How many relationships have you guys been in? I've only been in 4, and I am just sorta dfrifting around right now looking for a girl. So what about you guys?

[quote="1=2"]One with my computer. That's it.[/quote]

Same, we are engaged (me and my computer). We are very fond of each other. I must say it was love at first sight. As I gazed upon her empty screen, my heart sped up and my body temperature rose until I was positively fanning myself with the brochure. Then I saw her name, ASUS, and I thought to myself, 'that's the name of an Angel, I am more in love than Romeo when he first saw Juliet.' So I took out my wallet, but only found $ 3$ lonely dollars. I was shattered, I must have her, no matter what it takes. So in the dead of the night, I crept out of bed and put on my Ninja constume made of conductive polymer (polyethyne, due to the conjugated $ \pi$ electron system). My tricycle travelled at close to the speed of light and within seconds I was outside the store where I first saw my ASUS. Once again, I found sweat trickling down my neck despite the cold midnight breeze. Then, as though a sign from God, lightning struck and a tree fell. Thre apple hanging loosely on the branches travelled in projectile motion and struck the glass and for some unknown reason (where Classical Mechanics and Material Physics break down), the glass shattered. Millions of tiny pieces zoomed at me but my love for ASUS was so strong that physical pains did not stop my as I ventured towards the shop. I saw ASUS again, and a funny feeling stirred up in my stomach. I grabbed her and we were off to my place..........

And the tale goes on in a rather pointless manner.

[quote="BanishedTraitor"]

Same, we are engaged (me and my computer). We are very fond of each other. I must say it was love at first sight. As I gazed upon her empty screen, my heart sped up and my body temperature rose until I was positively fanning myself with the brochure. Then I saw her name, ASUS, and I thought to myself, 'that's the name of an Angel, I am more in love than Romeo when he first saw Juliet.' So I took out my wallet, but only found 3 lonely dollars. I was shattered, I must have her, no matter what it takes. So in the dead of the night, I crept out of bed and put on my Ninja constume made of conductive polymer (polyethyne, due to the conjugated \pi electron system). My tricycle travelled at close to the speed of light and within seconds I was outside the store where I first saw my ASUS. Once again, I found sweat trickling down my neck despite the cold midnight breeze. Then, as though a sign from God, lightning struck and a tree fell. Thre apple hanging loosely on the branches travelled in projectile motion and struck the glass and for some unknown reason (where Classical Mechanics and Material Physics break down), the glass shattered. Millions of tiny pieces zoomed at me but my love for ASUS was so strong that physical pains did not stop my as I ventured towards the shop. I saw ASUS again, and a funny feeling stirred up in my stomach. I grabbed her and we were off to my place..........
[/quote]

So what are the babies going to be? PDAS?

Wow, too fast man. Who said anything about babies?

[quote="pianoforte"]Sorry, but I've penetrated your computer's firewall with a trojan.[/quote]

She's dying..........Wohoo, I can get a new one! (This sounds incredibly offesive to females in the given context).

[quote="hunter34"] Well. You went to your crib after you stole her. So Babies are soon to com![/quote]

................ :|

I want a blackberry. Can you make one with the ASUS, or are you going to get another girl compy?

Wait, is it just me or are we simply ignoring the original question (which is.....)?

With the topic it is a good idea to stray away.

[quote="BanishedTraitor"][quote="pianoforte"]Sorry, but I've penetrated your computer's firewall with a trojan.[/quote]

She's dying..........Wohoo, I can get a new one! (This sounds incredibly offesive to females in the given context).[/quote]

 :rotfl:

One with my computer, one with my math materials, and one with my school friends.

[quote="BanishedTraitor"]I have been in a relationship with:

$ \sin, \cos, \tan, \sinh, \cosh, \tanh, \sin^{ \minus{} 1}, \cos^{ \minus{} 1}, \tan^{ \minus{} 1}, \sinh^{ \minus{} 1}, \cosh^{ \minus{} 1}, \tanh^{ \minus{} 1}$ and the notorious $ \theta$.[/quote]

DIE TRIG

Trig is cool (or is it are?) but anyway, they are my life long companion and I have known them as long as I have lived (man that means I haven't been living for very long).

I likes compy. I likes math. I likes Texas Instruments.

[quote="Nerd_of_the_Ages"]One with math
One with SSBM
One with procrastinating (trying to break up, but it's kinda' hard  :( )[/quote]

Man, 3 at once? It's gotta be hard to concentrate on anything.

yeah, actually it is
easy to see why procrastination would make it hard
And when I have free time it's often hard to decide whether I wanna' sit out and play SSBM for the whole night or do math (dont ban me  :o )


Is there any way to make it so that contributors can't? I'm thinking of adding a contributor or two to my blog, but I don't know if I can trust them with this sort of power.  :D

If you can't trust them, it might be a bad idea to include them in your blog to begin with.


Substitute every variable with its reciprocal, then try to use AM-GM

Use Muirhead
but first - you should say $ x_i\in \mathbb{R}^\plus{}$ because if not - for example $ x_{1\leq i\leq n}\equal{}\minus{}1$ and $ x_{n\plus{}1}\equal{}\pm n\Rightarrow$ the ineqality is wrong.


ARML
AMC
USA(J)MO
USAMO
geometry
USAMTS
summer program

Okay team, you're awesome, and i am so proud of you.  i couldn't have asked for a better result, and i'm already looking forward to next year!

brief results:
215 Lehigh
197 SFBA
196 million way tie

way to go team..

We own!  And Lehigh valley Fire is totally on steroids...215??

Now it remains to prove that $ 2 \equal{} 1000000$?

wow, congratulations to everyone!

Are the full results available anywhere yet? They don't seem to be on arml.com

Ted: see the link that gtf posted in post #32 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279622&start=20]here[/url].

You and we were neck and neck until the relays; the relays sunk us.

hmm you guys don't actually completely suck at relays (or individual obviously) but the power round leaves something to be desired (what are all those usamo qualifiers doing? we had only 11 on A1) :P

[quote="DataBox"]^Haha, I guess 3 Blue MOPpers and a billion Red MOPpers wasn't enough for the Power round =)[/quote]

Actually, I don't think THAT many Red MOPpers were at ARML,.  None of the 3 Red MOPpers from our school participated in ARML despite me constantly pestering them to, for example.

Speaking of MOP, shouldn't you change your sig, Ubey?

[quote="DeltaX"]Good job everyone, especially A team![/quote]

Good job A2 team as well!  you did at least as well as last year's B team, and it looks like we had the third best secondary team in the country.  Well done; it looks like we have a strong future..

Yeah, here are what I'd call "Division A2" standings: the standings of all teams that are not the top team in their region.

Thomas Jefferson B1     156
SoCal A2                    152
Bay Area A2                148
Georgia A2                  146
Lehigh Ice                  144
New York City N           132
San Diego Sun             130
New York City Y           129

I call this a very impressive performance from the A2 team!

Here's the counterpart to Si's table for best [i]third[/i] teams of organizations - a category SFBA didn't participate in this year:

NYC Y: 129
T. Jeff B2: 120
Mongomery D: 116 (Montgomery C was at 109)
Chicago C: 113
Florida B2: 104
Georgia B1: 103
SoCal B1: 102
E. Mass B1: 99
San Diego Fish Tacos: 95
Lehigh Lightning: 93
Ohio B2: 93

[quote="Scrambled"]Well, I heard that due to some confusion with power round scoring, Exeter actually got 199 (beating our score?)
I don't know if this is accurate or not.[/quote]
Yep; it's true  :lol: except that it's 201 points (they mixed up Exeter and Andover)

[quote="CircleSquared"][quote="Ubemaya"]What's wrong with my sig?[/quote]
It's outdated.

Also, why do I remember we do have a 3rd team?  A1, A2, B <-- third
Was there even a B team or did I somehow brainwash myself?

Congratulations to all, also.[/quote]
There is no B team. We had 5 people not on A1 or A2 this year.

Wait, for A1 team when do we get the code to get Geometer's Sketchpad?

My 1st copy of gsp, 3rd copy of Mathematica, and 3rd copy of TIN-Spire.

wait where are you supposed to get your mathematica?

[quote="timwu"][quote="Scrambled"][quote="junggi"]yay my 4th copy of gsp and my 5th copy of mathematica and my 2nd copy of nspire[/quote]
O.o I have none of these things. How can I claim a copy?[/quote]
gsp is free online.[/quote]

It wasn't free legally when I last checked a month ago...

I believe you have two free choices;

1) Get the demo free
2) Get version 3.1 (1998) free

I just checked, the gsp I downloaded last time was 4.03. I don't know if that's legal, however.

i bought my gsp #1, amp 07 gave me gsp #2, amp 08 gave me gsp #3
mathematica #1 from arml 08, #2 from usamts 08, #3 from mop 08, actually this is my 4th, but i will get another from mathcamp i think
both nspires from arml
yay

wait wait wait you got gsp from amp?
why? how?

maybe im mistaken but im pretty sure they just sent me a copy of gsp each year
and some silly guide on how to use it

You're going to Mathcamp?! YAAAY!!! (You'll love it. And yes, you'll get another copy.)

Back to question: How to get Mathematica?

Back to thread:
Join SFBA ARML next year

[quote="timwu"][quote="james4l"]Back to question: how to download GSP?[/quote]

Learn chinese. Install anti-virus.

Good luck.[/quote]

I'm Chinese, you know...

Rephrased: how to download the legitimate copy of GSP that we were supposed to have received for being second (well now third) place ARML team?

My assumption was that they were going to send you a copy.


calculus
integration
function
inequalities
real analysis
real analysis theorems

I've read the following proof for the theorem below (which is quite easy to prove if we consider $g(x)-f(x)$ as a new function), and I don't believe in it! Can someone clarify it for me?

Assume that $\alpha\nearrow$ on $[a,b]$. If $f,g\in R(\alpha)$ on $[a,b]$and if $f(x)\leq g(x)\forall x\in [a,b]$, then we have $\int_{a}^{b}f(x)d\alpha(x)\leq\int_{a}^{b}g(x)d\alpha(x).$
Proof. .... $S(P,f,\alpha)\leq S(P,g,\alpha)$... .

But why?! this is only necesarily true when we are choosing the same arbitrary point for $f, g$ for each interval of the partition $P$.

If $f$ is integrable, then its Riemann-Stieltjes sums converge to the integral of $f$ [b]regardless of the choice of sample points[/b]. Same for $g$. Thus, to prove an inequality between the integrals, it suffices to establish an inequality between Riemann-Stieltjes sums with sample points of our choice.

I know this, and I still find this kinda proof loose when we can prove it more clearly, easily and rigorously defining a new function.

Well, but first you have to prove clearly, easily and rigorously that $\int_{a}^{b}g(x)d\alpha(x)-\int_{a}^{b}f(x)d\alpha(x)=\int_{a}^{b}(g(x)-f(x))d\alpha(x)$

Of course we don't have to, if we follow the order of the theorems that have come in the book. :wink:


calculus
derivative
trigonometry
function
LaTeX
calculus computations

How do you write $ \cos y$ in terms of $ \sin y$?

oh,
so
cos y=1/sin y
and then dy/dx=1/(1/sin y)) which is sin y and in turn, x

so the final answer is x?
That sound right?  :)

[quote="iamhe"]cos y=1/sin y[/quote]
Are you sure?

[quote="Patterns_34"]Sorry but it's not $ x$.
[hide="Big Hint"]Consider inverse functions. In other words, $ f(y) \equal{} x \implies y \equal{} f^{ \minus{} 1}(x)$. So what's the inverse function of $ \sin$, and how can we use it in that last step substitution?[/hide][/quote]

So how do you do this problem? I'm a little confused now :huh:

Hold on,
is the inverse of sin $ sin^{ \minus{} 1}$?

If it is, then it would be $ 1/(cos(sin^{ \minus{} 1} x))$

P.S. I just did my first (and second) Latex images!!! :lol:

[quote="vicmiclovich"][quote="iamhe"]Let x=sin y.
Find dy/dx in terms of x.
i have managed to get to this step.
[hide]

x=sin y
dy/dx (x) = dy/dx(sin y)
1=(cos y)* dy/dx
dy/dx=1/cos y
[/hide]

i need a way to find y in terms of x.[/quote]....
from your above question....(the answer's just below!):....
I never really understood your question...finding y in terms of x...but will just assume finding dy/dx
given x = sin(y)
we could just write the equation differently!
 sin(y) = x
then
 d[sin(y)]/dx = dx/dx
cos(y)dy/dx = 1
****therefore; dy/dx = 1/cos(y)

implicit differentiation was applied in the expression " d[sin(y)]/dx"[/quote]

Yeah, but from the question that i'm reading, I have to find $ dy/dx$ in terms of x

[quote="iamhe"]would be $ 1/(cos(sin^{ \minus{} 1} x))$[/quote]
This is correct, but it's a little easier than this.  First, write \cos and \sin instead of cos and sin; they look nicer.  Second, recall that $ \sin^2 y \plus{} \cos^2 y \equal{} 1$.

ahhh,
i believe i have finally gotten the correct answer.

$ dy/dx \equal{} 1/ \sqrt {1 \minus{} x^{2}}$

I'm so slow!!!

P.S. how do you make fractions have a distinct numerator and denominator in latex, like make the fraction vertical instead of horizontal? thanks.

To make fractions, all you need to do is write the following code [code]\frac{a}{b}[/code] to get $ \frac{a}{b}$, and so forth.

Also, here are some [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]codes for symbols[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands]commands[/url] if needed.


who knows if and how and when the national competition will be on air or something?


I was trying to sort posts in a topic by author, only to discover that nothing happened when I punched the checkmark next to the sorting fields.  I was wondering if someone could help me solve this problem.  Any help is appreciated.  (I have IE7 on Windows Vista with the AoPS skin.)

It works for me, so it's not a global failure.

Has it worked for you before? If so, there's a good chance you've got a bad version of some file cached.

Yes, it has worked for me before.
[quote="jmerry"]If so, there's a good chance you've got a bad version of some file cached.[/quote]
At the risk of sounding dumb, what exactly does this mean, and how can I fix it?

Depends a little bit on your Browser, but you want to delete your Cache. For FireFox, this is done under "Delete private data" or something like this.

Well, as I mentioned earlier, I have IE, and I don't see an option quite like that.   :(
I have deleted the history of websites visited, and that has not fixed the problem.  (A clarification that might help:  This problem only occurs when I'm trying to use the sorting features in individual topics.  I can use the sorting features with a list of topics in a forum without problem.)

To delete your cache form IE, follow these Instructions.

Go to "Tools" --> Internet Options --> Under Browsing History, click "Delete", and there.  Just delete the ones you need to.

OK, I did that, and that did not fix the problem.


Given $0<a_{0}<a_{1}<...<a_{k}\in Z$
Difine the function f(x) as $f(x)=\sum_{i=0}^{k}\frac{x^{i}}{(1+x)^{a_{i}}}$
Prove that f(x) is a decreascent function

Should "decreascent" be "decreasing"?
I think that $f(x)$ is decreasing for $x>-1$.


AMC
AMC 12
AIME
geometry
analytic geometry
conics
parabola

This is transcript from Friday, 4/16/04 and I don't have a solution to last problem so please post here.

Here it is:

jwec05: I welcome everyone to the AoPS discussion
js1527k: do I invite them?
jwec05: who? tomorrow?
js1527k: nevermind.  Caroline is away and NAS is offline.
js1527k: I never found out white_horse_kings
js1527k: IM name
js1527k: anyway continue....
Tare111111111111: white_horse_king88
jwec05: As I said, I welcome everyone to this discussion and thank yo all for coming
jwec05: Before we start
jwec05: I'd like to explain few rules
jwec05: Please respect everone in discussion
jwec05: and eevryone should ask other people for help if they have a question
Tare111111111111: ok...
jwec05: there will be no instructor, but everyone will take turns and give away questions
Tare111111111111: nice parody of Mr. Rusczyk btw :-)
js1527k: heh
jwec05: Levels will be between MC to AMC and maximum of AMC12
jwec05: It means that no AIME, please
jwec05: After question is announced
jwec05: the questioner will at least give a minute to let people to solve a problem
jwec05: and after a minute
Tare111111111111: only a minute?
Tare111111111111: lol
Tare111111111111: nvm
Tare111111111111: go on
jwec05: in level of the question
jwec05: if no one has answer
jwec05: we'll give more
Tare111111111111: questions?
Tare111111111111: lol
Tare111111111111: jkjk
Tare111111111111: ignore me
jwec05: after question is given and at least a minute is passed
js1527k: time
jwec05: the questioner will type "Answer" to
jwec05: let people to know it's time to call the answer
jwec05: So, everyone ready?
Tare111111111111: nope
Tare111111111111: jking! lol
js1527k: yes
Tare111111111111: yessir
js1527k: lol Tare
Tare111111111111: arg
Tare111111111111: can we add one more rule?
Tare111111111111: No matter how annoying a person is being
Tare111111111111: you can't warn him until he can't be on AIM anymore
Tare111111111111: please?
Tare111111111111: lol
js1527k: LOL
jwec05: okay, unless the person is swering
Tare111111111111: ok
Tare111111111111: fine
js1527k: I will not warn you (unless you swear or anything like that)
Tare111111111111: fear enough
js1527k: Ok so who starts, silverfalcon?
jwec05: LEt me start with easy question from my level
jwec05: Find the center of the circle Q with a diameter whose endpoints are (-3,10) and (7, -2).
Tare111111111111: (2, -4)?
jwec05: Answer is disqualified and it's incorrect
js1527k: remember your supposed to wait for answer tare!
jwec05: try to wait for minute
Tare111111111111: lol
Tare111111111111: ok
js1527k: hehe
Tare111111111111: I wasn't listening about the whole explanation thing
jwec05: Answer?
Tare111111111111: lol
js1527k: (2,4)
Tare111111111111: yup
jwec05: That's correct
jwec05: How did you do it?
Tare111111111111: wow
Tare111111111111: *golf clap*
Tare111111111111: lol
js1527k: Center is at the middle of the diameter
Tare111111111111: mmm?
Tare111111111111: what?
jwec05: Yes, that's correct
js1527k: So you use the midpoint formula
Tare111111111111: lol
Tare111111111111: jkjk
jwec05: yes
Tare111111111111: oooh
Tare111111111111: I have a problem
jwec05: can you tell the formula?
Tare111111111111: but someone else go while I get it...
js1527k: yes
jwec05: what do you mean by "but someone else go while I get it?"
Tare111111111111: nvm
Tare111111111111: I got it
jwec05: okay
Tare111111111111: So can I go?
jwec05: Sure
js1527k: ((x1+x2)/2 , (y1+y2)/2 )
jwec05: Yes
js1527k: Is the midpoint of (x1,y1) and (x2, y2)
js1527k: well the segement w/those endpoints
jwec05: Yep
Tare111111111111: ready to move on?
jwec05: sure
js1527k: Yes
Tare111111111111: "There is a chord 26 cm long drawn in a circle. The length between the chord and the center is 5. What is the area of the circle?"
Tare111111111111: It's quite easy
Tare111111111111: draw a diagram
Tare111111111111: I'll give you a hint if no one gets it...
Tare111111111111: other than Arnav and Simon that is :wink:
Tare111111111111: um...
Tare111111111111: is it a minute yet? lol
jwec05: when you are ready please type "Answer
Tare111111111111: okey...
jwec05: 144pie
js1527k: sqrt(194)
Tare111111111111: lol
Tare111111111111: nicety
Tare111111111111: um...
js1527k: wait whoops
Tare111111111111: 144pie is the correct answer
Tare111111111111: lol
js1527k: got it lol
Tare111111111111: explanation?
js1527k: sorry
jwec05: Sure
Tare111111111111: I meant "Will anyone offer an explanation since I'm too lazy to type it up?"
joml88 has left the room.
Tare111111111111: lol
Tare111111111111: I'm too lazy even to type that up! lol
jwec05: center point is in the middle of the chord
jwec05: or something like that
Tare111111111111: ?
Tare111111111111: you mean midpoint?
joml88 has entered the room.
jwec05: um yes
Tare111111111111: go on
jwec05: and it's 5 cm away from point
jwec05: so we have triangle with 5, half of the legnth of chord , and x length
jwec05: half of 26 is 13
jwec05: Then we know x is 12
js1527k: a right triangle
jwec05: Yes
jwec05: Forgot those...
joml88: ahh
js1527k: just clarifying...
joml88: i did it a different way
jwec05: ANd 12 is radius
Tare111111111111: how did you do it?
jwec05: square it and pie
js1527k: tell us.
joml88: but got the same thing
jwec05: 144pie
jwec05: go on
js1527k: jwec you generally right 3.14159... as "pi" =)
Tare111111111111: lol
Tare111111111111: tis a joke
Tare111111111111: :-)
jwec05: oh
js1527k: yes I am just as funny as tare.
Tare111111111111: lol
Tare111111111111: ok
Tare111111111111: got another one
js1527k: Wait for joml
Tare111111111111: a little more harder
Tare111111111111: ok
Tare111111111111: joml?
js1527k: (his explanation)
joml88: ok
joml88: i drew the two chords
jwec05: oh, i forgot to point out the chat is for 1 hour
jwec05: keep going
joml88: and used the theorem with intersecting chords
joml88: i cant remember how to state it correctly
js1527k: a*b=c*d
joml88: yeah that pretty much sums up what it says
js1527k: like the segements they cut each other into?
jwec05: can u clarify what a, b, c, and d is?
jwec05: length
Tare111111111111: oy
Tare111111111111: you mean power of point?
Tare111111111111: that works I guess
Tare111111111111: The Two-Chord Power Theorem
joml88: but the triangle is much much easier
js1527k: Wait so the other chord you drew was the diameter?
joml88: yeah
jwec05: length of 20?
jwec05: nvr mind
Tare111111111111: um...
jwec05: what's the lenght of the second chord?
Tare111111111111: Does everyone here know Power of Point theorems?
jwec05: i'm not
Tare111111111111: Should we go over them?
jwec05: in short definition, if possible
Tare111111111111: ok
js1527k: I dont know what "power of point" means but I prob know the theorems...
joml88: they are kind of hard to explain w/out a picture
Tare111111111111: um
Tare111111111111: let me try:
js1527k: find a website?
js1527k: Ok
js1527k: go on
Tare111111111111: well mathworld's good
Tare111111111111: but here's Two-Chord Power Theorem in under 200 words:
jwec05: okay
Tare111111111111: Let there be a circle and two chords within the circle.
Tare111111111111: Let the chords be AB and CD and their intersection X.
Tare111111111111: Then AX*BX=CX*DX
Tare111111111111: The proof is abbreviated, but it just uses similar triangles and the Basic Proportionality Theorem
Tare111111111111: yup
jwec05: do length of two chords same?
joml88: no
jwec05: then how do you figure out them?
Tare111111111111: they should be given
Tare111111111111: or you can deduce them from othe stuff
js1527k: make your question clearer.  You are given three segments and you find the fourth.
Tare111111111111: right
jwec05: okay for this question
js1527k: I am still a little confused about what joml did
js1527k: He has x*y =169
js1527k: x+y=diameter
js1527k: ????
Tare111111111111: nope
Tare111111111111: r^2=12*12
Tare111111111111: think about it
Tare111111111111: he drew in the diameter
Tare111111111111: and he knew that the diameter bisects the other chord
Tare111111111111: so the four length are the radius, radius, 12 (24/2), and another 12
Tare111111111111: so solve for r
Tare111111111111: or rather leave it as r^2 and just stick a pi in
joml88: ok is everyone good?
js1527k: got it
jwec05: I think I better look at the Two-Triangle Theorem bby myself
jwec05: then I'll get ti
jwec05: go on
jwec05: js1527k, you can post the question
js1527k: Okay sure. Let me get one.
Tare111111111111: um...
js1527k: The coordinates of the vertex of the parabola P are (t,0).  If A(0,t) and B(6,t) are on the parabola P, find the equation of parabola P.
Tare111111111111: hello?
Tare111111111111: oh
Tare111111111111: hi
Tare111111111111: lol
js1527k: hi
jwec05: ah, is in conics?
js1527k: Silver I know you are just in Alg II now and might not have covered this yet.  I will explain.
js1527k: Yes, simple conics (ie parabola)
jwec05: i think i can get it but i need some time
jwec05: okay
js1527k: sure thing
Tare111111111111: me too
StellarSupremacy: whoa
StellarSupremacy: there are still people here?
Tare111111111111: lol
js1527k: lol
Tare111111111111: we're actually doing math!
Tare111111111111: And I'm participating!
joml88: can we answer now?
jwec05: Yes,....
Tare111111111111: APOCALYPSE is comign!
jwec05: yes
Tare111111111111: lol
Tare111111111111: I'm still doing it
Tare111111111111: lol
js1527k: yea go ahead
jwec05: i don't htink i can do it but...
joml88: y=(x-3)^2
jwec05: i know what everything question meant
jwec05: but just can't connect them
jwec05: is his answer correct?
js1527k: joml you are close, but incorrect.  check for silly mistakes
joml88: k
Tare111111111111: oy!
Tare111111111111: darn
jwec05: :-)
Tare111111111111: woe is me
Tare111111111111: oy
Tare111111111111: !
StellarSupremacy: hm, this seems organized. Much congratulations.
Tare111111111111: lol
Tare111111111111: is it like the y-coordinate of the vertex -b/a or somat?
Tare111111111111: and the x is like I forgot
Tare111111111111: lol
jwec05: okay, while someone try to get the answer, may I just ask and try to figure the answer?
StellarSupremacy: are you doing the parabola one?
Tare111111111111: yes
jwec05: yes
js1527k: Try using the standard form of the equation of a parabola:  y=a(x-h)^2 + k
StellarSupremacy: no
StellarSupremacy: not necessary
jwec05: h, k form?
StellarSupremacy: we have that (0,t) and (6,t) are on the parabola
jwec05: ah
Tare111111111111: I'm trying to solve a system of equations...
StellarSupremacy: thus the vertex must be (3,something)
StellarSupremacy: we are given that it is (t,0)
StellarSupremacy: thus t=3
StellarSupremacy: so (0,3), (6,3), and (3,0) are on the parabola
Tare111111111111: voila!
joml88: y=1/3(x-3)^2?
Tare111111111111: Arnav saves the day yet again!
Tare111111111111: lol
StellarSupremacy: shift all this over by 3 units to the left
js1527k: right
js1527k: lol he is quick
Tare111111111111: um...
js1527k: I used a system of equations too
StellarSupremacy: (-3,3), (3,3), (0,0) are on the parabola
Tare111111111111: or maybe we're all slowpokes :P
Tare111111111111: jkjk
Tare111111111111: lol
StellarSupremacy: this new parabola will have the equation y=ax^2
StellarSupremacy: simple checking yields a=1/3
StellarSupremacy: thus the shifted parabola has equation y=x^2/3
jwec05: hmm... how did you know vetex was (3, something)?
StellarSupremacy: and the orginial parabola has equation (x-3)^2/3=y
StellarSupremacy: I really didn't
jwec05: huh?
StellarSupremacy: I assumed the parabola was oriented so that the directrix was parallel to the x-axis
StellarSupremacy: This would be normal in most easy problems.
Tare111111111111: mmm
Tare111111111111: *acts like he gets this stuff*
Tare111111111111: lol
js1527k: Yea um dont get into rotations, its just a normal parabola
jwec05: um.. still bit confused
StellarSupremacy: ok good
jwec05: i only went to directrix by myself yesterday
StellarSupremacy: I assumed the parabola was like y=x^2, i.e. if you stretch it a bit and shift it around in the plane
StellarSupremacy: they'll be the same
StellarSupremacy: in other words, we won't have to rotate the parabola
Tare111111111111: matrix?
Tare111111111111: lol
jwec05: rotate the parabola?
StellarSupremacy: yeah
StellarSupremacy: y=x^2 can be rotated
Tare111111111111: y--x^2?
StellarSupremacy: and it'll still be a parabola
StellarSupremacy: but the equation will be a lot uglier
Tare111111111111: or is it x=y^2
Tare111111111111: ?
js1527k: guys it is a normal parabola
StellarSupremacy: x=y^2 is an example, but it is a simple example
StellarSupremacy: ok good
StellarSupremacy: anyway
js1527k: y=a(x-h)^2+k
StellarSupremacy: if it's a normal parabola
jwec05: yes
StellarSupremacy: and (0,t) and (6,t) are on the parabola
StellarSupremacy: their midpoint
StellarSupremacy: i.e. (3,t)
jwec05: okay, and they are in same line
jwec05: oh
StellarSupremacy: yeah
jwec05: then the midpoint is in axis of symmetry?
StellarSupremacy: must be on x=k
StellarSupremacy: yeah
jwec05: oh
jwec05: that clears up
jwec05: nice question
StellarSupremacy: how hard are y'all going?
js1527k: got it?
jwec05: to AMC12
jwec05: in maximum
StellarSupremacy: argh, I shouldn't give you these questions then
js1527k: By silverfalcon's rules, nothing beyond AMC12
jwec05: yep, I can't handle AIME
Tare111111111111: lol
jwec05: since we have about 10 minutes
joml88: i ll post the next question
jwec05: yes
Tare111111111111: okey
jwec05: last question
StellarSupremacy: :-[I was thinking USAMO, sorry
Tare111111111111: lol
js1527k: Well why must it only be 1 hour?
Tare111111111111: cuz
js1527k: Ok joml
Tare111111111111: I'm bored
Tare111111111111: are you?
Tare111111111111: lol
StellarSupremacy: go on as long as you want to
js1527k: no I like math
js1527k:  =)
Tare111111111111: mmm
StellarSupremacy: next problem, please?
Tare111111111111: you and 30 otehrs...
Tare111111111111: lol
jwec05: yes, next problem
js1527k: Joml is getting it
jwec05: oh
joml88: if a+b=3 and a^2+b^2=7  what does a^4+b^4 equal?
StellarSupremacy: good problem.
Tare111111111111: um...
Tare111111111111: why is this problem familiar to me?
Tare111111111111: wasn't this in AoPS???
joml88: i dont know
StellarSupremacy: 47?
Tare111111111111: ya
jwec05: hold on
jwec05: i want to do, too
Tare111111111111: chapter on manipulation
joml88: i didnt get it from aops but it probably was
js1527k: Is there an easier way than substitution?
Tare111111111111: I got 41
Tare111111111111: yup
StellarSupremacy: yes, manipulation is usually easier than substitution
Tare111111111111: square stuff
jwec05: i don't really get this question
StellarSupremacy: If you find yourself solving for a and b, consider trying a different method.
js1527k: hmmm
jwec05: can anyone explain it?
jwec05: i got ab = 1
StellarSupremacy: good.
Tare111111111111: oy
jwec05: it is right?
Tare111111111111: lol
StellarSupremacy: I believe so.
jwec05: then what do i do next?
StellarSupremacy: Now, you are given a^2+b^2, and you want to find a^4+b^4
StellarSupremacy: How can you make what you are given (a^2+b^2) look like what you want?
js1527k: ohh i got it
jwec05: it doesn't factor
StellarSupremacy: what can you do to the equation a^2+b^2=7 to help you get the terms (a^4+b^4) somewhere in there.
jwec05: then i don't know
Tare111111111111: um...
Tare111111111111: ******-dancing
Tare111111111111: you know, country dance
StellarSupremacy: square?
StellarSupremacy: oh
Tare111111111111: not line-dancing but...
Tare111111111111: lol
joml88: the first time you see this it seems hard
Tare111111111111: aw Arnav...
StellarSupremacy: oops
joml88: but you have to think outside the box
StellarSupremacy: I didn't even notice that.
jwec05: wait. I still don't get what StellarSupremacy meant
StellarSupremacy: call me bob
StellarSupremacy: what part don't you get?
Tare111111111111: call him Arny
Tare111111111111: :P
Tare111111111111: lol
Tare111111111111: jkjk
jwec05: like making sipmler
jwec05: simpler
js1527k: square both sides
StellarSupremacy: Indeed.
jwec05: which two sides?
StellarSupremacy: We have a^2+b^2=7
jwec05: yes
StellarSupremacy: and we want to get a^4+b^4 somewhere in there.
js1527k: a^2+b^2=7
jwec05: got it
StellarSupremacy: The simplest thing to do to get a^4 from a^2 and b^4 from b^2 is to square the corresponding terms.
Tare111111111111: One more?
StellarSupremacy: So we square both sides of a^2+b^2=7
StellarSupremacy: now what happens?
jwec05: wait slow down
js1527k: (a^2+b^2)^2 = a^4+2a^2b^2+b^4
jwec05: got that
js1527k: a^4+2a^2b^2+b^4=49
jwec05: correct
js1527k: ab=1
jwec05: yes, but-
js1527k: 2a^2b^2 = 2(ab)^2
js1527k: =2
js1527k: ?
Tare111111111111: yup
jwec05: equals 2?
js1527k: plug in 1 for ab
js1527k: a^4+b^4=49-2
js1527k: =47
Tare111111111111: one more?
Tare111111111111: anyone?
Tare111111111111: lol
jwec05: oh
jwec05: i think it's enough for today
js1527k: oh come on that was short
StellarSupremacy: pfft.
Tare111111111111: lol
jwec05: i'll arrange one more tomorrow at same time
joml88: ok
Tare111111111111: one more!
Tare111111111111: I have a good one!
jwec05: patience
Tare111111111111: lol
Tare111111111111: ah nvm
StellarSupremacy: ok, post it tare.
js1527k: let him post
jwec05: whatever
jwec05: yes
jwec05: post
jwec05: the last
jwec05: question
Tare111111111111: "There is a sphere. Plane A intersects the sphere and the intersection is a circle. The area of the circle is 144pie. The distance between the center of the sphere and plane A is 9. What is the surface area of the sphere?"
Tare111111111111: ok fine maybe it's not so goode :P
Tare111111111111: lol
StellarSupremacy: pi, not pie
Tare111111111111: I know
js1527k: heh
Tare111111111111: it's a continuation of the last proble
Tare111111111111: m
StellarSupremacy: 900pi?

I just copied directly from discussion and I won't identify the each person in here unless they want to.  For the last question, please post the correct answer with explanation here. Thanks!

Silver, people asked a few more questions after you left.  Please PM me if you want the rest of the transcript.  We could also put it online for refrence.

[quote]"There is a sphere. Plane A intersects the sphere and the intersection is a circle. The area of the circle is 144pie. The distance between the center of the sphere and plane A is 9. What is the surface area of the sphere?"
[/quote]

The radius of the circle is 12 because A= :pi: r :^2:    144 :pi: = :pi: r :^2:   solve for r and you get 12.  You're looking for the radius of the sphere.  So you set up a right triangle one leg is 9 and another 12.  So pythagorean theorem or fact that it is 3-4-5 gives you 15 for the radius.  Now, Surface area = 4 :pi: r :^2:   Plugging 15 in gives you 900 :pi:

Okay, then try to post it in this thread.
I kind of don't feel like to receive them in mail because I have enough emails coming to me so..
Please post the questions or exact transcript like joml88 kinda did..
Thanks

I suggest you clean up your transcripts because reading through "whatever" and "lol" is not what people intend to read.

basically, just get rid of all the stuff not pertaining to da maths.

[quote="Silverfalcon"]It's "to the math" not "to da math."

I agree on changing all them to neat form of questions and replies but well, I can't sit in my copmuter whole day and deleting all those useless replies, you know that.  I did erase for greeting part of the discussion but I don't think I feel necessary to erase all those unnecessary ones.

But thanks for suggestions, though[/quote]

Hmm.. I don't have much of a life.  ;)  If it would help people, you could send me the transcripts and I could clean them up.

Alright, I tried to clean up this transcript, mostly because I was bored/wanted to procrastinate writing an essay, but partly because it may be helpful.  Here you go.. (breaks indicate new problems)

jwec05: As I said, I welcome everyone to this discussion and thank you all for coming
jwec05: Before we start
jwec05: I'd like to explain few rules
jwec05: Please respect everyone in discussion
jwec05: and everyone should ask other people for help if they have a question
 jwec05: there will be no instructor, but everyone will take turns and give away questions
jwec05: Levels will be between MC to AMC and maximum of AMC12
jwec05: It means that no AIME, please
jwec05: After question is announced
jwec05: the questioner will at least give a minute to let people to solve a problem
jwec05: and after a minute
jwec05: if no one has answer
jwec05: we'll give more time
jwec05: the questioner will type "Answer" to  let people to know it's time to call the answer
jwec05: So, everyone ready?
jwec05: LEt me start with easy question from my level

----------------------------------------------------------------------

jwec05: Find the center of the circle Q with a diameter whose endpoints are (-3,10) and (7, -2).
Tare111111111111: (2, -4)?
jwec05: Answer is disqualified and it's incorrect
js1527k: remember your supposed to wait for answer tare!
jwec05: try to wait for minute
 jwec05: Answer?
 js1527k: (2,4)
 jwec05: That's correct
jwec05: How did you do it?
 js1527k: Center is at the middle of the diameter
 jwec05: Yes, that's correct
js1527k: So you use the midpoint formula
 jwec05: can you tell the formula?
 js1527k: ((x1+x2)/2 , (y1+y2)/2 )
 js1527k: Is the midpoint of (x1,y1) and (x2, y2)
js1527k: well the segement w/those endpoints
jwec05: Yep
Tare111111111111: ready to move on?
jwec05: sure
js1527k: Yes

----------------------------------------------------------------------

Tare111111111111: "There is a chord 26 cm long drawn in a circle. The length between the chord and the center is 5. What is the area of the circle?"
 jwec05: when you are ready please type "Answer
Tare111111111111: okey...
jwec05: 144pie
js1527k: sqrt(194)
 Tare111111111111: 144pie is the correct answer
 Tare111111111111: explanation?
 jwec05: center point is in the middle of the chord
jwec05: or something like that
 Tare111111111111: you mean midpoint?
 jwec05: um yes
Tare111111111111: go on
jwec05: and it's 5 cm away from point
jwec05: so we have triangle with 5, half of the legnth of chord , and x length
jwec05: half of 26 is 13
jwec05: Then we know x is 12
js1527k: a right triangle
jwec05: Yes
jwec05: Forgot those...
 js1527k: just clarifying...
joml88: i did it a different way
jwec05: ANd 12 is radius
Tare111111111111: how did you do it?
jwec05: square it and pie
js1527k: tell us.
joml88: but got the same thing
jwec05: 144pie
jwec05: go on
js1527k: jwec you generally write 3.14159... as "pi" =)
 joml88: i drew the two chords
 joml88: and used the theorem with intersecting chords
joml88: i cant remember how to state it correctly
js1527k: a*b=c*d
joml88: yeah that pretty much sums up what it says
js1527k: like the segements they cut each other into?
jwec05: can u clarify what a, b, c, and d is?
jwec05: length
 Tare111111111111: you mean power of point?
Tare111111111111: that works I guess
Tare111111111111: The Two-Chord Power Theorem
joml88: but the triangle is much much easier
js1527k: Wait so the other chord you drew was the diameter?
joml88: yeah
jwec05: length of 20?
jwec05: nvr mind
Tare111111111111: um...
jwec05: what's the lenght of the second chord?
Tare111111111111: Does everyone here know Power of Point theorems?
jwec05: i'm not
Tare111111111111: Should we go over them?
jwec05: in short definition, if possible
 joml88: they are kind of hard to explain w/out a picture
 Tare111111111111: let me try:
js1527k: find a website?
js1527k: Ok
js1527k: go on
Tare111111111111: well mathworld's good
Tare111111111111: but here's Two-Chord Power Theorem in under 200 words:
jwec05: okay
Tare111111111111: Let there be a circle and two chords within the circle.
Tare111111111111: Let the chords be AB and CD and their intersection X.
Tare111111111111: Then AX*BX=CX*DX
Tare111111111111: The proof is abbreviated, but it just uses similar triangles and the Basic Proportionality Theorem
 jwec05: do length of two chords same?
joml88: no
jwec05: then how do you figure out them?
Tare111111111111: they should be given
Tare111111111111: or you can deduce them from other stuff
js1527k: make your question clearer. You are given three segments and you find the fourth.
Tare111111111111: right
jwec05: okay for this question
js1527k: I am still a little confused about what joml did
js1527k: He has x*y =169
js1527k: x+y=diameter
js1527k: ????
Tare111111111111: nope
Tare111111111111: r^2=12*12
Tare111111111111: think about it
Tare111111111111: he drew in the diameter
Tare111111111111: and he knew that the diameter bisects the other chord
Tare111111111111: so the four length are the radius, radius, 12 (24/2), and another 12
Tare111111111111: so solve for r
Tare111111111111: or rather leave it as r^2 and just stick a pi in
joml88: ok is everyone good?
js1527k: got it
jwec05: I think I better look at the Two-Triangle Theorem bby myself
jwec05: then I'll get ti
jwec05: go on

----------------------------------------------------------------------

jwec05: js1527k, you can post the question
js1527k: Okay sure. Let me get one.
 js1527k: The coordinates of the vertex of the parabola P are (t,0). If A(0,t) and B(6,t) are on the parabola P, find the equation of parabola P.
 jwec05: ah, is in conics?
js1527k: Silver I know you are just in Alg II now and might not have covered this yet. I will explain.
js1527k: Yes, simple conics (ie parabola)
 joml88: y=(x-3)^2
jwec05: i know what everything question meant
jwec05: but just can't connect them
jwec05: is his answer correct?
js1527k: joml you are close, but incorrect. check for silly mistakes
joml88: k
 Tare111111111111: is it like the y-coordinate of the vertex -b/a or somat?
Tare111111111111: and the x is like I forgot
Tare111111111111: lol
jwec05: okay, while someone try to get the answer, may I just ask and try to figure the answer?
StellarSupremacy: are you doing the parabola one?
Tare111111111111: yes
 js1527k: Try using the standard form of the equation of a parabola: y=a(x-h)^2 + k
StellarSupremacy: no
StellarSupremacy: not necessary
jwec05: h, k form?
StellarSupremacy: we have that (0,t) and (6,t) are on the parabola
jwec05: ah
Tare111111111111: I'm trying to solve a system of equations...
StellarSupremacy: thus the vertex must be (3,something)
StellarSupremacy: we are given that it is (t,0)
StellarSupremacy: thus t=3
StellarSupremacy: so (0,3), (6,3), and (3,0) are on the parabola
Tare111111111111: voila!
joml88: y=1/3(x-3)^2?
  StellarSupremacy: shift all this over by 3 units to the left
js1527k: right
 js1527k: I used a system of equations too
StellarSupremacy: (-3,3), (3,3), (0,0) are on the parabola
 StellarSupremacy: this new parabola will have the equation y=ax^2
StellarSupremacy: simple checking yields a=1/3
StellarSupremacy: thus the shifted parabola has equation y=x^2/3
jwec05: hmm... how did you know vetex was (3, something)?
StellarSupremacy: and the orginial parabola has equation (x-3)^2/3=y
StellarSupremacy: I really didn't
jwec05: huh?
StellarSupremacy: I assumed the parabola was oriented so that the directrix was parallel to the x-axis
StellarSupremacy: This would be normal in most easy problems.
 js1527k: Yea um dont get into rotations, its just a normal parabola
jwec05: um.. still bit confused
StellarSupremacy: ok good
jwec05: i only went to directrix by myself yesterday
StellarSupremacy: I assumed the parabola was like y=x^2, i.e. if you stretch it a bit and shift it around in the plane
StellarSupremacy: they'll be the same
StellarSupremacy: in other words, we won't have to rotate the parabola
 jwec05: rotate the parabola?
StellarSupremacy: yeah
StellarSupremacy: y=x^2 can be rotated
 StellarSupremacy: and it'll still be a parabola
StellarSupremacy: but the equation will be a lot uglier
 js1527k: guys it is a normal parabola
StellarSupremacy: x=y^2 is an example, but it is a simple example
 StellarSupremacy: anyway
js1527k: y=a(x-h)^2+k
StellarSupremacy: if it's a normal parabola
jwec05: yes
StellarSupremacy: and (0,t) and (6,t) are on the parabola
StellarSupremacy: their midpoint
StellarSupremacy: i.e. (3,t)
jwec05: okay, and they are in same line
jwec05: oh
StellarSupremacy: yeah
jwec05: then the midpoint is in axis of symmetry?
StellarSupremacy: must be on x=k
StellarSupremacy: yeah
jwec05: oh
jwec05: that clears up
jwec05: nice question

----------------------------------------------------------------------

 joml88: i ll post the next question
 Tare111111111111: okey
jwec05: last question
 joml88: if a+b=3 and a^2+b^2=7 what does a^4+b^4 equal?
StellarSupremacy: good problem.
 StellarSupremacy: 47?
 js1527k: Is there an easier way than substitution?
Tare111111111111: I got 41
 StellarSupremacy: yes, manipulation is usually easier than substitution
Tare111111111111: square stuff
jwec05: i don't really get this question
StellarSupremacy: If you find yourself solving for a and b, consider trying a different method.
 jwec05: can anyone explain it?
jwec05: i got ab = 1
StellarSupremacy: good.
 jwec05: it is right?
 StellarSupremacy: I believe so.
jwec05: then what do i do next?
StellarSupremacy: Now, you are given a^2+b^2, and you want to find a^4+b^4
StellarSupremacy: How can you make what you are given (a^2+b^2) look like what you want?
js1527k: ohh i got it
jwec05: it doesn't factor
StellarSupremacy: what can you do to the equation a^2+b^2=7 to help you get the terms (a^4+b^4) somewhere in there.
jwec05: then i don't know
 joml88: the first time you see this it seems hard
joml88: but you have to think outside the box
 jwec05: wait. I still don't get what StellarSupremacy meant
 StellarSupremacy: what part don't you get?
 jwec05: like making simpler
js1527k: square both sides
StellarSupremacy: Indeed.
jwec05: which two sides?
StellarSupremacy: We have a^2+b^2=7
jwec05: yes
StellarSupremacy: and we want to get a^4+b^4 somewhere in there.
js1527k: a^2+b^2=7
jwec05: got it
StellarSupremacy: The simplest thing to do to get a^4 from a^2 and b^4 from b^2 is to square the corresponding terms.
 StellarSupremacy: So we square both sides of a^2+b^2=7
StellarSupremacy: now what happens?
jwec05: wait slow down
js1527k: (a^2+b^2)^2 = a^4+2a^2b^2+b^4
jwec05: got that
js1527k: a^4+2a^2b^2+b^4=49
jwec05: correct
js1527k: ab=1
jwec05: yes, but-
js1527k: 2a^2b^2 = 2(ab)^2
js1527k: =2
js1527k: ?
 jwec05: equals 2?
js1527k: plug in 1 for ab
js1527k: a^4+b^4=49-2
js1527k: =47

----------------------------------------------------------------------

 Tare111111111111: "There is a sphere. Plane A intersects the sphere and the intersection is a circle. The area of the circle is 144pie. The distance between the center of the sphere and plane A is 9. What is the surface area of the sphere?"
Tare111111111111: ok fine maybe it's not so goode
Tare111111111111: lol
StellarSupremacy: pi, not pie
Tare111111111111: I know
js1527k: heh
Tare111111111111: it's a continuation of the last proble
Tare111111111111: m
StellarSupremacy: 900pi?


vector
superior algebra
superior algebra unsolved

Hi,

I've just read that the group $(\mathbb{R}^{n},+)$ cannot be a field for $n>2$. Easy or not ?

This assumes that it has $\mathbb{R}$ as a subfield, and the scalar multiplication as a vector space is compatible with field multiplication. In that case, we would have a finite-dimensional vector space also a field, and therefore it would be an algebraic extension of $\mathbb{R}$. Since $\mathbb{C}$ is algebraically closed and has degree two over $\mathbb{R}$, no algebraic extension of $\mathbb{R}$ can have degree greater than two.

If you let that additive group stand alone, the statement is false.

Doesn't the following work¿:
Take some countable subset and define the operations of $\mathbb{Q}$ on it.
Then take a transcendental base $\mathfrak{X}$ of suitable cardinality and consider $\mathbb{Q}(\mathfrak{X})$.

[quote="Schoppenhauer"][quote="alekk"]On every infinite set $E$, can one find operations $+$ and $.$ such that $(E,+,.)$ is a field ?[/quote]

Let ${\cal L} = S \cup \{+, \cdot, e\}$ be the language Fields can be formalized.
Let $\Phi = \{ \forall x \forall y : x + y = y + x, \forall x \forall y : xy=yx, \forall x \forall y \forall z (x+y)+z = x+(y+z), \forall x \forall y \forall z: (xy)z = x(yz), \forall x xe= x, \forall x \exists y xy=e \}$ be the set of axioms for fields. It is easy to see  ;) that $|\Phi| < \aleph _ 0$.

From this, the Löwenheim-Skolem-Tarski-Theorem says, that for every infinite Cardinality $\kappa > |\Phi|$ which implies $\kappa \ge \aleph_0$ there exists a Model of $\Phi$ with cardinality k.
Hence, there exist Fields of any infinite cardinality k.[/quote]
I'm confusing... so do you say that I can define a field structure over $\mathbb{R}^n$ for every $n$? For example for $n=3$? I've read somewhere that it is only possible for $n=1,2,4,8$, the text refered to some "famous" Adams' theorem (but I can't find this theorem on the web right now :( ), but I'm a beginner, so maybe I'm wrong... Can someone
explain that point clearly?  :?

[quote="ZetaX"]There are only finite division-algebras over $\mathbb{R}$ for the dimensions you mentioned ($1,2,4,8$).
But the existence of a field having some cardinality has nothing to do with that. Indeed, $\mathbb{R}$ and $\mathbb{R}^n$ have the same cardinality, so take some bijection between them and define a field by using the field properties of the real numbers.
By this, you will loose your canonical additional group of $\mathbb{R}^n$ as a vectorspace.[/quote]
Thank's, in fact it was stupid not to consider the cardinality argument..., but what is a  finite division-algebra?

Here, a finite dimensional algebra (over the reals) is a ring (not necessary commutative) defined on a fnite dimensional vectorspace (over the reals here) with $a(xy)=(ax)y=x(ay)$ for all reals $a$ and elements $x,y$ from that algebra. It's called a division algebra, if the equations $ax=b,yc=d$ can be solved, too.

[quote="jmerry"]This assumes that it has $\mathbb{R}$ as a subfield, and the scalar multiplication as a vector space is compatible with field multiplication.[/quote]

Why do you mean by "compatible" in this case ?


real analysis
complex analysis
complex analysis theorems

[quote="julien_santini"]I was wondering why Rudin would prefer to use Parseval's formula to deduce Liouville's theorem, the Maximum Modulus Theorem and Cauchy's estimates (instead of going the easy way using only the development in power series).[/quote]
I don't think so.


inequalities
linear algebra
matrix
induction
linear algebra unsolved

$A \in M_n(R)$ and $B \in M_n(R)$ are two symmetric positive definite matrices.
$A_k \in M_{n-1}(R)$ is the matrix obtained from $A$ when deleting the
$k$-th row and the $k$-th column of $A$. Define $B_k$ similarly.

Prove that: $\frac{\det(A+B)}{\det(A_i + B_i)} \geq \frac{\det(A)}{\det(A_i)} +\frac{\det(B)}{\det(B_i)}$

Great problem!! We will see how it reduces to a very beautiful inequality that we have already discussed. But till then, some observations:
   The most  important idea is I think to see that $\frac{detA}{detA_i}=\frac{1}{(A^{-1})_{ii}}$. Indeed, it suffices to look at the element in position $(i,i)$ in the equality $^{t} Com(A)=det A\cdot A^{-1}$. Thus our nice problem becomes $\frac{1}{(A^{-1})_{ii}}+\frac{1}{(B^{-1})_{ii}}\leq \frac{1}{((A+B)^{-1})_{ii}}$. Now, since $A,B$ are both symmetric and positive, there exists $X$ such that $ A=X^{*}DX$ and $B=X^{*}FX$, with $D,F$ diagonal, having positive entries. This is not difficult, it is classical result. Replacing this in the inequality we have to prove and changing a little bit the notations, we see that we must prove in fact that $\frac{1}{(X^{*}DX)_{ii}}$ $+\frac{1}{(X^{*}FX)_{ii}}\leq \frac{1}{(X^{*}(D^{-1}+F^{-1})^{-1}X)_{ii}}$. Now, use the fact that $ (X^{*}UX)_{ii}=\sum_{k}{u_k x_{ki}^2}$ if $U$ is diagonal with entries $u_i$.  Practically, after making $ x_k=d_k\cdot x_{ki}^2$ and $y_k=f_k\cdot x_{ki}^2$ we have to prove that ${\frac{1}{x_1+...+x_n}+\frac{1}{y_1+...+y_n}\leq\frac{1}{\frac{x_1y_1}{x_1+y_1}+...+\frac{x_n y_n}{x_n+y_n}}}$. If we write it in the form $\frac{x_1y_1}{x_1+y_1}+...\frac{x_ny_n}{x_n+y_n}\leq\frac{(x_1+...+x_n)(y_1+...+y_n)}{x_1+...+x_n+y_1+...+y_n}$ this invites to induction and it really works easily by induction.


[hide]I think 1.a bone, 2.a man, 3.a chair and 4.a dog.[/hide]

[quote="K Sengupta"]Two legs sat upon three legs,
With one leg in his lap;
In comes four legs
And runs away with one leg;
Up jumps two legs,
Catches up three legs,
Throws it after four legs,
And makes him drop one leg.

Determine the item/ items referred to in the puzzle[/quote]

this one is similar to the bilbo-gollum riddles:

2 legs sits on 3 legs, no legs lays on 1 leg, and 4 legs gets some...
[hide="solution"]
a man sits on a 3legged stool, eats a fish on a 1legged table, and the cats gets some too[/hide]

[hide]One leg: chicken leg
Two legs: man
Three legs: stool
Four legs: cat[/hide]

That's awesome!!

Before I knew the different terms were meant to represent different entities, the riddle sounded like some sick joke...

[hide] 1 leg-chicken leg, 2 leg-man, 3 leg-stool,4 leg-dog[/hide]
what a cute puzzle!!


modular arithmetic
number theory
greatest common divisor

Have $8x=27k+1$
$27k+1\equiv 3k+1 mod 8$
So, $3k+1=8r$. The only value of $k$ which is an integer from $0$ to $7$ which satisfies this relation is $k=5$.
Then, $27k+1=136$ so $x=17$.
So, $x\equiv 17 mod 27$ is the solution.

[quote="MathStudent2"]Solve the congruence.

8x ≡ 1 mod (27)

Thanks![/quote]
If $ax\equiv b\pmod{m}$, where $\gcd{(m,a)}=1$ then you need find $a'$ such that $aa'\equiv 1\pmod{m}$.  :wink:

Yay, I'm currently learning how to solve linear congruences:

Since $\gcd(8, 27)=1$, we know there will only be one solution (plus the numbers congruent to this solution mod 27).

Anyway, just solve $8u+27v=1$, which can be done with the Euclidean Algorithm. $(u,v)=(-10, 3)$.

Then the solution is $x=\frac{1\cdot-10}{1}=-10$, which is 17 mod 27.

$x\equiv 17 \pmod{27}$ describes all the solutions.

[quote="MathStudent2"]Thanks everyone.  The problem I am still having is determing what different values are in different mods.  In this problem where -10 is 17 mod (27).  I understand the mod (27) part, I just don't see how -10 turns into 27 and how that is determined?  I have looked through my text and they don't provide any good examples and skip a lot of steps.  I understand the gcd part and that gives you the number of solutions you will eventually have.  I just don't see the one part yet.  Thanks.[/quote]

When you are working mod (say 27) then you can add and subtract the modulos (27 in this case) as many times are you want.

So if you have,
$2x\equiv 50(\mod 27)$
Subtract 27 once and once again to get,
$2x\equiv-4 (\mod 27)$
No divide by 2 because $\gcd(2,27)=1$ to have,
$x\equiv-2 (\mod 27)$
But it consider athesically pleasing when you have an answer as a remainder (meaning between 0 and 26).
So add 27 to get,
$x\equiv 25 (\mod 27)$

(Of course you could have divided by 2 from the very beginning but I was just showing you how this stuff works  :) )

Thank you sylow_theory!  I see what you mean.


By what common fraction does $ 0.\overline{81}$ exceed $ 0.81$?

Or we could note that their difference is $ .00\overline{81}$

Thus x-81/100=x/100 (where x is .8181...)

99/100(x)=81/100

x=81/99=9/11. We are looking for x/100 so our answer is 9/1100.


Hi everyone -- anyone on here ever take one of the Jeopardy Contestant Preliminary Tests (kids or teens)? -- either online or in person, to qualify for that show? If so, could you please PM me?  Did you take the test recently? What type of categories/questions are asked? Any way to prepare for such a test? Did anyone on here ever qualify for the second round or even make it onto the show?

i believe there is another jeapordy topic on these forums... make a search. it could be worthwhile.

If anyone knows where on this forum I can find more Jeopardy info, please let me know!!

Yes, I searched too!  I couldn't find anything.  :o

[quote=aryanpadarthi]This is the biggest bump in history of AOPS  :help:[/quote]

It's not, and that was unnecessary

[url=https://artofproblemsolving.com/community/c3t767494f3h19668_find_that_letter_challenge]EDIT- Here is an example

[quote=aryanpadarthi]This is the biggest bump in history of AOPS  :help:[/quote]

Lol imagine having nothing to do so that you decide to flex by bumping an old thread (that's not even the oldest).

What a minute.........................I CAN'T DELETE MY POSTS


geometry
parallelogram
quadratics
geometric transformation

1. A positive integer is called [i]good [/i] if it can be written as the sum of two distinct integer squares. A positive integer is called [i]better[/i] if it can be written in at least two was as the sum of two integer squares. A positive integer is called [i]best[/i] if it can be written in at least four ways as the sum of two distinct integer squares.

a) Prove that the product of two [i]good[/i] numbers is [i]good[/i].

b) Prove that $ 5$ is [i]good[/i], $ 2005$ is [i]better[/i], and $ 2005^2$ is [i]best[/i].

2. In each box of a $ 1 \times 2009$ grid, we place either a $ 0$ or a $ 1$, such that the sum of any $ 90$ consecutive boxes is $ 65$. Determine all possible values of the sum of the $ 2009$ boxes in the grid.

3. Show that if $ h_A, h_B,$ and $ h_C$ are the altitudes of $ \triangle ABC$, and $ r$ is the radius of the incircle, then

$ h_A \plus{} h_B \plus{} h_C \geq 9r$

4. The point $ M$ is chosen inside parallelogram $ ABCD$. Show that $ \angle MAB$ is congruent to $ \angle MCB$, if and only if $ \angle MBA$ and $ \angle MDA$ are congruent.

5. The weird mean of two numbers $ a$ and $ b$ is defined as $ \sqrt {\frac {2a^2 \plus{} 3b^2}{5}}$. $ 2009$ positive integers are placed around a circle such that each number  is equal to the the weird mean of the two numbers beside it. Show that these $ 2009$ numbers must be equal.

hello, for 3) i think it must be
$ h_A\plus{}h_B\plus{}h_C \geq 9r$
Sonnhard.

2. Denote by $ a$ the word made by the first $ 29$ boxes in the grid, and by $ b$ the word made by the next $ 61$ boxes in the grid. Then the word representing the grid is $ g \equal{} (ab)^{22}a$. Since the sum of the numbers in $ a$ may be any integer value between $ 0$ and $ 29$, the sum of the numbers in the grid may be any integer value between $ 22\cdot 65 \equal{} 1430$ and $ 22\cdot 65 \plus{} 29 \equal{} 1459$.

5. Expression $ \sqrt {\frac {2a^2 \plus{} 3b^2} {5}}$ is called weird mean for a good reason: indeed $ \min(a,b) \leq \sqrt {\frac {2a^2 \plus{} 3b^2} {5}} \leq \max(a,b)$ (in effect it is a pondered quadratic mean).
Take $ x$ (one of) the least number(s) written on the circle, and $ a, b$ its neighbors, so $ x \leq a$ and $ x \leq b$. According with the above it follows $ x \equal{} a \equal{} b$. Continue around the circle to get all numbers equal to $ x$.

[hide="solution to problem 3 "]

$ \frac{1}{r}\equal{}\frac{1}{h_a} \plus{}\frac{1}{h_b} \plus{}\frac{1}{h_c}$

By Cauchy Schwartz

$ (h_a \plus{}h_b \plus{}h_c) \cdot \frac{1}{r}  \geq 9  \implies h_a \plus{} h_b \plus{} h_c \geq 9r$

[/hide]

1a is a direct consequence of the [url=http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity]Brahmagupta-Fibonacci Identity[/url].

[hide="4"]
Ah, this idea helped me go to JBMO xD

Translation! Translate M to N, $ \vec{MN}\equal{}\vec{AB}$. Now MCNB is cyclic, easy angle chasing and we're done

[/hide]

[quote="modularmarc101"]4. The point $ M$ is chosen inside parallelogram $ ABCD.$ Show that $ \angle MAB$ is congruent to $ \angle MCB,$ if and only if $ \angle MBA$ and $ \angle MDA$ are congruent.[/quote]
Let $ X \equiv AM \cap CD$ and $ Y \equiv CM \cap AD.$ Note that quadrilateral $ ACXY$ is cyclic $ \Longleftrightarrow$ $ \angle DAX \equal{} \angle DCY$ $\Longleftrightarrow$ $ \angle YXM \equal{} \angle MCA$ and $\angle XYM \equal{} \angle MAC.$ Since $ \angle MXD \equal{} \angle MAB$ and $ \angle MYD \equal{} \angle MCB,$ it follows that $ MYDX$ and $ MABC$ are similar $ \Longleftrightarrow$ $ \angle MBA \equal{} \angle MDA.$


function
inequalities
inequalities unsolved

by AM-AG we have :$ \sum 2 \frac {4a^2 \plus{} b^2 \plus{} c^2}{\sqrt {3}(2a^2 \plus{} b^2 \plus{} c^2)}$ $ \ge$$ \sum 4\frac {a (\sqrt {a^2 \plus{} b^2 \plus{} c^2})}{(2a^2 \plus{} b^2 \plus{} c^2)}.$
now we prove that $ \sum2\frac {4a^2 \plus{} b^2 \plus{} c^2}{\sqrt {3}(2a^2 \plus{} b^2 \plus{} c^2)}\le 3\sqrt {3}$
let $ a^2 \equal{} x,b^2 \equal{} y$and $ c^2 \equal{} z$ and suppose that $ x \plus{} y \plus{} z \equal{} 1$ and consider the function $ f(x) \equal{} \frac {4x}{x \plus{} 1}$ it's easily to see that$ f$ is concave so by jensen's inequality we have  :$ \sum\frac {4a^2}{2a^2 \plus{} b^2 \plus{} c^2}\le 3$.
and we are done .


Show that it cannot be the case that $ U\equal{}X\plus{}Y$ where $ U$ is uniformly distributed on $ [\minus{}1,1]$ and $ X$ and $ Y$ are independent and identically distributed.
It has been puzzling me for a whole day. :(

take a look there:
http://www.mathlinks.ro/viewtopic.php?t=213903


i recently proved that "every integer > 3 is a sum of more than one prime." can this be useful to prove goldbachs conjecture?

It is possible to prove that every number is amount different prime

I think that I recall in Courant's "What is Mathematics?" that it can be shown that every sufficiently large integer is the sum of at most four primes.


function
trigonometry
algebra unsolved
algebra

If a, b, c, are real numbers such that a+b+c=abc. prove that any one of a, b, c is atleast 1.7

couldn't (a, b, c) be (1, 2, 3) as well?

Olorin was just providing counterexamples when $a,b,c$ are any reals. Note that the magnitudes of each number in the counterexamples are less than $1.7$.

$1,2,3$ satisfies the conditions because $3>2>1.7$ :wink:

Furthermore it is easy to see that there is no condition of the form
[i]... [b]all[/b] or a,b,c are at least k[/i]
because when $a$ is an arbitrarily small constant then we have a function in $b,c$ : $a+b+c-abc$ and this is positive at $b=c=0$ but for sufficiently large $b,c$ it is negative since the r.h.s has larger degree than the l.h.s so there is always some triplet which satisfies this regardless of $a>0$

but it is not given that, a,b,c are positive.

Please read!
[quote="olorin"]$(a,b,c)=({1\over 2},{1\over 2},-{4\over 3})$ or $(a,b,c)=(0,0,0)$

If $a,b,c> 0$, then ${abc\over 3}={a+b+c\over 3}\geq \sqrt[3]{abc}$
and $\max\{a,b,c\}\geq \sqrt[3]{abc}\geq \sqrt{3}>1.7$.[/quote]

Let $0<a\le b\le c$. Then $c=\frac{a+b}{ab-1}, \frac{\partial c}{\partial b}=\frac{-1-a^{2}}{(ab-1)^{2}}<0.$
Therefore $min c=b(a): ab^{2}=2b+a$ or $b(a)=\frac{1+\sqrt{1+a^{2}}}{a}$ and $\frac{\partial b}{\partial a}=\frac{-1-\sqrt{1+a^{2}}}{a\sqrt{1+a^{2}}}<0$. Therefore $c_{min}=b_{min}=a_{min}$ and $c_{min}^{2}=3$ or $c_{min}=\sqrt 3 .$

Another approach : (If $a,b,c > 0$)
By trigonometry substitution, we have $a = tan \alpha, b = tan \beta, c = tan \gamma$ where $\alpha+\beta+\gamma = \pi$ and $0 < \alpha, \beta, \gamma < \frac{\pi}{2}$.
Then, $max \{ a,b,c \}\geq tan \frac{\pi}{3}= \sqrt{3}> 1.7$ as desired..


Ring Theory
abstract algebra
superior algebra
superior algebra unsolved

I essentially know nothing about commutative algebra, and I came across this when reading a text,

[quote]L is a Lie algebra. There exists a unique maximal solvable ideal, called the radical of L and denoted Rad L. In case Rad L = 0, L is called semisimple.[/quote]

Now I remember from p.661 of Serge Lang's Algebra, there is an exercise saying that for an Artinian Ring R, the Jacobson radical = 0 iff R is semisimple.

Then I have two questions:
1. In the Lie algebra case, does the maximal solvable ideal = Jacobson radical?
2. Does the concept of "semisimple" in the Lie algebra case coincide with the "usual one"? (direct sum of simple algebra)

This is probably Humphrey's book, right? Keep reading, and he'll elaborate! It turns out that definition is equivalent to being a direct sum of simple (nonabelian) Lie algebras.

I suppose that is actually the intersection of the maximal ideals of a Lie algebra, which you can see using the description I just gave. To see this, note that it's clear in the semisimple case, and you can mod out by the radical to reduce to that case.

Thank you! Yes it's Humphreys' book, and I actually reached the coincidence of this semisimple and the usual "semisimple" thing yesterday. I'll look into Jacobson radical soon as I'm in a hurry.


find a 2 =b 2 +c 2
  prove that 5/4 \sum a 2 +2 \sum  \sqrt a  \leq
         3/2 \sum ab + \sum  \sqrt (a+b-c) :D  :D

Can you tell me the source of the problem ?
I think you should tell more clearly about it .
If you don't know the source , write Unknown.


long winded, confusing, and badly explained
[hide][quote="pkothari13"]$\text{Prove that }1(1!) + 2(2!) + ... + n(n!) = (n+1)! - 1$[/quote]
a combinatoriacal solution follows: digits: with a group of n+1 different, there are n+1! ways to make a letter string. n(n!) is the amount you can make a letter string such that a certain letter isnt the first. (n-1)(n-1)! is the amount of ways you can make with a fixed first letter (which is the one restricted the previous time) and a restriction on the second such that its not a certain letter.  with (n-2)(n-2)! is the amount of ways you can make it with a fixed first and second letter (the second letter is the one that was restricted for the previous time) and a restriction on the third letter. Going all the way down, we get that we have all the letter strings except for the last one, which we have restricted to anything but a certain letter, and there will only be one more for the certain letter we have restricted, giving that an extra one to be added.
so 1(1)!+2(2!)+...+n(n!)+1=(n+1)![/hide]
does it make any sense at all?


calculus
integration
limit
inequalities
complex analysis

I am having trouble with this step in the evaluation of Fresnel's Integrals ($ \int_{0}^{\infty} \cos(x^2)  \ dx$): Show that
\[ \lim_{R \rightarrow \infty} \int_{0}^{\frac{\pi}{4}} e^{\minus{}R^2e^{2\theta i}}Rie^{i\theta} \ d\theta
\]

Looks to me:

\[ \left|e^{\minus{}R^2(\cos(2\theta)\plus{}i\sin(2\theta))}\right|_{0\leq\theta\leq \pi/4} \leq e^{\minus{}R^2}\]

and so the limit is zero.

I think you have the inequality reversed? For example, taking $ \theta \equal{} \pi/4$ the LHS becomes $ |e^{\minus{}iR^2}|\equal{}1$ and the RHS can obviously be made arbitrarily large by an appropriate choice of $ R$.

[quote="Phelpedo"]I think you have the inequality reversed? For example, taking $ \theta \equal{} \pi/4$ the LHS becomes $ |e^{ \minus{} iR^2}| \equal{} 1$ and the RHS can obviously be made arbitrarily large by an appropriate choice of $ R$.[/quote]

Ok.  Didn't notice that.  Sorry.

Integrate by parts. The integrand becomes $ \frac{\text{bounded}}{R}$.


Has anybody had difficulties using the USAMTS LaTeX templates on a Macintosh?

One of our users is having some difficulties using the usamts.tex and solutions.tex files under OzTeX on a Mac, and we're trying to reproduce the error.  Unfortunately, here at AoPS World Headquarters, we don't have any Macs.

So if you're getting errors with the usamts.tex file on a Mac, please post them here (and, hopefully, how you fixed them).

Thanks for your help.

The file works for me, but I use TeXShop, not OzTeX.

Wow is there really an AoPS World Headquarters???

Do you guys have jobs other than AoPS or do you moderate these boards all day?  Just wondering

AoPS is our job, but it's much more than moderating these message boards.  Take a look around the site & you'll see.

I was just wondering what you guys do
I have seen so far:
teaching classes
writing books
moderating forums
selling books
finding and adding competitions to the resource section
doing USAMTS
answering questions
Is that all?

[quote="mustafa"]They do the Mandelbrot too, I think.[/quote]Actually, we don't.  [url]http://www.mandelbrot.org[/url]

I am not ready for this level math (yet) but I do have a mac. I tried veiwing some of the files and it works fine for me.. My macbook is new and it can run all windows files. Hope this helps!

Works fine here. (I use [url=http://www.tug.org/mactex/]MacTeX[/url] and run [b]pdflatex[/b] from the Terminal.)


How many weeks will it take for Shrek 2 to beat out Starwars for the number 2 all time box office earnings?

Facts:
Current Earnings: Shrek 2:$397,100,000 Star Wars: $460,998,007

Shrek made $10,500,000 this weekend and $1,958,000 on thursday (representing 1 of 4 weekdays). Shrek is currently showing in 2937 theaters.

Oh yeah, I forgot to mention: I think it will take 8 weeks and 3 days.

Never leads 3-2 which seems really improbable as Shrek made a total of 18 million last week and only need another 63 million total. considering it will be out at least another 20 weeks, it would need to average 5 million a week for the next 10 weeks (3 million on the weekends and 2 million total on 4 weekdays). and then average a bit over 1 million for the next 10 weeks. Personally, I think its decline will be about 2 million a week until it reaches 6 million a week, where it will run for quite a time. Could some "nevers" post their thoughts on why?

i was split between 7 and 8 so i picked the later one, 8


superior algebra
superior algebra unsolved

Let $A$ be a commutative ring with $2n+1$ elements , $n > 4$. Suppose that if $a$ is not inversable, then $a^{2}\in \{-a,a\}$. Prove that $A$ is a field.

the assumption $n>4$ excludes the case $A = \mathbb{Z}/3 \times \mathbb{Z}/3$, so we really have to make it. ;)

I've proven the claim for $\mathbb{Z}/k$ and deduced that $A$ has prime characteristic. any other ideas? :-)

Try proving the following result:

A finite ring with unit and with no non-zero nilpotent elements is a finite direct product of fields.

This will make the problem in this topic almost trivial, and it should prove to be a more interesting exercise. Assume apriori that the ring is commutative, if you prefer, but this is not necessary; the conclusion holds without assuming commutativity.


Two factors have a product of $ 3 \cdot 9^{2x} \minus{} 6 \cdot {27^x}$. One of the factors is $ 3^x \minus{} 2$. What is the other factor.

hello, we have $ 3\cdot9^{2x}\minus{}6(3^3)^x\equal{}3\cdot3^{4x}\minus{}6\cdot3^{3x}\equal{}3\cdot3^{3x}(3^x\minus{}2)\equal{}3^{3x\plus{}1}(3^x\minus{}2)$ so the other factor is $ 3^{3x\plus{}1}$.
Sonnhard.


geometry
perimeter
trigonometry
angle bisector
congruent triangles
China

Given a square $ABCD$ whose side length is $1$, $P$ and $Q$ are points on the sides $AB$ and $AD$. If the perimeter of $APQ$ is $2$ find the angle $PCQ$.

Let the excircle of triangle APQ opposite to the vertex A have center C' and touch the sideline AP of the triangle at the point B'. Then, $C^{\prime}B^{\prime}\perp AP$, and the distance AB' equals the semiperimeter of triangle APQ. Since the perimeter of triangle APQ equals 2, its semiperimeter must be 1; thus, we have AB' = 1. On the other hand, we know that AB = 1 (in fact, the square ABCD has sidelength 1). Since the points B and B' both lie on the ray AP, we thus can conclude that the points B and B' coincide. Hence, $C^{\prime}B^{\prime}\perp AP$ becomes $C^{\prime}B\perp AP$; in other words, $C^{\prime}B\perp AB$. Hence, the point C' lies on the perpendicular to the line AB at the point B. But since ABCD is a square, the perpendicular to the line AB at the point B is the line BC. Hence, the point C' lies on the line BC. Similarly, the point C' lies on the line DC. As the lines BC and DC have only one common point, namely the point C, it follows that the point C' coincides with the point C. Since we have defined the point C' as the center of the excircle of triangle APQ opposite to the vertex A, it follows that the point C is the center of the excircle of triangle APQ opposite to the vertex A. Thus, this point C lies on the external angle bisector of the angle APQ. Hence, $\measuredangle CPQ=90^{\circ}-\frac{\measuredangle APQ}{2}$. Similarly, $\measuredangle CQP=90^{\circ}-\frac{\measuredangle AQP}{2}$. Hence, by the sum of the angles in triangle PCQ, we have

$\measuredangle PCQ=180^{\circ}-\measuredangle CPQ-\measuredangle CQP=180^{\circ}-\left(90^{\circ}-\frac{\measuredangle APQ}{2}\right)-\left(90^{\circ}-\frac{\measuredangle AQP}{2}\right)$
$=\frac{\measuredangle APQ+\measuredangle AQP}{2}$.

But since the triangle PAQ is right-angled at A (this is because the quadrilateral ABCD is a square), we have < APQ + < AQP = 90°, and thus $\measuredangle PCQ=\frac{90^{\circ}}{2}=45^{\circ}$.

  darij

Find the point R in line AB but not in segment AB, which satisfies that BR=DQ. Then we can show triangle CDQ and triangle CBR are congruent triangles very easily by using S.A.S.  SO we can say that angle PCQ is equal to the half of right angle RCQ.

Hi,

How did you know that PCQ was exactly half of RCQ?  I followed you with extending AB and marking point R, and the triangle congruence, but I'm lost after that.

Thanks

using trigo it;s easy, but how do we prove that if $ (tgx+1)(tgy+1)=2$ then $ x+y=45$?

[quote][color=darkred][b][u]An easy extension.[/u][/b] Let $ ABCD$ be a quadrilateral for which $ AB=AD$, $ BC=CD$ and $ B=D=90^{\circ}$, $ m(\widehat{PAQ})=2\phi$
([b]a right deltoid[/b]). Let $ P\in (AB)$, $ Q\in (AD)$ be two points so that $ PQ=PB+QD$. Show that $ m(\widehat{PCQ})=90^{\circ}-\phi$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Observe that the semiperimeter of the triangle $ APQ$ is equal to $ AB$, $ CB\perp AP$ and the point $ C$ belongs to the bisector of the angle $ \widehat{PAQ}$. Therefore, the point $ C$ is the $ A$- exincenter of the triangle $ APQ$. Generally, $ m(\widehat{PCQ})=\frac{1}{2}\cdot \left(m(\widehat{APQ})+m(\widehat{AQP})\right)$, i.e. $ \boxed{m(\widehat{PCQ})=90^{\circ}-\phi}$.[/color]

[hide="Another Solution"]
Let $AP=x$ and $AQ=y$.  Then $PB=1-x$ and $QD=1-y$.  Next, note that

\begin{align*}\tan\angle PCQ&=\cot(\angle PCB+\angle QCD)=\dfrac{1-\tan\angle PCB\tan\angle QCD}{\tan\angle PCB+\tan\angle QCD}\\&=\dfrac{1-(1-x)(1-y)}{(1-x)+(1-y)}=\dfrac{x+y-xy}{2-x-y}.\end{align*}
Now turning to the condition that the perimeter of $\triangle APQ$ is $2$, since $PQ=\sqrt{x^2+y^2}$ we have $x+y+\sqrt{x^2+y^2}=2$.  This means that $\sqrt{x^2+y^2}=2-(x+y)$.  Squaring both sides gives $x^2+y^2=4-4(x+y)+(x+y)^2$, and simplification gives $2-x-y=x+y-xy.$  Hence $\tan\angle PCQ=\dfrac{x+y-xy}{2-x-y}=1\implies\angle PCQ=\boxed{45^\circ}$.  $\blacksquare$[/hide]

[quote="Tigerli"]You can then prove that CPQ and CPR are congruent triangles because
PQ=2-AQ-AP=(1-AQ)+(1-AP)=DQ+BP=BR+BP=PR. :)[/quote]

The 2 triangles are congruent as per s.s.s. criterion.

Best regards,
sunken rock


calculus
integration
function
real analysis
real analysis unsolved

Prove that if the real-valued function $f$ on the interval $[a,b]$ is integrable on $[a,b]$ then so is $f^{2}$.

I guess the problem narrows down proving that the product of two integrable functions is also integrable.

Can anyone show this?

This is a Riemann integration question, right? (The Lebesgue question, for not-necessarily-bounded functions, would be false: the product of measurable functions is measurable, but $L^{1}\not\subset L^{2}.$)

Kent, yes, it's a Riemann integration problem. Sorry I forgot to put that.

I'd prove instead that $g\circ f$ is Riemann integrable whenever $f$ is Riemann integrable and $g$ is Lipschitz on the range of $f$; $|g(y)-g(z)|\le C|y-z|$ for some real $C$ whenever $y,z$ are in the range of $f$.
Any Riemann integrable function $f$ is bounded, and thus $g(x)=x^{2}$ is Lipschitz on the range of $f$.

On any partition, the difference between the upper and lower sums of $g\circ f$ is at most $C$ times the difference between the upper and lower sums of $f$. That makes $g\circ f$ integrable; we can make this difference small with the same partitions as for $g$.

With Lebesgue tools, we can weaken the Lipschitz condition to continuity (on the range of $f$) and boundedness of $g$.
I thought about trying something with uniform continuity, but I'm not sure where that fits.

We can derive products from squares: if $f$ and $g$ are integrable, $f+g,(f+g)^{2},f^{2},g^{2}$ and finally $fg=\frac12((f+g)^{2}-f^{2}-g^{2})$ are integrable.


Ring Theory
number theory
prime factorization
superior algebra
superior algebra unsolved

Let R be a UFD, a in R*\U(R) that is a is non-zero, non-unit in R,

a =p_1^(a_1).......p_t^(a_t) for distinct non-associated primes p_j, (a_j) >= 1, 1=<j=<t.

If d|a the show that d~p_1^(b_1).....p_t^(b_t) for some 0=<b_j=<a_j, 1=<j=<t.

Show that there are (1+a_1)....(1+a_t) distinct ideals (d) such that (a)=< (d). Deduce that R satisfies ascending chain condition (ACC) on principal ideals.

I dont know how to do this question, please help

Many thanks

[quote="billwgd"]HINT for principal ideals "contains" = "divides", i.e. (a) > (b) <=> a|b. But, being a UFD, divisors must have said form.[/quote]

Thank you for your reply

But I still don't understad how to do this question. Can you show me more specifically, please?

The lattice of principal ideals containing $ (a)$ is in 1-1 correspondence with the lattice of divisors of $ a$ (mod units). Can you count the divisors based on the prime factorization?


THe area of a right triangle is 30. If the median to they hypotenuse is length 6, find b^2, where b is the length of the longer leg of the triangle.

thx


A circle of radius $ 1$ is internally tangent to two circles of radius $ 2$ at points $ A$ and $ B$, where $ AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the gure, that is outside the smaller circle and inside each of the two larger circles?
[asy]size(200);defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;

pair B = (0,1);
pair A = (0,-1);

label("$B$",B,NW);label("$A$",A,2S);

draw(Circle(A,2));draw(Circle(B,2));

fill((-sqrt(3),0)..B..(sqrt(3),0)--cycle,gray);
fill((-sqrt(3),0)..A..(sqrt(3),0)--cycle,gray);
draw((-sqrt(3),0)..B..(sqrt(3),0));
draw((-sqrt(3),0)..A..(sqrt(3),0));

path circ = Circle(origin,1);

fill(circ,white);
draw(circ);
dot(A);dot(B);

pair A1 = B + dir(45)*2;
pair A2 = dir(45);
pair A3 = dir(-135)*2 + A;

draw(B--A1,EndArrow(HookHead,2));
draw(origin--A2,EndArrow(HookHead,2));
draw(A--A3,EndArrow(HookHead,2));

label("$2$",midpoint(B--A1),NW);
label("$1$",midpoint(origin--A2),NW);
label("$2$",midpoint(A--A3),NW);[/asy]$ \textbf{(A)}\ \frac {5}{3}\pi \minus{} 3\sqrt {2}\qquad \textbf{(B)}\ \frac {5}{3}\pi \minus{} 2\sqrt {3}\qquad \textbf{(C)}\ \frac {8}{3}\pi \minus{} 3\sqrt {3}\qquad\textbf{(D)}\ \frac {8}{3}\pi \minus{} 3\sqrt {2}$
$ \textbf{(E)}\ \frac {8}{3}\pi \minus{} 2\sqrt {3}$

This is very easy for an AMC 25.... It might have been a problem on my teachers test in 7th grade.
Break the intercetion of the two circles into two sectors, call the intersection points of the circles C and D and the center of the small circle E. By drawing AE and EC, we form a 30-60-90 right triangle and thus arc CD is 120 degrees, now we use the fact that the area of each sector is the region between AD,AC and the circle minus the area of triange ACD, then we have to subtract the area of the inner circle, then answer is B.

No i do not think that is the answer. Well we have a two 60degree arcs. So that is 1/3 of the circles area. 4/3pi. but there is still extra which is the area of an equilateral triangle with side lenth 2. Area is sqrt3. So the extra region is 2 times 1/6 the area of the circle minus the equilateral triangle. so it is 2(2/3pi-sqrt3). 4/3pi+4/3pi-2sqrt3. 8/3pi - 2sqrt3 which is E.

[quote="sunehra"]A circle of radius $ 1$ is internally tangent to two circles of radius $ 2$ at points $ A$ and $ B$, where $ AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the gure, that is outside the smaller circle and inside each of the two larger circles?
[asy]defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;

pair B = (0,1);
pair A = (0,-1);

label("$B$",B,NW);label("$A$",A,2S);

draw(Circle(A,2));draw(Circle(B,2));

fill((-sqrt(3),0)..B..(sqrt(3),0)--cycle,gray);
fill((-sqrt(3),0)..A..(sqrt(3),0)--cycle,gray);
draw((-sqrt(3),0)..B..(sqrt(3),0));
draw((-sqrt(3),0)..A..(sqrt(3),0));

path circ = Circle(origin,1);

fill(circ,white);
draw(circ);
dot(A);dot(B);

pair A1 = B + dir(45)*2;
pair A2 = dir(45);
pair A3 = dir(-135)*2 + A;

draw(B--A1,EndArrow(HookHead,2));
draw(origin--A2,EndArrow(HookHead,2));
draw(A--A3,EndArrow(HookHead,2));

label("$2$",midpoint(B--A1),NW);
label("$1$",midpoint(origin--A2),NW);
label("$2$",midpoint(A--A3),NW);[/asy]$ \textbf{(A)}\ \frac {5}{3}\pi \minus{} 3\sqrt {2}\qquad \textbf{(B)}\ \frac {5}{3}\pi \minus{} 2\sqrt {3}\qquad \textbf{(C)}\ \frac {8}{3}\pi \minus{} 3\sqrt {3}\qquad\textbf{(D)}\ \frac {8}{3}\pi \minus{} 3\sqrt {2}$
$ \textbf{(E)}\ \frac {8}{3}\pi \minus{} 2\sqrt {3}$[/quote]

Connect the point of intersection of the two large circles, name them X and Y. Furthermore, connect B to X and Y and A to X and Y.

Noting that XY passes through the centre of the smaller circle, connect B to the centre of the small circle, thus we have a right angled triangle.

$ \angle AXB \equal{} \sin{1}{2}$

$ \therefore \angle AXB \equal{} 30^{\circ}$

$ \therefore \angle XBY \equal{} 120^{\circ}$

To determine the area of sector XAY, we shall use the previous information and the formula:

$ A \equal{} \frac{1}{2}r^2\left( \theta \minus{} \sin{\theta} \right)$

$ \equal{} \frac{1}{2}2^2 \left( 120^{\circ} \minus{} \sin{120^{\circ} } \right)$

$ \equal{}2 \left( \frac{2 \pi}{3} \minus{} \frac{\sqrt{3}}{2} \right)$

Furthermore, note that sector XBY is the same as sector XAY, hence their combined area is:

$ \equal{}2 \cdot 2 \left( \frac{2 \pi}{3} \minus{} \frac{\sqrt{3}}{2} \right)$

$ \equal{} 4 \left( \frac{2 \pi}{3} \minus{} \frac{\sqrt{3}}{2} \right)$

$ \equal{} \frac{8 \pi}{3} \minus{} 2\sqrt{3}$

The area of the small circle is $ \pi$

$ \therefore A \equal{} \frac{8 \pi \minus{} 3\pi}{3} \minus{} 2\sqrt{3}$

$ \therefore A \equal{} \frac{5\pi}{3} \minus{} 2\sqrt{3}$

$ \therefore \mbox{B}$

Wow this is ridiculously easy for a #25.

[hide="Alternate solution"]Draw in a couple of lines, $ AB$ and $ AC$ where $ C$ is the intersection of the two circles (either one), then we can find the area of a small arc thing cut off by equilateral triangle $ \triangle ABC$ to be $ \frac{1}{6}\cdot 4\pi\minus{}4\cdot\frac{\sqrt{3}}{4}\equal{}\frac{2\pi}{3}\minus{}\sqrt{3}$.

Four of those plus two equilateral triangles is $ \frac{8\pi}{3}\minus{}2\sqrt{3}$. Subtracting away the little circle gives $ \frac{5\pi}{3}\minus{}2\sqrt{3}\equal{}\boxed{\textbf{(B)}}$.[/hide]

varunrocks: Your solution is a bit hard to follow (well so is mine  :lol:). First off, they're $ 120^\circ$ arcs. After that you kind of lost me... but I think you're forgetting to subtract off the little inner circle, which would make your answer correct.

Oh I forgot to subtract the circle with area pi.


Find all the positive integers $a,b,c$ which are greater or equal to $2$ such that : $a^b+1=(a+1)^c$

this is an even more general problem:
http://www.kalva.demon.co.uk/short/soln/sh00n4.html


Shouldn't it just reduce to (a*b)?

because a*b shows up on both side of the + we can call it q and get:

q + (~c*q)

which must be true when q is true from the left side
and must be false when q is false because the left must be false and the right is made false by q.

Therefore the value of c has no impact so the answer is just q:

Substitute for a*b

What about this one.

[b]F(x,y,z) = ~x + ~(zy + ~x)(x + ~z(x + ~y))[/b]


Could someone help me with these please

1. If sin(theta) = 3/4 and pi/2<theta<pi

find the exact values of

(a) sin2(theta), cos2(theta) and tan2(theta)
(b) sin3(theta), cos3(theta) and tan3(theta)

2. Find, in simplest surd form, the value of cos75(theta), tan15(theta), sin105(theta) and tan22 degrees 30minutes.

[hide="Hint"]Use the addition formulas

$\sin\left(\alpha\pm\beta\right)=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha$

$\cos\left(\alpha\pm\beta\right)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

From these you can find multiple angle formulas.  Also use $\sin^{2}\theta+\cos^{2}\theta=1$.  Your answers should directly follow.[/hide]

But how do you use them for question b? where it has sin3(x)

i know how to do  question 1a and question 2......
for e.g. sin2(x) is just 2sin(x)cos(x)

but what happens when its sin3(x)? ?

it also might be easier to use $\sin(3x)=\sin(4x-x)$ since you can just use the double angle formulas


I am interested studying mathematics with computer science. I know that Harvard has a good math department and MIT has a good computer science department. I have one question

Is it possible to cross register and take all the CS courses at MIT while being a math major at Harvard?

Any information regarding this question would be greatly appreciated.

It should be added that MIT has a good math department and Harvard has a good CS department too.

I think it's better if you directly contact the colleges. I don't see why you wouldn't want to just stay in one college but if you really want to do what you said, I suggest talking to both schools and see if they can schedule you something.

Harvard specifically mentions on a Web page for its math majors that cross-registration at MIT is a good way to get a good combinatorics course.

it is possible to cross register between the two schools. i know that for MIT you can't take more than 2 classes at Harvard and that only some fraction of your major could be done away form MIT.... not sure about Harvard though!

You might also want to consider that the two schools have different academic schedules (MIT starts earlier, for one).

I believe that cross-registration is pretty easy.  A lot of people think that they want to do it, and then they get to college and realize that it's more trouble than it's worth.  "I can roll out of bed and get to class in under 5 minutes, or I can drag my lazy ass onto the 1 bus."  A larger practical issue is that it can be hard to schedule the courses you want if you have an awkward commute in the middle of the day to a school that does not run on the same schedule as your own.

By the way, the main reason for a math/CS person to cross-register is that MIT offers a lot more CS courses.  Cross-registering for a course that you can take at your home institution is just plain silly, unless the prof is known to be REALLY bad.


from Elementary Number Theory, Burton

1. Prove or disprove: If $a|(b+c)$, then either $a|b$ or $a|c$.
2. Prove that the product of any three consecutive integers is divisible by $6$; the product of any four consecutive integers is divisble by $24$; the product of any five consecutive integers is divisible by $120$.
3. Prove that if $a$ and $b$ are both odd integers, then $16|a^{4}+b^{4}-2$

p.s. Is the divisibility notation ok?

1.  False.  Counterexample: $(a,b,c)=(2,3,3)$.
2. $a(a+1)\hdots (a+n-1)=n!\binom{a}{n}\implies n!|a(a+1)\hdots (a+n-1)$.
3. Since $a\equiv b\equiv 1\pmod 2$, $a^{4}\equiv b^{4}\equiv 1\pmod{16}$.  As a result, $a^{4}+b^{4}-2\equiv 0\pmod{16}$.

[hide="1."]A counter example is a=3, b=5, and c=7[/hide]

[hide="2."]a.

a(a-1)(a+1). Since one of those has to be even, and one of those has to be a multiple of 3, the product ihas to be divisible by 6.

Same argument for n integers and n!.[/hide]

[hide="3."] For odd integers, a^4 is 1 mod 16.

1+1-2=0[/hide]


The number of ordered pairs of integers $ (m,n)$ for which $ mn \ge 0$ and \[m^3 \plus{} n^3 \plus{} 99mn \equal{} 33^3\] is equal to

$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 3\qquad
\textbf{(C)}\ 33\qquad
\textbf{(D)}\ 35\qquad
\textbf{(E)}\ 99$

Consider the case m,n>=0
Take the equation mods 3 and 11, respectively. We must solve $ m^3\minus{}n^3\equal{}0$. In both these mods, however, the only cube root of -1 is -1. Thus, $ m\equiv \minus{}n\mod{33}$(after CRT)
Now we bound: $ 99mn<33^3$ so $ \max(m,n)<19$. As $ 33|m\plus{}n$, $ m\plus{}n\equal{}0,33$. Toss out the first b/c this implies m=n=0. Now for the second:
\[ m^3\plus{}n^3\plus{}99mn\equal{}33^3\iff 33(m^2\plus{}n^2\minus{}mn)\plus{}99mn\equal{}33^3\\ \iff (m\minus{}n)^2\equal{}33^2\]
Thus we see only (33,0) and (0,33) work.

Now take m,n<=0
\[ (\minus{}m)^3\plus{}(\minus{}n)^3\plus{}33^3\geq 99mn\] by $ Am\minus{}Gm$ in three variables, with equality only when $ m\equal{}n\equal{}\minus{}33$.

Hence 3 solutions \[ \boxed{\textbf{(B)}}\]

[hide="Where did I go wrong?  Youre probably right..."]
Let $ m\plus{}n\equal{}s$.  Then $ m^3\plus{}n^3\plus{}3mn(m\plus{}n)\equal{}s^3$.  Subtracting the given equation from the second yields $ s^3\minus{}33^3\equal{}3mns\minus{}99mn$.  Thus, $ (s\minus{}33)(s^2\plus{}33s\plus{}33^2\minus{}3mn)\equal{}0$, so $ s\equal{}33$ or $ (m\plus{}n)^2\equal{}33(m\plus{}n)\plus{}33^2\minus{}3mn\equal{}0$.  The second equation is equivalent to $ (m\minus{}n)^2\plus{}(m\plus{}33)^2\plus{}(n\plus{}33)^2\equal{}0$, whose only solution is $ (\minus{}33,\minus{}33)$, and that works.  However, the solutions satisfying $ m\plus{}n\equal{}33$ and the required conditions are $ \{ (0,33),(1,32),(2,31),\ldots,(33,0) \}$, and there are $ 34$ of those.  Thus, the answer is $ 1\plus{}34\equal{}35$, or $ \boxed{\textbf{(D)}}$.
[/hide]

No, you're right. My solution had two typos, one fatal.
First: At the beginning when I say "solve m^3-n^3=0" i clearly meant $ m^3\plus{}n^3\equal{}0$
non-fatal...
but later on, I said:
\[ 33(m^2\plus{}n^2\minus{}mn)\plus{}99mn\equal{}33^3\iff (m\minus{}n)^2\equal{}33^2\]
where I meant to say
\[ 33(m^2\plus{}n^2\minus{}mn)\plus{}99mn\equal{}33^3\iff (m\plus{}n)^2\equal{}33^2\]
yup, those - and + signs

So all 34 solutions work

according to the solutions manual, the answer is [b]D[/b].

[quote="sunehra"]The number of ordered pairs of integers $ (m,n)$ for which $ mn \ge 0$ and

$ m^3 + n^3 + 99mn = 33^3$

is equal to

$ \textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 33\qquad \textbf{(D)}\ 35\qquad \textbf{(E)}\ 99$[/quote]
[hide]My solution:
We have: $ (m + n)^3 - 33^3 + 3\mn\(33 - m - n) = (m + n - 33)\(m^2 - 4mn + n^2) = 0$
Therefore m+n=33, which gives $ 34$ solutions(since m and n are the same-signed variables) or $ m^2 - 4mn + n^2 = 0$, which means $ (m - 2n)^2 = 3n^2$, which gives $ m = 0$ and $ n = 0$ is another solution.
So the answer is [b]D[/b]
How can I put the box around my answer?[/hide]

[quote="ghjk"]How can I put the box around my answer?[/quote]

Use \fbox{}.

Cool. I should start using that more, especially for fractions. \fbox{} goes crazy when you use \frac{}{} in it.

[quote="unimpossible"]How is m = 0 and n = 0 a solution? it doesn't fit the equation we are given.

[hide]

I think the best way to do this problem is to use the factorization $x^3+y^3+z^3-3xyz = (x+y+z)\frac{1}{2}((x-y)^2+(y-z)^2+(x-z)^2)$

so here, x = m, y = n, and z = -33.  so we have $m+n-33 = 0 $ or $\frac{1}{2}((m-n)^2+(m+33)^2+(n+33)^2)=0$. The first equation gives 34 solutions, and the second equation gives $m= -33$, $n=-33$.

The answer is D

[/hide][/quote]

Could you have used $x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ and still come up with the same answer?

@pylert: factor 1/2 out, "complete the squares" on the second factor and you'll get the same one as unimpossible's

@hsm174: Click "Contests" on the top bar and go to bottom of page, click "USA", click "AMC/AHSME", click "1999". Easy, right? You could've done all this instead of asking this in the thread.


trigonometry geometry perimeter rotation

Ok, by examining the following graphs, how would you describe the graphs, and how would you summarise your results when you change the variables (a, b and k)...

r = k + a sin (bx)
r = k + a cos (bx)

thanks guys

I hope you mean \[r = k+a\sin{b \theta}\]  \[r = k+a\cos{b \theta}\] The change of variable doesn't really affect the problem, it is just a matter of covention.

If it was just $r=k$, then you have a circle of radius $k$. Now by adding the sine wave to the circle, you are putting a wave like texture on the perimeter of the circle for small enough $a$.  If $a$ is larger than $k$, then you will have intersections on the interior of the circle of radius $k$. It has a radial period of $\frac{2\pi}{b}$, so note that non integer $b$ values yield a path that would need to be plotted over more than just $2\pi$. Effectively, the second graph is just a rotated version of the first graph.

I'm not sure how I would describe the graphs, maybe with excess verbage. The changes I can describe.

Increasing $a$ yields a greater range in radial values and if $a$ is large enough, there will be intersection at different $\theta$. $a = 0$ is just a circle. Increasing $b$ decreases the wavelength of the wave on the circle. Increasing $k$ increases the average radius of the graph.

cheers, that helps heaps!
exactly what i wanted


[quote="henryyangrui"]CAN SOMEONE HELP ME WITH THIS?
I AM RUNNING FOR PRESIDENT IN MY MATH CLUB, WHAT CAN I TALK ABOUT IN MY SPEECH?[/quote]

Talk about some really clever math problems, make them bow down to your awesomeness.  :D

yea, you are right, but ............ the thing is that i suck at math compared to them............ :D  :blush:

what you do is you run for vice-president when nobody else does, thus automatically making you vice-president. then, you wait for something to happen to the president, like quitting or graduating. that's how i became president. got it in the middle of 10th grade and i am still currently president.

Yup. If you're vice president, you'll become president if the president ever has a little "accident".

Not that I'm implying anything...

Speak on the importance of sanitation, making it sound like war propaganda.

There is almost a universal liking among mathy people for corny math jokes...

Haha, at my school, whoever was the most popular would be voted in as president, regardless of knowledge or anything else. However, if your school has a legit club system, then I would describe any other leadership positions I've had and relate them to how I would run this club.

[quote]You can talk about Hitler/Stalin/blah and how similar you guys are, you can also relate his charisma to yours too.[/quote]

Haha excellent advice  :)

I suggest a countdown show among the candidates in front of the audience before the voting.

Oh all of you dont know how henryyangrui really sucks at math...... He is hoping for a miracle.

Tell the members what you will contribute as president. From what I hear (read above, that is), you're not that great at math, but you should have some ideas about recruiting new members or preparing for/participating in/organizing an interschool competition or something like that.
If you don't intend to do anything as president, don't run. Don't wreck the club when someone more experienced, dedicated, whatever, could really take it to the next level and make it a worthwhile experience for everyone. (I don't mean any offence to you.)

[quote="Bictor717"] From what I hear (read above, that is), you're not that great at math,  (I don't mean any offence to you.)[/quote]

He is superb in math: 2nd in the state of Florida for Mathcounts, and 15th in the nation; USAMO qualifier. I believe he will make the MOP.

Well if you're that good, why do you have to cheat on a math contest by posting problems on AoPS? Both you and weiyangmiami have posted many of the problems from this year's CPSC in the forums. I have notified an administrator, so hopefully they wil be deleted soon. But the point of that contest is just having fun and working as a team! You shouldn't lie about the source of problems and post them to get help from other schools just in the hope of getting a better score.


Let $ n\equal{}4444^{4444}$, let $ A$ be the sum of the decimal digits of $ n$, let $ B$ be the sum of the digits of $ A$, and let $ C$ be the sum of the digits of $ B$. Find $ C$.

It can be done! And rather quickly in fact.


probability
rotation
geometry
geometric transformation
factorial
LaTeX

(please move this problem if it is in the wrong section)

Hey all you AoPSers! I've been working through the Introductory C & P book by David Patrick, but there's this problem that's been bugging me...but the solution to it is sort of confusing to me as in why it works.

Here is the problem as stated in the book: (3.4.3b)

"A Senate committee has $ 5$ Democrats and $ 5$ Republicans. In how many ways can they sit around a circular table...
...if all the members of each party all sit next to each other?"

Here's the solution:

[hide]"Choose any 5 consecutive seats in which to place the Democrats - it doesn't matter which $ 5$ consecutive seats that we choose, since we can rotate the table. Then there are $ 5!$ ways ways to place the Democrats in their seats, and $ 5!$ ways to place the Republicans in their seats, for a total of $ 5!\times5!\equal{}\boxed{14400}$ arrangements. "[/hide]
Now what confuses me is the fact that this chapter happens to be over how to [i]not overcount[/i] problems such as this one, and I can't seem to see why we don't do so here.  Let's call the Democrats letters A, B, C, D, and E; and the Republicans numbers 1, 2, 3, 4, and 5.  Shouldn't we have to divide by 10 or something since we can rotate the table and still have the same alignment, such as ABCDE12345 and BCDE12345A ?

Thanks to all the moderators for all the work they've done.

[quote="zhh"]
"Choose any 5 consecutive seats in which to place the Democrats - it doesn't matter which $ 5$ consecutive seats that we choose, since we can rotate the table."[/quote]

That part addresses what you're getting confused on.

If you still don't quite understand, consider a committee with only 2 democrats and 2 republicans. It should be clear to you why there are only $ 2!\times 2!\equal{}4$ arrangements in this case.

The rotations don't matter; all the solution is counting is the way in which the block of Democrats and the block of Republicans can each be independently arranged.

Otherwise Consider Democrats as one person and Republicans as another person .
You know no. of ways to seat n persons around a circular table is (n-1)!
Therefore no. of ways = $ 1 . 5! .5!$(5! for arranging the two parties among themselves)

Yeah...I think I get it now.

So basically we never counted those "rotation" possibilities in the first place, because of the method used in the solution.  Therefore, we have $ 5!\times5!\equal{}\boxed{14400}$ since we can rearrange each of the Democrats and Republicans in their groups of five.

madness, what are those decimal factorials you were talking about?

The decimals probably stand for multiplication.


C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)  ΔH=-1,370 kJ

For the combustion of ethyl alcohol as described in the above equation, which of the following is true?

I. The reaction is exothermic.
II. The enthalpy change would be different if gaseous water were produced.
III. The reaction does not involve oxidation-reduction.
IV. The products of the reaction occupy a larger volume than the reactants.

A. I, II
B. I, II, III
C. I, II, IV
D. III, IV
E. Only 1

I know that statements I and II are true (the change in ethalpy is negative, so the reaction is exothermic, and gaseous water has a different enthalpy value than liquid water, so it would change the overall change in ethalpy of the reaction).  But I'm not sure about III and IV.  I don't think that III is correct because there is a free element in the reaction (O2 gas).  I have no clue how to determine if statement IV is correct.  I know that the answer is either A or C.

Please help.

Carbon in the alcohol gets oxidized, and O2 gets reduced, so statement III is false.
We can look at the coefficients of the products and reactants that are gases. There are three moles of gas on the reactants side and 2 moles of gas on the products side. Therefore, the reactants occupy a larger volume than the products.

So the answer is A (only I and II are right?).