geometry
LaTeX
algorithm
inequalities
number theory

Why dont you people try these sums at all. People like soumya, chappli, vihang, pardesi can all try these sums no? Why dont you do it guyz?  :maybe:
See these sums  :D

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194424

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194423

And also, there are many more such problems in the olympiad forum, why dont you guys try those?

If only we indians tried, we would be able to prove ourselves  :)

Uhmm ... that's number theory ...  :maybe:
Now if you have some good Combinatorics or Geometry problem, ask me  :D

Besides I don't come to AoPS very frequently and I am too lazy to post the solutions even if I solve the problem.  :P

see the olympiad forums vihang! seriously the sums here are good  :lol:  esp for olympiad practise .
You neednt post the full answer but can give the method, that's sufficient

Here are the forums for olympiads:

http://www.artofproblemsolving.com/Forum/index.php?f=217

Hope it helps you guyz

PS: I knw that chappli is tooooo lazy to even see  :D

EDIT 1: [b]HOW DID I EVER FORGET BRUTEFORCE AND ASHWATH AND SANJITH AND MANY MORE![/b]!
EDIT 2: [b]AND OFC MYTHILI[/b]  :P

Solved one - too sleepy to try the other. Will go to class, sleep some more and then come back and try the other.

Yes saw that Nice!  :)  Well i dont mean these two sums only man! i mean the sooooooooo many sums in the forum as well. It isnt like solving only these sums  :)

Look da, people will solve if they are interested and if they have the time. Everybody is busy with their own stuff and people are way too lazy. I don't actually blame them. Its more of a challenge to post your solution nicely in $ LaTeX$ than it is to think of them, unless you are an expert like pardesi. None of us have got that much time to spend hours and hours on AOPS.

I actually like trying some of the combinatorics or recursion problems here.... I may not get many of them but thats the only area of higher maths, that I can really try. :D

However, I agree with madness. AOPS is a fantastic resource for people who are aspiring to get into the IMO team and higher....... There are people here like Darij , Valentin and Peter Scholze who have represented their countries more than once and got medals. (Peter has got 3G and 1S), which is more than what any Indian as got. I've been telling chappli to use AOPS for a long, long time. The problems found here are varied and great.

But in the end, it all boils down to the person. If you are interested, you do it, otherwise you dont.

[quote="madness"]yes but as i said, people can outline the method n not latex is required like that  :D
and ofc, no restrictions on what you should try   :)
And yes, here you can get the advantage of the expertise of many people. People like soumya have solved problems before but chappli, i think its time you do too :maybe:[/quote]

The charm of a solution is lost when you provide the algorithm, unless its CSC. I doubt chappli will listen to you all of a sudden. And for some people the computer is not a very conducive learning atmosphere especially when you have put certain things down in pen and paper.

Dei i mean you can say that use CS or sth like that instead of latexifying! What charm is lost there?  :huh:  Infact if you've worked it out fully on paper, then charm is gained  :D
Who said that they have to work on the comp? they can pen down the ideas n then type the algo  :maybe:

Why do you have to bring CS into this? I thought we were discussing maths. And besides, I'm sure a lot of us solve problems but don't post solutions on AoPS. So as long as we're making use of the resources, I don't see any problem. @Shreyas: After reading Darij's beautiful writings in geometry, one feels a bit embarrassed to post in the same forum as he does. His style of writing is so amazing that it simply bowls you over.

not only darij, t0rajir0u's assertiveness, JBL's domineering nature, fedja, jmerry, Carcul and Kent's thoroughness is impressive!  :D
Not to forget Altheman and The Zuton Force, scorpius and all.  :)

Alex is really original and is very inspiring. JBL is pretty awesome at combinatorics and yeah, jmerry is almost unbeatable at Algebra. Fedor Petrov and Virgil Nicula are pretty cool too. But at the end of the day, Darij and Grobber remain my favorites. God.. this is becoming like some sort of art discussion :).

See this too. Nice :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150855

My favourite anyday is Nadejda Petrovna.. :blush:

dont spam man :D

Yes and people seem to be in love with Vasc in the inequalities forum :D :D

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21786]this[/url] esp Darij's long post :roll: Mind blowing  :10:  :wow:

Fabolous and crystal clear. I like the way he explains his thoughts clearly. Many a time, many of us are unable to do that, especially on pen and paper and thats why we lose marks for rigority.


can anyone tell me of a site that lists how certain (new SAT) scores compair to other college-bound seniors? collegeboard is not providing this info, to date (that i can find), and its not on my score report. thanks!

I hate to bump this, but I promised my friend someone on AoPS would have a clue where to look. Can I suppose that the answer to this is no, you don't know where I could find that, since no one has answered me?

[b]After edit:[/b] The first information about 2005 college-bound seniors was just released today.

The report is called

[url=http://www.collegeboard.com/prod_downloads/about/news_info/cbsenior/yr2005/2005-college-bound-seniors.pdf]2005 College-Bound Seniors: Total Group Profile Report[/url].

Read and enjoy, and then we can discuss. I just found out about the posting of this.


In California, what is the second shortest month of the year?

[hide="BTW"]This will not be true in 2007.[/hide]

My answer:
>
>
>
>
>
.
.
>
>
.

No months are short.  They have all of the days in them that they're supposed to have.

That's not quite right.

Hint: This puzzle will be invalid by 2007.

[quote="nateharman1234"]is it whatever month loses an hour to day light savings?[/quote]
[hide]Correct, the month is April.

This puzzle will be invalid in 2007 because DST will start in March instead. The California part of the puzzle was necessary because many parts of the world already do this.[/hide]


integration
trigonometry
linear algebra
matrix
quadratics
real analysis
real analysis unsolved

$u_0,u_1$ are reals fixed $u_{n+2}=\int_{u_n}^{u_{n+1}}e^{-x^2}dx$

1/ Prove that $u_n$ is convergent to $0$

Pierre do you remember this problem ?

2/ What can you say about the monotony of $u_n$ ?
3/ An equivalent of $u_n$ when n tends to infty

Sure I remember it because I created it (hmm, only part a) i think)! Sometimes I have strange ideas :maybe:

Pierre.

I have some of it. I thought I could prove that $u_n\to0$ but now I'm not sure. The sequence cannot be monotone; in fact it is eventually nearly periodic of period 6, looking almost (at least in the short run) like $a\sin\left(\frac{\pi n}3-b\right)$ for some $a$ and $b$. I don't have the answer to question 3, other than to say that the rate of convergence to zero is very slow.

In what follows, let $F(x)=\int_{-\infty}^xe^{-t^2}dt.$ We can write the recursion as $u_{n+1}=F(u_n)-F(u_{n-1}).$

As an aside unrelated to the solution of the problem, I can prove that the series $\sum u_n$ has bounded partial sums. We can show that

$\sum_{k=2}^nu_k=F(u_{n-1})-F(u_0)$ from which we can see that

$\left|\sum_{k=2}^nu_k\right|<\sqrt{\pi}.$

The main technique is to visualize this a dynamical system given by a mapping from $\mathbb{R}^2$ to $\mathbb{R}^2.$ This mapping takes $(x,y)$ to $(y,F(y)-F(x)).$ We check this mapping for fixed points. The only possible fixed points occur when $x=y=F(y)-F(x).$ This can only happen when $x=y=0.$

Now consider the Jacobian matrix of this mapping. ... continued tomorrow.

I still don't have the result. The statement about looking similar to a sequence with period 6 was based on the value of the Jacobian matrix at the fixed point for the mapping above:

\[M=\left[\begin{array}{cc}0&1\\-1&1\end{array}\right]\]

This matrix satisfies $M^6=I,$ which is where the period-6 behavior is coming from.

I tried some numerical simulation, using a partial sum of the power series for $F(x)$. The pattern of 3 positive values followed by three negative values is clear, and the simulation suggests the possibilty that $u_n=O(n^{-1/2})$ as $n\to\infty.$

This still needs proof. One avenue I'm trying is to examine the quadratic form $Q_n= u_n^2-u_nu_{n+1}+u_{n+1}^2.$


A male bee has only one parent (a mother) and a female bee has two parents (a mother and a father). Assuming all ancestors are distinct, how many great, great, great, great grandparents does a male bee have?

[quote="littlereddevil"]On Ftw, it says the answer is 12. it is clearly 13, so should someone change it?[/quote]
The next time someone reviews that problem, they should click the report error with problem button.

[quote="pinkmuskrat"][hide]We can quite easily just go up the family tree. Let's make sure we correctly follow the great great great great grandparents part.

1 bee(male)
1 parent(female)
2 grandparents (1M, 1F)
3 great's (2F, 1M)
5 great great's

It's the awesome fibonacci sequence!
8 great great great
[b]13 [/b]quadruple greats

[/hide][/quote]
[quote="littlereddevil"]On Ftw, it says the answer is 12. it is clearly 13, so should someone change it?[/quote]
Actually, we don't include the male bee in our count.  This makes the answer $13-1=\boxed{12}$.


function
integration
calculus
calculus computations

The definition of the Gamma Function seems a bit strange (at least to the extent I have used in in Differential Equations).  I know that the definition is $\Gamma\left(n\right)=\int_{0}^{\infty}e^{-x}x^{p-1}\,dx$.  However, whenever I am able to apply the Gamma Function to something, I always encounter it in the form of $\Gamma\left(n+1\right)$.  Why wasn't Gamma n defined in the same way as Gamma of n+1 rather than the "p-1" on the x term?  What I mean is, why isn't the Gamma function defined as $\Gamma\left(n\right)=\int_{0}^{\infty}e^{-x}x^p\,dx$?

Thanks for your help :).

1. Tradition, going all the way back to Euler.

2. This places the rightmost pole in the complex plane at the origin, and we can say that near $z=0$, we have $\Gamma(z)\approx\frac1z.$

3. $\frac{dx}x$ is the Haar measure for the multiplicative group of positive reals.


limit
algebra
binomial theorem
calculus
calculus computations

$ \lim_{x\to \infty} \frac{\sqrt{x^2\plus{}1}\minus{} \sqrt[3]{x^3\plus{}1}}{\sqrt[4]{x^4\plus{}1}\minus{} \sqrt[5]{x^4\plus{}1} }$

Please give its solution. Its answer is $ 0$. Thanks a lot.

[quote="Spribo"]Try principle De l' Hospital: nominator as well as denominator converge to $ \infty$.[/quote]

Can you please show its solution? I could not solve it using L Hospital's Rule.

$ lim(\sqrt {x^2 \plus{} 1}(1 \minus{} \frac {\sqrt [3]{x^3 \plus{} 1}}{\sqrt {x^2 \plus{} 1}})) \equal{} \plus{} \infty$. The same is true for the denominator. Try now breaking the fraction in two fractions. Then find the limits of the inverses of the two fractions using De L'Hospital. You 'll find $ \infty$ and $ \infty$. So obviously you'll have 0+0=0.
Edit: Don't read that, I've made mistakes.

As I've made some mistakes, and as you must have got confused, here's a solution without  De L'Hospital:
$ \lim_{x\to\infty}\frac {\sqrt {x^{2} \plus{} 1} \minus{} \sqrt [3]{x^{3} \plus{} 1}}{\sqrt [4]{x^{4} \plus{} 1} \minus{} \sqrt [5]{x^{5} \plus{} 1}} \equal{} \lim_{x\to\infty}\frac {\sqrt {x^2 \plus{} 1}}{\sqrt [4]{x^{4} \plus{} 1} \minus{} \sqrt [5]{x^{5} \plus{} 1}} \plus{} lim_{x\to\infty}\frac {\sqrt [3]{x^3 \plus{} 1}}{\sqrt [4]{x^{4} \plus{} 1} \minus{} \sqrt [5]{x^{5} \plus{} 1}}$.Invert those fractions, break them, and you have (for example, and as $ x > 0$):
$ lim_{x\to\infty}\frac {\sqrt [4]{x^4 \plus{} 1}}{\sqrt {x^2 \plus{} 1}} \equal{} \lim_{x\to\infty}\frac {x\sqrt [4]{1 \plus{} \frac {1}{x^4}}}{x\sqrt {1 \plus{} \frac {1}{x^2}}} \equal{} 1$. Do the same for all your fractions, invert, and you'll have 1-1+1-1=0.
Note: in the fifth root in the denominator you should have $ x^5 \plus{} 1$, otherwise the limit gets $ \minus{} \infty$...

Well, I should 've been more careful :oops:

I'd rather use the binomial theorem.

$ \lim_{x\to\infty}\frac{x\sqrt{1\plus{}x^{\minus{}2}}\minus{}x\sqrt[3]{1\plus{}x^{\minus{}3}}}{x\sqrt[4]{1\plus{}x^{\minus{}4}}\minus{}x^{5/4}\sqrt[5]{1\plus{}x^{\minus{}4}}}\equal{}\lim_{x\to\infty}\frac{(1\plus{}x^{\minus{}2})^{\minus{}1/2}\minus{}(1\plus{}x^{\minus{}3})^{\minus{}1/3}}{(1\plus{}x^{\minus{}4})^{\minus{}1/4}\minus{}x^{1/4}(1\plus{}x^{\minus{}4})^{1/4}}$

$ \equal{}\lim_{x\to\infty}\frac{1\minus{}\frac12x^{\minus{}2}\plus{}O(x^{\minus{}4})\minus{}1\plus{}\frac13x^{\minus{}3}\plus{}O(x^{\minus{}6})}{1\minus{}\frac14x^{\minus{}4}\plus{}O(x^{\minus{}8})\minus{}x^{1/4}\plus{}\frac14x^{\minus{}15/4}\plus{}O(x^{\minus{}31/4})}$

The numerator is $ O(x^{\minus{}2})$ and tends to zero from the negative side. The denominator is dominated by the $ \minus{}x^{1/4}$ term and tends to $ \minus{}\infty.$ Hence the limit is zero.

With that said, what are the chances that this is all a typo and the problem was really meant to be this?

$ \lim_{x\to\infty}\frac{\sqrt{x^2\plus{}1}\minus{}\sqrt[3]{x^3\plus{}1}}{\sqrt[4]{x^4\plus{}1}\minus{}\sqrt[5]{x^5\plus{}1}}$

I'll attack that in the same way.

$ \lim_{x\to\infty}\frac{x\sqrt{1\plus{}x^{\minus{}2}}\minus{}x\sqrt[3]{1\plus{}x^{\minus{}3}}}{x\sqrt[4]{1\plus{}x^{\minus{}4}}\minus{}x\sqrt[5]{1\plus{}x^{\minus{}5}}}\equal{}\lim_{x\to\infty}\frac{(1\plus{}x^{\minus{}2})^{\minus{}1/2}\minus{}(1\plus{}x^{\minus{}3})^{\minus{}1/3}}{(1\plus{}x^{\minus{}4})^{\minus{}1/4}\minus{}(1\plus{}x^{\minus{}5})^{1/5}}$

$ \equal{}\lim_{x\to\infty}\frac{1\minus{}\frac12x^{\minus{}2}\plus{}O(x^{\minus{}4})\minus{}1\plus{}\frac13x^{\minus{}3}\plus{}O(x^{\minus{}6})}{1\minus{}\frac14x^{\minus{}4}\plus{}O(x^{\minus{}8})\minus{}1\plus{}\frac15x^{\minus{}5}\plus{}O(x^{\minus{}10})}$

$ \equal{}\lim_{x\to\infty}\frac{\minus{}\frac12x^{\minus{}2}\plus{}O(x^{\minus{}3})}{\minus{}\frac14x^{\minus{}4}\plus{}O(x^{\minus{}5})}\equal{}\lim_{x\to\infty}2{x^2}\plus{}O(x)\equal{}\plus{}\infty.$

The binomial theorem says that for $ u<1,$

$ (1\plus{}u)^{\alpha}\equal{}1\plus{}\alpha u\plus{}\frac{\alpha(\alpha\minus{}1)}{2}u^2\plus{}\frac{\alpha(\alpha\minus{}1)(\alpha\minus{}2)}{3!}u^3\plus{}\cdots,$

which I shortened to $ 1\plus{}\alpha u\plus{}O(u^2)$ as $ u\to0.$


LaTeX
inequalities proposed
inequalities

Let a,b.c be positive real numbers.
Prove or disprove     \sum (a^4b^2+a^3b^3+3a^2b^2c^2) \geq \sum (2a^3b^2c+3a^4bc)

[quote="Mathermatics are beautifu"]How to use Latex ?[/quote]
Like this:
[quote="Mathermatics are beautifu"]Let a,b.c be positive real numbers.
Prove or disprove     $ \sum (a^4b^2 \plus{} a^3b^3 \plus{} 3a^2b^2c^2) \geq \sum (2a^3b^2c \plus{} 3a^4bc)$[/quote]
It's wrong for the cyclic sum and for the symmetric sum.
Try $ a\rightarrow\plus{}\infty.$ :wink:

[quote="arqady"][quote="Mathermatics are beautifu"]How to use Latex ?[/quote]
Like this:
[quote="Mathermatics are beautifu"]Let a,b.c be positive real numbers.
Prove or disprove     $ \sum (a^4b^2 \plus{} a^3b^3 \plus{} 3a^2b^2c^2) \geq \sum (2a^3b^2c \plus{} 3a^4bc)$[/quote]
It's wrong for the cyclic sum and for the symmetric sum.
Try $ a\rightarrow \plus{} \infty.$ :wink:[/quote]

Sorry, I mean symetric sum.


I want you to say the name of the best actors that you know.(and if you can explain about him/her it will be so nice.) :D

i love $Russel$ $crew$.(you should excuse me if i wrote his name wrong) :D

[quote="Chocoboom"]Ashton Kutcher. Awesome in Butterly Effect.[/quote]
can you say about him/her more,because i didn't hear his/her name.where is he/she come from?


Alright there has been some confusion on my part, so I will try to clear some things up.  Silverfalcon is hosting a math chat room, and he has asked me to make the chatroom.

At his request you can ask any problem as long as it dosent go above AMC12 level (IMO the first few AIME problems are easier than the last 5 AMC 12 problems, but I suppose this involves your good judgement).  If you are going to ask a hard problem just be prepared to explain it.

The format is like: One person asks a question, they wait a min or so, then type "answer" the first person to answer it gives the explanation, answers questions, and suggests the next problem.

I will make the name of the chatroom MathChat, so that if I forget to invite you or you want to join at whatever time, you can invite yourself.

Okay so people who have signed up so far are:
Tare
JS1527
Silverfalcon
Caroline
joml88
NAS

If you don't sign up and want to join just invite yourself to the chatroom at the designated time/date.

The designated time/date... hmmm... Um okay there have been two dates/times floating around:

(1) this friday, 6 EST
(2) this saturday, some time (NAS?  what time do you want?)

So please do three things:
(a) post your name if you want me to invite you and your not already on the list
(b) vote, and if you vote for saturday or both, what time on saturday?

I am definitely open tomorrow but on Saturday I have a little bit of a schedule.  Well all I have is Chemistry but it will last 5-6 hours and is a couple hours from my house.  So I will probably be home only in the evening around 5-6.  If you guys do something on saturday feel free to do it at any time because I don't want to hold anyone back because I don't have time.

Basically, anytime after 4:00 on Saturday is good for me.. that is, 4:00 EST.  (Which is also after 4 PST, so it doesn't really make a difference).  Would people want to do.. umm.. 8 EST = 5 PST?  (I chose that kind of randomly; it seems like a time when people won't have too much going on except for maybe dinner.  Unless people actually go out on Saturday night..)

White Horse King -- please post your AIM name.

A life, qui moi? seriously tho Sat at 5:00 PST is cool with me

Thanks JS1527 for this thread.

And as she said, to join in the chat or discussion, you must sign up with your AIM sn or you won't get a chance.
I'm sharp on this because I don't want to waste anytime on getting people in and out.  I want to start this sharp at 6:00 EST if possible and last time (actually the first) many people joined but we had problem in lagging and it went kind of wrong than I expected.

For people who wonder why I set level to AMC, I did that because this is math discussion and everyone has to learn or teach something everytime.  It's not like countdown practice where you have fun by beating someone up.  In addition to that, not all people had took AMC tests or took AMC and qualified to AIME.

To finish this message, I'd like to thank everyone who wants to join in this.

Thanks.

Silverfalcon

Okay Im thinking saturday at 5 EST.

joml88 -- just invite yourself to the chatroom when you get home or IM me.

Before we set anything in stone we should probably see what silverfalcon has to say...

[quote="Silverfalcon"]
To finish this message, I'd like to thank everyone who wants to join in this.

Thanks.

Silverfalcon[/quote]

Thanks for organizing =D .. As for the chats themselves, are you looking to discuss actual AMC questions (i.e. questions that have appeared on the exams) or for us to devise our own questions of a similar level?

Oh, there he is...  OKay so the final time is

6:00 on SATURDAY

Ok, sorry, is that EST or PST please put that whenever you post a time (causes less confusion).

[quote="JS1527"]I think its just math questions in general.  If you have a good question from AMC then post it.  If you have a good question that you made up yourself, post it. =)[/quote]

Sounds great, thanks =)

EST.  Every time that I say will be in Eastern time b/c as you can see, Im from Rhode Island.

Just to remind you, 6 EST=3PST.. this might not work for people from the West coast (if there are any) if they have stuff to do during the day (sports practice, math team practice, whatever).

Everybody seems to be forgetting that at this time of the year almost the whole United States (and Canada too, I think) is on daylight-saving time, so that the abbreviations are "EDT," "PDT," or whatever. We have a few participants in Indiana (and maybe a few in Arizona) where daylight-saving time is not observed, but it's observed in most of the places mentioned in the thread. Oh well, look at your watch, and have a good chat. Isn't April full of changes?

OKay see you all tomm. at 6 my time (EST).  If you are late or have things to do or didn't sign up but want to be in it, just invite yourself -- the name of the chatroom will be MathChat.

I'm not quite sure who's on today or who's on tomorrow so try to stay on computer and turn on your AIM so I can contact you.

Follwoing people should hopefully in computer(I just posted names of people joining in either days)

JS1527 (she's making chatroom)
Tare
white_horse_knight (you have to tell us your sn first)
Caroline
NAS
joml88

I think that's about it...

Another one at 6:00 EST on Saturday, 4/17 today.

For more upcoming, I have no idea yet.

Special thanks to js1527k, for her additional help

No problem Silverfalcon.  See you all tonight.

White Horse King please tell us your AIM name.

Thanks for organizing these, guys.

For future reference - it might be advantageous of people who plan on participating prepare some questions/have some questions on hand, as well as know good solutions =)


geometry
algorithm
function
number theory
relatively prime
Euclidean algorithm
greatest common divisor

This is all elementary number theory and are from a book. The following problems I believe are from the BAMO (bay area math olympiad), but please correct me if I am mistaken.

1) Prove that among any 12 consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (the proper divisors of a positive integer n are all positive integers other than 1 and n which divide n. For example, the proper divisors of 14 are 2 and 7).

2)Since $ 24 \equal{} 3 \plus{} 5 \plus{} 7 \plus{} 9$, the number $ 24$ can be written as the sum of at least two consecutive odd positive integers.
(a)Can 2005 be written as the sum of at least two consecutive odd positive integers? If yes, give an example. If no, prove it.
(b) Can 2006 be written as the sum of at least two consecutive odd positive integers? If yes, give an example. If no, prove it.

3) Prove that any integer greater than or equal to 7 can be written as teh sum of two relatively prime integers, both greater than 1. For example, 22 and 15 are relatively prime, and thus $ 37 \equal{} 22 \plus{} 15$ represents the number $ 37$ in the desired way.

Please provide the most elementary number theoretic solution possible! Thanks!

Solution to 3 (not known if it is the most elementary...):

[hide]Case 1: $ x$ is even. $ x \equal{} 2k$; and $ k \ge 4$, so either $ k\plus{}1$ and $ k\minus{}1$ are relatively prime or $ k\minus{}2$ and $ k\plus{}2$ are relatively prime, since one of the two pairs must consist of solely odd numbers and, via the Euclidean algorithm, either the greatest common divisor of the pair is $ 1$ or even. Note that $ k\minus{}2 > 1$, so $ k > 3$, or $ k \ge 4$.

Case 2: $ x$ is odd. $ x \equal{} 2k\plus{}1$, and $ k \ge 3$. $ k$ and $ k\plus{}1$ are obviously relatively prime by the Euclidean algorithm. $ k > 1$, so $ k \ge 2$.

Therefore, $ x \ge 8$ for $ x$ even and $ x \ge 5$ for $ x$ odd, so $ x \ ge 7$.[/hide]

Solution to 1

[hide]Among 12 consecutive numbers, there must be a multiple of twelve.  We must prove that all multiples of 12 are abundant.

We have the sum of divisors function is $ (1\plus{}p_1\plus{}p_1^2\plus{}\ldots p_1^{e_1})(1\plus{}p_2\plus{}p_2^2\plus{}\ldots p_2^{e_2})\ldots(1\plus{}p_k\plus{}p_k^2\plus{}\ldots p_k^{e_k})$ if $ n\equal{}p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}$.  (Note that these include $ n$ and 1.)

We have that $ p_1\equal{}2, p_2\equal{}3, e_1\geq 2, e_1\geq 1$.  We plug these in to get:

Sum of divisors $ \geq(2^{e_1\minus{}2}\plus{}2^{e_1\minus{}1}\plus{}2^{e_1})(3^{e_2\minus{}1}\plus{}3^{e_2})(p_3^{e_3})(p_4^{e_4})\ldots(p_k^{e_k})$

$ \equal{}(1\plus{}\frac{1}{2}\plus{}\frac{1}{4})(1\plus{}\frac{1}{3})2^{e_1}3^{e_2}p_3^{e_3}p_4^{e_4}\ldots p_k^{e_k}$

$ \equal{}\frac{7}{3}n\equal{}n\plus{}(n\plus{}\frac{n}{3})$.  Now if we take out the $ n$ and 1, because these are not divisors, we get $ n\plus{}\frac{n\minus{}3}{3}\geq n$ because $ n>3$.[/hide]

Solution to 2

[hide]Notice that $ 24 \equal{} 6\cdot4 \equal{} 5^2 \minus{} 1^2 \equal{} (1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} 9) \minus{} 1 \equal{} 3 \plus{} 5 \plus{} 7 \plus{} 9$.

This argument can be extended to any number that is the difference of two squares, or all numbers not 2 mod 4.

For 2005, this gives:

$ 2005 \equal{} 5\cdot401 \equal{} (203 \plus{} 198)(203 \minus{} 198) \equal{} 203^2 \minus{} 198^2 \equal{} (1 \plus{} 3 \plus{} 5 \plus{} 7\ldots405) \minus{} (1 \plus{} 3 \plus{} 5 \plus{} 7\ldots395) \equal{} 397 \plus{} 399 \plus{} 401 \plus{} 403 \plus{} 405$.

For 2006 we use the following argument: the sum of any two consecutive odd numbers is divisible by 4, so if there is an even number of odd numbers in the sum then the sum is divisible by 4, and otherwise it is odd.  Thus consecutive odd numbers cannot add to 2006.[/hide]

[quote="theax"]

2)Since $ 24 \equal{} 3 \plus{} 5 \plus{} 7 \plus{} 9$, the number $ 24$ can be written as the sum of at least two consecutive odd positive integers.
(a)Can 2005 be written as the sum of at least two consecutive odd positive integers? If yes, give an example. If no, prove it.
[/quote]

I know in (a) and (b) you're being consistent by using the phrase
"the sum of at least two," but it would be be outright making
sense in (a) to ask instead the following because two odd integers
can never add up to 2005.

"Can 2005 be written as the sum of at least THREE consecutive
odd positive integers?"

Thanks for all the great solutions!
:D

But can someone either explain tenniskidperson3's solution to 1 or give their own? I chiefly do not understand the sum of divisors function and how someone would solve the problem if they didn't know what it was.

Thanks!

[quote="dgreenb801"]For #1, you can just say n, 2n, 3n, 6n, and 4n are divisors of 12n, QED. In fact for any 6 consecutive numbers where the lowest is at least 7, there is an abundant number, as 1,n,2n,3n all divide 6n.[/quote]

Do you not understand this solution? Among any $ 12$ consecutive integers, one must be divisible by $ 12$, and $ 12n < 16n \equal{} n \plus{} 2n \plus{} 3n \plus{} 4n \plus{} 6n\le\text{sum of the proper divisors of }12n$.

ohh... thanks!

just a quick note: for $ n\equal{}1$, the proper divisors is $ \equal{} 2 \plus{} 3\plus{} 4 \plus{} 6 \equal{} 15 \ne 16$. But other than that, thanks!

Whoops. :blush: Well yeah, you can just use $ 2n\plus{}3n\plus{}4n\plus{}6n\equal{}15n$ instead.


algebra
polynomial
superior algebra
superior algebra unsolved

When is $X^{m}+Y^{n}+Z^{2}$ irreducible over $\mathbb{F}[X,Y,Z]$? ($\mathbb{F}$ is an arbitrary field).

hey hi,
I feel the polynomial is reducible only if $char F =2$ and $m,n$ are even.

suppose our polynomial is reducible say $F(X,Y,Z)=A(X,Y,Z)B(X,Y,Z)$,then upon putting $X,Y=0$ we get $A(X,Y,Z)=Z+Xf_{0}+Yf_{1},B(X,Y,Z)= Z+Xg_{0}+Yg_{1}$.Now taking the product of $A,B$ we get the following:
$. f_{0}f_{1}= X^{m-2},g_{0}g_{1}=Y^{n-2}$
$. f_{0}+f_{1}=0,g_{0}+g_{1}=0$
$.f_{0}g_{1}+f_{1}g_{0}= 0$
from the first two equations we get $(f_{0})^{2}=-X^{m-2},(g_{0})^{2}=-Y^{n-2}$and from second and third equations we get $2f_{0}g_{1}= 0$ and so the conclusion ......
perhaps my proof is totally absurd  :lol:


Calculate $ 5\cdot5! \plus{} 4\cdot4!\plus{}4!$.

Pull out a $ 4!$ from all of them to get: $ 4!(5\cdot 5\plus{}4\plus{}1)\equal{}4!(30)\equal{}720$.


Suppose a function $f$ satisfys the equation $f(x)+f(2-x)+2=0$ for all $x\in\mathbb R$, then which point $(x_{0},y_{0})$ is the planar curve $y=f(x)$ symmetric about?


search
number theory unsolved
number theory

$ a,b,c\in\mathbb{Z^{\plus{}}}$
$ 2^{a}\minus{}1|2^{b}\plus{}2^{c}\plus{}1$  $ \implies$
find all $ (a,b,c)$


26. Let $x^{*}=x-\lfloor x \rfloor$, where $\lfloor x \rfloor$ equals to the greatest integer $y$ such that $y \leq x$. How many numbers $b$ are there such that $(b^{*})^{2}-b^{2}$ is an integer and $1 \leq |b| < 5$ \[\text{(A) }20\qquad \text{(B) }40\qquad \text{(C) }49\qquad \text{(D) }50\qquad \text{(E) }58\] I think there are only 48 possible values.

I didn't check your work, but what about $b =-5$ and $b = 5$?

What was the proposed answer?

Hmm, -5 and 5 doesn't work because the problem originally had 1<=abs(b)<5

[quote="13375P34K43V312"]I thought it was 1<|b|<=5?[/quote]

That doesn't matter, either $\pm 1$ works or $\pm 5$ works, still same number of solutions.

[quote="13375P34K43V312"]I thought it was 1<|b|<=5?[/quote]

I copied the question word for word from the test

I like how it says $1 < |b| \leq 5$ in the problem.

yo feiqi

The answer is 48. I heard the MAML people are going to throw out this question.

The contest director is aware of the problem, so the only question now is how it will be handled.  My guess is that everyone will have the question marked correct, regardless of how they answered it (though to me an answer of 49 is a lot better than an answer of 20 or 58, while not answering seems better than an answer of 49).

Since this is a forum... I think everyone should get it wrong. Because 48 is the right answer, and noone chose that... hehe maybe, i just don't want tim lu to beat my 146 from last year... He got a 150, if they mark 26 right, so yea, let's just make 26 wrong

Haha, I think credit should be granted or not granted to everyone, as I know of people who didn't put down an answer because 48 wasn't an option, and some people who thought 48 was an answer put down 59. Ah well. . .


superior algebra
superior algebra solved

$R$ is a ring. It has only two left ideals: 0 and itself.
Prove that $R^2=0$ or $R$ is a division ring.

I have a long proof of it...  :(
I'd like to see an elegant one...

First thought: $Rx$ is a left ideal for any element $x$. Thus either $Rx=R$ or $Rx=0$.

Furthermore, if $R$ has a multiplicative identity, then for $x\ne 0$ we have $Rx\ne 0$, meaning that for all $x\ne 0,\ Rx=R$. Each non-null $x$ has, then, a left inverse. From here it's easy to deduce that $R$ is a division ring.

I guess we're left with the case when $R$ has no multiplicative identity.

The outline of my proof is (assume that $R^2\neq 0$).

(1) there exist $a\neq 0$ s..t $Ra=R$.
(2) hence there exist $ea=a$.
(3) $e$ is a right (multiplicative) identity
(4) $e$ is the unique right identity
(5) $e$ is the (multiplicative) identity
(6) $R$ is a division ring.

Following the outline of Liyi's proof:
(1) Assume $R^2\neq 0$ there exist $a,b\in R\setminus \{0\}$ such as $ba\neq 0$ this leads to $Ra\neq 0$ so $Ra=R$.
(2) hence there exists $e\in R$ such that $ea=a$.
(3)$(ae-a)^2=0$. Let $u=ae-a$.
$Ru\in\{0,R\}$.
If $Ru=R$ then there exists $t\in R$ such as $tu=e$ so $t(ae-a)=e$ so $t(ae-a)^2=e(ae-a)=eae-ea=ae-a$. So $u=0$. This contradicts $Ru=R$ and $R^2\neq 0$.
So $Ru=0$. This means $rae=ra$ for all $r\in R$. Because $Ra=R$ it is easy to see that $xe=x$ for all $x\in R$ so $e$ is a right identity.

(4) Let $f\in R$ such as $xf=x$ for all $x\in R$. Then $r(e-f)=re-rf=r-r=0$ for all $r\in R$. So $R(e-f)=0$. Then $I=\mathbb{Z}(e-f)$ is a left ideal of $R$. $I^2=0$ so $I\neq R$. Thus $I=0$ leading to $e=f$.
So $e$ is the unique right identity.

(5) Let $x\in R$. Then $r(ex-x)=(re)x-rx=rx-rx=0$ for all $r\in R$, so $R(ex-x)=0$. Same as before, (ex-x)=0 leading to $ex=x$. $x$ was chosen arbitrarily in $R$, so $e$ is the multiplicative identity for $R$.

(6)As grobber did: Let $x\neq 0$. Then $x\in Rx$ so $Rx\neq 0$. Therefore $Rx=R$. So there exists $y\in R$ such as $yx=e$. $xyx=x\Rightarrow xy\neq 0$ and $(xy)^2=(xy)$. $xy\neq 0$ so $xy$ has left inverse so $(xy)^2=xy\Rightarrow xy=e$.
Therefore, for all $x\in R%Error. "setimus" is a bad command.
\{0\}$ there exists $y\in R$ such as $xy=yx=e$. So $R$ is a division ring.


function
algebra
polynomial
functional equation
IMO Shortlist
periodic function

Let $ f$ be a function from the set of real numbers $ \mathbb{R}$ into itself such for all $ x \in \mathbb{R},$ we have $ |f(x)| \leq 1$ and

\[ f \left( x \plus{} \frac{13}{42} \right) \plus{} f(x) \equal{} f \left( x \plus{} \frac{1}{6} \right) \plus{} f \left( x \plus{} \frac{1}{7} \right).\]

Prove that $ f$ is a periodic function (that is, there exists a non-zero real number $ c$ such $ f(x\plus{}c) \equal{} f(x)$ for all $ x \in \mathbb{R}$).

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
let $ a\equal{}\frac 16,b\equal{}\frac 17$ so $ a\plus{}b\equal{}\frac{13}{42}$ and:

$ f(x\plus{}a\plus{}b)\plus{}f(x)\equal{}f(x\plus{}a)\plus{}f(x\plus{}b)$

now in this equation,if we plug in $ x\plus{}a,x\plus{}2a,\ldots ,x\plus{}5a$ instead of $ x$,and then sum the equations up,we get that:

$ f(x\plus{}1\plus{}b)\plus{}f(x)\equal{}f(x\plus{}1)\plus{}f(x\plus{}b)$

now in this equation,plug in $ x\plus{}b,x\plus{}2b,\ldots ,x\plus{}6b$ instead of $ x$,and then sum them up to get that:

$ f(x\plus{}2)\minus{}f(x\plus{}1)\equal{}f(x\plus{}1)\minus{}f(x)\equal{}c$

hence for every natural number $ n$,we have:

$ f(x\plus{}n)\minus{}f(x)\equal{}nc$

now note that if $ c\neq 0$,then $ f(x\plus{}n)$ would tend to $ \infty$ as $ n$ gets bigger,so according to $ |f(x)|\leq 1$ we must have $ c\equal{}0$,so $ f(x\plus{}1)\minus{}f(x)\equal{}0$.i.e. $ f(x\plus{}1)\equal{}f(x)$ for all $ x\in\mathbb{R}$.

QED[/quote]

I claim that for all real $a$, $f(a+1) = f(a)$. If we define $f_a(x) = f(x-a)$, we see that $f_a$ trivially satisfies the functional equation of $f$, and that $f_a(0) = f_a(1)$ would imply that $f(a) = f(a+1)$.

Let $c_i = f_a\left(\frac{i}{42}\right)$ for all $i \geq 0$. We will show that $c_{n+42} = c_n$ for all integers $n$. $\{c_i\}$ satisfies $|c_n| \leq 1$ and $c_{n+13} = c_{n+7} + c_{n+6} - c_n$ for all integers $n$. Its characteristic polynomial is $x^{13} - x^7 - x^6 + 1 = (x^7 - 1)(x^6 - 1)$, which has 1 as a double root and the 6th and 7th roots of unity not equal to 1 as single roots. Let $r_1, r_2, \ldots, r_{11}$ be the 6th and 7th roots of unity not equal to 1, in some order. $c_n$ must satisfy $c_n = an + a_01^n  + a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n$, where $a, a_0, a_1, \cdots, a_{11}$ are some complex numbers. If $a \neq 0$, then we can pick some integer $n$ with $|an| > 1 + |a_1| + |a_2| + \cdots + |a_{11}|$, whence
\begin{align*}
|c_n|
&= |an + a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n| \\
&\geq |an| - |a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n| \\
&\geq |an| - (|a_1| + |a_2| + \cdots + |a_{11}|) \\
&> 1,
\end{align*}
a contradiction. Since $r_i^{42} = r_i$, $c_{n+42} = c_n$. In particular, $c_{42} = c_0$, so $f_a(1) = f_a(0)$, i.e., $f(a+1) = f(a)$, as desired.


advanced fields
advanced fields unsolved

Prove that the language {a^(n!) | 0 <= n} is not context free.

Use the pumping lemma. (Proof by contradiction):
Suppose A (your language) is context free.  Then let $p$ be the pumping length.  Let $s = a^{p!}$ clearly $s \in A$ and $|s| > p$.  Then we can break $s$ up into 5 parts: $s = uvxyz$ where $|vy|>0$, $|vxy| < p$, and $s$ can be "pumped": i.e. $uv^{i}xy^{i}z \in A$ for all integers $i \geq 0$. Okay, so now we need a contradiction, so look at pumping it to $i = 2$.  Then we have that $|vy| < p$ (from $|vxy| < p$), so $|uv^{2}xy^{2}z| = p!+k, \, k<p$, but since $|s| = p!$, we need something bigger still in the language: $|uv^{2}xy^{2}z| = p!+k = (p+1)!$ (or larger). This implies $p! = p!(p+1)-k$, but $k < p$ so this cannot be possible. Hence, a contradiction and our assumption that A was context free is wrong.


conics
hyperbola
geometry proposed
geometry

There is a right angle whose vertex moves on a fixed circle and one of it's sides passes a fixed point. What is the curve that the other side of the angle is always tangent to it.

the tnagency locus will be:

we have two points $A,A'$. The point $P$ is such that $PA-PA'=\text{const}$.wanted Curve is locus of $P$.

I don't know name of the locus in English.Can anyone help?

This is called hyperbola in English.
btw, try to generalize this problem ;)


What are the chances of flipping a penny four times and getting at least two tails?

[hide] We can calculate this as $ 1$ minus the probability of getting no tails or $ 1$ tail. The probability of getting $ 0$ tails is $ (\frac12)^4\equal{}\frac{1}{16}$. The probability of getting one tail is $ \binom{4}{1}(\frac12)^4\equal{}\frac{4}{16}$. So the probability of getting two or more tails is $ 1\minus{}\frac{1}{16}\minus{}\frac{4}{16}\equal{}\boxed{\frac{11}{16}}$.[/hide]


function
inequalities
calculus
calculus computations

Let $ f$ be a continuous and strictly increasing or decreasing function on positive reals and $ g \equal{} f^{ \minus{} 1}$. Let $ a_{1} < a_{2} < \cdots < a_{n}$  be $ n$ positive reals, then we define the Mean value of these numbers in regards to $ f$ as: $ M_{f} \equal{} g\left(\frac {1}{n}\sum^{n}_{i \equal{} 1}f\left(a_{i}\right)\right)$
$ \text{1.}$ Prove that $ f\left(M_{f}\right) \equal{} \frac {1}{n}\sum^{n}_{i \equal{} 1}f\left(a_{i}\right)$.
$ \text{2.}$ Show that $ a_{1} < M_{f} < a_{n}$.
$ \text{3.}$ If $ h(x) \equal{} af(x) \plus{} b$ where $ a\neq0$, prove that $ M_{h} \equal{} M_{f}$.

1. since f is strictly monotone its inverse is as well, so they are well defined and then we can just apply f to both sides of the definition and we are done

2.$ a_1< M_f <a_n$

assume f increasing, then $ f(a_1) \equal{}\frac{1}{n} \sum_{i\equal{}1}^n f(a_1) < f(M_f) < \frac{1}{n} \sum_{i\equal{}1}^n f(a_n)\equal{}f(a_n)$

taking g of all three gives us the desired inequality (since g is increasing too).

if f is decreasing then $ f(a_1) \equal{}\frac{1}{n} \sum_{i\equal{}1}^n f(a_1) > f(M_f) > \frac{1}{n} \sum_{i\equal{}1}^n f(a_n)\equal{}f(a_n)$

taking g of all three gives us the desired inequality (since g is decreasing and switches the signs).

3. $ h(x)\equal{}af(x)\plus{}b \to h^{\minus{}1} (x)\equal{} g\left( \frac{x\minus{}b}{a}\right)$

then $ M_h\equal{}h^{\minus{}1} \left( \frac{1}{n}\sum_{i\equal{}1}^n h(a_i)\right)\equal{}h^{\minus{}1} \left( \frac{1}{n}\sum_{i\equal{}1}^n [af(a_i)\plus{}b]\right)$
$ \equal{}g\left( \frac{\frac{1}{n}\sum_{i\equal{}1}^n [af(a_i)\plus{}b]\minus{}b}{a} \right)\equal{}g\left(\frac {1}{n}\sum^{n}_{i \equal{} 1}f\left(a_{i}\right)\right)\equal{}M_f$


induction
geometry
geometric transformation
homothety
ratio
real analysis
real analysis unsolved

Let $f(A)$ be the set of accumulation points for a set $A$. Can we find a subset $A$ in $R$ such that $A, f(A), f(f(A)),...$ are all different?

Yes.

Firstly we construct a sequence of countable subsets of $[0;1]$ $(A_n)$ sucht that $A_0=\nil$ and $A_n\neq A_{n+1}$, $A_n \subset A_{n+1}$ and $f(A_{n+1})=A_n$ by induction on $n$.
Put $A_1=\{0\}, A_2=\{0\}\bigcup \{\frac 1{2n}|n \in \mathbb{N}\}$.
Next, let's do the induction hypothesis:
Let $B_k=A_k \setminus A_{k-1}$. Since the elements in $B_k$ are not accumulation points of $A_k$ for every $X \in B_k$ we may find a $R_X$ s.t. no other element in $A_k$ lies in the interval $(X-R_X,X+R_X)$.
Then $r_X=\frac {R_X}2$ satifies the same condition just with the additional assumption that the intervals $(X-r_X,X+r_X)$ do not intersect.
Now let $W_X=\{X+\frac {r_X}{2n}|n \in \mathbb{N}\}$. Let $B_{k+1}=\bigcup_{X \in B_{k-1}} W_X$ ($B_k$ is countable) and let $A_{k+1}=A_k \bigcup  B_{k+1}$. Then since $X$ is an accumulation point of $W_X$ but the points in $W_X$ are not we just get $f(A_{k+1})=A_k$ and the induction is finished.
Now denote by $t_r(A)$ the translation of set $A$ of reals by a real $r$.
Then the seta $A=\bigcup_{i \in \mathbb{N}} t_{2i} (A_i)$ is our desired set.

Note: by using not only translation, but also homothety we may find a bounded set $A$ to respect our conditions:
If we let $h_r(X)$ be the image of the set $X$ under homothety centered at origin and ratio $r$ we see that
$[0;1] %Error. "superset" is a bad command.
A=\bigcup_{i \in \mathbb{N}} t_{\frac 1{2i+1}} h_{\frac 1{(2i+1)(2i+2)}} (A_i)$ also fullfills our requirements.

P.S. What is Miklos Albert paper with exercises?

Very nice. I found Miklos Abert paper with exercices on the web (don't ask me where, I forgot  :blush: ) and I think these problems were to be solved in seminars or something like this. Anyway, I find them very interesting and I will post more.


I am having a pretty hard time differentiating between the difference between these two things:

Mutually Exclusive, and Independant. When do you know which formula to use?

The same thing applies with *And* and *Or* problems.

I understand how to use the formulas, but I'm having a hard time deciding which category a problem actually goes into. Without knowing which category it goes into all my actualy knowledge is pretty much useless! Please help me, I have a math test on this tommorow.

for and, you multiply,
for or, you add.
 :)

Do you just add for or?  I always used $1-(1-A)(1-B)$ where A and B are probabilities of independent events.  Although, this means that both happening is acceptable, perhaps yours is for only one of the two happening.

Wait a minute.  If you have two probabilities, say 70% and 60%, adding is over 100%...

no, i was too General on that
 lets say, the chance of raining today is 70% and the chance of rainig tommorow is 60% then the chance that either today or tommorow would rain would be:
$1$ minus the chance that it would rain in both days, minus the chance that it wont rain in both days so you get:
$1-(0.3*0.4)-(0.6*0.7)=0.46$


[quote="epatjn"]What is the best way to teach someone how to approach this question:

Aviva is picking out chocolates for Father’s Day. The box holds 20 candies, and she wants 8 dark chocolates, 7 milk chocolates and 5 white chocolates. However, she wants each of the 20 candies to be different. The store carries 12 kinds of
dark chocolate candies, 10 kinds of milk chocolate candies and six kinds of white chocolate candies. How many distinct combinations can she make for her 20-piece box?[/quote]
teach the person permutations/combinations and they will get it quickly

of course i can't-i am terrible at explaining things!
but this is a website(in the hidden content) that always looked decent for the sort of mathcounts things i notice they have series and sequences but aops volume 1 is good for the sort too, it might be good for permutations  :fool:

[hide]http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm[/hide]

[quote="epatjn"]What is the best way to teach someone how to approach this question...[/quote]
I think two key concepts are required to teach this:

1.  When independent events are combined, the number of choices are multiplied.

2.  The "choose" function.

Can you take it from here?

[quote="epatjn"]I am not sure what you mean, so I cannot take it from there..[/quote]

I suppose Choose = Combinations = nCr =$\frac{n!}{r!(n-r)!}$

For teaching, I would first teach permutations by listing possible solutions in a branch diagram. For example, there are 5*4*3=60 ways to list 3 things from a total of 5 things (actually, I could figure out that multiplication trick as I was listing everything in a diagram, thanks to my step-by-step math study guide) Then combination is rather easy to see, since for each (a,b,c), there are $3 \cdot 2 \cdot 1=3!=6$ arrangements possible!  And this is how I learned both :)

See [url]http://regentsprep.org/regents/math/permut/Lperm.htm[/url] for short explanation

The choose function would be helpful here because it helps calculate the total number of ways she can have every type of chocolate different. For example, she has to choose eight different types of dark chocolate out of twelve. We can use nCr (out of n things, choose r) to find out the total number of ways. Then we can multiply (I think, I forgot the problem) the final ones to get the answer.

Thanks [b][u]kohjhsd[/u][/b].

That explanation and link to a website are very useful.

[quote="vamathletes"][hide]
If there are 12 dark chocolate, she wants 8, it is 12 choose 8 or
$12*11*10*9*8*7*6*5$
Do the same for the other chocolates, then multiply the numbers together.
Teach that "choose" is start with the first number and put lesser numbers until the total number of numbers is the second number (like 8).
[/hide][/quote]

12 choose 8 is definatley not that.  its should be that divided by 8!.

There are a couple of things that help for this problem:

1. Knowledge of combinations $\binom{n}{r}$

2. Multiplication of independent events

3. The identity $\binom{n}{r}=\binom{n}{n-r}$

The first 2 are more important though.


conics
parabola
geometry
function
geometry proposed

Problem: [i]For a parabola, the feet of the perpendiculars from the focus to the tangents all lie on one line.[/i]

[hide="Solution: "]Let the parabola, $ \mathcal{P}$ have focus $ F$ and let circle $ \mathcal{O}$ be the reciprocal  [in a projective sense; pole to polar, etc] of the parabola through a circle, $ \mathcal{C}$, center $ F$ that intersects the parabola at two points. Let $ A$ be diametrically opposed to $ F$ in $ \mathcal{O}$. Let $ l$ be the tangent to $ \mathcal{P}$ at the vertex. Note that $ l$ is the polar of $ A$. It follows that $ l$ and $ \mathcal{C}$ are inversive images about circle $ \mathcal{C}$.

Claim 1:
If $ t$ is a tangent to $ \mathcal{P}$ and $ t\cap l=P$, then $ t\perp PF$.

Proof:
Let $ t$ be a tangent to the parabola and let $ l\cap T=P$. Let $ P'$ be the inversive image of $ P$ in $ \mathcal{C}$. Clearly $ P'\in\mathcal{O}$ since $ \mathcal{C}$ and $ l$ are inversive images of each other in $ \mathcal{C}$ [this is obvious since $ \mathcal{C}$ passes through $ F$ so it maps to a line, they are symmetric about $ AF$.]

Now consider the polar [i.e reciprocal] of $ P'$. This is a line, through $ P$, perpendicular to $ FP'$, but since $ OP'P$ is a line, $ t\perp FP$.
--

Clearly the point that is the foot of the perpendicular from $ F$ to $ t$ is unique for any tangent, $ t$. Hence by Claim 1, these points lie on $ l$. The problem is proven.[/hide][hide="Comment:"]We can strengthen the claim of the problem to say that $ l$ is the locus of points, $ P^{*}$, such that the line, $ t^{*}$, perpendicular to $ P^{*}F$ at $ P^{*}$ is a tangent to $ \mathcal{P}$.

To prove this, we merely say that $ P^{*}'$ is the inversive image of $ P^{*}$ in $ \mathcal{C}$, then $ P^{*}'\in \mathcal{O}$ as before. Since
i) $ \mathcal{O}$ is the reciprocal of $ \mathcal{P}$, and
ii) the polar [i.e. reciprocal] of any point $ \in \mathcal{C}$ is a tangent to $ \mathcal{P}$
we have that the polar of $ P^{*}'$ is tangent to $ \mathcal{P}$. But by the inversion condition, we know that $ P^{*}$ is in the polar of $ P^{*}'$.[/hide]

I found this proof to be very involved for an exercise of Geometry Revisited. Since I have not had any actual teaching in this subject, I am not sure if I am folling a correct proof structure for reciprocal/pole/polar proofs. I would appreciate any feedback about my proof. I also welcome alternate solutions.

Sorry if some of the terminology is unintelligible. I am just using the words from Geometry Revisited. If necessary, I can explain/discuss this post further.

Sorry, here is a diagram. [It is a nice problem as to how to construct it!] My sketchpad has a function that fits a conic given 5 points. Can you figure out how to construct the figure given $ \mathcal{C}$ and $ \mathcal{O}$?


geometry
circumcircle
incenter
rectangle
trapezoid
geometric transformation
reflection

Dear Mathlinkers,
Let ABC be a triangle, 0 the circumcircle of ABC, O the center of 0, A' the antipode of A wrt O, M the midpoint of the arc BC which doesn't contain A, Q the point of contact of the A-excircle of ABC with BC, I the incenter of ABC, Z the meetpoint of the parallel to BC through I with the perpendicular to BC from A'.
The points Z, Q, and M are collinear.
Sincerely
Jean-Louis

$ D, E, F$ are contact points of the incircle $ (I)$ with $ BC, CA, AB$ and $ r$ is inradius. The A-altitude cuts the circumcircle $ (O)$ again at $ K$ and parallel to $ BC$ through $ I$ cuts $ AK$ at $ X.$ Inversion in $ (I)$ takes the circumcircle $ (O)$ into 9-point circle $ (O^*)$ of the intouch $ \triangle DEF,$ $ A$ into the midpoint $ A^*$ of $ EF,$ $ M$ into intersection $ M^*$ other than $ A^*$ of the perpendicular bisector $ AM$ of $ EF$ with $ (O^*),$ and the line $ MD$ into a circle $ \odot(M^*DI).$  This circle intersects $ IX, BC$ and $ U, V.$ Since it is centered on the perpendicular bisector of $ ID,$ the segments $ IV, DU$ are its diameters, $ IDVU$ is a rectangle, $ UV \perp BC$ and $ VM^* \perp IM^*.$ $ T$ is foot of the D-altitude of the intouch $ \triangle DEF$ $ \Longrightarrow$ $ IM^* \parallel DT$ (both $ \perp EF$). $ ID \equal{} r$ is circumradius of the intouch $ \triangle DEF,$ $ M^*T \equal{} r$ is diameter of its 9-point circle  $ \Longrightarrow$ the trapezoid $ IM^*DT$ is isosceles, cyclic. Consequently, $ UT \perp DT$ and $ UA^* \perp IA^*$ $ \Longrightarrow$ $ IU$ is diameter of a circle centered on $ IX$ and passing through $ I, A^*,$ which is inversion image of the A-altitude $ AX$ $ \Longrightarrow$ $ U \equiv X^*$ is the inversion image of $ X.$ Since the lines $ M^*V \perp IM, BC \perp ID, X^*V \perp IX$ are concurrent at $ V,$ their poles $ M, D, X$ with respect to $ (I)$ are collinear. As $ AK, AA'$ are isogonals with respect to the angle $ \angle CAB,$ $ KA' \parallel BC$ and $ Q, Z$ are reflections of $ D, X$ in the perpendicular bisector $ OM$ of $ BC.$

Dear Yetti,
thanks for you inversive proof. I research a direct one.
Let D the point of contact od the incircle of ABC with BC.
We know that MD and A'O intersect on the circumcircle of ABC... I think that there is away with Pascal's theorem...
Sincerely
Jean-Louis

I think that you mean that $ MD, A'I$ intersect on the circumcircle. It is obvious by inversion in the incircle $ (I)$ or in a circle $ (M)$ with radius $ MB \equal{} MC \equal{} MI$ ...
[hide]
(1): $ A^*$ is the midpoint of $ EF$ and $ T$ foot of the D-altitude of the intouch $ \triangle DEF,$ both on its 9-point circle $ (O^*).$ If $ IT$ cuts the circumcircle $ (O)$ of the $ \triangle ABC$ at $ L,$ then $ A^*, T$ are inversion images of $ A, L$ and the angles $ \angle ILA \equal{} \angle IA^*T$ are right $ \Longrightarrow$ $ IL$ cuts $ (O)$ at $ A'$ diametrically opposite to $ A.$ If $ M^*$ is the inversion image of $ M,$ then $ IM^*DT$ is isosceles cyclic trapezoid $ \Longrightarrow$ $ L \in MD.$

(2): A-altitude cuts the circumcircle $ (O)$ again at $ K.$ $ MD$ cuts $ (O)$ again at $ L$ and $ IL$ cuts $ AO$ at $ A'.$ By inversion in $ (M),$ $ MI^2 \equal{} MD \cdot ML$ $ \Longrightarrow$ $ \angle MLA' \equal{} \angle MLI \equal{} \angle MID \equal{} \angle MAK \equal{} \angle MAO \equal{} \angle MAA'$ $ \Longrightarrow$ $ MLAA'$ is cyclic, $ A' \in (O),$ $ AA'$ is a diameter of $ (O).$[/hide]
Knowing this, let $ L \in (O)$ be the intersection of $ A'I \cap MD.$ Let $ MM \parallel BC$ be the circumcircle tangent at $ M.$ The A-altitude of the $ \triangle ABC$ cuts $ (O)$ again at $ K.$ $ AA', AK$ are isogonals with respect to the angle $ \angle CAB$ $ \Longrightarrow$ $ A'K \parallel BC \parallel MM.$ By Pascal theorem for the cyclic hexagon $ ALKMMA',$ the intersectionx $ I \equiv AM \cap A'L,\ X \equiv AK \cap ML,\ J \equiv A'K \cap MM$ are collinear. But $ A'K \parallel MM,$ $ J$ is at infinity $ \Longrightarrow$ $ IX \parallel A'K \parallel MM \parallel BC.$

Hello yetti and jayme,
It's a nice result and I have a more elementary proof:
[hide="Proof"]Let D be the tangency point of the incircle (I) with BC, H,E be the foot of the A-altitude, A-bisector angle on BC, N be the midpoint of BC and K be the intersection of MD and AH. From $ ID\parallel{}AH\parallel{}MN$ and $ \triangle MBE\sim\triangle MAB$, we have that
$ \frac {EN}{DN} \equal{} \frac {EM}{IM} \equal{} \frac {EM}{BM} \equal{} \frac {BE}{AB} \equal{} \frac {IE}{IA} \equal{} \frac {DE}{HD}\Rightarrow \frac {HD}{DN} \equal{} \frac {DE}{EN}$$ \Rightarrow \frac {KH}{MN} \equal{} \frac {ID}{MN}\Rightarrow KH \equal{} ID\Rightarrow IK\parallel{}BC$
Thus K and Z are symmetric wrt the perp bisects of BC. By the symmetry, the collinearity of K,D,M follows the collinearity of Z,Q,M.[/hide]
Best regards, mr.danh

Another way round.

Reflect Q, Z w.r.t. line OM to get Q', Z' respectively.

Then by simply checking AZ' is perpendicular to BC and Z' on IZ while Q' is the point where the incircle touches BC.
As $ IQ' \perp BC$ and $ AZ' \perp BC$, so $ IQ' // AZ'$
So to prove Z, Q, M collinear $ \Leftrightarrow$ Z', Q', M collinear $ \Leftrightarrow \bigtriangleup IQ'M \sim  \bigtriangleup AZ'M \Leftrightarrow \frac{IM}{AM} \equal{} \frac{IQ}{AZ}$

Notice that $ IM \equal{} BM \equal{} CM$, and by Ptolemy's theorem yields $ AM \times BC \equal{} (AB\plus{}AC) \times IM$, i.e. $ \frac{IM}{AM} \equal{} \frac{BC}{AB\plus{}AC}$.

On the other hand, $ AZ \equal{} h_A \minus{} r \equal{} \frac{2S}{BC} \minus{} \frac{S}{s} \equal{} \frac{2S}{BC} \minus{} \frac{2S}{AB\plus{}BC\plus{}CA}$
and $ IQ \equal{} r \equal{} \frac{2S}{AB\plus{}BC\plus{}CA}$, so $ \frac{IQ}{AZ} \equal{} \frac{1}{\frac{AB\plus{}BC\plus{}CA}{BC} \minus{} 1} \equal{} \frac{BC}{AB\plus{}AC} \equal{} \frac{IM}{AM}$, so the result follows.


calculus
derivative
trigonometry
integration
real analysis
real analysis solved

Proof that $cos(arcsin x) = \sqrt{1-x^2}$
Just calculating their derivative is not enough!

I'm interested in how you would proof it.

$\frac{d}{dx}\cos(\arcsin x) = -\sin \arcsin x \cdot \frac{d\arcsin x}{dx} = -\frac{x}{\sqrt{1-x^2}}, \quad x\in (-\frac{\pi}{2},\frac{\pi}{2})$

and

$\frac{d}{dx}\sqrt{1-x^2} = -\frac{x}{\sqrt{1-x^2}}$

Let $f(x) = \cos(\arcsin x) - \sqrt{1-x^2}$ then $f'(x) = 0$.
hence $f(x)$ is a constant.

We see that $f(0)=0$. Hence $f(x)=0$ for all $x\in(-\frac{\pi}{2},\frac{\pi}{2})$.

Put arcsinx = t then x = sint and cos t = cos(arcsinx) = $\sqrt{1-x^2}$
Is very interesting the object cos(n arccosx)!

I found two different way's:

1) It is clearly that the $\int \sin(y)dy$ plus $\int \arcsin(x)dx$ is equal to $xy$.
When you calculate the integral of $arcsin(x)dx$ you wil find that it is different when you use partio integration. This leads to the conclusion that $cos(arcsin x) = \sqrt{1-x^2}$.

by partio integration:
$\int arcsin(x)dx = x\arcsin(x)+\sqrt{1-x^2}$
by using $\int \sin(y)dy + \int \arcsin(x) dx = xy$
$\int arcsin(x)dx = x\arcsin(x)+\cos(arcsin(x))$

2) another way is to use the rule $cos^2 x = 1 - sin^2 x$. filling in $a = arcsin x$ for $cos(a)$ leads to the same conclusion: $cos(arcsin x) = \sqrt{1-x^2}$

[quote="rhynoceros"]2) another way is to use the rule $cos^2 x = 1 - sin^2 x.$ filling in $a = \arcsin x$ for $\cos(a)$ leads to the same conclusion: $\cos(\arcsin x) = \sqrt{1-x^2}$[/quote]
Strictly speaking, what you have actually proved is that $|\cos(\arcsin x)|=\sqrt{1-x^2}.$ To finish this, you need to supply an argument that the sign is correct.


did you guys already take it?  how'd you do?

im taking the tuesday one.

jeez... why do you need to worry about psat's now?  I would think that it'd throw off your timing for the SAT.

*sigh*...i took the one yesterday...i thought it was remarkably harder than the one last yr for some reason...im guessing i got somewhere between 215-230 but not sure yet...

im taking it on tuesday as well...

yea, a lot of people i know said the saturday test was harder than average...i dunno

what grade are you miths?  215-230 is very good.  215 is probablly semifinalist, depending on where you live.  It's tough for Massachusetts, though, since the cutoff is around 22.  That's actually the highest in the nation.

I took it last saturday. did you guys think it was hard? i thought it was easier than others i have taken. my score was probably around 220. last year the cutoff was maybe 207 in MI.

so i just got home, how did you guys do?

I know I got 1 wrong in math, unfortunately because of bubbling the wrong answer...

I got 1 verbal wrong at least so far, and i dont know writing.  I thought writing was pretty tough, so was the reading comprehension part of verbal.

yeah, 3 hourish tests can do that to you. just think of the usamo. but at least you can get up if you have to

i thing i got exactly one off in math because of stupid mistakes, reading off calc display wrong. i think i did well in verbal and writing. why did you guys find writing hard?

*shrug* Took the Saturday one, pretty sure I have a perfect for math...verbal wasn't too bad - the dual compres had one tricky qn, otherwise it was okay - but I think the writing will kill my score. Oh well. 216 was the CA cutoff last year.

yeah, i took the tuesday one, the math was sadly easy, although i missed one because of a dumb careless error (I realized that I had screwed it up while I was doing the writing skills section)

youll prbly get a perfect score anywho though

i strongly doubt it...the curve on the math is like: if you get perfect, you get a 80, but if u miss even one, that moves you down to the 76-78 range...*sigh* i missed a math as well...

i think its slighly harder to get a perfect score on psat then sat. given, there is an extra math and verbal section on the sat, but the psat can be trickier. the extra variety of the writing section also makes it difficult. i go to a school where weve had perfect scores on sat and act every year, but in the last 6 years only one psat perfect score.

*nods* Of the seniors in my school this year, there are 5 1600s but only 2 240s...the writing section tends to be where most people trip up.

2 240's ?  that's really good.

Someone i know got a 237, and he said that that was like tied for 16th overall in MA.  (i guess theres some ranking of the scores, but i didnt know that.)

I just hope i get 222 for semifinalist, but that seems tough to do.

I took the test on Saturday morning. The double passage was hard... I guessed an uncomfortable amount. I found the passage too boring and long. Anyway, I hope I qualify for National Merit Scholarship--- which is hard b/c I live in Mass : (

Getting a merit scholarship also depends a lot on what college you go to.  I could have had one had I gone to the University of Chicago (which I was seriously considering, although not for that reason).

I took the one Saturday morning too...the double passage was pretty bad...everyone that took it complained about its difficulty level. Somehow, I even ran a lil short on time cause of it :-( . In any case, writing and math were pretty easy.

Personally, who cares about what 2 journalists think about their impact on the public?? (U probly wouldn't get this unless you took the PSAT last Sat.)

i think "socialist fever" was the answer, at least thats what most of my school got. i was more worried by the question asking what is different about the two authors opinions.s

There were too many questions on that double passage is my compliant...

Also, it was awful hard to understand...

they should make the passages more interesting

yes, i think so too...the main problem is that we realli dont care what they're talking about...

If that was solved, then perhaps us math ppl would do better...

When do you get the scores back?  What is the conversion for SAT to PSAT?

scores u get back in a couple of months...

in all honesty there is no real "conversion"...u just kinda have to take both since the PSAT has somewhat easier problems than the actual SAT and the SAT has more sections and bleh bleh bleh...for a rough estimate, juss subtract the writing score from ur PSAT and add a 0 at the end of it for SAT score...

PSAT gets back in december.

I dont think PSAT questions are easier than SAT.  They're supposed to be the same, but your score may (probablly will) be different from your PSAT score, since there are other factors (length/#of sections, can get more wrong on sat than on psat, etc.)

*nods* PSATs don't really do a good job of gauging your SAT score...though with the new PSAT and SAT that might change. I highly doubt it, though.

Maybe it's just me that thinks the PSAT questions are easier than SAT...or maybe its just that the SAT seems to go on and on and on...

but yes, with the new tests coming we can definitely expect change...thats for sure...


Let $a,\ b,\ c$ be real numbers such that $a\geq 1,\ b\geq 1,\ c\geq 1$ and $a^{2}+b^{2}+c^{2}=12.$

Prove that $\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq 3.$

We know \[LHS\leq\sqrt{3(a+b+c-3)}\] .Now we have from the condition mentioned, \[a+b+c\leq\sqrt{3(a^{2}+b^{2}+c^{2})}=6\] .Combining these two relations gives us the requested inequality.
PS:Sorry to say this,but it really is an easy one,unless I`ve made some mistake

[quote="kunny"][quote="andyciup"]We know \[LHS\leq\sqrt{3(a+b+c-3)}\] [/quote]
Sorry for that I can't understand it, because I am a beginner for inequality.[/quote]
It is just $\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\sqrt{3(x+y+z)}$,for$x=a-1$ etc
Also,the problem can be restated as: \[x,y,z\geq 0,\]  \[x^{2}+y^{2}+z^{2}+2(x+y+z)=9\] ,Prove: \[\sqrt{x}+\sqrt{y}+\sqrt{z}\leq 3\] I must say that I also have a problem with the condition$x^{2}+y^{2}+z^{2}+2(x+y+z)=9$(well,mine is $\leq9$) :)

Like the other "I don't know Muirhead" just take care of one variable :D
$(a-2)^{2}\geq 0$
$a^{2}-4a+4\geq 0$
$a^{2}+4\geq 4a$
$\frac{a^{2}}{4}+1\geq a$
$\frac{a^{2}}{4}\geq a-1$
$\sqrt{\frac{a^{2}}{4}}\geq \sqrt{a-1}$
$\sum\sqrt{\frac{a^{2}}{4}}\geq \sum\sqrt{a-1}$

And using QM-AM we have $3=\sqrt{3\sum\frac{a^{2}}{4}}\geq\sum\sqrt{\frac{a^{2}}{4}}$


geometry conics parabola ratio function calculus derivative

Let $ A,B$ be different points on a parabola. Prove that we can find $ P_1,P_2,\dots,P_{n}$ between $ A,B$ on the parabola such that area of the convex polygon $ AP_1P_2\dots P_nB$ is maximum. In this case prove that the ratio of $ S(AP_1P_2\dots P_nB)$ to the sector between $ A$ and $ B$ doesn't depend on $ A$ and $ B$, and only depends on $ n$.

By affine transformation, we may suppose the parabola is $ y \equal{} x^2$. Suppose $ A \equal{} (a, a^2), B \equal{} (b, b^2)$. The area of the sector is $ \frac{(b \minus{} a)^3}{6}$. We would like to select $ a \le t_1 \le t_2 \le \cdots \le t_n \le b$ such that the total area of the polygon is maximized. Splitting up the polygon by triangles $ AP_1B \plus{} P_2P_2B \plus{} \cdots \plus{} P_{n\minus{}1}P_nB$, we arrive at the expression $ \frac{b\minus{}a}{2} \Big((b\minus{}t_1)(t_1 \minus{} a) \plus{} (b \minus{} t_2)(t_2 \minus{} t_1) \plus{} \cdots \plus{} (b \minus{} t_n)(t_n \minus{} t_{n\minus{}1})\Big)$ for the area. This is clearly a continuous function on a compact region and thus achieves a maximum. The maximum clearly never occurs at the boundary since, at the boundary, the maximum we could obtain would be the maximal area with one fewer point, which we will show is not as large as the maximum we find in the interior. To find the maximum in the interior, all partial derivatives in the $ t_i$ vanish, so $ a \plus{} t_2 \equal{} 2t_1, t_1 \plus{} t_3 \equal{} 2t_2$, and so on. This has the unique solution when all the points $ a, t_1, \cdots, t_n, b$ are equally spaced, in which case all the areas are indeed proportional to $ (b \minus{} a)^3$, as desired.


limit
integration
calculus
calculus computations

guten morgen !  :D

find the limit

$ \lim_{m\to\infty}\;\int_0^\infty \; \sum_{n\equal{}1}^\infty \;\;\; 16\left\{\dfrac {2\,z^3\,\plus{}\,125\,n^\alpha}{(z^m\plus{}n^m)^{\beta\,\plus{}\,m}}\right\}\;\;\;\textbf dz$

$ (where \qquad \alpha, \beta  \quad  \in\mathbb R)$

[hide]
$ answer: \qquad \text{ our present year}$
[/hide]

You're going to want to captalize "Guten Morgen" or else it's wrong. :wink:


The scales show the masses of different collections of cylinders, cubes, and spheres. Weights given by the scales are in kilograms. One cylinder and one cube weigh 17 kg, two cubes and one sphere weigh 18 kg, and one cube, one cylinder, and one sphere weigh 21 kg. Assuming that the same shapes weigh the same amount, what is the number of kilograms in the mass of a cylinder?
[asy]import olympiad; size(300); import geometry_dev; import graph; defaultpen(linewidth(0.8));
void drawScale(pair blCorner){
 pair A = blCorner, B = A + (10,0), C = B + (-1,3), D = C + (-8,0);
 draw(A--B--C--D--cycle);
 draw(Arc(blCorner + (5,-3),4.5,50,130));
 draw(Arc(blCorner + (5,-3),5.5,50,130));
 draw((blCorner + (5,-3) + 4.5*dir(130))--(blCorner + (5,-3)+5.5*dir(130)));
 draw((blCorner + (5,-3)+4.5*dir(50))--(blCorner + (5,-3)+5.5*dir(50)));
 draw((blCorner + (5,-3)+4.5*dir(60))--(blCorner + (5,-3)+5.5*dir(60)),EndArrow);
 draw((D + (3,0))--(D + (3,1))--(D + (5,1))--(D + (5,0)));
 pair A2 = A + (0,4.5), B2 = B + (0,4.5), C2 = C + (0,1), D2 = D + (0,1);
 fill(A2--B2--C2--D2--cycle);
}
void drawCyl(pair blCorner,real h, real rad){
 draw(ellipse(blCorner + (0,rad),rad*0.5,rad));
 draw(ellipse(blCorner + (h,rad),rad*0.5,rad));
 draw(blCorner--(blCorner + (h,0)));
 draw((blCorner + (0,2*rad))--(blCorner + (h,2*rad)));
}
void drawCube(pair blCorner,real side){
 draw(blCorner--(blCorner + (0,side))--(blCorner + (side,side))--(blCorner +(side,0))--cycle);
 pair A = blCorner + (side,side);
 draw(A--(A + (0.2*side,-0.2*side))--(blCorner + (1.2*side,0)));
}
void drawSphere(pair center, real radius){
 draw(Circle(center,radius));
 draw(ellipse(center,radius,0.2*radius));
}
drawScale(origin); drawScale((15,0)); drawScale((30,0));
drawCyl((1,4.5),3,1); drawCube((6,4.5),2);
drawCube((15,4.5),2); drawCube((18,4.5),2); drawSphere((22,5.5),1);
drawCube((30,4.5),2); drawCyl((34,4.5),3,1); drawSphere((39,5.5),1);[/asy]

Add them all together to get:

2 cylinders + 4 cubes + 2 spheres = 56.

Cylinder + 2 cubes + Sphere = 28.

It is given that 2 cubes + Sphere = 18.

Thus, (Cylinder + 2 cubes + Sphere) - (2 cubes + Sphere) = 28 - 28 = $ \boxed{10}$.

I'll do it with variables. Slight bit cleaner in my opinion. :P

Let $x$= the weight of the cylinder, $y$= the weight of the cube, and $z$=the weight of the sphere.

1) $x+y=18$

2) $2y+z=19$

3) $x+y+z=21$

Substitute equation $1$ into $3$.

$18+z=21$

$z=3$

Plug $z$ back into equation $2$ and find $y$. Then plug the $y$ value into equation one to get $x=\boxed{10}$


Prove that the georgenne point of a triangle is the symmedian point of its Georgenne triangle

[hide]So this follows trivially from the following lemma:

Let the tangents to the circumcircle of $ \triangle{ABC}$ at $ B$ and $ C$ meet at $ D$.  Then $ AD$ is a symmedian of $ \triangle{ABC}$.

Proof: Let the circle centered at $ D$ going through $ B$ and $ C$ intersect $ AB$ and $ AC$ again at $ E$ and $ F$.  Then $ DE\equal{}DB \implies m\angle{DEB}\equal{}m\angle{DBE}$.  Let $ X$ be a point on ray $ DB$ so that $ B$ is between $ D$ and $ X$.  Then $ m\angle{DEB}\equal{}m\angle{DBE}\equal{}m\angle{ABX}\equal{}m\angle{ACB}$.  Similarly, $ m\angle{AFC}\equal{}m\angle{ABC}$.  Therefore, $ E, D, F$ are collinear, and triangles $ ABC$ and $ AFE$ are similar.  Because the two triangles are in opposite orientation, it follows that because $ AD$ is a median in $ \triangle{AFE}$, it is a symmedian in $ \triangle{ABC}$, as desired.

It is well-known and easy to prove that the symmedians of a triangle concur (at the isogonal conjugate of the centroid, the symmedian point).[/hide]


Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=6568&highlight=concours+generall

Pierre.


inequalities
function
geometry
3D geometry
rearrangement inequality
inequalities unsolved

Hello Friends , I am New to Rearrangement (Inequality) and Want your help in this Matter .
See The Following Questions and Reply :

If This Assumption is incorrect then How can I solve this Inequality using REARRANGEMENTS Only !!

 :?:

Hello, I think that you can answer this question for yourself. I ask you to investigate: in most inequalities that you see on this board, people will assume that $ a\ge b\ge c$. Why is this assumption justified? That is to say, what property does the inequality have that allows us to do that. I will say that you cannot make that assumption in this problem.

I don't think inequality holds...
try $ n\equal{}3$
we should prove:

$ (1\plus{}a)(1\plus{}b)(1\plus{}c) \geq (1\plus{}a^2b)(1\plus{}b^2c)(1\plus{}c^2a)$

But that is just wrong. Try $ a\equal{}b\equal{}c\equal{}2$
$ LHS \equal{} 27$
$ RHS \equal{} 9^3$  :!:

Then CAn't W SOlve this Inequality Using Rearrangements???
Any Other Method??

Aman I repeat Altheman's post here again.
see, u just cant assume like that for the sake of using Re arrangement Inequality.Always is it possible to assume $ a\geb\gec$? Investigate when it is possible. It is very simple. Depending upon the inequality u decide where to use which., especially Re arrangement inequalities.

[quote="amangupta2"]So If we consider the Question Below, it involves Only 3 Varieables a,b,c,

Can we COnsider here that a>=b>=c...???

I Hope that The Answer is Yes !!
 :([/quote]
Hmm I think no, its not symmetric :oops_sign:
But you can use weight AM-GM for it :wink:

Oh No !!! :(

I Learnt using Reaarragements The Day Before Yesterday, I was so happy that i could solve many Inequalities Instantly Usinf Rearrangements, But , I was Wrong :noo:  :wacko:

hey since i'm quite new to the inequalities
can someone tell me what are symmetric equations and how can we check if a certain eq is symmetric?
can someone tell me what are convex eqs and how can we check if a certain eq is convex?(for applying jensen,etc..)

Google is your friend.

http://en.wikipedia.org/wiki/Convex_function

Anyway, this seems a good link: http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf.

The first inequality is simple from Schur's but here is a solution from AM-GM:

a^3 + a^3 + b^3 >= 3a^2.b
b^3 + b^3 + c^3 >= 3b^2.c
c^3 + c^3 + a^3 >= 3c^2.a

Adding the three we have the result.

For second one, you would need to use Holder's inequality. These problems are from this year's INMOTC (Delhi).

You can use rearrangement though for that one :P

$ [a,b,c] and [a^2,b^2,c^2]$ are similarly sorted since $ f(x)\equal{}x^2$ is an increasing function over the non-negative reals.

Hence, $ a*a^2\plus{}b*b^2\plus{}c*c^2\ge a*b^2\plus{}b*c^2\plus{}c*a^2$

mathias is correct.  to see this, write $ a,b,c$ as $ a_1,a_2,a_3$.  Let $ \sigma$ be the permutation so that $ a_{\sigma(1)}\geq a_{\sigma(2)}\geq a_{\sigma(3)}$.  then certainly $ a^2_{\sigma(1)}\geq a^2_{\sigma(2)}\geq a^2_{\sigma(3)}$ as well.

Now, for any permutation $ \tau$ of $ \{1,2,3\}$, we have that
\begin{eqnarray*} a_1^3 + a_2^3 + a_3^3 & = & a^3_{\sigma(1)} + a^3_{\sigma(2)} + a^3_{\sigma(3)} \\
& = & a^2_{\sigma(1)}a_{\sigma(1)} + a^2_{\sigma(2)}a_{\sigma(2)} + a^2_{\sigma(3)}a_{\sigma(3)} \\
& \geq & a^2_{\sigma(1)}a_{\tau(\sigma(1))} + a^2_{\sigma(2)}a_{\tau(\sigma(2))} + a^2_{\sigma(3)}a_{\tau(\sigma(3))} \\
\end{eqnarray*}

by the rearrangement inequality!  now define $ \tau$ by $ \tau(1) = 2,\tau(2) = 3,\tau(3) = 1$.  this is kind of a silly post...

[quote="amangupta2"][quote="Mathias_DK"][quote="antiparticle"]We can't assume a>=b>=c as the inequality is not symmetric. Therefore, rearrangement is not applicable.[/quote]
The rearrangement inequality says something like the following.
Let there exists three sequences: $ a_1, a_2, ..., a_n$, $ b_1, .., b_n$, and $ c_1, c_2, ..., c_n$, such that $ c$ is a permutation of $ b$. If $ b$ and $ a$ are similar sorted then $ a_1b_1 \plus{} ... \plus{} a_nb_n \ge a_1c_1 \plus{} ... \plus{} a_nc_n$. This is also easy to prove.
So we can indeed use rearrangement here.
Further more he doesn't assume that $ a \ge b \ge c$ or anything like that. Read his post once again please.[/quote]

Are you  sure , we can use it ere, coz everyone above has posted a negative answer  !!! So Please , Tell me are you relaly sure i can  !!!(in the 2nd question involvinf a,b,c)

what about the  first question (involving a1,a2........an), can we apply rearrangement there!!!
 :?:[/quote]
I will strongly recommend you to go through the rearrangement inequality, and prove it yourself. When you have done that prove that $ a^3 \plus{} b^3 \plus{} c^3 \ge ab^2 \plus{} bc^2 \plus{} ca^2$. Post both proofs here, and I'll check them, so you're sure you understood :)

I final note: When in doubt if you can apply a theorem, It's always because you didn't understand something fundemental. (If you had, you could have just double-checked and you wouldn't be i doubt) So find the thing you don't understand and start to understand it :P

well !!  I first Learnt about Rearrangements from Engel's Problem solving strategies, it has 2-3  solved examples that illustrate its application , so i used exactly the same approach as they did .  But now i discover that many people disourage the use of Rearrangement inequality in solving this problem.
anyways i will try to post my approach as soon as possible !!

 :|

OH My GODDDDDDDDDDDDDDDD!!!
 I am really sorry , I am such a big FOOLLLLLLLL. I am sorry  :(
[b]Ronald is correct[/b], it's all cubes in the Left hand side.

Sorry once again.

[b]So ronald, now can I solve this using REARRANGEMENTS??[/b]

Please Reply Urgently , I need Help

Just use Holder's In this one:

$ (1\plus{}a_1^3)(1\plus{}a_1^3)(1\plus{}a_2^3) \ge (1\plus{}a_1^2a_2)^3$

Multiply similar inequalitites.
Note power of each term then will be 3 and taking the cube root we'll have the result we'll get the result.


[hide]E.[/hide]sorry, im a bit lazy to put a solution. i will later. but i didnt use a calculator or somat like that.

okay, now im ready for the solution. oh yeah, and to mathfanatic. you can delete my 2 posts, if you think other people should have a try, but thats the point of the spoiler font. but still. whatever floats your boat

solution:[hide]basically, find the pattern thingie. (10:^2:-10)/9=10, (10:^3:-10:^2:)/9=10:^2:, etc...(10^12-10^11)/9=10^11.[/hide]

[quote="Syntax Error"]oh yeah, and to mathfanatic. you can delete my 2 posts, if you think other people should have a try, but thats the point of the spoiler font. but still. whatever floats your boat[/quote]

Nonono.  I'll only delete something if it's rude or potentially offensive.  Nice solution, Syntax and Tare...

[quote="rippledance"]Bah. Same method as Tare.[/quote]
I thought it was "moo" :)

I think the easiest way... without using pattern methods... is to know that you can factor out a 10^11 on the numerator... thus it'll be 10^11 * (10-1) all divided by 9. So the 9's cancel each other and you're left with 10^11. simple alg 1.

yup. I think the finding a pattern method by Syntax is perfectly fine, but takes too much time.

yeah, but it didnt take any more than 20 seconds for me. but its just me, and it doesnt always work like that for all prblems

Sure.  :D
I guess whatever works for us individually is good.

The important thing is the answer (on multiple-choice),


(i) The equation of the circle $ C$ is

$ (x \minus{} 2t)^{2} \plus{} (y \minus{} t)^{2} \equal{} t^{2}$,

where $ t$ is a positive number. Show that $ C$ touches the line $ y \equal{} 0$.

Let $ \alpha$ be the acute angle between the $ x$-axis and the line joining the origin to the centre of $ C$. Show that $ \tan 2\alpha \equal{} \frac {4}{3}$ and deduce that $ C$ touches the line $ 3y \equal{} 4x$.

(ii) Find the equation of the incircle of the triangle formed by the lines $ y \equal{} 0$, $ 3y \equal{} 4x$ and $ 4y \plus{} 3x \equal{} 15$.

If the center of the circle is $ (2t,t)$, how can $ \tan\alpha\equal{}\frac{4}{3}$? Wouldn't it be $ 1/2$?

Sorry J. Mayers, it was a typo. I've edited it: it is $ \tan 2 \alpha \equal{} \frac{4}{3}$.


http://www.artofproblemsolving.com/Wiki/index.php/1989_AIME_Problems/Problem_9

So for the solution, I see up to the part where 4=n(mod5), but then I got lost...
Would anyone help me out from then on?

$ 133 \equiv 1$ modulo $ 3$, so $ 133^5 \equiv 1^5 \equiv 1$ modulo $ 3$.

$ 110 \equiv \minus{}1$ modulo $ 3$, so $ 110^5 \equiv (\minus{}1)^5 \equiv \minus{}1$ modulo $ 3$.

The same is true for $ 84^5$ and $ 27^5$, which are equivalent to $ 0$ modulo $ 3$.

By FLT, $ n^2 \equiv n$ modulo $ 3$. Therefore, $ n^5 \equiv n$ modulo $ 3$. so $ n^5 \equiv 0$ modulo $ 3$ implies $ n \equiv 0$ modulo $ 3$. We know that $ n \equiv 4$ modulo $ 5$, and since $ n$ must be even, $ \boxed{n\equal{}144}$. $ n \equal{} 174$ is too large because we note that $ n^5 < 4 \times 133^5$.

[quote="darkdieuguerre"]By FLT, $ n^2 \equiv n$ modulo $ 3$.[/quote]

You mean $ n^2\equiv 1\bmod 3$.


This is AHSME problem and it was like #20 or so. This may be very hard for some MC participants so moderators, if you consider this too hard, feel free to move it to Getting Started.

Question:

A wooden cube with edge length n units (where n is an integer > 2) is painted black all over. By slice parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit edge length. If the number of smaller of cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is n?
A) 5
B) 6
C) 7
D) 8
E) None of these

[hide="If_you_knew_this"]

This problem is quite tough if you're trying to understand everything it says in there.

Basically, the cube is divided into $n^3$ cubes and # of no paint = # of one painted.

This is same as saying:

$(n-2)^3 = 6(n-2)^2$

Then:

$(n-2) = 6\\
n = 8$

So the answer is 8.[/hide]

I thought it was a neat problem.

The number of cubes on each side with only one face painted is a square with side length n-2. The number of cubes in such a square is (n-2)^2 There are six faces.

The number of cubes with no faces painted is the cube that includes all of the small cubes except the outermost ones. It's side length is n-2, so the number of cubes it includes is (n-2)^3

Bruno Holanda:Can you use hide next time?

[hide]$(n-2)^3=6(n-2)^2$, so $n-2=6$, so n=8[/hide]

Actually it isn't that bad, just needs thinking about.

Billy


Aratati ca intr-un triunghiu ascutitunghic are loc urmatoarea inegalitate:
$(1+tgAtgB)(1+tgBtgC)(1+tgCtgA)\geq 64$

Notez $x=\frac{1}{tgA\cdot tgB}$ ,$y=\frac{1}{tgB\cdot tgC}$ si $x=\frac{1}{tgC\cdot tgA}$ ,$x,y,z>0$ atunci avem de demonstrat ca
$(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z})\geq 64$ , dar $x+y+z=1$
Fie $f(x,y,z)=(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z})-64$
$f(x,y,z)\geq f(x,\frac{y+z}{2},\frac{y+z}{2})$,$(*)$, $\Leftrightarrow$ $(1+\frac{1}{y})(1+\frac{1}{z})\geq \left(1+\frac{2}{y+z}\right)^{2}$ $\Leftrightarrow$ $\frac{1}{y}+\frac{1}{z}+\frac{1}{yz}\geq \frac{4}{y+z}+\frac{4}{(y+z)^{2}}$ ,relatie care este evident adevarata.
Deoarece relatia $(*)$ are loc ,este suficient sa aratam ca $f(x,y,y)\geq 0$ $\Leftrightarrow$ $\left( 1+\frac{1}{1-2y}\right)\left( 1+\frac{1}{y}\right)^{2}\geq 64$ $\Leftrightarrow$ $63y^{3}-33y^{2}+y+1\geq 0$ $\Leftrightarrow$ $(y-\frac{1}{3})(63y^{2}-12y-3)\geq 0$ ,semnul acestei ultime expresii se poate determina cu usurinta si se constata ca relatia este adevarata.

cred ca de la relatia $(1+\frac{1}{x})(1+\frac{1}{x})(1+\frac{1}{x})\geq 64$ te-ai complicat putin:P

se poate folosi doar AM-GM

$1+\frac{1}{3x}+\frac{1}{3x}+\frac{1}{3x}\geq 4 \sqrt[4]{\frac{1}{3^{3}x^{3}}}$

Ideea de a folosi astfel inegalitatea mediilor o sugereaza conditia de egalitate :wink:

Dupa inmultirea inegalitatilor analoage totul revine la $\frac{1}{3^{3}}\geq xyz$ care este clara din conditia $x+y+z=1$, folosind AM-GM


A certain math professor announces that he will select three out of 8 theorems to put on an exam, and that each student will choose one of the three theorems to prove.

a) A student decides to study only four of the theorems. What is the probability that at least one of the four theorems will be on the exam?

b) What is the least number of theorems the student should learn in order to be certain of being able to prove a theorem on an exam?

Make sure to hide your answers.

[hide="a"]
I will try to find the complement, the probability that none of the 4 are on the exam

$ \frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{2}{5}\equal{}\frac{1}{14}$

the probability that at least one is on the exam is the complement, or $ \frac{13}{14}$
[/hide]

[hide="b"]
Only 3 of the 8 theorems will be on the exam. That means that 5 of the 8 won't be on the exam. To be sure he will get at least one that is on the exam, he has to study more than 5, so the least number is $ \boxed{6}$
[/hide]


a disk of radius $R$ is located in the $x-y$ plane with its centere at the origin.  The diest has four disk shaped holes of radius $R/4$ which are centered at $(R/2,0),(0,R/2),(0,-R/2),(-R/2,0)$ as shown below.  The disk has mass $M$ and uniform density.  The moment of inertia of this dist about the $z$ axis is $CMR^2$ where $C$ is a constant.

a.  The dist starts from rest at the top of an incline of height $H$ and rolls without slipping to the bottom.  What is the speed of its centere of mass at the bottom?  Express your answer in terms of $H,C$ and fundimental constants.

b.  A nail of negligible mass is glued to the top of the disk at $(0,R)$.  the nail points in the $z$ direction and is mounted ina frictionless pivot so that the disk hangs in the vertical plane.  The disk is displaced by a small angle from its equilibrium postion and released from rest.  What is the period of the oscilliation of the dist in terms of $R,C$ and fundimental constants

c.  Find the numerical value of constant $C$.

Hints:

**********************************
a. Conservation of energy.
b. pendulum.
c. negative density methods.
***********************************

Nice problem, especially part C.

yes, part a is just conservation of angular momentum
part B is just $2\pi \sqrt{\frac{R}{g}}$

part c is the one that i couldn't figure out...

Are you sure for part B, cuz i get:

the moment of inertia about the center of mass is CMR^2, so by the parallel axis theorem the moment of inertia about the point (0, R) is CMR^2 + MR^2

Our torque equation becomes

(CMR^2 + MR^2)* alpha = -Mg sin(theta) R

cancelling like terms and using the small angle approximation, we get

alpha = - (g/[C(R+1)])*theta

g/[C(R+1)] is our W^2 in the equation 2* pi * f = W, so the period becomes

2pi*sqrt[C(R+1)/g]

The equation 2pi*sqrt(R/g) is only valid when you have a point mass on a string of length R.

For part C (this is tedious, so I'm not going to do it), imagine a large circle of density rho, and subtract from it the moments of inertia of 4 small circles of density rho using the parallel axis theorem to account for the fact that the circles are not concentric. The pain is finding the density by setting M = density * area of big circle - 4 (desnity * area of small circles), then using the density you get in your calculations. I messed up on arithmetic about 5 times...

you are correct, your answer for part B is correct.
mine is not :)
(although i would still like to see a ful solution for part c)

[quote="amirhtlusa"]
c.  Find the numerical value of constant $C$.[/quote]
[color=blue]My calculation gives the answer that $C=\frac{61}{128}$
I wonder if I'm wrong.[/color]

[color=blue]The moment of inertia of this disc without 4 holes about the $z$ axis is $I_0=\frac{1}{2}MR^2$
The moment of inertia of one hole with mass $m=\frac{M}{16}$ (because the disc has uniformed density)about the $z$ axis is $I=\frac{1}{2}mr^2+m(\frac{R}{2})^2=\frac{3}{512}MR^2$
So, the moment of inertia of this disc with 4 holes about the $z$ axis is $I'=I_0-4I=\frac{61}{128}MR^2$
$\Longrightarrow C=\frac{61}{128}$[/color]

but the M they give you in the problem is the mass of the disk with the 4 holes in it. This means that the mass of the large disk without the holes is 4/3M and thus its moment of inertia without the holes is 2/3MR^2. The moment of inertia of each of the small disks is 1/12M(1/2)((R/4)^2) + 1/12M(R/2)^2, making C = 55/96. You can reason out that 61/128 doesnt work intuitively by the fact that I for this arrangement should be bigger than I for a disk of equal mass since more mass on this figure is concentrated farther from the axis of rotation.

[color=blue]Oh yeah, you're right. I really misunderstood [/color]

[quote="AndersoN"]Does anyone know how Mathlinks transforms Latex code into images?[/quote]
[color=blue] Why do you want to know that?[/color]

[quote="psychicalchemist"]The moment of inertia of each of the small disks is 1/12M(1/2)((R/4)^2) + 1/12M(R/2)^2, making C = 55/96. [/quote]
how did you get 55/96... that simplifies to 3/128*M*R^2
i know im being stupid...

[quote="qweretyq"]how did you get 55/96... that simplifies to 3/128*M*R^2
i know im being stupid...[/quote]
[color=blue]You have to multiply $\large\frac{3}{128}$ with $4$ because there're 4 holes.
Ok?[/color]

I believe for part B you can use the formula $T = s \pi \sqrt{\frac{I}{mgh}}$

Using the parallel axis theorem we can determine I to be CMR^2 + MR^2

m = M

g = gravity

h (distance from suspension to the center of mass) = R

thus $T = s \pi \sqrt{\frac{CMR^2+MR^2}{MgR}}$

Yeah, but MEMORIZERS DIE in physics. Its better to just know concepts and re-derive stuff in 5-10 seconds. Plus, it looks more impressive in a solution.

You're the PHYSIC ALCHEMIST, remember?  How do you know who dies in physics?  :P

(Actually, I agree that trying to remember too many formulas is insane)


This is esy but I want to see different solutions.
Prove that two lines of Simson which correspond at antidiametric points are perperdicular.

Silouan ,

 it is meangless if it is easy( it is not !) or not , because it is a [b]THEOREM[/b] and you can use it without proof

  ( see ΝΤΑΝΗΣ , ΕΠΙΠΕΔΟΜΕΤΡΙΑ, in greek).As a theorem  , we are not interesting for many solutions but only for the best and more didactic. This is the reason for the problem , not to be answered untill now. Of course  , do not worry for that. For all other problems you got nice solutions. I could send you a solution , but I' m afraid it will be the same as yours.

Babis

Mr. stergiou grom ntanis I gave the problem. there it is an excelent solution. But another day someone said me that there exist a better solution. And because this problem is good I posted here.

Ges Silouan ,

    I' d like  , too , to see one more and simpler  proof to this theorem. Ask  your friend to post it ( the other solution) here for all linkers.

   Babis


When 0.050g of sodium was added to a 100cm^3 water the temperature rose by 1.24 degrees C. (assume that 1cm^3 of an aqueous solution is 1g)

a) Calculate the heat change in the reaction
b) How many moles of sodium were used?
c) Calculate the enthalpy change for the reaction in KJ per mole.

Note: We've done so many different things that involve enthalpy change, when this cropped up in class I got very confused. Please help me to understand.

q = m x cp x deltaT

(energy transfered = mass of water (g) x specific heat capacity of the water x temperature change)

specific heat capacity of water is stated as being 4.18 J per gramme per Kelvin.  I know that with this question I've got to work backwards, but I'm not sure where to start.

Moles = mass divided by molecular mass

Moles = volume X concentration (If it's a liquid)

Enthalpy change = q / moles of reactant, where q is theenergy transfered.

Please get back to me with any help as soon as you can, thanks a lot.
 :|

First, you will use $\Delta q=mc\Delta t$ to calculate the heat change (remember, $\Delta q$ is the heat change).  Since the amount of sodium used ($0.05\text{g}$) is so small compared to the amount of water ($100\text{g}$), it is acceptable to only consider the water in the equation.

To calculate moles of $\text{Na}$, just use the molar mass of sodium from the periodic table ($22.99\frac{\text{g}}{\text{mol}}$).

Finding $\Delta H$ in $\text{kJ/mol}$ is straightforward after the first two steps.

Hope that helps.


vector
geometry
3D geometry
inequalities
linear algebra
linear algebra unsolved

I encountered a problem in a book with a proof given. But I am a bit skeptic about it. I hope someone can help shed some light.

Let $\{g_{i}\}$ be a set of vectors and imagine a cone defined as $K = \left\{v \,\bigg|\, v =-\sum_{i}\lambda_{i}g_{i}, \textup{ where }\lambda_{i}\geq 0 \ , \forall i \right\}$.

Let $f \notin K$ and let $u \in K$ be the closest point to $f$. Obviously $u$ is the projected point of $f$ onto $K$. The objective is to prove that if $d = u-f$, then $g_{i}^\top d \leq 0, \, \forall i$. (Note that $d \neq 0$.)

The proof given is by contradiction: Suppose that is not true, that is, $\hat{g}_{i}^\top d = s_{i}$ for some scalar $s_{i}> 0, \, \forall i$, where $\hat{g}_{i}= g_{i}/\|g_{i}\|$. It is not difficult to see that $(u-s_{i}\hat{g}_{i}) \in K, \, \forall i$, i.e., it remains in the cone even by small or large perturbation on the vector $u$. Now, we shall show the perturbed point has smaller distance. Indeed this is the case since for any $i$,

\begin{eqnarray*}\|(u-s_{i}\hat{g}_{i})-f \|^{2}&=& \|(u-f)-s_{i}\hat{g}_{i}\|^{2}= \|(u-f)\|^{2}-2 s_{i}\hat{g}_{i}^\top (u-f)+s_{i}^{2}\|\hat{g}_{i}\|^{2}\\ &=& \|d\|^{2}-2s_{i}\hat{g}_{i}^\top d+s_{i}^{2}\\ &=& \|d\|^{2}-2s_{i}^{2}+s_{i}^{2}\\ &=& \|d\|^{2}-s_{i}^{2}\leq \|d\|^{2}, \end{eqnarray*}

which contradicts with the assumption that $u$ is the nearest point in $K$ to $f$ -- done!!!.

All looks good, however if I let $\hat{g}_{i}^\top d = t_{i}$ for which the scalar $t_{i}< 0,\, \forall i$ but sufficiently close to 0 such that $(u-t_{i}\hat{g}_{i}) \in K$ for any $i$, then using the same derivation I arrive at $\|(u-t_{i}\hat{g}_{i})-f \|^{2}= \|d\|^{2}-t_{i}^{2}\leq \|d\|^{2}$ too! This means it can contradict even for $g_{i}^\top d < 0$. I now question the validity of this proof. I welcome your comment.

[quote="kaos"]...such that $(u-t_{i}\hat{g}_{i}) \in K$ for any $i$[/quote]
This simply isn't true for negative $t_{i}$. Look at the definition of $K$.
Without that step, the argument falls apart.

I agree that it does not work for arbitrary $t_{i}$, but for sufficiently small value (in term of magnitude) it does work assuming the respective $\lambda_{i}\neq 0$.

How about this (to make things clearer):
Let $\bar{\lambda}= \min \{\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\}$, where $\lambda_{i}\neq 0$ for all $i$ come from $u$. Then let $-\bar{\lambda}\leq t_{i}< 0$ for all those $i$. For all $j$ of which $\lambda_{j}= 0$ in $u$, we set $t_{j}= 0$.

For each $i$, either $\lambda_{i}$ is zero or $u-f$ is orthogonal to $g_{i}$. This is consistent, and it's what you've really proved.

$u-f$ is orthogonal to $u$, but $u-f$ may not be orthogonal to $g_{i}$ -- recall that $u \in K$ and is not any one particular $g_{i}$.

That's irrelevant. Look at the picture in a managable number of dimensions- $u-f$ is orthogonal to [i]all[/i] of the $g_{i}$ that contribute to $u$.

Thank you for the insight. Although the original proof is not faulty, I still think the author did not make enough effort to strengthen its correctness.

Below is my version which I just thought of and which I think is better. I will give a direct proof. Consider the distance of a perturbed point of $u$ relative to the point $f \notin K$, which is given by

\begin{eqnarray*}\|(u-s_{i}\hat{g}_{i})-f \|^{2}&=& \|u-f\|^{2}-2 s_{i}\hat{g}_{i}^\top (u-f)+s_{i}^{2}\|\hat{g}_{i}\|^{2}\\ &=& \|d\|^{2}-2 s_{i}\hat{g}_{i}^\top d+s_{i}^{2}\end{eqnarray*}

Since $d$ is the minimum distance, hence

\begin{eqnarray*}\|d\|^{2}-2 s_{i}\hat{g}_{i}^\top d+s_{i}^{2}&=& \|(u-s_{i}\hat{g}_{i})-f \|^{2}\geq \|d\|^{2}\\ s_{i}\hat{g}_{i}^\top d &\leq& \frac{s_{i}^{2}}{2}. \qquad \qquad (1) \label{equ: distanceOfPerturbedPoint}\end{eqnarray*}

Therefore, for $s_{i}> 0$, the preceding inequality yields $\hat{g}_{i}^\top d \leq \frac{s_{i}}{2}$. Since it must be true for all positive $s_{i}$, hence

\begin{eqnarray*}\hat{g}_{i}^\top d &\leq 0, \qquad \textup{ for the case of }s_{i}> 0. \qquad \qquad (2) \end{eqnarray*}

On the other hand, for $0 > s_{i}\geq-\lambda_{i}$, the inequality (1) yields $\hat{g}_{i}^\top d \geq \frac{s_{i}}{2}$. Since it must be true for all negative $s_{i}$, hence

\begin{eqnarray*}\hat{g}_{i}^\top d &\geq 0, \qquad \textup{ for the case of }-\lambda \leq s_{i}< 0. \qquad \qquad (3) \end{eqnarray*}

Let $I(u)$ be the set of indices for which $g_{i}$, where $i \in I(u)$, contributes to $u$. Note that $\lambda_{i}> 0,\,\forall i \in I(u)$ and $\lambda_{j}= 0,\,\forall j \notin I(u)$. The perturbed point $(u-s_{i}\hat{g}_{i})$ is in $K$ only if $s_{i}\geq-\lambda_{i}$ for all $i$. By the virtue of these, for all $i \in I(u)$, $\hat{g}_{i}^\top d$ must satisfy both (2) and (3) and thus $\hat{g}_{i}^\top d = 0$. As a side result, this implies the vector $d$ is orthogonal to all $g_{i}$ that contribute to $u$. Meanwhile, for all $i \notin I(u)$, $\hat{g}_{i}^\top d$ only needs to satisfy (2) and hence $\hat{g}_{i}^\top d \leq 0$. By combining these results, it yields the result that is sought for, namely, $\hat{g}_{i}^\top d \leq 0$ for all $i$.


prove that

$ \frac{f(a)\plus{}f(b)}{2}\equal{}f ( \frac{a\plus{}b}{2} )\plus{} \frac{(b\minus{}a)^2}{8}\cdot f {''}(c)$

You forget many thigs $ f\in D^{(2)}[a,b],$  then exist $ c\in (a,b)$


The minimum value of $ x(x \plus{} 4)(x \plus{} 8)(x \plus{} 12)$ in real numbers is ?

$\textbf{(A)}\  \minus{} 240 \qquad\textbf{(B)}\  \minus{} 252 \qquad\textbf{(C)}\  \minus{} 256 \qquad\textbf{(D)}\  \minus{} 260 \qquad\textbf{(E)}\  \minus{} 280$

$ x(x\plus{}4)(x\plus{}8)(x\plus{}12) \equal{} (u\minus{}6)(u\minus{}2)(u\plus{}2)(u\plus{}6) \equal{} (u^2 \minus{} 36)(u^2 \minus{} 4) \equal{} u^4 \minus{} 40u^2 \plus{} 144, u \equal{} x \plus{} 6$

Minimum for that would be at $ u \equal{} \pm\sqrt{20}, f(\sqrt{20} \minus{} 6) \equal{} \minus{}256$

Setting $ x$ equal to $ y\minus{}8$, the expression becomes $ (y\minus{}8)(y\minus{}4)(y)(y\plus{}4)$ or $ (y^2\minus{}8y)(y^2\minus{}16)$.

This expression is equivalent to $ (y^2\minus{}4y\minus{}16)^2 \minus{} 16^2$ by difference of squares.
Verifying that $ y^2\minus{}4y\minus{}16$ has real roots, we find the minimum value of the expression is $ \minus{}256$.