LaTeX
inequalities
inequalities proposed

Just look at this one :

         If for the positive real x,y,z  is true xy+yz+xz=2 then provethe following enequality:

                        [b] symsum[{xy}^2]+symsum[{xyz}^2]>= symsum[(x^3)(y^2)z]+symsum[(x^2)yz][/b]

[quote="Arne"]If $x,y,z >0$ and $xy + yz + zx = 2,$ show that \[ \sum x^2y^2 + \sum x^2y^2z^2 \geq \sum x^3y^2z + \sum x^2yz.  \][/quote]
The inequality is equivalent to $\dfrac{1}{2}(\sum x^3y^3+\sum_{sym}x^3y^2z)+3x^2y^2z^2\ge \sum_{sym}x^3y^2z+\dfrac{1}{2}(\sum_{sym}x^3y^2z+3x^2y^2z^2)\Leftrightarrow\sum x^3y^3+3x^2y^2z^2\ge 2\sum_{sym}x^3y^2z$ that is incorrect for $x=y=z$ but correct the version $\sum x^3y^3+3x^2y^2z^2\ge \sum_{sym}x^3y^2z$ which is just Shur for substitution $xy=a,yz=b,zx=c$

[quote="amanin"][quote="Arne"]If $x,y,z >0$ and $xy + yz + zx = 2,$ show that \[ \sum x^2y^2 + \sum x^2y^2z^2 \geq \sum x^3y^2z + \sum x^2yz.  \][/quote]
The inequality is equivalent to $\frac{1}{2}(\sum x^3y^3+\sum_{sym}x^3y^2z)+3x^2y^2z^2\ge \sum_{sym}x^3y^2z+\frac{1}{2}(\sum_{sym}x^3y^2z+3x^2y^2z^2)\Leftrightarrow\sum x^3y^3+3x^2y^2z^2\ge 2\sum_{sym}x^3y^2z$ that is incorrect for $x=y=z$ but correct the version $\sum x^3y^3+3x^2y^2z^2\ge \sum_{sym}x^3y^2z$ which is just Shur for substitution $xy=a,yz=b,zx=c$[/quote]
why the inequality is incorrect for x=y=z?

After the substitution a=xy, b= xz, c=yz we get using the equality a+b+c=2 and

[b]$a^3+b^3+c^3+3abc=2(a^2+b^2+c^2-ab-ac-bc)+6abc$[/b]

a shur's  inequaliti and the inequality is proved


I can type the word munchkins in 72 minutes. How many minutes will it take for me to type the word floccinoccinihilapilaphacation?
 :starwars:

Assuming constant rate of typing per letter, it takes $ 8$ minutes to type one letter.  There are $ 30$ letters, so $ 8 \times 30\equal{}\boxed{240 \text{ minutes}}$.

Since you can type 8 min per one letter. there are thirty letters in the word floccinoccinihilapilaphacation so $ 8\times 30\equal{}\boxed{240\text{ minutes}}$. I did not look at the answer.

dude, don't post a solution unless it's different from the person above or u think the guy above (AIME15  :P ) is wrong!

EDIT: do u really think AIME15 is wrong?

according to my calculation I am right and thanks for the tip :furious:  :no:  :cursing:

munchkins has 9 letters
72 divided by nine is 8
Flo-blabla-somtin-somtin-tion has got 30 letters
8x30=240 min or 4 hours

ps: the problem has 130 character (with spaces) so it took you 17 hours and 20 minutes to write this.  :sleep2:

I think we already know the answer and there is no need for anymore solutions.


probability
combinatorics unsolved
combinatorics

Hi every body
could any one help me to solve this problem
i did not know where to write the topic
[img]http://img409.imageshack.us/img409/4158/22rw1.jpg[/img]
REGARDS
Please help me

For the second question:
The probability of rolling a $6$ is $\frac{1}{6}$ and the probability of rolling anything but a $6$ is $\frac{5}{6}$.  So for the first part of that question there are $\binom{4}{1}= 4$ ways of getting exactly one $6$.  So we have  $4\cdot\left( \frac{1}{6}\right) \cdot\left(\frac{5}{6}\right)^{3}$.  That will give you the probability of exactly one - see if you can use that to figure out the prob of at least one.

For the first question
P(1 works | system is functioning)
=P(1 works and system is functioning) / P(system is functioning)
=1/2 * (1 - (1/2)^(n-1)) / (1- (1/2)^(n-1))


inequalities
vector
linear algebra
matrix
linear algebra solved

Let $P,Q,R \in M_n(K)$. Prove that $rang(PQ)+rang(QR) \leq rang(Q)+rang(PQR)$

It is Frobenius inequality:
Firstly let's prove useful inequality,considering our matrices as mapping on vector space over $K$.Let $U \subset V$ and $X$:$V \to W $. Then $dim(KerX|_{U})\leq dimKerX=dimV-dimImX$ (1). If we use inequality (1) for the case $U=ImQR$,$V=ImQ$,$X=P$ we obtain that $dim(KerP|_{ImQR})\leq dimImQ-dimImPQ$  (2). On the other hand it is readily that $dim(KerP|_{ImQR})=dimImQR-dimImPQR$ (3). So, applying (2) and (3) we get our inequality

A different approach.

\[
\begin{pmatrix}
PQ & 0\\
Q & QR
\end{pmatrix}
\]
Multiply the first column by $-R$ to the right and add it to the second column, then multiply the second column by $-1$,
\[
\begin{pmatrix}
PQ & PQR\\
Q & 0
\end{pmatrix}
\]
Multiply the second row by $-P$ to the left, and add it to the first row
\[
\begin{pmatrix}
0 & PQR\\
Q & 0
\end{pmatrix}
\]

$rank(PQ)+rank(QR)=rank
\begin{pmatrix}
PQ & 0\\
0 & QR
\end{pmatrix}
\leq rank
\begin{pmatrix}
PQ & 0\\
Q & QR
\end{pmatrix}
=rank
\begin{pmatrix}
0 & PQR\\
Q & 0
\end{pmatrix}
=rank(PQR)+rank(Q)
$

where $rank(M)=rang(M)=r(M)$ denotes the rank of a matrix $M$.


Right, here is the news story.
http://news.yahoo.com/s/ap/20070410/ap_on_sp_co_ne/bkc_texas_durant
Comments?

Expected.

He is the next Kevin Garnett (but with a better shooting touch)

UT could mean University of Tennessee...
but yeah
He's a spectacular freshman prospect from the University of Texas.

A great guy that's going to wow lots of people.

Anyone think that he's going to get picked first in the draft?
I think it's highly reasonable, but it definitely depends on the team.

Alot of teams will want this kid
He's a swingman (means he can play Shooting Guard and Small Forward) and could play Power Forward if he develops his body into it.
Only teams stacked at Shooting Guard, Small Forward, and Power Forward won't pick him
Teams like:
Dallas Mavericks (Jason Terry, Josh Howard, Jerry Stackhouse, Dirk Nowitzki)
Detroit Pistons (Richard Hamilton, Tashyuan (sp?) Prince, Rasheed Wallace, Chris Webber, Antonio McDeyss (sp?)
Chicago Bulls (Ben Gordon, Luol Deng, Tyrus Thomas, P.J. Brown)
San Antonio (Manu Ginobli, Bruce Bowen, Tim Duncan)
aren't going to pick Durant
but these teams are all Playoff Contenders

so yeah
HE WILL GO NUMBER ONE! :D

Agreed. All the teams with decent big men, minus center, are strong playoff contenders... and all the teams headed for the draft are just the opposite. So yeah, Kevin Durant for #1.

this should go to G&FF

[i][Moderator says: So it is written, so it shall be done.][/i]

[quote="besttate"]Oden will go first. He has much more franchise value than Durant. Good centers are hard to come by, and swingmen aren't. Even if Durant is better, he won't get picked first based on how much he can improve a team over Oden.[/quote]

Are you even a sports fan?
Oden will have to develop his game
He gets WAY TOO many fouls during games
This renders him useless in the NBA
Good centers will not win games
San Antonio Spurs don't have a good center
Miami has an old center that's now a shadow of what he use to be
Detroit had a undrafted Center
Oden will never equal Shaq's dominance
Swingman can easily improve teams
Centers take TOO LONG to develop

Durant's franchise value > Oden's franchise value

I agree with besttate. I think Durant is more athletic and a better fit for the NBA, but that shouldn't take away any credit from Oden.
Plus, teams like the Celtics or the Hawks who might get top-3 lotto picks could definitely use Oden.
On a side note, DJ Augustin went to my school last year, and man, he was hoss.  :) People went to the games just to see him play (he was an All-American!). Our basketball team is consistently good, so I'm surprised we didn't get further in the playoffs last year. Wonder where he will end up in 3 years! Anyway, go UT! w00t.

eh...dont blame really good players to go pro early.  same with lendale white and a bunch of other football players.

[quote="Aznwithabrain"][quote="Art of Owna"][quote="mustafa"]Oden should stay in college to refine his skillz[/quote]

lol guess you didnt see Oden in the championship game-and he was playing against two lottery picks and another possible NBA player...

plus yea big guys are hard to come by, even if Durant is more talented than Oden, you still have to go with Oden because its pretty rare to find a seven footer with that much talent, as Oden can change a game without even touching the ball-he will draw double teams that can free up his teamates and hes a lot better defensive player than Durant, as he will make people change how they will shoot the ball when driving for a layup because those people dont want to get the shot blocked (if you look at the history of the NBA draft, most of the first overall picks have been guy over 6'10"-Bariangi (sp?), Bogut, Yao, Kwame Brown, Shaq, Elton Brand, Hakeem, Ewing.. and so one..not all of them have worked out but Oden is a lot better than people like Kwame Brown and stuff..)[/quote]

1) Oden was playing against Horford and Noah
2) Noah was HORRIBLE in the championship game. Noah will never survive in the NBA
3) Horford is undersized
3) Oden gets in too much foul trouble so the double teams won't come that often and the block shots won't come often
4) Oden will never be able to match up with other playoff team big men and make a difference.
5) Durant plays decent defense

If Oden stays in college, he will become a better player than durant
but that's only if

Durant will become remarkable player[/quote]

But Oden is leaving for the NBA (announced today), but hey look at Hakeem Olajuwan..he was pretty raw (just came from Nigeria where he never played basketball) like Oden and he would also get in foul trouble and didnt have that much stamina....when he came out for the NBA draft in 1984..hes one of the greatest centers 25 year later looking back on it (voted one of the top 50 NBA greatest players), but then again Michael Jordan was the 3rd pick that draft...hmm could Kevin Durant be the MJ of this draft..

[quote="Art of Owna"][/quote][quote="Aznwithabrain"][quote="Art of Owna"][quote="mustafa"]Oden should stay in college to refine his skillz[/quote]

lol guess you didnt see Oden in the championship game-and he was playing against two lottery picks and another possible NBA player...

plus yea big guys are hard to come by, even if Durant is more talented than Oden, you still have to go with Oden because its pretty rare to find a seven footer with that much talent, as Oden can change a game without even touching the ball-he will draw double teams that can free up his teamates and hes a lot better defensive player than Durant, as he will make people change how they will shoot the ball when driving for a layup because those people dont want to get the shot blocked (if you look at the history of the NBA draft, most of the first overall picks have been guy over 6'10"-Bariangi (sp?), Bogut, Yao, Kwame Brown, Shaq, Elton Brand, Hakeem, Ewing.. and so one..not all of them have worked out but Oden is a lot better than people like Kwame Brown and stuff..)[/quote]

1) Oden was playing against Horford and Noah
2) Noah was HORRIBLE in the championship game. Noah will never survive in the NBA
3) Horford is undersized
3) Oden gets in too much foul trouble so the double teams won't come that often and the block shots won't come often
4) Oden will never be able to match up with other playoff team big men and make a difference.
5) Durant plays decent defense

If Oden stays in college, he will become a better player than durant
but that's only if

Durant will become remarkable player[/quote][quote="Art of Owna"]

But Oden is leaving for the NBA (announced today), but hey look at Hakeem Olajuwan..he was pretty raw (just came from Nigeria where he never played basketball) like Oden and he would also get in foul trouble and didnt have that much stamina....when he came out for the NBA draft in 1984..hes one of the greatest centers 25 year later looking back on it (voted one of the top 50 NBA greatest players), but then again Michael Jordan was the 3rd pick that draft...hmm could Kevin Durant be the MJ of this draft..[/quote]

You do make some good points like:
Oden's foul troubles are similiar to Hakeem Oajuwan
Durant could be a bit similiar to Jordan

I agree with 3/4 of it
but
I'm pretty sure Olajuwan didn't come out of college in his freshman year
and
I'm certain Olajuwan developed his game in college
I remember reading numerous articles of him practicing in gyms with Houston Rocket players in collge
I think he came out in his junior or senior year

Don't forget:
Jordan came out as a Junior (certain)
None of those two came out as freshman

7 years later durant second draft pick and oden first draft pick durant went to the finals already and has a mvp award

Don't bump seven year old threads... that is an example of how NEVER to use the forums...


calculus
integration
calculus computations

Evaluate

\[ \int \frac{x^2 e^{\arctan x}}{\sqrt{1\plus{}x^2}}\,dx\]

I first differentiated

$ \sqrt{1\plus{}x^{2}}e^{\arctan x }$ to get $ \frac{(x\plus{}1)e^{\arctan x}}{\sqrt{1\plus{}x^{2}}}$

and then

$ x\sqrt{1\plus{}x^{2}}e^{\arctan x}$ to get $ \frac{(2x^{2} \plus{} x \plus{} 1)e^{\arctan x}}{\sqrt{1\plus{}x^{2}}}$.

Subtracting, these we have

$ \frac{2x^{2}e^{\arctan x}}{\sqrt{1\plus{}x^{2}}}$

and so $ \int \frac{x^{2}e^{\arctan x}}{\sqrt {1\plus{}x^{2}}} dx \equal{} \frac{ (x\minus{}1)\sqrt{1\plus{}x^{2}}e^{\arctan x}}{2} \plus{}c$.


Given $ 9^6 \equal{} 531441$, how would you represent $ 531440$ in base 9?

[hide="Answer"][b]Answer:[/b] $ 888888_9$[/hide]
I thought I had this one.. :(

[quote="Quickster94"]just like 10^3=1000, 1000-1=999.

So, when ur in base 9, one less then 9^6 should equal 88888.

Not sure how good the explanation is...[/quote]

It should have 6 8s; I'm not sure if that's a typo or not...


calculus
limit
trigonometry
function
derivative
complex analysis
calculus computations

And this? how calculate?

$ \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} \minus{} \frac{1}{{\cos x \cdot senx}}} \right)$

Thanks

Antonio

$ \lim_{x\to 0} \left ( \frac{1}{x} \minus{} \frac{1}{\sin x\cos x} \right )$
$ \equal{}\lim_{x\to 0} \left ( \frac{\sin x\cos x}{x\sin x\cos x} \minus{} \frac{1}{\sin x\cos x} \right )$
Use $ \lim_{x\to 0} \frac{\sin x}{x}\equal{}1$
$ \equal{}\lim_{x\to 0} \left ( \frac{\cos x\minus{}1}{\sin x\cos x} \right )$
Multiply by $ \frac{\cos x\plus{}1}{\cos x\plus{}1}$
$ \equal{}\lim_{x\to 0} \left ( \frac{\cos ^2x\minus{}1}{\sin x\cos x(\cos x\plus{}1)} \right )$
$ \equal{}\lim_{x\to 0} \left ( \frac{\minus{}\sin ^2x}{\sin x\cos x(\cos x\plus{}1)} \right )$
$ \equal{}\lim_{x\to 0} \left ( \frac{\minus{}\sin x}{\cos x(\cos x\plus{}1)} \right )$
$ \equal{}\frac{0}{1(2)}$
$ \equal{}0$

[quote="EastyMoryan"]$ \lim_{x\to 0} \left ( \frac {1}{x} \minus{} \frac {1}{\sin x\cos x} \right )$
$ \equal{} \lim_{x\to 0} \left ( \frac {\sin x\cos x}{x\sin x\cos x} \minus{} \frac {1}{\sin x\cos x} \right )$
Use $ \lim_{x\to 0} \frac {\sin x}{x} \equal{} 1$
$ \equal{} \lim_{x\to 0} \left ( \frac {\cos x \minus{} 1}{\sin x\cos x} \right )$
.
.
.
.
[/quote]

I have a doubt in this ... [b]when you apply the limit, shouldn't it be done for the entire function and not just a part of the function[/b]?

This is a wonderful problem for the Taylor series approach: $ \frac {1}{\cos x} \equal{} 1 \plus{} O(x^2)$.  $ \frac {x}{\sin x} \equal{} 1 \plus{} O(x^2)$.  So we have $ \lim_{x \to 0} \frac {1}{x} \minus{} \left(\frac {1}{x} \plus{} O(x)\right) \equal{} \lim_{x \to 0} O(x) \equal{} 0.$

If you replace $ \sec x$ with a function $ f$ such that $ f(0) \equal{} 1$ and $ f'(0) \equal{} a$ (I don't think I need any more niceness for $ f$ than the existence of this derivative at 0 -- is that right?) then the answer becomes $ \minus{}a$.

Can you please explain how is $ \frac {1}{\cos x} \equal{} 1 \plus{} O(x^2)$? Sorry I m weak in this kind of approximations.

The linear approximation (degree-1 Taylor series approximation) of $ \sec x$ at $ x \equal{} 0$ is $ \sec 0 \plus{} \sec' 0 x$, and this is just equal to 1, so $ \sec x \equal{} 1 \plus{} O(x^2)$.

$ \equal{} \lim_{x\to 0}\left (\frac {\sin x\cos x}{x\sin x\cos x} \minus{} \frac {1}{\sin x\cos x}\right )$
$ \equal{} \lim_{x\to 0}\left (\frac {\sin x\cos x}{x\sin x\cos x} \minus{} \frac {1}{\sin x\cos x}(1)\right )$
$ \equal{} \lim_{x\to 0}\left [\frac {\sin x\cos x}{x\sin x\cos x} \minus{} \frac {1}{\sin x\cos x}\left (\lim_{x\to 0}\frac {\sin x}{x}\right )\right ]$
$ \equal{} \lim_{x\to 0}\left (\frac {\sin x}{x}\right )\left (\frac {\cos x \minus{} 1}{\sin x\cos x}\right )$
$ \equal{} \lim_{x\to 0}\left (\frac {\cos x \minus{} 1}{\sin x\cos x}\right )$

We could also use L'Hopital's Rule:

$ \lim_{x\to 0} \frac {\sin x\cos x \minus{} x}{x\sin x\cos x}$
$ \equal{} \lim_{x\to 0} \left (\frac {\sin 2x \minus{} 2x}{x\sin 2x}\right )$
Here:
$ \equal{} \lim_{x\to 0} \left (\frac {2\cos 2x \minus{} 2}{2x\cos 2x \plus{} \sin 2x}\right )$
And here:
$ \equal{} \lim_{x\to 0} \left (\frac { \minus{} 4sin2x}{2(\cos 2x \minus{} 2x\sin 2x) \plus{} 2\cos 2x}\right )$
$ \equal{} \left (\frac {0}{4}\right )$
$ \equal{} 0$


Which online subject classes are extremely useful to take in preparation for the amc 12/aime. I will probably only be able to take about 1-3 subject classes. Also, I will have completed the aops series and part of art and craft of problem solving after.

Go for your weakest areas. That's the general advice.

I think the Problem Series may help you the most though if your goal is, for example, to qualify for USAMO.


geometry
3D geometry
pyramid
summer program
PROMYS

I learned about this game at (of all places) a computer camp in Colorado...

Anyways, I propose that we play it here, on AoPS. However, I'm not sure how we could do it on a forum, since people may want to join in mid-game, or forget to login for days.

Whatever.

[b]Blue[/b]!

Edit: For those of you who don't know how to play, here are the rules:
-On your turn, you must either say a color or country.
-If you say the name of a country, you lose.

Oooh, tough one.... hm...

[color=green]Green[/color]!

If this is anything like Mornington Crescent...then I'm out.

Oh are we? Very well then, I'm back in with turquoeise!

Should be.
Oh you have reached hodge conjecture I see.

Anyway pinkish white.

Egypt.......................ian pyramid tan!

Whew, close call  :)

Maceroni-and-cheese (I've seen it on crayola crayons) :D

[color=darkblue]Deep sea blue![/color]

I have lost 0 times so far! Wooo!

[quote="not_trig"][quote="Klebian"]magentaish chartruce[/quote]

I know this is an old post, but it's chartreuse  :D

The color of a dead man's face[/quote]

Haha, now that I think about it, I can't understand why I put a 2nd c in there, I know that at that time I couldn't remember how to spell it, but I know it doesn't have a c, and I've always known that...

Hmm... Europe! Yay!

that's neither a color nor a country. you have to go again.

The colour of Ivory Coast on the page entitled "Africa - Political" in the first standard school atlas you can lay your hands on.

[quote="nutz_for2.718281828"]The color of the colors of the American flag mixed together![/quote]
That would make a light purple. Maybe lavendar?


geometry
incenter
rectangle
cyclic quadrilateral
geometry proposed
Angle Chasing

If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$, $BCD$, $CDA$, $DAB$ are the vertices of a rectangle.

By angle chasing.

You can show that the incentres of ABC (K), BCD (L), B and C are cyclic because angle CLB = 180 - DCB/2 - DBC/2 = 90 + BDC/2, and similarly, angle CKB = 90 + BAC/2. But BAC = BDC. Thus CLB = CKB and so B,C,K,L are concyclic. Also if we call the incentre of ADC (M), D,C,L,M is cyclic. angle KLC = 180 - angle LBC, angle MLC = 180 - MDC, thus angle KLM = angle KBC + angle MDC = (ADC+ABC)/2 = 90, which is what we want.

PS: Can someone give me a crashcourse on latex? or point to me where?

It can be proved by complex numbers... See also

  [url=http://www.math.fau.edu/yiu/Geometry.html]Paul Yiu, [i]Notes on Euclidean Geometry[/i][/url], paragraph 10.2.6 (page 154);
  [url=http://groups.yahoo.com/group/Hyacinthos/message/8154]Barry Wolk, [i]Hyacinthos message #8154[/i][/url],
  [url=http://groups.yahoo.com/group/Hyacinthos/message/8178]Alexey A.Zaslavsky, [i]Hyacinthos message #8178[/i][/url].

  darij

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=77259]There is the converse of this fact[/url] (from Russian Summer TST 2005):
Let $ABCD$ be a quadrilateral. The incentres of the triangles $ABC, BCD, CDA, DAB$ form a rectangle. Proove, that $ABCD$ is cyclic.
It's very interesting and rather difficult problem.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Le%20rectangle%20de%20Ryokan%20Maruyama.pdf

Sincerely
Jean-Louis

We use complex numbers. Plot the points on the unit circle. Then we can choose $a,b,c,d$ such that $A=a^2,B=b^2,C=c^2,D=d^2$, and \[I_{ABC}=-bc-ca-ab,I_{BCD}=-cd-db-bc,I_{CDA}=da+ac-cd,I_{DAB}=-ab+bd+da.\] Then we can compute \[I_{ABC}+I_{CDA}=-ab-bc-cd+da=I_{BCD}+I_{DAB}\] and
\begin{align*}
\left|I_{ABC}-I_{CDA}\right|^2&=\left|-bc-2ca-ab-da+cd\right|^2=\frac{1}{abcd}\left(-bc-2ca-ab-da+cd\right)\left(-ad-2bd-cd-bc+ab\right)\\
&=\frac{1}{abcd}\left(-cd-2db-bc-da+ab\right)\left(-ab-2ac-ad-bc+cd\right)=\left|-cd-2db-bc-da+ab\right|^2=\left|I_{BCD}-I_{DAB}\right|^2
\end{align*}
so the incenters form a rectangle.

Labels are on the diagram.
By Fact 5 $AMNB$ is cyclic. And so is $AMQD$.
$\angle NMQ=\angle B/2 +\angle D/2=90^\circ$. And similarly the others.

i only could find those right angles on the rectangle by drawing external tangents
and since it is a quadrilateral
it must be a square or rectangle
and  by length chasing interestingly we could say that that square/rectangle is definitely composed of rectangles

We use complex numbers on this one.
We just easily prove this a parallelogram and then just show perpendicularity once or prove diagonals equal in length. :blush:

My solution at [url]https://artofproblemsolving.com/community/c6t48f6h24483_abcd_cyclic_gt_incenters_of_bcd_etc_form_rectangle[/url]
I hope you will enjoy it!

Best regards,
sunken rock


geometry
circumcircle
incenter
Euler
projective geometry
cyclic quadrilateral
geometry proposed

Let $ w$ be the circumcircle of $ \triangle ABC$ and let $ M\in w$ so that $ BC$ separates $ A$ , $ M$ .

Denote $ \|\begin{array}{ccc}
P\in (AB) & , & \widehat {AMP}\equiv\widehat {BMP}\\\
N\in (AC) & , & \widehat {AMN}\equiv\widehat {CMN}\end{array}$ . Prove that the incenter $ I\in NP$ .

Let $ MP,MN$ meet $ w$ at $ X,Y$ respectively. Then $ I\equal{}BY\cap CX$. By Pascal theorem, $ P,I,N$ are collinear.

This problem was a test IMAR in 2005 (Romania) and it is well-known, generally,
with the short and nice [b]Leonhard Euler[/b]'s proof. Exists another proof of it ?
Without the [b]Pascal[/b]'s theorem ? I know one.

A [b]Pascal[/b]-less proof for dear [b]Virgil Nicula[/b]
let B' and C' be midpoints of arcs AC and AB which doesn't contain B and C respectively.
now let $ MP\cap BI \equiv Y\ ,\ MN\cap CI\equiv Z$ we have that $ MYIZ$ is cyclic.
hence $ \angle IYZ \equal{} \angle IMZ \equal{} \angle B'IN\ ,\ \angle IZY \equal{} \angle IMY \equal{} \angle C'IP$ hence $ IN\parallel{}YZ\ ,\ IP\ \parallel{}YZ$ hence $ N,I,P$ collinear.

[color=darkred][i][b]Leonhard Euler[/b][/i] and [i][b]Amir S.[/b][/i] , here is [u]another[/u] proof without [b]Pascal's theorem[/b] :[/color]

[color=darkblue][b][u]Proof.[/u][/b] Apply the [b]theorem of bisector[/b] in  $ \{\begin{array}{cc}
\triangle\ AMB\ : & \frac {PB}{PA}=\frac {MB}{MA}\\\\
\triangle\ AMC\ : & \frac {NC}{NA}=\frac {MC}{MA}\end{array}\ \ (1)$ .

Apply the [b]Ptolemeu's theorem[/b] to the cyclic quadrilateral $ ABMC$ :

$ b\cdot MB+c\cdot MC=a\cdot MA$ $ \implies$ $ b\cdot\frac {MB}{MA}+c\cdot\frac {MC}{MA}=a$ . Using

the relations $ (1)$ obtain $ b\cdot\frac {PB}{PA}+c\cdot\frac {NC}{NA}=a$ , what characterizes that $ I\in NP$ .[/color]


Here all of us can introduce ourselves and kind of give a little "math profile". You do not need to give your real name if you don't want to. I'll start. I am kstan013. I am from Mosinee, WI and have taken 3rd place and 1st place at MathCounts at state in recent years. At nationals I have placed 121st and 78th. I also happen to know picounts and mustafa personally.

9th grade, Madison
I know mustafa and picounts, and have met kstan...

I'm from Madison, I know kstan, ciaccona, and picounts, I got 3rd at mathcounts state last year, and 63rd at nats. I also got the best score in wisconsin on the AMC 8 last year, but my school is the only school I've heard of taking the AMC 8, so I don't know if that really counts as anything special at all :roll: .

Well I am from Michigan, right next door, sort of.

Detroit area.

so mustafa was 3rd and kstan was 1st in state of Wisconsin.

Then Mustafa placed 63rd, higher than kstan during the Nats.

kstan got to first from 4th or 5th during the countdown round, on the written, I got 2nd and kstan was 4th or 5th.

I am in 8th grade from Appleton, WI.  I believe I will get to know kstan....being on the WI team.

I am in 8th grade from Mosinee, WI.  I believe I will get to know football....being on the WI team. :rotfl:  :D

Too bad nobody from Madison made it, then I could have come to your practices.

But all of us won't have to drive to Madison now..


[quote="phamvanquy"]Hi, I have problem so that. Find x, y if:

$\left\{ \begin{array}{l} 3x^2+2xy +y^2= 11 \\ x^3 + 2xy + 3y^2=17\end{array} \right.$

Oh! it's very easy if we change $x^3$ by $x^2$. But it's exact $x^3$.
Can you hepl me? Thanks you very much!
                                                         :wallbash_red: :help:[/quote]
So :
$9x^2+6xy+3y^2=33$
$x^3+2xy+3y^2=17$

Subtracting, we get $x^3-9x^2+16=4xy$ and, since $x=0$ is never solution, $y=\frac{x^3-9x^2+16}{4x}$

Plugging this in first equation, we get $x^6-10x^5+57x^4+32x^3-336x^2+256=0$

$x=1$ is a trivial root, and so $y=2$ and a first solution $\boxed{\text{S1 : }(x,y)=(1,2)}$

It remains to solve $x^5-9x^4+48x^3+80x^2-256x-256=0$

I'm sure your teacher tought you how to easily solve this quintic. Mine did not.

Waiting for a clever olympiad level solution by anybody, numeric calculus easily gives :

$\boxed{\text{S2 : }(x,y)=(\sim 2.34277416990797, \sim -2.19171663906611)}$

$\boxed{\text{S3 : }(x,y)=(\sim -2.04113089637945, \sim 3.7440041721502)}$

$\boxed{\text{S4 : }(x,y)=(\sim -0.909043992934478, \sim -2.14828785654138)}$


this one should be fun. I have a vexlar that I use for ice fishing ( it basically tells the depth and if there are any fish under you by showing a little suspended blip on the readout). Anyway, here is my question..

The transducer reads the depth. Lets say the depth is 10 feet and the angle it reads is 12 degrees, what is the equasion to find the surface area that it is covering at a depth of 10 feet (the area the vexlar can pick up fish movement on the bottom of the lake if it reads 12 degrees)..?

if you do not understand what I am trying to say I can put it like this. if the transducer is sitting on top of a traffic cone what is the surface area at the bottom of that cone given the cone under the transducer reads 12 degrees.

[hide="answer"]
$4.944\pi$

This is pretty easy...

Using Trigonometry......

$tan 12=\frac{r}{10}$  ,  $0.212*10=r$ , $r=2.12$,  then
$\pi(2.12)^2=4.944\pi feet squared$
[/hide]

[quote="ajai"][hide="answer"]
$44\pi$

This is pretty easy...

Using Trigonometry......

$tan 12=\frac{r}{10}$  ,  $0.212*10=r$ , $r=2.12$,  then
$\pi(2.12)^2=4.944\pi feet squared$
[/hide][/quote]

how comes the two answers you gave aren't consistent.
i guess the "4.9" part was edited out by accident

Please don't post trig problems in the middle school forums.


algebra
polynomial
number theory theorems
number theory

hi

can someone please explain about cyclotomic polynomials and how they are used in solving problems involving primes.

Perhaps you could start with http://mathworld.wolfram.com/CyclotomicPolynomial.html
or http://perso.wanadoo.fr/yves.gallot/papers/cyclotomic.html

HTH

A useful fact about Zp: The roots of phip-1(x) in Zp are exactly the primitive roots mod p (p prime).


Hello all!

I need some idees for the following thing. My english teacher said that once a week, durring the english class we have to be "the teacher"  :D. I think that's a good way for us to experss ourselfs better and that could help us at the oral exams. Some of my classmates already did that but they picked some topics that weren't so exciting and I say that by the way the rest of my collegues responded to the choosen subject. If you have any ideeas it would be nice of you. I thought to teach some math but they don't pay attetion sometimes at the math class itself :(. Thanks!

Some Fibonacci or golden ratio always helps    :)

Force them to pay attention!   :lol:   Our class had to do that in english to in my last year at high school, and me and my friend stubbornly wanted to teach english grammar (which is hated because over its abundance of rules about the choice of suitable tense).  The teacher really became anxious and in the days before it, he was quite nervous to learn when we were gonna say our real subject.  However i copied some notes to be filled in, and most of the class thought they were somehow forced to fill them in.

Seriously, there a few basic things about teaching math even I have encountered already :

Your audience is never up to speed as you would wish them to be.
You always go slower than you thought you would.

Also what I (as a movie lover) would really appreciate is that when you talk about the subject , you point out where it pops up.  For instance adjacency matrices pop up in Good Will Hunting, taylor approximations (somewhat) are used to beat a chinese store owner in eternal sunset...

[quote="José"]What about Pithagorean Theorem proof??? Quite simple, but it'd work...[/quote]

I'm in 11th grade  :lol: We learn now Analysis and Linear Algebra.

Well, I'd like to choose a original topic. Not math only, in fact no math at all. My teacher is allergic to math so it would be improper. I know you're smart guys so c`mon  :D !!!

Maybe you should try something connected with Computer Science. Perhaps talk about how a game is created (you can learn about that on Google :) ) ...

Well, an entire world of more knowledge opened up with [url]http://www.wikipedia.org[/url] for me

You could teach about some not well known historic event, like the third indochina war or the us grenada intervention, meiji revolution

or you could make a nice copyright infringing explanation of historic pictures (always nice)  :

[url]http://en.wikipedia.org/wiki/Category:Memorable_photographs[/url]

i'm just brainstorming, but i like recent history

 is also nice : [url]http://en.wikipedia.org/wiki/Shibboleth[/url]

Explain something interesting, such as current events.  Just bring along a newspaper!


You are at a game show.  There are three doors in front of you.  You know that one has treasure behind it and the other two have nothing behind them, but you don't know which one has the treasure.  You want to get the treasure.  So you pick a door.  But, before looking behind it, the game show host opens one of the two other doors and you find that the room behind it is empty.  The game show host now gives you the option of switching to the third remaining door.  The question is, would you have a better chance of getting the treasure by switching?

Just a small recomendation for K81o7.

Please, try to call problems in order that everyone could understand its essence from the title.
Also you can read http://www.mathlinks.ro/viewtopic.php?p=140098

Thanks

Oh...sorry about that.  I didn't know that there was a name for this problem...

[hide="Hint"]Think about the show host's options.  He has to open a door with nothing behind it right?  And that selection has to be made from the two doors remaining after you choose.[/hide]

[hide="Answer"]Well think about it.  If you chose a door with nothing behind it (the chance of doing so is 2/3), then the other door with nothing behind it is revealed, and so the third door has the treasure.  If you chose the one with treasure behind it, then either of the other two doors could have been opened but then switching would get you a door with nothing behind it.  Overall, when you switch, your chance of getting the treasure ends up as 2/3(as opposed to 1/3 from not switching).  So you would want to switch. [/hide][/hide]


[hide="Hint"]The recurrence formula telescopes in products[/hide]

An important condition is

$ a_{n}\equal{}a_{n\minus{}6}$


1) Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\leq\sqrt {\sum_{cyc}(5a^2 \plus{} 4ab)}
\]
For the collection.
With the same conditions prove that:
2)
\[ \sum_{cyc}\sqrt {4a^2 \plus{} ab \plus{} 4b^2}\leq\sqrt {\sum_{cyc}(22a^2 \plus{} 5ab)}
\]
3)
\[ \sum_{cyc}\sqrt {9a^2 \minus{}2ab \plus{} 9b^2}\leq2\sqrt {\sum_{cyc}(13a^2 \minus{}ab)}
\]

[quote="arqady"]1) Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\leq\sqrt {\sum_{cyc}(5a^2 \plus{} 4ab)}
\]
For the collection.
With the same conditions prove that:
2)
\[ \sum_{cyc}\sqrt {4a^2 \plus{} ab \plus{} 4b^2}\leq\sqrt {\sum_{cyc}(22a^2 \plus{} 5ab)}
\]
3)
\[ \sum_{cyc}\sqrt {9a^2 \minus{} 2ab \plus{} 9b^2}\leq2\sqrt {\sum_{cyc}(13a^2 \minus{} ab)}
\]
[/quote]
1) By Cauchy Schwarz Inequality, we have
\[ LHS^2 \le 2(a \plus{} b \plus{} c)\sum \frac {a^2 \plus{} ab \plus{} b^2}{a \plus{} b} \equal{} 2\sum (a^2 \plus{} ab \plus{} b^2) \plus{} 2 \sum \frac {c(a^2 \plus{} ab \plus{} b^2)}{a \plus{} b} \equal{} 4\sum a^2 \plus{} 6\sum ab \minus{} 2abc\sum \frac {1}{a \plus{} b}
\]
It suffices to prove that
\[ \sum a^2 \plus{} 2abc\sum \frac {1}{a \plus{} b} \ge 2\sum ab
\]

\[ \Leftarrow \sum a^2 \plus{} \frac {9abc}{a \plus{} b \plus{} c} \ge 2\sum ab
\]

\[ \Leftrightarrow \sum a(a \minus{} b)(a \minus{} c) \ge 0
\]
which is true by Schur's Inequality. :)

CYH is so strong! :)

P/s: 2) and 3) can be proved by a "similar" way. :)

[quote="can_hang2007"][quote="arqady"]1) Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\leq\sqrt {\sum_{cyc}(5a^2 \plus{} 4ab)}
\]

[/quote]
1) By Cauchy Schwarz Inequality, we have
\[ LHS^2 \le 2(a \plus{} b \plus{} c)\sum \frac {a^2 \plus{} ab \plus{} b^2}{a \plus{} b}
\]

[/quote]
Yes! :lol:
It's also my proof.

[quote="dduclam"][quote="arqady"]1) Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\leq\sqrt {\sum_{cyc}(5a^2 \plus{} 4ab)}
\]
[/quote]

Arqady, do you think the following inequalities are also true with the same conditions ?

1/ $ \sum_{cyc}\sqrt [3]{(a^2 \plus{} ab \plus{} b^2)^2}\le\sqrt [3]{3\left(\sum_{cyc}(2a^2 \plus{} bc)\right)^2}$

and

2/ $ \sum_{cyc}\sqrt [3]{(a^2 \plus{} ab \plus{} b^2)^2}\le\sqrt [3]{9\sum_{cyc}(2a^2 \plus{} bc)^2}$[/quote]
Problem 1.
\[ \sum_{cyc}\sqrt [3]{(a^2 \plus{} ab \plus{} b^2)^2}\le\sqrt [3]{3\left(\sum_{cyc}(2a^2 \plus{} bc)\right)^2} \leftrightarrow \sum_{cyc}\sqrt [3]{3(a^2 \plus{} ab \plus{} b^2)^2}\le \sqrt [3]{9\left(\sum_{cyc}(2a^2 \plus{} bc)\right)^2}
\]
By Holder ;
\[ \sum_{cyc}\sqrt [3]{3(a^2 \plus{} ab \plus{} b^2)^2} \leq \sqrt [3]{9(\sum_{cyc}(a^2 \plus{} ab \plus{} b^2))^2}
\]
It's remain to prove
\[ \sqrt [3]{9(\sum_{cyc}(a^2 \plus{} ab \plus{} b^2))^2} \le \sqrt [3]{9\left(\sum_{cyc}(2a^2 \plus{} bc)\right)^2}
\]
which is obvious.(because $ (\sum_{cyc}(a^2 \plus{} ab \plus{} b^2))^2 \equal{} \left(\sum_{cyc}(2a^2 \plus{} bc)\right)^2$)
Problem 2.
\[ \sum_{cyc}\sqrt [3]{(a^2 \plus{} ab \plus{} b^2)^2}\le\sqrt [3]{9\sum_{cyc}(2a^2 \plus{} bc)^2} \longleftrightarrow \sum_{cyc}\sqrt [3]{3(a^2 \plus{} b^2 \plus{} c^2)^2} \le 3\sqrt [3]{\sum_{cyc}(2a^2 \plus{} bc)^2}
\]
By Holder ;
\[ \sum_{cyc}\sqrt [3]{3(a^2 \plus{} b^2 \plus{} c^2)^2} \le \sqrt [3]{(9)(\sum_{cyc}(a^2 \plus{} ab \plus{} b^2))^2}
\]
It's remain to prove that
\[ \sqrt [3]{9(\sum_{cyc}(a^2 \plus{} ab \plus{} b^2))^2} \le 3\sqrt [3]{\sum_{cyc}(2a^2 \plus{} bc)^2}
\]

\[ \longleftrightarrow (\sum_{cyc}(a^2 \plus{} ab \plus{} b^2))^2 \le 3\sum_{cyc}(2a^2 \plus{} bc)^2
\]
which is obviously true by Cauchy-Schwarz ; $ 3\sum_{cyc}(2a^2 \plus{} bc)^2 \geq (\sum_{cyc}(2a^2 \plus{} bc))^2 \equal{} \sum_{cyc}(a^2 \plus{} ab \plus{} b^2)^2$
Edit : arqady is faster  :blush:

[quote="manlio"][quote="arqady"]With the same conditions prove that:
2)
\[ \sum_{cyc}\sqrt {4a^2 \plus{} ab \plus{} 4b^2}\leq\sqrt {\sum_{cyc}(22a^2 \plus{} 5ab)}
\]
3)
\[ \sum_{cyc}\sqrt {9a^2 \minus{} 2ab \plus{} 9b^2}\leq2\sqrt {\sum_{cyc}(13a^2 \minus{} ab)}
\]
[/quote]

I cannot prove them using the same method of can_hang2007. :(

A hint for me please, Arqady.

Thank you very much.[/quote]
The hint to the $ 3).$
By Cauchy Schwartz for all $ k\geq0$ we obtain:
$ \sum_{cyc}\sqrt {9a^2 \minus{} 2ab \plus{} 9b^2}\leq\sqrt {\sum_{cyc}\frac {9a^2 \minus{} 2ab \plus{} 9b^2}{a \plus{} b \plus{} kc}\cdot\sum_{cyc}(a \plus{} b \plus{} kc)}.$
Hence, it remains to find a value of $ k$ for which the following inequality is true.
$ (k \plus{} 2)(a \plus{} b \plus{} c)\sum_{cyc}\frac {9a^2 \minus{} 2ab \plus{} 9b^2}{a \plus{} b \plus{} kc}\leq 4\sum_{cyc}(13a^2 \minus{} ab).$
An equality in the original inequality holds also when $ a \equal{} b \equal{} 1$ and $ c \equal{} 0,$ which gives $ k \equal{} \frac {1}{2}.$ :wink:

[quote="manlio"]$ (k \plus{} 2)(a \plus{} b \plus{} c)\sum_{cyc}\frac {9a^2 \minus{} 2ab \plus{} 9b^2}{a \plus{} b \plus{} kc}\leq 4\sum_{cyc}(13a^2 \minus{} ab).$

Do you solve it by full expansion and pqr method, Arqady?

Thank you very much. :)[/quote]
For $ k\equal{}\frac{1}{2}$ after full expansion we obtain an inequality, which is true by Schur and Muirhead.

[quote="dduclam"][quote="arqady"]dduclam, the first is Jensen and the second follows from the first. :wink:[/quote]

I see, arqady  :lol:  And I think the following inequality is also true and hard :

$ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\le 2\sqrt3(\sum_{cyc}\frac {a^4}{b^3 \plus{} c^3})$

Do you think so ?[/quote]
It is still very easy, my friend ;)
Since we can easily prove that (by Cauchy Schwarz)
\[ LHS \le 3\sqrt{a^2\plus{}b^2\plus{}c^2} \le RHS\]
:)
Maybe you meant:
\[ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\le 2\sqrt3(\sum_{cyc}\frac {a^4}{a^3 \plus{} b^3})\]
?

[quote="can_hang2007"]
Maybe you meant:
\[ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\le 2\sqrt3(\sum_{cyc}\frac {a^4}{a^3 \plus{} b^3})
\]
?[/quote]

Oh, yes. I'm sorry for my wrong type :(

Besides, the following inequalities also seems correct :

$ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\le 2\sqrt3(\sum_{cyc}\frac {a^6}{a^5 \plus{} b^5})$

$ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\le 2\sqrt3(\sum_{cyc}\frac {a^7}{a^6 \plus{} b^6})$

And  $ \sum_{cyc}\sqrt {a^2 \plus{} ab \plus{} b^2}\le 2\sqrt3(\sum_{cyc}\frac {a^8}{a^7 \plus{} b^7})$ wrong allready   :)


Let $ a,b,c\geq 0$, s.t. $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \frac{1}{6\minus{}ab}\plus{}\frac{1}{6\minus{}bc}\plus{}\frac{1}{6\minus{}ca}\leq \frac{3}{5}$.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \frac {1}{6 \minus{} ab} \plus{} \frac {1}{6 \minus{} bc} \plus{} \frac {1}{6 \minus{} ca}\leq \frac {3}{5}$.[/quote]
$ <\equal{}>16(ab\plus{}bc\plus{}ca) \plus{}a^2b^2c^2 \le 36\plus{}13abc<\equal{}>$
$ a^2(b\plus{}c)\plus{}b^2(c\plus{}a) \plus{}c^2(a\plus{}b)\plus{}\frac{3}{4}a^2b^2c^2 \le a^3\plus{}b^3\plus{}c^3\plus{}\frac{15}{4}abc$( which is true by Schur and $ abc \le 1$); :)


In a 12-hour period of time, for how many minutes is at least one ``6'' displayed on a digital clock?

A "6" only occurs at x:06, x:16, x:26, x:36, x:46, or x:56.  Or when the first digit is 6.

So we have 11*6=66, add 60=126 minutes.

There are 59 times with a 6 in the hours, and $ 5 \cdot 12\equal{}60$ times for the minutes. However, there are $ 5$ times when we overcount, so $ 59\plus{}60\minus{}5\equal{}114$.

I think you're wrong - for one, there's 60 times with a 6 in the hours: 6:00-6:59.

Also, it's not 5*12, its 6*12, b/c we have x:06, x:16, x:26, x:36, x:46, and x:56.

So using your method, we have 60+72-6=126 minutes.

I thought a digital clock shows seconds too?...

It depends on which digital clock you're using, I suppose.  Mine doesn't.


What is the probbility that each player has a hand containing an ace when the 52 cards of a standard deck are dealt to four players , i.e. each player has 13 cards.

but isnt there $4!$ ways of distributing the aces?

but anyway how did you get what you got?

$\frac{24\cdot 13^4}{52\cdot 51 \cdot 50 \cdot 49}$

[hide] Yeah, I think he forgot to include the order of the aces. Consider it using permutations. There are a total of $52!$ ways to distribute the cards. For each player, there are $13$ possible slots to put the ace which gives $13^4$, and then a total of $48!$ ways to distribute the rest.

probability then equals: $\frac{4!\times13^4\times48!}{52!}=\frac{24\times13^4}{52\times51\times50\times49}$.
[/hide]


Find all continuous functions $ f: \mathbb{R} \to \mathbb{R}$ such that \[ f\left( {\frac {{x \plus{} y}} {2}} \right) \equal{} \sqrt {f(x)f(y)},\] for all reals $x$ and $y$.

[quote="Xaenir"]Find all continuous functions $ f: \mathbb{R} \to \mathbb{R}$such that $ f\left( {\frac {{x \plus{} y}} {2}} \right) \equal{} \sqrt {f(x)f(y)}$      , $ \forall x,y$[/quote]

If we can find $ x$ such that $ f(x)\equal{}0$, then put it into the equation we get $ f(x)\equal{}0$ for all $ x$ as a solution.

If not, then put $ p\equal{}\frac{x\plus{}y}{2}$ we get $ f(p)$ is positive for all $ p$

Now $ \sqrt{f(x\plus{}y)f(0)}\equal{}f(\frac{x\plus{}y}{2})\equal{}\sqrt{f(x)f(y)}$

So $ f(x\plus{}y)f(0)\equal{}f(x)f(y)$
$ \Leftrightarrow In (f(x\plus{}y))\plus{}In (f(0))\equal{}In (f(x))\plus{}In(f(y))$

Now let $ g(x)\equal{}In (f(x))\minus{}In (f(0))$ the equation becomes:

$ g(x)\plus{}g(y)\equal{}g(x\plus{}y)$

As $ g(x)$ is continuous (because $ f(x)$ is continuous), $ g(x)\equal{}cx$ (as Cauchy) , where $ c$ is a constant.

Let $ b\equal{}f(0)$, we have $ f(x)\equal{}e^{cx\plus{}In(f(0))}$

Let $ k\equal{}e^c$, $ t\equal{}f(0)$

We get $ f(x)\equal{}tk^x$ is the solution after checking, where $ t$ is any real constant and $ k$ is any non-zero constant.

[quote="hqthao"]you can take log two side, then you have Cauchy equation[/quote]

You must first show that $ f(x)\neq 0$

And, btw, this is exactly what Stephencheng did.  :wink:

[quote=pco][quote="Xaenir"]Find all continuous functions $ f: \mathbb{R} \to \mathbb{R}$such that $ f\left( {\frac {{x \plus{} y}} {2}} \right) \equal{} \sqrt {f(x)f(y)}$      , $ \forall x,y$[/quote]

We clearly have $ f(x)\geq 0$
If $ f(x_0)\equal{}0$ for some $ x_0$, we have obviously $ f(x)\equal{}f(\frac{x_0\plus{}(2x\minus{}x_0)}{2})\equal{}\sqrt{f(x_0)f(2x\minus{}x_0)}\equal{}0$ $ \forall x$

Consider now $ f(x)>0$ $ \forall x$

$ f(\frac{(x\plus{}y)\plus{}0}{2})\equal{}\sqrt{f(x\plus{}y)f(0)}$ and so $ f(x\plus{}y)\equal{}\frac{1}{f(0)}f(x)f(y)$

This is a classical equation whose the only continuous solutions are $ f(x)\equal{}a\cdot e^{bx}$

Plugging back in the original equation, we get the solutions :

$ f(x)\equal{}a\cdot e^{bx}$ $ \forall a\geq 0$[/quote]

what happen if x0 =0 ??? then we cannot divide by f(0) to get solution that you have arrived....

[quote=Understandingmathematics]what happen if x0 =0 ??? then we cannot divide by f(0) to get solution that you have arrived....[/quote]
I never divided by $f(x_0)$. I just trivially proved that $f(x_0)=0$ implies solution $f(x)=0$ $\forall x$ (whatever is $x_0$)

Read more carefully.
There are two parts in my proof :
Part 1 : If $\exists x_0$ such that $f(x_0)=0$
Then solution is $f\equiv 0$

Part 2 : If no such $x_0$ and so $f(x)>0$ $\forall x$ ..
Then ...
And in this part, I can divide by $f(0)$ since we know that $f(x)>0$ $\forall x$

this means that there are only one solution $f(x)=0$ or $f(x)$=a e^(bx)

[quote=Understandingmathematics]this means that there are only one solution $f(x)=0$ or $f(x)=$a$e^(bx)$[/quote]
This means that we have :
Either $f(x)=0$ $\forall x$
Either $f(x)>0$ $\forall x$ and then $f(x)=a.e^{bx}$, whatever are $a>0$, and $b$

And I merged the two solutions by allowing $a>0$ in the second to be $a\ge 0$

And so general solution $f(x)=a.e^{bx}$ $\forall x$, whatever are $a\ge 0$ and $b$


[quote="Ubemaya"]3 distinct unit fractions are added together to make $\frac{6}{7}$. What is the least possible sum of the denominators?

This was a #30 on the sprint, it prevented me from getting a 22 on the sprint :( .[/quote]
[hide]$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
We need to minimize x y and z. Try x=2 and this is now $\frac{1}{y}+\frac{1}{z}=\frac{5}{14}$ Try x=3. We got $\frac{1}{42}$. This is kinda big, I really doubt it minimizes. [/hide]
I'm about 85% sure that this is wrong.

That works, but the problem asks for unit fractions, which are of the form $\frac{1}{x}$ where x is an integer.

[hide]Use the Greedy Algorithm.

First go with $\frac{1}{2}$

Then go with $\frac{1}{3}$

Then the last one is $\frac{1}{42}$

So the answer is $47$[/hide]

[quote="mysmartmouth"]$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{6}{7}$
$\frac{xy+yz+xz}{xyz}=\frac{6}{7}$

Well, by analysis, $xyz$ in the denominator will have a factor of 7, so let $x=7$.  Then:

$\frac{7y+7z+yz}{7yz}=\frac{6}{7}$
$\frac{1}{z}+\frac{1}{y}+\frac{1}{7}= \frac{6}{7}$
$\frac{1}{z}+\frac{1}{y}= \frac{5}{7}$
Simon's favorite factoring trick won't work on this...

So try $x=14$ the next factor of 7.  And keep trying x of the form $x=7r$ until something works, and that will be the minimum???

Is that a good plan?  It seems kind of long-winded for a sprint.[/quote]
Oh, you can use simon's favorite factoring trick on that.
$(y-1.4)(z-1.4)=1.96$
$(5y-7)(5z-7)=49$
Clearly 2.8/2.8 won't work. Neither with the other thing. Both Neil and my solution are the same, but I have doubts about it being the smallest.

[hide="Solution"]Assume that the denominators $x$, $y$, and $z$ are in the order $x < y < z$.  Choosing $x = 1$ makes the sum too big, because $1 > \frac{6}{7}$.  Choosing $x \ge 3$ makes the sum too small, because
\[\frac{1}{3}+\frac{1}{4}+\frac{1}{5}= \frac{47}{60}< \frac{6}{7}\, . \]
Thus $x = 2$.  That means
\[\frac{1}{y}+\frac{1}{z}= \frac{5}{14}\, . \]
Choosing $y \ge 6$ makes the sum too small, because
\[\frac{1}{6}+\frac{1}{7}= \frac{13}{42}< \frac{5}{14}\, . \]
Thus $y$ is either 3, 4, or 5.  If $y = 3$, then $z = 42$.  If $y = 4$, then $z = \frac{28}{3}$, which is not an integer.  If $y = 5$, then $z = \frac{70}{11}$, which is not an integer.

Therefore the only triple that works (up to reordering) is $(2, 3, 42)$.  The sum is indeed $\boxed{47}$.[/hide]

the answer sheet indeed says that it is 47.


hello,
Prove that $ (x,y)\equal{}(2,1)$ is the only solution of the system : $ \quad \large (x,y) \in {\mathbb R}^2 ,\quad \left\{\begin{array}{l} y(x^3 \minus{} y^3) \equal{} 7 \\
y(x \plus{} y)^2 \equal{} 9 \end{array} \right.$

hello, from equation (2) we get $ x_1\equal{}\minus{}\frac{y^{\frac{3}{2}}\minus{}3}{\sqrt{y}}$ and $ x_2\equal{}\minus{}\frac{y^{\frac{3}{2}}\plus{}3}{\sqrt{y}}$, inserting $ x_1$ in equation (1) and simplifying we get
$ \minus{}2y^{\frac{11}{2}}\plus{}9y^4\minus{}27y^{\frac{5}{2}}\plus{}27y\minus{}7y^{\frac{3}{2}}\equal{}0$. This equation has only the real root $ y\equal{}1$ and from here we get $ x\equal{}2$.
By inserting $ x_2$ in equation (1) we get
$ \minus{}2y^{\frac{11}{2}}\minus{}9y^4\minus{}27y^{\frac{5}{2}}\minus{}27y\minus{}7y^{\frac{3}{2}}<0$ for $ y>0$, hence $ x\equal{}2,y\equal{}1$ is the only solution.
Sonnhard.


Let $ a,b,c,d\in [0,1]$. Find the maximum value of A
  $ A\equal{}\frac{a}{1\plus{}bcd}\plus{}\frac{b}{1\plus{}cda}\plus{}\frac{c}{1\plus{}dab}\plus{}\frac{d}{1\plus{}abc}$
 :D

[quote="toanIneq"]Let $ a,b,c,d\in [0,1]$. Find the maximum value of A
  $ A \equal{} \frac {a}{1 \plus{} bcd} \plus{} \frac {b}{1 \plus{} cda} \plus{} \frac {c}{1 \plus{} dab} \plus{} \frac {d}{1 \plus{} abc}$
 :D[/quote]
We have:
$ bcd\plus{}1 \geq abcd\plus{}1$
$ cda\plus{}1 \geq abcd\plus{}1$
$ dab\plus{}1 \geq abcd\plus{}1$
$ abc\plus{}1 \geq abcd\plus{}1$
=>$ A \leq \frac{a\plus{}b\plus{}c\plus{}d}{abcd\plus{}1} \leq \frac{1\plus{}ab\plus{}1\plus{}cd}{1\plus{}abcd}\leq \frac{1\plus{}1\plus{}1\plus{}abcd}{1\plus{}abcd} \leq \frac{3\plus{}3abcd}{1\plus{}abcd} \equal{}3$
Equality occurs when $ a;b;c\equal{}1;d\equal{}0$ and cyc..

Similar problem:
Let $ a,b,c,d;e \in [0,1]$. Find the maximum value of M
  $ M \equal{} \frac {a}{1 \plus{} bcde} \plus{} \frac {b}{1 \plus{} cdea} \plus{} \frac {c}{1 \plus{} deab} \plus{} \frac {d}{1 \plus{} eabc}\plus{}\frac{e}{1\plus{}abcd}$


geometry
circumcircle
ratio
cyclic quadrilateral
geometry proposed

Let $ABC$ be a triangle and $K$ be its symmedian point. Let $A', B', C'$ be the intersection of $AK, BK, CK$ respectively and the circumcircle. Let $U$ be the intersection of the tangent at $A'$ and the tangent at $B$. Call $M$ and $N$ the intersections of $UK$ with the circumcircle. Prove that $M, N, C, C'$ form a harmonic quadrilateral.

What is a harmonic quadrilateral?  In- and circumscriptible?

This site http://www.2dcurves.com/line/lineq.html says that it is a cyclic quadrilateral for which the products of opposite sides are the same.

That's probably not the most fortunate description here. Another definition is: a cyclic quadrilateral $ABCD$ s.t. the tangents in $A,C$ to the circumcircle meet on $BD$ (and the same can be said if the pair $A,C$ is interchanged with the pair $B,D$).

I'll just sketch it: if $V$ is the intersection between the tangents in $A,B'$ to the circumcircle, then $MN=UV$. We must thus prove that the tangents in $C,C'$ meet on $UV$, or, equivalently, since the polaes of the tangents in $C,C'$ are $C$ and $C'$ respectively, we must show that $P=AB'\cap BA'$, which is the pole of $UV$, lies on $CC'$.

This follows from the following general fact: given $XYZT$, a circumscribed quadrilateral, which touches its incircle in $W_1,W_2,W_3,W_4$ (these points lie on $XY,YZ,ZT,TX$ respectively), $W_1W_4,W_2W_3,YT$ are concurrent. We apply this to the circumscribed quadrilateral determined by the tangents to the circumcircle in $A,B,A',B'$, by also using the fact that the tangents in $A,B$ meet on $CC'$, and so do those in $A',B'$, so $CC'$ is one of the diagonals of this quadrilateral.

Sorry for the lack of detail.

[quote="blahblahblah"] [harmonic quadrilateral is] a cyclic quadrilateral for which the products of opposite sides are the same.[/quote]

Ah!  That explains the word "harmonic", because it is equivalent to the cross-ratio of the 4 vertices (under projection to any line) being equal to -1.


Let $ n$ be a positive integer and let $ K$ be a convex and compact subset of $ \mathbb{R}^{n}$. Is it true that $ K$ is always homeomorphic to a $ k$-dimensional closed disc?

If the interior of $ K$ is nonempty we can even state an [i]explicit[/i] homeormorphism between $ K$ and $ \mathbb{D}^{n}$. What can be said in general? :)

Links and pdf's are welcome, too.

The general case can be reduced to the special one:
wlog we can assume that $ 0\in K$. Now we consider the linear subspace $ V$ spanned by $ K$. It equals $ \bigcup_{n\in\mathbb{N}} n(K\minus{}K)$ as one can show. Now $ K$ has non-empty interior with respect to $ V$. So we can also assume that $ int K \neq \emptyset$ and if suffices prove that $ (K,\partial K)\cong (D^n,\partial D^n)$. And this one seems to be clear to you if I understand you correctly.

I have two questions on the notations in your post, Gockel.

1. What do you exactly mean by [b]K-K[/b]?
2. Could please rewrite the last line of your post? I'm not sure I followed you there... Especially when you say [i]if suffices prove that[/i]... Also, what do you mean by the pair [b](K,dK)[/b]?

Thanks again.

By $ K\minus{}K$ I meant the set of all differences: $ K\minus{}K: \equal{}\{k\minus{}k' | k,k'\in K\}$

I wanted to write "[i]it[/i] suffices [i]to[/i] prove that ..."
Sorry. :blush:

And the pair $ (K,\partial K)$ means exactly what it says: the pair $ (K,\partial K)$.
A homeomorphism $ (K,\partial K)\to (D^n,\partial D^n)$ is a homeomorphism of topological pairs, i.e. a bijective map $ f: K\to D^n$ which is continouos in both directions and maps $ \partial K$ to $ \partial D^n$.
Now that I think a bit more about the subject I think the construction is even stronger than that. There is a homeomorphism of $ \mathbb{R}^n$ onto itself which maps $ K$ to $ D^n$ (and hence $ \partial K$ to $ \partial D^n$).

[quote="Gockel"] ... prove that $ (K,\partial K)\cong (D^n,\partial D^n)$.[/quote]

Shouldn't it be $ \displaystyle (\mathrm{K}, \partial \mathrm{K}) \cong (\mathrm{D}^{\dim \mathrm{V}}, \partial \mathrm{D}^{\dim \mathrm{V}})$? ;)

That's the same... $ V\equal{}\mathbb{R}^n$.

[quote="Gockel"]$ V \equal{} \mathbb{R}^n$.[/quote]

Always? What if, for example, $ K\equal{}\{0\}$ and $ n\equal{}1$? Does it follow that $ V \equal{} \langle K \rangle  \equal{} \mathbb{R}$? :(

But $ \{0\}$ has empty interior. $ (K,\partial K)\cong(D^n,\partial D^n)$ holds iff $ int(K)\neq\emptyset$.

This is the post I'm referring to, dear [b]Gockel[/b]... ;)

[quote="Gockel"]The general case can be reduced to the special one:
wlog we can assume that $ 0\in K$. Now we consider the linear subspace $ V$ spanned by $ K$. It equals $ \bigcup_{n\in\mathbb{N}} n(K \minus{} K)$ as one can show. Now $ K$ has non-empty interior with respect to $ V$. So we can also assume that $ int K \neq \emptyset$ and if suffices prove that $ (K,\partial K)\cong (D^n,\partial D^n)$. And this one seems to be clear to you if I understand you correctly.[/quote]

I know... what is unclear about it? If $ K$ has non-empty interior then $ V\equal{}\mathbb{R}^n$ and $ (K,\partial K)\cong (D^n,\partial D^n)$. That's exactly what I wrote...

[quote="Gockel"] What is unclear about it? If $ K$ has non-empty interior then $ V \equal{} \mathbb{R}^n$...[/quote]

I thought Riquelme had told us to forget about that case... I also thought that in your first post you were taking K to be arbitrary, possibly with an empty interior.  Weren't you trying to show us how to proceed in that [i]scenario[/i]?

And I did...
Construct $ V$ as I did. Then $ K$ has non-empty interior in $ V$ and wlog we can assume that $ V\equal{}\mathbb{R}^n$. So we're again in the case which is clear to Riquelme.


(a) A polynomial $ P$ has degree at most $ n \minus{} 10\sqrt {n}$. Prove that

$ |P(0)|\leq\frac {1}{10}(\binom{n}{1}|P(1)| \plus{} \binom{n}{2}|P(2)| \plus{} ... \plus{} \binom{n}{n}|P(n)|)$

(b) Prove that for any sufficiently large integer $ n$, there exist a polynomial $ P$ with degree at most $ n\minus{}10\sqrt{n}$ such that

$ |P(0)|>\frac {1}{10}(\binom{n}{1}|P(1)| \plus{} \binom{n}{2}|P(2)| \plus{} ... \plus{} \binom{n}{n}|P(n)|)$

Indeed, how stupid I am.

Correct version for (b):

(b) Prove that for any sufficiently large integer $ n$, there exist a polynomial $ P$ with degree at most $ n \minus{} \frac{1}{10}\sqrt {n}$ such that

$ |P(0)| > \frac {1}{10}(\binom{n}{1}|P(1)| \plus{} \binom{n}{2}|P(2)| \plus{} ... \plus{} \binom{n}{n}|P(n)|)$


geometry
geometric transformation
rotation

I have one doubt.All bodies in the universe from electrons and quarks to galaxies and stars are rotating about some axis.Why do all the bodies rotate?What sets them into rotation?Also some bodies rotate faster than others.Can anyone explain this?Is it due to the magnetic field possessed by the body?

This is a very fundamental question, something perhaps no one can answer. It is akin to asking, "Why do objects move? Once moving, why do they stay moving?" They move because they are pushed into motion, just as things rotate because they are turned. Newton's first law states that things tend to stay moving when they are already moving, and Newton's first law for rotation says the same for rotating bodies. This is an experimental fact, deduced by observation, and cannot be derived from first principles.

Just as some objects move faster than others, some can rotate faster than others. If they are turned harder, they will move faster. This is also an experimental finding, and gives rise to Newton's second law for rotation.

Rotation, in general, has nothing to do with magnetic fields. A non-charged, non-magnetic object can still rotate. The concept of rotation does not stem from magnetism; rather, it is the other way round, that rotation often generates magnetism due to moving charges.

Then why do neutron stars and similar bodies with strong magnetic fields rotate so fast?

You've got cause and effect mixed up, I think.
Neutron stars rotate very fast, due to conservation of angular momentum. Hence, the moving charges inside them create extremely large magnetic fields.


$A_{n}^{a}=n(n-1)(n-2)...(n-a+1)$ for every integer $a\leq n$


calculus
integration
quadratics
number theory proposed
number theory
Diophantine equation

Determine all non-negative integral pairs $ (x, y)$ for which
\[ (xy \minus{} 7)^2 \equal{} x^2 \plus{} y^2.\]

let's rewrite the equation as $ 48\plus{}(x^2\minus{}1)(y^2\minus{}1)\equal{}14xy\qquad (*)$.

If $ x\equal{}0$ we get $ y\equal{}7$. putting aside the solutions $ (0,7),(7,0)$ we can assume that $ x,y$ are positive.

If $ x\equal{}1$ it turns out that $ y\equal{}\frac{14\pm\sqrt{61}}{3}\not\in \mathbb{Z}$

if $ x,y$ are both at least $ 5$ then $ x^2\minus{}4x\plus{}1$ and $ y^2\minus{}4y\plus{}1$ are both positive so from $ (*)$ it follows that

$ 14xy\equal{}48\plus{}(x^2\minus{}1)(y^2\minus{}1)>4x\cdot 4y\equal{}16xy$ a contradiction.

So let's assume $ x\in \{3,4\}$

If $ x\equal{}3$ we get $ y\in \{4,\frac54\}$

If $ x\equal{}4$ we get $ y\in \{3,\frac{11}{5}\}$

So in conclusion the only solutions are $ (x,y)\in \{(0,7),(7,0),(3,4),(4,3)\}$

$ (xy\minus{}7)^2\equal{}x^2\plus{}y^2$

$ x^2y^2\plus{}49\minus{}12xy\equal{}(x\plus{}y)^2$

$ (xy\minus{}6)^2\plus{}13\equal{}(x\plus{}y)^2$

$ 13\equal{}(x\plus{}y\minus{}xy\plus{}6)(x\plus{}y\plus{}xy\minus{}6)$

Now there are 4 cases, which solved give the solutions.

Let us Rearrange the Equation as
$x^2(y^2-1)-(14y).x+(49-y^2)=0$ which is a quadratic in $x$
Hence , $\Delta =(2y^2-1)+195=t^2$ and the rest is trivial to figure out  :lol:

[quote=Rijul saini]Can anyone provide an approach using pythagorean triplets?[/quote]

Seeing as no one gave a solution using Pythagorean triplets
The question undoubtedly boils down to finding the number of triangles for which $x,y and (xy-7)$ are sides of a given triangle

Now, we may or may not know the algorithm given by pythagoras himself that states that the sides of the Pythagorean triangle are of the form $2n^2 + 2n, 2n+1 and 2n^2 + 2n + 1$
Plugging the given for the given values
Such that $x = 2n^2 + 2n$
$y = 2n+1$
And $(xy - 7) = 2n^2 + 2n +1$

We find that $n^3 + n^2 = 2$
Or $n^2 \cdot (n+1) = 2$
This gives us $n$ as $1$ only (for positive integral values)

Thus this gives $x =4$
$y= 3$
And thus $xy-7 = 5$
Thus since the pairs are ordered we have $(3,4)$ and $(4,3)$

Also since they say “non negative”
Plugging in 0 in place of $x$ or $y$, we again get the solutions $(7,0)$ and $(0,7)$

Hence the solutions are $(3,4),(4,3),(7,0)$ and $(0,7)$

[quote=Syngates]Now, we may or may not know the algorithm given by pythagoras himself that states that the sides of the Pythagorean triangle are of the form $2n^2 + 2n, 2n+1 and 2n^2 + 2n + 1$[/quote]

What about the Pythagorean triple $(8,15,17)$? This doesn't seem to be of that form.

The approach using Pythagorean triples:
Let assume $x , y > 0$, so $x=a^2-b^2; y=2ab; xy-7=a^2+b^2$, where $a>b>0$. Obviously $a \ge 2, b \ge 1$
Thus, $2ab(a^2-b^2)=a^2+b^2+7$
$2ab = ab + ab \ge a + b + 1$ and $a-b \ge 1$
So $a^2+b^2+7=2ab(a-b)(a+b) \ge (a+b+1)(a+b)$
$7 \ge 2ab + a + b \ge 2ab + 2 + 1$
$2 \ge ab$
Thus $a=2, b=1$. This gives us $x=3, y=4$.

Storage

\begin{align*}&(xy-7)^2=x^2+y^2 \\ \Longleftrightarrow \;& x^2y^2-12xy+49=x^2+2xy+y^2 \\ \Longleftrightarrow \;& (x+y)^2-(xy-6)^2=13\\ \Longleftrightarrow \;& (x+y-xy+6)(x+y+xy-6)=13\end{align*}

$\mathbf{Case\; 1:}$
$$ (x+y-xy+6)=13 \quad \quad \text{and} \quad \quad  (x+y+xy-6)=1$$which gives $x+y=7$ and $xy=0$ which then yields the solutions $(7,0),(0,7)$

$\mathbf{Case\; 2:}$
$$ (x+y+xy-6)=13 \quad \quad \text{and} \quad \quad  (x+y-xy+6)=1$$which gives $x+y=7$ and $xy=12$ which then yields the solutions $(4,3),(3,4)$

Hence all the solutions are $\boxed{(7,0),(0,7),(4,3),(3,4)}$

Our aim is two transform the equation to the desirable form.\\
$(xy-7)^2=x^2+y^2 \; \Rightarrow \; (xy)^2+49-14xy=x^2+y^2$\\
$\Rightarrow (xy)^2+49-12xy=x^2+y^2+2xy \; \Rightarrow \; (xy-6)^2+13=(x+y)^2$ \\
$\Rightarrow 13 \cdot 1=(x+y)^2-(xy-6)^2 \; \Rightarrow \; 13 \cdot 1=(x+y+xy-6)(x+y-xy+6)$ \\
Now it becomes mere calculation... There are two cases to look on.\\
$\mathtt{Case\; 1:}$\\
$(x+y-xy+6)=13 \quad \quad \;  \quad \quad (x+y+xy-6)=1$\\
Solving for $(x+y)$ and $(xy)$ gives us $7$ and $0$ respectively which henceforth yields $(0,7)$ and $(7,0)$ as Solution.\\\\
$\mathtt{Case\; 2:}$\\
$(x+y+xy-6)=13 \quad \quad \;  \quad \quad (x+y-xy+6)=1$\\
Solving for $(x+y)$ and $(xy)$ gives us $7$ and $12$ respectively which henceforth yields $(3,4)$ and $(4,3)$ as Solutions.\\\\
Therefore, $(3,4)$, $(4,3)$, $(0,7)$, $(7,0)$ are the integral Solutions to the equation.

First, we rewrite with algebra: $(xy-7)^2=x^2+y^2 \leftrightarrow x^2y^2-14xy+49=x^2+y^2 \leftrightarrow (xy-6)^2-(x+y)^2=-13 \leftrightarrow [xy-6-(x + y)][xy-6+(x + y)] = -13$.

Now, we can break into cases:

[i]Case 1 $(xy-x-y-6=-1, xy+x+y-6=13)$[/i]:
Adding the equations and simplifying, we get $xy=12$. Subtracting the equations and simplifying, we get $x+y=7$. Solving the system, we get the solutions $(4,3), (3,4)$.

[i]Case 2 $(xy-6-(x + y) =-13, xy-6+(x + y)=1)$[/i]:
Manipulating as above, we get $xy=0$ and $x+y=7$. Solving yields $(0,7), (7,0)$ as solutions.

In summary, our solutions are $\boxed{(4,3),(3,4),(0,7),(7,0)}$.

\( (xy - 7)^2 = x^2 + y^2\Rightarrow x^2y^2-14xy+49=x^2+y^2 \Rightarrow x^2+y^2+12xy+36+13=x^2+y^2+2xy \Rightarrow (xy-6)^2-(x+y)^2=-13\)
\(\Rightarrow (x+y+xy-6)(x+y-xy+6)=13\). Let $(x+y+xy-6)=a$, $(x+y-xy+6)=b$. Thus we get 4 cases:

$(i)$ Case $1$: Observe that $a<0 \Leftrightarrow b<0$. If $a<0 \Rightarrow b<0 \Rightarrow a+b<0 \Rightarrow 2(x+y)<0$, which is not possible as $x,y$ are non-negative integers. Thus, this case is not possible.

$(ii)$ Case $2$: If $a=1$, $b=13$ $\Rightarrow a+b=2(x+y)=14 \Rightarrow (x+y) =7$, $a-b=2(xy-6)=-12 \Rightarrow xy=0 \Rightarrow x=0$ or $y=0$, yielding the solutions $(0,7)$ and $(7,0)$ from this case.

$(iii)$ Case $3$: If $a=13$, $b=1$ $\Rightarrow a+b=2(x+y)=14 \Rightarrow (x+y)=7$, $a-b= 2(xy-6)=13 \Rightarrow xy=12$. We form a polynomial $p(x_0)= x_0^2-7x_0+12=0$. The roots of this equation are $x$ and $y$. By factorising, $p(x_0)=(x_0-4)(x_0-3)$. Thus the roots are $4$ and $3$. We get one more pair of solutions: $(4,3)$ and $(3,4)$.

Solutions: $\boxed{(7,0),(0,7),(3,4),(4,3)}$


Does anyone know how to make a link to reveal hidden content?? I have seen numerous people do this, but I have no earthly idea how yet.

 :roll:

Write (hide=text you want as the title) stuff you want hidden (/hide).
Except [] instead of (), becomes [hide="text you want as the title"] stuff you want hidden [/hide].
Also, if you just go (hide) stuff you want hidden (/hide), the text will be "Click to reveal hidden content."

If you want to do something you see someone else writing, just quote them to see the code. :D

Please ask questions such as this in the questions forum, next time.  Thanks.

Anyway, to create hide codes, use the following code:

[code][hide=text or smilies to click here]YOUR TEXT HERE
MORE TEXT
EVEN MORE TEXT[/hide][/code]

does this feature work on other kinds of forums too? or just AoPS?

just out of curiousity

Depends on the forum, there is no general answer. Just try.

Some forums have the features. Others don't. There is a PHPBB mod to create that, but most don't to answer your question. Some do.  :D


algebra
polynomial
induction
algebra unsolved


[hide=" Have I missed something?"]

$ \frac{M}{N}\equal{}\frac{6}{5}$

$ 5M\equal{}6N$

$ 2N\equal{} \frac{5M}{3}$

$ 3M\plus{}2N\equal{} 3M \plus{} \frac{5M}{3}\equal{} \frac{14M}{3}$.

[/hide]

hello, we get
$ M\equal{}\frac{6}{5}N$
and plugging this in the term above we get
$ \frac{18}{5}N\plus{}\frac{10}{5}N\equal{}\frac{28}{5}N$
Sonnhard.

$ \frac {14M}{3} \equal{} \frac {14}{3}\cdot \frac{6N}{5} \equal{} \frac {28N}{5}$.

I still think there's something sri340 forgot to tell us.


Prove $(1^{k}+2^{k}+...+n^{k})$ is divisible by $\frac{n(n+1)}{2}$ where $k$ is an odd positive integer.

[I posted this earlier under a different post here but didn't get any response.]

We have

\begin{eqnarray*}2(1^{k}+2^{k}+\dots+n^{k}) &=& (1^{k}+n^{k})+(2^{k}+(n-1)^{k})+\dots+((n-1)^{k}+2^{k})+(n^{k}+1^{k})\\ &=& (n+1)p_{1}(n)+(n+1)p_{2}(n)+\dots+(n+1)p_{n}(n)\\ &=& (n+1)p(n)\end{eqnarray*}

where $p_{i}(n)$ are some polynomials and $p(n)$ is their sum. (We're using the fact that $a+b|a^{m}+b^{m}$ for an odd natural $m$.) Therefore, $2(1^{k}+2^{k}+\dots+n^{k})$ is always divisible by $n+1$. Subsequently, $2(1^{k}+\dots+(n-1)^{k})$ is always divisible by $n$, meaning that $2(1^{k}+\dots+(n-1)^{k})+2n^{k}$ is also divisible by $n$. Hence, $2(1^{k}+\dots+n^{k})$ is divisible by both $n$ and $n+1$, but as those two factors are coprime, it must be also divisible by their product:

$2(1^{k}+\dots+n^{k})=n(n+1)q(n)$

where $q(n)$ is some polynomial, but since $n(n+1)$ is always even, we can write

$1^{k}+\dots+n^{k}=\frac{n(n+1)}{2}q(n)$, hence the conclusion.

[quote="boxedexe"]You can easily derive that \[\sum_{i=1}^{n}i^{k}=\sum_{i=1}^{k}S(k,i)\binom{n+1}{i+1}i!\] Then you easily see that $\frac{n(n+1)}{2}| \sum_{i=1}^{k}S(k,i) \binom{n+1}{i+1}i!$.

Q.E.D.

Masoud Zargar[/quote]What is the notation $S(k,i)$?

Stirling number of the second type.  $S(n,k)$ is the number of ways to partition an n-set into k parts.

[quote="boxedexe"]You can easily derive that \[\sum_{i=1}^{n}i^{k}=\sum_{i=1}^{k}S(k,i)\binom{n+1}{i+1}i!\] Then you easily see that $\frac{n(n+1)}{2}| \sum_{i=1}^{k}S(k,i) \binom{n+1}{i+1}i!$.

Q.E.D.

Masoud Zargar[/quote]
[b]boxedexe,[/b] I cannot "easily" derive what you just wrote and I am sure very few people in this community can derive the same easily unless they have seen the formula beforehand, in which case the formula is obviously not easy.

Also, when you talk about Stirling numbers, I would appreciate it if you direct others like me to an online article that can explain what these numbers are. Without such help, it becomes very hard for me to comprehend what you write.

You have strange criteria for "easy", boxedexe... Probably we should beware of any "easy" problems you post in the future :wink:

[quote="boxedexe"]Well then I am sorry for not doing so.

1. [url=http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html]Stirling Numbers of the Second Kind[/url]
2. The proof of the formula is counting in two ways; trying to represent the number of k+1-tuples such that the k+1-th term is $x_{k+1}>\max_{i<k+1}x_{i}$ in $(x_{1},...,x_{k+1})$.  Then you may use combinatorics, and then followed by induction.

Masoud Zargar[/quote]
If this is easy -- something I was supposed to have seen by myself, I think I am gonna stop doing math from now on.  :rotfl:

Come on, guys, chill. I guess he just meant that it's easy when you know those notations etc. If that's his way to solve his problem, fine.

[quote="boxedexe"]You can easily derive that \[\sum_{i=1}^{n}i^{k}=\sum_{i=1}^{k}S(k,i)\binom{n+1}{i+1}i!\] Then you easily see that $\frac{n(n+1)}{2}| \sum_{i=1}^{k}S(k,i) \binom{n+1}{i+1}i!$.

Q.E.D.

Masoud Zargar[/quote]
Proof in math generally means more than  statments like "You can easily derive ..."  and putting  some formula which, to put it mildly, does not even bring you any closer to what you want to prove.

And how do you show, that  $\frac{n(n+1)}{2}| \sum_{i=1}^{k}S(k,i) \binom{n+1}{i+1}i!$.? And by "proof" I mean    explaining it in a    mathematically expected way ,  not  by merely saying ..."one can easily see"...

... and see if you can show that without a circular argument. ...

... Besides,  [b]what you say isn't  true!  all the time  [/b]
(that is $\frac{n(n+1)}{2}| \sum_{i=1}^{k}S(k,i) \binom{n+1}{i+1}i!$.? )  is false for, say, k=2

(Notice, you do not mention, k is odd here).  :)

You can see one  example,  just the post above your post, to see what a  clear proof looks like.

Hope this helps.

I'm not in a battle of proving myself here, so I'll say this, and leave it to that.

1.  You cannot judge someone's proofs by one or a few posts.

2.  Sorry, I did not mention k being odd.  The casework was not present due to the fact that I did not spend more than 2 minutes thinking about the problem.  However, that proof can be completed.

3.  I thought the notation was known since it had been posted in the forums before.

4.  The poof of the formula is pretty simple if you are familiar with the notation, which I thought you were.

5.  I agree that the proof was somewhat incomplete and it was not explained well enough.  I know how to write proofs that are "rigorous."  I simply did not have the time to explain.

Masoud Zargar

yeah...stop bombing this poor guy... :rotfl:.  we all have times when we glimpse at a problem and think we had an outline, and just type it out due to laziness to go in depth...  Anyways, I read this in PSS, and can't really get how he put the Stirling numbers in this context.  Maybe you can work the first few cases out (or post a link) and explain more thoroughly (at least better than PSS...)???

[quote]
4.  The poof of the formula is pretty simple if you are familiar with the notation, which I thought you were.
[/quote]
I believe when you say "pretty simple" you mean:
[quote]
Then you easily see that $\frac{n(n+1)}{2}| \sum_{i=1}^{k}S(k,i) \binom{n+1}{i+1}i!$.[/quote]

How do you see that?  For example,  is it 'easier' to see this in the above form rather than to  see this in the original  form:
$\frac{n(n+1)}{2}| \sum_{i=1}^{n}i^{k}$

(And you can assume, I am familiar with the notations)

( Calling   $\sum_{i=1}^{n}i^{k}= X$  and [b]merely [/b]saying it is true for $X$ is no proof.     ($X$ ,of course, here is $\sum_{i=1}^{k}S(k,i) \binom{n+1}{i+1}i!$, but that is not the point - point is you still have to show it is true for $X$ .. all I am trying to get is how do you show this for $X$)

PS -  I have no intention of "bombing"  or having battle of any kind with anyone ,just trying to help here...)

Sorry, [b]boxedexe[/b]! I wasn't trying to bomb you.  :) The whole connection between Stirling numbers and the problem I mentioned, well, I just couldn't see the way you saw it. Sometimes, you just make me feel stupid. But, I am sure you are a nice person. Cheers!

It`s possible use unduction to prove that $\large{\frac{n(n+1)}{2}\,|\,(1^{k}+^{k}+...+n^{k})}$ . Are there someone to do it ?


The sum of two positive numbers multiplied by the first of the two is 84. The difference of the two multiplied by the second is 16. Find the largest and smallest possible sums for the two numbers.

It's not that easy. Check your second equation.

Solution take 2:
[hide]
$a(b+a)=84$
$b(b-a)=16$
Sum them.
$a^2+b^2=100$
The maximum sum is when a=b, then $a=b=\sqrt{50}$ and $a+b=10\sqrt{2}$
The minimum comes as a or b approaches 0 (integers are not specified, only "numbers") So as one approaches 0 the other approaches 10, and the minimum sum is just greater than $10$ because neither can ever equal 0 and be positive.[/hide]

The problem is ambiguous.  Is the first number larger than the second?  Or the second larger than the first?  It affects the way we set up the second equation.

[quote="chesspro"]True, it may affect the second equation. However, since $b(a-b)=-b(b-a)$, we can just subtract instead of add, which will still result in the equation that bthings_2000 got after adding.

As a side note:
[hide]Instead of $a^2+b^2=100$ we could say $x^2+y^2=100$. What is this graphically?[/hide][/quote]

$x^2+y^2=c$ is a circle


linear algebra
matrix
algebra
polynomial

Let $ A,B \in M_n(\mathbb{C})$ Such that $ rank(A^{n\minus{}1})>rank(A^{n}),rank(B^{n\minus{}1})>rank(B^n),AB\equal{}BA.$
Prove that : $ (A\plus{}B)^n\equal{}0$
--------- :P ----------

$ dim(ker(A^{n\minus{}1}))<dim(ker(A^n))$ ; moreover the sequence $ dim(ker(A^{k\plus{}1}))\minus{}dim(ker(A^k))$ is decreasing then $ dim(ker(A^n))\equal{}n$ and $ A^n\equal{}0$.
After it is not difficult.

Glancing at the problem, i'm assuming one must prove that A^n-kB^k=0 - but, I'm stuck here. Can anyone elaborate the proof?

$ A,B$ are nilpotent ; $ AB\equal{}BA$ implies that $ A,B$ are simultaneously triangularizable. Then we may assume $ A,B$ strictly upper triangular. Thus $ A\plus{}B$ is nilpotent.
The total proof is valid for $ A,B\in\mathcal{M}_n({K})$ ($ K$ a field).

What is this fuss all about? If $ A$ and $ B$ are nilpotent and commute with each other, then $ A \plus{} B$ is nilpotent (this is a standard fact from commutative algebra). If a $ n\times n$ matrix $ X$ is nilpotent, then $ X^n \equal{} 0$ (this follows from rank considerations, or from Cayley-Hamilton and minimal polynomials, or is simply well-known). No need for additional conditions like $ rank(A^{n\minus{}1})>rank(A^{n}),rank(B^{n\minus{}1})>rank(B^n)$ (well, loup blanc didn't use them either).

  darij

the rank considerations were necessary to prove that a,b were nilpotent. The rest of the problem follows as you pointed out darij

Also Darij, it is not enough to prove that A+B is nilpotent, we must have it be nilpotent of order n. I'm pretty self taught and the only proof I know that commutative rings admit a nilradical doesn't preserve orders: i.e., if A^n=B^n=0 then A+B is nilpotent but not necessarily (A+B)^n=0

edit: i just looked up the "standard" proof of the existence of the nilradical for the first time and it too only shows that A^n=B^m=0 implies (A+B)^(n+m-1)=0

So, your cited result doesn't make progress on the problem unless you are able to prove that if A+B is nilpotent, then its order is not greater than n.

Hi me@home,
a free lesson for you:
let $ A\in\mathcal{M}_n(K)$ s.t. $ A^k\equal{}0$; then $ spectrum(A)\equal{}\{0\}$ . We can triangularize $ A$ over $ K$. Then we may assume that $ A$ is a strictly upper triangular matrix. Now calculate $ A^2,A^3,\cdots$ and you will see that $ A^n\equal{}0$.

me@home: it seems you need new glasses, too ;)

[quote="darij grinberg"]If a $ n\times n$ matrix $ X$ is nilpotent, then $ X^n \equal{} 0$ (this follows from rank considerations, or from Cayley-Hamilton and minimal polynomials, or is simply well-known).[/quote]

loup blanc: triangulation is great, but let us kill harder problems using it ;)

  darij


[hide]The LHS is equivalent to $(d+1)(a+b+c-1)-a^2-b^2-c^2$. This, in turn, can be simplified to $(d+1)(a+b+c-1)+d^2-1=(d+1)(a+b+c+d-2)$. By Cauchy-Schwartz, $(a^2+b^2+c^2+d^2)*4\ge (a+b+c+d)^2$ so $2\ge |a+b+c+d|$. Then subtracting two from a+b+c+d we have a nonpositive. However, since $|d|\le 1$, d+1 is nonnegative, so the given inequality must be true.[/hide]

Here is another approach of the proof for $a+b+c+d\leq 2.$

$\left(a-\frac{1}{2}^\right)^2+\left(b-\frac{1}{2}^\right)^2+\left(c-\frac{1}{2}^\right)^2+\left(d-\frac{1}{2}^\right)^2\geq 0.$

$\Longleftrightarrow a^2+b^2+c^2+d^2-(a+b+c+d)+1\geq 0.$

$\Longleftrightarrow a+b+c+d\leq 2.\ Q.E.D.$

Nice

Ok kunny, maybe this indeed was a bit easy, like you said. What about this? (another self-made one.)  :D

For $a,b,c \in \mathbb{R}^+$ with $a\geq b \geq c$
so that $a+b+c=1$

Prove that $ab(3a-1)+ac(3b-1)+bc(3c-1) \geq 0$

[quote="Jan"]
For $a,b,c \in \mathbb{R}$ so that $a+b+c=1$
Prove that $ab(3a-1)+ac(3b-1)+bc(3c-1) \geq 0$[/quote]

Jan, sorry are you sure this inequality? I think may be $ab(3a-1)+bc(3b-1)+ca(3c-1) \geq 0.$

kunny

No, I'm pretty sure about my inequality, but there may be indeed a problem with it.
I forgot an extra thing: $a \geq b \geq c$.
I'm not very familiar with these sort of inequalities, so it's possible that this is not needed in the inequality, but just to be sure, i'll add it in the question.
Sorry

No problem, Jan. I thought your ineqality hasn't the symmetry of $a,b,c.$
That is to say $ab\rightarrow bc\rightarrow ca.$

kunny

Yes, I know, it's not symmetric, so you can not assume that $a \geq b \geq c$, thus it needs to be given.
Pity, now there is a hint in the question, else it would be much more difficult.  :P

anyone has any ideas for solving this problem? I'm very curious.

That's true Arne, but it's not necessary.
I had something else in mind when I created the problem.  :D

[hide="my solution"]
Expand and re-arrange the inequality to obtain
$-a^2 - b^2 - c^2 + (d + 1)a + (d + 1)b + (d+1)c - (d + 1) \leq 0$
Now substitute $a^2 + b^2 + c^2 = 1 - d^2$ to get
$d^2 - 1 + (d + 1)a + (d + 1)b + (d+1)c - (d + 1) \leq 0$
dividing through by $d + 1$, (which we can do WLOG because $d \geq -1$, and the inequality is obviously satisfied if $d=1$) we get
$d - 1 + a + b + c - 1 \leq 0 \Rightarrow a + b + c + d \leq 2$
Applying Cauchy-Schwartz, we get
$(a^2 + b^2 + c^2 + d^2)(4) \geq (a + b + c + d)^2$
and the result follows.
[/hide]

[quote="Jan"]Nice

Ok kunny, maybe this indeed was a bit easy, like you said. What about this? (another self-made one.)  :D

For $a,b,c \in \mathbb{R}$ with $a\geq b \geq c$
so that $a+b+c=1$

Prove that $ab(3a-1)+ac(3b-1)+bc(3c-1) \geq 0$[/quote]
I don't think this inequality holds.  For example, try $a = 3, b = -1, c = -1$, and you find $ab(3a-1)+ac(3b-1)+bc(3c-1) = -16 < 0$.

?!? Wow, you're right. It's wrong. Can somebody tell me what's wrong with my method?

Chebyshev inequality tells us that \[ \frac{a+b+c}{3}.\frac{ab+ac+bc}{3} \leq \frac{a^2b+abc+bc^2}{3} \]
\[ \iff \frac{1}{3}.\frac{ab+ac+bc}{3}\leq \frac{a^2b+abc+bc^2}{3} \]
\[ \iff ab+ac+bc \leq 3a^2b+3abc+3bc^2 \]
\[ \iff 0 \leq ab(3a-1) + ac(3b-1) + bc(3c-1) \]

Looks right to me... As far as I know Chebyshev holds for real numbers.

First I also thought the use of Chevichev's inequality, but why can we use that?

[quote="Jan"]?!? Wow, you're right. It's wrong. Can somebody tell me what's wrong with my method?

Chebyshev inequality tells us that...[/quote]
Chebyshev's inequality only applies with two series that both have $a_1 \geq a_2 \geq ... \geq a_n$.  So, for that inequality to hold, you'd have to have $ab \geq ac \geq bc$, which doesn't necessarily follow from $a \geq b \geq c$.  As you'll note, my counter-example doesn't follow this inequality, and so Chebyshev's Inequality breaks down.

I believe if you added the condition that $a,b,c \geq 0$, then your inequality would hold.

You are right zanttrang, I didn't look very careful when I thought $ab \geq ac \geq bc$
Of course it is not always true.
This is something I should have seen. But the inequality isn't lost, I only have to say that $a,b,c$ are positive. Then the inequality holds and the problem is solvable.

Sorry all, I corrected the question.
But, now the part I was curious of, does anyone have proofs without Chebyshev?

For example, as Arne suggested, homogenizing:

We rewrite the equality as \[ 3a^b+3abc+3bc^2 \geq ab+bc+ac \]
Now we homogenize: \[ 3a^b+3abc+3bc^2 \geq ab+bc+ac (a+b+c) = a^2b+ab^2 +3abc + b^2c+bc^2+a^2c+ac^2 \]
So we get \[ 2a^2b+2bc^2 \geq ab^2 + b^2c+ac^2+a^2c \]

How do we prove that's true?

Any other solutions?

[quote="Jan"]
So we get \[ 2a^2b+2bc^2 \geq ab^2 + b^2c+ac^2+a^2c  \]

How do we prove that's true?

Any other solutions?[/quote]
Split this up in the following way:

$(ba^2 + bc^2 - ab^2 - ac^2) +(ba^2 + bc^2 - ca^2 - cb^2) \geq 0$

We can prove that each of these two terms is greater than or equal to zero separately, through two applications of the re-arrangement inequality.  Note that by the re-arrangement inequality,

$a(ab) + b(c^2) \geq b(ab) + a(c^2)$

and

$b(a^2) + c(bc) \geq c(a^2) + b(bc)$

So, both of the terms are greater than or equal to zero, and our inequality is proven (this wouldn't work if some of the terms were negative, because I rely on the fact that $ab \geq c^2$).  This is the first time I've actually used the re-arrangement inequality, so I may have used it incorrectly  :blush:
Also, because Chebyshev's Inequality is proven using the re-arrangement inequality, this is probably functionally equivalent to using Chevyshev.

[quote="Jan"][quote]$(ba^2 + bc^2 - ab^2 + ac^2) +(ba^2 + bc^2 - ac^2 - cb^2) \geq 0$[/quote]
I'm sorry, but this is not the same, you should have $(ba^2 + bc^2 - ab^2 - a^2c) +(ba^2 + bc^2 - ac^2 - cb^2) \geq 0$[/quote]
I made a typesetting error - the second $ac^2$ should really be $a^2 c$, and the first $ac^2$ should have a negative sign in front of it; then my solution should proceed normally (I should have checked what I wrote more closely - however, I believe I used the right form in the rest of my solution).

But in fact, your idea to use rearrangement is good.

So i'll write down your idea for a solution:
\[ 2a^2b+2bc^2 \geq ab^2 + b^2c+ac^2+a^2c  \]

By rearrangement, we get: \[ a(ab)+b(c^2) \geq ac^2 + ab^2  \]
And: \[ a^2(b) + bc (c) \geq b^2c + a^2c  \]
Adding these two inequalities gives us \[ 2a^2b+2bc^2 \geq ab^2 + b^2c+ac^2+a^2c  \]

QED  :D

Other ways to prove it?


can enyone tell me what is the fermat point in geometry.
thanks for helping.

Do you mean the Fermat point?

Can someone move this to the appropriate section?


If the college I want to go to does not recognize any AP credits, is it worth it to spend time self-studying one or two AP subjects and then take the exams?

Yes, it is worth it to learn something and show that you did.

if you don't know for sure that you will end going to that college, it's worth taking APs in case you end up going somewhere else that does give AP credit. Also, some colleges use AP scores for placement (esp calc), even if they don't use them for credit.

also, APs look good while you're applying- they show you've taken the initiative to take harder classes when you didn't have to!

Thanks for all the help!  :)

I still got one question here:
My school only offers AP Calculus and not any other courses. I am planning to take the AP Statistics exam this year and then AP Biology in Gr. 12. If I am not enrolled in an AP class, how can I show them that I am taking the AP Biology exam in May?

Colleges will want you to list your current courses when applying, so you can list "AP biology by self-study" if you like.

The other nice thing about AP classes is the GPA boost. I'm going to take as many as possible... I [b]need[/b] that GPA boost... :D

How high school grade-point averages are calculated depends on the arbitrary policies of each high school. In SOME high schools, a grade of A in an AP course counts for more than a grade of A in some ordinary course. Other high schools do not have this policy. If yours does not, don't worry about it. College admission officers have their own separate ways of evaluating an applicant's high school record to see if an applicant took challenging courses in high school. Each college may have a slightly different way of doing this, and no college I am aware of publishes its methodology for comparing high school courses.

[quote="satyam"][quote]The other nice thing about AP classes is the GPA boost. I'm going to take as many as possible... I need that GPA boost... [/quote]

Can anyone please explain how AP classes help in universtiy GPA boost?

Also, I am planing to take AP Calculus and may be AP Statistics next year when I am in grade 12. You think those two AP credits will make it easier for me to get into a well known university like MIT? A lot of students that I know take AP courses in grade 11 while I am planning to take them in grade 12, and also I am only going to take 2 AP course; do i need to take more AP courses to get into MIT?[/quote]

Last year my son attended a meeting with several recruiters from top colleges. I was talking to an Alumni recruiter from MIT. He said that three AP scores were baseline among applicants, and they even had applicants with 20+ AP Scores. I shot back with "Do you have any idea how elitist that is??" He immediately backed down - and told me that a kid who only has a couple because their school only offers a couple will look just as good as a kid who has many because they go to an excellent prep school.

BTW, if you are taking Calc BC, that counts as two - you get an AB and BC score.

W/R/T to the GPA boost - on a scale of 0-4 GPA, you get an extra point for an AP class, so if you got an A in an AP course, it is basically 5 out of 4.  I can average out that B in English :-)

[quote="lfm"]
BTW, if you are taking Calc BC, that counts as two - you get an AB and BC score.
[/quote]

I don't think this is true... you do get an AB subscore, but AFAIK it does not count as an [i]additional[/i] AP credit. Not that the sheer number of AP credits is a big deal. Also, I'm pretty sure most top colleges only accept BC anyway.


Find all composite integers M such that the product of 2005xM has exactly eight divisors

[hide="hint"]
$2005=5*401$ thus it has $4$ devisors itself.
now, find all numbers in a form of :$M=pq$

edit: but doesn't this have infinite number of answers?
you can choose any two primes to be $p,q$
[/hide]

[hide]Well any prime which isn't already a factor would work because that would give us 2*2*2 factors, but it has to be composite, so...
$25$ or $160801$
This would make the prime factorization $5^3*401$ or $5*401^3$ making the number of factors 2*4=8[/hide]

oh, I didnt realize those were the same problem.  My solution is in the other thread if anyone wants to read it :D


Books Pham Kim Hung - Secrets in Inequalities
Pages 23

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \leq2$
Equality holds for $ a \equal{} b \equal{} 1,c \equal{} 0$

But...
when $ a \equal{} b \equal{} 1,c \equal{} 0$
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} 1$

I think it can't take $ 2$

$ a \plus{} b \plus{} c \equal{} 2$
we assume that $ max(a,b,c) \equal{} a$
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \leq a^2b^2 \plus{} 2a^2bc \plus{} a^2c^2 \equal{} a^2(b \plus{} c)^2 \equal{} a^2(2 \minus{} a)^2$

$ \leq (\frac {a \plus{} b \plus{} c}{2})^2 \equal{} 1$

Written inequality is obviously wrong. You just have to put 1 instead 2 on the RHS.
In thesolution you gave us it's also proved that it's lower or equal than 1 but not 2.  :wink:


Six distinct symbols are transmitted through a communication channel. A total of 18 banks are to be inserted between the symbols with at least two blanks between every pair of symbols. In how many ways can the symbols and blank arranged?

[hide="Hint1"] the problem is equivalent to: Six distinct symbols are transmitted through a communication channel. A total of EIGHT blanks are to be inserted between the symbols with at least ZERO blanks between every pair of symbols. [/hide]
[hide="Hint2"] Equal to the number of ways of arranging 8 blanks and 4 separations [/hide]