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Valentin was here

How do we know that you were no 10000? Your profile number is 10153 (roll over username and see the status bar)

well here it says he was 9999th...
http://www.artofproblemsolving.com/Forum/memberlist.php?mode=joindate&order=ASC&start=9950

strange.

I'm sorry 10000th user, your name is really cool but as they say, you are just 9999

the 10000th user is conway, he joined the same day, and made one post immediately to disappear again

Can anyone understand mathI, he joined just one day after Creation by Valentin, waited until december [b]2004[/b] to disappear forever then??

I somehow think that the $9999$ comes from banning/deleting one user or somewhat like that...

Banning?  I thought that that was extremely rare on mathlinks?

However one suffices, and that it is sad to take that pride away,  :o  :(  sorry 10000th user

I just discovered I'm a pretty early bird, I am user 359 according to that list, I joined in july 2003. :roll:

[quote="bubka"]How do you know that math1 has disappeared forever? He may turn up again now. Maybe he makes a post every one and a half year!

And why does it show 78 when I roll over your name?[/quote]

 :huh:  :?:   No idea, I am number 359 according to the member list

you show 17356, while you are in fact 17201

Possibly one counts the number of registrations, the other one the number of existing users.
Or somewhat like that.

PS: yes, bans are very rare, but one out of $9999$ in about one year isn't much.

Hmm, I heard about a member who got agressive about Iranian-Israeli relations, ultimately ending in a ban, but Valentin accepted his pleas to come back to matlinks.  I don't if I got that story completely, but CAN he do that?  When you are 'killed', is your existence wiped off of this forum?  (your private messages, number of posts, your list of posts)

And now one other thing , if microsoft decides to prey on me, and I lose my hotmail account... are my mathlink days over then.  Or will I be able to come back with another email account, but the same username??

Well, there is a difference between ban and deletion, the first one makes the forum just unaccessible for you, the second one deletes everything like  e.g. PMs (but posts get just "Guests" or somewhat like that, not sure). At my last post, I meant more or less deletion instead of ban.
And as much as I know, there where two users banned in these Iranian-Israeli relations, only one of them got back (but they both really only got banned I think).
But I know of at least one user that was really deleted.

For the second, you can change your mail-address mathlinks uses as long as you have the MathLinks-password. When you loose both, you get into troubles... (but there should be a way, too)

Just to clear a bit of the mystery here :P Math1 is a test account that I occasionally use to check stuff up :D

[quote="fredbel6"]interesting :

some questions :

1) can valentin see our passwords?  And our email accounts (that are hidden if you refuse to display them)?

2) do banned/deleted people get warnings before they are thrown off?

3) can moderators ban/delete as well or is that decision only up to valentin himself?

4)can mathlinks do anything against double account people?
I really hate that, people who get into arguments and nobody agrees with them... except one other guy that doesn't say anything else , or banned people who refuse to let the site be (you should the IMO article on wikipedia, it's pathetic on the talk page)[/quote]

I think he can see your email but not your password, and while the other questions are really for moderators to answer, I would say it depends on the circumstances.

[quote=Potato2017]oh my a 14 year old thread

look at how far aops has come; it now has like 500k useres i think[/quote]

I think 600k now

[quote=10000th User]Well, people were talking about me when I first joined, so let's see who will be the next person to hit 20000th user :D[/quote]

Rias Gremory is the OG of AoPSers!

Currently the newest user is s128190 which is user #687863 (I found him)

https://artofproblemsolving.com/community/user/687863

[quote=Valentin Vornicu]Haha, I know the secret identity of the user #20000 (with the username #20000) that is and it's not user #10000 :)

About admins: no you need not 42 points in the IMO. There is no condition that can be fulfilled that guarantees us needing you in the team or anything like that. There are though conditions that, if not fulfills, automatically disqualify any attempt: being very fluent in English is one of them for example.[/quote]

Who-

bruh how do u have a space in ur username

that's the 2nd user with there being no first

[quote=StarshipTigerTheNext]Who will be the 1,000,000th user? I hope they're active. [tip=nothing to see here dotted]they will probably be more orzed than dotted and cents[/tip][/quote]

I see what you did there  :-D

Almost all of you in the top posts died anyway.


$ABCDEF$ is a cyclic haxegon with $AB=CD=EF$. $G, H, I$ are the midpoints of $BC, DE, AF$.

Is $\triangle{GHI}$ an equalateral one? Prove your answer.

[quote="shobber"]$ABCDEF$ is a cyclic haxegon with $AB=CD=EF$. $G, H, I$ are the midpoints of $BC, DE, AF$.

Is $\triangle{GHI}$ an equalateral one?[/quote]

This holds only when AB = CD = EF is the radius of the circumcircle of ABCDEF. This fact is known under the name "asymmetric propeller theorem". Check out http://www.mathlinks.ro/Forum/viewtopic.php?t=16908 and http://www.mathlinks.ro/Forum/viewtopic.php?t=45874 .

  Darij


If the nth term of a product is $ {(4(2k\minus{}1)^4)}\plus{}1$ as numerator and $ {(4(2k)^4)}\plus{}1$ as denominator find the product up to 6 terms.
This is a star question

i guess using sophie germain identity we will get a telescopic cancellation but with me that got too computational

[hide]By the Sophie Germain Identity, we have $ 4x^4\plus{}1\equal{}(2x^2\minus{}2x\plus{}1)(2x^2\plus{}2x\plus{}1)$. Also note that $ 2x^2\minus{}2x\plus{}1\equal{}2(x\minus{}1)^2\plus{}2(x\minus{}1)\plus{}1$. Hence:
\[ \frac{4\cdot 1^4\plus{}1}{4\cdot 2^4\plus{}1}\cdot\frac{4\cdot 3^4\plus{}1}{4\cdot 4^4\plus{}1}\cdots\frac{4\cdot 11^4\plus{}1}{4\cdot 12^4\plus{}1}\equal{}\frac{2\cdot 1^2\minus{}2\cdot 1\plus{}1}{2\cdot 12^2\plus{}2\cdot 12\plus{}1}\equal{}\frac 1{313}.\][/hide]

i was plainly lazy to do that calculation


geometry
geometric transformation
rotation
geometry proposed

The plane is covered with network of regular congruent disjoint hexagons. Prove that there cannot exist a square which has its four vertices in the vertices of the hexagons.

[i]Gabriel Nagy[/i]

We can even assume the lattice is triangular: for each hexagon, add a point in its center and connect it to the six vertices of the hexagon. The conclusion holds for this larger lattice as well:

Assume WLOG that we can find points $ A,B $ belonging to this lattice s.t. if $ O $ is the origin of the plane, $ OB $ is obtained from $ OA $ through a rotation of angle $ \frac\pi 2 $ around $ O $. This is equivalent to finding integers $ a,b,c,d $ s.t. $ \frac{a+b\varepsilon}{c+d\varepsilon}=i $, where $ \varepsilon $ is a solution to $ x^2+x+1=0 $ ($ a+b\varepsilon $ represents the point $ A $, while $ c+d\varepsilon $ - the point $ B $). In turn, this implies that we can find rationals $ u,v $ with $ u+v\varepsilon=i $. This is claerly absurd, since eliminating $ u $ between $ u+v\varepsilon=i,u+v\bar\varepsilon=-i $ gives an irrational value for $ v $.


Prove that $ (\sin x \plus{}\cos x)^4 \leq 4$

[quote="enndb0x"]Prove that $ (\sin x \plus{} \cos x)^4 \leq 4$[/quote]

$ \Leftrightarrow (\sin \ x \plus{} \cos \ x)^2\le 2\Leftrightarrow \sin^2\ x \plus{} \cos^2 \ x \plus{} 2\sin\ x\cos \ x\le 2\Leftrightarrow 2\sin \ x\cos\ x\le$ $ 1\Leftrightarrow \sin \ 2x\le 1$

better for intermediate forum .

[quote="enndb0x"]Prove that $ (\sin x \plus{} \cos x)^4 \leq 4$[/quote]

$ \sin x \plus{} \cos x \equal{} \sqrt{2} \ \sin (x\plus{} \frac{\pi}{4})$

$ ( \sin x \plus{} \cos x)^4 \equal{} 4 \ \sin ^4 (x \plus{} \frac{\pi}{4}) \leq 4$


[hide="hint"]Let $y=x\sqrt{2}$. The polynomial looks nicer in terms of $y$.[/hide]

You would need to convert all of the $x$'s to $y$'s in order to make the substitution useful.

This is $2(y^2-6y+9)$, which can be factored further, of course.

After factoring completely, rewrite in terms of $x$.

so it's

2(y^2 -6y+9)
[b]2(y-3)^2[/b]

but when you expand it again it doesn't work out back to the original equation? because the surd is in the first section too.. sorry i must seem really dumb.

2(x√3-3)^2
2(x√3-3)(x√3-3)
2(2x^2√3 - 6√3 +9)
4x^2√3- 12√3 +18

[quote="laurenluvspink"]Just doing some homework and I came across an equation that I can't seem to factorise. The surd has confused me  :oops: .

The equation:
[b]4x^2 - 12√2x + 18[/b]

This is what I have done for the next step but I'm not sure if it's right:
2(2x^2 - 6√2x +9)

Can anyone please help?[/quote]

4X^2-12√2X+18=0.

Root (Only one): (3/2)√2.

Then, 4X^2-12√2X+18=4(X-3/2√2)(X-3/2√2).

How do you type those square-root thingies without $\LaTeX$?

It's a special character. You can go into a text formatter (MS Word) make one, copy/paste it, or copy/paste his into the text field in which you reply with.

[quote="JavaMan"]square roots that cant be simplified into an integer, i think[/quote]

Specifically, square roots that are irrational--can't be simplified to rational numbers


I understand long division but synthetic i dont like...so can someone solve this division problem using synthetic division....thanks

$x^3-4x^2-19x=9/x+3$

thanks

-3[_1__-4___-19___0___
 ______  -3 ___21 __-6
____1 _ -7____2 __ -6

X^2 - 7X + 2 Remainder -6
Sorry for the sloppyness I'm not sure how to use latex. Do you see what I did?

:starwars:

[hide]
You use 3 for the divider, but make it -3.
You take the coefficents of $x^3-4x^2-19$
Then, you bring down the one, multiple -3 by the number and put it below the next number and subtract:

-3
  1             -4             -19
                 -3               3
_____________________
  1             -1              -22
So: $x^2-x-\frac{22}{x+3}$
Ummm... I think that's how you do it.
The -3 is supposed to be under the -4, and the 3 under the -19.[/hide]


geometry
angle bisector
area of a triangle

Prove that it is impossible to have a triangle in which the trisectors of an angle also trisect the opposite side.

[hide="Hint"]Begin by showing that if an angle-bisector in a triangle is also the median to the opposite side, it must be perpendicular to that opposite side, and the triangle is isosceles.[/hide]

The area of a triangle is $ \frac {1}{2}absin\theta$.

By trisecting the side, you now have 3 triangles with equal area. Label the sides going from the vertex $ p,q,r,s$ from left to right.

$ \frac{1}{2}pq*sin\theta \equal{} \frac{1}{2}qr*sin\theta \equal{} \frac{1}{2}rs*sin\theta$

Or $ pq \equal{} qr \equal{} rs$ so From the first equality, $ p \equal{} r$ from the second $ q \equal{} s$.

Note that they can't all be equal, since there can be at most 2 segments from a point to a line with equal length.

Now, it is a simple argument. Ask yourself if $ p > q$ or $ p < q$ can you have the segments in that order? Hint, consider the foot of the altitude from the vertex, and the distances from the bases of the segments to the foot of the altitude.


[quote="papu24"]The sum of the first n terms of the series 1^2+2.(2^2)+3^2+2.(4^2)+5^2+2.(6^2)...... is {n(n+1)^2}/2, when n is even.What will be the sum when n is odd?[/quote]

Do you mean:

$ 1^2 \plus{} 2(2^2) \plus{} 3^2 \plus{} 2(4^2) \plus{} 5^2 \plus{} 2(6^2)\equal{}\frac{n(n \plus{} 1)^2}{2}$

? Please use $ LaTeX$ in the future.

[hide]
$ 1^2 \plus{} 2(2^2) \plus{} 3^2 \plus{} 2(4^2)...\plus{} 2(n^2) \equal{} \frac {n(n \plus{} 1)^2}{2}$

We are attempting to find the value of N+1. We currently have the value of N. Set $ m \equal{} n \minus{} 1$, so that we can work with the odd series without being confused

$ 1^2 \plus{} 2(2^2) \plus{} 3^2 \plus{} 2(4^2)...\plus{} (m^2) \equal{} \frac {(m\minus{}1)(m)^2}{2}\plus{}m^2$

$ \equal{} \frac {(m\minus{}1)m^2\plus{}2m^2}{2}$

Factor out m^2

$ \equal{} \frac {(m\minus{}1\plus{}2)m^2}{2}$

$ \equal{} \frac {(m\plus{}1)m^2}{2}$[/hide]

Uh, in your second line, don't you mean set $ m\equal{}n\plus{}1$?


geometry
circumcircle
angle bisector
congruent triangles
geometry solved

Let ABC be a triangle. M is a point in it's plane. The bisectrix of angles BMC, AMC,AMB intersect BC,AC,AB in A1,B1,C1.
Show that:
a) AA1 \cap BB1 \cap CC1=P
b) PA/PA1*PB/PB1*PC/PC1=8 if and only if M is the circumcenter of ABC

Do you mean that M lies inside ABC ? If so, the first part is quite obvious.

We have t(a) = CA1/BA1 = CM/BM by the angle bisector theorem. Similarly, t(b) = AB1/CB1 = AM/CM and t(c) = BC1/AC1 = BM/AM. Now t(a)t(b)t(c) = (CM/BM)(AM/CM)(BM/AM) = 1 so AA1, BB1 and CC1 are concurrent at P by Ceva's theorem.

Now we have to show that PA*PB*PC = 8*PA1*PB1*PC1 if and only if M is the circumcenter. Now PA/PA1 = AB/A1B, PB/PB1 = BC/B1C, PC/PC1 = AC/A1C. So we should prove that AB*BC*CA = 8*A1B*B1C*C1A if and only if M is the circumcenter. The if-part is obvious. If M is indeed the circumcenter, then triangles MAB, MBC, MCA are isosceles so A1, B1, C1 are the midpoints of BC, CA, AB. In that case AB = 2*C1A, BC = 2*A1B, CA = 2*B1C so indeed AB*BC*CA = 8*A1B*B1C*C1A. Now the "only if" -part remains.

Here's the proof of the "only-if" part. So we want to prove that if 8*A1B*B1C*C1A = AB*BC*CA, then M is the circumcenter.
Now note that A1B*B1C*C1A = A1C*C1B*B1A (by Ceva).
So A1B*B1C*C1A = \sqrt(A1B*A1C)*\sqrt(B1C*B1A)*\sqrt(C1A*C1B).
Then \sqrt(A1B*A1C) \leq 1/2(A1B + A1C) = BC/2 with equality only if A1B = A1C, i.e. if A1 is the midpoint of BC.
Similarly, \sqrt(B1C*B1A) \leq AC/2 and \sqrt(C1A*C1B) \leq AB/2.
Multiplying, we have
8\sqrt(A1B*A1C)*\sqrt(B1C*B1A)*\sqrt(C1A*C1B) \leq AB*BC*CA
with equality only if A1, B1, C1 are the midpoints of BC, CA, AB.
So if 8\sqrt(A1B*A1C)*\sqrt(B1C*B1A)*\sqrt(C1A*C1B) = AB*BC*CA then A1, B1, C1 are the midpoints of BC, CA, AB. This means M is the circumcenter (this is obvious, it can be proven by congruent triangles for example.)

It is very easy to see, that PA/PA_1=AB_1/B_1C+AC_1/C_1B= MA/MC+MA/MB.

(MA/MC+MA/MB)(MC/MA+MC/MB)(MB/MC+MB/MA)\geq 8  and equality holds iff MA=MB=MC


...and $ f(r),g(r)$ is increasing functions. No one interested?

[quote="Sunjee"]If $ r\geq 0$ and $ \infty > f(r),g(r)\geq 0$ then
\[ \left(\frac {w_1f^r(r) \plus{} w_2g^r(r)}{w_1 \plus{} w_2}\right)^{\frac {1}{r}}\geq \left(f^{w_1}(r)g^{w_2}(r)\right)^{\frac {1}{w_1 \plus{} w_2}}
\]
here $ w_1,w_2 > 0$[/quote]Is this what you mean really ? If so, simplily put $ f(r)\equal{}: a$, $ g(r)\equal{}: b$, $ \frac {w_1}{w_1\plus{}w_2}\equal{}: \lambda$, $ \frac {w_2}{w_1\plus{}w_2}\equal{}: \mu$,  and your ineqality becomes

$ (\lambda a^r\plus{}\mu b^r)^{\frac1r}\ge a^\lambda b^\mu$. Weigthed power mean.  :lol: :rotfl:

[quote="Amunaho"][quote="Sunjee"]If $ r\geq 0$ and $ \infty > f(r),g(r)\geq 0$ then
\[ \left(\frac {w_1f^r(r) \plus{} w_2g^r(r)}{w_1 \plus{} w_2}\right)^{\frac {1}{r}}\geq \left(f^{w_1}(r)g^{w_2}(r)\right)^{\frac {1}{w_1 \plus{} w_2}}
\]
here $ w_1,w_2 > 0$[/quote]Is this what you mean really ? If so, simplily put $ f(r) \equal{} : a$, $ g(r) \equal{} : b$, $ \frac {w_1}{w_1 \plus{} w_2} \equal{} : \lambda$, $ \frac {w_2}{w_1 \plus{} w_2} \equal{} : \mu$,  and your ineqality becomes

$ (\lambda a^r \plus{} \mu b^r)^{\frac1r}\ge a^\lambda b^\mu$. Weigthed power mean.  :lol: :rotfl:[/quote]

 $ f(r)\equal{}a,g(r)\equal{}b$ this is not true


analytic geometry
complex numbers
complex analysis
complex analysis unsolved

What are the Cauchy-Riemann Equations in polar coordinates FOR the modulus and the argument?

The Cauchy-Riemann equations state that:

$ \frac{\partial u}{\partial x}\equal{}\frac{\partial v}{\partial y}$ and $ \frac{\partial v}{\partial x}\equal{}\minus{}\frac{\partial u}{\partial y}$

That's if $ z\equiv x\plus{}iy$

If $ z\equiv re^{i\theta}$ we get

$ \frac{\partial u}{\partial r}\equal{}\frac{1}{r}\frac{\partial v}{\partial \theta}$ and $ \frac{1}{r}\frac{\partial u}{\partial \theta}\equal{}\minus{}\frac{\partial v}{\partial r}$

which is what you were asking.

Thank you for your answer, let me tell you why Im confused.

We know that if  $ f(z) \equal{} u(x,y) \plus{} i v(x,y)$ and $ f$ is analytic then the real and the imaginary part are connected by the
Cauchy-Riemann equations:

$ \frac {\partial v}{\partial x} \equal{} \minus{} \frac {\partial u}{\partial y},$ and $ \frac {\partial u}{\partial x} \equal{} \frac {\partial v}{\partial y}.$

If $ f(z) \equal{} u(r,\theta) \plus{} i v(r,\theta)$ then we have the Cauchy-Riemann equations in polar coordinates:

$ \frac {\partial u}{\partial r} \equal{} \frac {1}{r}\frac {\partial v}{\partial \theta},$ and  $ \frac {1}{r}\frac {\partial u}{\partial \theta} \equal{} \minus{} \frac {\partial v}{\partial r}.$

Now if we put $ f(z) \equal{} R(x,y)e^{i \phi(x,y)}$ then the modulus $ R$ and the argument $ \phi$ are connected by the relations

$ \frac {\partial R}{\partial x} \equal{} R \frac {\partial \phi}{\partial y}$ and $ \frac {\partial R}{\partial y} \equal{} \minus{} R \frac {\partial \phi}{\partial x}$.

So my questions is: The latter are the Cauchy-Riemann equations FOR the modulus and the argument? If so, then I guess I can deduce them in polar coordinates, in the same way we can do it for the real and the imaginary part.

I mean if $ f(z) \equal{} R(r,\theta)e^{i \phi(r,\theta)}$ what are the equations relating the modulus and the argument.

Thank you


Let $ a,b,c$ be positive real numbers such that $ abc\equal{}a\plus{}b\plus{}c\plus{}2$.Prove that:
$ \frac{ab}{(a\plus{}c)(b\plus{}c)}\plus{}\frac{bc}{(b\plus{}a)(c\plus{}a)}\plus{}\frac{ca}{(c\plus{}b)(a\plus{}b)}\le \frac{3(abc\minus{}4)}{8\plus{}abc}$

[quote="vo thanh van"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} a \plus{} b \plus{} c \plus{} 2$.Prove that:
$ \frac {ab}{(a \plus{} c)(b \plus{} c)} \plus{} \frac {bc}{(b \plus{} a)(c \plus{} a)} \plus{} \frac {ca}{(c \plus{} b)(a \plus{} b)}\le \frac {3(abc \minus{} 4)}{8 \plus{} abc}$[/quote]
IT's easy because $ (abc)^2\geq \frac{9}{8}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \geq 64$! :P


Prove that none of the digits 2,4,7,9 can be the last digit of a number equal to the sum $1+2+3+ . . . +n$  where $n$ is an arbitrary natural number.

$\sum_{i=1}^{n}{n}=\frac{n(n+1)}{2}$ , after this just check remainders on 10.

[hide]It is easily verifiable for n = 1, 2, 3, 4....., 10
So we will prove it by induction for n+10

Assume true for n=k, ie (n+1)(n)/2 not equal no 2,4,7,9 (mod 10)
Subbing in k+10 and simplifing we get (n+1)(n)/2 + 55 + 10n
therefore by putting in k +10 we either add 5 or 0 (mod 10) so we end up that once again, (k+11)(k+10)/2 does not end in 2,4,7,9... and by the princable or induction, is true for all n. (This is a quick, not reall detailed proof, but not to hard to verify yourself...)[/hide]

[hide="Even simpler is"]
mod 5 rather than 10 :) But of course, in the few seconds it takes you to think of that, you could probably have finished the mod 10 way..
[/hide]

[quote="seamusoboyle"]Curses!I hate when someone puts in their message while your typing yours! :)[/quote]
I guess that "someone" don't know when you are typing your message. :)


How many positive multiples of 13 are three-digit integers?

There are 900 positive 3-digit integers, from 100~999. So, our answer is 900/13, or 69.23..., and you ignore the decimal, so 69.
P.S. do not make the mistake of rounding...for example, when 900/7, it is NOT 129.

[quote="isabella2296"]Hmm, actually first you do $ 99 \div 13 \approx 69$ and then $ 100 \div 13 \equal{} \approx 8$. $ 69 \minus{} 8 \plus{} 1 \equal{} \boxed{62}$.[/quote]
I think isabella2296 meant $\lfloor\frac{999}{13}\rfloor=76$ and $\lceil\frac{100}{13}\rceil=8$.  Therefore, the number of 3-digit multiples of $13$ is $76-8+1=68+1=\boxed{69}$.


Warum gelten reelle Formeln für $ sinx$, $ cosx$, $ sinhx$, $ coshx$, $ e^x$ automatisch auch im Komplexen? Wie ist die Situation fuer Funktionen, die nicht ueberall definiert sind?($ arctan x$, $ log x$(!) unw.)

Fitim

Diese Funktionen, wie auch alle daraus resultierenden, sind lokal holomorph/analytisch (d.h. zu jedem Punkt im Definitionsbereich gibt es eine offene Umgebung, in dem die Funktion komplex differenzierbar ist/in dem die Funktion durch eine Potenzreihe gegeben ist; beide Begriffe stimmen faktisch überein). Nun gibt es den Eindeutigkeitssatz für holomorphe Funktionen:
$ f,g: U \to \mathbb C$ seien holomorph auf einem (offenen, zusammenhängenden) Gebiet $ U$. Enthalte weiter $ M \subset U$ einen seiner Häufungspunkte. Dann gilt: Stimmen $ f,g$ auf $ M$ überein, so auf ganz $ U$.

Nimmt man also einen zusammenhängenden Definitionsbereich in dem genannten Funktionen holomorph sind, so gilt eine "Formel" bereits dann, wenn sie z.B. auf einem Intervall einer positiven Länge gilt.


inequalities
function
inequalities theorems

[quote="Pascual2005"][b]Triangles relation for inequalities:[/b]

Hi: I just wanted to know from you, some of the relations between triangle things, specially, I would like to find expressions in terms of $R, r ,s (ABC)$, for the symetric functions of the sides $a,b,c$. If someone knows something, I will be very pleased if he could pot it here...[/quote]

It's well known that the following three identities hold in every triangle.
$a+b+c=2s$;
$ab+bc+ca=s^2+4Rr+r^2$;
$abc=4Rrs$.
With them, we can ratiocinate all other symetric functions of the sides $a,b,c$.
For example,
$a^3+b^3+c^3=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=2s(s^2-6Rr-3r^2)$.


I post this not to receive a solution, because this seems to be rather impossible, but because the results is magnificient:
  There is a constant $C$ such that there are no numbers of the form $ y^n$ with $y>1$ and $n>C$ in the Fibonacci sequence.

Harazi: I have an idea that i cant finish...

Suppose no such constant exist, first we can prove we can find  a composite index $m$, such that $F_m=y^k$, now i think we can extend this lemma [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=27757[/url] to the field $a+b\sqrt{5}$. using it we can prove that $m=ty$, and that $F_t=y^{k-1}$ but then $F_m/F_t=y$ wich fails for big enough $k$ I dont know, it is just some crazy idea.... ;)  and i have not been trainig lately ( i am on VACATION!!)


Can two triangles have two equal sides and three equal angles, and still be noncongruent?

If yes, then how? ;)

No, of course they have to be congruent (unless I misunderstood what you meant  :? ). If any 2 angles and 1 side of the triangles are equal, they are congruent, and here more than 1 side and more than 2 angles are equal.

[quote="Altheman"]well we know that they are similar, but we could have it so the corresponding sides are not equal[/quote]

I don't think that's true, because consider that you have a $3,4,5$ triangle, and using trig we see that the angles are $36.9, 53.1, 90$ degrees, let's say we have another triangle with the same angles and sides $4$ and $3$ but let's say $4$ is the hypotenuse, but $3$ can't be another side because it goes against the ratio of the sides and law of sines.


What is the maximum number of regions that 9 planes can divide a box?

I'm guessing 81 (which is 9 squared). I think there may be more, but I can't find that solution.  :blush:

never mind (edit), what the heck am I thinking?

[hide]
1 plane-2
2-4
3-8
4-14?

pattern

9planes-58
[/hide]

I think with four cuts, you can have 16 pieces, just cut into the shape of an 8-sided asterisk.

So now I think it's 2^9, which is 512.

I got fifteen regions for four cuts now...

1-2
2-4
3-8
4-15?

differences-2,4,7...
11,16,22,29,37

terms-
5-26
6-42
7-64
8-93

9-130

EDIT:
I just noticed,

points cutting a line (points, section)
1-2
2-3
3-4
4-5...difference is 1

lines cutting a plane (lines, regions)

1-2
2-4
3-7
4-11...differences are the ammount of sections from "points cutting a line"

planes cutting a space (planes, regions)

1-2
2-4
3-8
4-15
5-26...differences are the ammount of regions from "lines cutting a plane"

[quote="pianoforte"]I got fifteen regions for four cuts now...

[/quote]

Strive for excellence: try to find one more  :D

[hide]
I think each plane divides the current space into 2 parts, so 9 planes would divide this into 2^9=512 parts??!!
[/hide]

[quote="236factorial"][quote="pianoforte"]I got fifteen regions for four cuts now...

[/quote="pianoforte"]

Strive for excellence: try to find one more  :D[/quote]

I can't find one more, and my answer fits the pattern.  :?

[quote][hide]EDIT:
I just noticed,

points cutting a line (points, section)
1-2
2-3
3-4
4-5...difference is 1

lines cutting a plane (lines, regions)

1-2
2-4
3-7
4-11...differences are the ammount of sections from "points cutting a line"

planes cutting a space (planes, regions)

1-2
2-4
3-8
4-15
5-26...differences are the ammount of regions from "lines cutting a plane"[/hide][/quote][/quote]

EDIT: I don't see how the forth plane could split all of the existing 8 planes, at least one has to be left out :?

sorry piano forte, I was thinking of something else. How did you get 8 for three cuts?

One plane going side to side and up and down,
another going side to side and forward and back,
the third going up and down, and forward and back, :P

like stacking eight cubes into a bigger cube, four on top, four on bottom.


What makes a good math teacher/tutor/professor?

I believe that a good math teacher is someone who can break the material down to the evel of students.

Your view?

I want several replies.

I believe a good math teacher is fair and appreciates everyone's skill in math, whether good or bad.

A good math teacher should not give a lot of homework lol

a good math teacher is one who only cares about kids who do math competitions and doesnt teach the other people.

[quote="funcia"]one that lets you go fast if you are getting it but understands when you don't[/quote]

one who believes that mistakes do happen and should be less strict about grades and more eabout learning the material itself

Anyone can take something and make it more complicated. It takes true insight and understanding to take something that's complicated and make it simple. It takes a truly good teacher to do this with mathematics.

Building on that, mathematics teachers (before college, anyway) are notorious for making math seem like a collection of formulas. This goes with taking something simple and making it more complicated. A good mathematics teacher can show the relationships between concepts and unify ideas.

A good math teacher is willing to deviate from the curriculum a bit and explore different concepts. Most math teachers will not teach things like Ptolmey's, finite differences, etc.

A good math teacher does not teach to the test. Most math teachers teach to their own tests, and a lot of AP teachers teach to the AP test. I find this very annoying.


function symmetry conics parabola calculus

Function $y=kx^2-4x-1$ is decreasing on $[3,6]$. Find the range of $k$.

[hide="Solution"]We have $y'=2kx-4$ and it must be negative on $\lbrack 3,6\rbrack$. The derivation is a linear function, so its maximal value is in the left boundary for $k<0$ or in the right boundary for $k>0$:

\begin{eqnarray*}1^\circ\quad k<0 \land y'(3)\leq 0\implies k<0 \land 6k-4\leq 0\implies k<0\land k\leq{2\over 3}\implies k<0\end{eqnarray*}

\begin{eqnarray*}2^\circ\quad k>0\land y'(6)\leq 0\implies k>0\land 12k-4\leq 0\implies k>0\land k\leq{1\over 3}\implies k\in(0,{1\over 3}\rbrack\end{eqnarray*}

$3^\circ\quad k=0\implies y=-4x-1$, so the function is decreasing on the whole $\mathbb{R}$.

Hence, the answer is $k\in(-\infty,{1\over 3}\rbrack$[/hide]

[hide="solution"]

For negative $k$ , we have $y=k(x-\frac{2}{k})^2-1-\frac{4}{k}$ . So the symmetry line of the parabola is negative value and hence over $[3,6]$ , $y$ definitely decreases .

For positive $k$ , since the symmetry line is at $x=\frac{2}{k}$ . So in order for $y$ to decrease over $[3,6]$ , the symmetry line must be at least at $6$ which means $\frac{2}{k}\geq 6$ or $k\leq \frac{1}{3}$

Hence combine both cases to get the range for $k$ to be $(-\infty,\frac{1}{3}]$
[/hide]


what proportion  of the area of a circle lies inside an inscribed regular hexagon?

[hide="My sorution"]
Divide the hexagon into 3 equliteral triangles with side $s$. All of these have area $\frac{\sqrt{3}}{4}s^{2}$. So the area of the hexagon is
\[\frac{\sqrt{3}}{4}s^{2}\cdot 6=\frac{3\sqrt{3}}{2}s^{2}\]
Therefore, the proportion of the area of the circle that lies inside the hexagon is
\[\frac{\frac{3\sqrt{3}}{2}s^{2}}{\pi s^{2}}=\frac{3\sqrt{3}}{2\pi}\]
[/hide]


The ratio of the perimeters of two similar pentagons is 3 : 4.  If the length of the shortest side of the smaller pentagon is 48, what is the length of the shortest side of the larger pentagon?

The choices are:

(A) 24

(B) 36

(C) 64

(D) 72

This should be in Classroom Math...

[hide]$ \frac{48}{3}=\frac{x}{4}$

$ x=64$[/hide]

He has the right to post questions, but they have to be in the right forum.

I am sharkman not interval.  Maybe this interval person was a real pain but I am tired of being compared to someone who probably lives on the other side of the globe.

I am sharkman!!  Why sharkman???

I like sharks, the fish.

Get it?

Thank you i_like_pie for your defense but I have no idea what these people are talking about.

He used similarity. You know, if we say:

$ \frac{1}{x}= \frac{2}{3}$

then $ x = \frac{3}{2}$.

Similarity is a crucial part of geometry. Good luck learning it.


What is the sum of the numbers less than 200 that have exactly 9 divisors?

[quote="usaha"][hide]36,100,196 so 332[/hide][/quote]
How did you get those numbers?

[quote="usaha"][hide]36,100,196 so 332[/hide][/quote]

[hide]A number can only have an odd number of factors if it is a perfect square. Now to find the amount of factors a number has you add 1 to each exponent and multiply the sums. So this can equal 9 only when the number equals one exponent which has to be eight or, two exponents which are 2 and 2. [/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

[color=violet][size=134]An abelian group $ M$ is called indecomposable if it can't be writen by  a direct sum of two non-zero subgroup .

Is any indecomposable abelian group  $ M$    torsion-free(no finite order elements)?  except the trival situation:$ Z[1/p]/Z,Z_{p^{n}}$

in other word ,if an indecomposable abelian  group $ M$   is not torsion-free, it must be $ Z[1/p]/Z$, or $ Z_{p^{n}}$??[/size][/color]

[size=150]i don't know whether this problem has been solved ?

but  in the condition that $ M$ has only finite maximal subgroups we have got it !!

thanks very much for who gives  advices ![/size]

This is theorem 10, page 22 of Kaplansky's textbook on Infinite Abelian Groups: an indecomposable abelian group is either Z[1/p]/Z, Z/p^nZ, or torsion-free.

thanks
but i don't have read this book.
but in another  book there is this result.

INFINITE ABELIAN GROUPS
Ld.szlo Fuchs
Ttclane University
New Orleans, Louisiana
V O L U M E I

"Theorem 3.1. A group C is cocyclic if and only if CgZ(pk) with
k = 1, 2, - - .o r 00.

Corollary 27.3 (Kulikov [ l ] ) . If a group contains elements of fitiite order,
then it contains a cocyclic direct summand.
Corollary 27.4 (Kulikov [l]). A directly indecomposable group is either torsion-free or cocyclic."

so the guess is right....but it isnot proved by me.... :(


:pilot:  :censored::  Let a,b,c be posive numbers such that $ a\plus{}b>c, b\plus{}c>a, c\plus{}a>b$ and $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that      $ \sum  \sqrt{2\minus{}a^2}\geq a\plus{}b\plus{}c$

[quote="thegod277"]:pilot:  :censored::  Let a,b,c be posive numbers such that $ a \plus{} b > c, b \plus{} c > a, c \plus{} a > b$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that      $ \sum \sqrt {2 \minus{} a^2}\geq a \plus{} b \plus{} c$[/quote]
I think, it's wrong. Try $ a\equal{}1.4$ and $ b\equal{}c\equal{}\sqrt{0.52}.$ :wink:


[quote="mikejones"]could someone help me factor these:
 -x^3+8x^2-8x-32
and -x^3+3x+2[/quote]

Wrong difficulty level, I think. ;)

But as for your problems, think Rational roots. ;)


calculus
integration
analytic geometry
trigonometry
Gauss
calculus computations

Calculate the integral

$\int \int_V \int \frac{1}{x^2 + y^2 + z^2} \text dx \text dy \text dz$

over the volume $V$ defined by $x^2 + y^2 + (z - 1)^2 \le 1$.

I am in a hurry right now, but let's see. I suppose that the best way to do it is make a change of coordinates to spherical coordinates, and so, If I am not mistaken (As I said, right now I don't have time to review my calculations), the integral becomes,

\[\int_{0}^{2\pi}\int_{-\frac\pi2}^{\frac\pi2}\int_{0}^{2\cos\phi}\frac{1}{\rho^2}\;d\rho\;d\phi\;d\theta\]

What is easier to compute. Sorry, gotta go. Bye,

[quote="fedja"]In hurry you forgot to multiply by the Jacobian $\rho^2\sin\varphi$. ;)  If you do that the integral becomes really easy...[/quote]

It was not just the missing jacobian $r^2 \sin \theta$, but also the integration limits. If you go to spherical coordinates $r, \phi,  \theta$

$x = r \cos \phi \sin \theta$

$y = r \sin \phi \sin \theta$

$z = r \cos \theta$

the integration limits are not constant for either of $r, \phi$ or $\theta$. If you go to spherical coordinates $r, \phi,  \theta$

$x = r \cos \phi \sin \theta$

$y = r \sin \phi \sin \theta$

$z = r \cos \theta + 1$

then the integration limits are constant and the integral becomes

$\int \int_V \int \frac{1}{x^2 + y^2 + z^2} \text dx \text dy \text dz = \int_0^{2 \pi} \int_0^{\pi} \int_0^1 \frac{r^2 \sin \theta}{r^2 + 2r \cos \theta + 1} \text dr \text d\theta \text d\phi = 2 \pi \int_0^{\pi} \int_0^1 \frac{r^2 \sin \theta}{r^2 + 2r \cos \theta + 1} \text dr \text d\theta$

This can be integrated in a rather laborious way by expanding the integrand into a power series $\sum_{n = 0}^{\infty} a_n(\cos \theta) r^n$, integrating term by term and then figuring out how to sum the results. But there is another solution, much simpler than going to spherical coordinates.

[quote="yetti"] If you go to spherical coordinates $r, \phi,  \theta$

$x = r \cos \phi \sin \theta$

$y = r \sin \phi \sin \theta$

$z = r \cos \theta$

the integration limits are not constant for either of $r, \phi$ or $\theta$.

[/quote]
I think djimenez used
$x = \rho \sin \phi \cos \theta$

$y = \rho \sin \phi \sin \theta$

$z = \rho \cos \phi$

Then his limits of integration are correct ;)

There are too many different angle conventions for spherical coordinates here.

djimenez is using the convention (standard in calculus books) that $\theta$ is longitude, $\phi$ is colatitude, and the radius is $\rho$, but his limits are slightly wrong as well as the Jacobian: it should be $\int_0^{2\pi}\int_0^{\frac\pi2}\int_0^{2\cos\phi}\sin\phi\, d\rho\,d\phi\, d\theta$
(The entire ball is in the "north" half-space $0\le\phi\le\frac\pi2$. The southern half-space would be $\frac\pi2\le\phi\le\pi$, and I don't know where $\frac{-\pi}2$ came from.)
This is a simple integral- easier than yetti's method.

yetti is using the convention (standard in physics books) that $\theta$ is colatitude, $\phi$ is longitude, and the radius is $r$.

Spherical coordinates centered at the origin make the integrand nice. Since the limits aren't too bad, it works better than spherical coordinates centered at the ball of integration.

[quote="jmerry"] it should be $\int_0^{2\pi}\int_0^{\frac\pi2}\int_0^{2\cos\phi}\sin^2\phi\, d\rho\,d\phi\, d\theta$
[/quote]
Oops  :blush:  jmerry is right as to the limits. As to the power of $\sin\phi$, I disagree:
the volume element is $d\rho\cdot \rho \,d\phi \cdot \rho\sin\phi \,d\theta$, so the integral becomes $\int_0^{2\pi}\int_0^{\frac\pi2}\int_0^{2\cos\phi}\sin\phi\, d\rho\,d\phi\, d\theta$. Seems like we all are not at our best today  :P

Oops... I was confused about a physical interpretation when I edited that in. $\sin^2\phi$ comes from the $z$-component of an inverse-square force.

[quote="fedja"]
Then his limits of integration are correct ;)[/quote]

Not entirely. It should be

$\int_0^{2 \pi} \int_0^{\frac \pi 2} \int_0^{2 \cos \phi} \frac{r^2 \sin \phi}{r^2} \text dr \text d\phi \text d\theta$

What I had in mind was to substitute $\frac{1}{r^2} = \text{div}\left( \frac{\vec r}{r^2} \right)$ and then use Gauss' theorem

$\int \int_V \int \frac{1}{r^2} \text dV = \int \int_V \int \text{div}\left( \frac{\vec r}{r^2} \right) \text dV = \int_S \int \frac{\vec r}{r^2} \cdot \text d\vec S$

Substituting $\vec \rho = \vec r - \vec k$, the integration surface becomes a unit ball $\Sigma$ and

$r^2 = |\vec \rho + \vec k|^2 = \rho^2 + 2\rho \cos \theta + 1 = 2(1 + \cos \theta)$

$\vec \rho \cdot \text d\vec \Sigma = \text d\Sigma$ [color=white].[/color] and [color=white].[/color] $\vec k \cdot \text d\vec \Sigma = \cos \theta\ \text d\Sigma$

$\int_{\Sigma} \int \frac{(\vec \rho + \vec k)}{|\vec \rho + \vec k|^2} \cdot \text d\vec \Sigma = \int_{\Sigma} \int \frac{1 + \cos \theta}{2(1 + \cos \theta)} \text d\Sigma = \frac 1 2 \int_{\Sigma} \int \text d\Sigma = 2 \pi$

[quote="fedja"]In hurry you forgot to multiply by the Jacobian $\rho^2\sin\varphi$. ;)  If you do that the integral becomes really easy...[/quote]

Upsss :blush:  :blush: , as I told, I was in a hurry, and well, you already saw the resoult of that  :oops: .

[quote="djimenez"]..., as I told, I was in a hurry, and well, you already saw the result of that  :oops: .[/quote]

It was very nice and quick solution. At first, I could not get it, because of the incorrect limits $-\frac \pi 2, \frac \pi 2$. This gave me the impression that the missing jacobian must be $\rho^2 \cos \phi$ and than it did not add up.

yetti


geometry
algebra
polynomial
quadratics
absolute value

Let y = ax^2 + bx + c be a second degree polynomial with real roots, where a < 0. Find the area enclosed by the x axis and the quadratic in terms of a,b,c .

And don't give me Archimedes' 2/3bh ! I want it in terms of a,b,c.

why must you know this??

x1 = (-b -  :sqrt: (b^2-4ac))/2a
x2 = (-b +  :sqrt: (b^2-4ac))/2a

y = ax^2 + bx + c

ax^3 / 3 + bx^2 / 2 + cx from x1 to x2
that's assuing x2 is to the right of x1, if u get a negative answer... just absolute value it.

area ( a/3*(x2)^3 +b/x*(x2)^2 +c(x2) ) -
( a/3*(x1)^3 +b/x*(x1)^2 +c(x1) )

i think that's right, i dunno, its a ridiculous formula, not too bad when you use actual numbers though.

You must go further! hehehe don't worry, keep working on it. It actually works out to a pretty nice formula


The area of a square is 64 square centimeters. What is the number of square centimeters in the area of a new square made by decreasing each side of the original square by $ 75\%$?

Each of the sides of the square has length $ \sqrt{64}\equal{}8$. So, $ 8(1\minus{}.75)\equal{}2$. Hence, the new area is $ 2^2\equal{}\boxed{4}$.


Interested in what IPO is about? Here is the [url=http://www.philosopiad.org/]site[/url]. Enjoy :)

I actually didn't know that such an olympiad exists until now !!  :( And anyway, it might be sometime before India begins to participate...


If the word REPETITION is repeatedly written until there are 1000 letters, what percent of the letters are E's?

Well, there are exactly $ 10$ letters, so you can divide the 1000 letters by 10 and by 10 again.  Thus, there are only $ \boxed{20}$ percent E's.

No, ernie is correct. There are 2 e's. We write the word REPETITION 100 times. We have 20 e's. Thus the percent that we want is $ \boxed{\%20}$.

Edit: Oh...

AIME15 wasn't stating an answer, just stating an observation.

Sorry if I'm being ignorant but I don't quite understand what AIME15's observation was about.  What does it have to do with the problem?

I just looked at the problem and saw it was 2 e's out of 10. no matter what multiple of 10, it will always be 20%.


hey friends!

going thru many resources i hav come across the symmetric and cyclic summation ....symmetric mostly...

the expression of a symmetric summation is generally done in a single step as if it were on their tips...

i find it tedious and irritating to find these summations by the beginners' method..

so if any of u r aware of any ways tricks or shortcuts to find these expressions please post them here..

sfsreporter

does no one want 2 help or is it that there not any known methods 2 do this?

i think u got me wrong i have problems expanding an expression in the symmetric\cyclic forms

for example---

http://www.mathlinks.ro/Forum/viewtopic.php?t=32470&start=20

post #22

how did this man got the expression in a single step...... fine he must have done some computations on paper beforehand ...but this seems 2 be genius.

even in assignments like Kiran Kedlaya and Mildrof they do this saying "after some [b]simple computations[/b] we get....."

i keep trying these "simple computations" and it seems that i am a fool!! who finds this difficult.

so i wanted 2 know if there is any method to do all this by "SIMPLE COMPUTATIONS"???

By "simple computations" they mean something that, say, a computer could do for you. It isn't hard to keep applying the distributive law over and over and over and over and over (as long as you don't make a mistake in the middle).

It doesn't mean that they found a really fast way to do it by magic. It means that they actually sat down and expanded and expanded and expanded, and didn't feel like writing all of the expanding down so they skipped right to the end of all the expanding in their explanation for us.

An analogy: somebody asks you what 534/716 is to eight decimal places. You tell them, "after some simple calculations, I got 0.74581005." They say, "ohmigod how did you figure that out, I kept guessing 0.7457 or maybe 0.74583??? How did you do it in one step??" The truth is you spent three hours with a pencil and paper doing long division...

sorry cant edit but i meant consolation...not consolidation

[quote="sfsreporter"]thx for that kindda consolidation.

i feel better now..

but then if i get Iran 1996... i mean a similar problem in an olympiad ...so do i need to expand it all sitting dere in da examination hall??

is there no other possible way?[/quote]
First of all: Skip the abbreviations. It is very difficult to read what you have written, when you use such bad english.
In general there is a lot of other ways than expanding.
Let's take an example:
$ \sum_{cyc} \frac {x \plus{} z }{2y} \ge \sum_{cyc} \frac {2y}{x \plus{} z}$ for positive $ x,y,z$.. Now you could multilply with $ xyz(x \plus{} y)(y \plus{} z)(z \plus{} x)$ expand it and then use muirhead and schur or something like that.
A easy way to solve this is to notice that $ x \plus{} z \ge 2\sqrt {xy}$ and thus $ \frac {y}{\sqrt {xz}} \ge \frac {2y}{x \plus{} z}$.
But the inequality $ \sum_{cyc} \frac {x \plus{} z}{2y} \ge \sum_{cyc} \frac {y}{\sqrt {xz}}$ is very easy to prove. It comes from muirhead since $ (1,0, \minus{} 1) \succ (1, \minus{} \frac {1}{2}, \minus{} \frac {1}{2})$. Also $ \frac {y}{x} \plus{} \frac {y}{z} \ge \frac {y}{\sqrt {xz}}$, and so on with all three variables. Addition yields the desired.
Sometimes you can also make a minor change in the inequality like i did, and then expand. It comes from experience:)

[quote="sfsreporter"][quote]Skip the abbreviations. It is very difficult to read what you have written, when you use such bad english. [/quote]

sorry but i was in a hurry.. i think it was not that much of sms language..it was quite moderate but its fine...i'll prefer the other way now..

i'm afraid but u hav made a typo in the problem ...cud you plz edit it so that i can have a better look at the thing.

thanks![/quote]
I sound more arrogant than I actually am, when I'm writing. I just get confused when I see "R" instead of "are" :)

Anyway. The example is fixed.


I am sorry for asking so much but:
How many positive two-digit integers are there in which each of the two digits is prime?
I got 4 but the answer says 16...

akhil0422

There are $ 4$ prime digits: $ 2$, $ 3$, $ 5$, and $ 7$.  Thus, there are $ 4$ choices for the first prime digit, and $ 4$ choices for the second prime digit.  Multiplying yields $ 4 \times 4 \equal{} \boxed{16}$.


MATHCOUNTS
AMC
AMC 8
email
articles
search

Sigh...you people who left early to have time to trick-or-treat or whatever missed out on sooo much. Oh well. I hope you have the info necessary to do the homework. If not I can explain stuff here I guess. If you stayed, thanks for being patient. Either way, discuss the problem sets here. Last week's answers will be up eventually.

Oh yeah--AMC-8 NEXT TUESDAY! Sign up! Yay.

hey, I don't think you're doing a good job of making the mc kids come here.
this forum is like, pretty inactive. No one(from jls) posts here cept susan, linda and that other slt person that I dont know.
and do everyone who sign up for AMC8 get to take it? that's nice....

susan, linda, slt don't post here anymore. this forum is dead. now i can't post since there aren't any new posts. also can i take the amc 8 for fun?

Huh, I think I'll start demanding that everyone post at least one solution to a problem beginning the week after next. I don't know if anyone will listen to me but whatever. David, we've only got 30 tests. I strongly suggest you not waste one that could have been used by a JLSer. Though I suppose you could go in case less than 30 people show up.

Wait a sec...the AMC-8 is NEXT week?! I thought it was today! Darn it. That means I shouldn't have gone last week, when I did, and should have gone today, when I didn't. I'm so easily confused. Darn it. I hope Ms. Beddoes didn't wait too long for me. Better email her about this.

Uh... As amusing as that was...no more spam.

It's a good thing that the AMC wasn't today: now I can recruit that former 6th grader that totally pwned math class to math club. :)

[quote="ccy"]It's a good thing that the AMC wasn't today: now I can recruit that former 6th grader that totally pwned math class to math club. :)[/quote]

Who's that? Does (s)he go to Terman?

[quote="moogra"][quote="junggi"][quote="moogra"]according to ccy, its jeffrey wang's brother[/quote]
and according to you, you're a retard[/quote]

no thats you. ur the only one here trying to screw around. im answering questions.[/quote]
no you weren't.
you were giving answers that made no sense.
according to you, ppl throw helmets at each other when theyre a tiny bit pissed. of course that's complete nonsense. even if that was true, you still haven't justified your action, because from what I saw from your face before you threw the fing shiit, you weren't just pissed. You were EXTREMELY angry. Just because i threw a piece of paper. wtf is that? (Remember that you tried to punch my head off before you threw that helmet.)

just to remind you what happened, bc, given your intelligence level, I doubt that you remember.

I tried to get you to come to mandelbrot because I thought you would have liked to(because you spend a lot of time on mathcounts and all that. and of course I was wrong to think that you would still have interest left in math. oh and btw, you coming to this forum is a complete waste of your time, because what do you do? YOU POST SOLUTIONS FOR MATHCOUNTS AND MOEMS PROBLEMS. I'd be fine with it if you were trying to teach them something, but no, you post either wrong solutions, or completely inelegant solutions. I wouldn't be saying this if posting solutions there actually helped you. But it doesn't. You only do the trivial problems. You'd be better off playing Maple story or some other games for retards.). Then you keep insulting me, and how I'm superlame because I like math and that bs.
So I get pissed, like everyone would when theyre insulted. So I throw a piece of paper. Then you suddenly turn into this beast thingy with an extremely angry face on, and you try to punch my head off. Then you stop for some reason, so I try to pick up the paper. Then when I had my head down, you, like an fing coward, pick up the helmet, and threw it at my head. WTF

You do have a point. and i never said math was lame, and i don't post wrong solutions. Also good job on the essay. It was interesting from your point of view.
-edit-
(even if I do post wrong solutions (might be 1 somewhere), i don't try to and that shows I actually tried to do the problems)

er...hi...am I interrupting anything??????

OK. SO bookaholic. I'm gonna drop out of Analysis soon. Another thing. Tell me. Just HOW do you avoid idoticly retarded mistakes, like the one I did on the practice Chapter Round and the five retardedly idiotic mistakes I made on the Analysis test?

By the way, junggi, nice er...essay...you have there....

Something else I forgot........

[quote="moogra"]Yes this is weird.  :D This scares everyone away[/quote]
ok f u, retard.
and seriously, stop coming here, and play maple story or something

[quote]By the way, junggi, nice er...essay...you have there.... [/quote]
lol yea did you actually read that? of course not lol

[quote="Bluestalk"][quote]Quote:
By the way, junggi, nice er...essay...you have there....

lol yea did you actually read that? of course not lol
[/quote]

WRONG junggi, I actually read it. Shows how bored I got. 0.0

BOOKAHOLIC WHENEVER YOU'RE ON I NEED SOME AMC 8, 10. and 12 PRACTICE PROBLEMS!!! (Er...I don't exactly know where to find them so.... :?:  :| )[/quote]
................................................................................
TRY THE AOPS FORUMS? :?:  :?:
(THAT MEANS MATHCOUNTS,AND HSB FORUMS)

[quote="Bluestalk"]
I'm gonna drop out of Analysis soon.
[/quote]
  :o whoa i didnt see that before...
as for stupid mistakes, this might help
[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php[/url]
[quote="moogra"]and i never said math was lame, and i don't post wrong solutions. Also good job on the essay. It was interesting from your point of view.
[/quote]
dude, stop lying.
I clearly remember you saying "YOURE TRYING TO MAKE TO COME TO A [i]MATH [/i]CONTEST! HOW LAME CAN YOU GET?" and similar stuff
[quote="moogra"]
(even if I do post wrong solutions (might be 1 somewhere), i don't try to and that shows I actually tried to do the problems)
[/quote]
dude, I saw you posting SUPER inelegant(means retarded soln in case you didnt know) solns to trivial problems. Also, PLEASE dont post solns after 1028084107 people already posted the exact same sonl you have

[quote="bookaholic"]Drop out of Analysis?! Why?! You're an eighth grader, your grades this year won't be recorded. Even if you de-skip a grade, shouldn't you at least wait until it actually matters? Or is it taking too much time or something?

As for stupid mistakes...I'm not sure. I think there's an article about eliminating stupid mistakes in the Resources section. You should try the ideas in there. I didn't find it very helpful but that's probably just me. My carelessness is too ingrained in my nature. Um. If the problem is longish and complicated, I would suggest running through the solution in your head or even writing it out (very abbreviated of course) if you have time, and stopping at each step to make sure it's logical. Keep your scratch work very organized if possible. Make sure you're reading the problems right, I think misreading is the most preventable of stupid mistakes. What else...if you're making arithmetic mistakes or something, that should theoretically get better with practice. I don't know if it actually does.

Oh. Practice problems? They're all over AoPS. http://www.artofproblemsolving.com/Forum/resources.php?c=182 is the official page for all the former AMC series contest problems, but the unofficial problems in the forums are almost as useful. If you search in the AMC forum, people have been writing a lot of mock AMCs and AIMEs. I think there have been mock MathCounts and AMC-8 contests in the MathCounts forum. You can also buy some from http://www.unl.edu/amc . They're not really all that expensive.[/quote]

Uh bookaholic...  Next time you see bluestalk, ask her about the retarded (for the lack of a better word) way our teacher grades tests and quizzes.  I got points knocked off because I didn't round correctly :( and for some other trivial mistakes I can't remember.
[hide="testimony by junggi"][20:05:37] me: do you agree that beck grades retardedly?
[20:05:43] junggi: of course
[20:05:49] junggi: she takes off 3 pts for one mistake
[20:06:00] me: thanks im quoting you
[20:06:05] junggi: ok[/hide]

@bluestalk: Here's a nice resource for AMC practice problems: http://www.unl.edu/amc/mathclub/04,1-problems.html (click the IAP links).  What will you do if you drop analysis?  School service?  :maybe:

Gee, people, FYI, I'm stupidder than you think. Stupidder. Stupider. Which one...?

OK, thanks for all those links....I'm only allowed like 10 minutes on the computer at home, so that helps. =P

It's probably stupider. But at least you probably have never written "2n^2-2n+1=2n^2-1" on a contest problem sheet...

(See, the thing is, I distributed 2n(n-1)+1 as 2n^2-2+1, so I wrote 2n(n-1)+1=2n^2-2+1=2n^2-1 on my paper. Later I realized I'd left out an n and scribbled it into the expression in the middle but FORGOT that that made the expression on the right invalid and just used it anyway. Argh.)

Er...OK.....I do that periodically...

Bookaholic, I pmed you about some Mock AMC questions I didn't get.....

If you have the time, could you post solutions? Thanks!!

OH YEAH. One of the orchestra flute players is the president of PALY Math Club. Adrian Sanborn or somethinglikethat...anyway he says you're good at math. =P Don't deny it....

Deny it in your head if you must then....


We have a regular polygon with 12 sides. How many edges can we draw between 2 of the vertex in such a way that there are no triangles???

Do you mean triangles whose vertices are vertices of the 12-gon? Or whose vertices are allowed to be intersection points of some diagonals?

Pierre.

I mean triangles whose vertices are vertices of the 12-gon.

So, the maximum is 12 2 /4 = 36 as a direct consequence of Turan's theorem on graph without triangle.

Pierre.


modular arithmetic
number theory unsolved
number theory

Let $ n$ be a natural number, $ n \geq 2$.Prove that if $ \frac{a^n\minus{}1}{a\minus{}1}$ is a prime power for some positive integer $ a$, then $ n$ is a prime number.

Assume that n is a composite number. We can write it as n=x.y x and y are positive integers greater than 1. (Not neccesarily distinct.) W.l.o.g. we may assume that x is greater than or equal to y. (a^(x.y) - 1)/(a-1) = p^k for some prime p and positive integer k. We can rewrite this equation as (a^x - 1)((a^x)^(y-1) + (a^x)^(y-2) + ... + 1)/(a-1). For the sake of the proof we will use the following notation. a^x = r. (a^(x-1) + a^(x-2) + ... + 1)(r^(y-1) + r^(y-2) + ... + 1) =p^k. r=1 (mod a^(x-1) + ... + 1). (Since r=(a-1).(a^(x-1)+ ... 1) +1.) So the first is smaller than the second. a^(x-1) + a^(x-2) + ... + 1 divides r^(y-1) + ... +1. r=1 (mod a^(x-1) + ... +1) So r^(y-1) + ... + 1= y (mod a^(x-1) + ... + 1) So a^(x-1) + ... + 1 should divide y. But at first, we assumed that x is greater than or equal to y. a^(x-1) + ... + 1 is greater than x. (It can be shown by using Bernoulli's Inequality.) So our assumption was wrong hence n must be a prime.  :P

it can be easily proved by the Lemma $ v_p(a^n\minus{}1)\equal{}v_p(a\minus{}1)\plus{}v_p(n)$  :)  :D


A coin is flipped, a 6 sided die numbered 1-6 is rolled, and a 10 sided die numbered 0-9 is rolled.
What is the probability that the coin is heads and the sum of the numbers on the faces is 8?
[hide]
probability coin is heads = 1/2
there are 7 ways to have a sum of eight: (1,7) (2,6) (6,2) (3,5) (5,3) (4,4)
the chance of roling each of these is 1/6 x 1/10 = 1/60
the chance of rolling an eight is 6/60, multiply that by 1/2 to get 6/120
so the answer is [b]1/20[/b][/hide]

Six possibilites on the first die, ten on the second. Sixty in all. Only six combinations sum to eight, so we have $ p_2\equal{}\frac{1}{10}$. Of course, $ p_1\equal{}\frac{1}{2}$. Looks like the probability is $ \frac{1}{20}$.

I wonder why people post solutions on the first post...

Wouldn't that just allow people to cheat off of it? :maybe:

Hmmm... :|


Perhaps more realistic. Which individual do you think will win at nationals?
Argh, spelled Neil Gurram wrong, can a mod correct that, please?

[size=75][color=green][LadyKn1ght: Fixed the spelling][/color][/size]

BIff actually does have a good chance of making it. Have you seen some of his recent scores on nats rounds.

Obviously he can't include all prospective winners. However, I think the major favorites are Andrew Ardito, Neal Wu, and Nathan Benjamin, for their performance last year.

[quote="kyyuanmathcount"][quote="Ignite168"]Someone else from CA, TX, MI, or IN <- That's condescending considering Washington tied MI in cutoff and OK tied TX for cutoff.

I voted Nathan btw but I think these people should also be included in your poll

Sachin Shinde (Indiana)
Anand Oza (Maryland)
Matthew Vengalil (Michigan)
Daniel Li (Virginia)[/quote]

You were just taking the first places from those states right?

Matthew Vengalil wrote 7th in Michigan, and used countdown to gain first place, so there probably isn't a really high chance he'll make countdown, unless that was a choke.[/quote]

No, Sachin Shinde got 4th, Anand Oza got 3rd or 4th. Daniel Li keeps getting 30 on practice sprints. If I was just taking 1sts, I'd put first place in Texas up there or even 2nd as Kevin Chen was 3rd.

Wow, thanks for putting me on your poll.

Where is all your self-esteem, Klebian? I see that you did not put yourself up there. (this doesn't include someone else from IN, TX...)

[quote="Ignite168"]
Matthew Vengalil (Michigan)
Daniel Li (Virginia)[/quote]
^^^ know them in real life.

They are both very good compeditors. If Matthew V. makes it to CD, he will surely go far. I don't know Daniel Li's ability in CD, but I am sure he is proficient at it.

Yeah, this is true.
I am also pretty good at countdown, but I just choked at state countdown.  Usually, I do best for some reason when I am 3rd Place or 4th Place.

Intrepid Math, did you qualify for nats?

[quote="samath"]I feel so happy to be included in the poll! haha....I only have 2 votes though  :(

It's going to be an interesting year...[/quote]

Now you have 3 :P I just voted for you.

Thanks.  Yay now I'm in 4th place...not counting other.

I'm both None of the Above and Person from MI, TX,indiana so I have 8 votes.

Actually, since you are from TX, IN, MI, etc, you can't be none of the above, since that category is "above".

Actually as far as Daniel Li in VA, I've seen him the past 2 years at state.  I don't deny his written round abilities, but I personally beat him in the countdown round at state this year. (And I'm not half as good as some of you guys!)

[quote="aznness"]SOMEONE ACTUALLY VOTED FOR ME!!!

Oh wait that was me.

I think that we are overrating Neal and Nathan. I know Neal has godly speed, but nathan really needs to sharpen up if he wants to win. I figure i have a decent chance as long as i stop misreading. So, my money's on Neal. Sorry, gurram, never even heard of you. I'd like to meet you though.

[size=75][color=orange][Around here we don't worship Aphrodite and Zeus and Hermes. And we consider references to their *cough* practices rather uncalled-for. Would you say OM** to a girl you liked in front of your pastor and principal? -Billy][/color][/size][/quote]

Overrating? In 7th grade, one of them got #2 written and the other was National Champion. There is no question that they are the big favorites heading into the competition.

I voted Gurram because Michigan people are KO0o0OLer than everybody else.

P.S. How many of those people up there are AoPSers?

Everyone but Neal Wu. :D
Nathan Benjamin (I_Would_Say)
Sam Keller (samath)
Andrew Ardito (hexahedron)
Kevin Chen (bionuniquineist (sp?))
Neil Gurram (mathgeniuse^ln(x))
Kevin Yang (aznness)

[quote="jrshoch"]Hey just wanted to let everyone know they shouldn't base 8th grade performance very much on 7th grade performance..... speaking from personal experience  :oops: I got 31st as a 7th grader (5th best for 7th grade), and then got 48th as an 8th grader.  It was a bad day... and harder problems than I was expecting.

This could also serve as a warning to those contending for countdown round (don't get less than 3 hours of sleep the night before the competition, even if it means using sleeping pills)

But don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]

Absolutely. Always sleep :blush: before the competition. And if you make Nats in 7th grade, you need to know that not many people improve in 8th, judging from Mathcounts and Word Power.

All of these examples are from another competition - NOT MC! : From the first year of going to nats to the second (in another competition), one person went from 4th to 7th, 12th to 13th, 7th to 13th, and 11th to not making nats at all. How many people improved enough to get in the top 10 the second year who had returned? Nobody. (Of course, 3rd year people, having gotten disappointed the second year, generally do much much better; in that competition, two different people have made it to nats one year, didn't make the finals or nats the second, and won 3rd the third year.) Remember, this ISN'T from MC results.

So don't be overconfident. You are not the best in the world.

I would also point out this goes for state competition as well. Last year, a 7th grader placed 5th at our State Competition and was there as an individual. Top 4 were all 8th graders. (including jrshoch who won) This year, that 7th grader's team was there as well and he placed 6th while 7th graders who placed 8th and 11th went to 3rd and 1st respectively and the two other people who made the team were seventh graders.

[quote="pgpatel"][quote="solafidefarms"][quote="jrshoch"]Hey just wanted to let everyone know they shouldn't base 8th grade performance very much on 7th grade performance..... speaking from personal experience  :oops: I got 31st as a 7th grader (5th best for 7th grade), and then got 48th as an 8th grader.  It was a bad day... and harder problems than I was expecting.

This could also serve as a warning to those contending for countdown round (don't get less than 3 hours of sleep the night before the competition, even if it means using sleeping pills)

But don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]

Absolutely. Always sleep :blush: before the competition. And if you make Nats in 7th grade, you need to know that not many people improve in 8th, judging from Mathcounts and Word Power. More examples of falling: From the first year of going to nats to the second (in another competition), one person went from 4th to 7th, 12th to 13th, 7th to 13th, and 11th to not making nats at all. How many people improved enough to get in the top 10 the second year who had returned? Nobody. (Of course, 3rd year people, having gotten disappointed the second year, generally do much much better; in that competition, two different people have made it to nats one year, didn't make the finals or nats the second, and won 3rd the third year.)

So don't be overconfident. You are not the best in the world.[/quote]
I agree, however you are ignoring people who do well after returning. Looking ONLY at 04 and 05 I see that both david and sergei were in lower parts of countdown. Pardha got 30th in 04. Mark Zhang got 40th in 04.
Almost half of last years countdown made the top 57 in 04. Half of the people making it into countdown are returning. You only named 4 drops in the past many years.[/quote]

Ooops, I didn't make it clear I wasn't talking about MC at all in those examples.

[quote="solafidefarms"][/quote][quote="pgpatel"][quote="solafidefarms"][quote="jrshoch"]Hey just wanted to let everyone know they shouldn't base 8th grade performance very much on 7th grade performance..... speaking from personal experience  :oops: I got 31st as a 7th grader (5th best for 7th grade), and then got 48th as an 8th grader.  It was a bad day... and harder problems than I was expecting.

This could also serve as a warning to those contending for countdown round (don't get less than 3 hours of sleep the night before the competition, even if it means using sleeping pills)

But don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]

Absolutely. Always sleep :blush: before the competition. And if you make Nats in 7th grade, you need to know that not many people improve in 8th, judging from Mathcounts and Word Power. More examples of falling: From the first year of going to nats to the second (in another competition), one person went from 4th to 7th, 12th to 13th, 7th to 13th, and 11th to not making nats at all. How many people improved enough to get in the top 10 the second year who had returned? Nobody. (Of course, 3rd year people, having gotten disappointed the second year, generally do much much better; in that competition, two different people have made it to nats one year, didn't make the finals or nats the second, and won 3rd the third year.)

So don't be overconfident. You are not the best in the world.[/quote]
I agree, however you are ignoring people who do well after returning. Looking ONLY at 04 and 05 I see that both david and sergei were in lower parts of countdown. Pardha got 30th in 04. Mark Zhang got 40th in 04.
Almost half of last years countdown made the top 57 in 04. Half of the people making it into countdown are returning. You only named 4 drops in the past many years.[/quote][quote="solafidefarms"]

Ooops, I didn't make it clear I wasn't talking about MC at all in those examples.[/quote] :idea:

[quote="jrshoch"]
But don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]

I like coffee.  :D  Yeah, my friend and I both went to the nationals last year as 7th graders, but this year, my friend got 7th place I think at states. And I dropped from 2nd to 3rd at states. :oops: Oh well.

Thanks for not including my name in your list of other potential winners :( Just because MathCounts was mean and didn't post my 44th placing last year on their returning competitors list doesn't mean i didn't get 44th :( :( :(. (Btw, Daniel Li got 47th. I beat him on tiebreaker :D).

Anyway, both his written and CD skills are good, but I've beaten him at countdown more times than he's beaten me (if I remember correctly), so as long as my written round is good enough to get into CD i'll have a good shot, but i mainly wanna get top 4 for masters round and top 12 for being on TV ;)

A few other points:
1) Daniel's only getting 30s on practice sprints cuz he's done em all recently and remembers them. There's no way of knowing how he'll do on a new thing considering he's doing better at these nats competitions than the 2006 state one.
2) Everyone from VA is getting 2x on all of their practice sprints with 14/16 on targets.
3) Written isn't everything.

I voted for someone else meaning someone from VA.

[quote="lotrgreengrapes7926"]I like coffee.  :D  Yeah, my friend and I both went to the nationals last year as 7th graders, but this year, my friend got 7th place I think at states. And I dropped from 2nd to 3rd at states. :oops: Oh well.[/quote]

Same in Washington.  I won again, but he dropped from 3rd to 7th and so he's not on my team.  It's because he's not an AOPSer, I bet  :D


AMC
USA(J)MO
USAMO
trigonometry
inequalities
inequalities proposed

I think it is a USAMO question.

Subsitute $a = 2\cos{\alpha}$, $b = 2\cos{\beta}$ and $c = 2\cos{\gamma}$ where $\alpha, \beta, \gamma$ are the angles of an acute triangle. It is then very easy using the Hardy-Littlewood-Polya majorization inequality. I haven't checked it properly but it seems the values are 2 and 3, am I right?

I doubt if the min is 2.
I don't think equality case can be attained.

[quote="minhkhoa"]Yes!!
$2 \le a+b+c \le 3$
sory, Arne! I'd like to know Hardy-Littlewood-Polya majorization inequality :)[/quote]

minhkhoa, here's majorization inequality, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14975
may i know when a+b+c=2?

[quote="minhkhoa"]Dear Suihochung!
(a,b,c)=(2,0,0) :)[/quote]

ah! stupid me...  :blush:

[quote="Arne"]
Subsitute $a = 2\cos{\alpha}$, $b = 2\cos{\beta}$ and $c = 2\cos{\gamma}$ where $\alpha, \beta, \gamma$ are the angles of an acute triangle.[/quote]

I'm not exactly a stranger to trigonometric identities, but I didn't know this one.

Are you saying that for a triangle,  $\sum cos^2(\alpha) + 2 \Pi cos(\alpha) = 1$?

(so that triangles parametrize the $a,b,c$ satisfying the identity)

[quote="fleeting_guest"]I'm not exactly a stranger to trigonometric identities, but I didn't know this one.

Are you saying that for a triangle,  $\sum cos^2(\alpha) + 2 \Pi cos(\alpha) = 1$?[/quote]

Yes, exactly. We have:

[b]Theorem 1.[/b] Given a triangle with angles A, B, C, then, denoting, x = cos A, y = cos B, z = cos C, we have $x^2+y^2+z^2+2xyz=1$.

There exists a kind of converse of this theorem for positive numbers and acute-angled triangles:

[b]Theorem 2.[/b] If x, y, z are three positive reals such that $x^2+y^2+z^2+2xyz=1$, then there exists an acute-angled triangle whose angles A, B, C satisfy x = cos A, y = cos B, z = cos C.

I have proven this theorem in http://www.mathlinks.ro/Forum/viewtopic.php?t=23888 post #8.

Now, using Theorem 2, we can prove the original inequality:

[quote="minhkhoa"]Let $a,b,c$ be non-negative numbers such that
                   $a^2 + b^2 + c^2 + abc = 4$
Find the least and the greatest values of $a+b+c$[/quote]

In fact, we will show that $2\leq a+b+c\leq 3$. Equality in $2\leq a+b+c$ is obtained for a = 2, b = 0, c = 0, and equality in $a+b+c\leq 3$ is obtained for a = 1, b = 1, c = 1; thus, the problem of finding the least and the greatest values of a + b + c will be solved.

In order to prove $2\leq a+b+c\leq 3$, we consider two cases:

[b]Case 1.[/b] All three numbers a, b, c are positive.
[b]Case 2.[/b] At least one of the numbers a, b, c is 0.

(In fact, we know that the three numbers a, b, c are nonnegative, so that each of them is either positive or 0.)

In Case 1, define the numbers $x=\frac{a}{2}$, $y=\frac{b}{2}$, $z=\frac{c}{2}$. Then, a = 2x, b = 2y, c = 2z, and thus, the equation $a^2 + b^2 + c^2 + abc = 4$ rewrites as $\left(2x\right)^2+\left(2y\right)^2+\left(2z\right)^2+2x\cdot 2y\cdot 2z=4$, what simplifies to $x^2+y^2+z^2+2xyz=1$. Furthermore, since the three numbers a, b, c are positive, the numbers $x=\frac{a}{2}$, $y=\frac{b}{2}$, $z=\frac{c}{2}$ are positive, too. Thus, after Theorem 2, there exists an acute-angled triangle whose angles A, B, C satisfy x = cos A, y = cos B, z = cos C. Now, a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21188]really well-known inequality[/url] yields that $1<\cos A+\cos B+\cos C\leq\frac32$; since $\cos A=x=\frac{a}{2}$, $\cos B=y=\frac{b}{2}$ and $\cos C=z=\frac{c}{2}$, this rewrites as $1<\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\leq\frac32$. Multiplication with 2 yields $2<a+b+c\leq 3$, and thus, our inequality $2\leq a+b+c\leq 3$ is proven.

In Case 2, we cannot apply Theorem 2, but this case is really easy. At least one of the numbers a, b, c is 0; we can WLOG assume that a = 0. Then, the condition $a^2 + b^2 + c^2 + abc = 4$ rewrites as $b^2+c^2=4$. Hence, $\left(b+c\right)^2=b^2+c^2+2bc\geq b^2+c^2=4=2^2$, so that $b+c\geq 2$; in other words, $2\leq b+c$. On the other hand, the AM-QM inequality yields $\frac{b+c}{2}\leq\sqrt{\frac{b^2+c^2}{2}}=\sqrt{\frac{4}{2}}=\sqrt2$, so that $b+c\leq 2\sqrt2$, and since $2\sqrt2<3$, we thus get $b+c<3$. Combining this with $2\leq b+c$, we get $2\leq b+c<3$. Since a = 0, we can rewrite this as $2\leq a+b+c<3$. Thus, again, the inequality $2\leq a+b+c\leq 3$ is proven.

Problem solved.

  darij

[quote="darij grinberg"] [In $\Delta ABC$], x = cos A, y = cos B, z = cos C, we have $x^2+y^2+z^2+2xyz=1$.  [/quote]

Thank you.  Any nice proof or interpretation?

I think there must be some elegant ways, but direct substitution works in most cases.
\[
\cos C =  - \cos A\cos B + \sin A\sin B
\]
Substitute this into x^2+y^2+z^2+2xyz
\[
\begin{array}{l}
 \cos ^2 A + \cos ^2 B + \cos ^2 C + 2\cos A\cos B\cos C \\
  = \cos ^2 A + \cos ^2 B + \left( {\cos A\cos B - \sin A\sin B} \right)^2  - 2\cos A\cos B\left( {\cos A\cos B - \sin A\sin B} \right) \\
  = \cos ^2 A + \cos ^2 B - \cos ^2 A\cos ^2 B + \sin ^2 A\sin ^2 B \\
  = \cos ^2 A + \cos ^2 B - \cos ^2 A\cos ^2 B + \left( {1 - \cos ^2 A} \right)\left( {1 - \cos ^2 B} \right) \\
  = 1 \\
 \end{array}
\]

Nice solution Darij Grinberg !!!
Here is the main idea of my solution(i'll post it later)
$\sqrt{4-a^2} \le b+c \le 2\sqrt{2-a}$

[b]for the least value[/b]
From $a^2 + b^2 + c^2 + abc = 4$ we get $a,b,c \le 2$
Hence,
$4-a^2 = b^2 + c^2 + abc \le b^2 + c^2 + 2bc = (b+c)^2 $

$\Rightarrow b+c \ge \sqrt{4-a^2}  $

$\Rightarrow a+b+c \ge \sqrt{4-a^2}+a = \sqrt{(2-a)(2+a)}-(2-a)+2 $
$= \sqrt{2-a}(\sqrt{2+a}-\sqrt{2-a})+2 \ge 2  $

[b]for the greatest value[/b]
It is easy to prove that $b^2+c^2+abc \ge \frac{1}{2}(b+c)^2 + \frac{a}{4}(b+c)^2$
Hence,
$4=a^2 + b^2 + c^2 + abc \ge a^2 + \frac{1}{2}(b+c)^2 + \frac{a}{4}(b+c)^2$

$\Rightarrow   (b+c)^2 \le 4(2-a)$

$\Rightarrow   b+c \le 2\sqrt{2-a} \le 1+2-a   (AM-GM)$

$\Rightarrow   a+b+c \le 3$
it's not so clear!!! ;)

[quote="Arne"]Sorry, this is not a USAMO question.

The USAMO question I had in mind is similar:

If $a, b, c$ are positive reals such that $a^2 + b^2 + c^2 + abc = 4$, prove that $0 \leq ab + bc + ca - abc \leq 2$.[/quote]
The ineq. is nice!
Where can i find its solution?

[quote="i am a chinese"][quote="darij grinberg"]http://www.kalva.demon.co.uk/usa/usoln/usol013.html .

  darij[/quote]
but why i can not open it[/quote]

Because you are a chinese  :(  and for some reason, the chinese "firewall" blocks Kalva's domain.

I hope Kalva has nothing against me quoting his solution:

[quote="Kalva"]30th USAMO 2001
------

[b]Problem A3[/b]

Non-negative reals x, y, z satisfy $x^{2}+y^{2}+z^{2}+xyz=4$. Show that $xyz\leq xy+yz+zx\leq xyz+2$.

[b]Solution[/b]

Assume $x \geq y \geq z$. If z > 1, then $x^{2}+y^{2}+z^{2}+xyz > 1+1+1+1 = 4$. Contradiction. So $z \leq 1$. Hence $xy+yz+zx \geq xy \geq xyz$.

Put x = u + v, y = u - v, so that $u, v \geq 0$. Then the equation given becomes $u^{2}\left(2+z\right)+\left(2-z\right)v^{2}+z^{2}= 4$. So we we keep z fixed and reduce v to nil, then we must increase u. But $xy+yz+zx-xyz = \left(u^{2}-v^{2}\right)\left(1-z\right)+2zu$, so decreasing v and increasing u has the effect of increasing xy + yz + zx - xyz. Hence xy + yz + zx - xyz takes its maximum value when x = y. But if x = y, then the equation gives $x = y = \sqrt{2-z}$. So to establish that $xy+yz+zx-xyz \leq 2$ it is sufficient to show that $2-z+2z\sqrt{2-z}\leq 2+z\left(2-z\right)$. Evidently we have equality if z = 0. If z is non-zero, then the relation is equivalent to $2\sqrt{2-z}\leq 3-z$ or $\left(z-1\right)^{2}\geq 0$. Hence the relation is true and we have equality only for z = 0 or 1.[/quote]

  darij


find all values of $ x/x \plus{} y \plus{} y/y \plus{} z \plus{} z/x \plus{} y$,x,y,z>0

Do you mean $ \frac{x}{x\plus{}y}\plus{}\frac{y}{y\plus{}z}\plus{}\frac{z}{x\plus{}y}$?

You should do the same in the Inequalities Marathon .

\[ \sum_{cyc}\frac{x}{x\plus{}y}\ge \sum_{cyc}\frac{x}{x\plus{}y\plus{}z}\equal{}1\]

and \[ \sum_{cyc}\frac{x}{x\plus{}y}\plus{}\sum_{cyc}\frac{y}{x\plus{}y}\equal{}3\] and use the fact that \[ \sum_{cyc}\frac{y}{x\plus{}y}\ge 1\]


calculus trigonometry parameterization quadratics algebra

A driver in a car is in the desert, 10 km from the nearest point, $ N$, on a long, straight road.  The driver can cover 30 km/h in the desert, and 50 km/h on the road.  What is the shortest time for the driver to reach a point on the road that is:

(a) 20 km from $ N$?
(b) 30 km from $ N$?

I would imagine the road is perpendicular to the path from the driver to $ N$?

Yes, since the $ N$ is the nearest point on the road, 10 km is the perpendicular distance from the driver to the road.

[hide="Ugly Solution"]While on the desert, the path will contain only lines, because the shortest distance between any two points is a line, and once we hit the road, it is the most efficient to continue to the finish line by the Triangle Inequality. Now assume that the path while on the desert is not just one line, but a composition of them. Start from the last two lines; we can make this shorter with only one line by the Triangle Inequality. We can continue like this to show that the shortest desert path to some given point on the road is always a line.

So let the point that the desert path ends be $ x$ away from $ N$, and thus $ 20 \minus{} x$ from the finish point. The time taken would then be $ \frac {\sqrt {x^2 \plus{} 100}}{30} \plus{} \frac {20 \minus{} x}{50} \equal{} \frac25 \plus{} \frac1{150}(5\sqrt {x^2 \plus{} 100} \minus{} 3x)$. Now we want to find the maximum $ k$ such that $ 5\sqrt {x^2 \plus{} 100} \minus{} 3x\ge k$ for all $ 0\le x\le20$. So moving $ 3x$ to the RHS and simplifying yields $ (16x \minus{} 3k)^2 \plus{} 25(1600 \minus{} k^2)\ge0$. Keeping in mind that $ k$ is positive, if $ k \equal{} 40$, $ 0\le x \equal{} \frac {120}6 \equal{} \frac {15}2\le20$ is where the minimum occurs, if $ k < 40$, the minimum is not achievable for any $ x$ since $ 1600 \minus{} k^2 > 0$ so certainly not $ 0\le x\le20$, and if $ k > 40$, then for $ 0\le x \equal{} \frac {15}2\le20$, $ 5\sqrt {x^2 \plus{} 100} \minus{} 3x < k$, which is a contradiction. Thus the minimum time occurs when $ x \equal{} \frac {15}2$, which makes the time $ \frac23$ of an hour or $ 40$ minutes.

Similarly, for the second part, we want to minimize $ \frac35 \plus{} \frac1{150}(5\sqrt {x^2 \plus{} 100} \minus{} 3x)$. But this obviously also occurs when $ x \equal{} \frac {15}2$, giving a time of $ \frac {13}{15}$ of an hour or $ 52$ minutes.[/hide]

Ahh there should be a nicer solution...

[quote="math154"] and if $ k > 40$, then for $ 0\le x \equal{} \frac {15}2\le20$, $ 5\sqrt {x^2 \plus{} 100} \minus{} 3x < k$, which is a contradiction. [/quote] How was this contradiction proved (without circular reasoning of $ 5\sqrt{x^{2}\plus{}100}\minus{}3x < k$ for $ k>40$ since the maximum $ k$ is $ k\equal{}40$)?

OK, so if $ k > 40$, then by definition, $ 5\sqrt {x^2 \plus{} 100} \minus{} 3x\ge k > 40$ for all $ 0\le x\le20$. But $ x \equal{} \frac {15}2$ (which is between $ 0$ and $ 20$) yields $ 5\sqrt {x^2 \plus{} 100} \minus{} 3x \equal{} 40$, which is less than $ k$, contradicting the role of $ k$.

OK I see... for some strange reasons I misunderstood the definition of $ k$... :huh:

Now we want to find the maximum $ k$ such that $ 5\sqrt {x^2 \plus{} 100} \minus{} 3x\ge k$ for all $ 0\le x\le20$.

This is the first sentence of the solution that I don't understand.  Could you please explain it differently?  Among other things, I'm not sure why the sentence calls for a maximum when we're trying to minimize the driving time.

Thanks.

[quote="math154"]Now we want to find the maximum $ k$ such that $ 5\sqrt {x^2 \plus{} 100} \minus{} 3x\ge k$ for all $ 0\le x\le20$.[/quote]

This is the first sentence of the solution that I don't understand.  Could you please explain it differently?  Among other things, I'm not sure why the sentence calls for a maximum when we're trying to minimize the driving time.

Thanks.

Of course. What if we picked, say $ k\equal{}\minus{}100$? Clearly all $ 5\sqrt{x^2\plus{}100}\minus{}3x\ge k$, but it's not [i]achievable[/i], which is what we need for $ k$ to really be a minimum. So we need to find $ k$ such that $ 5\sqrt{x^2\plus{}100}\minus{}3x\ge k$ for all $ 0\le x\le20$, and so that for some $ x$ in the range $ 0\le x\le20$, $ 5\sqrt{x^2\plus{}100}\minus{}3x\equal{}k$. This corresponds to finding the maximum possible $ k$ with these conditions. For example, here $ k\equal{}40$ is the maximum possible lower bound, achievable when $ 0\le x\equal{}7.5\le20$, because if $ k<40$, then there wouldn't exist any $ x$ in the range achieving the minimum, and if $ k>40$, then when $ x\equal{}7.5$, we would have $ 5\sqrt{x^2\plus{}100}\minus{}3x\equal{}40<k$, but that contradicts the definition that $ 5\sqrt{x^2\plus{}100}\minus{}3x\ge k$ for all $ 0\le x\le20$!

[b]math154[/b]'s argument clearly shows how the problem is really to find the most economical way of getting to the highway given which direction on the highway the driver ultimately wants to go. That is, we want to minimize the time required to get to the highway minus the highway time saved. It is also clear that we should take a straight path to the highway.

I would instead suggest setting up the problem using angles/trigonometry, since the problem scales.

Suppose the driver going directly to the highway means an angle of zero, and the driver traveling in the direction of the highway means an angle of 90, and generally, the angle the driver makes from the normal route to the highway an angle of theta.

We're a distance $ d$ away. Let the velocity on the road be $ 1$, and the relative velocity on the sand be $ v \in [0,1]$. We're interested in the time required to get to the highway minus the highway time saved. This is given by:

$ \frac{d}{v \cos \theta} \minus{} d \tan \theta \equal{} \frac{d}{v} \left( \frac{1 \minus{} v \sin \theta}{\cos \theta} \right)$

[hide="Minimizing"]The trick is now to minimize $ \frac{1 \minus{} v \sin \theta}{\cos \theta}$ without calculus. I would suggest using the parametrization $ \sin \theta \equal{} \frac{1\minus{}t^2}{1\plus{}t^2}$ and $ \cos \theta \equal{} \frac{2t}{1\plus{}t^2}$, with the order of the two strategically chosen.

We thus get $ \frac{(1\minus{}v)\plus{}(1\plus{}v)t^2}{2t} \equal{} \frac{(1\minus{}v) t^{\minus{}1} \plus{} (1\plus{}v) t}{2}$

By AM-GM, we get this to be minimized when $ (1\minus{}v) t^{\minus{}1} \equal{} (1\plus{}v) t \implies t^2 \equal{} \frac{1\minus{}v}{1\plus{}v}$

Sticking this value of $ t^2$ back into our expression for $ \sin \theta$, we find that:

$ \sin \theta \equal{} \frac{1\minus{}\frac{1\minus{}v}{1\plus{}v}}{1\plus{}\frac{1\minus{}v}{1\plus{}v}} \equal{} v$

So now, we know exactly what angle to approach the road. If the desired destination is at a larger angle than $ \arcsin v$, then we should go in at an angle of $ \arcsin v$ and take the highway. If the desired destination is at a smaller angle, we should just go directly to that point.[/hide]

Refer to the diagram. The point C is the present location of the car in the desert, D is the point to be reached with $ ND \equal{} d$. Let the car first go to point P ($ NP \equal{} x$) and then continue on the highway. Let $ u$ be the velocity of the car in the desert and $ v$ on the highway. Also, let $ CN \equal{} a$. The time taken is
$ t \equal{} \dfrac{CP}{u} \plus{} \dfrac{PD}{v} \equal{} \dfrac{\sqrt {x^2 \plus{} a^2}}{u} \plus{} \dfrac{d \minus{} x}{v} \equal{} \dfrac{v\sqrt {x^2 \plus{} a^2} \minus{} ux}{uv} \plus{} \dfrac{d}{v}$
In order to minimize $ t$, we need to minimize $ v\sqrt {x^2 \plus{} a^2} \minus{} ux$. Let
$ y \equal{} v\sqrt {x^2 \plus{} a^2} \minus{} ux$
$ \Rightarrow\ y^2 \plus{} u^2x^2 \plus{} 2uxy \equal{} v^2x^2 \plus{} a^2v^2$
$ \Rightarrow\ (v^2 \minus{} u^2)x^2 \minus{} 2uxy \plus{} a^2v^2 \minus{} y^2 \equal{} 0\qquad(1)$
Since $ x$ must be real, the roots of the above quadratic equation in $ x$ must be real. That happens if the discriminant of (1) is non-negative; i.e.
$ 4u^2y^2 \minus{} 4(v^2 \minus{} u^2)(a^2v^2 \minus{} y^2)\ge 0$
i.e.
$ y\ge a\sqrt {v^2 \minus{} u^2} \equal{} y_\mathrm{min}$
When the equality is attained, the discriminant of (1) is zero, and the double root becomes
$ x_m \equal{} \dfrac{uy_\mathrm{min}}{v^2 \minus{} u^2} \equal{} \dfrac{au}{\sqrt {v^2 \minus{} u^2}}$
As such the minimum time is
$ t_\mathrm{min} \equal{} \dfrac{y_\mathrm{min}}{uv} \plus{} \dfrac{d}{v} \equal{} \dfrac{a\sqrt {v^2 \minus{} u^2}}{uv} \plus{} \dfrac{d}{v} \equal{} \dfrac{d \plus{} a\sqrt {(v/u)^2 \minus{} 1}}{v}$

Thanks to all of you.  These are all superb explanations, and I'm trying to work through them now.  I'm very grateful for your help.


You have $f(x)>1$ for all $x$. What condition does this impose on $m$?


vector
function
complex analysis
complex analysis unsolved

Could someone help me with this problem:

f is a conformal mapping from the unit disk onto a convex set.
Show that the image of any disk centered at the origin is convex.

[url=http://www.mathlinks.ro/viewtopic.php?t=280607]this[/url] is one of the most beautiful problems in complex analysis I have ever seen.

The image of the circle |z|=r is convex iff u(z)= Re(1+zf"(z)/f'(z)) is non-negative on this circle.
(This is an easy computation: parametrize the circle by r.exp(it) and then write the condition
that the argument of the tangent vector to the image of the circle is increasing as a function
of t.)
Now u is a harmonic function (because f'(z) is never 0), so if the image of some circle |z|=r
is convex then the image of any smaller circle |z|=r'<r is also convex, by Maximum Principle.

Pym.

P.S. This class of functions is called convex univalent; it was studied a lot, see, for
example Hayman's Multivalent Functions, chap. I


geometry
geometric transformation
reflection
vector
trigonometry
rhombus
angle bisector

oxy: angle
od: bisector
AB=A'B'
AM=MA'
BN=NB'

PROVE THAT: MN//od

THERE IS AN IMAGE IN OREDER TH HELP:[url=http://img236.imageshack.us/my.php?image=graphij5.png][img]http://img236.imageshack.us/img236/5430/graphij5.th.png[/img][/url]

No hard.
Let K on AB' so KA=KB' and .... can you do continue . :lol:

I DON;T GET IT.  :wink:  ANY CHANCE FOR FURTHERMORE INFORMATION? :)

Denote the points $ M',$ $ N',$ as the intersection points of $ AA',$ $ BB'$ respectively, from the angle bicector of $ \angle BOB'$ and as $ C,$ $ C',$ the intersection points of $ BB',$ from the lines through $ A,$ $ A'$ resprctively and parallel to $ ON'.$

By [b][size=100]Thales theorem[/size][/b] and the [b][size=100]angle bisector theorem,[/size][/b] we can now to prove that $ BC = B'C'$ and then $ NC = NC'.$

Hence, we conclude that $ MM'\parallel AC\parallel A'C'\parallel ON'$ and the proof is completed.

Kostas Vittas.

PS. I will post here later more details.

[quote="Virgil Nicula"][color=darkred][b][u]Lemma[/u] (well-known).[/b] Let $ ABC$ be a triangle and let $ M\in AB$, $ N\in AC$ be two points so that $ BM=CN$.
Denote the middlepoints $ X$, $ Y$ of the segments $ [BC]$, $ [MN]$ respectively. Prove that :
[b]1.[/b] If the line $ BC$ doesn't separate the points $ M$, $ N$ then the line $ XY$ is parallel with the internal bisector of the angle $ \widehat{BAC}$.
[b]2.[/b] If the line $ BC$ separates the points $ M$, $ N$ then the line $ XY$ is parallel withe the external bisector of the angle $ \widehat{BAC}$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote $ \{D,D'\}\subset BC$ so that the rays $ [AD$, $ [AD'$ are respectively the internal and external bisectors of the angle $ \widehat{B AC}$. Define
the reflection $ P$ of the point $ N$ w.r.t. the point $ X$. Observe that $ MP\parallel XY$, $ PB\parallel AC$, $ MB=NC=PB$, i.e. $ \triangle BMP$ is $ B$- isosceles.

$ \{\begin{array}{cccccccc}1\blacktriangleright & m(\widehat{MBP})=B+C & \implies & m(\widehat{BMP})=\frac{A}{2}& \implies & MP\parallel AD & \implies & XY\parallel AD\ .\\\\ 2\blacktriangleright & m(\widehat{MBP})=A & \implies & m(\widehat{BMP})=\frac{B+C}{2}& \implies & MP\parallel AD' & \implies & XY\parallel AD'\ .\end{array}$

[b]Remark.[/b] Let $ M\in AB$, $ N\in AC$ be two mobile points so that $ BM=CN$.
Then the geometrical locus of the midpoint of the segment $ [MN]$ is $ AD\cup AD'$.[/color]

[quote="Virgil Nicula"][color=darkred][b][u]Lemma[/u] (well-known).[/b] Let $ ABC$ be a triangle and let $ M\in AB$, $ N\in AC$ be two points so that $ BM=CN$.
Denote the middlepoints $ X$, $ Y$ of the segments $ [BC]$, $ [MN]$. Prove that :
[b]1.[/b] If the line $ BC$ doesn't separate the points $ M$, $ N$ then the line $ XY$ is parallel with the internal bisector of the angle $ \widehat{BAC}$.
[b]2.[/b] If the line $ BC$ separates the points $ M$, $ N$ then the line $ XY$ is parallel withe the external bisector of the angle $ \widehat{BAC}$.[/color][/quote]
Nice one...I suppose the nicest way to do this is what cuopbientcd said?  Create the point $ K$, use the midsegment theorem, and then use vectors?

There is also an easy solution using complex numbers. Let $ A=e^{-i\theta}$, and $ A'=a e^{i\theta}$, and $ B=b e^{-i\theta}$. A short calculatin gives $ N-M=2(b-1)\cos\theta$, which is real.

[quote="Virgil Nicula"][color=darkred][b][u]Lemma[/u] (well-known).[/b] Let $ ABC$ be a triangle and let $ M\in AB$, $ N\in AC$ be two points so that $ BM=CN$.
Denote the middlepoints $ X$, $ Y$ of the segments $ [BC]$, $ [MN]$ respectively. Prove that :
[b]1.[/b] If the line $ BC$ doesn't separate the points $ M$, $ N$ then the line $ XY$ is parallel with the internal bisector of the angle $ \widehat{BAC}$.
[b]2.[/b] If the line $ BC$ separates the points $ M$, $ N$ then the line $ XY$ is parallel withe the external bisector of the angle $ \widehat{BAC}$.[/color][/quote]
[color=darkblue][b]Proof I.[/b] Denote $ \{D,D'\}\subset BC$ so that the rays $ [AD$, $ [AD'$ are respectively the internal and external bisectors of the angle $ \widehat{B AC}$. Define
the reflection $ P$ of the point $ N$ w.r.t. the point $ X$. Observe that $ MP\parallel XY$, $ PB\parallel AC$, $ MB=NC=PB$, i.e. $ \triangle BMP$ is $ B$- isosceles.

$ \{\begin{array}{cccccccc}1\blacktriangleright & m(\widehat{MBP})=B+C & \implies & m(\widehat{BMP})=\frac{A}{2}& \implies & MP\parallel AD & \implies & XY\parallel AD\ .\\\\ 2\blacktriangleright & m(\widehat{MBP})=A & \implies & m(\widehat{BMP})=\frac{B+C}{2}& \implies & MP\parallel AD' & \implies & XY\parallel AD'\ .\end{array}$

[b]Remark.[/b] Let $ M\in AB$, $ N\in AC$ be two mobile points so that $ BM=CN$.
Then the geometrical locus of the midpoint of the segment $ [MN]$ is $ AD\cup AD'$.

[b]Proof II.[/b] Let $ \{\begin{array}{c}\{M,M'\}\subset AB\\\ \{N,N'\}\subset AC\end{array}$ be four different points so that $ \{\begin{array}{c}M\in (AB)\ ,\ N\in (AC)\\\ BM=BM'\ ,\ CN=CN'\end{array}$. Denote the midpoints $ X$, $ Y$, $ U$, $ V$ $ Z$ of the segments $ [BC]$, $ [MN]$, $ [M'N]$, $ [MN']$, $ [M'N']$ respectively. Prove easily that the quadrilateral $ UYVZ$ is a rhombus ($ UY\parallel VZ\parallel AB$, $ YV\parallel UZ\parallel AC$) and $ X\in UV\cap YZ$. Thus, $ \overline{UXV}\perp \overline{YXZ}$, i.e. $ \overline{YXZ}\parallel AD$ and $ \overline{UXV}\parallel AD'$.

[b]Remark.[/b] Let $ ABCD$ be a convex quadrilateral so that $ E\in AB\cap CD$ and $ AB=CD$. Denote the midpoints $ X,Y,U,V$ of the segments $ [BC]$, $ [AD]$, $ [AC]$, $ [BD]$ respectively. Then the lines $ XY$, $ UV$ are parallel respectively to the internal and external bisectors of the angle $ \widehat{AED}$.
[/color]

I'M LOOKING FOR A SIMPLE SOLUTION (WITH DETAILS IF IT'S POSSIBLE) THANKS :)

I think that Virgil Nicula's proof is quite very simple and very clear. Try to read his proof then.

[b]HINT(nothing the new !!!)[/b]

Well,

 take two segments $ MK,ML$ equal to $ AB$ and parallele to $ AB, A'B'$respectively. Then $ MK = ML$ and $ BKB'L$ is a parallelogram.But then $ KL$ passes through N and MN is the bisector of $ \angle KML$ in the isosceles triangle $ MKL$. Obviously $ MN$ is parallele to the bisector of angle $ BOB'$  too ect.

[quote="vittasko"]....
By [b][size=100]Thales theorem[/size][/b] and the [b][size=100]angle bisector theorem,[/size][/b] we can now to prove that $ BC = B'C'$ and then $ NC = NC'.$
...... [/quote]

From $ AC\parallel ON',$ we have that $ \frac{BA}{BO}= \frac{BC}{BN'}$ $ ,(1)$

From $ A'C'\parallel ON',$ we have that $ \frac{B'A'}{B'O}= \frac{B'C'}{B'N'}$ $ ,(2)$

From $ (1),$ $ (2),$ $ \Longrightarrow$ $ \frac{BA}{BO}\cdot \frac{B'O}{B'A'}= \frac{BC}{BN'}\cdot \frac{B'N'}{B'C'}$ $ ,(3)$

From $ (3)$ $ \Longrightarrow$ $ \frac{B'O}{BO}= \frac{BC}{B'C'}\cdot \frac{B'N'}{BN'}$ $ \Longrightarrow$ $ \frac{BC}{B'C'}= 1$ $ ,(4)$ $ ($ because of $ \frac{B'O}{BO}= \frac{B'N'}{BN'}$ by [b][size=100]angle bisector theorem[/size][/b] $ ).$

So, from $ (4)$ $ \Longrightarrow$ $ BC = B'C'$ $ \Longrightarrow$ $ CN = C'N$ $ ,(5)$

From $ (5)$ we conclude that $ MN\parallel AC\parallel A'C'\parallel ON'$ and the proof is completed.

Kostas Vittas.

transfer $ AB$ with $ \overrightarrow{AM}$ and $ A'B'$ with $ \overrightarrow{A'M}$.let $ B\rightarrow B_{1}$ and $ B'\rightarrow B_{1}'$.also we know $ \overrightarrow{AB}+\overrightarrow{A'B'}=2\overrightarrow{MN}$ and $ MB_{1}=MB_{1}'$ then $ MN\parallel \text{angle bisector of}\ \angle A$


algebra
polynomial
vector
real analysis
real analysis unsolved

Given a polynomial $ P \in \mathbb{R}[X]$, find a norm $ \|. \|$ on $ \mathbb{R}[X]$ such that $ \|X^n \minus{} P \| \rightarrow 0$ when $ n \rightarrow \infty$

Without extra assumptions, it is easy: given any algebraic basis $ e_1,e_2,\dots$ in an infinite dimensional vector space, we can always define a norm such that $ \|e_n\|\to 0$ (say, by $ \Bigl\|\sum_j \alpha_j e_j\Bigr\|\equal{}\sum_j\frac{|\alpha_j|}j$). It remains to note that $ 1,x,\dots,x^m,x^{m\plus{}1}\minus{}P, x^{m\plus{}2}\minus{}P,\dots$ is a basis in $ \mathbb R[X]$ where $ m$ is the degree of $ P$. Here is a more challenging question: can we find such a norm with the extra property that $ \|QR\|\le\|Q\|\cdot\|R\|$ for any polynomials $ Q,R$? ;)

thanks fedja, I'll think about your question  :wink:


Yeah... I dunno how this could be applied to real life, but maybe one day, someone will make use to it... i don't know how, but I hope someone considers my theory and give feedbacks...

To be honest, I thought mathematicians are always open to new ideas, and not become dependent on the facts that they already know.  I always thought that mathematicians are ready to analyze the unthinkable, and consider whether it's valid or not.

Maybe how I worded my idea, may throw people off... I agree that infinity is a literally defined value (representing the largest value), but its value remains undefined (despite that no one knows the largest value) (oddest oxymoron).  Every contradiction stated in this forum thus far is relied on what I believe to be "lies."

If I offend you even more, I'll take it all back... Honestly, I should have not stated this, because apparently no one is considering my theory, which is backed up with true information as well...  I should have find a better place to post this -.-;

sorry Peter, may you delete this... I pushed the wrong button -.-;


在梯行ABCD中 AB平行CD
過A點在BC邊上取一點E
連接AE及DE 使角AED為直角
請問如何證明得到三角形ADE面積為梯形面積的一半

Translate:

In a trapezium $ ABCD , AB//CD ,$ take a point $ E$  on $ BC$ such that $ \angle AED$ is a right angle.

Prove that the area of $ \triangle ADE$ is as half as that of trapezium $ ABCD$ .

The problem is obviously wrong: basically there are 2 points $E$ on $BC$ with $AE\bot DE$ and, as per your 'problem' we need each of them to have area half of that of the trapezoid, i.e. $AD\parallel BC$, or $ABCD$ a parallelogram.

Best regards,
sunken rock

[color=darkred][b][u]PP[/u].[/b] Let $ ABCD$ be a trapezoid, where $\left\{\begin{array}{c}
AB\parallel CD\\\
AB\perp AD\end{array}\right|$ so that  exists unique $E\in (BC)\ ,\ EA\perp ED$ . Prove that $\frac {[ADE]}{[ABCD]}\le\frac 12$ .[/color]