linear algebra
matrix
vector
superior algebra
superior algebra unsolved

Prove that any endomorphism $f$ of $M_n(\mathbb C)$ such that
\[f(X\cdot Y)=f(X)\cdot Y+X\cdot f(Y)\]
for any matrices $X,Y$ is of the following form: $f(X)=AX-XA$ for a fixed matrix $A$.

[i]Edited by Myth[/i]

Such $f$ is called derivation, there is way to solve this using
Kronecker product of matrices

$A\otimes B = \left[\begin{array}{ccc}a_{11}B&...&a_{1n}B\\...&...&...\\a_{n1}B&..&a_{nn}B\end{array}\right]$

I don't know if there is another way to solve it.

what's this proof with Kronecker product ?

To each matrix $A$ we define a vector
$vec A=(a_{11},...,a_{n1},a_{12},...,a_{n2},...,a_{1n},...,a_{nn})^{t}$

The condition on $f$ can be written like

(1) $vec(f(X,Y))=vec(f(X)Y)+vec(Xf(Y))$

We use the

Lemma:$g:M_n(C)\rightarrow M_n(C)$ linear map, then there exista unique matrix
$K(T) \in M_{n^2}(C)$ such that $vec(f(X))=K(T)vec(X)$ for all $X\in M_q(C)$

Use also $vec(AB)=(I_n\otimes A)vec(B)$

(1) becomes

$K(T)(I_n\otimes X)vec(Y)=(I_n\otimes f(X))vec(Y)+(I_n\otimes X)K(T)vec(Y)$

(2) $K(T)(I\otimes X) - (I\otimes X)K(T)=(I\otimes f(X))$ for all $X\in M_n(C)$

use the block decomposition  $K(T)=(K_{ij})$ wher $K_{ij}\in M_n(C)$

(2) implique

$f(X)=K_{ii}X-XK_{ii}$
$f(X)=K_{ij}X-XK_{ij}$  for $i\neq j$

$i=1$ give $f(X)=AX-XA$, $A=K_{11}$

Matrix Analysis
Horn & Johnson
Cambridge University Press

Theorem4.3.4


Does anyone know how many of the semifinal questions you need to get to camp? What were last year's figures?

Whoa, I don't think there's a need to start so many little threads on the USAPhO.  From a casual glance, there were already at least seven on the first page of these forums (not counting this one).  Please do some searching: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199609.


Last three digits of a 6-digit number forms a 3-digit number which is 1 more than a 3-digit number which is formed by the first 3 digits of the same 6-digit number, eg. 499500;613614 etc.

Find all such 6-digit numbers that are perfect squares.

I'm thinking about this:

$ 7\cdot11\cdot13k \equal{} (n \minus{} 1)(n \plus{} 1)$, But I don't know what to do now...

If the first 3-digit number is k, then the whole number is $ 1001k\plus{}1\equal{}n^2$ which is exactly the same as yours, Bachukas. Note that then $ n^2\equiv 1(mod7)$ and same for 11 and 13, so that means $ n\equiv \pm 1 (mod7)$ and so for the 11 and 13. Now find all $ n$, and then all numbers.

8 situations - much work  :oops:

However thank you #5-1Mole  :)

No problem :)
There are problems with more than just 8 situations, I don't like these either.


probability
geometry
search
expected value

Can someone help we with these tough probability questions?

1. People in this room are asked their birthday month and day, not their year. What are the chances that two ppl in the room have the same birthday?

2. The Ace Bubble Gum Factory includes on picture in each of its packages of gum. THe company printed the following numbers of pictures and randomly put them packages of gum:

45,000 of Bill Cosby, 45,000 of Michael J Fox, 90,000 of Bart Simpson, 180,000 of Alf the ALien. How many packages would you probably have to buy in order to obtain a complete set of all 4 pictures?

3. Design 2 identical circular spinners. Each spinner is divided into 4 equal sections. THe sections are numbered 1, 2, 3 and 3. Each spinner will be spun and the 2 numbers added together. WHich sum will occur most often?

4. Roll a regular 6-sided die as often as you wish. Your score is your total of all numbers rolled, but if roll a 1, you lose all the points for that round. What strategy would maximize your total score?

5. A man leaves for work at a random time b/w 7:00AM and 8:00Am. His paper arrives at his house at a random time b/w 6:30 AM and 7:30 AM. What is the probability that he has his newspaper before he leaves for work?

I don't understand how to set the geometric regions up for this problem. I have a horizontal axis for the man and a vertical axis for the newspaper => each axis goes from 6:30, 7:00, 7:30, 8:00 => I shade in both areas for the newspaper and the man, and isn't their intersection the probability? I get 1/9 for the area of the intersection over the total area, but I've asked many people and they all say the prob. is 1/8 Can someone show me why?

Thanks to all who help out!

Search for #1, #2 is pigeon hole -ish, #3 and #4 are multiplication, and #5 your graph doesn't make any sense to me, I think you should look at it in a different way.

mod note, not pre-olympiad.

Hints:
1.  Consider instead the probability that they don't have the same birthday.
3.  There is a small enough number of cases to consider each (but also think which seems logical?)
4.  Hmm, i think making an expected value tree diagram works here.
5.  The way you drew your diagram might be misleading you, are you keeping the original conditions in mind?  Like if the paper arrives at 6:30, he can't leave at 6:31, gotta wait till 7.  Try just having a square, 6:30 to 7:30 on x, 7:00 to 8:00 on y, shade the desired (and note, 1/8 is actually the probability that that does NOT happen).


Hello..
could you solve the first question of the third mathlinks...
If you send it to me. ;)

[quote="pbornsztein"]Do you mean the first question of the 3 round of mathlinks contest spring 2004?

Yes I can.... :D  :D  :D

Pierre.[/quote] Not only that he could solve it, but he also made it up :P :P

I want to post my solution.

Let $S$ be a set with $2004$ point and supose it only determinates $91$ distances.
Let $N$ be the number of ocurrences of a point of $S$ in the perpendicular trought the midpoint of other two points in the set. Obviously for every bisector, there are at most $7$ points on it, so there are at most $7\binom{2004}{2}$ ocurrences. also, if a point is equidistant from $k$ other points, it appears in $\binom{k}{2}$  bisectors. Since every poiint is at one of $91$ possible distances from the other $2003$ points, every point appears in at least $91\binom{22}{2}$ bisectors. So we have the following:

$7.\binom{2004}{2}>N>2004.91.\binom{22}{2}$ wich is obviously a contradiction.

What do you think Myth?


calculus
derivative
algebra
polynomial
geometry
3D geometry
calculus computations

[b][b](1) If y^2 = p(x) here p(x) is a polynomial of degree 3
Then find the value of d/dx(y^3*d2y/dx2).
means y cube  multiplied by d square y divided by d x square.

ans must be in the form of p(x) or p'(x) or p''(x) or p'''(x). [/b]

I want a less complex solution[/b]

Less complex than what? Show us what you've done.

Start with $ y^2 \equal{} p(x)$ and differentiate implicitly thrice (3 times) w.r.t. x and divide by 2 to get

\[ 3\frac{dy}{dx}\cdot \frac{d^2y}{dx^2}\plus{}y\cdot \frac{d^3y}{dx^3} \equal{}{\scriptstyle \frac{1}{2}} p^{\prime\prime\prime}(x)\]

where the right-hand side is better known as some constant, but I'll keep it in terms of $ p(x),p^\prime (x),\ldots$ .

We wanted to simplify

\[ \frac{d}{dx}\left( y^3\cdot\frac{d^2y}{dx^2}\right) \equal{} 3y^2\cdot\frac{dy}{dx}\cdot\frac{d^2y}{dx^2} \plus{} y^3\cdot\frac{d^3y}{dx^3} \equal{} y^2\cdot {\scriptstyle \frac{1}{2}} p^{\prime\prime\prime}(x) \equal{}\boxed{{\scriptstyle \frac{1}{2}} p(x)\cdot p^{\prime\prime\prime}(x) }\]


If a sentence says:

One force is pushing an object in a direction 50 degrees south of east with a force of 25 N.

Then if I want to resolve this into components, is it just

25*cos(50)i - 25*sin(50)j ??

Yes, that is correct.  Technically, you should be taking the cosine and sine of $ \minus{}50^{\circ}$ but since cosine is even and sine is odd, it is equivalent to what you have.

If the question is :

A second force is simultaneously pusihing the object in a 70 degrees north of west with a force of 60 newtons, then is it:

60 * cos(110) i +  60 * sin(110)j


Let $AK,BL$ and $CM$ be the altitudes of $\triangle ABC$ , and $P$ arbitrary point of the plane which does not belong to the lines which contain these altitudes. Prove that the circles circumscribed about $\triangle AKP,\triangle BLP$ and $\triangle CMP$, beside the point $P$ have one more common point.

Since $HA\cdot HK=HB\cdot HL=HC\cdot HM$ (denote their common value by $\alpha$), it means that $H$ is the radical center of $ABC$. This means that the point $P'$ situated on $(PH$ s.t. $HP\cdot HP'=\alpha$ must belong to all the circles.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Can someone give me an example of a countable group with uncountably many subgroups.

Since I'm pretty new to grouptheory: why would that particular example fulfil the requirements? :)

Thanks in advance

Consider the group $\mathbb{Z}^{(\mathbb{N})}$ of those sequences $(x_{1}, x_{2}, ...)$ with only finetely many entries $\neq 0$ and coordinatewise addition.
For each $n \in \mathbb{N}$, we define $a_{n}= (0,0,...,0,1,0,...)$ with the $1$ on position $n$.
Then for any subset $N$ of $\mathbb{N}$, we can define $A_{N}$ as the subgroup generated by those $a_{n}$ with $n \in N$.
Since there are uncountable many (lol, I was told in school that "many" is used if the objects can be counted  :lol: ) subsets $N$ of $\mathbb{N}$, we just have to show that all these subgroups are different, but this should be clear.


Let $ k$ be a circle and $ k_{1},k_{2},k_{3},k_{4}$ four smaller circles with their centres $ O_{1},O_{2},O_{3},O_{4}$ respectively, on $ k$. For $ i \equal{} 1,2,3,4$ and $ k_{5}\equal{} k_{1}$ the circles $ k_{i}$ and $ k_{i\plus{}1}$ meet at $ A_{i}$ and $ B_{i}$ such that $ A_{i}$ lies on $ k$. The points $ O_{1},A_{1},O_{2},A_{2},O_{3},A_{3},O_{4},A_{4}$ lie in that order on $ k$ and are pairwise different.

Prove that $ B_{1}B_{2}B_{3}B_{4}$ is a rectangle.

It is true regardless of the order of $ O_i$,$ A_i$ on $ k$
(directed angles were used in what I hope is a correct way)

$ \angle B_1 A_1 A_2$
$ \equal{} \angle B_1 A_1 O_2 \minus{} \angle A_2 A_1 O_2$
$ \equal{} \left( 90^{\circ} \minus{} \frac {1}{2} \angle A_1 O_2 B_1 \right) \minus{} \angle O_2 A_2 A_1$
$ \equal{} 90^{\circ} \minus{} \angle A_1 A_2 B_1 \minus{} \angle O_2 A_2 A_1$
$ \equal{} 90^{\circ} \minus{} \angle O_2 A_2 O_1$
Also,
$ \angle A_4 A_1 B_1$
$ \equal{} 90^{\circ} \minus{} \angle O_2 A_4 O_1$
$ \equal{} 90^{\circ} \minus{} \angle O_2 A_2 O_1$
$ \equal{} \angle B_1 A_1 A_2$

Thus,
$ \angle B_1 A_1 A_2 \equal{} \frac {1}{2} \angle A_4 A_1 A_2$
Similarly,
$ \angle A_2 A_3 B_3 \equal{} \frac {1}{2} \angle A_2 A_3 A_4$

Therefore,
$ \angle B_1 B_2 B_3$
$ \equal{} \angle B_1 B_2 A_2 \plus{} \angle A_2 B_2 B_3$
$ \equal{} \angle B_1 A_1 A_2 \plus{} \angle A_2 A_3 B_3$
$ \equal{} \frac {1}{2} \angle A_4 A_1 A_2 \plus{} \frac {1}{2} \angle A_2 A_3 A_4$
$ \equal{} 90^{\circ}$

$ (QED)$

[quote="weiquan"]
$ \equal{} 90^{\circ} \minus{} \angle A_1 A_2 B_1 \minus{} \angle O_2 A_2 A_1$
$ \equal{} 90^{\circ} \minus{} \angle O_2 A_2 O_1$
[/quote]

I just realised I forgot to justify that $ O_1 B_1 A_2$ are collinear, which I proved it first and conveniently assumed that it was given :P

$ \equal{} 90^{\circ} \minus{} \angle A_1 A_2 B_1 \minus{} \angle O_2 A_2 A_1$
$ \equal{} 90^{\circ} \minus{} \angle A_1 O_2 O_1 \minus{} \angle O_2 A_2 A_1$
$ \equal{} 90^{\circ} \minus{} \angle A_1 A_2 O_1 \minus{} \angle O_2 A_2 A_1$
$ \equal{} 90^{\circ} \minus{} \angle O_2 A_2 O_1$


[quote="Michael"]Hello!

Let $a,b,c$ be three real numbers such that $abc=1$. Prove that:
\[ {{1}\over{a^2+ab}}+{{1}\over{b^2+bc}}+{{1}\over{c^2+ca}}\geq {3\over 2}.  \][/quote]
I think $a,b,c$ are positive real numbers.;)
Let $a=\frac{y}{x},$ $b=\frac{z}{y},$ $c=\frac{x}{z}.$
Hence, $\sum_{cyc}\frac{1}{a^2+ab}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}\frac{x^2}{y^2+xz}\geq\frac{3}{2}.$
$\sum_{cyc}\frac{x^2}{y^2+xz}=\sum_{cyc}\frac{x^4}{x^2y^2+x^3z}\geq\frac{(x^2+y^2+z^2)^2}{x^2y^2+x^2z^2+y^2z^2+x^3z+y^3x+z^3y}.$
But $\frac{(x^2+y^2+z^2)^2}{x^2y^2+x^2z^2+y^2z^2+x^3z+y^3x+z^3y}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}(2x^4+x^2y^2-3x^3z)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(8x^4+4x^2z^2-12x^3z)\geq0\Leftrightarrow\sum_{cyc}(7x^4+z^4+4x^2z^2-12x^3z)\geq0.$
But $7x^4+z^4+4x^2z^2\geq12\cdot\sqrt[12]{x^{28+8}z^{4+8}}=12x^3z.$ :)

Thank you very much arqady. Yes, you are right, the real numbers $a,b,c$ are positive and verify $abc=1$.

Does your solution apply also for more than three variables (for examples for four real positive numbers $a,b,c,d$ such that $abcd=1$, or more generally for $n$ reals). I think that in this case the RHS of the inequality should be $n/2$.

Michael.

[quote="arqady"][quote="Michael"] I think that in this case the RHS of the inequality should be $n/2$.
Michael.[/quote]
This is very nice! :)
I had difficulties already for n=5. :([/quote]

Did you solve it for $n=4$? :)

[quote="Michael"]
Did you solve it for $n=4$? :)[/quote]
Yes. Perheps to prove it like for the case $n=3.$ :)

[quote="Vasc"]The more general statement is valid:
Let $p$ and  $a,b,c$ be positive numbers such that $abc=1$. Prove that:
${{1}\over{a(a+pb)}}+{{1}\over{b(b+pc)}}+{{1}\over{c(c+pa)}}\geq{3\over{1+p}}$.[/quote]
For the proof we can use the following your inequality.
\[(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x)\]

[quote="Michael"]Hello!

Let $a,b,c$ be three real numbers such that $abc=1$. Prove that:
\[ {{1}\over{a^2+ab}}+{{1}\over{b^2+bc}}+{{1}\over{c^2+ca}}\geq {3\over 2}. \][/quote]
An easier method is using AM-HM.
$(\frac{1}{a(a+b)} +\frac{1}{b(b+c)} +\frac{1}{c(c+a)})(a(a+b) + b(b+c) + c(c+a)) \ge 9$
$a^2 +ab +b^2 +bc +c^2 +ac \ge 6$
From which the result follows

[quote="Unknown-6174"]
An easier method is using AM-HM.
$(\frac{1}{a(a+b)} +\frac{1}{b(b+c)} +\frac{1}{c(c+a)})(a(a+b) + b(b+c) + c(c+a)) \ge 9$
$a^2 +ab +b^2 +bc +c^2 +ac \ge 6$
From which the result follows[/quote]
Doesn't work - need $a^2 +ab +b^2 +bc +c^2 +ac \le 6$

$x^4+y^4+z^4\ge xy^3+yz^3+zx^3$ is true by Rearrangement though so it is an easy fix.

[quote="arqady"]I think your last step is not so good.[/quote]
Dear arquady! Why? Can you tell me more okay?

[quote=Michael]Hello!
Let $a,b,c$ be three real numbers such that $abc=1$. Prove that:
\[ {{1}\over{a^2+ab}}+{{1}\over{b^2+bc}}+{{1}\over{c^2+ca}}\geq {3\over 2}. \][/quote]
Let $a,b,c$ be three real numbers such that $abc=1$. Prove that:
\[ \frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}.\]

[url=https://artofproblemsolving.com/community/c6h183916p1010959]Zhautykov Olympiad, Kazakhstan 2008[/url]

I have a solution (I think) with MA - MH

[quote=sqing]
Let $a,b,c$ be three real numbers such that $abc=1$. Prove that:
\[ \frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}.\]
[url=https://artofproblemsolving.com/community/c6h183916p1010959]Zhautykov Olympiad, Kazakhstan 2008[/url][/quote]
Let $a,b,c$ be positive real numbers such that $abc=1$ . [url=https://artofproblemsolving.com/community/c6h1064117p4863125]Prove that[/url]$$ \frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq\frac{3}{2}.$$

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove that$$ \frac{2}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq2\sqrt 2-1.$$
Equality holds when $a=\sqrt 2+1,b=1,c=\sqrt 2-1.$
$$ \frac{3}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq2\sqrt 3-\frac{3}{2}.$$Equality holds when $a=2\sqrt 3+3,b=1,c=\frac{2}{\sqrt 3}-1.$
$$ \frac{k}{a(b+1)}+\frac{k}{b(c+1)}+\frac{1}{c(a+1)}\geq2\sqrt k-\frac{1}{2}.$$
Where $k> \frac{1}{4}.$ Equality holds when $a=1,b=2\sqrt k-1,c=\frac{1}{2\sqrt k-1}.$

[quote=sqing]Let $a,b,c$ be three real numbers such that $abc=1$. Prove that:
\[ \frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}.\][/quote]
Let $ a, b, c$  be three real numbers such that  $ abc = 1$. Prove that$$ \frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)} \geq \frac{3}{4}$$
[hide=S]$x= \frac{1}{a(b+1)},y=\frac{1}{b(c+1)},z=\frac{1}{c(a+1)}$
$$(x+y+z)^2\geq 3(xy+yz+zx)$$[/hide]
[url=https://artofproblemsolving.com/community/c6h2155588p15929956]h[/url]

[quote=sqing]
Let $ a, b, c$  be three real numbers such that  $ abc = 1$. Prove that$$ \frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)} \geq \frac{3}{4}$$[/quote]


logarithms
function
inequalities unsolved
inequalities

Given a>b>1:
    Prove that a^(b^a) > b^(a^b)

$a>b>1,\text{ show }a^{b^{a}}>b^{a^{b}}$
Take $ln$ of the sides twice:
$b^{a}\ln a>a^{b}\ln b$
Since $\ln$ is increasing, it suffices: $b^{a}>a^{b}$
$a\ln b>b\ln a \Longleftrightarrow \frac{a}{\ln a}>\frac{b}{\ln b}$
Consider $f(x)=\frac{x}{\ln x}$. $f'(x)=\frac{\ln x-1}{(\ln x)^{2}}$ uh oh, this function is only increasing when $x>e$ so lets try new approach:
$b^{a}\ln a>a^{b}\ln b\implies a\ln b+\ln \ln a>b\ln a+\ln\ln b$ now we can use fact that $e>a>b>1$ here but it gets too messy
$b^{a}\ln a>a^{b}\ln b\implies \ln\log_{b}a>b\ln a-a\ln b$
---------------------------------------
new way completely
$f(a,b)=a^{b^{a}}-b^{a^{b}}$
Derivate w.r.t a,b, respectively:
$b^{a}a^{b^{a}-1}\ln b b^{a}-b\ln a b^{a^{b}}b a^{b-1}$
same except vars flipped
It can easily be shown that this is always positive

I'm sorry for such a messy and not nice solution... do you have a better one in mind?

your new way:
    How can you see f'(a,b)>0 ?

Disregard what I wrote, I ruined the problem :(
I believe that b^a>a^b must hold


real analysis
function
algebra
polynomial
integration
inequalities
triangle inequality

Let $ f: [a,b]\rightarrow \Re$ be a Lebesgue integrabe function, and let $ \epsilon> 0$.
Prove that there exists a polynomial $ p$ such that $ \int_a^b |f(x)\minus{}p(x)| dx <\epsilon$

approximate $ f$ by a continuous function, then use stone-weierstrass.

it's a standard result that for every $ f\in L^1([a,b])$ and every $ \varepsilon>0$, there exists $ g\in C([a,b])$ with $ \|f\minus{}g\|_{L^1([a,b])}<\varepsilon/2$.  By stone-weierstrass, we can find a polynomial $ h$ with $ \|g\minus{}h\|_{L^\infty([a,b])}\leq \varepsilon/(2(b\minus{}a))$.  by holder's inequality, $ \|g\minus{}h\|_{L^1([a,b])}<\varepsilon/2$.  by the triangle inequality, $ \|f\minus{}h\|_{L^1([a,b])}<\varepsilon$.


algebra
polynomial
special factorizations

Hi everyone,

Can someone remind me how to complete the square with an equation like -4y^2+4y+6? Do I need to factor out a 2 first?

Thank you very much

Well, assuming that the original equation is equal to 0, you start with
$ \minus{}4y^{2}\plus{}4y\plus{}6 \equal{} 0$. I think it is necessary to divide by $ \minus{}4$ in this case, because the coefficient of the first term must be 1, but I'm not positive. Dividing the whole equation by $ \minus{}4$ gives $ y^{2}\minus{}y\minus{}\frac{3}{2}\equal{} 0$
Now, you move the constant to the other side by adding, giving you $ y^{2}\minus{}y \equal{}\frac{3}{2}$
Now, you take $ \frac{1}{2}$ of the coefficient of the middle term and square it, then add this number to
both sides, giving you $ y^{2}\minus{}y\plus{}\frac{1}{4}\equal{}\frac{3}{4}$.
Now the LHS is a perfect square, so it can be rewritten as $ \left(y\minus{}\frac{1}{2}\right)^{2}\equal{}\frac{3}{4}$

Taking the square root of both sides and manipulating it, you get $ y \equal{}\frac{1\pm\sqrt{3}}{2}$.

Solution

$ \minus{}4y^{2}\plus{}4y\plus{}6$

$ \equal{}\minus{}4(y^{2}\minus{}y)\plus{}6$

$ \equal{}\minus{}4\left(y\minus{}\frac{1}{2}\right)^{2}\plus{}4\left(\frac{1}{2}\right)^{2}\plus{}6$

$ \equal{} 7\minus{}4\left(y\minus{}\frac{1}{2}\right)^{2}$

Completing the square just means you add/subtract a number to both sides of an equation to make a polynomial expression a perfect square polynomial.

In the future, please don't post things about simple algebra like completing the square in this forum. I think you should post that in Classroom Math. (under Middle School) or under High School Basics.

I knew I was screwing something up. You don't divide by -4, just factor it out :P


$ n$ is of the form $ 3k \plus{} 1$. We start writing the nos. $ 1,2,\cdots,n$ in any order.At any instant, the sum of the nos. written should not be divisible by 3. In how many ways can these nos. be written?

In mod 3, this is a sequence of $ k$ 2's, $ k$ 0's and $ k \plus{} 1$ 1's in some order. Consider a sequence satisfying the given, and take out all the 0's. If the remaining sequence started off with 22, then it must then alternate between 121212... However, at any point in this sequence, there are more 2's to the left of the $ i$th term than 1's, so this is not possible. Thus, the remaining numbers must start with 11 and then alternate between 2121...., with the right most number a 2. We may then put $ n$ 0's in the sequence where ever we wish, as long as it is to the right of the leftmost 1. Thus chopping off the leftmost 1, and considering the remaining $ 3k$ numbers as 0 or nonzero, we see we must place the $ k$ 0's into $ 3k$ total positions, so the answer is $ \dbinom{3k}{k}$.

Does anyone know how to generalize this problem for not divisible by $ a$?


Hello.

I'm having a lot of trouble with the following problem. Any help would be appreciated.

For each integer $ n >\equal{} 1$, let $ A_{n} \equal{} (1 \minus{} \frac{1}{2^1})(2 \minus{} \frac{1}{2^2})(3 \minus{} \frac{1}{2^3})...(n \minus{} \frac{1}{2^n})$ and $ B_{n} \equal{} (1 \plus{} 2^1)(2 \plus{} 2^2)(3 \plus{} 2^3)...(n \plus{} 2^n)$.

Verify that $ (n\plus{}1)A_{n} <\equal{} B_{n}$.

Explain.

[hide]
Note that $ A_n < n!$ since each factor is less than each consecutive integer

Then see that $ B_n > (n\plus{}1)!$ since each term is greater than one more than the consecutive integers from 2 to n+1

Pulling it all together

$ (n\plus{}1)A_n < (n\plus{}1)! < B_n$
And we're done
[/hide]


calculus
integration
analytic geometry
calculus computations

Can anyone help me with this double integral

I^2 = inegr.\integr.|0,oo| e^[-(x^2 + y^2] .dx dy

Show that I = sqrt.(pi) /2 using rectangular coordinates.
I was asked to do this problem in polar coordinates, and a hint was given :

to let u = /sqrt.(x^2 + y^2)

Any help would be appreciated.

Switching to polar coordinates, you get $I^{2}=\int_{0}^{\pi/2}\int_{0}^{\infty}e^{r^{2}}r\,dr\,d\theta$, which can be evaluated using the substitution $u=r^{2}$.


[quote="Quevvy"]If $f(x)=3x+9$, what is $f^{-1}(42)$?

How would you do negative powered functions?[/quote]
[hide]
$f^{-1}(x)=\frac{x}{3}-3$
$f^{-1}(42)=\frac{42}{3}-3=14-3=11$
to do negative powered fcns, just reverse everything that's done of its positive power(if it passed the horizontal and vertical line test)

[/hide]

[quote="ProtestanT"]to do negative powered fcns, just reverse everything that's done of its positive power(if it passed the horizontal and vertical line test) [/quote]
Then wouldn't that mean that you would do
$f^{-1}(x)=-3x-9$?
I'm confused... :huh:

Well, it's not really a negative power function, but rather a function inverse. That is, it's the operation that you perform to "undo" the function. So, in this problem, you're really just looking for the value of $x$ when $f(x)=42$. So, $42=3x+9$ and $x=\frac{42-9}{3}= 11$.

[quote="Quevvy"]If $f(x)=3x+9$, what is $f^{-1}(42)$?

How would you do negative powered functions?[/quote]

[hide]Okay, the definition of the -1 is to denote an inverse function.

An inverse function in this case has to deal with the number that you inputted given the output.

Opposed to the function where you are given the input and find the output.

So you want to find the value of x such that 42=3x+9

This is 11

$11$[/hide]

[quote="mathgeniuse^ln(x)"][quote="ProtestanT"]$f^{2}(x)=3(3x+9)+9=9x+36$, and take the inverse of that, which is $f^{-2}(x)=\frac{x}{9}-4$
$f^{-2}(42)=\frac{2}{3}$
either its like that or theres no such thing as $f^{2}$ or $f^{-2}$
i think thats how u do it. im not too sure[/quote]

But we want f^{-1}, not f^{-2}.

That would be right if it was -2 instead of -1.[/quote]
[quote="Quevvy"]Then, using the same $f(x)$, how would you do $f^{-2}(42)$?[/quote]

Sorry.

In this case, you could just find the value that gives the value that gives the value of 42.

$f^{2}(x)=f(f(x))$
$f^{-2}(y)$ is the value of $x$ such that $f(f(x))=y$.


How is the score needed to qualify for USAMO calculated?
if you can't tell me that then if i'm getting around the 130's on the AMC 10's then how many correct would i need on the AIME to qualify for USAMO?

The USAMO index is typically around 210 and is calculated as AMC + 10*AIME.

Another general rule of thumb is that if you get $ x$ on practice AMC 10's then you'll get $ x\minus{}15$ on the real thing. So you have to raise your practice AMC 10 score to at least 135 before you think about making AIME.

So if you get 135 on practice AMC's, then you'll get 120 on the AMC 10A and you'll need 9 on the AIME to qualify.

I'm not so sure about the x-15 part...

I was scoring about 105 on AMC 10's, and ended up with 114.

I was scoring about 90 on AMC 12's, and ended up with 97.5

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1221302#1221302]USAMO index cutoffs[/url] in recent years.

[quote="pythag011"]depends. I always do WAY worse on practice AMC 10s then the real thing.[/quote]You remind me of myself- except that at that age, I basically didn't know anything about math contests.

There is one very important difference between practice and the real thing: adrenaline. Some people suffer from this test anxiety, and do worse under pressure. Others react well, and actually do better.
I'm one of those people who does better in the adrenaline rush, thinking faster and more clearly.

There's no way to know where you are on the scale except by experience, but it should affect your strategy in preparing. If you react negatively to the adrenaline, you should practice to make the test as routine as possible, minimizing disruption. If you react positively, you should be familiar with the format so you don't waste time, but most preparation effort should go into the knowledge you need rather than the details of the test.

i also disagree with yongyi. take someone who got a 136.5+ on AMC 10. they probly took practice tests. so are you saying that these people got over perfects on practice AMC 10's. for me, in 7th grade, i would get like 114's and 111's (occasionally 120) and stuff on the practice tests but on the actual test i got a 117. although yongyi was kinda right about my 8th grade year. I was getting 138+ on AMC 10's for practice, but on the real thing i got a 126. so blah.

@poincare: try to do as well as you can on AMC 10 (try to get 140+) to give yourself some wiggle room on the AIME in case you have a bad day or the test doesn't work for you.

thank you everyone.

So, right now i am getting around 140 on untimed AMC 10's but i average just over 75 minutes when i finish test. if i am doing this right, then i probably need like a 7 or 8 on the AIME. that's pretty hard. so, i am going to study!

@jmerry
i seem to be doing a lot better with the time pressure. i don't know why.

Also, for you, there's no rush, really.  You're in what, 5th or 6th grade?  Don't stress out if you don't make it.

i am in 6th grade. I don't really do it cause i want to win the competitions, but i do it cause its fun. So, I just wanted to know how far the people who made USAMO are from me.

The one of the important things to keep in mind is that because your grade is lower than 11th, what you do in AMC 10 is irrelevant. All that matter is floor value, which varies each year. But I say 8-9 is always safe score (it goes lower than that most times but there were times when the floor value was 8).

Again, because you're so young, don't worry about the score. Just keep practicing and enjoy.

[quote="cognos599"]actually, now there is a required index in addition to the floor score for <11th graders. this year it was 202.5 or something...[/quote]

Hmm, interesting. I never heard of this information. Thanks.

[quote="buzzer11"]I haven't heard it either. Is there a website or something that can confirm this?[/quote]

I have heard it, check the AMC website...

I got the floor score in AIME this year, but my AMC score was wayyy too bad to get in...and i'm in 7th grade, didn't make USAMo.

The two-part system has been on the books the whole time, but it has only mattered the last two years. If they had used the floor score alone these last two years, they would have had far too many qualifiers, overwhelming their resources. This activated the secondary cut of the index score, which is a bit unfair in that it counts AMC 10 scores and AMC 12 scores equally. In 2008, the index score was high enough that the floor didn't matter at all.

yes, if last year, someone received the floor score of 6, they would need at least a 144 on AMC 10/12 in order to qualify. but i do know of many people who got 144/6 and qualified.

The point isn't that 144 was needed with a 6- it's that $ \ge 152.5$ was needed with a $ 5$, and that's impossible.

But to qualify for USAMO, you must achieve the floor score and achieve the required index. Take '07 for example. I think the cutoff for <11th graders was 183.5 or something. It is very possible that someone got a 133.5+/5 and did not make it to USAMO.

You're not actually disagreeing with me. Look at the post I linked earlier.


After restarting my 17 powerbook g4 (leopard) when it had auto-updated, I left it only to find that it only shut off, but didn't turn back on. I turned it on manually and it worked (i heard the startup chime and the fan turned on and it made it to the gray loading screen with the rotating dial), but it got stuck at that loading screen. The dial keeps rotating, so I know it's not frozen.

So i force shut it down and started it up like 5 times and the same thing happened every single time. I tried starting it up in Safeboot, but it didn't start up again.

does anyone know if there's anything i can do to start up my computer?

Were you having any problems before? Every time I've run into problems like that, it's been a hardware issue.
Can you start it off the system CD?

That particular stopping place seems like either a hard drive problem or a corrupted system installation- and the first could cause the second. I'd recommend taking it to an Apple store to get looked at. You may have to pay for any actual repairs since the standard warranty is one year, but the diagnosis at least is free.
If it is the hard drive, you want to do this soon so you have a chance at recovering your data.

thanks for the help!

i ran the hardware check and none of the hardware was corrupted.

i guess it seems strange that the hard drive is malfunctioning because it's been working perfectly fine for about 2 years.

Don't trust that diagnostic. That just means it can't find a problem, not that there isn't one. I remember trying to run it and having the computer crash, then successfully running it and finding nothing.

When my hard drive went bad, I lost all data- by the time I took it in, it was possible to mount the drive on another machine, but not to copy due to corruption. A reformat and reinstall bought me a few days, but when everything recurred, I went for the replacement. Save as much as you can now, before drastic measures.
These things can happen suddenly- my hard drive died about eight months in, and I didn't have any real warning before the crashes started. They did get worse fairly rapidly from there, with successful startup progressively less likely.

my hard drive suffered a head crash two months from purchase :dry:

replaced for free cause of warranty, though I lost all data.. now I keep paranoid backups :D

mwpl, like I said in the email, you might try resetting the PMU.

you should get hard drives from Hitachi :) my dad has like five of them, and none of them has failed after three years of usage.

fyi, I got my 250GB SATA drive here: [url]http://eshop.macsales.com/item/Hitachi/0A53329/[/url], though you may need an IDE drive for a powerbook.. not sure.

EDIT: get a Linux live CD like the one here: [url]http://www.ubuntu.com/getubuntu/download[/url] and boot your mac off of it to back up your data first.. I smell a potential Linux convert :D

[quote="undefined117"]EDIT: get a Linux live CD like the one here: http://www.ubuntu.com/getubuntu/download and boot your mac off of it to back up your data first.. I smell a potential Linux convert [/quote]

if i do that will i be able to restore all the information i already have stored on my computer?

the only information that i really need is my adobe cs3 serial number. unfortunately, its written on a sticky in my dashboard. that will probably be wiped by now, right?

[quote="undefined117"]Linux live CDs are loaded onto RAM, so no it won't mess with any data.

the stickies database should be here: ~/Library/StickiesDatabase

save that file elsewhere/upload it to some file host if you're going to do a clean install..[/quote]So you're convincing others to put Linux on their Macs now... :)

But yeah, like he said; LiveCDs don't use virtual memory.  They load as much of the OS as is necessary into RAM; they don't touch the hard drive at all.

mwpl: Diagnose your mac, don't switch to using Linux unless you have strong feelings about doing so and want to promote open source development, increase efficiency through command line, etc. People usually don't switch only for a more stable OS.  But if you like what you see after you run the (Ubuntu?) LiveCD and are comfortable, then go for it.

oh and remember don't judge by looks alone! You can skin your Linux system however you want :)

Debian's claims of stability holds true.

after two days of desperation, IT FINALLY WORKS!!! on my third time reinstalling an OS, i finally decided to downgrade and reinstall tiger instead of leopard (my ppc came with tiger). so, ya. it worked.

and guess what? IT DIDN'T DELETE ADOBE CS3!!! i lost the product key, all my photos, all my files, all my music, but none of my apps, so its fine with me. internet still doesn't work and most of my apps are severely outdated (tiger versions, so much so they won't even open), but as long as i don't have to restart the computer again, it'll hopefully be okay.

now just watch me turn all paranoid like undefined and back up my entire computer every two seconds :ninja: .

:D I'll go first:

1. A Mac just works.
2. Solidity of UNIX + GUI simplicity.
3. Apple machines are not only visually appealing, but also comes with high-quality hardware.
4. For switchers, they can boot both Windows and Mac OS on a Mac if need be.
5. Applecare = awesome.
6. Everything is tightly integrated.. and who doesn't use an iPod? :)
7. No viruses, literally.
8. Most Windows programs have a Mac version, and there are good - or better - alternatives for those that don't.
9. Don't say you can't right-click - you can plug in a standard two (or more) -button mouse and it'll work perfectly.
10. Support the underdog.

but then again.. Linux on Mac = ultimate pwnage

I must admit, undefined, that your reasons were pretty good (and comical), particularly because you used the word "simplicity" in a sentence (few do that).

Who's next?

[quote="MathAndKnowledge"]So now that your experience is over, mwpl11, I offer a solution that would have been far simpler:

Don't use Macs.

I'm sure that at this point, MustIt will provide 10 reasons why I'm wrong (of course, I still want to hear them!).  Yeah, I dislike Macs.

I anticipate that with the above statement, I will start some sort of war (me vs. everyone else on the Indiana forum) about Macs.  Oh well...[/quote]Pretty much exactly what undefined said, and because you wanted me to:

1. Don't get me wrong, I tend not to prefer Macs becasuse of their lack of developer community, which means no open source apps, whatsoever.  But since we're not really talking about the apps made for it, I'll disregard that.

2. It uses an OS based on Unix, and they even admit that it's based on Unix. :) This means STABLE OS! (but not necessarily stable hardware :) )

3. Most people, probably including you, think that Mac OS has a nice GUI.

4. Though it's not easy to install 3rd party hardware, Apple makes EVERYTHING for it, which means no need for cluttering your OS with outdated driver installations, etc.  Mac can handle all of that for you.

5. Going along with 2 and 4, there is no room, really, compared to Windows, for viruses to find their way into your computer.

6. Apple tends to look into graphics before anything else. :)

7. Mac OS comes with a simpler boot loader.

8. No registry. yay!

9. You may not be smRt enuff to know what to put on your computer.  Apple puts everything.

10. Don't support the OS selling more, support the one which is going to suck more money out of you!

Ok so that list was kinda making fun of undefined, and most of those can be said as bad things.  Anyway, just to note, I'm working on a Win2k box from 8 or 9 years ago.  It hasn't crashed since it's been using win2k (used to run 98 a long, long, long, long time ago) the only thing I don't particularily like is that anti-virus takes up pretty much all the free memory there is. (but who cares, i work with everything online anyway :) )

mhmm yeah this is all basically personal preference anyway.. I like pretty GUIs :) Linux offered a pretty GUI AND awesome performance/configurability, which is why I switched. I was thinking of it even before I'd bought my macbook anyway, since I'd tried Debian.

My excuse for their high prices is that Apple needs the money :P I think it's still within reasonable limits, though.

But I don't like how Apple is both the software and hardware supplier.. it means non-Apple software or hardware isn't totally guaranteed to work with the Mac OS, which kinda keeps you in a box provided by Apple when you first buy your Mac.

Which is why while Linux on a Mac might be uber awesome, getting it installed and functioning at near-perfection is a nasty ordeal for the beginner.

You could get a Dell XPS (MustIt seems to worship this, so I'll take his word for its magnificence) and put Linux on it.. would be a lot cheaper/easier :)

EDIT: either way, Windows is only still here because users are reluctant to switch. Mac systems are appealing for the beginner since it's so simple to start using it out of the box. A beginner wouldn't care about all the software configurability or whatever, so they wouldn't know what they're missing. They'd see the sleek machines with pretty GUI and be all oooh ahhhh :w00t:

An advanced user, on the other hand, would know better than to use Windows. They'd use Linux, of course. No further explanation needed :)

[quote="undefined117"]An advanced user, on the other hand, would know better than to use Windows. They'd use Linux, of course. No further explanation needed :)[/quote]Uhh many advanced users are fine with Windows and Cygwin :D

People develop apps for Windows.  Corporations develop apps for Macs.  Clear enough? :rotfl:

Dell XPSs are NOT CHEAP.  At all, but they are pretty much the best branded PCs out there, in my opinion. (in service too) Unfortunately, you're going to have to get one with Vista now. (We got an XPS 210 RIGHT when it came out, which means RIGHT when Core 2 Duo came out. Blazing fast. :) )

[quote="mwpl11"]well, from my past experience with windows (i had a windows laptop that crashed like every week without warning; it didn't care if i was working on a major paper or anything before it just decided to kill the os an:dry: ), many more tears were shed in the two years i was using my windows than this one incident i've had in the two years i've been using this mac.

for all the purposes i use a computer, a mac fits my needs perfectly. now we just need to convert schools to mac so that no one will have compatibility issues anymore!

and undefined: how would you get linux and how is it different from, say, leopard?[/quote]Well, our school is using Windows, for well...compatibility reasons. :D  Also, I wouldn't be surprised if you were using a DOS based version of Windows.

And any version of Linux cannot really be compared to Leopard.  Well, in terms of stability, they're probably about the same.  GUI can be customized any way you want (literally). But Linux is targeted towards advanced users.  There is a learning curve which is much more than "what are the differences?"

As for getting Linux.  There are three methods.  One is to download the LiveCD, preferrably off a torrent (but can usually be done off their website as well), and to burn that iso onto a CD.  The second is to do the same with the command line installation; useful if you need to install on a slower computer, but much more complicated. The last is to request a CD be sent to you.  Ubuntu (Canonical), I know sends out CDs for free if you request them, but it takes months before they arrive.

..well anyway like I said, Windows is only here because of compatibility and people's reluctance to change.

The international school I took the AMC at is all Mac.. I was surprised :lol:


Please help me with this problem, solving:

z^2 + 13iz - 51 - 3i = 0
where z=a+ib

Thanks!

You may apply the Quadratic formula. It works with complex numbers as well.


calculus
integration
function
analytic geometry
graphing lines
slope
derivative

f(t)=6+cos(t/10)+3sin(7t/40)

and f(t) is the velocity of a plane.

according to this, what is the average velocity of the plane in miles per mintue over the time interval 0<=t<=40.

according to the answer its just 1/40*integral of f(t) from 0 to 40, but isnt that finding the average value in miles?  or am i mistaken.  or are they the same thing??  or is the average value in miles when f(t) is the position instead of the velocity?

also what does f(40)-f(0)/(40-0) find?? the average rate of change?  whats the difference.

thanks.

The units of the answer to the integral are miles, and the 40 that you divide by is 40 minutes. So the units of the answer are miles per minutes. In general the units of the average of a function must be the same as the units of the function.

The average rate of change of a function is what you gave as $ \frac{f(40)\minus{}f(0)}{40\minus{}0}$, which is the slope of the secant line. This is the definition. Let $ f$ have a derivative $ f'(x)$. Then it isn't unreasonable to call $ f'(x)$ the rate of change of $ f$. But then the average of $ f'$ is the average of the rate of change of $ f$ which is already given as $ \frac{f(40)\minus{}f(0)}{40\minus{}0}$. Now apply the fundamental theorem to obtain the usual integral for average. You get that the average of $ f'(x)$ on interval a to b is $ \frac{\int_a^b f'(x)\,dx}{b\minus{}a}$.

Good luck on the test.

You know that \[ \int^b_a f(t)dt\equal{} F(b)\minus{}F(a)\] then \[ \frac{1}{b\minus{}a}\int^b_a f(t)dt\equal{}\frac{F(b)\minus{}F(a)}{b\minus{}a}\]
So no, you're not finding the average value in miles.
You're finding the average slope of $ F(x)$ over the interval $ [a,b]$.

k thanks guys

i just have a tough time understanding the basic concepts as i get them mixed up.

and just one more question:

is there any other way of determining which function goes with a slope field besides just trying to sketch out the function on the slope field and matching it to one?

like im given a slope field and the question asks this slope field most liekly goes with what?
a. y=cos x
b. y=sin x.
c. y=lnx
...
and so forth?  beccause i sometimes dont remeber what lnx and like e^(-x) look like.

Honestly, it's those two functions and a few other elementary ones that are MOST critical to know.
If you don't know what a $ \cos(x)$ looks like by now, you're in deep water.
Same goes for a standard exponential graph, and a polynomial up to degree three.

Elementary anti-derivatives are good to know, as well.

ok yea sorry i was kinda exaggerating, i can pretty much brute force those functions but I was thinking there might be an easier way (to check my answer)

oh well i think im prepared...good thing you can get a 70/108 on this test and basically be guareenteed a 5.

Slope fields:
Best method I've found to deal with the ones the AP gives you:
They'll usually give you a field, and five possible functions which might dictate the field.

Test a point in each quadrant (usually $ \pm1, \pm1$) in each function. Check whether the resulting value is positive or negative, and compare it to whether the slope line in the given quadrant is positive or negative. Usually you can eliminate 3/5 of the answers simply using the positive/negative check. The other two answers usually vary by some exponent which is pretty easy to check.

k thank you.

hehe i should be studying instead of being here on AoPS.


How come when i click submit with an attachment the site goes back to the message screen where i'm supposed to type up my message/ topic and it says something like "extension bmp is not allowed" or something? Is it my computer?

No, it's just that the extension bmp is not allowed. It's too big. Use MS Paint (Start >> Programs >> Accesories >> Paint) tool to create a .PNG file (which is much smaller) and attach that image.

Thanks.
But now it says
"Sorry, you have reached your maximum Upload Quota Limit of 2 MB"


geometry
rectangle
circumcircle
conics
ellipse
geometric transformation
reflection

Let a rectangle $ ABCD$ is insribe in a circle $ (O)$, $ M$ move on circle $ (O)$. The perpendicular bisector of $ OD$ cut $ MA$ at $ E$, perpendicular bisector of $ OC$ cut $ MB$ at $ F$. Prove that: $ EF$ is always tangent with a fix circle.

see my solution here also: http://www.mathlinks.ro/viewtopic.php?t=279085

The proposed problem can be generalized.

$ ABCD$ is a convex cyclic quadrilateral with circumcircle $ (O,R).$ Its diagonals $ AC,BD$ meet at $ P.$ $ M$ is a variable point on $ (O)$ and the perpendicular bisectors of $ PD$ and $ PC$ cut $ MA$ and $ MB$ at $ E,F,$ respectively. Then  $ EF$ is always tangent to the ellipse $ \mathcal{E}$ with foci $ O,P$ and major axis $ R.$

[quote]$ ABCD$ is a convex cyclic quadrilateral with circumcircle $ (O,R).$ Its diagonals $ AC,BD$ meet at $ P.$ $ M$ is a variable point on $ (O)$ and the perpendicular bisectors of $ PD$ and $ PC$ cut $ MA$ and $ MB$ at $ E,F,$ respectively. Then $ EF$ is always tangent to the ellipse $ \mathcal{E}$ with foci $ O,P$ and major axis $ R.$[/quote]
Let $ A'$ and $ B'$ denote the reflections of $ P$ on $ MB,MA.$  Let $ H$ be the orthocenter of  $ \triangle MAB$ and $ T$ the Anti-Steiner point of $ HP$ with respect to $ \triangle MAB.$ $ B_0 \equiv BH \cap (O),$  different from $ B,$ is the reflection of $ H$ across $ MA$ $\Longrightarrow$ $ T \equiv B'B_0 \cap (O).$ Since $ \angle B'PD \equal{} \angle B_0BD \equal{} \angle B_0TD$ $ \pmod\pi,$ it follows that $ T \in \odot(PDB')$ and similarly circle $ \odot(PCA')$ goes through $ T.$ Therefore, perpendicular bisectors of $ PD,PC$ cut  $ MA,MB$ at the centers $ E,F$ of the circles $ \odot(PDB')$ and $ \odot(PCA')$ $ \Longrightarrow$ Line $ EF$ is the perpendicular bisector of $ PT.$ Let $ Q \equiv OT \cap EF.$ From the isosceles triangle $ \triangle PQT$ with apex $ Q$ we obtain $ QO \plus{} QP \equal{} QO \plus{} QT \equal{} R$ $ \Longrightarrow$ $ Q$ moves on the ellipse $ \mathcal{E}$ with foci $ O,P$ and major axis $ R.$ Since $ EF$ becomes the external bisector of $ \angle PQO,$ then $ EF$ is tangent to $ \mathcal{E}$ at $ Q$ and the proof is completed.

There is even more general.

[i] Let $p$ and $q$ be arbitrary lines and $\mathcal{C}$ circle.
Let $A$ and $B$ be fixed point on $\mathcal{C}$ and $T$ variable point on it.
Let line $AT$ cuts $p$ at $X$ and $BT$ cuts $q$ at $Y$.

Then $XY$ is tangent to fixed conic.[/i]


What is Presidential Prayer Team? What ever happened to separation of church and state?

A possible explanation for the Time statistics is that the smart college graduates with conservative leanings disproportionately end up becoming CEOs and investment bankers, while the ones with liberal leanings go into academia.
Furthermore, at colleges there are a limited number of positions available in teaching no-nonsense liberal arts courses such as "American Literature", but there is a unlimited amount of possible politicized courses in areas such as "Gender Studies".

The "Presidential Prayer Team" sounds pretty ridiculous to me, but it is legal under the constitution as long as Bush is running it as a private citizen, rather than using the government apparatus to fund or advertise it. The first amendment guarantees free practice of religion in addition to prohibiting the federal government from establishing an official church.

[quote="schulmannerism"]Furthermore, at colleges there are a limited number of positions available in teaching no-nonsense liberal arts courses such as "American Literature", but there is a unlimited amount of possible politicized courses in areas such as "Gender Studies".[/quote]

Where in the world did you get this idea?

Rep123max, where did you get this idea?  For one, the Presidential Prayer Team is not affiliated President Bush or the White House.  It is run by a private foundation and is being handled by a company called The Envoy Group.  The organization is by no means sponsored by government money (or by the President's private funds, either).

See the [url=http://www.snopes.com/rumors/prayer.htm]Urban Legends Reference Page[/url] on this one.

Also, on the Presidential Prayer Team website, the FAQ says:

[quote="http://www.presidentialprayerteam.org/faq.php"]Is this effort affiliated with any political party, elected official or governmental agency?

No. The Presidential Prayer Team is a spiritual movement of the American people which is not affiliated with any political party or official. It gains no direction or support, official or unofficial, from the current administration, from any agency of the government or from any political party, so that it may be free and unencumbered to equally serve the prayer needs of all current and future leaders of our great nation.

How is the effort funded?

The Presidential Prayer Team is an independent, nonprofit, non-partisan organization. It is funded by those who wish to support its ongoing efforts. We receive no government or political party financial support.

Did the President start this effort?

The current President supports and welcomes our prayer effort, however he did not start it and his specific endorsement or commissioning is not desired due to inherent constitutional issues. The Presidential Prayer Team gains no financial support from the government or any agency of the government.
[/quote]

Even when you find an official-looking source, you have to watch out.  For instance, [url=http://www.whitehouse.org/dof/prayer.asp]this page[/url] seems official, but look on the bottom of the page, and you see "a chickenhead productions parody".

And for the record, the $1.5 million figure that Rep123max noted is the Presidential Prayer Team's annual budget; as before, it is not the amount of money the government gives it, because the government doesn't give it any!

For those of you who are interested, the relevant time articles that were mentioned in the original post are [url=http://www.time.com/time/magazine/article/0,9171,1101040531-641149,00.html]here[/url] and [url=http://www.time.com/time/magazine/article/0,9171,1101040531-641136,00.html]here[/url].

JBL - I'm not saying that anything more than a fraction of courses offered are on these sort of topics. However, I'm sure if you look at the course catalog in 1960, 1970, 1980, 1990, 2000, there are more of what could be called "politicized" courses in each successive decade.

I do have to say that this is starting scare me more and more.

I mean, a group of people worshipping President Bush?  That's what it almost sounds like, and that is definitely a disturbing thought.  This would inherently place the separation of church and state in jeopardy.

[quote="Rep123max"]And Mathfanatic, you are correct on one thing, I did misinterpret the 1.5 million dollars and I admit to my mistake. However, I still believe the Presidential Prayer Team is disgusting and anti-American. Perhaps it is because I am Jewish and not very religious, but nonetheless, it is wrong.[/quote]

Since we have the issue about whether or not Bush sponsors this movement straightened out, I won't address that. However, no matter what you think of it Max, it is a freedom provided by the first admendment. I think someone like you who is so inclined to Kerry's idea of giving people rights about their babies being born would understand the idea of "rights". Furthermore, although you may no find it appealing, you can bet there are millions of people that pray for the president on a daily basis, whether or not they're part of this team.

Btw, it's in no way anti-American. In fact, it's pro-American if anything.

You may not support it, but there's no way you can argue that it's illegal or wrong or corrupt or something Bush did for his reelection efforts or anything like that.

[quote="Rep123max"]But heres my take on it. I am sure that all of us are biased but the feeling I get from the president and his administration (I do not at all think that all Republicans are like this, a Kerry-Mcain ticket is what I want). This is what they do. They let guys decide whether a woman can get an abortion. Straight people decide whether gay people can marry. And Christian people decide what faith everybody should be. Even I will admit that my last statement about them could be better worded that they spread Christianity, but I just thought that it shows how they decide how other people should live.[/quote]
You've drawn more false conclusions in this paragraph than any other I've seen recently.

[i]They let guys decide whether a woman can get an abortion[/i]
First, the Representatives and Senators are elected to their positions by the populace, and the percentage of the Representatives and Senators that support abortion is roughly equal to the percentage of the populace that supported it at the last election. That's the way this system works - we elect people based on their stances on issues. Secondly, there are many female Senators and Representatives - Hillary Clinton is one example.

[i]Straight people decide whether gay people can marry[/i]
I'm willing to wager that there are at least a few gay Senators and/or Representatives. But further, the decision will be made by the populace, which includes both straights and gays. Assuming there are no confounding variables like a higher precentage voter turnout of one group over the other, if there are x% gays and 100-x% straights in the population, then x% of the vote will be cast by gays. This is exactly how it should be - surely you don't think it's fair to deprive the 100-x% of the population of the right to vote on some issue so the minority will get what it wants, do you?

[i]Christian people decide what faith everybody should be.[/i]
WHERE do I begin? Should I begin by pointing out that the government sponsors no particular religion? Should I begin by pointing out that while the Christians try to convert us all to Christianity, the Muslims are trying to bring us to worship Allah and read the Qur'an, the Jews are trying to get us into their synagogues, the Mormons are trying to coax us into the Church of Jesus Christ of Latter Day Saints by telling us about Joseph Smith and his golden plates, the Seventh Day Adventists are trying to convince us the end is near, the scientologists are trying to convince us the E. Ron Hubbard was correct, and the athiests are trying to tell us that there is no God at all? Or should I go back to the argument that there are Christians, Muslims, Jews, Atheists, and probably numerous other groups are all represented in our Congress?

Sigh. I see no validity in your arguments.

[quote="confuted"]Jews are trying to get us into their synagogues.[/quote]

Oh really?

[quote="Rep123max"]btw- I am Jewish.[/quote]
Yes, I suppose you can claim that you are Jewish by birth. I'll give you that.

But as far as I am concerned, birth does not make you Jewish in anything other than name. From what you have said, it seems that you fall into this "non-practicing Jew" category:
[quote="Rep123max [in this thread]"]
...being a nonreligious person...
...that they call Pray the Vote, which is going to make me barf...
...Perhaps it is because I am Jewish and not very religious, but nonetheless, it is wrong.[/quote]
If prayer makes you barf, somehow I doubt that you are a Jew by anything but birth. It bothers me a bit that Judaism is passed on by birth, and one doesn't need to practice or even believe in order to say [i]"I am Jewish"[/i].

Well, on the issue of gay marriage, it definitely should be allowed.

Assume it should not be allowed.  It immediately follows then that

sanctity of marriage :ge: equality for gays

multiplying both sides by law and applying the idea principle (which states that law is the multiplicative identity for the set of all ideas not written in law)

sanctity of marriage :ge: constitution

According to the corollary to the Constitution theorem (also known as the common sense postulate),

For a [tex]\in[/tex] law, b :ge: a iff b [tex]\in[/tex] law.

Since clearly constitution [tex]\in[/tex] law, then sanctity of marriage [tex]\in[/tex] law must also be true.

However, by the Constitution theorem, the intersection of religion and law is the empty set.

But by Religion's property

sanctity of marriage [tex]\in[/tex] religion
thus,
sanctity of marriage [tex]\notin[/tex] law, giving us contradiction.

QED

[quote="Rep123max"]Nice post probability. Praying does not make me barf. Praying that is somewhat related to the government (no matter what, you cannot deny the strong ties to this administration and Christianity, and the Presidential Prayer Team). I could respond to you calling me a Jew by birth (even though I got bar mitzvahed), you are correct, I do not go every week to temple, or in fact, I dont even go once a month. I probably now go once a year. But I will be the bigger person and stop there and not say what I want to say right now.[/quote]
So millions of people praying for the future of their nation makes you feel ill. Fascinating. I'm only going by what you've said here. You said you were not very religious - I do not see how I am insulting you by repeating that, but I suppose words can be twisted to mean any variety of things. Just look at John Kerry, who actually did vote for the $87 billion before [he] voted against it" - whatever that means.

[quote="confuted"]Why is sanctity of marriage [tex]\in[/tex] religion ? Why not sanctity of marriage [tex]\in[/tex] time honored tradition?[/quote]

By definition, sanctity [tex]\in[/tex] religion, and it's also true that most subsets of religion are [tex]\in[/tex] time honored tradition.

[quote]It's not that I can't find something; it's that I do not care to spend my time looking when I could be learning multivariable calculus and physics.[/quote]

I congratulate you for not wasting your time with this debate! On the other hand, you can't argue in favor of Bush unless you atually present your arguments.

[quote]I believe that going to war with Iraq qas the right thing to do because the U.S. Had recieved many threats from Iraq about terrorism. [/quote]

Iraq itself never threatened any country. Also, threats of terrorism belong to terrorist organizations as opposed to nations. If you think that there were many terrorist groups present in Iraq, you should think about Iran or Saudi Arabia. There exists much more evidence of terrorist activities in these countries. Why didn't the US invade them? Its due to the fact that they are stronger and have more impressive international reputations. Bush simply needed a weak country that he could easily topple and Iraq provided this. Unfortunately, its turned into a real nightmare.

[quote]They also had intelligence that turned out to be wrongbut they could not be sure, about WMD.[/quote]

There is impressively strong evidence, infinitely more impressive than that of WMD in Iraq, that WMD are being built by North Korea. It has been shown by intelligence of several countries that this country is in the active process of building a nuclear missile. So, if WMD is a reason for going into Iraq, why not go to North Korea as soon as possible.

[quote]We can all agree I'm pretty sure that what Sadam was doing was wrong.[/quote]

So? I'm sure we can agree that what Kadaffi or Idi Amin had done was wrong too. Or, again, Kim Jung Il. Why not topple their goverments? Some of these people have caused more damage than Saddam.

[quote]We warned Sadam to leave the country and he wouldn't attack.[/quote]

You mean we wouldn't attack. Well, if say Osama bin Laden decided to tell Bush "you have two days to leave the White House or we will attack it," do you think that Osama's actions are justified? Of course not! Ultimatums never justify actions.

[quote]Sadam didn't heed the warning. Bush had no choice really to follow through on his threat.[/quote]

Why didn't he have a choice? I can admit the fact that I have made a mistake. Bush can't? Or is it a matter of politics, that is, it would take away from his hard-liner image?

[quote="tetrahedr0n"]
There is impressively strong evidence, infinitely more impressive than that of WMD in Iraq, that WMD are being built by North Korea. It has been shown by intelligence of several countries that this country is in the active process of building a nuclear missile. So, if WMD is a reason for going into Iraq, why not go to North Korea as soon as possible.
[/quote]
But in iraq, there was no danger of being attacked by WMD as there is in north korea.

if we go to war with North Korea, our allies be nuked.  Besides the US worries that the north koreans already have a unit capable of launching an ICBM (though they arent sure).  From what i heard, they launced a capsule into space some time ago so there is no reason why they cant launch a missile.

[quote]Look. The goverment watches North Korea closely. They have been watched for centuries.[/quote]

So by the same logic, we could have watched Iraq right? Why did we invade Iraq but not North Korea? What is watching them going to help in preventing North Korea from developing WMD? And I must seriously doubt that two centuries ago, the US was in a position to "watch" any country.

[quote]But in iraq, there was no danger of being attacked by WMD as there is in north korea.
[/quote]

Let me get this straight. We invaded Iraq because they had WMD but there was no danger of them using WMD? What country develops WMD and doesn't use them when under attack? You claim we invaded Iraq because they had WMD. If we were right, wouldn't they use the WMD on our miltary? Seems like very poor military logic to me.

[quote]if we go to war with North Korea, our allies be nuked. [/quote]

Same logic: Had Iraq had WMD, which we believe they had, since we invaded, our allies could have been attacked with chem/bio weapons. We still went in.

[quote]Besides the US worries that the north koreans already have a unit capable of launching an ICBM (though they arent sure). [/quote]

They also were not sure of Iraq having bio/chem weapons. They still invaded.

[quote]From what i heard, they launced a capsule into space some time ago so there is no reason why they cant launch a missile.[/quote]

North Korea has not launched a capsule into space as far as I know. I do know that they did test missiles capable of carrying nuclear warheads. This makes North Korea more dangerous, and by Bush's logic, we should invade them in order to prevent their accumulation and eventual use of logics.

[quote="tetrahedr0n"].

[quote]But in iraq, there was no danger of being attacked by WMD as there is in north korea.
[/quote]

Let me get this straight. We invaded Iraq because they had WMD but there was no danger of them using WMD? What country develops WMD and doesn't use them when under attack? You claim we invaded Iraq because they had WMD. If we were right, wouldn't they use the WMD on our miltary? Seems like very poor military logic to me.

[quote]if we go to war with North Korea, our allies be nuked. [/quote]

Same logic: Had Iraq had WMD, which we believe they had, since we invaded, our allies could have been attacked with chem/bio weapons. We still went in.

[quote]Besides the US worries that the north koreans already have a unit capable of launching an ICBM (though they arent sure). [/quote]

They also were not sure of Iraq having bio/chem weapons. They still invaded.[/quote]
\
What i meant for no danger of WMD is for them to attack the US.  Anyway, i believe chemical weaopons rained on Isreal
For the last point, i dont get what you are saying.  I meant that the US doesnt attack because then nukes will go up.

[size=75][The message has been deleted by CrystalCrown.][/size]

[quote="confuted"][quote="BHorseMath"]An invasion of North Korea is being looked at.[/quote]
Unfortunately, it is not being seriously considered, and perhaps almost as unfortunately, "at" is a preposition and doesn't go at the end of a sentence.[/quote]

sorry confuted...  Thats not what my english teacher told me this year though.  Hmm I will have to look that up.  Thank you for correcting it though.

heh, unfortunately, many English teachers nowadays tend to make things up :-)

Some mistakes that English teachers (esp. elementary school) sometimes make:

"Mr./Mrs. *****?  That's me!"

"Can you give me and ***** some help?"
"[i]***** and I[/i]"

correct scambled.  Thats exacly what I meant.

[quote="Scrambled"][quote="confuted"][quote="BHorseMath"]An invasion of North Korea is being looked at.[/quote]
Unfortunately, it is not being seriously considered, and perhaps almost as unfortunately, "at" is a preposition and doesn't go at the end of a sentence.[/quote]
i believe that in this case "at" is an adverb describing "is being looked"[/quote]

No, "at" is not an adverb.  If it were modifying "is being looked," then I can kill it and try to say "is being looked," which does not work.  I think "look at" is just one of those messed up verbs that makes the following hold:

English < min(all other languages) :-)

I know what you mean probability.  English is one of the hardest laguages in the world.  I have problems with it.


Point $ M(4,4)$ is the midpoint of $ \overline{AB}$. If point A has coordinates (8,4), what is the sum of the coordinates of point B?


I'm somewhat liberal also. I took a test in history class that said that my views were very similar to Gandhi's

[quote="ursula"]In order to vote, I would need some kind of definition of the possibilities in the poll. I'm sorry for my ignorance, but what does somewhat liberal means, or very liberal or conservative. I don't know

Ursula[/quote]
"Conservative" and "liberal" are very politically-charged terms in the USA. I am not sure if they are used in the same way in other countries because of this. I believe these were the [url=http://en.wikipedia.org/wiki/American_conservatism]conservative[/url] and [url=http://en.wikipedia.org/wiki/American_liberalism]liberal[/url] to which the poll means to refer.

(I know my choice of wiki links is very USA-ethnocentric but I just feel that this is a very USAish way of classifying people's social and political views. Someone can challenge me if he or she likes.)

Xevarion, I also consider myself slightly socially liberal but very economically conservative, and thus somewhat conservative overall.

I have a problem with the idea of taking more money from the rich, because they can afford it.  That's communism in my opinion.  Along the same lines we could even argue that we should tax the poor at HIGHER rates so that we can get as much money from them as we get from the rich.

I do believe that stem cell research needs to be done, on adults stem cells rather than embryos, however.  I would like to call myself socially conservative as well, but there are a few key issues I cannot reconcile.

Ursula,
I really had a problem understanding and remembering the difference between conservative and liberal.  They don't mean strictly Republican and Democratic, but Bush is very conservative, while Michael Moore is pretty liberal.  A conservative is someone who wants to keep things they way they are.  They favor lower taxes, and usually less government.    A good explanation is that liberals value rights over traditional values, while conservatives value traditional values over rights.  That is not entirely unbiased, but it is a good explanation.  Just remember that Republicans are generally conservative, while Democrats are generally liberal.

Also, take the [url=http://www.okcupid.com/politics]OKCupid Politics test[/url].  It will tell you EXACTLY what you are, and its the best, most accurate self-test quiz I have ever seen.  My results:
http://seansoni.blogspot.com/2006/07/my-political-views.html

I am supposedly close to Donald Trump, and somewhat of a capitalist.  Go take the [url=http://www.okcupid.com/politics]test[/url]!

EDIT: Also note that liberal/conservative are political parties in the UK, and are TOTALLY different there.  For example, Tony Blair was a liberal (Labour Party?) but he was definitely an ally of Bush, and conservative by US standards.

You are a

Social Liberal
(68% permissive)

and an...

Economic Liberal
(30% permissive)

You are best described as a:

Democrat

You exhibit a very well-developed sense of Right and Wrong and believe in economic fairness. loc: (68, -75)

I took it twice.

  You are a

Social Conservative
(38% permissive)

and an...

Economic Liberal
(10% permissive)

You are best described as a:

Strong Democrat

You exhibit a very well-developed sense of Right and Wrong and believe in economic fairness. loc: (-44, -150)

You are a
Social Liberal
(61% permissive)

and an...
Economic Liberal
(18% permissive)

You are best described as a:

Socialist

You exhibit a very well-developed sense of Right and Wrong and believe in economic fairness.

I was right at the very bottom of the scale, right in the center. I different test that I took, which I trust a bit more, put me right with Gandhi around the middle of the bottom left quadrant.

[quote="Xevarion"]I'm very liberal socially but pretty conservative economically... Not crazy enough to vote Libertarian though...[/quote]
Maybe I should be voting Libertarian. The OKCupid test put me at 76% socially permissive (liberal) and 83% economically permissive (conservative), well into Libertarian territory.

You are a

Social Liberal
(61% permissive)

and an...

Economic Liberal
(28% permissive)

You are best described as a:

Democrat

You exhibit a very well-developed sense of Right and Wrong and believe in economic fairness. loc: (43, -82)

You are a

Social Liberal
(60% permissive)

and an...

Economic Liberal
(35% permissive)

You are best described as a:

Centrist

(though a corner of my square is sticking out into Democrat area)

[quote]what does that imply about mathematicians?[/quote]
:D

[quote="ccy"]Ignite, you really like make threads into flamewars, don't you?[/quote]

Please explain how this is a flame war? I was merely explaining how some of my views are contrary to what the test shows and that the test may not be that an accurate gauge of a person's views. If people disagree with my views, then they should me mature enough to 1) Not bring this thread offtopic with posts like the one I just quoted 2) Debate rationally without flaming.

I`m a fairly strong liberal on most issues.

If anyone is interested in the test, these are my results:
[quote]You are a
Social Liberal
(61% permissive)

and an...
Economic Liberal
(11% permissive)

You are best described as a:

Socialist

You exhibit a very well-developed sense of Right and Wrong and believe in economic fairness. [/quote]

[quote="SoxFan34"]"Happy holidays" just doesn't cut it with me.[/quote]
What would you prefer? "Happy/merry Christmas, Hanukkah, Ramadan, Kwanzaa, Saturnalia, etc., etc., etc."? Where everything has to be spelled right, which it probably isn't right now, and the etc.'s had better be replaced by every single other winter holiday somebody somewhere in the U.S. might celebrate? I don't get it. Why not just say "Happy holidays"? What harm does it do to acknowledge all holidays, instead of just the Christian one or just the Abrahamic ones?

[quote="SoxFan34"]I am somewhat conservative, but that doesn't mean i think Dubya is a good prez. I think religion in the united states should be distinguished and free, not melded into one big stew of confusement by John Kerry and John edwards. "Happy holidays" just doesn't cut it with me.[/quote]

When a religion becomes distinguished? I'm asking this question because I don't know the meaning in this context of the word distinguished. How does John Kerry saying 'happy holidays' contradicts your idea of free religion? What is Dubya? I'm guessing it means Bush, because I'm guessing prez stands for president. Sorry, sometimes the forum is like the test, you don't get it if you are not from the USA, also I might be growing old and unaware of modern virtual language.

[quote="Xevarion"]My (need-based) scholarship covers more than 90% of my annual tuition and I still pay far more than you do[/quote]

People tend to go to public universities in many countries in Europe (UK would be a big exception, I think). Also related is the fact that public university is widely regarded as better in many countries, most notably Germany where, at least, until very recently, I think you paid about 30 euros a year or something. Correct me somebody if I'm wrong. I also think this is changing right now and they will pay something like the amount fredbel6 cites.

[quote="fredbel6"]You have to know that -regardless of who is in power - we are pretty leftwing compared to USA. Half of what you earn can go to taxes. You are forced to get health insurance (there is a socialist, christian and liberal - corresponding to those three parties- mutuality). A visit to a normal doctor costs about 20 euros (about 25 US dollars)... but then the mutuality ends up paying you back about 15 euros (about 18 US dollars). [/quote]

I've never been in the US, but I shall say that this does seems accurate. Take France. Chirac is rightwinger and in the past elections, the left aligned couldn't believe it when they saw themselves voting Chirac because they couldn't risk the other option, LePen, unconceivable!. However, in the international arena, Chirac is consistently to the left of the US. Actually I don't think there is much of a left party in the US. In Europe, the democrats would be at least centrists. Issues as the war on Iraq serve as examples of dubious positions of the democrats, along the whole conflict while what is called left in Europe was evidently and actively against. In the other hand, I don't understand quite why this is so. Why in the US people are politically allocated in general in other sectors that the rest of the first world?. Or I should say Europe,  Australia is quite rightwinged too I think. Of course I have my theory but it is in itself incomprehensible.

Ursula

[quote="ursula"] What is Dubya? I'm guessing it means Bush, because I'm guessing prez stands for president. [/quote]
People call Bush "Dubya" because he is George W. Bush. It is to distinguish him from George H. W. Bush and also a play on his Southern accent (affectionate for some, derogatory for others :P).

[quote="ursula"]People tend to go to public universities in many countries in Europe (UK would be a big exception, I think). Also related is the fact that public university is widely regarded as better in many countries, most notably Germany where, at least, until very recently, I think you paid about 30 euros a year or something. Correct me somebody if I'm wrong. I also think this is changing right now and they will pay something like the amount fredbel6 cites.
[/quote]
Well even public universities in the US are many times more expensive than what [b]fredbel6[/b] pays. (In fact, I would get less scholarship money from a public university, so I personally would pay more for a public university than for Princeton. And as I said I still pay several times what he does.)

I don't want to get into a discussion about attitudes towards education in the USA, but I think this is the reason behind it.

[quote]  You
are a

Social Liberal
(61% permissive)

and an...

Economic Moderate
(56% permissive)

You are best described as a:

Centrist

You exhibit a very well-developed sense of Right and Wrong and believe in economic fairness.[/quote]
I'm between Donald Trump, Adam Sandler and John Kerry.

I'm pretty conservative, except when it comes to things like stem-cells research and funding science. Then I say: tax and spend!

[quote] Sorry, sometimes the forum is like the test, you don't get it if you are not from the USA, also I might be growing old and unaware of modern virtual language.[/quote]
Ursula, I don't know where you are from but you are definitely right about that.  There are many people on this board who are still learning English and trying very hard to be clear.  USA-ers and other anglophonics should respect their efforts, be grateful for their own "not having to translate" comfort and do at least some effort to write clearly.  After all, this forum was established by ROMANIANS. :|

I cringe when I see stuff like "I dont no we're I'm go in too nite but it wont be da house"

"prez" comes from "president"
"Dubya" comes from "Double U", which is the English name for the letter "w", a reference to George W. (=William Bush)
LOTR = "Lord of the Rings"
HPGOF="Harry Potter Goblet of Fire"
IMO = "In my opinion"

....

I guess it's just the endless "abbrev(iation) compulsion" in USA.  It got over the top when Guantanamo became "Gitmo". :huh:

[quote]People tend to go to public universities in many countries in Europe (UK would be a big exception, I think).[/quote]
I don't think there are decent non public universities in Flanders. :!:

[quote]I also think this is changing right now and they will pay something like the amount fredbel6 cites.[/quote]
As I said, that's the full amount, when your family is not wealthy enough (lots of kids) things can still be arranged.  The university also owns homes in Ghent that students can rent.  The price depends on the income of your parents.  It's possible you are denied a place when that income is too high.  However, their newest homes are located on the campus itself and has a bathroom and internet connection for each student... so they are in fact much more comfortable than the privately owned homes. :o

[quote="fredbel6"]You have to know that -regardless of who is in power - we are pretty leftwing compared to USA. Half of what you earn can go to taxes. You are forced to get health insurance (there is a socialist, christian and liberal - corresponding to those three parties- mutuality). A visit to a normal doctor costs about 20 euros (about 25 US dollars)... but then the mutuality ends up paying you back about 15 euros (about 18 US dollars). [/quote]

[quote]I've never been in the US, but I shall say that this does seems accurate. Take France. Chirac is rightwinger and in the past elections, the left aligned couldn't believe it when they saw themselves voting Chirac because they couldn't risk the other option, LePen, unconceivable!. However, in the international arena, Chirac is consistently to the left of the US. Actually I don't think there is much of a left party in the US. In Europe, the democrats would be at least centrists. Issues as the war on Iraq serve as examples of dubious positions of the democrats, along the whole conflict while what is called left in Europe was evidently and actively against. In the other hand, I don't understand quite why this is so. Why in the US people are politically allocated in general in other sectors that the rest of the first world?. Or I should say Europe,  Australia is quite rightwinged too I think. Of course I have my theory but it is in itself incomprehensible.[/quote]
I think I have a theory.  In France the president is elected in two rounds : if no candidate obtained 50%  the two best scoring candidates face each other in a second round.  During the last election in 2002, Chirac got most of the votes, then far right winger Le Pen, and then socialist Jospin(you yourself probably know all this :wink: ).    That way the socialist voters still had a chance to stop Le Pen.  That system is GOOD.
In the USA that doesn't exist.  That way any vote for a leftist president is a republican vote.  That's why nobody dares to vote left, and that's what keeps USA right forever.
In the Netherlands 150 parliament members are elected into the Second Chamber (comparable to the French National Assembly or the House of Representatives in the USA)  The government is then a "coalition" : a set of parties the sum of their number of seats exceeding 75.  Several coalitions are possible.  That way the government is not made up of members from the biggest party alone.  In fact, a coalition can be formed without the biggest party.  That way you aren't politically immobile.
(I could have used my own country as an example...but Belgium is DOUBLE FEDERAL and thus a bit more complicated than the unitary Netherlands. :roll: )

[quote="fredbel6"]After all, this forum was established by ROMANIANS. :| [/quote]
Um I don't think so. It's a forum made by Americans merged with a forum made by Romanians.

[quote="Ignite168"][quote="mysmartmouth"]Interesting how AoPS is so liberal, except for the one college student, where college students (or at least professors) are notoriously liberal...what does that imply about mathematicians?[/quote]

Except I'm firmly opposed to f*gs marrying, in favor of death penality, think illegal immigrants need to get the - out of this country and be deported, think affirmative action sucks, and think women should stay in the kitchen and not get jobs.[/quote]
Time for a flamewar. Can we get a mod to split this?

While we're restricting freedoms, let's ban inter-racial dating!
Or maybe we should have the government randomly dump people's ISP logs. If they don't have anything to hide, they shouldn't care.

1) On marrying, I see it as an archaic, government endorsed religious cermony, with tax benefits. But while we're at it, why shouldn't gays be allowed to do what ever they dang well please? How does it affect your life if we allow polygamy or gay marriage? While I prefer my partners to be girls, why does it matter if Bob down the street likes guys? Why would this relationship be at all inferior, or any less deserving of being formalized?

2) On killing people: taking the life of anyone is murder. This is only justified if they pose an imminent and deadly risk to someone else. If I happen to be walking down a dark alley, and some thug pulls a knife and starts going for me, I would pull my (hypothetical) gun and pop that sucker. Without a second thought. I would of course call 911 and try to make sure he doesn't die, but if he does, it was justified if he was trying to kill me. But if you have someone inprisioned for the rest of his or her life, why should we kill them? And how do you justify it, morally? Why should we kill him instead of letting him rot in prison?

3) On the immigrant thingy: I don't care. The current laws suck, and they need to be changed.

4) Why should women stay in the kitchen and not get jobs? Are they inferior creatures?

More generally, why should me restrict the freedom of someone when they are not hurting anybody? Should I be able to prance down mainstreet in pantyhose, wearing a shirt with naked girls on it and smoking weed?

(Hint: Yes)

[quote="PenguinIntegral"]
 Should I be able to prance down mainstreet in pantyhose, wearing a shirt with naked girls on it and smoking weed?

(Hint: Yes)[/quote]
There's this place called Amsterdam. :wink:
Trust me, even Flemings (like me) can be surprised about the totally free attitude over there.

And why isn't Ignite referring to "f*gs" as homosexuals (or possibly bisexuals) which is the proper term?
I'll assume Ignite is from the USA, I find it very generous that he wants to kick almost all 300 million USA-ers out in favour of the native Indians.  Sometimes people forget the history of their nation.

I voted strong conservative and...You are a

Social Conservative
(30% permissive)

and an...

Economic Conservative
(61% permissive)

You are best described as a:

Republican

I was right!

Is this considered reviving if the topic is on the very first page?

[color=green][i]Moderator responds: Yes.[/i][/color]

You are a

Social Moderate
(43% permissive)

and an...

Economic Liberal
(16% permissive)

You are best described as a:

Strong Democrat

You exhibit a very well-developed sense of Right and Wrong and believe in economic fairness. loc: (-25, -125)

I thought I was a Republican...

And [i]why[/i] am I near Darth Vader?

[quote="Xevarion"]Related: does anyone know who are the two people in the upper right corner? I mean the sketch of the guy (which looks really familiar, I'm going to be mad at myself when I hear his name) and the 18th century-looking guy?[/quote]
Since this thread has already been bumped, I'll answer my own question. The sketch in the upper left (anarchist) corner is an old police sketch of Ted Kaczynski, the Unabomber.

I pay 60 USD a semester to go to university  :D. I used to have a scholarship so that I lived for free in my dorms too, but now I have to pay 390 USD a month (9 months a year). My smokes also cost like 160 USD a month.

Hmmm...Centrist but close to totalitarian and republican, when I always though I was perfectly republican. Oh well, I voted what I thought, so I guess I'm a centrist, but I still support Republicans :)


There is a triangle ABC. The bisector of (angle A) cut BC in a point  $ P$ . In the middle line AP is taken a point E such

$ BC \equiv AE$ and a point F such $ AC \equiv AF$ . Compare BF and  CE.

Ohh .. man I relly need a solution of this.

The thing is... there is no specific solution... it depends on the relative lengths of $ BC$ and $ AC$... so like if $ BC>AC$, then $ BF>CE$, and vice versa for the other case. You'll get equality iff $ BC\equal{}AC$.


ratio
number theory proposed
number theory

Let $ s(n)$ be the sum of the decimal digits of positive integer $ n$. Is the sequence $ \frac{{s(n)}}{{s(n^2 )}}$ bounded?

If $ n$ has $ m$ digits then $ n^2$ has $ 2m \minus{} 2$, $ 2m \minus{} 1$ or $ 2m$ digits.

Thus $ s(n) \leq 9m$ (if all digits were 9) while $ s(n^2) \geq 2m \minus{} 2$

And thus the desired ratio is $ \leq \frac {9}{2 \minus{} \frac {2}{m}}$ which for $ m > 2$ is bounded. Thus we are left to check finitely many cases $ < 100$ so it is indeed bounded

How you got $ s(n^2) \geq 2m\minus{}2$¿  :wink:

If each digit was 1  (stunning realisation hits, curse you zetaX for constantly finding flaws in my solutions!)


Given $ a, b, c$ are the reals satisfy $ \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\equal{}2$.
Prove that: $ \frac {1}{2a^2 \plus{} 1} \plus{} \frac {1}{2b^2 \plus{} 1} \plus{} \frac {1}{2c^2 \plus{} 1} \leq\ 2$
 :)

[quote="caubetoanhoc94"]
We need prove that:
$ 4(1 \minus{} q)^2 \plus{} \frac {4q^2}{3} \ge 2$
$ < \equal{} > (4q \minus{} 3)^2 \ge 0$
[/quote]
Are you sure for this step???
Or you want to say $ 4(1 \minus{} q)^2 \plus{} \frac {4q^2}{3} \ge 1$ :?:

[quote="caubetoanhoc94"][quote="nguoivn"]Given $ a, b, c$ are the reals satisfy $ \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \equal{} 2$.
Prove that: $ \frac {1}{2a^2 \plus{} 1} \plus{} \frac {1}{2b^2 \plus{} 1} \plus{} \frac {1}{2c^2 \plus{} 1} \leq\ 2$
 :)[/quote]

Nice.
My soulution :)
$ \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \equal{} 2$
So:$ ab \plus{} bc \plus{} ca \equal{} 1 \minus{} 2abc$
$ \frac {1}{2a^2 \plus{} 1} \plus{} \frac {1}{2b^2 \plus{} 1} \plus{} \frac {1}{2c^2 \plus{} 1} \leq\ 2$
$ < \equal{} > 16a^2b^2c^2 \plus{} 4\sum a^2b^2 \ge 1$-expand
Setting:$ q \equal{} ab \plus{} bc \plus{} ca$
$ 16a^2b^2c^2 \plus{} 4\sum a^2b^2 \equal{} 4(1 \minus{} q)^2 \plus{} 4\sum a^2b^2 \ge 4(1 \minus{} q)^2 \plus{} \frac {4q^2}{3}$
We need prove that:
$ 4(1 \minus{} q)^2 \plus{} \frac {4q^2}{3} \ge 1$
$ < \equal{} > (4q \minus{} 3)^2 \ge 0$
It's always true  :P
P/s:Đến 9 h tối 1/6 mới hết ban của em anh à[/quote]

Sr.I edited


A circle of area 20 is centered at point $D$. Point $A,B,C$ are in the circumference of this circle, and $\triangle ABC$ has area of 8. If $\angle ADB = \alpha, \angle BDC = \gamma, \angle ADC = \beta$, find the value of $\sin \alpha + \sin \beta + \sin \gamma$.

[quote="kunny"]Let $R$ be the radius of circumcircle, we have $\pi R^2=20$.

$\triangle ABC=\triangle DAB+\triangle DBC+\triangle DCA$

$\Longleftrightarrow \frac{1}{2}R^2(\sin \alpha +\sin \beta+\sin \gamma)=8$

$\Longleftrightarrow \sin \alpha +\sin \beta +\sin \gamma =\frac{16}{R^2}=\boxed{\frac{4}{5}\pi}$[/quote]

That's correct, kunny. The "key" to this problem was to see area formula $\sin \theta a b \frac{1}{2}$ and since $a = b$ for all three triangles in this case (this length is the radius), we can find $\sin \alpha + \sin \beta + \sin \gamma$.


hi every one please write solution this inequality!
 i know this inequality has some topic which i forgot :?

a,b,c - positive real numbers, sides of triangle

a + b + c = 1
 prove that:

\[ \sum\limits_{SYM} {\sqrt[n]{{a^n + b^n }}} \le 1 + \frac{{\sqrt[n]{2}}}{2}  \]


Okay I'm stumped.

How many odd perfect square factors does 2^4 x 3^6 x 5^10 x 7^ have?

I get 12.  The book says the answer is 120.  Is this a typo?

thanks,

Um, 7 to the what?

I got more than 12 already, but I'm not sure since you didn't give an exponent to the 7.

Even if we don't know the exponent of 7, we have enough information to figure it out(assuming it is an even integer by pattern).  Let's say 7's exponent is n.  We know that the odd prime divisors of said number cannot have a 2 in its prime factorization, so we ignore the 2.  As for the 3, we know that 3^0, 3^2, 3^4, and 3^6 are possibilities, for a total of 4 ways.  Similarly, there are 6 ways for the 5 term, and n/2+1 ways for the 7 term.  Hence, we have:
4 X 6 X ([n/2] +1) ways, or 24(n/2)+12 ways=12n+12 ways.  Since this equals 120, we have that n=8.  However, it can also equal nine since [9/2]=[8/2].  So, I guess that term is 7^(8 or 9).

[quote="brendajay"]OOPs it's

2^4 x 3^6 x 5^10 x 7^9

my bad typing.  It's from the 2003 book.[/quote]
Ok so anything multiplied by two is even right?
So you need to find out the number of odd factors.
So now, we could either have $ 3^0$,$ 3^2$,$ 3^4$,$ 3^6$
So for 5, we could have
$ 5^0$,$ 5^2$,$ 5^4$,$ 5^6$,$ 5^8$,$ 5^{10}$,
For 7, you could have
$ 7^0$,$ 7^2$,$ 7^4$,$ 7^6$,$ 7^8$,
So $ 4\times6\times5 \equal{} 120$


"The total interest $ u$ paid on a home mortgage of $ P$ dollars at interest rate $ r$ for $ t$ years is given by:

$ u \equal{} P\left[\frac {rt}{1 \minus{} \left(\frac {1}{1 \plus{} \frac {r}{12}}\right)^{12t}} \minus{} 1\right]$

A $ \$$$ 120000$ mortgage is made at $ 9\frac{1}{2}$ percent.

Approximate the length of the mortgage when the total interest paid is the same as the size of the mortgage."

Show all work please. I have no idea how this interest formula was created. It makes no sense at all.

well, i dont want to show my work right now. But all you do is just plug in and solve. plug in what you know and make sure you set the u and the mortgage equal. Did you make up that formula? lol  :)


let $a_1  ,a_2 $ be given positive constant,then we define a sequence$a_3,a_4,a_5,...$by :$a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}$ show the $Lim_{n\rightarrow \infty }a_n $ exists and find this limit.

I asked sam-n to tell me the solution, but meanwhile I think I managed to solve it, although it won't be complete.

Earlier this year I encountered this problem at a contest:

Given $\displaystyle t_i\ge 0,\ i\in \overline{1,k},\ \sum_{i=1}^kt_i=1$ and a bounded sequence (we only need it to be bounded from below) $\{a_n\},n\ge 1$ s.t. $a_{n+k+1}\le t_1a_{n+1}+t_2a_{n+2}+\ldots+t_ka_{n+k},\ \forall n\ge 0$, show that $a_n$ converges.

I won't prove it here, but we can use it on this problem:

Let $x_n=\frac {\sqrt 2}{a_n}$ (I like it better when sequences tend to $1$ instead of $\sqrt 2$ :)). Then $x_{n+1}=\frac 2{x_n+x_{n-1}}$. This means that $x_{n+1}=\frac 2{\frac 2{x_{n-1}+x_{n-2}}+\frac 2{x_{n-2}+x_{n-3}}}=\frac {(x_{n-1}+x_{n-2})(x_{n-2}+x_{n-3})}{x_{n-1}+2x_{n-2}+x_{n-3}}\le \frac 14x_{n-1}+\frac 12x_{n-2}+\frac 14x_{n-3}$.

I think you can easily see why this problem is a particular case of the one above: k=4, t_1=t_3=\frac 14,t_2=\frac 12,t_4=0$. After we show the convergence, the limit is easy to compute.

I think the proof of this lemma is harder than original problem ,for k=2 is the problem vojtech jarnik april 1997, any way I tried to prove that lim sup =lim inf i difintly sure that it works :maybe:

Well, we might try to prove it only for this particular case. Anyway, I think that lemma is a really cool thing :). It was proposed by Vasile Popa for that contest I wa talking about (I just thought it wouldn't do any harm if I mentioned it :)).

Could you post your proof, please?

[quote="grobber"]
Given $  t_{i}\ge 0,\ i\in \overline{1,k},\ \sum_{i=1}^{k}t_{i}=1$ and a bounded sequence (we only need it to be bounded from below) $ \{a_{n}\},n\ge 1$ s.t. $ a_{n+k+1}\le t_{1}a_{n+1}+t_{2}a_{n+2}+\ldots+t_{k}a_{n+k},\ \forall n\ge 0$, show that $ a_{n}$ converges.
[/quote]

This is not true in general. For example $ a_{n+3}\leq a_{n+1}$ is obviously not enough to
ensure the convergence. We need $ t_{i}>0$ or what?

Yes, I think the problem I mentioned had the stronger condition $ t_{i}>0,\ \forall i$.


Is the sequence
\[ \frac{1}{1\plus{}n(nx^2\minus{}n\minus{}1)^2}\]
uniformly convergent on $ (1,2)$? Could you please give me some advice how to find the answer?

It converges pointwise to $ 0$; $ n(x^2 \minus{} 1) \minus{} 1$ goes to $ \infty$ everywhere but $ 1$, and we're squaring and multiplying by $ n$. Thus, if it were to converge uniformly, the limit would be $ 0$. To disprove this, it is sufficient to find points $ x_n$ so that $ f_n(x_n)$ is bounded away from zero.


function
trigonometry
quadratics
algebra
quadratic formula

What positive acute angle(s) satisfies the equation of all 6 squared trigonometric functions equaling 31?

or Sin squared x + cos squared x + tan squared x + cot squared x + csc squared x + sec squared x = 31

Thanks

Why do you need a response so fast?

[hide]Hint: Use the Pythagorean identities and put it all in terms of sine and cosine.

Also,

$2\sin{x}\cos{x}=\sin{2x}$[/hide]

This isn't an extra credit problem for your class is it?  :huh:

Using the identities and simplifing to cos/sin I came upwith

1 + (sin squared over cos squared) + (cos squared over sin squared) + (1 over cos squared) + (1 over sin squared) = 31

Do I multiply everything by sin squared cos squared?

Substitute $\sin^{2}x=1-\cos^{2}x$, then let $\cos^{2}x=a$, which leads to some pretty straightforward algebra.

I think we need to have a giant sticky at the top of the forum: DO NOT POST YOUR HOMEWORK HERE, WE WILL NOT DO IT FOR YOU, or some such thing.

Yeah. I like to give people the general idea to solve their HW, but not the whole solution as that's a breach of academic integrity.

[quote="DiscreetFourierTransform"]Complete Solution:
[hide="Solution"]
Equation:
$\cos^{2}x+\sin^{2}x+\tan^{2}x+\sec^{2}x+\csc^{2}x+\cot^{2}x = 31$

Rewrite in terms of sin and cosine:
$cos^{2}x+\sin^{2}x+\frac{\sin^{2}x}{\cos^{2}x}+\frac{1}{\cos^{2}x}+\frac{1}{\sin^{2}x}+\frac{\cos^{2}x}{\sin^{2}x}= 31$

Use fundamental pythagorean identity $\sin^{2}x+\cos^{2}x = 1$ to rewrite in terms of sin^2.

$(1-\sin^{2}x)+\sin^{2}x+\frac{\sin^{2}x}{1-\sin^{2}x}+\frac{1}{1-\sin^{2}x}+\frac{1}{\sin^{2}x}+\frac{1-\sin^{2}x}{\sin^{2}x}= 31$

Let $\sin^{2}x = A$:
$1-A+A+\frac{A}{1-A}+\frac{1}{1-A}+\frac{1}{A}+\frac{1-A}{A}= 31$

Which simplifies algebraically to
$\frac{2a^{2}-2a+2}{a(1-a)}= 30$

and further down to
$32A^{2}-32A+2 = 0$
$16A^{2}-16A+1 = 0$
Then solve with quadratic formula

$A = \frac{16 \pm \sqrt{256-(4)(16)(1)}}{32}$
$A = \frac{16 \pm \sqrt{256-64}}{32}$
$A = \frac{16 \pm \sqrt{192}}{32}$
$A = \frac{16 \pm 8\sqrt{3}}{32}$
$A = \frac{2 \pm \sqrt{3}}{4}$

So, we have $\sin^{2}x = \frac{2 \pm \sqrt{3}}{4}$

So as not to give the entire solution for free, I'll let you finish it up.

[/hide][/quote]
I got something much nicer than that...

Will post on Monday.


1.The list 2,3,5,6,7,10... consists of all positive integers which are neither squares nor cubes in increasing order. What is the 2005th number in this list?
2. The distance from a point inside a quadrilateral to the four vertices are 1,2,3 and 4. Determine the maximum value of the area of such a quadrilateral.
3. The positive integers a.b and c are such that a+b+c=20=ab+bc-ca-b^2 Determine all possible value of abc.
4.  The sum of 49 positive integers is 624. Prove that three of them are equal to one another.

[hide="sum1"]
1. note that till the number 2006, the number of squares is 43 and the number of cubes is 11 and the number of 6th powers is 2 .
So by Inclusion Exclusion, no. of numbers omitted till 2006 = 54-2 = 52.
So 2006 is the 1954th term. The next square is 2025 and the next cube is 2197 and the next 6th power is 4096. So 2026 is the 1973th term and then the 2005th term is 2058 :)

Therefore the answer is $ \boxed{2058}$

PS : Hope there arent calculation mistakes :)
[/hide]

[hide="3"]since 20=ab+bc-ac-b^2=(b-c)(a-b) we obtain (b-c,a-b)=(1,20),(20,1),(4,5),(5,4),(2,10) or (10,2)
if (b-c,a-b) = (1,20) or (20,1) then b-c+a-b=a-c=21 since then a+b+c > 20 contradiction .
if (b-c,a-b) = (2,10) then b-c+a-b = 12 -> a=c+12 , since a+b+c=20 we have c+12+c+2+c=20 which means c=2
if (b-c,a-b) = (10,2) similarly we obtain a=c+12  ; a+b+c=c+12+c+10+c=20 which shows c=8/3 which is not intergers .
if (b-c,a-b) = (4,5) then b-c+a-b = 9 -> a= c+9 then a+b+c=c+9+c+4+c =20 gives c = 7/3 not satisfying .
if (b-c,a-b) = (5,4) similarly we obtain a=c+9 , a+b+c=c+9+c+5+c = 20 gives c=2
in conclusion (a,b,c)=(14,4,2) or (11,7,2) which means abc = 112 or abc=154 [/hide]


geometry
rectangle
projective geometry
angle bisector
geometry proposed

Let $ABCD$ be a rectangle and denote by $M$ and $N$ the midpoints of $AD$ and $BC$ respectively. The point $P$ is on $(CD$ such that $D\in (CP)$, and $PM$ intersects $AC$ in $Q$. Prove that $m(\angle{MNQ})=m(\angle{MNP})$.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55017]Rectangle & an angle bisector[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14246]geometry stuffs[/url].

Let $ O $ be the centroid of recyangle. We draw a parallel through $O$ to $BC$ that intersects $QN$ at point $K$. We have: $\frac{QM}{PM} = \frac{QO}{OC}$ , but $\frac{QO}{OC} = \frac{QK}{KN}$  ($KO\parallel BC$).
Now we see that:
$\frac{QM}{PM} = \frac{QO}{OC} = \frac{QK}{KN}$ $\Longrightarrow$ $KM\parallel NP$ ,so $\angle MNP \equiv \angle KMO$ , but $\triangle MKN $ is right-angled so, $\angle KMO \equiv \angle QNM $.

Dear Mathlinkers,
1. R is the point of intersection of AB and NQ
2. by a converse of Pappus theorem applied to the frontiers PR and AQC to the degenerated hexagon
PQRA (line AD) (line CD) P, the Pascal line MN being parall to AB (or CD), we have PR // AD.
And we are done
Sincerely
Jean-Louis