ie gerefe sadeie G ba n ras darim.
ras haie an ra tori rang amizi mikonim ke hich do rasi ke be ham ertebet darand iek rang nabashand.(rang amizie mojaz)
farz mikonim adade rangi rasi in graph(kamtarin tedad rang mored niaz baraie erzaie sharaiet masale) k bashad.[X(G)=k]
hamin kar ra baraie motameme G iani G' ham anjam midahim. farz konid adade rangi rasi motamme G iani G', k' bashad.[X(G')=k']
rabeteie bein k va k' ra biabid?
 :o  :( :ninja:  :wink: :?:

behtare soal ro iejoore dige matrah konam ke ghabele fahm tar bashe.
farz konim G iek geraph va G' mokamele an bashad.
adade rangi rasi G va G' ra betertib ba X(G)=k va X(G')=k' neshan midahim.
1:behtarin karane bala va behtarin karane paiin,bar hasb v(G)[tedad ras haie G] ra biabid.
2:agar v(G)=n va pasokhe ghesmate ghabl be shekle
f(n)<=|k-k'|<=g(n) bashad angah hameie geraph haii ra ke
|k-k'|<=f(n) va |k-k'|<=g(n) ra biabid.(ba zekre dalil)
3:ghesmate 1 va 2 ra baraie k+k' bejaie |k-k'| hal konid.
 :huh:  :(  : :idea:  :maybe:

nodet to  hamayesh riazie mash had oomade boodi?:D

khanoom / aghaye soal konande!

adade rangi, tooye graph ha ye masleye pichdeyiye,
in hame zoor zadan, baraye graph haye mosatah
adade rangi, dar avordan.

mitooni motmaen bashi ke beyni X(G), va X(G')
dar halat kolli hich rabeteyi bargharar nist.

rabete ke ghatan hast.
in soalo marhum reza sadeghi hal kardan.
midooni ke ki hast?avvale keabe nazarie adade mirzakhani esmesh hast.
pas nabaiad naomid shod.
dar har hal mamnoon ke fek mikonin.


suppose $(C)$ is a circle and $F$ is a fixed point on it.$A,B$ are to variable points which

move clock wise,with the property of:

$FA.FB=K$(fixed)

the tangents to circle at points $A,B$, meets each other at $M$.

what is the locus of $M$.

[hide]i only could find the locus isnt circle[/hide]

suppose $L$ is a line and $F$ is a fixed point not on $L$. A,B are two variable points on $L$ inwhich:

$FA.FB=K$(fixed)

a circle is tangent to $L$ at $A$ and also goes through $F$.another one is tangent to $L$ at $B$,and goes through

$F$. name the second intersection of two circles $M$.

what is the locus of $M$?

[hide]i only could find the locus isnt line[/hide]

Actually the second problem is inverse of first problem wrt $F$ .

The locus isn't a line and a circle either.

[quote="amir2 pofiuz"]Yes I agree with MathX.

Why do you think on this problem when you know the locus isn't a Line and a circle Ashegh?

from the famous relation we have $FA.FB=2R.h_F \longrightarrow h_F=\text{Constant}$[/quote]

if u read  it carefully u could find why.

i said one of them isnt circle and the other one isnt a line.

the first one might be line and the second might be circle...

ok thanks dears... ;)

[quote="Ashegh"][quote="amir2 pofiuz"]Yes I agree with MathX.

Why do you think on this problem when you know the locus isn't a Line and a circle Ashegh?

from the famous relation we have $FA.FB=2R.h_F \longrightarrow h_F=\text{Constant}$[/quote]

if u read  it carefully u could find why.

i said one of them isnt circle and the other one isnt a line.

[/quote]

 :lol:  :D so when one of them isn't a circle so it's inverse isn't a circle either .


geometry
geometric transformation
reflection
algebra solved
algebra

Thirteen birds arrive and sit down in a plane. It's known that from each 5-tuple of birds, at least four birds sit on a circle. Determine the greatest $M \in \{1, 2, ..., 13\}$ such that from these 13 birds, at least $M$ birds sit on a circle, but not necessarily $M + 1$ birds sit on a circle. (prove that your $M$ is optimal)

(1) Suppose there are 5 points A1,A2,A3,A4,A5 and circle w, which contains these points.
Suppose B and C are not situated on w. Consider 5-tuple A1,A2,A3,B,C. We can discard from with 5-tuple only A1, A2 or A3 (otherwise, (A1,A2,A3,B \in w1 ==> w=w1 and B \in w1) or (A1,A2,A3,C \in w1 ==> w=w1 and C \in w1)). WLOG let w1 contains A1,A2,B and C. Then consider 5-tuple A3,A4,A5,B,C ==> WLOG w2 contains A3,A4,B and C. Finally consider A1,A3,A5,B,C ==> w3 contains {B,C and A1 and Aj (Aj not in w1)} or {B,C and A3 and Aj (Aj not in w2)} ==> w3=w1 or w3=w2 ==> w3 contains three points from {A1,A2,A3,A4,A5} ==> w3=w ==> B,C \in w.

Thus if (1) is true ==> 12 points necessarily are situated on a circle.

(2) Suppose there are no 5 points on the same circle.
Suppose A1,A2,A3,A4 \in w, A5,A6,A7,A8,A9,A10,A11,A12,A13 not \in w.
Consider all 5-tuples which contains three points from {A1,A2,A3,A4}, one point from {A5,A6,A7,A8,A9,A10,A11,A12} and A13.
We must discard from such 5-tuples one point which lies on w, otherwise w contains additional point from {A5,A6,A7,A8,A9,A10,A11,A12,A13}. So for each of C(4,3)*8=32 5-tuples exists circle wi, which contains A13 and intersects w in two points; and each wi is counted 4 or less times, otherwise there are 5 points on the same circle (verify it! for example, wi can arise from 4-tuple A1,A2,A13,Aj and from 4-tuple A3,A4,A13,Aj ==> it can't arise from A1,A3,A13,Aj or something like this) ==> 8 or greater distinct circles. Set {A1,A2,A3,A4} contains only 6(<8) two-points subsets ==> for some i,j circle wi contains the same points from w as wj, additionally A13 \in wi and A13 \in wj ==> wi=wj - it is contradiction.

Thus (2) is false and (1) is true ==> 12 points necessarily are situated on a circle, so M=12.

Excuse me my english.

Arne! Is my answer correct? I hope you have elegant solution.

Well, Myth ... that's not bad I must say :D
Your answer is entirely correct !
In fact, the official solution was even a bit more complicated I think ...
I'm not sure anymore, I will look it up.

Did you enjoy the problem ?

I didn't enjoy. This problem is too straightforward as my solution.

Do you mean you found it an easy problem ?

Ummm... How to explain? A simple problems is a problem without alternatives. My reflections moved in the following direction:
1) to construct nontrivial example - I failed;
2) to find any proper M
3) naturally I begun to examine M=5 and I realize that if M=5 then M=12
4) after all I must prove that there are 5 points on the same circle - I used well known idea that two circle intersects in two or less points


If $ a^{2k} \plus{} a^kb^k \plus{} b^{2k}$ can be factorized (in $ \mathbb{R}$), what do we know about $ k$?

I moved it:

http://www.mathlinks.ro/Forum/viewtopic.php?p=1711939#1711939

a^2k+b^2k+a^k b^K

(a^k+b^k)^2 - (ab)^k

this can be factorized when

k=2n form


1. We have a funtion $ f: D \to R$ such that $ \left\{\begin{array}{l} f(1) \equal{} 1 \\
f(n \plus{} 1) \equal{} f(n) \plus{} 2n \plus{} 1 \\
\end{array} \right.$ with $ n \equal{} 1,2,3........$ .

Prove that $ f(n) \equal{} n^2$

2. We have a funtion $ f: R \to R$ such that:
$ \left\{\begin{array}{l} f(1) \equal{} 1 \\
f(1) \plus{} f(2) \plus{} f(3) \plus{} ............ \plus{} f(n) \equal{} n^2 f(n) \\
\end{array} \right.$ with $ n \equal{} 2,3,4......$

Prove that $ f(n) \equal{} \frac {2}{{n(n \plus{} 1)}}$

********
[b][color=red]Note: Not use induction method! [/color][/b] :D  :D  :D

[hide="1"]This is begging for induction, but if you want to get around it you can say that by repeated use of the recursion, $ f(n)\equal{}1\plus{}3\plus{}\cdots \plus{}2n\minus{}1\equal{}n^{2}$ as desired.[/hide]

[hide="2"]We have

$ n^{2}f(n)\plus{}f(n\plus{}1)\equal{}(n\plus{}1)^{2}f(n\plus{}1)$
$ f(n\plus{}1)\equal{}\frac{n^{2}}{(n\plus{}1)^{2}\minus{}1}\cdot f(n)$
$ f(n\plus{}1)\equal{}\frac{n}{n\plus{}2}\cdot f(n)$

Now repeated use of the recursion gives a telescoping product:

$ f(n)\equal{}\frac{n\minus{}1}{n\plus{}1}\cdot\frac{n\minus{}2}{n}\cdots\frac{1}{3}\equal{}\frac{2}{n(n\plus{}1)}$[/hide]

[quote="matt276eagles"][hide="1"]This is begging for induction, but if you want to get around it you can say that by repeated use of the recursion, $ f(n) \equal{} 1 \plus{} 3 \plus{} \cdots \plus{} 2n \minus{} 1 \equal{} n^{2}$ as desired.[/hide][/quote]

Can you give me a detail solution!!! :)  :)

[quote="matt276eagles"][hide="2"]We have

$ n^{2}f(n) \plus{} f(n \plus{} 1) \equal{} (n \plus{} 1)^{2}f(n \plus{} 1)$
$ f(n \plus{} 1) \equal{} \frac {n^{2}}{(n \plus{} 1)^{2} \minus{} 1}\cdot f(n)$
$ f(n \plus{} 1) \equal{} \frac {n}{n \plus{} 2}\cdot f(n)$

Now repeated use of the recursion gives a telescoping product:

$ f(n) \equal{} \frac {n \minus{} 1}{n \plus{} 1}\cdot\frac {n \minus{} 2}{n}\cdots\frac {1}{3} \equal{} \frac {2}{n(n \plus{} 1)}$[/hide][/quote]

That's intelligent solution!!! :)  :)


function
geometry
3D geometry
Asymptote
calculus
derivative
algebra

[quote="king_23"]My question is how to find the criticals numbers for this function (solve for x for this particular function) and find the intervals of increasing and decreasing?

x^2 - 1/x[/quote]

Well,

$x^{2}-1/x = 0$
$x^{3}-1 = 0$

Difference of cubes $x^{3}-1 = (x-1)(x^{2}+x+1)$

Since the discriminant of $x^{2}+x+1$ is negative, the only root of this function is $x=1$, so the graph will cross the x-axis only there.

So we just have that function, except there is an asymptote at $x=0$, since that would cause the function to be undefined.  So this means the function will be split into two parts.  Also, this function will not be symmetrical to the y-axis, since it is an odd function.

If you want to find where a function is increasing and decreasing, you need calculus (I think).  But we can find where this range of function is positive and negative, by plugging in values and seeing where they go from positive to negative.

Thx for the help i managed to get the c.n i just have a question on how you get the intervals of increasing and decreasing. When you said plug in values im just wondering which values and which equation, the derivative of the function or the original equation?

Since you want increasing and decreasing, you're going to need calculus (I think).  I don't know much calculus yet, so you probably need someone else's help.


modular arithmetic
number theory open
number theory

Let $ k$ be a positive Integer, $ q_i,p$ be distinct primes

If $ \prod_{i\equal{}1}^{k}q_i \equiv a \mod(p^2)$  then

$ \prod_{i\equal{}1}^{k}(q_i \minus{}1) \not\equiv a \mod(p)$

[quote="wauwau"]Let $ k$ be a positive Integer, $ q_i,p$ be distinct primes

If $ \prod_{i \equal{} 1}^{k}q_i \equiv a \mod(p^2)$  then

$ \prod_{i \equal{} 1}^{k}(q_i \minus{} 1) \not\equiv a \mod(p)$[/quote]

Wrong : choose $ k\equal{}2,p\equal{}7,q_1\equal{}3,q_2\equal{}5$

$ q_1q_2\equal{}3\cdot 5\equiv 15\pmod{49}$
$ (q_1\minus{}1)(q_2\minus{}1)\equal{}2\cdot 4\equal{}8\equiv 15\pmod 7$

[quote="wauwau"]Are there any a for which this could be true?
(e.g. a=2 or  1<a<p) ?[/quote]

If $ a\equal{}0$, then proposition "$ \prod_{i \equal{} 1}^{k}q_i \equiv a \mod(p^2)$  implies  $ \prod_{i \equal{} 1}^{k}(q_i \minus{} 1) \not\equiv a \mod(p)$" is true (when $ q_i,p$ are distinct primes)  :D

You can always solve the equations $ xyz\equal{}a$ and $ (x\minus{}1)(y\minus{}1)(z\minus{}1)\equal{}a$ in $ \mathbb Z / (p)$. And more isn't necessary if you take a closer look.


geometry
algebra
polynomial
Vieta
perimeter
number theory
relatively prime

The lengths of the sides of a triangle are the roots of the equation $ 7x^{3}+35x=28x^{2}+13$. If the square of the area of the triangle is $ \frac{p}{q}$, where $ p$ and $ q$ are relatively prime positive integers, what is $ p+q$?

[hide]
By heron's theorem:

$ A^{2}= s(s-a)(s-b)(s-c)$ in a triangle

expanding we have

$ s[s^{3}-(a+b+c)s^{2}+(ab+bc+ac)s-abc]$

$ s[s^{3}-\frac{s^{3}}{2}+(ab+bc+ac)s-abc]$

$ s[\frac{s^{3}}{2}+(ab+bc+ac)s-abc]$

back to our cubic polynomial
$ 7x^{3}-28x^{2}+35x-13$ with roots a,b,c

using vieta's,
$ abc = \frac{13}{7}$
$ ab+bc+ac = \frac{35}{7}= 5$
$ a+b+c = \frac{28}{7}= 4 \implies s = 2$

dropping all that good stuff in

$ 2[\frac{8}{2}+(5)2-\frac{13}{7}]$

$ 28-\frac{26}{7}$

$ \frac{170}{7}$

so $ \boxed{p+q = 177}$
[/hide]

[hide]From Vieta's formulas, we know that $ a+b+c=-\frac{-28}{7}=4$, so thus, the semi-perimeter ($ s$) of $ \triangle ABC$ is 2.
We also know that:
$ ab+ac+bc=\frac{35}7=5$
$ abc=-\frac{-13}7=\frac{13}7$

By Herons formula,

$ \noindent[\triangle ABC]^{2}=2(2-a)(2-b)(2-c)=16-8(a+b+c)+4(ab+ac+bc)-2abc=16-32+20-\frac{26}7=\frac27$

Thus, $ p+q=2+7=9$[/hide]

@chickendude:  $ a+b+c\neq\frac{s}2$

Note that $ p(s)=7*(s-a)(s-b)(s-c)$ so you can just plug $ s$ into the function.


How would I draw a free body diagram for the below?

[img]http://i27.tinypic.com/2d0nfol.jpg[/img]

The rope is connected to the circumference of the sphere and is massless

You have the tension, which have the direction of the rope, the weight of the sphere (vertical), and the reaction of the wall (horizontal).

i used paint :)

cut the pciture - save as jpeg - http://www.tinypic.url
upload the pic and paste here


I'm not sure if this is totally latex-related, but how would I latex a lot of .tex files at a time? I tried [i]latex *.tex[/i] from the terminal, but it only latexed one file..

Er, which operating system are you using as that makes a lot of difference? In Windows the DOS command would look something like
On the command line:
[code]for %f in (*.tex) do latex %f[/code]but in a batch file
[code]for %%f in (*.tex) do latex %%f[/code]See [url=http://www.computerhope.com/forhlp.htm]Microsoft DOS for command[/url] for example and lots of other places on the net.
If it's not Windows then look up whatever shell you are using in that operating system.

Oh, and wouldn't you prefer to use pdflatex to get pdf files?

[url=http://www.cyberciti.biz/faq/bash-loop-over-file/]Bash Shell Loop Over Set of Files[/url] amongst lots of other places.


[color=purple]Let ABC is an acute triangle and define its element normally.Prove that
  $\sqrt{ab+bc+ca}+\frac{1}{4}(|a-b|+|b-c|+|c-a|)\ge 3R$
  Have fun ! :D [/color]

[color=purple] I think it's quite hard but any one can post the solution ?
   P/S:  It's of Le Dinh Huy a VietNamese [/color]


MIT college Princeton Stanford trigonometry calculus AMC

Hey.  I'm going to be a junior in highschool this year, and I'd like some advice about college.
I've been thinking about colleges that I may want to attend, and some of these colleges include prestigious ones, such as MIT, Princeton, Stanford, ect.  Considering my chances at getting into such a college, I believe that academically i could qualify, but I am unsure about my extracurricular activities.  I don't have any serious commitments to school activities (I've only been a math tutor, in french club, and in science club).  Although my grades are almost all 100's, I don't think I have any distinguishing qualities.
However, this past year I have become almost obsessively interested in mathematics and took trigonometry and calculus in the summer.
As I look at the situation now, I still have time to do some more extracurricular activities... the only problem is i wouldn't enjoy them.  I would much rather be spending my extra time learning math.  I suspect someone will say that I should try to make USAMO or something, but I doubt I could this year since I am probably far behind most people who make it.  Making USAMO senior year is feasible, but that won't really appear on my college applicatioin.

So what I'm really asking is should I do what I like to do or sacrifice time to improve my college application?

the two arent necessarily mutually exclusive.  i think it helps a lot to have a passion for something, whatever it is, and pursue it.  if it's math, make it clear to the college that you love math and go out of your way to find related activities/contests.  Do lots of math and prepare for usamo, you never know you might surprise urself and if nothing else at least you'll get a decent aime score to show colleges.  dont bother doing some extr acurricular that you dont really care about, its a waste of ur time and colleges can probablly see through that.

not everyone that gets into MIT, etc are USAMO qualifiers. not even the ones that are passionate about math. in FACT, [b]NEITHER[/b] of the guys i know that got into MIT were USAMO qualifiers. both of them made AIME at least twice (and may it be noted that as a freshman, i beat both of their sorry butts) and both of them did well. they took every upper-level math course our school offered. they performed well in local math team competitions, and neither of them went out-of-state to perform. one of them was also passionate about aerospace engineering and did the NASA SHARP program. the other was also into tennis, and was a large factor in his tennis program in his hometown. one held a lot of leadership positions within the school. the other held a few, but that wasnt really his thing. he did, however, attend the PROMYS program (loved it loved it loved it). let it also be noted that while one's major passion was DEFINITELY mathematics, he still made a B in AP calculus one year (and not because he couldn't do it). volunteerism was also a really big thing with both guys- tutoring, cleaning places up, even soup kitchen sometimes- they logged a lot of free hours.

so, in summary, from these two case studies:
1. have one or two things you are really passionate about, and GO OUTSIDE of your school to do them. do internships or camps or programs over the summer- this really shows how much you want to do whatever it is you like.
2. you can have slip-ups on occasion. nobody's perfect, don't have a stess attack.
3. volunteer!!!!! bonus points if you can do #3 AND #1 at the same time.

because you're a junior, even if you do qualify for USAMO senior year, it'll be too late for that to affect you college application. also, you don't really need an extraordinary number of extracurricular activities, you just have to show a lot of comittment to something.

MIT has actually revised its application form to reduce the number of extracurricular activities you can list on it, all to show that it cares about what YOU are passionate about, not how many little activities you can cram onto a laundry list. Dare to be yourself, and you will stand out from a lot of other college applicants. Too many applicants spend their senior years, junior years, sophomore years, and even earlier worrying only about what looks good on a college application, instead of thinking, "Who am I? What do I really enjoy doing well?"

Good luck on your applications. It sounds like you are off to a good start.

if you dont think you have the time to work to make usamo, something you might wish to consider is doing math research and entering it into a competition such as siemen's westinghouse.

Whoa a lot of respones already....
I think the general response confirms my idea that I should not waste time on things I don't care about.

To tokenadult and cats, thanks for the encouraging respones.

bleumoose, it seems i have number one of your points covered, as I plan to continue studying math outside of school and either taking some interesting classes or going to a camp next summer.  About volunteering, I suppose I could do a little, but it wouldn't really be because I am a good person that wants to do it; it'd only be to help with college.  Although I'll consider the idea, I don't really see it happening.

h78reg, I did happen to mention that I may be able to make USAMO senior year- at the same time i mentioned that I knew it wouldn't affect my college application, as you said.  I still think it would be a good experience, even if it wouldn't help with college.

And finally, ffdbzathf, your suggestion sounds interesting and I will look into such competitions.

My advice would be to enjoy your (remaining) high school years, not live them as a 2-year-long college application.  I thought very little about my applications until just before they were due.  Now I'm not encouraging procrastination--I do regret not starting to fill them out a little earlier--but the beginning of your junior year is way too early, in my opinion, to be thinking about what looks good on college apps.

Although I've learned so much about the college process through the 2 years that ive been paying attention to it, I dont have much to offer because I've pretty much learned it all from this site, and what bluemoose just said is probably the best post ive ever seen on it, encompassing everything you need to know.
One thing to keep in mind though is that doing good research takes a lot more time, is a lot harder, and unless you find something that you're really interested in, a lot less enjoyeable than preparing to make the USAMO. Even if you dont make the usamo, the things that you learn in the process, the hardships that you would have gone through, and the passion you'd develop would all have made it worthwhile.

IMO, if you don't like to do community service and can't find a way to do it in a way that you would like, then suck it up and just do it. Sometimes you have to do thing that you don't want to do, no matter how much people want to say 'do what you like' it doesn' take away from the fact that you are competing with thousands of other well qualified applicants, and I'm not saying that you need to go out and do a million extra curriculars and make the USAMO, I'm just saying that you've got to prove to the college that you deserve to go there. If you feel that your current achievements are good and like them that way, then stay, but sometimes you just have to do things that you don't want to do, I wish I could get into a great college just having a passion for math, but that's not how the real world works.

and if it helps any, i always have way more fun volunteering than working.

are sports valued about the same as other extracurricular activities?

Usually sports require more time and dedication than other extracurriculars.  Unless it is something like math team where you have shown great dedication the sport will be more greatly valued.

I'm actually quite surprised how much sports seem to be valued by MIT in recent years.


A game is played as follows:

There are 3 piles of coins, each pile having $ a$, $ b$, and $ c$ coins, respectively. Two players take turns taking away coins from the piles. They can either take away 1 coin, or they can take away all of the coins in one pile. The player who takes the last coin loses. For which $ (a, b, c)$ does player 1 win?

You haven't made it clear who the winner of the game is...

i sopose both players do the best move possible....

player 1 will lose if  all a,b and c are odd...

player 2 would only have to make sure that the number of coins that are left when he starts the turn isn't odd...

....

about your question ... i don't know  :( ... i'm still thinking... :lol:

It is a little easier if you solve first the one pile version of this game, then the two pile version, then the three pile version.  Also, it might be a little easier to characterize the winning positions for the second player (which of course is equivalent; Wichking has some, but not all of them).

hm ... player 2 can win if any of piles has odd number coins

No: starting with $ 1, 1, 0$ is a sure win for player 1.

Uh, for starters,

If $ b \equal{} c \equal{} 0$, then if $ a$ is even, 1 wins. If $ a$ is odd, 2 wins.

If $ c \equal{} 0$, then if $ a,b$ is even, odd, AND odd,odd, 1 wins (just remove a pile to leave one odd pile).

If $ a,b$ is even,even, 2 wins (if you remove one pile, you lose.  If you remove one, he'll remove the other pile, and you lose).

For $ a,b,c \neq 0$, if there are two even piles, 1 should remove one whole stack to guarantee win (because it reduces down to the $ c \equal{} 0$ case with even,even and he becomes player 2.) [this takes care of even even even, odd even even]

When you have odd,odd,odd, you're guaranteed a lose, because if you take one pile, he'll win (look at $ c \equal{} 0$), and if you take one, he'll just continue to remove one, until you remove the last piece from one column, then he wins (look at $ c \equal{} 0$ for odd,odd).

When you have odd, odd, even,  removing one from the even guarantees a win.

Uh...I haven't double checked this, AND, if you say something like, what about $ (1,1,1)$, removing one is the same as removing a whole stack.

I'll finish this once I finish dinner. brb.


According to my textbook, there are two types of hyperbolas:

(1) for horizontal transverse axes

(2) for vertical transverse axes

The math book provides the following two different hyperbolas equations:

(x - h)^2/a^2 - (y - k)^2/b^2 = 1

OR

(y - k)^2/a^2 - (x - h)^2/b^2 = 1, IF

c = sqrt{a^2 + b^2}

In the math book we find this data:

length of transverse axis = 2a.

length of conjugate axis = ab.

The distance from the center to a focus is c.

The center, of course, being the point (h, k).

Here is a question that the author of the book did not explain very well.

[b]A hyperbola whose foci are (0, 3 - sqrt{10}) and (0,3 + sqrt{10}) has a conjugate axis of length 2.  Write the equation of the hyperbola in standard form and graph it.[/b]

This is how the author of the book explained the SOLUTION:

"The center of the segment whose endpoints are the two given foci is (0, 3). So, h = 0 and k = 3."

My question is:

How did he find that h = 0 and k = 3 from the given foci?

He did not explain this at all.

He went on to say:

"Since you need to to go either sqrt{10} units up or down from the center to reach one of the foci, c = sqrt{10}."

I do not understand how c = sqrt{10}.

Can someone explain this to me another way?

Of course, we need to find a, b, c, h and k to be able to plug all this stuff into one of the hyperbolas given above.

This would give me the equation that I am looking for.

Please, explain and solve.

=========================================

Also, let's PRETEND the question provided two vertices (not two foci), the length of the transverse axis and/or the length of the conjugate axis.

We are told to write an equation given this data.  How would I find the correct hyperbola equation then?

Well it's sort of self-explanatory.  You should know the equation of a circle is given by $ (x\minus{}h)^{2}\plus{}(y\minus{}k)^{2}\equal{}r^{2}$ and you should know that $ (h,k)$ is the center of the circle.  Well in the case of the hyperbola, it's similar.  The point $ (0,3)$ is the center of the hyperbola given because both foci are centered around that point, separated by some distance $ c$.  That is, foci=$ (h\pm c,k)$ or $ (h,k\pm c)$.  In your case, we have the latter, which happens to be $ \sqrt{10}$.  I hope I clarified it a little.

The center is the midpoint of the conjugate axis, by symmetry. So to find the center, you need to find the midpoint of the conjugate axis or the midpoint of the transverse axis.
By the definition you gave, $ c$ is the distance from either focus to the center. If you know where the foci and center are, you can calculate $ c$.


thank you arqady;
I think :

if $ 0<a<b$ , prove that:
$ a^3 \minus{} 3a \minus{} 2 \le b^3 \minus{} 3b \plus{} 2$

[quote="mathson"]if $ a < b$ , prove that:
$ a^3 \minus{} 3a \minus{} 2 < b^3 \minus{} 3b \plus{} 2$[/quote]
Try $ a\equal{}\minus{}1$ and $ b\equal{}1.$ :wink:

[quote="arqady"][quote="mathson"]if $ a < b$ , prove that:
$ a^3 \minus{} 3a \minus{} 2 < b^3 \minus{} 3b \plus{} 2$[/quote]
Try $ a \equal{} \minus{} 1$ and $ b \equal{} 1.$ :wink:[/quote]

pravo arqady;
now look to the problem again.

[quote="mathson"]if $ a < b$ , prove that:
$ a^3 \minus{} 3a \minus{} 2 \leq b^3 \minus{} 3b \plus{} 2$[/quote]

OK. Let $ b \equal{} a \plus{} x.$ Then $ x > 0$ and $ a^3 \minus{} 3a \minus{} 2 \leq b^3 \minus{} 3b \plus{} 2\Leftrightarrow$
$ x\left(a^2 \plus{} a(a \plus{} x) \plus{} (a \plus{} x)^2\right) \minus{} 3x \plus{} 4\geq0\Leftrightarrow3xa^2 \plus{} 3x^2a \plus{} x^3 \minus{} 3x \plus{} 4\geq0.$
Thus, it remains to prove that $ 9x^4 \minus{} 12x(x^3 \minus{} 3x \plus{} 4)\leq0,$ which is equivalent to
$ \minus{} 3x(x \plus{} 4)(x \minus{} 2)^2\leq0,$  which is true.
The equality holds when $ x \equal{} 2$ and $ a \equal{} \minus{} 1.$


In how many ways can you locate 8 married couples in line with the condition that every husband is next to his wife?

Is it correct: 8! * 2!

In how many ways can you locate 8 married couples around a circular table with the condition that every husband is next to his wife?

Answer: 7! * 2!

The university has to cover 10 algebra courses each with a professor and they have 10 professors

In how many ways can you distribute them?

Answer: 10!

Are this answers correct? What do you think?

[quote="sanchez45"]In how many ways can you locate 8 married couples in line with the condition that every husband is next to his wife?

Is it correct: 8! * 2![/quote]
[hide]Consider each couple as a single object. There are $ 8!$ ways of placing the couples in a line. There are two possible arrangements (wife, husband; husband, wife) for each couple, so multiply by $ 2^8$. The total number of arrangements is $ \boxed{8!\cdot2^8}$.[/hide]
[quote="sanchez45"]In how many ways can you locate 8 married couples around a circular table with the condition that every husband is next to his wife?

Answer: 7! * 2![/quote]
[hide]Same as the above question, divided by $ 8$. Total arrangements: $ \boxed{7!\cdot2^8}$.[/hide]
[quote="sanchez45"]The university has to cover 10 algebra courses each with a professor and they have 10 professors

In how many ways can you distribute them?

Answer: 10![/quote]
[hide]There are $ \boxed{10!}$ ways to permute $ 10$ professors.[/hide]


If cosx + sinx = 2/3, find sin2x.

Heres how I did it:

\[ (\sin x \plus{} \cos x) \equal{} (\frac{1\plus{}\sqrt{3}}{2})^2\]
\[ \sin^2 x \plus{} 2\sin x\cos x \plus{} \cos^2 x \equal{} \frac{1\plus{}2\sqrt{3} \plus{} 3}{4}\]
\[ 1 \plus{} 2\sin x\cos x \equal{} \frac{4\plus{}2\sqrt{3}}{4}\]
\[ 1 \plus{} \sin 2x \equal{} \frac{2\plus{}\sqrt{3}}{2}\]
\[ \sin 2x \equal{} \frac{\sqrt{3}}{2}\]

However, by solving the equation on the calculator, x = 30 degrees.  \[ \cos 30^\circ \plus{} \sin 30^\circ\] does not equal \[ \frac{2}{3}\].

What am I doing wrong?  Thanks.

$ (\sin x \plus{} \cos x)^2 \equal{} (2/3)^2$

$ \sin^2 x \plus{} \cos^2 x \plus{} 2\sin x \cos x \equal{} 4/9$

Since $ \sin^2 x \plus{} \cos^2 x \equal{} 1$ and $ 2\sin x \cos x \equal{} \sin 2x$, we have

$ 1 \plus{} \sin 2x \equal{} 4/9$

$ \therefore \sin 2x \equal{} \boxed{ \minus{} 5/9}$

I don't know where you got the $ \frac{1 \plus{} \sqrt{3}}{2}$ from.

[quote="modularmarc101"]$ (\sin x \plus{} \cos x)^2 \equal{} (2/3)^2$

$ \sin^2 x \plus{} \cos^2 x \plus{} 2\sin x \cos x \equal{} 4/9$

Since $ \sin^2 x \plus{} \cos^2 x \equal{} 1$ and $ 2\sin x \cos x \equal{} \sin 2x$, we have

$ 1 \plus{} \sin 2x \equal{} 4/9$

$ \therefore \sin 2x \equal{} \boxed{ \minus{} 5/9}$

I don't know where you got the $ \frac {1 \plus{} \sqrt {3}}{2}$ from.[/quote]

wooooooow /facepalm for looking at the question below.  happens when you have 40 questions squished on a page

Thanks :P.


whould the red line be considered a diagonal?

A line segment connecting two nonadjacent polygon vertices of a polygon.

According to this description, yes, the red line would be considered a diagonal.

Remember, a quadrilateral must have 2 diagonals.

Yes, this would be considered a diagonal.

In the diagram, there are two diagonals. One of them "bisects" the quadrilateral.

[quote="4everwise"]That has already been stated.[/quote]
oh, sorry i didn't look at the previous posts even though i should have.  Sorry.

[quote]this is a diagonal. It connects two non-adjacnet verticies.[/quote]

but another property of a diagonal is that it has to be inside of convex polygons...

so which one's correct :?:

[quote="4everwise"]That has already been stated.[/quote]
First of all that was  :spam:

Yes. A diagonal is any line that connects 2 nonajacent veticies of a polygon.

[i]When writing word problem use Albatross and Bobert.[/i]


If $ a$,$ b$,$ c$ are positive integers such that $ 0 < a^{2} \plus{} b^{2} \minus{} abc \le c$,show that $ a^{2} \plus{} b^{2} \minus{} abc$ is a perfect square !

This problem is sloved by Vieta Jumbling
View http://www.mathlinks.ro/viewtopic.php?p=849193#p849193 :P


Let $n=1667$.  Then the first nonzero digit in the decimal expansion of $\sqrt{n^{2}+1}-n$ reading from left to right is?

Here is what I have:
[hide]
$\sqrt{n^{2}+1}-{\sqrt{n^{2}}}$
$=\frac{1}{\sqrt{n^{2}+1}+\sqrt{n^{2}}}$
$\approx\frac{1}{2n}= \frac{1}{3334}$

Recall that $\frac{1}{3.333333...}= 0.3$
Divide this by 1000 to get $\frac{1}{3333.33333333...}= 0.0003$

Since $3334>3333.3333...$, $\frac{1}{3334}< 0.0003$

Repeat this process only with $\frac{1}{5}$ to know that $\frac{1}{3334}> 0.0002$

Therefore, $0.0002 < \frac{1}{3334}< 0.0003$, and the first nonzero digit is $2$.

edit: To clarify, we can safely approximate $=\frac{1}{\sqrt{n^{2}+1}+\sqrt{n^{2}}}$ as $\frac{1}{2n}$ because the 1 under the radical only adds an insignificant value to the denominator, while you would need a significant amount of increase to the denominator to make $=\frac{1}{\sqrt{n^{2}+1}+\sqrt{n^{2}}}< 0.0002$[/hide]


oh, thanks!

and one more:

how can i connect a point on coordinate (1,2,3) and the triangle's vertex?

Point projections often give you the "camera is too close to the object" error. I'm not sure how exactly they are defined (the manual is rather vague at this point) but it seems that they just project to the plane perpendicular to the direction to the "center" of the picture (whatever that means) at the distance somehow related to the picture size and the distance to the picture, and, if the picture intersects this plane, you get an error (this description may be inaccurate in details but is, probably, adequate as far as the general idea is concerned). So, just move your point (9,3,4) away a bit and it'll work.


Let $ \overrightarrow{a},\ \overrightarrow{b}$ be vectors such that  $ |\overrightarrow{a}\plus{}2\overrightarrow{b}|\equal{}1,\ |\minus{}3\overrightarrow{a}\plus{}\overrightarrow{b}|\equal{}1$, then find the maximum value of $ |\overrightarrow{a}\plus{}\overrightarrow{b}|.$

\[ |\overrightarrow{a} \plus{} \overrightarrow{b}| \equal{} | \scriptstyle{\frac {4}{7}} \left( \overrightarrow{a} \plus{} 2\overrightarrow{b}\right) \minus{} \scriptstyle{\frac {1}{7}} \left( \minus{} 3\overrightarrow{a} \plus{} \overrightarrow{b}\right)|\le \scriptstyle{\frac {5}{7}}\]

Yeh, that's a super solution using triangle inequality. :lol:


Short for robot, or in this case some sort of automated user. Bots are generally only for spam.

They come installed on the site. They do the functions assigned to them.

The bots listed as online users here are search indexing programs; they read threads to index them, but never post.


from usamts website:

"As opposed to most mathematics competitions, the USAMTS allows students a full month to work out their solutions."

and the deadline for the first one is october 14, so we have two months for the first one?  am i hallucinating [uh, august, september, october... okay i counted right], or are we all blessed with the gift of lots of time? :)

yeah, the first round typically gives you more time.


Let $a,b,c,d$ be positive real numbers such that $abcd=1$

Prove that

$\frac{1}{a(1+b)} +\frac{1}{b(1+c)} +\frac{1}{c(1+d)} +\frac{1}{d(1+a)} \geq 2$

Put $a=\frac{x}{w}, b=\frac{y}{x}, c=\frac{z}{y}, d=\frac{w}{z}$.
The given ineq becomes
$\sum_{cycl}\frac{w}{x+y} \ge 2 ,$
which can be easily proved using Cauchy-Schwarz.


function
algebra
domain
trigonometry
calculus
integration
algebra unsolved

I'm just wondering when a function f has an infinite number of zeros, for example if f(x^2) = 0 for x any positive integer, does it necessarily mean f is the zero function? :?

Edit:
This question is from Canadian MO 2002
Let X be the set of non-negative integers. Find all functions f: X → X such that x f(y) + y f(x) = (x + y) f(x^2 + y^2) for all x, y.

Can we just say since f(x^2) = f(0) for all x in the domain of f, then f is the constant function?

$f(x)=\sin x$.  If you want it to take on integral values at the integers, $f(x)=\sin \pi x$.  If you want it to take on nonnegative values at the integers, $f(x)=\vert \sin \pi x\vert$

Ah I must've thought all functions are polynomials. :blush:   Thanks.  :)


Evaluate

$\displaystyle\lim_{x\to\infty}\frac{tan^3(2x)}{x^3}$

and

$\displaystyle\lim_{x\to{0}}\frac{tan^3(2x)}{x^3}$

[quote="maokid7"]$\displaystyle\lim_{x\to\infty}\frac{tan^3(2x)}{x^3}$
[/quote]

This limit doesn't exist, as $\lim_{t\to\infty}{\frac{\tan t}{t}}$ doesn't exist.


[hide]First, let us express .76767676...  as x. The repeating pattern is a 2 digits (76), so you multiply by 10^2 (100). Therefore, 100x=76.767676... 100x-x=76, so 99x=76 and x=76/99.[/hide]

[quote="mysmartmouth"]Hey,

What is .767676... expressed as a fraction?

Also,
I just got GMAIL from a very generous person, and I have 50 invites to spread.  I noticed almost everyone already has GMAIL here, but if you need GMAIL, let me know.[/quote]

[hide="Answer"]
76/99
[/hide]
And for gmail, i have 150 invites. :P

[hide="Answer but no solution cuz I am way too lazy"]
76/99
[/hide]

I think the whole point of gmail by invitation only is to keep out auto-created accounts. (You know how you have to copy a string of random letters/digits to make a hotmail account?  same thing).  Because anyone who wants it can get GMAIL, by posting a request on Google, or posting a message on any board.

That was g-mails intended purpose.  They may never release it to the public, when they do...All of those new people will more than likely start out with a very low amount of webspace that gradually increases over a period of time.

If you look G-Mail now is up to 2.6 gigs....And it is increasing every second!

What I really dont get is why it's there. I mean look at it!

They even give free pop3 access.  So I can just junk this other crappy email I have...

I think you meant 2.06 (It's 2.068 GB now)

[quote="mathgeek2006"] 1 GB=1024 MB.[/quote]

I never knew that! I always assumed it would be 1000 MB.

[quote="bob123"][quote="mathgeek2006"]BTW, you got it wrong.  1 GB=1024 MB.  so it is 2.0205078125GB.[/quote]

I know it is, just that most people use it as 1000 so it's easier.[/quote]

Well, most people on this frum are fairly aquainted with the powers of 2.

1
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384
32768
65536.........

As far as I have memorized........


let k : positive integer
Prove that
$\frac{\sqrt{k}}{\sqrt{k}+\sqrt{k-1}}+\frac{\sqrt{k}}{\sqrt{k}+\sqrt{k+1}}\geq\frac{1}{4k}$
thanks

Inequality is equivalent to $\sqrt{k}(\sqrt{k}-\sqrt{k-1})+\sqrt{k}(\sqrt{k+1}-\sqrt{k})\geq\frac{1}{4k}$ or $\sqrt{k+1}-\sqrt{k-1}\geq \frac{1}{4k\sqrt{k}}$ or $8k\sqrt{k}\geq \sqrt{k+1}+\sqrt{k-1}$ or $32k^{3}-k\geq \sqrt{k^{2}-1}$. This is trivial by $32k^{3}-k>k>\sqrt{k^{2}-1}$.


compare the two aimes, which one is easier?

I definitely thought 2003 was easier(I solved the first 9 on AIME I in 50 minutesin in practice), while I only solved 9 total on 2005 here :(

Actually, there was only 2 problems that I couldn't have solved(#10, 14), everything else that I didn't get were  pretty easy(#9, 12, 13, 15), well maybe not #13 because it took me like 10 recursion graphs to get the right answer.

2005 was one of the hardest AIME ever :) , on average I solve 11 questions on old practice AIME's.

lol... everybody is saying that 2005 is easy... but ppl always do better on practice than in the real thing.

[quote]2005 was one of the hardest AIME ever, on average I solve 11 questions on old practice AIME's.[/quote]
Is that a typo? People ALWAYS do better in practices... there is less pressure and you're doing the test when you feel like doing it: thus, you're in the zone...

I just glanced over older AIMEs and they are MUCH harder.. take for example even the "easy" 2004 exams.

On mathcounts practices (I hav states tomorrow) I score a 29 or 30 sprint nearly every time, but when the real thing comes, my scores have never been that good. It's just much easier to take anything but the SATs (since they're so monotonous to take at home) if not everything at home... and there's no pressure when you're practicing.

[quote="hockeyadi23"][quote]2005 was one of the hardest AIME ever, on average I solve 11 questions on old practice AIME's.[/quote]
Is that a typo? People ALWAYS do better in practices... there is less pressure and you're doing the test when you feel like doing it: thus, you're in the zone...
[/quote]

Right, my question is: how to not let pressure affect you? I couldn't deal with it on Tuesday unfortunately :(

[quote="hockeyadi23"][quote]2005 was one of the hardest AIME ever, on average I solve 11 questions on old practice AIME's.[/quote]
Is that a typo? People [b]ALWAYS[/b] do better in practices... there is less pressure and you're doing the test when you feel like doing it: thus, you're in the zone...
[/quote]

ALWAYS, eh?  [b]NONE[/b] of my practice AIME's even came close to my score on the real thing....

[quote="darktreb"]Just because you were pressured doesn't make 2005 the hardest AIME of all time.[/quote]

Yeah, I regret saying it was, I just realized actually it was pretty easy(easier than last year's). Ah first year doing real AIME, I didn't realize how much pressure real AIME test is. Hope I'll do better next year.

[quote]On mathcounts practices (I hav states tomorrow) I score a 29 or 30 sprint nearly every time, but when the real thing comes, my scores have never been that good. It's just much easier to take anything but the SATs (since they're so monotonous to take at home) if not everything at home... and there's no pressure when you're practicing.[/quote]
thats funny.
back in the day I always did better on contest day than practices

well mebbe thats cuz i dont concentrate.  :D

[quote]ALWAYS, eh? NONE of my practice AIME's even came close to my score on the real thing....[/quote]

More indication that it was the easiest AIME of all time, as I have said several times.

to me this is easier..although i only get 6,but mainly it's due to my stupid mistake in 3,6,8,10 and12....i usually can only solve 7-8 question on practice,but this time 11 out of 15,at least i am on the right track.....

Hmm...I recall scoring...well...not very well on the 2003 AIME. But in terms of diffculty, I think they're wow...roughly the same? This one produced higher scores though, imo.

[quote="soulzmischief"]My pratices always come out worse than the real thing. I did 12 this time (11c, 1w). On practices, I usually get about 12-13 problems, but I make a lot of stupid mistakes, because there's little motivation for me to do well, and I usually miss about 2 or 3 problems.[/quote]

True. I haven't really done a real AIME at full level yet, but during practices, I usually just nod off after an hour and a half and start listening to music or playing games as I check or finish some of the problems. But in past math contests, I always check my work to make sure I did everything correctly and focus in well.

i'm not too good at AIME-style questions yet i felt this year's was a little easy. actually, what made it easy was that so many of the questions could be solved by brute force instead of finding the elegant solution(#'s 4,5, and 13 for example). i managed to get a 4 and i should have gotten a 5 but i couldn't count straight on #13  :wacko:

Hahaha, I tend to spin my pencil and play around with my ruler as well.

For some reason 2003 seemed much harder when I took it (well... it was 2 years ago).  In addition, the AIME cutoff was an 8 that year.  I think this year's will be much higher than that.

Incidentally, the 2003 AIME was a lesson to those people that don't bring a protractor to the test.  I talked to a lot of people who got the angle problem (#9?) just by measuring it on the diagram.

This year it wasn't that way. People who used graph paper and a protractor to do #14 got 960 and 963. Finally they learn not to make it drawn to scale. Numerous AMC 12 questions this year were drawn to scale so many people were able to just measure them using their pencil, eraser, etc.

[quote="qweretyq"]practice avg: 3-4
score 2005:  7 :)[/quote]

For me, even more drastic:
practice: 1-3
score 2005: 8  :D
Hmm.. I seem to have developed an adrenaline-handling mechanism over the years since Mathcounts.
I haven't been mathematically stimulated at all since then... that might be why.

Another thing... on long tests ya maybe people start to just doze off after the first hour, wondering who won the playoffs or something, or how bored you are... but if you break the test up into 1 or 1 1/2 hour segments then you'll really try. I know what some of you mean, for example I rarely like to check while practicing, b/c It's a bit hard to concentrate... but I found that practicing checking helps...

I take back the always thing, but if you honestly do the test and don't pull a yifan or something  :)  then ya you'll probably do better.


hallo,
prove that the equation \[x^{2}+y^{2}+z^{2}=59^{n}\] has infinite solutions for every $n \in N$.  :)

There's something wrong with the problem, because when the $n$ is fixed it's pretty obvious that the number of solutions is finite.

The 3 squares theorem kills it (assumed that $n$ is allowed to vary).
We can even see $\mod 16$ that $x,y,z$ are all $\neq 0$ for $n \equiv 1 \mod 4$.

[quote="aviateurpilot"]$nombre\ of\ solutions\ is\ <(1+59^{\frac{n}{2}})^{3}$[/quote]
hallo,
i actually mean that for every natural number $n$ you can find solutions. because there are infinite natural numbers you can find infinite solutions.  :)

[quote="modeler"]The equation definitely has infinite solutions, because x, y, z are not bounded to be integers! :D[/quote]
hallo,
o sorry i forgot to say that $x,y,z \in Z$  :oops:


group theory
abstract algebra
algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Let $ L=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $ M=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$. Note that $ L\subseteq M$.

a)Prove that $ \sqrt{5}\not\in L$.
b)Prove that $ [M: \mathbb{Q}]=8$, that $ M$ is a normal extension of $ \mathbb{Q}$, and that every element of $ Aut(M/\mathbb{Q})$ has order $ 1$ or $ 2$.
c)Dertemine the number of distinct subfields $ F$ of $ M$ such that $ [F: \mathbb{Q}]=4$.

Please post your solution!  :wink:

a) Suppose $ \sqrt{5}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then $ \sqrt{5}=a\sqrt{2}+b\sqrt{3}$ where $ a,b\in\mathbb{Q}$. But this means that $ 5=2a^{2}+3b^{2}+2ab\sqrt{2}\sqrt{3}$ or that $ \sqrt{2}\sqrt{3}=\frac{5-2a^{2}-3b^{2}}{2ab}$ is rational. A contradiction.

b) Can we use the Tower Theorem? If so, then $ [M: \mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}): \mathbb{Q}(\sqrt{2},\sqrt{3})]\cdot [\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})]\cdot [\mathbb{Q}(\sqrt{2}) : \mathbb{Q}]$ and each of those can be seen to be $ 2$ by examining the order of the minimal polynomial. So the whole order has extension degree $ 8$.

Your solution for part a) is not necessarily correct. $ L$ also contains $ \sqrt{6}$, so it wouldn't have to be that $ \sqrt{5}=a\sqrt{2}+b\sqrt{3}$. I would do this instead: it is not difficult to show that if $ p,q$ are distinct primes, then $ \mathbb{Q}(\sqrt{p})\neq\mathbb{Q}(\sqrt{q})$. It follows that $ |\mbox{Aut}(L/\mathbb{Q})|=4$. If $ \sqrt{5}\in L$, then $ L$ has at least four distinct subfields of index $ 2$. This includes $ \mathbb{Q}(\sqrt{6})$, which is distinct from the others because $ 6,5,3,2$ are all squarefree. Since $ L$ is a normal extension, this would mean that $ \mbox{Aut}(L/\mathbb{Q})$ contains $ 4$ distinct subgroups of order $ 2$. This is not possible, since it only contains $ 4$ elements, one of which must be the identity.

b) follows simply from a). We know that $ [L\colon\mathbb{Q}]=4$, and so we have $ [M\colon\mathbb{Q}]=[L(\sqrt{5})\colon L]\cdot 4=2\cdot 4=8$. The fact that $ [L(\sqrt{5})\colon L]=2$ follows from a), since $ \sqrt{5}\notin L$. The fact that it is a normal extension is also simple: $ L$ is a normal extension of $ \mathbb{Q}$, being a normal extension of $ \mathbb{Q}(\sqrt{2})$, which is itself a normal extension of $ \mathbb{Q}$. And $ M$ is a normal extension of $ L$, since it is the splitting field over $ L$ of the polynomial $ p(x)=x^{2}-5$.

If $ \phi\in\mbox{Aut}(M/\mathbb{Q})$, then $ \phi$ fixes all rationals and permutes the roots of $ p(x)$. It also permutes the roots of $ q(x)=x^{2}-3$ and of $ r(x)=x^{2}-2$. Each of these polynomials has two roots. So if $ |\phi|\neq1$, let $ x\in M$. If $ x\in\mathbb{Q}$, then obviously $ \phi^{2}(x)=x$. If not, assume without loss that $ x$ is a root of $ p$ and that $ \phi$ does not fix $ x$. Then $ \phi(x)$ is the other root of $ p$, and so $ \phi^{2}(x)=x$, since $ \phi$ cannot fix the other root of $ p$, being an automorphism. Thus $ |\phi|=2$.

As for c), this is the same as finding the number of subgroups of $ G=\mbox{Aut}(M/\mathbb{Q})$ of order $ 2$. This is due to the proof of the fundamental theorem of Galois theory, where it can be shown that if the intermediate field $ F$ and subgroup $ H$ correspond to each other, then $ [M\colon F]=|H|$. Since $ [F\colon\mathbb{Q}]=4$, $ [M\colon F]=2$. So you want to find all subgroups $ H\subset G$ which are generated by a single non-identity element of $ G$, since each of these must have order $ 2$. There are $ 7$ of these.

[quote="N.T.TUAN"]Let $ L=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $ M=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$. Note that $ L\subseteq M$.

a)Prove that $ \sqrt{5}\not\in L$.
[/quote]
HilbertThm made a mistake on this one. The error was that $ \sqrt{5}= a\sqrt{2}+b\sqrt{3}$ for $ a,b\in \mathbb{Q}$ is not correct, because the basis is $ \{ 1,\sqrt{2}, \sqrt{3}, \sqrt{6}\}$.  :wink:

Here is how I would do it. We can view $ \mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2})(\sqrt{3})$.
And hence if $ \sqrt{5}\in \mathbb{Q}(\sqrt{2})(\sqrt{3})$ then $ \sqrt{5}= a+b\sqrt{3}$ for $ a,b \in \mathbb{Q}(\sqrt{2})$. Hence $ 5 = a^{2}+2ab\sqrt{3}+3b^{2}$. But then $ \sqrt{3}= \frac{5-a^{2}-3b^{2}}{2ab}\in \mathbb{Q}(\sqrt{2})$ which is a contradiction*.

*)Actually this takes some more works, because if $ a=0$ or $ b=0$ then we have division by zero  :wallbash_red:
But we can easily disregard those cases.

[quote="r2sputin"]The fact that it is a normal extension is also simple: $ L$ is a normal extension of $ \mathbb{Q}$, being a normal extension of $ \mathbb{Q}(\sqrt{2})$, which is itself a normal extension of $ \mathbb{Q}$. And $ M$ is a normal extension of $ L$, since it is the splitting field over $ L$ of the polynomial $ p(x)=x^{2}-5$.[/quote]
If $ M|L$ and $ L|K$ is normal, we don't necessarily get that $ M|K$ is normal.

Yes, I should have just said that it is the splitting field of $ p,q,r$ over $ \mathbb{Q}$, where $ p(x)=x^{2}-5,q(x)=x^{2}-3,r(x)=x^{2}-2$.

Darn, I realized my mistake half-way through my day, but the error had already been pointed out.  :(

I have a question about notation.
What is the meaning of $ \mbox{Aut}(E/F)$? I never seen this in my book.

That that supposed to be the group $ \mbox{Gal}(E/F)$, same thing? As in the group of all automorphisms of $ E$ that have $ F$ fixed?

Thanks :)

[quote]b) Prove that $ [M: \mathbb{Q}]=8$ and that each element has order 1 or 2[/quote]
Im gonna assume what I asked just above about notation is true.

We dont need to be fancy here with all the theorems. We can use basic observations.
Let $ \iota$ be the identity automorphism so $ \iota \in \mbox{Gal}(M/\mathbb{Q})$.

And $ \tau_{1}= \psi_{\sqrt{2},-\sqrt{2}}, \tau_{2}=\psi_{\sqrt{3},\sqrt{3}}, \tau_{3}= \psi_{\sqrt{5},-\sqrt{5}}$ be the [i]conjugation[/i] automorphism of $ M$, we immediately see that $ \tau_{1}, \tau_{2}, \tau_{3}\in \mbox{Gal}(M/ \mathbb{Q})$

And so is $ \tau_{4}= \tau_{1}\tau_{2}, \tau_{5}=\tau_{1}\tau_{2}, \tau_{6}= \tau_{2}\tau_{3}$.

And finally let $ \tau_{7}=\tau_{1}\tau_{2}\tau_{3}$.

It is not difficult to show that $ \mbox{Gal}(M/\mathbb{Q}) = \{\iota, \tau_{1},...,\tau_{7}\}$.

So the automorphism group is of 8 elements and its easy to see that this is the complete group by looking at $ \{1,\sqrt{2},\sqrt{3},\sqrt{5},....,\sqrt{30}\}$ and seeing how it acts on the elements of the basis. Basic computation shows that each non-identity element has order 2.

In fact, this a group of low order so its pretty easy to classify it up to isomorphism. I just dont know cuz I didnt make a group table but I am guessing its $ \mathbb{Z}_{2}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}$. Would anyone confirm please?


The sequence $ a_n$ of positive integers is such that (1) $ a_n \leq n^{\frac32}\forall n$, and (2) $ m\minus{}n$ divides $ a_m \minus{} a_n\forall m > n$. Find $ a_n$.

Have a look here
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1274252063&t=84909

Pierre.


real analysis
inequalities
real analysis unsolved

Let $ A,B \subset \mathbb{R}^n$ Lebesgue measurable set's,and $ \lambda(A)\equal{}a,\lambda(B)\equal{}b$. Let $ A\plus{}B\equal{}\{x\plus{}y|x \in A,y\in B\}$. How large  can be at most and at least $ \lambda(A\plus{}B)$.($ \lambda$ is Lebesgue measure)

the upper bound is clearly $ \infty$, take $ A$ to contain a dense subset of $ \mathbb{R}^n$ and $ B$ to be an open ball, then $ A\plus{}B$ is all of $ \mathbb{R}^n$.

the lower bound seems harder though.  If we assume that $ a,b>0$, then it seems like the measure of $ A\plus{}B$ should be smallest when $ A$ and $ B$ are balls, so you can compute how big the measure of $ A\plus{}B$ should be.  But I am not sure how to prove this (maybe some fooling around with the isoperimetric inequality?)

[quote="blahblahblah"]...  (maybe some fooling around with the isoperimetric inequality?)...[/quote]
 they are not compact .
if we suppose that they are compact $ A$ and $ B$ then $ A\plus{}B$ is also compact and one can use Brunn-Minkowsky inequality in $ \mathbb{R}^{n}$
$ Vol(A\plus{}B)^{1/n} \geq Vol (A)^{1/n}\plus{}Vol(B)^{1/n}$

If  $ \lambda(A)=0,\lambda(B)=0$  and the two sets are in [0,1] , could the sum $ A+B=\{x+y|x \in A,y\in B\}$ be [0,1] ?

Yes. Moreover, $A, B$ could be chosen to have Hausdorff dimension zero, which is stronger.
http://www.artofproblemsolving.com/community/c7h607102


[color=darkred]Α)[/color]
Εστω τριγωνο $ABC$ και $I$ το εγκεντρο του.
Αν για καθε σημείο $M$ του επιπέδου του ισχύει $\vec{MA}+\vec{MB}+\vec{MC}=3\vec{MI}$
νδο ειναι ισόπλευρο.

[color=darkred]B[/color]
Εστω τριγωνο $ABC$ και $H$ το ορθοκεντρο του.
Αν για καθε σημείο $M$ του επιπέδου του ισχύει $\vec{MA}+\vec{MB}+\vec{MC}=3\vec{MH}$
νδο ειναι ισόπλευρο.

 :)

[quote="pontios"][hide]

Έστω $G$ το κέντρο βάρους

$\vec{MA}+\vec{MB}+\vec{MB}=$

$\vec{MG}+\vec{GA}+\vec{MG}+\vec{GB}+\vec{MG}+\vec{GC} =$

$3\vec{MG}+(\vec{GA}+\vec{GB}+\vec{GC}) =$

$= 3\vec{MG}$

$\text{a) }\vec{MA}+\vec{MB}+\vec{MB} = 3\vec{MI} \Rightarrow 3\vec{MG} = 3\vec{MI} \Rightarrow$

$\vec{MG} = \vec{MI} \Rightarrow G \equiv I$

$\text{b) }$ ομοίως,  $G \equiv H$

Σε κάθε περίπτωση, το τρίγωνο είναι ισόπλευρο[/hide]

[u]Ερώτηση:[/u]
Δεν αρκεί να δοθεί ότι ισχύει για ένα μόνο σημείο Μ;

Π.χ. "υπάρχει σημείο Μ του επιπέδου για το οποίο ισχύει ισχύει η ισότητα ..."[/quote]

Έχεις δίκιο, Λουκα.
Η αρχική ιδέα φαίνεται πως ήταν να βαλουμε Μ=Α μετα Β .... και τελος Γ

 :D


equation
inequalities
recurrence relation
algebra
Integer sequence
IMO Shortlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]let $ a_1,a_2,\ldots ,a_n$ be such numbers,we have $ a_{k \minus{} 1} < a_k$ hence for $ k \equal{} 2,3,\ldots ,n \minus{} 1$ we get that $ 2\leq k$.
now we can rewrite the inequality $ (a_{k \plus{} 1} \minus{} 1)a_{k \minus{} 1}\geq a_k^2(a_k \minus{} 1)$ as:

$ \frac {a_{k \minus{} 1}}{a_k} \plus{} \frac {a_k}{a_{k \plus{} 1} \minus{} 1}\leq\frac {a_{k \minus{} 1}}{a_k \minus{} 1}$

now if we write this inequality for $ k \equal{} i \plus{} 1,i \plus{} 2,\ldots ,n \minus{} 1$ and add them up,according to the obvious inequality $ \frac {a_{n \minus{} 1}}{a_n} < \frac {a_{n \minus{} 1}}{a_n \minus{} 1}$,we'll get that:

$ \frac {a_i}{a_{i \plus{} 1}} \plus{} \frac {a_{i \plus{} 1}}{a_{i \plus{} 2}} \plus{} \ldots \plus{} \frac {a_{n \minus{} 1}}{a_n}\leq\frac {a_i}{a_{i \plus{} 1} \minus{} 1}$

now if we use the above inequality for $ i \equal{} 0$,according to the given equation,we would have:

$ \frac 1{a_1}\leq\frac {99}{100} < \frac 1{a_1 \minus{} 1}$

hence $ a_1 \equal{} 2$.

similarly for $ i \equal{} 1$ we get that:

$ \frac 1{a_2}\leq\frac 1{a_1}\left(\frac {99}{100} \minus{} \frac 1{a_1}\right) < \frac 1{a_2 \minus{} 1}$

which yields $ a_2 \equal{} 5$.

in a similar way we conclude that $ a_3 \equal{} 56$ and $ a_4 \equal{} 25\times 56^2 \equal{} 78400$.

but for $ a_5$ we would end up with:

$ \frac 1{a_5}\leq\frac 1{a_4}\left(\frac {99}{100} \minus{} \frac 12 \minus{} \frac 25 \minus{} \frac 5{56} \minus{} \frac {56}{25\times 56^2}\right) \equal{} 0$

which is impossible.

also we can easily see that $ a_1 \equal{} 2,a_2 \equal{} 5,a_3 \equal{} 56,a_4 \equal{} 25\times 56^2$ is a solution,and also this solution shows that this set of numbers are the only valid answer.[/quote]

[quote=Twoisaprime]First, I shall show by reverse induction that $ \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \dots - \frac{a_{i - 1}}{a_i} \leq \frac{a_i}{a_{i + 1} - 1} $ for all nonnegative integers $ i < n $.  For the base case, letting $ i = n - 1 $ it suffices to show that $ \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \dots - \frac{a_{n - 2}}{a_{n - 1}} = \frac{a_{n - 1}}{a_n} \leq  \frac{a_{n - 1}}{a_n - 1} $ which is trivial.

Now, assume that for some positive integer $ m < n $ we have that $ \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \dots - \frac{a_{m - 1}}{a_m} \leq \frac{a_m}{a_{m + 1} - 1} (*) $.  I shall show that $ \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \dots - \frac{a_{m - 2}}{a_{m - 1}} \leq \frac{a_{m - 1}}{a_m - 1} $.

It clearly suffices to show that $ \frac{a_{m - 1}}{a_m} \leq \frac{a_{m - 1}}{a_m - 1} - \frac{a_m}{a_{m + 1} - 1} $, because adding this to $ (*) $ then yields the desired result.  But upon expansion this is equivalent to the following inequality: $ (a_{m + 1} - 1)a_{m - 1} \geq a_{m}^2(a_m - 1) $ which we know to be true.  Therefore the induction hypothesis holds.

Now, since $ \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \dots - \frac{a_{i - 1}}{a_i} = \frac{a_i}{a_{i + 1}} + \frac{a_{i + 1}}{a_{i + 2}} + \dots + \frac{a_{n - 1}}{a_n} \geq \frac{a_i}{a_{i + 1}} $ we have that $ \frac{a_i}{a_{i + 1}} \leq \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \dots - \frac{a_{i - 1}}{a_i} \leq \frac{a_i}{a_{i + 1} - 1} (**) $ for all nonnegative integers $ i < n $.

This inequality clearly implies that all of the $ a_i $ are uniquely determined.  Letting $ i = 0 $ in $ (**) $ we find that $ a_1 = 2 $.  Plugging in $ i = 1 $ we find that $ a_2 = 5 $.  Continuing in this fashion we find that the only solution is given by $ (a_0, a_1, a_2, a_3, a_4) = (1, 2, 5, 56, 25^2 \cdot 56) $.

I want to provide some motivation for this solution.  The idea for finding $ (**) $ comes from the greedy algorithm... by applying it to the problem we immediately get a solution, and then it is pretty likely that we want to show that this is the only solution.  The greedy algorithm works by finding an $ a_i $ that satisfies $ (**) $ every time we plug in a new $ i $ so it is natural to try to prove that $ (**) $ must always hold, and then induction becomes apparent.  Despite its similarity to the above solution posted by BaBaK Ghalebi, I wanted to post this because his solution seems to magically derive an identity from the given inequality that trivializes the problem seemingly without motivation.[/quote]


Prove that$\forall a,b,c>0$we have
$\sum\frac{a^2-bc}{b^2+c^2+2a^2} \ge 0$     (1)    :)

[quote="pvthuan"]That is a good inequality, come on, nthd[/quote]
Solution of nthd  nice too:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=403193&highlight=#403193

Yet another proof: By Cauchy-Schwarz,

\[ 3-2\sum_{cycl}\frac{c^2-ab}{2c^2+a^2+b^2}=\sum_{cycl}\frac{(a+b)^2}{a^2+b^2+2c^2} \le \sum_{cycl}\Lp\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\Rp = 3 \] from which the result follows.

The following inequality is also true.
Let $a,$ $b$ and $c$ are real numbers such that $ab+ac+bc\neq0$ and $0\leq k\leq2.$ Prove that:
\[\frac{a^2-bc}{ka^2+b^2+c^2}+\frac{b^2-ac}{a^2+kb^2+c^2}+\frac{c^2-ab}{a^2+b^2+kc^2}\geq0\]

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3480335#p3480335


Find the coefficient of $ a^3b^4c^5$ in the expansion of $ (bc\plus{}ca\plus{}ab)^6$.

[quote="thomaskurian89"]Find the coefficient of $ a^3b^4c^5$ in the expansion of $ (bc \plus{} ca \plus{} ab)^6$.[/quote]

anwer is 60.
[hide="Solution"] $ (ab\plus{}bc\plus{}ac)^6\equal{}(a(b\plus{}c)\plus{}bc)^6\equal{} \binom{6}{0}a^6(b\plus{}c)^6\plus{}\binom{6}{1}a^5(b\plus{}c)^5bc\plus{}\binom{6}{2}a^4(b\plus{}c)^4b^2c^2\plus{}\binom{6}{3}a^3(b\plus{}c)^3b^3c^3\plus{}\binom{6}{4}a^2(b\plus{}c)^2b^4c^4\plus{}\binom{6}{5}a(b\plus{}c)b^5c^5\plus{}\binom{6}{6}b^6c^6$
So we can observe only $ \binom{6}{3}a^3(b\plus{}c)^3b^3c^3$.
 $ \binom{6}{3}a^3(b\plus{}c)^3b^3c^3\equal{}20a^3b^3c^3(b^3\plus{}3b^2c\plus{}3bc^2\plus{}c^3)\equal{}20a^3b^6c^3\plus{}60a^3b^5c^4\plus{}60a^3b^4c^5\plus{}20a^3b^3c^6$
So therefore coefficient  of $ a^3b^4c^5$ is $ 60$[/hide]

[quote="SaYaT"][quote="thomaskurian89"]Find the coefficient of $ a^3b^4c^5$ in the expansion of $ (bc \plus{} ca \plus{} ab)^6$.[/quote]

anwer is 60.
[hide="Solution"] $ (ab \plus{} bc \plus{} ac)^6 \equal{} (a(b \plus{} c) \plus{} bc)^6 \equal{} \binom{6}{0}a^6(b \plus{} c)^6 \plus{} \binom{6}{1}a^5(b \plus{} c)^5bc \plus{} \binom{6}{2}a^4(b \plus{} c)^4b^2c^2 \plus{} \binom{6}{3}a^3(b \plus{} c)^3b^3c^3 \plus{} \binom{6}{4}a^2(b \plus{} c)^2b^4c^4 \plus{} \binom{6}{5}a(b \plus{} c)b^5c^5 \plus{} \binom{6}{6}b^6c^6$
So we can observe only $ \binom{6}{3}a^3(b \plus{} c)^3b^3c^3$.
 $ \binom{6}{3}a^3(b \plus{} c)^3b^3c^3 \equal{} 20a^3b^3c^3(b^3 \plus{} 3b^2c \plus{} 3bc^2 \plus{} c^3) \equal{} 20a^3b^6c^3 \plus{} 60a^3b^5c^4 \plus{} 60a^3b^4c^5 \plus{} 20a^3b^3c^6$
So therefore coefficient  of $ a^3b^4c^5$ is $ 60$[/hide][/quote]

Thank you.  :)

Easier to use Multinomial theorem

[hide]
The coefficient of $ x^p y^q z^r$ in $ (x\plus{}y\plus{}z)^n$ with p+q+r = n is $ \frac{n!}{p!q!r!}$

In $ (ab\plus{}bc\plus{}ca)^6$, the general term will look like $ (ab)^p (bc)^q (ca)^r \equal{} a^{p\plus{}r} b^{p\plus{}q} c^{q\plus{}r}$

Comparing exponents of like terms, we get p =1, q = 3 and r = 2

Hence the coefficient is $ \frac{6!}{3!2!1!} \equal{} 60$[/hide]

Where can I find a proof for the Multinomial theorem?


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let $ A_1, A_2$ be rings, and $ A_1 \times A_2$ is their direct product. What are prime ideals of the ring $ A_1\times A_2$?

The ideals of $ A_1 \times A_2$ are exactly the sets $ I_1 \times I_2$ with ideals $ I_1 \subseteq A_1$ and $ I_2 \subseteq A_2$.
Let $ I_1 \times I_2$ be a prime ideal, meaning that $ (A_1 \times A_2) / (I_1 \times I_2) \cong (A_1 / I_1) \times (A_2 / I_2)$ is an integral domain. This can only happen of each $ A_k / I_k$ is an integral domain and then one of them is the trivial ring. Conversely, this condition is sufficient.
Thus the prime ideals are of type $ P_1 \times A_2$ or $ A_1 \times P_2$, $ P_k \subseteq A_k$ a prime ideal.

Why one of them should be trivial? I think we can just take two prime ideals $ I_1, I_2$ and $ I_1\times I_2$ will be prime by the criteria you wrote: $ (A_1/I_1)\times(A_2/I_2)$ is integral.

[quote="ZetaX"]The ideals of $ A_1 \times A_2$ are exactly the sets $ I_1 \times I_2$ with ideals $ I_1 \subseteq A_1$ and $ I_2 \subseteq A_2$.[/quote]

unfortunately, this is false :)
just take $ ((1,1))\subset \mathbb{Z}\times\mathbb{Z}$: it's the diagonal, and it's not of the form you say it should be :)
anyway, it turns out that prime ideals are actually of the form you state.

ZetaX is right (of course). (1,1) is a unit and thus generates the whole ring ... you probably thought of groups?

ehm.. lol :blush:
is it possible for you all to forget my previous post? :blush:

erm, which post are you talking about?  :wink:


inequalities
Ramsey Theory
graph theory
combinatorics solved
combinatorics

Prove that within a group of $ 23$ actors that there are three who know each other or seven who don't know each other.

Let R(a,b) be the Ramsey number (that is the minimal n such that if we 2-color, say red and blue, the edges of the complete graph K_n then there exist a complete red subgraph K_a or a complete blue subgraph K_b).

The following results are classical :
- R(a,b) = R(b,a).
- We have R(2,b) = b.
- We have R(a,b)  \leq R(a-1) + R(a,b-1).
- The preceeding inequality is strict if both R(a-1,b) and R(a,b-1) are even.

It follows easily that R(3,7)  \leq 23.

Note that it is known that R(3,7) = 23 (Greenwood and Gleason, 1955).

Pierre.

[quote="pbornsztein"]...

It follows easily that R(3,7)  \leq 23.

...[/quote]

Although it is true that $ R(3, 7) = 23,$ I think the above inequality doesn't follow so "easily."
Actually, if you use the inequality $ R(a, b) \leq R(a-1, b)+R(a, b-1)$ recursively, you'll end up with 28 as an upper bound.
And, even if you take the value of $ R(3, 6)$ for granted (which is 18), you get 25 as an upper bound.


geometry
projective geometry
vector
analytic geometry
superior algebra
superior algebra unsolved

hi

i know this is not completely algebra but there is a link

it is well known that desargues is not true in all axiomatic projective planes
however, if the three concurrent lines in three dimensional projective space
are not coplanar i can prove it easily as all three 'would be collinear' points
are in the unique intersection line of two planes

however, it's not so easy for me to see that it also holds if the three
are coplanar

which seems more than just a little hole, because it would show that two
dimensional desargues in a plane, imbedded in three dimensional space, is
provable

any ideas

fred

To prove the coplanar case, project a non-coplanar case into that plane.

If Desargues' theorem fails, it must be impossible to embed our 2-dimensional projective plane into a 3-dimensional projective geometry.

i'm sorry i'm so slow, how exactly project it?

we are working in a projective space right, constructed with in fact subspaces of 4D vector space?

Definition of projection onto a space of dimension one less than the whole:

Let the subspace be $P$. Choose a point $x$ not in that space. For each line $l$ passing through $x$, map all points other than $x$ on $l$ to the intersection of $l$ and $P$.

Suppose we have a Desargues configuration $ABCDEFGHIJ$ in $P$ such that the triples $ABE$, $ACF$, $ADG$, $BCH$, $BDI$, $CDJ$, $EFH$, $EGI$, $FGJ$ are each collinear. We want to prove that $HIJ$ are collinear.
To do so, choose $D'\not\in P$ such that $xD'D$ are collinear and let $G'$ be the intersection of $xG$ and $AD'$. Let $I'$ be the intersection of $BD'$ and $EG'$, then let $J'$ be the intersection of $CD'$ and $FG'$.
Now $ABCD'EFG'HI'J'$ is a noncoplanar Desargues configuration, so by the theorem you have already proved $HI'J'$ are collinear. These points project onto $HIJ$ and the line projects to a line, so $HIJ$ are collinear.

[quote="fredbel6"]we are working in a projective space right, constructed with in fact subspaces of 4D vector space?[/quote]
Actually, I'm working in axiomatic projective geometry. No coordinates needed.
If I want coordinates, I'll trim to an affine space (unique parallel axiom) and define the operations geometrically. With Desargues' theorem, the points on an affine line should be a division ring.

hi
forgive me for getting back to this, but i actually do have some more questions about this

you work axiomatic? but what exactly are the axioms of let 's say three dimensional projective space?

i suppose you start with sets (A,B,C,I) with I incidence relation

but what are all the conditions : a unique throught two distinct points
a unique plane through three noncollinear points
when a line is not incident with a plane, there is a unique point incident with both...

how much more axioms are needed, and does it get even more difficult for higher dimensions

also, in class, we reasoned like this : hilbert proved that in a plane desargues implies coordinatizable over a skew field
and in higher dimensions desargues is always true---> all higher dimensional projective spaces are over a skew field

does it go that fast, i mean hilbert only did it for a plane?

fred

This is going to be long...

The basic idea is to let the points of a line be the field, and then define the operations geometrically. To read and understand this, you probably need to draw some diagrams.

It's easier to work with an affine plane here; if you started with a projective plane, remove a line. A line at infinity can alwalys be added back later. The axioms of an affine plane:
(I1) Any two points lie on exactly one line.
(I2) Any line contains at least two points.
(I3) There exist three noncollinear points.
(P) Through any point not on a given line $\ell$, there is exactly one line parallel to $\ell$. (parallel = does not intersect)

There is a special case of Desargues' theorem I will use heavily here, and call Little Desargues: if the corresponding vertices of two triangles lie on parallel lines and two pairs of corresponding sides are parallel, the third pair of sides is parallel. Conversely, if the corresponding sides are parallel and two pairs of corresponding vertices are on parallel lines, the third pair of vertices lies on a line parallel to the other two. This special case is weaker than the full theorem.
On the other hand, the full Desargues theorem is needed here as an extra axiom- we can infer it from a longer list of incidence axioms if our plane is part of a higher-dimensional affine space.

Choose a line $\ell$ and points $0,1$ on it.
Let $F$ be the set of points of $\ell$.

Once we know the definition works, we create a coordinate grid to complete the description of this affine geometry as the Cartesian plane over $F$. To create the grid, choose a line $\ell'$ passing through $0$ to be the second axis. The point $x$ on the first axis becomes $(x,0)$. Choose an arbitrary point $(0,1)$ on $\ell'$. For all $x$, let the line through $(x,0)$ parallel to the line through $(0,1)$ and $(1,0)$ intersect $\ell'$ at $(0,x)$. Let the parallel to $\ell'$ through $(x,0)$ and the parallel to $\ell$ through $(0,y)$ intersect at $(x,y).$

For $a,b\in F$, define $a+b$ as follows:
Choose $x\not\in F$ arbitrarily. Let $\ell_x$ be the line parallel to $\ell$ through $x$. Let $p$ be the intersection of $\ell_x$ and the line through $b$ parallel to $0x$. Let $a+b$ be the intersection of $\ell$ and the line through $p$ parallel to $xa$. Now we need to prove that $+$ satisfies the appropriate axioms:

$+$ is well-defined: Let $x,p$ be as above, and also choose $y,q$ similarly.
The vertices of the triangles $0xy$ and $bpq$ lie on parallel lines, and two of the corresponding sides are parallel. By Little Desargues, $xy$ and $pq$ are parallel. Now, consider the triangle with vertices $p,q$ and sides parallel to $xa$ and $ya$. By Little Desargues, the third vertex is on the line through $a$ parallel to $\ell_x$ and $\ell_y$. Therefore, the third vertex is $a+b$, and the two definitions agree.

$+$ is commutative ($a+b=b+a$): Let $x,p$ be as above. Choose $y$ such that $0y\parallel xa$ and $ay\parallel x0$. Let $\ell_y$ be the parallel to $\ell$ through $y$ and let $q$ be the intersection of the parallel to $0y$ through $a$ with $\ell_y$. Then $b+a$ is the intersection of the parallel to $yb$ through $q$ with $\ell$.
By Little Desargues for $xap$ and $0yb$, $ap\parallel yb\parallel q(b+a)$. By Little Desargues for $0xp$ and $ya(a+b)$, $0p\parallel y(a+b)$. By Little Desargues for $0ap$ and $yq(a+b)$, $ap\parallel q(a+b)$. Since $q(b+a)$ and $q(a+b)$ are parallel to the same line, they are one line, and $a+b$ must coincide with $b+a$.

$+$ is associative ($a+(b+c)=(a+b)+c$): Choose $x,y$ randomly. Let $p$ be the intersection of $\ell_x$ and the parallel to $0x$ through $b$. Let $q$ be the intersection of $\ell_y$ and the parallel to $0y$ through $c$. $a+b$ is the intersection of $\ell$ and the parallel to $xa$ through $p$. $b+c$ is the intersection of $\ell$ and the parallel to $yb$ through $q$. Let $r$ be the intersection of $\ell_x$ and the parallel to $0x$ through $b+c$. $a+(b+c)$ is the intersection of $\ell$ and the parallel to $xa$ through $r$. $(a+b)+c$ is the intersection of $\ell$ and the parallel to $y(a+b)$ through $q$.
By Little Desargues on $pby$ and $r(b+c)q$, $py\parallel rq$. By Little Desargues on $py(a+b)$ and $rq(a+(b+c))$, $q(a+(b+c))\parallel y(a+b)$. Since $q((a+b)+c)\parallel y(a+b)$ by definition, the two lines coincide and $a+(b+c)=(a+b)+c$.

It is obvious by the definition that $0$ is an identity for $+$. To construct inverses, choose $x$ arbitrarily. Let $p$ be the intersection of $\ell_x$ and the parallel to $ax$ through $0$. Let $-a$ be the intersection of $\ell$ and the parallel to $0x$ through $p$. It is then trivial to show that $a+(-a)=0$.

Choose $x$ arbitrarily. Let $p$ be the intersection of $0x$ and the parallel to $1x$ through $b$. Let $a\cdot b$ be the intersection of $\ell$ with the parallel to $xa$ through $p$.

$\cdot$ is well-defined: Let $x, p,a,b,a\cdot b$ be as above. Also choose $y$ randomly and let $q$ be the intersection of $0y$ with the parallel to $1y$ through $b$. By Desargues' theorem for triangles $1xy$ and $bpq$, $xy\parallel pq$. By Desargues' theorem for $axy$ and $(a\cdot b)pq$, $(a\cdot b)q\parallel ay$ and $\cdot$ is well-defined.

$\cdot$ is associative ($a\cdot(b\cdot c)=(a\cdot b)\cdot c$): Choose $x,y$ arbitrarily. Let $p$ be the intersection of $0x$ and the parallel to $1x$ through $b$. Then $a\cdot b$ is the intersection of $\ell$ and the parallel to $xa$ through $p$. Let $q$ be the intersection of $0y$ and the parallel to $1y$ through $c$. Then $b\cdot c$ is the intersection of $\ell$ and the barallel to $yb$ through $q$. Let $r$ be the intersection of $0x$ with the parallel to $1x$ through $b\cdot c$. Then $a\cdot(b\cdot c)$ is the intersection of $\ell$ with the parallel to $xa$ through $r$. Also, $(a\cdot b)\cdot c$ is the intersection of $\ell$ with the parallel to $y(a\cdot b)$ through $q$.
By Desargues for $pyb$ and $rq(b\cdot c)$, $py\parallel rq$. By Desargues for $py(a\cdot b)$ and $rq(a\cdot(b\cdot c))$, $p(a\cdot b)\parallel r(a\cdot(b\cdot c))$ and $a\cdot(b\cdot c)=(a\cdot b)\cdot c$.

Distributivity, part 1 ($(a+b)\cdot c=a\cdot c+b\cdot c$): Choose $x$ arbitrarily. Let $p$ be the intersection of $0x$ and the parallel to $1x$ through $c$. Choose $y$ arbitrarily on the parallel to $xb$ through $0$. Then $a\cdot c$ is the intersection of $\ell$ with the parallel to $xa$ through $p$, and $b\cdot c$ is the intersection of $\ell$ with the parallel to $xb$ through $p$. Let $z$ be the intersection of $0y$ with the parallel to $ay$ through $a\cdot c$. Let $\ell_y$ and $\ell_z$ be the parallels to $\ell$ through $y$ and $z$ respectively. Let $q$ be the intersection of $\ell_z$ with the parallel to $0z$ through $b\cdot c$ and let $r$ be the intersection of $\ell_y$ with the parallel to $0y$ through $b$. Then $a+b$ is the intersection of $\ell$ with the parallel to $ya$ through $r$ and $a\cdot c+b\cdot c$ is the intersection of $\ell$ with the parallel to $z(a\cdot c)$ through $q$. Also, $(a+b)\cdot c$ is the intersection of $\ell$ with the parallel to $x(a+b)$ through $p$.
By our definitions, $p(b\cdot c)q$ and $xbr$ are straight lines parallel to each other. By Desargues for $xya$ and $pz(a\cdot c)$, $pz\parallel xy$. By Desargues for $pzq$ and $xyr$, $0qr$ are collinear. By Desargues for $pq(a\cdot c+b\cdot c)$ and $xr(a+b)$, $p(a\cdot c+b\cdot c)\parallel x(a+b)$ and $(a+b)\cdot c=a\cdot c+b\cdot c$.

Distributivity, part 2 ($a\cdot(b+c)=a\cdot b+a\cdot c$): Choose $x$ arbitrarily. Let $p$ and $q$ be the intersections of $0x$ with the parallels to $1x$ through $b$ and $c$ respectively. Then $a\cdot b$ and $a\cdot c$ are the intersections of $\ell$ with the parallels to $xa$ through $p$ and $q$ respectively. Let $\ell_p$ be the parallel to $\ell$ through $p$, and let $r$ and $s$ be the intersections of $\ell_p$ with the parallels to $0p$ through $a\cdot c$ and $c$ respectively. Then $b+c$ is the intersection of $\ell$ with the parallel to $pb$ through $s$ and $a\cdot b+a\cdot c$ is the intersection of $\ell$ with the parallel to $p(a\cdot b)$ through $r$. Let $y$ be the intersection of $0x$ and $s(b+c)$; since $s(b+c)\parallel x1$, $a\cdot(b+c)$ is the intersection of $\ell$ with the parallel to $xa$ through $y$.
By Little Desargues on $yrs$ and $q(a\cdot c)c$, $yr\parallel q(a\cdot c)\parallel p(a\cdot b)\parallel r(a\cdot b+a\cdot c)\parallel y(a\cdot(b+c))$, so we must have $a\cdot(b+c)=a\cdot b+a\cdot c$.

It is obvious from the definition that $1$ is a multiplicative identity. To construct inverses of nonzero elements, choose $x$ arbitrarily. Let $p$ be the intersection of $0x$ with the parallel to $ax$ through $1$. Let $a^{-1}$ be the intersection of $\ell$ with the parallel to $1x$ through $p$. It is easy to show that $a\cdot a^{-1}=a^{-1}\cdot a=1.$

This completes the proof that an affine plane with Desargues' theorem is the Cartesian plane over a division ring. Since Desargues' theorem is true in any affine or projective space of more than two dimensions, all such spaces can be coordinatized.

jmerry you really are took kind
i didn't know it was possible to construct that division ring that 'fast' (i was told it took ten pages?)

however, what exactly did you do know : prove that desargues implies in a affine plane that it is over a division ring, or did you prove that in all affine spaces desargues implies that

and forgive me for getting back to my question : you gave the axioms for affine plane
could it be that that axiom that every line has at least two points is implied by the rest? (we have a minicourse projective (and affine ) planes in our course of projective geometry, and we only had three axioms)

and could you state the axioms for like a three dimensional affine or projective space
i know that they are all created over division rings, but then again i can't understand the theorem that in higher dimensions all spaces are over such ring , if i don't know what a space in general is

thx anyway however,

you really DO know much about geometry :ninja: !

[quote="fredbel6"]i didn't know it was possible to construct that division ring that 'fast' (i was told it took ten pages?)
[/quote]
That would be ten pages if I included the diagrams, or if I expanded the print.

The incidence axiom (I2) is probably redundant with (P), but redundant axioms don't actually hurt. This labeling comes from a treatmemt of Euclidean geometry, in which we might leave out (P) to allow non-Euclidean possibilities.
For this process, I would state the extra axiom as
(D) If $AB\parallel DE$, $BC\parallel EF$, and $CA\parallel FD$, then either $AD,BE,CF$ are parallel or $AD,BE,CF$ intersect at $G$.

I'm not sure exactly what a nice set of the axioms of an affine space are, but any subspace will satisfy the lower-dimensional axioms. We can build the division ring in a subspace of dimension two, then construct as many axes as we need. Since the incidence axioms in at least three dimensions imply (D), we don't need anything extra in that case.

A set of axioms for three-dimensional affine space (which may be redundant)

(I1) Any two points lie on exactly one line.
(I2) Any line contains at least two points.
(I3) There exist three noncollinear points.
(I4) Through any line and any point not on that line, there is exactly one plane.
(I5) The intersection of two distinct planes which are not parallel is a line.
(I6) There exist four noncoplanar points
(P3) Through any point not on a given plane $p$, there is exactly one plane parallel to $p$.
Definition: two lines are parallel if they lie in a common plane and do not intersect.
(P2) Through any point not on a given line $\ell$, there is exactly one line parallel to $\ell$.


Let $ x,y,z \in R$ and $ \begin {cases} x^2 \plus{} y^2 \plus{} z^2 \equal{} 8 \\
xy \plus{} yz \plus{} zx \equal{} 4 \end{cases}$. Find maximum and minimum of $ z$.

[hide="Hint"]
$ (x\plus{}y\plus{}z)^2\minus{}2(zy\plus{}yz\plus{}zx)\equal{}x^2\plus{}y^2\plus{}z^2 \ \ \text{i.e.} \ \ x\plus{}y\plus{}z\equal{}\pm4$
Let $ t\in\{x, y, z\}, \ xyz\equal{}c$
$ t^3\mp4t^2\plus{}4t\minus{}c\equal{}0$

[hide="Answe"]$ \minus{}\dfrac{8}{3} \le z \le \dfrac{8}{3}$[/hide]
[/hide]


quadratics
inequalities
vector
complex numbers
algebra
quadratic formula
triangle inequality

This is the problem.

x^2 + x + 1 > 0

Since the roots are complex numbers, I'm not sure how to go about doing this problem.
All real numbers are included in x, but otherwise I'm confuzzled.
Is i greater than 0? Can the two be numerically compared?
What if the real portion and the imaginary portion of the complex number have different signs?

do you want the solution set for $ x$?

we know that $ x^2\plus{}x\plus{}1\equal{}(x\plus{}\frac{1}{2})^2\plus{}\frac{3}{4} >0$, so it's true for all $ x\in\mathbb{R}$.

We can't compare complex numbers (e. g. say $ i >0$) because complex numbers are not ordered, so we define the magnitude of a complex number $ z\equal{}a\plus{}bi$ and $ |z| \equal{} \sqrt{a^2\plus{}b^2}$ to convert complex numbers into a metric space.

Well, the assumption is that we're only talking about real numbers for this type of problem.  There simply isn't a way to order the complex numbers as nicely as we can the reals.

Since the roots are non-real numbers, the value of the expression is positive for all reals.  We just stop there.

Since there's no truly nice way to order the complex numbers, we can't really say a priori whether i is greater than 0.  There certainly is a way they can be compared, but we actually have to define what it means for one complex number to be greater than another first.

For instance, we could define it this way:
$ a \plus{} bi > a \plus{} ci$ if and only if $ b > c$ and $ a \plus{} bi > c \plus{} bi$ if and only if $ a > c$.  However, what did we give up in defining it like this?  Well, we can't compare $ 0$ and $ 1 \minus{} i$, for one.

Or we could define it this way:
$ a \plus{} bi > c \plus{} di$ if and only if either (1)$ b > d$ or (2)$ b\equal{}d$ and $ a>c$.
But then we lose a rather nice structure of the reals.

My point, in the end, is that there is no standard way to compare complex numbers, so we have to define what being "greater than" would mean first.  Only after that can we answer, for instance, whether i is greater than 0.

[quote="Mathnerdmo"]
Since the roots are non-real numbers, the value of the expression is positive for all reals.  We just stop there.
[/quote]

This is not true; what you can say is that a quadratic with real coefficients and non-real roots must be either always positive or always negative. This can be proven simply by noting that, WLOG, if the quadratic was positive at $ x \equal{} a$ and non-positive at $ x \equal{} b$, then, since $ f$ is continuous, there would exist $ c\in [a,b], c\in \mathbb{R}$ such that $ f(c) \equal{} 0$, a contradiction.

[quote="sunehra"]So if the inequality was $ x^2 \plus{} x \plus{} 1 < 0$ would there be no solutions?[/quote]

Correct.

[quote="7h3.D3m0n.117"]To solve $ x^2 \plus{} x \plus{} 1 > 0$, you can use the [url=http://www.artofproblemsolving.com/Wiki/index.php/Quadratic_formula]Quadratic Formula[/url] on the LHS.

This will give you the roots you seek.

Imaginary numbers cannot be compared with real numbers. There is no value for $ i$.

And it is possible to have opposite signs in a complex number. Are you familiar with the [url=http://www.artofproblemsolving.com/Wiki/index.php/Complex_plane]Complex plane[/url]?[/quote]

I know how to find the roots, it's just that I wasn't sure about how to deal with comparing complex numbers to zero, coz zero is real. Like I said in the first post, I know that all real numbers are included in x, I just wanted to know if I can go further.

Yes, I am. I know that there can be opposite signs, but my point was that if a complex number with both positive signs, such a 3 + 2i could be considered positive and greater than zero (I was very unsure about this, which is why I made the thread, now I have my answer), then what would happen if the signs were opposite. I should have been more clear.

[quote="DiscreetFourierTransform"]One measure of determining how "big" a complex number is is the magnitude, defined (for $ a \plus{} bi$) to be $ \sqrt {a^2 \plus{} b^2}$. It's usually written (for $ z \equal{} a \plus{} bi$) as $ |z|$.
This allows us to compare them, in a sense. This is really just the Euclidean Norm of a vector, however, which is a different topic.[/quote]

Can you explain this to me more thoroughly, or point me toward a good tutorial for explaining it?
Please and thank you. :)

If you think about complex numbers as vectors in the complex plane (from the origin to the point) then the complex magnitude (sometimes called the complex modulus) is the distance from the origin to the point. This is known as the norm of a vector.

An interesting thing about norms:
They satisfy the triangle inequality
$ |a \plus{} b| \le |a| \plus{} |b|$

He's referring to finding the distance from a complex point to 0 using the Pythagorean Theorem.


Let $a,b,c>0$ that satisfy $abc\leq 1$. Prove that:
\[ \frac{a+b}{a^2b^2}+\frac{b+c}{b^2c^2}+\frac{c+a}{c^2a^2}\geq 6 \]

It isn't hard. I think.
Put $a= \frac1x$ $b= \frac1y$ $c= \frac1z$ then $abc \leq 1$ we have $xyz \geq 1$
Then $\sum \frac{a+b}{a^2b^2}=\sum{xy(x+y)}$ from here we AM-GM for 6 numbers  ($xyz \geq 1$ ) and done!

He he . I inspire from you sols.
We have $\sum \frac{a+b}{a^2b^2}=\sum{(\frac{1}{ab^2}+\frac{1}{a^2b})}\geq 6 \sqrt[6]{\frac{1}{a^7b^7c^7}}\geq 6$ :D

[quote="Cezar Lupu"]I think this is too easy for this topic, so I'll move to Pre-Olympiad. :)[/quote]

Yeah, right. Sorry :oops:


Seven zeroes and a $1$ are placed on the vertices of a cube. Each move, you can add $1$ to the two numbers at the end of a random chosen side.
Can you make all the numbers equal? Can you make all the numbers divisible by $3$ ?

[hide="Vertices Equal"]
No, because the total of the numbers at the vertices will always be odd (starts at $1$, and every turn you add $2$ to the total). If all of the vertices were the same (let's say $k$), then the total would be $8k$, which is even.
[/hide]

Divisibility by 3.

[hide]Orientate the cube somehow. On the top face, let the corners be abcd, and on the bottom, efgh, such that e is under a, f under b, g under c, and h under d.

Now notice that $a-b+c-d-e+f-g+h$ is invarient. If all the corners were divisable by 3, so would this quantity. But in the beginning its equal to $+-1$ contradiction.[/hide]


induction
probability
invariant
real analysis
probability and stats

Let $ X_1,\dots, X_k\sim U([0,x])$. Order the first $ k \minus{} 1$ variables, and denote
\[ A_1 \equal{} [0,X_{(1)}], A_{i \plus{} 1} \equal{} [X_{(i)},X_{(i \plus{} 1)}], i \equal{} 1,\dots,k \minus{} 2,A_{k} \equal{} [X_{(k \minus{} 1)},x].
\]
Prove $ \mathbb{P}(X_k\in A_i) \equal{} \frac {1}{k},i \equal{} 1,\dots,k$.

I tried with induction, but I get tangled in the process :oops: .

isn't it almost obvious when you say that with words: what is the probability that $ X_k$ is at the $ i$-th position among $ X_1, X_2, \ldots, X_k$ ?

Yes, the claim is obvious, but I don't understand you comment.

This is how I went:

$ 1\equal{}\mathbb{P}(X_k\in [0,x])\equal{}\mathbb{P}(X_k\in \cup_{i\equal{}1}^{k}A_i) \equal{}\sum_{i\equal{}1}^{k}\mathbb{P}(X_k\in A_i)$,
 we know that $ \mathbb{P}(X_k\in A_1)\equal{}\mathbb{P}(X_k\in A_k)$ (cause $ A_k$ can be obtained via $ A_1$ under some transformations that are invariant under Lebesgue measure), so we need to prove that
$ \mathbb{P}(X_k \in A_1)\equal{}\mathbb{P}(X_k \in A_l)$ for some $ l\equal{}2,\dots,k\minus{}1$.
Here come the problems, as I said I went with induction, but my reasoning isn't very precise...

I would be glad for a further hint, or an explanation of your comments.


I would know how to find formula of $C_{artin}$:
\[C_{artin}=\prod_{p\in\mathcal{P}} \left(1-\frac{1}{p(p-1)}\right)\]

Thanks

What do you mean? Do you want closed expression?
Did you see http://mathworld.wolfram.com/ArtinsConstant.html ?

Yes I saw this page. I want to know how we find (the logic reasonement to obtain that) this formula since the product on primes p is ... mystic for me.


geometry
trapezoid
rectangle
ratio
number theory
relatively prime
AMC

Let $ ABC$ be a triangle with sides $ 3,4,$ and $ 5$, and $ DEFG$ be
a $ 6$ by $ 7$ rectangle.  A segment is drawn to divide triangle
$ ABC$ into a triangle $ U_1$ and a trapezoid $ V_1$, and another segment is drawn to divide rectangle $ DEFG$ into a triangle $ U_2$
and a trapezoid $ V_2$ such that $ U_1$ is similar to $ U_2$ and $ V_1$ is similar to $ V_2$.  The minimum value of the area of $ U_1$ can be written in the form $ m/n$, where $ m$ and $ n$ are relatively prime positive integers.  Find $ m\plus{}n$.

How do you find the lengths of triangle $ U_2$?  Let $ s_1$ be the line segment drawn in triangle $ ABC$. How do you know that "
were $ s_1$ parallel to the side of length $ 4$, the corresponding acute angles in $ V_1$ and $ V_2$ would not be congruent, therefore trapezoids $ V_1$ and $ V_2$ would not be similar"?

Thanks!

[quote="myc"]Let $ s_1$ be the line segment drawn in triangle $ ABC$. How do you know that "were $ s_1$ parallel to the side of length $ 4$, the corresponding acute angles in $ V_1$ and $ V_2$ would not be congruent, therefore trapezoids $ V_1$ and $ V_2$ would not be similar"? [/quote]

In $ V_1$, the acute angle is less than $ 45^\circ$. In the rectangle $ DEFG$, a $ 45^\circ$ angle cannot be the acute angle in a trapezoid. (Draw it!)

You know that:
1_ Only a segment that goes through exactly one of the vertices can divide DEFG into a triangle and a trapezoid. A right triangle and a trapezoid with two right angles result.
2_ ABC is a right triangle.
3_ A segment must be parallel to one of the legs of ABC to split it into a right triangle (similar to ABC) and a trapezoid with two right angles.

After making the proper sketches, think of all the acute angles and of the length of the legs of the triangles and how they are all related.

How are the acute angles in ABC, U1 and V1 related? Which are congruent and which are complementary? How are the acute angles in U2 and V2 related?

Further hint: If you know the similarity ratio (k) of two figures, the ratio of the areas is k squared.


[quote="Franz Joseph"]Show that:
$1\times2\times3\times4\times......\times p+2\times3\times4\times5\times...........\times(p+1)+..........+n(n+1)+.........+(n+p-1)=\frac{n(n+1)(n+2).....(n+p)}{p+1}$

for all Natural  Value  of  $p$.

I know it is really easy to prove by means of Mathematical Induction but what I really want is a Non-Inductive Proof. I don't care how ungly or monstrous it is. :lol:[/quote]

divide both sides by $p!$, then it becomes

$\binom{p}{p}+\binom{p+1}{p}+...+\binom{n+p-1}{p}=\binom{n+p}{p}$ which is true (using catalyst or induction, etc) with pascal's identity


$w=e^{\frac{pi}{3}}$ then it's OBVIOUSLY -1.


Same approach as last one...what don't you understand about solving these equations?

on this one, all three lines have variables. The last one didn't.

Well, how far can you get? Do you not know how to get started?

[hide="If that is the case"]Try eliminating variables by subtracting equations from other equations, substitution, etc.[/hide]

[hide="If you are still stuck"]
Try subtracting 3 times the 3rd equation from the 2nd one. Then, subtract 5 times the last equation from the 1st one. Do you see a general technique for eliminating variables?[/hide]

Try the same thing with the last equation and the 1st equation, then you have a system of equations in 2 variables.

-11y-9z=11
14y+11z=14

So this is what i got, but i still don't know how to solve the rest of the way.

If you want the old substitution way... Make it so that the y-coefficients are the same...

$ \minus{} 11y \minus{} 9z \equal{} 11$ <multiply by 14
$ 14y \plus{} 11z \equal{} 14$ <multiply by -11

becomes

$ \minus{} 154y \minus{} 126z \equal{} 154$
$ \minus{} 154y \minus{} 121z \equal{} \minus{} 154$

Magically subtract...

$ \minus{} 5z \equal{} 308$
$ z \equal{} \minus{} \frac {308}{5}$

Solve from there.

HEY WAIT: I think you messed up on your last post...  :o

-11y-9z cannot be both 29 and 11. There is a problem with TPotter26's post.

[quote="TPotter26"]
3x+2y-3z=20
3x-9y+6z=9

-11y-9z=29

I don't understand what to do from this point...[/quote]
Okay.
So we will get:

11y-9z=11
-11y-9z=29

woah i think i messed up.

Didn't you post the exact same problem in Classroom Math?

Oh he did.
It's [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225835[/url]

It is not really Highschool Basics, it is more like Intermediate Middle School Topics.

You mean Intermidiate High School?  :wink:

No, it is too easy for highschool, it should be in the MOEMS section, the Classroom math section, or the MC section.


function
algebra
functional equation
algebra unsolved

Find every function $ f$ that is continous on $ \mathbb{R}$ and satisfy $ f(x) \equal{} x \plus{} f(x \minus{} f(x)),\forall x\in \mathbb{R}$.

Two clear solution is $ f(x) \equal{} x$ or $ f(x) \equal{} \minus{}x$.
Let $ g(x) \equal{} x \minus{} f(x)$.
Clearly, $ g(x)$ is continuous, $ Im(\mathbb{R})$ is an interval, and that for all of this interval, $ f(x) \equal{} \minus{}x$
[b]Case 1[/b]: Suppose the interval formed, is a point, i.e. the function is in the form of $ f(x) \equal{} x \plus{} c$, it can be easily checked that $ c \equal{} 0$.
[b]Case 2[/b]: Suppose the interval is bounded, and the upper bound and lower bound is $ a$ and $ b$ respectively. Clearly, for this interval of $ x$, $ f(x) \equal{} \minus{}x$. Hence, we get $ x \minus{} f(x)$ can attain another interval, from 2a to 2b, it is impossible for this interval to be a subset of the image interval, and leads to contradiction.
[b]Case 3[/b]: Suppose the interval is not bounded, but if this interval covers the whole $ \mathbb{R}$, the result would be trivial.
[b]Case 3.1[/b]: Suppose the interval has a lower bound, is the lower bound is smaller than zero, it can be proved by the similar tricks in Case 2, that the interval enetually covers the whole range. So, suppose the lower bound is larger than zero. However, I still don't have any idea on this part up till now...

Hope someone, can help me to finish this proof...

i wish you finish your proof , coz i really enjoy it  :D  ,  am just asking if an other approach... like using sequences would be simpler  :blush:
for example : $ \forall x$ we state $ u_0(x) \equal{} x$ and $ u_{n \plus{} 1}(x) \equal{} u_n(x) \minus{} f(u_n(x))$
from the functional equation we get : $ u_{n \plus{} 1}(x) \plus{} f(u_{n \plus{} 1}(x)) \equal{} 0$ so : $ 2u_{n \plus{} 1}(x) \equal{} u_{n \plus{} 2}(x)$
so $ \forall n\ge1$ we have $ u_n(x) \equal{} 2^{n \minus{} 1}u_1(x) \equal{} 2^{n \minus{} 1}g(x)$

I think you should wath your solution, the functions $ f(x)\equal{}|x|$ and $ f(x)\equal{}\minus{}|x|$ are also satisfy our condition :D

Let $ A\equal{}\{x\minus{}f(x)|x\in\mathbb{R}\}$.
If $ a\in A$, $ f(a)\equal{}\minus{}a\Rightarrow 2a\equal{}a\minus{}f(a)\in A$.
Since $ x\minus{}f(x)$ is continuous, $ A$ has to be interval or just one point.
If there is positive $ a\in A$, $ 2^na\in A$ implies $ A$ has no upper bound.
If there is negative $ a\in A$, $ 2^na\in A$ implies $ A$ has no lower bound.

Case 1:
$ A$ is one point. $ f(x)\equal{}x\minus{}c$, $ f(c)\equal{}\minus{}c\equal{}c\minus{}c\equal{}0$. So $ f(x)\equal{}x$

Case 2:
$ A$ has both positive and negative. $ A\equal{}(\minus{}\infty,\infty)$. $ f(x)\equal{}\minus{}x$.

Case 3:
$ A$ has positive, no negative. $ \inf_xA\equal{}a\geq0$. Since $ x\minus{}f(x)$ is continuous, $ a\in A$. So $ A\equal{}[a,\infty)$.
Then $ f(x)\equal{}\minus{}x$ when $ x\geq a$; $ f(x)\leq x\minus{}a$ when $ x<a$. It is easy to check the condition holds as long as continuity holds.

Case 4:
$ A$ has negative, no positive. $ \sup_xA\equal{}\minus{}a\leq0$. Since $ x\minus{}f(x)$ is continuous, $ \minus{}a\in A$. So $ A\equal{}(\minus{}\infty,\minus{}a]$.
Then $ f(x)\equal{}\minus{}x$ when $ x\leq \minus{}a$; $ f(x)\geq x\plus{}a$ when $ x>a$. It is easy to check the condition holds as long as continuity holds.