typical american arrogance :D (no offense to my american friends)
http://gprime.net/video.php/foxoncanada

of course, there is also this classics "talking to americans" on cbcs this hour has 22 minutes by rick mercer... check it out!!

http://home.comcast.net/~wwwstephen/americans/

everyone from Harvard Students to Jerry Springer to George W Bush

[quote]
"Congratulations Canada on getting FM radio."

"Congratulations Canada on getting running water in all five states."
[/quote]

oh yeah, if you didnt know, Jean Chretien (not Jean Poutine) is the former prime minister of canada for i think ten years or so.

of course, as much as I enjoy the videos, I dont support making fun of average american citizens.  Only making fun of politicians should be allowed ;)

also see a wiki entry: http://en.wikipedia.org/wiki/Talking_to_Americans

I have notice that you say

$\displaystyle i \equiv 5 (mod 13)$ where $i^2=-1$. You know that it is false, don't you!, at least it is false in $\mathbb{Z}[i]$, because there is no element $x\in\mathbb{Z}_i$ such that $x(5-i)=13$, and that is quite easy to show.

I suppose that your argument is that $5^2=25\equiv-1=i^2(mod 13)$, but that  is not correct, otherwise we would have $6\equiv7(mod 13)$!!

Best,

really? I didnt know that.  thanks for pointing it out!!  as you can see, we all are ignorant to something  :D   but i remember from The Art of Craft of Problem solving that 'i' is congruent to something which i forgot.  they used it to solve equations like x^2+25y^2 or something like that

Hahaha I LOVE the "TAlking to Americans!"

Its sooo funny, not true but still hilarious.  "Congratulations Canada on getting your first University!"  And the seal hunt in Saskatchewan too...

Good show

what do you mean by not true?  i think they are real interviews... :)
but mercer's way of deplicit average american citizens is too harsh, he should only target harvard students and politicians :D

Oh I know they are true interviews, but this is like....1%, maybe not even, of America's population.

Still hilarious!

Do you Canadians do any math on those bobsleds your always on?  :D  :D

It would only seem logical, after seeing that at last years IMO one of the 4 perfect scores was a 10th grader from Canada  :?  ;)  :D

Somehow.. I've never seen one of those interviews, but I've heard many things about it.. ;)

Are Americans really THAT ignorant?  :D
(Even though I'm sure it's like 1 in 10000.. )

Also wondering: My school offers an american history course.. Does any american high schools offer Canadian history course?
That might help in acquiring the knowledge of who the prime minister of Canada is.. (Hint: it's not Tim Horton!!!  :lol: )

[quote="jpark"]That might help in acquiring the knowledge of who the prime minister of Canada is.. (Hint: it's not Tim Horton!!!  :lol: )[/quote]
Mmmmm....donuts....

We have Tim Horton's here in the States now, but not quite at the density that they exist at in Ontario.  And not at all here in California  :(

Wait, so Tim Horton retired as Prime Minister?  :D

I find Canada really nice, so I go there about twice or so a year. It's one of my top three favorite vacation places. Maine is first though.  :)

well lets see if Americans knows their own cities/states well enough :D:D:D Trivia question: which place is south of Detroit (notice how I did not say Michigan :D)?  no peeking at the map!  :D  hide your answer please :D :D :D

Well, there's a ton of places south of detroit. But the one your talking about may be... [hide]chicago? or is it ann arbor?[/hide]

There's like a ton of places south of Detroit!

Okay, here is a full fledge quiz  :D :D
[moved to a new thread]

Besides question 2, I fail to see the relevance of the rest of this test with US geography :D

:D okay, geography around the US :D

btw the quiz is courtesy of Soroban :D

[quote="11th dimension"]Wow, we are the only forum in North America that currently has posts...LOL[/quote]

There are posts in the other forums, it's just that you have to be a part of their group in order to view the posts. Apparantly they want privacy. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22123]here[/url]for more details.

That's a small "bug" :) (0 posts thing). Well it's not actually a bug, since most users don't have access but to 0 threads in those forums, thus they have 0 topics :)

In plus it seems that Billzhao "artificially" increased the post count in the Canada forum over night :D :D :D

[quote="jpark"]Nice to see that Canada is FINALLY the first in something..
(Wait, we're the forum with most posts, right?  :D )[/quote]...  :? ... you are not :)

Aww... Well, we'll catch up, right canadians??

(we'd better!!  :D  ;)  )

Hehe now that Sarah's back in action after a week, There'll be a lot more spam!!!

[quote="bugzpodder"]So I assume only ppl with canadian flags could visit this part?  :O   :D[/quote]I visit, and I don't have a Canadian flag :D ... :P

[quote="Valentin Vornicu"]I visit, and I don't have a Canadian flag :D ... :P[/quote]

Ooooh red alert, outsider in the boards!  Hehehe.

Michigan is practically part of canada... eh?

is this forum private or not??   :?  :(  :?

[quote="bugzpodder"]is this forum private or not??   :?  :(  :?[/quote]

Not currently. I feel that if we make it private, we may block out potential Canadian members as well (especially since some people find this forum through google, which would not be possible if the forum were private).

This may change however, if we gain sufficient popularity in the future. But for now, we'll just have to rely on the honour system to keep the outsiders out.


In Math class today, we learned about Conditional Proofs.  Our homework was about it too, but there are two problems which I don't understand at all.  can anyone help me with it?

$ 1.      \text{Premise:   }\\
\text{                     P}\\
\text{                     Q}\rightarrow \text{R}\\
\text{Conclusion: (P}\rightarrow\text{Q)}\rightarrow\text{R}$

$ 2.$  $ \text{Premise: }$
           $ \text{                                      P}\rightarrow\text{(Q}\rightarrow\text{R)}$
           $ \text{                                      R}\rightarrow\text{S}$
           $ \text{                                      Q}$
$ \text{Conclusion: P}\rightarrow\text{S}$

I need to explain my answers using two-coloumn proofs.  [b]I am not asking for you to do my homework. Even not explaining answers in two coloumn format is fine.[/b]

Thanks.

oh snap i dont even know these wierd words but i will try to go through a run dow of what you need to do
first one:
you know p is true and r is true when q is true
so if the statement "if p then q" is true with the above information, then q will be true (using the detachment law blah) and r will be true again using detachment law junk

second one (actuly easier)
suppose p is true
then "if q then r" is true
and we already know q is true, so r must be true, then using the 2nd given, you can say s is true

im sorry if i am not following some rules for proving these problems


function
inequalities
algebra unsolved
algebra

[quote="perfect_radio"]Solve $2^{1-x} + 2^{\sqrt{2x-x^2}}=3$
in $\mathbb{R}$

$x=0$ and $x=1$ verify.
What about $x \in (0,1)$? How can you demonstrate that $2^{1-x}+2^{\sqrt{2x-x^2}}>3$?
[/quote]
Well, it's a high school problem, use Roll's Theorem, because only exists a point $x_0$ where the function $F(x)=2^{1-x} + 2^{\sqrt{2x-x^2}}-3$ has an extremal point between $(0,1)$ then as $x=0$ and $x=1$ are zeroes of $F(x)$ and $F(x)$ is continuous, then the function has not a sign change in the interval, and if $x->0^+$ then it's easy to show that $F(x)>3$. And that's all.

this was given at an olympiad for 10th grade. i think the proof shouldn't imply roll's theorem, which isn't studied in tenth grade.
i didn't really understand your proof.

Let him sleep. I will try to post soon a proof for 10-th grade.

Let $t=1-x$ which is also between 0 and 1. Suppose t is not 0 or 1. If $ 5t>3$ then apply the AM-GM inequality with the three terms $ 2^{t-1},2^{t-1}, 2^{\sqrt{1-t^2}}$ and prove next that $ 2(t-1)+\sqrt{1-t^2}$ is greater than 0. If $ 5t$ is at most 3, apply the Am-GM for the numbers $2^t,2^{\sqrt{1-t^2}-1}, 2^{\sqrt{1-t^2}-1}$ and prove next that $ t+2(\sqrt{1-t^2}-1)$ is greater than 0. That's all.


$ o\equal{}b\minus{}20$
$ 7b\equal{}9o$

Therefore, $ 7b\equal{}9(b\minus{}20)$

$ 7b\equal{}9b\minus{}180$

$ 180\equal{}2b$

$ b\equal{}90$

A banana has 90 cals.

[quote="ackter78692"]the orange has 20 fewer calories than a banana. if 7 bananas have the same #of calories as 9 oranges, how many calories r in a banana[/quote]

o = b - 20
7b = 9o

There is one more solution to this, and it is the elimination method. We can multiply the first equation by 9 to get;

9o = b - 20
9o = 7b

Subtract the top equation from the bottom equation to get;

0 = 6b + 20

Then you can just solve for b, and then substitute that value in to solve for o.

You need to multiply both sides of the equation by 9, wickedestjr.

You should actually get

9o=9b-180
9o=7b.

But this is just making more work for yourself. Elimination is a viable method to use here, but substitution is the easiest, especially because one equation is already in terms of a variable.

Oops!!! Sorry about that mistake. When you multiply one side by 9, you have to multiply the other one by 9 also.


Given n letters $a_1, a_2, \cdots$ find the length of the minimal string such that each letter is next to each other letter.
For example, for $a_1, a_2, a_3$, the shortest such string is $a_1a_2a_3a_1$ so the length is 4.

this is a very difficult question...it has been posted before...they are called brujin cycles:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=school&t=72392[/url]

(note that the last solution there does not really make sense)


How is a cyclic thioacetal group deprotected with I2 in aq. acetonitrile?
How is the thioacetal activated? I don't think it is by the formation of an S-I bond.
Isn't acetonitrile simply a solvent?
Thanks for your ideas!

In my opinion,Acetonitrile is just used as a polar solvent and also it's somewhat inert. We use such mixed solvent to adjust the appropriate condition for this reaction . :?:
Indeed. When we treat thioacetal with I2 ,sometimes oxidation occur;subsequently disulfide bond forms and protected substance become alkane.


inequalities
geometry
geometric transformation
rotation
triangle inequality

In triangle $ABC$,$AB$ is the longest side.Prove that if $P$ is a point on the interior of the triangle, then $PA$ $+$ $PB$ $>$ $PC$.
Isn't this trivial from Pompieu's theorem? Please help.

Could you state that theorem?  I don't know what you're talking about.

[hide="Anyway..."]
It's not too hard using the triangle inequality and some logic.

Let $AB=c, BC=a, CA=b$.  Now, it's not too hard to see that $PC\le \max\{a,b\}$.  Also, $PA+PB\ge c$.  So \[PA+PB \ge c > \max\{a,b\}\ge PC.\] [/hide]

Oops! Anyway Pompieu's Theorem is apllicable only for equilateral triangles.

All I can find for Pompieu's Theorem:

[quote="Google Search"]This result is intimately related to Pompieu's areolar derivative theorem which states that lim f c' f(z)dz = 2/~[f(z)] /y--,o ~r where ~ is the area ...[/quote]

[b]Pompeiu`s Theorem[/b] says:

If $P \in Int ABC$, where $ABC$ is an echilateral triangle, we have the result which says that:

$PA , PB , PC$ can be sides of a triangle.

[hide="Proofs"][b]$1$st Proof:[/b] $\rightarrow$ we make a classical rotation around $C$

[b]$2$nd Proof:[/b] $\rightarrow$ we take $3$ paralel lines from $P$ to the sides of our triangle ($AB,BC,CA$) and we obtain $3$ sides which are exactly as long as $PA, PB, PC$ and have their extremities on the sides of our triangle.

[b]$3$rd Proof:[/b] $\rightarrow$ it follows with the famous triangle inequality.

And there are many other proofs of this theorem :)

[b]PS:[/b] sorry for my english, i`m really tired.[/hide]


[img]http://img48.imageshack.us/img48/7366/mickeypp0.png[/img]

The firure shows three circles (representing head and ears of Mickey Mouse) where $a,\ b,\ c$ are respective radii, and $d$ is the length of the exterior tangent. Prove that: $x^{2}=\frac{a^{2}\cdot d^{2}}{(a+b)\cdot (a+c)}$

Wow, wow, wow

I think I forgot something...

Without trig.

 :rotfl:  :rotfl:


For $n$ sets $A_1,A_2,..A_n$ satisfying:
$|A{}_i{}_1\cup A{}_i{}_2 \cup...\cup A{}_i{}_k|\geq k \forall 1\leq i_1<i_2<...<i_k\leq n$.Show that:there exists $a_i\in A_i \forall 1\le i\le n$such that:$a_i\neq a_j\forall 1\le i<j\leq n$

This is just Hall's Marriage Theorem. Google "hall condition", "hall theorem", "perfect matching", "system of distinct representatives", etc. You'll find lots of information, and I do mean lots :).

But I can't find out its solution in MathWorld....


greatest common divisor
algorithm
number theory
Euclidean algorithm

What is the greatest common factor of 252 and 96?

OR, use the Euclidean Algorithm :D .

$ 252 \equal{} 2* 96 \plus{} 60$

$ 96 \equal{} 1 * 60 \plus{} 36$

$ 60 \equal{} 1*36 \plus{} 24$

$ 36 \equal{} 1*24 \plus{} 12$

$ 24 \equal{} 2 * 12$

Hence, the greatest common factor is $ \boxed{12}$.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

help me with this problem:
If R is commutative ring with 1, then either R is an integral domain or every non-zero element of P (prime ideal of R) is a zero divisor.

is $P$ fixed, or do you assert that there exists such a $P$ if $R$ is no integer?

P is not fixed.  So, if R is not an integral domain, there is a prime ideal P such that....

if $R$ is a ring, a standard application of zorn's lemma shows that the set of ideals consisting only of zero-divisors (incl. $0$) has a maximal element, and that this has to be a prime ideal $P$.

so you don't have to assume that $R$ has zero divisors $\neq 0$. but in this case, $P \neq 0$, and otherwise $P = 0$.

Thanks for your help, but how do we know that maximal element has to be a prime ideal?

assume $P$ is not a prime ideal, i.e. there are $x,y \in R$ s.t. $xy \in P$ and $x,y \notin P$. then $P+Rx$, $P+Ry$ are greater ideals than $P$, so that they have to contain elements $p+rx$, $p'+r'y$ which are no zero divisors. then also their product

$(p+rx)(p'+r'y)=pp'+pr'x+p'rx+rr'xy$

is no zero divisors, but it lies in $P$ - a contradiction.

I understood the statement as "If $R$ is commutative ring with $1$, then either $R$ is an integral domain or every non-zero element of [b]any[/b] prime ideal $P\subset R$ is a zero divisor." :maybe: Is this too strong to be true?


Factor $5^{1985}-1$ into three integers each of which is greater than $5^{100}$.

[quote="pbornsztein"]This one is from the IMO short-list 1985.

Pierre.[/quote]

... and can also be found in one of the Kvant issues... But anybody may post a solution which will become part of the ISL/ILL project.  :)

That is the question...  :D

There are many similar problems.

Are there any tricks or standard techniques for factorizing ???

Who knows...  :lol:
Each new problem is a bit new technique.

Heres one.

For any base b, exponent x:

bx - 1 can be factored into (b-1) (bx + bx-1 + bx-2 + ... + b 2 + b + 1).

Theres one for bx + 1. Anyone know what that is?

Heres [b]correct[/b] one.  ;)
For any base $b$, $x\in\mathbb{N}$:
$b^x- 1$ can be factored into $(b-1) (b^{x-1} + b^{x-2} + b^{x-3}+ ... + b^2 + b + 1)$.

For any base $b$, $x\in\mathbb{N}$, $x$ is odd:
$b^x+1$ can be factored into $(b+1) (b^{x-1} - b^{x-2} + b^{x-3}- ... + b^2 - b + 1)$.


trigonometry
geometry
circumcircle
power of a point

In quadrilateral ABCD, $\angle{ABC}=60^{o}$, $\angle{BAD}=\angle{BCD}=90^{o}$, AB=2, CD=1, the intesection of AC and BD is O. Find $\sin{\angle{AOB}}$.  (1996 Beijing Math Contest)

It can be done by trig, but the computations are horrible. Is there another way :?:

[hide="hint"]it is a cyclic quadrilateral[/hide]

[quote="4everwise"]... We also know that $2\angle BCA=\angle CAD,$ because the former subtends an arc twice the size of the latter. ...[/quote]
Why, [i]4everwise[/i] ? [u]Here $AB$ is the length of a segment and isn't the length of an arc ![/u]

I don't know any good ways to solve it. The answer is pretty ugly: $\frac{15+6 \sqrt{3}}{26}$. Here is one solution:

[hide]Since $\angle{BAD}= \angle{BCD}=90^{o}$, so A,B,C,D are concyclic. Let BA and CD meet at P, the $\angle{ADP}=\angle{ABC}=60^{o}$.

Let $AD=x$, then $AP= \sqrt{3}x$, $DP=2x$. By the Power of a Point theorem, there is: \[(2+\sqrt{3}x) \cdot \sqrt{3}x = 2x(1+2x)\] Thus $AD=x=2 \sqrt{3}-2$, $BC= \frac{BP}{2}=4-\sqrt{3}$.

In quadrilateral ABCD, we use Ptolemy's theorem: \[BD \cdot AC=(4-\sqrt{3})(2\sqrt{3}-2)+2=10\sqrt{3}-12\] Since $S_{ABCD}=\triangle{ABD}+\triangle{BCD}= \frac{3\sqrt{3}}{2}$,so $\frac{1}{2}\cdot (10\sqrt{3}-12) \cdot \sin{\angle{AOB}}= \frac{3\sqrt{3}}{2}$, thus $\sin{\angle{AOB}}=\frac{15+6 \sqrt{3}}{26}$.[/hide]

Oops, wasn't thinking.  :blush: Yeah, an answer of 1 doesn't seem right either.

Nice solution Shobber.

[color=darkblue]Denote the circumcircle $w=C(I,r)$ of the quadrilateral $ABCD$ and $m(\widehat{CBD})=x$. Thus,
$m(\widehat{ACB})=30^{\circ}+x$ $\Longrightarrow$ $m(\widehat{AOB})=30^{\circ}+2x$. Therefore, $AB=2r\sin\widehat{ACB}\ ,\ CD=2r\sin\widehat{CBD}$ $\Longrightarrow$
$1=r\sin (30^{\circ}+x)=2r\sin x$ $\Longrightarrow$ $\sin (30^{\circ}+x)=2\sin x$ $\Longrightarrow$ $\tan x=\frac{4+\sqrt 3}{13}$ $\Longrightarrow$ $\boxed{m(\widehat{AOB})=30^{\circ}+2\arctan\frac{4+\sqrt 3}{13}}\ .$
[b]Remark.[/b] $2\arctan\frac{4+\sqrt 3}{13}=\arctan \frac{24+7\sqrt 3}{33}\in \left(\frac{5\pi}{12},\frac{\pi}{2}\right)$. Verify if $\sin \left(30^{\circ}+2\arctan\frac{4+\sqrt 3}{13}\right)=\frac{15+6\sqrt 3}{26}\ .$[/color]


function
calculus
derivative
calculus computations

Show that the fuction$ f(x) \equal{} \frac{1}{x}\sin(x)$ for $ x\neq0$ can be extended to a continuous function on all of $ \mathbb{R}$.

Also, if I wanted to show that $ f(x)$ is uniform continuous, could I show it is by the Mean Value Theorem?

Does it have a limit as $ x\to0?$ If it does, then that tells you what the extension has to be.

Yes you can use the Mean Value theorem to show that it is Lipschitz, hence uniformly continuous. That will require some careful work on estimating the size of the derivative near zero - unless you want to go a little further and claim the the extended function is not merely continuous but in fact infinitely differentiable (even entire).


analytic geometry
geometry
inequalities
rectangle

How many square units are in the area of the region determined by the system of inequalities? Express your answer as a decimal.

\[ x\minus{}y\leq 4\\
x\geq \minus{}1\\
x\leq 3\\
y\geq \minus{}4\\
y\leq 5
\]

[hide]Ans:31.5 (if this answer is wrong, don't read on)
First, you draw out everything besides x-y is less than or equal 4    equation
You get a rectangle. Yay!.
Start by drawing the line, x-y=4. That is equal to x-4=y.
This intersects the right part of the rectangle  (with the equation, x=3) at (3,-1)
This intersects the bottom part of the rectangle( with the equation, y=-4) at (0,-4)
These points form a right triangle with the lower right vertex of the rectangle (at (3,-4))
Since the equation is: y is greater than or equal to x-4. Then the answer is the area above the line but in the rectanglel. This is also the area of the rectangle - the area of the triangle just formed.
area of rectangle:4*9=36
Area of triangle: 3*3/2=4.5
36-4.5=31.5 YAY!![/hide]


Is is true that given a positive integer n we can find a set of n natural numbers such that the sum of the elements in any subset is a perfect square? I think the answer is no, but I cannot prove that.

Well, the answer is clearly no, because as far as I know, it is already an open problem to know if there exist 4 squares whose sums in pairs are squares.

In another hand, I think this argument cannot be accepted to be a solution of the initial question, especially in a test in Math sup  ;)  (even at Louis-le-Grand).

Pierre.

Yes, this is what I also knew and I was just amazed by this problem. Well, probably the teacher whoi gave this problem solved that open question, who knows?  ;)


There was a cowboy. And he went into a town on Wednesday to settle some "unfinished business". It took him four days to finish that other cowboy up, then, satisfied with himself, he rode back on Thursday.

How is this possible?

A 10-year-old told me this so if you get it, remember that I know how easy it is.

I think that 10-year-old kid is possbly cleverer than I am, my answer should be completely wrong. :blush:

I guess the cowboy rode back on Thursday [b]a week after[/b] the "unfinished business".

I hope I have not embarrassed myself too badly with this answer! :P

[quote="Kev-Eile"]I love this one....

[hide]His horse's name was Wednesday[/hide][/quote]

Very well done Kev-Eile!  :P, I love this puzzle as well.

By the way, that 10-year-old kid is possbly cleverer than I am... :blush:

Then why did he ride back on "Thursday"? Did he steal the other guy's horse too? :huh:

grr I saw this exact same problem in a book except the name was thursday.....

yes easy once u think about it

[hide]No, he rode back on horse Wednesday, day Thursday[/hide]

[hide="classic"]horse's name was thursday[/hide]

[hide="Possibly"]
He finished 2 cowboys up
[/hide]

this is that type of thing that 10 year olds tell each other when i was 10 i heard this.....

well Imean, it's a popular teaser and many people know about it


So, how do you guys keep yourselves awake when you pull all nighters? I mean, Morrison n Boyd tends to get a little groggy after sometime.........

This post is a tribute to Shre and myself...

Tea and coffee FTW

anytime COFFEE
i hate tea.....but have to drink it once a day my mother loves it :(
agian what's FTW :rotfl:  :rotfl:

but not at night i normally like sleeping early morn and get up late at around 9:00 AM...except on days of pujai....where i ahve to :)

Lol.. I love coffee and shre loves tea.. But other than that minor difference, we both are caffeine ~almost~ addicted :)

Vodka heh? Lol.... Perhaps a bloody mary then :D

[quote="SOUMYASHANT NAYAK"]nothing less than vodka will do...
(It helps me relieve tension of nighters by driving me to sleep)
 :P[/quote]
and how could u forget ....what u need the most at nights a nice,wonderful,sweet,beautiful,little(may or may not be)[hide]SLEEP :P  :P  :P [/hide]

not for Soumya:Adaka da :P

What pervert? Ans it seems someone has finished nving bio fully  :P

DIAGRAMS... They are the bane of my existence... Havent started them.. Left them for some late night torture..... Have to keep the candles burning na.... Jag has become a night time reading favourite.... But yea, to answer your question finished NVing

soumya might feel insulted by that..he deserves much more than that  :P

was that sarcasm...or was that 'seriously' funny :)

i 'drink' music to stay awake...nothing of these energizers....
on an average just 2 cups of STRONG Coffee(and i f i am at chennai which has better qulaity cofee podi i drink 3-4 :D)
hey the link doesn't lead to anything useful

3-4 a, my mom will kill me. I drink 1 coffee and 2 teas on holidays and generally 2 teas on school days. It should work now da.

BTW, are you planning to come to Chennai anytime soon??

may be at best after another 4 months :D ....
don't worry we all will surely meet once in our lives....Surely :)

nice try da semma talent  :D  :D chinna vayasu  :P

@GV ada pavi sollavella :P
soumya ke bikini spelling kude theriyadhu :noo:
avan nejamave 'chinna payen semma talent :D  :D '

[quote="SOUMYASHANT NAYAK"]

Don't you know that dog is man's best friend??
And students are men..

[/quote]
shruthi.....is a girl(women) and a student :P

I am sorry, you may call it male chauvinism or sth....

But 'man' also refers to human beings in language though presently there are attempts to remove such words...

A female chairman is a woman and not a man...(not a chairwoman)

'Practice maketh a man perfect'

So do you think it doesn't refer to women... :wink:

[quote="Cyber-Shreyas"]I am a big fan of both Tea and Coffee. Infact I like Tea better (and also tea is healthier cause it contains bioflavanoids).. I drink Tea everynight to stay awake till ...whenever I sleep.

Ofc I try to maintain a limit everyday. Pardesi, how much do you drink per day (in terms of cups)...

Shruthi loves Naploean brandy, do you all know that.

And this is a tribute to rockesh and rohit.
[img]http://upload.wikimedia.org/wikipedia/commons/thumb/d/d8/Caffeine.svg/175px-Caffeine.svg.png[/img]

This was my quiz question but it also came in FIITJEE. :D
http://chronotron.wordpress.com/2007/09/22/the-quiz-code-2/[/quote]
My key to keeping awake is the FM ringing in my ears. It keeps jarring and keeps me awake.

@ shreyas: "TOO MUCH OF CAFFIENE IS BAD FOR UR HEALTH". So drink T bcs it has bioflavanoids(English NV).

[quote="Vihang"][quote="SOUMYASHANT NAYAK"]A student's best friend? : Tea or coffee??
Is it a question worth asking??
Stupid people...

Don't you know that dog is man's best friend??
And students are men..

=> Dog is student's best friend...

So throw away your coffee and tea cups ...

Have 10 hotdogs...(varies proportionally to your mass..)[/quote]

Saddam's shop FTW !  :D
Seriously ... I survived the camp only on Hotdogs ...[/quote]

you put it the best way...
Saddam's shop is the only solution..

[quote="Chapplikka"]Hey,
was is it called the saddam's shop?? Doubtless, the hot dogs there r simply world class...[/quote]
The owner of the shop in 2004 looked like Saddam ... apparently he made the best hotdogs, the quality has deteriorated a bit over time ...

Ada Paavi, Vignesh... I thought you were veg.. Well atleast by definition hot dogs are not veggie. :D Though veggie hot dogs may have existence.

Rak: Even tea has caffeine da.

Ama: Endha [url]http://www.mathlinks.ro/album.php?mode=attach&id=12297[/url] Moonjikku Bikini babes kekkudha?? (Note: The guy standing in the left wearing a reddish, maroon shirt)

Edit: Endha Moonjikkum. [url]http://www.geocities.com/astronomyolympiad/images/iao10/iao10.jpg[/url] (3rd from the right, am I right??)

ella theveye ellai.....yaradha theriyum naan enga amuchu vaye...treat ke badule edhu kuditha podhum

Wow... Do you know that caffeine is in the list banned drugs in the olympic committee.

Everyone's obsessed with bikini babes from Miami :D Lol, dei Shre Napoleon Brandy is ok man, but nothing like a bloody mary :D Thats a modified form of vodka...

And I usually drink 2-3 cups of hot stronnnng coffee everyday :D Used to drink only 2 but with study hols, its gone up...

I def think Miami babes is more interestin than nv'ing who Hargobind Khorana's life partner(I meant partner for his life in working with genetics) was...:D

Remember I was mentioning to the topic Miami babes and not the miami babes themselves... Jus like how colloid is a state of substance and not a substance by itself, here Miami babes is a topic and not the actual ppl...

[quote="Cyber-Shreyas"][quote="Chapplikka"]I def think Miami babes is [/quote]

Not 'is' its 'are'. Remember you are talking to the height of english grammar nv.. (not me, shru)[/quote]

Shre ur list seems to be smaller than usual, why not add more? Confused after chem?  :maybe:   :wink:   :rotfl:


trigonometry
limit
calculus
calculus computations

If $x$ is rational, there is some $M$ such that $m > M$ implies $m!x$ is an integer, so $\cos^{2n}(\pi m!x) = 1$ for all $n$ and so the limit is 1.  If $x$ is irrational, for every $m$ $|\cos(\pi m! x)| < 1$ and so the limit in this case is 0.


A square of sidelenght $ 7$ contains $ 51$ points .Show that at least three of these points lie inside a circle with radius $ 1$

[hide="Solution"]Split the square into a grid containing $ 25$ squares of sidelength $ \frac{7}{5}$. Then by the pigeon-hole principle, at least one of these squares contains at least 3 points.

Now isolate this square. The radius of the circle required to cover the whole square is equal to half the diagonal length. In a square of sidelength $ \frac{7}{5}$ this is ${ \frac{\sqrt{\frac{7}{5}^2+\frac{7}{5}^2}}2}$.

This is equivalent to $ \frac{7}{5\sqrt{2}}$ which is less than $ 1$. So a circle with radius $ 1$ can cover the entirety of one of the squares in the grid. Since one of these squares must contain at least 3 points, the circle can cover 3 points.[/hide]


Euler
geometry
incenter
circumcircle
projective geometry
analytic geometry
geometry unsolved

[quote="Moonmathpi496"]Prove that, the Euler line of the triangles $ IBC,ICA,IAB$ are concurrent where $ I$ is the incenter of $ \triangle ABC$.[/quote]
[b]Lemma:[/b] Given a triangle ABC. P lies in triangle ABC. $ O_a, O_b,O_c$ are the circumcenters of triangle $ PBC, PCA, PAB$ then the Euler line of triangle $ PBC, PAC, PAB$ are concurrent if and only if $ AO_a, BO_b, CO_c$ are concurrent. (*)
[b]Proof:[/b] (figure 1)
By [b]Desargues[/b]'s theorem we get 3 Euler lines are concurrent if and only if $ (G_bG_c\cap O_bO_c), (G_cG_a\cap O_cO_a), (G_aG_b\cap O_aO_b)$ are collinear.
$ \Leftrightarrow \frac {A''G_b}{A''G_c}.\frac {B''G_c}{B''G_a}.\frac {C''G_a}{C''G_b} \equal{} 1 (1)$
Because $ G_bG_c//BC$ then $ \frac {A''G_b}{A''G_c} \equal{} \frac {A'B}{A'C}$ ...
So $ (1)\Leftrightarrow \frac {A'B}{A'C}.\frac {B'C}{B'A}.\frac {C'A}{C'B} \equal{} 1$
$ \Leftrightarrow AO_a, BO_b, CO_c$ are concurrent ([b]Desargues[/b]'s theorem)
Return to this problem:
Let $ O_a, O_b, O_c$ are 3 circumcenter of triangle $ BOC, COA, AOB$. It's easy to see that $ O_a, O_b, O_c$ is the midpoint of arcs $ BC, CA, AB$, respectively.
Then $ AO_a, BO_b, CO_c$ are concurrent.
Applying lemma (*)->QED

Further, they concur on the Euler line of $ \triangle ABC.$ We use barycentrics WRT ABC.

Coordinates of centroid $G_c$ of $ \triangle ABI$ are $ G_c (2a \plus{} b \plus{} c: a \plus{} 2b \plus{} c: c)$

Coordinates of the orthocenter $H_c$ of $ \triangle ABI$ are $ H_c  ( b \minus{} c: a \minus{} c: c)$

$\Longrightarrow \ell_c \equiv  H_cG_c \equiv  c(b \plus{} c)x \minus{} c(a \plus{} c)y \minus{} (a^2 \minus{} b^2)z \equal{} 0$

Cyclically, equations of Euler lines $\ell_a,\ell_b$ of $ \triangle BCI$ and $ \triangle CAI$ are given by

$ \ell_a \equiv (b^2 \minus{} c^2)x \plus{} a(c \plus{} a)y \minus{} a(b \plus{} a)z \equal{} 0$

$\ell_b \equiv \minus{} b(b \plus{} c)x \plus{} (c^2 \minus{} a^2)y \plus{} b(a \plus{} b)z \equal{} 0$

Hence, $\ell_a,\ell_b,\ell_c$ concur at  $ X_{21} \left (\frac {a \minus{} b \minus{} c}{b \plus{} c}: \frac {b \minus{} a \minus{} c}{c \plus{} a}: \frac {c \minus{} a \minus{} b}{a \plus{} b} \right)$

Which is the Schiffler's point of $ \triangle ABC$ lying on its Euler line.

[quote="livetolove212"]By [b]Desargues[/b]'s theorem we get 3 Euler lines are concurrent if and only if $ (G_bG_c\cap O_bO_c), (G_cG_a\cap O_cO_a), (G_aG_b\cap O_aO_b)$ are collinear.
$ \Leftrightarrow \frac {A''G_b}{A''G_c}.\frac {B''G_c}{B''G_a}.\frac {C''G_a}{C''G_b} \equal{} 1 (1)$
[/quote]

[quote="livetolove212"]So $ (1)\Leftrightarrow \frac {A'B}{A'C}.\frac {B'C}{B'A}.\frac {C'A}{C'B} \equal{} 1$
$ \Leftrightarrow AO_a, BO_b, CO_c$ are concurrent ([b]Dersaques[/b]'s theorem)
[/quote]
Could you please explain these two parts? Unfortunately I don't understand them  :maybe:

I let A' be the intersection of BC and $ O_bO_c, A''$ be the intersection of $ G_bG_c$ and $ O_bO_c$.
[quote="livetolove212"]By [b]Desargues[/b]'s theorem we get 3 Euler lines are concurrent if and only if $ (G_bG_c\cap O_bO_c), (G_cG_a\cap O_cO_a), (G_aG_b\cap O_aO_b)$ are collinear.
$ \Leftrightarrow \frac {A''G_b}{A''G_c}.\frac {B''G_c}{B''G_a}.\frac {C''G_a}{C''G_b} \equal{} 1 (1)$
[/quote]
This part is Menelaus's theorem for triangle $ G_aG_bG_c$.
[quote="livetolove212"]So $ (1)\Leftrightarrow \frac {A'B}{A'C}.\frac {B'C}{B'A}.\frac {C'A}{C'B} \equal{} 1$
$ \Leftrightarrow AO_a, BO_b, CO_c$ are concurrent ([b]Dersaques[/b]'s theorem)
[/quote]
Fom $ \frac {A'B}{A'C}.\frac {B'C}{B'A}.\frac {C'A}{C'B} \equal{} 1$ we get $ A',B',C'$ are collinear then applying Dersargues's theorem we get $ AO_a, BO_b, CO_c$ are concurrent.

When P is altitude of triangle, your solution isn't right because $ O_bO_c$ doesn't tough BC.
Can U slove it when $ O_bO_c$//BC?

No, my solution is still right if we note that $ O_bO_c$ intersects $ BC$ at the infinity point.

Not right for high school students! Can U slove it for all school student?


function
induction
algebra unsolved
algebra

The sequence $ (x_n)_{n \geq 1}$ is defined by:
$ x_1\equal{}1$
$ x_{n\plus{}1}\equal{}\frac{x_n}{n}\plus{}\frac{n}{x_n}$
Prove that $ (x_n)$ increases and $ [x_n^2]\equal{}n$.

[quote="Thjch Ph4 Trjnh"]$ x_{n + 1} > x_n$.
$ \rightleftarrows x_n^2 + n^2 > nx_n^2$.
$ \rightleftarrows x_n^2 < \frac {n^2}{n - 1}$.
Using induction proof:
$ \sqrt {n} \leq x_n < {\sqrt {n + 1}}$.[/quote]
Shorter !: Using simple proof, $ x_n$ is increasing and $ [x_n^2]=n$

More seriously, could you develop your proof for us, please ?

I'm sorry to insist, but could you kindly give us your solution, Thjch Ph4 Trjnh, please ?
I dont succeed in showing your so simple induction.

Thanks.

[quote="Thjch Ph4 Trjnh"]Assume holds for $ n$:$ \sqrt {n}\leq x_n\leq \sqrt {n \plus{} 1}$.
Function $ f(x) \equal{} \frac {x}{n} \plus{} \frac {n}{x}$ increase on $ [\sqrt {n}, \plus{} )$.
Thus $ f(\sqrt {n})\leq x_{n \plus{} 1} \equal{} f(x_n) < f(\sqrt {n \plus{} 1})$.
$ f(\sqrt {n}) \equal{} \sqrt {n} \plus{} \frac {1}{\sqrt {n}}\geq \sqrt {n \plus{} 1}$,
$ f(\sqrt {n \plus{} 1}) < \sqrt {n \plus{} 2}$.[/quote]

Ok, thanks (I really did not see this  :blush: ).

Notice that $ f(\sqrt {n \plus{} 1}) < \sqrt {n \plus{} 2}$ is not true for the lowest values of $ n$ so the start of induction needs to be taken about $ 3$.


Hello  :) ,
I have a book and its state the interior area if the circular sector in a strange way  :(. Its written as $ A\equal{}90r^2\theta/\pi$ as this format. Do you see how its should be written as?

And what is the parameter of the circular sector??? Its written as $ B\equal{}2r(1\plus{}90\theta)/\pi$ but I still can't get it  :(

Maybe this will help?
http://en.wikipedia.org/wiki/Circular_sector


[quote="pluricomplex"][quote="Fermat -Euler"]Find non-negative integers $x, y, z, t$ such that $5^x + 6^y + 7^z + 11^t = 2005$[/quote]
Can i use the very quite slow computer man?  :D[/quote]
but i want to solve it without computer  :)

I think $(x,y,z,t)=(4,1,2,3)$ (gives 2011) and $(x,y,z,t)=(4,2,1,3)$ (gives 1999) are the closest you can get.

$5^x + 6^y + 7^z + 11^t = 2005$. Take the expression modulo 5
\[ 1 + 2^z + 1 = 0 ( \mbox{mod } 5 )  \]
Thus $2^z = - 2 = 3 ( \mbox{mod } 5 )$, and this means that $z = 4 k + 3$ for some $k$ positive integer (check !). Now take the original expression modulo 6
\[ ( - 1 )^x + 1 + ( - 1 )^t = 5 ( \mbox{mod } 6 )  \]
So that $( - 1 )^x + ( - 1 )^t = - 2 ( \mbox{mod } 6 )$, so $x$ and $t$ are odd (1). Now observe that for any $k \geqslant 1$ we have $4 k + 3 \geqslant 7$, and $7^7 > 2005$, thus
\[ 2005 < 7^7 < 7^{4 k + 3} < 5^x + 6^y + 7^{4 k + 3} + 11^t = 2005  \]
A contradiction. So $k = 0$ and $z = 3$. This means that
\[ 5^x + 6^y + 11^t = 2005 - 7^3 = 1662  \]
Take the expression modulo $3$, then
\[ ( - 1 )^x + ( - 1 )^t = 0 ( \mbox{mod } 3 )  \]
Homever this means that $x$ and $t$ cannot be both odd or even, which contradicts the remark from (1). There is no positive integer $x, y, z, t$ such that $5^x + 6^y + 7^z + 11^t = 2005$.


Let be $ n\in N$ and $ z\in C\minus{}R$ such that $ z^n$ and $ (z\plus{}1)^n$ are reals. Find the values of $ z$.

[quote="Adriana N."]Let be $ n\in N$ and $ z\in C \minus{} R$ such that $ z^n$ and $ (z \plus{} 1)^n$ are reals. Find the values of $ z$.[/quote]       z=[e^(ikπ/L)]/[2cos(kπ2/L)] $ \frac{\minus{}n}{4}<k<\frac{n}{4}$

$\text{Since }(i)^4 \text{ and } (1+i)^4 \text{ are both real numbers, we conclude that }z=\boxed{\pm i}\text{.  }$

Visit The Following Link For A More Logical And Proofed Answer:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2631644#p2631644[/url]

Consider the fact that $\frac{z+1}{z}$ are the $n$th roots of unity.


geometry
circumcircle
angle bisector
geometry unsolved

Let ABC be a triangle. Let D be the midpoint of side BC. Let E be a point on
BC such that AE is the angle bisector of \BAC. The circumcircle of DEA intersects
AB at X and AC at Y . Prove that BX = CY .

Let $ \Gamma$ be the circumcircle of $ DEA$.
The power of $ B$ with respect to $ \Gamma$ is $ BE \cdot BD \equal{} BX \cdot BA$.
The power of $ C$ with respect to $ \Gamma$ is $ CD \cdot CE \equal{} CY \cdot CA$.

Dividing these equalities we have $ \frac{BD}{CD} \frac{BE}{CE} \equal{} \frac{BX}{CY} \frac{BA}{CA}$.

From the bisector theorem we have $ \frac{BE}{CE} \equal{} \frac{BA}{CA}$. Also, since $ D$ is the midpoint of $ BC$, $ \frac{BD}{CD}\equal{}1$.
Hence $ \frac{BX}{CY} \equal{} 1 \rightarrow BX\equal{}CY$.

$ \mbox{Let } \Omega \mbox{ circumcircle of triangle ABC and } \Gamma \mbox{ circumcircle of triangle ADE. }\\
 \mbox{If } \Omega \cap \Gamma \equal{} \{A; F\} \mbox{, then prove that: } DF\perp BC.$


I took around 45 mins to figure out this cheap sum.....

Prove that :    n(n+1)(n+2)(n+3)  is not a perfect square.

dont laugh at me :?  :? ,i know this is pretty common,but i got it after doin cheap stuff...(i'll tell u what cheap stuff after somone's given the solution)

[hide="Answer"] It's impossible for $-n(n+1)(n+2)(n+3)$ to be a perfect square becuase negative numbers cannot be perfect squares.  We can prove this assersion as follows:

  Either a number is positive, negative, or 0 (we will not consider 0, since 0 squared is not condsidered a square).  The square of a positive number is positive, likewise, the square of a negative number is postive.  Thus, there are no negative perfect squares. [/hide]

G-UNIT, you must also prove that

[hide]
the number is indeed negative; keep in mind that he didn't say that $n$ was $\ge 0$.
[/hide]

I'm not sure if that was intended to be negative. It would be much clearer if Latex was used in the original post...

U guys,thats just a dash,nothins negative :rotfl:  :rotfl:
Sorry for the inconvienience chess(u kow u've deleted around 4 comments of mine ;)  :lol: )

Cmon guys,give the sol already!!Ps-the people who have SEEN this problem somewhere are requested not to post and let the others try and no hints.This includes eminem.(cuz i spent more than 30mins doin this question ;) )

ummm..ches64 did u read the section of my earlier post where I mentioned that we do no need to consider terms that are 0 because there product will be 0, and $0^2$ is not considered a perfect square.....?

  lol, shady you really ought to learn latex I spent over 3 hours coming up with that :P

  But keep in mind, apart from mathcounts, acuraccy is much more important than speed!

  Anyways, this is a pretty well-known problem (at least for me) and is usualy put in the "generalized" form:

 Prove that the product of three or more consecutive positive integers cannot
 be a perfect square.

Assuming n>0  and an integer etc.. (Read it only you want to .. it will give away the solution ..)
[hide]$n(n+1)(n+2)(n+3) = n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2) $
Now if you think $n^2+3n$ as a number $k$ and realize that
product  of two consecutive odd (or even)  numbers can  never be a square: .... because

[hide]Because  $k^2 < k(k+2) <(k+1)^2$  there is no integer  between $k$ and $k+1$
[/hide]
You are done!

Btw, one can prove, (though it is little harder) that  product of any consecutive numbers(>0)  can  never be a square (or a power)

There is an interesting proof (or a few proofs)  here for 5 numbers .. Here [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=kabootar&t=30932]here[/url]

or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6337&postorder=asc&highlight=kabootar&start=20]here ..[/url][/hide]

okay

I think you might be able to drive out some thoughts from the fact that the numbers are consecutive

since the numbers are consecutive

$n(n+1)(n+2)(n+3)$:arrow: $(n-1)(n+1)n(n+2)=(n^2-1)n(n+2)$

this equation means $n(n+2)$ has to equl $(n^2-1)$ in order to be the perfect square( which can't be true )

well, it doesn't so it'll never be the perfect squre

I think this is the easiest solution than any other's mentioned here

and also I know that someone's going to argue with me...

best wishes from
Chris Kang

Not necessarily. Remember, $2\cdot 8=16=4^2$. Also, $n^2-1=n^2+2n$ for $n=-\frac{1}{2}$, but we're looking for integers, so that doesn't work. However, you can still prove that it doesn't work.

exactly

they are all under assumtion that they are integers

Since everyone's given their sols,heres mine.....

I tried this question for 2 days,then i got sick of it and just to pass time,i put in values fer n and then realised somethin....it was always one less than a perfect square!!!Then i added one to the equation and proved it to be a perfect square!!!Hence,the original thing could not be a perfect square,hence proved i guess.

I liked gyan's sol,the simplest. :lol:

[quote].it was always one less than a perfect square!!!T[/quote]
Exactly! (this is why trying a few special cases, are often very helpful in  finding a solution.  So,  are two  consecutive odd (or even) numbers ($(x-1)(x+1)=x^2-1$ is $1$ less than a perfect square ..and the product of 4 consecutive numbers can be written as product of two consecutive odd (or even) numbers)

[quote=Gyan wrote]Exactly! (this is why trying a few special cases, are often very helpful in finding a solution. So, are two consecutive odd (or even) numbers ((x-1)(x+1)=x^2-1 is 1 less than a perfect square ..and the product of 4 consecutive numbers can be written as product of two consecutive odd (or even) numbers)[/quote]

isn't basically same as mine?

yup, thats basically the solution JF, the longer solution was suggested by shady who mentioned the pattern of $k^2-1$  that combined with noticing theres a direct proportion with the squared term and the sequence yields:

$n(n+1)(n+2)(n+3)={k}^2 - 1^2= [{(n+1)(n+2)-1]}^2 - {1}^2$

$n(n+1)(n+2)(n+3)=(n+1)(n+2) \cdot [(n+1)(n+2)-2]$

$n(n+3)=(n+1)(n+2) - 2$

$n^2 + 3n= n^2 + 3n$

  Thus: $n(n+1)(n+2)(n+3)=k^2 - 1^2$

  So the condition holds (there equal) and it can never be a perfect square.

I don't know why he put it here, but it isn't [i]too[/i] hard... :P


inequalities
invariant
triangle inequality
inequalities proposed

Let a,b,c and d be positive real numbers.
Prove that

\[|\max(a,b)-\max(c,d)|\leqq|a-c|+|b-d|\]

We have to prove that $\left|\max\left(a,b\right)-\max\left(c,d\right)\right| \leq \left|a-c\right|+\left|b-d\right|$. We will prove a stronger assertion (stronger since all three absolute values | max(a, b) - max(c, d) |, | a - c | and | b - d | are nonnegative): We will prove that

$\left|\max\left(a,b\right)-\max\left(c,d\right)\right| \leq \max\left(\left|a-c\right|,\left|b-d\right|\right)$.

We will distinguish the following 4 cases:

[b]Case I.[/b] We have $a\leq b$ and $c\leq d$.
[b]Case II.[/b] We have $a\leq b$ and $c\geq d$.
[b]Case III.[/b] We have $a\geq b$ and $c\leq d$.
[b]Case IV.[/b] We have $a\geq b$ and $c\geq d$.

Since Cases [b]III[/b] and [b]IV[/b] result from Cases [b]II[/b] and [b]I[/b] after interchanging (a, b) and (c, d) (this operation clearly leaves the above inequality invariant, since |-x| = |x| for every real number x), we need to consider the Cases [b]II[/b] and [b]I[/b] only. In Case [b]I[/b], the inequality is trivial, since max(a, b) = b and max(c, d) = d, so that it remains to prove $\left|b-d\right| \leq \max\left(\left|a-c\right|,\left|b-d\right|\right)$, what is trivial. So Case [b]II[/b] is the only one we have to consider yet.

In Case [b]II[/b], we have b = max(a, b) and c = max(c, d), so that we have to show

$\left|b-c\right| \leq \max\left(\left|a-c\right|,\left|b-d\right|\right)$.

Now, let x = c - d and y = b - a; then, $x\geq 0$ and $y\geq 0$ since $c\geq d$ and $a\leq b$. But a - c = (b - c) - (b - a) = (b - c) - y, and b - d = (b - c) + (c - d) = (b - c) + x. Thus, the inequality we have to prove rewrites as

$\left|b-c\right| \leq \max\left(\left|\left(b-c\right)-y\right|,\left|\left(b-c\right)+x\right|\right)$.

Now, if $b-c\leq 0$, then $\left|b-c\right| \leq \left|\left(b-c\right)-y\right|$, and if $b-c\geq 0$, then $\left|b-c\right| \leq \left|\left(b-c\right)+x\right|$. In both cases, we get $\left|b-c\right| \leq \max\left(\left|\left(b-c\right)-y\right|,\left|\left(b-c\right)+x\right|\right)$, so that the proof is complete.

  Darij

Thank you for your reply, darji.

You are right.Here is my solution.

$\max (a,b)=\frac{a+b+|a-b|}{2},\max (c,d)=\frac{c+d+|c-d|}{2}$ . From triangular inequality,

$|\max (a,b)-\max (c,d)|=\frac{\left|(a+b)-(c+d)+|a-b|-|c-d|\right|}{2}
\leqq \frac{|(a-c)+(b-d)+|a-b|-|c-d||}{2}$

$\leqq\frac{|a-c|+|b-d|+|a-b|-|c-d|}{2}$

On the other hand, $|a-b|=|(a-c)+(c-d)+(d-b)|\leqq |a-c|+|c-d|+|d-b|$, thus

$|a-b|-|c-d|\leqq|a-c|+|b-d|$ ,so we are done.

kunny

[quote="kunny"]From triangular inequality,

$|\max (a,b)-\max (c,d)|=\frac{\left|(a+b)-(c+d)+|a-b|-|c-d|\right|}{2}
\leqq \frac{|a-c|+|b-d|+|a-b|-|c-d|}{2}$[/quote]

How do you conclude this from the triangular inequality? I even think this conclusion is wrong (e. g., it is wrong for a = 1, b = 1, c = 2, d = 1). In fact, using the triangle inequality, you can get $\left|\left(a+b\right)-\left(c+d\right)+\left|a-b\right|-\left|c-d\right|\right|\leq \left|a-c\right|+\left|b-d\right|+\left|a-b\right|+\left|c-d\right|$, but not $\left|\left(a+b\right)-\left(c+d\right)+\left|a-b\right|-\left|c-d\right|\right|\leq \left|a-c\right|+\left|b-d\right|+\left|a-b\right|-\left|c-d\right|$.

  darij

Oh! Yes. You are right, darij. I made a mistake. :blush:
I will try again!

kunny


Prove vieta's formulas for quadratic equation without the quadratic formula.

*sigh*

I thought someone would ask this.  :P

Check this out:

http://mathworld.wolfram.com/VietasFormulas.html

by the factor theorem, a quadratic $ax^2 + bx + c$ can be written as $a(x-r_1)(x-r_2)$ (where $r_i$ is a root of the polynomial), and expanding that, we get $ax^2 - a(r_1+r_2)x +ar_1r_2$ and since the roots cannot have variables in them, we can compare coefficients of the x's, and we get that $\dfrac{(-1)^1a(r_1+r_2)}{a}=s_1$ and  $\dfrac{(-1)^2ar_1r_2}{a} = s_2$
which is vieta's theorem, and we are done


geometry
circumcircle
inradius
incenter
geometry solved

I couldn't draw a valid figure...

Given two circles internally touching each other, construct a triangle which has the smaller of the two as incircle and the bigger of the two as nine point circle.

i didn't understand the problem- :? -what does it mean. can u explain it a little bit better after CONSTRUCT A TRIANGLE

The problem is underdefined, infinitely many such triangles do exist. The 9-point circle radius is half of circumradius. Once you know the circumradius $R$ and the inradius $r$, you can calculate and construct the distance $d = OI$ between the circumcenter and the incenter:

$\frac 1 r = \frac{1}{R + d} + \frac{1}{R - d}$

$d = \sqrt{R^2 - 2rR}$

Connect the incenter $I$ and the 9-point circle center $N$. Let the line $IN$ intersect the 2 circles at points $T, U, V$, where $T$ is their tangency point. Draw a circle with center $V$ and radius $UV = TV - TU = R - 2r$ intersecting the line $IN$ at another point $W$ different from $U$. Find the midpoint $M$ of the segment $TW$ and construct a circle with center $M$ and radius $MT = MW = \frac{TW}{2}$. Erect a normal to the line $IN$ at the point $V$, intersecting the circle $(M)$ at a point $X$. The distance $VX$ is equal to $VX = d$. Draw a circle with center $I$ and radius $d$. The circumcenter can be anywhere on this circle, each point corresponding to a triangle with different angles. If you want just any triangle, construct an isosceles one. Select the circumcenter $O$ on the line $IN$ and construct the circumcircle with radius $R = TV$. Construct a tangent to the incircle perpendicular to the line $IN \equiv OI$, touching the incircle at the point $T$ (which is on this line) and intersecting the circumcircle at points $A, B$. Draw the remaining tangents to the incircle from these points, which intersect at a point $C$ on the circumcircle, thus completeng an isosceles triangle $\triangle ABC$.

Sol: Take any point X on the inner circle. Draw a tangent at X to the inner circle to cut the outer circle in M and D. One of these M and D is midpoint of the base and the foot of altitude. Take M as the midpoint of the base and draw a perpendicular to the tangent line at D.
Ley Y be a point on the opposite side of M to X such that MY=MX. Let T be be the point diametrically opposite to X on the inner circle.Draw YT and produce to cut the perpendicular at D, at a point A. From A, draw tangents to the inner circle to meet the base line at B,C. Clearly by constraction(i love this word)
the inner circle is the inner circle of triangle.
The last step is to prove that the outer circle is the nine point circle.(it's easy)

Y and X are symetrically placed M. If X is the point of contact of the incircle with base, Y is the point contact of the corresponding excircle by homothety. Hence M is the midpoint of the base BC. D is the foot of the altitude from A to the base. The outer cicle passes through the foot of the altitude and the midpoint of the base. Hence it's nine point circle.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ R \equal{} K[X,Y]$. Prove that every ideal $ I$ of $ R$ is isomorphic as an $ R$-module to an ideal $ J$ of $ R$ such that $ dim_K\frac RJ < \infty$.

[edit] The problem changed from $ n$ variables to $ 2$ variables.

Are you sure this is true? I can't see what's wrong with this counterexample :maybe: :

Take $ n \equal{} 3$ (so $ R \equal{} k[x,y,z]$) and $ I \equal{} (x,y)$. The ideal $ I$ can obviously be generated as an $ R$ module by two elements, but it cannot be generated by one (it is a prime ideal of height $ 2$). This is preserved by an $ R$-module isomorphism, so it must hold for $ J$ as well. Now since $ J$ is a proper ideal generated by two elements (proper because it is not principal), its height is at most $ 2$. We can thus find a prime ideal $ \frak p$ containing $ J$ and of height $ \le 2$. The dimension of $ R/\frak p$ is now at least $ 1$, contradicting the fact that $ R/\frak p$ is supposed to be a quotient of the Artin ring $ R/J$.

The assertion is clearly true for $ n \equal{} 1$ (when $ I$ is non-zero, of course), and it may be true for $ n \equal{} 2$; I haven't checked the latter.

I guess I have to check with the person that used this result in proving or citing the proof of some theorem.
There seems to be nothing wrong with your proof.


Support
integration
topology
real analysis
real analysis unsolved

Let $ u\in H^1(\mathbb{R}^n)$ have a compact support and weak solution of :$ \minus{}\Delta u\plus{}c(u)\equal{}f$ in $ \mathbb{R}^n$,where $ f\in L^2(\mathbb{R}^n)$ and $ c: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is smooth ,with $ c(0)\equal{}0,c'\geq 0$.prove that $ u\in H^2(\mathbb{R}^n)$.

Try to understand Thm 1 in $ \S$6.3.1 p.309. If you cannot, I will help.

Let $ \Omega$ be a bounded open set that contains $ \mathrm{supp}\,u$, then $ u \in H_0^1(\Omega)$. Since $ u$ is a weak solution of our problem, we have
\[ \int_\Omega \nabla u \cdot \nabla v = \int_\Omega fv - \int_\Omega c(u)v
\]
for any $ v \in H_0^1(\Omega)$. Take $ v = D_i^{ - h}D_i^h u$, where
\[ D_i^h u(x) = \frac {u(x + he_i) - u(x)}h.
\]
Then
\[ \int_\Omega \nabla u \cdot \nabla v = \int_\Omega |D_i^h \nabla u|^2
\]
and
\begin{align*} \left| \int_\Omega fv \right| \le \frac 12 \int_\Omega |D_i^{ - h}D_i^h u|^2 + \frac 12 \int_\Omega |f|^2 \le \frac 12 \int_\Omega |D_i^h\nabla u|^2 + \frac 12 \int_\Omega |f|^2. \end{align*}
Hence
\begin{align*} \int_\Omega |D_i^h \nabla u|^2 \le \int_\Omega |f|^2 - 2 \int_\Omega c(u) D_i^{ - h}D_i^h u \le \int_{\mathbf R^n} |f|^2 - 2 \int_\Omega D_i^h c(u)D_i^h u \end{align*}
It suffices to show that
\[ \text{(*)} \qquad \sup_{k\in\mathbf N} \int_\Omega |D_i^{h_k}\nabla u|^2 < \infty
\]
for some sequence $ h_k\downarrow0$ (and any $ i = 1,\ldots,n$). In particular, one may replace (*) with stronger condition:
\[ \limsup_{k\to\infty} \int_\Omega |D_i^{h_k}\nabla u|^2 < \infty.
\]
The "obvious" idea is as follows: since $ c'\ge0$,
\begin{align*} \limsup_{k\to\infty} \int_\Omega |D_i^{h_k}\nabla u|^2 & \le \int_{\mathbf R^n} |f|^2 - 2 \liminf_{k\to\infty} \int_\Omega D_i^{h_k} c(u)D_i^{h_k} u \\
& \le \int_{\mathbf R^n} |f|^2 - 2 \int_\Omega c'(u) |D_i u|^2 \\
& \le \int_{\mathbf R^n} |f|^2. \end{align*}
But I don't see how to prove the legitimacy of
\[ \liminf_{k\to\infty} \int_\Omega D_i^{h_k} c(u)D_i^{h_k} u \ge \int_\Omega c'(u) |D_i u|^2.
\]
:blush:

[quote="sam-n"]Let $ u\in H^1(\mathbb{R}^n)$ have a compact support and weak solution of :$ \minus{} \Delta u \plus{} c(u) \equal{} f$ in $ \mathbb{R}^n$,where $ f\in L^2(\mathbb{R}^n)$ and $ c: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is smooth ,with $ c(0) \equal{} 0,c'\geq 0$.prove that $ u\in H^2(\mathbb{R}^n)$.[/quote]

Oh no, in whole $ \mathbb R^N$ we have a short approach. For each $ h \in \mathbb R^N, h \ne 0$ fixed, we denote
\[ \left( {D_h u} \right)\left( x \right) \equal{} \frac{{u\left( {x \plus{} h} \right) \minus{} u\left( x \right)}}{h}.
\]
By a weak solution $ u$ of $ \minus{} \Delta u \plus{} cu \equal{} f$ in $ \mathbb R^N$ we mean that
\[ \int_{\mathbb{R}^N } {\nabla u\nabla \phi dx}  \plus{} \int_{\mathbb{R}^N } {cu\phi dx}  \equal{} \int_{\mathbb{R}^N } {f\phi dx} \quad , \quad \forall \phi  \in C_0^\infty  \left( {\mathbb{R}^N } \right).
\]
Since $ u \in H^1$ then we can put $ \phi  \equal{} D_{ \minus{} h} \left( {D_h u} \right)$ which yields
\[ \int_{\mathbb{R}^N } {\left| {\nabla D_h u} \right|^2 dx}  \plus{} \int_{\mathbb{R}^N } {c\left| {D_h u} \right|^2 dx}  \equal{} \int_{\mathbb{R}^N } {fD_{ \minus{} h} \left( {D_h u} \right)dx} .
\]
Therefore
\[ \left\| {D_h u} \right\|_{H^1 }^2  \leqslant \left\| f \right\|_{L^2 } \left\| {D_{ \minus{} h} \left( {D_h u} \right)} \right\|_{L^2 } .
\]
On the other hand, it is known that
\[ \left\| {D_{ \minus{} h} v} \right\|_{L^2 \left( \Omega  \right)}  \leqslant \left\| {\nabla v} \right\|_{L^2 \left( {\mathbb{R}^N } \right)} ,\forall \Omega  \subset  \subset \mathbb{R}^N ,\forall h
\]
which gives
\[ \left\| {D_{ \minus{} h} v} \right\|_{L^2 \left( {\mathbb{R}^N } \right)}  \leqslant \left\| {\nabla v} \right\|_{L^2 \left( {\mathbb{R}^N } \right)} ,\forall v \in H^1 \left( {\mathbb{R}^N } \right).
\]
Thus $ \left\| {D_h u} \right\|_{H^1 }  \leqslant \left\| f \right\|_{L^2 }$. This allows us to deduce that
\[ \left\| {D_h \frac{{\partial u}}
{{\partial x_j }}} \right\|_{H^1 }^{}  \leqslant \left\| f \right\|_{L^2 } ,\forall j \equal{} \overline {1,N} .
\]
Thus $ u \in H^2$.

In general $ u \in H^{m\plus{}2}$ provided $ f \in H^m$. When $ m > \frac{N}{2}$ we also obtain $ u \in C^2$.

You treated $ c$ as a constant while it is a [i]nonlinear function.[/i]


geometry
geometric transformation
reflection
articles
abstract algebra
function
percent

Personally I like traditional chinese more :D

The one thing I notice about "simplified" Chinese characters is that because they have fewer strokes per character on average, they tend to be a bit less distinct from one another than are "traditional" characters, and thus perhaps marginally harder to read in some typefaces. But most simplified character typefaces I have seen in printed books are at least a little bit better looking than traditional character typefaces I have seen in printed books, so this is a minor difference.

All I ask is the publishing companies hire proofreaders  :(  because I have seen some amazing typographical errors in Chinese-language magazines (especially) over the years, and those are becoming more commonplace with modern computerized methods of production.

Isn't it great to have AoPS forums working again?

P.S., after edit: The poll should have an "I don't care" option, to reflect the choice of people like me.

I tried to add that, but it seems that the poll cannot be edited anymore after somebody has voted for it :(

I like simplified. Speed and simplicity matter more to me.

I prefer traditional Chinese, because I'm learning that in school :P .

Traditional Chinese is sort of more similar to Korean, and plus it has pronounciation clues.

For example, the character "ge" as in "chang ge" (which means sing) includes the character "ge" as in older brother. I haven't seen simplified Chinese before, but I doubt it has clues like that.

[quote="Treething"]Traditional Chinese is sort of more similar to Korean, and plus it has pronounciation clues.

For example, the character "ge" as in "chang ge" (which means sing) includes the character "ge" as in older brother. I haven't seen simplified Chinese before, but I doubt it has clues like that.[/quote]

Actually that exists in simplified too: 唱歌 and 哥哥.

i'd prefer simplified, more ppl are learning it...   traditional is  basically the same  as simplified and it takes more time.
it might look better but i think simplified is just simpler.

I have to say simplified because, well, I can't read traditional AT ALL. Plus I personally think simpler is prettier in this case. And it's faster.

[quote="RobinHood3000"]Personally, I prefer Traditional.  It presents more of a challenge, and just feels more Chinese, I guess.  Simplified seems to me to be the equivalent of Internet-speak in English, I suppose.[/quote]

I kno u didn't mean it that way, but don't b dissin my language. Simplified is just as grammatically correct as traditional. Plus, the vast majority of people use it. I'd say traditional is analogous to old english in english, not the number of charectors but the complexity of the structure of a sentence.

Most people these days learn simplified, including a vast majority of students in China. My personal view is that the purpose of learning a language is to communicate with the other people who know it, not because it looks pretty or sounds nice. To that extent, since simplified Chinese is becoming the contemporary standard, that's my preference.

tokenadult, I noticed that too while skimming some articles and whatnot. :-) But it's definitely explanable. There's new systems of typesetting these days that make typing chinese faster than, say, English and by a significant a difference too. So, it's obviously very easy to make mistakes like that. However, that's hardly an excuse and many major newspapers (e.g. People's Daily) seldom make such mistakes because they pay attention. Also, the myriad of homophones in Chinese make it extremely easy to make a mistake (especially when typing fast) and some are so subtle that even well-educated adults proofreading the text might miss it.

:) Well, I am from the mainland of China, I learn the simplified Chinese at school, I have learnt nothing about the traditional Chinese, but I can still read the whole traditional Chinese, and I can also write some of the traditional Chinese words, most young students have the same situation with me. I think the reason is that Chinese people have a natural feeling on the Chinese words no matter they are simplified or traditional. :D
  To me, I think the traditional Chinese looks more beautiful, but the simplified Chinese might be more easy to learn and be faster to write.

Simplified chinese is simpler in that the words are less complex.  Also, simplified to traditional isn't one-to-one.  Multiple words in traditional are written the same way in simplified.

[quote="Valentin Vornicu"]What is the big difference between traditional and simplified Chinese? Is it just in the way things are written / read? Or are some words changed as well?[/quote]

The way things are said is the same, although there are multiple dialects in China, each with a different way of pronouncing certain words.

A really good book to look up on the nature of Chinese writing is John DeFrancis's book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0824808665/learninfreed]The Chinese Language: Fact and Fantasy[/url], which gets into the facts around several of the issues being discussed in this thread. Professor DeFrancis has been a proficient speaker of Chinese since the 1930s and has long experience in researching writing reform in China.

[quote="JBL"]So, in other words, you're defending the efficiency of Chinese as compared to English by pointing out that we can use English to quickly render Chinese?  This seems highly ironic to me...[/quote]

You don't use English, you use transcripted Chinese  :) It is like writing in Russian using  Latin alphabet by the means of transliterator program. BTW, this is similar to "on-the-fly" compression of txt files: you write a word just how you say it and it is immediately compressed to take less space (up to 2 times). But while you cannot read compressed txt file without decompression by some program, you can easily read "compressed Chinese text"   :)

I must detalize, that of course, I don't believe you can reach the speed of 150-200 [b]Chinese characters[/b] p/m using pinyin method (because you should type 700-800 Latin characters p/m to do that). I believe you can reach the speed of 150-200 [b]English characters[/b] of the translated Chinese text  p/m (because 1-2 Chinese characters correspond to the whole word in English). I.e. you can type the same text as fast as English people would type it in English. But once again, Chinese will take less space. On the other hand, this is true for the [i]average text-writer[/i]. I know that Russian journalists use the method of speed writing called [i]stenography[/i], which allows typing the text much faster than the [i]average text-writer[/i] should, but before you must learn the system  :D  which is much like the separate language. I think, that using the method in which a Latin character stands for the certain part of Chinese character is the Chinese equivalent of stenography: you can be faster, but before you must learn it.

[quote="orl"] A reason that the Chinese system of writing is not abolished is the following fact: There are lots of different dialects in Chinese and lots of Chinese coming from different regions won't understand each other. But pointing to the characters often resolves the problem for them and they both know what they are talking about. For me it seems that Chinese is kind of XML document and people from different parts of the country have implemented different stylesheets, e.g. XSLT, in order to pronounce/render their characters accordingly.  :D[/quote]

Are you smart in Scottish Gaelic, [b]orl[/b]?  :D

[quote="nonIMO"]
Are you smart in Scottish Gaelic, [b]orl[/b]?  :D[/quote]

I know some basic phrases in Gaelic. There is even a Gaelic TV channel here. But it is hard to grap some phrases as most young Scottish don't know it because it is useless for them unless you are deeply connected with the culture and history with it. Thus especially just elder people know it. And I must say it is quite a weird language and I could explicitely name several reasons why it is an uncommon language.  :)

So, not only China has the dialect problem  :?  But basically I must agree that for China it is more large-scale problem than for other countries because of its big population. According to the latest [url=http://news.xinhuanet.com/english/2004-12/27/content_2383913.htm]research[/url] only 53 percent of population can communicate using standard dialect - Mandarin. There are many wide-spread dialects, mostly in the south-eastern part of the PRC. And if I should ask those Chinese, whose native dialect is Mandarin, if they can understand Chinese speaking Wu (Shanghaiese dialect) or Yue (Cantonese dialect), I should hardly hear: "Yes, we can"  :D

[quote="nonIMO"]According to the latest [url=http://news.xinhuanet.com/english/2004-12/27/content_2383913.htm]research[/url] only 53 percent of population can communicate using standard dialect - Mandarin.[/quote]

Wow! I had no idea the figure for Mandarin speakers in China ascertained by Chinese social scientists was still that low. The comparable figure for Taiwan has been higher for a generation--which illustrates the huge socioeconomic gap between Taiwan and China. Thanks for sharing the link.

[quote="kueh"]yeah... even in uk, the chinese people are normally be split into those speaking chinese and those speaking catonese. although those speaking catonese onyl come mainly from hongkong.[/quote]
People in Guangdong speak Cantonese. Cantonese in Chinese literally means "Guangdong language"...
They do not speak Mandarin well...

[quote="nonIMO"]According to the latest [url=http://news.xinhuanet.com/english/2004-12/27/content_2383913.htm]research[/url] only 53 percent of population can communicate using standard dialect - Mandarin.[/quote]
This percent might be increasing. I don't know the overall situation but in my city fewer and fewer people speak the local dialect. Many students at my age (including me) don't speak the local dialect but we can understand it. The situation is: people speak local dialects to local people, and they speak Mandarin to people from other places.

[quote="liyi"]I don't know the overall situation but in my city fewer and fewer people speak the local dialect. Many students at my age (including me) don't speak the local dialect but we can understand it. The situation is: people speak local dialects to local people, and they speak Mandarin to people from other places.[/quote]

I think this is not a good thing, because that means the local culture is going to die out. Luckily, because I live in rural areas, so I could still speak my local tongue very well while fewer and fewer urban students at my age could speak their local tongue

[quote="tokenadult"] The comparable figure for Taiwan has been higher for a generation--which illustrates the huge socioeconomic gap between Taiwan and China.[/quote]
You constantly broach off-topic subjects, [b]tokenadult[/b]  :D Just remind you that the topic is: "Tradtional Chinese Vs Simplified Chinese"  :)  I have some arguments in defense of mainland China's economics and I feel it would be long-long discussion, but I don't want to go into it neither here nor anywhere else. As concerns social situation, I think you should admit, that to teach 700-800 million people some language, especially in rural areas is not that easy  :)

What I am saying is that there is a known model for taking a population with fewer than 10 percent Mandarin speakers (that's a generous estimate) in 1949 and increasing the percentage of proficient Mandarin speakers to a much higher percentage, a good bit above the 53 percent figure you so kindly shared with the thread. (By the way, was that post of yours strictly on-topic in relation to the thread title?   :D  )  One aspect of that successful model, which is closely related to earlier messages in this thread, is the pervasive use of sound-indicating writing (the "bo po mo fo" characters of the zhuyin fuhao system) in elementary school textbooks and books published for children in Taiwan. My nieces were quite proficient in reading and writing anything they could speak in Mandarin, in that system, at first-grade age. The great increase in general literacy in China since 1949 has much to do with use of Hanyu pinyin alphabetic spelling in elementary reading instruction since the 1950s, but linguists who have compared the situation in the two countries note that textbooks (as I can verify) shift to all-Chinese-character writing at a much earlier age in China than in Taiwan, reducing the degree to which the standard national language is reinforced by school instruction.

Of course people don't learn the standard national language of a country only in school. I will give an example: my maternal grandparents were both born in the United States, but each received schooling only in the German language. They learned English, which is a language they could use with great proficiency, from their neighbors in their community, and in later years by listening to radio broadcasting. Similarly, one aspect of the amazing spread of Mandarin in Taiwan from a tiny minority language to the most interethnically useful common language is the use of that language in movies and TV broadcasting (radio broadcasting has always been dominated by Taiwanese) and in daily use contexts in ethnically diverse neighborhoods in Taiwan. Watching TV and watching movies, of course, requires being able to afford to pursue those recreations. Furthermore, internal travel and a broad availability of telephones (which tends to unify a national speech community) also happened much earlier and more pervasively in Taiwan, and that is how earlier economic development hastened the spread of Mandarin in one place rather than another.

[quote="tokenadult"]What I am saying is that there is a known model for taking a population with fewer than 10 percent Mandarin speakers (that's a generous estimate) in 1949 and increasing the percentage of proficient Mandarin speakers to a much higher percentage, a good bit above the 53 percent figure you so kindly shared with the thread. (By the way, was that post of yours strictly on-topic in relation to the thread title?   :D  )  [/quote]
I feel my fault too  :)  but it seems to me that this off-topic branch has its root in your statement that alphabetical system should be better for China.

I like Traditional because that's how I learned it.

But I guess I will still have to learn Simpliefied for colledge

After reading the extraordinarily long discussions, I have a few things to note:

(1)Japanese is strongly influenced by Chinese. Many of the characters used in Japanese nowadays come from Chinese, and some of them even get the exactly same meaning as Chinese. Of course there are some exceptions, like ?供, which means child in Japanese.
(2)Cantonese was once a major dialet in China long times ago(I think in the Song Dynasty).
(3)It's not a surprise that Guangdong people speak Putonghua badly. Putonghua is originally a dialet that Northern Chinese speak. Since Guangdong is located at the very south of China, the dialet used there naturally has a number of differences compared to Putonghua. Consequently Guangdong people cannot speak Putonghua well.


probability
geometry
3D geometry
pyramid

I posted this question once before but the right answer didn't come up.
3 numbers are chosen at random between 0 and 1. What is the probability that the difference between the greatest and the least is less than $\frac{1}{3}$.

Last question in AoPS probability chapter. I believe this will be helpful: [hide]geometric probability[/hide]

[hide]I thought I had a different idea than the AOPS book II, but is seems like my idea is the same, but the book puts it into educated terms to people like me can't understand...At least I figured it out...

Call the three numbers A,B,C  First consider is A<B and A<C

If A is between 0 and 2/3, there is a (1/3)^2 chance B and C are less than 1/3 away from it.

If A is between 2/3 and 1, then we need to find the average of all the numbers of 0-1/3 squared.  So, if we set up a square pyramid with base of area (1/3)^2 and a height of 1 we get a volume of 1/27.

So adding up all this casework we get (1/3)^2*(2/3)+(1/27)*(1/3)=7/81

But we only considered the case if A is the smallest, if we include if B and C is the smallest we multiply 7/81 by 3

7/27[/hide]


Find inverse map
a)$ f: R\to R$
$ f(2x)\equal{}3x\minus{}5|x|$
b)$ R\to [\minus{}1;1]$
$ f(x)\equal{}\frac{2x}{x^2\plus{}1}$

[hide="My hint"]
$ f(x)\equal{}\frac {2x}{x^2\plus{}1}$
$ x^2.f(x)\plus{}f(x)\equal{}2x$
$ x^2.f(x)\plus{}f(x)\minus{}2x\equal{}0$
$ x\equal{}\frac {\minus{}2\plus{}\sqrt {1\plus{}8x^3}}{2x^2}$
[/hide]

You are wrong! They don't have inverse map! :rotfl:  :rotfl:


induction
advanced fields
advanced fields solved

See here for more info:

[url]http://www.artofproblemsolving.com/Wiki/index.php/Mathematical_induction[/url]

[quote="James Bond007"]Its a pretty big topic. Submit Problems instead[/quote]Did you notice the topic was from January 2006?


So if we compute the inside of the parentheses, we get:
$ (5.5)^2\minus{}(4.5)^2$

And using the special form:
$ a^2\minus{}b^2\equal{}(a\plus{}b)(a\minus{}b)$, we plug it in:

$ (5.5\plus{}4.5)(5.5\minus{}4.5)$, and we simplify to

$ 10*1$, so it's 10!

Or just $ (a\plus{}b)^2\minus{}(a\minus{}b)^2\equal{}4ab\implies4\cdot5\cdot0.5\equal{}\boxed{10}$.


algebra
polynomial
linear algebra
matrix

I'm a little confused about how to do the following so any help would be appreciated.

Find the characteristic polynomial p(x) of G, the following matrix:

(2 0
-1 1)

Use p(x) to determine the eigenvalues of G, and use these to determine the eigenvectors/eigenspaces of G.  Are the eigenvectors corresponding to different eigenspaces linearly independent?

Okay, so I found the characteristic polynomial by doing

[X 0
 0 X]

minus
[2 0
-1 1],

which yielded

[x-2   0
  1    x-1]

(x-2)(x-1) - 0 = x^2 -3x +2 <- the characteristic polynomial

Therefore, the eigenvalues are 2 and 1.

However, I'm confused about how to determine the eigenvectors/eigenspaces of G.

For each eigenvalue $\lambda$, look at the null space of $\lambda I-A$; that's the eigenspace for $\lambda$.


Hey guys.
here are a few problems I came across and I'm not sure how exactly to solve them.

"A cricle has radius 3. Find the sum of the squares of the distances from any point on the circle to the vertices of an inscribed square."

There are also a few problems where there is a semicircle and a line, which is perpendicular to the diameter, goes to some point on the circle, then it wants to know what the length is.
I know it needs a picture but maybe my description can help:

"Semicricle with BD is perpen. to AC at B. AB=6, BC=24." Find BD.

Another problem I know it has come up in mathcounts, is the problem where there is a triangle and a small square at the base of it, and you have to find the length of one of the points on the triangle to one of the points on the square.

An easier one, I think:
"one diagonal of a rhombus is 4rt5, and the perimeter is 40. Find the other diagonal.)

That's it for now, thanks in advanced!

I think your problems require diagrams... at least... I need them.  :oops:

[hide="Rhombus Problem Solution"]The side length of the rhombus is obviously $ 10$. We want $ 2x$, where $ x^2 \plus{} (2\sqrt {5})^2 \equal{} 10^2$. $ x^2 \equal{} 80$, so $ x \equal{} 4\sqrt {5}$, so $ \boxed{2x \equal{} 8\sqrt {5}}$.[/hide]

well, i agree, your problems are the kind that almost always require a picture, but here goes nothing..

assuming that AC is the diameter of the circle, and B is a point on AC, and the intersection of the line perpendicular to AC and the circle is D, then that means that triangles ABD and CBD are similiar, so triangle ACD is a right triangle, with hypotenuse AC. If BC is 24, and AB is 6, then AC is 30 by the Segment Addition Postulate. So, by using Similiar Triangle Proportions, we get BC(the longer leg of the right triangle; Length 24)/CD(labeled x) = CD(the longer leg of triangle ACD; and labeled x)/AC(length 30). So, we can solve for CD, which will end up being the square root of 720. Then, assuming that all the work up until now was correct, you can solve for BD, which i believe will end up being the square root of 236 (?). i can't be quite sure because i'm doing all of this in my head really fast... so if i'm wrong with the calculations, let me know. And if my interpretation of the drawing is incorrect, then please give us an accurate drawing for the problem. or atleast a link. i hope this helped! :P