vector
algebra
function
domain
calculus
calculus computations

This is really a topological question about the domain ([url=http://en.wikipedia.org/wiki/De_Rham_cohomology]de Rham cohomology[/url]). If that domain is all of $\mathbb{R}^{3}$ (and the function is differentiable everywhere), the dimension is zero. The answer would be different on some other sets, such as $\mathbb{R}^{3}\setminus\{0\}$.

Actually, on $\mathbb{R}^{3}\setminus\{(0,0,0)\},$ as with any simply connected set, the dimension is zero. For something different, try $\mathbb{R}^{3}\setminus\{(0,0,z): z\in\mathbb{R}\}.$

Right. $\mathbb{R}^{3}\setminus\{0\}$ would be appropriate if we were looking at divergence and curl rather than curl and gradient.

sorry, i had a typo. yes the domain is not just R^3
fixed!

[quote="e^pi"]$V=\{\vec{F}: \mathbb{R}^{3}-X\rightarrow\mathbb{R}^{3}\text{s.t }\nabla\times \vec{F}=0\}$
$W=\{\vec{F}=\nabla g\}$

what is $\dim{(V/W)}$?[/quote]

basically this depends on X itself. for example if the domain is what Kentmerryfield said, then the domian is not 1-connected which implies that curl=0 is not a sufficient condition to show that F has a potential.

this problem, more complicated version of a classical physics problem: determining whether a static electric field necessarily has a potential function. If the vector field is allowed to be infinite or simply nonexistent at certain points (collectively indicated by X), the question becomes physically unrealistic but mathematically very rich; which is why most physics books say that vanishing curl implies existence of a potential.

For the particular case I mentioned, omitting the entire $z$ axis, we will have $\text{dim}(V/W)=1.$ Since amirhtlusa wants to talk physics, the generator of that quotient space will be the magnetic field of an infinitely long straight wire carrying a steady current.

With that modification to the problem- this is definitely [i]not[/i] elementary vector calculus. The theory applicable to this problem is at the advanced undergraduate/early graduate level- I learned it myself in a first-year graduate course. Follow the link in my first post for more information.

The question also doesn't have a simple answer.

can someone explain what the question is asking? :maybe:
so V is set of all vector fields with domain R^3-X and range R^3 s.t. its curl is the zero vector.
whats W?

what does $\dim(\frac{V}{W})$ mean?

$W$ is the set of vector fields on the same set which are gradients of something. Both $V$ and $W$ are (very large) vector spaces with $W\subset V$; $V/W$ is the algebraic quotient, which is also a vector space (finite-dimensional in many cases).

If you're just taking vector calculus now, you shouldn't understand this.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14070[/img]

 :(

A stay on the perpendicolar line r to HaMa on Ha. Call D the feet of the perpendicolar from I to HaMa, and E the symmetric of D w.r.t. Ma. DI meet the incircle on F, then $ A: EF \cap r$. Draw the tangent form A to the incircle that meet HaMa on B and C.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14073[/img]

 :thumbup:


Does anyone know anything about majoring in math at Georgia Tech (georgia tech)? I've heard that you can only major in applied math and discrete math for undergraduate. Does that mean that you can't take classes like undergraduate number theory or heavily theoretical calculus courses?

Check Georgia Tech's math department's

[url=http://www.math.gatech.edu/academic/undergraduate/degreePrograms/UGprograms.php]Undergraduate Programs[/url]

Web page, where indeed it appears that there is not much pure math on offer.


The sum of the digits of a positive two-digit integer is 45 less than the integer.  What is the tens digit of the integer?

The greatest sum of the digits of a two digit number is 9+9=18.

45 greater than 18 is 63.  So test 63.  63-9=54.  Fail.

A higher tens digit obviously won't help here, so try a lower one.

59-14=45.  I win. Tens digit is 5.

An algebraic way would be:

For a two-digit number $ ab$, with $ a$ and $ b$ representing the tens and units digits, respectively,
$ 10a\plus{}b\minus{}(a\plus{}b)\equal{}45$
$ 9a\equal{}45$
$ a\equal{}5$
Tens digit is 5.

Note that this works no matter what the units digit is.


geometry
3D geometry
pyramid
AMC
AIME
probability
USA(J)MO

What was your score?

109.5

I got #2 wrong because of some stupid bubbling error, and I got a lot of blanks: wicked geometry.

I got a 126, which was not at all what I was hoping for, although it is a 1.5 point improvement over the A (but still a 4.5 point drop from last year . . .  :( )

I made a stupid mistake on #20, and I wasted way too much time on #17 (almost 10 minutes  :o ).

This was really not a good AMC year for me . . .  I need to do better on the AIME.

Pretty sure I got 144... I thought the B was harder than the A this year though

Edit: I'm also unsure of #20 :/

124.5 (22 answered, 2 careless mistakes)... I thought B was harder also

138-144 :D
not sure for No.20  :P
What is the correct answer for No.20?  :maybe:

126

Hmm... I don't necessarily know that it was harder than the 12A, on which I got a 135.  The numbers, in this case, do lie.  At least numbers 1-24 (I didn't look at 25 yet) were very doable.  I personally didn't like number 20, the one with the truck and the person walking.

My story is this.  Up to 17, I was fine.  Then, for some reason, I couldn't solve 18.  I starting panicking, and it blend into the rest of the test.  I ended up missing 20, 21, 23, and 25.

For number 23, I should have had it.  I had $ \frac{n^2(n\plus{}1)}{2}\equal{}792$ instead of $ \frac{n(n\plus{}1)^2}{2}\equal{}792$.  Something wasn't working, so I guessed 12 instead of 11.

Number 21 I just misread.  Flat out misread.  When I finally read the problem correctly, I solved it in about 2 minutes.

Number 20 was a stupid mistake.

My favorite question on the test was number 24.  Nice pattern, nice question.  That was the one with  the graph of $ y\equal{}\sqrt{x}$ and the equilateral triangles and $ A_n$ and $ B_n$ and whatnot.

I think for #20 you could make it a lot more organized by drawing position vs. time graphs for both the person and the truck and count the intersections... but during the test I attempted #20 for 5 minutes and got really confused so I gave up.

I got a 145.5  :D - got #1-23 and #25 right, left #24 blank.  Yeah, I know, that makes me a huge loser for not guessing on the last one, but I'm pretty happy with my score.

When I was taking the test I didn't open it all the way, so after I finished #6, I accidentally went right to #20, the annoying garbage truck problem.  I thought it was #7 so I was expecting a really easy solution, but I wasted a lot of time failing to find one.  Finally I realized it was #20, solved it, and went back to #7.  Also, for some reason #17, the right triangle/parabola problem, was really hard for me.  Maybe if that didn't take me 10 minutes I would have had more time for #24.  :P

108 (I think, may have bubbled something wrong, probably not though).
Argh, I was being stupid on #23 and set stuff up wrong.

108. I failed  :(

I got #1 and #6 marked off for some freak reasons, even after I chekced them...also misread the pyramid one, #18, thinking the two sides were adjacent and couldn't solve. I wasted too much time on #20 (I guessed at last) and didn't get enough time to do the last five. In panic I skipped #22, which was fairly easy, and spent all the rest of my time on #23, and kept getting wrong answers because of a careless mistake at the beginning. So I ended up with guessing on #23 and leaving blank on 24 and 25.

I was expecting something between middle 120s to low 130s; had I got #1 and #6 right, I could get at least a 120. Anyways there is always next year, and also AIME for this year. The lessons I learnt were: 1, never guess; 2, get good sleep before the test!

As for the test it self, when I went over it after the test, I thought 24 was really nice; to my surprise, 25 was fairly easy in comparison. 20 was hard, or at least I was bad at this type of questions; I heard it was 25 on AMC 10B.

argh

127.5

missed the parabola/line parallel to x-axis one... because I forgot to subtract 1... argh

and then I missed the garbage truck one

I did, however, get lucky on the two intersecting circles probability thing

lol I think i got a 124.5 not sure... I forgot that $ |a\plus{}bi|\equal{}\sqrt{a^2\plus{}b^2}$, not $ a^2\plus{}b^2$... -_-... stupidness for the win...

I had 1 minute left and I almost had #24 -_-... oh well. #25 was a nice question in my opinion. Very simple if you knew what to do.  #20 =... I forgot to include the time they intersect in the middle  :wink:

Like last time, I got the first 23, but left the last two blank this time, for a 141. I actually thought the B was easier, I solved 1-23 much more quickly, and 25 was surprisingly easy to bash (which I found after the test).

i'm a grade 9 with 114 on 12b

what do you think my chances are of making usamo

please be honest :maybe:  :)

108

Somehow I guessed 6 questions right. Hmm...strange. Anyways, congrats to everyone who made AIME.

Ugh, 106.5.  Lowest I've ever gotten.  Answered 18, 2 stupid mistakes.  Didn't multiply by 4 after finding the area of a triangle in 15, and then messed up #20 like everyone else.

At least the no calc thing didn't do much to me.  After hardly any practice at all (only 2 tests the day before), the only difference from last year was 1 more wrong (last year was a 112.5).  And I'm the only person from my school to make AIME this year :stink:

21 people from my school qualified on the A, and a few more on the B.   :lol:

132, improvement of 19.5 points from last year.
missed my goal of 141 :(
eh, w/e

Hmm, I think I did about 30 points worse on this exam compared to last year.

Also, someone has the same avatar as me; it's blowing my mind.

[quote="zolojetto"]AGGGHHH...I didnt take AMC 12 just 10B (8th grd) 115.5 (unofficial) what were the answers!!!  and what will it take to make AIME for the B like top 1%[/quote]

The answers are posted on this website:

[url]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2008-12a/07-2-1012-Answers.shtml[/url]

ARGGGGGH i got a 105  :blush:
probably one of the worst scores

i went home and did them, and thought "HOW EASY!!"

overall i think it was easy, on contrary with my score

PS: my score was low because I was disturbed at the thought of someone chasing a garbage truck...
and didn't read the next few problems.

[quote="chris579"]I don't know why you guys had so much trouble on #20...
I only got a 99 (partially because of stupid mistakes) on th 12B,
but I got that one in less than a minute.  I thought it was fairly easy.[/quote]

many people thought they was only meet at the pails (which would be 4).

91.5 as a freshman but I know I did bad.  Any chance for me of  getting in the AIME?

Hey has any school gotten their AMC B results email?

The emails will begin going out today, Monday, March 10, 2008.  Postal mailed hard copies, will also start today.

Steve Dunbar
MAA Director, American Mathematics Competitions

93 on the AMC 12B only because I wasted too much time on one problem (the cone underwater one) and because I made a stupid mistake on #1 and #16. :mad:  :blush:


calculus derivative function trigonometry Asymptote conics parabola

This is just repeated application of chain rule (after a quotient rule.) It is really tedious - do you want to go through all the steps or is this enough?

[quote="holywrath"]Can some one show me how to find the first and second derivative of:

(cos((x+3)/x^2))^2+(sin(3x))^4-5
---------------------------------    (division bar)
4x^4+12x^3-8x^2+16x

thx[/quote]

holywrath, do you know the Lord? Go to [url=http://library.wolfram.com/webMathematica/Education/WalkD.jsp]this website[/url] and cut and paste the function below into the first field and click "DO IT".

(cos[ 1/x+3/x^2 ]^2+sin[3*x]^4-5)/(4*x^4+12*x^3-8*x^2+16*x)

So much for graded calculus homework. Well, I don't grade homework anyway, but give in-class quizzes based on it.

This example does appear unreasonably hard, so it may be that the teacher who assigned it deserves nothing better than a printout from wolfram.com  :ninja:

Thanks everyone.
i understand the quotient rule, but i don't get how the chain rule is applied and how did it get simplified down to the one they had up there.

and yea, my teacher loves to do this to my class...lol

I'll take one piece of the function: $\cos((x+3)/x^{2})$. This is a composition of two simple functions: $g(x)=\frac{x+3}{x^{2}}$ and $f(x)=\cos(x)$. In a formula, $f(g(x))=\cos((x+3)/x^{2})$. The Chain Rule says that the derivative of this composite function is the product of $f'(g(x))$ with $g'(x)$. So you compute $f'(x)=-\sin x$, insert $g$ into it to get $f'(g(x))=-\sin((x+3)/x^{2})$, and multiply all this by $g'(x)=-\frac{1}{x^{2}}-\frac{6}{x^{3}}$. (I did not want to use the Quotient Rule to compute $g'$, so I wrote $g(x)$ as $x^{-1}+3x^{-2}$.)

Try these [url=http://archives.math.utk.edu/visual.calculus/2/chain_rule.2/index.html]drill problems[/url]

omg so much easier when u explain it to me... i hope ure my teacher lol...
now i have hard time finding the asymptotes, i know 0 are both horizontal and verticle, but when i graph it with graphic calculator, i got another verticle asymptotes, i have no idea where that came from.. also the maxiume and minimum, i set the first derivative to 0, and then i make it 0/1, then i crossed multiply, left with the numerator = 0.. then i couldn't solve it...

[url=http://imageshack.us][img]http://img183.imageshack.us/img183/4418/untitledkq0.png[/img][/url]

Vertical asymptotes appear at the points where the denominator $4x^{4}+12x^{3}-8x^{2}+16x$ turns into $0$. One such point is $x=0$; others are the roots of the cubic equation $4x^{3}+12x^{2}-8x+16=0$. Unfortunately, the roots are not easy to find -- you actually need [url=http://en.wikipedia.org/wiki/Cubic_equation]Cardano's formula[/url] for them. Can't help you with critical points: finding the zeros of that derivative appears a hopeless task.

it is okay u helped me a ton already :P
by the way, are those maxium and minimum points on the graph i attached on my last pose?

Those bumps [b]look[/b] like maximum and minimum points, but there are so many of them that I suspect they may be a result of computational errors of your graphing software.  :maybe:

hum...not sure, but i though rectangluar parabola does not have maxium or minium...

Those bumps are real, and there's infinitely many of them, coming from that cosine term.

but they are not absolute right, the are only local?

can someone explain how cardano's method works please?

man, how did all the sin and cos dropped out after simplify...is there a step by step simplifying website or somthing...


Find all complex numbers $z$ such that :

$(3z+1)(4z+1)(6z+1)(12z+1)=2$

May I ask what's Horner's principle?  :)

[quote="leepakhin"]
P.S. I also don't know what is Horner's principle. :?[/quote]

I just found that : http://en.wikipedia.org/wiki/Horner_scheme :)


Fie numerele reale $ x_1,x_2,...,x_n$ care satisfac $ 0<x_1<x_2<...<x_n<1$ astfel incat $ x_0\equal{}0,x_{n\plus{}1}\equal{}1$. Presupunem ca aceste numere satisfac sistemul de ecuatii:

$ \displaystyle\sum_{j\equal{}0,j \not\equal{} i}^{n\plus{}1}\frac{1}{x_i\minus{}x_j}\equal{}0$ unde $ i\equal{}1,2,...,n.$

Demonstrati ca $ x_{n\plus{}1\minus{}i}\equal{}1\minus{}x_i$ pentru $ i\equal{}1,2,...,n$.

Fie $ P(x)\equal{}(x\minus{}x_0)(x\minus{}x_1)...(x\minus{}x_n)(x\minus{}x_{n\plus{}1})$. Atunci:

$ P'(x)\equal{}\displaystyle\sum_{j\equal{}0}^{n\plus{}1}\frac{P(x)}{x\minus{}x_j}$ si $ P''(x)\equal{}\displaystyle\sum_{j\equal{}0}^{n\plus{}1}\displaystyle\sum_{k \not\equal{} j}^{} \frac{P(x)}{(x\minus{}x_j)(x\minus{}x_k)}.$

Asadar, $ P''(x_i)\equal{}2P'(x_i)\displaystyle\sum_{j \not\equal{} i}^{}\frac{1}{(x_i\minus{}x_j)}$ pentru $ i\equal{}0,1,...,n\plus{}1,$ si conditia data implica $ P''(x_i)\equal{}0$ pentru $ i\equal{}1,2,...,n.$ In consecinta, $ x(x\minus{}1)P''(x)\equal{}(n\plus{}2)(n\plus{}1)P(x)$. $ (1)$

Este usor de vazut ca exista un singur polinom monic de gradul $ n\plus{}2$ care satisface ecuatia diferentiala $ (1)$. Pe de alta parte, polinomul $ Q(x)\equal{}(\minus{}1)^n P(1\minus{}x)$ de asemenea satisface aceasta ecuatie, este monic si grad $ Q$=$ n\plus{}2$. Asadar, $ (\minus{}1)^n P(1\minus{}x)\equal{}P(x)$, si astfel rezultatul se deduce.


Prove that nroot{3}{25 + 10sqrt{5} }  + nroot{3}{25 - 10sqrt{5} }  is rational


how much can you do until you get bored?

There is no 'getting bored' while doing math...

I mean, at least to mathematicians... :wink:

Yes, and Mr. New Member, you will certainly quickly see there are TONS of ppl TONS TONS who do math every second of their lives. Me addicted? yes Me love it? yes Me have other life? Yes, but their are some here who dont. coughcough

Anyway, have fun here at aops

ok

i like math but i guess i can get tired.
bored is the wrong word

[quote="pinkmuskrat"]Me have other life? Yes, but their are some here who dont. coughcough[/quote]
Are you implying that not having life is a negative thing?

How so?

Besides, I strongly believe that doing math is one way of having life.

I never get tired. In fact, I have NOTHING else to do. Without math I would get bored and tired.

This is pretty much my typical day: get up, eat breakfast, go on AoPS, eat lunch, go on AoPS, do problems, go on AoPS, think about a problem, eat dinner, think about how to prove something I saw on AoPS, sleep. Of course, I would go to school if I had school that day. I am currently on summer break.

Answering pinkmuskrat's problems:
Am I addicted (notice the grammar)? Yes
Do I love it? Yes
Do I have another life? NO
Can I live without it? NO

I think you get the point.

I have to agree with everyone's responses here. I do math for the most part of my day. Now...I have my stupid, pointless, boring finals coming up...so I have to study for it.  But, without math my life would have been divided by zero (as in undefined ;))

Umm...shouldn't this be moved to The Round Table or something?

[quote="limac"]Umm...shouldn't this be moved to The Round Table or something?[/quote]

More like Fun Factory?

I [i]wish[/i] all I had to do was math... Unfortunately, I have finals coming up... AND my parents think that doing just math is unhealthy(so I watch basketball :rotfl: )I do have a life, though. I have a ton of friends and a facebook.

Luckily for me, it's almost summer!

You can't get bored. There isn't such thing as too much Math.  :idea:

I like math but I don't do it 24/7. I'm on summer break (have been for almost 2 weeks) and I'm working on math to prepare for next year's math class. But I also have a life...I hang out with friends, email, text, and do other things like practice piano, and listen to music, and play Wii!

Math is fun but it depends what kind of math:

Arithmetic- stuff you get in school- not fun
Geometry- not the kind in school- fun

You know? :yup:

I am a junior member  :oops: , but still I get a feeling that no [b]Aspiring Mathematician[/b] can get bored with sums.Previously I had a problem similar to [b]xsquare[/b] but when I came to know about Olympiads, I was immediately motivated.

We are all hungry for problems(the difference is only in the level of Maths we know) at AoPS.So, there is no question of me getting bored. :)

This is my experience with math.

In junior high, I did the math club for both years then I grew to HATE it and I quit for a year. After a year of not doing math, I came to this site and realized how much I missed doing it. So in the end, math and I parted for a year, but we were never truly seperated. I guess math was never boring for me, I just needed a break.

achieving stuff through math is a great feeling, whether it's a competition or a really annoying proof. that's mainly why i like math.

also, as mentioned above, my talents are more balanced(i.e. soccer, writing(lol yea), swimming, taekwondo)

I completely agree with louiethechewy123 that this website touches a tip of the iceberg.I have a suggestion:

This website is simply amazing.I realized that [b]I am a nobody [/b]after I came to the AoPS.Here we can interact with so many people.Why do we not drastically expand this website, and open forums for  all topics upto PhD level.That will keep us glued to site all our lives!
As for math being fun, I can say that I cannot do without it at all( like you people.).Of course I am not abnormal,it is so strange that some people think that people interested in maths are supposed to be mad.

Of course, people here are in a  better position to comment than somebody as ordinary as me.

Don't take math taught in schools as a good example of how fun math can be, studying on your own is much more interesting :lol:

Yeah, math has definitely totally changed how I view things. I didn't really start competitive math until just a few months ago, and it's really become a major hobby. Of course, I have others, like programming, physics, and some computer games, but I love math.

shouldn't this topic be in the Fun Factory?

Anyhow, who on AOPS DOESN'T like math?  :D

Of course math is fun! Not only that, but it can also teach you important skills to help you in your life, such as logical thinking, and problem solving skills

Math is basically addicting after a while (when you appreciate the complexity and elegance of problems and math itself.) And even though I sometimes want to do something else, this 'earnest' desire/addiction keeps me messing around with ideas in math. This addiction, in my opinion is like a game. It's just like the feeling you get when you accomplish something out of the ordinary. This feeling is what I really enjoy and relish. Thus, math is extremely fun to me.

[hide="Proof that math is fun"]A fun game = Very addicting
Math = Very addicting $ \implies$ Math = A fun game = Very addicting $ \implies$ Math is a fun game that is very addicting.
$ Q.E.D$
[/hide]

[quote="batteredbutnotdefeated"][hide="a better proof :lol: "] fun game=f, math=m, tough=t, addicting=a, c=cool

f=t^2+a^2
m=(c*a)+(c*t)
c=t=a

m=f

QED

[/hide][/quote]

f=ma

so a=1. :lol:

This thread is starting to lose its purpose. I think xsquare already got what (s)he wanted.

Moderators, please lock.

thanks i recently started taking the algebra class online
and yyes it does feel good to finish a hard problem :lol:


geometry
parallelogram
rectangle
Pythagorean Theorem

Diagrams are usually made to scale.  It doesn't state what $ FG$ could be (as in any irrational).  This probably means that $ FG$ is a whole number.

By Pythagorean theorem, $ BG$ is $ 6$.

The only Pythagorean triple with hypotenuse $ 26$ and integer side lengths is $ 10$-$ 24$-$ 26$.

$ FB$ is clearly longer than $ FA$.  So $ FB$ should be $ 24$.

Thus, the answer is $ 24 \plus{} 6$ or $ \boxed{30}$.

I think the question should state $ BC \equal{} AF \equal{} 10$, not $ BC \equal{} AD \equal{} 10$. That is because the original statement is trivial (it is a parallelogram), and also, they didn't state $ AB \equal{} CD \equal{} 26$, only $ AB \equal{} 26$ (so why would they do it for $ BC \equal{} AD$?).

How do you report a question...

[quote="ernie"]By Pythagorean theorem, $ BG$ is $ 8$.[/quote]

It is 6, by Pythagorean theorem. 10^2-8^2=36.

And yes, this question does not make sense.

???


What is the value of the following expression, where all the natural
numbers 1-99 are included and the repeated operations are add,
add, subtract?
\[ 1 \plus{} 2 \minus{} 3 \plus{} 4 \plus{} 5 \minus{} 6 \plus{} 7 \plus{} 8 \minus{} 9 \plus{} 10 \plus{} 11 \minus{} 12 \plus{} \ldots \plus{} 97 \plus{} 98 \minus{} 99\]

When combined, we are adding $ 3\plus{}9\plus{}15\plus{}21\plus{}\cdots\plus{}195\equal{}3(1\plus{}3\plus{}5\plus{}7\plus{}\cdots\plus{}65)$ and subtracting $ \minus{}3\minus{}6\minus{}9\minus{}\cdots\minus{}99\equal{}\minus{}3(1\plus{}2\plus{}3\plus{}\cdots\plus{}33)$.  The first sum is equal to $ 3(33^{2})\equal{}3(1089)\equal{}3267$ and the second is equal to $ \minus{}3(561)\equal{}\minus{}1683$.  Hence we have $ 3267\minus{}1683\equal{}1584$.


vector
advanced fields
advanced fields unsolved

I have a problem with calculating explicitly the Casimir operator for $ \mathfrak{sl}(n, \mathbb{C})$. I choose the following basis for the algebra: ($ H_{i}, X_{\alpha}, X_{\minus{}\alpha}$), where $ X_{\alpha}$ are root vectors $ E_{kl}. k > l$ and $ H_{i}$ are given by:
$ H_{1} \equal{} diag(1,\minus{}1,0,\dots,0). H_{2}\equal{}diag(0,1,\minus{}1,0,\dots,0), \dots, H_{n}\equal{}diag(0, \dots, 0, 1, \minus{}1)$

The Casimir operator is given by:
\[ C \equal{} \sum_{i} Y_{i}Y^{i}\equal{} \sum_{i,j} g_{ij}Y_{i}Y_{j}\]
where $ g_{ij}\equal{}B(Y_{i},Y_{j})\equal{}2n \ tr(Y_{i}Y_{j})$ is the Killing form for $ \mathfrak{sl}(n, \mathbb{C})$ and we sum over all basis vectors $ Y_{i}$.

After calculating $ g_{ij}$ fot this particular basis I get the following formula:
\[ C \equal{} 2n \sum_{\alpha > 0} (X_{\alpha}X_{\minus{}\alpha} \plus{} X_{\minus{}\alpha}X_{\alpha}) \plus{} 4n \sum_{i\equal{}1}^{n\minus{}1} H_{i}^{2} \minus{} 4n \sum_{i\equal{}1}^{n\minus{}2} H_{i}H_{i\plus{}1}\]

Now, $ C$ is supposed to be a multiple of identity; however the above formula doesn't seem to be right (the 1-st and n-th diagonal entries are too small). Where is the error?

(btw, is there any simple orthonormal basis for the algebra generated by $ H_{i}$, i.e. Cartan subalgebra?)

The mistake was using $ g_{ij}$ instead of $ g^{ij}\equal{}g_{ij}^{\minus{}1}$ - a trivial mistake, sorry for wasting your time.


[size=150][b]RtΔABC.AB=AC.An ellipse passes through point A and B.One of its focal points is C,
the other one is in RtΔABC.Try to find the eccentricity of the ellipse.[/b][/size]

Thanks for helping. :)

Rt means ∠A=90

Could anybody give simple solution?

Why not give us a img of it?
i m a little lazy... :lol:

[quote="castigioni"]If eccentricity is c/a where c=semi focal distance and a=semi major axis then e=6^.5-3^.5
I consider that the other foci is on leg AB[/quote]

[b]Thanks but why  the other foci is on leg AB :?: [/b]

The other foci is on AB because the problem ask for it.Another position is the foci on BC but I didn't yet

The eccentricity is 2-2^.5 if the other foci is on BC

To determine the ellipse using R.e.C or Cinderela :
1.Right isosceles ABC with A=90 ,B on horizontal and C on vertical
2.Choose T on AB
3.Circle center B radius BT cut line CB at M
4.Circle center A radius AT cut line CA a N
5.Trianle CMN must be isosceles with CM=CN  and T(foci) is determined until the MN bissector go through C

Then the ellipse can draw by this way:Circle center C radius CM. Choose a point P on it .The bissector ofTP cut yhe line CP in point J.This point is on the ellipse.Moving P along the circle you will have the ellipse

The point T on AB as item 2 is the tangency point of the excircle


[quote="toanviet"]Find x in the sequence: 1, 20, 189, 1520, x

Find y in the sequence: 1, 10, 91, 730, 5113, y

Have A Great Day![/quote]

this looooooooooooooooooooooooooooks hard

yeah, well that's spam too

and this post


Fie $(a_n)_{n>0} $ un sir de nr. reale pozitive ,convergent la a>0.
Demonstrati ca sirul $x_n=sin(n\pi\sqrt[n]{{a_n }})$ are limita daca si numai daca $lna \in Z$.

$a=e^k$ where $k \in Z$

An asymptotic expansion gives
$\sqrt[n]{a_n}=1+\frac{k}{n}+\frac{k^2}{2n^2}+\frac{1}{n^2}g_n$

with $\lim g_n =0$

$x_n=\sin (\pi(n+k)+\frac{\pi k^2}{2n}+\frac{1}{n}h_n)=(-1)^{n+k}\sin(\frac{\pi k^2}{2n}+\frac{1}{n}h_n)$  with $\lim h_n=0$

$x_n$ tends to zero


Let $ S $ be a set of $ n $ concentric circles in the plane. Prove that if a function $ f: S\to S $ satisfies the property
\[ d( f(A),f(B)) \geq d(A,B) \] for all $ A,B \in S $, then $ d(f(A),f(B)) = d(A,B) $, where $ d $ is the euclidean distance function.

[quote="julien_santini"]Consider the two circles $ D $ and $ E $ such that $ d(D,E) $ is maximal. Then $ d(f(D),f(E))=d(D,E) $, and either $ f(D)=D, f(E)=E $ or $ f(D)=E, f(E)=D $. Now assume for two circles $ A,B $, $ d(f(A),f(B))>d(A,B) $. Then $ d(f(D),f(E)) = d(D,E) \leq d(D,A)+d(A,B)+d(B,E) < $ $ \min(d(f(D),f(A))+d(f(A),f(B))+d(f(B),f(E)), $ $ d(f(D),f(B))+d(f(B),f(A))+d(f(A),f(E))) = d(f(D),f(E)) $.
Absurd.[/quote]

I don't think this argument works unless you can show that $f$ is continuous.  Anyway, I think we just first show that the function $f$, restricted on the largest circle, is an isometry.  Then, get rid of this circle, and proceed inward to the smallest circle.

(Due to Chung Ping Ngai, HK IMO team member) First off, f must be bijective, for otherwise there exists two distinct elements (say, X, Y) in S which are mapped to the same element, so that d(f(X), f(Y)) = 0 < d(X, Y), a contradiction.

Note that since S is finite, for all A, B in S we have $f^p(A)=A$ and $f^q(B)=B$ for some positive integers p, q. Then $f^{pq}(A)=A$ and $f^{pq}(B)=B$. But $d(f^{pq}(A), f^{pq}(B)) \geq d(f^{pq-1}(A), f^{pq-1}(B)) \geq ... \geq d(A, B)$, so all $\geq$ signs must become equal signs. QED.

Ohhh.  OK.  I thought $S$ was the set of points of $n$ circles.  Apologies to the first reply.


hello, calculating the left-hande side of the equation we get
$ \frac{24\cdot12}{9}\equal{}2^{x\plus{}1}$ simplifying and factoring $ 2^5\equal{}2^{x\plus{}1}$, from here we get the solution $ x\equal{}4$.
Sonnhard.


algebra
function
domain
search
logarithms
algebra unsolved

Determine all maps $f: \mathbb{N}\rightarrow [1,+\infty)$ that satisfy the following conditions:
[list]
[*]$f(2)=2$,
[*]$f(mn)=f(m)f(n)$, for all $m,n \in \mathbb{N}$, and
[*]$f(m) < f(n)$ if $m < n$.
[/list]

Hello Jan

[quote="Jan"]Determine all maps $f: \mathbb{N}_{0}\rightarrow [1,+\infty[$ that satisfy:
[list]
[*]$f(2)=2$
[*]$f(mn)=f(m)f(n)$, $\ \forall \ m,n \in \mathbb{N}_{0}$
[*]$f(m) < f(n)$ if $m < n$[/list][/quote]

First, we can ignore the requirement $f(2)=2$
Obviously, $f(m^{r})=f(m)^{r}$
Let a,b two integers > 1 and n,p two integers > 0 such that $\frac{n}{p}\geq\frac{ln(a)}{ln(b)}$
Then, $b^{n}\geq a^{p}$ hence $f(b^{n})\geq f(a^{p})$ then $f(b)^{n}\geq f(a)^{p}$ and $\frac{n}{p}\geq\frac{ln(f(a))}{ln(f(b))}$

So : Let a,b two integers > 1 and n,p two integers > 0 : $\frac{n}{p}\geq\frac{ln(a)}{ln(b)}$ ==> $\frac{n}{p}\geq\frac{ln(f(a))}{ln(f(b))}$

But, it is always possible to find two integers p and q such that $\frac{n}{p}$ be as near as possible and above $\frac{ln(a)}{ln(b)}$
==> $\frac{ln(a)}{ln(b)}\geq\frac{ln(f(a))}{ln(f(b))}, \forall a,b$ integers > 1
Since we also have then $\frac{ln(b)}{ln(a)}\geq\frac{ln(f(b))}{ln(f(a))}$

The conclusion is $\frac{ln(a)}{ln(b)}=\frac{ln(f(a))}{ln(f(b))}$, hence $\frac{ln(f(a))}{ln(a)}=\frac{ln(f(b))}{ln(b)}$

Hence $f(x)=x^{c}$

But f(2)=2

Hence f(x) = x

--
Patrick

$f(1)=(f(1))^{2}\Longrightarrow f(1)=1$. Let $a$ be an integer $a > 2$ . It's easy to see that $f(a^{n})=(f(a))^{n}$ .
Let $u_{n}$ be an  integer such that $2^{u_{n}-1}\leq a^{n}\leq 2^{u_{n}}$ so $2^{\frac{u_{n}-1}n}\leq a\leq 2^{\frac{u_{n}}n}$ so  $\frac{u_{n}}n$  is convergent to $\frac{\ln (a)}{\ln (2)}$
We can write :
$2^{u_{n}-1}\leq a^{n}\leq 2^{u_{n}}\Longrightarrow (f(2))^{u_{n}-1}\leq (f(a))^{n}\leq (f(2))2^{u_{n}}$$\Longrightarrow 2^{\frac{u_{n}-1}n}\leq f(a)\leq 2^{\frac{u_{n}}n}\Longrightarrow f(a)=2^{\frac{\ln (a)}{\ln (2)}}=a$
Of course this proof is verry known ..
 :cool:


geometry incenter circumcircle inradius trigonometry inequalities vector

Let incenter of$\triangle{ABC}$ is $I$ circumcenter is $O$.$({O}\not={I})$ And $D$, $E$ are points on rays $[AC)$ and $[BC)$ respectively such that $AD=BE=AB$. Prove that $DE$ is perpendicular to $OI$

[quote="Litlle 1000t"][color=darkred]Let incenter of $\triangle{ABC}\,\left(AB=\min\{AB,\,BC,\,CA\}\right)$ is $I,$ circumcenter is $O.$ And $D$, $E$ are points on $AC$ and $BC$ respectively such that $AD=BE=AB.$ Prove that $DE$ is perpendicular to $OI$[/color][/quote]

[color=indigo][b]Proof.[/b] Let $M,\,N$ be the projections of $O$ on $AC,\,BC$ respectively; $H,\,K$ be the projections of $I$ on $AC,\,BC$ respectively.
We have:
\begin{eqnarray*}\overrightarrow{OI}\cdot\overrightarrow{DE}&=&\overrightarrow{OI}\left(\overrightarrow{CE}-\overrightarrow{CD}\right)\\ &=&\overline{NK}\cdot\overline{CE}-\overline{MH}\cdot\overline{CD}\\ &=& \left(\overline{CK}-\overline{CN}\right)\overline{CE}-\left(\overline{CH}-\overline{CM}\right)\overline{CD}\\ &=&\left(p-c-\frac{a}{2}\right)(a-c)-\left(p-c-\frac{b}{2}\right)(b-c)=0 \end{eqnarray*}
Hence: $OI\perp DE$

[b]Remark.[/b] See also for another property:[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=19133

[color=red][b]P.S.[/b] Are you sure your problem is still true when $AB\neq\min\{AB,\,BC,\,CA\}?$[/color]

we know that $ DE \perp OI \iff DO^{2}-DI^{2}=EO^{2}-EI^{2}$

so its sufficient to show that:

$ IE^{2}-DI^{2}=EO^{2}-DO^{2}$

we know that the power of points $ E,D$ wrt circle $ (I)$ respectively is equal to:

$ P^{E}_{(I)}=IE^{2}-r^{2}$

$ P^{D}_{(I)}=ID^{2}-r^{2}$

where $ r$ is the inradius,now we get that:

$ IE^{2}-DI^{2}=P^{E}_{(I)}-P^{D}_{(I)}$

now let the incenter touch the sides $ AC,BC$ at $ P,Q$ respectively,now we have:

$ P^{E}_{(I)}=EQ^{2}$

$ P^{D}_{(I)}=DP^{2}$

so we get that:

$ IE^{2}-DI^{2}=EQ^{2}-DP^{2}$

but we know that $ EQ=p-a,DP=p-b$ where $ p$ is the semiperimeter of $ \triangle ABC$...(because we have $ DP=DA-AP=c-(p-a)=c+a-p=p-b$)

so we have:

$ \boxed{EQ^{2}-DP^{2}=ac-cb (I)}$

also we know that the power of points $ E,D$ to the circle $ O$ respectively are equal to:

$ |P^{E}_{(O)}|=|OE^{2}-R^{2}|=R^{2}-OE^{2}=EB.EC$

$ |P^{D}_{(O)}|=|OD^{2}-R^{2}|=R^{2}-OD^{2}=DA.DC$

where $ R$ is the circumradius of $ \triangle ABC$

hence we get that:

$ \boxed{OE^{2}-OD^{2}=DA.DC-EB.EC=c(a-c)-c(b-c)=ca-cb (II)}$

from $ (I),(II)$ we get the desired resault...

and also there is another similar problem:

[b]assume that $ ABCD$ is a convex quadrilateral which $ AB=AC=BD$,now let $ P$ be the intersection of $ AC,BD$ and denote $ O,I$ as the incenter and circumcenter of $ \triangle ABP$ respectively,prove that if $ O\not= I$ then $ OI \perp CD$[/b]

[quote="April"][color=red][b]P.S.[/b] Are you sure your problem is still true when $ AB\neq\min\{AB,\,BC,\,CA\}?$[/color][/quote]

Of course it doesn't require that $ AB=\min\left\{AB,\ BC,\ CA\right\}$ as long as the points $ D$ and $ E$ are supposed to lie on the rays $ AC$ and $ BC$, respectively (and not on the segments $ AC$ and $ BC$). See http://www.mathlinks.ro/Forum/viewtopic.php?t=14655 for more solutions.

  Darij

there is also another solution which wasnt in the links given above (at least I didnt find it) here it goes:

let $ M$ be the midpoint of arc $ BC$ of circumcircle,and let $ K$ be the midpoint of side $ BC$ and let $ N$ denote the
intersection point of angle bisector $ AM$ and $ BD$.and as said above assume that $ R$ is the circumradius of $ \triangle ABC$

we know that $ MI=MB$ and also $ MB=2R.sin \angle BAM$

$ \Rightarrow \boxed{MI=2R\sin{\frac{A}{2}}(I)}$

now in $ \triangle ABD$ we have $ AB=AD$ so $ N$ is the midpoint of $ BD$ hence we have:

$ \boxed{BD=2BN=2AB.\sin{\frac{A}{2}}(II)}$

now according to $ (I),(II)$ we get that:

$ \frac{BD}{MI}=\frac{2AB.\sin{\frac{A}{2}}}{2R\sin{\frac{A}{2}}}=\frac{AB}{R}$

and also we have:

$ \frac{BE}{OM}=\frac{AB}{R}$

now by the above two equations we get that:

$ \frac{BD}{MI}=\frac{BE}{OM}(*)$

now in the triangles $ OIM,EDB$ we have:

$ OMN \perp BE$

$ IM \perp DB$

yielding $ \angle IMO=\angle DBE$

so according to $ (*)$ we get that:

$ \triangle IMO \sim \triangle DBE$

now in the similar triangles $ IMO,DBE$ two sides are pariwise perpendicular to each other,so the third sides must be perpendicular...

so $ OI \perp DE$

and also as a reasult of the above solution we get that:

$ \triangle IMO \sim \triangle DBE \Rightarrow \frac{DE}{OI}=\frac{BE}{OM}=\frac{AB}{R}$

now according to $ \sin$law in triangle $ ABC$ we have:

$ \frac{AB}{R}=2\sin C$

$ \Rightarrow \frac{DE}{OI}=2\sin C (1)$

now let $ R'$ be the circumradius of $ \triangle CDE$,now according to $ \sin$law in $ \triangle CDE$ we get that:

$ \frac{DE}{R'}=2\sin C (2)$

$ (1),(2) \Rightarrow \frac{DE}{OI}=\frac{DE}{R'}\Rightarrow OI=R'$

[quote="April"][quote="Litlle 1000t"][color=darkred]Let incenter of $ \triangle{ABC}\,\left(AB =\min\{AB,\,BC,\,CA\}\right)$ is $ I,$ circumcenter is $ O.$ And $ D$, $ E$ are points on $ AC$ and $ BC$ respectively such that $ AD = BE = AB.$ Prove that $ DE$ is perpendicular to $ OI$[/color][/quote]

[color=indigo][b]Proof.[/b] Let $ M,\,N$ be the projections of $ O$ on $ AC,\,BC$ respectively; $ H,\,K$ be the projections of $ I$ on $ AC,\,BC$ respectively.
We have:
\begin{eqnarray*}\overrightarrow{OI}\cdot\overrightarrow{DE}&=&\overrightarrow{OI}\left(\overrightarrow{CE}-\overrightarrow{CD}\right)\\ &=&\overline{NK}\cdot\overline{CE}-\overline{MH}\cdot\overline{CD}\\ &=&\left(\overline{CK}-\overline{CN}\right)\overline{CE}-\left(\overline{CH}-\overline{CM}\right)\overline{CD}\\ &=&\left(p-c-\frac{a}{2}\right)(a-c)-\left(p-c-\frac{b}{2}\right)(b-c)=0\end{eqnarray*}
Hence: $ OI\perp DE$

[b]Remark.[/b] See also for another property:[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=19133

[color=red][b]P.S.[/b] Are you sure your problem is still true when $ AB\neq\min\{AB,\,BC,\,CA\}?$[/color][/quote]

how did you get from the first line to the second line?

$ N,K$ are the projections of points $ O,I$ on $ BC$ respectively...
so $ \overrightarrow{OI}\equal{}\overrightarrow{NK}$...

[quote="BaBaK Ghalebi"]$ N,K$ are the projections of points $ O,I$ on $ BC$ respectively...
so $ \overrightarrow{OI}=\overrightarrow{NK}$...[/quote]

so could you explain it fully,

generally $ \overrightarrow{AB}\dot\overrightarrow{AC}\leq AB\dot AC$ so when do we have inequality, thanks

I cant understand what youre trying to say...  :maybe:

note that $ \overrightarrow{v}\cdot\overrightarrow{w}\equal{} |v|.|w|cos\alpha$ where $ \alpha$ is the angle between $ v,w$ and $ |v|,|w|$ denote the length of $ v,w$ respectively...

so $ v\perp w$ implies that $ \alpha \equal{} 90$ which implies that $ \overrightarrow{v}\cdot\overrightarrow{w}\equal{} 0$...

also in [b]April[/b]`s solution,he is trying to prove that $ \overrightarrow{OI}\cdot\overrightarrow{DE}\equal{} 0$ which implies that $ OI\perp DE$

now we know that $ \overrightarrow{DE}\equal{}\overrightarrow{CE}\minus{}\overrightarrow{CD}$ also as I mentioned in my last post
$ \overrightarrow{OI}\equal{}\overrightarrow{NK}\equal{}\overrightarrow{MH}$

also we know that

$ \overrightarrow{x}\cdot(\overrightarrow{y}\plus{}\overrightarrow{z}) \equal{}\overrightarrow{x}\cdot\overrightarrow{y}\plus{}\overrightarrow{x}\cdot\overrightarrow{z}$

so as April said we have:

$ \overrightarrow{OI}\cdot\overrightarrow{DE}\equal{}\overrightarrow{OI}\cdot(\overrightarrow{CE}\minus{}\overrightarrow{CD}) \equal{}\overrightarrow{OI}\cdot\overrightarrow{CE}\minus{}\overrightarrow{OI}\cdot\overrightarrow{CD}$

now we can replace $ \overrightarrow{OI}$ by $ \overrightarrow{NK}$ and $ \overrightarrow{MH}$ to get that:

$ \overrightarrow{OI}\cdot\overrightarrow{DE}\equal{}\overrightarrow{NK}\cdot\overrightarrow{CE}\minus{}\overrightarrow{MH}\cdot\overrightarrow{CD}$

but we know that $ N,K,C,E$ lie on $ AC$ so the angle between $ NK,CE$ is equal to $ 0$ so the $ \cos$ of this angle is equal to $ 1$ so we get that:

$ \overrightarrow{NK}\cdot\overrightarrow{CE}\equal{} |NK|.|CE|$

and similary we get that:

$ \overrightarrow{MH}\cdot\overrightarrow{CD}\equal{} |MH|.|CD|$

thus the last equality is equivalent to this:

$ |NK|.|CE|\minus{}|MH|.|CD|$

but we know that:

$ NK \equal{} CK\minus{}CN \equal{} p\minus{}c\minus{}\frac{a}{2}$

and also

$ MH \equal{} CH\minus{}CM \equal{} p\minus{}c\minus{}\frac{b}{2}$

also

$ CE \equal{} a\minus{}c,CD \equal{} b\minus{}c$

thus the last equality can be written as:

$ |NK|.|CE|\minus{}|MH|.|CD| \equal{} (|CD|\minus{}|CN|)|CE|\minus{}(|CH|\minus{}|CM|)|CD|$

$ \equal{} (p\minus{}c\minus{}\frac{a}{2})(a\minus{}c)\minus{}(p\minus{}c\minus{}\frac{b}{2})(b\minus{}c) \equal{} 0$

so we get that:

$ \overrightarrow{OI}\cdot\overrightarrow{DE}\equal{} 0$

so the angle between these two lines is equal to $ 90$ as wanted...

thank you for your detailed explanation, one more thing:

is it true if O $ \in$ AC, and when does $ \over arrowhead{AC}\dot\over arrowhead{AB}\leq |AC\parallel AB|$

[quote="BaBaK Ghalebi"]$ N,K$ are the projections of points $ O,I$ on $ BC$ respectively...
so $ \overrightarrow{OI}=\overrightarrow{NK}$...[/quote]

could you give a proof

for your first question:

$ \overrightarrow{AB}\cdot\overrightarrow{AC}= |AB|.|AC|.\cos\alpha$

but we know that $ \cos\alpha\leq 1$ so we have:

$ |AB|.|AC|.\cos\alpha\leq |AB|.|AC|$

so we have:

$ \overrightarrow{AB}\cdot\overrightarrow{AC}\leq |AB|.|AC|$

and the equality occurs when $ \cos\alpha = 1\iff\alpha = 0$ i.e. $ AB\parallel AC$...

and for your second question,that was a stupid typo of mine sorry,here is the corrected version:

note that $ N,K$ are the projections of points $ O,I$ on side $ BC$,now let $ I'$ be the point such that $ NI' = OI$ and $ NI'\parallel OI$ in other words $ \overrightarrow{NI'}=\overrightarrow{OI}$,now in the right triangle $ \triangle NI'K (\angle I'KN = 90)$ we have:

$ NK = NI'\cos NI'K = OI\cos\NI'K$

but note that $ OI\parallel NI'$ so the angle between $ OI$ and $ BC$ is equal to the angle between $ NI'$ and $ BC$ which is $ \angle NI'K$...
so if we denote $ \alpha$ as the angle between $ OI$ and $ CE$ then $ \alpha =\angle NI'K$ so we get that:

$ NK = NI'\cos NI'K = OI\cos\alpha$

so we get that:

$ CE.OI\cos\alpha = NK.CE$

thus we get that:

$ \overrightarrow{OI}\cdot\overrightarrow{CE}= |NK|.|CE$...

similary we get that:

$ \overrightarrow{OI}\cdot\overrightarrow{CD}= |MH|.|CD|$...

and the conclusion follows...

P.S:sorry again for the stupid mistake of mine...
 :oops:

[quote="BaBaK Ghalebi"]for your first question:

$ \overrightarrow{AB}\cdot\overrightarrow{AC}\equal{} |AB|.|AC|.\cos\alpha$

but we know that $ \cos\alpha\leq 1$ so we have:

$ |AB|.|AC|.\cos\alpha\leq |AB|.|AC|$

so we have:

$ \overrightarrow{AB}\cdot\overrightarrow{AC}\leq |AB|.|AC|$

and the equality occurs when $ \cos\alpha \equal{} 1\iff\alpha \equal{} 90$ i.e. $ AB\perp AC$...

[/quote]

 Please correct a typo :

 $ \cos\alpha \equal{} 1\iff\alpha \equal{} 0$ i.e. $ AB\parallel AC$...

 and not  $ AB\perp AC$

[quote="BaBaK Ghalebi"]thanks [b]stergiu[/b]...   :oops:
its correct now...  :D[/quote]

 :wink: Nice !!!

so generaly let A', B' be projections of AB onto CD. Then vector AB=vector A'B'

no D.P.L,you can say that only when $ AB\parallel CD$...
in the general case if $ A',B'$ are the projections of $ A,B$ on $ CD$ we can say that:
$ \overrightarrow{AB}\cdot\overrightarrow{CD}\equal{}|A'B'|.|CD|$
because of the reason that I used in my previews post...

[quote="Litlle 1000t"][color=darkred]Let $ ABC$ be a triangle with the circumcircle  $ w = C(O,R)$ and the incircle $ C(I,r)$. Prove that $ \{\begin{array}{c}D\in (AC\ ,\ E\in (BC\\ \\ AD = BE = AB\end{array}\|\implies OI\perp DE$.[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ p_{w}(X)$- the [b]power of a point $ X$ w.r.t. the circle $ w$[/b]. I'l use some  simple rules : $ \{\begin{array}{c}|\overline{XY}| = XY\\ \\ \overline{XY}+\overline{YZ}=\overline{XZ}\end{array}$.

$ 1\blacktriangleright\{\begin{array}{c}p_{w}(D)\equiv OD^{2}-R^{2}=\overline{DA}\cdot\overline{DC}=\overline{DA}\cdot (\overline{DA}-\overline{CA}) = c(c-b)\\ \\ p_{w}(E)\equiv OE^{2}-R^{2}=\overline{EB}{\cdot\overline{EC}=\overline{EB}\cdot (\overline EB}-\overline{CB}) = c(c-a)\end{array}\|$ $ \implies$ $ OD^{2}-OE^{2}= c(a-b)$.

$ 2\blacktriangleright\{\begin{array}{c}ID^{2}= XD^{2}+XI^{2}= (\overline{XA}-\overline{DA})^{2}+r^{2}= (p-a-c)^{2}+r^{2}\\ \\ IE^{2}= YE^{2}+YI^{2}= (\overline{YB}-\overline{EB})^{2}+r^{2}= (p-b-c)^{2}+r^{2}\end{array}\|$ $ \implies$ $ ID^{2}-IE^{2}= c(a-b)$.

Therefore, $ OD^{2}-OE^{2}= ID^{2}-IE^{2}= c(a-b)$, what means that $ \boxed{DE\perp OI\ }$.[/color]

[color=darkred][b][u]Remark I.[/u][/b] In the case $ (c-a)(c-b)\ge 0$ there is the relation $ DE^{2}= c^{2}\cdot(1-\frac{2r}{R})$, i.e. $ \boxed{\ DE = 2\cdot OI\cdot\sin C\ }$.

[b]Proof.[/b] Observe that $ \{\begin{array}{c}CD = |c-b|\\ \ CE = |c-a|\end{array}$ and apply the [b]PYthagoras's theorem[/b] for the side $ [DE]$ in the triangle $ DCE$ :

$ DE^{2}= (c-a)^{2}+(c-b)^{2}-2(c-a)(c-b)\cos C =$ $ a^{2}+b^{2}-2c(a+b)+2c^{2}-2(c-a)(c-b)\cos C =$

$ 2ab\cos C+c^{2}-2c(a+b-c)-2ab\cos C+2c(a+b-c)\cos C =$ $ c^{2}-4c(p-c)+4c(p-c)\cos C =$

$ c^{2}-8c(p-c)\sin^{2}\frac{C}{2}=$ $ c^{2}\cdot\frac{abc-8(p-a)(p-b)(p-c)}{abc}=$ $ c^{2}\cdot\frac{4Rpr-8pr^{2}}{4Rpr}$ $ \implies$ $ DE^{2}= c^{2}\cdot(1-\frac{2r}{R})$.

From the [b]Euler's relatiom[/b] $ OI^{2}= R^{2}-2Rr$ obtain the shorter form of this result :

$ DE^{2}=(\frac{c}{R})^{2}\cdot(R^{2}-2Rr) = (2\sin C)^{2}\cdot OI^{2}$ $ \implies$ ${ DE = 2\cdot OI\cdot\sin C}$. [/color]

[color=darkred][b][u]Remark II.[/u][/b] Let $ ABC$ be a triangle with the circumcircle  $ w = C(O,R)$ and the incircle $ C(I,r)$.

Let the points $ D$ and $ E$ so that $ \{\begin{array}{c}A\in (CD)\ ,\ B\in (CE)\\ \\ AD = BE = AB\end{array}$ Then $ \boxed{\ F\in OI\ ,\ \overline{FI}= 3\cdot\overline{FO}\implies FD = FE\ }$.[/color]
[color=darkblue][b]Proof.[/b]

$ 1\blacktriangleright\{\begin{array}{c}p_{w}(D)\equiv OD^{2}-R^{2}=\overline{DA}\cdot\overline{DC}=\overline{DA}\cdot (\overline{DA}+\overline{AC}) = c(c+b)\\ \\ p_{w}(E)\equiv OE^{2}-R^{2}=\overline{EB}\cdot\overline{EC}=\overline{EB}\cdot(\overline{EB}+\overline{BC}) = c(c+a)\end{array}\|$ $ \implies$ $ OD^{2}-OE^{2}= c(b-a)$.

The incircle touches the sides $ (AC)$, $ (BC)$ in the points $ X$, $ Y$ respectively. Thus,

$ 2\blacktriangleright\{\begin{array}{c}ID^{2}= XD^{2}+XI^{2}= (XA+AD)^{2}+r^{2}= (p-a+c)^{2}+r^{2}\\ \\ IE^{2}= YE^{2}+YI^{2}= (YB+BE)^{2}+r^{2}= (p-b+c)^{2}+r^{2}\end{array}\|$ $ \implies$ $ ID^{2}-IE^{2}= 3\cdot c(b-a)$.

Therefore, $ 3\cdot(OD^{2}-OE^{2}) = ID^{2}-IE^{2}$ $ \Longleftrightarrow$  $ 3\cdot DO^{2}-DI^{2}= 3\cdot EO^{2}-EI^{2}$. From the [b]Stewart[/b]'s relation

applied to the points $ D$, $ E$ and the line $ \overline{OFI}$, obtain : $ \{\begin{array}{c}3\cdot DO^{2}-DI^{2}= DF^{2}-\frac{3}{2}\cdot OI^{2}\\ \\ 3\cdot EO^{2}-EI^{2}= EF^{2}-\frac{3}{2}\cdot OI^{2}\end{array}\|$ $ \implies$ $ FD = FE$.[/color]

[size=117][color=darkred][b]Attention ![/b][/color][/size] [color=darkblue]An interesting extension of this problem you can find at http://www.mathlinks.ro/Forum/viewtopic.php?t=165597 .[/color]


Show that for positive reals $ a,b,c$ ;
\[ \sum_{cyc}\frac{a^3}{(2a^2\plus{}b^2)(2a^2\plus{}c^2)} \leq \frac{1}{a\plus{}b\plus{}c}\]

[quote="apollo"]Show that for positive reals $ a,b,c$ ;
\[ \sum_{cyc}\frac {a^3}{(2a^2 \plus{} b^2)(2a^2 \plus{} c^2)} \leq \frac {1}{a \plus{} b \plus{} c}
\]
[/quote]

Here

$ \left( {2a^2  \plus{} b^2 } \right)\left( {2a^2  \plus{} c^2 } \right) \ge \left( {a^2  \plus{} ab \plus{} ac} \right)^2$


I don't know what forum this belongs in, but since I want people at the intermediate level to see it, I'll post it here.

I've heard a lot about summer programs focusing on research, or certain people choosing to go to a university in order to do undergraduate research. I think that people on this forum could engage in collaborative research as well. There's definitely enough talent and all it would take is a few people to thoughtfully participate. I know there is some type of research in the Olympiad section, but most of that stuff is above people on the intermediate thm. for now.

We could choose some topic to be explored further, accessible to people on the intermediate level, then get a few people to start working on it. Then we could post here about any progress or whatever. If people value research so much (like at summer programs) why not take advantage of this forum to do so? Just some thoughts. Let me know if you think its actually possible.

Hmm, research as a high school level is difficult, we need at least an experienced enough mentor to lead us for the research. There are a few points not mentioned in your posts like the level or type of research. Undergraduate research is way beyond a lot of people in high school level (it's undergrad after all!). What can a high school student possibly research that is at the intermediate level of this forum? Topics can be hard without the proper knowledge to tackle them. Perhaps we can work on some of the Olympiad problems and try to develop some propositions of our own or prove some old theorems from scratch, which can be fun. Or, perhaps we can look up some stuff like those POWER questions from ARML that seems like research at the intermediate level :) Just my 2 cents.

(This should be in The Round Table Forum)

The problem, as user said, is that is extremely difficult to research at this level. I mean, there are a lot of MathLinkers that are excellent solving problem, but one thing is solving a problem, when you know that there is a solution, and nother thing, completely different, is to research, when you don't know how to find, I mean, you're just going around somthing without knowing if you are going to find something, and you don't know the kind of "something" you will find. Moreover, what kind of themes will you research at a high school level????

well, the way that i do "research" is that i make aime and above level problems, you just start from scratch, and it is amazing how you can make up questions that work out so nicely

Based on Altheman's comments, this opens up more options to this collaborative research: we have seen examples of this like Mildorf's Mocks, who's problems are fun and I'm sure he did a lot of research and problem solving to develop those elegant problems.

I'm beginning to get the idea that drunner2007 is hinting at: perhaps we can develop our researching and problem solving skills at the same time while discussing methods and techniques for attacking certain problems. This way we can develop some results (probably results that are already proven elsewhere but who cares :P) and we can practice our researching skills. I'm not sure if you guys understand me but ... meh.

yeah, that's kind of the idea. I think that collaboratively working on olympiad/usamo problems, arml power rounds, or making up usamo/aime problems is a good idea.

Then, maybe the idea is to generalize problems to be proposed at a certain olympiad

I found a problem that some people can start working on. I think its fairly hard but not impossible.

An $n$-digit positive integer $N$ is a [i]Kaprekar number[/i] if the sum of the number formed by the last $n$ digits in $N^2$, and the number formed by the first $n$ or $n-1$ digits in $N^2$ equals $N$. For example, $297$ is a Kaprekar number since $297^2=88209$ and $88+209=297$. Find all three-digit Kaprekar numbers.


Hint: Using the Newton's second law.

[hide]My calculation gives the answer $N_{14}=N_{41}=110N$ and $N_{24}=N_{42}=70N$[/hide]

yes it is the right answer
but how did you get that,can you give the explanation
i think you did 120-10=110
but why didnt you take the resultant force
is there any concept of contact forces on the net

[quote="bkunchanapalli"]three blocks of mass 1,4,2 kg are held together(not by string).on the side

of 1kg block 120N are applied and on the side of 2 kg block 50N are

applied.wat is the normal force acting between 1kg block and 4kg block.

is there any site for this topic[/quote]
The net force on the system is $120-50=70\,\textrm{N}$, so the acceleration is $10\,\textrm{m}/\textrm{s}^{2}$.  Therefore you have \[120\,\textrm{N}-F_{14}=10\,\textrm{N}\implies \boxed{F_{14}=110\,\textrm{N}}.\] Similarly you have \[F_{42}-50\,\textrm{N}=20\,\textrm{N}\implies \boxed{F_{42}=70\,\textrm{N}}.\] As stated above, these are obtained by noting that \[\sum_{n}F_{n}=F_{\textrm{net}},\] so if you simply draw a free body diagram, the answer should be apparent.

But for normal force between 1kg body and 4kg body

why dont we take $F_{14}$=net force - 10N

=70-10=60

can you please explain this,and why we only take 120N instead of 70N

Once you find the net force, forget about the other block.  You only have two that matter now.  One is pushing on the other (which is the force you are looking for) and then there is an external force pushing backwards against it.  By Newton's Second Law, the net force must be the sum of all of the forces acting on the object.  In this case, you only have two such forces acting on the block in question.  One will be negative of course because it is acting in the opposite direction of the net force.  Adding these together gives you an expression which you can use to solve for the desired force.


The three side lengths of a triangle are $a$, $b$ and $c$. Given the area is $\frac{1}{4}$ and the circumradius  is $1$.
Let $m=\sqrt{a}+\sqrt{b}+\sqrt{c}$, $n=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Which of the following must be true?
(A) $m \geq n$
(B) $m=n$
(C) $m \leq n$
(D) can't be determined

Since $S=\frac{4R}{abc}$, so $abc=1$. Because $2\sqrt{a}=2\sqrt{\frac{1}{bc}}\leq \frac{1}{b}+\frac{1}{c}$, so $m \leq n$.


abstract algebra
linear algebra
matrix
group theory
superior algebra
superior algebra unsolved

Let $ 0 \to A \to B \to C \to 0$ be a short exact sequence of abelian groups.
If $ C$ is a free abelian group then the above sequence is split.

Choose a basis $ \mathcal B_C$ of $ C$ and define a section of $ B\to C$ by extending an appropriate mapping $ \mathcal B_C\to B$ to $ C$.

or use the universal property of projective objects, after all free abelian groups are just free $ \mathbb{Z}$-modules.

$ \begin{matrix}
 & & C &\\
 & \swarrow^{\exists h} & \downarrow & \\
B & \rightarrow & C & \rightarrow 0
\end{matrix}$


The inequality for n=2 is the inequality of Yuong (I'm sorry if I wrote the name incorrectly)
It will be the base of induction
Then

*)  $\displaystyle \sum_{i=1}^{n} \lambda {}_ix{}_i =\lambda_1x1 + \sum_{i=2}^{n}\lambda_i\sum_{i=2}^{n} \frac{\lambda _i}{\sum_{i=2}^{n}\lambda_i} x_i \geq x_1^\lambda_1*(\sum_{i=2}^{n}\frac{\lambda_i}{\sum_{i=2}^{n}\lambda_i}x_i)^\sum_{i=2}^{n}\lambda_i$ under our inequality

After that we have

$\displaystyle \sum_{i=2}^{n} \frac{\lambda _i}{\sum_{i=2}^{n}\lambda_i} =1$

by the assumption of induction we have

$\displaystyle \sum_{i=2}^{n} \frac{\lambda _i}{\sum_{i=2}^{n}\lambda_i}x_i \geq \prod _{i=2}^{n} x_i^{\frac{\lambda_i}{\sum_{i=2}^{n}\lambda_i}}$

so

$\displaystyle (\sum_{i=2}^{n} \frac{\lambda _i}{\sum_{i=2}^{n}\lambda_i}x_i)^ {\sum_{i=2}^{n}\lambda_i}\geq \prod _{i=2}^{n} x_i^{\lambda_i}$

and in inequality *) we have  $\displaystyle \sum_{i=1}^{n} \lambda {}_ix{}_i \geq x_1^\lambda_1*... \geq \prod _{i=1}^{n} x_i^{\lambda_i}$

[quote="seamusoboyle"]If $x_i \geq 0$ and $\lambda _i \geq 0, \sum_{i=0}^{n} \lambda _i=1$

then:

$\displaystyle \sum_{i=1}^{n} \lambda _ix_i \geq \prod _{i=1}^{n} x_i^{\lambda_i}$[/quote]
This is generalization of AM-GM ineq.
Let $\lambda_i=\frac{\alpha_i}{S}$ with $S=\sum_{i=1}^n\lambda_i$ then, try to use AM-GM.
 :)  :D

[quote="Lovasz"]
In my country, AM-GM is called Cauchy's ineq.
[/quote]

Just like in my country  ;)  I just didn't understand the name "AM - GM" , but now it's clear for me  :)
Thanks a lot  :)

[quote="Lovasz"][quote="GHAVR"][quote="shyong"]Isnt that just weight AM-GM ? :ninja:[/quote]

What is AM-GM ?  :blush:[/quote]
AM-GM ineq (very famous) :)
Let $a_i$be positive reals, $i=\overline{1,n}$ then:
$\sum_{i=1}^na_i \ge n.\sqrt[n]{\prod_{i=1}^na_i}$
In my country, AM-GM is called Cauchy's ineq.
 :?  :?[/quote]

in my country we speak of cauchy, when speaking of:
\[ \left(\sum_{k=1}^n a_k^2\right)\cdot \left(\sum_{k=1}^n b_k^2\right) \ge \left( \sum_{k=1}a_kb_k \right)^2 \] or in terms of vectors:
\[ \forall a,b \in \mathbb R^n: a^2\cdot b^2 \ge \left( a\cdot b\right)^2 \]

(i guess if anybody of my country says something different, i am wrong, saying "in my country we speak of cauchy, when speaking of [...]" ;) )

[quote="shyong"]Isnt that just weight AM-GM ? :ninja:[/quote]

i'd call it that way too, but it needs a prove anyway :P

[quote="seamusoboyle"]I have a beautiful proof using jensens and the fact $e^x$ is convex.[/quote]

Now I know this proof too :thumbup:


analytic geometry
symmetry
geometry
3D geometry
icosahedron
Euler
group theory

I was thinking about the following simple problem:

Prove that the 600 cell has 600 cells.

More precisely, prove that if a convex, four-dimensional, polytope has cells that are all regular tetrahedra of equal size, and exactly five tetrahedra meet at each edge, the polytope must have exactly six hundred cells.

Obviously this is a well known mathematical object so this is not exactly an unsolved problem.  However, it seems that most books in describing the 600-cell simply introduce it, give its definition, and state without proof that it has 600 cells.  There are often accompanying diagrams that are usually less than edifying (to say the least) for someone who might wish to actually count the cells.

Clearly, one can accomplish a proof simply by exhibiting the coordinates of the 120 vertices and 600 cells.  I wrote a little Java program that starts with a group of five regular tetrahedra that meet at an edge, and then adds vertices and tetrahedra in the same basic structure until no more vertices or tetrahedra can be added.  There
are, indeed, 120 vertices and 600 cells when you are done, so I am quite persuaded that this object does have 600 cells.

But is there a simple, elegant, argument to prove that this object, which has a fairly simple definition, has 600 cells without actually computing them all?  I know that there is a correspondence between the vertices of the 600 cell and the rotational symmetry group of the regular icosahedron--is there any kind of argument based on this?  Is
there a simple elementary argument?

(I apologize if I am posting to the wrong forum.  I am posting to the group theory forum because group theory has been used to attack this problem but I am hoping there is a more elementary solution someone knows about.)

This seems like as good a place as anyway, and it fits this site's definition of "unsolved" well enough- the problem has a solution, but you don't know it.

A simple method of combinatorial counting:
Let the number of cells be $N$. Each two-dimensional face is on two cells, and each cell has four faces, so the number of faces $F$ is $\frac{N\cdot 4}2$. Each cell has six edges, and each edge is part of six cells, so the number of edges $E$ is $\frac{N\cdot 6}5$.
To count the vertices, we take a look at the structure around one vertex. Let $v$ be a vertex, and $A$ be the set of other vertices which share an edge with $v$. The members of $A$ form a polyhedron in their own right, using only the edges and faces that are also edges and faces of the original polytope. This polyhedron has triangular faces, with five such faces meeting at each vertex $w$,one from each of the five tetrahedra around the edge between $v$ and $w$. This should sound familiar- we already know that that polyhedron has to be an icosahedron.
This icosahedron has 20 faces, so 20 cells meet at $v$ and there are $\frac{4\cdot N}{20}$ total vertices.

Now we'd really like something like the Euler formula. Unfortunately, the basic generalization $V-E+F-C$ simply has value zero, for this or any four-dimensional polytope. That doesn't help calculate $N$, although it does confirm that we've got the right ratios.
I'll see about what the group can do for us (the symmetry group of an icosahedron/dodecahedron is isomorphic to $A_5\times \mathbb{Z}_2$, which has 120 elements)


[quote="kky1638"][b]21982145917308330487013369[/b] is thirteenth power of some natural number
what is that natural #?[/quote]

Quite a large number. The reduired number ends in 9. So, I have helped others now they can solve. HiHi... :rotfl:

The number $n$ must end in a $9$ because of the patterns of powers, however, this number has $26$ digits, but if you look at a few possible answers $9^{13} \Longrightarrow 13 \text{digits}, 19^{13} \Longrightarrow 17 \text{digits}, 29^{13} \Longrightarrow 20\text{digits}, 39^{13} \Longrightarrow 21 \text{digits}$,but I don't know if this is a good way since the possible answers are $84-99$ so either $99$ or $89$ could be the answers

THe method I used to find digits for $n^{13}$ was $13 \cdot \log{n}+1$ and round but I have no idea how to do it any other way, perhaps mods?

[quote="Hamster1800"][hide]the number ends in 9 and is 2 digits and close to 100 because $100^{13} = 100,000,000,000,000,000,000,000,000$ (27 digits)

99 is too high, so the answer is $\boxed{89}$[/hide]

btw, the actual way to get the number of digits is $\lfloor 13\cdot\log n + 1 \rfloor$

Try plugging 1 into your formula before. It says it has 0 digits![/quote]

Right, I forgot to add $1$ and round, so $\boxed{89}$ is the correct answer, btw, does anyone know how to do this w/o calculator?

[quote="varun"][quote="Hamster1800"][hide]the number ends in 9 and is 2 digits and close to 100 because $100^{13} = 100,000,000,000,000,000,000,000,000$ (27 digits)

99 is too high, so the answer is $\boxed{89}$[/hide]

btw, the actual way to get the number of digits is $\lfloor 13\cdot\log n + 1 \rfloor$

Try plugging 1 into your formula before. It says it has 0 digits![/quote]

Hamster1800, what is the formula to find the number of digits.
Is it for, $n^{x}$ equal to $\lfloor n\cdot \log n + 1 \rfloor$[/quote]

Yes and then round your answer to the nearest integer :)

[quote="varun"][quote="Hamster1800"][hide]the number ends in 9 and is 2 digits and close to 100 because $100^{13} = 100,000,000,000,000,000,000,000,000$ (27 digits)

99 is too high, so the answer is $\boxed{89}$[/hide]

btw, the actual way to get the number of digits is $\lfloor 13\cdot\log n + 1 \rfloor$

Try plugging 1 into your formula before. It says it has 0 digits![/quote]

Hamster1800, what is the formula to find the number of digits.
Is it for, $n^{x}$ equal to $\lfloor n\cdot \log x + 1 \rfloor$
What is the base of the log, is it 10?
Do we write "ln" for log base to 'e'?

Varun[/quote]

acutally, the formula is $\lfloor x\cdot\log n + 1 \rfloor$ The reason is that we are actaully doing $\log{n^x} = x\cdot\log n$. And yes, the log is base 10, and yes, we write ln for base e.

Can anyone do this without a calculator, because I think this problem was intended to be solved without one. ;)

$n^{13} \equiv n \mod{13}$

You need that big number mod 13, then you can decide if its 89 or 99

Mod 10 tells you it ends in 9.
$\mod 9 \Leftrightarrow \ \text{sum of digits}$ tells you it is 8 mod 9.

Numbers ending in 9 that leave a remainder of 8 when divided by 9: 89, 179, 269, 359,... But we can easily see that 100 is an upper bound.  Thus 89 is the only possibility left.

How I solved it was the following.  By the casting out nines rule in number theory, I said
\[ \text{(The sum of all the digits in the number)}\equiv 8\mod 9 \].
We then take a look at numbers $a_1,...,a_n$ that have a remainder of 8 and satisfy $89\le a_1,...,a_n<100$.  We started with 89 because $89\equiv 8 \mod 9$. Numbers  $a_1,...,a_n$ could be 89 or 98, while satisfying $89\le a_1,...,a_n<100$. Since, $13\log98+1=26.89\approx 27>13\log89+1=26.34\approx 26$, we say that 89 is better.  We can do this without a calculator by just saying that since $98^{13}$ is very close to $100^{13}=1\times 10^{27}$, 89 is the answer.  $\therefore 89^{13}=\text{The huge number}$. :)

Masoud Zargar

The problem in your signature is THE (common) definition of $\exp x$

lol, people have been PMing me abtou this. :)


My parents are always nagging me to sleep. How important is sleep to you guys? How important is it overall? I here middle schoolers should get 7-9 hours, but I only get barely 7, so I'm always sleepy - that is why I am sleeping 2 hours earlier today - but I'll probably end up reading 1984 just because I'm really not sleepy - but for the Math Counts cause I will today and tomorow.

I wish I could sleep a lot more than I am able to. Usually, I finish my homework about 9-10, and I'm falling asleep by then. Anyhow, I usually can't get to bed until around 12. I voted that sleep is very important because I like to sleep!

Very: It can make you thinner, study better, grow taller, and all sorts of marvelous things like not making you crazy, so if you don't want to act like me, then [b]go to sleep[/b]!

I didn't have that much trouble with sleep deprivation until college. I typically got about 7 hours of sleep a night, which was enough since I never had to do anything particularly demanding. When I got to college, I started going to sleep at 1AM and getting up at 7AM. It didn't work. In a drastic attempt to stay awake for my classes, I completely changed my sleep schedule. Now I go to sleep at 9:30PM and get up at 4:30AM. One of the great things about the morning is that there are far fewer distractions, so it's much easier to get work done. Also, I don't have to put up with my roommate's noise pollution, er, I mean music, that sounds so terrible half the time.

it's weird... usually i feel better with 3 or 4 hours of sleep instead of 6 or 7.  I've been trying to go to sleep earlier, but before i used to get about 3 or 4 hours of sleep 3-4 times during the week, and performance in school or wahtever didn't change much when i started getting more sleep.  But i always try to get a lot of sleep before big tests (sat's or math stuff).

i get about 8 sleep at 11 and wake up at 7. i use to sleep 9 hours however the only major impact i am the slowest swimmer ever and my runnings getting a little worse.

I only sleep at 11 or 12 o clock. Otherwise, Im working. [color=white]Jarey I might need some help for mathcounts[/color]
i try till the end, but, it always doesnt turn out that way :mrgreen:

Sleep is very important.  When you don't get enough sleep your reasoning goes way down.  If I'm well rested I get a significantly more number of math problems right than when I'm just slightly sleep deprived.  Also, you can't just get a lot of sleep the day before your exam.  It will help-but basically it just makes up from lost sleep the days before.  In order to be totally alert, and able to do your best you need to start getting extra sleep at least a few days in advance.  The first AMC tests are coming up Tuesday!  Start getting to sleep early now!

I've never had enough to make a judgment.  I bet that if I did get more than what I do get, I would significally do better in maths competitions. Bah

Meh, well... I get a moderate amount of sleep. I thought it was little until I read how much yall get. I go to bed at 9 and wake up at 6. Thats about 9 hours. Sometimes my mom lets us sleep in until 6:30 if we are lucky. Sleep is somewhat important however, it is not the most important. To me my grades are more important than sleep. I have to continue to keep up my grade in order to stay in all the extra crics Im in and to maintain the position of President of the National Junior Honor Society. So, sleep is not as important as school is to me.

Yep, it's set up just like our government except with less positions just your 5 main officers and a planning commitee and a judgment commitee. All of the officers are involved with everything and I like being the president because I even have athourity over my principle sometimes. I have the right to be late to class or be exempt from an assignment if I am doing something for NJHS. I have the choice on everything however it has to go through the planning commitee first and if it passes there it has to go on to the judgment commitee.

what is njhs i've never heard of it at my school (or is it for senior high).

njhs=National Junior Honour Society

njhs is for middle school/junior high where as nhs (national honour societ) is for high school

It doesn't take effort to be in njhs.  I somehow managed a treasurer position after writing my application in 5 minutes during homebase and then bsing my interview.  Yes, and I get around 4 hours of sleep most of the time.  Stop complaining.  8th grade requires no effort at all.

8th grade is the time to work on your maths competitions.

Im in 9th grade at AAST, I typically sleep at 12:30 to 1:00, and wake up at 5:30-ish so I can catch the bus at 6:30... You just kinda have to get used to lack of sleep.. I'm fine without it...

Ok that was treasurer this is president. I dont know how it was at your school but at mine I have to plan the meetings, Im in charge of making sure all the members are there, I have to make all announcments for the njhs, that in turn means I have to be at all meetings and events that we sponser, it's kind of hard to concentrate on math competitions when youre busy all the time. Im also doing an idependent study and I just dropped out of hadit (hornets against drugs in Texas.) Hornets are our mascot if youre wondering. Thats alot when youre doing all honors classes too and choir then Im in math team and dropped out of athletics. I hate being so busy I think though so Im not so busy in college Id rather take those extra crics and get extra credit and recorded community service hours. Now syntax tell me that you balanced all that on one plate please. This is alot of work and Im not complaining I like doing all this stuff I just hate being under so much stress. Its overwhelming sometimes and any who have done this much at one time will know what Im talking about. Im just happy today since its my b-day even though it was one of those days where I had to do everything because of iflirtz (a valantines deal we do at school.) Its retarded but hey its community service hours and Im all for that.

Okay maybe at your school but not at mine. We have to maintain a 92 or above average and it increases as time goes on so by the end of the year it could be somewhere around a 95. I did not run for this position. The kids dont vote you in the teachers do. You dont run for any positions at my school they just stick you there. It's not fair. At my school some of the advanced classes are easy but not english. Almost everyone fails her class and I normally have the highest grade in there which is a 92 to a 94. My history teacher is the same way I have the highest grade in class which is a 96. All the other classes are pretty easy but you run out of time to do everything. There is so much to do in a small town it's crazy!The only competiton teams we have here are the math and science team, chess club, and athletics.

Serious? Mmm can you post like a sample problem since we have no idea what the standards are in your school...

P.S. The above is a typical example to disprove the statement "There's spam only in Fun and Games," also proves whitehorsemath's point elsewhere

P.P.S. Course the above result is understandalbe since [b]The Round Table is part Etc.[/b]

PPPS the above is a typical example to prove that Tare doesn't know the definition of spam

You don't know the definition of spam, the above is under the category "off-topic posts" and is subject to either splitting or deletion.

P.S. It really doesn't matter what you think spam is

P.P.S. You're post is also spam (off-topic)

P.P.P.S. I stand thoroughly corrected on my statement "the only time TripleM spammed"

Tare are you in a bad mood tonight please fill us in on what happened if something did. You are never in this type of mood.

[size=150]Hey![/size] --

all of you, quit acting like babies.  TripleM, Tare, and everyone else who keeps up this nonsense: go warn each other on AIM or something.  I'm tired of reading all your trash-talking complaining nonsense.

"You're spamming!"
"No, you're spamming!  And deleting things!"
"No, I'm not!"
"Yes, you are!  And your forum stinks!"
"No it doesn't!  No I'm not!  My forum is great!  And you're spamming!"
"Shut up, spammer!"
"Spammy spam spam.  Go spam yourself.  Stop spamming!"

^-- that is what you all sound like!  Cut it out!  Grow up!

[quote="Syntax Error"]your grades are pass or fail going into high school anywho...[/quote]

Maybe for you.  That's certainly a local decision, though, and could be done differently in different places.  For example, at my highschool, you could get grades for advanced classes you took in Jr. high if you brought them a transcript, but the school never even saw any of your other grades.

[quote="bugzpodder"]i sleep for 8-9 hrs and i feel like i need 10...  i am always tired in class...  maybe somethings wrong with me[/quote]

Thanks for bringing the thread back on topic. Do you get much physical exercise. I notice that I--and my kids, at all ages--sleep much better if I get an adequate amount of exercise, which, alas, I don't get every day. Today we had perfect weather for an outdoor snowball fight here in Minnesota, so the three boys were out for two hours while I stayed in to watch my one-year-old daughter, who took a nap part of the time. I think my oldest son will sleep reasonably well tonight, before we get up early (for us, on a Saturday) for tomorrow's Mathcounts chapter meet.

I hope everybody gets the rest they need before the big tests and competitions this year.


ca nu-s apt la ora asta de problema lui maky, mai bag si eu o problema de geometrie de data asta  :blush:

Cercul C trece prin punctele B, C si prin centrul cercului inscris triunghiului ABC.Cecul C intersecteaza a doua oara dreapta AB in B1 si pe AC in C1. Sa se arate ca BB1 = CC1. :)

[hide="solutie"]pentru inceput observam ca $B_{1}$ si $C_{1}$ sunt pe segmentele $(AB)$, $(AC)$ (pornim constructia de la cercul $\mathcal{C}$, apoi alegem doua puncte $B,C$ si $I$ astfel incat $\angle IBC$ e obtuz, apoi il construim pe $A$ si observam ca $BA$ si $CA$ intersecteaza cercul in semiplanul determinat de $A$ si $BC$)
apoi $\angle B_{1}IB=\angle B_{1}CB=C/2$, si deci $\angle BB_{1}I=180-(B/2+C/2)$, deci si angle $CC_{1}I=180-(B/2+C/2)$, deci arcele $BB_{1}$ si $CC_{1}$ sunt egale, deci si coardele sunt egale.[/hide]

super misto rezolvarea...eu o facusem astfel :) postez diseara :P


Peter said:"The day before yesterday I was 10, but I will turn 13 in the next year."
                   IS THIS POSSIBLE?

If today were the first of January 2009 and Peter was born on the 31st of December 1997, then the day before yesterday was the 30th December 2008 when Peter was 10. He will be 11 on the 31st of December 2008 and 13 on the 30th of December 2010.


If a,b,c are in interval (0,1) prove:

$ 3\plus{}4(ab\plus{}bc\plus{}ca)\ge4(a\plus{}b\plus{}c)$

[quote="Bugi"]If a,b,c are in interval (0,1) prove:

$ 3 \plus{} 4(ab \plus{} bc \plus{} ca)\ge4(a \plus{} b \plus{} c)$[/quote]

[hide="Solution"]$ 3 \plus{} 4(ab \plus{} bc \plus{} ca)\ge4(a \plus{} b \plus{} c)\ge4\sqrt {3(ab \plus{} bc \plus{} ca)}$
Let $ \sqrt {ab \plus{} bc \plus{} ca} \equal{} t (0 < t < 3)$, we have an equivalent inequality:
$ (2t \minus{} \sqrt3)^2\ge 0$ [/hide]

[quote="chrischris"]:maybe:

but ... when a = b =0 , c =1
the inequality doesn't hold[/quote]
Well $ a,b,c\in (0,1)$ :P

um.. I don't understand..

does the notation mean a, b and c lie between 0 and 1?

When a and b are close to 0, c close to 1, the inequality doesn't hold.

[quote="Bugi"]If a,b,c are in interval (0,1) prove:

$ 3 \plus{} 4(ab \plus{} bc \plus{} ca)\ge4(a \plus{} b \plus{} c)$[/quote]

There must be something missed in the question.And Kyoshiro didn't prove the inequality. He only proved a weaker result.

Just put $ a\equal{}0.3, b\equal{}0.7$ we know that the inequality is wrong.

My mistake, it is wrong.

@chrischris: (0,1) is a interval which excludes 0 and 1. [0,1] is a interval which includes 0 and 1. (0,1] is a interval which includes 1, excludes 0? Get it?

I think that the condition should be $ a,b,c\in[\frac{1}{2},1]$ :maybe:


For the three identical squares shown, points A, B and C are vertices, and AB = 2 square root 5 cm. What is the length of AC, in centimeters? Express your answer in simplest radical form.

problem no 39 please see the figure in  the following pdf link

http://www.mnspe.org/mathcounts/2007tests/510-07_C_Countdown.pdf

Hi.

If the side-length of each square is $ x$, then, by Pythagoras,  $ AD^{2} \plus{} DB^{2} \equal{} AB^{2}$ , where $ D$ is the left-hand corner of the square.

That is, $ x^{2} \plus{} (2x)^{2} \equal{} (2\sqrt{5})^{2}$.

Therefore $ x\equal{}2$.

Now $ AC^{2} \equal{} AD^{2} \plus{} DC^{2} \equal{} 2^{2} \plus{} 6^{2} \equal{} 40$,

so $ AC\equal{}  \sqrt{40} \equal{} 2\sqrt{10}$.

(Or, we could use the cosine rule in triangle $ ABC$.)


modular arithmetic
least common multiple
number theory
superior algebra
superior algebra unsolved

On a test I just got out of, there was the following problem (some of the details might be a little bit off):

Let $ \beta ^2 = (1 3 5 7 9)(2 6 8)$.   Find $ \beta $.

Well, what popped into my head was that $ \beta ^2 $ must have order 15, since 15 is the least common multiple of the length of the disjoint cycles.  Does this imply that $ \beta $ has order 30?  This is where I got stuck, because that means that $ \beta $ can be written as disjoint cycles of lengths 5 and 6, 2 and 15, or 1 and 30, but each of these are of course impossible since there are only 10 symbols (actually, this wasn't stated explicitly...could it have been a trick?).  I think I may be confusing concepts here or misunderstanding a theorem.  Any help would be appreciated.

Look at the disjoint cycles separately. Any non-trivial 5-cycle is the square of another 5-cycle, and any nontrivial 3-cycle is the square of another 3-cycle.

Look at it this way: $\beta^2=\alpha\gamma,$ where $\alpha\gamma=\gamma\alpha,\,\alpha^5=e,$ and $\gamma^3=e.$

Then $(\alpha^3)^2=\alpha$ and $(\gamma^2)^2=\gamma.$

So let $\beta=\alpha^3\gamma^2.$

I'll let you work out what that looks like in cycle notation. And the order of $\beta$ is 15.

An alternative that's even easier to think of: since the order of $\beta^2$ is 15, just solve the number theory problem $2x\equiv1\pmod{15},$ getting $x\equiv8\pmod{15}.$

So let $\beta=(\beta^2)^8.$

It turns out that this is $\alpha^8\gamma^8=\alpha^3\gamma^2,$ so it is the same as the previous computation.

We hope there aren't too many more symbols; the answer is not unique if there are at least 11 objects to permute.


inequalities
number theory theorems
number theory

[quote="Raminder"]Prove that d(n)<2(square root of n).[/quote]

Let $ n\equal{}p_1^{a_1}p_2^{a_2}....p_k^{a_k}$.

Then it becomes to prove $ [(a_1\plus{}1)(a_2\plus{}1)...(a_k\plus{}1)]^2<4p_1^{a_1}p_2^{a_2}....p_k^{a_k}$

Clearly, for $ p_i \ge 5$,
$ (a_i\plus{}1)^2 \le 5^{a_i} \le p_i^{a_i}$ (because $ (\frac{x\plus{}2}{x\plus{}1})^2<5$ for $ x \ge 0$)

Similarly we have $ (a_2\plus{}1)^2<2 \times 3^{a_2}$
and $ (a_1\plus{}1) \le 2 \times 2^{a_1}$

Multiply these three inequalities we have $ [(a_1\plus{}1)(a_2\plus{}1)...(a_n\plus{}1)]^2 < 4p_1^{a_1}p_2^{a_2}....p_k^{a_k}$, done.

Thanx a lot. But I need some more explainations. :roll:

Well the solution above is quite elegant except for one point that perhaps needs some clarification the reasoning of 5^a>(1+a)^2. Anyway the main trick of the sum would be that following the above statement we just have to manipulate the 4 and a1 and p1 and p2 to justify the inequality.

[hide="Aliter"]

If $ d_1 d_2 \equal{} n$, then one of them is necessarily less than $ \sqrt n$. Thus we can pair up all divisors this way. Hence there will be at most $ 2 (\sqrt n \minus{} 1)\plus{}1 < 2 \sqrt n$ divisors

[/hide]


[hide]Notice that the possible units digits of $3^a$ mod 10 are $3,9,7,1$. The possible units digits for $7^b$ are $7,9,3,1$. Therefore, we just need to find how many of these 16 possible sums pair up to a units digit of $8$.

1,7 pair, 7,1 pair, and 9,9 pair, for a 3/16 probability   [/hide]

[quote="4everwise"]First $a$ is chosen at random from the set $\{1,2,3,\ldots,99,100 \}$, and then $b$ is chosen at random from the same set. The probability that the integer $3^a+7^b$ has units digit $8$ is

$\text{(A)} \ \frac1{16} \qquad \text{(B)} \ \frac18 \qquad \text{(C)} \ \frac{3}{16}\qquad \text{(D)} \ \frac15 \qquad \text{(E)} \ \frac14$[/quote]
[hide]I got $(C)\frac{3}{16}$.
The units digit of $3^a$ from ${1,2,3,\ldots,99,100}$ is $3,9,7,\ldots,3,9,7,1$, since 100 is divisible by four. For $7^b$, it is basically the same: $7,9,3,1,\ldots,7,9,3,1$. If three is chosen from the first set then nothing can be chosen from the second to get a units digit 8 (added together). If 9 is chosen, then 9 from the second set can be chosen to get units digit 8. If 7 is chosen, 1 can be chosen from the second set to get units digit eight, and if 1 is chosen, 7 can be chosen to get units digit 8. So the probability is $3\displaystyle\left(\frac{1}{4}\cdot\frac{1}{4}\right)=\boxed{\frac{3}{16}}$.[/hide]
EDIt: too late... :P


In triangle $ ABC$, $ D$ is the midpoint of $ AB$ and $ E$ is the point of trisection of $ BC$ nearer to $ C$. Given that $ \angle ADC \equal{} \angle BAE$ find $ \angle BAC$.

See here:  http://www.mathlinks.ro/viewtopic.php?t=42760

Best regards,
sunken rock


can someone please explain the physics behind the functioning of a battery(bot ideal and non ideal)

well taht's ok about the internal battery mechanism...
what i wanted was how electrons move in a conducting wire that is attached to the battery and why this field that makes them move is so insignificant( :maybe: ) out side the conducting wire


deos genetics have to do anything with being smart? :?:

Yes, it does.  ;)

However, I think perseverence can get you further than just being creative.

Genetics can probably give a slight advantage although I have never seen scientific evidence, but, honestly -

"Smart" is such a general term. There are so many different kinds of "smart" - my guess being that some have more to do with genetics than others - that you can't really give a straight answer to this questions.

I think that genetics helps you somewhat, but there are other factors. If you're parents are smart, it doesn't necessarily mean you've got good genetics, you could just of been exposed to knowledge that your parents have. I think perseverance is the most important factor.

In addition - you could get either a blended form of intelligence from your parents or just from one parent it seems. I have a friend who isnt very smart.. low classes of everything, but her father is a genius, and her mother isnt.. really. It could also just be influences (ex. shes distracted for whatever reason)


[quote="Stephen"]$ x, y, z$ are intergers.

If $ 2x^2 \plus{} x^2y^2 \plus{} y^4 \equal{} z^2$, prove that $ x \equal{} 0$.[/quote]

Oh, sorry, it was an error. The question is:

$ 2x^4 \plus{} 2x^2y^2 \plus{} y^4 \equal{} z^2$.

the equation is equivalent to: $ x^4 \plus{} (x^2 \plus{} y^2)^2 \equal{} z^2$

this is Pytago type equation


Prove that 2c/((-b :pm:  :sqrt:) b :^2:+4ac) = ((-b :pm:  :sqrt:) b :^2:+4ac)/2a
EDITED

I don't understand that.  Your typing is a little sloppy could you fix it up a little please.  It looks like you said prove:

[tex]\displaystyle{\frac{2c}{-b\pm\sqrt{b^2+4ac}}=\frac{2a}{-b\pm\sqrt{b^2+4ac}}}[/tex]

[quote]Prove that 2c/((-b  ) b +4ac) = ((-b  ) b +4ac)/2a [/quote]

[tex]\frac{2c}{-b\pm\sqrt{b^2+4ac}}=\frac{-b\pm\sqrt{b^2+4ac}}{2a}[/tex]

Right?

In the  :pm: are the signs opposite each other? If so, just [hide]cross multiply and expand?[/hide]

Cross multiplying does 'work' but since this is a proof, should "cross-multiplying" be allowed?  Note that cross multiplying is a shortcut for applying the "multiplication property of equality".

Ie since we are trying to prove that these two expressions ARE equal, a vital step of the proof cannot require the "multiplication property of equality".    This is analogous to the rules for proving trig identities - work on either or both sides of the identity, but do not 'move' things left and right.

Challenge:  find a proof without using properties of equality.

btw, this problem yields a second version of the quadratic formula.  From a computation or problem solving purpose, is this second version ever preferable to the original version?

Basically to make what Yooper said a little clearer, you can't cross multiply b/c you don't know A=B until you prove it.

blahblah =? somethingelse

Now you can manipulate blahblah and you can manipulate somethingelse, you can multiply both sides by a constant/variable, but you cant do something like:

blahblah - somethingelse =? 0

Since you don't know that blahblah=somethingelse.

This probably was not too helpful, especially because I used "blahblah" instead of a variable.

That was actually pretty clear, once I figured out "blahblah" meant "some expression" instead of the AoPS member with a similar name.  I disagree with yooper, however.  Because every step, including that cross-multiplying, is reversible, you can start with A = B and derive something true.  Granted, it's bad form, but it is a valid proof nonetheless.

[quote="bob123"][quote="nr1337"]I think this should be

[tex]
\frac{2c}{-b \pm \sqrt{b^2 - 4ac}} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
[/tex][/quote]

No, it's this because when you simplify it, you want the 4ac to be positive in order to make both sides of the equation equal.

[tex]
\frac{2c}{-b \pm \sqrt{b^2 + 4ac}} = \frac{-b \pm \sqrt{b^2+4ac}}{2a}
[/tex][/quote]
it is minus.

Wouldn't you be able to go ahead and cross-multiply? If it turns out then that a=b, then wouldn't you stll be able able to use that inforrmation?

Only if each of your steps is reversible.  In this case, yes.

What exactly do you mean by "reversible"? (sorry, it's been a while since I did even minutely complex algebra.)

Well, a step is reversible if it is true in both directions.  For example, if we subtract something from both sides of an equality, we have made a reversible step: each side implies the other.

That is, x - 3 = 7 [b]if and only if[/b] x = 10.

Squaring, however, is not a reversible step, since

x = 3 implies x2 = 9, but
x2 = 9 does [i][b]not[/b][/i] imply x = 3.  So, you cannot start from what you are trying to prove and then square it to get a true statement in order to prove something.  However, you can start from what you're trying to prove and multiply by a non-zero constant on both sides to get something true and thus prove your result.

Since in our problem, we are multiplying by denominators, we cannot be multiplying by things equal to zero.  Since that is a reversible step, it means that when we arrive at a true statement, we have in fact proven what we were trying to.


linear algebra
matrix
vector
linear algebra unsolved

We define a [i]pseudo-inverse[/i] $B\in \mathcal M_n(\mathbb C)$  of a matrix $A\in\mathcal M_n(\mathbb C)$ a matrix which fulfills the relations
\[ A = ABA \quad \text{ and } \quad B=BAB. \]
a) Prove that any square matrix has at least a pseudo-inverse.

b) For which matrix $A$ is the pseudo-inverse unique?

[i]Marius Cavachi[/i]

The creator of this notion of pseudo-inverse is Penrose. Here is a paper with a lot of applications of this concept. ;)

Can anyone kindly translate the article?  :(

Let A = PDQ, D is I_k direct sum with zero matrix, P,Q invertible (obtained from gaussian elimination)

(1)Let $B = Q^{-1}DP^{-1}$
(2)If A is invertible, obviously B is unique, and it's $A^{-1}$, otherwise, take B = $Q^{-1}DP^{-1}$ or $Q^{-1}(2D-I)P^{-1}$, they are distinct, so we are done.

Soarere, can you send me please this article to the following e-mail adress: lupucezar@yahoo.com

[quote="RealZeta"]I am not able to fully understand the part about Soarer's solution where he says "D is  I_k direct sum with zero matrix", it'd be awesome if somebody could explain that part.

Thanks[/quote]
I mean a matrix with only diagonal entries 1,...1,0,...0

[quote="Soarer"](2)If A is invertible, obviously B is unique, and it's $A^{-1}$, otherwise, take B = $Q^{-1}DP^{-1}$ or $Q^{-1}(2D-I)P^{-1}$, they are distinct, so we are done.[/quote]
Are you sure that $B = Q^{-1}(2D-I) P^{-1}$ as a pseudo-inverse works? My computations show otherwise.

[quote="Soarer"]Can anyone kindly translate the article?  :(

Let A = PDQ, D is I_k direct sum with zero matrix, P,Q invertible (obtained from gaussian elimination)

(1)Let $ B \equal{} Q^{ \minus{} 1}DP^{ \minus{} 1}$
(2)If A is invertible, obviously B is unique, and it's $ A^{ \minus{} 1}$, otherwise, take B = $ Q^{ \minus{} 1}DP^{ \minus{} 1}$ or $ Q^{ \minus{} 1}(2D \minus{} I)P^{ \minus{} 1}$, they are distinct, so we are done.[/quote]

In (2) your second example doesn't work. One can take
\[ L \equal{}  \left(  \begin{array}{ccc}
I_k & 0 \\
* & 0 \end{array} \right) \]
instead of $ 2D \minus{} I$ to fix it. Then $ LD \equal{} L$, $ DL \equal{} D$ so $ LDL \equal{} LD \equal{} L$ and $ DLD \equal{} D^2 \equal{} D$, as desired. Note that here we assume that $ A\neq 0$ in which case the pseudo-inverse is also unique $ B \equal{} 0$.

I had another solution using Jordan Normal Form. Just a sketch here: WLOG we may assume that $ A$ is in JNF, then that $ A$ has only Jordan Blocks with eigenvalue $ 0$, then that there is only one such block and in such case one may take
\[ B \equal{}  \left(  \begin{array}{ccc}
v & 0 \\
I_{n\minus{}1} & 0 \end{array} \right) \]
where $ v$ is any row vector of length $ n \minus{} 1$.


prove that the algorithm of [color=red]sqrt(x)[/color], x real number is correct!

Which algorithm?  There are [url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots]quite a few[/url].

I was refering to digit by digit calculation, in base 10( the ones we do at school on the blackboard)
The link to the topic is quite useful, thank you!
If you have other ideas/observations please tell them!
                                                                        A.I.


In the ninth grade class, each of 130 students attends exactly one of the five English classes.  Each class has fewer than 30 students.  What is the least possible number of students in any one of these five classes?

A.10	 B.14	 C. 20	 D.23	  E.26

[hide] In order to minimize one class, we must maximize the others. The maximum in any one class is $ 29$, so $ 130\minus{}29(4)\equal{}14$. Hence, the answer is B.[/hide]


The 2 pairs of red triangles and the 2 red-white pairs use all $ 2(2)\plus{}2\equal{}6$ of the red triangles, so there are no blue-red pairs.  The 3 pairs of blue triangles leave $ 2(5)\minus{}2(3)\equal{}4$ blue triangles to be pairs with white triangles, so $ 4$ white triangles are used in these pairings.  Two of them were already used in the red-white pairs, so there are $ 8(2)\minus{}(4\plus{}2)\equal{}10$ white triangles left.  These are paired with each other, so there are $ 10\div 2\equal{}\boxed{5}$ coinciding white pairs.


I wasn't sure where else to put this. What does "watching a forum" mean? At the bottom of the forums there's this little icon that says something like "Watch this forum".

huh. im not sure but i think AoPS gives you updates via email about the forum?!?! idk

Oh, never mind. I just figured it out. I clicked "Watch this Forum" on the FTW Problems and Discussion. You get an e-mail every time there's a new post. Now I gotta go clean my inbox...it's flooded!


Hello fellow math mates :wink:

Given that:
$ a \plus{} b \equal{} ab \equal{} a^b$
Prove that:
$ a \equal{} b \equal{} 2$

I know it isnt that hard, but I just dont get how to start on it. Any solutions?
Thanks in advance! :blush:

Edit: I might have posted this in the wrong forum. Sorry.

What happens if you subtract $ a\plus{}b$ from both sides of the first equality, then add 1?

Now what happens if you take the second equality and divide both sides by $ a$, then take the $ b\minus{}1$'th root of both sides?

Can you solve for $ a$ in both equations, then set them equal and solve for $ b$?

[quote="lolbits"]Given that:
$ a \plus{} b \equal{} ab \equal{} a^b$
Prove that:
$ a \equal{} b \equal{} 2$
[/quote]

[hide]
-----------------------------------------------------

$ ab \equal{} a^b$

$ \frac {ab}{a} \equal{} \frac {a^b}{a}$

$ b \equal{} a^{b \minus{} 1}$ . . . (**)

$ log(b) \equal{} log(a^{b \minus{} 1})$

$ log(b) \equal{} (b \minus{} 1)log(a)$ . . . (*)

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

Solve for $ a$ in $ a \plus{} b \equal{} ab$.

$ ab \minus{} a \equal{} b$
$ a(b \minus{} 1) \equal{} b$

$ a \equal{} \frac {b}{b \minus{} 1}$ . . . (**)

To substitute the expression in (**) into (*), first take the logarithm
of each side will be taken in (**):

$ log(a) \equal{} log(\frac {b}{b \minus{} 1})$

$ log(a) \equal{} log(b) \minus{} log(b \minus{} 1)$

Make the substitution into (*):

$ log(b) \equal{} (b \minus{} 1)[log(b) \minus{} log(b \minus{} 1)]$

$ log(b) \equal{} (b \minus{} 1)log(b) \minus{} (b \minus{} 1)log(b \minus{} 1)$

$ (b \minus{} 1)log(b) \minus{} log(b) \equal{} (b \minus{} 1)log(b \minus{} 1)$

$ [(b \minus{} 1) \minus{} 1]log(b) \equal{} (b \minus{} 1)log(b \minus{} 1)$

$ (b \minus{} 2)log(b) \equal{} (b \minus{} 1)log(b \minus{} 1)$

$ log(b^{b \minus{} 2}) \equal{} log[(b \minus{} 1)^{b \minus{} 1}]$

$ b^{b \minus{} 2} \equal{} (b \minus{} 1)^{b \minus{} 1}$

When $ b \equal{} 2$, the equation becomes:

$ 2^0 \equal{} 1^1$

In (**), $ a \equal{} 2$

Can it be shown that $ a \equal{} b \equal{} 2$ is the only solution?

EDIT:  I claim that the following could answer it.

Contrapositive:

If it is not the case that $ a \equal{} b \equal{} 2$, then at least one of the following
is not true:
$ a \plus{} b \equal{} ab, a \plus{} b \equal{} a^b, ab \equal{} a^b$.

If it is not the case that $ a \equal{} b \equal{} 2$, then
either $ a \equal{} 2$ and $ b \ne 2$, or $ a \ne 2$ and $ b \equal{} 2$, or $ b \ne 2$ and $ a \ne 2$.

Without loss of generality, suppose $ b \ne 2$.

(i) Let $ b \equal{} 2 \minus{} \epsilon$, where $ \epsilon > 0, \epsilon$ belongs to the real numbers.  Then by (**), $ a \equal{} \frac{2 \minus{} \epsilon}{1 \minus{} \epsilon}$.

(ii) Let $ b \equal{} 2 \plus{} \epsilon$.  Then by (**), $ a \equal{} \frac{2 \plus{} \epsilon}{1 \plus{} \epsilon}$

In either case, $ a$ would equal a value other than $ 2$, while $ b \ne 2$.

[/hide]

Well.. I dont get it? Doesnt a equal to b?

[hide="@Arrange_your_tan"]I'm not too sure if that counterexample (sort of, i'm not sure how to classify it) is accurate.

I think you would still have to also use (*) because you would get multiple (infinite) answers for an equation that has two variables.

$ \log {b} \equal{} (b\minus{}1)\log{a}$

$ \log{a} \equal{} \frac{\log{2\minus{}\epsilon}}{2\minus{}\epsilon\minus{}1}$

$ \log{a} \equal{} \log{(2\minus{}\epsilon)^{2\minus{}\epsilon\minus{}1}}$

$ a \equal{} (2\minus{}\epsilon)^{2\minus{}\epsilon\minus{}1}$

which begins to become very messy.
Sorry, I'm not too sure how to prove that $ a \equal{} b \equal{} 2$ is the only solution.[/hide]

[hide="@lolbits"]Arrange your tan was trying to prove that $ a \equal{} b \equal{} 2$ is not the only solution. [/hide]

Ok, I have tried, and here are the results:

[hide="Solution"]
$ ab \equal{} a^b$
$ \rightarrow b \equal{} a^{b\minus{}1}$

$ a \plus{} b \equal{} ab$
$ 1 \plus{} \frac {b}{a} \equal{} b$
$ \rightarrow b\minus{}1 \equal{} b/a$

Combining both:
$ a^\frac {b}{a} \equal{} b$
or: $ a^b \equal{} b^a$

Use logarithmic function:
$ b * log(a) \equal{} a * log(b)$
or:
$ \frac {log(a)}{a} \equal{} \frac {log(b)}{b}$

Since $ \frac {log(b)}{b}$ is strictly increasing (can't log negatives), so is $ \frac {log(a)}{a}$.  Meaning the function is non-recurrent, thus $ a \equal{} b$.

so now that's proven that a = b:
$ ab \equal{} a \plus{} b$
$ \rightarrow a^2 \equal{} 2*a$
$ \rightarrow a^2/a \equal{} 2$
or: $ a \equal{} 2$
thus $ b \equal{} 2 \equal{} a$[/hide]
Please, do comment back if I have accidentally forgotten something / done something wrong.

Cheers

[quote="lolbits"]Since $ \frac {\log(b)}{b}$ is strictly increasing[/quote]  Nope: increasing for $ 0 < b < e$, decreasing thereafter ($ \lim_{b \to \infty} \frac {\log b}{b} \equal{} 0$).  By the way, note \log (and similarly \sin, \tan, \lim, etc.) and the multiplication dot \cdot in LaTeX.  (This looks like some potential progress, though.)


This may be a bit easy.

You have 12 blocks in a train. How many rectangles can you make?

[hide] 2 lines makes 1 rectangle
$\frac {13!}{2!11!} = 78$.[/hide]

[hide]Well, I get 78. You can have: 1 train-12, 2 trains-11, 3 trains- 10, ... until you get to 12 trains -1 .

So the total number of trains you can make is 12 + 11 + 10....3+ 2+ 1

And an easy way to find the first n consecutive positive numbers is n(n+1)/2.

So you have 12(13)/2 = 78.[/hide]

Can anyone of you explain how this works though?

[hide]If you pick any two vertical lines from the train, you have a rectangle. Since there are 13 lines then you just do 13 choose 2.[/hide]


Find the next number of the following sequence:
$1,11,21,1211,111221,312211,?$.

[hide="Hint"] Say each digit out loud[/hide]

[hide="Answer"]
one three, one one, two two, two one

13112221
[/hide]

It's the famous "look and say" sequence.

The first term is 1.

Now you say this: There is one one.

The second term is what you pronounced : 11

Now repeat: Two ones.

Third term is 21.

Repeat.

each term is the previous term "pronounced".

I first saw this problem from Art and Craft of Problem Solving book.


logarithms
real analysis
real analysis unsolved


Hello. I've recently begun considering taking some online classes such as the courses on EPGY, especially some at the college level (not just in mathematics, but in other subjects as well). I was wondering if, beside the personal benefit I'd garner from these classes, they would be helpful in for making my college application look better. Would they make any difference? I was considering courses such as

EPGY (especially this one)
BYU (http://ce.byu.edu/is/site/)
Others that I haven't heard of but that you may know of

Any and all comments would be much appreciated!

Being resourceful and using the internet to further your education beyond what your high school offers defininetly looks good to colleges!

[quote="horselvr101"]

EPGY (especially this one)
BYU (http://ce.byu.edu/is/site/)
Others that I haven't heard of but that you may know of

Any and all comments would be much appreciated![/quote]

Try this site

http://ocw.mit.edu/OcwWeb/index.htm

You wont have a transcript, or a grade. Speaking as a homeschooling parent, all that is highly overrated. Admission committees do know how to handle this sort of item on your application.


Hi there,
Does anyone have any clue approximately how long after submitting all application materials will the admission decision be mailed/emailed???
Hope to see some of you there :-)

As was explained in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87536]last year's thread[/url], it varies a lot and can be slow.  People who need a decision by a certain day or something can always try to request that by email.