analytic geometry
algebra
polynomial
algebra unsolved

[u][b]Problem1: [/b][/u]Is there any quadrilateral whose tangents of the interior angles are equal to each other?

 [u][b]Problem2:[/b][/u] Given two points whose coordinates are integer on the graph of a polynomial with integer coefficient. Prove that if the distance between these points is an integer then the line connecting the given points is parallel to x-axis.

[u][b] Problem3:[/b][/u] Drawn the squares ABB_1A_2, BCC_1B_2 and CAA_1C_2 on the outside of triangle ABC. If the squares A_1A_2A_3A_4 and B_1B_2B_3B_4 are drawn on the line segments A_1A_2 and B_1B_2 (outside of the triangles AA_1A_2 and BB_1B_2) then prove that A_3B_4 is parallel to AB.
                                                                                fatih sahin

[u][b] Problem4:[/b][/u] Given the sequence a_n = 1 + 2^n + 3^n  + 4^n + 5^n. Is there any 5-consequtive terms is divisible by 2005.

[u][b]Problem5:[/b][/u] Given 200 points in the space. Each two point is connected by the line segments (not intersecting each other). For k-different colors, 1st player paints one of the line segments then 2nd player paints each point with the same color. If there is two points and line segments between these points painted with the same color then 1st player wins otherwise 2nd player wins. For these conditions,  prove that 1st player can absolutely win;
              a) if k=7
              b) k=10

fatih sahin

Solution4: Assume it's possible for n=k, k+1 , k+2, k+3, k+4... then
        2005 divides the 5 consecutive terms  a_k, a_(k+1), a_(k+2),a_(k+3),a_(k+4)...
Hence,
        2005 divides also the sum S = a_k + a_(k+1) +  a_(k+2) + a_(k+3) + a_(k+4)
And,
                       S = 5+ 31. 2^k +121. 3^k +341. 4^k +781. 5^k

The last digits of the powers of 2,3,4,5:
                 2 = 2                  3=3             4=4                 5=5                 (mod 10)
             2^2 = 4              3^2=9         4^2=6             5^2=5                 (mod 10)
             2^3 = 8              3^3=7         4^3=4             5^3=5                 (mod 10)
             2^4 = 6              3^4=1         4^4=6             5^4=5                 (mod 10)

So it's clear that the sum S isn't a multiple of 5...
Hence,
      2005 cannot divide S and at least, one of a_k, a_(k+1), a_(k+2),a_(k+3),a_(k+4) cannot be divisible by 2005...

fatih sahin

[b][u]Problem6:[/u][/b] Given two sets A={a_1,...,a_n}  and B={b_1,...,b_n} of rectangular parallelepipeds. It is known that it is possible to place all parallelepipeds from A into rectangular box C; and for all  i=1,2,...,n one of the sides of b_i  is smaller than in a_i, and two remaining sizes are the same as in a_i. Is it true that we can reduce one of the sides of C in such a way that it will be possible to place all parallelepiped from B into this reduced box?

solution3: Since AB = BB_1, BC = BB_2  and < B1BB2 = pi - <ABC, so |ABC|=|BB1B2| .
 Analogically, |B1A2A3| = |AA1A2| = |ABC|=|BB1B2| =|B1A2B4|.Hence  A3B4|| A2B1||AB.

Take the coordinates (x1, p(x1)) and (x2, p(x2)), where x1 ≠ x2 . The distance between the coordinates is:
[(x1-x2)^2 +[p(x1)-p(x2)]^2]^0.5 = |x1-x2| . [1+[( p(x1)-p(x2))/ (x1-x2)]^2]^0.5

Since x1 ^n –x2 ^n is divisible by x1 − x2, where n is a natural number.
a= [p(x1)-p(x2)]/ (x1-x2) is integer (since the coefficients of the polynomial p(x) are integers)
(1+a^2)^0.5 is an integer for only  a=0 then  p(x1) = p(x2), it verifies the given condition.


Ν. δ. ο υπάρχουν απειρες καμπυλες $ C_l: (x \plus{} y)^3 \equal{} 3xy(2y \plus{} x) \plus{} l$ οπου $ l$  θετικος ακεραιος που εχουν στην γραφικη τους παρασταση τρια τουλαχιστον σημεια $ A,B,C$ με ακεραιες συντεταγμενες και το βαρυκεντρο του τριγωνου $ ABC$ να ειναι η αρχη των αξονων.

Επιλέγουμε αριθμούς της μορφής $ l \equal{} k^3$ οι οποίοι προφανώς είναι άπειροι σε πλήθος
 Τότε

$ C_k: \ x^3 \plus{} y^3 \minus{} 3xy^2 \equal{} k^3$

Τα σημεία $ (x,y) \equal{} (0,k),\ ( \minus{} k, \minus{} k),\ (k,0)$ είναι τα ζητούμενα σημεία A, B, C   :)

Ωραια Sil.
Ας αποδειξουμε και την παρακατω .
Αν η $ C_l$ για συγκεκριμενο θετικο ακεραιο $ l$ εχει ενα σημειο με ακεραιες συντεταγμενες τοτε να αποδειξουμε παλι οτι υπαρχει τετοιο τριγωνο

Την γράφουμε πάλι ως  $ x^3 \minus{} 3xy^2 \plus{} y^3 \equal{} l$  (1)

Τότε παρατηρούμε ότι αν $ (x_0,y_0)$ είναι μία λύση της (1) τότε οι $ (y_0 \minus{} x_0, \minus{} x_0)$ και $ ( \minus{} y_0,x_0 \minus{} y_0)$ είναι λύσεις
και ικανοποιούν το ζητούμενο.

Τώρα θέτω  το εξής ερώτημα για το οποίο έχω φτάσει σε ένα ικανοποιητικό σημείο αλλά δεν το έχω τελειώσει .
Για ποια $ l$ έχει η εξίσωση (1) λύσεις στους  ακεραίους ?

Βρείτε μια κλειστή μορφή για αυτές τις λύσεις  :wink:


[hide]well i imagine this is the system

$3^{x+y}=81$
$81^{x-y}=3$

so...if you apply a certain rule that i don't remember the name of then you get

$3^{x}+3^{y}=81$
$81^{x}-81^{y}=3$

...then im stuck  :blush:  lol well the systems i've solved never actually involved exponents of any degree...

hmm...do you exponentiate both sides or something????[/hide]

Hi 7h3,

Yes, that's the system.

Thanks though for trying! Any further help would be appreciated!

[quote="catalan"]Hello everyone,

Could you please tell me how I could solve this system, without solving it for me?

3^(x + y) = 81

81^(x – y) =3

Thank you very much in advance!

Patrick[/quote]

[hide="first step"]so 3^4=81, so x+y=4.

81^(1/4)=3, so x-y=1/4.[/hide]

[hide="second step"]x+y+x-y=2x=17/4
x=17/8[/hide]

[hide="solution"]y=4-17/8=15/8[/hide]

[quote="catalan"]Hello everyone,

Could you please tell me how I could solve this system, without solving it for me?

3^(x + y) = 81

81^(x – y) =3

Thank you very much in advance!

Patrick[/quote]
[hide]$3^{x+y}=81$ and $81^{x-y}=3$

We know $3^{4}=81,$ so $3^{x+y}=3^{4},$ and $x+y=4.$

Also, $81^{\frac{1}{4}}=3,$ so $81^{x-y}=81^{\frac{1}{4},}$ and $x-y=\frac{1}{4}.$

Now, we solve for $x$ and $y.$

$\left\{\frac{17}{8},\,\frac{15}{8}\right\}$[/hide]


[quote="nguoivn"]Given $ a, b, c > 0.$ Prove that:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {21(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \geq\ 10$
It's well-known as a hard problem which not easy to find a nice proof.
But now, I can say it's very easy.
My hint: Using Am-Gm, we have: $ 7.017 > 7$
Please think about it  :)
Besides, from it, we are easily to see that:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq\ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}$
This ineq is also well-known.
See also here for them:
http://www.mathlinks.ro/Forum/viewtopic.php?t=231645[/quote]
It's just my old result, nguoivn. ;) In 2 years ago, I have proved that
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{10(a^2\plus{}b^2\plus{}c^2)\minus{}ab\minus{}bc\minus{}ca}{(a\plus{}b\plus{}c)^2}.\]
;)

[quote="nguoivn"]I saw your proof in this link:
http://www.mathlinks.ro/Forum/viewtopic.php?t=231645
It's longer than mine  :wink:[/quote]
Of course, I have also another proof with SS, which is very short, but I dont like SS too much. That's why I only post the solution by SOS. ;)

P/s: I think your AM-GM solution is to use AM-GM after rewriting the inequality into SS form. Am I right? ;)

Yes, you are right. I think we have the same hint  :oops:


calculus videos blogs HMMT derivative function inequalities

As time goes by, I notice that I start to forget more and more of what I've learned years ago. It is mostly because I don't to related problems to, e.g., ODEs, hence I forget how to solve some type of differential equations.

So how do you guys keep refreshing your knowledge in previous subjects? Is it mostly by reading here at the forum (various problems at different forums) or do you have some other neat tricks?

Drinking lots of green tea is supposed to help. :wink: Maybe I don't drink enough because I had to read my calculus textbook to recall how to solve 1st order linear ODEs.

It's best if you reviewed a little. Don't wait til the last minute to cram something in for a final. Take a little bit of your time and just briefly look over your notes.

The sad truth is that whatever is not being constantly used tends to get forgotten. I hardly remember anything from advanced courses in algebra or the same differential equations I took when I was a student. The bright side is that once you learned something for the first time (even if it happened  long ago), the time needed to "relearn" it for the second time if you wish to refresh your knowledge is considerably shorter than the time you had to spend learning it in the first place.

As to "neat tricks", I know none :(. Whenever I have to refresh my knowledge of something, I have to go back to my student textbooks and notes (thank God, I was clever enough to make reasonably good notes of most things I was taught) and to reread them. Another thing that helps a lot is teaching high level courses (especially, so called "topics courses", where you are essentially free to choose any subject). They really make you remember things at least as far as the basics are concerned.

As to the Forum, it is always a good idea to spend as much time solving assorted mathematical problems as possible. I don't know if that it facilitates remembering particular things, but, at least, it keeps your brain in a good shape.

Oh, yes, the last advice: breath clean air, eat nutritious food, and do enough physical exercise :P

I am not old enough to say that I am beginning to forget things that I have learned a [i]long[/i] time ago.  However, if you are trying to remember mathematical concepts, then I think the best way (at least for me) is to develop (from memory) some parts of the theory from time to time.  Sometimes on your free time, you can think about you have previously learned and try to refresh your memory.  Of course, eating nutritious food, doing physical exercises (this is suggested, but I cannot provide deeper insight), and drinking green tea could help you.

well..
i don't know..
like in korea, i had to learn chinese characters, but i forgot them all...
i would say read books on subjects that you learned a long time ago..
for me, i took algebra last year, so it's pretty sharp, but i always read algebra books once in a while...
i know there's a video set that explains mathematics for older people who forgot basic math (as in algebra).

I tape soliloquies that I've memorized to my wall (there aren't very many, so this is feasible). When I look up, I review a few words.

Perhaps, write model problems (ones that involve many concepts) on notecards and hide them in places? For example, among your socks, between cans of food, in obscure compartments of your lunchbox, between books, etc. You review once when making the problems, and another time when doing them.

This is a great blog by one of most knowledgeable brain experts. He specializes in memory and learning. There are some great past articles.

[url=http://thankyoubrain.blogspot.com/]Thank you brain[/url]

With the Algebra involved in nearly every single one of my units, its hard to avoid it. I forgot how to find the variable for numerators, denominators, etc, but there is an online textbook that's available.

In terms of math, I find it difficult to forget concepts that I use often (and that's a pretty broad set, considering I've been studying like Christ for the AIME).

But man, forgetting sucks; at HMMT, I bashed a problem out for around five minutes, and when I got near the end, I couldn't even remember how to take the derivative of a function of the form $ \frac{f(x)}{g(x)}$. I had to go all the way back to the beginning of the problem and find an elegant geometric method of solving the problem. It was horrible.

Seriously, though, I find reviewing past material a great way to keep concepts relatively fresh in my head. Even fifteen minutes every other week helps a lot.

If you forgot it, it's probably because it's not useful and you aren't interested in it otherwise you'd think about it and remember it. So do you really need to remember it?

I've come to the conclusion that most things aren't worth remembering.
If you can always go back to textbooks for them, why take the pains?
It is as useless as keeping record of all your contacts in your head and using no diary for that.

Very probably, I don't remember the Cauchy Schwarz inequality, I don't remember the poisson mass, I don't remeber the gamma density, I don't remember how to prove common continuous functions continuous and common differentiable functions differentiable, I've forgotten almost all sophisticated results from integration, I don't remember much of monotonous convergence theorem, law of large numbers, central limit theorem, and I hardly care for all that - because I just need five minutes in a decent library or the internet to find any of them.

Concentrate only on things you find quite interesting at present. The rest doesn't matter.

Well, sorry to say this but I've never had this problem. Usually if I learn something it sticks forever. On that note I've nevered ever studied for a test. I can just seem to remember everything I've heard in class.


For each real number $ r$, $ [r]$ denotes the largest integer less than or equal to $ r$, e.g. $ [6] \equal{} 6, [\pi] \equal{} 3, [\minus{}1.5] \equal{} \minus{}2$. Indicate on the $ (x,y)$-plane the set of all points $ (x,y)$ for which $ [x]^2 \plus{} [y]^2 \equal{} 4$.

I'm not quite sure what the problem means by "on the plane" (like quadrants or what?). So I decided to use inequality.

[hide="Solution"]

The key part is to note that $ [k]$ for real number $ k$ is always an integer. This means that there are only selected ways for $ [x]^2 \plus{} [y]^2 \equal{} 4$. This is done by $ x \equal{} 0, y \equal{} 2; x \equal{} 0, \minus{}2; x \equal{} 2, y \equal{} 0; x \equal{} \minus{}2, y \equal{} 0$.

Now that this is done, the actual answer is easy to find.

The solution set for $ x,y$ are:

$ (0 \leq x < 1,  2 \leq y < 3)$
$ (0 \leq x < 1, \minus{}2 \leq y < \minus{}1)$
$ (2 \leq x < 3, 0 \leq y < 1)$
$ (\minus{}2 \leq x < \minus{}1, 0 \leq y < 1)$ [/hide]

The solution is correct, but I think you should write
instead of this: $ x \equal{} 0, y \equal{} 2; x \equal{} 0, \minus{}2; x \equal{} 2, y \equal{} 0; x \equal{} \minus{}2, y \equal{} 0$
this: $ [x] \equal{} 0, [y] \equal{} 2; [x] \equal{} 0, \minus{}2; [x] \equal{} 2, [y] \equal{} 0; [x] \equal{} \minus{}2, [y] \equal{} 0$
Nothing important, but we have to be typical :D
I hope I'm correct! :blush:


limit
trigonometry
logarithms
function
LaTeX
calculus
calculus computations

1. $\ln\left(x^{\sin x}\right)=(\sin x)(\ln x).$

As $x\to0^{+},$ $\sin x$ looks like $x,$ while $\ln x$ tends to infinity slower than any power of $x.$ Hence the product tends to zero, and hence, the original limit is $1.$ (That is, if it is a one-sided limit from the right. The function is not defined to the left of zero.)

2. Now we're looking at $\frac{\ln x}{1-\ln x}$ as $x\to e.$ The numerator tends to $1$ and the denominator tends to zero. For $x>e,$ this quotient is negative, and for $x<e,$ this quotient is positive. Hence:

$\lim_{x\to e^{-}}\frac{\ln x}{1-\ln x}=\infty$ hence $\lim_{x\to e^{-}}x^{\frac1{1-\ln x}}=\infty$

$\lim_{x\to e^{+}}\frac{\ln x}{1-\ln x}=-\infty$ hence $\lim_{x\to e^{+}}x^{\frac1{1-\ln x}}=0$

[quote="shyong"]Find

1)$\lim_{x\to 0}x^{\sin x}$

2)$\lim_{x\to e}x^{\frac{1}{1-\ln x}}$

3)$\lim_{x\to \frac{\pi}{2}}\frac{\ln\tan x}{\ln\tan 2x}$[/quote]In other words, all these limits, as they are written in the above quote, do not exist? :?

this problem is given to me by my friend and I also think that the limit doesnt exist, just that Im not very sure how to prove it .

As for the 3rd problem , I also think that limit doesnt exist because

$\lim_{x\to \frac{\pi}{2}}\ln\tan 2x$ is not defined ??

Thanks Prof.Kent .

The first lim exists why not? It is actually a one-sided limit from the right. It's 1
I could send you a complete solution in Word if you want
The second obviously doesn't exist
The third is a one-sided limit from the left because the function is not defined to the right of π/2

To me the limit definition is that $f(x)$ should be close to $A$ as long as $|x-a| < \delta$ AND $x \in D_{f}$, where $D_{f}$ is the set on which $f$ is defined.

So, I don't agree with Kent that the first limit only exists as a right limit. Although it's arguably only a choice of semantics in this case I think the definition I use is better in for example cases where the denominator vanishes infinitely often since it allows a limit to exist there. I think there once was a thread with a discussion about exactly this.

I probably should have said it more clearly in the first place. On the semantic issue, I agree with thatin and Kalle. All of us are in agreement on the substance in any case.

As for the third problem - shyong doesn't go far enough. As originally stated, there is no punctured neighborhood of $\frac{\pi}2$ on which the original function is defined - on each side of that point, you're taking one logarithm of a negative number.

To make it an interesting problem, let's make it

$\lim_{x\to\frac{\pi}2}\frac{\ln|\tan x|}{\ln |\tan 2x|}$

Then it's an $\frac{\infty}{\infty}$ indeterminate case, which I'll leave to someone else to sort out. (I will concede that once in a while L'Hôpital's rule has a few practical uses.)

When x->π/2- then tanx>0 and tan2x<0
When x->π/2+ then tan2x>0 and tanx<0
I used De L'Hopital's rule, then "broke" the lims seperating tanx,tan2x and (cosx)^2, (cos2x)^2 then used De L'Hopital's rule at tans and I finally ended to lim(cosx)^2/(cos2x)^2
in which I used again De L'Hopital's rule
I will write my solution later at word (didn't learn latex yet)
I think it's correct
I didn't calculated them to the end so I don't know if finally the lim exists (I am in a hurry)

[quote="shyong"]Find : $\lim_{x\to 0}x^{\sin x}$ ; $\lim_{x\to e}x^{\frac{1}{1-\ln x}}$ ; $\lim_{x\to \frac{\pi}{2}}\frac{\ln\tan x}{\ln\tan 2x}$[/quote]$\blacktriangleright 1.\ \lim_{x\to 0}x^{\sin x}=\lim_{x\searrow 0}x^{\sin x}=\lim_{x\searrow 0}\left(x^{x}\right)^{\frac{\sin x}{x}}=1\ .$
$\blacktriangleright 2.$ [color=darkblue]The second limit has a sense and don't exist : $\lim_{x\nearrow e}x^{\frac{1}{1-\ln x}}=\infty$ ; $\lim_{x\searrow e}x^{\frac{1}{1-\ln x}}=0\ .$
$\blacktriangleright 3.$ The third limit hasn't sense ([b]very nice ![/b]) because for any $x\in \left(\frac{\pi}{4},\frac{3\pi}{4}\right)\ ,\ x\ne\frac{\pi}{2}$ we have $\tan x\cdot\tan 2x<0\ !$
[b]Another examples :[/b] the limit $\lim_{x\to\infty}\arcsin\frac{x+1}{x}$ hasn't sense and $\lim_{x\to 1}\frac{\ln \left(x^{2}-3x+2\right)}{\ln \left|4x^{2}-8x+3\right|}=\infty$.[/color]

[quote="Virgil Nicula"][quote="shyong"]Find : $\lim_{x\to 0}x^{\sin x}$ ; $\lim_{x\to e}x^{\frac{1}{1-\ln x}}$ ; $\lim_{x\to \frac{\pi}{2}}\frac{\ln\tan x}{\ln\tan 2x}$[/quote]$\blacktriangleright 1.\ \lim_{x\to 0}x^{\sin x}=\lim_{x\searrow 0}x^{\sin x}=\lim_{x\searrow 0}\left(x^{x}\right)^{\frac{\sin x}{x}}=1\ .$
$\blacktriangleright 2.$ [color=darkblue]The second limit has a sense and don't exist : $\lim_{x\nearrow e}x^{\frac{1}{1-\ln x}}=\infty$ ; $\lim_{x\searrow e}x^{\frac{1}{1-\ln x}}=0\ .$
$\blacktriangleright 3.$ The third limit hasn't sense ([b]very nice ![/b]) because for any $x\in \left(\frac{\pi}{4},\frac{3\pi}{4}\right)\ ,\ x\ne\frac{\pi}{2}$ we have $\ln x\cdot\ln 2x<0\ !$
[b]Another example :[/b] the limit $\lim_{x\to\infty}\arcsin\frac{x+1}{x}$ hasn't sense.[/color][/quote]
To Virgil Nicula:
I have never seen the following notations before,
$\lim_{x\nearrow e}(\bullet) = \lim_{x\to 0^{-}}(\bullet)$ and $\lim_{x\searrow e}(\bullet) = \lim_{x\to 0^{+}}(\bullet)$. Are they standard notations? Where/how did you learned about them?

To everyone:
Isn't there a more straightforward way to evaluate these limits? (or limits in general for that matter)? All this analysis seems to be a bit speculative don't you think.


Come on, Beat!! It follows immediately from this well-known indentity:

                             \[ a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca). \]

 I think that you should have posted this at getting Started. ;)

Don't be. That's ok. Nobody is perfect. ;)

[quote="Beat"]:blush:   How didn`t i see this :what?:... thanks anyway

Question: Can the equation $a^3+b^3+c^3=3abc$ be solved a) in natural, b) in integer numbers[/quote]
a)in natural numbers
            applying AM-GM $a^3+b^3+c^3\ge3abc$ with equality iff $a=b=c$
only Solution$(a,a,a)$
b)in integers this takes longer. Iwill post later

[quote="Beat"]Prove that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$.[/quote]

Since the equation is omogeneus, we can divide it by $c^3$; let's call $m: =a/c$ and $t: =b/c$ then $(m+n)(m^2+n^2-mn)=J$ but $m+n=-1$ so $(m+n)^2=1$, so $-m^2-n^2+mn=-1+3mn$

You can solve the equation $a^3+b^3+c^3=3abc$ directly in $\mathbb {R}^3$; just use the identity Cezar Lupu has already mentioned and obtain either $a+b+c=0$ or $a^2+b^2+c^2-ab-ac-bc=0$. In the second case use the identity
$a^2+b^2+c^2-ab-ac-bc=\frac 1{2}[(a-b)^2+(a-c)^2+(b-c)^2]$
and get $a=b=c$. Solutions: $\{(a,a,a) /a\in\mathbb{R}\}\ U\ \{(a,b,-a-b)/a,b\in \mathbb {R}\}$.


Wait, what does it mean when two events are mutually exclusive? Is it the same as independent?

Please explain and give an example... Thanks!~

Ok, thanks for the help!

Just a side note: Someone told me today that Wikipedia is fake?? huh!?

Maybe they mean that it isn't always true. This is because the content can be written by anyone, although they do have moderators. I once saw a whole section deleted and replace by a bunch of swear words, but after a few minutes, it was undone.

[quote="pascal12"]Because the content is written and edited by users, you definitely need to be wary and cautious of the information presented on wiki. There are moderators, but they can't reach every single vandalized article, so I would just double-check with a second source just in case.[/quote]
i've found, however, that most of the math articles are pretty accurate.
i always check with 3 other sites if i use Wiki, though

What other websites for math do you use?

Would the Intro to Counting and Probability answer my question?

Thanks.

Snowhawk try these websites: http://mathcounts.saab.org/mc.cgi http://regentsprep.org/

The second website helps with math itself more than problem solving.

Thanks for the help!

Mods, please lock this thread now.


[quote="anthony19891019"]n is a integer; n>=2;
k is a positive real number;
Find the minimum of this function: $y=\frac{x^n}{x-k}$;
x is a postivie real number that bigger than k.
[/quote]

Put $x-k=a$ to get $f(a)=\frac{(a+k)^n}{a}$
By AM-GM  $\displaystyle a+(n-1)\frac{k}{n-1}\ge n\sqrt[n]{a\cdot \left(\frac{k}{n-1}\right)^{n-1}}\Rightarrow f(a)\ge n^n\left(\frac{k}{n-1}\right)^{n-1}$


namosavie zir ra baraye hameye adade mosbate $a_i,b_i,i=1,2,...,n$ saabet konid.
$(\sum_{i=1}^{n} a_i)(\sum_{i=1}^{n} b_i)\geq(\sum_{i=1}^{n} a_i+b_i)(\sum_{i=1}^{n} \frac{{a_i}{b_i}}{a_i+b_i})$

in dige akharish bood.baraye in in hame ro post kardam ke man mikham ye chand mahi beram mosaferat va nemitoonam biyam....bye

I have a nice solution for this one, but I'm afraid that I've used an equality I am not sure about it's correctness.

Define $f(x_{1},x_{2}) = \frac{x_{1}x_{2}}{x_{1}+x_{2}}$ and one can see that $f$ is concave on $\mathbb{R}^{+}^{2}$ because it's second partial derivatives are negative. Therefore because of Jensen's unequality for concave functions (in fact I'm not sure about correctness of Jensen's for two-dimensional functions) we have $f(\frac{x_{1}+x_{2}+\ldots+x_{n}}n) \geq \frac{f(x_{1})+\ldots+f(x_{n})}n$.

Now put $x_{i}=(a_{i}, b_{i})$ and the requested unequality follows!

This problem is from Columbia2005 :wink:


A magician designed an unfair coin so that the probability of getting a Head on a flip
is $ 60\%$. If he flips the coin three times, what is the probability that he flips more
Heads than Tails? Express your answer as a common fraction.

Case 1: He gets 2 heads.

$ 3 \times \frac{3}{5} \times \frac{3}{5} \times \frac{2}{5}\equal{}\frac{54}{125}$

Case 2: He gets 3 heads.

$ \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5}\equal{}\frac{27}{125}$

Total:  $ \frac{81}{125}$.

[quote=ec04]Why do you multiply Case 1 by (3 choose 2) or 3?[/quote]

Something Pascal's triangle.

Yeah, I think it has to do with how many ways the coins could be ordered for the three coins to have two land on heads, which would use 3 choose 2.


algebra polynomial function conics hyperbola puzzles

the 2 pt. by 2 pt.  system in a square lattice?  Imagine the points
as {(0,0),(0,1),(1,0),(1,1)} for convenience.  A "curve" HERE is defined as one or more discrete points, one or more disconnected or connected line segments, or a combination of one or more discrete points WITH one or more line segments.  Nothingness (lack of any points or line segments) does not constitute a curve.  The same shape repeated, but shifted and/or flipped counts as a different item towards the total.  You may use a computer, OR paper & pencil/pen to come up with the total.

2)  Unknown:  Let there be an n pt. by n pt. system in a square
lattice also in the plane.   Because the total number of "curves" (that is, involving points and line segments) is growing very fast, I don't see anyone getting close to an expression for the total #
of curves, call it f(n), for n pts. by n pts.

Question:  If f(n) is a polynomial, can you state its degree?
                  Or if f(n) is an exponential function, can you state
                  its leading term?

This post is welcome to any poster who has not announced that
any of my other posts have been "ridiculous."

Good luck!

Err, what have we here... let's all go home and be happy campers.

And the problem seems to be pretty interesting, though I'm not quite sure about how a curve is actually defined... any set of points and lines? So they don't even need to be connected? Maybe I'm wrong, but that just seems sort of strange, so if someone could perhaps clarify this definition... and what if line segments overlap each other? Can they both be part of a "curve?"

Any set of points and line segments as described.  Line segments CAN overlap.  They don't need to be connected (think parts of one hyperbola for just one instance).  The two or more pieces that overlap
make up ONE curve when you go to do a count.

Ex.  Line segment from (0,0) to (1,1) AND the line segment from
       (1,0) to (0,1) when graphed AT THE SAME TIME make up ONE
       curve for this problem.

would be 1) the set of discrete points (0,1) and (1,0) and 2) the set of discrete points (0,0) and (1,1).  They are to be counted as separate items.


Let $\lambda\in\mathbb{R}$ and $P\in R[X]$ a polynom with the following propriety: $P(1)+P(2)+...+P(n)=n^7+\lambda$ with $n\in\mathbb{R}^{*}$. Find $P(\lambda)$.

Yes it has. For $n=1$ we have $P(1)=1+\lambda$ and so on... but $P(\lambda)=?$  :?

[hide="experiments with possible solution"]

Setting $n = 1,2,\dotsc$,

$P(1) = 1^7 + \lambda$
$P(2) + P(1) = 2^7 + \lambda \implies P(2) = 2^7 - 1^7$
$P(3) + P(2) + P(1) = 3^7 + \lambda \implies P(3) = 3^7 - 2^7$
$P(4) + P(3) + P(2) + P(1)= 4^7 + \lambda \implies P(4) = 4^7 - 3^7$

It makes sence that $P(x) = x^7 - (x-1)^7$, except for $P(1) = 1 + \lambda$, because that way $\displaystyle\sum_{i=1}^n P(i)$ will always equal $x^7 + \lambda$.

Then it seems that $\lambda$ could be any real number.  So $P(\lambda) = \lambda^2 - (\lambda - 1)^2$

But this doesn't seem right... :roll:

[/hide]

[quote="calculuslover800"][hide="experiments with possible solution"]
It makes sence that $P(x) = x^7 - (x-1)^7$, except for $P(1) = 1 + \lambda$[/hide]
[/quote]

[hide]You're on the right track calculuslover. Just put these two together to find what $\lambda$ equals and you're there.[/hide]

[hide="Well... to evaluate that..."]$\lambda ^2 - (\lambda -1)^2=2\lambda -1$?[/hide]

His answer says squared... I was kind of wondering about that.

[hide="Fixed"]$\lambda ^7 - (\lambda -1)^7=\lambda ^7-\lambda^7 +7\lambda ^6-21\lambda ^5+35\lambda ^4-21\lambda ^3+7\lambda ^2-\lambda$
$=7\lambda ^6-21\lambda ^5+35\lambda ^4-21\lambda ^3+7\lambda ^2-\lambda$[/hide]

Yes i got the same by this way:

$P(1)+P(2)+...+P(\lambda-1)+P(\lambda)=\lambda^7+\lambda$ (1)

$P(1)+P(2)+...+P(\lambda-1)=(\lambda-1)^7+\lambda$ (2)

Now
\begin{eqnarray*}P(\lambda) &=& P(1)+P(2)+...+P(\lambda-1)+P(\lambda)-[P(1)+P(2)+...+P(\lambda-1)] \\
&=& \lambda^7+\lambda-[(\lambda-1)^7+\lambda]\\
&=& \lambda^7-(\lambda-1)^7\end{eqnarray*}

My question is: Is this the final form?

The problem was multiple choice type and the possible answers were:
$(a)\ -1\ \ \ \ \ \ (b)\ 0 \ \ \ \ \ \ (c)\ 1\ \ \ \ \ \ (d)\ \lambda^7+7\ \ \ \ \ \ (e)\ \lambda+7$

You guys are so close. You have the right form for the polynomial, but $\lambda$ has to have a [i]specific[/i] value! Look closely at the information.

So one answer form the multiple choice answers is correct :)?

Yes, one of the answer choices is correct.

:trampoline: I think is correct :D.
And $P(1)=1^7-(1-1)^7$ so $\lambda=0$ ;)


modular arithmetic
quadratics
number theory unsolved
number theory

Solve the equation $x^5+31=y^2$ in positive integers, I think.
I think modulo something small doesn't work, unless we missed something

First if x is even then y^2=3(mod 4)
If x =3(mod 4) then y^2=2(mod 4)
So x=1(mod 4)
Then x+2=3(mod 4)
There must be a prime p|x+2 with p=4k+3 form
then x^5+2^5=y^2+1
y^2+1=0(mod p)
which obviously yields no solution of y by euler's criterion.

just a tip, when we are going to use modulo something and we need to evaluate $x^k$ and we are so lucky that 2k+1 is prime, wi should obviously try it!

In our problem $x^5=+-1$ so we would need $-1=y^2$, or $8=y^2$ both of them imposible.

(oops i forgot x=0)

or some Brute force will also do: $x^2+4=(x+2i)(x-2i)$

[quote="vess"][quote="Peter Scholze"]similar, but more interesting(this one cannot be done via some modulo computations): solve $x^5=y^2+4$ in integers.[/quote]

No, it is also trivial modulo $11$: $x^5-4$ must be one of $6,7,8 \pmod{11}$, and all of these are quadratic non-residues.  ;)

Another reason that this is trivial is that $y^2+6^2 = x^5+2^5=(x+2)(x^4-2x^3+4x^2-8x+16)$ must have a prime factor of the form $4k+3$.

--Vesselin[/quote]

how can u find mod 11? :?
trial and error?
any systematic way to find 11?
Thanks. :D

[quote="siuhochung"]how can u find mod 11? :?
[/quote]

Pascual already gave an answer to this above: if you have $x^k$ and $2k+1$ happens to be some prime $p$, then it might be a good idea to look modulo $p$, because $x^k$ takes only very few values, namely $0, \pm 1$. Here we have $k=5$.

In general, if $(k,p-1)=1$, then $x^k$ takes all possible values $\mod{p}$, so it won't make any sense to reduce modulo $p$. To have a chance, $k$ must have a large common divisor with $p-1$. For $k=5$, the smallest such prime is $11$, so it's natural to check $p=11$ first (and totally meaningless to look modulo a smaller prime). If things don't work out, then the second prime to look at is $p=31$ (no smaller prime will help at all).

Hope this helps,
--Vesselin

Pascual, I know the method! The first attempt I did is checking the equation modulo $11$ but if $11|x$ we get $y^2=9$ $)mod$ $11)$ obviously possible.

I know Iura, you are one of the top solvers in this forum so i am sure you tried it, I also saw i made a mistake becuase i didnt considered that possibility.

I'm sorry, couldn't we do this wth Siuchong's method?
I mean we can prove $x,y$ are odd then not divisible by $3$.
Then since $y^2+36$ is the sum of two squares so is $x^5$ and hence $x$ and we get $x=4k+1$ so $x-2=4k-3$.
Then $y^2+4=x^5-32=(x-2)(\cdots)$ and we produce a contradiction since $y^2+4$ is again sum of two squares

Totally unrelated, but I reminded by what my mathematics teacher said when asked why use a so-and-so mod. He said he was just lucky.  :D


[quote="1/(ln x)"]What prep books would be good for SAT II Chemistry and US History?[/quote]

McGill SAT Chem is the best.
Don't know much about US History.

for US History, get either Kaplan or Barrons.

I got an 800 on Chemistry with the Kaplan book.

What books would be good for SAT II physics and biology?

I used PR for physics

I think a friend used Barrons for biology and he said it was pretty good (he got 720 w/o ever taking a bio class and only two weeks of studying so...must be somewhat effective)

again, Barrons or Kaplan. The Princeton Review is also pretty good.


[hide="Solution"]
By just brute force, the answer is 73.

The easiest way is I think using logic.

The last digit cannot be 0,2,4,5,6,8. That eliminates a lot.

What is left, just filter the prime numbers that end in 1,3,7 or 9.

37 and 73 work, because both digits have to be 1,3,7,9 (since the end has to be one of those).

So, 13, 31 is too small, 17, 71 is too small, 19, 91 is too big. 37,73 is right.

After that, only 6 possibilies exist. Then, the answer is 37,73. The question asks for the bigger number, which is $\boxed{73}$.[/hide]

[hide]42 is equal to 10*4 + 2

So we can write:

(10a + b) + (10b + a) = 110

11(a+b) = 110
a+b = 10

Since we're working with Primes here, you have: 1,9, and 3,7 and that's it. 91 is not prime so the answer is 73.[/hide]

[quote="Stargirl240"]I just realized that 91 isn't prime. That's why I got this problem wrong! And since the sum of this pair of primes ends in a zero, the primes have to end in 1, 9, or 3,7. That eliminates a lot of primes. Are there any other ways of doing this problem besides brute force?[/quote]

You should memorize primes.

[quote="Stargirl240"](copied from '93 Chapter) The number 13 is prime. If you reverse the digits you also obtain a prime number, 31. What is the larger of the pair of primes that satisfies this condition and has a sum of 110? I know that this problem is really easy, but are there any shortcuts instead of running through the primes in your head?[/quote]
[hide="answer"]
just look for the primes that form 10s when its tens and ones place integers are added. also the ones place cant be even.
this should be like a chapter countdown, it might set a record for the fastest answered ever.
73.
 [/hide]

[quote="InsanelyBored"]
this should be like a chapter countdown, it might set a record for the fastest answered ever.
[/quote]

It was #9. #9's should be that easy

[hide="answer"]I just used brute force and after several minutes, I got 73.( I just wrote down all the primes over fifty and reversed them and added them together.)[/hide]

[quote="jackdillon"]*clip* after several minutes *clip* [/quote]

jackdillon, I don't think you will have that much time in a mathcounts question (this sprint should be done in ~1 min if you want to finish)

but it probably should be done by an aopser in a lot less than that, just saying

jorian

Yeah, but I was doing it in the back of math class during a test, and I got bored. I could of used algebra but chose not to.


rotation
integration
calculus
calculus computations

Hey all

I have a physics problem that I need some help setting up the problem:

A bar of length 2L rotates with angular velocity $ \omega$, which points in the +y direction. The bar makes an angle $ \alpha$ with the y axis, with its centre corresponding to the origin. The bar has linear mass density $ \rho(r) \equal{} \frac {b}{L^2}r^2$, where r is the distance along the rod measured from the origin. At the instant that the bar is in the xy plane, find the angular velocity.

Now, I am having trouble computing the moments of inertia. Using the inertia tensor notation, and since $ \omega \equal{} (0,\omega,0)$, it is clear that only the $ I_{xy}, I_{yy}, I_{zy}$ components will appear in the solution. For those who might not know,

$ I_{xy} \equal{} \minus{} \int{xydm}$
$ I_{yy} \equal{} \int{(x^2 \plus{} z^2)dm}$
$ I_{zy} \equal{} \minus{} \int{zydm}$

Now I know that $ dm \equal{} \rho(r)dr$ but I have no idea how to get $ \rho(r)dr$

The class has as yet not covered double integrals, so there must be a way to reduce these three to a single variable integration. Note that z=0 throughout the problem.

Any help setting up would be appreciated, and if anything in unclear I'll try to make it clearer.

Something is missing in the question; seems the angular velocity is a constant (why should it change ?)...

What for are you needing the inertial tensor? The problem doesn't mention any force acting on the bar.


Support
function
trigonometry
topology
real analysis
real analysis unsolved

Which real sets are the compact supports of some continuous function?

The support of a function is the closure of the set where it's non-null. Compact support means a support which is compact :). And the "s" at the end of "supports" comes from the plural of support. Does it make more semse now? :)

Yes, those are the sets, but you've only shown the necessity. We must also show that every such set is the support of a continuous function (this is not hard either :)).

Was that a reply to yourself, grobber? :?

Of course not :). When I wrote that reply, there was a post by ilyasu, in which he gave the answer. I don't know what happened to that post (there's also the possibility that there was no message from ilyasu, and I'm going crazy :)).

Yes, I posted the reply but I thought that my answer is incorrect, so I erased it.

Sorry,  :blush:

I think the answer is a bounded set, which is the union of at most countably many disjoint closed intervals.

Let $E = \{x:f(x)\neq 0\}$. Since $f$ is continuous, $E$ is an open set, and since $f$ has a compact support, $E$ must be bounded. Thus $%Error. "supp" is a bad command.
f = \overline{E}$ is a union of most countably disjoint closed intervals.

It is not difficult to construct a continuous function from a given such set $F = \bigcup_{i=1}^\infty [a_i,b_i] (a_i<b_i)$.

Let $f(x) = \sin \frac{\pi}{b_i-a_i}(x-a_i)$ on $[a_i,b_i]$ and $f(x)=0$ otherwise.


[quote="matheus"]when will the results be available for mathlinkers?[/quote]
I'd like to know it either,just out of curiosity ...


mine in C++, but Java and C# is pretty cool also

I don't like Java very much.  Too bad the AP people had to switch form C++.

Java is pretty cool if you get to know it more.

[quote="Peter VDD"]PHP!! :)[/quote]
I had a list of 40 programming languages, but the stupid poll wont take that many options so i had to cut it down to 10 :S

Durn.  Cocoa isn't on there.  So I voted C/C++/Objective C/VC++, but I haven't programmed VC++ since I have a mac.

in case you don't already know about it, check out the free programming contests and training material at http://www.usaco.org. The fall contest schedule should be posted soon (probably right after the int'l competition in mid-Sept.)

is there a limit to one's programming skills?  I can only go as far as section 4.1 in USACO training before getting stuck and I have taken USACO green for three years and the most i've ever gotten is about 400 points (just once)... and the rest of the contests i think i only score at most 250   :?

I've only done a few USACO contests, but it seems that orange and green have gotten noticeably more difficult with the addition of the blue division. Is this just a coincidence?

For those who haven't done USACO contests, the green division is the hardest, followed by orange division and then blue for those just getting started.

[quote="bugzpodder"]is there a limit to one's programming skills?[/quote]
In my opinion, I think it's similar to chess or martial arts.  There is this barrier that you keep hitting your head against over and over and over until for no reason whatsoever, you suddenly break through.

That's what I think, anyway.

[quote="countersixte"][quote="texas137"]The fall contest schedule should be posted soon (probably right after the int'l competition in mid-Sept.)[/quote]
The 2004 IOI (International Olympiad in Olympiad) is in Athens, Greece this year and takes place from Sep. 11 - Sep. 18.

I can't wait for contest season to start!  Last year I didn't do so well.  This year, however, I'm going to be ready! ( I think...  :maybe: )

btw, the training pages were very helpful in improving my programming skills.  Anybody who is interested should definitely try it![/quote]

As a fellow CS- student I wish you all the luck in the world at the IOI.

[quote="h_s_potter2002"]When doing USACO stuff, I like using C, mainly because it works, and is the right language for whaat they want you to program.  But I mainly like Java, becuase it's a whole lot cleaner with it's object-oritented stuff, but that's just my opinion.[/quote]

C++ is Object Oriented.

Rexx - Interpretive language.

PL/I,   - For scientific stuff.

IBM Assembler - for really machine level stuff.

Prob. written > 10^4 lines of code in last few decades..

 (Though I have prob. written thousands of line of code in FORTRAN, and some what in ALGOL, APL and host of other languages..._

OK, I have learned a little more. I now know a "real" programming language - Ruby. Really popular in Japan, though I am not japanese. Quite simple! Just Google for it.

Billy

Java because it is also a kind of coffee and an island!

Blitz Basic is the best way to program games :D

[quote="sponge008"]Hey, an open-source freak at my school also is trying to introduce me to Ruby! Perhaps I should try it?[/quote]

Go to webdeveloper.com. They have an article on how to get started. All taht you need to see is the download link. Then open the help book(REALLY THOROUGH and logical too!) and get started!

java is the coolest because
its the only one i know
its a coffee
and its the capital of indonesia*

Mine used to be perl, then java, then c++;

my all time favorite line of c++ code:

#define f(x,y) for (int x = 0; x < y; ++x)

you can probably see why i liked perl;

Ever tried FORTRAN ?? It's fun .......... :D

my favorite is VB, because that's the only one i know

[quote="probability1.01"]I think Java is too high level.  C++ is still the best balance of level of language and flexibility/ease of use.  But maybe that's just because I can never get used to Java.[/quote]
Hmm, most people think C++ is harder than Java.

i don't know that much.  I know TI a little bit.

Basic and VB. Basic because everything is like raw commands (PRINT "blah blah") and VB because its the same but with a GUI.

I hate Java. I took Comp Sci A and couldn't get a single question.


(i'm so sorry because my last problem had a mistake in writing but here is the corect one)
 Let $ a,b,c >0$ prove that
\[ \frac{a^{n}}{b^{n\minus{}1}}\plus{}\frac{b^{n}}{c^{n\minus{}1}}\plus{}\frac{c^{n}}{a^{n\minus{}1}}\geq \frac{a^{n\minus{}1}}{b^{n\minus{}2}}\plus{}\frac{b^{n\minus{}1}}{c^{n\minus{}2}}\plus{}\frac{c^{n\minus{}1}}{a^{n\minus{}2}}\]

[quote="M.A"]
 Let $ a,b,c > 0$ prove that
\[ \frac {a^{n}}{b^{n \minus{} 1}} \plus{} \frac {b^{n}}{c^{n \minus{} 1}} \plus{} \frac {c^{n}}{a^{n \minus{} 1}}\geq \frac {a^{n \minus{} 1}}{b^{n \minus{} 2}} \plus{} \frac {b^{n \minus{} 1}}{c^{n \minus{} 2}} \plus{} \frac {c^{n \minus{} 1}}{a^{n \minus{} 2}}
\]
[/quote]
It's true for all reals $ n\geq1.$ :wink:


modular arithmetic
number theory
prime numbers

We are given that $100!$ is a $n$ digit number ending in $4$ followed by $24$ zeros... so here are some questions:

$Question 1$
Delete these Zeroes and the $4$; leaving a $n-25$ digit number. Add up the digits and for a multi digit sum; add them up again ....
Go on till you are left with a single digit number.

What will that be?

$Question 2$
Can this $n-25$ digit number have prime factors greater than $100$?
If not why not ?
If yes ; a bit of explanation.

$Question 3$
Which $n!$ when treated as before ; ( deleting all Zeroes at the end , and the Least Non Zero significant digit ) ; when repeatedly added as before , will result in the same last digit number deleted from its end ?

[hide="Q1"]
Taking digital sums in this manner preserves the value of the number modulo 9. Since $9|100!$, (and dividing by a power of 10 does not alter this) 9 divides our single digit number, which must therefore be $9$.[/hide]

[hide="Q2"]
There are no prime factors>100 that divide this number, which we call $m$. Since you get $100!$ from $m$ by adding zeros to the end, $100!=10^{25}m$ and $m|100!$. So if $p|m$, $p|100!$. But then $p|1$, $p|2$, ... or $p|100$. (By Euclid's Lemma). Hence $p<100$.[/hide]

[quote="tim1234133"][hide="Q1"]
Taking digital sums in this manner preserves the value of the number modulo 9. Since $9|100!$, (and dividing by a power of 10 does not alter this) 9 divides our single digit number, which must therefore be $9$.[/hide][/quote] Remember you are taking the 4 off the end also.

Which  when treated as before ; ( deleting all Zeroes at the end , and the Least Non Zero significant digit ) ; when repeatedly added as before , will result in the same last digit number deleted from its end ?

For this the answer is no such value of n exists
Reason is that..If the number ends in a non-zero significant digit as x...then we wnt (9k+x)*10^n + x mod9=0

that is to say 2xmod9=0..this means x is either 0 or 9...bt we hav it as the last nonzero digit..so x is not 0..then if x is 9..which isn possible coz nonzero digit of n! shud be even...so this isnt possible...

[quote="matt276eagles"][hide="1"]The $n-25$ digit number is equal to $\frac{100!}{10^{24}}-4$.  Since $\frac{100!}{10^{24}}\equiv 0\pmod{9}$, the $n-25$ digit number is $\equiv 5\pmod{9}$.  As we continually take the sum of the digits, the number $\pmod{9}$ is invariant.  Therefore the single digit number left at the end is 5.[/hide]

[hide="A start for 2"]Suppose that the $n-25$ digit number has no prime factors greater than 100.  The number is equal to $\frac{100!}{10^{24}}-4$.  Notice that $100!$ is divisible by all primes less than 100 and no primes greater than 100, and has more factors of 2 than 5.  There are 24 factors of 5 because we are given that $100!$ ends in exactly 24 zeros.  Therefore $\frac{100!}{10^{24}}$ is divisible by all primes less than 100 except 5, and no primes greater than 100.

From this, we know that $\frac{100!}{10^{24}}-4$ is not divisible by any of the primes that divide $\frac{100!}{10^{24}}$, except 2, since $2|4$.  Therefore, the only prime factors of our $n-25$ digit number are 2 and/or 5.[/hide][/quote]
for Q2. why not more prime numbers greater than 100? only 5 and 2?


1)Find $a,b$ so that $x^3+ax^2+11x+6$ and $x^3+bx^2+14x+8$ may have a common factor of the form  $x^2+px+q$

2)Solve for $\mathfrak{x,y,z}$

$x+y+5+\sqrt{xy+3x+2y+6}=39$

$x^2+y^2+7x+8y+xy+9=741$

1)
[hide]Use the division algorithm for polynomials to divide $x^3+ax^2+11x+6$ by $x^2+px+q$, giving $x+1$ and a certain remainder. Then, use synthetic division to divide the two given polynomials by $x+1$, yielding remainders of $a-6$ and $b-7$, respectively. Hence, $a=\framebox{6}$ and $b=\framebox{7}$. Checking this, we see that by substituting the values we found for $a$ and $b$ into their respective polynomials, the polynomials factor to $(x+1)(x+2)(x+3)$ and $(x+1)(x+2)(x+4)$, respectively, with a common factor in the desired form of $x^2+3x+2$.[/hide]

[quote="shadysaysurspammed"]1)Find $a,b$ so that $x^3+ax^2+11x+6$ and $x^3+bx^2+14x+8$ may have a common factor of the form  $x^2+px+q$[/quote]

For this to be true, then $x^2+px+q | x^3+ax^2+11x+6 - (x^3+bx^2+14x+8) = (a-b)x^2-3x-2$.

Also, one may say that $x^2+px+q | x^3+bx^2+14x+8 - (x^3+ax^2+11x+6) = (b-a)x^2+3x+2$.

Therefore, any $a$ and $b$ will work, so long that $|a-b| = 1$.

arjunlandes, your solution is not necessarily wrong, but it only gives one answer. (I think)

[quote="pkothari13"][quote="shadysaysurspammed"]1)Find $a,b$ so that $x^3+ax^2+11x+6$ and $x^3+bx^2+14x+8$ may have a common factor of the form  $x^2+px+q$[/quote]

For this to be true, then $x^2+px+q | x^3+ax^2+11x+6 - (x^3+bx^2+14x+8) = (a-b)x^2-3x-2$.

Also, one may say that $x^2+px+q | x^3+bx^2+14x+8 - (x^3+ax^2+11x+6) = (b-a)x^2+3x+2$.

Therefore, any $a$ and $b$ will work, so long that $|a-b| = 1$.[/quote]

huh $k(x^2+px+q)=(a-b)x^2-3x-2\iff k=a-b,kp=-3,kq=-2$ where $k\in\mathbb{Z}$

so $k=\pm 1$, $p=\mp 3$, $q=\mp 2$

[quote="Adunakhor"][hide="2"]
$x+7+5+\sqrt{xy+3x+2y+6}=39$

$x^2+y^2+34^2+2(xy-34x-34y)=xy + 3x + 2y + 6$

$x^2+y^2+x - 71x - 70 y + 1150 = 0$

Subtract the second equation from this equation to get:

$-78x - 78y = 1882$
$x+y = \displaystyle -\frac{1882}{78}$

[/hide][/quote]

You didn't finish the solution.  You still need to plug that back in and solve for x and y...

[quote="Adunakhor"][hide="2"]
$x+y+5+\sqrt{xy+3x+2y+6}=39$

$x^2+y^2+34^2+2(xy-34x-34y)=xy + 3x + 2y + 6$

$x^2+y^2+x - 71x - 70 y + 1150 = 0$

Subtract the second equation from this equation to get:

$-78x - 78y = 1882$
$x+y = \displaystyle -\frac{1882}{78}$

[/hide][/quote]
How can the intersection of two conics in the plane be a line?

Yeah  :blush: I wasn't finished with my solution when I had to leave, and I forgot to put the disclaimer that it wasn't done when i left. I'll edit it after I eat dinner  :P

[quote="pkothari13"][quote="chess64"]
huh $k(x^2+px+q)=(a-b)x^2-3x-2\iff k=a-b,kp=-3,kq=-2$ where $k\in\mathbb{Z}$

so $k=\pm 1$, $p=\mp 3$, $q=\mp 2$[/quote]

yep that's essentially what i said.  :) i think you just introduced a new variable $k$.[/quote]

err sorry misread

Are you sure the second equation on #2 isn't supposed to be $+19$?  It would make the answer a lot nicer :D

[hide="2"]
The first equation can be rewritten

$(x + 2) + (y+3) + \sqrt{ (x + 2)(y + 3) } = 39$

The second equation can be rewritten

$(x + 2)^2 + (y + 3)^2 + (x + 2)(y + 3) = 751$

Let $a = x + 2, b = y + 3$.  Then

$a + b + \sqrt{ab} = 39$
$a^2 + b^2 + ab = 751$
$ab = (39 - (a + b))^2$

Let $u = ab, v = a + b$.  Then

$u = (39 - v)^2$
$v^2 - u = 751$
$v^2 - (39 - v)^2 = 751$
$(39)(2v - 39) = 751$
$v = \frac{ \frac{751}{39} + 39}{2} = \frac{1136}{39}$

$u = (39 - v)^2 = \frac{385^2}{39^2}$

...

Seriously, the final solution to this is annoyingly ugly :(
[/hide]


[quote="ch1n353ch3s54a1l"][quote="nebula42"]What is the minimum number of minutes after 3:00 such that the minute and hour hand are $20^\circ$ apart?[/quote]
[hide]
After $m$ minutes $6m$ degrees pass on the minute hand and $.5m$ degrees on the hour hand, or $5.5m$. So then we have $|90-5.5m|=20$ and for the two cases, the lesser $m$ is $140/11$.[/hide][/quote]

I don't even think you need the absolute value.

[hide]There are two cases, when the minutes are less than 15 and greater than 15.  Since less than 15 would be the minimum, we will find that.  15 min to 0 min is $90^\circ$.  So, for each minute out of 15 you go, $\frac{90}{15}= 6^\circ$ will pass by.  Also, the hour hand will go $\frac{1}{2}^\circ$.  So, the equation becomes $90-\frac{11m}{2}= 20$, where m is minutes.  Solve and $m = \boxed{\frac{140}{11}}$.[/hide]

[hide]Let m be the number of minutes that pass by.  This means that the minute hand has gone by $6^\circ$ and the hour hand has gone by $\frac{1}{2}^\circ$.  So the angle measure changes by $\frac{11}{2}^\circ$.  So, since the angle measure is decreasing, we just have to solve the equation.

$90-\frac{11m}{2}=20$, so $m=\frac{140}{11}$

$\frac{140}{11}$[/hide]

I got the same equation as mathgeniuse^ln(x) :

Final Ans:
 140/11 :D


Every once in a while, I would notice that my rating goes up/down or one of my posts is rated with no reason. So I'd like to suggest a new rating system:

1) The rater would rated the user one a scale of 1-6 cubes as usual.
2) There would be a pop-up or a new window of some sort, which would be something like:

[i]Was this post good or bad? (Choose one)
- Good
- Bad
Any comments about why this post was good/bad: (Optional I guess)
<textbox>
Would you like to leave your username here? (Y/N)[/i]

3) The rated user would have a new section, maybe in the profile, that would show rated topics and such (Includes the topic link, the rating, comments, and rater (possibly anonymous).

Rating is weighted as usual and such.

Now I'd have to find a way to encourage people to rate.

Good idea, but I think the biggest problem is that people are rating spam rather than helpful posts (just an observation)

That's an observation by someone having a rating of less than 2 :wink:

Since this forum is built on the phpbb software (I'm pretty sure it is), you can't really add that sort of stuff. :maybe:

^ I've seen another phpBB board that has something similar, not quite same. I'd just say the difference is rating 1-6 and maybe changing some wording/lists.

And for those of you who don't see how this would help:

1) You get commented for good explanations/ideas/etc.
2) Those annoyed by not knowing who rated them might know.
3) Those that get annoyed by not knowing what they did wrong will likely to know what they did wrong.
4) You can visit the spammy posts and look at how to improve them.
5) Less flaming/lecturing about forum etiquette in forums.

Comments/constructive criticism anyone?

EDIT: About number 5, maybe after a lot of unchecked bad rated posts, you will receive some kind of warning/consequence.

General question:

List as many things why you think rating is important. I don't care even a little about my rating, although I try to post productive.

I second Bugi's suggestion.But, as far as I'm concerned, there doesn't exist a necessity for most people to carp about their low rating. I also used to fret about my rating but once I observed that most people (on AoPS) have a severe predisposition to discern and rate "unproductive posts" than "productive posts" -- I just stopped carping!

But it's important to contribute to the forum as much as possible by posting solutions to puzzles/problems, and composing relevant, productive posts. (Even when it doesn't have any effect on your rating)

Well, from so many users who are complaining about bad rating (no, it's not just dragon), no one listed one thing why he/she thinks rating is important. But, topics like this are going to continue.

Listen to my advice: Don't care about your rating except it's extremely low. My rating now is 3,4 some may say it's high, some may say it's low, but I don't care. Neither should you.


trigonometry
logarithms
calculus
integration
geometric sequence
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]
Determine whether the series converges conditionally,converges absolutely or diverges.

$ \sum_{n\equal{}2}^{\infty}\frac{\sin (n\plus{}\frac{1}{n})}{\ln (\ln n)}$

[b]3. The attempt at a solution[/b]

$ \frac{\sin (n\plus{}\frac{1}{n})}{\ln (\ln n)}\equal{}\frac{\sin n\cos \frac{1}{n}\plus{}\sin \frac{1}{n}\cos n}{\ln (\ln n)}\equal{}\frac{\sin n(1\plus{}O(\frac{1}{n^2}))\plus{}\cos n(\frac{1}{n}\plus{}O(\frac{1}{n^3}))}{\ln (\ln n)}$

$ \sum_{n\equal{}2}^{\infty}\sin n,\sum_{n\equal{}2}^{\infty}\cos n$ are bounded

From Dirichlet's Test we can deduce that this series converges

$ |a_n|\equal{}\left|\frac{\sin n(1\plus{}O(\frac{1}{n^2})}{\ln (\ln n)}\right|\plus{}\left|\frac{cos n(\frac{1}{n}\plus{}O(\frac{1}{n^3}))}{\ln (\ln n)}\right|$

if we look for example at $ \left|\sum_{n\equal{}2}^{\infty}\frac{cos n}{n\ln (\ln n)}\right|>\left|\sum_{n\equal{}2}^{\infty}\frac{cos n}{n\ln (n)}\right|$

0<|cosn|<1 and $ \sum_{n\equal{}2}^{\infty}\frac{1}{n\ln (n)}$ diverges by Integral Test
So this series converges conditionally!!

A slight repackaging of the argument:

By Taylor's Theorem, $ \sin\left(n\plus{}\frac1n\right)\equal{}\sin n\plus{}\frac{\cos n}n\plus{}O(n^{\minus{}2}).$

$ \sum_n\frac{\sin n}{\ln\ln n}$ converges conditionally.

$ \sum_n\frac{\cos n}{n\ln\ln n}$ converges conditionally.

Finally, $ \sum_n\frac{O(n^{\minus{}2})}{\ln\ln n}$ converges absolutely.

So the original series converges conditionally.

I guess this is the night of unobvious series for me:

Why is it clear that the series doesn't converge absolutely?

I'd better go to bed before I come up with any more (probably point-missing) questions...

i can explain  absolutely by many ways:
peoples who know well rational aproximation to $ \pi$ problems like this is obviusly:
we consider :
$ \sum |\sin n \plus{} \frac {\cos n }{n}| \frac {1}{\ln \ln n }$
and
this serie $ \sum \frac {1}{n^{1 \plus{} \alpha}\ln \ln n }$ will convergens if and only if when $ \alpha > 0$ it's well known and easy)
if we consider sequance
$ A_{n} \equal{} |\sin n \plus{} \frac {\cos n }{n} |$
numbers $ n_{j}$ for which $ |\sin n_{j} \plus{} \frac {\cos n_{j} }{n_{j}}|\approx\frac {1}{n_{j}^{1 \plus{} \beta}}$ or $ n_{j}\approx 0(mod \pi)$ is very few (it's proved that it's like geometric progression ) therefore those numbers can't help for convergent serie .
by this way can explain peoples who know well about aproximation and study this part of math.
for Us it's must be some hint to explain why serie absolutely wiil divergent :!:
if we suppose that $ A_{n_{0}}$ is very closely to zero for fix $ n_{0}$ say $ A_{n}$~$ 0$ then let consider $ A_{n \plus{} 1}$ and prove that it will big or not closely to $ 0$. this is Idea.
I will post sketch solution ( because don't want to explain some easy step)
let $ n$ is very "big" :  $ |\sin n \plus{} \frac {\cos n }{n} |\approx 0$ or
then
$ A_{n \plus{} 1} \equal{} |\sin (n \plus{} 1) \plus{} \frac {\cos (n \plus{} 1) }{(n \plus{} 1)}|\approx |\cos 1 (\sin n \plus{} \frac {\cos n}{n}) \plus{} \sin 1 (\cos n \minus{} \frac {\sin n }{n})) \plus{} O(n^{ \minus{} 2})|\approx \sin 1|\cos n \plus{} O(n^{1 \plus{} \gamma}) |$
and if $ n > 10^{100}$ for which   $ |\sin n \plus{} \frac {\cos n }{n} | < \frac {1}{10^8}$ then $ |\cos n | > \frac {1}{10^{10}}$ for all like this $ n$.


floor function
probability
trigonometry
geometry
function
complex numbers
algebra

Due to the fact that I'm *very* busy preparing for the new COP the 4th, I didn't get chance to compile my solution and answers. It's somewhere around my desk and I'm bit busy to find it. So, here's a deal:

Everybody can post the answers and when all 15 questions are answered, then we'll take who is the winner. I don't want to post the COP the 4th now anyway so it's going to take sometime as well.

I'll now and then come help the answers.

Below are the answers which two contestant send me (RC-7th did not send the answers)

[hide]
Farazy's answer
1. 2482
2. 652
3. 1207
4. 16665
5. 18
6. 2(34) where 34 is the units digit...guess
7. guess...17
8. guess...1
9. 55
10. 1-$\frac{\sqrt{10}}{10}$..guess
11. 1/2...it can either happen...or not happen
12. There are none...
13. 60
15. guess...7[/hide]

[hide]
Jimmy's answer
1) 2482
2) 391
3) 1207
4) 5000 (guess)
5) 2997/163 (guess)
6) 4 95
7) $2\pi$
8) 144
9) 40
10) 23/36
11) 1/16 (guess)
12) $\frac{\sqrt{5}-1}{2}$
13) $30\sqrt{2}$
14) $\pi/6$, $\pi/4$, $\pi/2$, $3\pi/4$, $5\pi/6$
15) 807[/hide]

#1

Since $z_1$ and $z_2$ are complex numbers that their sum is NOT complex number, it is easily figured out that they must be conjugate to each other, and further more, if they're in form of:

$a+bi$ and $a-bi$

$a=7$ because those add to 7.

So, we have:

$(7+bi)(7-bi) = 49 + b^2 \Rightarrow b = 2$.

So, putting them in the value:

$(7+2i)^4 + (7-2i)^4 \Rightarrow 2482$

Please post out here so we can get result soon! :)

#10
Find the exact value of $1-\left(\frac{1}{10}+\frac{1}{18}+\frac{1}{28}+\frac{1}{40}+\frac{1}{54}+\frac{1}{70}+...\right)$.
This has a closed form expression of $1-\sum_{n=1}^\infty \frac{1}{(x+1)(x+4)}$.
Using partial fractions on this sum, we have $\sum_{n=1}^\infty \frac{1}{(x+1)(x+4)}=\sum_{n=1}^\infty \left(\frac{1}{3(x+1)}-\frac{1}{3(x+4)}\right)$.
If we write out the terms, we see that this is a telescoping sum.
$\left(\frac{1}{6}-\frac{1}{15}\right) + \left(\frac{1}{9}-\frac{1}{18}\right) + \left(\frac{1}{12}-\frac{1}{21}\right) + \left(\frac{1}{15}-\frac{1}{24}\right)$ with all all terms cancelling except $\frac{1}{6}+\frac{1}{9}+\frac{1}{12}=\frac{13}{36}$.
Thus the answer is $\frac{23}{36}$.

#12
Find all value of $x$ such that $x=1-x+x^2-x^3+x^4-x^5+...$.
First we have, $|x|<1$ in order for the RHS of the equation to converge.
$x-1=-x(1-x+x^2-x^3+x^4-x^5+...)=-x^2$.
$x^2+x-1=0, |x|<1$.
$x=\frac{\sqrt{5}-1}{2}$.

Wow...What you did for #1 was a lot quicker then what I did...

#1
I did this with two manipulations!

$z_1 \cdot z_2 = 53$
$z_1 + z_2 = 14$

$z_1^4 + z_2^4=?$

Let $z_1=x$ and $z_2=y$

$(x+y)^4=14^4$
$x^4+4x^3y+6x^2y^2+4xy^3+y^4=38416$
$6x^2y^2=16854$
$x^4+4x^3y+4xy^3+y^4=21562$
$x^4+y^4+4xy(x^2+y^2)=21562$

$(x+y)^2=14^2$
$x^2+2xy+y^2=196$
$2xy=106$
$x^2+y^2=90$
$4xy=212$

$x^4+y^4+212(90)=21562$
$x^4+y^4=2482$

#2
$[CODABP]=CX(OP)+5(OA)+21(AX)= \beta$

CX=20(cos 15)
OP=20(cos 15)+20(sin 15)
OA=20(cos 15)-20(sin 15)
AX=20(sin 15)

Plug these values in to the original equation, yank out a calculator
and you get the $\lfloor{\beta}\rfloor$ giving you [b]652[/b]

#3
$p_1=p_2=4$
$p_2005=?$

$p_1=4=4^1$
$p_2=4=4^1$
$p_3=16=4^2$
$p_4=64=4^3$
$p_n=4^{n-1}$

$p_2005=4^2004$
To find the number of digits,
(log base)xPower and round up

$(log 4) \times 2004 \approx 1206.52$
Round up: [b]1207[/b]

#4
Sum of all numbers with digits r,s,t,u:

1. $6666r+6666s+6666t+6666u=x$
2. $6666(r+s+t+u)=x$

Thus the sum MUST be divisible by 6666

Since we know the sum of all numbers except rstu we need to subtract 1000r+100s+10t+u

3. $5666r+6566s+6656t+6665u=59871$

We need to find rstu plus the largest formation of r,s,t,u
From our first equation and our third we can find rstu as

4. $x-59871=rstu$

x must be a value that is a multiple of 6666 and when 59871 is subtracted from x it results in a number in the thousands. So we need to find a multiple of 666 such that
$1000 \le x-59871 \le 9999$

The only value for x that this holds true for is 66660, plugging this value in we get that
rstu is 6789

So 6789 + the highest formation is 6789+9876=[b]16665[/b]

#9

10 people chose 6.

Two people must be larry and jane so now we have

 8 people choose 4=70

but laura and John cannot both be chosen

So now subtract the number of cases in which they are together

6 choose 2 =15

70-15=55

#12

There are no values for y that makes this equation true.

Consider the follwoing case

$y=1-y+y^2-y^3+y^4-y^5+...$

when y=1

1=1-1+1-1+1-1+...
if we organize this equation as follows:

1=1+(-1+1)+(-1+1)...
1=1
which is true

but if we organize it like this

1=(1-1)+(1-1)...
1=0
which is false...

Showing that the equation is faulty.

#13

_$\cdot$_$\cdot$_$\cdot$_$\cdot$_$\cdot$_$\cdot$_$\cdot$_

There are 4 possible values for the first one (7,8,9,9)
There are 3 for the second
2 for the third
1 for last but we divide by two since 9 and 9 are the same

the forus has 3 values, the fifth has 10, the sixth has 1 the last has 5

2x3x2x1x3x10x1x5
=30 :sqrt: 2

#4 doesn't seem to work.
[hide]It can be said that each digit $r$, $s$, $t$, $u$, appears in each place value 6 times. Then the total sum of all of the arrangements of $rstu$ would be $6666r + 6666t + 6666s + 6666u = 6666(r + s + t + u)$. The problem says that $rstu$ was taken out, so $59871 + 1000r + 100s + 10t + u = 6666(r + s + t + u)$. By trial an error I think Farazy found that 6789 was $rstu$ ... but it doesn't work here. If 6789 worked then the total sum would be $6666(30) = 199980$ but when 6789 is added to the sum of the 23 other numbers as said in the problem, 59871, we only get 66660. That would mean $r + s + t + u$ is equal to 10, but that isn't the case.[/hide]

11
[hide]Probability that David wins on first try is $\frac 18$. Then second try $\frac 18 * \frac 18$... and so on. Since it's an infinite geometric sequence we plug the numbers in: $\frac {\frac 18}{\frac 78} = \frac 17$.[/hide]

#7
$r(x)=\frac{\csc^4(x)-2\csc^2(x)+1}{\csc{x}(\csc(x)-\sin(x))+\frac{\sin(x)-\cos(x)}{\sin(x)}+\cot(x)} = 0 =\frac{\cos^4(x)}{\sin^2(x)}$
$\cos(x)=0$
$x=\frac{\pi}{2}$ or $x=\frac{3\pi}{2}$.
Thus the sum of the two possible values of $x$ is $2\pi$.

#8
Each term of $S(z)$ can be though of as the hypotenuse of a right triagle with legs of lengths $c_k$ and $d_k$.  Putting all of these terms together yields a staircase. $S(z)$ is at a minimum when the staircase is a straight line, meaning $S(z)=\sqrt{\sum_{k=1}^{n} c_k^2 + \sum_{k=1}^{n}d_k^2} = \sqrt{17^2+z^2}$.
We know that $S(z)$ and $z$ are integers, so we can solve this using decompostion.
$S(z)^2-z^2=17^2$
$S(z)-z=1$ and $S(z)+z=17^2$
$S(z)=145$ and $z=144$.

I do remember some problems.. So, I'll post the ones that I'm pretty sure are correct.

1. 2482
3. 1207
7. $2\pi$.
8. 144
10. 23/36
11. 1/7
12. $\frac {\sqrt 5 - 1}{2}$
15. 807

Others are not sure.. They seem right but they're combinatorics.. I have to check #7.... I'm not so confident about them..

Ones that need to be checked:

2,4-6,9,13,14

I know 15 is correct because it was Mildorf who sent me the problem.. Unless he's wrong, it seems right! :)

11) Since it continues infinately, the probabiltiy that it happens vs not happening is 1/2

#7

Your answer is correct, Jimmy. Although way you got it is not exactly what I have. ;)

[quote]$\displaystyle r(x) = \frac {\csc^4 x - 2 \csc^2 x + 1}{\csc x (\csc x - \sin x) + \frac {\sin x - \cos x}{\sin x} + \cot x}$[/quote]

Top:

$\csc^4 x - 2 \csc^2 x + 1 \Rightarrow (\csc^2 - 1)^2 \Rightarrow (\cot^2 x)^2 = \cot^4 x$

Bottom:

$\csc x (\csc x - \sin x) + \frac {\sin x - \cos x}{\sin x} + \cot x\\
\csc^2 x - 1 + \frac {\sin x - \cos x}{\sin x} + \cot x\\
\cot^2 x + \cot x + \frac {\sin x - \cos x}{\sin x}\\
\frac {\cos^2 x}{\sin^2 x} + \frac {\cos x}{\sin x} + \frac {\sin x - \cos x}{\sin x}\\
\frac {\cos^2 x + \cos x \sin x + \sin^2 x - \cos x \sin x}{\sin^2 x}\\
\frac {1}{\sin^2 x} \Rightarrow \csc^2 x$

There was no need to do the bottom part, however. Since to have 0, we only need to look for the top. :)

$\cot x$ is 0 on $\frac {\pi}{2}, \frac {3\pi}{2}$. So, adding them together yield $2\pi$.

#4

tarquin, nice observation. :) Although what Farazy did was the method to get the answer, when we plug the value in, it doesn't work. :D So, which means this problem has $\text {no solution}$.

#2

There are actually 5 triangles..

Let's call them A,B,C,D, and E, respectively. Their area is given as:

$A = 20 \cdot 20 \cdot \sin (30) \cdot \frac {1}{2}\\
B = 10 \sqrt {3} \cdot 10 \cdot \frac {1}{2}\\
C = 10 \cdot 10 \cdot \frac {1}{2}\\
D = 10 \sqrt {2} \cdot 10 \cdot \frac {1}{2}\\
E = \sin (15) \cdot 20 \cdot 2 \cdot 21 \cdot \frac {1}{2}$

Adding them gives you the large polygon, and this value is when used for the floor function, $\fbox {416}$.

#5 (Sorry for irregular order)

Let's call that angle ACR is composed of angle $x$ and angle $y$.

By Tangent Identity, we have:

$\tan (x+y) = \frac {\tan x + \tan y}{1 - \tan x \tan y}$

In this case, we had to see the word "altitude." So:

$\frac {333}{652} = \frac {\frac {5}{x} + \frac {13}{x}}{1 - \frac {65}{x^2}}$

Solving for $x$, we get $x = 37$. (Other answer is negative).

Putting in for x, we have $[CAR] = 18 \cdot 37 \cdot \frac {1}{2}$ and take the square root and the floor function value is $18$.

#6

Hmm.. I'm not sure about this one.. I have 95 as answer.. But I got them from the problem where there was 9 points..

That case, the number of intersection was four-element subset of 9 which is defined by $\binom {9}{4}$.

If that's the case here, then it's $\binom {12}{4}$. But I'm not so confident.. So, I'll exclude this one from point value. If anyone knows clear solution, please post here.

Thanks.

#13

There are $\binom {4}{2}$ ways for the first four. There are 3! options for the fifth one. For sixth digit and seventh, it is just 10 since it doesn't matter which one the seventh is, but only matters on what sixth digit is. There are 5 optoins for eighth. So, it's $\sqrt {1800}$ which is equal to $30 \sqrt {2}$.

#14

The answers are:

$\frac {\pi}{2}, \frac {3\pi}{2}, \frac {\pi}{4}, \frac {3\pi}{2}, \frac {\pi}{6}, \frac {5\pi}{6}$

See if you can find a neat way to solve this. ;) I like this problem a lot even I made it.. :P

Since all solutions are posted, let's call the winner.

By the way, the answer to #9 was 55, like Farazy stated.

Jimmy is the winner with score of 8 points out of 15 points. :)

Congratulation Jimmy for the Champion of the COP 3rd.

[hide="To Jimmy"]
:first: [/hide]

Also congratulation to Farazy for making the second placer.

[hide="To Farazy"]
 :cool: Nice job! (That icon's my favorite) [/hide]

Both of the name will appear up on the COP Hall of Fame post. That thread is in About COP thread. Check it out!


It's a Wallis product. It can be rewritten as the limit of

$ 2\cdot\frac{2n}{2n+1}\cdot\frac{2n}{2n-1}\cdot\frac{2n-2}{2n-1}\cdots \frac45\cdot\frac43\cdot\frac23\cdot\frac21$


I'm not sure if there's anymore..

[hide="View"]

$(5,4)$ works.

I found this by using: $y = \frac{x^2+3}{x+2}$. The problem I see here is that denominator and numerator will be in different parity at least for positive integers and zero. I didn't check negatives though.[/hide]

[quote="mna851"]...to the equation $\ x^2 + 3=y(x+2) $[/quote]

i did it this way

[hide]i did it by long division
$y=\frac{x^2+3}{x+2}=x-2+\frac7{x+2}$ and $y-x+2=\frac7{x+2}$ must be integer so $x+2$ must divide 7. 7 is prime and so $x+2=\pm 1,\pm 7$. we get respective $x=5,-1,-3,-9$ and $y=4,4,-6,-12$[/hide]

[quote="mna851"]Palytoxin I believe your solution is correct, I did it another way though.[/quote]
can you show us the "another way"? i want to learn it too

I like palytoxin's way, it has helped me at so many competitions.


function
calculus
derivative
analytic geometry
vector
quadratics
calculus computations

Let Z be a function of the n variables $ X_i$. I am reading a text that says the following: the second order variation of Z is

$ \delta _2 Z \equal{} \frac{1}{2} \sum_{i} \sum_{k} \frac{\partial ^2 Z}{\partial X_i \partial X_k} \delta X_i \delta X_k$.

Can someone explain to me what a "second order variation" is and how it arises? It seems that it is simply the extension of the total differential concept to second derivatives, because of its form and the 1/2 factor. And it seems to arise by differentiating the total differential. Is this the case? Also, why is used the notation $ \delta X_i$ instead of $ d X_i$?

It is exactly the second differential term in the Taylor formula. The terminology comes from the fact that if $ \delta X_j$ are thought to be of certin size $ s$, the resulting expression is of order $ s^2$. Don't bother too much about $ \delta$ versus $ d$, it is just historical. If you know the general Taylor formula and are comfortable with it, you can easily translate the language of the text you are reading into more familiar one of Taylor expansions and you'll lose nothing in the process.

Thank you so much. So we can write the above as

$ d_2 Z \equal{} \frac{1}{2} \sum_{i} \sum_{k} \frac{\partial ^2 Z}{\partial X_i \partial X_k} dX_i dX_k$

So if I have understood this right, we have

$ dZ \equal{} \sum_i \frac{\partial Z}{\partial X_i} dX_i$

and something similar for $ d_3Z$. But now a doubt: in the Taylor formula for a function in the neighbourhood of a point (a, b, c, ...) we have terms like $ (X_i \minus{} a)$ instead of $ dX_i$. So, I can say that in an infinitesimal neighbourhood of the point (0,0,0, ...) (or another point?) we can write $ Z(X_i)$ as

$ Z(X_i) \equal{} Z(0,0,0,...) \plus{} \sum_i \frac{\partial Z}{\partial X_i} dX_i \plus{} \frac{1}{2} \sum_{i} \sum_{k} \frac{\partial ^2 Z}{\partial X_i \partial X_k} dX_i dX_k \plus{} ...$

Is this right?

Usually by $ dX_i$ (or $ \delta X_i$) in such formulae one already means the increment of $ X_i$, i.e., $ X_i\minus{}a$. A bit more accurate language would be to define $ dX_i$ as the linear form that returns the $ i$-th coordinate of a vector and $ d^n Z$ as the $ n$-th degree form given by
\[ d^nZ\equal{}\sum_{i_1,\dots,i_n}\frac{\partial^n Z}{\partial X_{i_1},\dots,\partial X_{i_n}}\,dX_{i_1}\dots dX_{i_n}\]
so the Taylor expansion of $ Z$ near $ X$ becomes
\[ Z(Y)\equal{}Z(X)\plus{}\sum_{n\ge 1}\frac 1{n!}(d^n Z)(Y\minus{}X).\]
where $ (d^n Z)(Y\minus{}X)$ means the value of the form $ d^n Z$ on the increment vector $ Y\minus{}X$.
Looks familiar, doesn't it :)

Yes, that looks familiar, thanks. Two more questions:

1) Why can we write $ d^2 Z \equal{} \frac {1}{2} \sum_{i} \sum_{k} \frac {\partial ^2 Z}{\partial X_i \partial X_k} dX_i dX_k \equal{} \frac {1}{2} \sum_i dx_i dX_i$,

where $ x_i \equal{} \frac {\partial Z}{\partial X_i}$?

2) So I have a function $ Z(X_1, X_2, ..., X_n)$ and I am interested in the conditions for which Z is a minimum; therefore, the following two conditions must hold,

$ dZ \equal{} \sum_i \frac {\partial Z}{\partial X_i} dX_i \equal{} 0$

and

$ d^2Z \equal{} \frac {1}{2} \sum_{i} \sum_{k} \frac {\partial ^2 Z}{\partial X_i \partial X_k} dX_i dX_k > 0$

and in this last case, we must require that the Hessian determinant is > 0, right?

1) Notation, notation, notation. Some people prefer to say "second order variation", write $ \delta$ and include $ 1/2$ in the definition, some other people prefer to say "second differential", write $ d$ and have $ 1/2$ as an explicit factor. Some even prefer to use it both ways simultaneously. As long as you clearly understand what is meant and can easily translate from one language to another, you shouldn't bother about such things at all. It seems that $ \delta$ is more common in books on theoretical mechanics and $ d$ more common in mathematical ones but who really cares? :roll:

2) Be careful here: what we must require is that the Hessian is non-negative definite (as a necessary condition) and positive definite (as a sufficient condition). This implies non-negativity (positivity) of the determinant but is not implied by it.

[quote="Fedja"]2) Be careful here: what we must require is that the Hessian is non-negative definite (as a necessary condition) and positive definite (as a sufficient condition). This implies non-negativity (positivity) of the determinant but is not implied by it.[/quote]

So, in other words, the Hessian must be $ \geq 0$ and not just > 0? But I have read that, when finding the minimum of a several variable function, if the Hessian is zero than we can make no decision. Also, my text says that the Hessian determinant must be positive, and so also the principal minors, but when translating into mathematical notation the text writes $ \geq 0$ instead of just > 0?. Why is that?

Read again what fedja wrote.

If that Hessian has an eigenvalue $ < 0$, then that point is at best a saddle point. Hence, if that point is a minimum point, then all eigenvalues are $ \geq 0$. However, the converse is not true. When we have some eigenvalues equal to $ 0$, and the others $ > 0$, we must use higher order terms from Taylor's expansion.

What is true is that if all eigenvalues are $ > 0$, then the point is a minimum point (because the quadratic form $ x^T H x$ is always $ > 0$).


on a contest i took today (got a perf), the last problemw as the one that tripped up most people (actually all but two of us), and when my friend asked me how to do it, i told him to use the Principal of inclusion exclusion.... when he asked me to explain why it works on this problem, i coudlnt for the life of me do it....

So can someone please say why it works for this...

9x + 16y = n... find n, where n is the greatest number that cant be made with non-negative integer solutions to x,y

One way I like to think about this problem is to say that you need both x and y to be positive. So we get 9(x+1)+16(y+1)=n, and we want a large n such that there are no solutions. We can see easily that 9 16s can be replaced with 16 9s, but we can't switch anything smaller. Thus 9*16 is not achievable. But if 9(x+1)+16(y+1)=9*16=144, then 9x+16y=144-9-16=119. That's not a proof that it's the largest, but it's true.

yeh that really isnt a good enuff explanation i dont think. Im not looking for something rigorous, just something to tell this person so they understand why it works.

Just noting that this was the mathleague, which I also got a perfect on, and that the third question was really weird cuz the percent was messed up.

Yeah, the wording on the percent problem was confusing.

i didnt think they'd have some irrational number like that as an answer.

Isn't it well-known that if a,b are positive integers with gcd(a,b) = 1, then

ab-a-b

is the greatest positive integer which can't be written as

xa+yb

for positive integers x, y ?

A proof of this useful fact can be found at

http://www.kalva.demon.co.uk/imo/isoln/isoln833.html

Taking a = 16, b = 9 we get the result.

yeah, i missed the percentage problem... sigh.... so stupid. but it had  a very messed up answer.

[quote="Arne"]
ab-a-b

is the greatest positive integer which can be written as

xa+yb

for positive integers x, y ?
[/quote]

You mean can't :)

Yeah, its been mentioned here a lot that problem. Its famous enough to just say that formula really..

what the.. I had the same contest for the "Canadian" National Math League

right.  It's all the same contest, just different name.  They're all run by http://www.mathleague.com .

thanx for all the replies. I guess illl just translate the imo solution for him.

I got a look at the official solutions today (wewt!) Here's a synopsis.

9x+16y=n
If x :ge: 7 then n+1 can be achieved with a multiple of 16.
For example:
9(7)+16(0)=63
9(0)+16(4)=64

If y :ge: 5 then n+1 can be achieved with a multiple of 9.
For example:
9(0)+16(5)=80
9(9)+16(0)=81

Therefore, it can be established that if x:ge:7 and y:ge:5, there exists another pair of positive integers (x,y) for which 9x+16y=n+1. So all
n:ge:9(7)+16(5) can be solved for positive integers (x,y)

So to get the largest number that CAN'T be expressed as 9x+16y, x<7 and y<5. So x=6, y=4. 6*9 + 4*16 = 118, but this number can be expressed, so add 1 to get 119.

As a quick check...
(in the form n = x9 + y16)
120 =   8*9 + 3*16
121 =   1*9 + 7*16
122 = 10*9 + 2*16
123 =   3*9 + 6*16
124 = 12*9 + 1*16
125 =   5*9 + 5*16
126 = 14*9 + 0*16
127 =   7*9 + 4*16
128 =   0*9 + 8*16
129 =   9*9 + 3*16
130 =   2*9 + 7*16
And so on.  (Hey, if you look at the values of x and y there, specifically for every other n, you'll start to see patterns.)

Anyway, I guess we'll all have to remember how to do this problem.

yeh um... i was leafing through the diophantine section of aops 2 today and it mkaes a lot of sense. I was able to derive the formula using the table of mods idea it presents.

And through that, I have come to the conclusion that this isnt really PIE.... just something that looks a LOT like it.... and another conclusion is that i am very very happy that it still worked. I guess I must have remembered this formula from soemthing, and at the same time, i was remembering PIE, so i attached the name to it.


The noon temperatures for seven consecutive days were $ 80^{\circ}$, $ 79^{\circ}$, $ 81^{\circ}$, $ 85^{\circ}$, $ 87^{\circ}$, $ 89^{\circ}$, and $ 87^{\circ}$ Fahrenheit. What is the mean noon temperature, in degrees Fahrenheit, for the week?

The average of any collection of numerical data is the sum of the pieces of data divided by the number of pieces of data. $ \frac{80\plus{}79\plus{}81\plus{}85\plus{}87\plus{}89\plus{}87}{7}\equal{}\boxed{84}$.


limit
logarithms
induction
calculus
calculus computations

[color=darkblue]Consider the recurrent sequence $ x_1 > 1$ and $ e^{x_{n \plus{} 1}} \equal{} e\cdot x_n$ for any $ n\in \mathbb N^*$ .

Prove that $ x_n\rightarrow 1$ . Ascertain $ \lim_{n\to\infty}x_n^n$ and $ \lim_{n\to\infty}n\cdot \left(x_n\minus{}1\right)$ .[/color]
[hide="Answer."]
$ \lim_{n\to\infty}x_n^n\equal{}e^2$ and $ \lim_{n\to\infty}n\cdot \left(x_n\minus{}1\right)\equal{}2$ .[/hide]

[color=darkblue]Here is a solution for Japanese highschool students who don't usually study [/color]$ \epsilon - \delta$.

$ x_n > 1\Longrightarrow e^{x_{n + 1}} > e$. $ \therefore x_{n + 1} > 1$ and $ x_1 > 1$,  Inductively $ x_n > 1$ for $ \all n\in\mathbb{N}$.

Now $ e^{x_{n + 1}} = e\cdot x_n\Longleftrightarrow x_{n + 1} = 1 + \ln x_n$.

Let $ f(x) = 1 + \ln x$ and $ x_{n + 1} = f(x_n)$, we have $ x_{n + 1} = f(x_n),\ 1 = f(1)$. Thus

By Mean Value of Theorem,

$ \displaystyle |x_{n + 1} - 1| = |f(x_n) - f(1)| = \left|\frac {f(x_n) - f(1)}{x_n - 1}\right \parallel{}x_n - 1|\ (1 < c < x_n)$

$ = |f'(c)\parallel{}x_n - 1|$

$ = \frac {1}{c}|x_n - 1|$.

Let $ r = \frac {1}{c}$, $ 0 < r < 1$, we obtain $ |x_{n + 1} - 1| < r|x_n - 1|$.

$ \therefore |x_n - 1| < r|x_{n - 1} - 1| < r^2|x_{n - 2} - 1| < \cdots < r^{n - 1}|x_1 - 1|$.

$ \therefore 0 < |x_{n} - 1| < r^{n - 1}|x_1 - 1|$. By $ \lim_{n\to\infty} r^{n - 1}|x_1 - 1| = 0$, we have

$ \lim_{n\to\infty} |x_n - 1| = 0\Longleftrightarrow \lim_{n\to\infty} x_n = 1$.

$ x_n^n \equal{} e^{n(x_{n \plus{} 1} \minus{} 1)}$

$ \lim x_n^n \equal{} e^{\lim (n \plus{} 1)(x_{n \plus{} 1} \minus{} 1) \minus{} \lim (x_{n \plus{} 1} \minus{} 1)}$

$ \lim x_n^n \equal{} L_1, \lim n(x_n \minus{} 1) \equal{} L_2, \lim x_n \minus{} 1 \equal{} 0$

$ L_1 \equal{} e^{L_2}$

edit: in the sense that existence of one implies existence of the other and this relationship.

That is not a proof.
Why does $ \lim_{n \to \infty} x_n^n$ exist?

A first try (partial result):
Let $ k \in \mathbb{N}$ be such that $ x_1 > 1 \plus{} \frac {2}{1 \plus{} k}$.
Statement 1:
$ x_n > 1 \plus{} \frac {2}{n \plus{} k}$ for all $ n$.
Induction:
It is true for $ n \equal{} 1$.
$ x_{n \plus{} 1} \equal{} 1 \plus{} \ln(x_n) > 1 \plus{} \ln(1 \plus{} \frac {2}{n \plus{} k}) > 1 \plus{} \frac {2}{n \plus{} 1 \plus{} k}$

So, $ n(x_n \minus{} 1) > \frac {2n}{n \plus{} k}$ for all $ n$ and
$ \liminf_{n \to \infty} n(x_n \minus{} 1) \ge 2$


geometry
circumcircle
trigonometry
geometric transformation
reflection
power of a point
radical axis

In a  triangle $ \triangle ABC $ , let  $H $ the orthocenter , $AM $ a median  and $ HE \perp AM $ ( E is point on AM) . If  $ N $ is the symmetric point of  $E $ with respect to $ BC $, prove that $ N $ belongs to the circumcircle of $\triangle ABC $.

Babis

 [i]( Years 1950 - 1978 were the golden age for  geometry in Greece. In that perod were written the best geometry books. Now , nobody solves geometry here  - only  a few exseptions .)[/i]

I've seen a very similar problem before. It's obvious that it's actually the same question (just look carefully :)). It's from some old Kurschak competition, I am copy-pasting my solution from a document I wrote, a long time ago (so I'm sorry for the trig here, I wasn't such a "pure" geometer back then :D):

[b]Problem 17.[/b] Let $ABC$ be a triangle and let $M$ be the midpoint of $BC$. Let the line $AM$ meet the circumcircle of $ABC$ again at $A_1$ and let $A_2$ be the reflection of $A_1$ in $M$. Define the points $B_2$ and $C_2$ similarly. Prove that the orthocentre of $ABC$ lies on the circumcircle of $A_2B_2C_2$.

[b]Solution.[/b] A stronger result is true. In fact, the points $A_2$, $B_2$ and $C_2$ lie on the circle with diameter $GH$, where $G$ and $H$ are the centroid and orthocentre of $ABC$ respectively. It is therefore sufficient to prove that $\angle{GA_2H} = 90^{\circ}$. Let $AH$ meet $BC$ at $P$. Since $A_2$ lies on $AG$, and $\angle{APM} = 90^{\circ}$, it is sufficient to prove that $PMA_2H$ is cyclic, which is equivalent to $AA_2 \cdot AM = AH \cdot AP$. Hence we focus on proving this equality. Notice that $AA_2 \cdot AM = (AM - A_1M)\cdot AM = AM^2 - A_1M\cdot AM = AM^2 - BM \cdot CM$ (by power of a point) which equals $\frac{1}{4}\cdot\left(2b^2 + 2c^2 - a^2\right) - \frac{1}{4}a^2$ (by the formula for the length of a median), which is equal to $\frac{1}{2}\left(b^2 + c^2 - a^2\right) = bc \cdot \cos \alpha$ (by the cosine rule). (As usual we write $a = BC$, $b = CA$, $c = AB$). Let $BH$ meet $AC$ at $Q$. It is easy to see that $PHQC$ is cyclic and hence $AH \cdot AP = AQ \cdot AC = (c \cdot \cos \alpha) \cdot b = bc \cdot \cos \alpha$. It follows that $AH \cdot AP = AA_2 \cdot AM$ and we are done.

It's equivalent to $\angle BNC=\pi-\angle BAC$, or $\angle BEC=\angle BHC$, which is another way of saying that $BHEC$ is cyclic.

If $A',B',C'$ are the feet of the perpendiculars through $A,B,C$ and $M$ is the midpoint of $BC$, then the inversion of pole $A$ and power $AH\cdot AA'$ turns $B,C,H,E$ into $C',B',D,M$, which are concyclic (they all lie on the nine-point circle of $ABC$). This means that $B,C,H,E$ must also be concyclic, and we're done.

[quote="Arne"][b]Problem 17.[/b] Let $ABC$ be a triangle and let $M$ be the midpoint of $BC$. Let the line $AM$ meet the circumcircle of $ABC$ again at $A_1$ and let $A_2$ be the reflection of $A_1$ in $M$. Define the points $B_2$ and $C_2$ similarly. Prove that the orthocentre of $ABC$ lies on the circumcircle of $A_2B_2C_2$.
[...] In fact, the points $A_2$, $B_2$ and $C_2$ lie on the circle with diameter $GH$, where $G$ and $H$ are the centroid and orthocentre of $ABC$ respectively. It is therefore sufficient to prove that $\angle{GA_2H} = 90^{\circ}$.[/quote]

Thanks for reminding me about this ;) . Indeed, this was discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=384 and Grobber gave a synthetic solution. More solutions can be found in [url=http://groups.yahoo.com/group/Hyacinthos/messages/8744?xm=1&viscount=7&m=e]Hyacinthos messages #8744, #8747, #8748, #8751[/url].

The above result would also follow from the generalization of the Fuhrmann circle, but I don't have a synthetic proof of this generalization yet.

  darij

Hehe, I remember this was one of the first "harder" problems I "discovered" when I started my training two years ago :)  [I couldn't solve it back then...] Since then, I've always remembered this problem as one of my favourite geometry problems - geometrical facts. I even used it in my geometry lectures in training camps for IMO 2005.

This nice problem becomes very easy problem if the student knows the following remarkable property of a triangle $ABC$ w.r.t. its circumcircle $w=C(O,R)$ :
"The reflection of the orthocenter $H$ in the line $BC$ and the reflection $A'$ of the same point $H$ in the midpoint $M$ of the side $[BC]$ belong to the circle $w$ $(A'\in AO\cap w)$.

[b]Solution.[/b] I note $\{ F,A\}\subset AM\cap w$. From the relations $MH=MA',\ HE\perp AM,\ FA'\perp AF\equiv AM$ results $HE\parallel FA',\ HE=FA',\ MF=ME=MN\Longrightarrow  MF=MN\Longrightarrow $ the point $N$ is the reflection of the point $F$ w.r.t. the diameter which passes through $M$, i.e. the point $N$ belongs to the circle $w.$

Here is my solution (?) :

  It is easy to see and feel that point E is the orthocenter of triangle A'BC,

  where A'  is on the circle ABC and A'E _|_ BC.

  Thank you.

  M.T.

For the initial problem: we need to show that $L$ belongs to the circle with diameter $AH$. And to prove this, the “inversion is incredible”: the image of this circle in the inversion of pole $A$ with power $AH\cdot AD$ is the line $BC$: let ${E} \equiv BH \cap AC$, ${F} \equiv CH\cap AB$, and ${D}\equiv AH \cap AD$. Obviously, $AF\cdot AB=AC\cdot AE=AH\cdot AD$, but from power of $A$ to the circle of diameter $BC$  we have $AE\cdot AB=AM^2-MB^2$, so we have to prove that $AL\cdot AM=AM^2-BM^2$.
As already shown, $AM$ is the radical axis of the 2 circles tangent to $BC$, hence: $BM^2=AM\cdot ML$. Consequently, $AL\cdot AM=AM\cdot AM-AM\cdot ML=AM^2-AM\cdot ML=AM^2-MB^2$, and our assumption is proved to be correct.

Best regards,
sunken rock


projective geometry
geometry solved
geometry

Let $\Gamma$ be a circumference, and $A,B,C,D$ points of $\Gamma$ (in this order).

$r$ is the tangent to $\Gamma$ at point A.
$s$ is the tangent to $\Gamma$ at point D.

Let $E=r \cap BC,F=s \cap BC$.
Let $X=r \cap s,Y=AF \cap DE,Z=AB \cap CD$
Show that the points $X,Y,Z$ are collinear.

Note: assume the existence of all above points.

let $XY\cap AD=M,XY\cap BC=N,AD\cap BC=P$

use ceva and menelaus in triangle $XAD$ you can reach that $(AD,MP)=-1 $

so,$XY$ is polar of $P$   $\longrightarrow$$ (BC,NP)=-1$ now you can get $Z$ is on polar of $P$. therefore $X,Y,Z$are collinear.

[quote="Johann Peter Dirichlet"]Let $\Gamma$ be a circumference, and $A,B,C,D$ points of $\Gamma$ (in this order).

$r$ is the tangent to $\Gamma$ at point A.
$s$ is the tangent to $\Gamma$ at point D.

Let $E=r \cap BC,F=s \cap BC$.
Let $X=r \cap s,Y=AF \cap DE,Z=AB \cap CD$
Show that the points $X,Y,Z$ are collinear.[/quote]

This is a purely projective problem.

Consider the degenerate hexagon BCDDAA inscribed into the circle $\Gamma$. Applying the Pascal theorem to this hexagon, we see that the points $BC\cap DA$, $CD\cap AA$ and $DD\cap AB$ are collinear. Hereby, the lines AA and DD have to be understood as the tangents to the circle $\Gamma$ at the points A and D; in other words, these lines AA and DD are simply our lines r and s. Hence, $BC\cap DA=EF\cap DA$, $CD\cap AA=CD\cap r=XE\cap ZD$ and $DD\cap AB=s\cap AB=FX\cap AZ$. Hence, we obtain that the points $EF\cap DA$, $XE\cap ZD$ and $FX\cap AZ$ are collinear. Thus, by the converse of the Desargues theorem, the triangles XEF and ZDA are perspective, i. e. the lines XZ, ED and FA are concurrent. In other words, the line XZ passes through the point $ED\cap FA$. But $ED\cap FA=AF\cap DE=Y$; thus, the line XZ passes through the point Y. In other words, the points X, Y, Z are collinear. Problem solved.

  darij

Let AP and DQ be the polars of E and F, where $P,Q\in\Gamma$. Now it suffices to prove that $AP\cap FD,DQ\cap AE,AD\cap BC$ are collinear, which is equivalent to triangles XAD and SPQ are perspective, where $S=AP\cap DQ$. This can be done by Brianchon's Theorem.

Let $N=AC\cap BD$.

Then $X$, $N$ and $Z$ are (by well known theorem) collinear.

Let $Y_1 = XZ\cap DE$ and $Y_2=AF \cap XZ$.

Then we have:

$(F,C,D,E) = (AF,AC,AB,AE)=(AY_1,AN,AZ,AX)=(Y_1,N,Z,X)$
and
$(F,C,D,E)= (DF,DC,DB,DE)=(DX,DZ,DN,DY_2)=(X,Z,N,Y_2)$

Now we have $(X,Z,N,Y_2)=(Y_1,N,Z,X)$ which means $Y_1=Y_2=Y$
So $X,Y,Z$ are collinear.


abstract algebra
superior algebra
superior algebra solved

Let A be a simple R-module. Prove that Hom_R (A, A) is a division ring.


Let ABC be an acute triangle and (O) be its circumcircle. The tangents from
A and B to (O) meet the tangent from C to (O) in points D and E respectively.
Let AE meet BC at P and BD meet AC at R. Points Q and S are
respectively the midpoints of AP and BR. Prove that <ABQ = <BAS.

This "infamous" problem has been posted before. For instance, see

[url]http://www.artofproblemsolving.com/viewtopic.php?t=275380[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207440[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19806[/url]


conics
ellipse
analytic geometry
graphing lines
slope
special factorizations
algebra unsolved

$ x^3\plus{}y^2\equal{}2$
$ x^2\plus{}xy\plus{}y^2\minus{}y\equal{}0$

This is a easy system of equation You let equation (1) -equation(2)
we have :(x^2-y)(x-1)=0

On subtracting the two equations, we get (x^2 - y)(x - 1) = 2 (and not 0).

Substituting y = +or- sqrt(2- x^3) from the first equation into the second, we get
x^6 - x^5 + 3x^4 + 5x^3 - 6x^2 -4x + 2 = 0,
giving (x,y) = (1, 1), (1, -1), (-1, sqrt3), (-1, -sqrt3).
Of these, only (1, -1) satisfies both equations.
There is another root between x = 0 and x = 0.5 but I haven't got a calculator to hand.

Found a calculator:

x is about 0.371 and so y is about +/- 1.396, both of which satisfy the first equation but not the second. So the only solution is (x,y) = (1, -1).

I divided the sextic by (x+1), (x+1) and (x-1) to get the cubic f(x) = x^3 -2x^2 +6x - 2.
f(0) = -2 and f(1/2) = 5/8.

There must be a neater way (which may be (by working backwards) how the original pair of equations was derived).

It seems that there are no real solutions to this pair of equations.

A rough sketch shows that the second equation represents an ellipse passing through (0,0) and (0,1), whereas the first equation represents a curve passing though (0, sqrt2) and missing the ellipse completely.

(I must try to think before I rush into posting.)

I still think that there must be a neat way of doing this.

By differentiating, we see that the slope of the first curve is less steep than that of the ellipse for 0<x<2/3 and x = 2/3 is to the right of the ellipse since by completing the square y^2 is negative when x = 2/3.

Had the 2 in the first equation been 0, then we have (x,y) = (0,0) or (-1,1).

[color=darkblue]Hi! duongptnkvn.

Your system of equtions is nice.

This is solution:

System $ \Longleftrightarrow \begin {cases} x^3 \plus{} y^2 \equal{} 2 \\
x^3 \minus{} x^2 \minus{} xy \plus{} y \equal{} 2 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^3 \plus{} y^2 \equal{} 2 \\
(x \minus{} 1).y \equal{} x^3 \minus{} x^2 \minus{} 2 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^3 \plus{} y^2 \equal{} 2 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$

$ \Longleftrightarrow \begin {cases} x^3 \plus{} \frac {(x^3 \minus{} x^2 \minus{} 2)^2}{(x \minus{} 1)^2} \equal{} 2 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6 \minus{} x^5 \minus{} x^4 \minus{} 3x^3 \plus{} 2x^2 \plus{} 2x \plus{} 2 \equal{} 0 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$

$ \Longleftrightarrow \begin {cases} (x^3 \minus{} 2)(x^3 \minus{} x^2 \minus{} x \minus{} 1) \equal{} 0 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$

Hence we get two systems:
$ \begin {cases} x^3 \minus{} 2 \equal{} 0 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$ ; $ \begin {cases} x^3 \minus{} x^2 \minus{} x \minus{} 1 \equal{} 0 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$

Then it's easy.[/color]

--------------------------------------------------------------------------------
My school site: http://www.chuyenquangtrung.com.vn

Fist system: $ \begin {cases} x^3 \minus{} 2 \equal{} 0 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$

$ \Longleftrightarrow \begin {cases} x^3 \equal{} 2 \\
y \equal{} \frac { \minus{} x^2 }{x \minus{} 1} \end {cases}$

$ \Longleftrightarrow \begin {cases} x \equal{} \sqrt [3]{2} \\
y \equal{} \frac { \minus{} \sqrt [3]{4} }{\sqrt [3]{2} \minus{} 1} \end {cases}$

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My school site: http://www.chuyenquangtrung.com.vn

Second system: $ \begin {cases} x^3 \minus{} x^2 \minus{} x \minus{} 1 \equal{} 0 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$

$ \Longleftrightarrow \begin {cases} x^3 \minus{} x^2 \equal{} x \plus{} 1 \\
y \equal{} \frac {x \minus{} 1 }{x \minus{} 1} \equal{} 1\end {cases}$

But $ x^3 \plus{} y^2 \equal{} 2$ then we get $ x^3 \equal{} 1 \Longleftrightarrow x \equal{} 1$, but $ x^3 \minus{} x^2 \minus{} x \minus{} 1 \equal{} 0$ so $ x \equal{} 1$ is't root.

Hence second system $ \begin {cases} x^3 \minus{} x^2 \minus{} x \minus{} 1 \equal{} 0 \\
y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1} \end {cases}$ has not root.

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My school site: http://www.chuyenquangtrung.com.vn

[quote="ith_power"][quote="AndrewTom"]Nice, but you've changed the second equation.[/quote]
its not changed we get it upon substracting the 2 equations.[/quote]

Yes. Let $ (1) \minus{} (2)$ we get $ x^3 \minus{} x^2 \minus{} xy \plus{} y \equal{} 2$

Then $ \begin {cases} x^3 \plus{} y^2 \equal{} 2 \\
x^2 \plus{} xy \plus{} y^2\minus{}y \equal{} 0 \end {cases} \Longleftrightarrow \begin {cases} x^3 \plus{} y^2 \equal{} 2 \\
x^3 \minus{} x^2 \minus{} xy \plus{} y \equal{} 2 \end {cases}$

AndrewTom is right, there is no real solutions.
We find $ x^6\plus{}2\minus{}x^5\minus{}x^4\minus{}3x^3\plus{}2x^2\plus{}4x\equal{}0$; this equation has no real solutions in $ x$.