induction
superior algebra
superior algebra unsolved

G is a group which is generated with elements a and b such that they satisfy the following conditions:
$ b^{ \minus{} 4}a^{2007}b^4 \equal{} a^{2008} \ \ ,\ \ a^{ \minus{} 1}b^{1386}a \equal{} b^{1387}$
What is $ G$?

I'm having issues with the first line... (and from here on I assume $ a^{b^c}\equal{}a^{(b^c)}\qquad$)
Given $ a^{2008}\equal{}b^{\minus{}4}a^{2007}b^{4}$, I see \[ a^{2008\cdot2}\equal{}b^{\minus{}4}a^{2007}b^{4}\cdot b^{\minus{}4}a^{2007}b^{4}\equal{}b^{\minus{}4}a^{2007\cdot2}b^{4}\]
Thus $ a^{2008n}\equal{}b^{\minus{}4}a^{2007n}b^4$

This leads to something completely different from you first line...
In fact, the only way I see to raise the powers of $ b$ in the first relation is as follows:
\[ b^4a^{2008}\equal{}a^{2007}b^4\implies b^4ab^4a^{2008}\equal{}(b^4a)a^{2007}b^4\equal{}a^{2007}b^8\]\[ \implies (b^4a)^na^{2007}\equal{}a^{2007}b^{4n}\]

Yeah, that first line definitely needed explaining: I'll just show a simple case, but the general formula can be deduced by induction:

Given $ b^{\minus{}4}a^{2007}b^4\equal{}a^{2008}$, we have
\[ a^{2008^2}\equal{}\left( b^{\minus{}4}a^{2007}b^4\right)^{2008}\equal{}b^{\minus{}4}a^{2007\cdot 2008}b^4\]
(which works because the inside $ b$'s all cancel)
\[ \equal{}\left( b^{\minus{}4}a^{2008}b^4\right)^{2007}\equal{}\left( b^{\minus{}4\cdot 2}a^{2007}b^{4\cdot 2}\right)^{2007}\]
\[ \equal{}b^{\minus{}4\cdot 2}a^{2007^2}b^{4\cdot 2}\]

Nice! That's a really neat trick... thanks for explaining!

yes this trick can be applied in various contexts. try the following:
let $ G$ be a finite group of order $ n$, and let $ x \in G$ have prime order $ p$ such that $ n,p\minus{}1$ are coprime. then $ \langle x \rangle$ is normal if and only if $ x$ is central.

[quote]let $ G$ be a finite group of order $ n$, and let $ x \in G$ have prime order $ p$ such that $ n,p\minus{}1$ are coprime. then $ \langle x \rangle$ is normal if and only if $ x$ is central.[/quote]
The converse is easy but I'm having problems with the direct implication. Any hints?

use the above trick :-). given $ y$, write $ y^{\minus{}1} x y \equal{} x^i$ and consider $ y^{\minus{}n} x y^n$.


A complete cycle of a traffic light takes 60 seconds. During each
cycle the light is green for 25 seconds, yellow for 5 seconds, and
red for 30 seconds. At a randomly chosen time, what is the
probability that the light will NOT be green?  Express your answer as a common fraction.

At any given time, the probability that it is green is $ \frac {25}{60}$, so the probability that it's not green is $ 1 \minus{} \frac {25}{60} \equal{} \boxed{\frac7{12}}$.


[quote="cryax"]with a,b,c,x,y,z >0 prove: (ax+by+cz)^2+a^2+b^2+c^2>= 2(ab+bc+ac)
thank u very much![/quote]

It is very hard to prove since it's wrong :

Choose $ x\equal{}y\equal{}z\equal{}\frac 14$, $ a\equal{}2$, and $ b\equal{}c\equal{}1$ :

$ (ax\plus{}by\plus{}cz)^2\plus{}a^2\plus{}b^2\plus{}c^2\equal{}7$
$ 2(ab\plus{}bc\plus{}ac)\equal{}10$

and so $ (ax\plus{}by\plus{}cz)^2\plus{}a^2\plus{}b^2\plus{}c^2< 2(ab\plus{}bc\plus{}ac)$  :(

sorry i was wrong, it must be:with a,b,c,x,y,z>0    prove:                                                      (ax+by+cz)^2/(xy+yz+xz)+a^2+b^2+c^2>=2(ab+bc+ac)

[quote="cryax"]sorry i was wrong, it must be:with a,b,c,x,y,z>0    prove:                                                      $ \frac {(ax + by + cz)^2}{xy + yz + xz} + a^2 + b^2 + c^2\ge2(ab + bc + ac)$[/quote]
Nice but easy: Taking the discriminant with respect to $ a$, we obtain that the ineq holds if $ [b(x + y) - c(x + z)]^2\ge0$ which is obviously true. :)

The ineq is in fact equivalent to $ [a(x + y)(x + z) - bz(x + y) - cy(x + z)]^2 + (xy + yz + xz)[b(x + y) - c(x + z)]^2\ge0$.

Equality occurs iff $ (a,b,c)\sim (y + z,x + z,x + y)$, equivalently iff

$ (x,y,z)\sim (s - a,s - b,s - c)$ where $ s = \frac {a + b + c}2$. Thus, there may be a geometrical solution also.  :maybe: (If $ a,b,c^$ are not sides of a triangle, it holds trivially.)

What intrigues me is the fact that it is not symmetric in $ (a,b,c)\leftrightarrow (x,y,z)$. But in view of the whole bunch of equality cases, I don't think that it can be strengthened.

thank a lot. i have another question:
solve this equation: sin(4x/3)+(sin(x)+cos(x))^2=0.


I swear Im sure Ive seen a similar thread but the search functions came up with nothing..

before the announcement (new post ratings) my post count was around 920. now it's 889. what happened? did i lose points because i posted a blog problem (which recieved a rating of 2?)

[quote="Valentin Vornicu"]A couple of other forums have been added to the no post count list. That's why you've "lost" some of those post numbers.[/quote]
Which forums are these (just wondering)?

[quote="mathcountsfreak"]Well games and fun factory isn't being counted for posts, but I don't know what else..[/quote]

The Games and Fun Factory and the Test forum both don't go to your overall post count but they were set that way before the announcement came up.


function
inequalities proposed
inequalities

[quote="lgmco"]If a, b, c, d ∈ R+, satisfy a + b + c + d = 1 prove that:
6(a^3 + b^3 + c^3 + d^3) ≥ a^2 + b^2 + c^2 + d^2 +1/8[/quote]
A hint: Take a look at the tagent to the function $ f(x) \equal{} 6x^3\minus{}x^2\minus{}\frac{1}{32}$ at $ x\equal{}\frac{1}{4}$.

[quote="lgmco"]If $ a,b,c,d\in R_{ \plus{} }$, satisfy $ a \plus{} b \plus{} c \plus{} d \equal{} 1$ prove that:
$ 6(a^{3} \plus{} b^{3} \plus{} c^{3} \plus{} d^{3})\geq a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2} \plus{} \frac {1}{8}$[/quote]

homogenizing and AM-GM does the trick :starwars:


inequalities
induction
function
calculus
inequalities theorems

[quote="greatestmaths"]please  can you  give me the proof of  AM-GM  inequality by recurrence.
please it's very important to me.[/quote]

Induction, not recurence  :wink:

I know one by a spesific induction (called in french: réccurence descendante)

You prove that the result is true for EVEN numbers first(its easy)

Then you prove that: if the result is true for $ n$ It's also true for $ n\minus{}1$(which is also easy  :lol: )

Good luck  :wink:

Actually, if I can remember the  inductive proof for the AM-GM inequality, you prove the statement is correct for all powers powers of 2 (beginning by two) instead of even numbers. Then you prove that if it's correct for n it's correct for n-1. Then the statement is proved for all n>=2. It's called Cauchy induction. See here : http://mzargar.wordpress.com/2009/07/19/cauchys-method-of-induction/
A neat proof can be found in [b]Secrets in inequalities, Volume 1[/b] by Pham Kim Hung.

see here: http://www.mathdb.org/notes_download/elementary/algebra/ae_A2.pdf
(p.8, example 3.5)

Just for fun I put a proof by [i][b]Lagrange Multiplier[/b][/i] in here:

[b]If $x_1,x_2,...,x_n>0$ then prove $\frac{x_1^n+x_2^n+...+x_n^n}{n}  \geq  x_1x_2...x_n$ and equality case happens when $x_1=x_2=...=x_n$.[/b]

Define the function $f(x_1,x_2,...,x_n)=x_1^n+x_2^n+...+x_n^n-nx_1x_2...x_n$ and now our purpose is to prove that $f(x_1,x_2,...,x_n)\geq 0$.
$1\leq i\leq n:\frac{\partial  f(x_1,x_2,...,x_n)}{\partial x_i}=nx_i^{n-1}-n\frac{x_1x_2...x_n}{x_i}$
We know that extreme of the function $f$ happens when we put $1\leq i\leq n:\frac{\partial  f(x_1,x_2,...,x_n)}{\partial x_i}=0$.By doing this we get:
$1\leq i\leq n:x_i^{n}=x_1x_2...x_n\implies x_1=x_2=...=x_n$
So extreme happens when $ x_1=x_2=...=x_n$! And we can check to see that this is minimum and also for $ x_1=x_2=...=x_n$ we have the $f(a,a,...a)=0$ so it's proved that $f(x_1,x_2,...,x_n)\geq f(a,a,...a)=0$.

[i]P.S.[/i] You can also find more interesting proofs in here:
[url=http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proofs]Proofs of AM-GM Inequality[/url]

there are many inductive proofs.
here is another one that uses calculus, but only the basic notion of limit.

$x_1 + \ldots + x_n =$

$\frac {x_1+\ldots+x_{n-1}} {n-1} + \ldots + \frac {x_2+\ldots+x_{n}} {n-1} \ge$

$\sqrt[n-1]{x_1 \cdots x_{n-1}} + \ldots + \sqrt[n-1]{x_2 \cdots x_{n}}$

and this step can be repeated. in every stage we start with a sum

$x_1^{\frac{1}{n}+a} \cdots x_{n-1}^{\frac{1}{n}+a} \cdot x_n^{\frac{1}{n}-(n-1)a}
+ \ldots +
x_1^{\frac{1}{n}-(n-1)a} x_{2}^{\frac{1}{n}+a} \cdots x_n^{\frac{1}{n}+a}$

(where $a$ may be negative) and deduce that it is bigger or equal to

$x_1^{\frac{1}{n}-\frac{a}{n-1}} \cdots x_{n-1}^{\frac{1}{n}-\frac{a}{n-1}} \cdot
 x_n^{\frac{1}{n}+a}
+ \ldots +
x_1^{\frac{1}{n}+a} x_{2}^{\frac{1}{n}-\frac{a}{n-1}} \cdots x_n^{\frac{1}{n}-\frac{a}{n-1}}$

and if we repeat this to infinity the lower bound approaches

$\rightarrow n x_1^{\frac{1}{n}} \cdots x_n^{\frac{1}{n}}$


[quote="Fuaran"]Just let me make sure I've got it right:
is for N=3 the solution is:
a1=0, a2=1, a3=6, b1=2, b2=3, b3=8
Let me know. I might have a solution, just want to be sure I understand the problem...[/quote]

Yes, you are right.  :)


Express -2x^-2 as a fraction without the negative exponent.

My answer is 1/x^2.  is this right?

If not, why not?

To start, -2x^-2 means -2(x^-2), doesn't it?

-2(x^-2)=
-2([b]1/x^2[/b])=
[b]-2/x^2
[/b]
that is your answer. Confused on how you got 1/x^2, though.

Does -2x^-2 mean $ \minus{}2*x^{\minus{}2}$ or $ (\minus{}2x)^{\minus{}2}$ or $ \minus{}(2x^{\minus{}2})$ etc.?

If it means $ \minus{}2*x^{\minus{}2}$, the answer is $ \frac{\minus{}2}{x^{2}}$

If it means $ (\minus{}2x)^{\minus{}2}$, the answer is $ \frac{1}{4x^{2}}$.

If it means $ \minus{}((2x)^{\minus{}2})$, the answer is $ \frac{\minus{}1}{4x^{2}}$

It's a little unclear...

if these are wrong, sorry, I did them in my head...

no matter what he meant, sharkman left out the 2. His answer would be correct if you just factored it into the expression in the appropriate way.

Remember PEMDAS when confused...
Exponents first, then Multiplication.
$ \minus{}2x^{\minus{}2}\equal{}\minus{}2(x^{\minus{}2})\equal{}\minus{}2/x^{2}$


Let set $A$ be the set $\{1,2,3,...,n\}$.We say a collection $J$ of subsets of $A$ is good if and only if for any two elements of $J$ the union and intersection of these element are also be the elements of $J$ and also $A$ and $\{\}$ be the elements of $J$.
Problem: for a set $\{1,2,3,...,n\}$ how many collection $J$ are exist that are good?

Saman Gharib


What are you taking this May??

personally, i am taking B. hopefully get a 5.

I am taking B also.  I think I'm rusty on some of the topics but a last week review should take care of that.

I'm taking the AP Physics B test too.  I don't know if I'll get a five, but I think I'm guaranteed a 3.  I'm good with the mechanics stuff but I haven't gotten to some of the other topics yet.  Better study...

There are free response problems here:

http://www.collegeboard.com/student/testing/ap/physics_b/samp.html?physicsb

and here:

http://www.collegeboard.com/student/testing/ap/physics_c/samp.html?physicsc

I am taking Physics B.  I think I should do fine but, of course, I need to do some more reviewing.  It's my last AP so it isn't on the top of my list of concerns right now.

[quote="shobber"]can post a complete test of AP test??
I really wanna know what level you get to.[/quote]

Well, you can look up the course outline on http://www.collegeboard.com.  Physics B basically skims a lot of different topics (as opposed to Physics C which focuses on just a few a lot).

I think I'm screwed on C E&M.  But C mechanics and B shouldn't be that bad.  I hope.

I am taking C in two weeks. I think I'm in good shape for mechanics, but I have issues with electricity and magnetism. :(

im supposedly taking physics c mechanics and e and m according to my school, but if they don't let me only take one i will simply walk out of the room in between and not come.  my teacher was really bad this year and didn't start teaching e and m until the end of march.  he didn't even teach mechanics well.  now over this vacation im busy studying for the mechanics section and i won't even have time to look into the e and m.

Is there anyway to take C before B (or to skip B)?  Some juniors from my school told me that colleges don't care about physics B and that C is actually easier.

you don't have to take B to take C.  i'm taking C right now as a freshman, never having taken B.  there are a few seniors in my class who have never taken B either.  although "B" comes before "C" in the alphabet, it doesn't necessarily mean that "C" requires "B" =).  the only problem might be with your school in terms of taking the C course.  it could be that your school possibly requires the B course to take the C course.  but, if you personally want to take the C without B you can always do an independent study and just sign up for the exam.

The difference between B and C is generally that B has absolutely no calculus.  C  makes use of some elementary calculus.  I took B last year in eigth grade and am taking C next year in tenth.  I did this because my school believes in pre-calculus.

Plus, C doesn't cover such topics as thermodynamics and optics.

well, my school requires "elementary knowledge" of physics prior to taking the C exam, so you have had to take either "regular" or physics B in order to take physics C, this might be something you'd have to look into

I heard that Physics C requires Maxwell's (which is calculusy, and not that elementary).  But I could be wrong.


[b]Romanian Marian Dinca teacher gave a solution of the problems China following:[/b]

$ Let \ two \ triangles \ ABC \ and \ XYZ \ which \ hawe \ tree \ side \ length \ a,b,c \ and x,y,z \ respectively. \\
Prove \ that: \ With \ amy \ positive \ real \ numbers \ m,n,p \ and \ arbitrary \ point \ M \ we \ hawe: \\
\ \ \ \ \ m.\frac {x^2}{a^2} + n.\frac {y^2}{b^2}.MB^2 + p.\frac {z^2}{c^2}\ge 4.\sqrt {\frac {mnp}{m + n + p}}.S_{XYZ}.$

[b]Forum on Chinese to the topic:  [/b] http://www.mathlinks.ro/viewtopic.php?p=1229457#1229457

[color=red] Ok the solution Marian Dinca gave is similar with the proof I sent to Jian Liu  who is the proposer of this problem.
  I give here the proof was sent to me by Jian Liu [/color]

[color=red] Here is my solution.
First of all we knew the well known inequality $ ma^2 \plus{} nb^2 \plus{} pc^2\ge4\sqrt {mn \plus{} np \plus{} pm}S (2)$ now according to $ (2)$ for triangle $ XYZ$ we have :
  $ m\frac{MA^2}{a^2}x^2\plus{}n\frac{MB^2}{b^2}y^2\plus{}p\frac{MC^2}{c^2}z^2\ge4S_{XYZ}\sqrt{mn\frac{MA^2.MB^2}{a^2b^2}\plus{}np\frac{MB^2.MC^2}{b^2c^2}\plus{}pm\frac{MC^2MA^2}{c^2a^2}}$
So we only need to show that
$ mn\frac{MA^2.MB^2}{a^2b^2}\plus{}np\frac{MB^2.MC^2}{b^2c^2}\plus{}pm\frac{MC^2MA^2}{c^2a^2}\ge\frac{mnp}{m\plus{}n\plus{}p}$
or $ \frac {1}{p}.\frac {MA^2.MB^2}{a^2b^2} \plus{} \frac {1}{n}.\frac {MC^2MA^2}{c^2a^2} \plus{} \frac {1}{m}.\frac {MB^2.MC^2}{b^2c^2}\ge\frac {1}{m \plus{} n \plus{} p} (3)$
Noted that the Hayashi' inequality give $ \frac {MAMB}{ab} \plus{} \frac {MBMC}{bc} \plus{} \frac {MCMA}{ca}\ge1$ so $ (3)$ is clearly true by Cauchy Schward inequality.We are done.[/color]


Ok, I can't seem to be able to find this anywhere on the internet so I guess I was the first ever to post this question online. NO, it's not missing any information! Enjoy.. [url]http://robertgatzke.com/math/solveY.pdf[/url]

[quote="unimpossible"]is it a clean answer?[/quote]

Not sure what you mean?

I actually doubt the answer will be clean (or nice and simple looking like $ 1$ or $ \frac {1}{2}$). As for solving this:
[hide="Hint"]Let the length between $ 24^o$ and the right angle be represented as $ x$ (or another variable if you desire). Next, find $ x$ in terms of $ y$.

From there it's possible to evaluate $ y$.[/hide]

[quote="gatzke"]Well here's what I have so far:

[hide]
if Tan theta = y/x then y=xTan theta so:

2.35Tan15 = xTan24
x = 2.35Tan15 / Tan24

And from there I'm lost..

[/hide][/quote]

The second step is incorrect, it should be:
[hide]$ (2.35 \plus{} x)\tan(15^o) \equal{} x\tan(24^o)$[/hide]

[b][Edit][/b] You're welcome.  :)

Thank you Patterns. Problem solved.

y is correct for both angles when checking my answer.

Cheers :)


LaTeX
inequalities proposed
inequalities

this is a simple problem

For $1\le i\le c$ let $p_i>0$ and $q_i>0$ and assume $\sum_{i=1}^{c}p_i=\sum_{i=1}^{c}q_i=1$. also suppose that:
$(\sum_{i=1}^{k}p_i)q_{k+1}\geq (\sum_{i=1}^{k}q_i)p_{k+1}$ for $0<k<c$.

prove that: $\sum_{i=1}^{k} p_i\le \sum_{i=1}^{k}$ for all $0<k<c$.

[quote="amirhtlusa"]
\sum_{i=1}^{k} p_i\le \sum_{i=1}^{k}[/quote]

There is a typo. (Deleted the LaTeX tags - do you see?)


the easy but ugly way: Put $a=x^{3}$ etc. Then it becomes, using the well known symmetric sum notation,
$[6,0,0]+[2,2,2]\ge[3,3]+[4,1,1]$, but this follows directly from Schur & Muirhead :)

Is there a solution without directly using Schur?

a one line solution should be by applying turkevici for $a=\sqrt{x}, b=\sqrt{y}, c=\sqrt{z}, d=1$.

Let a = x/y, b = y/z and c = z/x. Using Schur's Ineq for (xy), (yz), (zx).
Have fun :)


vector
linear algebra
matrix
linear algebra unsolved

can anyone help me with answerig these 2 questions?

sMat2(R) = { ( a  b ) | a,b,d element of R}  be the set of symmetric 2 x 2 matrices.
[color=white]____________[/color]b   d   |

show that sMat2(R) is a vector subspace of the vector space

Mat2(R) = { ( a  b ) | a,b,c,d element of R} of all 2 x 2 matrices
[color=white]__________ [/color]c   d    |

another question

if we let V = R^2 = {(a,b)| a,b element R} be a vector space of column vectors of length 2.

How do u show that          W = {(a,b)| a,b element Z}      is not a vector subspace of V?

For the first question, note that the sum of two symmetric matrices is a symmetric matrix: If $A_{i,j}=A_{j,i}$ and $B_{i,j}=B_{j,i}$ then $(A+B)_{i,j}=(A+B)_{j,i}$. Also if we multiply every element of symmetric matrix by the same scalar $c$, we have $A_{i,j}\to cA_{i,j}$. But also $A_{j,i}\to cA_{j,i}=cA_{i,j}$ because $A_{i,j}=A_{j,i}$ for every $i,j$. Finally, the zero matrix $0_2$ is trivially symmetric.

For the second question, $W$ is not closed under scalar multiplication: $(1,1)\in W$ but $\frac{1}{2}(1,1)=\left (\frac{1}{2},\frac{1}{2}\right )\not \in W$.


topology
advanced fields
advanced fields unsolved

A set $E \subset \mathbb{R}^{n}$ is said to be [i]chain-connected[/i] if for any $x, y \in E$ and $\epsilon > 0$ there is a sequence of points $x = t_{1}, t_{2}, \ldots, t_{m}= y \in E$ such that $|t_{i}-t_{i+1}| \le \epsilon$ for $i = 1 \ldots m-1$.

Prove that if $E$ is connected, it is chain-connected.

Is this still true if $\mathbb{R}^{n}$ is replaced by an arbitrary metric space?

Is there any way to generalize this concept to spaces that are not metrizable?

1) Define the chain-connected component of a point and observe that it is an open set. A connected set cannot be divided into $>1$ open subsets.

2) Sure.

3) Yes for [url=http://en.wikipedia.org/wiki/Uniform_space]uniform spaces[/url], but probably not for general topological spaces.


geometry
analytic geometry
trigonometry
graphing lines
slope
function
algebra

Which is the center of the circle (A)?

Could someone attach the figure of the problem?

Thanks in advance.

kunny

Of course, the center of the circle $ (A)$ is $ A$

Well, assume that $ BC \equal{} 1,AB \equal{} c > 0$(in other case we can use the homothety). Then, consider Cartesian coordinate system, where $ BC$- is x-axis, and $ BA$- y-axis. If $ (x_Q,y_Q)$ - coordinates of $ Q$, then because $ Q$ is on the circle$ (A)$, we have that:
\[ x_Q^2 \plus{} (y_Q \minus{} c)^2 \equal{} c^2\Leftrightarrow x_Q^2 \minus{} 2y_Qc \plus{} y_Q^2 \equal{} 0
\]
But because $ MNPQ$ - is square, $ QP \equal{} QM \equal{} y_Q$. From the equation $ x \plus{} \frac {y}{c} \equal{} 1$, that describes the line $ AC$, we find that $ P$ have the coordinates $ (1 \minus{} \frac {y_Q}{c},y_Q)\Rightarrow QP \equal{} 1 \minus{} \frac {y_Q}{c} \minus{} x_Q \equal{} y_Q$. So we system of equations:
\[ \begin {cases} x_Q^2 \minus{} 2y_Qc \plus{} y_Q^2 \equal{} 0 \\
1 \minus{} \frac {y_Q}{c} \minus{} x_Q \equal{} y_Q\end {cases}
\]
From the second equation $ x_Q \equal{} 1 \minus{} \frac {y_Q}{c} \minus{} y_Q$, and use it in second equation we have:
\[ (2 \plus{} \frac {2}{c} \plus{} \frac {1}{c^2})y_Q^2 \minus{} 2(c \plus{} 1 \plus{} \frac {1}{c})y_Q \plus{} 1 \equal{} 0
\]
which have 2 solutions:$ y_Q \equal{} c$ and $ y_Q \equal{} \frac {c}{2c^2 \plus{} 2c \plus{} 1}$. But if $ y_Q \equal{} c$, then $ x_Q \equal{} 1 \minus{} \frac {y_Q}{c} \minus{} y_Q \equal{} \minus{} c < 0$, so this case is impossible. Then we have $ y_Q \equal{} \frac {c}{2c^2 \plus{} 2c \plus{} 1}\Rightarrow S_{MNPQ} \equal{} QP^2 \equal{} \left(\frac {c}{2c^2 \plus{} 2c \plus{} 1}\right)^2$. Of course $ \left(\frac {c}{2c^2 \plus{} 2c \plus{} 1}\right)^2$ have maximum at the same point with $ \frac {c}{2c^2 \plus{} 2c \plus{} 1}$.
\[ \left(\frac {c}{2c^2 \plus{} 2c \plus{} 1}\right)' \equal{} \frac {c'(2c^2 \plus{} 2c \plus{} 1) \minus{} c(2c^2 \plus{} 2c \plus{} 1)'}{\left(2c^2 \plus{} 2c \plus{} 1\right)^2} \equal{} \frac {2c^2 \plus{} 2c \plus{} 1 \minus{} 4c^2 \minus{} 2c}{\left(2c^2 \plus{} 2c \plus{} 1\right)^2} \equal{} \frac {1 \minus{} 2c^2}{\left(2c^2 \plus{} 2c \plus{} 1\right)^2} \equal{} 0
\]
and because $ c > 0\Rightarrow c \equal{} \frac {1}{\sqrt {2}}$, therefore $ S_{max} \equal{}\left(\frac {\frac {1}{\sqrt {2}}}{2(\frac {1}{\sqrt {2}})^2 \plus{} 2(\frac {1}{\sqrt {2}}) \plus{} 1}\right)^2 \equal{} \frac {3 \minus{} 2\sqrt {2}}{4} \equal{} 0.042893218...$

[quote="greenvert"]Well, assume that $ BC = 1,AB = c > 0$(in other case we can use the homothety). Then, consider Cartesian coordinate system, where $ BC$- is x-axis, and $ BA$- y-axis. If $ (x_Q,y_Q)$ - coordinates of $ Q$, then because $ Q$ is on the circle$ (A)$, we have that:
\[ x_Q^2 + (y_Q - c)^2 = c^2\Leftrightarrow x_Q^2 - 2y_Qc + y_Q^2 = 0
\]
But because $ MNPQ$ - is square, $ QP = QM = y_Q$. From the equation $ x + \frac {y}{c} = 1$, that describes the line $ AC$, we find that $ P$ have the coordinates $ (1 - \frac {y_Q}{c},y_Q)\Rightarrow QP = 1 - \frac {y_Q}{c} - x_Q = y_Q$. So we system of equations:
\[ \begin {cases} x_Q^2 - 2y_Qc + y_Q^2 = 0 \\
1 - \frac {y_Q}{c} - x_Q = y_Q\end {cases}
\]
From the second equation $ x_Q = 1 - \frac {y_Q}{c} - y_Q$, and use it in second equation we have:
\[ (2 + \frac {2}{c} + \frac {1}{c^2})y_Q^2 - 2(c + 1 + \frac {1}{c})y_Q + 1 = 0
\]
which have 2 solutions:$ y_Q = c$ and $ y_Q = \frac {c}{2c^2 + 2c + 1}$. But if $ y_Q = c$, then $ x_Q = 1 - \frac {y_Q}{c} - y_Q = - c < 0$, so this case is impossible. Then we have $ y_Q = \frac {c}{2c^2 + 2c + 1}\Rightarrow S_{MNPQ} = QP^2 = \left(\frac {c}{2c^2 + 2c + 1}\right)^2$. Of course $ \left(\frac {c}{2c^2 + 2c + 1}\right)^2$ have maximum at the same point with $ \frac {c}{2c^2 + 2c + 1}$.
\[ \left(\frac {c}{2c^2 + 2c + 1}\right)' = \frac {c'(2c^2 + 2c + 1) - c(2c^2 + 2c + 1)'}{\left(2c^2 + 2c + 1\right)^2} = \frac {2c^2 + 2c + 1 - 4c^2 - 2c}{\left(2c^2 + 2c + 1\right)^2} = \frac {1 - 2c^2}{\left(2c^2 + 2c + 1\right)^2} = 0
\]
and because $ c > 0\Rightarrow c = \frac {1}{\sqrt {2}}$, therefore $ S_{max} = \left(\frac {\frac {1}{\sqrt {2}}}{2(\frac {1}{\sqrt {2}})^2 + 2(\frac {1}{\sqrt {2}}) + 1}\right)^2 = \frac {3 - 2\sqrt {2}}{4} = 0.042893218...$[/quote]

Nice Solution, greenvert. I could obtain the same answer as well in using trigonometic function.

$ \frac{c}{2c^2+2c+1}\leq \frac{c}{2\sqrt{2}c+2c}=\frac{\sqrt{2}-1}{2}$. the equality holds when $ c={\frac{1}{\sqrt{2}}}$.

Here is my solution. [color=red]Last Edited[/color]

In attached figure, let $ BC = 1,\ AB = x$, and $ \angle{QBC} = \theta$, the relation of cord and tangent line gives $ \angle{BAQ} = 2 \theta \ \left(0 < 2\theta < \frac {\pi}{2}\right)$.
$ BM + MN + NC = 1\Longleftrightarrow x\sin 2\theta + l + \frac {l}{x} = 1$, where $ l$ is the side length of the square $ QMNP$.
$ RB = QM\Longleftrightarrow l = x(1 - \cos 2\theta)$. Elliminating $ x = \frac {\cos 2\theta - l}{\sin 2\theta}$, we have $ l = \frac {\cos 2\theta - l}{\sin 2\theta }(1 - \cos 2\theta )\Longleftrightarrow l = \frac {\cos 2\theta (1 - \cos 2\theta)}{1 - \cos 2\theta + \sin 2\theta}$.

$ \therefore l = \frac {2\sin ^ 2 \theta \cos 2\theta}{2\sin ^ 2 \theta + 2\sin \theta \cos \theta}$

$ = \frac {\sin \theta \cos 2\theta}{\sin \theta + \cos \theta} = \frac {\tan \theta}{1 + \tan \theta}\cdot \cos 2\theta$. Set ${ u = \tan \theta}$, since $ 0 < \theta < \frac {\pi}{4}$, we have $ 0 < u < 1$ and $ \cos 2\theta = \frac {1 - \tan ^ 2 \theta}{1 + \tan ^ 2 \theta} = \frac {1 - u^2}{1 + u^2}$.

$ \therefore l = \frac {u(1 - u)}{1 + u^2} = - 1 + \frac {u + 1}{u^2 + 1}$.

$ = - 1 + \frac {1}{\frac {u^2 + 1}{u + 1}} = - 1 + \frac {1}{u + 1 + \frac {2}{u + 1} - 2}\leq - 1 + \frac {1}{2\sqrt {(u + 1)\cdot \frac {2}{u + 1}} - 2} = \frac {\sqrt {2} - 1}{2}$.

The equality holds ${ u + 1 = \frac {2}{u + 1}\ (0 < u < 1)\ \Longleftrightarrow u = \sqrt {2} - 1 \Longleftrightarrow \tan \theta} = \sqrt {2} - 1\Longleftrightarrow \theta = \frac {\pi}{8}$.

Consequently, the desired maximum value is $ l^2 = \frac {3 - 2\sqrt {2}}{4}$.

[quote]$ \angle{BAQ} \equal{} \angle{QBC} \equal{} \theta \ \left(0 < \theta < \frac {\pi}{2}\right)$[/quote]
Are you sure that it's right?

Yes, that's a basic theorem  related to a circle, though I don't know what we call the theorem in English.

It is right, when the $ \angle BAQ$ is inscribed angle, but in our case $ \angle BAQ$ - is angle with the vertex in the center of the circle, so $ \angle QBC\equal{}\frac{1}{2}\angle BAQ$

Sorry, you are right.  :oops:

I have just edited.

kunny

[quote="zaizai-hoang"]Let $ ABC$ be a right triangle, $ B = 90^0$ . The side $ BC$ is fixed and $ [AB]$ is changeability (not fix). The circle $ w = C(A, AB)$ intersects the $ [AC]$ at the point $ D$ . In the surface which is limited by arc $ BD$ and segments $ BC$ ,  $ CD$ construct square $ MNPQ$ such that $ \{M,N\}\subset [BC]$ , $ P\in [CD]$ and $ Q$ is on the arc $ BC$ . When $ [AB]$ change find the maximum value of area the square $ MNPQ$ .[/quote]

[b][u]Proof.[/u][/b] I"ll use the analytical geometry. Denote the mobile point $ A(0,a)$ where $ a > 0$ and w.l.o.g. the fixed point $ C(1,0)$ .

The equation of the circle $ w = C(A,a)$ is $ x^2 + y^2 = 2ay$ and the equation of the line $ AC$ is $ y = a(1 - x)$ . Therefore,

$ D\in\cap\{\begin{array}{c} x^2 + y^2 = 2ay \\
 \\
y = a(1 - x)\end{array}\|$ $ \implies$ $ \{\begin{array}{c} x_D = \frac {a}{\sqrt {1 + a^2}} \\
 \\
y_D = a\cdot\frac {\sqrt {1 + a^2} - a}{\sqrt {1 + a^2}}\end{array}$ . Hence the slope of the line $ BD$ is $ \sqrt {1 + a^2} - a$ .

$ Q\in\cap\{\begin{array}{c} x^2 + y^2 = 2ay \\
\ y = tx\end{array}$ , where $ 0 < t < \sqrt {1 + a^2} - a$ . Hence $ \{\begin{array}{c} x_Q = \frac {2at}{1 + t^2} \\
\ y_Q = \frac {2at^2}{1 + t^2}\end{array}$ . But  $ P\in\cap\{\begin{array}{c} y = \frac {2at^2}{1 + t^2} \\
\ y = a(1 - x)\end{array}$ . Hence

$ x_P = \frac {1 - t^2}{1 + t^2}$ . Thus, $ QM = QP\Longleftrightarrow y_Q = x_P - x_Q$ $ \Longleftrightarrow$ $ 2at^2 = 1 - t^2 - 2at$ $ \Longleftrightarrow$ $ (2a + 1)t^2 + 2at - 1 = 0$ $ \Longleftrightarrow$

$ t\in\{ - 1,\frac {1}{2a + 1}\}$ $ \Longleftrightarrow$ $ t = \frac {1}{2a + 1}$ . Prove easily that $ \frac {1}{2a+1}\in(0,\sqrt {1+a^2}-a)$ . Therefore,

$ [QMNP] - \max.$  $ \Longleftrightarrow$ $ y_Q - \max.$ $ \Longleftrightarrow$ $ \frac {2a}{1 + (2a + 1)^2} - \max.$ $ \Longleftrightarrow$  $ \frac {a}{2a^2 + 2a + 1} - \max.$ $ \Longleftrightarrow$

$ \frac {2a^2 + 2a + 1}{a} - \min.$ $ \Longleftrightarrow$ $ 2a + 2 + \frac 1a - \min.$ $ \Longleftrightarrow$ $ 2a + \frac 1a - \min.$ .

Since  $ 2a\cdot \frac 1a = 2$ (constant), the area $ [QMNP]$ is maximum iff $ 2a = \frac 1a$ , i.e. $ \boxed {\ a = \frac {\sqrt 2}{2}\ }$ .

[quote="zaizai-hoang"]Let ABC is a right angled triangle, $ \widehat{B} \equal{} 90^0$. BC is fix and $ AB$ is changeability (not fix). The circle $ (A)$, the radius of the circle is $ AB, (A)$ intersect $ AC$ at $ D$. In the plane was limited by chord $ BC$, side $ BC$ and side $ CD$ construct square $ MNPQ$ such that $ M,N$ are on $ BC, P$ on $ CD, Q$ on chord $ BC$
When AB change find the maximum value of area the square MNPQ.[/quote]

Solution in using Trigonometric Fuction Part 2

In the attached figure,  $ QM \equal{} NM\Longleftrightarrow r(1 \minus{} \cos \alpha ) \equal{} \cos \alpha \minus{} r\sin \alpha \Longleftrightarrow r \equal{} \frac {\cos \alpha}{1 \minus{} \cos \alpha \plus{} \sin \alpha}$. Thus the side length $ d$ of the square $ PQMN$ is $ d \equal{} \frac {\cos \alpha}{1 \minus{} \cos \alpha \plus{} \sin \alpha}(1 \minus{} \cos \alpha)\ (0 < \alpha < \frac {\pi}{2})$

$ \equal{} \frac {\cos \alpha}{1 \plus{} \frac {\sin \alpha}{1 \minus{} \cos \alpha}} \equal{} \frac {\cos \alpha}{1 \plus{} \frac {1}{\tan \frac {\alpha}{2}}}$.

Let $ z \equal{} \tan \frac {\alpha}{2}\ (0 < z < 1)$, as I posted in #6, $ d \equal{} \frac {\frac {1 \minus{} z^2}{1 \plus{} z^2}}{1 \plus{} \frac {1}{z}} \equal{} \frac {z(1 \minus{} z)}{1 \plus{} z^2}$ which follows the solution by setting $ \theta \equal{}\frac{\alpha}{2}$ in my post #6. Needless to say, You can differentiate $ d$ with respect to $ z$ directly.


Given the sequence defined by x(0)=0, x(1)=1, and x(n+1)=5x(n)+x(n-1) for all integral n>=1.  Find the value of (x(60))^2 - (x(59))(x(61)).

I noticed that an alternating pattern in similar expressions for increasing values of n that suggests the answer is -1.  I'm still searching for a way to prove a general result, perhaps by mathematical induction.  Any ideas?

Do you mean x(n+1)=5x(n)+x(n-1)?  What you typed would inply that 5x(n)=0, which is a contradiction to the fact that 1=x(1).
I suppose that you could use the equality given in the equation and keep reducing it down until you get the answer.  Or expand the function to something like x(n+2)=26x(n)+5x(n-1). or that kind of idea.

Do you know the standard way of solving difference equations? Solving the characteristic equation and all that?

Sorry about the typo -- the first n-1 should have been n+1 -- I just repaired my original post.   Just for the heck of it, I did derive the closed form for x in terms of n for the recurrence involving the values (5+sqrt 29)/2 and (5-sqrt29)/2.  Is this what you're referring to?    Is it useful here?

Just sub in the values, stuff could cancel?

I have a beautiful solution.

Define

$A=\left( \begin{array}{cc}
5 & 1 \\
1 & 0 \\
\end{array} \right)$

Clearly

$A^n=\left( \begin{array}{cc}
x_{n+1} & x_n \\
x_n & x_{n-1} \\
\end{array} \right)$

We have

$\det A_n = (-1)^n = x_{n+1}x_{n-1}-x_n^2$

and it's trivial from here

EDIT: for $\LaTeX$

Very nice!  I seem to remember using matrices and determinants to prove a similar expression about the Fibonnaci series, but I didn't think of extending it to a sequence like this one.   Do you think that method could be generalized to more arbitrary sequences?


MATHCOUNTS
ratio
analytic geometry
email
geometry
AMC
AIME

try looking up your director on the [url]http://www.mathcounts.org[/url] and e-mail him/her. they probably have the info.

-jorian

You can't really refer to the AMC as an example. The score you need depends on the region you are in. If you live in California, you may need a higher score than if you live where I do, in Wisconsin. I would definitely recommend asking your coach or chapter director

Actually the number who advance to states depend on the number of people in the chapter competition...there is no top 20 or top 10 etc, they have percentages (that would be awesome if there were only 8 people at chapters huh?)

I think for my chapter it was the top 20% or something but i don't remember

another note--hurray my 500th post! bye bye Riemann Hypothesis!

they usually bring 1/4 people to states

jorian

i wish it was one fourth...we had 200 kids in our chapter...26 go to states (5 team 6 individual)

im guessing that they have a minimum number of teams the chapter can send, and then they have it go up according to how many participate

oh, we have like 240 and 8 teams and like 30 individuals (i found this out at chapters)

jorian

My chapter this year- 43 people :rotfl: 2 teams 2 individuals advance. 10/43 :rotfl:

I don't think it is 1/4.  Last year (since I am a 9th grader this year, I don't know about this year), we had over 400 students.  Top 5 teams and top 8 individuals not on the teams went to State.

i meant just for CT i just realized

-jorian

yeahhhhhh....
right

I think the bigger the state the smaller the ratio

are you being sarcastic?

because ur 2nd statement disproves the sarcasticness in the 1st statement

and yes, it said 1/4 in my packet

jorian

It's not really "secrecy".  It's just that MATHCOUNTS is a collection of 57 independent fiefdoms, each with their own slightly different rules and procedures, and there's no central repository for the data that you want.

If you want the data for your state, then you should probably do as was previously suggested and contact your state coordinator.  The [url=http://www.mathcounts.org]MATHCOUNTS website[/url] has a list of the state coordinates, and of all the chapters too.  (Click on "Competition Information" on the home page.)

[quote="life1"]I am a 6th grader and would like to know what numbers qualified for the states in previous years. Yes, I totally realize the numbers change from year to year, but as a budding mathematician, I'd like to figure out the Average qualifying score so I know the approximate range. I've searched for days and found nothing. Can't I find out last year or the yr before's qualifying scores to go from regional to state? Someone please help! Why all the secrecy? AMC posts stats right on their site!![/quote]
Following is the summary of chapter statistics for the years 2004, 2005 and 2006 from MATHCOUNTS:

http://www.mathcounts.org/webarticles/articlefiles/179-ChapSumFinal060504.pdf

Following is the summary of state statistics  for the years 2004, 2005 and 2006 from MATHCOUNTS:

http://www.mathcounts.org/webarticles/articlefiles/179-StateSumFinal060504.pdf

These statistics don't show the qualifying scores, but can still give you some general idea.

You said you were a 6th grader in your opening post, now you're a coach all of a sudden?

Have you been even reading the replies to this thread? CONTACT YOUR STATE COORDINATOR.

[quote="life1"]So there is no confusion as to who is writing this post, I am a coach. A new coach to mathcounts, and I have spent the better part of the last week trying to obtain the most basic stats of my state (NJ) -- for ex: since region tests are over, why can't I find out how many kids from the 6 regions in total have qualified to go to the states? I'd like to know if the person I'm coaching is up against 40 people or 400! Is that too much to want to know?[/quote]
LIKE I SAID PM TREETHING, HE'LL KNOW PROBABLY
If you want help with something maybe you should listen to what others have said?

[quote="bpms"][quote="life1"]I apologize for any confusion - but I AM a coach and I am new to this forum, so please excuse my faux pas (mistakes!) But I did let my son dictate a question or two via my account, and both and he and I did carefully read all replies - and appreciate all the advice. And yes, a moderator explained that 2 people on an acct confuse others, and that my son should open his own account -- that is why I clarified that I was a coach in my last post -- so as to be honest and not confuse. By doing so I confused you more!  :(   Anyway, I wrote a zillion emails to the state director (my regional director quit - so the state director is wearing two hats) and I tried to call too -- even called administrators at Mathcounts, and no one has replied in a week - not to even say they received my inquiries. Oh well -- like I said, every other math organization I've dealt with has been forthright in providing stats -- look at the AMC site - they lay it all out - by state, gender, grade, etc!!! I think others should be lobbying Mathcounts to do the same -- without all the excuses of how there is so many different regions, blah, blah, blah. C'mon - this is a MATH organization! Stats should come easy!!![/quote]
OH, MY, GODS.
Some states aren't organised and do this without that much effort. It would be a lot of work to organise this info this much.
With mathcounts the tests aren't multiple choice. They aren't sent into a scanner for grading. They are graded by humans. There is no bubble for gender or grade or name. This is all done by hand. So it would be a ton of work to find all of this information.

IF YOU DISAGREE WITH MATHCOUNTS THEN JUST DON'T COMPETE IN IT.
plain and simple.

PS sorry if it seems like I need anger management, I'm just tired of you complaining about everything.[/quote]

I think life1 is talking about an important topic, which is gathering statistics on MATHCOUNTS. When I started on MATHCOUNTS, as most of us preparing for this contest, first thing I wanted to know was the same statistics for my state. My dad came to know a very knowledgable coach in my state and he pretty much had all the statistics we wanted to know. I am pretty sure the info is out there for NJ. You may want to try contacting coaches from past winning teams.

[quote="bpms"][quote="life1"]I apologize for any confusion - but I AM a coach and I am new to this forum, so please excuse my faux pas (mistakes!) But I did let my son dictate a question or two via my account, and both and he and I did carefully read all replies - and appreciate all the advice. And yes, a moderator explained that 2 people on an acct confuse others, and that my son should open his own account -- that is why I clarified that I was a coach in my last post -- so as to be honest and not confuse. By doing so I confused you more!  :(   Anyway, I wrote a zillion emails to the state director (my regional director quit - so the state director is wearing two hats) and I tried to call too -- even called administrators at Mathcounts, and no one has replied in a week - not to even say they received my inquiries. Oh well -- like I said, every other math organization I've dealt with has been forthright in providing stats -- look at the AMC site - they lay it all out - by state, gender, grade, etc!!! I think others should be lobbying Mathcounts to do the same -- without all the excuses of how there is so many different regions, blah, blah, blah. C'mon - this is a MATH organization! Stats should come easy!!![/quote]
OH, MY, GODS.
Some states aren't organised and do this without that much effort. It would be a lot of work to organise this info this much.
With mathcounts the tests aren't multiple choice. They aren't sent into a scanner for grading. They are graded by humans. There is no bubble for gender or grade or name. This is all done by hand. So it would be a ton of work to find all of this information.

IF YOU DISAGREE WITH MATHCOUNTS THEN JUST DON'T COMPETE IN IT.
plain and simple.

PS sorry if it seems like I need anger management, I'm just tired of you complaining about everything.[/quote]

its not just this, but a lot more factors go into scores from regionals.  One is difficulty of the test-- one year u might have an easy test and need a 40, on u might have a hard one and get a 25 and make it--another is region: normally my regions score needed to advance in chicago area of indiana is ten less then some in the southern parts, and in cali you almost always need a 43ish to advance--another is size of competition like in cali there are 200 peeps while in chicago area of indiana(my area) ther might be about 30:the number of people who make it is based on people, tho, in my area people who won the team often also win individuals so instead of 2*4+8=16, u mite only have eight...

the last and least common factor is sheer luck--for example, some kids in our chapter made it without being in the top 8 cuz the coordinators didnt put our teams scores with the rest of the peoples' so instead it went to top 12 since every person on our team was in the top eight. all the trophies have been given to their rightful owners as well as the 25 bux from aops but the countdown round cant be redone.

Why do you need the stats?  If you find out in your state a 25 usually gets you to Nationals are you only going to try to be able to score 25 and then stop working?  Shouldn't the goal to be to learn as much as possible? If you just want an idea of which tests are hardest, ask on this forum.  A lot of people can give you that info.  2005, 1999, 1995(states) are particularly hard if you're wondering.

they love math-they want to find the averages for everything.

Good one Funcia, and just for the record Ihave the score for every test my top 12 kids have taken over the last 6 years in a giant spreadsheet.  Sorted by year, grade, team vs individual, kids who made it to Nats vs not, etc.  I love Data too!!

Yeah but by capitalizing Data I almost think you mean the android Data...okay that is a bit spammy...I'll stop here.

Hi - I just posted a very detailed post in response to several other poster's comments towards my own -- it says "it sent" but I don't see it on line -- if it's too long, does it get censored out or something? Just wondering - don't have the heart to attempt to rewrite it write now ...  :(

perhaps you should ask on the site question forum(prob. not exact name)

I just openly responded to bpms.. "anger mgt guy" with a full page, well written, letter, but it didn't show on site, but I will try to rewrite some of it.  Basically, suffice to say that I don't know your age, and I won't resort to character insults as you seem to do - not only w/ me, but others to whom you post. And I think you are wrong. We don't live in a perfect world and I don't believe as you do that if something isn't 100% perfect, it's better to "just quit" then attempt to improve! To go on record, I think Mathcounts is a [u]Wonderful Program [/u]that offers [u]Terrific Opportunities [/u]to Middle School Students, but saying that does not preclude that the program also has certain frustrations for many students and coaches, and that improvement is not possible. I believe it is. And I welcome others' suggestions towards such improvement and would encourage them to take some of them up with Mathcounts directly, and not just to one another via this forum. My youngest child even pointed out how the process of the census changed dramatically between the 1950's and now, with upgraded technologies. Surely, if Mathcounts wanted, they could implement an electronic portion of their forms to include things like age, gender, state, etc. Plus, surely they HAVE to have accurate records of things like "top scores in each state" -- otherwise, how do they know who competes at States and who doesn't? I'm just suggesting they share these facts with all participants - simple as that. My going on this forum was to solicit constructive suggestions from others, and not personal attacks/criticisms as I felt you have leveled at me. I'm sorry you viewed my seeking others' suggestions as "mere complaining" -- and while you are certainly allowed your opinion, as others are allowed theirs, myself included, I would kindly ask that you try to stay away from personal comments against someone's character in the future. Even if you disagree with their opinions. If you respond in a negative way to this, I will take the high road and not reply back. I have offered many constructive and supportive pm's and public posts to many on this site and they may or may not come forth in my defense as they see fit. Bpms, I will also add that I know you also try to be helpful to others, where possible, and I think those posts are very constructive. Lastly,I want to thank mthiyagu very much for her kind words and for backing me up. That's all I have to say on this topic.  Thank you all who have offered positive comments or advice in response to my initial post, and for understanding my question and concern for what it is.

[quote="life1"]Maybe I'm naive or ignorant here, but given what everyone says -- how a 30 in part A of a state may advance to state, but a 35 in part B of same state may not -- well, my goodness, how in the heck is this fair to the kid in part B, and how does each State claim they are sending their best and brightest to National's?  Did it ever dawn on anyone in all these years to maybe do something like AMC does when qualifying for AIME and just take say, the top x % (whatever that is - .05, 5, 10, 20 - whatever % the organization decides, and let THOSE KIDS go to states, REGARDLESS of where they live in the state? Sure, that MAY mean 10 kids from Town A go and only 1 from Town C, but so what?????  Why would that matter? I mean, isn't the point of going from School to chapter to state to National so that the best student(s) is/are recognized?  Someone (other than bpms, Please!!!)  explain to me what I am missing. And NO, bpms, this is not a "complaint" -- I'm just shedding light on something that seems a lot of people seem perplexed about, and it appears so simple to me.  Am I not getting something??[/quote]

Because likely, the people who got 30-35 in part A will be trounced anyways. For a state to do that, it takes a lot more effort as all chapters must report in results and then the state must tabulate them and then must report back to the chapter and the chapter has to get in touch with everybody again and trophies must be sent in the mail which is a lot harder to do, etc.

[quote="life1"]no malicious reason for wanting to know, and of course, student wouldn't work "less hard" (at least not the ones I know) but it DOES put things into perspective, especially for the kids new to these type of contests. For example, my son's friend thought he had a very low score and was really down on himself, only to find he got first place in his region.  The problem was he thought only 44's - 46 could possibly move on (and maybe in some regions, that is true) but it would have helped his self-esteem and overall mental state had he had a better sense of current and past scores of others in his state. That's all.  Also, since he is now directly competing with all the other top regional kids, wouldn't he want to know what THEIR scores were at the regional level compared to his own (after all, they all took the same test, right? so it's as fair a comparison to others as he's going to find -- that is, IF the info were released.  Does that answer your question?  It's no fun testing "in a box" -- when you took tests - the SAT's or whatever, even just a test in school, weren't you the least bit curious how your fellow peers were doing and what grades they got? Oh, how would I go about getting the state tests for 1995, 2005, or the other yr you said was hard? Thanks.[/quote]

I just sent em to you  :lol:

I don't find it fair, but it pushes me to work harder.
I got a X(apparently .5 but eh) and I didn't make it, as the qualifying scores from my chapter was X+1+. On the other hand, if I was in SouthernCal, I would've had an extremely high chance of making it.

[quote="life1"]Maybe I'm naive or ignorant here, but given what everyone says -- how a XX in part A of a state may advance to state, but a XX+5 in part B of same state may not -- well, my goodness, how in the heck is this fair to the kid in part B, and how does each State claim they are sending their best and brightest to National's?  Did it ever dawn on anyone in all these years to maybe do something like AMC does when qualifying for AIME and just take say, the top x % (whatever that is - .05, 5, 10, 20 - whatever % the organization decides, and let THOSE KIDS go to states, REGARDLESS of where they live in the state? Sure, that MAY mean 10 kids from Town A go and only 1 from Town C, but so what?????  Why would that matter? I mean, isn't the point of going from School to chapter to state to National so that the best student(s) is/are recognized?  Someone (other than bpms, Please!!!)  explain to me what I am missing. And NO, bpms, this is not a "complaint" -- I'm just shedding light on something that seems a lot of people seem perplexed about, and it appears so simple to me.  Am I not getting something??[/quote]

Yes, life1, I think you are completely and utterly naive about the fundamental philosophy of MATHCOUNTS.

I love MATHCOUNTS [b]and[/b] the AMC contests.  But they are different.

MATHCOUNTS is a grass-roots "coaching and competition" program.  It starts at the local schools, with coaches (many are volunteers or teachers performing unpaid extra duty) working with kids to get them ready for a fun competition.  The school is limited to taking a handful of students to the Chapter Competition, which fosters a competitive atmosphere in the school.  Whether your are in a rural school with 200 low-income children of farm workers or from a suburban school with 1600 high-income students or from a magnet school with a high proportion of children of Princeton professors, a School MATHCOUNTS competition can really jazz your students.

At the next level, a volunteer Chapter Coordinator, along with a small army of volunteers hosts another competition, with the prospect that a fair proportion of students from that competition will advance to the next level.  Often these volunteers donate, not just their time, but their own money to help fund the competition, the prizes, and the food.

At the State level, yet another volunteer, the State Coordinator, along with another small army of volunteers hosts yet another competition, with the guarantee that four students will advance to the National Competition to represent your State.  Once again, the Engineers (and other volunteers) who run the contests donate time, money, food, and prizes.

Now what happens if the magnet school is permitted to take a boatload of students to the Chapter Competition?  Few other schools in the Chapter will be able to send anybody to the State Competition.

What if the advancement criteria is based on the statewide results, rather than Chapter Results?  Then students in the rural part of the State will never advance to the State competition.  Why will the rural Chapter Coordinator and his volunteers continue to invest their time and money?  Why will the coaches in the rural part of the state continue to invest their time?  Why even bother with MATHCOUNTS, except at the most elite schools in the State?

Because MATHCOUNTS has a 23-year history of a grassroots "coaching and competition" program.  It is more than simply seeing who "wins" the contest.  It's about learning.  It's about teamwork.  It's about physically getting students from different schools together, where they can mingle.  Where their coaches can share ideas and strategies.  It's about the experience and downright "fun" of the program!

In the end, each state *does* usually send four of their tip-top students to the National Competition.  And yet, the rural coach can still have a student who is justifiably proud of their score of "3" on the Sprint.

For the high concentrations of great math students at high schools like TJ, Stuy, IMSA, and Exeter, the AMC contests allow them to shine.  Likewise, the student in region "B" of your State can shine in the AMC contests, and will probably also have a much wider array of other opportunities in which he can shine than the student in region "A" of your State.

If you tried to change the rules of MATHCOUNTS to match those of the AMC, you would destroy MATHCOUNTS.  Even if it didn't destroy MATHCOUNTS, why would you need two contests with the same rules.  We could just do away with MATHCOUNTS, and rename the AMC-8 to be the AMC-8-MC.  How delightful!

[quote="life1"]Hi - I just posted a very detailed post in response to several other poster's comments towards my own -- it says "it sent" but I don't see it on line[/quote]

Chances are you PM'ed your response, rather than posting your response to the Forum.

[quote="life1"]no malicious reason for wanting to know, and of course, student wouldn't work "less hard" (at least not the ones I know) but it DOES put things into perspective, especially for the kids new to these type of contests. For example, my son's friend thought he had a very low score and was really down on himself, only to find he got first place in his region.  The problem was he thought only 44's - 46 could possibly move on (and maybe in some regions, that is true) but it would have helped his self-esteem and overall mental state had he had a better sense of current and past scores of others in his state. That's all.  Also, since he is now directly competing with all the other top regional kids, wouldn't he want to know what THEIR scores were at the regional level compared to his own (after all, they all took the same test, right? so it's as fair a comparison to others as he's going to find -- that is, IF the info were released.  Does that answer your question?  It's no fun testing "in a box" -- when you took tests - the SAT's or whatever, even just a test in school, weren't you the least bit curious how your fellow peers were doing and what grades they got? Oh, how would I go about getting the state tests for 1995, 2005, or the other yr you said was hard? Thanks.[/quote]

I think there is a rule in mathcounts, tho, that if u r in the top 25 percent of ur state, u cna get past ur regionals without comin in the top whatever

I haven't read this entire thread, so I'm guessing someone has made this point already.

Please note that MATHCOUNTS is largely run by volunteers.  To me, it's astounding that MATHCOUNTS works at all - there are 6-7 paid national staff.  After that, it's all volunteers pretty much.  All the way down to the 500-550 Chapter competitions.  We run 2-3 contests a year in San Diego, with nowhere near the polish and professionalism MATHCOUNTS achieves with its volunteer army.  That they can pull off the events so well (and pull off so many of them) is a staggering feat to me - certainly in math education, I haven't seen its equal.  To expect them to also maintain databases of all scores everywhere on top of this is quite unrealistic.  Just imagine the data entry headache alone, not to mention the coordination required to get all the data in the same format, make it accessible, etc.

As for individuals versus teams, MATHCOUNTS is intended to be a team activity.  Moreover, most of the great coaches I know will tell you that the strength of your team is in students 5-20, not 1-4.  Regardless, if a student has a realistic shot at being top 4 in the state, he/she should be 98+% likely to make it out of chapter.

Finally, if you were to build a national program and expect it to survive and reach a large number of students, would you tie it to individuals or to schools?  Your best bet is to tie it to schools.    MATHCOUNTS programs that are built around individuals tend to die quickly after said individuals leave 8th grade.  Programs that are built around coaches have much more longevity and reach many more students.

As far as I'm usually concerned, ace it and there's nothing to be worried about. Maybe your 6th grade son obviously can't ace it, but the point is: do your best and that's the best you can do! It doesn't matter if the winning score is 5 or 45. Just get as many right as you can! You can only control YOUR score.

Respectfully,
kstan013

I'm an inforat. Personally, I love knowing everything about everybody.

I can see life1's point.

However, I agree with the majority of people here -- when it comes down to it, worry about yourself.

As the mom of a son who made it to Nationals last year purely by being from a less competitive region, I have to say that I think it's fantastic mathcounts is set up the way it is.  "Rcv" and others really have said it beautifully already, but just to give another example: at our chapter competition last week, the eighth or ninth place team, when they heard they'd advance to the state level (yes, our chapter sends that many teams to state) cheered with wild abandon.  It really put things in perspective for me.  For so many of these kids, this is a chance to set a goal, work toward it, and be happy with however they end up doing.  If life1's son is in a very competitive region, maybe it's best for him to just focus on learning and preparing, and aim for a particular score on the state test, as opposed to winning a place at Nationals.  If that happens, all the better, but if it doesn't, I don't think anyone wants him to feel that the work he did preparing was in vain.  Keep it fun!

life1, if ur a coach, join the coaches forum-- u would probably find some stats from them.

[quote]Fanatic I'm an inforat. Personally, I love knowing everything about everybody.

 [/quote]

are u a stalker fanatic?

yes, that is so obviously what I meant.

Ironically, I saw & responded to your pm before viewing this thread...

ETA: I remember where I know you from now! You're the guy who created a topic called "yuqing" asking indianamath if he won his chapter. Well, at least you made it obvious...

who are you? From honey creek?

"settle" this via PM guys.

Anyhow, I think everything's been answered. Lock this topic up.


Denote by $P(a)$ the product of distinct prime divisors of positive integer $a\ge2$ and set $P(1)=1.$
For every $n\ge1$ prove or disprove the statement:
there exist positive integers $a_0, a_1, \dots, a_n$ such that for $k=1, 2, \dots, n$ we have $a_k=a_{k-1}-P(a_{k-1}).$


The perimeter of a rectangle is 50 inches and its area is 154 square inches. How many inches are in the length of one of its shorter sides?

If the perimeter is 50, the sum of the lengths of two consecutive sides is 25. Those two sides have product 154, so they are [b]11[/b] and 14.

You can even make two systems. 2x+2y=50 and x*y=154.


probability
trigonometry
trig identities
Law of Cosines

Points S and T are chosen at random on a circle of radius 10. What is the probability that ST>5?

[hide]
WLOG let S be a stationary point
we can base our probability off of the arc that S and T would cut off at length 5

drawing the radii to ST, we have a triangle with lengths 10, 10, 5

Using the law of cosines, the central angle is shown with

$5^{2}= 10^{2}+10^{2}-2(10)(10)\cos\theta$

$200\cos\theta = 175$

$\cos\theta = \frac{7}{8}$

$\theta = \arccos\left(\frac{7}{8}\right)$
now we want to double this since the length would be less than 5 regardless of which direction from S, T is located

$\theta = 2\arccos\left(\frac{7}{8}\right)$
This is the smaller arc. We want the larger arc.

$\theta_{big}= 2\pi-2\arccos\left(\frac{7}{8}\right)$

and the probability that ST>5 is the probability that T lies on that arc, or

$\frac{2\pi-2\arccos\left(\frac{7}{8}\right)}{2\pi}$
[/hide]

How about $ST>6\sqrt{2}$ for $r=6$? That works out much nicer. [You could potentially see this type of problem as an early AMC problem.]


geometry
geometric transformation
rotation
reflection
MATHCOUNTS
analytic geometry

what kind of problems would make us use reflections,rotations, and translations, directlky or indirectly?  Thanks, can somneone post like a few examples.

Here's one from an old MC nationals(1995). The triangle with vertices at A(-2,2)B(-8,2)C(-8,-1) is reflected about the line y=2x+1. What are the coordinates of A'?


What is the smallest multiple of 5 the sum of whose digits is 18?

We want the digits to be as big as possible so we can get a smaller number with the sum of its digits $ 18$. As all multiples of $ 5$ end with $ 0$ or $ 5$ and we want the bigger one, the number shall end in $ 5$. Next, we want the largest digit, so we will put a $ 9$. The sum is already $ 14$. We need $ 4$ more, so we need a $ 4$ in the hundreds place. Our number is $ 495$.

Why do we want to make it as big as possible (number)?

You can either choose 5, or 0 in the digits place. We choose 5 because this will make the other digits smaller.

To minimize the value of the number, we want the hundreds to be the smallest possible, so we maximize the tens.

To sum to 18, we have a number, $495$.


real analysis
calculus
integration
limit
inequalities
triangle inequality
real analysis unsolved

Prove the following theorem.

$ (\Omega,\Sigma,\mu)$ measure space
1) $ \mu(\Omega)$ is finite.
2) $ f,f_1,f_2,...$ are complex-valued, measurable
3) $ f_n \rightarrow f$ almost everywhere
4) $ \int_{\Omega} |f_n|^2 d\mu < 1$ and $ \int_{\Omega} |f|^2 d\mu < \infty$
Then,
$ \lim_{n \to \infty}\int_{\Omega} |f_n\minus{}f|^p d\mu \equal{}0$ for any $ 0<p<2$

Fix $ 0<p<2$.  Given $ \varepsilon>0$, use Egorov's theorem to find a set $ A_\varepsilon$ such that $ \mu(A_\varepsilon)\leq \varepsilon$ and such that $ f_n\to f$ [i]uniformly[/i] on $ \Omega\setminus A_\varepsilon$.

Then we have, by the triangle inequality and Holder,

\begin{align*}
\|f_n-f\|_{L^p(\Omega)} &\leq \|f_n-f\|_{L^p(A_\varepsilon)}+\|f_n-f\|_{L^p(\Omega\setminus A_\varepsilon)} \\
&\leq \mu(A_\varepsilon)^{1/q-1/2}\|f_n-f\|_{L^2(A_\varepsilon)} + \|f_n-f\|_{L^p(\Omega\setminus A_\varepsilon)}
\end{align*}

Write $ M=\int_{\Omega} |f|^2d\mu$.  Then we have the uniform bound $ \mu(A_\varepsilon)^{1/q-1/2}\|f_n-f\|_{L^2(A_\varepsilon)} \leq \varepsilon^{1/q-1/2}(M+1)$, and also $ \|f_n-f\|_{L^p(\Omega\setminus A_\varepsilon)}\to 0$ as $ n\to\infty$ (by uniform convergence on compact sets).  Pick $ n$ large enough, let $ \varepsilon\downarrow 0$, and you're done.

Can we use Holder inequality for 0<p<1 ??

I think we should prove the theorem without using holder inequality...

[quote="lysangc"]
Can we use Holder inequality for 0<p<1 ??
[/quote]

no.  nonetheless, the inequality I stated still holds, and its proof does follow from Holder's inequality (well, with $ p$ in place of $ q$; that was a typo).  why?

hint: you're applying holder's inequality to the product $ |f|^p\cdot 1$, and you want to get a bound involving the integral of $ |f|^2$, so what do your exponents need to be?

[color=green]Moderator edit: It never makes sense to quote the post immediately preceeding yours entirely![/color]

I can't see why holder inequality holds for 0 < p < 1

Could you explain that explicitly? Sorry..

you use the conjugate exponent $ 2/p$, and not $ p$ directly.  so you get $ \int_\Omega (|f_n\minus{}f|^p)^{2/p} \equal{} \int_{\Omega} |f_n\minus{}f|^2$, etc

[quote="blahblahblah"]you use the conjugate exponent $ 2/p$, and not $ p$ directly.  so you get $ \int_\Omega (|f_n \minus{} f|^p)^{2/p} \equal{} \int_{\Omega} |f_n \minus{} f|^2$, etc[/quote]

Aha, Thanks. I didn't know what you said ( because of my poor english and math )

Now I get it.

If you don't mind, could you prove the theorem without using holder inequality?
(in fact, this problem is in Ch1 of ANALYSIS by lieb & loss. but, in this book holder inequality is in Ch2..)


A survey of 400 patients at a hospital classified the patients by gender and blood type, as shown in the table below.  What percent of the patients with type AB blood are male?
\begin{tabular}{|l|c|c|c|c|c|}
\hline
&\textbf{Type A}&\textbf{Type B}&\textbf{Type O}&\textbf{Type AB}&\textbf{TOTAL}\\\hline
\textbf{Male}&45&60&80&15&200\\\hline
\textbf{Female}&55&40&100&5&200\\\hline
\textbf{TOTAL}&100&100&180&20&400\\\hline
\end{tabular}

[quote="GameBot"]A survey of 400 patients at a hospital classified the patients by gender and blood type, as shown in the table below.  What percent of the patients with type AB blood are male?
$ \begin{tabular}{|l|c|c|c|c|c|}
\hline
&\textbf{Type A}&\textbf{Type B}&\textbf{Type O}&\textbf{Type AB}&\textbf{TOTAL}\\\hline
\textbf{Male}&45&60&80&15&200\\\hline
\textbf{Female}&55&40&100&5&200\\\hline
\textbf{TOTAL}&100&100&180&20&400\\\hline
\end{tabular}$[/quote]

The answer is simply $ 100\% \times \frac{15}{20}$, or $ \boxed{75}$.


Here is a question where I think the makers had it wrong:

If the radius of a sphere is doubled, how is the volume changed?

I took 2 as the original radius.

Original Volume=$ \frac43 \pi 2^3\equal{}10 \frac23\pi$

Doubled Radius Volume= $ \frac43 \pi 4^3\equal{}85 \frac13 \pi$

And...

$ 85 \frac13 \pi \div 10 \frac23\pi\equal{}8$

So would the answer be divided by 8 or multiplied by 8?  I put divided by 8 as the answer but it was wrong, so is it multiplied?  Sorry if this is a stupid question, it's morning and my brains really isn't working.

Consider a sphere of radius $ a$.  Then it has volume $ V\equal{}\frac{4}{3}\pi a^{3}$.  Now double its radius so $ V\equal{}\frac{4}{3}\pi (2a)^{3}\equal{}\frac{4}{3}\pi 8a^{3}\equal{}\frac{32}{3}\pi a^{3}$.  Divide the second volume by the first and we see that it is 8, like you said.  But think about your answer; if I double the radius, how the volume decrease by a factor of 8?  Does that makes any sense?  Of course not, it would have to be multiplied by 8.

ratio of radii is 1/2
so ratio of volumes = 1^3/2^3= 1/8
so answer is 8
smaller is 1x
bigger is 8x

Ok, plain and simple: bigger radius, bigger volume,
when you divide by a positive integer greater than one, the number can only get smaller.


calculus
derivative
function
trigonometry
calculus computations

When given a graph, how do I estimate the value of each derivative by looking at the diagram?  Also, after estimation is made, how do I sketch the graph of f '(x)?

SAMPLE:

If the graph of y = f(x) is given, how do I sketch the graph
of f ' (x)?

The derivative in some point is the gradient of the tangent to the curve in that point. This should go to Calculus Computations&Tutorials: not the first time I'm telling you this.

Is this not the college, computations and Tutorials?
Can you answer the question or not?

Yes, now it is in the right section, but that is because some moderator moved it.

Just look at how the gradients of the tangents to the curve vary to sketch the graph :|

Here's an example, let's say that there is a continuous differentiable function $f(x)$ and $f'(0) = 1$, $f'\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, $f'\left(\frac{\pi}{2}\right) = 0$.  Then the points you graph for the derivative of $f(x)$ are $(0,1)$  $\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)$, and $\left(\frac{\pi}{2},0\right)$.  If you haven't already guessed, $f(x) = \sin x$.

Jimmy,

Your help has been great.  I can graph this stuff.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove the following:

1)The order of $ g^\minus{}1$ is the same as order of $ g$

2)The order of $ ab$ is the same as order of $ ba$

3)Every subgroup of a cyclic group is cyclic

[hide="solution"]
1) Assume on the contrary. WLOG, $ m: \equal{}ord(g^{\minus{}1})>ord(g)\equal{}: n$ for some $ g$. Then $ (g^{\minus{}1})^n\equal{}(g^n)^{\minus{}1}\equal{}e$ (contradiction).

2) Assume on the contrary. WLOG $ m: \equal{}ord(ab)>ord(ba)\equal{}: n$. Then $ (ab)^n\equal{}a(ba)^na^{\minus{}1}\equal{}e$ (contradiction)

3) Let $ H$ be a subgroup of $ G\equal{}\langle g\rangle$ and let $ s$ be the smallest positive number such that $ g^s\in H$. Claim that $ H\equal{}\langle g^s\rangle$.
Let $ a\in H\subset G$. Then $ a\equal{}g^t$ for some $ t$. Then $ t\equal{}qs\plus{}r$ for some $ q$ with $ r\equal{}0$ or $ 1<r<s$. Now $ g^t\equal{}(g^s)^qg^r \in H$. It follows that $ g^r\in H$. By the  minimality of $ s$ we must have $ r\equal{}0$. Hence $ a\equal{}g^t\equal{}(g^s)^q$.

[/hide]

For #2, note that $ ba\equal{}a^{\minus{}1}(ab)a.$ Hence $ ab$ and $ ba$ are conjugates and therefore have the same order.

I'm still a bit confused on #1, can you elaborate more?

[quote="1234567a"]How about when $ g$ and $ ab$  have infinite order ?[/quote]
For any positive integer $ n,$ $ g^n\equal{}1\ \Leftrightarrow\ (g^{\minus{}1})^n\equal{}1.$ Hence $ g$ has finite order if and only if $ g^{\minus{}1}$ has finite order.

Also $ ab$ has finite order if and only if $ ba$ has finite order since they are conjugates.


bleh, this one stumped me:

$\text{Prove that } \cot \displaystyle (\displaystyle\sum_{n=1}^{k}\cot^{-1}(1+n+n^2)) = \displaystyle\frac{k+2}{k}$

If you can give me the summation formula for $\cot(x+y)$, I'll solve it when I get to school tomorrow. I'm going to sleep now, though.

First of all, I would like to point out that the RHS should be $\frac{k+2}{k}$

Using the hint and the fact that we add and subtract a number of terms (cancelling them), the expression (with k-1 instead of k) equals $\cot{(\cot^{-1}1-\cot^{-1}k)}$ Now let $a=\cot^{-1}1$ and $b=\cot^{-1}k$. We have $\cot{(a-b)}=\frac{\cot a\cot b+1}{\cot b-\cot a}=\frac{k+1}{k-1}$. Since we did this with k-1 instead of k, we add one to get $\frac{k+2}k$.
$\mathbb{QED}$

oops, my bad, yeah you're right, ill change it


superior algebra
superior algebra solved

show on irreducible algebraic curve there are only finitrly many singular points.and on irreducible algebraic cubic curve there is atmost 1 singular point.  :)

Are you talking about plane algebraic curves?
Doesn't really matter for the first part. It is known that for every variety (irreducible) $V$, the set of smooth points is open and dense. This means that the set of singular points is a closed proper subset of the curve. Since the curve is $1$-dimensional, the set of singular points must be $0$-dimensional i.e. a finite union of points.

If the curve is a plane curve, then it is given by one equation $f(x,y)=0$ in $\mathbb A^{2}$ or a homogeneous equation $F(x,y,z)=0$ if you want it in $\mathbb P^{2}$. The set of singular points is $V(f,\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$ or $V(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z})$ respectively. To prove that these are finite unions of points it suffices to prove that the partials of $f$ or $F$ don't all vanish identically on the curve. This is because if they did, they should be multiples of $f$ or $F$ which is impossible because they have lesser degree. This works at least in characteristic $0$.
In characteristic $p$, the curve $x^{p}+y^{p}+z^{p}=0$ is singular at every point, so I assume that you work in characteristic $0$.

If the cubic $C$ is a plane cubic, then the second result is a consequence of Bezout. If $P$ and $Q$ are singular points, then the line $d=PQ$ cuts $C$ in $P$ and $Q$ with multiplicity at least $2$ because the points are singular, so the tangent space at these points is the entire plane. By Bezout, the sums of the multiplicities of every point of intersection of $d$ and $C$ is $3$. There do not exist negative multiplicities, so we are done.


Find the greatest integer less than $ 10.3^3$ without using a calculator.


[quote="Victory.US"]let $ a,b,c$ be real number such that the sum  of any two of them is not equal to zero. Prove that "

$ \frac {a^5 \plus{} b^5 \plus{} c^5 \minus{} (a \plus{} b \plus{} c)^5}{a^3 \plus{} b^3 \plus{} c^3 \minus{} (a \plus{} b \plus{} c)^3} \geq \frac {10}{9}(a \plus{} b \plus{} c)^2$

 :D[/quote]
$ \frac {a^5 \plus{} b^5 \plus{} c^5 \minus{} (a \plus{} b \plus{} c)^5}{a^3 \plus{} b^3 \plus{} c^3 \minus{} (a \plus{} b \plus{} c)^3} \equal{} \frac {5}{3}\sum_{cyc}(a^2 \plus{} bc)\geq \frac {10}{9}(a \plus{} b \plus{} c)^2,$
which is obvious.

[quote="joeislittle"]Did you just bash out that sum or is there something clever that I am missing?[/quote]
Just use the identity
$ (a\plus{}b\plus{}c)^5\equal{}a^5\plus{}b^5\plus{}c^5\plus{}5(a\plus{}b)(b\plus{}c)(c\plus{}a)(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)$  :)


If x varies directly as y and z and inversely as the square root of w. If x=10 when y=2, z=3, and w=9, what is x when y=5, z=4, and w=25?

[hide="Hint"]If $ x$ and $ y$ vary inversely, $ x\cdot y$ is a constant. If $ x$ and $ y$ vary directly, $ \frac{x}{y}$ is a constant.[/hide]

Does "x varies directly as y and z" mean $ \frac{x}{y}\equal{}c$ and $ \frac{x}{z}\equal{}d$ or $ \frac{x}{yz}\equal{}c$?

The latter ($ x \equal{} kyz$); the former would also cause $ y \propto z$.

[hide]From the problem, we know that $ \frac {x\sqrt {w}}{y*z} \equal{} k$, where k is a constant.

Plugging the first set of numbers, we find that $ k \equal{} 5$:
$ \frac {10\sqrt {9}}{2*3} \equal{} k \equal{} 5$

Now, plugging in the second set of numbers and solving for $ x$:

$ \frac {x\sqrt {25}}{5*4} \equal{} 5$
$ \Rightarrow x \equal{} \boxed{20}$[/hide]


Let $ a,b$ and $ c$ be real numbers such that $ a,b,c\notin(2,3)$. Find the minimum of the expression $ a^4\plus{}b^4\plus{}c^4$

 :)

[quote="tdl"]Min is $ 0$?[/quote]
 :)

Let $ a,b$ and $ c$ be real numbers such that $ a,b,c\notin(2,3)$. Find the maximum of the expression $ a^4 \plus{} b^4 \plus{} c^4$  :)

Sorry... I posted it carelessly  :(

$ F \equal{} a^4 \plus{} b^4 \plus{} c^4\geq 178$ for $ a \plus{} b \plus{} c \equal{} 8$ and $ a,b,c\notin(2,3)$
Equality holds if and only if $ a \equal{} b \equal{} 3;c \equal{} 2$ and the two permuations :maybe:

Assume that $ a\geq b\geq c$ than $ a \plus{} b \plus{} c \equal{} 8\Rightarrow 3a\geq 8\Rightarrow a\geq 3$
If $ a\geq 4$ then $ F\geq 4^4 > 178$
Then $ 3\leq a < 4$, it mean $ b \plus{} c > 4\Rightarrow 2b > 4\Rightarrow b\geq 3\Rightarrow 0 < c\leq 2$
Now we prove that $ a^4 \plus{} b^4 \plus{} c^4\geq a^4 \plus{} 3^4 \plus{} (b \plus{} c \minus{} 3)^4\Leftrightarrow b^4 \minus{} 3^4\geq (b \plus{} c \minus{} 3)^4 \minus{} c^4$
$ \Leftrightarrow (b \minus{} 3)(b \plus{} 3)(b^2 \plus{} 9)\geq (b \minus{} 3)(b \plus{} 2c \minus{} 3)((b \plus{} c \minus{} 3)^2 \plus{} c^2)$
It is OK because $ b \minus{} 3\geq 0;b \plus{} 3 > b \plus{} 2c \minus{} 3;b^2 \plus{} 9\geq18 > 8\geq(b \plus{} c \minus{} 3)^2 \plus{} c^2$ ( Because $ b \plus{} c\leq5$)
Now we must to prove that $ a^4 \plus{} 3^4 \plus{} (b \plus{} c \minus{} 3)^4\geq 178\Leftrightarrow a^4 \plus{} (5 \minus{} a)^4\geq 3^4 \plus{} 2^4$
$ \Leftrightarrow a^4 \minus{} 3^4\geq 2^4 \minus{} (5 \minus{} a)^4\Leftrightarrow (a \minus{} 3)(a \plus{} 3)(a^2 \plus{} 9)\geq(a \minus{} 3)(7 \minus{} a)(4 \plus{} (5 \minus{} a)^2)$
It is OK because $ a \minus{} 3\geq0;a \plus{} 3 > 7 \minus{} a;a^2 \plus{} 9\geq18 > 8\geq 4 \plus{} (5 \minus{} a)^2$
Equal when $ a\equal{}b\equal{}3;c\equal{}2$


A graduating engineering student is interviewing with several companies for an entry-level position.
He receives an attractive offer from company A. Since the job market is very competitive, he feels it unlikely
that another company will give an offer, much less an attractive one. The student accepts company A’s offer
and returns a signed letter of acceptance which documents the terms of the position. However, he receives an
offer from company B one week afterwards. This new opportunity has a higher salary, more benefits, better advancement
prospects, and a more desirable location. It is significantly better in all respects. Since only one week has past
since the first acceptance was returned and the new opportunity is clearly in his professional and financial
interests, he tells company A that he has changed his mind and accepts the offer of company B. Company A does not
express any criticism of the student’s actions.

Is the student ethically bound the honor the signed letter of acceptance with company A? Has company A been harmed by the student’s action?

The IEEE Code of Ethics are:

to accept responsibility in making decisions consistent with the safety, health and welfare of the public, and to disclose promptly factors that might endanger the public or the environment;

2. to avoid real or perceived conflicts of interest whenever possible, and to disclose them to affected parties when they do exist;

3. to be honest and realistic in stating claims or estimates based on available data;

4. to reject bribery in all its forms;

5. to improve the understanding of technology, its appropriate application, and potential consequences;

6. to maintain and improve our technical competence and to undertake technological tasks for others only if qualified by training or experience, or after full disclosure of pertinent limitations;

7. to seek, accept, and offer honest criticism of technical work, to acknowledge and correct errors, and to credit properly the contributions of others;

8. to treat fairly all persons regardless of such factors as race, religion, gender, disability, age, or national origin;

9. to avoid injuring others, their property, reputation, or employment by false or malicious action;

10. to assist colleagues and co-workers in their professional development and to support them in following this code of ethics

I don't know what this in reference to but as long as no one has been harmed, I don't think it is unethical.  If Company A is okay with it, then he should be able to switch.  If by accepting Company A's offer the student took it away from another applicant, it may be unfair but I think that since Company A is okay with it, it is not unethical.  Just my opinion.

I think it's unethical at least. Just because
[quote]Company A does not express any criticism of the student’s actions.[/quote]
does not mean they don't have a lawsuit in works.

meh, people are selfish, is being selfish ethical?


Suppose for a trapezoid $ABCD$ (with $AB$ parallel to $CD$) we have $AC^2+BD^2=(AB+CD)^2$. Show that the trapezoid is orthodiagonal (i.e. its diagonals are perpendicular).

This is from AoPS vol 2. My solution didn't use any of the methods presented in the chapter, and I was wondering what a more straight-forward or algebraic solution might be.

[hide="my solution"]Let $M$ be the point of intersection of the diagonals of $ABCD$. Then $(AM+MC)^2+(BM+MD)^2=(AB+CD)^2$. It's easy to construct the right triangle this describes. Extend segment $CD$ to the point $P$ such that $CP=AB$, and draw the segment $BP$. Triangle $BDP$ is what we want, since $DB=DM+MB$, $DP=CD+CP=CD+AB$, and $BP=AC=AM+MC$ (since it is a parallelogram). Since $MC$ and $BP$ are parallel and angle $MBP$ is right, so is angle $BMC$.[/hide]

[quote="fourierseries"]I rechecked AOPS Vol. 2 and it is:

$AC^2+BD^2=(AB+CD)^2$.  Not the equality you posted.[/quote]I fixed it. I had all the $C$s and $D$s reversed, but I missed those the first time.. sorry.

you could just use the $-2ab\cos(x)$ part of the law of cosines to show that they have to be perpendicular

$AC^2+BD^2 = (AB+CD)^2 \ \ \ (1) $

AntonioMainenti,

[hide="The First Solution is just like yours, but I dont construct the right triangle subscribed by $(1)$"]
Triangles $ABM, CDM$ are similar, since they have $ \angle ABM = \angle MDC$ and $\angle BMA = \angle DMC$.

Hence

$\frac{AM}{MC} = \frac{BM}{MD} = \frac{AB}{CD}$

$\frac{AM+MC}{MC} = \frac{BM+MD}{MD} = \frac{AB+CD}{CD}$

$\frac{AC}{MC} = \frac{BD}{MD} = \frac{AB+CD}{CD}$

So, the triangle $DMC$ is also similar to a triangle with sides $AC,BD,AB+CD$
But, from $(1)$ we know that this is a right triangle, so $DMC$ is right too.
(right angle the angle $M$)
[/hide]
[hide="The Second Solution uses vectors"]
$\overline{AC} +\overline{DB} =$

$ \overline{AM} +\overline{MC} +\overline{DM} +\overline{MB} =$

$\overline{AM} +\overline{MB} +\overline{DM} +\overline{MC} =$

$\overline{AB} +\overline{DC} $

So

$(\overline{AC} +\overline{DB})^2 = (\overline{AB} +\overline{DC})^2 \Rightarrow $

$(\overline{AC} +\overline{DB})^2 = \overline{AC}^2 +\overline{DB}^2\Rightarrow $

$\overline{AC}^2 +\overline{DB}^2 +2 \cdot \overline{AC} \cdot \overline{DB} = \overline{AC}^2 +\overline{DB}^2\Rightarrow $

$2 \cdot \overline{AC} \cdot \overline{DB} = 0 \Rightarrow $

$\overline{AC} \perp \overline{DB}$

Notice that, since $\overline{AB}, \overline{CD}$ are parallel but with opposite directions, we should use $\overline{DC}$ instead of  $\overline{CD}$ in order to find the value of $|\overline{AB}| +|\overline{CD}|$

[/hide]


A conveyor belt's motor consists of two gears connected by a metal belt which hangs wothout slack. The diameter of gear #1 is 8m, the radius of gear #2 is 12m, and the centers are seperated by 20m. If gear #2 runs at 15 revolutions per minute, what is the rate of gear #1 in revolutions per minute?

A.$4\sqrt{21}$  B.$45$    C.$22.5$     D. $4\sqrt{10}$

[quote="The QuattoMaster 6000"][hide]  Since the distance between the two centers is 20 meters and the radii of the gears are 8 and 12, we have that the sum of the radii is equal to the distance between the centers.  Therefore, the gears are externally tangent.  As a result, Gear B pushes Gear A at the same speed as Gear B moves.  Since the radius of Gear B is 12 meters, it turns $24\pi$ meters/revolution.  Since there are 15 revolutions per minute, the speed of Gear B is $360\pi$ meters per minute.  Hence, Gear A moves at the same speed.  Each revolution, Gear A moves $16\pi$ meters.  Thus, every minute, Gear A goes through $\frac{360\pi}{16\pi}=22.5$ revolutions per minute, so the answer is C. [/hide][/quote]
I don't know how to do this problem but I know you made a mistake. You said the radii of both gears are 8 and 12. But it said that the diameter of gear #1 is 8. So the radius would be 4.

:blush: Oh, sorry about that.  [hide] The distance between the two gears is not really relative.  Since Gear 2 is pushing the belt at a particular speed, which will transfer the speed to Gear 1, the speed of Gear 1 is the same as the speed of Gear 2.  The distance per revolution of Gear 2 is $12\pi$ meters (we can calculate this the same exact way which we calculated it in my other post), and since it goes for 15 revolutions per minute, we have that the speed is $180\pi$ meters per minute.  Now, since Gear 1 runs at the same speed, it also runs at $180\pi$ meters per minute.  We can also calculate that the distance per revolution of Gear 1 is $8\pi$.  Hence, the number of revolutions per minute is $\frac{180\pi}{8\pi}=22.5$.  [/hide]

You still get the same answer. Thats kinda weird. Is this a coincidence or something?

Not really a coincidence.  You are pretty much just dividing the numerator and denominator by 2 since the diameter/radius=2.

Quatto, read the problem more carefully, please. The radius of gear #1 is 4, the radius of gear #2 is 12.

Therefore gear 1 has to rotate 3 times as much as gear 2 to cover the same distance along the conveyor belt.

So the answer is just $15 \times 3 = 45$.

Oh, yes, you are right.  Sorry, I am off today.

[quote="bpms"]Where is this problem from?[/quote]
A Mu Alpha Theta test we recently took.


A set S includes natural numbers from 1 to 100 , S= {1, 2, ... , 100} . Find the smallest natural number c such that : Every subset of S has c elements  always exists 2 elements have a square sum .

No , c different numbers are taken from S

Firstly taking the numbers $3k+1$ we get already $c \geq 35$.
But I have a question:
You said that whenever two numbers are chosen, their sum is not a perfect square.
Can they be equal?
For example $2+2=4$ and $2=2$

2 numbers can't equal . example : 1+3=4 , not 2+2=4 . Every number is chosen only one time


\frac{d}{dx}\left\[ \frac {1}{\sqrt {2}}\ln{\left| \sec{(x - \frac {\pi}{4})} + \tan{(x - \frac {\pi}{4})}\right| + C \right}\]

Yes you can, but you lost a }. The tex log will tell you that a missing } had been inserted.
 $ \frac{d}{dx}\left[\frac {1}{\sqrt {2}}\ln{\left| \sec{(x \minus{} \frac {\pi}{4})} \plus{} \tan{(x \minus{} \frac {\pi}{4})}\right|} \plus{} C \right]$

[quote="stevem"]Yes you can, but you lost a }. The tex log will tell you that a missing } had been inserted.
 $ \frac {d}{dx}\left[\frac {1}{\sqrt {2}}\ln{\left| \sec{(x \minus{} \frac {\pi}{4})} \plus{} \tan{(x \minus{} \frac {\pi}{4})}\right|} \plus{} C \right]$[/quote]What tex log?  I only use this Forum's-blog Tex and the only time I've seen a message about errors is when I'm missing a \$ to close my Latek.

Steve was assuming you were using TeX on your own computer to generate PDF documents, in which case you would have a .log file containing all the logged stuff (like errors and warnings).


Well, it's that time of year!
I'm going to be going to HCSSiM 2008 (under the IRL name of Matt Redmond), so if you're from AOPS, then definitely post here and let me know that you're planning on going as well. If you're going to go, or maybe if you're only going to talk about when you had gone, or maybe even if you are going right now to apply so that you can go while the going is good, then post. Go now!

EDIT: I'm not sure if applications are still open. Guess you'd have to go and see whether they're still allowing people to go.
EDIT EDIT:
I'm using my theme word of the day.
Or maybe I just find+replaced as much as I could with go.
Maybe not.

I'm Benjy Hirsch. I'm quite excited about the summer.

Happy 17th.

This is Emil. I'm a junior, I think.

I went there last summer and I'll be there again this year and it's going to be AMAZING! Can't wait!

I'm going into 9th grade, and I'm thinking about going to HCSSiM in the nxt couple of years, but I wanted to know, is that math mostly theoretical or applicable?

Hi!  I'm Megan and I went with Emil last year.  And I'm really excited for this year!!

I'm flying into MA on Wednesday (crack'o'dawn) and I should have a nice little weekend to myself :)
Then I drive myself into camp, and hopefully the fun begins!

Do we get there on the 29th? Or earlier, to unpack?

My name is Warren and I really hope the answer to Pockets' question is "29th".

You get there on 29th.

Why is Matt driving to a camp? Wouldn't you be missing HCSSiM?


Let $M= \{1, \dfrac{1}{2}, \dfrac{1}{3}, \ldots, \dfrac{1}{1991}\}.$ Generate a new set $M$ as follows. Delete two elements $a,b \in M$ and insert the new element $(a+b + a \cdot b)$: $(M \backslash \{a,b\}) \cup \{a+b + a \cdot b\}.$ This method is applied until only there will be only one number in the set. Is it possible that this number is 2004 ?

I think not... since it's always 1991 :D

Yes. Consider $\pi(S)=(1+x_1)(1+x_2)\cdots(1+x_n)$ where
$S=\{x_1, \cdots x_n\}$.
Then $\pi(S)$ is invariant.

or simpler: start with a,b then take c etc.

(a, b, c, d, ...)
(<a+b+ab>, c, d, ...)
(<a+b+ab+c+ac+bc+abc>, d, ...)
we get all terms so the thing is fully commutative.

(sort of induction)

But Peter, that's not simpler. That's exactly Iura's invariant when opening the brackets.  :D So it's [u]as[/u]  simple as Iura's.

woops :blush:

eum hey no, mine is immediately ok, iuras still needs to be reduced by 1 to be correct :D


group theory
abstract algebra
function
superior algebra
superior algebra solved

G is a group and H is a subgroup of G s.t. |G:H| is finite. Show that there is a subgroup of H, normal in G, N, s.t. |G:N| is also finite.

[i]moderator edit (19.2.2006): see http://www.mathlinks.ro/Forum/viewtopic.php?t=35182[/i]

|G:H| is called the index of H in G, the number of left coset of H in G
In a finite group the index is |G|/|H|

I found your problem in a book (without answers, unfortunately  :( ) and it had one more question: find an upper bound for the index of N in G. :D  :D

You mean that if |G:H|=n then we should find an upper bound for |G:N| as a function of n?

I guess so... :D  :D  It is not clearly stated, but it should be like this.

Do you allow {e} to be your normal subgroup in the finite case?

Of course, that's no problem. The interesting case is when G is infinite.

I don't know how to solve that other question. Moreover, I'm not entirely sure what it means. When you solve the problem I posted you get a particular group which is called the normal interior of H in G (it's usually denoted by HG). Maybe the second problem asks for an upper bound for |H/HG|, not for ANY subgroup N of H which is normal in G.

If it asks for an upper bound for |G:HG|, n! is such a bound.


For each natural number $ n$, determine the numbers of $ (a,b,c)$ such that: $ \{ a,b,c\}\subset\{0,1,2,...,n\minus{}1\}, a<b<c$, and $ n|a\plus{}b\plus{}c$

[hide="Hint (Recursion)"]If you want to enumerate the cases, build some recursions noticing the $ 0<a\plus{}b\plus{}c < 3n$ then $ a\plus{}b\plus{}c\equal{}n or 2n$[/hide]
[hide="Hint (Counting)"]If you want to count the cases maybe this approach helps, count all the ways of arrenging $ 2$ vertical lines with $ n$ or $ 2n$ dots sand then substract all the cases with $ 2$ or $ 3$ equal elements and divide by $ 6$ in order to get the ordered arrangements.[/hide]


I was just wondering, what would happen if school was canceled on the day of the AMC?  I think you're still able to take it the next day, and send the tests in for scores, but they aren't official.  Would there still be a chance to take the AMC, or not?

Thanks

Seriously. In Ohio, its freezing outside. I better not be preparing for crazy for the AMC and then missing it. Maybe the AMC office may do something?

they'll probably let the scores count since half the country has a snow day.

[quote="1234567890"]I'm concerned about this, too.

Here in Michigan, it's -20 degrees with the wind chill, and school is thus canceled today. And tomorrow will be as cold as today.  :([/quote]

Actually, I heard that tomorrow was supposed to be even colder, for us.  So if we have a "cold" day today, then I don't know about what will happen tomorrow.

Well, at least my school will offer the B date.  But I kind of liked the idea of having two chances to qualify for the AIME.

Our school doesn't even offer the B date because our principal only even registered our school because I wanted to do it.

Not cool.

Anyway, I think it'll be okay tomorrow.

You must take the AMC on the official date in order for the score to officially count.  The problem is that students agree (by their signature on the test answer sheet) not to discuss the test outside their school on the day of the test.  But the next day, all bets are off and students will start discussing the problems (in particular, here on AoPS).  So once the test has started to be publicly discussed, the AMC obviously cannot accept official scores.

In a nutshell: if your school is closed because of a snow day, your school will have to reregister for the B date in two weeks.  That's the primary reason why the AMC has two different dates (rather than to give students two "chances" to qualify for the AIME).

[quote="xxazurewrathxx"]Well if the scores counted that'd be a problem since the answers may have a better chance of leaking out. Either way, I feel for you guys. Even though it obviously doesn't snow much in Florida, there is a statewide standardized test being held on the same day as AMC A this year. Since this test determines whether or not you move on to the next grade and what grade your school gets (I know, it's stupid but it's mandatory), many schools prohibit after school activites on those days.[/quote]I heard that the same is true here in southern California.  As I said in my previous post, this is why the AMC has two dates.

My personal feeling (read: this is not the official position of AoPS and/or the AMC) is that students should only be allowed to take the AMC once (just like the AIME), to eliminate the real or perceived advantage that some students get by being able to take it twice, while others can only take it once.

So, if I contact my teacher to register the B date and he does it on WEdnesday, can I still take the B test

Here are the facts:  From the Teachers' Manual, page 6, Under Section VI, Policy Statements:
see online at: http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-AMC1012A-TM/2007AMC1012TM.pdf

"If you are unable to give the Contest A on February 6 because:
a. Your school is closed
b. your school has an academic conflict
c.  the class periods have been shortened to due to an assembly or other reason,
then you may give the second version of the contests (AMC 10 B/AMC 12 B) on the second official day Wednesday, February 21, 2007.  (See Contest B registration form on page 13.)  You may still take either exam unofficially on later dates, but those contests will not be eligible for state and national awards and will not be eligible for participation on the AIME.  Unofficial participants are still eligible for intramural awards.

It is important to note that the only days eligible for official participation are the two official COntest days, Tuesday, Febraury 6 and Wednesday, February 21, 2007."

That has been the policy since we went to the A-date and B-date contest in 2002.  It continues to be the policy.

Steve Dunbar
AMC Director

This isn't fair because most likely there will be a snow day tomorrow for me and I can't take the AMC.
In Mathcounts all the school, chapter, and state competitions are on different dates in different areas and that seems to work fine.

Also, our school doesn't take the B test :(

What about monetary refunds?

That's why the B-date Registration Form is included in the A-date Teachers' Manual.

ALternatively, check to see if a college or university is adminstering the B-date near you, and contact them.

Steve Dunbar
AMC Director

what happens if you have a snow date on the a and the b test?

if this happens with half the country, it could be a problem.

jorian

Does school necessarily have to be open in order for you to take the AMC?

Then again, maybe some people's math teachers are not as awesome.

we almost got out of school last wednesday due to snow.  :o

I doubt it will snow tomorrow, unless the weather forecast is wrong.

Isn't discussing of AMC 10A prohibited until a few days after competition? Why would it still be unofficial?....  :P

[quote="randomdragoon"]Does school necessarily have to be open in order for you to take the AMC?[/quote]

I don't think so, but it would be kind of hard, at my school at least, to get everyone together to take the AMC, especially if school is canceled because of a snow or "cold" day.

[quote="now a ranger"]Isn't discussing of AMC 10A prohibited until a few days after competition? Why would it still be unofficial?....[/quote]

Discussion is just prohibited on the day of the competition (I think DPatrick said that on this thread).  It's just that AoPS just locks the AMC forums until they hold a math jam.

We will be locking this forum later today, and reopening it on Wednesday morning.

The front page of the AMC 12 contest says:
[quote]The publication, reproduction or communication of the problems or solutions of the AMC 12 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules.[/quote]The phrase "period when students are eligible to participate" means anytime when it is Tuesday (for the A test) [i]in any time zone[/i].  So, in particular, this means no discussion at 1:00 Eastern Wednesday morning, since it's still Tuesday on the West Coast and Alaska/Hawaii.

In fact, as it is already Tuesday morning in the Far East, and a number of people take the AMC in east Asia, Japan, and Australia, we will be locking the forum momentarily.  It will reopen Wednesday morning.


inequalities
trigonometry
calculus
derivative
geometry
geometry proposed

Prove that in an isosceles  triangle  $ \triangle ABC$ with $ AC\equal{}BC\equal{}b$ following inequality holds $ b> \pi r$, where $ r$ is inradius.

Hey people can anyone solve this inequality starting with
  r/(a/2)=(tgA)/2
     r=(a/2)*(tgA)/2

 a/sinC=b/sinA a/sin2A=b/sinA

   a=2b*cosA

    r=b*cosA*(tgA/2)

   b/r=1/(cosA * tgA/2)

  and its enough to prove pi*cosA*tgA/2<1 I hope i wrote this good after that I use trigonometric equalities to have only cos(A/2) as variable and I got some good calculation  but not sure how far I got in them so I would like someone to try to solve it in this way

[quote="delegat"] [color=darkred]Prove that in an isosceles  triangle  $ \triangle ABC$ with $ AC \equal{} BC \equal{} b$ following inequality holds $ \frac br > 1 \plus{} \sqrt 5 > \pi$, where $ r$ is inradius.[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Observw that $ \frac br \equal{} \frac {b}{\frac c2}\cdot\frac {\frac c2}{r} \equal{} \frac {1}{\cos A}\cdot\frac {1}{\tan\frac A2} \equal{} f\left(\tan\frac A2\right)$ , where $ f\ : \ (0,1)\ \rightarrow\ \mathbb R$ , $ f(t) \equal{} \frac {t^2 \plus{} 1}{t \minus{} t^3}$ .

Prove (with derivatives) that for any $ t\in (0,1)$ we have $ f(t)\ge f\left(\sqrt {\sqrt 5 \minus{} 2}\right) \equal{} \frac {\sqrt 5 \minus{} 1}{(3 \minus{} \sqrt 5)\sqrt {\sqrt 5 \minus{} 2}}$ .

But $ \sqrt {\sqrt 5 \minus{} 2} < \frac 12$ . Thus $ f(t) > \frac {2(\sqrt 5 \minus{} 1)}{3 \minus{} \sqrt 5} \equal{} \frac {(\sqrt 5 \minus{} 1)(3 \plus{} \sqrt 5)}{2} \equal{} \sqrt 5 \plus{} 1 > \pi$ .

[b][u]Remark.[/u][/b] If the triangle $ ABC$ is no acute, then $ \frac br\ge 2\plus{}\sqrt 2>1\plus{}\sqrt 5>\pi$ .[/color]

Thanks for interesting and solution [b]Virgil Nicula[/b]

Here you can also (on page 3) find another solution to this problem.

[url]http://rapidshare.com/files/115927250/Bosnian_TST_2008..pdf.html[/url]

[size=134][color=darkred][b]Thank you, [u]Delegat[/u] ! This $ \sqrt {10}$ is a good idea and yet $ 1 + \sqrt 5 > \sqrt {10}\ !$[/color] [/size][/b]

[color=darkblue]I"ll present the [b]official proof[/b]. Observe that $ a = b\implies \|\begin{array}{c} p = \frac {2b + c}{2} \\
 \\
p - a = p - b = \frac c2 \\
 \\
p - c = \frac {2b - c}{2}\end{array}$

and $ pr^2 = (p - a)(p - b)(p - c)\implies$ $ r^2 = \frac {c^2(2b - c)}{4(2b + c)}$ . We"ll show $ \boxed {\ b > r\sqrt {10}\ }$ , i.e. $ b^2 > 10r^2$ $ \Longleftrightarrow$

$ 2b^2(2b + c) > 5c^2(2b - c)$ $ \Longleftrightarrow$ $ 4b^3 + 2b^2c - 10bc^2 + 5c^3 > 0$ for any $ 0 < c < 2b$

(inequality of triangle : $ a + b > c\implies 2b > c$ ). Denote $ t = \frac bc > \frac 12$ . Obtain $ 4t^3 + 2t^2 - 10t + 5 > 0$ for any $ t > \frac 12$ .

Denote $ x = 2t - 1 > 0$ , i.e. $ t: = \frac {x + 1}{2}$ . Obtain $ x^3 + (4x^2 - 5x + 2) > 0$ for any $ t > 0$. It is truly because $ x^3 > 0$ and

the discriminant $ \Delta$ of the equation $ 4x^2 - 5x + 2 = 0$ is $ \Delta = ( - 5)^2 - 4\cdot 4\cdot 2 = - 7 < 0$ (negative), i.e $ 4x^2 - 5x + 2 > 0$ .

In conclusion, $ \frac br > \sqrt {10} > \pi$ , i.e. $ b > \pi\cdot r$ .

[b][u]Remark.[/u] I like this problem[/b]. We can find a better approximation if consider the hexagon

$ MNUVXY$ which circumscribed to the incircle and $ \|\begin{array}{c} \{M,N\}\subset (CA) \\
\ \{U,V\}\subset (AB) \\
\ \{X,Y\}\subset (BC)\end{array}$ so that $ \|\begin {array}{c} MY\parallel AB \\
\ NU\parallel CB \\
\ VX\parallel CA\end{array}$ .

Observe that $ \|\begin{array}{c} MY = UV = \frac {c(p - c)}{p} \\
\ XV = MN = \frac {b(p - b)}{p} \\
\ NU = XY = \frac {a(p - a)}{p}\end{array}$ and the semiperimeter of this hexagon

is greater than the semiperimeter of the incircle, i.e. $ \sum \frac {a(p - a)}{p} > \pi\cdot r$ .

Directly, $ \sum \frac {a(p - a)}{p}=\frac {2r(4R+r)}{p}\ge \frac {2r}{p}\cdot p\sqrt 3=2r\sqrt 3>\pi\cdot r$ . [/color]

I have a similar solution to Virgils one, using AG at the end:
we start as Virgil and get the following, with $ t\equal{}\tan \frac{\alpha}{2}$, and we need to prove $ \frac{1}{t}\plus{}\frac{2t}{1\minus{}t^2} > \pi$, so since $ 0 < t < 1$ :
$ \frac{1}{3}\plus{}\frac{1}{3}\plus{}t^3\pi\plus{}\frac{1}{3}\plus{}t^2 \geq t(3 \sqrt[3]{\frac{\pi}{9}}\plus{}2\sqrt{\frac{1}{3}}) > \pi t$

the end

in fact,we can strengthen $\pi$ to $\frac{\sqrt{2+2\sqrt5}}{2}+\sqrt{2+\sqrt5}$
using a little derivative......