calculus
integration
logarithms
trigonometry
calculus computations

i was trying to prove that $ \int_0^{\frac\pi2}\ln(\sin x)\,dx < \infty.$

i have for $ 0\le x\le\frac\pi2$ is $ \sin x\le x,$ and takin' logs from there, but this doesn't work, since the integrand is negative for $ 0 <x\le\frac\pi2,$ so comparison test doesn't apply.

does anyone have a better approach?

$ I\equal{}\int_{0}^{\frac{\pi }{2}}\ln (\sin x)\,dx\equal{}\allowbreak \int_{0}^{\frac{\pi }{2}}\ln \left( \cos x\right) \,dx$

$ 2I\equal{}\int_{0}^{\frac{\pi }{2}}\ln (\sin 2x)\,dx\minus{}\frac{\pi }{2}\ln 2$

$ \int_{0}^{\frac{\pi }{2}}\ln (\sin 2x)\,dx\equal{}\allowbreak \int_{\minus{}\frac{1}{2}\pi }^{\frac{1}{2}\pi }\frac{1}{2}\ln \left( \cos x\right) \,dx\equal{}\int_{0}^{\frac{1}{2}\pi }\ln \left( \cos x\right) \,dx\equal{}I$

Hence $ 2I\equal{}I\minus{}\frac{\pi }{2}\ln 2$

Therefore $ I\equal{}\minus{}\frac{\pi }{2}\ln 2$

That's the computation, but it's only valid if you can prove that $ I$ is finite, so we still need what Null asked for: the estimate that says that it's finite.

By the way, since $ \ln(\sin x)<0$ on $ (0,\pi/2),$ what we need to prove shouldn't be phrased as asking that the integral is $ <\infty.$ Instead, we should say that the integral is $ >\minus{}\infty.$

Note that there is some $ a>0$ such that $ \sin x>ax$ for $ 0<x<\frac{\pi}2.$ (It's not hard to get a specific value for $ a$ but we don't really need that.) Then:

$ \int_0^{\frac{\pi}2}\ln(\sin x)\,dx>\int_0^{\frac{\pi}2}\ln(ax)\,dx$

and we can specifically compute that integral and show that it is finite.

Another argument. Let $ 0 < a < \frac{\pi}{2}$. Then $ \int_0^{\pi /2} \ln(\sin x)\,dx \equal{} \int_0^a \ln(\sin x)\,dx \plus{} \int_a^{\pi /2} \ln(\sin x)\,dx$. The second integral on the right is not improper. For the first, consider the change of variables $ \sin x \equal{} u$ and let $ b \equal{} \sin a$: we get

$ \int_0^b \frac{\ln u}{\sqrt{1 \minus{} u^2}}\,du$,

and the problem is near zero. But what can you say about $ \frac{1}{\sqrt{1 \minus{} u^2}}$ for $ u \approx 0$?

[hide]We have $ \frac{1}{\sqrt{1 \minus{} u^2}} \equal{} 1 \plus{} \frac{u^2}{2} \plus{} O(u^4)$. Now what is the nature of $ \int_0^a x^k \ln x\,dx$, for $ k \geq 0$?[/hide]


[quote="Crissy"]Are you looking for the coefficient of any particular term?[/quote]

sorry, ive changed the question, it should be the coefficient of $ x^{16}$

The coefficient of $ x^{16}$ in

the first 1.)

$ \frac{10!}{7!0!2!1!}+\frac{10!}{6!2!0!2!}+\frac{10!}{5!3!2!0!}= 2^{2}\cdot 3^{2}\cdot 5 \cdot 23$

[hide="Solution to the first 1.)"]
More generally, the expansion of $ (1+x^{2}+x^{5}+x^{6})^{10}$ is the sum of all the terms of the form $ \frac{10!}{a!b!c!d!}1^{a}x^{2b}x^{5c}x^{6d}$, where $ a,b,c,d$ range over all possible sets of non-negative integers satisfying $ a+b+c+d = 10$.  We desire also that $ 2b+5c+6d = 16$.   Fix a value of $ a$, then $ b$, thus solving the 2x2 system in $ c$ and $ d$.
[/hide]

the second 1.)

$ \frac{10!}{7!1!1!1!}+\frac{10!}{5!4!1!0!}= 2^{2}\cdot 3^{2}\cdot 5 \cdot 11$

[hide="Solution to the second 1.)"]
Same as the first but we need to solve with $ 2b+5c+6d+3 = 16$.
[/hide]

[quote="Crissy"]The coefficient of $ x^{16}$ in

the first 1.)

$ \frac{10!}{7!0!2!1!}+\frac{10!}{6!2!0!2!}+\frac{10!}{5!3!2!0!}= 2^{2}\cdot 3^{2}\cdot 5 \cdot 23$

[hide="Solution to the first 1.)"]
More generally, the expansion of $ (1+x^{2}+x^{5}+x^{6})^{10}$ is the sum of all the terms of the form $ \frac{10!}{a!b!c!d!}1^{a}x^{2b}x^{5c}x^{6d}$, where $ a,b,c,d$ range over all possible sets of non-negative integers satisfying $ a+b+c+d = 10$.  We desire also that $ 2b+5c+6d = 16$.   Fix a value of $ a$, then $ b$, thus solving the 2x2 system in $ c$ and $ d$.
[/hide]

$ \frac{10!}{7!1!1!1!}+\frac{10!}{5!4!1!0!}= 2^{2}\cdot 3^{2}\cdot 5 \cdot 11$

[hide="Solution to the second 1.)"]
Same as the first but we need to solve with $ 2b+5c+6d+3 = 16$.
[/hide][/quote]

could you please explain further, sorry i already forgot this stuff

First, let me apologize, my solution is wrong (I am missing two terms in the first (1)), but the method is correct.  Thank you for being polite.  I see that you have been working with binomials.  This is very similiar, but is under the guise of binomials and multiplication principle.  Here is a rather lengthy explanation.


superior algebra
superior algebra solved

How do you prove that [b]R[/b][x]/<$x^2 + 1$> is a field.

By proving that $x^2+1$ is irreducible, or equivalently, the ideal generated by $x^2+1$ is maximal.


calculus
function
inequalities
calculus computations

Please help me with this exercise...
function f:R  R and |f(x)-e^x|<=x^2.Both f' and f'' exist in R and f'' continuous at 0.Prove that:f''(0)<=3.
I am not sure about the terms in english.With f'' continuous at 0 i mean that limf(x)=f(0)
                  x
:arrow 0

 :maybe:

[quote="Brambila"]This is not difficult. If you prove that g(x)=|f(x)-e^x| is differentiable you just differentiate your inequality and you can open the modulo to avoid any problem with that and you will get the answer very easily[/quote]
Warning, $g(x)=|f(x)-e^{x}|$ is not allowes differentiable . Of course $h(x)=f(x)-e^{x}$ is differentiable .. There is $-x^{2}\leq h(x)\leq x^{2}$
In general you can't differentiate an inequality !!
You can write $h(x)=h(0)+xh'(0)+\frac{x^{2}}2 h''(c(x))$ , with $|c(x)| \leq |x|$ .. and so on ..

:cool:


group theory
abstract algebra
number theory
least common multiple
linear algebra
matrix
superior algebra

Let $ SO(\mathbb Q) \equal{} \{A\in \mathcal M_n(\mathbb C)|A^TA \equal{} I,\det A \equal{} 1\}$. Prove that for each $ n\geq2$, the multiplicative group $ SO_n(\mathbb Q)$ is not finitely generated.

Here is an idea.
Any finetely-generated subgroup of $ SO_n(\mathbb{Q})$ is contained in $ SO_n(\mathbb{Z}[1/k])$ where $ k$ is the lcm of he denominators of the matrix entries of the generators and their inverses.


int x^m*P_n(x) dx=0 where integration is from (-1) to (+1).Given m<n

I took integrand F(x) and saw that F(-x)=(-1)^(m+n)*F(x)
should that help anyway?

Here is a useful tip for the integration of legendre polynomials:
$(2n+1)P_{n}=\frac{d}{dx}(P_{n+1}-P_{n-1})$


group theory
abstract algebra
superior algebra
superior algebra unsolved

A normal subgroup H of a group G is said to be a direct factor if there exists a (normal) subgroup K of G such that $ G \equal{} H \times K$.
Prove: If H is a direct factor of K and K is a direct factor of G, then H is normal in G. (Hungerford I.8.12)

Attempted solution:

Let $ K \equal{} H \times I$ and $ G \equal{} K \times J$. where I is normal in K and J is normal in G.

Now, by the previous exercise in Hungerford $ K \equal{} H \times I$ implies that hi=ih for all $ i \in I$ and $ h \in H$ and that every nonidentity element of K can be written uniquely as h,i,or ih where $ e \neq i \in I$ and $ e \neq h \in H$. And we can say analogous things for the product $ G \equal{} K \times J$.

Now, fix $ g \in G$. Then $ h^{\minus{}1} g h \equal{} h^{\minus{}1} k j h  \equal{} ?$ and this is pretty much where I am stuck. We could insert h h^{-1} somewhere or write k = h' i but I don't see how to deal with that j, which we basically know nothing about...

i think the solution is much simpler, if we can show $ H$ is normal in $ G$ then we have the following:

$ G \equal{} K \times J$ and $ K \equal{} H \times I$, so we have that $ G \equal{} H \times I \times J$.

Since the direct product of normal subgroups is normal, we take $ N \equal{} I \times J$. Then $ G \equal{} H \times N$ where $ N$ is normal, hence $ H$ is a direct product.

[hide="to show $ H$ is normal in $ G$."]
as you said, ever nonidentity element of $ G$ satisfies $ kj\equal{}jk$, particularly for $ k \equal{} h$. so $ h \equal{} jhj^{\minus{}1}$.[/hide]

[quote="shallow"]
$ G \equal{} K \times J$ and $ K \equal{} H \times I$, so we have that $ G \equal{} H \times I \times J$.

[/quote]

I was going to use that but I convinced myself it (i.e. associativity of direct products) wasn't true for some reason but now that I think about it follows immediately from the definition of a direct product.

[quote="shallow"]
Since the direct product of normal subgroups is normal, we take $ N \equal{} I \times J$. [/quote]

I am kind of confused about this. You are saying that N is normal because it is a direct product of normal subgroups? Can you be more precise? I haven't seen a theorem like that... Also remember I is a normal subgroup of K, not G...

i think i'm oversimplifying the problem, taking direct product might not be associative (found two different sources one claiming each, *edit* personally, i think it is, what makes you think otherwise?).

and you're right about $ I$, i forgot it was a subgp of $ K$, but you can show that it is normal in the same way that $ H$ is normal.

as for the last part about $ N$, this is true. i think what necessary is an isomorphism between $ IJ$ and $ I\times J$ where we know the former is normal, i believe the mapping

$ \phi: IJ \longrightarrow I\times J$
$ \phi(ij) \longmapsto (i,j)$

although i've made so many mistakes already, i'm doubting this as well


AwesomeMath
summer program
number theory

We are very fortunate to have Dr. Zuming Feng present to the Metroplex Math Circle this coming Saturday.

Dr. Zuming Feng, Phillips Exeter Academy

Zuming Feng received his Ph.D. from Johns Hopkins University with emphasis on Algebraic Number Theory and Elliptic Curves. He teaches at Phillips Exeter Academy. Zuming also served as a coach of the USA IMO team (1997-2006), was the deputy leader of the USA IMO Team (2000-2002), and an assistant director of the USA Mathematical Olympiad Summer Program (1999-2002). He has been a member of the USA Mathematical Olympiad Committee since 1999, and has been the leader of the USA IMO team and the academic director of the USA Mathematical Olympiad Summer Program since 2003. Zuming is also co-founder and academic director of the AwesomeMath Summer Program (AMSP) since 2006. He received the Edyth May Sliffe Award for Distinguished High School Mathematics Teaching from the MAA in 1996 and 2002.

http://metroplexmathcircle.wordpress.com/2009/02/02/february-7-2009-dr-zuming-feng/

Um, shouldn't this be in the Texas forum?


inequalities
calculus
derivative
inequalities proposed

The following are stronger:

$\sqrt\frac{1+a}{2}+\sqrt\frac{1+b}{2}+\sqrt\frac{1+c}{2} \ge 3$
and
$\sqrt\frac{1+2a}{3}+\sqrt\frac{1+2b}{3}+\sqrt\frac{1+2c}{3} \ge 3$.

Of course, the solutions all of them are the same, even the more general :

Find the best constant $k$ st:

$\sqrt{k+a}+\sqrt{k+b}+\sqrt{k+c} \ge 3\sqrt{k+1}$

Which can be solved the same.

But I Choose the first because It's more becautiful : ;)

[quote="hungkhtn"]
Find the best constant $k$ st:

$\sqrt{k+a}+\sqrt{k+b}+\sqrt{k+c} \ge 3\sqrt{k+1}$

Which can be solved the same.

But I Choose the first because It's more becautiful : ;)[/quote]
But the first may be solved by AM-GM ineq., because
$(3+a)(3+b)(3+c) \geq 64$.  :lol:

[quote="campos"]in the same conditions, does this inequality holds:

$9(a+b+c)+abc\geq 28$ ?[/quote]

I think so. :lol:

in my post i asked if that inequality can be proven... if it's true, then hungkthn's inequality can be proved by am-gm, as vasc said.

[quote="campos"]$9(a+b+c)+abc\geq 28$ ?[/quote]
<=>
$81(a^2+b^2+c^2+6)\geq (28-abc)^2$
<=>$81(\sum a^2)+56abc\geq 298+a^2b^2c^2$
Using the following ineq.:
$a^2+b^2+c^2+2abc+1\geq 2(ab+bc+ca)$(Schur's Ineq) and AM-GM we are done!

[quote="zhaobin"][quote="pvthuan"][quote="hungkhtn"]Prove that if $ab+bc+ca=3$ then :

$\sqrt{3+a}+\sqrt{3+b}+\sqrt{3+c} \ge 6$[/quote]
One year ago, I created a similar one:

$\text{Let}\; x,y,z \; \text{be positive real numbers such that }\; xy+yz+zx=3,\;\text{prove that}$

$\sqrt{1+x}+\sqrt{1+y}+\sqrt{1+z}\geq3\sqrt2.$

We have a long proof using calculus. Have you got a solution without derivative?[/quote]
I think my solution also using calculus.
and here it is:
let $\sqrt 2a=\sqrt{1+x}$,and $b,c$ similar:
we have $a,b,c \ge \frac{\sqrt 2}2$
it is easy to see the inequality is equal to when:
$a+b+c=3,a,b,c \ge \frac{\sqrt 2}2$
then:
$\sum_{cyc}(2a^2-1)(2b^2-1) \le 3$   $(1)$

$(1)$ is equal to $a^2b^2+b^2c^2+c^2a^2 \le a^2+b^2+c^2$
see the solution in
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=68165]http://www.mathlinks.ro/Forum/viewtopic.php?t=68165[/url]
I think we can find a solution.[/quote]

THis solution is really the same as mine. This method can be use to solve the more difficult of Vasc

$(2-ab)(2-bc)(2-ca) \le 1$ if $a^2+b^2+c^2=3$

I think you reversed the sign,hungkhtn.Take $a=b=\frac{1}{\sqrt2},c=\sqrt2$

Yes, it should be

$(2-ab)(2-bc)(2-ca) \ge 1$


Hi there,
Could all of us Mac users please get some help with latex?
Thanks,
silvercharcoal5

Are you wondering how to install latex or use latex?

There's mactex available. Google is your friend. :)

Hi silvercharcoal5,

welcome to the board!
Have a look here for software: [url=http://texblog.net/latex-link-archive/distribution-editor-viewer/mac-osx/]Distributions, Editors and Viewer for Mac OS X[/url]. I recommend MacTeX and TeXShop.
Further you could look at [url=http://tug.org/mactex/faq/]MacTeX Frequently Asked Questions[/url].

Stefan

Thanks guys,
I'm more concerned on exactly how to use Mactex. I would really like to be able to create my own code without having to consult the website and copy+paste.
-silvercharcoal5

If you're wondering how to actually use $ \text{\LaTeX}$, [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]AoPS has its own excellent guide[/url].


A goods train left Moscow at x hrs y mins and arrived in Saratov at y hrs z mins. The journey took z hrs x mins. Find all possible values of x.

I have a start of a solution.
[hide]
Represent everything in minutes and make an equation,
$ 60x\plus{}y\minus{}(60y\plus{}z)\equal{}60z\plus{}x$
$ 59x\minus{}59y\equal{}61z$
As long as x and y are integers, taking the LHS mod 59 gives a residue of 0.
So, we need to find two numbers x and y such that 59x-59y is divisible by 61. In other words,
$ (59)(x\minus{}y)$ must be divisible by 61 and in more other words, x-y must be divisible by 61 (59 is prime). But, if x and y have to be less than 60, (they're used as minutes), how is this possible?
[/hide]

(the question is inside the solution)

I did the same way you haev, except putting 60y+z-60x-y
therefore obtaining 61(z+x)=59y
Thus 61|y, and y can only be 0.
x,y,z have to be under 24 too, since hours are used.
Therefore, I conclude the train had traveled at warp speed, and x=y=z=0.
Damn the russians have good technology...

[quote="jackfeng6"]I did the same way you haev, except putting 60y+z-60x-y
therefore obtaining 61(z+x)=59y
Thus 61|y, and y can only be 0.
x,y,z have to be under 24 too, since hours are used.
Therefore, I conclude the train had traveled at warp speed, and x=y=z=0.
- the russians have good technology...[/quote]
No solutions... That's what I thought :D

http://web.archive.org/web/20040405010511/www.kalva.demon.co.uk/russian/rus95.html
^ there's the link

Thanks

Suppose $ x > y$ and it went past midnight/noon.

Then $ 60(y \plus{} 12) \plus{} z \minus{} 60x \minus{} y \equal{} 60z \plus{} x\Rightarrow59y \minus{} 59z \plus{} 720 \equal{} 61x$

Mod 59 this gives $ 2x\equiv 12\mod 59\Rightarrow x\equiv 6\mod 59$

So let $ x \equal{} 6$.  Then $ 59(y \minus{} z) \plus{} 720 \equal{} 61(6)\Rightarrow 59(y \minus{} z) \equal{} \minus{} 354\Rightarrow z \minus{} y \equal{} 6$.

We can see that if the train left at 6:01 am and got there at 1:07 pm, then it took 7 hours and 6 minutes, as required.

Now we try it in 24-hr time.

We have $ 60(y \plus{} 24) \plus{} z \minus{} 60x \minus{} y \equal{} 60z \plus{} x\Rightarrow 59y \minus{} 59z \plus{} 1440 \equal{} 61x$

Mod 59 this gives $ 2x\equiv 24\mod 59\Rightarrow x\equiv 12\mod 59$

Since it is 24 hr time, $ x$ must be 12.  This gives $ 59(y \minus{} z) \plus{} 1440 \equal{} 61(12)\Rightarrow y \minus{} z \equal{} \minus{} 12\Rightarrow z \minus{} y \equal{} 12$

So starting at 12:05 and going until 5:17 the next day is a 17 hour 12 minute journey.

Thus if 24-hr time is required, the only value for $ x$ is 12.  If 12-hr or 24 hr time is allowed, then 6 and 12 both work.

But, what if it doesn't refer to the next day?  Maybe 2 days in the future?  I'll leave this case to you guys.


Τρεις πράξεις μόνο επιτρέπονται:

1) Διαλέγεις ένα πολυώνυμο στην τύχη.
2) Εναλλάσεις το συντελεστή του μεγιστοβάθμιου όρου με το σταθερό όρο.
3) Θέτεις όπου χ το χ+τ όπου τ πραγμαικός αριθμός.

Έστω ότι διαλέξαμε το πολυώνυμο χ^2-χ-2. Ποιος είναι ο μικρότερος αριθμός πράξεων (μεταξύ των 2, 3) που απαιτείται ώστε να το μετατρέψουμε στο χ^2-χ-1?

loipon oraio problimataki etsi protonoie :wink:
loipon paratiriste oti oi 2 praxeis den metaballoun tin diakrinousa tou trionimou!!!!! :o
simperasma epeidi exoume  arxikasto 1o polionimo diakrinousa 9 ki sto teliko polionimo diakrinousa 5 einai adinato me tis epitreptes praxis na kanoume to zitoumeno :D
xaireto olous sas


Ring Theory
quadratics
algebra
function
domain
superior algebra
superior algebra unsolved

I'm trying to learn how to factor ideals by doing this problem. Please help if you know how, thanks : /

Suppose the ring of integers O of a quadratic field K is a Principal Ideal Domain (PID). Let p be a prime in Z. Factor pO as a factor of prime ideals.

Now let K be Q(sqrt(-6)), factor 3O as product of prime ideals.

Does it help to know that in a PID, factoring $ xO$ into prime ideals is exactly the same as factoring $ x$ into a product of prime elements of $ O$?

sorry, but what would factoring the ideal look like?
how could i be sure after righting pO as a product of ideals that these ideals are actually prime?

$ (ab)O\equal{}(aO)(bO)$.  Furthermore, if $ p$ is a prime element in $ O$, then $ pO$ is a prime ideal, almost by definition.  (consider what it means for a product of elements to be in $ pO$)  These two facts mean that factoring ideals in a PID is the same, up to units, as factoring elements.

This doesn't fully answer the question, of course, as the ring of units in $ \mathbb{Q}(\sqrt{\minus{}6})$ is not a PID, but it should alleviate some confusion on what is meant by "factorization of ideals".

So now I'm trying to factor 3 in Z[sqrt(-6)] which is not a UFD I know.  But I cannot factor it, is 3 irreducible in Z[-6]?

Yes, $ 3$ is irreducible in $ \mathbb{Z}[\sqrt{\minus{}6}]$.  Can you prove it?

But in order to finish the problem, you need to answer a more difficult question: is $ 3O$ a prime ideal?  I believe that it's the same question as: is $ 3$ prime in $ O$?

By the way, you can show that $ \mathbb{Z}[\sqrt{\minus{}6}]$ is in fact the ring of integers in $ \mathbb{Q}(\sqrt{\minus{}6})$, right?

suppose 3=ab

N(3)=9=N(ab)=N(a)N(b)

Suppose N(a)=3

a=c+f*(sqrt(-6))
thus N(a)= c^2 + 6*f^2=3 which is impossible.

same thing for b, thus either N(a) or N(b) is 1 which implies that either a or b is a unit and 3 is irreducible.

how do i know if 3 is prime though? Z[(sqrt(-6)] is Noetherian.  so I know everything can be factored into irreducibles.

thus i think i need to show that if  3|ab then 3|a or 3|b.

(also, i've already shown that if d is not 2,3 mod 4, then Q(sqrt(d)) has ring of integers Z(sqrt(-6)).)

[quote="Keno20"]thus i think i need to show that if  3|ab then 3|a or 3|b.[/quote]

In fact, this is the definition of a prime.

If you can show that, then you are done.  If it's false, then the counterexample will tell you how to factor $ 3O$.

By the way, $ \minus{}6$ [i]is[/i] 2 mod 4, so I'm a little confused about your last comment.

sry. I meant to say that if d is 2,3 mod 4 then Z[sqrt(-6)] is the ring of integers.

15= (3+sqrt(-6))(3-sqrt(-6))

3|15, but 3 does not divide either of these.

So far you're on the right track.

I said something misleading before, as I'm not used to thinking about factorization of ideals.  Even though $ (3)$ is not prime (as you showed) it is possible that $ (3)$ does not factor as a product of prime ideals.  I think I've worked out the details, (though I won't give them all away, especially if this is homework) and looking at the alternate factorizations for 15 that you've written down seems to be an appropriate way to analyze whether this is the case.


I dont think you should post it here,let post it in unsolved forum :wink:
withyour problem,we only let $ a \plus{} b \plus{} c \equal{} 3$,and some small comment
FOr a example $ \frac{1}{2a^2\minus{}6a\plus{}9}\le \frac{1}{5}\minus{}\frac{2(a\minus{}1)}{25}$,we will easyly prove it
I remember,it is a Japan MO 2002

set a+b+c=1 anh then easy to use jensen or chebysev! :oops:

[quote="Allnames"]Can you detail it,i need the proof which be used Jensen :lol:[/quote]
Oh , I think Jensen is useful . It 's here
http://www.mathlinks.ro/viewtopic.php?t=28678


linear algebra
matrix
vector
analytic geometry
probability and stats

Afternoon all,

I'm looking for a solution to the following problem.

I have a Nx3 matrix, called "XYZ" where the columns represent the x,y, and z axes and the individual row vectors are coordinates in a 3 dimensional space.  I'm looking for a way determine the least squares fit plane for that 3D dataset.  The tricky thing that I'm hung up on is that the resulting plane must contain a reference point.

Hopefully, a worked solution exists so that I can model it in Octave or Matlab.

I apologize if this is the wrong subforum, as I'm new and unsure where this topic belongs.  Please move accordingly.

Thank you for your time,
Mike.

Hello [b]mj-barton[/b]!

Not sure whether I understand your problem properly (hint: it's a good idea to give not only desctiption, but also some example at what is the input of your problem and what output do you want). Have a look at Sostools package ([url=http://www.cds.caltech.edu/sostools/]link[/url]) for Matlab. Is this what you are looking for?

Sergey Markelov


[color=darkblue]Fie o multime finita $ M$ de numere reale nenule cu $ n$ elemente. Exista o functie injectiva $ f\ : \ M\ \rightarrow\ M$  care verifica relatia $ f(x)\plus{}f^{\minus{}1}(x)\equal{}x$  pentru orice $ x\in M$ . Sa se arate ca $ n$ este multiplu de $ 6$ si sa se arate ca ipoteza este consistenta, adica sa se construiasca un exemplu de functie cu proprietatea mentionata in ipoteza.[/color]

Am gasit o functie, la partea cu divizibilitatea inca mai lucrez.
Luam $ M\equal{}\left\{\minus{}3,\minus{}2,\minus{}1,1,2,3\right\}$ si $ f(\minus{}3)\equal{}\minus{}2,f(\minus{}2)\equal{}1,f(\minus{}1)\equal{}\minus{}3,f(1)\equal{}3,f(2)\equal{}\minus{}1,f(3)\equal{}2$.


integration
function
inequalities
algebra
binomial theorem
combinatorics proposed
combinatorics

[quote="Rafal"]ok, try to proove combinatorically my favourite one :) :

$\sum_{i=0}^{n}\frac{(-1)^i}{2i+1}{n \choose i}=\frac{(2n)!!}{(2n+1)!!}$

[color=red]Edited by Megus: next time post problems in new topics - this post was taken from http://www.mathlinks.ro/Forum/viewtopic.php?t=41694&sid=4bdc912db5da86feabefc93dcb2eccfe [/color][/quote]

What is $n!!$? I used to know what it is, but I forgot!!! :(

Remark $\frac{1}{2i+1}=\int_0^1 x^{2i} dx$ and use binomial theorem

$\sum_{i=0}^{n}\frac{(-1)^i}{2i+1}{n \choose i}=\sum_{i=0}^n \int_0^1 x^{2i}(-1)^i {n\choose i} dx$

$=\int_0^1 \sum_{i=0}^n {n\choose i} (-x^2)^i dx=\int_0^1 (1-x^2)^n dx$

which follows [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=38551[/url]

kunny

Didn't Rafal said "prove combinatorically"? The proof doesn't seem to be combinatorical. :D

What an ugly function! I don't like this !! thingy. :(

Well actually it can be faster obtained $\int_0^1 (1-x^2)^n dx=\int_0^{\frac{\pi}{2}} sin^{2n+1}dt=\frac{(2n)!!}{(2n+1)!!}$ (taking $x=cost$), where the last inequality is from the recurence :
$\int sin^nx dx=-\frac{1}{n}sin^{n-1}cosx+\frac{n-1}{n}\int sin^{n-2}xdx$, where $\int_0^{\frac{\pi}{2}} sinx dx=1$.


Let ABCD be a convex quadrilateral such that
(1) AD = AC and AB = BC ;
(2) DAC = 20 and ABC = 100.
Show that AD = BC + CD.

I love this kind of problems (with 20, 40, 80 degree angles, where you need some constructions to help you solve them). Here's my soln:

Let E be on AD s.t AE=AB=BC. Since angle BAE=40+20=60 it means that the triangle ABE is equilateral, so AB=AE=BE=BC. EBC=100-60=40, so BCE=(180-40)/2=70 deg. Since BCA=40 deg, we get ACE=30 deg. ECD=ACD-ACE=80-30=50 deg. Because ADC=EDC=80 deg we get DEC=ECD=50, so the triangle EDC is isosceles, so ED=DC. This means that AD=AE+ED=AB+DC=BC+CD, what we wanted.

Yep, I love this one also ... I'm proud of this one :mrgreen:.
I found already FIVE different solutions !

I also added another selfmade problem in the intermediate section.

[hide]My other solutions are :

(1) Purely trigonometric  (SimPson and stuff like that) ;

(2) Cosine rule ;

(3) Notice that ABCD is cyclic and prove that sin20 + sin40 = sin80 ;

(4) Take a point E and DC such that CE = CB ;

(5) Sth with circles (too complicated in fact for this simple problem.   [/hide]

Another solution:
Extend $(BC$ with $CE=AC-BC\ (\ 1\ )\ (\text i.e. BE=AC)$; we now know that $\angle CEA=30^\circ$, i.e. $AE\bot CD$, but $\triangle CDE$ is isosceles (due to symmetry) and has $\angle DCE=60^\circ$, so $CE=CD$; with $(1)$ we are done.

Best regards,
sunken rock


Let $\mathbb N_{0}$ be the set of non-negative integers. Suppose that the function $f: \mathbb N_{0}\to\mathbb N_{0}$ satisfy
\[f(m+n)-f(m)-f(n)=0\text{ or }1 \]
for all $m,n\in\mathbb N_{0}$ and $f(8888)=2222$.
(1) Prove that $k-1\geq f(\sum_{i=1}^{k}n_{i})-\sum_{i=1}^{k}f(n_{i})\geq 0$.
(2) Find $f(4)$.
(3) Find $f(2006)$.

This is almost exactly the same as IMO 1982, #1.


induction
strong induction
number theory unsolved
number theory

Does there exist an integer power of two so that by changing the order of its digits you could get another power of two? (Remember, the power of two we start with might have a few zeros in its decimal representation, therefore these two powers of two may not have the same number of digits.)

[hide="start of proof I was typing before I realized it did not help..."]Unless it fails in some mod, it seems like we need to rigorously define switching two numbers: \[\text{Let }2^{a}=\underline{a_{k}}...\underline{a_{3}}\ \underline{a_{2}}\ \underline{a_{1}}\text{ where each }a_{i}\text{ is a digit in base 10, and }a_{k}\not= 0\] To switch $a_{i}$ and $a_{j}$, we add $10^{i}(a_{j}-a_{i})+10^{j}(a_{i}-a_{j})=(a_{j}-a_{i})(10^{i}-10^{j})$ to our number, $2^{a}$.
After performing all the switches we want, we have only changed $2^{a}$ by a multiple of $9$. By looking at the number of digits we have to switch around, the largest possible power of $2$ we can make is $2^{a+3}$. Therefore, the only possible powers of $2$ that are larger are $2^{a+1},2^{a+2},2^{a+3}$. Subtracting $2^{a}$ from each of these, we find that the amount of change needed is never a multiple of $9$. Therefore, we cannot switch around digits to make a larger power.
Now we are ready to try to prove this by induction.[/hide]
Now I am thinking more of strong induction: assume no power below $2^{n}$ can switch into any other power. Then prove that $2^{n}$ can't switch into another power.
Or a slightly different approach:
Assume no power below $2^{n+1}$ can switch into any other power below $2^{n+1}$. Then prove that no power below $2^{n+2}$ can switch into any other power below $2^{n+2}$. (Base case is easy for both of these)

[quote="K81o7"]Well, the best I can do is prove that one power of 2 is the other times a power of $2^{6}$ by considering each mod 9...but that's it...[/quote]

This is useful.  It means that at the least, we need at least one $0$ (and if we start with an initial digit of anything other than $1$ we need at least two).  Now, the distribution of $0$s is difficult to capture, but they are not too common.

Can we show that there must exist a power of $2$ with an arbitrary number of $0$s in its decimal representation?

Even so, it is not enough.  It seems as if there may be an answer, but that it will be large.

[quote="pbornsztein"]To be honest, the original problem from the Tounament of the Towns asked for two powers of $2$ with the same number of digits, which made the problem not so difficult.

Pierre.[/quote]
This problem is from the Tournament of Towns year 1988 and it does not say that the number of digits has to be the same; however in the solution it solves it only for powers with the same number of digits.

Well two such powers of two cannot have the same number of digits...  They are congruent mod 9, and are both units mod 9.  2 has order 6 when taken mod 9, so their quotient is a power of 64...


calculus
integration
function
logarithms
calculus computations

$ \int{\frac {1}{x + \sqrt {x^2 - x + 1}}dx}$
my code is different from yours

hello, setting $ \sqrt{x^2\minus{}x\plus{}1}\equal{}t\plus{}x$ we get $ x\equal{}\frac{1\minus{}t^2}{2t\plus{}1}$ and $ dx\equal{}\minus{}2\frac{(t^2\plus{}t\plus{}1)}{(2t\plus{}1)^2}\,dt$ and our integral is$ \minus{}2\int\frac{t^2\plus{}t\plus{}1}{(2t\plus{}1)(2\plus{}t)}\,dt$. This gives $ t\plus{}2\ln|2\plus{}t|\minus{}\frac{1}{2}\ln|2t\plus{}1|\plus{}C$.
Sonnhard.


An ordinary pack of cards is shuffled, and 2 cards are dealt face up. Find the probablity that at least one of these is a spade.

There are $ 13$ spade cards, and there are $ 52$ cards total. So there are $ 49\cdot48$ ways to chose something that is not a spade, so the probability that at least one of them is a spade is $ \frac {52\cdot51 - 49\cdot48}{52\cdot51} = \frac {300}{2652} = \frac {25}{221}$... I think....  :maybe:

Edit: Ah, fail..... Good job Rofl..

Incorrect.

The probability that we get at least one spade is one minus the probability that neither are spades. That probability is $ \frac{39}{52}\cdot\frac{38}{51}\equal{}\frac{19}{34}.$ The final answer is $ 1\minus{}\frac{19}{34}\equal{}\boxed{\frac{15}{34}}.$


Let $F_{n}$ denote the $n^{th}$ Fibonacci number. Prove that if $P>5$ is a prime, then $F_{p}=$legendre symbol p/5 $(mod p)$
<=> if $p=1,4(mod 5)\to F_{p}=1(mod p)$ , $p=2,3(mod 5) \to F_{p}=-1(mod p)$

Hint: Use Binet's formular and Binomial theorem


algebra
polynomial
floor function
LaTeX
modular arithmetic
trigonometry
linear algebra

I obtained $P(3n+1)=(-1)^{\lfloor\frac{n}{2}\rfloor}\sqrt{3}^{3n}+1$ for $n$ even, and $P(3n+1)=(-1)^{\lfloor\frac{n}{2}\rfloor}\sqrt{3}^{(3n+1)}+1$ for $n$ odd.  Therefore, $n=4$. Is this correct? If so, I'll post the solution.

Masoud Zargar

[quote="boxedexe"] $P(3n+1)=\sqrt{3}^{3n}+1$ for $n$ even, and $P(3n+1)=(-1)^{\lfloor\frac{n}{2}\rfloor}\sqrt{3}^{(3n+1)}+1$ for $n$ odd.
Masoud Zargar[/quote]

How did you come up with that??

EDIT: Sorry boxedexe i didnt read that you said you'd post the solution if it was correct..

[quote="boxedexe"]I obtained $P(3n+1)=\sqrt{3}^{3n}+1$ for $n$ even, and $P(3n+1)=(-1)^{\lfloor\frac{n}{2}\rfloor}\sqrt{3}^{(3n+1)}+1$ for $n$ odd.  Therefore, $n=4$. Is this correct? If so, I'll post the solution.

Masoud Zargar[/quote]
Yes, it is correct. Please post your solution.

I'll post the solution tomorrow since I have to go do some other work, and I don't feel like typing the LaTeX now.  Please be patient. :)

although it is nice to come up with the general form, but its not needed.
take a look here.

http://www.kalva.demon.co.uk/usa/usoln/usol845.html

There is another solution that involves "guessing" the polynomial using roots of unity. I will post that one if boxeded's is different.

[b]Solution[/b] We will consider the third root of unity, that is, $\zeta=\mathrm{e}^{i\frac{2\pi}{3}}$.  Let $Q(x)=P(x)-1$.  Since $\zeta^{p}=\zeta^{p\pmod 3}$, the third roots of unity are periodic with period 3.  Due to this periodicity,
\[\\ Q(m)=\frac{2\sqrt{3}}{3}\Im((\mathrm{e}^{i\frac{2\pi}{3}})^{2m+1}). \]
By [i]Newton's Binomial Theorem[/i], the fact that $\deg Q=\deg P$, and $1+\zeta+\zeta^{2}=0$, we have the equivalent equality
\[Q(m)=\frac{2\sqrt{3}}{3}\Im(\zeta\sum_{k=0}^{3n}\binom{m}{k}(-1)^{k}(\zeta+2)^{k})\\ \implies Q(3n+1) =\frac{2\sqrt{3}}{3}\Im(\zeta((-1-\zeta)^{3n+1}-(-\zeta-2)^{3n+1}))\\ =-\frac{2\sqrt{3}}{3}\Im(i^{3n+1}\zeta^{3n+2}\sqrt{3}^{3n+1})\\ =-2\sqrt{3}^{3n}\Im(i^{3n+1}(\cos\frac{2(3n+2)\pi}{3}+i\sin\frac{2(3n+2)\pi}{3}))\\ =(-1)^{n+2}2\sqrt{3}^{3n}\Im(i^{n+1}(\frac{1}{2}+i\frac{\sqrt{3}}{2}))\\ =\{\begin{matrix}(-1)^{\frac{n}{2}}\sqrt{3}^{3n},\ \text{if n even}\\ (-1)^{\lfloor\frac{n}{2}\rfloor}\sqrt{3}^{3n+1},\ \text{if n odd}\end{matrix} \]
Since $P(x)=Q(x)+1$,
\[P(3n+1)=\{\begin{matrix}(-1)^{\frac{n}{2}}\sqrt{3}^{3n}+1,\ \text{if n even}\\ (-1)^{\lfloor\frac{n}{2}\rfloor}\sqrt{3}^{3n+1}+1,\ \text{if n odd}\end{matrix} \]
When $P(3n+1)=730$, $n=4$.  As a result, $P(13)=730$.$\blacksquare$

Masoud Zargar


probability
integration
limit
probability and stats

this is more of a general technique that I'm asking about, but for the sake of concreteness, lets apply it to a specific example.  Let $ X_{i}$ ($ 1\leq i\leq 2n+1$) be uniformly distributed on $ [0,1]$ and let $ M_{n}$ be the median of the $ X_{i}$'s.  We want to show that $ M_{n}\to 1/2$ in probability and a.s.

A straightforward computation shows that
\[ P(a\leq M_{n}\leq b) = \int^{b}_{a}\frac{(2n+1)!}{n!n!}t^{n}(1-t)^{n}dt \]

So the first half is not that hard.  A little bit of work shows that
\[ P(|M_{n}-1/2|>\epsilon) = 2\int^{1}_{1/2+\epsilon}\frac{(2n+1)!}{n!n!}t^{n}(1-t)^{n}dt \]

We note that $ \frac{(2n+1)!}{n!n!}\sim 4^{n}\cdot \sqrt{\pi n}\cdot (2n+1)$ (e.g. by stirling's formula), and hence we get

\[ P(|M_{n}-1/2|>\epsilon) < 2\cdot \sqrt{\pi n}\cdot (2n+1) \cdot (1-4\epsilon^{2})^{n}\to 0 \]
This is convergence in probability.

I'm completely mystified on how we would move to a.s. convergence though :(  Am I just being too optimistic in assuming that this approach will lead to a proof of a.s. convergence and do I need to be looking for stronger tools/estimates?

[quote="mlok"]The first [url=http://en.wikipedia.org/wiki/Borel-Cantelli_lemma]Borel-Cantelli Lemma[/url] should help.[/quote]

I think I see what you're getting at.  Since $ P(|M_{n}-1/2|<\epsilon)\to 0$ exponentially quickly, it follows that $ \sum_{n\geq 1}P(|M_{n}-1/2|<\epsilon)<\infty$ for each $ \epsilon$, that is, $ P(|M_{n}-1/2|<\epsilon \ i.o.) = 0$ for each $ \epsilon>0$ (by Borel-Cantelli).  Now we have that
\begin{eqnarray*}P(\lim M_{k}\not=1/2) &=& P(\bigcup_{n\geq 1}|M_{k}-1/2|<n^{-1}\ i.o.) \\ &=& \bigcup_{n\geq 1}P(|M_{k}-1/2|<n^{-1}\ i.o.) \\ &=& 0 \end{eqnarray*}
which shows a.s. convergence.  Am I right?

Correct, except for sloppy formulas. Another way to put it: if $ \lim\ne 1/2$, then either $ \limsup>1/2$ or $ \liminf<1/2$, and both these events have probability zero.


topology
advanced fields
advanced fields unsolved

Prove that for a space $X$ the following conditions are equivalent:
a). $\pi_{i}(X)=0$ for $i \leq n$, i.e. $X$ is $n$-connected.
b). Any continuous mapping $f : S^{i}\to X$ can be extended to a mapping $\tilde{f}: D^{i+1}\to X$ for $i \leq n$.

a->b We know that $f$ is nullhomotopic, i.e. there is a continuous map $F\colon S^{i}\times [0,1]\to X$ such that $F(\cdot,1)=f$ and $F(\cdot, 0)$ is constant. Define $\tilde(x)=F(x/\|x\|,\|x\|)$.

b->a The extension $\tilde f$ gives a homotopy between $f$ and a constant map, as above.


Show that the coefficient of $x^n$ in the expansion of $\frac{1}{{\left( {1 - x} \right)\left( {1 - x^2 } \right)\left( {1 - x^3 } \right)}}$
is $\frac{{\left( {n + 3} \right)^2 }}{{12}} - \frac{7}{{72}} + \frac{2}{9}\cos \frac{{2\pi n}}{3} + \frac{{\left( { - 1} \right)^n }}{8}$


this appears in elementary number theory charles vanden eynden


So how did every1 do? based on projected score for R4 and R1,2,3. I know I got a Silver, wtb every1 else? any golds

13 (:blush:) / 24 / 19 / $ x\equal{}56\plus{}x$

If $ x\geq 19$, then I get silver.
If $ x<19$ then I get bronze.

Scores:

25/25/21 (:()/ x

if x = 25, I get gold
if x < 25, I get silver...

I got all the problems correct so I'm hoping I get a 25...

6th grade:

13/<lost>/18/probably 14=45+<lost>, so likely a bronze.

[quote="Yongyi781"]13 (:blush:) / 24 / 19 / $ x \equal{} 56 \plus{} x$

If $ x\geq 19$, then I get silver.
If $ x < 19$ then I get bronze.[/quote]
haha yea what happened on R1 yongyi

I left problem #1 blank (:o)*. Yeah. 3-D combinatorics is my least favorite type of problem.

* Well actually I didn't leave it blank. I wrote:

It trivially follows that the answer is 30.

 :rotfl:

16/22/20/(22)=80 => silver

The first round really hit me hard. It was not too hard yet by far my lowest score. I had never written a proof before and was obviously not up to par with the rigor, etc. But, this was clearly part of the learning experience.

11/20/12/x=43+x

Probably a few points short of bronze...epic fail :(

17/7/9/x=33+x

In order to get bronze, I must score 27 on Round 4

[quote="BOGTRO"]GL with that!  :lol:[/quote]

Thanks BOGTRO :)

I'll do my best

Wait...what is the smallest number of points you need to get a T-shirt? Because I totally dig just getting that.

All people who were ranked more than Honorable Mention (i.e. 40)

Heh, I was going for that too :D

Yes! One more shirt for my small closet! That isn't from MathCounts! (It's kind of embarrassing to wear a MathCounts shirt as a high schooler, so....)

I have 23/25/8/n
So I have 56 points so far. I need 19 more points for silver.
I can get bronze if i get 4 more points... and best I can get is silver :)

20/13/14/x

I need x>=13 to get bronze. I only did #1, #2, and #4 on R4...

This is really embarassing  because I'm a freshman...

Yeah, not too "research"-oriented.

But at least better than last year, where I looked at the first round, said **** this, and got a 0.

[quote="dnkywin"]25+24+25+25?=gold  :lol:[/quote]

Same here.

Does anyone know if Wolfram will be donating student licenses for Mathematica again?  That would be great.

[quote="benzi455"]Does anyone know if Wolfram will be donating student licenses for Mathematica again?  That would be great.[/quote]Yes.  We will also have a few new books available as prizes, as well as many of the books from last year.  All this will be announced officially next month.

fail .. I should stop sending in my solutions so early .. 25/25/25/25(?) ..  :P

[quote="Temperal"]bleh that's the same thing in my world.[/quote]
yeh both miss bronze

0/10/0/16ish  = ~26

I win nothing  :D
I wish I'd done rounds 1 and 3 (and actually read the questions more than once for rounds 2 and 4), but I had no time...

[quote="gauss1181"][quote="Temperal"]Hmph; 7/12/18/~22=~49, missing bronze by a bit. Huh, my scores are increasing almost linearly. Funny, because while I barely glanced at round 1, I spent the most time on round 2, not very much time on round 3, and almost no time on round 4.[/quote]
eh temperal your math computation was wrong...it should have added to 59 not 49. :wink: Just saying.[/quote]

I wonder why he did so poorly. :P

Mieh, 16/14/21/~19=70. Yay, i achieved my goal maybe

[quote="buzzer11"]22/18/12/0 = 52.
I think that's bronze?
haha i just stopped trying as the year went on.[/quote]

Nope, that's honorable mention.

Heh, wow, you almost had my scores in reverse?

DARN.
what's bronze again?

haha that's funny. you got better while i got suckier.

Bronze is 60-79 (?), I think? Not sure of the maximum score for bronze.

[quote="Brut3Forc3"]Might change this year because of round 2...[/quote]
lower or higher?

[quote="Yongyi781"]I think they will have separate cutoffs for those who lost their #2's and those who didn't. :)[/quote]
yay then I'm fine  :D

I didn't do too well... too much stress on my part lol

18 / 14 / 11 / x = 43 + x
x > or  = 17 ===> bronze
x < 17 ====> honorable mention.

This is worse than my 18 / 16 / 13 / 20 last year (Bronze = 67)

I got Round 4, Questions 1, 4, 5 right (hopefully getting 5's for each) and 2 and 3 I should get partial credit.  There is still some chance for me to get bronze :)

I got 2, 3, 4, and 5 and missed 1. :(
18/10/18/hopefully >14

#5 was bashtastic.

This is a dumb question, but when is the first day we are able to view year 2009-2010 round 1 problems?
I don't know much about USAMTS... (I knew it when it's too late to participate in this year, i.e.,year 2008-2009)  :(

[quote="mathwizarddude"]This is a dumb question, but when is the first day we are able to view year 2009-2010 round 1 problems?[/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218708]This topic[/url] was posted on August 4, 2008. While there are probably reasons why it won't be exactly that date next time, "first week in August" would make a plausible first guess.


function
algebra
rational function
complex analysis
complex analysis unsolved

Suppose that $f,g$ are entire functions which for each $z\in \mathbb C$ we know that $|f|\geq |g|$. Prove that there exist $c\in \mathbb C$ which $|c|\leq 1$ and $g=cf$

Liouville theorem applied to $h=\frac{g}{f}$ should do the trick: it is a rational function, bounded by $1$ almost everywhere. Then it is a bounded entire function, thus constant to $c$.

Well but you must consider the case which $f(z_0)=g(z_0)=0$. For this case you see there exist $m,n\in \mathbb N$ and entire functions $F,G$ which $f(z)=F(z)(z-z_0)^m,g(z)=G(z)(z-z_0)^n$ and $F(z_0)\neq 0, G(z_0)\neq 0$ and also $m\leq n$. Now you can define $\frac gf$ on the point $z_0$.


Let  be a convex polygon with $n >5$ vertices and area $1$. Prove  that there exists a convex hexagon inside the given polygon with area at least $\dfrac{3}{4}$

There's already a solution by Myth, I posted this problem second time because the wording was quite different, also I'm waiting for more geometrical approach.

http://www.mathlinks.ro/Forum/viewtopic.php?p=143505#p143505


Why do you think it is important to receive a well-rounded education in an increasingly diverse world?

I'm interested in your response.

People seem to be so broken up into their separate worlds... education is one of the few times when you're forced to be exposed to people, ideas, and things that you wouldn't necessarily choose for yourself.

I think it's almost impossible to focus on one or just a few things in such a diverse world, so maybe the emphasis on a well-rounded education can in fact be counterproductive by encouraging this indecision. However, at the same time, it's scary to think how ignorant people are of things outside their immediate sphere... so I don't know.

Ha, I'm broken, the world's broken, and that isn't always such a disadvantage... But probably the clearest benefit of a well-rounded education is it exposes you to different options so that you don't have to do what your parents did before you.

(Am I the only one overwhelmed by the sheer volume of knowledge and experience in the world, and the way the total dwarfs any one person's reach?)

A Well-Rounded education has been deprioritized for many good reasons in the last 100 years or so (or so it would seem to me... I have no super-awesome-kickass evidence for this statement)

The many good reasons have to do with society's more pragmatic, relativist tendencies..... knowledge is no longer viewed as some sort of ultimate end of existence by as many people.  Most people now are focused on knowledge and even intelligence and understanding as TOOLS.

The reason these are good reasons is because from an individual perspective, most people were never all that interested in learning in the first place (what percentage of people you know are REALLY intellectually curious and "learn for the sake of learning" and all that stuff).  So, the degredation of a social standard requiring people to be "rennaissance" people is more or less a good thing, from the academically uninvoled's perspective.  Consider the fall of the liberal-arts education, and the dramatic rise of engineering and trades schools.  In Europe, they have this too a greater degree in a lot of places, I think(than in Amurika).......  Additionally, it does not make academic sense to the average person to be educated in areas in which they will not exercise their educatedness.

So, is this a fundamentally good trend from an objective perspective?

I don't know.

There are a lot of ups and downs to either side.  I tend to think that a well-rounded education is benefitial in two ways/cases:

1.  For the individual who has afforementioned "academic curiosity"  This should be more or less a solid point.  I think the shear diversity of ideas and different conceptual lenses which one aquires through a wide-spread, rounded education is an extremely valuable tool (almost) regardless of the task.

2.  For a basic level of education and "learning to learn" & "learning to think".  There are some basic things which everyone in the world should know.  There are some basic things everyone in America/Canada/France/Russia/&c. should know (laws and history...) There are some things which everyone in a minorly skilled job should know; there are some things everyone in a certain skilled profession should know (law, medicine, engineering, yeah......) etc. etc.  This is especially important in America, I think, being that it is a socially and politically powerful democracy.  Having the most powerful political and social identitiy being controlled, largely, by un-educated or under-eduicated people leads to bad stuff....

There are two problems relating to this, I think in American education.  One is that the economic benefits of a college or high school degree still make it profitable to go through all of high school or go to college and, in general, to seek an excesive amount of education for some who do not benefit proportionally from the additional time spent in school or from learning the things outside their field which they'll never use.

The other problem is on the opposite side.  I've used this example already, and I'll use it again for lack of a better one:  Engineers.  There are a number of colleges and programs designed specifically to train a person to be an engineer in the most efficient way possible and thereby neglecting other academic areas.  In many cases this is super.  However, it can be an example of sacrificing quality for quantity(more like efficiency, whatever..); the tools of a well-rounded education can be extremely, even critically useful for many professions in which they are viewed as unneccessary.

In my personal case, I want a "well-rounded" education (what couldn't you tell??) but I think for many (even most) people it is frivoulous to complete an entire liberal-arts college education.  It often seems (despite the particualrly ludicrous example of categorization which this is...) that there is an ON/OFF switch on teh back of people's heads, actually, a whole set of them, one general one and a bunch of specific ones.  Ideally, everyone gets the general one flipped; how much time/energy this takes veries greatly... some get it switched on in grade school, some not until graduate school.... as an end note, if your going to be an engineer, it'd be good to hae your engineer switch flipped on too.....

Yeah.
Aha.
Touche.
Nice.

[quote="capybaralet"]
In my personal case, I want a "well-rounded" education (what couldn't you tell??) but I think for many (even most) people it is frivoulous to complete an entire liberal-arts college education.  It often seems (despite the particualrly ludicrous example of categorization which this is...) that there is an ON/OFF switch on teh back of people's heads, actually, a whole set of them, one general one and a bunch of specific ones.  Ideally, everyone gets the general one flipped; how much time/energy this takes veries greatly... some get it switched on in grade school, some not until graduate school.... as an end note, if your going to be an engineer, it'd be good to hae your engineer switch flipped on too.....

Yeah.
Aha.
Touche.
Nice.[/quote]
Unfortunately, the engineer switch seems to be incompatible with the others, especially the has social life one or cares about stuff he won't used.


Hi ,every body!

 I have a inequality: (Created by my)
 Let $a,b,c,n  \geq 0$
 Prove that:
[tex]\displaystyle \frac{na+b}{na+c}+\frac{nb+c}{nb+a}+\frac{nc+a}{nc+b} \leq   \frac{a}{b}+\frac{b}{c}+\frac{c}{a}[/tex]

@ It's a general from MathLinks Contest

------------------------------------------
My E is very bad!!! :blush:

here is my approach,but I haven't complete it :?
let $f(n)= \frac{na+b}{na+c}+\frac{nb+c}{nb+a}+\frac{nc+a}{nc+b}$,
$f'(n)= \frac{ac-ab}{(na+c)^2}+\frac{ba-bc}{(nb+a)^2}+\frac{cb-ca}{(nc+b)^2}$
$f'(n) \leq 0$,the same as,$ \frac{ac}{(na+c)^2}+\frac{ba}{(nb+a)^2}+\frac{cb}{(nc+b)^2} \leq  \frac{ab}{(na+c)^2}+\frac{bc}{(nb+a)^2}+\frac{ca}{(nc+b)^2}$
but I don't how  to do for this,anybody want to complete it?

[quote="siuhochung"]i guess it's not true for all $n \ge 0$.[/quote]
 I don't think so  ;)

No.It's a real number :)

------------------------------
It isn't difficult but nice  :P

Some Interesting Problems from It

 [b]Problem:[/b]
 a,b,c>0.Prove that:

 a) $\displaystyle \frac{1+b^2c}{1+bc^2}+\frac{1+c^2a}{1+ca^2}+\frac{1+a^2b}{1+ab^2} \le \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$

 b) $\displaystyle \frac {a^2+b}{a^2+c}+\frac{ab+c}{ab+a}+\frac{ac+a}{ac+b}\le \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$

 $\cdots$

-----------------------------
My E bad.Help me!!!  :blush:
(Thanks alot  :P  :P  :P  )

Does anybody have the solution?

 My solution is very simply  :P  :P  :P

 (You can see three solution in MathLinks Contest for n=1
but I think my solution is better  ;)  )

And here is my soulution  :D  :D  :D

 Put
       $X=\displaystyle\frac{\frac{a}{b}+n}{n+1}$

       $Y=\displaystyle\frac{\frac{b}{c}+n}{n+1}$

       $Z=\displaystyle\frac{\frac{c}{a}+n}{n+1}$

 It's easy to prove $\displaystyle XYZ \ge 1$ (1)

 We have

 $\displaystyle\frac{a}{b}-\frac{nc+a}{nc+b}=\frac{nc(a-b)}{b(nc+b)}
                               =\frac{nc}{nc+b}\cdot(\frac{a}{b}-1)$
 $\displaystyle =\frac{n}{(n+1)Y}\cdot(n+1)\cdot\frac{\frac{a}{b}+n-(n+1)}{n+1}=\frac{n(X-1)}{Y}$
 Then the inequality equivalent
 $\displaystyle\frac{X}{Y}+\frac{Y}{Z}+\frac{Z}{X} \ge \frac{1}{X}+\frac{1}{Y}+\frac{1}{Z} $ (*)

 But by Cauchy we have

$\displaystyle\frac{X}{Y}+\frac{X}{Y}+\frac{Z}{X} \ge 3\sqrt[3]{\frac{XZ}{Y^2}}
=3\sqrt[3]{\frac{XYZ}{Y^3}} \ge \frac{3}{Y}$
(Because $\displaystyle XYZ \ge 1$)

The same we have

$\displaystyle 2\frac{Y}{Z}+\frac{X}{Y} \ge \frac{3}{Z}$

$\displaystyle 2\frac{Z}{X}+\frac{Y}{Z} \ge \frac{3}{X}$

Then We have (*)

Finish  :P

 :D  :D  :D  :D


I was told by my friends that the recent years' AIMEs are way harder than the 80's, and a little bit harder than the 90's since they stopped to reuse the problems and took away those basic problems.  But I thought that they are almost identical difficulty wise, what's your opinion?

i thought this year's AIME was slightly easier compared to the years before; but then, I wasn't as good of a problem solver as i am now back then.
The questions this year, however, had more room for errors and some required decent calculation skills.

There was a good combination of geometry and combination problems on this year's AIME.


Let n be a natural number >=3. Show that n^(n^(n^n))-n^(n^n)
is divisible by 1989.

I substituted 3 into the equation and got 7625597465304 which wasn't divisible by 1989... (It came out to 3833885100.705)

I don't think you can handle a number the size of 3^3^3^3 with a calculator or computer program. I put in 3^3^3=3^27 into NuCalc, and it responded with 7.625597e+012. When I tried 3^3^3^3, it spit out infinity. This problem seems to be hard. Where did it come from?

1989 = 9 * 13 * 17

A try to prove 17 | n^n^n^n - n^n^n which may be extending (I don't see how for now) for the other cases.

Factoring n^n^n at first, and after n^n in the exponent :

n^n^n^n - n^n^n = n^n^n (n^(n^n^n - n^n) - 1) = n^n^n (n^(n^n * (n^(n^n - n)) - 1) - 1)

If 17 | n it's obvious.

Otherwise :

. if n even, n^n = 0 mod 16.
By Fermat Little Theorem x^(16*y) = 1 mod 17 if gcd(x,17) = 1.
So (n^n * (n^(n^n - n)) - 1) = 1 mod 17 and it's good for this case.

. if n is odd, 4 | n^n - n, 16 | n^(n^n-n) - 1 and it's good also.

Unfortunately it doesn't seem to work nicely for 9 or 12 but it may be a good start.

Hey! Vialli, I'm a member of the 'happy community' of mathlinks.go.ro   :lol: Just like you (I've seen you're a member there). Anyway, this problem was posted there today by Alekk, and I posted a soln there, but it's kind of long, and that forum doesn't seem to be working :( After it starts working again I'll post it here too. By the way, Alekk said it was an unused problem for the Balkan olympiad in ... guess what year: 1989.

Try to prove that 12/n^n*(n^(n^n-n)-1). If you prove this it works for 13. It's not hard for 9 either.


Let $a,b,c$ real numbers and $-2<p\le 2$. Prove that
$(a^2+pab+b^2)(b^2+pbc+c^2)(c^2+pca+a^2) \ge \frac{(2+p)^3}{27}(ab+bc+ca)^3$

[quote="xtar"]Let $a,b,c$ real numbers [/quote]

so (a+b)(b+c)(c+a)/8 \ge (a+b+c)(ab+bc+ca)/9 is wrong.

sorry.
I think the inequality in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61543]that post[/url] which also xtar posted helps.but no one post a solution there. :)


function
real analysis
real analysis unsolved

This is a problem from the text (This text is extremely poorly written):

Let $a,b \in R, a<b$ and let $f$ be a continuous real-valued function on $[a,b]$. Prove that if $f$ is one-to-one then $f([a,b])$ is $[f(a),f(b)]$ or $[f(b), f(a)]$, whichever expression makes sense.

What do they mean by $[f(a),f(b)]$ or vice versa? Do they mean the interval?

I think this is a well known result (theorem?)

Yes, $[f(a), f(b)]$ is an interval. Note that if $f(a) = 2$ and $f(b) = 1,$ then the interval $[2,1]$ doesn't make sense but $[1,2]$ does.


(A.Zaslavsky, 9--10) Given four points $ A$, $ B$, $ C$, $ D$. Any two circles such that one of them contains $ A$ and $ B$, and the other one contains $ C$ and $ D$, meet. Prove that common chords of all these pairs of circles pass through a fixed point.

It's the reverse of this problem

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1995Problem3


limit
number theory proposed
number theory

For natural numbers $ a, b , a\ge 2 ,b\ge 2 , $ let  ${ S(a,b):= \sum\limits_{k=1}^{\infty} \frac{1}{a^{{k}^b}}} \; . $ [b]For which (a,b) the number $ S(a,b) $  is irrational  ? [/b]

REFERENCES :
[1-a] G. Cantor , ,,Ueber die einfache Zahlensysteme ", Zeitschrift fuer Mathem.und Physik 14 (1869)=   Gesammelte  Abhandlungen , Seite  35.
[1-b] G. Cantor , ,,Zwei Saetze ueber eine gewisse Zerlegung der               Zahlen in unendliche  Produkte " , Zeitschrift fuer    Mathem.und Physik 14 (1869)Gesammelte  Abhandlungen , Seite 43.
[2] J. Lueroth , ,, Ueber eine eindeutige Entwicklungen von Zahlen  in eine unendliche Reihe ", Math. Annalen 21 (1883) .
[3] O.Perron , ,,Die Lehre von der Kettenbruechen", Leipzig, 1913.
[4] O.Perron , ,,Irrationalzahlen" , Goeschens Lehrbuecherei, Band 1,                Walter de Gruyter & Co., dritte Auflage, 1946.
[5] J.J. Sylvester , ,,On a point in the theeory of vulgar fractions",                Amer. J. Math. 3(1980).
===============
(,, Ancient books= interesting books ! " , e.g. see   [4] )
==============

All of them.

More generally, if $(p_n)$ and $(q_n)$ are two infinite sequences of positive integers such that :
a) The sequence $(p_n)$ is bounded.
b) The sequence $( \frac {q_n} {q_{n+1}} )$ has limit $0$
c) For each $n$, $q_n$ divides $q_{n+1}$
then
the serie $ \sum \frac {p_i} {q_i}$ is convergent and its limit, say $\alpha$, is an irrational number.

I will prove it later if non-one does it (I'm in a hurry now).

Pierre.

Here we go :

From b), if $0 < e < 1$, there exists an integer $N > 0$ such that, for $n \geq  N $ we have $q_n < e \cdot q_{n+1}$, that is $\frac 1 {q_{n+1}} < \frac e {q_n}$.
It follows that, if $n \geq N$, then (with  $p_i \leq  M$, for all $i$) :
$S_n = \sum_{i=1}^n \frac {p_i} {q_i} \leq S_{N-1} + \frac M {q_N}(1 + e + e^2 + ...) \leq c + \frac {c'} {(1-e)}$, where $c,c$' are two constants
Therefore, the sequence of the partial sums is increasing and bounded, thus it converges to a limit, say $L$.

Now, we prove that $L$ is irrationnal :
Suppose, for a contradiction, that ${L = \frac p q}$, where $p,q$ are two positive integers.
Then, for all $N$, we have :
$\frac p q = S_{N-1} + \sum_{i geq N} \frac {p_i} {q_i}$
But, from c), for all $n$, we have $q_{n+1} = a_{n+1}q_n$, where $a_{n}$ is a positive integer (with $a_1 = q_1$)
From b), we deduce that $\lim a_n = + \infty$
Then, for $N > 1$ :
$\frac p q = \sum_{i=1}^{N-1} \frac {p_i} {a_1 \cdots a_i} + \sum_{i \geq N} \frac {p_i} {a1 \cdots a_i} .$
Multiplying by $q \cdot a_1 \cdots a_{N-1}$ it leads to
$p \cdot a_1 \cdots a_{N-1} = q \cdot \sum_{i=1}^{N-1} p_i \cdot a_{i+1} \cdots a_{N-1} + q \cdot \sum_{i  \geq N} \frac {p_i} {a_N \cdots a_i }$
Let $A = q \cdot \sum_{i=1}^{N-1} p_i \cdot a_{i+1}\cdots a_{N-1}$,  which is an integer. Then :
$ B = p \cdot a_1 \cdots a_{N-1} - A = q\cdot \sum_{i \geq N} \frac {p_i} {a_N \cdots a_i}.$
Thus $B > 0$ is an integer, and from a), :
$B \leq qM \cdot \sum_{i \geq N} \frac 1 {a_N \cdots a_i}$.

Let $0 < e < 1$. Since $\lim a_n = + \infty$, there exists $N> 1$ such that if $i \geq N$ then $\frac 1 {a_i} < e.$
For such a $N$, we then have
$B \leq qM \cdot \sum_{i \geq N} \frac 1 {e^{N-i+1}} = \frac {qMe} {1-e}$.
Thus, if initially, we choose $e$ such that $\frac {qMe} {1-e} < 1$, we deduce that  $B$ is an integer such that $0 < B < 1$, a contradiction.

Pierre.


A farmer wishes to build a rectangular sheep pen with a 160m fence, using one side of an existing shed as the side. The shed is also rectangular of measurement 20m x 20m.
(a) Find the measurement and the area of the sheep pen if one side of the existing shed is used as the width of the sheep pen.
(b) By using the fence of the same length and using the width of the shed as part of the width of the sheep pen,find the length and width of the sheep pen with maximum area.
(c) If two sheep pens of the same size are built adjacent to each other, find (i)the length of the new sheep pen if one side of the shed is used as the width of the sheep pen.(ii)the length and width of the sheep pen, which is adjusted so that the enclosed area is maximum.
(d) Find the maximum enclosed area if the same fence is used to construct three rectangular sheep pens.
FURTHER INVESTIGATION
If n rectangular sheep pen of width 20m are constructed (one side of the existing shed is used as part of the pen) from a fence of length x m, find in terms of x and n:(iii)the length of the sheep pen (iv)total area of the sheep pens
*State at least 2 approaches*verify the suggested approaches by showing relevant steps*use correct mathematical notations, correct terminology, relevant pictorial presentation(tables,graphs or pictures)

There are many problems like this solved. Go and find a book about optimization. It can help you with this problem.


Prove: If {v_1,v_2,…,v_n} is a basis for V and {w_1,w_2,…,w_n} are vectors in W, not necessarily distinct, then there exists a linear transformation T:V→W such that T(v_1 )=w_1, T(v_2 )=w_2,… T(v_n )=w_n.

Define the function $ T: V \to W$ such that $ T \left(\sum_{j \equal{} 1}^n \alpha_j v_j \right) \equal{} \sum_{j \equal{} 1}^n \alpha_j w_j$. Observe that:

I. $ T$ is indeed a function, since $ \forall v \in V$, $ T(v)$ exists and is unique. In fact, if $ v \in V$ then there are scalars $ \alpha_1, ..., \alpha_n$ such that $ v \equal{} \sum \alpha_jv_j$, and the uniqueness of those scalars imply the uniqueness of $ T(v) \equal{} \sum \alpha_jw_j$.

II. $ T$ is a linear transformation, as is easy to see.

III. $ T(v_j) \equal{} w_j$.

IV. $ T$ is unique.


algebra
modular arithmetic
AMC 10
recursion
Sequences

The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the 2005th term of the sequence?

$ \textbf{(A)}\ 29\qquad
\textbf{(B)}\ 55\qquad
\textbf{(C)}\ 85\qquad
\textbf{(D)}\ 133\qquad
\textbf{(E)}\ 250$

[hide="Answer"]Looking at the first few terms, we have:

$2005\Rightarrow 2^3+5^3=133$
$133\Rightarrow 1^3+2(3^3)=55$
$55\Rightarrow 2(5^3)=250$

This has the same two non-zero digits as $2005$, so the pattern will repeat every three values. $2005\equiv 1\bmod{3}$, so we take the first value. $\boxed{E}$[/hide]

[hide="Jesusfreak,"]the $1 \pmod 3$ term in the sequence is 250, not 133.  (133 is the $2 \pmod 3$ term) The answer is $(E)$[/hide]

[hide]Continuing the sequence, we get 2005, 133, 55, 250, 133, 55, 250, 133...

It repeats every three terms. Since 2005 leaves a remainder of 1 when divided by three, it must be the same term as term 4 (also a remainder of 1), which is 250. [/hide]


[color=blue]I need to know how to do coin toss problems similar to:

If a fair coin is tossed 10 times, what is the probability of getting 7 heads and 3 tails.  I could look up the answer, but I would like to know how to do this for future reference.  Can anyone explain how to do these problems in general?[/color]

Wait is it the first 7 heads and next 3 tails?? If it is.. then i think the answer is 1/1024, because its 1/2 * 1/2 * 1/2*.... Ten times. 2^10 = 1024, and only one works, so its 1/1024. Im not sure if this is how you do it though...my probability is very weak.

is the answer [hide="answer"]
15/128?[/hide]
if so, i'll post a solution

[hide]Its 15/128. Use Pascal's triangle. The fourth (number of + 1)  term in the 11th row (number of tosses + 1) Is the probability (out of 1024) is 120. THat reduces to 15/128[/hide]

i HATE probability problems, cuz i can also never figure out WHEN to use factorials or permutations or whatever........ :(
i just need to know IN GENERAL how to do em


Euler function quadratics algebra polynomial

For both let $ n\equal{}41$. Expanding gives $ 41\cdot41\plus{}41\plus{}41\rightarrow43\cdot41$ and $ 41\cdot41\minus{}41\plus{}41\rightarrow41^2$

Euler proved that for any function in $ x$ of the form $ f(x) \equal{} x^2 \plus{} x \plus{} a$ for all $ a \in \mathbb{Z}$, $ f(x)$ is not always prime. I'm not sure whether there is a proof of the non-existence of a function $ g(x)$ for which $ g(x)$ is prime for all $ x$.

[quote="johnnash"]Prove why does the equations n^2+n+41 and n^2-n+41?
not always give you a prime number??please show the steps[/quote]

[hide="First Equation:"]If we factor out $ n$ from $ n^2\plus{}n$, we get $ n(n\plus{}1)\plus{}41$. Now for there to be a non prime solution, we need $ x$ such that $ x|n(n\plus{}1)$ and $ 41$. $ 41\equal{}x$ is a simple answer.
[/hide]

Similar method for second equation.

I personally think that involving number makes it seem more complicated. Assume you're dealing with a standard quadratic equation $ f$ with integer coefficients; $ f(n)\equal{}an^2\plus{}bn\plus{}c$. We now observe that for all such functions \[ f(c)\equal{}c(ac\plus{}b\plus{}1)\] is guaranteed to be composite when neither $ c$ nor $ ac\plus{}b\plus{}1$ have a value of one.

y're right igiul. i totally agree with u. no need to introduce numbers here.

[quote="igiul"]I personally think that involving number makes it seem more complicated. Assume you're dealing with a standard quadratic equation $ f$ with integer coefficients; $ f(n) \equal{} an^2 \plus{} bn \plus{} c$. We now observe that for all such functions
\[ f(c) \equal{} c(ac \plus{} b \plus{} 1)
\]
is guaranteed to be composite when neither $ c$ nor $ ac \plus{} b \plus{} 1$ have a value of one.[/quote]

What if they both have a value of one?

then also, it is not prime. 1 is neither prime nor composite.

igiul's argument does not rule out the possibility that $ c\equal{}1$ and $ a\plus{}b\plus{}1$ is prime. There are instances where his argument fails but it does prove that a 'majority' of quadratics, or in fact, a class of polynomials that do not always yield primes; but this is far from being 'complete' to show that there exist NO polynomials of a single variable that would yield primes.

Google Matijasevich, I remember that his work was based on the existence of multivariable polynomials that would yield primes, a deep result.


geometry
email
incenter
inradius
geometry solved

I. Construct a bicentric quadrilateral, given a side and the two adjacent angles.

 II. Construct a bicentric quadrilateral, given two consecutive sides and the angle between them.

[quote="sprmnt21"]I. Construct a bicentric quadrilateral, given a side and the two adjacent angles.[/quote]

So assume that we have to construct a bicentric quadrilateral ABCD knowing the side AB and the angles at the vertices A and B. In other words, we know the lines g and h passing through the points A and B, respectively, on which the remaining vertices D and C of the required bicentric quadrilateral ABCD lie. Now we must find the exact positions of the points D and C on these two lines. Well, the incircle of our bicentric quadrilateral ABCD must touch the lines AB, AD = g and BC = h; these lines are known, and hence we can construct the incircle of our bicentric quadrilateral ABCD. Call i this incircle and I its center.

Let an arbitrary circle through the points A and B meet the lines g and h at the points F and E, respectively (apart from the points A and B). The perpendicular to the line EF through the point I meets the circle i at two points; the one lying farther from the line AB will be called P. The tangent to the circle i at the point P meets the lines g and h at the points D and C. Then, the quadrilateral ABCD is our required bicentric quadrilateral.  ;)

To prove this construction, one has to show that the quadrilateral ABCD obtained in this way is bicentric, and that conversely, any bicentric quadrilateral ABCD with vertices D and C on the lines g and h can be obtained by the above construction. I will prove the first part only (the second one can be showed by reversing the proof of the first one): In fact, it's clear that the quadrilateral ABCD obtained in this way has an incircle (since the line CD is the tangent to the circle i at the point P); it remains to show that this quadrilateral ABCD also has a circumcircle. Well, since I is the center of the circle i, the tangent to the circle i at the point P is perpendicular to the line IP; but the line IP is perpendicular to the line EF. Hence, the tangent to the circle i at the point P is parallel to the line EF. In other words, the line CD is parallel to the line EF. Hence, < ADC = < AFE. On the other hand, < AFE + < ABE = 180 since the points A, B, E, F lie on one circle. Hence, < ADC + < ABE = 180, or, in other words, < ADC + < ABC = 180. Hence, the quadrilateral ABCD has a circumcircle. Therefore it is bicentric. Qed.. As I have said, the second part, namely to show that any quadrilateral ABCD with vertices D and C on the lines g and h can be obtained by the construction I gave above, can be proved by reversing the proof of the first part: First show that the line CD must be parallel to EF, hence the line CD is perpendicular to IP, and since the line CD must touch the circle i, it follows that this line touches the circle i at the point P, etc..

[quote="sprmnt21"]II. Construct a bicentric quadrilateral, given two consecutive sides and the angle between them.[/quote]

This is equivalent to constructing a bicentric quadrilateral ABCD given three vertices A, B, C. This was Problem 5 of the IMO 1962; see a [url=http://www.kalva.demon.co.uk/imo/isoln/isoln625.html]solution on the Kalva site[/url]. Note that I had sent Kalva an email about this solution:

[quote="Darij Grinberg"][b]Subject:[/b] IMO 1962 Problem B2
[b]From:[/b] Darij Grinberg
[b]To:[/b] John Scholes

Dear John Scholes,

A note of thanks and appreciation for the wonderful problem collection at first.

A friend of mine has found a few typos in your solution of the B2 problem of the IMO 1962: In the very beginning of the solution, the "< ABI + < BCI = 90" should be "< BAI + < BCI = 90" [color=red][this is now corrected - Darij][/color]. Then, when you say "< AIC = 270 - < ABC", you should note that this is true for the angle < AIC of the quadrilateral ABCI but not for the angle < AIC we see in an acute triangle ABC. Then, "< AXC = 90 + < ABC" should be "< AXC = 90 - < ABC".

A few remarks by myself: As a synthetician, I find it unnecessary to invoke continuousity in the last argument. Just perform the construction and you will easily see that the resulting quadrilateral ABCD has an incircle.

And finally, here is an alternative (I don't say simpler, but it is rather symmetric) construction of the circle such that the chord AC suspends the chordal angle 90 - < ABC: Construct either the point E where the perpendicular to BC at C meets AB or the point F where the perpendicular to AB at A meets BC. The points E, F, C, A lie on one circle - this is the required circle. Feel free not to include this comment if you don't find it useful - it is indeed rather a remark from triangle geometry than a part of the problem solution.

  Sincerely,
  Darij Grinberg[/quote]

  Darij

[quote="darij grinberg"]
Let an arbitrary circle through the points A and B meet the lines g and h at the points F and E, respectively (apart from the points A and B). The perpendicular to the line EF through the point I meets the circle i at two points; the one lying farther from the line AB will be called P. The tangent to the circle i at the point P meets the lines g and h at the points D and C. Then, the quadrilateral ABCD is our required bicentric quadrilateral.  ;)
[/quote]

It seems very similar to my solution(but I have to check better; I have some doubt about arbitrary circle).
Here is my reasonment:

 if ABP is the triangle constructed from the given data and I its incenter let Q be the intersection between AB and the ortogonal line to PI from I. The line tangent from Q to the incircle of ABP cuts PA and PB in the points C and D we are looking for.

[quote="sprmnt21"]II. Construct a bicentric quadrilateral, given two consecutive sides and the angle between them.[/quote]

This problem one can prove being equivalent to that wich require to construct a triangle of wich are given a side the ooposite angle and the difference between the other two sides. Wich in turn is equivalente to the problem of cunstruct a triangle of wich we know two sides and an angle (obtuse) opposite wrt one of the two sides knwon.

Only for the part I:
We know:for any bicentric quadrilateral, the sum of the squars of the inverses of the distances from incenter to two opposite vertices is equal to square of inverse of inradius.
So if we know the side $AB$ and the angles at vertices $A$ and $B$ of our bicentric quadrilateral $ABCD$, then we will find it with the next construction: draw the bisector of $A$ and $B$ st the cut point is $I$(incenter), then we know $IA$ and $IB$, draw the perpendicular from $I$ to $AB = r$(inradius of $ABCD$) also draw a perpendicular to $IB$ at $I$ st the cut point with $AB$ is $F$, then $FIB$ is a right triangle,so: $\frac{1}{IB^2}+\frac{1}{IF^2}=\frac{1}{r^2}$.Now we have $IF=ID$, with center at $I$ draw a arc of circle with radius $IF=ID$ st the cut point with the line through vertice $A$ at $D$.Finally from $D$ we draw a tangent to circle (incircle) with radius r st the cut point with the line through vertice $B$ is $C$ point, so we have founded $ABCD$ as bicentric quadrilateral.


:blush: the line joining the mid-points of the diagonals of a trapezoid has length  3 units.if the longer base is 97 units,then the shorter base is- :?:

[b][u]The Answer: 91
Formula: B1 = B2-2S where B1 is the shorter base, B2 is the longer base and S is the segment connecting the midpoints of the diagonals.[/u]
[/b]

I was wondering about your question and will apologize for my means of answering it. The way I came about my answer was by
construction on Cabri II. I'm not aware of any easier ways to construct it but this is what I did (I'm including it so you can reproduce it and/or check for mistakes).

[u][b]The Construction[/b][/u]
1)   Construct a line.
2)   Construct a line parallel to the first.
3)   Construct a quadrilateral with two vertices on each of these lines and only one pair of parallel opposite sides.
4)   Connect the points.
5)   Make sure that the quadrilateral only has one pair of parallel sides.
6)   Construct the diagonals and their midpoints.
7)   Construct a line segment through the midpoints of the diagonals.
8)   Extend either base until its length is 97.
9)   Extend the other base until the length of the segment through the midpoints of the diagonals is equal to 3.
10) You will find this length to be 91 and your answer.

[hide]Draw trapezoid ABCD. Let the diagonals AC and BD intersect at X, and let the midpoints of AC and BD be M and N, respectively. Since AB = 91 and MN = 3 and triangles ABX and MNX are similar, we can let AX be 91y and BX be 91z. Consequently, MX will be 3y and NX will be 3z. Then, CM is (91 + 3)y = 94y and DN is 94z. The side lengths of triangle DCX are then 97y, 97z, and 97.

So 97 is correct.[/hide]

 Hopefully it will make sense once you draw it xD


inequalities
inequalities solved
3-variable inequality
AM-GM

If $a,b,c \geq 0$ then we have $a^3+b^3+c^3+abc \geq \frac{1}{2}(a+b)(b+c)(c+a)$

The inequality is equivalent to $2\sum a^3 + 2abc \geq 2abc + \sum (a^2b +ab^2)$.
but $(a-b)^2(a+b) \geq 0 \implies a^3+b^3 \geq a^2b+ab^2$ and adding all three inequalities of this type the problem is solved.


Find the product $ (n \plus{} 1)(n \plus{} 2)(n \plus{} 3)\cdots(n \plus{} n)$

[hide="Answer"]$ 2^n\cdot1\cdot2\cdot3\cdots\cdot(2n \minus{} 1)$[/hide]

[quote="triplebig"]Find the sum $ (n \plus{} 1)(n \plus{} 2)(n \plus{} 3)\cdots(n \plus{} n)$

[hide="Answer"]$ 2^n\cdot1\cdot2\cdot3\cdots\cdot(2n \minus{} 1)$[/hide][/quote]

A smilar question

Find the exact value of

$ (1 \plus{} \frac{1}{n})(1 \plus{} \frac{1}{n \plus{} 1})(1 \plus{} \frac{1}{n \plus{} 2})\cdots(1 \plus{} \frac{1}{2n \minus{} 1})$

Not so much, all you do is sum and they will all cancel. How is it similar?

For Stephen's question, simply write with a common denominator and cross out to get the answer $ n$.

Triplebig's is pretty obvious to figure out algebraically, but the expression is equal to $ n!\binom{2n}{n}$ so there might be a nice combinatorial argument; however, I can't find it.

Also "find the sum"? Do you mean product?

Yes, I did mean product, thank you for correcting.

I did not see it factorial-wise, it is pretty obvious, and thanks for the tip.


geometry
circumcircle
incenter
search
geometry proposed

hint: Inversion pole $ I$ (incenter of $ \triangle ABC$ ) power $ IM^2$

[quote="efoski1687"]The incircle of a triangle ABC is tangent to BC,CA,AB at M,N and P, respectively. Show that
the circumcenter and incenter of △ABC and the orthocenter of △MNP are collinear.[/quote]

http://www.mathlinks.ro/viewtopic.php?search_id=240517078&t=6228


King Kamehameha wants to slay a dragon. He can go on the mission alone, or he can take some number of court jesters with him. If he has $n$ distinct court jexters, let $f(n)$ be the number of ways in which Kamehameha can form an entourage from these court jesters, choosing none or one or more. Calculate $\log_{2}f(1)+\log_{2}f{2}+...+\log_{2}f(100)$.

[hide]Since ${n\choose 0}+{n\choose 1}+{n\choose 2}\ldots+{n\choose n}=2^{n}$, we can simplify the problem to:

$\log_{2}2^{1}+\log_{2}2^{2}+\log_{2}2^{3}\ldots \log_{2}2^{1}00=1+2+3\ldots+100=\frac{100(101)}{2}=\boxed{5050}$
I hope I didn't mess up  :maybe: [/hide]


probability geometry MATHCOUNTS FTW AMC AIME USAMTS

algebra is my strongest and counting&probability are my weakest.

Yes somebody who agrees with me in this post.

Too bad! My strongest is geometry. I wonder if that's in my genes...

For me, Algebra is my strongest. It looks like this is the case for most people, so far.

It's probably because MathCounts algebra problems tend to be simple equations and soon.

look at my other post. Too lazy to copy and paste.

My List:

Strong:
Algebra-Geometry tie (not too good at Geo or Alg in Chinese texts  :o ...)
Probability

Weak:
NT
Counting

[quote="Yongyi781"]1. Tie between number theory and geometry
3. Algebra
4. Probability
5. Counting[/quote]

Yay! Agrred.

Since everyone else is making a list so will I.

1.geometry
2.algebra
3.number theory
4.counting
5.probability

I agree that geometry and algebra in china is wicked hard. But strangely the counting and probability isn't that hard.(or is it just me?)

DEFINITELY GEOMETRY FTW! It's pwnage for me on Mathcounts!

lol, i don't have a strongest topic.  My weakest:  all of 'em.

For MC level stuff, I'm best at Number Theory and C&P (thanks to the classes xD).  Then probably algebra, and I really stink at geometry

[quote="Wickedestjr"]Algebra is my strongest, but I am pretty good at Number Theory also. I am actually pretty good at Counting and Probability also. The only one I really have a lot of trouble with is Geometry.[/quote]
Didn't you already say that?

[quote="Wickedestjr"]Here's my countdown of strongest topics;

1 Algebra - Strongest
2 Number Theory
3 Counting
4 Probability
5 Geometry[/quote]

[quote="Wickedestjr"]
Algebra is my strongest, but I am pretty good at Number Theory also. I am actually pretty good at Counting and Probability also. The only one I really have a lot of trouble with is Geometry. [/quote]

1. Algebra
2. Geometry
3. Counting/Probability
4. Proofs

My answer is similar to Nick's; Algebra is my top, maybe followed by geometry (possibly). However, I stink at the rest.

Gahhhhh proofs < combinatorics.

Geometry>Algebra.

[quote="numbermaster"]Geometry :winner_first:
Algebra :winner_second:
Probability :winner_third:[/quote]
You totally agree with me!

But only on Mathcounts.

On other contests, especially olympiad ones, Probability>Algebra. :D

The only topic that I am not really good at is geometry. I do okay at the others.

2 best i think are probability and number theory. Geometry is a close 3rd, followed by brute force ;]

and i make list for fanfan

1. CALCULUS
2. CALCULUS
3. CALCULUS
4. CALCULUS

etc.

I fail at everything, but I'm definitely best at geometry, algebra 2nd best, counting/probability semi-ok, and number theory...really bad. That's probably not a good thing, since most AMC questions are about algebra and number theory, with probably just up to 6 geometry and a couple counting and probability. :D

Algebra is my strongest, then geometry, then probability. Only on Mathcounts, though. My rankings change is tougher competitions(question-wise).

Pointlessness and getting spammy ergo locked (and * some people * find it important to post multiple times in this topic, which confuses me; cannot one state ones' likes in one post? If it takes multiple posts to state ones' likes, one is spamming.)


geometry
3D geometry
sphere
function
calculus
topology
analytic geometry

Let $S$ be the set of interior points of a sphere and $C$ be the set of interior points of a circle. Find, with proof, whether there exists a function $f:S\rightarrow C$ such that $d(A,B)\le d(f(A),f(B))$ for any two points $A,B\in S$ where $d(X,Y)$ denotes the distance between the points $X$ and $Y$.

[i]Marius Cavachi[/i]

Use \to for a simple right arrow in LaTeX.  Also, the last instance of $ f$ in your problem should of course be $ d$.[color=#FF0000] [mod: this has been edited] [/color]

Presumably the sphere and circle are meant to have the same radius?  If so, an equilateral triangle inscribed in a great circle (= equator) of the sphere must be sent to an equilateral triangle inscribed in the circle.  Now choose the north pole relative to the selected equator and it's clear there is nowhere for it to go.  Or, if you don't want me to use points on the boundary, just begin with a slightly smaller sphere inside our larger one.

If the circle can have arbitrarily larger radius, it's not clear what the answer should be (but in this case the problem probably belongs to the Calculus and Analysis forum, not here).

I think it doesn't belong to analysis forum and is a fine Olympiad problem. It can be solved (for any radius of the circle) without ideas of continuity, homeomorphism etc.

Consider a cube of side length $a$ in the interior of the sphere. Clearly, all points from the cube are mapped to points on the given circle (say with diameter $D$). Now we set our coordinate system such that the origin is at a corner of the cube and the positive $x,y,z$ axes are along three adjacent sides of the cube. For a positive integer $n$, consider the set \[S_n=\{\left(\frac{ax}{n},\frac{ay}{n},\frac{az}{n}\right)|\ 0\le x,y,z\le n\}\]This set is a subset of the said cube. Let $f(S_n)$ denote the set of images of points in $S_n$ under $f$. Each point in $S_n$ is at least at a distance $d=a/n$ from all other points in $S_n$. So same must hold for $f(S_n)$. Now we consider disks of diameter $d$ centered at the points in $f(S_n)$. Clearly they are disjoint to one another. And they are all contained within a circle of diameter $D+d$ which is concentric with the given circle of diameter $D$. So the total area of those $(n+1)^3$ disks should be less than the area of that circle. i.e.\[(n+1)^3\frac{\pi}{4} \left(\frac{a}{n}\right)^2\le \frac{\pi}{4} (D+\frac{a}{n})^2\]\[\text{But we have }na^2<(n+1)^3\left(\frac{a}{n}\right)^2\le (D+\frac{a}{n})^2\le (D+a)^2\implies n<\left(1+\frac{D}{a}\right)^2\]But this must be true for all $n$, which is impossible. Hence no such function exists.


In acute $ \triangle RST$, $ TA \perp RS$ at $ A$, and $ M$ is the midpoint of $ RS$. $ A$ lies between $ R$ and $ M$. $ \overline{TA} \equal{} \frac 45 \overline{TM}$ and $ \triangle ATM$ has perimeter $ 24$ feet. $ \triangle TMS$ has perimeter $ 36$ feet. Find the perimeter of triangle $ RST$. (give answer in exact form)

Since AT=(4/5)TM and the perimeter of triangle AMT is 24, we figure out that AM=24-(9/5)TM.  Using triangle TAM, we can find TM using the pythagorean theorem:
TM^2=[(4/5)TM]^2+[24-(9/5)TM]^2
Solving this for TM, we find that it equals 10.
Thus, AT=8 and AM=6.
From triangle TMS, we see that MS+TS=36-10=26.
Letting MS be m, in right triangle ATS, we have that TS^2=(6+m)^2+(8)^2=100+12m+m^2.  Hence, we know that TS=sqrt(100+12m+m^2). Plugging this into MS+TS=m+TS=26, we find that m=MS=9.  From triangle TAR, we see that TR^2=(8)^2+(m-6)^2=m^2-12m+100.  Thus, the perimeter of triangle RTS is:
RS+RT+TS=
2m+sqrt(m^2-12m+100)+TS=
m+sqrt(m^2-12m+100)+m+TS=
9+sqrt(73)+26=
[u]35+sqrt(73)[/u]


[hide]
It can be proved that for even $n$, $n^4 +4 \equiv 0 \pmod {10}$ and for odd $n$, it is congruent to 5. (I'll post the proof as soon as I finish it... :P )

Of course a prime cannot end in a $0$ because it must contain a factor of $5 \cdot 2$.

A number that ends in 5 must be a multiple of 5, unless it is 5 itself. This case is eliminated by the fact that $n>1$, so $n^4+4 > 5$.
[/hide]

[hide="Proof - the hard part"]
Base case: $n=2$.

$2^4+4 \equiv 20 \equiv 0 \mod {10}$

Inductive step: $n=k+2$.

$(k+2)^4+4 \equiv k^4+8k^3+24k^2+32k+5 \equiv f(k)=k^4+8k^3+4k^2+2k+5 \mod {10}$

Nvm, I'm doing this wrong... time to go to sleep :sleeping:
[/hide]

[quote="chess64"]
...... and for odd $n$, it is congruent to 5. (I'll post the proof as soon as I finish it... :P )

[/quote]
Not true .. check when n=5.

The question has been asked before here (mulltiple times) .
Big hint : do not look,  unless you want to look at the solution..

 [hide]Can you factor it what if  ..[hide]you add $4n^2$ and subtract $4n^2$[/hide] [/hide]

[hide]
$n^4+4$ can be factored by either adding and substracting $4n^2$ or using Demoivre's Theorem. After factoring it into (n^2+2n+2)(n^2-2n+2), we can tell that it can't be prime because both factors are greater than 1.[/hide]

[quote="white_horse_king88"]Show that no prime numbers in the form $n^4 + 4$ exist for $n \; \epsilon \; \mathbf{Z}$ where $n > 1$.[/quote]

I think that you wanted to use \in as opposed to \epsilon, i.e. $\in$ as opposed to $\epsilon$. ;)


[hide]Using the Pythagorean Theorem:

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

$2x^{2}=4$

$x=\sqrt{2}$

The side length of the square is $\sqrt{2},$ so the area of the square is $2.$[/hide]

Some ppl in my class was grabbing calculators lol.....

if you remeber the lengths for an isocoles triangle
you can just say the lengths are root(x)

well, since it's a square, and all sides are going to be equal,
$x^{2}+x^{2}=4$
$2x^{2}=4$
$x=\sqrt{2}$

so, that is the sidelength, and the area would be, $(\sqrt{2})^{2}$ which is equal to $4$.  :10:

8parks11: you should learn LaTeX

so it would be $(\sqrt{2})^{2}= 2$

 :blush:

[quote="i_like_pie"][quote="klutzycherry14"]the area would be, $(\sqrt{2})^{2}$ which is equal to $4$.  :10:[/quote]
$\left(\sqrt{2}\right)^{2}=2$ :roll:[/quote]

sorry...my bad...that was a typo! lol...that is sooo me! :blush:  :roll:

[hide]To find to area of a square with a diagonal of $x$, do $\frac{x^{2}}{2}$ In this case, it would be $\frac{2^{2}}{2}$ $=2$[/hide]

[quote="Necrowarrio0"]Here is a easy question (Lots of ppl got it wrong in my class lol)

A square has a diagonal of 2m, what is the area?[/quote]

[hide]If a square has a diagonal $2m$, then it has sides $\sqrt{2}m$. Therefore, its area is $2m^{2}$[/hide]

[quote="jessica_chuang"]The answer is $2m^{2}$ right? :|[/quote]
That is correct. :)

@ anirudh, Ihatepie - The [b]amount of units[/b] in the area of the square is $2,$ but the units are $m^{2},$ not $m.$

[quote="i_like_pie"]@ anirudh, Ihatepie - The [b]amount of units[/b] in the area of the square is $2,$ but the units are $m^{2},$ not $m.$[/quote]

I fixed that...typo in latex...soon after my posting.  :roll:

Yup it is 2m^2 cause it's area of a 2D shape[/quote]


Equation of a straight line passing through the point of intersection of [b]x-y+1=0 [/b] and [b]3x+y-5=0[/b] and perpendicular to one of them is

[b](a) x+y+3=0
(b) x+y-3=0
(c) x-3y-5=0
(d) x-3y+5=0
[/b]

I think that the answer can be (b) or (d) both.But the answer given is (d). Please tell me why (b) is not the answer. And how can I know that the line is perpendicular to which of the two lines? Can't I assume that the line is perpendicular to any one of the both given lines?

Thanks a lot for your help.

Let 2 equation of a straight line : $ a_1x\plus{}b_1y\plus{}c_1\equal{}0$, $ a_2x\plus{}b_2y\equal{}c_2\equal{}0$.

parallel : $ \frac{a_1}{a_2} \equal{} \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
coincide : $ \frac{a_1}{a_2} \equal{} \frac{b_1}{b_2} \equal{} \frac{c_1}{c_2}$
perpendicular : $ a_1a_2 \plus{} b_1b_2 \equal{} 0$

Thanks a lot. I got that. I didn't notice it in this case, so it became a problem for me.Thank you so much.You solved my problem within a minute.


A sector is cut out from a circular sheet of paper, and the two edges of the cut are  joined together (with no overlap) to form a cone. The cone formed has a radius of 6 and a volume of 96pi. What is the number of degrees of the sector that was cut off?

[hide="What I did"] Since the area of a cone is $ \frac {1} {3} \pi r^2 h$ and r = 6, h=8. The radius of the sector is the slant height of the cone (visualize it), so we make a right triangle with legs of the radius and the height and hypotenuse of the slant height. This is a 6, 8, 10 right triangle, thus the radius of the sector is 10. The arc of the sector is the circumference of the base of the cone (again, visualize it). Since the base radius is 6, the circumference is 12pi. So we have a sector with radius 10 and arc of 12pi. The circumference of the circle from which the sector is cut is 20pi, so the sector is $ \frac {12\pi} {20\pi} \equal{} \frac {3} {5}$ of the cone. The measure of degrees of the sector would be $ \frac {3} {5} (360) \equal{} \boxed {216}$. [/hide]

They said the answer is...
[hide]144[/hide]

Can someone help explain how they arrived at this answer? Thanks.

I think $ 144^\circ$ is the angle of the other part of the circular piece of paper and, unless I'm doing something silly, I agree with your answer.

Ae you sue it is not "what is the number of degees in the sector of the circle [b][i]not[/i][/b][i]  cut off?

B/C  got 216 as well...[/i]

The question reads exactly:

"A sector is cut out from a circular sheet of paper, and the two edges of the cut are joined together (with no overlap) to form a cone. The cone formed has radius 6 and volume 96pi . What is the number of degrees in the sector that was cut off?"


invariant
induction
function
group theory
superior algebra
superior algebra unsolved

1)Let G be a group of odd order. Show that [b]x [/b]is not conjugate to [b]x[/b]^(-1) unless [b]x[/b]=e.

2)Let [b]g, h [/b]in S10. [b]g[/b]= (12) (345) (678) and [b]h[/b]= (13)(249) (6 8 10). How many solutions (for[b] x[/b] in S10) are there to the equation [b]x[/b][b]gx[/b]^(-1) =[b]h[/b]??? Write down one of them.

-------------------------------------------------------------------

For 1)
I said the situation is equivlent to: there exist a [b]g[/b] in G such that [b]gxg[/b]^(-1)=[b]x[/b]^(-1)    <=>  [b] x[/b]=e
so (<=) is extremely easy which I've done already.
for (=>), I said     [b]gxg[/b]^(-1)=[b]x[/b]^(-1)   implies [b]gxg[/b]^(-1)[b]x[/b]^(-1) = [b]x[/b]^(-2)
ie, <[b]x[/b]^(-2)> = <[b]x[/b]^2> is in the commucator of G,
since G is a group of odd order, implies that <[b]x[/b]^2> = <[b]x[/b]>
but then I don't know how to continue....or am I going on the wrong way???

For 2)
All I can do is the second part, [b]x[/b]=(23) (59) (7 8 10) is an example.
And I noticed that if I write [b]g[/b] or[b] h [/b]differently, like [b]g[/b]= (12)(678)(345)...etcetc....I would have different[b] x[/b]. But is it the way I should count for [b]x [/b](the number of ways to write [b]g[/b] times number of ways write [b]h[/b])??? Is there a better way?

the missing part on your 1)-proof: $ gxg^{\minus{}1}\equal{}x^{\minus{}1}\Longrightarrow g^2xg^{\minus{}2}\equal{}x$, ie. $ g^2$ and the cyclic group generated by it will centralize $ x$, but since $ g$ has odd order we have $ g\in <g^2>\Longrightarrow x\equal{}gxg^{\minus{}1}\equal{}x^{\minus{}1}\Longrightarrow x^2\equal{}1\Longrightarrow x\equal{}1$.

As for 2) let $ \Omega(g)\equal{}\{\{1,2\},\{3,4,5\},\{6,7,8\},\{9\},\{10\}\},\Omega(h)\equal{}\{\{1,3\},\{2,4,9\},\{6,8,10\},\{5\},\{7\}\}$ which are the invariant sub-sets decompositions of the action of $ g$ and $ h$ respectively.

Then $ xgx^{\minus{}1}\equal{}h$ implies for all $ \omega\in\Omega: x(\omega)\equal{}h(x(\omega))$, i.e. we must have for the fixed points the identification

(i) $ \{x(9),x(10)\}\equal{}\{5,7\}$ and then we can conclude
(ii) $ \{x(1),x(2)\}\equal{}\{1,3\}$ and finally,
(iii) $ \{\{x(3),x(4),x(5)\},\{x(6),x(7),x(8)\}\}\equal{}\{\{2,4,9\},\{6,8,10\}\}$

Generally, if two cycles $ c_1\equal{}(i_1,\ldots,i_k)$ resp. $ c_2\equal{}(j_1,\ldots,j_k)$ are conjugated by some element $ x$ with $ x(i_1)\equal{}j_1$, then we must have $ \forall s\equal{}1,\ldots, k: x(i_s)\equal{}j_s$ (this follows from induction, since modulo k in the index we have $ x(i_{s\plus{}1})\equal{}xc_1(i_s)\equal{}c_2x(i_s)\equal{}c_2(j_s)\equal{}j_{s\plus{}1}$) or in other words conjugation preserves "ordering of cycles".

Now any function determined by (i),(ii),(iii) is a well-defined permutation and iff the cycle ordering is preserved, it will solve $ xgx^{\minus{}1}\equal{}h$. This means, there are $ 72\equal{}2\cdot 2\cdot (6\cdot 3)$ different solutions (two choices for $ x(9)$ or (i), two choices for $ x(1)$ or (ii) and 6 choices to select $ x(3)$ and then 3 choices to determine $ x(6)$, which has to be in the "other" 3-cycle, i.e. 18 choices for (iii) ).

[quote]a little hint: notice that in $ G$ every element is a square, hence $ f(g)\equal{}g^2$ is a biyection.
[/quote]
what I did was: $ gxg^{\minus{}1}\equal{}x^{\minus{}1}$ then $ gx\equal{}x^{\minus{}1}g\equal{}\alpha$ then $ \alpha^2\equal{}gxx^{\minus{}1}g\equal{}g^2$ then $ gx\equal{}\alpha\equal{}g$ so $ x\equal{}1$.


1. The equation $ x^2\minus{}3x\plus{}1$ has two positive real roots, one smaller than 1 and one greater than 1.
Denote the smaller root by $ \beta\equal{}(3\minus{}\sqrt5)/2$.

2. Prove by induction:
If $ x_k\le\beta$ then $ x_{k\plus{}1}\equal{}(1\minus{}x_k)^2\ge(1\minus{}\beta)^2\equal{}\beta$.
If $ x_k\ge\beta$ then $ x_{k\plus{}1}\equal{}(1\minus{}x_k)^2\le(1\minus{}\beta)^2\equal{}\beta$.

3. Since $ 25$ is odd, the case $ x_1\le\beta$ after 25 steps implies $ x_1\ge\beta$,
and the case $ x_1\ge\beta$ after 25 steps implies $ x_1\le\beta$.

4. Hence $ x_k\equiv\beta$ for $ 1\le k\le25$ is the only solution.


Congrats to the Cdn IMO team this year!
What are the criterias for the Canadian IMO team? What kind of marks do you need to get on the CMO, APMO or the USAMO to be selected?

My quote from 'CMO result' topic.

[quote="lightrhee"][quote="Sunny"][quote="shobber"]well, can anyone tell me how does your selecting program work?[/quote]

It is basically based on the 3 contests APMO, CMO, and USAMO.  There may be other factors in there, but I don't know for sure.[/quote]
 In Canadian system, people decides who are going to be on IMO team, not like America where everything is strictly decided by numbers(scores on USAMO and so on). But the scores still plays important roll when it comes to choosing members of the team.
 APMO counts about 30%, CMO counts about 40%, and USAMO counts about 30%. They also look at MOCP(Mathematical Olympiads Correspondence Program) and winter camp.
 As I said, they do not use the percentages. They consider it about that much when choosing a team.[/quote]

Thanks

I got invited this year to USAMO, but not CMO or APMO.....

[quote="TheDreamer"]Really!?

That's really weird.

I did the exact reverse lol; did APMO and CMO but didn't do USAMO[/quote]

I believe that you have to write the AMC and then the AIMC in order to be invited to write the USAMO. not many school offer to write the AMC

[quote="opple"]I believe that you have to write the AMC and then the AIMC in order to be invited to write the USAMO. not many school offer to write the AMC[/quote]

You may be able to write the USAMO unofficially solely for the purpose of team selection.

On the other hand, qualifying USAMO officially does not necessarily put you into consideration for the team.

Yeah no school in our region writes AMC or AIME.

Is the best way to get it just to talk to my teacher?

Can someone tell me their homepage, so I can learn more information?

AMC and AIME are at http://www.unl.edu/amc