,e_k$ edges respectively. Note that $e_i\geq v_i-1$, so $14=\sum e_i\geq \sum (v_i-1)=20-k\implies k\geq 6$. Since the problem statement demands that every connected component contains an edge, we conclude. $\Box$

Consider a graph $G = (20, 14)$. From the degree sum formula, $\sum_{v \in V} \deg(v) = 2 \cdot E = 28$ where $V$ is the set of vertices. Suppose $k$ of the vertices have degree $1$. Then it follows that $\sum_{v\in V} \deg(v) = 28 \ge k + (20-k)\cdot 2 \Rightarrow k\ge 12$. Thus, at most we have $8$ vertices with degree that is greater than $1$. If we ignore $8$ such edges associated with these vertices then we have at least $6$ edges left and the degree of all remaining vertices is less than or equal to $1$. Hence, this is the $6$ games that occurs within $12$ distinct players, as desired.

[quote=JSGandora]We can represent this as a graph of $20$ vertices and $14$ edges.  For the sake of contradiction, assume in any set of $6$ edges, there is at most $11$ vertices.  Thus, the sum $S$ of the number of vertices over all sets of $6$ edges is
\[S \leq \binom{14}{6}\times 11.\]
Now each edge is represented $\binom{13}{5}$ times over all the sets of $6$ edges.  Every edge has two vertices, and therefore $S=\binom{13}{5}\binom{14}{1}\cdot 2$.  Thus,
\begin{align*}
\binom{13}{5}\binom{14}{1}\times 2&\leq \binom{14}{6}\times 11 \\
\cancel{\frac{13\times 12\times 11\times 10\times 9}{5\times 4\times3\times2\times1}\times\frac{14}{1}}\times 2&\leq \frac{\cancel{14\times13\times12\times11\times10\times9}}{6\times\cancel{5\times4\times3\times2\times1}}\times11 \\
2&\leq \frac{11}{6},
\end{align*}
a contradiction, therefore there must exist some set of $6$ edges containing $12$ vertices and we are done.[/quote]

I think that this solution is incorrect because when you get that $S=\binom{13}{5}\binom{14}{1}\cdot 2$ you are not counting  the same thing as here \[S \leq \binom{14}{6}\times 11.\] .At first you are counting sum of  all cardinalities of covering sets with $6$ edges ,and on later sum of degres of all covering sets(with $6$ edges).By covering set i mean the set of vertices such that ends of all chosen edges are in that set and no other edge is. By the way $S\leq\binom{13}{5}\binom{14}{1}\cdot 2$

[quote=anantmudgal09][quote=MithsApprentice]The 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players.[/quote]

Pick a maximal matching of $G$ with $m$ edges. Each edge of $G$ emanates from a vertex of this matching, so since each player plays at least once, we have $20-2m \ge 14-m \implies m \ge 6$ as desired.[/quote]

It should be $m\le6$!

[quote=tc1729]So consider a maximal matching. Suppose it contains $t$, with $t\le 5$ edges. Then there are $20-2t$ vertices not in this matching. Since the matching is maximal, those vertices span no edges. However we know that every vertex has a positive degree, so the number of edges from the remaining vertices is greater than or equal to $20-2t$. This requires a total that is greater than or equal to $(20-2t)+t=20-t\ge 15$ edges, which is too many.[/quote]

The number of edges associated with the remaining vertices might not be at least $20-2t$ as some of the edges may have both endpoints among the remaining vertices.


calculus
integration
analytic geometry
calculus computations

$\int_0^1 \int_0^\sqrt{1-y^2}$$\sqrt{1-y^2} dx dy$. why when i evaluate the double integral by changing it to polar form yield a different answer ( i get $\frac{1}{2}$ instead of $\frac{2}{3}$, when i evaluate without changing to polar form )? shouldn't it be the same? thanks in advance

[quote="du0_hell"]$\int_0^1 \int_0^1\sqrt{1-y^2} \sqrt{1-x^2} dx dy$. [/quote]

Is this the integral u are solving ?

no guys, i believe i've posted the question correctly.. ramanujan. the answer that u posted is the answer to the solution, but i think u mistyped the question. my question is: why when we use polar coordinates to integrate the double integral we get  $\frac{1}{2}$ instead?

Perhaps you neglected to add the extra $r$; $dxdy = rdrd\theta$.

i just realised i used a wrong substitution..  :blush: sorry guys.. but now i'm stuck on the polar form
$\int_0^1 \int_0^{\sqrt{1-y^2}} \sqrt{1-y^2}\, dx dy$ $= \int_0^{\frac{\pi}{2}} \int_0^1 r\sqrt{1-r^2sin^2\theta}\, dr d\theta$ $= -\frac{1}{3}\int_0^{\frac{\pi}{2}} \frac{(1-cos^2\theta)^\frac{3}{2}-1}{cos^2\theta}\, d\theta$  $= -\frac{1}{3}\int_0^{\frac{\pi}{2}} \frac{sin^3\theta-1}{cos^2\theta}\, d\theta$

how do i evaluate $\int_0^{\frac{\pi}{2}} \frac{sin^3\theta}{cos^2\theta}\, d\theta$ ?

This integral $\int_0^{\frac{\pi}{2}} \frac{sin^3\theta}{cos^2\theta}\, d\theta$ does  not converge. You need to evaluate $\int_0^{\frac{\pi}{2}} \frac{sin^3\theta-1}{cos^2\theta}\, d\theta$

For this you write

$-\frac{1}{3}\int_0^{\frac{\pi}{2}} \frac{sin^3\theta-1}{cos^2\theta}\, d\theta = -\frac{1}{3}\int_0^{\frac{\pi}{2}} \frac{sin\theta (sin^2\theta)-1}{cos^2\theta}\, d\theta =\\ \\ \\ -\frac{1}{3}\int_0^{\frac{\pi}{2}} \frac{sin\theta (1-cos^2\theta)-1}{cos^2\theta}\, d\theta =-\frac{1}{3}\int_0^{\frac{\pi}{2}} \left(\frac{sin\theta}{cos^2\theta} -sin\theta-\frac{1}{cos^2\theta}\right)\, d\theta= \\ \\ \\ -\frac{1}{3}\left[\frac{1}{cos \theta}+cos \theta -tan \theta \right]_{0}^\frac{\pi}{2}=\frac{2}{3}$

I hope it is correct  :)

thanks a lot for ur help ramanujan ...  :)


Hi, I am looking for a substance that is similar to rubber, but which works a bit like like playdough in that it will deform and keep its shape when put under pressure - when for example massaged in the hand. It must not however break up like playdough, nor must it be sticky, again like playdo (you shouldn't for example, be able to tear it and then stick the seperate pieces together again by just using the pressure of your hands. It should preferably be able to be produced with different 'stiffness' qualities', ranging from the consistancy (and stiffer) of playdough, to a very loose (almost watery) consistancy ( a bit like the flubber stuff that is being discussed here: http://homeschooling.gomilpitas.com/extras/Gak.htm) and preferably would have the same (or very similar) surface texture to to soft latex.it should also be able to regain it's own natural moulded shape if left alone and untouched for long enough. (Perhaps after an hour or two).

What are the most likely candidates for a material with these properties? Is it possible to produce a single material with all of these properties, or would it take two two different substances? Basically I initially  interested in looking at substances such as this because I wanted to make some inner ear moulds for my specialised Ipod earphones that would adjust to the the shape of my ear and provide a tight seal from external noise - but which could also deform and grow as my ears changed and grew through the years. But then I though of a few other interesting applications that a substance/s like this might have.

Any input would be appreciated.

GJ


I   search  for  a  good  book  in  combinatorics. Is  any  free  in  the  net ...........................

           please

Please, come here for some interesting e-books

http://www.cis.upenn.edu/~wilf/


[quote="phan hong son-11T-THPTNK QB"][i]Let[/i] $a,b,c \ge$ $0$[i].prove that[/i]
\[\underset{\text{cycl}}\prod(1+\frac{a(b+c)}{(a+b)(a+c)})\le\frac{27-11(\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a})^{2}}{8}\]  :wink:
[/quote]
@ all: bye bye see you again :P


limit
logarithms
calculus
calculus computations

Compute $ \lim_{x \to 0^\plus{}} x^{\left(x^{x^x} \minus{} x^x \plus{} 1\right)}$.

$ x^x\to 1\minus{}$

$ x^{x^x}\equal{}x^{1\minus{}}\to 0\plus{}$

$ x^{x^x} \minus{} x^x \plus{} 1\equal{}(0\plus{}) \minus{} (1\minus{}) \plus{} 1 \to 0\plus{}$

$ x^{0\plus{}}\to 1\minus{}$

The answer may be right, but the solution isn't.  [quote="Yuriy Solovyov"]$ x^{0 \plus{} }\to 1 \minus{}$[/quote] is false in general: consider something like $ \lim_{x \to 0^ \plus{} } x^{\frac { \minus{} 1}{\log x}}$.  We have $ \lim_{x \to 0^ \plus{} } \frac { \minus{} 1}{\log x} \equal{} 0$, $ \frac { \minus{} 1}{\log x} > 0$ for $ 0 < x < 1$, but the limit I've given is $ \frac {1}{e}$, not 1.  (Actually, $ x^{\frac{\minus{}1}{\log x}} \equal{} \frac{1}{e}$ whenever it's defined.)

OK, a closer look:
Take logarithms:
$ \exp\left(\lim_{x\to 0^\plus{}}\ln x\left(x^{x^x}\minus{}x^x\plus{}1\right)\right)$
$ \exp\left(\lim_{x\to 0^\plus{}}\ln x\left(x\cdot x^{x^x\minus{}1}\minus{}(x^x\minus{}1)\right)\right)$
We have $ x^x\minus{}1\equal{}\exp(x\ln x)\minus{}1\equal{}x\ln x\plus{}O(x^2\ln^2 x)$ near zero. Then $ x^{x^x\minus{}1}\equal{}1\plus{}x\ln^2 x\plus{}O(x^2\ln^3 x)$, and the limit becomes
$ \exp\left(\lim_{x\to 0^\plus{}}\ln x\left[x(1\plus{}x\ln^2 x\plus{}O(x^2\ln^3 x))\minus{}(x\ln x\plus{}O(x^2\ln^2 x))\right]\right)$
$ \exp\left(\lim_{x\to 0^\plus{}}\ln x\left(x\minus{}x\ln x\plus{}O(x^2\ln^3 x)\right)\right)$
The dominant term $ \minus{}x\ln^2 x$ goes to zero. Taking the exponential, the limit is $ 1$.


[i]Ούτε πια θυμάμαι που έχω αυτήν την άσκηση :( .Τώρα την πήρα από το αγγλικό φόρουμ.Εκεί έχει ελαφρά τριγωνομετρική λύση .Ας μαζέψουμε όμως μερικές λύσεις για να την έχουμε ολοκληρωμένη στο αρχείο για το βιβλίο. Έχει απλή και καλή εκφώνηση.[/i]

 Δίνεται οξυγώνιο τρίγωνο $ ABC$ και η διχοτόμος  $ AD$.
Ας είναι  $ E$,$ F$ οι προβολές του $ D$  στις  $ AC$,$ AB$  αντίστοιχα.Έστω ότι οι  $ BE$ και  $ CF$  τέμνονται στο  $ H$.
Να αποδείξετε ότι η  $ AH$ είναι κάθετη στην  $ BC$.

 Μπάμπης

Βάζω μια λύση αλλά χρσηιμοποιώ και εγώ τριγωνομετρία (εύχομαι να μην είναι ίδια με αυτή στην οποία αναφερθήκατε...)

Λοιπόν κατ' αρχάς ας πούμε $ K$ την τομή $ AH$,$ BC$. Τώρα έχουμε από Ceva:

$ \frac{BF}{FA}\cdot\frac{AE}{EC}\cdot\frac{CK}{KB}\equal{}1$.

Τώρα έχουμε ότι  $ \frac{BF}{FA}\equal{}\frac{FD\cdot\cot{B}}{FD\cdot\cot{\frac{A}{2}}}\equal{}\frac{\cot{B}}{\cot{\frac{A}{2}}}$
(όπου cot είναι η συνεφαπτομένη) και όμοια γαι το 2ο κλάσμα.

Αρα:

$ \frac{\cot{B}}{\cot{\frac{A}{2}}}\cdot\frac{\cot{\frac{A}{2}}}{\cot{C}}\cdot\frac{CK}{KB}\equal{}1\iff$

$ \frac{KB}{CK}\equal{}\frac{\cot{B}}{\cot{C}}$

Απ' την τελευταία ισότητα γίνεται φανερό ότι $ K$ είναι το ίχνος της καθέτου απ' το $ A$, αφού αν το ίχνος αυτό ήταν το $ K'$ θα ίσχυε $ \frac{KB}{CK}\equal{}\frac{\cot{B}}{\cot{C}}\equal{}\frac{K'B}{CK'}$ που σημαίνει ότι τα 2 σημεία ταυτίζονται :wink:

Νομίζω ότι άλλη απλούστερη απόδειξη είχα παλιότερα για την πρόταση αυτή, αλλά δεν τη θυμάμαι και δεν μπόρεσα να τη βρώ.

[hide]Γράφουμε τον κύκλο $ (O_{1}),$ με διάμετρο το τμήμα $ AD,$ ο οποίος τέμνει την πλευρά $ BC$ στο σημείο έστω $ K.$

Έτσι έχουμε $ AK\perp BC$ και έστω $ Q\equiv EF\cap AK$ και $ L\equiv (O)\cap DQ.$

Επειδή είναι και $ DL\perp AL,$ συμπεραίνουμε ότι το σημείο $ Q,$ είναι το ορθόκεντρο του τριγώνου $ \bigtriangleup PAD,$ όπου $ P\equiv BC\cap AL$ και άρα η ευθεία $ EF\perp AD$ $ ($ η $ AD$ είναι μεσοκάθετη του $ EF$ $ ),$ περνάει από το σημείο $ P,$ ως το τρίτο ύψος του $ \bigtriangleup PAD.$

$ \bullet$ Έχουμε τώρα δια του σημείου $ P,$ τις δύο τέμνουσες $ PKD,\ PLA$ του κύκλου $ (O_{1})$ και όπως είναι γνωστό, το σημείο $ Q\equiv AK\cap DL,$ είναι αρμονικό συζυγές του $ P,$ ως προς τα $ E,\ F.$

Η αρμονική σημειοσειρά $ P,\ F,\ Q,\ E,$ προβάλλεται από το $ A,$ στη σημειοσειρά $ P,\ B,\ K,\ C,$ που είναι επομένως, επίσης αρμονική.

Είναι εύκολο τώρα να δούμε ότι στο πλήρες τετράπλευρο $ BCEFAP,$ η ευθεία $ AQK$ περνάει από το σημείο $ H,$ τομής των διαγωνίων του $ BE,\ CF$ και έτσι έχουμε ότι το σημείο $ H$ ανήκει στην ευθεία $ AK\perp BC.$

Άρα  $ AH\perp BC$ και η πρόταση έχει αποδειχθεί.[/hide]
Κώστας Βήττας.

Έστω ότι η κάθετη από το $ A$ τέμνει την $ BC$ στο $ K$.

Αρκεί $ \frac{AE}{EC}\frac{CK}{KB}\frac{BF}{FA}\equal{}1$

Όμως $ AE\equal{}AF$ οπότε αρκεί:

$ \frac{CK}{CE}\equal{}\frac{BK}{BF}$

Επίσης $ BF\equal{}BD*cosB$ και $ CE\equal{}CD*cosC$

ή ισοδύναμα:

$ BF\equal{}BD*\frac{KB}{AB}$ και $ CE\equal{}CD*\frac{KC}{AC}$

οπότε αντικαθιστώντας είναι προφανές από το θεώρημα διχοτόμων.

Δεν τη βρήκα την απόδειξη που λέω πιο πάνω, αλλά μάλλον πρέπει να είναι αυτή :

[hide]Έστω $ AK,\ BL,\ CM,$ τα ύψη του τριγώνου  $ \bigtriangleup ABC$ και για να είναι το  $ H\equiv BE\cap CF$ σημείο της ευθείας $ AK,$ αρκεί να αποδειχθεί ότι ισχύει (αντίστροφο του [b][size=100]θεωρήματος Ceva[/size][/b]).

$ \frac {KB}{KC}\cdot \frac {EC}{EA}\cdot \frac {FA}{FB} \equal{} 1$ $ \Longrightarrow$ $ \frac {KB}{KC} \equal{} \frac {FB}{EC}$ $ ,(1)$ $ ($ γιατί $ EA \equal{} FA$ $ ).$

Από $ DF\parallel CM$ $ \Longrightarrow$ $ \frac {BF}{BM} \equal{} \frac {BD}{BC}$ $ \Longrightarrow$ $ BF \equal{} BM\cdot \frac {BD}{BC}$ $ ,(2)$ και από $ DE\parallel BL$ $ \Longrightarrow$ $ \frac {CE}{CL} \equal{} \frac {CD}{BC}$ $ \Longrightarrow$ $ CE \equal{} CL\cdot \frac {CD}{BC}$ $ ,(3)$

Από $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {FB}{EC} \equal{} \frac {BM}{CL}\cdot \frac {BD}{CD}$ $ ,(4)$

Από $ (1),$ $ (4)$ προκύπτει ότι αρκεί να αποδείξουμε ότι ισχύει $ \frac {KB}{BM}\cdot \frac {CL}{KC} \equal{} \frac {BD}{CD}$ $ ,(5)$ η οποία όμως αληθεύει γιατί

$ \frac {KB}{BM} \equal{} \frac {AB}{BC}$ $ ,(6)$ από όμοια ορθογώνια τρίγωνα $ \bigtriangleup KAB,\ \bigtriangleup MCB$ και $ \frac {CL}{KC} \equal{} \frac {BC}{AC}$ $ ,(7)$ από όμοια ορθογώνια τρίγωνα $ \bigtriangleup LBC,\ \bigtriangleup KAC.$

Επίσης, έχουμε $ \frac {BD}{CD} \equal{} \frac {AB}{AC}$ $ ,(8)$ από την $ AD,$ ως διχοτόμο της γωνίας $ \angle A.$

Άρα η ισότητα $ (1)$ αληθεύει και η πρόταση έχει αποδειχθεί.[/hide]
Κώστας Βήττας.


geometry rectangle perimeter quadratics ratio calculus limit

Let $ABCDE$ be a convex pentagon such that $A,B,C,E$ are the vertices of a rectangle and $C,D,E$ are the vertices of an equilateral triangle. The area of the pentagon $ABCDE$ has a given value $F$. Determine whether among these pentagons there exist a pentagon with minimum perimeter $u$. If so, determine for those pentagons the value $|AB|:|BC|$. If possible, try to find solutions with and without calculus.

[quote="orl"]... If possible, try to find solutions with and without calculus.[/quote]
This is like starting a fire with and without matches.

Denote $a = AB = CD = DE,\ b = BC = DA$. Let $p, S$ be the pentagon perimeter and area. Finding the minimum perimeter at a given area is equivalent to finding the maximum area at a given perimeter.

$p = 3a + 2b,\ \ S = ab + a^2 \frac{\sqrt 3}{4}$

$b = \frac{p - 3a}{2},\ \ S = \frac{a(p - 3a)}{2} + a^2\frac{\sqrt 3}{4}$

$\frac{6 - \sqrt 3}{4}a^2 - \frac p 2 a + S = 0$

This quadratic equation for $a$ has a real solution, iff the discriminant is greater than or equal to zero:

$D = \frac{p^2}{4} - (6 - \sqrt 3)S \ge 0,\ \ S \le \frac{p^2}{4(6 - \sqrt 3)}$

The area is maximum at equality and the quadratic equation for $a$ becomes

$\frac{6 - \sqrt 3}{4}a^2 - \frac p 2 a + \frac{p^2}{4(6 - \sqrt 3)} = 0$

$(6 - \sqrt 3)^2 a^2 - 2(6 - \sqrt 3) p a + p^2 = 0$

$[(6 - \sqrt 3) a - p]^2 = 0$

$p = (6 - \sqrt 3)a$

$b = \frac{6 - \sqrt 3 - 3}{2} a = \frac{3 - \sqrt 3}{2} a$

$\frac{AB}{BC} = \frac a b = \frac{2}{3 - \sqrt 3} = \frac{3 + \sqrt 3}{3} = 1 + \frac{\sqrt 3}{3}$

$\frac{S}{p^2} = \frac{ab + a^2 \frac{\sqrt 3}{4}}{(3a + 2b)^2} = \frac{x + x^2 \frac{\sqrt 3}{4}}{(3x + 2)^2}$

where $x = \frac a b$. The maximum ratio $\frac{S}{p^2}$ is then

$(\frac{S}{p^2})_{\text{max}} = \frac{1 + \frac{\sqrt 3}{3} + (1 + \frac{\sqrt 3}{3})^2 \frac{\sqrt 3}{4}}{(3 + \sqrt 3 + 2)^2} = \frac{\frac 3 2 + \frac 2 3 \sqrt 3}{2(14 + 5\sqrt 3)} =$

$= \frac{66 + 11 \sqrt 3}{12 \cdot 121} = \frac{6 + \sqrt 3}{132} \approx \frac{7.7}{132} = \frac{7}{120} = \frac{35}{600}$ [color=white].[/color] (within 1%)

Check with calculus:

$\frac{\text d}{\text dx}(\frac{S}{p^2}) =  \frac{(1 + x \frac{\sqrt 3}{2})(3x + 2)^2 - 6(x + x^2 \frac{\sqrt 3}{4})(3x + 2)}{(3x + 2)^4} = 0$

$(1 + x \frac{\sqrt 3}{2})(3x + 2) - 6(x + x^2 \frac{\sqrt 3}{4}) = 0$

$3x + 2 + \frac{3x^2 \sqrt 3}{2} + x \sqrt 3 - 6x - \frac{3x^2\sqrt 3}{2} = 0$

$(3 - \sqrt 3)x = 2,\ \ x = \frac{2}{3 - \sqrt 3} = \frac{3 + \sqrt 3}{3} = 1 + \frac{\sqrt 3}{3}$

$\lim_{x \rightarrow 0}(\frac{S}{p^2}) = \lim_{x \rightarrow 0} \frac{x + x^2 \frac{\sqrt 3}{4}}{(3x + 2)^2} = 0 < (\frac{S}{p^2})_{\text{max}}$

$\lim_{x \rightarrow \infty}(\frac{S}{p^2}) = \lim_{x \rightarrow \infty} \frac{x + x^2 \frac{\sqrt 3}{4}}{(3x + 2)^2} = \frac{\sqrt 3}{36} \approx \frac{29}{600} < \frac{35}{600} \approx \frac{6 + \sqrt 3}{132} = (\frac{S}{p^2})_{\text{max}}$

(within 1%)

Thus we have a maximum ratio $\frac{S}{p^2}$ at $x = \frac a b = 1 + \frac{\sqrt 3}{3}$.


[color=cyan]If   x + yz = -2,  x^2 + y^2 + z^2 = 3, and  y + z = 2, compute both possible values for the complex number x.[/color]

Solution:

[hide]if y+z=2, then y=2-z

Then, plug y into both of the other equations, to get

x+(2-z)z=-2 ----> x+2z-z^2=-2

x^2+(2-z)^2+z^2=3  ---->x^2-4z+2z^2=-1

Multiply first equation by 2: 2x+4z-2z^2=-4

Now add the two equations

You get x^2+2x=-5, and now you can easily solve for x.[/hide]

For a nicer solution, add twice the first equation to the second equation, and add one to both sides, getting x^2 + 2x + 1 + y^2 + 2yz + z^2 = 0, which simplifies to (x+1)^2 + (y+z)^2 = 0. Substitute y+z=2 to get (x+1)^2 = -4, hence x = -1 +/- 2i.

hmmmmmmm
interesting...

they're sorta the same, but urs has got a different twist to it.


I love the AoPS textbooks and intend to use them as primary curriculum for homeschooling my son.  The classes sound great as well; however, my son is a competitive gymnast and spends four hours every evening - including the class timeslot - at the gym.  What is the value of the introductory series classes above and beyond the textbook if you can only view the classes as transcripts after the fact?

You'd still get the benefit of having the teachers grade challenge problem sets, and homework problems, etc. However, other than that, I think the textbooks are better than the transcripts because of the structure unless you're going to actually participate in the classroom.

A small portion of students do use our online school in the manner you suggest, but usually it's best if you can actually attend the class.  The one other benefit in addition to what grn_turtle mentions is that the classes provide structure.  If you can provide that structure for your son, or if he's motivated enough to do it himself, he should be fine with the books alone.


Given a segment $\overline{AB}$ and two arbitary points $P,Q$ on it. What's the probability that the three segments$\overline{AP},\overline{PQ}$ and $\overline{QB}$ can form a triangle?
[img]http://i5.photobucket.com/albums/y194/jelly119/seg.jpg[/img]

this is a pretty well known problem, and there are many ways of solving the problem. here is one easy way:
clearly P and Q have to be on the both sides of the midpoint of AB, this has a probability of $\frac{1}{2}$ let the midpoint of AB have a cordinate m, let n<m. assum P has coordinates m-n, now consider points that $\frac{1}{2}\le P-Q$ now if you pair up all the m-n's and n's you can conclude that the probability of $\frac{1}{2}\le P-Q$ for P, and Q's on the oppposite side of the midpoint is $\frac{1}{2}$..
thus the probability that we want is:
$\frac{1}{2}\frac{1}{2}=\frac{1}{4}$


symmetry
function
algebra unsolved
algebra

Find all $ f: R\to R,g: R\to R,h: R\to R$ satisfying
1) $ f,g,h$ is continuous on $ R$
2) $ f(x^2) \minus{} f(y^2) \equal{} [h(x) \minus{} h(y)]g(x \plus{} y)$ with $ \forall x,y\in R$

Let $ y \rightarrow x$, the RHS approaches $ 0$. (since $ h,g$ are continuous). Hence, $ 2f(x^2) \rightarrow 0$ for all $ x$. This implies $ f(x)\equal{}0 \forall x \in \mathbb{R}_{\geq 0}$. For $ x\leq 0$, any definition of $ f$ so that it is continuous will satisfy.

This leads to $ \forall x,y \in \mathbb{R}: h(x) \equal{} h(y) \vee g(x\plus{}y)\equal{}0$. (this could also be deduced from symmetry of $ x$ and $ y$) So, $ h(x)\equal{}h(0) \vee g(x)\equal{}0 \forall x$. Any definition of $ h(x)$ and $ g(x)$ satisfying this condition (and the continuity of $ h$ and $ g$) will be fine.

I guess that your problem should be changed to
$ f(x^2) \minus{} f(y^2) \equal{} [h(x) \minus{} h(y)]g(x \plus{} y)$

[quote="mathangel"]Let $ y \rightarrow x$, the RHS approaches $ 0$. (since $ h,g$ are continuous). Hence, $ 2f(x^2) \rightarrow 0$ for all $ x$. This implies $ f(x) \equal{} 0 \forall x \in \mathbb{R}_{\geq 0}$. For $ x\leq 0$, any definition of $ f$ so that it is continuous will satisfy.

This leads to $ \forall x,y \in \mathbb{R}: h(x) \equal{} h(y) \vee g(x \plus{} y) \equal{} 0$. (this could also be deduced from symmetry of $ x$ and $ y$) So, $ h(x) \equal{} h(0) \vee g(x) \equal{} 0 \forall x$. Any definition of $ h(x)$ and $ g(x)$ satisfying this condition (and the continuity of $ h$ and $ g$) will be fine.

I guess that your problem should be changed to
$ f(x^2) \minus{} f(y^2) \equal{} [h(x) \minus{} h(y)]g(x \plus{} y)$[/quote]
maybe! Can you post solution?

I didn't find the solution. I think that there is no construction for all the functions satisfying the condition.

"I think that there is no construction for all the functions satisfying the condition."

Why?


A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form m/n, m n relatively prime. find M+N

[hide]The problem is just the same as 1 :le: a_1  :le:  a_2  :le: a_3 :le: a_4 :le: 6, find number of possible (a_1,a_2,a_3,a_4). We can imagine a possible sequence of 6 A's and 4 B's and the sequence must start with A. The 4 B's divides the sequence that the value of a_n correspond to the number of A's before the nth B (for example, AABABAABBA, a_1=2, a_2=3, a_3=5, and a_4=5. ABBAAABAAB, a_1=1, a_2=1, a_3=4, a_4=6) Since the sequence must start with A, so the answer is (10-1)C4=9C4. 9C4/6^4=7/72. so 7+72=79.  [/hide]

[hide]

Furthermore, it is quite easily seen that the number of such sequences is the number of ways we can pick 4 numbers from {1,2,3,4,5,6}, with replacement.

This is C(6+4-1,4) = C(9,4).

Divide this by 6^4 to get your answer.

Combinations with replacement: your friend on the AIME.

[/hide]


How many numbers are in the list $ 2008, 2003, 1998, \ldots, 8, 3?$

This sequence is $ a_n \equal{} 2008\minus{}5(n\minus{}1) \equal{} 2013\minus{}5n$.
So, $ 2013\minus{}5n\equal{}3 \iff 5n\equal{}2010 \iff n\equal{}402$.
Therefore $ a_{402} \equal{} 3$.
Thus, ｔhe length of the list is $ \boxed{402}$.

$ 2008, 2003, 1998, \ldots, 8, 3$

Alternatively,

Let's add 2 to all numbers in the sequence!

$ 2010, 2005, 2000, \ldots, 10, 5$

Now let's divide all by 5!

$ 402, 401, 400, \ldots, 2, 1$

It is clearly 402.

First, we reverse the list to become $3,8,\ldots,1998,2003,2008$. Now we add 2 to each number to get $5, 10,\ldots, 2000, 2005, 2010$. Then we divide each number by 5 to get $1,2,\ldots,400,401,402$, so there are $\boxed{402}$ numbers.

[quote name="X_star12321" url="/community/p19309408"]
can you not bump everything unless you are putting a solution
[/quote]

okok fine


linear algebra
matrix
superior algebra
superior algebra solved

Fie A=
(cosx  sinx)
(sinx  cosx) Sa se calculeze A^n
It means calculate A^n where A is the given matrix.
Gheorge Andrei

denote
x = 1/2 + 2 n-1  cos n2x
y = 1/2 - 2 n-1  cos n2x

A^n =
(x y)
(y x)

Here is what I think:

denote A=
x y
y x

We consider that A^n=
x_n y_n
y_n x_n

we get from the fact that A^(n+1)=A^n *A
x_(n+1)+y_(n+1)=(x_n+y_n)(x+y)=(x+y)^(n+1)  and
x_(n+1)-y_(n+1)=(x-y)^(n+1)

so A^n=
[(x+y)^n+(x-y)^n]/2  [(x+y)^n-(x-y)^n]/2
[(x+y)^n-(x-y)^n]/2   [(x+y)^n+(x-y)^n]/2

hope it's okay! :)
cheers! :D :D

notice that x+y=1... and (cos x)^2 - (sin x)^2 = 2 cos 2x

so you can simplify your answer... and it is exactly same to mine,


Three isosceles right triangles are erected from the larger side of a rectangle into the interior of the rectangle, as shown on the right, where M is the midpoint of that side. Five circles are inscribed tangent to some of the sides and to one another as shown. One of the circles touches the vertex of the largest triangle. Find the ratios among the radii of the five circles. Have a look at problem 5/39.

The figure is symetric hence....
Let the radii of the top two circles be [b][i]r[/i][/b]
Let the radii of the bottom two circles be R
Let the radius of the bottom center circle be r.

Let cot( \pi /8)=w=  \sqrt2+1
Let the length of the longer side be 'a' and the length of the smaller side be 'b'

Now we have
[b][i]r[/i][/b] = a/(2 \sqrt2(w+1)) = a( \sqrt2 -1)/4------(A)

a/ \sqrt2 + 2r =b       ----(1)
R(w+1)= b                ----(2)
4 \sqrt (Rr) + 2R = a ----(3)

which gives us
R=( \sqrt2+ \sqrt6-1)a/2 (B)
r= ( \sqrt2+1)\sqrt3 a/2  (C)

From eq (A),(B),(C)
We have the ratio of the radii of the circles.


A silly question, but I really need the answer..

In the common app., there is a part where you fill in info about your extracurrlcular activitites.. And there are some spaces titled Grade level/Post secondary. (your options are 9, 10, 11, 12 or PS). I'm not sure about the meaning of that..

Am I supposed to check off the years I participated in the activity, or do I indicate my level?
(For example, if I play the piano at an undergraduate level, do I check off PS, etc, or Do I check off 9, 10, 11 and 12 because I played piano through all high school?)

It's embarassing but I need someone's help, and this forum seems like a good place!

Thanks in advance!  :blush:  :blush:  :blush:  :blush:  :blush:

You should only check off the years of participation in each extracurricular activity. You level and talent should be expressed in the description or in your essays.


A square sheet was cut into 42 smaller squares, one of which has an area greater than 1 cm² and the others have over 1 cm². What is the measure on the leaf ?

A) 6 cm		B) 12 cm	C) 21 cm 	D) 19 cm 	E) 20 cm

How to do it ?

Hi Luis,

I can't follow the question; could you post the Portuguese version? Thanks.

AndrewTom.

Portuguese:

Uma folha quadrada foi cortada em 42 quadrados menores, dos quais um tem área maior do que 1 cm² e os demais têm área de 1 cm2. Qual é a medida do lado da folha?

Hi luis,

I translate this as

A square sheet (of paper) is cut into $ 42$ smaller squares, of which one has an area of more that $ 1$ $ cm^{2}$ and the rest (of them) have areas of $ 1$ $ cm^{2}$. What  is the measure of the side of the sheet?

Hi Luis, I make it $ (C): 21$.

I placed the large square, $ x$ by $ x$, in the top right-hand-corner, with $ y$ unit squares to its right. This gave me $ xy \plus{} y(x\plus{}y) \equal{} 41$.

$ y^{2} \plus{} 2xy \minus{}41 \equal{} 0$

$ y\equal{} \frac {\minus{}2x \pm \sqrt{4x^{2} \plus{} 164}}{2}$

$ 4x^{2} \plus{} 164 \equal{} 4(x^{2} \plus{} 41) \equal{} 4z^{2}$.

$ x^{2} \plus{} 41 \equal{} z^{2}$

$ z^{2} \minus{} x^{2} \equal{} 41$

$ (z\minus{}x)(z\plus{}x)\equal{}41$

$ x\equal{}20, z\equal{} 21$

$ y\equal{} \frac{\minus{}40 \pm \sqrt{1600 \plus{} 164}}{2} \equal{} 1$.

So we have a large $ 20 \times 20$ sqare in the top left-hand-corner with an L-shape of $ 41$ unit squares to its right and below it, giving a $ 21 \times 21$ square, that is, a square of side $ 21$


modular arithmetic
number theory proposed
number theory

Suppose that $ p$ is a prime number. Prove that the equation $ x^2\minus{}py^2\equal{}\minus{}1$ has a solution if and only if $ p\equiv1\pmod 4$.

http://www.mathlinks.ro/viewtopic.php?t=25234
(the case $ p \equiv 3 \mod 4$ is comparibly trivial)


[quote="Sylwek"]$ f : \mathbb{R} \rightarrow \mathbb{R} \\
f( f(x) \minus{} f(y) ) \equal{} (x \minus{} y)^2 f ( x \plus{} y )$[/quote]

i)put x=y in eqn,you get f(0)=0.then
ii)put x=x and y=0,so you get f(f(x))=x^2*f(x).
iii) put x=f(x),you get f(f(f(x)))=x^4*f(x).resubstituting x=f(x) again and again we get
    f^n(x)=x^(2*(n-1)).f(x)
iv)now put n=0,therefore 1=x^(-2).f(x)
    or simply f(x)=x^2.

i don't know how to use the code language and maybe few notations are wrong.
hope you are convinced by my solution.

$ f(f(x) - f(y)) = (x - y)^{2}f(x + y)$.  $ (*)$.
    Setting $ x = y$ in the given equation implies $ f(0) = 0$. Easy to check that $ f(x) = 0$ is the function we are looking for. Assume that there exists $ t$ for which $ f(t)$ doesn't equal to $ 0$. Setting in the given equality $ y = - x$ we get $ f( - x) = f(x)$, for any real $ x$. If we substitute $ y$ by $ - y$ in $ (*)$ , our equality becomes
 $ f(f(x) - f(y)) = (x + y)^{2}f(x - y)$.  $ (1)$. Easy to see that for any pair of real numbers $ (a;b)$ we can find $ x$ and $ y$ so that $ x - y = a$ and $ x + y = b$. From $ (*)$ and $ (1)$ we find out that $ a^{2}f(b) = b^{2}f(a)$ $ (2)$, for any real numbers $ a,b$. Assume there is number $ u$ such that $ u$ doesn;t qual to $ 0$ but $ f(u) = 0$. Setting $ (a;b) = (t;u)$ in $ (2)$ we get $ a^{2}f(t) = 0$, which is obviously impossible. It means that there doesn't exist such $ u$. Rewrite $ (2)$ as $ f(a)/{a^{2} = f(b)/b^{2}}$, $ (3)$, for any real numbers both different from $ 0$. $ (3)$ means that $ f(x)/x^{2}$ is constant, implying $ f(x) = kx^{2}$, for any real $ x$, but if we set $ x$ different from $ y$ in $ (*)$, we get $ k^{2} = k$, which leads to the conclusion that $ k = 1$. So, $ f(x) = x^{2}$ and $ f(x) = 0$ are the only solutions. Easy to verify that both of them is indeed solution.  Lasha Lakirbaia.

   P.S.  Sharath S I, your solution is obviously wrong. You got the equality by writing $ f(f(x)) = x^{2}f(x)$ $ n$ times, so you can't set $ n = 0$ in new equality, because you can't write equation $ 0$ times.  :D

Exactly! For lasha's solution: we don't know that $ f$ is injective. Actually, if $ f(x) \equal{} f(y)\neq 0$ then $ x^2 \equal{} y^2$

If there exists $ f(a)\ne0$, we can assume, WLOG $ f(a)<0$.
Then $ f(f(x)\minus{}f(a\minus{}x))\equal{}(a\minus{}2x)^2f(a)$.
Looking at the RHS, we see $ f$ can take any nonpositive value. $ f(x)\minus{}f(y)$ can take any value.
$ f(f(x)\minus{}f(y))\equal{}f(f(y)\minus{}f(x))$ implies $ f$ is even function.
Similarly for $ f(a)>0$.
Now lasha's argument should be OK.

Thank you for correcting my mistake. Great solution!


trigonometry
geometry
calculus
function
integration
trig identities
Law of Sines

How are you supposed to do these two problems (I found them in ten real SATs)?

(For the first one, I used sqrt(s(s - a)(s - b)(s - c)), and I used Law of Sines for the second one, but these two formulas aren't tested on the SAT are they? I can't think of an easier way to do them)

1.

One side of a triangle has length 6 and a second side has length 7. Which of the following could be the area of this triangle?

I. 13
II. 21
III. 24

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

2.

[quantitative comparison]

(m<QRS = 120 degrees, RS = 4, QS = 6)

===================          ==============
|  The area of triangle QRS |         |            12             |
===================          ==============

I think there might be ways around those formulas, but since those are the easiest you might as well as use them if you know them.
I'm pretty sure everything but Calculus is tested on the SATs, but I'm not sure (I don't remember trigonometry being necessary for SAT, though it's going to be very helpful.)

Okay, so for the first one I think we can agree that herons formula will suck to impliment.  Instead, do the following: consider the two edges "hinged" together at their common vertex, and slowly move them closer and farther apart.  The area will be a continuous function of the angle.  So, we can clearly get an area of 0 if we push them all the way together.  The biggest area we will get is if we make them perpendicular (as a result of K = ab*sin(C)/2, incidentally), which will give you 1/2 * 6 * 7 = 21.  So we can get any real number between 0 and 21.

As for the second one, I recommend exactly the same trick.  Try it around vertex S, see what happens.

This is a very common trick for the SAT's -- worth hanging on to.  Good time-saver.

[quote="JBL"]Instead, do the following: consider the two edges "hinged" together at their common vertex, and slowly move them closer and farther apart.  The area will be a continuous function of the angle.  So, we can clearly get an area of 0 if we push them all the way together.  The biggest area we will get is if we make them perpendicular (as a result of K = ab*sin(C)/2, incidentally), which will give you 1/2 * 6 * 7 = 21.  So we can get any real number between 0 and 21.
This is a very common trick for the SAT's -- worth hanging on to.  Good time-saver.[/quote]

Indeed. I must have used some kind of intuitive reasoning like this when I took the SAT back in the 1970s, because I didn't learn the tools of trigonometry at all back then (I knew what trigonometry was about, but I really couldn't solve most straight-up trigonometry problems), and I DEFINITELY didn't even know of the existence of Heron's formula until decades later, but I almost aced the SAT math section. I usually don't find problems like this very hard, and I do habitually think about "hinged" angles when facing problems of this nature. All the REQUIRED syllabus for the SAT math section (until the test is revised, the year after next) is plain-old high school "algebra 1," which is seventh-grade math in much of the world. But SAT problems, like some competition problems, take math that "everyone knows" and put in little twists that require mental agility. After all, the SAT attempts to be a test of REASONING, not of achievement obtained through school lessons. It's pretty clear that studying more math (inside or outside of school) rather than less helps prepare for the SAT, because people with more preparation have done more THINKING about math, but the number of strictly required tools and techniques for the SAT math section is remarkably small.

Thanks, JBL, for "thinking out loud" for us.

I'm just going to post what I would do for the 1st one quickly because I have to get to school.

If the sides are length 6 and 7, then you know the other leg must be anywhere between 2----12 (they only use integers don't they?). Calculate the area if the leg was 2, then again if it was 12. Find what number falls within this range....

I agree that JBL's hinge is an excellent method for these problems.  But you don't really need the trig formula.  A=bh/2 is sufficient.

For the first problem, keep one of the legs fixed as the length of the base, and allow the other leg to rotate.  "A" varies proportionally with the only remaining variable in the formula, "h".  What is the maximum possible "A"?  The minimum?

Similarly for the second problem.  Put a hinge at R, and pick one of the legs to be the fixed base.  If you can find a triangle that has A=12, perhaps you can decide whether your triangle will be smaller, equal, or larger, without having to compute its area.

scm, your solution has 2 problems.  The first is that the requirement is not for integers, so your starting assumption was wrong.  The second problem is that given two sides, the largest and smallest possible values for the third side will NOT be the largest and smallest possible values for the area.  To see why, take a triangle with two sides of length 20 and 11.  The largest (integral) value for the third side will be 30, while the smallest will be 10.  However, both of these triangles are skinny little buggers with very small area.  The maximal area will come when the third side is about 23.

OK well then you can just see that 1/2 (6*7) = 21 Which is an answer. Yeah I see what you eman, I didn't really think about it that much.


quadratics
number theory unsolved
number theory

Let $p$ be an odd prime. Let $A_{p}$ and $B_{p}$ be the sets of quadratic residues and non-residues respectively (modulo $p$). Let $\zeta$ be a primitive $p$-th root of unity and define $a = \sum_{k \in A_{p}}\zeta^{k}$ and $b = \sum_{k \in B_{p}}\zeta^{k}$. Show that $1-4ab = p \cdot \left(\frac{-1}{p}\right)$.


Several [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]MATHCOUNTS Math Jams[/url] have been scheduled for the next few months.  The Problem Discussion Hours give students a chance to seek help on the problems that frustrate them most.  The focus is on finding the best method for each problem.

The MATHCOUNTS Games are very popular and many state and national competitors participate and everyone has a good time.

You can view past Math Jams [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]here[/url].

2002 National MATHCOUNTS champion Albert Ni says he may host a Math Jam or two also.  Check the schedule for updates.

Albert Ni has just informed me that he will host a MATHCOUNTS Problem Discussion Hour Math Jam this Wednesday at 7 PM ET/4PM PT.  I will add it to the schedule tomorrow.

Two questions Mr. Crawford

I am a beginner student just starting to hit more complex algebra. If you are familiar with Saxon textbooks, I am just starting Algebra One. If you are not, then never mind. How do I know which MJs are for me and which aren't? Solafidefarms said something about #s denoting how hard a MJ is, but I cannot find those #s.

What is the link to the "games" you are talking about?

Numbers:
At the far right, you see a column that says difficulty, that's the number.

Games:
If you look under title of MJ, you'll see some that say MathCounts Game Mathjam. (There's 3).

I read the Algebra I, Geometry, and Algebra II Saxon books early in middle school.  I would say that are level 1 in regards to our Math Jam difficulty levels.  Most all of our material is more difficult.  However, the best way to find out how much you are ready for is to visit a Math Jam and see.  The MATHCOUNTS Math Jams and MOEMS Math Jams are a good place to start.


This morning I voted a 2._ rating user a 3 cube, and the rating went down.

Just a few minutes ago, I rated a post a 4 cube, and the overall rating for that specific post showed 2. So when I rate something it takes half of what i really rated?

I believe the rating system takes the average of all the ratings that a post has received.

We would need the post number to research this and determine if it is an issue.

[quote="james4l"]Happens with me too: rating a 1 to a post sometimes makes it registers a 1.5

It can possibly have to do with the weighting system and that the weighting system compares ratings in relation to the person (which can be flawed; 1 should = 1 for all cases) , with my post rating not especially great (3.3), and theirs may be a 4. Or a 1.5. I don't know the formula itself, so...

On the other hand, the weights should affect ratings in relation to each other, not to the rating itself.[/quote]

This happens to me too, I believe they average the rating of that post....

Also, I never see the rating for any of my posts though I'm sure some of them are rated.

I posted a similar issue earlier, with no response. A new user with a rating of 4.8 made a spam post, which I rated accordingly, and the rating actually went up to 5. This happened a few more times as well. At first I figured that perhaps someone had just  happened to rate of the user's other posts at the same time I was, but it's rather unlikely this would keep happening over and over again in such a short time period.

A slightly more dramatic example is something that happened this morning. I rated someone with a 4.3 post rating 4 cubes, and it shot down to 3.2.

Please provide post numbers when encountering such issues. We can not investigate without.

[quote="dragon96"]Actually that is not exactly true - ratings are weighted, so not exactly the average.

But I am positive that no one rated those posts before I.[/quote]If a post hasn't been rated a sufficient number of times before it will not show up as rated. Once a required amount of ratings is done on the post, the (weighted) average of those ratings is showed. So if $ n\minus{}1$ people before you voted 1, and you are the last to vote and you vote 4, the rating of the post would probably be 1.5

This sort of thing happens to a lot of posts. Like I rated a post higher than average and the rating went down. And sometimes the opposite.

Also, where do I find the post number?

[quote="AwesomeToad"]Also, where do I find the post number?[/quote]Click on the #2 or whatever post number, in the post's top-right side. You will get a link to the post. The last digits in the url are the post ID.


calculus
derivative
calculus computations

Let $x=f(u, v)$ and $y=g(u, v)$. Prove that
$\left({\partial x\over\partial u}\right)_{v}\left({\partial u\over\partial x}\right)_{y}= \left({\partial y\over\partial v}\right)_{u}\left({\partial v\over\partial y}\right)_{x}$ . :)

[quote="mlok"]I don't know what $\left(\frac{\partial u}{\partial x}\right)_{y}$ means. Would you mind working out a simple example like $x=uv$, $y=u/v$?[/quote]
The index of each partial differentiation tells you that, the index is assumed to be constant while computing the derivative. Your example leads to the $\frac12=\frac12$ identity.

[quote="JBL"]So, when you write "$\left({\partial x\over\partial u}\right)_{v}$" you mean, "take the derivative of $x$ with respect to $u$, treating $v$ as a constant?"  Because normally one just writes ${\partial x\over\partial u}$ for that -- implicit in taking a partial is that you hold other variables constant.

With that definition, I'm not sure where you get $\frac{1}{2}$ from.[/quote]
As far as I'm concerned this is a well known notation. $\frac12$ is obtained simply by combining the teo given relations when differentiating. You may obtain $1=1$ if applying the derivative directly.


algebra
binomial theorem
combinatorics unsolved
combinatorics

I know there is a way using combinatorics to solve this sumation and derive a formula . Could someone help me out as quick as possible. Thanks

Ben

CRACKED, its actually easier using Binomial Theorem. Thanks anyways:P

If I am not mistaken, these are known as Bernoulli Numbers.  Here is a link:

[url]http://en.wikipedia.org/wiki/Bernoulli_number[/url]

right... id like to keep it a little more simple for people to understand... its easier with binomial theorem to get it;)
Thanks for the comment

Ben

I wonder how you got what formula by simple binomial theorem...


There are 400 types of poisonous snakes in the world. Given that this number represents $ 13 \frac{1}{3} \%$ of the total number of snake varieties, how many varieties of snakes are in the world?

Set up a ratio: $ \frac{400}{x}\equal{}\frac{2}{15}\Rightarrow x\equal{}3000$


function
calculus
derivative
real analysis
real analysis unsolved

g:R-->R of classe C^1 such that
there exist A>0 for any x \geq A,  g(x) > 0 and g'(x) > 0
Prove that any solution of the differential equation y" + g(x)y = 0
is bounded in a neighboroud of +oo

Nice one! I took again all problems with differential equations, since I'm zero in this field and our chapter of study is now  differential equations. So please sorry if I made stupid mistakes, but I think that the function $ y^2(t)+\frac{(y'(t))^2}{g(t)}$ has negative derivative, thus is it decreasing and the conclusion follows.

nice proof!
what if we only suppose that $g(x)>a>0$ for all $x$, $g$ only continuous. I tried this morning to show that the same conclusion holds but i got nowhere. i don't even know if it is true.

in fact I don't think the conclusion holds with that weaker condition. I made some experimentations with Mapple and ploted solutions that seem to be unbounded but I don't know if it is numerical errors or not  :(


trigonometry
limit
calculus
calculus computations

Evaluate

\[\lim_{x\rightarrow +0} \left(\frac{1}{x^n}-\frac{1}{\sin ^ n x}\right)\ (n=1,2,3,\cdots)\]

[quote="zhaobin"]I think the answer is $- \infty$ when $n \geq 3$,
$0, n=1$
$-1/6, n=2$
right?[/quote]

Close (I think)

Just double check n=2

[quote="zhaobin"]I think the answer is $- \infty$ when $n \geq 3$,
$0, n=1$
$-1/6, n=2$
right?[/quote]

i'm new to calculus. can you explain more clearly the result?

just $\lim_{x\rightarrow +0} \left(\frac{1}{x^n}-\frac{1}{\sin ^ n x}\right)=\lim_{x\rightarrow +0} \left(\frac{sin^nx-x^n}{x^n\sin ^ n x}\right)$
and use $sinx=x-\frac{x^3}{6}+O(x^5),x \to +0$
 :) is it okay?

Can we compute $\lim_{x \to 0} \frac{\sin x - x}{x^3}$ without using l'Hospital / Taylor series?


A bug starts at one vertex of a cube and moves along the edges of the cube accoring to the following rule.  At each vertex the bug will choose to travel along one of the three edges emanting fromt hat vertex.  Each edge has an equal probability of being chosen, and all the choices are independant.  What is the probability that after seven moves, the bug will have visited every vertex exactly.

$ A\: \frac{1}{2187}\quad B \: \frac{1}{729} \quad C \: \frac{2}{243} \quad D \: \frac{1}{81}\quad E \: \frac{5}{243}$

Looking for this?
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2006&p=423925