Any Harry Potter fans here? I like the books, but I'm not a J.K. Rowling fanatic like a few people I know  :P. What are your favorite books and why? I liked Chamber of Secrets the best because of the awesome plot. What are your opinions?

for me, I too like chamber of secrets, but goblet of fire comes in a close 2nd. I really didn't like prisoner of askaban, despite loopy lupin (who was really cool).

I thought they've all been real good, can't wait for the next one.. liked the last one so far cos it was so long! Longer the better..

[quote="andy17null"]I think that Lupin is going to have a major role in OOTP (order of the Phoenix.) By the way, what are your opinions as to what OOTP will be about? My guess is that the Order of the Phoenix is an elite Hogwarts student group, union, society, or whatever.

-Isaac[/quote]

Actually, I think I read a thing about what it was going to be about and if I remember correctly, I believe the order of the phoenix is some elite wizard's organization, not hogwarts students... but I could be wrong.  I liked the 4th book simply because it had a rather (or at least the best a kid's book could be) complicated plot.

-Taylor

Well, my guess was pretty close, huh? I also liked the length of the 4th book... if they keep getting thicker, you won't be able to buy a boxed set. You'll buy a crated set. 16 days till Book 5 release...

-Isaac

Noticing the choice was not on the list...I don't really like the Harry Potter books. They scare me :shock::roll:!

wats wrong with Harry Potter? i think the books are great, especially the fourth one. when is the fifth coming out anyway?

9 days if I remember correctly...
Can't wait...

-Isaac

I don't remember...it was all so long ago...:cry:

I remember my friend told me it was June 21. (I'm not totally sure.)

OotP is definitely coming out on June 21st.

I am looking forward to it as well. I just reread the series so that I'd be all caught up for it. Amazon.com has a review that says it is even longer than Goblet of Fire. Hooray!

As a long time fan of sci-fi/fantasy both teen and adult, it is great to see the genre finally coing into its own and gaining a wider audience, both in print and film.

Book 5 came out an hour ago here...I wish I were at a bookstore.

:shock: must get Order of Phoenix  :shock:. My car is broken down today of all days. Post here when you finish the book, but no spoilers please. I heard that a major character dies in the book... my guess is Hagrid.

-Isaac

woooooooooooooot. i got order of the phoenix at midnight. woooooooooooooooooooooooot. its great. i would have finished it by now, but i didnt sleep the day before cause i was at a parte. but ill finish up today.

I got it at noon or so and finished reading it at 11:55 last night. Good book imo. see my sig?  :wink:

-Isaac

it was pretty cool. i kept reading it between jobs. i was actually sadder when cedric died than when ****** died, to tell you the truth. you guys?

Well, there's a chance that they'll be reunited I think. Remember Luna's quote when they are discussing that event? idk, but i hope so... Personally, I didn't really care about either character (cedric or ******) that much.

-Isaac

the eighth book is out is there a ninth?

can you please stop bumping posts from 15 years ago

When this post was created, in 2003, they had not been written yet.

[quote=AnimalLover111]is there going to be a 9th book?[/quote]

inb4 back in 2018 there was no 9 and 10

there's already an 8th(kind of). its a play.
my favorite book is the deathly hallows though


system of equations
algebra
IMO
IMO 1965

Find all sets of four real numbers $x_1, x_2, x_3, x_4$ such that the sum of any one and the product of the other three is equal to 2.

If I undestood the question correctly we have:

\[ x_1+x_2x_3x_4=2 \] and so on.So we have:

\[ x_1^2-2x_1+x_1x_2x_3x_4=0 \] And for that if we make \[ k=x_1x_2x_3x_4 \] a constant then \[ x_i \] with \[ i=1,2,3,4 \] can only equal one of the 2 rooths of the equation:

\[ x^2-2x+k=0 \]. And for that we have only 3 possibilitys: \[ x_1=x_2=x_3=x_4; or; x_1=x_2, and, x_3=x_4; or; x_1=x_2=x_3, and, x_4 \]. In the first possibility we have the equation \[ x+x^3-2=0 \] and the only real rooth to this equation is x=1 and we have the set \[ {1,1,1,1} \]

If \[ x_1=x_2=a;and;x_3=x_4=b \] we have that the rooths of \[ x^2-2x+a^2b^2=0 \] are a;b so we have \[ ab=(a^2b^2)/1 \] and ab=0 or 1. If ab=0 then a and b are the rooths of \[ x^2-2x=0 \] so (a,b)=(0,2) but {2,2,0,0} is not a set with the propriety.
If ab=1 then a and b are the rooths of \[ x^2-2x+1=0 \] so (a,b)=(1,1) and we have the set {1,1,1,1} again.
Now we only have to check the case that \[ x_1=x_2=x_3=a;and;x_4=b \]. We have that a and b are the rooths of \[ x^2-2x+a^3b=0 \] so a+b=2 and ab=a^3b or \[ a^2=1 \] and we have the sets {1,1,1,1} or {-1,-1,-1,3}. or ab=0. If ab=0 then a and b are rooths of \[ x^2-2x=0 \] so (a,b)=(0,2) but neither {2,2,2,0} or {0,0,0,2} are good sets.
So the sets are {1,1,1,1} and {3,-1,-1,-1}. If there are any mistakes in this solution please tell me.

I think you could speed up the reasoning if you consider two relations and then substract them. For example $(x_1-x_2)(1-x_3x_4)=0$.

Observe that subtracting the equations $x_1+x_2x_3x_4=2~(1)$ and $x_2+x_3x_4x_1=2~(2)$ yields $x_1-x_2-x_3x_4x_1+x_3x_4x_2=0$, which rearranges as
$$(x_1-x_2)-x_3x_4(x_1-x_2)=(x_1-x_2)(1-x_3x_4)=0\implies x_1=x_2\text{ or }x_3x_4=1.$$
In other words, subtracting equations $(1)$ and $(2)$ implies that $x_1=x_2$ or $x_3x_4=1$, and generally subtracting equations $(a)$ and $(b)$, where $1\leq a\neq b\leq 4$, implies that $x_a=x_b$ or $x_cx_d=1$, where $\{a,b,c,d\}=\{1,2,3,4\}$. It follows that $x_a=x_b$ or $x_cx_d=1$ for all such $a,b$.

Now, we consider how many of the $x_k$ are equal. If none of them are equal, we have that $x_cx_d=1$ for all $c,d$, which yields $(1,1,1,1)$, a contradiction. Similarly, if only $2$ of them are equal, we have that $x_cx_d=1$ for all $c,d$ except for one pair, but regardless we again have $(1,1,1,1)$, impossible. Furthermore, if $2$ of them are equal and the other $2$ are equal, WLOG let $x_1=x_2$ and $x_3=x_4$; then $x_2x_3=1$ since $x_1\neq x_4$ so $x_3=x_4=\tfrac{1}{x_2}$, but plugging into $x_1+x_2x_3x_4=2$ gives $x_1+\tfrac{1}{x_1}=2$, whose only real solution is $x_1=1$. Consequently we have $(1,1,1,1)$ again, impossible.

Hence, we have shown that either $3$ of them are equal and the other is distinct, or all $4$ of them are equal, the latter of which yields the solution $\boxed{(1,1,1,1)}$. In the former case, WLOG let $x_1=x_2=x_3$. Then since $x_1\neq x_4$, $x_2x_3=x_2^2=1$, so $x_1=x_2=x_3=\pm1$. Having $x_1=x_2=x_3=1$ yields $(1,1,1,1)$ once again, which is impossible, so we take $x_1=x_2=x_3=-1$, which yields $\boxed{(-1,-1,-1,3)\text{ and permutations}}$ upon substituting back into $x_1+x_2x_3x_4=2$. All cases have been exhausted, so we're done.

Subtracting the first $2$ equations gives us $(x_1-x_2)(1-x_3 \cdot x_4)=0$, so we have $x_1=x_2$ or $x_3 \cdot x_4 =1.$

Now we proceed with bashy casework. After doing this casework we get $(1,1,1,1)$ and $(-1,-1,-1,3)$ and permutations. $\square$


calculus
derivative
trigonometry
logarithms
calculus computations

$ f(x) \equal{} x \sin (i \log x)$; find $ f'''(x)$.

$ f(x) \equal{} x \sin (i \log x) \equal{} i x \sinh (\log x) \equal{} i x \frac{1}{2} \left( x \minus{} \frac{1}{x} \right) \equal{} \frac{i}{2} (x^2 \minus{} 1)$
So,
$ f '(x) \equal{} ix$,
$ f ''(x) \equal{} i$,
$ f '''(x) \equal{} \boxed{0}$.


quadratics
number theory unsolved
number theory

What are the primes in $ \mathbb{Z}[\sqrt{\minus{}2}]$? Correspondingly, what are the primes of the form $ x^2 \plus{} 2y^2$? What numbers can be written in the form $ x^2 \plus{} 2y^2$?

Quadratic reciprocity gives you necessary conditions.  [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]Are they sufficient?[/url]


modular arithmetic
number theory unsolved
number theory

[quote="MeKnowsNothing"]Prove that every year, there is at least one month such that the 13th day is on Friday.[/quote]

Between november 13th and december 13th, we have $ 30$ days so $ 2\pmod 7$
Between october 13th and december 13th, we have $ 61$ days so $ 5\pmod 7$
Between september 13th and december 13th, we have $ 91$ days so $ 0\pmod 7$
Between august 13th and december 13th, we have $ 122$ days so $ 3\pmod 7$
Between july 13th and december 13th, we have $ 153$ days so $ 6\pmod 7$
Between june 13th and december 13th, we have $ 183$ days so $ 1\pmod 7$
Between may 13th and december 13th, we have $ 214$ days so $ 4\pmod 7$

So exactly one of these $ 7$ days (may 13th, june 13th, july 13th, august 13th, september 13th, october 13th, november 13th) is a friday.


If $\imath^x+\imath^x=2$, what does $9(4x)^2$ equal?

A: 16
B: 256
C: 2304
D: 4953
E: 8327

Also, in your post, please hide the stuff AND tell me how to make the squares and square root symbols please!@

EDIT: Thank you for ccy on informing me on a huge mistake I made!

X has an infinite number of possibilities.  Choose the LOWEST NATURAL NUMBER X CAN BE for this problem only

Edit2: Added a misdone problem

Edit3: Fixed latex, this should be last edit, all that is left is for peeps to solve i guess.

does i refer to i as in the square root of -1.... if it does i think this problem might be too hard for the Basics.

hmmmmm sorry everyone!

SHould i repost in highschool one already and not wait for it to ge moved?

[quote="la mafioso"]If i to the x plus i to the x equals 2, what does 9(4x)squared equl?

A: 16
B: 256
C: 9216
D: 4953
E: 8327

Also, in your post, please hide the stuff AND tell me how to make the squares and square root symbols please!@[/quote]

i^2 -> $i^2$
\sqrt{i^2} -> $\sqrt{i^2}$

All in dollar signs.

[hide]
$1+1=2$ (lol)
$i^{x}=1$
There's no one solution to the problem.  $x$ can be 0, 4, 8, 12...
[/hide]

Hrm, I apologize.

In this question, for simplification if people are limited with pencil's and paper and calculators, we shall pretend that x is equal to THE LOWEST POSSIBLE NATURAL NUMBER IT CAN BE!

For la mafioso, to create $x^2$ terms, just put dollar signs around x^2.  For square root of something $\sqrt{x^2+y^2}$, you would have \sqrt{x^2+y^2} in dollar signs.  For more help, one of the administrators has a place to learn $\LaTeX$ in his signature.

[hide="Answer to Problem"]You can write the problem as $2\cdot\imath^x=2$.  Then, you get $\imath^x=1$.  From this, you get $\imath=4$.  Then, I'm not sure what your problem is asking.  If it means $9(4x)^2$, then the answer is 2304.  If it means $9(4)x^2$, then the answer is 144.  If it means $[9(4x)]^2$, then the answer is 20,736.  I don't know what else it could mean, and none of those solutions are choices.[/hide]

okay b-flat got it.

It was supposed to be

[hide=":"]c[/hide]

I screwed up so it should be

[hide=":"]2304[/hide]

I apologize for all the mess up's...it shall be edited asap

You might want to change the problem to look like $\imath^x+\imath^x=2$, which is \imath^x+\imath^x=2 in $\LaTeX$.  Also, $9(4x)^2$, which is 9(4x)^2 in $\LaTeX$.  It will make it easier for people to read.


limit
trigonometry
real analysis
real analysis unsolved

Find the limit :
$\lim_{n\to\infty} \sin^2(2\pi \sqrt[3]{n^3+n^2})$

Well, as it is stated it doesn't exist: $\sin^2( n (1+\frac{1}{n})^{\frac{1}{3}}) = \sin^2 (n(1+\frac{1}{3n}+o(\frac{1}{n})))=\sin^2(n+\frac{1}{3}+o(1))$ and it's limit when $n \to \infty$ surely doesn't exist

I am so sorry that i've made a typo :blush:

So, in this case using what i wrote above the limit is $\displaystyle \lim_{n \to \infty} sin^2( 2\pi n+ \frac{2 \pi}{3}+o(1))= \frac{3}{4}$

For an expression Exp of that form try comuting $\sin(n\pi-Exp)$


Find a marvellous solution for this admirable problem :

Let $a_1,...a_n,b_1,...,b_n$ be real numbers.
Find infinite number of real polynomial`s such that

$f(a_i)=b_i$,

The solution our teacher gave,(by his word) is due to lagrange .

I think your teacher used this: http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html
There are other ways to get a polynomial with the desired properties, e.g. by Newton.

But when you have found one such $f$, all other are given by $g(x)=f(x)+p(x) \prod_{k=1}^n (x-a_i)$ where $p$ is any polynomial.


Let $ A,B$ are max roots of $ 4x^3\minus{}2x^2\minus{}15x\plus{}9$ and $ 12x^3\plus{}6x^2\minus{}7x\plus{}1$. Prove that $ A^2\plus{}3B^2\equal{}4$

Have you ever tried using Cardano's formula? If not, please make a research on wiki or whatever and you see this problem is so easy! :)

Can somebody give a more simply solution? Solution using Cardano's formula is bit complex!


let (C) be a circle inwich its center is inaccessible.and let P be a point.

find point Q on (C),inwich PQ is tangent to (C).

What does it mean that "the center is accessible", I'm assuming this is important to the problem, or else there is an infinite number of answers.

ofcourse its important....

accessible means that we dont know where the center of (C)is.and we should solve the problem,with out using the

center of (C). :?

these problems are so nice.tell me where do you get these problems? :D

thanks ashegh.and there is one more thing can we draw it with only compas or ruler?(only one of them)

to iversonfan 2005:

i edited the problem.see it again plz :)

So the problem asks to find/construct $Q$ on the circle $(C)$ such that $PQ$ is tangent to the circle, where $P$ is a given point (outside of the circle), right?

If so, what are we allowed to use? Only compass (just like your other problems)?

see: http://www.mathlinks.ro/Forum/viewtopic.php?t=57006

jensen its usefull for u... ;)


Hi!

Since last night i'm playing the adventure game da vinci code!

I 'm stuck somewhere........

Has anyone made a lot progress?

 :)

I've only seen the movie and read the book.  What's the game like? What do you do?

it's an adventure. It has some puzzles as well. I like it.
But it's not too easy...

do you play pc-games?

 :)


function
integration
logarithms
complex analysis

Show that for $x>0$ we can write :

$\Gamma(x)=\sum^{\infty}_{n=0}\frac{(-1)^{n}}{n!}\frac{1}{n+x}+\sum^{\infty}_{n=0}c_{n}x^{n}$

where $c_{n}=\frac{1}{n!}\int^{\infty}_{1}t^{-1}e^{-t}(\log t)^{n}$

Note that:

$\Gamma(x)=\int^{\infty}_{0}e^{-t}t^{x}dt$


function
limit
real analysis
real analysis unsolved

The function $f(x)\ (-\infty <x<\infty)$ always take positive values and satisfies for all real numbers $x,y$ as follows .

\[3f(x+y)=f(x)f(y)\ \ f(1)=6\]

Evaluate

\[\lim_{n\to\infty} n\left[f\left(1+\frac{1}{n}\right)-f(1)\right]\]

[quote="kunny"]The function $f(x)\ (-\infty <x<\infty)$ always take positive values and satisfies for all real numbers $x,y$ as follows .

\[3f(x+y)=f(x)f(y)\ \ f(1)=6\]

Evaluate

\[L=\lim_{n\to\infty} n\left[f\left(1+\frac{1}{n}\right)-f(1)\right]\][/quote]

Change slightly the definition of $f$, by $g =f/3$ then $g$ satisfies
$ g(x+y) = g(x)g(y)$ for all $x,y$. Then there exists  $\lambda$ such that  $g(x) = \exp( \lambda x) $ for all $x$.
With the condition on $f(1)$, we get $f(x) = 3 \exp( \ln(2) x)$ for all $x$.
Then the differential of $f$ at $1$ gives : $L= 6 \ln(2)$.

Fabrice

[quote="fmathurin"]Then there exists $\lambda$ such that $g(x) = \exp( \lambda x)$  for all x. [/quote]
This isn't actually true without an extra assumption such as continuity. On the other hand, it is true for all [i]rational[/i] $x$ automatically, and the problem only uses the value of $g$ at rational $x$.


i kno this is a  pretty vague question still what are the general methods of carbon chain expansion/degradation

Yes, its a little vague because that depends on the particular structure of the molecule. Anyway, the most general carbon-carbon bond forming reactions are perhaps the following: aldol condensations and related reactions (Michael reactions, Robinson annulations), acetoacetic and malonic esther synthesis, Friedel-Crafts alkylations and acylations, nucleophilic substitutions with acetylides and cyanide anions, Wittig reaction, organometalic reagents (for example, organolithium reagents, Grignard reagents, and Gilman reagents), hydroboration, Diels-Alder reaction, and 1,3-dipolar cycloadditions.

The degradation of a carbon chain depends more strongly on the functional groups present in the molecule. I have provided some example of application of degradation in "Organic synthesis" problems.

thanks..yes ur synthesis problems are great ..just need some time to work out(and of course skill  :P )


geometry
ratio
function
rectangle
trigonometry
cyclic quadrilateral
trig identities

How can we tell whether two quadrilaterals are similar? Are there rules associated with this sort of similarity in the same way that there are rules with triangle similarity (SSS, SAS, etc..)?

Specifically, I was studying a geometry problem from Canada 1990:

The feet of the perpendiculars from the intersection point of the diagonals of a convex cyclic quadrilateral to the sides form a quadrilateral q. Show that the sum of the lengths of each pair of opposite sides of q is equal.

The solution can be found here:

http://www.kalva.demon.co.uk/canada/casoln/csol903.html

The problem involves two similar cyclic quadrilaterals with one pair of equal angles. Because the quadrilaterals are cyclic, their opposite pair of angles must be equal as well. However, even with these two equalities, why can we say that the two quadrilaterals are equal? Furthermore, once we have determined similarity, why can we say that side1/diameter1 = side2/diameter2 as Scholes does? Aren't we only allow to make this ratio when comparing pairs of sides (or quantities that can be expressed as a function of the sides)?

Any help would be appreciated. Thanks.

Obviously if all angles are congruent (we only need three, as the fourth will follow), the quadrilateral is similar...I g2g for now. Bbl with an updated post.

Thats not true. Easiest way to realise that - consider a rectangle :)

[quote="Fierytycoon"]How can we tell whether two quadrilaterals are similar? Are there rules associated with this sort of similarity in the same way that there are rules with triangle similarity (SSS, SAS, etc..)?[/quote]
i think SSAS will work

No, it doesn't. Suppose you fix the SAS part. That is a unique triangle. Then you can move the other S around in a circle.

You need just about everything to make it work.  By the by, we've discussed this elsewhere, I believe.  Anyhow, it's going to take you probably 2 sides and 3 angles, and order is very important.  Nothing really nice works, unfortunately.

Mmm...then how can Scholes make the assertion that he does?

This part:

[quote="Fierytycoon"] why can we say that side1/diameter1 = side2/diameter2 as Scholes does?[/quote]

comes from the Law of sines applied to the two triangles PAS and BAD.

Cyclicity is a very strong condition.  As for the question about similarity, anything well-defined is similar under similarity.  That includes any kind of curlicues on the inside of the triangle, so long as they will scale up or down.  So it also applies to the diameters.  (Or you could use Law of Sines, I suppose.)

So, I sat down for a little while, and as far as I can tell, you basically need 5 pieces of information about a quadrilateral, plus order, to determine similarity.  In other words, 2 sides (adjacent or opposite, I think) and 3 angles, 3 sides and 2 angles, or 4 sides and 1 angle, so long as you give them all names.  That is, if I just give you 4 lengths and 1 angle, you can form more than 1 quadrilateral.  I have to tell you which angle it is and what the order of the sides around it is in order to uniquely define it.

I don't know if this will help much, but I know we discussed it here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8141&highlight=polygons+similar

cause I had the same question as you.

Thanks for the responses.

I understand why we can say side1/diameter1 = side2/diameter2 now.

However, about the actual determining of the similarity...in this problem, we have ∠PAS = ∠BAD and thus, since both quadrilaterals are cyclic, ∠PXS = ∠BCD (2 opposite angles). But, err, isn't that all we have, or am I missing something...?

[quote="TripleM"]You have the fact that they're both cyclic :)[/quote]

Yeah, but what other similar sides/congruent angles does that give us?

It tells you everything. Since you know the angles, the cyclic quadrilateral is unique (other than similar ones). Think about it like this: Choose any chord of the circle as one of the quads sides.. Then draw the next side at the correct angle. Then draw the next side at the right angle. Then draw the last side at the right angle. Well, theres only one way its going to meet up..

I mean, theres probably a more rigorous method but..

Well, we know that ∠PAS = ∠BAD and ∠PXS = ∠BCD, but what about the other two pairs. All we know is that they the other two angles of either quadrilateral sum to 180 degrees between the quadrilaterals are cyclic, but what lets us know that they are equal?


What is the remainder if the greatest comman divisor of the these $ 5^{2007}\minus{}1$ and $ 5^{669} \minus{} 1$ numbers are divided by 7?

Where are the rest $ 5^{2007}\minus{}3$ numbers? We have $ (5^{2007},5^{669}\minus{}1)\equal{}1$ so the answer is $ 1$.

[quote="puuhikki"]Where are the rest $ 5^{2007} \minus{} 3$ numbers? We have $ (5^{2007},5^{669} \minus{} 1) \equal{} 1$ so the answer is $ 1$.[/quote] sorry -1 was missed from the quetion. added now.


integration
trigonometry
calculus
derivative
function
calculus computations

hello

I have the solution to a qn. but i cannot understand how or why the limits are suddenly changed from $ \pi$ to $ \frac{\pi}{2}$

[img]http://i44.tinypic.com/15x9y03.jpg[/img]

Ok i see it now. But for this question why did they change the limit from 2 to 0 ==> 1 to -1 ??

what has happened to $ \frac {1}{L}$ of the formula for $ a_n$ and $ b_n$part?

[img]http://i41.tinypic.com/eu1xmu.jpg[/img]

Changing the interval - always with length $ 2$ - in which calculate the $ a_m$s and $ b_m$s is just for simplicity.

You have the same result calculating $ a_m\equal{}\displaystyle\int_{0}^{2}{f(x)\,\cos({m\pi{x}})\,dx}\equal{}\int_{0}^{1}{x\,\cos({m\pi{x}})\,dx}\plus{}\int_{1}^{2}{({1\minus{}x})\,\cos({m\pi{x}})\,dx}$ or

$ a_m\equal{}\displaystyle\int_{\minus{}1}^{1}{f(x)\,\cos({m\pi{x}})\,dx}\equal{}\int_{\minus{}1}^{0}{({\minus{}1\minus{}x})\,\cos({m\pi{x}})\,dx}\plus{}\int_{0}^{1}{x\,\cos({m\pi{x}})\,dx}$.

The same for $ b_m$s.

Here's my calculation using weak derivatives:

$ f'(x)\equal{}\delta(x)\minus{}\delta(x\minus{}1)\plus{}\begin{cases}1&0<x<1\\\minus{}1&1<x<2\end{cases}$

$ f''(x)\equal{}\delta'(x)\minus{}\delta'(x\minus{}1)\plus{}2\delta(x)\minus{}2\delta(x\minus{}1)$

$ \widehat{f''}(k)\equal{}\frac12\int_0^2f''(x)e^{\minus{}\pi ikx}\,dx$

$ \equal{}\frac12\left(\pi ik\minus{}(\pi ik)e^{\minus{}\pi i k}\plus{}2\minus{}2e^{\minus{}\pi ik}\right)$

$ \equal{}\frac{\pi ik(1\minus{}(\minus{}1)^k)}2\plus{}(1\minus{}(\minus{}1)^k)$

$ \equal{}\begin{cases}\pi ik\plus{}2&k\text{ odd}\\0&k\text{ even}\end{cases}$

For $ k\ne 0$ we divide that by $ (\pi ik)^2$ to get $ \widehat{f}(k):$

$ \widehat{f}(k)\equal{}\begin{cases}\frac1{\pi ik}\minus{}\frac2{\pi^2k^2}&k\text{ odd}\\0&k\text{ even}\end{cases}$

$ \widehat{f}(0)$ (the average value of the function) must be computed separately, but it is zero. Hence we have:

$ f(x)\equal{}\sum_{\substack{k\equal{}\minus{}\infty\\ k\text{ odd}}}^{\infty}\left(\frac1{\pi i k}\minus{}\frac2{\pi^2k^2}\right)e^{\pi ikx}$

If we turn that back into sines and cosines, it will be the same as the series in the quote in bos1234's post.


number theory
least common multiple
group theory
superior algebra
superior algebra unsolved

Let $ \omega(G)$ be the set of element orders and $ \mu(G)$ the set of maximal elements of $ \omega(G)$ by divisible relation in finite group $ G$.  And denote by $ s_n$ the number of elements of order $ n$.

Q: Whether for every $ \mu\in\mu(Sym_n)$, we have $ n! \equal{} \mu\cdot s_{\mu}$? Where $ Sym_n$ is the symmetric group with degree $ n$.

For $ n$ not a power of a prime, the equality $ n! \equal{} n\cdot s_n$ doesn't hold: there are always $ (n \minus{} 1)!$ of the $ n$-cycles and for composite $ n \equal{} ab$, $ \gcd(a,b) \equal{} 1$, you get a lot of combinations of $ a$-cycles with $ b$-cycles.

But I think you made some mistake, in that equality, there is no occurence of any $ \mu$, so why defining all that stuff...  :wink:

Thanks, ZetaX.

Yes, you're right, for example $ n\equal{}6$. Let $ \mu\equal{}6$. Then $ s_6\equal{}2\cdot 5!$.

But if $ n$ is a power of prime, then whether for every $ \mu\in\mu(Sym_n)$, we have $ n!\equal{}\mu\cdot s_{\mu}$?

Every element is a product of disjoint cycles of length, say, $ a_1, ..., a_k$; then $ a_1 \plus{} ... \plus{} a_k \leq n$. Then the order of that element is $ \mathrm{lcm}(a_1, ..., a_k)$. Now if we want that order to be $ \equal{} n \equal{} p^s$, then all $ a_i$ are powers of $ p$, say $ a_i \equal{} p^{s_i}$. If there is more than one cycle ($ \iff k > 1$), then $ a_i < n$, thus $ s_i < s$, thus $ \mathrm{lcm}(a_1, ..., a_k) \equal{} \mathrm{lcm}( p^{s_1}, ..., p^{s_k}) \equal{} p^{\mathrm{max}(s_1,...,s_k)} < p^s \equal{} n$, so it cannot have the required order $ n$.
So the only way to get an element of order $ n \equal{} p^s$ is to have a single cycle of length $ n$, and there are $ (n \minus{} 1)!$ many.


Predictions:
Backho[WHITE]> Type-B
By.Fantasy> Stork[gm]
Yellow[Arnc] < Die.Jaedong
By.Flash < EffOrt

Go go go!

Post your predictions

woah there seem to be a lot of good players
also wait what happened to the other 8 guys

also, stork will clearly pull something out of his ass and rape fantasy

this is just the end of the ro16 (the last 4 games)

flash better qualify :( darn

I think...

backho > type-b
fantasy > stork
jaedong > yarnc
flash > effort

I hope, at least. flash has a chance~

dude leejaedong has a pretty long last name

[quote="junggi"]yarnc>gogo
leta>type b
fantasy>canata
jaedong>zero

yarnc>leta
fantasy>jaedong

yarnc>fantasy[/quote]

Yarnc > Gogo
Leta > Type B
Fantasy>Canata
Jaedong > ZerO

Leta > Yarnc
Jaedong > Fantasy

Jaedong 3-2 Fantasy
Yay!

darn flash > effort

yarnc > gogo
leta > type b
fantasy>Canata
jaedong > ZerO

leta > Yarnc
fantasy > jaedong

fantasy 3-2 leta

dude what now? YARNC VS JAEDONG WOOOOO although yarnc vs fantasy would've been way more exciting

NOW WE GET TO SEE JvZ and a golden mouse. hopefully.

Finals: Yellow[Arnc] vs n.die.Jaedong

Winner prediction; Son Chan Woong

oh dang hi there golden mouse

also umm dang... calm vs kwanro? lololol

[quote="Scrambled"][quote="not_trig"][quote="Scrambled"]NEW OSL SEASON!

BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHO BACKHooOOOOooOOOoOOOoOOOOoooOOOOOooooooOOOoOOOO!!!!![/quote]

er no sir

LEEEEE YOUNGGGGG HOOOOO

also do you have a fantasy pl team yet? *checks*[/quote]
BackHo (captain) Skyhigh Jangbi Pusan Crazy-hydra Movie KHAN

Anti: Guemchi Kal (instead of Calm, by the advice of phr0zen) 815[/quote]

How many points is that for main?

Its 16+pusan+crazy-hydra i think...

mine keeps changing

currently it's

yarnc (captain)
fantasy
skyhigh
movie
calm
ruby (iris I think would be kinda... OWAIT KWANRO TO WOONGJIN ok back to iris)
CJ

anti:
stork
justin
soo

[quote="Ubemaya"]hmm how does this anti thing work?[/quote]

dude maybe you should get on TL and find out

or you could be like anti >= 13 points, and you get negative(total points earned by everyone on anti) points


$ a,b,c\in \mathbb{R_{\plus{}}}$ and $ \sum_{cyc}\frac{a^{3}\plus{}2a}{a^{2}\plus{}1}\equal{}\frac{9}{2}$
prove that $ \sum_{cyc}\frac{1}{a}\geq 3$

[quote="Dumel"]$ a,b,c\in \mathbb{R_{ \plus{} }}$ and $ \sum_{cyc}\frac {a^{3} \plus{} 2a}{a^{2} \plus{} 1} \equal{} \frac {9}{2}$
prove that $ \sum_{cyc}\frac {1}{a}\geq 3$[/quote]
we have
$ \frac{9}{2}\equal{}\sum a\plus{}\sum \frac{a}{a^2\plus{}1}$ so we have $ \sum a \ge 3$

but $ \frac{9}{2}\equal{}\sum \frac{1}{a}\plus{}\sum \frac{a^4\plus{}a^2\minus{}1}{a^3\plus{}a}\equal{}\sum a\plus{}\sum f(a)$
where $ f(a)\equal{}\sum \frac{a^4\plus{}a^2\minus{}1}{a^3\plus{}a}$
but $ f''(a)\equal{}\frac{\minus{}2(6a^4\plus{}3a^2\plus{}1)}{a^3(a^2\plus{}1)^3}<0$ for every $ a>0$
so we have $ \sum f(a) \le 3f(\frac{a\plus{}b\plus{}c}{3})\le 3f(1)\equal{}\frac{3}{2}$
so we have $ \sum \frac{1}{a}\ge 3$
Q.E.D

Let $ a\plus{}b\plus{}c\equal{}u , \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}v$

$ \frac{a}{a^2\plus{}1}\plus{}\frac{b}{b^2\plus{}1}\plus{}\frac{c}{c^2\plus{}1} \geq \frac{9}{u\plus{}v}$ by Cauchy Schwarz inequality.

$ (\frac{9}{2}\minus{}u)(u\plus{}v) \geq9$

If $ v<3 \Longrightarrow$

$ (\frac{9}{2}\minus{}u)(u\plus{}3) >9 \Longleftrightarrow$

$ (2u\plus{}3)(u\minus{}3) < 0$ but $ \frac{a}{a^2\plus{}1} \leq \frac{1}{2} \Longrightarrow$

$ \frac{9}{2}\minus{}u \leq \frac{3}{2} \Longleftrightarrow u \geq 3$ is a contradiction.  :wink:

[quote="crazyfehmy"]
$ (\frac {9}{2} \minus{} u)(u \plus{} 3) > 9 \Longleftrightarrow$

$ (2u \plus{} 3)(u \minus{} 3) < 0$ [/quote]$ u \equal{} 4$?  :|

[quote="tuandokim"]
$ 3f(\frac {a \plus{} b \plus{} c}{3})\le 3f(1) \equal{} \frac {3}{2}$
[/quote]but $ f$ is increasing function...


geometry
circumcircle
trigonometry
conics
hyperbola
geometry proposed

[color=darkblue][u][b]Sorry[/b][/u], I don't remember if this problem was posted at least once. ![/color]

[color=darkred]Given are a fixed line $d$ and a fixed point $F\not\in d$ and two mobile points $M$, $N$ so that $m(\widehat{MFN})$ is constant. Prove that the circumcircle of the triangle $MFN$ are tangent to a fixed circle.[/color]

Very nice problem, but it isn't hard as I thought.
Let $O$ be circumcenter of triangle $FMN$  and $\widehat{MFN}=\alpha$ we have
$\frac{OF}{d(O,d)}=\frac{R}{\frac{MN}{2}\cot\alpha}=\frac{R}{R\sin\alpha\cot\alpha}=\frac{1}{\cos\alpha}=const>1$
Thus locus of $O$ is a hyperbola with a focus $F$ now let $F'$ be the second focus of this hyperbola.Let  $I\in OF'$ such that $OI=OF=R\Rightarrow I\in (O)$ but $O$ in hyperbol therefore $F'I=|OF'-OF|=a=const\Rightarrow I\in (F',a)$ Thus $O$ always tagent to fix circle $(F',a)$.

also we can use an inversion with the pole $O$.


geometry
circumcircle
geometric transformation
homothety
geometry proposed

A larger circle contains a smaller circle and touches it at N. Chords BA, BC of the larger circle touch the smaller circle at K, M respectively. The midpoints of the arcs BC, BA (not containing N) are P, Q respectively. The circumcircles of BPM, BQK meet again at B'. Show that BPB'Q is a parallelogram.

PS
perhaps this is easier than previous of the same contest  but seems to me very interesting.

First of all consider the homotety with center N that sends the smaller circle into the big one. Then K will go to Q , so N-K-Q collinear. Simillary N-M-P collinear. Also it's obvious from angles that K-B'-M collinear. So <BQB'=<BKB'=<BMK=<BPM=<QNP=180-<QBP so QBPB'- a parralelogram.

[quote="Sailor"]First of all consider the homotety with center N that sends the smaller circle into the big one. Then K will go to Q , so ...
[/quote]

Exactly what I did.

I think, but is just my opinion, you should (to better balance what you prove explicitely and what you consider immediate or triavia) explain way "K will go to Q".

Actually it's not that hard:
Under this homotety AB should go into a paralell line but also into a tangent to the big circle at point(say) Q'.We have that N-K-Q' collinear. But then we have by angles that Q' should be the midpoint of arc AB. Hence Q=Q'.
Similary for BC.


[quote="CCMath1"]HERE IS THE PROBLEM:   P is a point in triangle ABC PA' is perpendicular to BC ;use the same way to makr point B' and point C', point AA'' is perpendicular to B'C'  use the same way to make rays BB'' ,CC'';1)Prove that: AA'',BB'',CC'' meet at one point M;2)Denote the circumcenter of triangle A'B'C' is O prove that M,P,O are collinear and 3) MO=PO.[/quote]
1) easily if we use Cacno theorem ;)

It is easy to prove M,P are Conjugate Points of ABC
and according to the property of Conjugate Points M,O,P are collinear   O is midpoint of MP

[hide]hint:to prove this, you can project M on AB,BC,AC and the foot of six altitude are concyclic[/hide]


geometry
rectangle
induction
combinatorics proposed
combinatorics

A cute problem :):

We have an $m\times n$ recatngle tiled with $2\times 1$ bricks, except for a $1\times 1$ corner. If the small side of a brick is right near the hole, we can move the brick along its larger side so that it covers the existing hole and creates a new one. Prove that by performing this sort of move we can place the hole at any other corner of the rectangle.

anyone ? I find this pb interesting but I can't solve it ..

Grobber do you have any idea on the solution ? I tried too but I got nothing till now ...

@+

a closely related problem(in my opinion, they are the same) is that for any $2n-1\times 2n-1$ square, the number of domino tilings of this square minus a corner square, is the same as the number of spanning trees of an $n\times n$ grid graph.
basically, one can form from any spanning tree of the $n\times n$ grid graph, one can associate a tiling of the $2n-1\times 2n-1$ minus a corner square in a rather obvious way, if we regard the $n\times n$ square to be the one formed by the alternating squares of every alternate row.to go the reverse, note that for any tiling of the larger square, will give rise to a graph on the $n\times n$ grid. Since there are $n^2-1$ such squares that are covered by dominoes, the corresponding graph will have $n^2-1$ edges. now it is easy to argue that this graph cannot have cycles(parity arguments, or induction, whatever), and so that gives rise to a spanning tree of the smaller grid graph.

Thank you, seshadri. It is exactly what I did :).

[quote="seshadri"]
basically, one can form from any spanning tree of the $n\times n$ grid graph, one can associate a tiling of the $2n-1\times 2n-1$ minus a corner square in a rather obvious way, if we regard the $n\times n$ square to be the one formed by the alternating squares of every alternate row.to go the reverse, note that for any tiling of the larger square, will give rise to a graph on the $n\times n$ grid. Since there are $n^2-1$ such squares that are covered by dominoes, the corresponding graph will have $n^2-1$ edges. now it is easy to argue that this graph cannot have cycles(parity arguments, or induction, whatever), and so that gives rise to a spanning tree of the smaller grid graph.[/quote]

Hmm... I can't see how this implies what the problem asks for...

Well, the edges of this graph show how the empty square can "travel" along the graph. Since this graph has one edge less than it has vertices, plus it has no cycles, it must be a tree and hence a connected graph.  Thus the empty square can travel anywhere on the big square, including the other corners.

Also note that the fact that the empty square is initially in the corner is essential for proving that the resulting graph is cycle-free, since otherwise it is possible to give a tiling which gives cycles (and thus the underlying graph is not connected).

Jan


function
Gauss
number theory
prime factorization
number theory proposed

For all $n \in \mathbb N^*$ denote by $p(n)$ the product of all divisors of $n$.
Prove that : $p(m)=p(n) \Rightarrow m=n$

the function p(n) depends only the cannonical prime factorization of a positive integer n so if we have equality then m and n have the same prime divisors and it is obvious that it has to have the same prime factorization. there is really nothing to explain here.

it may be an easy problem, but i don't think it's exactly *obvious*...

well i guess not obvious but it is indeed trivial dont you think???

i think it's quite simple, but it's not yet obvious.
If \[
n = \prod {p_1^{k_1 } }
\]
, then the product of divisiors equals \[
\prod {p_1^{\frac{{k_1 \left( {k_1  + 1} \right)}}
{2}} }
\]
Assume there are m,n satisfying p(m)=p(n) and their exponents of p_1 are different, then
\[
p_1^{\frac{{k_1 \left( {k_1  + 1} \right)}}
{2}} M = p_1^{\frac{{l_1 \left( {l_1  + 1} \right)}}
{2}} N
\]
where M,N are not divisible by p_1
\[
p_1^x M = N
\]
Assuming M > N and x is some non-negative integers.
Since M,N are not divisible by p_1, x=0, that means the exponents of p_1 are the same.
So the exponents of all prime divisors are the same which means m=n.

[quote="jmerry"]Here's a simple approach.

Let $\tau(n)$ be the number of divisors of $n$. Then $p(n)=n^{\frac{\tau(n)}{2}}$. Clearly, we can only have $p(m)=p(n)$ if $n=m^r$ for some rational $r$. WLOG, $r>1$, so $m|n$. Then $\tau(n)>\tau(m)$ because every divisor of $m$ is a divisor of $n$, and $n$ is also a divisor of $n$. Then $p(n)=n^{\frac{\tau(n)}{2}}>m^{\frac{\tau(n)}{2}} >m^{\frac{\tau(m)}{2}}=p(m)$. This contradiction proves that $p(m)=p(n)$ if and only if $m=n$.

The formula for $p$ in the previous post was wrong- $p$ is not a multiplicative function.[/quote]

can you explain how you obtained the formula of p?

[quote="ma_go"]it's the same as gauss' idea for the sum of the first $n$ numbers...
for each $a|n$ there exist one (and only one) $a'|n$ such that $aa' = n$.
now, take $p(n) = \prod_{a|n} a = \prod_{a|n} a'$: squaring it, you get $p(n)^2 = \prod_{a|n} aa' = n^{\tau(n)}$, and it's done...
quite nice, by the way :)[/quote]

Very interesting! :lol:


inequalities
trigonometry
inequalities proposed

Prove that in any triangle $ABC$ the following inequality holds:

       \[ \sqrt{\sin\frac{A}{2}}+\sqrt{\sin\frac{B}{2}}+\sqrt{\tan\frac{C}{2}} > 1+\frac{1}{r}{R}. \]

:? your expression is not even defined for all triangles... and even for sharp angles it seems false to me ($A\rightarrow\frac\pi2,B\rightarrow\frac\pi2,C\rightarrow0$ causes $R\rightarrow k,r\rightarrow0$ with $k$ the longest sidelength)

Maybe you meant 3x tangens instead, in sharp triangles?

I thought so too. The problem was given to me by Kuba. I told him that there is something sospicious about this inequality. ;)


Could anyone suggest me some classic;challenging calculus textbooks? :maybe:

Its very hard to find tough calculus questions for high school syllabus. JEE questions are good but if you want more interesting try the calculus section of this forum.

have u tried CALCULUS by THOMAS & FINNEY.......

Try Amit M Agarwal unless you are tooooo advanced.

Well,I have only done a bit of limits from Hall and Knight and will not be doing much of calculus till next year,as I have got to take care of my ICSE. :weightlift:


what is the difference between
(a+b+c+d+.....)^2  and
|a+b+c+d+......|^2

so can I wright (a+b+c+d+....)^2=|a+b+c+d+.......|^2  ?

There is no difference. Consider, for example, (1-2-3)^2 and |1-2-3|^2.

Your problem quite easy.
$ |x|^2\equal{}x^2,\forall x\in R$
What is your question?

oke thx

I needed to know for sure

 :lol:


Find the value
1. $(x-a)(x-b)(x-c)...(x-z)$
2. $\frac{\sin{x}}{n}$

(Maybe well-known)

[hide]
1. 0
2. 6
[/hide]

shouldn't this be in one of the math forums? ;)

uhh...how are you getting these answer?  i don't get shobber's solutions.

dude, they are only math jokes.
the first one is $x$ minus $a$, time $X-b$, ... $(x-y)(x-z)$.
in between you have $(x-x)$ which is zero, so the whole thing is zero.
the second one. the $n$ in the $sinx$ cancelles with the denominator that leaves only $SIX$

lol, they are pretty well known

[hide]1. 0  2. 6[/hide]


geometry
geometric transformation
reflection

[quote="dgreenb801"]This problem is from Geometry from the Land of the Incas([url]http://agutie.homestead.com/[/url]).

You have right triangle ABC with angle B 90. D is on AC and E is on BC. Angle EAD is twice angle EAB. Angle AED=Angle BCA. Show DE=2BE. [/img][/quote]
[hide="Solution"]
Reflect $ E$ over $ AB$ to form $ F$ and say that $ \angle BAE \equal{} \alpha$.  Thus, we have that $ \angle EAD \equal{} 2\alpha$, $ \angle BAF \equal{} \alpha$, $ \angle AED \equal{} \angle C \equal{} 90 \minus{} \angle BAC \equal{} 90 \minus{} 3\alpha$, $ \angle ADE \equal{} 180 \minus{} \angle DAE \minus{} \angle AED \equal{} 90 \plus{} \alpha$, and $ \angle BFA \equal{} 90 \minus{} \angle BAF \equal{} 90 \minus{} \alpha$.  Therefore, $ \angle EFA \plus{} \angle EDA \equal{} 90 \plus{} \alpha \plus{} 90 \minus{} \alpha \equal{} 180$, so $ ADEF$ is cyclic.  Hence, $ \angle DFE \equal{} \angle DEA \equal{} 2\alpha \equal{} \alpha \plus{} \alpha \equal{} \angle BAE \plus{} \angle BAF \equal{} \angle EAF \equal{} \angle EDF$, so $ \triangle EDF$ is isoceles, and thus $ ED \equal{} FE \equal{} BE \plus{} FB \equal{} 2BE$ and we are done. [/hide]


square
algebra
Cubic
polynomial
cubic equation
IMO Shortlist

Real constants $ a, b, c$ are such that there is exactly one square all of whose vertices lie on the cubic curve $ y \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c.$ Prove that the square has sides of length $ \sqrt[4]{72}.$

[quote="Rofler"]Well, if such a curve exists, then translate the curve down s.t. the center of the square is at (0,0), to get another monic cubic curve. Such a transformation preserves length, so after we find the length of the side of such a square, we can just translate back to the original curve.

If the point (x,y) is a point, then so is (-y,x) (-x,-y) (y,-x).

$ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} y$
$ ( \minus{} x)^3 \plus{} a( \minus{} x)^2 \plus{} b( \minus{} x) \plus{} c \equal{} \minus{} y$

Summing the 2 up, we have $ 2ax^2 \plus{} 2c \equal{} 0$

$ ( \minus{} y)^3 \plus{} a( \minus{} y)^2 \plus{} b( \minus{} y) \plus{} c \equal{} x$
$ y^3 \plus{} ay^2 \plus{} by \plus{} c \equal{} \minus{} x$

Summing up the 2, we have $ 2ay^2 \plus{} 2c \equal{} 0$

Now, if a is not equal to 0, we have x,y are negatives of each other, or are equal to each other, since they are both solutions to $ 2at^2 \plus{} 2c \equal{} 0$

Therefore, $ (x,x)$ and $ (x, \minus{} x)$ would both have to be points, but since cubics are one-to-one, x=-x, so $ x \equal{} 0$, so no square exists. Hence, a=0, and therefore c=0.

The set of equations reduces to:

$ x^3 \plus{} bx \equal{} y$
$ y^3 \plus{} by \equal{} \minus{} x$

x=0 or y=0 give degenerate squares, so there has to exist a non-zero pair (x,y) which satisfy the system.

If b is positive then if x is positive, y is positive, so -x is positive, contradiction.
If b is positive then if x is negative, y is negative, so -x is negative, contradiction.

Hence, b is either 0 or is a negative number.

Anyone can finish?[/quote]
continue with Rofler's solution, now we have non-zero $x$, $y$

$ x^3 \plus{} bx \equal{} y$
$ y^3 \plus{} by \equal{} \minus{} x$

but we need to use the condition that there is "exactly one" square on the cubic curve,
which is equivalent to there is exactly one solution (x, y) in the 1st quadrant satisfy the above equations.
Plot the cubic curves we can figure out this implies the two curves tangent at (x, y).  then

$ (3x^2 \plus{} b)dx \equal{} dy$
$ (3y^2 \plus{} b)dy \equal{} \minus{} dx$

cross out we have $ (3x^2 \plus{} b)(3y^2 \plus{} b) \plus{} 1 \equal{} 0 $

now there are three equations and three variables $x,y,b$ so we can, in principle, solve the equations

Maybe a solution with a bit less algebra.

Shift the coordinate system so that the center of the square is the origin and let $P(x)=x^3+ax^2+bx+c$. Then $P(x)+P(-x)=0$ for the four vertices of the square. Since this is a quadratic, it must be identically $0$. Hence, $a=c=0$.

Thus, $y=x^3+bx$. Rotating this by $90^\circ$ around the origin leads to the curve $x=-y^3-by$. Since exactly one square is possible, the two cubics have to be tangent to another. Hence, by substituting $y=x^3+bx$ into the other equation (and dividing out the solution $x=0$), we must get a polynomial with four double roots. Hence, $0=x+y^3+by=x+(x^3+bx)^3+b(x^3+bx)=x(x^8+3bx^6+3b^2x^4+(b^3+b)x^2+b^2+1)$. Setting the second factor equal to $(x^4+a_3x^3+a_2x^2+a_1x+a_0)^2$ and comparing coefficients leads to $a_1=a_3=0$, $3b=2a_2, 3a_0=b^2+1,a_2^2+2a_0=3b^2$. This implies $b=\pm 2\sqrt 2$. Now $b<0$, otherwise $P$ would be increasing and the two curves would only intersect at $(0,0)$. Hence, $b=-2\sqrt 2$. Then the factorization becomes $(x^4-3\sqrt 2x^2+3)^2$. If the coordinates of the vertices are $(x_i,y_i),1\leq i\leq 4$ and the square has sidelength $a$, then by the quadratic formula $a^2=\frac 12(x_1^2+y_1^2+\hdots+x_4^4+y_4^2)=x_1^2+x_2^2+x_3^2+x_4^2=4\cdot\frac{3\sqrt 2}{2}=\sqrt {72}$, as desired.


Let $ a,b,c>0$ with $ abc\geq 1$.Prove that $ \frac{bc\minus{}a}{1\plus{}a}\plus{}\frac{ca\minus{}b}{1\plus{}b}\plus{}\frac{ab\minus{}c}{1\plus{}c}\geq 0$

Inequality become\[ \sum {\frac{1\plus{}bc}{1\plus{}a}}\ge3\] But \[ \sum {\frac{1\plus{}bc}{1\plus{}a}}\ge3\sqrt[3]{\prod {\frac{1\plus{}bc}{1\plus{}a}}},\]  so is suffices to prove that \[ \prod {(1\plus{}bc)}\ge \prod {(1\plus{}a)}\] The last inequality is equivalent with \[ (abc\minus{}1)(abc\plus{}a\plus{}b\plus{}c)\ge 0\] wich is cleary true.


Let $ABCD$ be a trapezoid with parallel sides $AB > CD$.
Points $K$ and $L$ lie on the line segments $AB$ and $CD$, respectively, so that
$\frac{AK}{KB}= \frac{DL}{LC}$. Suppose that there are points $P$ and $Q$ on the line segment $KL$ satisfying $\angle APB = \angle BCD$ and $\angle CQD = \angle ABC$.
Prove that the points $P, Q, B, C$ are concyclic.

See: http://www.mathlinks.ro/Forum/viewtopic.php?t=152249  :)

It is ISL 2006 G2, so please continue the discussions here: [url]http://www.mathlinks.ro/viewtopic.php?t=155705[/url]


N (N=100000) coins are tossed If you end up having H heads you get a score ((H*N))/(H+1) for H heads out of N coins. Thus You get score of 0 for 0 Heads.
What is the expected score for a single flip. Your answer should be of form X.abcdefghij (Exactly 10 decimal places )

Note: Single Flip signifies the expected score per coin.

This subforum is for [b]Number Theory[/b]! -> Moved.

isn't it H/(H+1) since the expectation is additive?

There is no $ H$. That answer doesn't make sense.

ok, thanks jmerry, I try again
I think it's
P(0 heads)*expected score(0  heads)+P(1 head)*expected score(1 head)=1/4


given f(0) =3, f(1)=-1, and f(n) = f(n-2)-f(n-1), find f(100) in terms of the fibonacci numbers.

[hide="solution"]Let, fibonacci sequence be denoted by $ f(n)$ and the given sequence by $ g(n)$
This sequence goes on like:
$ 3, \minus{} 1 , 4 , \minus{} 5.....$   and it can be proved that :
                         $ (i)$ it is a fibonacci like sequence whithout considering the signs $ (ii)$ the signs have form $ ( \minus{} 1)^{g \minus{} 1}$

 Next, as it is a fibonacci like sequence, observing from $ g(4)$ onwards(not considering sign): $ 9, 14, 23, 37 ....$
 and considering our fibonacci sequence $ f(6)$ onwards: $ 8, 13, 21, 34 ...$
  these differ by:$ 1, 1, 2, 3,....$ which is the fibonacci sequence again

   thus in general $ g(n) \equal{} f(n) \plus{} f(n \minus{} 3)$

  Applying signs we get $ g(n) \equal{} ( \minus{} 1)^{g \minus{} 1}f(n) \plus{} f(n \minus{} 3)$
   thus, $ \boxed{g(100) \equal{} \minus{} [f(100) \plus{} f(97)]}$[/hide]


Hi all,

Can anyone tell me what the latex sequence for the '=?' symbol (as used in conjectures, such as P =? NP, with the '?' appearing above '=') is? I have looked through many symbol tables and have been unable to find it. I have, however, seen a pdf (created using pdflatex) that displays this character.

Thanks,
Srikanth.

You could do
\stackrel{?}{=}
Which gives
$P\stackrel{?}{=}NP$
I don't know if that's a good idea though.

Personally, I prefer using \overset and \underset.

[code]P\overset ?= NP[/code]
gives
$P\overset ?= NP$

You can also have more than one character, for example
[code]\underset{bbb}{aaaaa}[/code]
gives
$\underset{bbb}{aaaaa}$


Hi, My friend gave me this problem, But I'm not sure this is the original statement. Here it is first :
Does there exist a function $s: \mathbb Q \rightarrow \{-1;1\}$ satisfying :
For all $x,y \in \mathbb Q$ such that $xy=1$ or $x+y=1$ we have $s(x)s(y)=-1$
I found a very easy solution Take $x=y=1$ we get $(s(1))^2=-1$ which is absurd.
But this can't be an olympiad problem  :? Maybe as I said above It's not the correct statement.
Does anyone have an Idea about that ??

I think it's like this:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30462[/url]

Unbelievable I see it for third time in a row. So popular problem!


quadratics
number theory proposed
number theory

prove that  $ a$ and $ b$ have the same  parity $ <\equal{}>$ there are  two  positive integers $ c$ and $ d$ such that
$ a^2\plus{}b^2\plus{}c^2\plus{}1\equal{}d^2$

Let $ a^2 \plus{} b^2 \plus{} 1 \equal{} 2n \plus{} 1$
Then we have: $ 2n \plus{} 1 \equal{} d^2\minus{}c^2 \equal{} (d\minus{}c)(d\plus{}c)$
Let $ d\minus{}c \equal{} 1$ and $ d\plus{}c \equal{} 2n\plus{}1$ Then we have $ d \equal{} n\plus{}1$ and $ c \equal{} n$.

So $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 1 \equal{} d^2$ if $ c \equal{} \frac{a^2\plus{}b^2}{2}$ and $ d \equal{} \frac{a^2\plus{}b^2}{2}\plus{}1$ (These are integers because $ a$ and $ b$ have same parity)

for the only if part...
suppose that $ a,$ and $ b$ have different parity then:
$ a^2\plus{}b^2\plus{}1 \equiv 2 (4)$ thus $ d^2\minus{}c^2 \equiv 2 (4)$
however the quadratic residues mod $ 4$ are $ 0,1$ contradiction


geometry
parallelogram
trapezoid
geometry proposed

We have a convex quadrilateral with no parallel sides. On each side $AB$ select a point $P$ following this procedure:

Draw lines from $A$ and $B$ parallel to the opposite sides and let $A'$ and $B'$ be the new points where these lines intersect the sides neighboring $AB$ and let $P$ be the point of intersection of $AB$ with $A'B'$. The four points selected in this way are the corners of a .............. :lol:

Discover by yourself the missing figure ;)

I didn't exactly undestand how to construct $P$.

If $ABCD$ is the convex quadrilateral, draw from $A$ and $B$the lines parallel to $CD$ (that is the side opposite to $AB$).The line for $A$ meet the line $BC$ (or its extension over $B$) in the point $B'$ while the line for $B$ meet the line $AD$ (or its extension over $A$ in the point $A'$). Now join $A'$ with $B'$ and draw the line $A'B'$. Finally call $P$ the intersection of $AB$ with $A'B'$.
Repeat this construction for the four sides and join the four points found. :)

I hope it is clear now.

Nice problem, Manlio!

[quote="Myth"]It is a parallelogram![/quote]

Indeed, it's a parallelogram with sidelines parallel to the diagonals of the quadrilateral ABCD. To prove this, it will be enough to consider only one side of this parallelogram. In other words, it will be enough to prove:

[b]Theorem 1.[/b] Let ABCD be a quadrilateral.
Let the parallel to the line AB through the point C meet the line DA at M, and let the parallel to the line AB thorugh the point D meet the line BC at N. Let the line MN meet the line CD at V.
Let the parallel to the line BC through the point D meet the line AB at K, and let the parallel to the line BC thorugh the point A meet the line CD at H. Let the line KH meet the line DA at U.
Then, UV || AC.

[i]Proof.[/i] Let the lines AB and CD meet at a point X, and let the lines BC and DA meet at a point Y.

It is well-known that in a trapezoid, the line joining the point of intersection of the diagonals with the point of intersection of the legs passes through the midpoints of the bases. Hence, for the trapezoid AHDK (this is a trapezoid since AH || BC and DK || BC yield AH || DK), the line joining the point of intersection U of the diagonals DA and KH with the point of intersection X of the legs DH and AK passes through the midpoints of the bases AH and DK. In other words, the line UX passes through the midpoints of the segments AH and DK. Denoting the midpoint of the segment AH by S, we thus see that the line UX passes through this point S. Now, if this line UX meets the line BC at a point K, then AH || BC yields with Thales BK : KC = AS : SH, while AS : SH = 1 because the point S is the midpoint of the segment AH. Thus, BK : KC = 1, and therefore, the point K is the midpoint of the segment BC.

Hence, we get an alternative construction of the point U: Take the midpoint K of the segment BC; then, the point U is the point of intersection of the line KX with the line DA. Similarly, if L is the midpoint of the segment AB, then the point V is the point of intersection of the line LY with the line CD.

Now, we will prove an important fact:

[b]Theorem 2.[/b] Let ABCD be a quadrilateral, let the lines AB and CD meet at a point X, and let the lines BC and DA meet at a point Y. Let K be an arbitrary point on the line BC, and let U be the point of intersection of the line KX with the line DA. Let L be an arbitrary point on the line AB, and let V be the point of intersection of the line LY with the line CD. Then, the lines AC, KL and UV concur.

[hide="First proof of Theorem 2"]
[i]First proof of Theorem 2.[/i] Applying the Pappos-Pascal theorem to the collinear points Y, L and V and to the collinear points X, U and K, we see that the points of intersection $LK\cap VU$, $VX\cap YK$ and $YU\cap LX$ are collinear. But $VX\cap YK = C$ and $YU\cap LK = A$; thus, the points $LK\cap VU$, C and A are collinear. In other words, the point $LK\cap VU$ lies on the line CA. Equivalently, the lines LK, VU and CA concur. This is exactly Theorem 2.[/hide]

[hide="Second proof of Theorem 2"]
[i]Second proof of Theorem 2.[/i] Have you already thought about what happens to the Pappos-Pascal theorem under projective duality? ;) [/hide]

Okay, now after having proved Theorem 2, we can apply it to our original problem and see that the lines AC, KL and UV concur. But the points K and L are the midpoints of the segments BC and AB, respectively; hence, the line KL is parallel to the line AC (as a midparallel in the triangle ABC). Thus, the point of concurrence of the lines AC, KL and UV must be an infinite point, so that we have UV || AC, and Theorem 1 is proved.

  Darij


The diagram shows masses of 8 kg and 12 kg lying on a smooth horizontal table and joined, by a light inextensible string, to a mass of 5 kg hanging freely. This string passes over a smooth pulley at the edge of the table.
The system is released from rest. Find:
a. the tension in the string connecting the 8 kg and 12kg masses
b. the tension in the string connecting the 12kg and 5kg masses
c.the acceleration of the system.

can you answer my question?

This question should be posted in the [url]http://www.artofproblemsolving.com/Forum/index.php?f=332[/url]


find such real p,q,a,b that for all x an equality is hold

$(2x-1)^{20}-(ax+b)^{20}=(x^{2}+p x+q)^{10}$

Does this really belong in high-school basics?  It seems like more advanced material.

It doesn't seem to be out of reach for HSB.

[hide="Solution"]The leading coefficient of the LHS is $2^{20}-a^{20}$, and the leading coefficient of the RHS is $1$. Hence $a^{20}=2^{20}-1\iff a=\pm\sqrt[20]{2^{20}-1}$.

Now put $x={1\over 2}$. The equation becomes $-\left({a\over 2}+b\right)^{20}=\left({1\over 4}+{p\over 2}+q\right)^{10}$. Since the LHS is non-positive and the RHS is non-negative, the equality can hold iff both are zero. Therefore $b=-{a\over 2}=\mp{1\over 2}\sqrt[20]{2^{20}-1}$

Now let's substitute $a$ and $b$ into the original equation:

\begin{eqnarray*}&&(2x-1)^{20}-\left(x\sqrt[20]{2^{20}-1}-{1\over 2}\sqrt[20]{2^{20}-1}\right)^{20}\\ &=& (2x-1)^{20}-(2^{20}-1)\left(x-{1\over 2}\right)^{20}\\ &=& 2^{20}\left(x-{1\over 2}\right)^{20}-(2^{20}-1)\left(x-{1\over 2}\right)^{20}\\ &=& \left(x-{1\over 2}\right)^{20}\\ &=& \left(x^{2}-x+{1\over 4}\right)^{10}\end{eqnarray*}

Therefore, $(a,b)=\pm(\sqrt[20]{2^{20}-1},-{1\over 2}\sqrt[20]{2^{20}-1}), (p,q)=\left(-1,{1\over 4}\right)$[/hide]


conics
analytic geometry
linear algebra
matrix
parabola
symmetry
geometry

Say I wanted the equation of a conic passing through a few points on the cartesian coordinate system. How would I go about finding this equation?

I know I can set up a determinant matrix and just expand it, but are there any other ways that do not require matrices?

A circle's equation can be determined with three points, plugging them into the general form of

$x^2+y^2+Dx+Ey+F=0$

Then solving the system of three equations and three unknowns.

A parabola is also determined by 3 points ;)

By the way, there's another way to find the equation of a circle that goes through 3 points.
Basically, you use the property that the perpendicular bisector of any chord goes through the center of the circle.
You find the equation of the perpendicular bisector of the chord twice, then find the intersection of these two perpendicular bisectors... The intersection is the center of the circle.

[quote="gotztahbeazn"]umm well i guess it depends on how many pts ur actually given... like if ur given enuff couldnt u just use the form like

$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$

and plug in for $x$ and $y$ 4 times to find the general equation? that seems a bit tedious but this is the only way i can think of if given random pts

now if the pts they give u are like the center/focus/endpts then its entirely different....[/quote]

The above, assumes that the major (or minor)  axis is parallel to x-axis.. in general you would need  5 points to fix a curve. ( not four). If you know it is a circle you  need only 3 points, if it is a parabola  you need 4 points. .. so  the following is [b]not [/b]quite correct:
([quote]A parabola is also determined by 3 points[/quote]

Anyway:

The general equation is $ax^2+bxy+cy^2+dx+ey+f= 0$  where not all the constants are zero  (so you can divide by that and you would have five independent coefficients)   just substitute (x,y) for 5 points, you get 5 equations  and thus can find the values of  the coefficients.

[quote="Gyan"]The above, assumes that the major (or minor)  axis is parallel to x-axis.. in general you would need  5 points to fix a curve. ( not four). If you know it is a circle you  need only 3 points, if it is a parabola  you need 4 points. .. so  the following is [b]not [/b]quite correct:
([quote]A parabola is also determined by 3 points[/quote][/quote]
Sorry, I wasn't clear about that.
What I was trying to say was, if we already know that the axis of symmetry is vertical (or horizontal), then we only need to know 3 points on the curve.
A parabola whose axis of symmetry is vertical, for instance, has no $xy$ terms and $y^2$ terms. Hence it is in the form $y=ax^2+bx+c$, and that's why we only need 3 points.

Thanks for the clarification, Gyan :)

[quote="jhcreinhardt"]For a purely general curve in general form, one needs 6 points (6 variables).[/quote]
No,  they are not independent (at least one of them is not zero, so divide all coefficients by that number and   now you have only five)

For example $Ax +By +C = 0$  is  general equation of a  line  (with 3 coefficients)  , but you need only 2 points  to fix a line.
(IOW if B is not zero, the equation can be written as y= mx+b , (where m= -A/B and b = -C/B)  and you have only two varibles m and b)

As mentioned before,   to fix a conic,   you need 5 points only.
Hope that helps.

How is it determined that a circle only needs 3, a parabola needs 4, etc?

As other people mentioned in the previous posts, it depends on the number of unknown coefficients.

For instance, given $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, if $A=C$ and $B=0$, then the equation represents a circle.
We can divide both sides by $A$ (or $C$) to get $x^2+y^2+D'x+E'y+F'=0$, where $D'$, $E'$ and $F'$ are some other constants.
Since we have 3 unknowns, we need 3 points, as jhcreinhardt said.

Okay, I see. Thank you.

Basically if you want to solve a system of equations with n variables, you need n equations?


If $ a \plus{} b \equal{} 2$ and $ a^{2}b \plus{} b^{2}a \equal{} 1$, what is the value of $ ab$?

hello, we have $ ab(a\plus{}b)\equal{}1$ setting $ a\plus{}b\equal{}2$ in our equation then we get the answer $ ab\equal{}\frac{1}{2}$.
Sonnhard.

We can factor $ a^{2}b \plus{} b^{2}a \equal{} 1$ as $ ab(a \plus{} b) \equal{} 1$

We can substitute the first equation into this to get $ 2ab \equal{} 1$

Solving gives $ \boxed{\frac{1}{2}}$

EDIT: Oops, beaten


My solution isn't very good  :mad:

Denote H be the intersection of PE and FC and K be the intersection of QT and DE
By Thales Theorem and calculating, we have $ HQ \equal{} HE \equal{} \frac{CE.AE}{AC}$. Moreover, QE || TD, which follows that F, C, K are collinear

[quote="hophinhan"](MOP 1997) Let ABC be a triangle, and D,E, F the points where the incircle touches sides BC,CA,AB, respectively. The parallel to AB through E meets DF at Q, and the parallel to AB through D meets EF at T. Prove that the lines CF,DE,QT are concurrent.[/quote]
Let $ K\equal{}DE\cap QT,K'\equal{}FC\cap DE, H\equal{} DE\cap AB$.
Because $ (ABFH)\equal{}\minus{}1$ then $ (EDK'H)\equal{}\minus{}1\Rightarrow (FE, FD,FC,FH)\equal{}\minus{}1$
Moreover $ EQ//DT//AB$ we get FC passes through the midpoints of QE and DT.
Applying the lemma about trapezium we obtain FC passes through K.

[quote](MOP 1997) Let ABC be a triangle, and D,E, F the points where the incircle touches sides BC,CA,AB, respectively. The parallel to AB through E meets DF at Q, and the parallel to AB through D meets EF at T. Prove that the lines CF,DE,QT are concurrent.[/quote]
[b]Another approach:[/b]
Let $ X\equiv CF\cap (I)$, where $ (I)$ is the incircle wrt $ \triangle ABC$. It is easy to see that $ FEXD$ is a harmonic quadrilateral. Thus, $ (FE,FD,FA,FX) \equal{} \minus{} 1$. Therefore, if $ CF\cap \{QE,TD\}\equiv \{M,N\}\Longrightarrow M,N$ will be the midpoints of $ QE,TD$ respectively. But $ QE\parallel{}TD$, hence, it is obvious that $ MN$ passes the intersection of $ QT,DE$. The result is lead as follow.
Our proof is completed then.


Hello,

I'm planning to do a simple document with a footer and a table of content.

I don't want to make you tired so if you know a guide or an e-book describing these things.

Your help is appreciated  :)

The [url=http://en.wikibooks.org/wiki/LaTeX]wiki[/url] and the [url=http://tobi.oetiker.ch/lshort/lshort.pdf]Not So Short Guide[/url] are both good.


IMO
combinatorial geometry
combinatorics
counting
convex quadrilateral
IMO 1969

Given $n>4$ points in the plane, no three collinear. Prove that there are at least $\frac{(n-3)(n-4)}{2}$ convex quadrilaterals with vertices amongst the $n$ points.

[hide]

Orient the points so that one of them corresponds to the origin (A), another lies on the y-axis (B), and all the others are in either quandrant I or IV. (In other words, you are creating a boundary line.) Select one of the points (C) which minimizes the angle BAC. You cannot have two of them because then they would be collinear. Also no points lie inside the area of ABC because if such a point D did exist, then the angle DAC would be less than BAC - contradiction. So there are none of the other n-3 points inside the area of ABC.

Now we will show that for any two of the remaining n-3 points, you can create a convex quadrilateral containing two of them and two of A,B,C.

Now select two arbitrary points from the remaining n-3. Call them E and F, and draw a line though them; call this line l. If l does not intersect ABC, then ABCEF is convex and you choose the quadrilateral ABEF.
The other case is that it separates ABC such that one of A,B,C is one one side of l and the other two are on the other side. Without loss of generality, A is on one side of l while B and C are on the other side. Consider the quadrilateral BCEF (or CBEF, depending on orientation.) We know that BC are on the same side of EF. Also, since no points are in the third quadrant, we know that EF are on the same side of BC. Now we can choose either BCEF or CBEF to be our convex quadrilateral.

So we now have that for the given ABC, all pairs of the other n points give a distinct convex quadrilateral. So we have n-3 points to choose from, and we must choose two of them. So this gives us at least C(n-3,2) = $\frac{(n-3)(n-4)}{2}$ convex quadrilaterals.

[/hide]

[edit] Slight technicality.

We can also proceed by showing that given 5 points with the given conditions, there is always a convex quadrilateral.
Then we can choose 5 points from the set of n points in $n\choose 5$ ways. But each quadrialteral is counted $n-4$ times, so the problem is equivalent to proving that ${n\choose5}\frac{1}{n-4}\geq{{n-3}\choose2}$, which is trivial.

[quote="Philip_Leszczynski"]
Now select two arbitrary points from the remaining n-3. Call them E and F, and draw a line though them; call this line l. If l does not intersect ABC, then ABCEF is convex and you choose the quadrilateral ABEF.
The other case is that it separates ABC such that one of A,B,C is one one side of l and the other two are on the other side. Without loss of generality, A is on one side of l while B and C are on the other side. Consider the quadrilateral BCEF (or CBEF, depending on orientation.) We know that BC are on the same side of EF. Also, since no points are in the third quadrant, we know that EF are on the same side of BC. Now we can choose either BCEF or CBEF to be our convex quadrilateral.
[/quote]
Sorry, I just noticed this. I don't think this is right. The critical case is actually whether l intersects AB or not. And it seems possible for line BC to intersect segment EF and line EF to intersect ABC: A=(0,0), B=(0,4), C=(2,2), E=(2,1), F=(4,1). Here the only convex quad is ABCE.

On the other hand, letting the C be the point that maximises BAC seems to solve the problem. Then all the points are between lines AB and AC, and lines AB and AC will not intersect segment EF.
And the only case when line EF manages to intersect both segments AB and AC is a case where segments EF and BC do not intersect. A convex quad with both E and F is found in all cases.


topology
function
real analysis
real analysis unsolved

I am trying to determine which parts of the problem below are false and which are true. For the false, I am trying to think of a counterexample, but none come to mind. Any help to guide me in the right direction would be much appreciated:

A map $ f: M \rightarrow N$ is open if for each open set $ U \subset M$, the image set $ f(U)$ is open in $ N$.

a) If $ f$ is open, is it continuous? [hide] I am pretty sure this is false. But I can't think of an open set that would inverse map into a closed set , especially if that closed set has to be contained in the open set $ M$, since $ f$ is open. The only example I tried was $ \mathbb{R}$ mapping into the $ \mathbb{Q}$, but I 'm not sure if this makes sense. [/hide]
b) If $ f$ is a homeomorphism, is it open? [hide] True, by defn, a homeomorphism is topological property preserving [/hide]
c) If $ f$ is an open, continuous bijection, is it a homeomorphism? [hide] false? [/hide]
d) If $ f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous surjection, must it be open?
e) If $ f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous, open surjection, must it be a homeomorphism? [hide] This cannot be a homeomorphism because the 1-1 condition is missing. [/hide]
f) If $ f: S_{n}(1) \rightarrow S_{n}(1)$ is a continuous, open surjection, must it be a homeomorphism? [hide] I want to say that this is false as well for the same reason as e), but how do I justify my guess? [/hide]

a) false - there is e.g. a function s.t. $ f(U)\equal{}\mathbb R$ for each open $ U$
b) you are right
c) true, because the inverse is continuous when $ f$ is open
d) of course false (there are a plenty of extremely simple examples - just think a little bit)
e) in general false, but for $ \mathbb R$ is true. $ f$ must be monotone (prove that on each interval it takes maximum
in one of the ends)
f) what is $ S_n(1)$?

[quote="mstrip"]a) Am I false or is the statement above false?
e)If f is not monotone, then this statement is false right? So I could take $ f(x) \equal{} x^{3}$ and that can serve as my counterexample?
f) $ S_{n}(1)$ is the unit ball.[/quote]
a) you are right
e) $ f(x) \equal{} x^{3}$ IS monotone

[quote="mstrip"]f) If $ f: S_{n}(1) \rightarrow S_{n}(1)$ is a continuous, open surjection, must it be a homeomorphism?[/quote]This is not true, Take $ z\mapsto z^k$ ($ k>1$) on $ S^1$.

Note that in general, if $ f$ is continuous and open, $ f$ is surjective.


geometry
3D geometry
pyramid
trigonometry
AMC

Consider a pyramid $P-ABCD$ whose base $ABCD$ is square and whose vertex $P$ is equidistant from $A,B,C$ and $D$. If $AB = 1$ and $\angle APB = 2 \theta$ then the volume of the pyramid is

$\text{(A)} \ \frac{\sin \theta}{6} \qquad \text{(B)} \ \frac{\cot \theta}{6} \qquad \text{(C)} \ \frac{1}{6 \sin \theta} \qquad \text{(D)} \ \frac{1 - \sin{2 \theta}}{6} \qquad \text{(E)} \ \frac{\sqrt{\cos \ 2 \theta}}{6 \sin \theta}$

It's a bunch of trig!

[hide]Okay so label the midpoint of AB as E.  We know $\angle BPE = \theta$ and $EB = .5$, so $PE = \frac{\cot \theta}{2}$.  Label the center of the square as O.  $EO = .5$, so the height $PO^2 = \frac{\cot^2 \theta}{4} - \frac{1}{4} = \frac{\cot^2 \theta - 1}{4} = \frac{\cos^2 \theta - \sin^2 \theta}{4\sin^2 \theta}.$  Using the identity $\cos^2 \theta - \sin^2 \theta = \cos 2\theta,$ this means the height OP of the pyramid is $\frac{\sqrt{\cos 2\theta}}{2\sin \theta}.$  Because the volume of a pyramid is a third of the product of the area of the base and the height, our answer is $\fbox{E}.$
[/hide]

[hide="Answer"]Call the altitude of $\triangle APB$ $\overline{PH}$.

$\angle APB=2\theta \Rightarrow \angle APH=\theta$
$\tan APH=\tan \theta =\frac{\frac{1}{2}}{PH}\Rightarrow \overline{PH}=\frac{1}{2\tan \theta}$

Call the altitude from $P$ to $ABCD$ $\overline{PE}$. $\overline{HE}=\frac{1}{2}$, so $\overline{PE}=\sqrt{\overline{PH}^2-\left(\frac{1}{2}\right)^2}=\sqrt{\frac{1}{4\tan ^2\theta}-\frac{1}{4}}=\frac{1}{2}\sqrt{\frac{2\cos ^2\theta -1}{\sin ^2\theta}}$. This simplifies to $\frac{\sqrt{\cos 2\theta}}{2\sin \theta}$, so then we multiply that by the area of the base and divide by $\frac{1}{3}$ for the volume of the pyramid to find $\frac{\sqrt{\cos 2\theta}}{6\sin \theta}\Rightarrow \boxed{E}$.[/hide]


Find the minimum value of $2x^{2}+2y^{2}+5z^{2}-2xy-4yz-4x-2{z}+15$ for real numbers x, y, z.

[quote="Ubemaya"][quote="10000th User"][hide]$(x-y)^{2}+(y-2z)^{2}+(x-2)^{2}+(z-1)^{2}+10$[/hide][/quote]

How did you come up with that factorization?[/quote]
There were a lot of squares and I tried using completing the squares and it worked.

[quote="bpms"][quote="sonny"][quote="10000th User"][hide]$(x-y)^{2}+(y-2z)^{2}+(x-2)^{2}+(z-1)^{2}+10$[/hide][/quote]

Well, you didn't answer the question completely but your thinking is right.  :)[/quote]
I think it was a hint :|
[hide="using the hint"]
x=2, y=2, z=1
$10$[/hide][/quote]

Oh oops. Okay, 10 is the correct answer.


Find the sum of the roots of $f(x) = x^{17} - 3x + 1$ to the 17th power.

[hide="hint"]
newton sums
[/hide][/hide]

Is this a trick question or something? Isnt it just 0.

To clarify, let the roots be $r_1,r_2,\ldots,r_{17}$, you need to find $r_1^{17} + r_2^{17} + \ldots r_{17}^{17}$

you actually don't need to know newton's sums...you just need vieta's sums and a little clever insight


AMC
USA(J)MO
USAMO
AMC 8
AMC 12
AMC 12 B
AMC 10

Yeah I came home and checked the mail. I happily saw a package from MAA, thinking it was a complementary problem book free with USAMO participation but to my sadness it was the results from AMC/AIME/USAMO.

In the book they honored the beloved, friend-to-all, Mathew Crawford <A perfect scorer on AMC '95.> Hopefully, I should be among the remembered in the perfect scorers section next year. Surprisingly, in a school where only three people took the test we actually placed among the top 5 schools for team scores in region 9... very chic indeed.

However, since they honored past AMC scorers why not did they mention past USAMOers (At least the top 12 winners)

What did you guys think of the book, or the AMC program overall?

There are probably less than 12 people who get a perfect on the AHSME; not necessarily the AMC 10 or 12 of course.  That's why they have the AHSME perfect scorers remember, I guess.  Good luck with your perfect score goal!

-interesting_move

Heh, had a very similar experience.  Of course, if I hadn't made that one stupid little mistake on the AMC8 (:-x), I woulda gotten "two donor mathematics books" (in the result book, p. 35) for a perfect score...but as it is, I'm pretty happy.  Found out I just have to get around a 144 on the '12 in my region, and I'll get something, be it subscription, plaque, or CAMC book... 8-)

But yeah, a perfect score would be nice.  I have a feeling I could achieve that by finishing the test in 45ish minutes (quite possible) and then checking to make sure I didn't make any stupid mistakes (quite difficult, due to my short attention span).  And the way I'm gonna do that... well, I'll have to spend 1 hour a day on competition math.  Just some ideas.

[quote]I have a feeling I could achieve that by finishing the test in 45ish minutes (quite possible) and then checking to make sure I didn't make any stupid mistakes (quite difficult, due to my short attention span). [/quote]

Hmm, I personally think that itd be easier to spend more time on each problem, but do each problem slowly so that you are sure of its correctness on the first go-around, so you don't have to spend time checking over everything.

Fierytycoon

[quote="Fierytycoon"]Hmm, I personally think that itd be easier to spend more time on each problem, but do each problem slowly so that you are sure of its correctness on the first go-around, so you don't have to spend time checking over everything.[/quote]

I KNOW that doing problems slowly will probably be more accurate.  But for me, over the past two years I've gone [i]way[/i] too slowly, showing all my work - which is fine for harder problems but pointless for really easy ones.  Besides, I was kind of kidding on that last one.  No need to get so worked up...

But seriously, 50 minutes for the entire test is a pretty good strategy for me, because I'm pretty accurate but when I do have a mistake, it takes a while for me to find it.

I think that's the difference between a 144 and a 150. I never got a 150, but not because I couldn't do all the problems. It wasn't even because I couldn't do the problems in time. Rather, I finished quite early, but I would only get 144. The problem is that if you miss one, you don't know which one it is, so you have to do the entire test over again, which is tough to do in time. (I did that on AMC12B in 2002 because the first time it took me only 35 minutes for some reason, but I still got 144.) The people who get 150 I imagine must work slowly and carefully the first time and (my personal failing that kept me from a 150 once) read the question extremely accurately.


number theory
greatest common divisor
algebra
polynomial

Let [tex]a_0, a_1,\cdots,a_n[/tex] and [tex]b_0,b_1,\cdots,b_m[/tex] be integers such that

[tex]\gcd(a_0,a_1,\cdots,a_n)=1[/tex] and [tex]\gcd(b_0,b_1,\cdots,b_m)=1[/tex].

Define [tex]a_i=0[/tex] for [tex]i>n[/tex] and [tex]b_j=0[/tex] for [tex]j>m[/tex] and define

[tex]c_k=a_0b_k+\cdots+a_kb_0[/tex] for [tex]k=0,1,\cdots[/tex]. Show that

[tex]\gcd(c_0,c_1,\cdots,c_{(m+n)})=1[/tex].

Some people might immediately recognize that this is related to ...

[quote="fuzzylogic"]

Some people might immediately recognize that this is related to ...[/quote]

Linear independence?

I think fuzzylogic recently posted another problem about the multiplication of polynomials that is equivalent to this problem about convolution.


1. Find all solutions of equation : $\sum^n_{k=1}a_k^2=n\prod^n_{k=1}a_k$

2. What condition must integers $p$ and $n$ satisfy such that the equation : $\sum^n_{k=1}a_k^2=p\prod^n_{k=1}a_k$ ,
    (a) has no solution ?
    (b) has infinitely many solutions ?

By Power Mean Ineq., $\displaystyle RMS \geq GM \Rightarrow \sqrt{\frac{\sum_{k=1}^n a_k^2}{n}} \geq \sqrt[n]{\prod^n_{k=1}a_k} = \sqrt[n]{P}$. By our equality, we have $\displaystyle RMS = \sqrt{P}$. Assume $P$ is positive because if it is not there is no equality since the sum of squares is always positive. So, by substitution $\displaystyle \sqrt{P} \geq \sqrt[n]{P} \Rightarrow P^{\frac{1}{2}-\frac{1}{n}} \geq 1$ Since $P$ is a product of integers, we know $P \geq 1$ which tells us $P^j \geq 1$ is only true if $j \geq 0$. So, we now have $\displaystyle \frac{1}{2}-\frac{1}{n} \geq 0 \Rightarrow n \geq 2$. Now, when $n \geq 2$ we have $RMS \geq GM$ in which equality holds only when ???? I have to jet, sorry I can't finish right now. Toodles.


logarithms
advanced fields
advanced fields unsolved

Let $ p\ge 2$ be an integer and $ a>0$ is a constant, solve the ordinary differential equation \[ y'\plus{}2y\frac{y^p\minus{}a}{(p\plus{}1)y^p\plus{}(p\minus{}1)a}\equal{}0,\] for $ y\equal{}y(x)$.
 :)

hello, Maple gives the following solution
$ y \left( x \right) \equal{}{{\rm e}^{2\, \left( x\plus{}{\it RootOf} \left( {
{\rm e}^{{\it \_Z}}}\minus{}{{\rm e}^{2\,{\frac { \left( x\plus{}{\it \_Z}\plus{}{\it
\_C1} \right) p}{p\minus{}1}}}}\plus{}a \right) \plus{}{\it \_C1} \right)  \left( p\minus{}1
 \right) ^{\minus{}1}}}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, Maple gives the following solution
$ y \left( x \right) \equal{} {{\rm e}^{2\, \left( x \plus{} {\it RootOf} \left( { {\rm e}^{{\it \_Z}}} \minus{} {{\rm e}^{2\,{\frac {\left( x \plus{} {\it \_Z} \plus{} {\it \_C1} \right) p}{p \minus{} 1}}}} \plus{} a \right) \plus{} {\it \_C1} \right) \left( p \minus{} 1 \right) ^{ \minus{} 1}}}$
Sonnhard.[/quote]

The form is complicated. I forgot to put the initial condtition $ f(0)\equal{}1$.  :blush:

hello, with this initial condition we get
$ y \left( x \right) \equal{}{\it RootOf} \left( 2\,xp\plus{}2\,\ln  \left( {{\it \_Z
}}^{p}\minus{}a \right) p\minus{}\ln  \left( {{\it \_Z}}^{p} \right) p\plus{}\ln  \left( {
{\it \_Z}}^{p} \right) \minus{}2\,\ln  \left( 1\minus{}a \right) p \right)$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, with this initial condition we get
$ y \left( x \right) \equal{} {\it RootOf} \left( 2\,xp \plus{} 2\,\ln \left( {{\it \_Z }}^{p} \minus{} a \right) p \minus{} \ln \left( {{\it \_Z}}^{p} \right) p \plus{} \ln \left( { {\it \_Z}}^{p} \right) \minus{} 2\,\ln \left( 1 \minus{} a \right) p \right)$
Sonnhard.[/quote]
What is $ \it \_Z$ in the expression?


inequalities
analytic geometry
graphing lines
slope
vector
trigonometry

[quote="kunny"]You can use Cauchy Schwarz inequality and triangular function.[/quote]

actually to be honest with you

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=418742#p418742

Its kinda of the same problem.  However, I didnt know how to do that problem, so I clicked on the solutions.  The only solution posted was cauchy schwartz.

However, I have no idea what that is and I know there is an analytic solution so when I posted this I was honesty looking for an analytic solution (beucase there wasnt one posted). The problem is almost like (the one posted now) that old amc problem.

wtf..this better be right..took me like 10 mins with the heavy computation..

$ \frac{1722+287\sqrt{41}}{41}$, slopes should do the trick.

I have found my post, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29545]here is[/url].
Anyway, Here is my solution.

Assumed that $ 4x+5y$ takes some value $ k$ for $ x,\ y$ satisfying $ (x-3)^{2}+(y-6)^{2}=49$,
The circle has at least one intersection point with the line $ 4x+5y=k$ ,that is to say, the distance of the center of the circle $ (3,\ 6)$ and the line

$ 4x+5y-k=0\leq \text{the radius 7}\Longleftrightarrow \frac{|4\cdot 3+5\cdot 6-k|}{\sqrt{4^{2}+5^{2}}}\leq 7$

$ \Longleftrightarrow |42-k|\leq7\sqrt{41}$, yielding $ 42-7\sqrt{41}\leq k\leq 42+7\sqrt{41}$.Thus the answer is $ 42+7\sqrt{41}.$ The rest duty we should do is to find $ x,\ y$ that gives the maximum value. This is left for you.

By the way, can you use the inner product of vector? If so, you can find the value very easily, especially for $ x,\ y$.

[quote="vikas"][quote="t0rajir0u"][quote="vikas"]The only solution posted was cauchy schwartz.

However, I have no idea what that is[/quote]

Then you should learn.  It's much easier than the analytic solution.  In two variables, the Cauchy-Schwarz inequality states that

$ (a^{2}+b^{2})(c^{2}+d^{2}) \ge (ac+bd)^{2}$

In two variables, it's exceedingly easy to prove thanks to the identity

$ (a^{2}+b^{2})(c^{2}+d^{2}) = (ac+bd)^{2}+(ad-bc)^{2}$

[hide="How can we apply it to this problem?"] $ ( (x-3)^{2}+(y-6)^{2})( 4^{2}+5^{2}) \ge (4(x-3)+5(y-6))^{2}\Leftrightarrow$
$ 49 \cdot 41 \ge ( (4x+5y)-42 )^{2}\implies$
$ 7 \sqrt{41}\ge (4x+5y)-42 \Leftrightarrow$
$ 42+7 \sqrt{41}\ge 4x+5y$

The equality case of Cauchy-Schwarz in two variables occurs when $ ad = bc \Leftrightarrow \frac{a}{b}= \frac{c}{d}$, that is, the vectors $ \left< a, b \right>, \left< c, d \right>$ are in the same direction.  This is consistent with the interpretation of Cauchy-Schwarz in terms of the dot product.  [/hide][/quote]

Thank you very much.  However, just one more question.  If it asked for the minimum value, would you then have to use analytic because I dont see how cauchy would work then.[/quote]

Not so.  Cauchy puts an upper bound not on $ 4x+5y$, but on $ (4x+5y-42)^{2}$.  I took the positive square root to achieve an upper bound; instead I can bound the negative square root and attain the result

$ 7 \sqrt{41}\ge 42-(4x+5y) \Leftrightarrow$
$ 4x+5y \ge 42-7 \sqrt{41}$

In agreement with kunny's result.

Thank you for clarification my answer.
The dot product's solution give us easy solution, isn't it? :lol:

[quote="vikas"]If an equation of a circle is $ (x-3)^{2}+(y-6)^{2}=49$, then what is the maximum value of $ 4x+5y$ where $ (x,y)$ is a point on the circle.[/quote]

I use the parametric representation of any point on the given circle.
Let $ x = 3+7\cos \theta ,\,y = 6+7\sin \theta$.

This gives $ 4x+5y = 42+7(4\cos \theta+5\sin \theta )$  whose maximum value, from basic trigonometry, is $ 42+7\sqrt{41}$

Um guys >_>

We know that $ 4x+5y$ is maximized at the point where a line $ 4x+5y=k$ is tangent to the circle, so all we need to do is find the point where $ 4y=5x$ intersects $ x^{2}+y^{2}=49$ and translate it 3 units right and 6 units up. A little bit of computation later, we get:

$ (\frac{28}{\sqrt{41}}+3,\frac{35}{\sqrt{41}}+6)$ as this point.

$ 4(\frac{28}{\sqrt{41}}+3)+5(\frac{35}{\sqrt{41}}+6) = \frac{287}{\sqrt{41}}+42 = 7\sqrt{41}+42$


Ψάχνω παίδες σημειώσεις λογισμόυ 1  για το πρώτο εξάμηνο
του μαθηματικού..
κανένας συνάδελφος μπορεί να με βοηθήσει?

[quote="Yperbatikos_e"]να μου πει πως ονομαζονται οι Δυναμοσειρές στα αγγλικα?[/quote]

Power series

[quote="Yperbatikos_e"]να μου πει πως ονομαζονται οι Δυναμοσειρές στα αγγλικα?[/quote]

Για την ανάποδη κατεύθυνση, μαθηματικούς όρους από τα αγγλικά στα ελληνικά,
ο συγγραφέας του δίνει δωρεάν από την ιστοσελίδα του
το εξής Λεξικό:

http://www.ucy.ac.cy/~georgios/bookfiles/dict1.pdf

Είναι έγκυρο και το συνιστώ.

Υπάρχει και συμπιεσμένο στο

http://www.thegreekz.com/forum/showthread.php?t=304404


calculus
inequalities unsolved
inequalities

Let a,b,c be real numbers such that
$ a \ge 2,b \ge 9,c \ge 1945,a \plus{} b \plus{} c \equal{} 2000$
Find the maximum of
$ A \equal{} abc$

hello, with the help of calculus we get
$ abc\le\frac{5069349}{2}$
and will be attained for $ a\equal{}\frac{51}{2}\equal{}b,c\equal{}1949$.
Sonnhard.


calculus
ratio
logarithms
geometric sequence

[quote="rob22"]I stumbled across this problem, and know the answer, but wasn't quite sure how you would determine what $ \displaystyle\prod_{x \equal{} 1}^{\infty} \frac { (4x \minus{} 2)^2 } { (4x \minus{} 3) (4x \minus{} 1) }$ is equal to.[/quote]

[hide]$ \frac{16x^2\minus{}16x\plus{}4}{16x^2\minus{}16x\plus{}3}\equal{}1\plus{}\frac{1}{(4x\minus{}3)(4x\minus{}1)}\equal{}1\plus{}\frac{\minus{}1/2}{4x\minus{}1}\plus{}\frac{1/2}{4x\minus{}3}$

Then it's telescoping sums.[/hide]

It's an infinite product and not an infinite sum.

[quote="jli"][quote="rob22"]I stumbled across this problem, and know the answer, but wasn't quite sure how you would determine what $ \displaystyle\prod_{x \equal{} 1}^{\infty} \frac { (4x \minus{} 2)^2 } { (4x \minus{} 3) (4x \minus{} 1) }$ is equal to.[/quote]

[hide]$ \frac {16x^2 \minus{} 16x \plus{} 4}{16x^2 \minus{} 16x \plus{} 3} \equal{} 1 \plus{} \frac {1}{(4x \minus{} 3)(4x \minus{} 1)} \equal{} 1 \plus{} \frac { \minus{} 1/2}{4x \minus{} 1} \plus{} \frac {1/2}{4x \minus{} 3}$

Then it's telescoping sums.[/hide][/quote]
I don't think it telescopes at all beside the fact that the expession is an infinite product...

Well, we could try some values for x and find what the resulting fraction is. Who knows? It might just telescope.

That does not follow in general and is wrong in this case, too: just as a trivial example, consider the product $ \prod_{n \geq 1} \frac {2 \minus{} \frac {1}{n \plus{} 1}}{2 \minus{} \frac {1}{n}}$.  Every term is bigger than 1, but it certainly converges.

Notice that every term is greater than one, yet multiplying them one by one surely does not guarantee divergence . . . . because each succeeding term then approaches 1.
Product = (4/3)(36/35)(100/99)(196/195)(324/323) . . . . .
I think it will converge to a certain value as [b]JBL[/b] pointed out.

For a product of the form $ \prod_{n \geq 1} (1 \plus{} a_n)$ with $ a_n > 0$, we have convergence if and only if $ \sum_{n \geq 1} a_n$ converges.  (The condition $ a_n > 0$ can be weakened substantially: see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204755]this thread[/url] for much more discussion.)  Since $ \frac{(4n \minus{} 2)^2}{(4n \minus{} 1)(4n \minus{} 3)} \equal{} 1 \plus{} \frac{1}{(4n \minus{} 1)(4n \minus{} 3)}$ and $ \sum_{n \geq 1} \frac{1}{(4n \minus{} 1)(4n \minus{} 3)}$ converges, the product converges.  But this tells us very little about its actual value.

The partial product $ \prod_{x \equal{} 1}^n \frac{(4x \minus{} 2)^2}{(4x \minus{} 1)(4x \minus{} 3)}$ can be rewritten as $ \frac{2^{2n} ((2n \minus{} 1)!!)^2}{(4n \minus{}1)!!}$ where $ (2m \minus{} 1)!! \equal{} (2m \minus{} 1)(2m \minus{} 3)(2m \minus{} 5)\cdots 3\cdot 1$.  Note that $ (2m \minus{} 1)!! \equal{} \frac{(2m)!}{2^m m!}$, so we can further rewrite this as
\[ \frac{2^{2n} \cdot \left(\frac{(2n)!}{2^n n!}\right)^2}{\frac{(4n)!}{2^{2n} (2n !)}} \equal{} \frac{2^{2n} \cdot (2n)!^3}{(4n)! n!^2}.\]
From here, the "natural" thing to do is use Stirling's approximation (by the way, this thread should be moved to Calculus Computations and Tutorials) to get that the given expression is asymptotic to
\[ \frac{2^{2n} \cdot \frac{(2n)^{6n}}{e^{6n}} (4 \pi n)^{3/2}}{\frac{(4n)^{4n}}{e^{4n}} \cdot \frac{n^{2n}}{e^{2n}} \cdot \sqrt{8 \pi n} \cdot 2\pi n}.\]

Simplify this (eyeballing it, it looks like it works out to $ \sqrt{2}$?) and you have your answer.

[quote]$ \prod_{x \equal{} 1}^{\infty}\frac { (4x \minus{} 2)^{2}}{ (4x \minus{} 3) (4x \minus{} 1) }$[/quote]
Since
$ sin( \frac {\pi}{4}) \equal{} \frac {\sqrt {2}}{2}$

Perhaps the power series representation of $ sin(x)$ could be manipulated to yield the product for which we are trying to solve.

[quote]
There is no product. The numerator is always greater than the denominator so the product will be infinitely large[/quote]

Perhaps you are thinking of the sum of a geometric sequence, where $ \left | r \right |< 1$. But, in this case, each succeeding term becomes closer and closer to one...logically, the series should converge. Only if the ratio of terms $ \frac{{a_{n\plus{}1}}}{a_{n}}> 1$ would the series not converge (as you would be multiplying by a bigger and bigger number).

[quote="RoFlLoLcOpT"]There is no product. The numerator is always greater than the denominator so the product will be infinitely large.[/quote]

Let $ a_1, \; a_2, \; \ldots$ be a sequence of positive numbers asymptotically approaching $ 1$. Suppose I want the value of $ P \equal{} \prod_{k\in \mathbb{N}} a_k$.

Clearly, $ P$ is not $ 0$ (well, ok, this isn't [i]that[/i] trivial, but convince yourself it's true) and $ P$ is not negative. Thus, if $ P$ exists, it must be positive, and thus must have a logarithm.

$ \ln P \equal{} \ln \left( \prod_{k\in \mathbb{N}} a_k\right) \equal{} \sum_{k \in \mathbb{N}} \ln a_k$

So, it depends on if the series $ \ln a_k$ converges. I can think of an example in which it does:

$ a_k \equal{} e^{1/2^k}$
$ \ln a_k \equal{} 1/2^k$
$ \ln P \equal{} \sum _{k \in \mathbb{N}} 1/2^k \equal{} 1$
$ P \equal{} e$


linear algebra
matrix
algebra
polynomial
number theory
greatest common divisor
relatively prime

$A\in M_n(N)$ such that $m\in N-\{0;1\}$ with $gcd(m;det(A))=1$

¨prove that $A^2+A+mI_n$ is invertible.

If $\alpha,\bar\alpha$ are the roots of $x^2+x+m$, then it's clear that they are complex, so if $P(\alpha)P(\bar\alpha)=0$ (where $P$ is the characteristic polynomail of $A$), then $P(\alpha)=P(\bar\alpha)=0$, meaning that $x^2+x+m$ is the characteristic polynomial of $A$, or, in other words, $A^2+A+mI_n=O_n$, which is impossible because of the condition $(m,\det A)=1$.

[quote="grobber"] meaning that $ x^2 \plus{} x \plus{} m$ is the characteristic polynomial of $ A$[/quote]

impossible the characteristic polynomial is $ det(A \minus{} xI_n)$ polynomial of degree the

size of the matrix $ A$ , $ n$ in the statement n can be >2

:) I don't know why I was thinking about matrices of order $2$. Terribly sorry.  :blush:

Some remarks

 1/We should say $A\in M_n(Z)$ with positive entries
not $A\in M_n(N)$ for ring properties

2/ Since $gcd(det(A),m)=1$ with Bezout theorem there exist $x,y$ integers such that

(1) $mx+ydet(A)=1$

If $det(A)=0$ the above equation becomes $mx=1$ no solution in integers since $m>1$

So $det(A)\neq 0$,  $A$ is invertible matrix in $M_n(Z)$ it is known that

$det(A)=1 \; or \; -1$

Let $p$ be the characteristic polynomial of $A$; $p$ has integer coefficients, and its constant term is relatively prime to $m$. The gcd of $p(t)$ and $t^2+t+m$, which is an integer polynomial, must have constant term $\pm1.$ If $|m|>2$, this only happens when the gcd is 1, which makes $A^2+A+mI$ invertible. We must exclude $m=2$ and $m=-2$ as well as $-1,0,1$ because they allow roots of $\pm1$ as well.

the equation $x^2+x+m=0$ has complex roots. Supposing that $(P_A(X), x^2+x+m)\ne 1$ then a root of $x^2+x+m=0$  is also a root of $P_A$ but if $\alpha\in\mathbb C$ is a root of $P_A$ then $\overline{\alpha}$ is also a root of $P_A$. If $\alpha,\overline\alpha$ are the roots of $x^2+x+m=0$ then thier product is $m$ and because they are roots of $P_A$ the following relation holds
$x_1\ldots x_{n-2}\alpha\overline\alpha=\det A$ (where $x_i$ are the other eigenvalues of $A$) which is equivalent to $km=\det A, k\in\mathbb N$. This is a contradiction with the fact that $(m,\det A)=1$. So the matrix $A^2+A+mI_n$ is invertible.


geometry
circumcircle
geometric transformation
homothety
search
trigonometry
power of a point

In $ \triangle ABC$ let $ AM (M \in BC)$ be a median and let $ CC_1$ $ (C_1 \in AB)$ and $ BB_1 (B_1 \in AC)$ be two altitudes. The line through $ A$ perpendicular to $ AM$ intersects lines $ BB_1$ and $ CC_1$ in points $ E$ and $ F$, respectively. Denote by $ k$ the circumcircle of $ \triangle EFM$. Let $ k_1$ and $ k_2$ be any two circles tangent to both $ EF$ and to
the arc $ EF$ on $ K$ not containing $ M$. If $ P$ and $ Q$ are the intersection points of $ k_1$ and $ k_2$, prove that points $ P, Q$, and $ M$ lie on a line.

[hide="Solution"]
[b]Lemma 1:[/b] $ PQ$ passes through the midpoint of major arc $ EF$ of the circumcircle of $ \triangle EMF$.

[i]Proof: [/i]Let the $ X$ be the tangency point of $ k_1$ and with the circumcircle of $ \triangle EMF$ and $ Y$ be that between $ k_1$ and $ EF$.  Then, the homothety that takes $ k_1$ to the circumcircle of $ \triangle EMF$ takes $ EF$ to a line that is tangent to the circle and is parallel to $ EF$.  Thus, this line passes through the midpoint of major arc $ EF$.  Hence, $ XY$ passes through the midpoint of this arc, say $ N$.  Now, we have that $ \angle FXN \equal{} \angle FEN \equal{} \angle YFN$, so $ \triangle NYF\sim \triangle NFX$, meaning that $ NY\cdot NX \equal{} NF^2$.  Similarly, the if the tangency points between $ k_2$ and the circumcircle of $ \triangle EFM$ and $ EF$ are $ T$ and $ U$ respectively, then $ N$, $ T$, and $ U$ are collinear and $ NT\cdot NU \equal{} NE^2 \equal{} NF^2 \equal{} NY\cdot NX$.  It follows that the powers of $ N$ with respect to $ k_1$ and $ k_2$ are equal to $ NF^2$, meaning that $ N$ is on the radical axis of $ k_1$ and $ k_2$.  Therefore, $ PQ$ passes through $ N$.

[b]Lemma 2:[/b] $ FA \equal{} EA$.

[b]Sublemma (well-known):[/b] Given a cyclic quadrilateral $ ABCD$ with $ AB$ and $ CD$ meeting at $ R$, $ AD$ and $ BC$ meeting at $ Q$, $ O$ being its circumcenter, and $ P$ being the intersection between $ AC$ and $ BD$, we have that $ RP\perp QO$.

[i]Proof of Lemma 2:[/i] Letting the orthocenter of $ \triangle ABC$ be $ H$, $ AH$ meet $ BC$ at $ A_1$, and the intersection between $ B_1C_1$ and $ BC$ be $ R$, the sublemma gives that $ RH\perp AM$ when applied to cyclic $ B_1C_1CB$.  Then, $ FE\parallel RH$.  Now, we have that $ (R, B, A_1, C) \equal{} \minus{} 1$, meaning that $ H(R, B, A_1, C) \equal{} \minus{} 1$, so the intersections between the mentioned harmonic bundle and $ EF$ form a harmonic division.  Thus, $ (\infty, E, A, F) \equal{} \minus{} 1$, meaning that $ FA \equal{} AE$, establishing Lemma 2.

[i]The Finish: [/i]Since $ FA \equal{} AE$ and $ MA\perp EF$, we have that $ MF \equal{} ME$.  Thus, $ M$ is the midpoint of the major arc $ EF$ of the circumcircle of $ \triangle MEF$.  Hence, $ M \equal{} N$, so $ P$, $ Q$, and $ M$ are collinear. [/hide]

[quote="ishfaq420haque"]can you prove the sub lemma in another way???i mean is there any other proof? :([/quote]
You can find the proof of the sublemma here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1826082895&t=219812

If you want another proof of Lemma 2 (perhaps without using harmonic divisions), then here:
[hide="Another Proof"]
We extend $ AM$ to meet the circumcircle of $ \triangle ABC$ at $ X$, the $ A$-symmedian to meet it at $ Y$, let the foot of the perpendicular from $ F$ to $ EB$ be $ K$, and let the foot of the perpendicular from $ E$ to $ FC$ be $ L$.  Now, since $ AB$ and $ EL$ are both perpendicular to $ CF$, we have that they are parallel.  Thus, $ \frac {FA}{FE} \equal{} \frac {AC_1}{EL}$ and, similarly, $ \frac {EA}{EF} \equal{} \frac {AB_1}{FK}$.  Now, for $ FA \equal{} AE$, we need $ \frac {AC_1}{EL} \equal{} \frac {AB_1}{FK}$.  This means that we need for $ \frac {AC_1}{AB_1} \equal{} \frac {EL}{FK}$.  Yet, $ BC_1B_1C$ is cyclic, so $ \triangle AC_1B_1\sim \triangle ACB$, meaning that $ \frac {AC_1}{AB_1} \equal{} \frac {AC}{AB}$.  Thus, we wish for $ \frac {AC}{AB} \equal{} \frac {EL}{FK}$.  We also have that $ \angle FKE \equal{} \angle FLE \equal{} 90$, meaning that $ FKLE$ is cyclic.  Thus, $ \frac {EL}{FK} \equal{} \frac {EH}{FH}$.  Furthermore, since $ \angle AC_1F \equal{} 90 \equal{} \angle MAF$, we have that $ \angle HFA \equal{} \angle XAB \equal{} \angle XCB$.  Similarly, $ \angle HEF \equal{} \angle XBC$, establishing that $ \triangle HFE\sim \triangle XCB$, so $ \frac {EH}{FH} \equal{} \frac {BX}{XC}$.  It suffices to show that $ \frac {BX}{XC} \equal{} \frac {AC}{AB}$.  Since $ \angle YAB \equal{} \angle XAC$, we have that $ BY \equal{} XC$, so $ BYXC$ is an isosceles trapezoid.  Then, $ BX \equal{} YC$.  Noting that $ \angle AYB \equal{} \angle ACM$, we have that $ \triangle ACM\sim \triangle AYB$.  Thus, $ \frac {BY}{MC} \equal{} \frac {AY}{AC}$.  Similarly, $ \frac {YC}{BM} \equal{} \frac {AY}{AB}$.  Dividing gives that $ \frac {AB}{AC} \equal{} \frac {BY}{YC} \equal{} \frac {CX}{XB}$, as desired. [/hide]

[b]Quatto[/b], could you show why CD*BD=DM*DP in April's proof of the sublemma?

Sorry for the confusion; [b]April[/b] does not prove the sublemma.  In the link I sent, I used the sublemma to solve the problem.  I prove the sublemma in post $ 7$ of the link.  If  you want the proof of $ CD\cdot  BD\equal{}DM\cdot DP$, then it is a property of harmonic divisions.  Say that $ (A, B, C, D)\equal{}\minus{}1$ is a harmonic division.  Letting $ M$ be the midpoint of $ BC$, we would have that $ D(R, B, M, C)\equal{}\minus{}1$ is harmonic, impling that $ \frac{DR}{DB}\equal{}\frac{DC}{DM}$.  Here is the proof of the property.  Letting $ RB\equal{}c$, $ BD\equal{}a$, and $ DC\equal{}b$, we get that $ MC\equal{}\frac{a\plus{}b}{2}$, so $ DM\equal{}\frac{b\minus{}a}{2}$.  This means that
\[ \frac{DR}{DB}\equal{}\frac{DC}{DM}\iff \frac{a\plus{}c}{a}\equal{}\frac{b}{\frac{b\minus{}a}{2}}\iff 2ab\equal{}ab\minus{}a^2\plus{}bc\minus{}ac\iff\]\[ ab\plus{}a^2\plus{}ac\minus{}bc\equal{}0\iff a(a\plus{}b\plus{}c)\equal{}bc \iff \frac{c}{c\plus{}a\plus{}b}\equal{}\frac{a}{b}\iff\]
\[ \frac{RB}{RC}\equal{}\frac{DB}{DC}\]
which follows from the fact that $ (R, B, D, C)\equal{}\minus{}1$.

My approach seems to be quite different from the one proposed by [b]The QuattoMaster 6000[/b].
[b]Solution:[/b]
Consider the circumcircle $ \Pi$ with diameter on $ BC$. Denote the intersection point of $ BC$ and $ B_1C_1$ by $ X$ and let $ H$ be the orthocenter of $ \triangle ABC$. Clearly $ M$ is the circumcenter of $ \Pi$. Therefore, $ AM$ is the polar of $ X$ and it means that $ HX\perp AM$, or $ HX\| EF$.
Consider the harmonic four $ \left(XY,BC\right)$ and the pencil of this division through point $ H$ and its projection on $ EF$.
Since $ HX\| EF$ it follows that $ \boxed{AF \equal{} AE}$ and therefore [i]$ M$ is a midpoint of arc $ EF$[/i].

Clearly for either of $ k_1,k_2$, theirs tangent points with $ EF$ and $ k$ passes through $ M$. So after inversion centered at $ M$, points $ P$ and $ Q$ will turn into themselves.  It precisely means that $ P$,$ Q$ and $ M$ are collinear.

Alternate proof of Lemma 2:
Obvious, B and C are equally apart of AM and, if note $ \angle BAM \equal{} \angle AFC_1\equal{}  A_1$ and $ \angle CAM \equal{} \angle AEB_1\equal{} A_2$, then
$ c.sin(A_1) \equal{} b.sin(A_2)$ and from the triangles $ AFC_1$ and $ AEB_1$ by calculation we get AE = AF. Or, even better, use two pairs of similar triangles, avoid trig, but same idea.

Best regards,
sunken rock