Ross Mathematics Program
real analysis
real analysis unsolved

This is rather simple, but I liked it :).

Show that a disk cannot be partitioned into two congruent subsets.

Let the unit disk $D$ be partitioned into sets $A$ and $B$.
Lemma: At least one of $\overline{A}$ and $\overline{B}$ has diameter 2, where $\overline{X}$ is the closure of $X$.
Proof: Let $A'$ and $B'$ be the intersections of $\overline{A}$ and $\overline{B}$ with the boundary $C$ of $D$. If either of $A'$ and $B'$ is all of $C$, we are done, since $C$ has diameter 2. Otherwise, since $C$ is connected and both $A'$ and $B'$ are closed, $A'$ and $B'$ intersect. Choose a point $x$ in the intersection; its antipode is in at least one of $A'$ and $B'$, and we are done.

Now, if $A$ and $B$ are congruent, both $\overline{A}$ and $\overline{B}$ must have diameter 2. Therefore, the only unit disk that covers $A$ is $D$, and the only unit disk that covers $B$ is $D$. Since exactly one of the sets contains the center of $D$, exactly one of the sets can be covered by a unit disk centered at a point in the set, contradicting congruence.

Nice solution jmerry!
I know a solution, completly elementary , which is presented in Mathematical Gems III (of Ross Honsberger)
I wont post this proof, but i would like to write some thoughts from this book...
Isn't it natural to ask if this holds for all figures in R^2 which have a center ?
Well, we can easily find a counterexample using an unbounded set. But does it hold for bounded figures ?
The answer is no,
So can you find such a bounded figure that has a center and it can be partition in two congruent sets ?

[There is a solution for this, in the book mentioned in top, not realy trivial but very clever! ]

[quote="Anto"]
So can you find such a bounded figure that has a center and it can be partition in two congruent sets ?
[/quote]

How about this:
Consider a circle centered at $-1$ of radius $1$ (call it $\mathcal C_1$), and a circle centered at $1$ of radius $1$ (call this one $\mathcal C_2$). On $\mathcal C_1$ we place the points $-1+z^k,k\ge 0$, while on $\mathcal C_2$ we place the points $1-z^k,k\ge 0$, where $z$ is a complex number of modulus $1$ whose argument is an irrational multiple of $\pi$. Additionally, our set also contains the centers of the circles, i.e. $\pm 1$.

The set is clearly symmetric wrt $0$, and the two congruent sets that partition it are $\{-1\}\cup\{-1+z^k|k\ge 0\}$ and $\{1\}\cup\{1-z^k|k\ge 1\}$. They are congruent through the correspondence $-1\mapsto 1,\ -1+z^k\mapsto 1-z^{k+1},\ \forall k\ge 0$.
I hope it's correct. My apologies if it isn't: it's about 4.00 AM here, and I'm tired and confused :).

[quote] [There is a solution for this, in the book mentioned in top, not realy trivial but very clever! ] [/quote]

About such a clever solution i was talking before.   :)

I found a nice problem related to the ones discussed here on [url=http://www.math.columbia.edu/~zare/prob_all.html]this page[/url]. It's problem number $12$:

Does there exist a plane set which can be partitioned into two subsets, both of which are congruent to the initial set?

Enjoy! :)

There is a solution to that problem in the book i'm mentioning before. (In the same chapter. )


Hi, I'm a high school student currently living in S.Korea and attending Daewon Foreign Language HS.
I've entered some mathmatical competitions here in Korea and won a price at the National Olympiad, but I didnt enter any international competitions until now.
Since I'm moving to New Jersey in March, I thought it would be good to take the AIME.

So here are the questions:
1.Should I try to enter this year's AIME? (the AMC is in February, before I move to NJ)
2.If I take the AMC12 in Korea and qualify for the AIME, am I able to take the AIME in the US?

please help me, I have no idea how the AMC & AIME works. (i've read the site, but couldn't get the info I needed)

you need to take the amc 10 or 12 and qualify for the AIME. you cant just register for it.

He's asking if he can take the AMC 12 in Korea and take the AIME in the USA.

AMC 12 should definitely be worth it. Seeing that you made nationals in a country like Korea, you're pretty good.

Read his post...he's asking if he's ALLOWED to take AMC 12 in Korea and AIME in USA...

Yeah, as long as you score in the top 1% for AMC 10 or top 5% for AMC 12, you can go international.

Your question has at least two distinct parts, I will restate them and answer them separately:

Are you allowed to register for the AMC 12 and take the AMC 12 in February in Korea?:

The answer is a qualified yes.  Specifically, your school registers for the contest, and administers the contest to you as a student at the school.  The AMC 12 is not an individual participation event.  So someone at your school, usually a math teacher, must register for the AMC 12.  The person at your school will have to do the regstration by postal mail or by fax.  (The on-line registration requires a CEEB number and for your school to be in our database.  Probably neither of those conditions apply.)   Schools in about 30 countries around the world, several from Korea, register for the AMC contests annually, and students at those schools participate fully in the AMC 12.

Will you be able to participate in the AIME, given that you will be in the United States in March?

The answer is again a qualified yes.  Assuming that you score at least 100 points, or are in the top 5% of scores on the AMC 12, you will qualify to take the AIME in March.  Normally, our procedures would automatically send the AIME and the participation instructions to your administering school.  That would be your school in Korea.  However,  you will not be at the school in March.  Therefore you will have to contact the AMC office and tell us where your AIME materials should be sent in New Jersey.  That means you will need to alert your school in New Jersey of the arrangements, alert your school in Korea of the arrangements, and alert us at the AMC office about the arrangements you make.  You should do this by writing an email to amcinfo@maa.org explaining the full situation, following up with a postal letter to the AMC office, (see our website for the postal address) and including a copy of the postal letter with the AMC 12 answer sheets that your school in Korea sends to us for scoring.

We have about 12,000 students in the United States and around the world who qualify for AIME, and we have 2 to 3 weeks to get all the AIME materials to them at their school.  Obviously this requires
automated procedures.  You are asking us to suspend our automated procedures to attend to one specific student with special circumstances.

The proper way to get answers to detailed administrative questions and special situations like this is to contact the AMC office.  Students on this forum simply do not have the inforrmation or experience necessary to appropriately and completely answer the question.

Steve Dunbar
MAA Director, American Mathematics Competitions


topology
real analysis
real analysis unsolved

$F \subset %Error. "mathh" is a bad command.
{R}^1$ is a closed set. $\{G_k\}$ is a sequence of open sets. $\overline{G_k \cap F} = F$ for all $k$. Prove that
\[\overline{G_0 \cap F} = F\]
where $G_0 = \bigcap_{k=1}^\infty G_k$

The details are rather murky, but it shouldn't be too hard if we use the theorem stating that an open set in $\mathbb R$ is a countable union of disjoint open intervals. First we prove that $\overline{G_1\cap F}=F,\overline{G_2\cap F}=F\Rightarrow \overline{G_1\cap G_2\cap F}=F$, then we repeat the process to show that $\overline{S_k\cap F}=F,\ \forall k\ge 1$, where $S_k=\displaystyle \cap_{i=1}^kG_k$. After this, it's easy (I think).

To prove $\overline{G_1\cap F}=F,\overline{G_2\cap F}=F\Rightarrow \overline{G_1\cap G_2\cap F}=F$ we use the fact that any $x\in F$ is the limit of some sequence $\{x_n\}\subset G_1\cap F$, and each $x_k$ is the limit point of a sequence $\{y_n^k\}\subset G_2$ together with the theorem mentioned above.


For n = 1,2,... two sequences {x_n} and {y_n} are defined recursively by x_1 = 1, y_1 = 2, x_(n+1) = 22*y_n - 15*x_n, y_(n+1) = 17*y_n - 12*x_n.

a.) Prove that x_n and y_n are not equal to zero for all n = 1,2,...

b.) Prove that each sequence contains infinitely many positive terms and infinitely many negative terms.

c.) For n = 1999 1945, determine whether x_n and y_n are divisible by 7.

I give up! I let someone another to solve THAT :D


complex analysis
complex analysis unsolved

hello.

I'm sorry for opening 2 topics in two days..but I have a test on monday and I need some help:

given:

$ f(z) \equal{}\frac{z^{2}\plus{}1}{2z}$

i need to find the laurent series about $ z\equal{}0$

I tried to "play" with this expression,but I got stuck

one more thing about $ f(z)$ how do I show it's one to one for $ |z| < 1$ ?

thank you very much  :)

[quote="t0rajir0u"]Note that $ f(z) \equal{}\frac{1}{2}z\plus{}\frac{1}{2z}$, which is already a Laurent series, to my knowledge.[/quote]

I noted that,but I then tried to expand $ \frac{1}{2z}$
and that's where I got stuck,why can't I expand this?

thank you

Why are you trying to "expand" it?  It's already a valid term of the Laurent series.  Specifically, we have $ f(z) \equal{}\sum_{n \equal{}\minus{}\infty}^{\infty}a_{n}z^{n}$, where $ a_{1}\equal{} a_{\minus{}1}\equal{}\frac{1}{2}$ and $ a_{i}\equal{} 0$ otherwise.

I don't know why I tried to expand it,stupid mistake.

thank you for your help.

can you help me with the other question about the one to one?

and one more question: when does Laurent series has a finite number of terms?


For each n there exist k such that kn=111........1110000......0000

if gcd(n,10)=1 than there exist such m that 10^m-1 devided by 9n.
(For instance m=\phi(9n), where \phi -- Euler function). Than
11...11 (1 -- m times) devided by n.

if n= 2^s5^t n_1 where gcd(n_1,10)=1. Than we can add at the end some zeroes.

Any alternative proof (my own solution):

Consider the n + 1 numbers
1
11
111
...
111...111 (n + 1 digits)
Two of them have the same remainder (mod n).
Hence n divides their difference. That is, n divides
111...111000...000.

Hey! What's going on here? Does the second problem (the one solved by Arne and Fedor) solve the first problem Arne posted?

Indeed, to solve the Yugoslavian problem you have to solve another one:

Prove that any positive integer has a multiple of the from 11...1100...00.

Or do you mean something else, Grobber?

What I mean is this: Look at your first post. Now look at your last post. It might be obvious, but I see absolutely no connection between the problems  :D  :D Please enlighten me.

[quote="Arne"]Does there exist a positive integer k which satisfies:

[i]None of the digits 3, 4, 5, 6 appears in the decimal representation of 2002!*k?[/i]

Yugoslavian math olympiad 2002[/quote]

Fedor started proving that for each n, there is a k such that

kn = 111...111000...000

using Fermat's little theorem. I proved this in an alternative way, using pigeonhole.

And when that's proved, the problem is obviously solved :D

Actually, I was that guest :)

When taking n = 2002! above and taking the appropriate k,
there is obviously no 3, 4, 5, 6 in the decimal representation (only 0 or 1).

Oh my god! That's so obvious! I don't know what i was thinking about.. I guess not paying attention to what you people wrote, combined with my being really hungry right now ( :D ) + the amount (I like to think it's just a drop  :D  :D ) of stupidity that lurks inside me yielded a deadly combination  :D  :D

Thx for the enlightenment. Bye! :D


linear algebra
matrix
function
integration
vector

Let $ \lambda_1\geq\lambda_2\cdots\geq\lambda_n > 0$,
$ c_{ij} \equal{} \frac {1}{\lambda_i \plus{} \lambda_j}$,$ C \equal{} [c_{ij}]_{n\times n}$.
Can you tell me why $ C$ is positive?

You know [url=http://en.wikipedia.org/wiki/Cauchy_matrix]this[/url]? Apply this to the minors of your matrix.

  darij

Consider the functions $ f_j(x)\equal{}x^{\lambda_j}$ in the inner product space in which the inner product is
\[ \langle f,g\rangle \equal{}\int_0^1f(x)g(x)\,\frac{dx}x\]
Then $ \langle f_i,f_j\rangle\equal{}\frac{1}{\lambda_1\plus{}\lambda_j}$ and thus the proposed matrix $ C$ is the grammian of this set.

If the $ \lambda_j$ are all distinct, then the matrix $ C$ is positive definite; if not, then $ C$ is positive semidefinite but not positive definite.

The general proof: let $ \{v_j\}$ be a set of vectors in a real inner product space and let $ c_{ij}\equal{}\langle v_i,v_j\rangle.$ Consider $ u\equal{}\sum_j\alpha_jv_j$ and write $ a$ for the column vector whose entries are $ \alpha_j.$ Then

$ \|u\|^2\equal{}\langle u,u\rangle \equal{}\sum_i\sum_j\alpha_i\alpha_j\langle v_i,v_j\rangle \equal{}a^TCa.$

Since $ \|u\|^2\ge 0,$ we have that $ C$ is at least positive semidefinite. If in addition, the set $ \{v_j\}$ is independent, then $ u\ne 0$ for nonzero $ a$ and thus $ C$ is positive definite.

It's "gramian" in Wikipedia, "grammian" in PlanetMath. The former spelling appears to be about 2-3 times more common.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=34537]here[/url] for a stronger version.


inequalities
trigonometry
real analysis
real analysis theorems

Dear everyone.

Prove $ \forall x(\in R), \cos{(\sin{x})} > \sin{(\cos{x})}$

Thanks for reading my post. :blush:

look at me

it is \[{ \Leftrightarrow sin(\frac{\pi}{2}-sin(x)})-sin(cos(x)=0\]
\[ \Leftrightarrow 2cos(\frac{\pi}{4}-\frac{\sqrt{2}}{2}cos(x+\frac{\pi}{4})sin(\frac{\pi}{4}-\frac{\sqrt{2}}{2}cos(x-\frac{\pi}{4})\]
latest pression it right because
\[ \frac{\pi}{4}-\frac{\sqrt{2}}{2} \leq \frac{\pi}{4}-\frac{\sqrt{2}}{2}cos(x+\frac{\pi}{4}) \leq \frac{\pi}{4}+\frac{\sqrt{2}}{2}\]
\[ \Rightarrow -\frac{\pi}{2} < \frac{\pi}{4}-\frac{\sqrt{2}}{2}cos(x+\frac{\pi}{4}) < \frac{\pi}{2}\]
\[ \Rightarrow cos(\frac{\pi}{4}-\frac{\sqrt{2}}{2}cos(x+\frac{\pi}{4}) >0\]
and
\[ 0 <\frac{\pi}{4}-\frac{\sqrt{2}}{2}cos(x-\frac{\pi}{4}) <\pi\]
\[ \Rightarrow sin(\frac{\pi}{4}-\frac{\sqrt{2}}{2}cos(x-\frac{\pi}{4}) >0\]

 :rotfl:


Could some one please help me with this question? I don't want you to solve it for me; a nudge in the right direction would be much appreciated.  :(

[b]A buffer solution is made by dissolving 10.3g sodium ethanoate, CH3CO2Na, in 0.1M ethanoic acid, CH3CO2H, and making up to 100cm3. Calculate the pH of this buffer given than Ka for ethanoic acid is 1.86 x 10-5 mol dm-3.[/b]

[Ethanoate = acetate. I'll use the trivial name below.]

One thing you'll need is the molecular weight of sodium acetate, so that you'll know how many moles 10.3 grams would be.

I'm not quite sure what "in 0.1M ethanoic acid" means here. Since it gives a more solid foundation for working the problem, I'll assume it means that you dissolve the sodium acetate in 0.1 [b]moles[/b] of acetic acid (in solution, at some concentration), and then you add pure water until the whole mixture occupies 100 ml of volume. That would make the concentration of the acetic acid that you added 1 mole per liter.

Anyway, you have $a$ moles per liter of the NaA salt combined with $b=1$ moles per liter of the HA acid solution. We track the concentrations [H+], [Na+], [HA], and [A-]. (To a first approximation, the concentration [OH-] will be negligibly small.)

We have the following equations:

$[\text{Na}^{+}]=a$
$[\text{HA}]+[\text{A}^{-}]=a+b.$
$[\text{Na}^{+}]+[\text{H}^{+}]=[\text{A}^{-}]$ (charge balance)
$\frac{[\text{H}^{+}][\text{A}^{-}]}{[\text{HA}]}= \text{K}_{\text{a}}$

Now, what can you do with those? If you can solve for $[\text{H}^{+}],$ then you will have the pH. In fact, you can take the negative (base 10) logarithm of the last equation to get that

$\text{pH}-\log[\text{A}^{-}]+\log[\text{HA}]= \text{pK}_{\text{a}}$


Let a,b,c,d positive real and (a+b)(b+c)(c+d)(d+a)=1 show that
(2a+b+c)(2b+c+d)(2c+d+a)(2d+a+b)(abcd) 2  \leq  1/16

my student has g good solution i will post it tomorrow

" My" solution:
    In the post " Nice inequality" source Crux, solved problems subforum, Namdung proved the following inequality:
    (a+b)^3(b+c)^3(c+d)^3(d+a)^3>=16(abcd)^2(a+b+c+d)^4. AM_GM for the LHS combined with this inequality solves the problem.

and the link is http://www.mathlinks.ro/viewtopic.php?t=1934 :)


Suppose V is any vector space over field F, g:V->F is a linear form on V, and for some vector u in V, g(u)=1. Prove that

span{u} intersection with ker(g) = {0}
span{u} + ker(g) = V

I can prve the first one, but I am stuck on the second one. This is an assignment, so no solutions please, just a nudge in the right direction would be nice. Is there some theorem saying that the dimension of the kernal of alinear form in 1 less than the vector space on which it acts? Because that would be really nice, but it is probably just wishful thinking ^_^

Also, let me know if you are unfamiliar with any notation here, I am assuming that what I am learning is pretty standard, but I could be wrong.

[quote="KBriggs"]Is there some theorem saying that the dimension of the kernal of alinear form in 1 less than the vector space on which it acts?[/quote]

Yes (unless the linear form is identically $ 0$), but this is usually proven using the exercise you're doing, so it is easier to solve the exercise first and then conclude this.

You want to prove that $ \mathrm{span}\left\{u\right\} \plus{} \mathrm{ker} g \equal{} V$. This means proving that for every $ v\in V$, there exists a $ \lambda\in F$ and some $ w\in\mathrm{ker} g$ such that $ v\equal{}\lambda u\plus{}g$. How would you construct these $ \lambda$ and $ g$?

  darij

More generally, for $ g : V \to W$, if $ U \subset V$ is a set of linear independent vectors such that $ g(U) \equal{} \{g(u) \ \mid \ u \in U\}$ spans $ \textrm{Im}(g)$, then $ \mathrm{span}\left\{U\right\} \plus{} \mathrm{ker} g \equal{} V$.

Hey cool I got it

It actually was an exercise in the book, but the exercise didn't make the connection to this or prove the dimension thing, so I only thought of it when you laid it out like that.

Thanks :)


[b][color=darkblue]Prove that:[/color][/b] $ \boxed {\frac {1}{3} < \sin{20^0} < \frac {7}{20}}$

$ \sin x \equal{} x \minus{} \frac{x^3}{3!} \plus{} \frac{x^5}{5!} \minus{} \frac{x^7}{7!} \plus{} \ldots$

x is in radians on the left

[quote="TZF"][hide="Hint to Solution 1"]Consider the taylor series of sin x, x in radians.[/hide]

[hide="Solution 1"]For positive x, note that x-x^3 < sin x < x
Then, note that 1/3 < (pi/9)-(pi/9)^3 /6 < sin pi/9 < pi/9 < 7/20[/hide]

[hide="Hint to Solution 2"]
sin a > sin b
iff. a > b
iff. cis a is counterclockwise of cis b
iff. cis na is counterclockwise of cis nb

If a and b are not very close or n is particularly large, it becomes difficult to really say what counts as "is counterclockwise of". But if we can be certain that |nb-na| < 180º, then it makes sense.[/hide]

[hide="Solution 2"]Read the hint above. Let $ n \equal{} 9$, so that $ \text{cis } 20n ^{\circ} \equal{} \minus{} 1$. We'll first prove the left part.

$ \sin \alpha \equal{} \frac {1}{3} \implies \text{cis } \alpha \equal{} \frac {2\sqrt {2}}{3} \plus{} \frac {1}{3} i$
$ \implies \text{cis } 6\alpha \equal{} ( \text{cis } \alpha)^6 \equal{} \minus{} \frac {329}{729} \plus{} \frac {460 \sqrt {2}}{729} i$

which has argument less than 180 degrees, so $ \frac {1}{3} < \sin 20^{\circ}$

Similarly, if $ \sin \beta \equal{} \frac {7}{20}$, then $ (\text{cis } \beta )^9 \equal{} \minus{} \frac {319318497 \sqrt {39}}{2000000000} \minus{} \frac {152945093}{2000000000} i$

which has argument greater than 180 degrees, so $ \sin 20^{\circ} < \frac {7}{20}$

We could have avoided some of the computation by, say, dropping the denominators and just looking at the arguments. But generally, this is a computation-heavy approach.

An similar approach that will get you the result more quickly if you know the numbers are really close together is to, given sin a, find cos a, and then sin 3a, and then cos 3a, and then sin 9a. You can assume the cosine will be near -1, and just consider the sign of the sine.[/hide][/quote]

Thank you very much. I like way use:  $ x \minus{} x^3 < \sin{x} < x$


function
inequalities
inequalities proposed

Let $ a,b,c,d>0$.Prove that:
$ (ac\plus{}bd)^5\plus{}(bc\plus{}ad)^5\leq (a\plus{}b)^5(c^5\plus{}d^5)$

[quote="Inequalities Master"]Let $ a,b,c,d > 0$.Prove that:
$ (ac \plus{} bd)^5 \plus{} (bc \plus{} ad)^5\leq (a \plus{} b)^5(c^5 \plus{} d^5)$[/quote]
Let $ f$ is a convex function and $ x \plus{} y \equal{} 1,$ where $ x\geq0,$ $ y\geq0.$ Then
$ xf(c) \plus{} yf(d)\geq f(xc \plus{} yd)$ and $ yf(c) \plus{} xf(d)\geq f(yc \plus{} xd).$
Hence, $ f(xc \plus{} yd) \plus{} f(yc \plus{} xd)\leq f(c) \plus{} f(d).$
After substitution $ f(x) \equal{} x^5,$ $ x \equal{} \frac {a}{a \plus{} b}$ and $ y \equal{} \frac {b}{a \plus{} b}$ we obtain your inequality.

[quote="Inequalities Master"][b]Dr Sonnhard Grabner wrote:[/b]
[quote]hello, your inequality is equivalent with
$ 5ab(c \plus{} d)(c \minus{} d)^2(a \plus{} b)(a^2c^2 \plus{} a^2d^2 \plus{} 2abcd \plus{} abc^2 \plus{} abd^2 \plus{} c^2b^2 \plus{} b^2d^2)\geq0$ which is obviously true.
Sonnhard.[/quote]

Ok, how did you prove that:
$ (ac \plus{} bd)^5 \plus{} (bc \plus{} ad)^5\leq (a \plus{} b)^5(c^5 \plus{} d^5)\ \iff 5ab(c \plus{} d)(c \minus{} d)^2(a \plus{} b)(a^2c^2 \plus{} a^2d^2 \plus{} 2abcd \plus{} abc^2 \plus{} abd^2 \plus{} c^2b^2 \plus{} b^2d^2)\geq0$ :?:[/quote]

Maybe he got it before used to CAS  (Computer Algebraic Scientist :D) Recently time there are a lot of inequalities can be factorred very nice by CAS   :D  For example:
http://www.mathlinks.ro/viewtopic.php?t=219632
http://www.mathlinks.ro/viewtopic.php?t=215971
And a famous problem was proposed in IMO 2008  :P
Althought we can make it by hand if you really want but wasting time to do like that. Sorry for spam. It is just a my opinion about the powerful of CAS in solving inequality  :P

Can U show me CAS? I haven't seen it yet. :D


Let $ x_{1},x_{2},x_{3},x_{4}$ be the four digits of a four digit number $ N\equal{}x_{1}x_{2}x_{3}x_{4}$

Find the maximum value of $ \frac{N}{x_{1}\plus{}x_{2}\plus{}x_{3}\plus{}x_{4}}$

Alternatively, let the fraction be A.

$ A\equal{}1\plus{}\frac{999x_1\plus{}99x_2\plus{}9x_3}{x_1\plus{}x_2\plus{}x_3\plus{}x_4}$

We see that for maximum A, we have $ x_4\equal{}0$

$ A\equal{}1\plus{}9\plus{}\frac{990x_1\plus{}90x_2}{x_1\plus{}x_2\plus{}x_3}$

Same logic, $ x_3\equal{}0$

$ A\equal{}1\plus{}9\plus{}90\plus{}\frac{900x_1}{x_1\plus{}x_2}$

$ x_2\equal{}0$

$ A\equal{}1\plus{}9\plus{}90\plus{}900\equal{}1000$, equality for $ x_2\equal{}x_3\equal{}x_4\equal{}0$


geometry
3D geometry
tetrahedron
analytic geometry
vector
trigonometry
algebra

How would you figure out the integer coordinates of the 4th vertex of a tetrahedron if you know the other three?
Ex. (0,1,2), (4,2,1), (3,2,5), (x,y,z).

I don't think it is possible unless it's given that the tetrahedron is regular or isosceles.  (I don't know very many properties of 3D polyhedra, but I don't see how else you could get the fourth vertex.)

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

Wolfram Mathworld has some amazing resources as well.

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

Hehe, sorry, forgot to say that it's [b]regular[/b]

Think about it.  Try to visualize it in your head - what is the definition of a regular tetrahedron?

[hide="hint"]The distance between the points remains constant.[/hide]

[quote="TachyonPulse"][hide="Hint to another solution"]
You could do a vector bash using the formula $ uv \equal{} |u\parallel{}v|\cos{\theta}$ for vectors $ u$ and $ v$. You will end up with a system of equations with three equations and three variables.
[/hide][/quote]

That seems interesting.  What is a vector bash?  Could you explain the formula you provided or redirect me to a link?


modular arithmetic
algebra proposed
algebra

Let $ F(n)$ be the $ n$-th Fibonacci number. Prove that $ (F(F(1998)))^2\plus{}(F(F(1999)))^2\equal{}F(F(1997))\cdot F(F(2000))$.

this follows from Catalan's identity, $ F(n)^2\plus{}(\minus{}1)^{n\minus{}m\plus{}1}F(m)^2 \equal{} F(n\minus{}m)F(n\plus{}m)$. letting $ n \equal{} F(1999), m \equal{} F(1998)$ produces the desired result. ($ n\minus{}m\plus{}1 \equal{} F(1997)\plus{}1$ is even.)

Please post proof of the Catalan's identity.

obviously that's not the identity we're talking about here, soruz. oricum, ce faci frate!

if you're interested in a proof of Catalan's Fibonacci number identity, there's no point in me writing one up since a proof appears here: http://planetmath.org/?op=getobj&from=objects&id=6389

Thank you very much. Now, i'll post my solution.

It is easy see that $ F(n) \equal{} \frac {\alpha^n \minus{} \beta^n}{\alpha \minus{} \beta}$, here $ \alpha \plus{} \beta \equal{} 1$ and $ \alpha\beta \equal{} \minus{} 1$. This statement called as Binet's fomular.

We need prove $ \left(\frac {\alpha^{F(1998)} \minus{} \beta^{F(1998)}}{\alpha \minus{} \beta}\right)^2 \plus{} \left(\frac {\alpha^{F(1999)} \minus{} \beta^{F(1999)}}{\alpha \minus{} \beta}\right)^2 \equal{} \left(\frac {\alpha^{F(1997)} \minus{} \beta^{F(1997)}}{\alpha \minus{} \beta}\right)\left(\frac {\alpha^{F(2000)} \minus{} \beta^{F(2000)}}{\alpha \minus{} \beta}\right)$

This is equivalent to $ \alpha^{2F(1998)} \plus{} \beta^{2F(1998)} \plus{} \alpha^{2F(1999)} \plus{} \beta^{2F(1999)} \minus{} 2(\alpha\beta)^{F(1998)} \minus{} 2(\alpha\beta)^{F(1999)} \equal{} \alpha^{F(1997) \plus{} F(2000)} \plus{} \beta^{F(1997) \plus{} F(2000)} \minus{} \alpha^{F(2000)}\beta^{F(1997)} \minus{} \beta^{F(2000)}\alpha^{F(1997)}$

Note that the sequence $ \{F(n)\pmod{2}\}$ is $ 1,1,0,1,1,0,\cdots$, therefore $ F(3k \plus{} 1)\equiv F(3k \plus{} 2)\equiv 1\pmod{2},F(3k \plus{} 3)\equiv 0\pmod{2}\;\;\forall k \equal{} 0,1,2,\cdots$. And $ F(2000)\equal{}2F(1998)\plus{}F(1997),F(2000)\plus{}F(1997)\equal{}2F(1999)$.

We're done.


integration
inequalities
triangle inequality
real analysis
real analysis unsolved

by the triangle inequality and Holder's inequality,
\begin{eqnarray*}
\left|\int_I (f_n - f)gd\mu\right| &\leq& \int_I |f_n-f\parallel{}g|d\mu \\
&\leq&\|f_n-f\|_{L^1(I,\mu)}\|g\|_{L^\infty(I,\mu)} \\
&\to& 0
\end{eqnarray*}
the hypothesis that the $ f_n$'s are uniformly bounded is unnecessary.  did you make a mistake in the statement?


Let x,y,z be positive real numbers satisfy : x+y+z=3. prove that x1/2+ y1/2 + z1/2 >= xy+yz+zx.
I've a solution, and you?

We have that xy+yz+zx=[(x+y+z)^2-(x^2+y^2+z^2)]/2 and our inequality becomes x^2+y^2+z^2+2(x^1/2+ y^1/2 + z^1/2)>=9. But by applying the AM-GM inequality we have that x^2+x^1/2+x^1/2>=3 and so on and the inequality is proved.

\[ \sum(x^{2}+\sqrt{x}+\sqrt{x})\geq\sum3x=\sum x^{2}.
   Then       \sum\sqrt{x}\geq\sum yz \]

This is the solution that my friend gave me

Oooops I made a little mistake :blush:
 \[   \sum(x^{2}+\sqrt{x}+\sqrt{x})\geq\sum3x=(\sum x)^{2}.\]

I have two solution for this problem :
 * The first solution :
    $ xy + yz + zx = \frac{9-x^2-y^2-z^2}{2} $
  the ineq is equivalent to $ (2\sqrt{x}+x^2) + (2\sqrt{y}+y^2) + (2\sqrt{z}+z^2) \geq 9 $
    It's true because $ (2\sqrt{x}+x^2) \geq 3x $
  * The second solution :
  Using mixing variables (ugly solution ! )


You are given $50$ intervals on the number line such that no $8$ of these intervals have a real number in common. Prove that for some $8$ of them, no $two$ have a real number in common.

First, let's state [url=http://mathworld.wolfram.com/DilworthsLemma.html]Dilworth's theorem[/url]:

Given a poset with $ab+1$ elements, we can either find a chain with $a+1$ elements, or an antichain with $b+1$ elements.

Now let's consider a special kind of poset: the elements will be (closed, say) intervals on the real line. We set $[x,y]<[z,t]$ whenever $y<z$. Clearly, this is a partial order. Notice that a set of intervals is a chain iff no two intervals in the set intersect, and it's an antichain iff every two intersect iff all of them intersect. Dilworth's theorem applied to this special situation yields the following:

Given $ab+1$ intervals on the real line, we can either find $a+1$ no two of which intersect, or $b+1$ all of which intersect.

Your problem is the special case of the statement above, with $a=b=7$.


C2 (IRN) Let k be a fixed positive integer. A company has a special
method to sell sombreros. Each customer can convince two persons to buy
a sombrero after he/she buys one; convincing someone already convinced
does not count. Each of these new customers can convince two others and
so on. If each one of the two customers convinced by someone makes at
least k persons buy sombreros (directly or indirectly), then that someone
wins a free instructional video. Prove that if n persons bought sombreros,
then at most n/(k + 2) of them got videos.

look at the .pdf file here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2005+shortlist+imo&t=101128[/url]

page 11. :)

mamnoon amir.
man yek soale dige ham daram. in  soalam yeki az soalaye shortliste 2005 man hads mizanam ke hamvare namosaviye $N^{2}\geq 2^{mn+1}M$ dorost va bargharare.
 shomaha chi fekr mikonid? :?:
8. C3 (IRN) In an m× n rectangular board of mn unit squares, adjacent
squares are ones with a common edge, and a path is a sequence of squares in
which any two consecutive squares are adjacent. Each square of the board
can be colored black or white. Let N denote the number of colorings of
the board such that there exists at least one black path from the left edge
of the board to its right edge, and let M denote the number of colorings
in which there exist at least two non-intersecting black paths from the left
edge to the right edge. Prove that $N^{2}\geq 2^{mn}M.$


geometry
3D geometry
number theory
prime numbers

6. What is the sum of the distinct prime numbers which divide 257^4 + 514^4 ?   :maybe:

7. Each face of a cube is assigned a different integer. Then each vertex is assigned the sum of the integer values on the faces that meet at the vertex. Finally, the vertex numbers are added. What is the largest number that must divide the final sum for every possible numbering of the faces?      :fool:

I can't explain #1 but No. 2 isn't that hard.

Assume numbers $ a, b, c, d, e,$ and $ f$ for the cube.

Take $ 3$ adjacent sides (that's how you get a vertex). Call the first $ a$. We can have $ a \plus{} b \plus{} c, a \plus{} b \plus{} d, a \plus{} c \plus{} e,$ and $ a \plus{} d \plus{} e$. The same for the remaining side $ f$.

$ f \plus{} b \plus{} c, f \plus{} b \plus{} d, f \plus{} c \plus{} e,$ and $ f \plus{} d \plus{} e$.

Add these all together to get $ 4a \plus{} 4b \plus{} 4c \plus{} 4d \plus{} 4e \plus{} 4f$. Obviously, this number must be divisible by $ \boxed{4}$.

For number $ 1$, we get
\begin{align*}
257^4+(2\cdot 257)^4 &= 1\cdot 257^4 + 16\cdot 257^4
&= (1+16)\cdot 257^4
&= 17\cdot 257^4
\end{align*}

17 and 257 are the primes (both are actually fermat primes), so $ 17+257=\boxed{274}$

ARGH! I'm so dumb!! I didn't realize that the second number was twice as much as the first one. It would've been so easy then!!!

(I need to calm down :P )

#2 can probably be done more logically. Each face has four vertices that it contributes to, so the total sum must be divisible by 4, as the number from each face is counted four times.

thnx 5849206328x
it makes since now. with the breaking down and factoring


If $ y\equal{}3x^2\minus{}2$, what does $ \frac{dy}{d(2x\minus{}3)}$ equal?

Transforming $ y\equal{}3x^2\minus{}2$ into $ y\equal{}\frac{3}{4}(2x\minus{}3)^2\plus{}\frac{9}{2}(2x\minus{}3)\plus{}\frac{19}{4}$, then \[ \frac{dy}{d(2x\minus{}3)}\equal{}\frac{3}{2}(2x\minus{}3)\plus{}\frac{9}{2}\equal{}3x.\]

From the chain rule:

\begin{eqnarray*} \frac {dy}{d(2x - 3)} & = & \frac {dy}{dx} \cdot \frac {dx}{d(2x - 3)} \\
& = & \frac {dy}{dx} \cdot \left( \frac {d(2x - 3)}{dx} \right) ^{ - 1} \\
& = & 3x \end{eqnarray*}


Let $f(x)=3^x-2x-1$,suppose $a_1,a_2,...,a_{17}\in [0,2] $ satisfying:
$a_1+a_2+...+a_{17}=24$.
Prove that $\exists a_i,a_j,a_k$ such that $f(a_i),f(a_j),f(a_k)$ be three sides of the triagle

$f(x)\leq 0$ if $x\leq 1$ and f is increasing between 1 and 2.
$f(1)=0$
$f(1,17)=0,2760...$
$f(1,28)=0,5205...$
$f(1,45)=1,0184...$
$f(1,69)=2,0222...$
$f(2)=4$
now, if three of the $a_{i}$ are simultaneously in one of the four intervals (1,17;1,28),(1,28;1,45),(1,45;1,69),(1,69;2), then their images by f are the lengths of a triangle, because the image of each of these 4 intervals is an interval of the form (a,b), with $2a>b$.
Suppose that all four intervals (1,17;1,28),(1,28;1,45),(1,45;1,69),(1,69;2) contain only 2 or less $a_{i}$. Then, (0,1,17) contains 9 or more  $a_{i}$. In that case the sum of all the $a_{i}$ is lower than 9*1,17+2*1,28+2*1,45+2*1,69+2*2=23,27. Therefore, the sum is not 24, which is absurd. That is why one of the four intervals contains at least three $a_{i}$. Their images are the lengths of a triangle.


Prove that for all positive integers $n,m$, with $m$ odd, the following number is an integer
\[ \frac 1{3^mn}\sum^m_{k=0} { 3m \choose 3k } (3n-1)^k. \]

Let us fix $n$, and let
\[
P_m = \sum^m_{k=0} \binom{3m}{3k} (3n-1)^k
\]
we are required to show that $3^m n$ divides $P_m$ when $m$ is a positive odd integer.

Now either this is a very computationally heavy problem, or I'm just not using the right methods.

Let $\omega = \frac{-1 + i\sqrt{3}}{2}$ be the 3rd root of unity. Consider the coefficient of $x$ in $(1+x)^{3m} + (1+\omega x)^{3m} + (1+\omega^2 x)^{3m}$. By noting that $1 + \omega + \omega^2=0$, we see that the coefficient of $x^i$ is $0$ if $3 \nmid i$, and $\binom{3m}{i}$ if $3 \mid i$. So
\[
(1+x)^{3m} + (1+\omega x)^{3m} + (1+\omega^2 x)^{3m} = 3\sum^m_{k=0} \binom{3m}{3k} x^{3k}
\]
so
\[
P_m = \frac{A^m + B^m + C^m}{3}
\]
where $A= (1+ \sqrt[3]{3n-1})^3$, $B= (1+ \omega \sqrt[3]{3n-1})^3$, $C = (1+ \omega^2 \sqrt[3]{3n-1})^3$.

Suppose that $m \geq 7$. Using the identity
\begin{eqnarray*}
A^m + B^m + C^m &=& (A^2 + B^2 + C^2)(A^{m-2} + B^{m-2} + C^{m-2}) \\
&& - (A^2B^2 + B^2C^2 + C^2A^2)(A^{m-4} + B^{m-4} + C^{m-4}) \\
&& + A^2B^2C^2(A^{m-6} + B^{m-6} + C^{m-6})
\end{eqnarray*}
we find that
\[
P_m = (A^2 + B^2 + C^2)P_{m-2} - (A^2B^2 + B^2C^2 + C^2A^2)P_{m-4} + A^2B^2C^2P_{m-6}
\]

Now, computing, we get
\begin{eqnarray*}
A^2 + B^2 + C^2 &=& 3^3(n^2+6n-2) \\
A^2B^2 + B^2C^2 + C^2A^2 &=& 3^5(n^4-18n^3+33n^2-18n+3) \\
A^2B^2C^2 &=& 3^6 n^6
\end{eqnarray*}
so
\[
P_m = 3^3(n^2+6n-2)P_{m-2} - 3^5(n^4-18n^3+33n^2-18n+3)P_{m-4} + 3^6 n^6 P_{m-6}
\]
It follows that if $P_{m-2}$, $P_{m-4}$, $P_{m-6}$ satisfy the result, then so does $P_m$. It suffices now to show the three base cases:
\begin{eqnarray*}
\frac{P_1}{3n} &=& 1 \\
\frac{P_3}{3^3n} &=& n^2+27n-9 \\
\frac{P_5}{3^5n} &=& n^4+150n^3+355n^2-270n+45
\end{eqnarray*}
and we are done by induction.

hey, I had posted a solution for this one before and it dissapeared!

We can solve this problem by induction.
Used $(1+\sqrt[3]{3n-1})^{3m}$

Although it was written in Vietnamese , I think you coud understand it clearly

[quote="billzhao"]
$ P_m \equal{} 3^3(n^2 \plus{} 6n \minus{} 2)P_{m \minus{} 2} \minus{} 3^5(n^4 \minus{} 18n^3 \plus{} 33n^2 \minus{} 18n \plus{} 3)P_{m \minus{} 4} \plus{} 3^6 n^6 P_{m \minus{} 6}$
[/quote]

Or we can use:
With $ x$ denote for $ A,B,C$ we have $ (\sqrt [3]{x} \minus{} 1)^3 \equal{} 3n \minus{} 1\Rightarrow x \minus{} 3\sqrt [3]{x^2} \plus{} 3\sqrt [3]{x} \minus{} 1 \equal{} 3n \minus{} 1$
$ \Rightarrow x \minus{} 3n \equal{} 3\sqrt [3]{x}(\sqrt [3]{x} \minus{} 1)\Rightarrow (x \minus{} 3n)^3 \equal{} 27x(3n \minus{} 1)$
$ \Rightarrow x^3 \minus{} 9nx^2 \plus{} (27n^2 \minus{} 81n \plus{} 27)x \minus{} 27n^3 \equal{} 0$
$ \Rightarrow P_{m \plus{} 3} \minus{} 9nP_{m \plus{} 2} \plus{} (27n^2 \minus{} 81n \plus{} 27)P_{m \plus{} 1} \minus{} 27n^3P_{m} \equal{} 0$
Then OK by induction!

PS: We also can calculate $ A \plus{} B \plus{} C,AB \plus{} BC \plus{} CA,ABC$ to have $ P_{m \plus{} 3} \minus{} 9nP_{m \plus{} 2} \plus{} (27n^2 \minus{} 81n \plus{} 27)P_{m \plus{} 1} \minus{} 27n^3P_{m} \equal{} 0$


What is the difference between the sum of the first 20 positive even integers and the sum of the first 20 positive odd integers?

[hide="obvious"]Each of the evens will be 1 more than 1 of the odds. So, it's 20.[/hide]

[hide="easy solution"]Note that each even integer is one more than the corresponding odd integer, for a sum of 20.[/hide]

[hide="slow solution"]The sum of the first 20 even numbers is $2\left(\frac{(20)(21)}2\right)=420$ and the sum of the first 20 odd numbers is $20^{2}=400$, so the difference is $420-400=20$.[/hide]


Find all pairs $ (a,b)$ of naturals such that $ a^b\minus{}b^a\equal{}0$

It's very well known and old. The solutions are $ (1,1),(2,4),(4,2)$

I know it is an old problem, but I've been looking for the complete solution. I couldn't find it , so I posted the problem again. By the way, yeah that are the solutions. In fact, if $ a\equal{}b$ there are infinitely many solutions: (1,1), (2,2).......

[quote="aleksandar"]Find all pairs $ (a,b)$ of naturals such that $ a^b \minus{} b^a \equal{} 0$[/quote]
when we replace the condition $ a,  b$  integer by condition  $ a,  b$ is rational number.
problem will be very had


[quote="The Tech"]
May Zhou ...04 Dead, Found in Car Trunk
Police Think Suicide May Be Cause of Death

By Nick Semenkovich
STAFF REPORTER

Mengyao “May” Zhou ’04 was found dead in the trunk of her car last Thursday, Jan. 25, from an apparent suicide. The City of Santa Rosa Police Department is waiting on toxicology and other test results before declaring an official cause of death.

Zhou, an electrical engineering graduate student at Stanford University, was reported missing to the Stanford Police on Jan. 21. She was last seen on the evening of Jan. 20 leaving her residence to run errands. At approximately 3:30 a.m. on Jan. 25, an officer from the Santa Rosa Junior College Police Department located her 2006 Toyota in one of the student lots, according to a Santa Rosa Police press release. A deceased woman, later identified as Zhou, was found in the trunk.

There are “no early indications of foul play” and “items in the trunk indicate the subject may have committed suicide,” according to the Santa Rosa Police press release.

The Sonoma County Coroner’s Office completed an autopsy Friday, Jan. 26 which was inconclusive. According to another Santa Rosa Police press release, there were “no outward signs of trauma to the body.” Toxicology studies will take several weeks to perform.

Wechung M. Wang ’04, a Stanford classmate of Zhou’s, said that she was shocked at the news. “She seemed happy, healthy, and having a normal day,” describing the Thursday before she went missing.

The Santa Rosa Police were not available for comment and did not respond to repeated phone messages.

Yitong Zhou, May Zhou’s father, said to The Santa Rosa Press Democrat that the police had not released details of the crime scene and case to him. “I asked, they don’t want to tell me,” Yitong Zhou said to The Democrat. “They don’t want to tell me details because they haven’t released anything to the outside, they still want to study.”

“We’re making progress toward trying to have a better understanding of what occurred,” Santa Rosa Police Sgt. Paul Henry also said to The Democrat. “It’s a suspicious death … if we find evidence of foul play in the future, the suspect will be the only one who knows what’s in there.”

Zhou’s father is convinced that his daughter did not commit suicide. “I don’t believe it, because I’m a parent,” Zhou said to the The San Jose Mercury News. “I know her. I don’t think this is what she would do.” Her family suspects an abduction because Zhou did not like to drive long distances. Santa Rosa Junior College, where Zhou’s car was found, is some 90 miles away from Stanford.

In an interview with The Tech, Yitong Zhou said detectives had spoken with May Zhou’s younger sister about academic pressures, asking her questions about how their father would have reacted if May Zhou received an A-minus in a class. “They asked all these strange questions,” he said to The San Jose Mercury News. “My daughter was a straight-A student … I think they’re going the wrong direction. They’re looking for clues to support their hypothesis.”

Zhou graduated from MIT in 2004 with a degree in Electrical Engineering and Computer Science and obtained an MEng in EECS a year later.

[i]This article originally appeared in The Tech, issue 64 volume 126. It may be freely distributed electronically as long as it includes this notice but cannot be reprinted without the express written permission of The Tech. Write to archive@the-tech.mit.edu for additional details.[/i][/quote]

What do you think?  I have some thoughts on this but I'll save them until after others post.  I do find it very tragic though.

No I did not hear about it.

Anyway, I don't find that this is anything new, the pressure of forcing your children to do well in school has led to suicide before, and like all things, pushing your child to succeed is a good thing, but it must be done in moderation.

very tragic. it makes me sad as well.

i wonder how many individuals with suicidal thoughts actually seek help.

it hasnt been proved that it was suicide...it was kinda local, so ive read about it...sad.

it was in the local newspaper for me. Still I think it's pretty tragic, but not that much. I've heard worse cases. I do think it's suicide, because poisons were found in the trunk of the car...

[quote="moogra"]Still I think it's pretty tragic, but not that much. I've heard worse cases.[/quote]

Please explain.

[quote="moogra"]I do think it's suicide, because poisons were found in the trunk of the car...[/quote]

Well, I think the lack of evidence for a homocide is more compelling for a suicide than the presence of substances that one can commit suicide with.

[quote="drunner2007"]i wonder how many individuals with suicidal thoughts actually seek help.[/quote]
This is somewhat tangential, but I wonder if the Internet is actually good for suicide prevention. I know that some people claim PostSecret saved them from committing suicide. There are tons of suicide forums and chatrooms, where people go when they're considering suicide, and maybe the information they find and people they talk to would dissuade them.

On the other hand, the Internet provides a lot of information and ideas about how to do it, and it provides whole new reasons for suicide (I recall a few years ago some dude killed himself because of some bad thing that happened to him in Everquest).

I don't really know what sort of discussion the OP was hoping for so if this is way off, I'm sorry.

A random comment about heavy academic pressure: whether this works is very disputable. I have a feeling that most of the geniuses I've read about were educated in a (relatively) low-pressure environment (like Terence Tao, for example). I personally wasn't pressured at all either, and I think I'm doing fine. I don't know if I'd be better at academic stuff if my parents had been tyrants, but I suspect not much. This is because I've developed a mentality of "I can do whatever I want if I try hard enough, and if I fail it's because I was lazy or unfocused." On the other hand, I get the feeling that people whose parents pressure them all the time have a mentality of "I will do whatever my parents think I should, because I must, and if I fail I am worthless." This kind of motivation doesn't sound as effective as mine (to me). Perhaps it diminishes laziness and carelessness, but it surely diminishes creativity and enthusiasm too.

I tend to like tangential discussions as long as they're interesting and, well, [i]tangential[/i] (i.e. not way off base).  And, actually, I was hoping there would be discussion of academic pressures and how they relate to suicide.

Now, my thoughts on it.  From my limited information, it seems likely that she committed suicide.  I don't know [i]why[/i] she did, but there can be many reasons.

I don't really like how the media is portraying her dad.  They suggest he might be a major cause of her suicide (at least that's the implication I feel it's leading toward).  While he might have been a main cause of her pressure (I don't know, but usually parents can cause a lot of pressure on their children), I don't think the blame goes on him.  If you want to blame someone, put most of the blame the person who committed suicide.  It was her decision.  I know that sounds rough, but that's what I think.  Also, she was in graduate school.  If her father were bothering her too much it shouldn't be too hard to get away from him.

As for suicide in general, I don't think you can always predict suicide.  Sometimes you can, if a person is clearly not doing well.  However, suicide is just one event.  It might not take much for someone to pull the trigger, literally or figuratively.

[quote="joml88"]I don't really like how the media is portraying her dad.  They suggest he might be a major cause of her suicide (at least that's the implication I feel it's leading toward). [/quote]

This theme was raised not by the media but by the dad.  It is understandable, if tasteless, that (apparently only) the MIT Tech pursued this angle given its parallels to some other MIT-affiliated suicides.

her dad was definitely protrayed negatively...it was like the stereotypical "asian dads are so demanding" kinda crud...i'm not saying that he isnt or is like that.

[quote="pieterminate"]Someone without academic pressure (according to stanford prof.) would drive 90 miles into some small random college, crawl into the trunk, and then commit suicide. Odd... :huh:  Need some real police people.[/quote]

well although it seems like it is probably due to academic pressure, you can't really be sure.

there are other things that can make like rough (family problems, self-esteem problems, financial problems,...)

[quote="joml88"]Hm?  I believe pieterminate was saying that a stanford prof said she didn't have much academic pressure and that the circumstances make a suicide seem doubtful.[/quote]

oh ok. i interpreted what you were saying incorrectly pieterminate.

[quote="star99"]her dad was definitely protrayed negatively...it was like the stereotypical "asian dads are so demanding" [/quote]

The media did not portray him as positive or negative, but allowed him to paint a picture of himself through his own statements.  It's hard to say how the newspapers should have handled this, because although shattered by the events and probably not in his right mind, the father was also publicly speculating that it was murder before the police have finished their work.


algebra
polynomial
limit
real analysis
real analysis unsolved

$ f(x)$ is a polynomical and $ Deg(f(x))\equal{}2k$ with k is a negative number . Prove that We have a real number $ y$ such that : $ f(y) \leq f(x)$ with all value of $ x$

What is a polynomial of negative degree?

A point of terminology: you meant a "natural number", or "positive integer".

This is also false as stated; we need to require that the leading $ x^{2k}$ term has a positive coefficient. With that, look for a minimum on some interval $ [\minus{}M,M]$ and show that the values $ f$ takes outside that interval are all larger for large enough $ M$.

Sorry Everyone for some mistakes in my post >"< . And true problem is :
$ f(x) \equal{} x^{n} \plus{} a_{n}x^{n \minus{} 1} \plus{} ..... \plus{} a_{2}x \plus{} a_{1}$ with n is a multiple of 2 , n > 0 . Prove that we have a real number $ y$ such that : $ f(y)\leq f(x)$ with all value of x

Let $ N$ be an arbitrary number in the image of $ f$. Since $ \lim_{|x| \to \infty} f(x) \equal{} \infty$, we can find an interval $ [\minus{}M,M]$ such that $ f(x) > N$ for $ x$ outside this interval. By continuity, $ f$ attains a minimum on the compact set $ [\minus{}M,M]$ which obviously is a global minimum for $ f$ on the larger set $ \mathbb{R}$.

[quote="CommandoGuard"] Since $ \lim_{|x| \to \infty} f(x) \equal{} \infty$, we can find an interval $ [ \minus{} M,M]$ such that $ f(x) > N$ for $ x$ outside this interval. [/quote]

Cam you explain it ? i think isn't an evident clause  :)

We have a simple fact about even-degree polynomials together with a definition of what it means to go to infinity.  But since your question is completely vague and imprecise, it's impossible to know which of these parts you find confusing.


http://www.mathcamp.org/parents-whattobring.php

This recommends we bring $100 -$150.

Edit: I'm not using LaTeX right now!

If financial matters aren't an issue, you can certainly bring more than that; I don't think anything was stolen last year the entire camp.

There are occasional trips into the city to shops and various other places, so you'll definitely want a supply of spending money. There will most likely be ATMs on campus, though, so if you don't want to have large amounts of cash on hand you don't need to.


function
trigonometry
algebra unsolved
algebra

Find all functions $ f: \mathbb{R}\to\mathbb{R}$ that

$ y(x\plus{}y)f(x)\sin{x}\plus{}x(x\plus{}y)f(y)\sin{y}\equal{}xyf(\minus{}x\minus{}y)\sin{(x\plus{}y)}$

for all $ x$, $ y\in\mathbb{R}$.

[quote="V.V."]Find all functions $ f: \mathbb{R}\to\mathbb{R}$ that

$ y(x + y)f(x)\sin{x} + x(x + y)f(y)\sin{y} = xyf( - x - y)\sin{(x + y)}$

for all $ x$, $ y\in\mathbb{R}$.[/quote]

Strange problem :)

[hide="My solution"]

Let $ x,y$ such that $ xy(x + y)\neq 0$. Then dividing the equation by $ xy(x + y)$, we get $ g(x) + g(y) = g( - x - y)$ with $ g(x) = \frac {f(x)\sin(x)}{x}$

Let then $ a\neq 0$ : $ (a)( - \frac {a}{2})(\frac {a}{2})\neq 0$ $ \implies$ $ g(a) + g( - \frac a2)$ ${ = g( -a + \frac a2})$ $ = g( - \frac a2)$ and so $ g(a)=0$

So $ \forall x\neq 0$, $ g(x) = 0$ $ \implies$ $ \forall x\neq 0$ : $ \frac {f(x)\sin(x)}{x} = 0$ $ \implies$ $ \forall x\neq 0$ : $ f(x)\sin(x) = 0$ $ \implies$ $ \forall x\neq k\pi$ : $ f(x) = 0$

And it's easy to see that $ f(k\pi)$ may have any value.

So the solutions are :
Let $ h(x)$ any function from $ \mathbb Z\to\mathbb R$ :

$ f(k\pi) = h(k)$ $ \forall k\in\mathbb Z$
$ f(x) = 0$ $ \forall x\neq k\pi$

And it's easy to check that this solution indeed fits the initial requirements.

[/hide]


MIT
college
geometry
geometric transformation
rotation
puzzles

In a lateral thinking puzzle, you are given a situation and asked to explain it, without being given all the information. You are allowed to ask any number of yes or no questions.

[size=150]Puzzle 1[/size]

In one town, there are four shoe shops, which sell the same brands of shoes. Each one puts shoes out on display, which are occasionally stolen. All four shops take the same security measures. However, shoes are stolen from one of the shops much more often than the other three. Why?

[hide]
Three shops display left shoes, while the fourth displays right shoes. The thieves need to make pairs, so they steal three times as much from the fourth than from the first three.
[/hide]

Problem 2

A man drinks water because he doesn't have water. If he had water he would drink milk. What is his occupation?

Question about Number 2: Does the state of the water (solid, liquid, gas, plasma) affect the outcome of the problem (e.g. if you changed the states would the problem still work)?   :?

[hide="hint"]
It wouldn't.
[/hide]

I never heard of water being plasma but....

do conundrums qualify as "lateral thinking puzzles"??

This sounds familiar, but I don't remember the answer. Some questions:
1) Does he drink water from a bottle?
2) Does he drink water from a body of water (lake, river, etc.)?
3) Does he have milk?
4) Does he use water for his occupation?
5) Does he use milk for his occupation?
6) Does not "having" water mean not having it in his body?

1 and 2. Whether he drinks water from a bottle or from a body of water is irrelevant.

3. We don't know whether he does or not. It depends on other things.

4 and 5. Yes to both. Very much to #4

6. No.

[hide="Perhaps?"]The guys a farmer. If he had enough water, he would raise cows, so he could drink milk. But he doesn't have enough water, so he drinks water.[/hide]

If I'm right, I'll post a problem.

Yep. RC-7th, you've changed your avatar.

Post #3.

Zetonian library's new buildings are finally ready. Zeta, the head librarian, is a creation of A+Math, and therefore is very lazy. But he's also very thrifty. What is the cheapest and easiest (in terms of work) way for him to get the books from the old library to the new one?

Note: he does of a staff, including Bobbert, Caddlebot, and others, but they get tired after awhile.

This is a pretty vague question... he could ask people who borrow his books to return them to the other library, but that might take a long time.  Along those lines, he could wait for magic pixie dust to gather on the books and teleport them....

Mark Antony and Cleopatra were found dead in their villa, with only a smashed bowl of water. No one was there at the time they were killed. How were they killed?

[hide]Mark Antony and Cleopatra were goldfish. [/hide]

The music stopped, and everyone in the tent watched as the blindfolded man fell to his death. Why did he die?

1. Is this a circus tent?
2. Was the end of the music some kind of cue?
3. Was he a tightrope walker?

We also know that the object was ONCE a living thing or part thereof, it is either composed of plant or animal parts but is not a plant or animal. It could be something wooden, leather, fur, etc. It could be food of some sort. We know it's not pills or medication....maybe he lost his cane? Could it be some sort of ointment/adhesive? He has a physical problem that is easily noticeable, perhaps he's handicapped? What kind of pain would he experience as a result of not having this object? Would it be physical pain, or would it be emotional pain?

Sorry to necro post, but I read through the whole thread and G-UNIT's puzzle is driving me nuts.

So, G-UNIT, wherever you are, come back and tell us the solution? Please? :|

Heh...this problem has been up for 4 months and over 7 pages.  People have probably mostly forgotten about it or gotten bored of it, and are just randomly guessing  In any case, I would like to know the answer so the show can go on.

Did he have a physical defect (like no legs etc.) ?
                Or
Was it some physical characteristic(like fat,thin etc.) ?

 :mad:  :mad:  :mad: Pleaseeeeeee help G-UNIT

[quote="mathnerd314"]New question:

A man is in a cabin in the middle of the woods, dead. There were no signs of violence. How did he die?[/quote]

1) He starved.
2) He had a heart attack (or something like that).
3) He was in the cabin of a crashed plane (probably what you want).

I agree with number three. Except that there would be the signs of violence of an exploded plane, which I think would be pretty violent.

Grrr. That's right. :( :)

Any other problems? Any answers from G-UNIT?

To G-units problem

Maybe the man was starved to death. There is some food that was stored under the couch and if he looked he could have ate them.

Sorry if this was repeated.

The thing is not man-made, that is very important, I think. At the same time somebody put the thing under the couch. That leaves very little possibilities of what the thing is.
The man was alone when he killed himself but not before. So, who was the person? This might allow us to find out what the thing was.
Also, because the thing is not human-made, the medical condition of the person is probably that he is crazy. WAIT! The condition is important. It's something specific. Like epilepsy.(That's probably not the one)
First thing: find out the condition.

Oh yeah, and he killed himself. Something was wrong with his brain.
DRUGS!!!!!(As in  marijuana, etc.):D :D :D
They are not man-made;
The man was addicted, without  gettting them he killed himself;
Somebody brought them to him;
They were under the couch, so nobody else would see them.
This is the answer unless the "couch" part is important.

Btw, if the thing was man-made and the man had no health issues, the thing could be
[hide="Hilarious"]ALARM CLOCK ringing :rotfl: :rotfl: :rotfl: [/hide]

ok i got one.

A man is running down a very long hallway with big lights on the sides. The lights flicker for a second. The man stops running and begins to walk. Why?


Shouldn't any problem with "infinitely differentiable" as a hypothesis be in "College Playground > Calculus - Analysis"?


We call the sum of any $k$ of $n$ given numbers (with distinct indices) a $k$-sum. Given $n$, find all $k$ such that, whenever more than half of $k$-sums of numbers $a_{1},a_{2},...,a_{n}$ are positive, the sum $a_{1}+a_{2}+...+a_{n}$ is positive as well.

k must be $\geq\frac{n}{2}$ because otherwise we may take one negative number all the others positive so that the sum is negative and exactly $\binom{n-1}{k}>\frac{\binom{n}{2}}{2}$ k-sums are positive.

[quote="Baronov"]k must be $\geq\frac{n}{2}$[/quote]

Proof that it works?


A plane is taking a round trip flight from town A to town B, and back to town A.  With no wind, the total trip takes 2 hours and 48 minutes.  When there is a 100 mph wind that blows from town A to town B, the round trip takes 5 hours.  How far is it from town A to town B?

I tried setting up equations, but didn't get very far.  Help please!

[hide="Solution"]
Suppose $ A$ and $ B$ are $ d$ miles apart. With no wind, the plane goes at $ \frac{5d}{14}$ miles per hour.
Taking wind into account, $ \left(\frac{5d}{14}\minus{}100\right)5\equal{}\frac{25}{14}d\minus{}500\equal{}d$.

Solving gives $ d\equal{}\boxed{\frac{7000}{11}}$.
[/hide]

Sorry for being dumb, but where does $ \frac{5d}{14}$ come from?

$ \frac{d\text{ miles}}{2\cdot 60\plus{}48\text{ minutes}\cdot\frac{1\text{ hour}}{60\text{ minutes}}}\equal{}\frac{d}{\frac{14}{5}}\text{ mph}\equal{}\frac{5d}{14}\text{ mph}$.

Guess I should have mentioned that ;)

Are you sure you're reading correctly?  The trip is round-trip, not one way.

[hide]We'll solve this in general variables, to illustrate the procedure, and then substitute the values later.

Denote the plane's speed by $ v$, the speed of wind by $ w$, the distance by $ d$, and the given times by $ t_1$ and $ t_2$ respectively.

When there's no wind, the plane's effective speed is $ v$, hence $ {2d\over v}\equal{}t_1$

When there IS wind, the plane's effective speed is $ v\plus{}w$ going from A to B and $ v\minus{}w$ going from B to A, hence $ {d\over v\plus{}w}\plus{}{d\over v\minus{}w}\equal{}t_2$

This yields

$ {d(v\minus{}w\plus{}v\plus{}w)\over v^2\minus{}w^2}\equal{}t_2$

$ {2vd\over v^2\minus{}w^2}\equal{}t_2$

From the first equation, $ d\equal{}{vt_1\over 2}$, so we plug that in:

$ {v^2t_1\over v^2\minus{}w^2}\equal{}t_2$

$ {v^2\over v^2\minus{}w^2}\equal{}{t_2\over t_1}$

$ {v^2\minus{}w^2\over v^2}\equal{}{t_1\over t_2}$

$ 1\minus{}{w^2\over v^2}\equal{}{t_1\over t_2}$

$ {w^2\over v^2}\equal{}1\minus{}{t_1\over t_2}$

$ {w\over v}\equal{}\sqrt{1\minus{}{t_1\over t_2}}$

$ v\equal{}\frac{w}{\sqrt{1\minus{}{t_1\over t_2}}}$

Then from $ d\equal{}{vt_1\over 2}$ we get

$ d\equal{}\frac{wt_1}{2\sqrt{1\minus{}{t_1\over t_2}}}$

Plugging $ t_1\equal{}{14\over 5}, t_2\equal{}5, w\equal{}100$, we get

$ d\equal{}\frac{100\cdot{14\over 5}}{2\sqrt{1\minus{}{14\over 25}}}\equal{}{100\cdot 7\over\sqrt{11}}\equal{}{700\over\sqrt{11}}$[/hide]

Solution:
You know that the distances in this problem are equal. So if distance from A to B is d, thus from B to A.
The distance can be calculated by multiplying rate and time: d = s * t
2d = s * (2 + 48 /60)
2d = s * 14 / 5
d = s * 7 / 5
Thus the rates are ( s + 100 ) for with the wind of 100 miles per an hour and ( s - 100) for against the wind at 100 miles per an hour.
From A to B
d / (s + 100) = t1
From B to A
d / (s – 100) = t2
t1 + t2 = 5
d / (s +100) + d / (s -100) = 5
[(s -100)d + (s + 100)d] / [(s +100)(s - 100)] = 5
2sd / (s^2 – 100^2) = 5
14s^2 / 5(s^2 – 100^2) = 5
14s^2 / (s^2 – 100^2) = 25
s^2 / (s^2 – 100^2) = 25 /14
(s^2 – 100^2) / s^2 = 14 /25
1 – 100^2 / s^2 = 14 / 25
100^2 / s^2 = 1 – 14 / 25
100 / s = sqrt(1 – 14 / 25)
From previous equation d = s * 7 / 5, thus s = 5d / 7
100 * 7/ 5d = sqrt(1 – 14 / 25)
100 * 7 / sqrt(11) = d

Answer: Distance from town A to town B is 700 / sqrt(11) miles.

To learn more on Word Problems, you may click the link below and go through video lesson on ‘Time and Distance’:
http://www.winpossible.com/lessons/Algebra_I_Other_common_types.html


ESPN's announcers repeatedly refered to world cup group E, which contains the Czech Republic, Italy, Ghana and the United States, as the "Group of Death."  There were also suggestions during the Argentina-Ivory Coast game that group C, which also is home to the Netherlands and Serbia and Montenegro, might be the "Group of Death."

Suppose that you had accurate information about the quality of every team in the World Cup.  How would you determine which group was the Group of Death?  My initial thought was that the group with the best third-best team would be a good candidate.  On further thought, I decided maybe we should take the quality of the second-best team into more serious account, so say we take as our measure of deathiness one part the quality of the second-best team, two parts the quality of the third-best team in the group.

What do you think?

For those who don't know, the first round of the World Cup has 32 teams, broken into 8 groups of 4 teams.  In each group, every team plays every other.  A win earns 3 points, a tie earns 1.  The best two teams advance.  In case of a tie, the head-to-head result is the first tie-breaker and goal differential is second.

Hmm, could you explain your reasoning for giving weight to the 2nd and 3rd best teams?  I can see why the top team would not be that crucial, but why not the 4th?

Because really the 4th team isn't going to qualify anyhow.  :)  Maybe this is not such a good reason, after all.

My original feeling for choosing just the third-best team's strength was that then I would be choosing the group which would deny a place to the strongest team.  But, of course, we might have teams ranked 15, 16, 17 among the group in which case the fact that the third team is very strong for a third team is totally offset by the fact that the first and second teams are very very weak for first and second teams.  So qualifying in this group would in fact not be so very hard, because basically you would need a win and a couple of ties against only middle-ranked teams.

Maybe the following is a better approach: create a function $f(a, b, c)$ which takes the skills of three teams and spits out the difficulty qualifying for a 4th team.  Then add up the four values this gives you when you apply it to each triple of teams in the group.  Of course, we'd still need a good idea of what a decent function $f$ is, but this is at least more "scientific" then my other proposals.

[quote="shadysaysurspammed"]Someone tell me the exact ankins of the teams JBL mentioned.Czech is like 3rd and italy isn't far behind.If 2 teams have a real close top-10 ranking,then they would be my group of death thing because maybe a lot of expectation is there on these teams rather than having 3 consistent ood btu not super good teams.[/quote]

Well, one problem with soccer is that the world rankings don't seem to be very reliable.  You can get the world rankings for a month of your choice via the [url=http://www.fifa.com/en/mens/statistics/index/0,2548,All-May-2006,00.html]FIFA website.[/url]  In May's ranking, Czechs are second, US is tied for fifth and Italy is 13th.

I don't think having two good teams is meaningful for choosing a hard group.  After all, look at group D, and suppose world rankings are good predictors of team success.  Then Mexico (4th) and Portugal (7th) would seem to have a tough group lined up.  But the other two teams, Iran and Angola (respectively 23rd and 57th) shouldn't stand a chance against either of them.  So there shouldn't really be any competition to qualify in that group, thus not a Group of Death.  Having two closely-ranked good teams will lead to some excellent first-round play, perhaps, but it doesn't necessarily mean the group itself is difficult.

oh yeah i forgot the teams play eaach other once in the 1st round,I think the 3rd team is decisive then.

Well, you call the group of the death when there are 3 teams that have good chances to pass, and the 4º one can be a surprise e.g. african teams....... for me the grop of the death is C

Well let see the Group of Death:

Argentina 6
Netherlands 6
Ivory Coast 0
Serbia and Montenegro 0

It didn't show us that it was THAT difficult (I mean, 6-0 in a death group)

And these group:

Italy 4
Czech Republic 3
Ghana 3
USA 1

Nice definition! I'll say that Italy and Czech Republic will make the 2º round (2 ties at the 2 games left)

If you're judging by World Rankings and the 2nd and 3rd teams only, then Group F would be an obvious Group of Death with Croatia and Japan and so far, looking at the results, it is because Australia is 1st, although they aren't very accurate as Australia got lucky in 8 minutes and Brazil was horrible.

The group of death the one of Brazil???? Come on!! Japan does not play well, and, in my opinion, they'll finish in 4º place. Then, Brazil is going to win the group, and between Croatia and Australia, there is the 2º place.... I mean, group F is not as difficult as E man, Australia and Japan have nothing, and Croatia just a bit to show....


it's really nice problem.  It's equivalent of:

Prove that \[ \sum_{i\equal{}1}^{n}\frac{i}{2^i} < 2 \]

[quote="matex (L)(L)(L)"]Could you explain more clearly?
$ \sum_{i\equal{}1}^{n}\frac{i}{2^i} < 2$[/quote]
It simply follows from, \[ S\equal{}\sum_{i\equal{}1}^\infty \frac{i}{2^i} \Longrightarrow 2S\minus{}S\equal{}S\equal{}\sum_{i\equal{}1}^\infty \left (\frac{i}{2^{i\minus{}1}} \minus{}\frac{i\minus{}1}{2^{i\minus{}1}}\right ) \equal{}\sum_{i\equal{}1}^\infty \frac{1}{2^{i\minus{}1}}\equal{}2 \]
So,$ \sum_{i\equal{}1}^{n}\frac{i}{2^i} < 2$

[quote="Enygma"]But why
\[ S = \sum_{i = 1}^\infty \frac {i - 1}{2^{i - 1}} )
\]
in this:
\[ 2S - S = S = \sum_{i = 1}^\infty  (\frac {i}{2^{i - 1}} - \frac {i - 1}{2^{i - 1}} )
\]
[/quote]
Here, what I did was nothing but telescoping the sum.
OK,here I am showing it clearly,
I took, $ S = \frac {1}{2} + \frac {2}{2^2} + \cdots$. Now, $ 2S = \frac {1}{2^0} + \frac {2}{2^1} + \cdots$

So,$ 2S - S = S = \frac {(1 - 0)}{1} + \frac {(2 - 1)}{2^1} + \frac {3 - 2}{2^2} + \cdots$
I hope that it is clear now :)

EDIT: and most importantly, $ S = \sum_{i = 1}^\infty \frac {i - 1}{2^{i - 1}} )$ does not change $ S$ as you see, at the first step, $ i-1=1-1=0$ :)


This program solves the "Your Ride is Here" problem, but how can I edit it so that it is submittable, i.e. it uses the provided inputs?
[hide][code]/*
ID: insert_id_here
PROG: ride
LANG: C++
*/

#include <iostream>
using namespace std;

int word_total()
{
  char in;
  int k, total=1;
  cin.get(in);
  while(!cin.fail() && in!='\n')
  {
    k = in - 'A' + 1;
    total *= k%47;
    total = total%47;
    cin.get(in);
  }
  return total;
}

int main()
{
  int word1, word2;
  word1 = word_total();
  word2 = word_total();
  if(word1 == word2)
    cout << "GO";
  else
    cout << "STAY";
  return 0;
} [/code][/hide]

your program needs to read from the file ride.in and write to the file ride.out

put #include<fstream> at the top of your program, then put the following in main():
[code]
ifstream fin("ride.in");
ofstream fout("ride.out");
[/code]
then use fin, fout instead of cin, cout.

This [i]is[/i] explained on the training pages...read the text "submitting solutions."


a) Calculate I+J
b) Calculate I-J (using cos2x=cos^2x-sin^2x)
c) Calculate I and J

[img][url=http://imageshack.us][img]http://img246.imageshack.us/img246/7457/int1cd6.gif[/img][/url][/img]
[img][url=http://imageshack.us][img]http://img356.imageshack.us/img356/1923/int2vq2.gif[/img][/url][/img]

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=141378]here[/url] for a formula that may be useful for b. The rest is trivial.

for c, 0.5((I+J)+(I-J))=I


Post a random word! (this is gonna get locked, but it is fun!)

OK, this is it. Math92 you will no longer be able to reply in this forum for 7 days. Hopefully that will cool you off a little bit from spamming randomly :)


#13, 2001 State Sprint

[img]http://i84.photobucket.com/albums/k25/Dim_04/13State2001Sprint.jpg[/img]

[hide]Sqrt2/2[/hide]

Let the width be x, the length be y.
Then, x/y = y/(x/2)
x/y = 2y/x
x^2 = 2y^2
y^2/x^2 = 1/2
y/x = 1/sqrt2
In SRF (simplest radical form) this is sqrt2/2.

I get srt2/2 because I go to your step where
x^2=2y^2 and divide both sides by y^2
to get x^2/y^2=2, and sqrt the entire equation.
therefore x/y=srt 2, and flipping them around gets you
y/x=srt2/2.  It's easier to do divide y^2 instead of skipping
four steps as you did.  Good job though.

[hide="LaTeX anyone?"]
For clarity, let the original length be $ l_{1}$ and the original width $ w_{1}$. Similarly, the folded rectangle's length can be $ l_{2}$ and the width $ l_{2}$. Since a folded rectangle can only be similar if its longest side is folded, we have the proportion

$ \dfrac{l_{1}}{w_{1}} = \dfrac{l_{2}}{w_{2}}$

The problem and diagram give the facts that $ l_{1} = 2w_{2}$ and $ w_{1} = l_{2}$. Substituting gives

(Note: Most of the above can be done by intuition, eliminating the need for extra variables, etc.)

$ \dfrac{2w_{2}}{l_{2}} = \dfrac{l_{2}}{w_{2}}$.

So

$ \dfrac{w_{2}^{2}}{l_{2}^{2}} = \dfrac{1}{2}$

Our answer is $ \sqrt {\dfrac{1}{2}}$

$ = \dfrac{\sqrt {2}}{2}$.[/hide]

myspacesucks, you make absolutely no sense. Your solution is no faster, and uldivad9's is not flawed (besides the fact that neither of you explained your original proportion very clearly).

I really don't understand how you got the two equations you said could be done by intuition. How could the second length and the first width be equal? Since the second length is the same as the first one, wouldn't that make the first rectangle a square? And if that's true, they couldn't be similar...

Mind explaining where I went wrong?

the two rectangles were similar, so the sides of each rectangle are proportional to one another. So if l is the length and w is the width of the original rectangle,

l/w=w/(1/2)l since the length is folded in half.


I was wondering, if I won't be able to make it to the time of the online class, would I have access to the transcripts of the session after class like one can in Math Jams? Thanks.


[color=green]Removed by mod on request[/color]

#1 is a very well-known problem.

[hide="A solution sufficient for the problem itself"] Let $ f(x) \equal{} \frac {\ln x}{x}$.  The problem is equivalent to showing that there exist an infinite number of ordered pairs $ (x, y)$ such that $ f(x) \equal{} f(y)$, which is obvious because the graph is concave down and has a maximum. [/hide]
[hide="A better solution"] The set of rational solutions is given by $ x \equal{} \left( 1 \plus{} \frac {1}{n} \right)^n, y \equal{} \left( 1 \plus{} \frac {1}{n} \right)^{n \plus{} 1}$ for $ n \in \mathbb{N}$.

The set of real solutions is given by $ x \equal{} k^{\frac {1}{k \minus{} 1} }, y \equal{} k^{\frac {k}{k \minus{} 1} }$ for $ k \neq 1$.  [/hide]

[hide="Lemma for 2"]Lemma: If we have a set of $ n \geq 3$ sticks, with the smallest length $ 1$, and the longest length less than or equal to $ F_n$, then $ 3$ of them can form a triangle. ($ F_n$ is the $ n$th Fibonacci number with $ F_1 \equal{} F_2 \equal{} 1$)

Proof: By contrapositive. Let the stick lengths be $ a_1, a_2, \ldots a_n$ with $ 1 \equal{} a_1 \leq a_2 \leq \ldots \leq a_n$. We show that if no triangle can be made, then $ a_n \geq F_n$. We know $ a_1 \geq 1 \equal{} F_1$ and $ a_2 \geq 1 \equal{} F_2$. Then all other sticks have length greater than $ a_1 \plus{} a_2 \geq F_1 \plus{} F_2 \equal{} F_3$. So $ a_3 > F_3$. Then all other sticks have length greater than $ a_2 \plus{} a_3 \geq F_2 \plus{} F_3 \equal{} F_4$. So $ a_4 > F_4$. In this way we prove $ a_k > F_k$ for $ 3 \leq k \leq n$, and we're done.[/hide]

But I think that lemma implies that the problem is incorrect as stated. I might be wrong.

Also prove that i) $ x<e<y$ and ii) $ xy>e^2$


I have the full MikTeX distribution installed, then when I tried to use \looparrowdownright, it requires the  mathabx package. When compiling gives the error: mathabx.sty not found.

I don't understand, if I have full MikTeX distribution installed, why this happens?

mathabx is not part of the MiKTeX distribution so isn't installed. There are thousands of packages out there and no LaTeX distribution can provide all of them.
You can find and install mathabx from [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=mathabx]here[/url].

Okay thanks, but I couldn't successfully install it. I downloaded mathabx.sty & mathabx.dcl, then I placed those files in \miktex\tex\latex\mathabx and I hit the "Refresh FNDB" button, but when compiling, it appeared lots of errors.

Can you tell me how to install it correctly?

You haven't installed the [url=http://tug.ctan.org/tex-archive/fonts/mathabx/source/]fonts[/url]. These are included in the download [url=http://tug.ctan.org/tex-archive/fonts/mathabx/]mathabx.zip[/url] so not sure why you ignored them. The readme file says:
[quote]Suggested TDS location for all ``.mf'' files:
---------------------------------------------
TEXMF/fonts/source/public/mathabx[/quote]


geometry
geometric transformation
reflection
geometry proposed

Let $A$, $B$, $C$, $D$, $E$, $F$ be six points on a circle such that $AE\parallel BD$ and $BC\parallel DF$. Let $X$ be the reflection of the point $D$ in the line $CE$. Prove that the distance from the point $X$ to the line $EF$ equals to the distance from the point $B$ to the line $AC$.

$AE\parallel{BD}$=>$AB=DE$
$BC\parallel{DF}$=>$BF=CD$=>$\angle DEC=\angle FEB$

$\angle FEX=\angle FEC-\angle XEC=\angle FEC-\angle DEC =\angle FEC-\angle FEB=\angle BEC=\angle BAC\Rightarrow \angle FEX=\angle BAC$

Let $B_1$ is feet of perp. from the point $B$ to the line $AC$ and $X_1$ is feet of perp. from the point $X$ to the line $EF$.
$\angle X_1EX=\angle FEX=\angle BAC=\angle BAB_1$=>$\angle X_1EX=\angle BAB_1$
But we else know that $AB=DE=EX$=>$\triangle{ABB_1}=\triangle{XEX_1}$=>$XX_1=BB_1$  :)


:blush:
last prob was easy yall got it right, rotflol.
Here's a bit harder one.

pi+po=20
pe+pp=35
pa=pi+3
a=sqrt(36)
p=5

Q: What is the value of ((E/P)(O+I)/O)+A???
please post answer below...

This might be better off in the classroom math forum...
[hide="Solution"]
Well, from $ p\equal{}5$, we have (line 2)
$ 5e\plus{}25\equal{}35$
$ e\equal{}2$
We also have $ a\equal{}6$, so from line 3...
$ 30\equal{}5i\plus{}3$
$ i\equal{}27/5$
Now... (line 1)
$ 27\plus{}5o\equal{}20$
$ o\equal{}3/5$
Now we can just plug in and get the answer (I'm too lazy to do it right now)[/hide]

The answer is 34/7.

The facts are:
"pi+po=20
pe+pp=35
pa=pi+3
a=sqrt(36)
p=5

Q: What is the value of ((E/P)(O+I)/O)+A???"

So right away we know that p=5.

We also know that a=6 since 6 is the square root of 36.

In the equation pe+pp=35, by plugging p as 5, the equation becomes 5e+25=35.  Therefore, 5e=10 and e=2.  B
y plugging p as 5 in the equation pa=pi+3, the equation becomes 5a=5i+3.  By plugging a=6 into the newly found equation, 30=5i+3.  Therefore, 5i=27 and i=27/5.

By plugging p as 5 in the equation pi+po=20, the equation becomes 5i+5o=20.  Dividing by 5, i+o=4.  By plugging in i=27/5 into the newly found equation, (27/5)+o=4.  Therefore, o=(20/5)-(27/5) and o=-7/5.

Back to the original question, the value of ((e/p)(o+i)/o)+a can be broken down by solving for e/p and (o+i)/o.
e/p=2/5
(o+i)/o=1+(i/o)=1+((27/5)/(-7/5))=1+(27/-7)=(7/7)+(-27/7)=-20/7.
Therefore, the equation becomes ((2/5)(-20/7))+a, or -40/35+6.
Finally, the equation becomes (-40/35)+(210/35)=170/35=34/7

Of course, I solved the problem in fewer steps than this, but I made it step-by-step in case anyone out there doesn't follow what I'm doing when I explain how I solved it.

Edit:  By the way, I think this problem should go to Classroom Math since it's not really a MathCounts problem.

I also noticed that the problem either has a major flaw or is an elementary school problem.
This is because when you stated the fact that a=sqrt(36), the value of a can be either 6 or -6, since either number squared is 36.
You also said "What is the value of..." in the question, which means that there is only one value/answer to the problem.  This is impossible since the variable a has two separate values which therefore results in two separate answers at the end of the problem.
Therefore, as I stated earlier, this problem either has the major flaw of ignoring the fact that a=-6 is possible, or the problem is an elementary school problem (which is understandable when the fact is considered that elementary school problems are so easy that the problem solver is not supposed to know that a=-6 is possible).

Gah  :ewpu: I didn't see the $ p\equal{}5$ and ended up going in circles...


S is an infinite set of points in the plane. The distance between any two points of S is integral. Prove that S is a subset of a straight line.

Maybe this can help you:
[url]http://mathcentral.uregina.ca/mp/previous2002/oct02sol.html[/url]


Sorry, I don't know how to use an equation editor or something in here, can anyone teach me?
The numbers beside the logs are their bases..

Find all x that satisfies:

log5 x log4 x log3 x = log5 x log4 x + log5 x log3 x + log4 x log3 x .

Thanks in advance~

Put your math around dollar signs.

[code]$\log_{5}x$[/code]
This produces $\log_{5}x$.

[quote="AMPIStiffler"][hide]$(\log{x})^{3}= (\log{x})^{2}\log{3}+(\log{x})^{2}\log{4}+(\log{x})^{2}\log{5}$

$\log{x}= \log{3}+\log{4}+\log{5}$[/hide][/quote] Careful! You missed a solution. :D

[hide]Yeah, the not so obvious but obvious "one" you missed[/hide]

[hide]

if $x=1$ then the equation would become $0\bullet0\bullet0=0\bullet0+0\bullet0+0\bullet0$

[/hide]


probability
number theory
least common multiple
modular arithmetic

How would u do this problem?

If aliens abduct someone from Spokane once every 3 days, someone from Seatlle, once every 8 days, and someone from Moses Lake once every 4 days, what is the probability that on a given day, someone is abducted from Spokane and Seattle, but not form Moses Lake?

[quote="amcavoy"][hide]We seek $x<\textrm{lcm}\left(3,4,8\right)=24$ such that $x\equiv 0\bmod 3$, $x\equiv 0\bmod 4$, and $x\not\equiv 0\bmod 8$.  The only such value is $x=12$.  Therefore the probability is \[{\textrm{P}=\frac{1}{\textrm{lcm}\left(3,4,8\right)}=\boxed{\frac{1}{24}}}.\] [/hide][/quote]

I think you misread the problem.  It should be [hide]$x<\textrm{lcm}(3,4,8)=24$ such that $x\equiv0\pmod3$, $x\equiv0\pmod8$, and $x\not\equiv0\pmod4$.  There are no values that fit these criteria, so the answer is $\boxed{0}$[/hide]

Yes I did, thanks for the correction.  My answer would give the probability that someone is abducted from Spokane and Moses Lake, but not Seattle.

[hide]The answer should be 0, since how can a number be a multiple of 8 without being a multiple of 4.[/hide]

Since this is a probability question, it was probably meant as "there is a $\frac{1}{3}$ chance of being abducted in ... "

This is more modular arithmetic than probability.

Yeah, but the modular arithmetic is so simple, you don't even really need to use the mods.  All you need to know is that there's no multiple of 8 that's not a multiple of 4...

Btw, lotrgreengrapes, the probability is $0$, what else could it be asking for?

Since it was labeled probability question, [hide="Possible Interpretation"] $\frac{1}{3}\cdot\frac{1}{8}\cdot(1-\frac{1}{4})=\boxed{\frac{1}{32}}$ [/hide]

[hide="answer"]The sample space is a set of 24 days. Every 24 days, the pattern of abductions repeats itself. So we just need to look at the probability for a set of 24 days.

Since Spokane and Seattle meet every day a multiple of the GCM of 8 and 3, then every 24 days these abductions intersect. However, on every 24th day, Moses Lake abduction occurs because 24 is a multiple of 4 days. So these events are not exclusive and thus the probability of them occuring exclusively is $\boxed{0}$.[/hide]

btw, to above post, I don't think that those probabilities can be treated as independent events because the chance of Moses lake occuring if seattle occurs is 1, not .25

Even if it was not intended to be a probability question, who says the aliens have to start abducting on the same day?  :P

How would you find an answer to that, lotrgreengrapes7926? Would the mods be like:
$1\equiv\pmod{4}$
$3\equiv\pmod{8}$

Wouldn't you have to use the CRT?

If they don't start abducting on the same day, we have to be told how many days apart they started abductions in each city, sor there isn't enough info to solve the problem.

The only logical way to go about this problem with the information given is to assume that the abductions started on the same day...


A train, traveling at constant speed, takes 20 seconds from the time it first enters a tunnel that is 300 meters long until it completely emerges from the tunnel. One of tthe stationary ceilng lights in the tunnel is directly above the train for 10 seconds. Find the length of the train.

This sounds like an aime or amc question, so this is just a quick solution.

[hide]let the length of the train be x meters. It takes 10 seconds for the train to pass one certain point (the light). So the speed of the train is $ \frac{x}{10}$ meters per second. Hence, $ \frac{x\plus{}300}{20}\equal{}\frac{x}{10}$ and multipling 20 on both sides, we have x=300 meters.[/hide]


function
integration
Functional Analysis
real analysis
real analysis unsolved

Let $ C([0,1])$ Banach space of functions $ f: [0,1]\rightarrow \mathbb{C}$ with $ \parallel{}\cdot\parallel{}_\infty$ norm. Let $ S\equal{}\{f\in C([0,1]): \int_0^{\frac{1}{2}} f(t)dt\minus{}\int_{\frac{1}{2}}^1 f(t)dt\equal{}1\}$. Prove that there doesn't exists function $ f_0\in S$ such that $ \parallel{}f_0\parallel{}\leq \parallel{}f\parallel{},\forall f\in S$.
Any hints are welcome.

[quote="Mr.Doe"]Let $ C([0,1])$ Banach space of functions $ f: [0,1]\rightarrow \mathbb{C}$ with $ \parallel{}\cdot\parallel{}_\infty$ norm. Let $ S \equal{} \{f\in C([0,1]): \int_0^{\frac {1}{2}} f(t)dt \minus{} \int_{\frac {1}{2}}^1 f(t)dt \equal{} 1\}$. Prove that there doesn't exists function $ f_0\in S$ such that $ \parallel{}f_0\parallel{}\leq \parallel{}f\parallel{},\forall f\in S$.
Any hints are welcome.[/quote]A minimizer in $ L^\infty(0,1)$ is given by
\[ f_0(x) \equal{} \begin{cases} 1 \quad (x < \frac{1}{2}) \\ \minus{}1 \quad (x > \frac{1}{2}) \end{cases}\]
But of course this function isn't continuous ... :wink:


logarithms AMC USA(J)MO USAMO probability AIME ratio

Hey guys, how many stupid mistakes did you make?

I made a total of four. One on the AMC 10 (used diameter instead of radius), and three on the AIME. Of course, I missed #1 with 807. I missed #4 (119, should've been 177) by a answer-to-answer-page transcription error. I missed #6 by 1 (413 instead of 412) because the fourth root of 625 doesn't work...

Which left me with an index of 195. So had I not messed up any of those, my index would be 231. Big difference.

Had I not mistranscribed #4........................

 :mad:  :mad:  :mad:

Hmm I said $ \log (1\plus{} something) \equal{} 1 \plus{} \log(something)$ for the sequence one... then I proceeded from there and got 65... :wallbash:
Oh and I forgot about the <9999 condition for #9 and forgot to subtract 3 from 15  :noo:

I said there were 24 even numbers from 0 to 49...

One of my mistakes was so stupid. For #8, I found N as the sum of all the sums, not differences...what a waste of time because I did this by patterns and brute calculation. I only realized after the test was over.

I didn't make any stupid mistakes, and I didn't have to catch any during the test (although I did spend some time checking my work anyway).  The key for me was that throughout the test, I told myself to read the problems carefully and work slowly.  I read the problems insanely carefully, too.  :lol:

The tradeoff is that I didn't get enough time to solve all the problems, but anyone going for USAMO or a more modest goal shouldn't care about having time for all the problems anyway.

Hmmm...i failed worst?

#3. i got probability=$ 8C3/p^3(1\minus{}p)^5$ instead of $ 8C3p^3(1\minus{}p)^5$ so everything was flipped...
#6. $ 4^4\equal{}6$ most definitely...
#7. I misread as $ 5^{(a_n\minus{}a_{n\minus{}1})}$ instead of $ 5^{(a_{n\plus{}1}\minus{}a_{n})}$

I didn't really improve my math skills much over the past year, but I cracked down on dumb mistakes big time. Only made one arithmetical error in adding all of the differences of powers in #8, but other than that, I'm surprised at the number of ones that I thought I had right that actually were. 7/8. I wasted too much time fuddling with 13 and 14. I had brain dying syndrome where I just couldn't think and my eyes were all out of focus from staring at the test for 3 hours straight and there were distracting people outside the conference room were I was taking the test yelling really loud and just being distracting. Then there was no clock in the room because I had to switch rooms the instant before the test started because there was a meeting in there later in the day so I go over to the adjacent conference room and then I fuddled too long with crazy multiplication with 15 digits and the proctor is an intern for the other one who is on maternity leave and probably forgot to mail it in because I took the test at 12:00 and she's like "The mail already went out I'll mail it tomorrow" but I doubt she mailed it today and then they didn't excuse me from any of my classes that day and it just didn't go well at least I was the only person from my school who took it because then the pain wasn't dually bad.

Thank you for your time that is all.

Stupid Mistakes?

138 on 10B, should have had 150.  I can't multiply anymore, apparently.  Nor can I read.

6 on AIME, should have had 8 or 9.  I fail at arithmetic.

So 198 index instead of 230+, which makes me sad.

I made 1 careless error on the AMC 12 and 3 on the AIME.[hide]
On #25 on the AMC 12, I got 23+12 is congruent to 23 mod 24. However, I was slow enough to not answer #25 before time was called, so it had no effect on my score.
On #3 on the AIME, I forgot to take the square root of 25.
On #9, I forgot about permutations of A, B and C.
On #12, I got (37^2)/3 instead of 37/3 for the length of the tangent from I to the circle.[/hide]
Oh well... I hope I still have a chance to make USAMO...

135/5=45

0 is not "nonnegative"

13 and 31 are the same things.

1/cot = cot

I'm still kind of mad, so I'll keep on posting in these topics.

#1: I had the correct geometric sequence (4,6,9) but wrote it as 469 instead of 964 because... well, that's why they're called stupid mistakes.
#2: Must've been calculation error. I ended up with 497, exactly 200 less than actual
#4: Read as $ \frac{AM}{MB}$ instead of $ \frac{AM}{AB}$, which resulted in a ratio of $ (3009 \plus{} 2*17)/17 \equal{} 179$, instead of 177.
#6: Calculation again, I had the same formula that TZF (?) came up with in the solutions
#7: Found the telescoping series, found the wrong n. Not as stupid as the other ones.
#14: I knew I'd miscalculate this one once I began working through the (identical) solution.

Sorry for whining.

#1, didn't read #9.

AMC 12, I got the birdfeeding one wrong. With the millet. Yeah. Don't ask. Please.

On the AMC 12 B, I got #1 wrong by solving for muffins instead of bagels.
It was such an awesome modular arithmetic solution, too.

And on the same AMC 12, I correctly solved the question about the girl and her twin brothers, and then forgot to add one of the brothers' ages into the total, and got 10 instead of 18.

If I hadn't missed those two, I'd have made AIME  :mad:

Any tips on not making stupid mistakes?  I read Mr. Rusczyk's article, so I used a separate sheet of paper for each AIME problem and wrote neatly.  Then, I failed.

[quote="feuxfollets"]Hmm I said $ \log (1 \plus{} something) \equal{} 1 \plus{} \log(something)$ for the sequence one... then I proceeded from there and got 65... :wallbash:
Oh and I forgot about the <9999 condition for #9 and forgot to subtract 3 from 15  :noo:[/quote]

I got 65 on that problem as well. However, I plugged in n+1 instead of n for $ \frac {1}{n \plus{} \frac {2}{3}}.$

And for #3, I did $ \frac {1}{p}$ and $ \frac {1}{1 \minus{} p}$ for some odd reason.

And then for #8 I was 6 off due to stupid addition errors.

[quote="Heero Kirashami"]I missed #4 (119, should've been 177)[/quote]

Wow. I put 119 as well (but then caught THAT mistake, but then wrote 178 assuming the fraction was a+b)

So -10.

Then on #8 I thought the sequence started with a0 instead of a1.

So -10.

On the AMC10B I got a perfect so I can't exactly lose points there, right?

Total pts lost to stupidity: 20.
Not too bad (last year was 46, year before was 42.5)

[quote="feuxfollets"][quote="mgao"]
3) #2 wrong, epic fail. I went like 656-41 = 615 (!).[/quote]

Last time I checked 656 - 41 = 615....[/quote]
yeah it does but it was supposed to be 656 + 41 which gave correct answer of 697 lol

haha worst is me.. 12 pts lost on 10b for retarded reasons.. and then 40 on aime because of computation

I kicked my stupid-mistakes habit this year (none/11 answered on the AIME, I don't remember any on the AMC12).  Basically, what I can suggest is:
1) Have an overall mindset of relaxed focus.
2) Read the question carefully. Make this a habit, even practice it if it's a problem for you.
3) Work through the problem in a deliberate manner. Flashes of inspiration are good, but make sure you go through them methodically and not miss anything.
4) Do all arithmetic calculations in two different ways. (e.g. 223 + 1242, you can do it by adding them up on paper, and check by doing it mentally like seeing that it's going to be 23+42 past 1400; I usually do it mentally first and write down what I got, then check on paper).
5) Once you have the answer, reread the problem. Work through your logic and make sure it's bulletproof; follow each step and ask yourself if there's anything you could have missed. Play the role of the devil's advocate trying to prove your solution wrong. If everything looks good, read the part of the question about what you're supposed to find, and write exactly that down at the bottom of your solution. There's your answer.

If at any point in the verification steps you find an error, check to see which parts of your solution (if not all) it impacted, and start over from whichever point you need to. If it's just an arithmetic problem, redo the calculation. Also - if you do something two different ways and get two different answers, try it a third way. Doing things in duplicate is very useful to avoid dumb mistakes.

Doing the process above, I only had time to do 11 problems on the AIME (although I might have done more had I not blown 30-45 minutes on #5), but all of those 11 answers were correct.

4 mistakes cost me exactly 36 points on my index score- try to figure out how many on which test.

Anyways, if you give up: [hide]
On the AMC 10, the millet one, I figured out when the birds would find half a pint of millet or something like that.  That was wrong.
On the AIME, the game show one, I forgot that 11113 was not allowed, and I permuted wrong, and I overcounted...
On the AIME, the minimization one, I didn't even use the fact that the sum of the $ a_k$'s was 513 except to show that there had to be at least 187 1's.
On the AIME, the difference of powers of 2's one, I found a formula that included multiplying a subanswer by 2, which I forgot to do later.
[/hide]

[quote="Waxing Rhapsodic"]Well, there was the matter of somehow missing the word "distinct" in #1...[/quote]
I missed that too at first...and I thought it was too easy...so I spent like 4 minutes rereading the problem and finally realized it had to be distinct.

[quote="Waxing Rhapsodic"]Well, there was the matter of somehow missing the word "distinct" in #1...[/quote]

Hint Hint: Signature :(

[quote="Waxing Rhapsodic"]Well, there was the matter of somehow missing the word "distinct" in #1...[/quote]

That's really got to suck.  :wallbash_red:

My horrible mistake was somehow getting #1 of the AMC 10B wrong.

Me too, I got 12 questions(1-5,7-13) but made 4 stupid mistakes which lead to only 8 marks...

All you guys complaining about stupid mistakes, that's factored into the test. Last year, I worked the whole time on the AIME, answered 9, got 6 right. I wasn't really confident in my answers, I used logic but didn't really have a thorough understanding. This year, I finished 10 problems in an hour and a half, then I checked and caught 5 mistakes (got these 10 right) then managed to do two more, of which I got one right. You need to know the material really well to have the time to check it, having a vague idea of the concept required for the problem after spending a long amount of time is not enough.

Lost 12 points on the AMC 12B due to general stupidity, and another 4 questions on the AIME to failure of basic arithmetic  :(

So instead of a 132 and a 9 I am left with a 120 and a 5....

The only stupid mistake was on #1 of AIME. I feel like I have been doing much better with avoiding stupid mistakes. Reducing stupid mistakes has really improved my scores on all my competitions.

AMC 12A: don't remember.
AMC 10B: none
AIME: On number two, I thought that the real part of $ \frac{x\plus{}iy}{x\plus{}n\plus{}iy}$ was $ \frac{x}{x\plus{}n}$. Biz Are.


Suppose $ a_n > 0, \forall n$ and $ \sum_{k \equal{} 1}^{\infty} a_k$ diverges.

Prove that $ \sum_{k \equal{} 1}^{\infty} \frac {a_k}{(1 \plus{} a_1 )(1 \plus{} a_2 ) \cdots (1 \plus{} a_k)}$ converges ,and find its sum.

[color=red]sorry,everybody.My typo mistakes and correctly edited.[/color]

The result doesn't appear to be true for $ a_k \equal{} 1$, unless I"m misreading it.

It is also not true for $ a_k \equal{} \frac1k$.

For $ a_k \equal{} 1$,

$ \sum\limits_{k \equal{} 1}^{\infty} \frac {1}{2^k} \equal{} \frac {1}{1 \minus{} \frac {1}{2}} \equal{} 2$?

Edit: Oh. He made a typo.

[quote="Extremal"][hide="hint"]
$ \frac {a_{k \plus{} 1}}{(1 \plus{} a_{1})...(1 \plus{} a_{k \plus{} 1})} \equal{} \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{k})} \minus{} \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{k \plus{} 1})}$
[/hide][/quote]

would you have details offered to post?

That's rather a lot of information there.  It's really a very nice way of looking at the problem: this is a generalization of the observation that $ 0.999... \equal{} 1$ applied to [url=http://en.wikipedia.org/wiki/Mixed_radix]mixed radix[/url] representation of real numbers (rather than integers).

[quote="vinskman"]

would you have details offered to post?[/quote]
Let's consider $ S_{N} \equal{} \sum_{j \equal{} 1}^{N}\frac {a_{j}}{\prod_{k \equal{} 1}^{j}(1 \plus{} a_{k})}$
since
$ \frac {a_{k \plus{} 1}}{(1 \plus{} a_{1})...(1 \plus{} a_{k \plus{} 1})} \equal{} \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{k})} \minus{} \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{k \plus{} 1})}$
then we can write :
$ S_{N} \equal{} \sum_{k \equal{} 1}^{N}\frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{k})} \minus{} \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{k \plus{} 1})} \equal{} 1 \minus{} \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{N \plus{} 1})}$
thus if $ \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{N \plus{} 1})} \to 0$ as $ N \to \infty$ then we can write :
$ \sum_{j \equal{} 1}^{\infty}\frac {a_{j}}{\prod_{k \equal{} 1}^{j}(1 \plus{} a_{k})} \equal{} \lim_{N \to \infty} S_{N} \equal{} \lim_{N \to \infty}1 \minus{} \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{N \plus{} 1})} \equal{} 1$
this is the solution of this prioblem if you can to prove the Next statement :
[hide="Lemma"]
$ a_{j} > 0$ and $ \sum a_{j} \equal{} \infty$
then
$ \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{N \plus{} 1})} \to 0$ as $ N \to \infty$
 :)
[hide]
let $ S_{N}\equal{} \sum ^{N} a_{k} \to \infty$ as $ N \to \infty$
but $ (1 \plus{} a_{1})...(1 \plus{} a_{N}) \equal{} 1 \plus{} S_{N} \plus{} b_{N}$ where $ b_{N} > 0$
hence
$ \frac {1}{(1 \plus{} a_{1})...(1 \plus{} a_{N })} \equal{} \frac {1}{1 \plus{} S_{N} \plus{} b_{N}} \to 0$ since $ S_{N} \to \infty$
[/hide]
[/hide]