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If $A$ and $B$ are subsets of $\mathbb{R},$ define $A+B$ by $A+B=\{a+b:a\in A, b\in B\}.$ Suppose that $A$ and $B$ are measurable sets of positive finite measure. Prove that $A+B$ contains an open interval.

Hi Kent!
Finally, I am at home!

Very nice problem.

Suppose contrary, i.e. $\frac{A+B}{2}$ doesn't contain intervals. Assume WLOG $A,B\subset [0,1]$, then $\frac{A+B}{2}\subset [0,1]$.
Let $n>1$ be natural number. Consider subintervals $\left(0,\frac{1}{n}\right)$, ..., $\left(\frac{n-1}{n},1\right)$. Each of these intervals contains point $c$ s.t. $c\not\in \frac{A+B}{2}$. It follows that $(2c-A)\cap B=\varnothing$.
So we have set $C\subset [0,1]$ s.t. $\overline{C}=[0,1]$ and $\forall\,c\in C:\ (2c-A)\cap B=\varnothing$.
Since $B$ is a mesurable set, we can find open set $\hat{B}$ s.t. $B\subset \hat{B}$ and $|\hat{B}\setminus B|\leq \epsilon$, where $\epsilon$ will be defined later. Any open set in $\mathbb{R}$ is a union of at most countable many pairwise disjoint intervals. So $\hat{B}=B_1\cup B_2\cup ...$, where $B_k$ is an interval, $B_k\subset [0,1]$.
For each interval $B_k$ we can find $c_k\in C$ s.t. $|(2c_k-A)\cap B_k|\geq \frac{1}{3}|B_k|\cdot |A|$ [b](!)[/b].
Thus we have
1) $\bigcup_{k=1}^\infty (2c_k-A)\cap B=\varnothing$;
2) $\left|\bigcup_{k=1}^\infty (2c_k-A)\cap \hat{B}\right|\geq \frac{1}{3}|\hat{B}|\cdot|A|$;
3) $|\hat{B}\setminus B|\leq \epsilon$.
It follows that
\[B''=\bigcup_{k=1}^\infty (2c_k-A)\cap \hat{B} \subset \hat{B}\setminus B=B',\]
so $|B''|\leq |B'|$. It yields
\[\frac{1}{3}|\hat{B}|\cdot|A|\leq \epsilon.\]
Take $\epsilon =\frac{1}{4}|B|\cdot|A|$ to obtain contradiction.

I think I also have another solution using characteristic functions and integrals.

[quote="Myth"]I think I also have another solution using characteristic functions and integrals.[/quote]
That would probably be the solution I had in mind. It will take me until tomorrow to process your first solution.

I am glad you will read it! :)

P.S. I hope you checked my solution for topic "Convex function" http://www.mathlinks.ro/viewtopic.php?p=129665 ;)

[quote="Myth"]For each interval $ B_k $ we can find $ c_k\in C $ s.t. $ |(2c_k-A)\cap B_k|\geq \frac{1}{3}|B_k|\cdot |A| $ (!).[/quote]
Although I think I see this, I'm not quite sure where the 1/3 is coming from. Could you expande the explanation of this a little, please?

I'll go ahead and post my proof. Convolution on $\mathbb{R}$ is defined by

\[f*g(x)=\int_{-\infty}^{\infty}f(x-y)g(y)dy\]
It can be shown that if $f\in L^1$ and $g\in L^{\infty}$, then $f*g$ is continuous (in fact, uniformly continuous and tending to $0$ at $\infty$, not that we'll use those parts.)

Consider $h(x)=\chi_A*\chi_B(x).$ From above, this is a continuous function. If $h(x)\ne 0$ then there must be some $y$ (in fact, a set of $y$ of positive measure) such that $x-y\in A$ and $y\in B$. In other words, $x\in A+B.$ However, $h$ is a nonnegative continous function whose integral is $|A|\cdot|B|$, a positive number. A nonnegative continous function with positive integral must be positive on some interval.

[quote="Kent Merryfield"][quote="Myth"]For each interval $ B_k $ we can find $ c_k\in C $ s.t. $ |(2c_k-A)\cap B_k|\geq \frac{1}{3}|B_k|\cdot |A| $ (!).[/quote]
Although I think I see this, I'm not quite sure where the 1/3 is coming from. Could you expande the explanation of this a little, please?[/quote]
Actually, it is easy to show that we can find $c\in[0,1]$ s.t. $ |(2c-A)\cap B_k|\geq \frac{1}{2}|B_k|\cdot |A| $ and $\frac{1}{3}$ is just "correction", because we search for $c\in C$, where $C$ is a dense set.

Your solution requires some nontrivial (?) convolution's features. I tried to find solution which appeal only to theory of measure itself.

[quote]Your solution requires some nontrivial (?) convolution's features.[/quote]
A walk through the features that it does require:

1. There are "nice" dense subspaces of $L^1(\mathbb{R}).$ One particular manifestation of this: the step functions are dense in $L^1.$ A path to this: you can approximate an $L^1$ function by a simple function, and then you can approximate each characteristic function that goes into the linear combination by a characteristic function of a finite union of intervals. This is a version of "measurable sets are nearly finite unions of intervals", which is an idea that Myth used rather more directly in his proof.

2. For $f\in L^1$, define $f_t(x)=f(x-t).$ Then $\lim_{t\to0}||f_t-f||_1=0.$ This is true by calculation for step functions, and we use the density from #1.

3. $|f*g(x-t)-f*g(x)|\le \int_{-\infty}^{\infty}|f_t(x-y)-f(x-y)||g(y)|dy$  $\le ||f_t-f||_1||g||_{\infty}.$ Hence the result about $f*g$ being uniformly continuous.

That's what was used.

A nice immediate corollary is that sets of positive measure contain hamel basis', the A - A contains a neighborhood of the origin analogue gives a proof of that any cantor set contain a hamel basis.

How do we jump from "any set with positive measure contains a Hamel basis" to "the Cantor set contains a Hamel basis" (since the Cantor set has measure $0$) ?

As has been discussed many times before on the forum (I remember seeing the problem at least three times; use the search function), if $C$ is the Cantor set, then $C+C=[0,2]$ (despite the fact that $C$ has measure zero; measure has nothing to do with this :)). This means that every real can be written as a rational multiple of a number in $C+C$, which, in turn, means that $C$ contains a system of generators of $\mathbb R$ over $\mathbb Q$. From any system of generators one can extract a basis, so that's that.

[quote="grobber"]if $C$ is the Cantor set, then $C+C=[0,2]$[/quote]

Oh, oh ... this sheds much light !

I was searching for posts on convolutions, and ran across this. I was excited to see that out of the two solutions given, the way I did this last week was not one of them.

There are $ a_0$ and $ b_0$ in $ A$ and $ B$ respectively with metric density 1. (Metric density at $ x$ is $ \lim_{r\to 0}\frac{m(E\cap B(x,r))}{m(B(x,r))}$). Since metric density is 1 at almost every point of a set, a sets of positive measure are guaranteed to have the points $ a_0$ and $ b_0$ as listed.

Choose $ \delta>0$ to be small. Then let $ c\equal{}a_0\plus{}b_0$. For each $ \epsilon$, define $ B_\epsilon\equal{}\{c\plus{}\epsilon\minus{}b: b\in B, \ |b\minus{}b_0|<\delta\}$.

So we now have $ B_\epsilon\subset(a_0\plus{}\epsilon\minus{}\delta, a_0\plus{}\epsilon\plus{}\delta)$. So proper choosing of $ \delta$ and $ \epsilon$ small enough we have $ A\cap B_\epsilon\neq\emptyset$ (by the metric density of 1), i.e. $ A\plus{}B\supset (c\minus{}\epsilon_0, c\plus{}\epsilon_0)$ for some $ \epsilon_0>0$.


geometry
geometric transformation
reflection
incenter
circumcircle
geometry unsolved

Let us call a triangle rational if each of its angles is a rational number when measured in degrees. Let us call a point inside triangle rational if joining it to the three vertices of the triangle we get three rational triangles. Show that any acute rational triangle contains at least three distinct rational points.

[hide]
if someone can show that there is one, then reflect the point across the heights you get other two.
the bad news is the 3 heights intersect at one point, but the point doesn't have such a reflection.
[/hide][/hide]

For non-equilateral triangle: incenter, circumcenter, orthocenter.
For equilateral triangle, there are infinitely many rational points on any one of the three altitudes.


Is it true that for any n we can find a,b>1 integers such that for any k=1,2,...,n we have a^k-1|b^k-1, but b is not a power of a?
  I do not have a solution to this one, but I like how it sounds.

I was just going through some old posts and found this one ... Well, trivially, just take
\[
b = 1 + \prod_{1 \leq j \leq n} (a^j - 1).
\]
Perhaps the condition that $a^n - 1 \mid b^n - 1$ for infinitely many $n$ will imply that $b$ is a power of $a$ (as a matter of fact, I [b]do[/b] know that this is true, as a corollary of a more general result), though I don't know how to show it. As for the initial problem (the one that treats the case of all $n$), it is an immediate corollary of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27052&highlight=]this[/url].


abstract algebra
Functional Analysis
real analysis
real analysis unsolved

If M is a finite-dim subspace in normed space Y, then how to show X
has a closed supplement space? (e.g. a closed subspace N of Y, such
 that M/\N={0} and M+N=Y)

You need Hahn-Banach here- there are so many possibilities that we need something close to the axiom of choice to find any.

Find a basis $v_{1},v_{2},\dots,v_{n}$ of the subspace $M$. For each $i$, let $L_{i}$ be the linear functional $L_{i}(av_{i})=a$; extend $L_{i}$ to $Y$ continuously. The kernel of the extension is a closed subspace complementary to $v_{i}$; let $N$ be the intersection of these kernels.


A Sudoku matrix is deﬁned as a $ 9\times9$ array with entries from $ \{1, 2, \ldots , 9\}$ and with the constraint that each row, each column, and each of the nine $ 3 \times 3$ boxes that tile the array contains each digit from $ 1$ to $ 9$ exactly once. A Sudoku matrix is chosen at random (so that every Sudoku matrix has equal probability of being chosen). We know two of the squares in this matrix, as shown. What is the probability that the square marked by ? contains the digit $ 3$?

$ \setlength{\unitlength}{6mm} \begin{picture}(9,9)(0,0) \multiput(0,0)(1,0){10}{\line(0,1){9}} \multiput(0,0)(0,1){10}{\line(1,0){9}} \linethickness{1.2pt} \multiput(0,0)(3,0){4}{\line(0,1){9}} \multiput(0,0)(0,3){4}{\line(1,0){9}} \put(0,8){\makebox(1,1){1}} \put(1,7){\makebox(1,1){2}} \put(3,6){\makebox(1,1){?}} \end{picture}$

[hide="Solution"]If there is a 3 in the first 3x3 square, in the bottom row, there is a $ 0$ probability. Otherwise, the probability is $ \frac16$ that there is a 3 in the question mark. Thus, the total probability that the ? is a 3 is $ \frac47\times\frac16\plus{}\frac37\times0\equal{}\boxed{\frac{2}{21}}$.[/hide]


[color=darkred]$ 1\blacktriangleright$ The equation $ 2^{x}\plus{}3^{x}\plus{}x\equal{}6^{x}$ has only two real zero : $ x_{1}\equal{}1$ and $ x_{2}<0$.

$ 2\blacktriangleright$ The equation $ 2^{x}\plus{}6^{x}\plus{}3^{x}\equal{}2x^{2}\plus{}6x\plus{}3$ has only three real zero :[/color] $ x_{1}\equal{}0$, $ x_{2}\equal{}1$ and $ x_{3}\in (\minus{}3,\minus{}2)$.

for the first equation see that for x>1 we have 6^x-2^x-3^x>x.similar whenx<0


In how many ways can n married couples be seated around a circular table in such a manner than there is always one man between two women and none of the men is next to his own wife? Derive a formula !

reference:

http://mathworld.wolfram.com/MarriedCouplesProblem.html
http://mathworld.wolfram.com/Inclusion-ExclusionPrinciple.html

We assume that the woman have been seated at alternate places. Call the ladies 1 to n. The problem amounts to placing the integers 1 to n on a circle with positions numbered 1 to n such that for all i the integer i is not in position i or in position i+1 (mod n). Let E_i be the set of seatings in which husband i is sitting next to his wife. We use the principle of inclusion-exclusion (PIE) and must therefore calculate in how many ways it is possible to seat r husbands incorrectly. Call this number A_r. We do this as follows. Consider a circular sequence of 2n positions. Put a 1 in position 2i-1 if husband i is sitting to the right of his wife; put a 1 in position 2i if he is sitting to the left of his wife. Put zeros in the remaining positions. The configurations which that we wish to count therefore are circular sequences of 2n zeros and ones, with exactly two ones, no two adjacent. Let A'_r be the number of sequences starting with a 1 (followed by a 0). By considering 10 as one symbol, we see that we must choose r-1 out of 2n-r-1 positions. To count the number A''_r of sequences starting with a 0, we place the 0 at the end, and then it amounts to choosing r out of 2n-r places. Hence:

A_r = A'_r + A''_r = Cn(2n-r-1,r-1) + Cn(2n-r,r) = 2n/(2n-r) * Cn(2n-r,r)

By PIE we find the number of ways to seat the men is:

\sum^(n)_(r=0) (-1)^r * (n-r)! * 2n/(2n-r) * Cn(2n-r,r)

Before you tackle the problem you may try the three less challenging problems by Kaplansky:

(1)

In how many ways one can mark in the sequence 1,2,3,...,n p elements such that between two marked elements there is at least one non-marked element ?

(2)

In how many ways one can mark in the sequence 1,2,3,...,n p elements such that between two marked elements there are at least s non-marked elements ?

(3)

In how many ways one can mark in the sequence 1,2,3,...,n in cyclic order p elements such that between two marked elements there are at least s non-marked elements ?

[quote="orl"][b](1)[/b] In how many ways one can mark in the sequence 1,2,3,...,n p elements such that between two marked elements there is at least one non-marked element ?[/quote]

I will solve another problem:
[i]In how many ways one can mark in the sequence 1,2,3,...,n [b]some[/b] elements such that between two marked elements there is at least one non-marked element?[/i]

Let $K_n$ be desired number of ways.
Suppose most left marked number is $m$, then $m+1$ is unmarked and we are to mark remaining $n-(m+1)$ numbers, so
\[K_n=\sum_{m=1}^{n} K_m\cdot K_{n-m-1},\]
where $K_{-1}=K_0=K_1=1$.

Anyway, I guess there is closed form of this recurrence. :)

What about creating a bijection from the sequences:
$(x_1,x_2,\ldots,x_p)$ to $(x_1,x_2+1,\ldots,x_p+p-1)$ wouldn't this be easier????

Yes, of course. So we obtain sum of binomial coefficients.
But I wonder is there closed form for such sum.

Ooopppss.  :blush:
I realized that I was solving another problem, because I didn't see that we need to choose exactly $p$ numbers.  :blush:  :blush:
So Dust's post completely solves problem.


[quote="#H34N1"]A roll of paper contains 440 sheets each 9cm $\times$ 9cm. The outside diameter of the roll is 14 cm and the inside diameter is 4 cm. Find the thickness of each sheet to the nearest hundredth of a centimeter. Use $\frac{22}{7}$ as an approximation for $\pi$.[/quote]

[hide=" :D ←Click this"]Let the thickness be $x$.
When seen from the side, there are two concentric circles with diameters 14cm and 4cm. The area of the tissues being shown is the area in between the two circles, which is equal to $7^{2}\pi-2^{2}\pi = 45 \pi$. Each of the 440 sheets have the surface area of $9x$ when seen on this side. From this,

$45 \pi = 440 \cdot 9x$

$45 \cdot \frac{22}{7}=440 \cdot 9x$

$x= \frac{45 \cdot \frac{22}{7}}{440 \cdot 9}\approx \boxed{0.04}$ (cm).[/hide]


Hello, the second admissions test for this summer is posted here: http://www.awesomemath.org/tests/TestB-2009.pdf The test runs from April 3rd to April 18th. Good luck!

--Alex

When will Awesome math tell you if you made it or not, and by what means?

Thanks, btw I just looked at the page again and found that exactly under the link for the test is says "April 22, 2009" and that's when they inform who got in or not.

lol kay :)
good luckkk XD amsp rocks =]]

[quote="shelly32494"]I got accepted maybe a week after i sent in my apps [it was pretty fast tho] & i got notified first by email , then they sent me my packet.[/quote]
Have you received your packet yet ? I got accepted on Feb 18. They said that they will send me packet, but I am still waiting, I know that at least two more kids have been waiting also ...

shtsxc12:
i went last year. not going this year T_T
but the packet came pretty late; cuz i remember worrying alot last year too. dont worry. it'll come :)

[quote="sjyoo93"]Hi guys...

I did the normal registration, and my packet arrived like, just now...

I guess just a heads-up, although it will probably arrive.[/quote]

did you receive an email about it?
i'm afraid their admissions letter (about accepting/denying) i may have deleted from my spam folder.

No, I didn't get any e-mails.. Although I am not sure if it was my mom's email or mine.. Think it was mine.

Just to make sure, the packet is all the forms and the student handbook (?) and the packing list and all that, right? If it is, I got it.. and..

Don't worry too much. The date on UTD says it will respond to its applicants by 22nd.. today?? Hope you got it.

And if you didn't, ask them about it, and really, don't panic. I probably got it early because again, my house is 30 minutes from UTD and where Dr. Andreescu (director) lives..

I got in.
Going to both of the camps, taking all 4 intermediate classes hopefully.
Don't worry guys, I haven't gotten a packet.

How long should it take to get the packets anyways?

Also, can the check be POSTMARKED May 1? Or do you have to GET IT BY May 1?

Also: I applied and got into this camp called COSMOS that's also at UC Santa Cruz. The last two days of COSMOS are the first two days of AwesomeMath. What would be the best way to resolve this issue?

:rotfl: Simply don't go to the last day in COSMOS and the first day in AwesomeMath  :P
By the way isn't that out of topic  :wink:


Homologation of Nitriles.

Devise methods for converting a nitrile to its next higher homolog,

$ RC \equiv N \longrightarrow RCH_{2}C \equiv N$,

and for converting a nitrile to its next lower homolog,

$ RCH_{2}C \equiv N \longrightarrow RC \equiv N$.

A good way for the first one might be:

RCN (hydrolysis) --> RCOOH (LiAlH4 for reduction) --> RCH2OH (PBr3 then NaCN) --> RCH2CN

Perfectly correct. How about the second part?


Let $ f,g: \left [ a,b \right ] \rightarrow \mathbb{R}$ be two functions such that:

$ a)$ $ g$ is increasing on $ \left [ a,b \right ]$

$ b)$ $ |f(x)\minus{}f(y)| \le |g(x)\minus{}g(y)|, \forall x,y \in \left [ a,b \right ]$

$ c)$ $ f(a)\equal{}g(a)$ and $ f(b)\equal{}g(b)$

Prove that $ f(x)\equal{}g(x), \forall x,y \in \left [ a,b \right ]$

From (b) we have $ f(x)\minus{}f(y) \le g(x)\minus{}g(y)$ for $ x>y$ because $ g(x)$ is increasing.

If we can find a $ k$ so that $ f(k)>g(k)$, then add it with $ f(a)\minus{}f(k) \ge g(a)\minus{}g(k)$ (the sign change because $ k>a$)
we get $ f(a)>g(a)$, contradiction.

Similarly if we can find a $ k$ so that $ f(k)<g(k)$, then add it with $ f(b)\minus{}f(k) \le g(b)\minus{}g(k)$
we get $ f(b)<g(b)$, contradiction again, done.


number theory
greatest common divisor
modular arithmetic
number theory unsolved

Show that 13 is the largest prime that can divide two successive integers of the form n^2+3

For $ n \equal{} 2k,n \plus{} 1 \equal{} 2k \plus{} 1$, $ \text{gcd}\,(4k^2 \plus{} 3, 4k^2 \plus{} 4k \plus{} 4)$ $ \equal{} \text{gcd}\,(4k^2 \plus{} 3, 4k^2 \plus{} 4k \plus{} 4)$ $ \equal{} \text{gcd}\,(4k^2 \plus{} 3, 4k \plus{} 1)$ $ \equal{} \text{gcd}\,(3 \minus{} k, 4k \plus{} 1) \equal{} \text{gcd}\,(3 \minus{} k, 13)$; and similarly for $ n \equal{} 2k \minus{} 1, n \plus{} 1 \equal{} 2k$. We can achieve this for $ n \equiv 6\pmod{13}$.


For each convex hexagon $A_1 B_1 C_1 D_1 E_1 F_1$, there exists a convex hexagon ABCDEF such that $A_1, B_1, C_1, D_1, E_1, F_1$ are the midpoint of $AB, BC, CD, DE, EF, FA,$ respectively. Prove your answer.
-------------------------
I think it is a nice pr. Can anyone solve it?

If we let $x_1$ be the complex number corresponding to vertex $X_1$ ($X$ is one of the letters $\{A,\ B,\ldots,\ F\}$) then we start from another point $A$ and draw its symmetric wrt $B_1$, $B$. We draw the symmetric of $B$ wrt $B_1$ and so on until we have drew $G$, the symmetric of $F$ wrt $F_1$.

The problem asks if $G=A$. Let $a$ be the complex number corresponding to $A$. The the complex number of $G$ is $g=2f_1-2e_1+2d_1-2c_1+2b_1-2a_1+a$. We can see that $g-a$ is constant when we move $a$ around, and it only depends on $a_1,\ b_1,\ldots,\ f_1$. At the same time it's obvious that $g=a\iff a_1+c_1+e_1=b_1+d_1+f_1\iff$ triangles $A_1C_1E_1$ and $B_1D_1F_1$ have the same center of gravity.

That's the necessary and sufficient condition for the hexagon to exist: $A_1C_1E_1$ and $B_1D_1F_1$ must have the same centroid (center of gravity).

[quote="grobber"]Oh, I didn't see the word "convex", but as we've seen, the hexagon exists for any position of $A$ we choose, so I think the hexagon can be made convex by a proper choice of $A$.[/quote]
yeah, if there was not word convex, this pr. would be done easily like you :D :D

[quote="grobber"]So the answer is wrong? Or is the sanswer right but with an incomplete solution?[/quote]
The answer is not completed, moreover this solution is one of very many step to solve this pr
ex. take a hexagon ABCDEF with A=B=c=D=E=F=$2\pi /3$ and AB=CD=EF, BC=DE=FA, and AB/BC $\rightarrow 0^+$


search
real analysis
real analysis unsolved

What is the cardinality of bijections on N, the natural numbers?

[hide="Solution"]$ 2^{|\mathbb N|}$ - asked here many times already. [/hide]

I wish I could find one of those many instances in the search to see a proof

Here's one, with a hint:

http://www.mathlinks.ro/viewtopic.php?t=219073

One can always think of partial fractions  :blush:

*I meant continued fractions, not partial fractions..  :oops:


Which of the following is larger:

2^2^2^...^2 (10 twos in all) or 3^3^3^...^3 (8 threes in all)  :huh:

I know if there are 9 threes then the tower of 3s is larger but I have been unable to prove either way for 8 threes. Any thoughts?

Correct me if I'm wrong, but aren't you comparing $ 2^{512}$ and $ 3^{2187}$?

No. He's comparing $ 2^{2^{2^{2^\dots}}}}$ and similarly for 3's, which is different.

This is equivalent to comparing:

$ 2^{2^{2^{2^{2^{2^{2^{16}}}}}}}$

and

$ 3^{3^{3^{3^{3^{3^{3^3}}}}}}$

both of which are power towers with 8 terms. Starting from the inside out, $ 2^{16} \equal{} 65536$ is greater than $ 3^3 \equal{} 27$, $ 2^{65536} > 3^{27}$, etc... so the power tower of 2's is greater.

That doesn't appear to be rigorous...you seem to be assuming that if $ x>y,2^{x}>3^{y}$, which is not true.

[quote="saillsha"][quote="Brut3Forc3"]How about taking the log log log log log log log log of both and comparing?[/quote]

Please elaborate more on that method. I'm not sure how it would work because if you do $ \log{\left(\log{x^{x^{x}}}\right)}$, it simplifies to $ \log{\left(x^{x}\log{x}\right)}$ and you can't go very far from there...[/quote]
[hide="Here you go"]Note: Refer to [url]http://en.wikipedia.org/wiki/Tetration[/url] for my notation.
We know that for positive $ a,b$, $ a>b \iff \ln a > \ln b$. Therefore, we can compare both of these values by comparing their natural logarithms. Or rather... their natural logarithms taken 8 times. :D

First, we start with $ ^{10}2$.
$ \ln\ln \ln\ln \ln\ln \ln\ln \,^{10}2\equal{}\\
\ln\ln \ln\ln \ln\ln \ln (\,^{9}2\ln2)\equal{}\\
\ln\ln \ln\ln \ln\ln (\, ^{8}2\ln2\plus{}\ln\ln2)$
Now, $ ^{8}2\ln2$ is an unimaginably high value, whereas $ \ln\ln2$ is approximately equal to $ \minus{}.3665$, so $ \ln\ln2$ can safely be dropped out. We then have
$ \ln\ln \ln\ln \ln\ln (\, ^{8}2\ln2)\equal{}\ln\ln \ln\ln \ln(\, ^{7}2\ln2 \plus{}\ln\ln2)\approx\ln\ln \ln\ln \ln (\,^{7}2\ln2)$
This pattern continues, and can be stopped at $ \ln\ln (\, ^{4}2\ln2)\equal{}\ln\ln (65536\ln2)$, after which it can be easily computed by a calculator. This is approximately equal to $ 2.3725$.

Similar methods for $ ^{8}3$ give
$ \ln\ln \ln\ln \ln\ln \ln\ln \,^{8}3 \approx \ln\ln\ln(\, ^{3}3 \ln3) \approx 1.2217$

Since $ \ln\ln \ln\ln \ln\ln \ln\ln \, ^{10}2 > \ln\ln \ln\ln \ln\ln \ln\ln \, ^{8}3, \, ^{10}2 > \, ^{8}3$.[/hide]

There's also something called the super-logarithm. :D
Blah, I'm bad at posting rigurous proofs, thus I must be bad at math in general. QED.


Did anyone go to this? I got 5th in Alg 2, and Matthew Crawford was at the awards ceremony to give out certificates to the top scorers (and to advertise for AOPS :D). They gave out flyers about AOPS to everyone after the test.

Congradulations on your 5th place finish!!! :-D

I've heard quite a bit of Vestavia Hills High School. They have a very active math team, and they travel a lot (sometimes very long distances) for math meets. I always wondered whether they get support from their school for traveling, or if they raise their own money somehow.

What is the format of the Vestavia Hills contest? Tell us a little more about it.

As for their funding, Mrs. Tipton always thanks the rotary club at the awards ceremony, because they donate huge sums to the Math and Debate teams (they got  $17, 400 this year...Wow. I don't think our team even gets any funding, lol).

The contest is very simliar to all other local contests here in Alabama. There is a 25-question, multiple choice 1-hour written test and four 5-question rounds of ciphering. The person who got 1st place in Alg2 got a 66 out of a possible 100, so the test is pretty difficult. The upperclassmen and some former students write the tests (which range from pre-algebra to comprehensive), but I'm not so sure that the faculty checks them (Last year's Alg2 test was absolutely impossible. I think the high score was somewhere in the 30s or 40s).

You're right about their math team. They have an excellent math team, but recently this area has become very competetive and they've been beaten a few times over the last couple of years. Still, they're very far ahead of everyone else, although my goal is to beat them by the time I graduate (:D).


the answer has to be an equation not an expression ;)

[quote="math92"]Form a quadratic equation with rational coefficients having $2-4(3)^{\frac{1}{2}}$ as one of its roots.[/quote][hide="solution"]$\xi=2-4\sqrt{3}\Longrightarrow\xi-2=-4\sqrt{3}\Longrightarrow\xi^2-4\xi+4=48\Longrightarrow\xi^2-4\xi-44=0$. Thus $\xi$ is a solution to $x^2-4x-44=0$.

By the way, I like your pickiness over "expression" versus "equation." I wish more people around here paid attention to those details :D. [/hide]

[hide]$x=2-4\sqrt{3}$ is one solution and the coefficients are rational, so the other root is $x=2+4\sqrt{3}$.

Sum of the roots: $(2-4\sqrt{3})+(2+4\sqrt{3})=4$
Product of the roots: $(2-4\sqrt{3})(2+4\sqrt{3})=4-48=-44$

$\therefore x^2-4x-44=0$[/hide]

[quote="chess64"]Why does everyone use those weird symbols instead of normal variables... :? $\xi$... :roll:[/quote]
i agree.  when i was looking at a math textbook the used a 0 with a horizontal line going through it when they could have just used x.

[quote="math92"]when i was looking at a math textbook the used a 0 with a horizontal line going through it when they could have just used x.[/quote]
Did you mean $\theta$?

[quote="math92"]Form a quadratic equation with rational coefficients having $2-4(3)^{\frac{1}{2}}$ as one of its roots.[/quote]

A quadratic equation has two solutions and since they are both rational, the other solution is $2+4\sqrt{3}$, thus knowing that the sum of the roots is $\frac{-b}{a}$ and by making the $a=1$, we can see that the sum is therefore: $4$ and the product being $\frac{c}{a}$, we can see that the product is $-44$, therefore the equation is:

$\boxed{x^2-4x-44}$

frt and Iversonfan's methods are just fine but what about if you want to find such equations that have roots like $\sqrt[3]{2-3\sqrt{5}}+3$?

[quote="AntonioMainenti"][quote="chess64"]Why does everyone use those weird symbols instead of normal variables... :? $\xi$... :roll:[/quote]In [i]An Introduction to the Theory of Numbers[/i] G.H. Hardy used that to represent the solution to a polynomial equation. If you are criticizing the choice of variable names of the great G.H. Hardy I shall have to ask you to step outside![/quote]

Well, he's a mathematician. And you're still a student. It does look cool though. $\xi$... wow! :D


Molly has seven U.S. coins with a total value of 88 cents. She
does not have any half-dollars. How many dimes does Molly
have?

Well... first off, you need 3 pennies (or 8, but that's too many).
85 cents for 4 coins.

It should be clear that 3 quarters + 1 dime, so 1 dime.

[quote="cookieboy123456789"]3*25+10*1+1*3=88
3+1+3=7[/quote]
However, the question is just asking the number of dimes.


AMC
USA(J)MO
USAMO
number theory
relatively prime

Let $d$ be a divisor of $n$. Let $d(n)$ be the number of divisors of $n$. Prove that if $n$ is not a product of two primes then, for every $d$, the number of divisors that are not relatively prime to $d$ is at least $\frac{d(n)}{2}$.

[hide]
Let $n=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{n}^{e_{n}}$ where the $p_{i}$ are prime.  It follows that $d(n) = (e_{1}+1)(e_{2}+1)\cdots(e_{n}+1)$.    Also let $d = p_{1}^{f_{1}}p_{2}^{f_{2}}\cdots p_{n}^{f_{n}}$.

The statement which we want to prove is equivalent to proving that the number of divisors relatively prime to $d$ is at most $\frac{d(n)}2$.

A divisor of $n$ can be relatively prime to $d$ only if at least one of the $f_{i}$ are 0.  So let $f_{k_{1}}, f_{k_{2}},\ldots f_{k_{j}}$ all be 0.  (Note that not all of the $f_{i}$ can be zero since taht would make $d=1$).  Then the number of divisors of $n$ which are relatively prime to $d$ is \[\prod_{i=1}^{j}(f_{k_{i}}+1).\] However, the total number of divisors of $n$ can be written as $d(n) = \prod_{i=1}^{n}(f_{k_{i}}+1)$.  We then have \[ \prod_{i=1}^{j}(f_{k_{i}}+1) \cdot \prod_{i=j}^{n}(f_{k_{i}}+1) = d(n)\Rightarrow \prod_{i=1}^{j}(f_{k_{i}}+1) = \frac{d(n)}{\prod_{i=j}^{n}(f_{k_{i}}+1)}.\] Since $f_{i}\ge 1$ we arrive at our desired result: \[\prod_{i=1}^{j}(f_{k_{i}}+1) \le \frac{d(n)}2.\] [/hide]

with this little lemma we can Solve USAMO 2005 #1 easily :)

I don't see why if $n$ is a product of two primes the claim fails... :(


arithmetic sequence
geometric sequence
AMC

The sequence 6,-9,x,y is such that the first 3 terms form an arithmetic sequence and the last three terms form a geometric progression. Find (x,y)

...too much math...can't breathe...*gasp*...*pant*...*monotone*

Anyways, [hide]since the first three forms an arithmatic sequence, x=-24 (-9-6=-15, -9-15=-24

The last three form a geometric sequence, so y=-64 (-24/-9=2+2/3, (2+2/3)*-24=-64)

Answer: x=-24, y=-64[/hide]

ya that was pretty easy

Are these questions too easy?

Not at all! Although I'd love it if you post them in MATHCOUNTS. It's like a ghost town and I'm like the deputy there. :) [size=75]I'm scared! Mommy![/size]


In a unit cube , skew diagonals are drawn in two neighboring faces.Find the minimum distance between them .

[quote="Albanian Eagle"]$\frac{\sqrt{3}}{3}$ of course  :)[/quote]

I also know the answer   :lol:  . But how did you came up to this result ?This is the reason I posted here

let the unit cube be $ABCDA_{1}B_{1}C_{1}D_{1}$ our two diagonals are $AC$ and$BC_{1}$ now since this two skew lines are on the two parallel planes $(ACD_{1})(BC_{1}A_{1})$ their distance is considered to be the distance between these two planes which in this case is :$\frac{\sqrt{3}}{3}$ :wink:


calculus
function
vector
conics
parabola
search
calculus computations

Problem 1.

-We want to build a paralelipipedical container with a volume of $V = 20 m^{3}$. The cost of the material for the lateral faces is $1 dolar/m^{2}$, for the bottom face the cost is $2 dolars/m^{2}$, and for the upper face the cost of material is $3 dolars/m^{2}$. What should be the dimensions of the container in order for the total cost to be minimum?

Problem 2.

Indicate (without solving) a system whose solutions allow to solve the following problem:

-In a certain region two rivers pass nearby each other. Supposing, to simplify, that the region is planar, we verify that the trajectories of the rivers can be approximately described by the equations $x^{2}-2x+y = 2$ and $2x^{2}-y+3 = 0$. We want to build a channel, the shortest possible, that connects both rivers. Between which points of the two rivers must the channel be build and what will be its lenght?

(These two problems are from a Calculus II class, where it is studied differential calculus of functions of two or more variables)

[b]Problem 1[/b]
Correct me if I am wrong, but in this case the Parallelepiped reduces to a Cuboid,
since it is more effective.
Let the cost of each surface be $a$, $b$ and $c$ per $\rm{m}^{2}$.
The problem reduces to find the minimum of the function
$f(x,y,z)=(a+b)yx+2az(x+y)$ under the constraint $g(x,y,z)=xyz=V=20$

Ok, for the first one:

[hide="Solution"]I have considered the bottom face as being in the xy plane, with one side having x meters and the other y meters. The height is z meters. So we have $V = xyz = 20 \Rightarrow  z = \frac{20}{xy}$.
The cost of the four lateral faces will be

$2xz+2yz = 2x \cdot \frac{20}{xy}+2y \cdot \frac{20}{xy}= \frac{40}{y}+\frac{40}{x}$.

The cost of the bottom one will be $2xy$, and the cost of the top face will be $3xy$. So the function $C(x,y)$ that describes the total cost of the container is

$C(x,y) = \frac{40}{x}+\frac{40}{y}+5xy$.

Now, first we have to calculate the critical points of $C(x,y)$, for which we have to solve the equations

$\frac{\partial C}{\partial x}= 0$ and $\frac{\partial C}{\partial y}= 0$.

Solving those we get $(2,2)$ as the only critical point. Now we must find if this point is indeed a minimum. For that, we start to make the following computations:

$\frac{\partial^{2}C}{\partial x^{2}}(2,2) = 10$, $\frac{\partial^{2}C}{\partial y^{2}}(2,2) = 10$, and $\frac{\partial^{2}C}{\partial x \partial y}(2,2) = \frac{\partial^{2}C}{\partial y \partial x}(2,2) = 5$.

Next, we calculate the Hessian of C in (2,2), which equals

$\frac{\partial^{2}C}{\partial x^{2}}(2,2) \cdot \frac{\partial^{2}C}{\partial y^{2}}(2,2)-\left[\frac{\partial^{2}C}{\partial x \partial y}(2,2)\right]^{2}= 10 \cdot 10-5^{2}= 75.$

Now, as both the Hessian and $\frac{\partial^{2}C}{\partial x^{2}}(2,2)$ are > 0, then (2,2) is indeed a minimum. So the dimensions that will minimize the total cost are $x = 2, y = 2, z = 5.$
[/hide]

I havent figured out which method is the best one, but
one would be to parametrixe the curves you described above
$y_{1}(x)=-x^{2}+2x+2$ and $y_{2}(x)=2x^{2}+3$ into
$r_{1}(t)=(t,y_{1}(t))$ and $r_{2}(t)=(t,y_{2}(t))$ (vectors)
Then the distance between the curves would be the length of the vector
$r_{1}-r_{2}$ for some certain points. Now we can put some constraint on this.
The minimum distance occurs when $(r_{1}-r_{2})\cdot r'_{1}=0=r'_{2}\cdot (r_{1}-r_{2})$
from this it should be possible to solve the two values for $t$

another method would be to find a distance function between the two curves
an minimize it. (One can consider the points ($x_{1},y_{1}$) and ($x_{2},y_{2}$) one each curve)

[quote="Zakary"]Does the second problem involve Calculus and more than 2 variables?[/quote]

Yes, the purpose is to use functions of two or more variables. However, if you can solve it in other way, you can post it.

Does anyone have some more thoughts about problem 2?

Why? Am I wrong or do you want it more detailed

These are the curves in [b]PROBLEM 2[/b]

Does anyone have an Idea or a easy method to solve problem 2?

Carcul: Do you?

I have solved it just like your second proposal: find the function that gives the distance between the two curves (as a funtion of two points) and then minimize it.

well i remeber seeing a theorem that [b]probably [/b] :maybe: stated the minimum distanec between the parabolas is the ditance between the endpoints along the line joining their focus,...just may be i am not sure i will search for atext though ..meanwhile googling may help

Let $P(s,-s^{2}+2s+2)\in y=-x^{2}+2x+2,\ Q(t,\ 2t^{2}+3)\in y=2s^{2}+3$, $\overline{PQ}$ is minimal $\Longleftrightarrow$ the tangent lines at $P,\ Q$ are perpendicular to the line $PQ$ simultaneously. That is to say, the circle with daiameter $PQ$ is tangent to these parabolas at a time.

Condition 1: The tangent lines at $P,\ Q$ are pararell $\Longleftrightarrow-2s+2=4t$.

Condition 2: The line $PQ$ is perpendicular to the tangent line at $P$ or $Q \Longleftrightarrow \frac{-s^{2}-2t^{2}+2s-1}{s-t}\cdot 4t=-1$.

Elliminating $s=1-2t$ from these equations gives the cubic equation $24t^{3}+3t-1=0$. I calculated twice, I got same result. How can we solve this cubic equation?


Who will win Men's 2009 French Open?  Who?

Man.. Tough call, but I'm picking Maria Sharapova to defeat Max Schindler in straight sets  :P
In reality I'm picking Nadal.. again.

Funny, how Max Schindler is not on the list of wild-card qualifiers nor is he in the ATP top 100... Yet he is a favorite to win Rolland Garros... I picked Nadal...

tough one eh....

i agree juggler, maria would destroy max. poor victor.. :(

i think runpeng will defeat nadal in a close match

Uh, resurrection:

notice that the question implied is for Men's.   :D

[quote="resurrection"]i think maria would own everyone cause she's russian. and in soviet russia. ya[/quote]

In America, you beat Max Schindler at tennis, but in Soviet Russia....

[quote="pythag011"][quote="james4l"][quote="pythag011"]I don't think that Max is playing in going to play in any Starcraft tournament in the near future, so why is max's name on there?[/quote]

Neither would David Yang. Zuming could theoretically be there, seeing how he beat several random MOPpers.[/quote]

Zuming is a noob cheeser (sorry :P.) He proxy gates a lot.[/quote]

Explain with a detailed thesis on his build order, attack strategy, amount of macro/micro, and give eighteen different instances (must be replays) of such.

[quote="james4l"][/quote][quote="pythag011"][quote="james4l"][quote="pythag011"]I don't think that Max is playing in going to play in any Starcraft tournament in the near future, so why is max's name on there?[/quote]

Neither would David Yang. Zuming could theoretically be there, seeing how he beat several random MOPpers.[/quote]

Zuming is a noob cheeser (sorry :P.) He proxy gates a lot.[/quote][quote="james4l"]

Explain with a detailed thesis on his build order, attack strategy, amount of macro/micro, and give eighteen different instances (must be replays) of such.[/quote]

[hide="Build Order"]

7 pylon, 9 gateway, 10 proxy pylon, 10 proxy gateway, 10 forge, get 12 zealots while getting +1 zealot attack, attack.[/hide]

Attack strategy:

Kill sunkens, then units, then eggs (lol...), then larva, then hatchery, then drones.

[hide="Macro Micro"]

5 macro and 13 micro.[/hide]

NO everyone knows that you attack drones before eggs

Zuming is out.

[quote="pythag011"][/quote][quote="james4l"][/quote][quote="pythag011"][quote="james4l"][quote="pythag011"]I don't think that Max is playing in going to play in any Starcraft tournament in the near future, so why is max's name on there?[/quote]

Neither would David Yang. Zuming could theoretically be there, seeing how he beat several random MOPpers.[/quote]

Zuming is a noob cheeser (sorry :P.) He proxy gates a lot.[/quote][quote="james4l"]

Explain with a detailed thesis on his build order, attack strategy, amount of macro/micro, and give eighteen different instances (must be replays) of such.[/quote][quote="pythag011"]

[hide="Build Order"]

7 pylon, 9 gateway, 10 proxy pylon, 10 proxy gateway, 10 forge, get 12 zealots while getting +1 zealot attack, attack.[/hide]

Attack strategy:

Kill sunkens, then units, then eggs (lol...), then larva, then hatchery, then drones.

[hide="Macro Micro"]

5 macro and 13 micro.[/hide][/quote]

blasphemy are you saying he does the infamous COMPUTER RUSH (with an added +1 and a random proxy!?)

so like what

if you proxy you better be attacking way before your 12th zealot

[quote="not_trig"][/quote][quote="pythag011"][/quote][quote="james4l"][/quote][quote="pythag011"][quote="james4l"][quote="pythag011"]I don't think that Max is playing in going to play in any Starcraft tournament in the near future, so why is max's name on there?[/quote]

Neither would David Yang. Zuming could theoretically be there, seeing how he beat several random MOPpers.[/quote]

Zuming is a noob cheeser (sorry :P.) He proxy gates a lot.[/quote][quote="james4l"]

Explain with a detailed thesis on his build order, attack strategy, amount of macro/micro, and give eighteen different instances (must be replays) of such.[/quote][quote="pythag011"]

[hide="Build Order"]

7 pylon, 9 gateway, 10 proxy pylon, 10 proxy gateway, 10 forge, get 12 zealots while getting +1 zealot attack, attack.[/hide]

Attack strategy:

Kill sunkens, then units, then eggs (lol...), then larva, then hatchery, then drones.

[hide="Macro Micro"]

5 macro and 13 micro.[/hide][/quote][quote="not_trig"]

blasphemy are you saying he does the infamous COMPUTER RUSH (with an added +1 and a random proxy!?)

so like what

if you proxy you better be attacking way before your 12th zealot[/quote]

This.

[quote="Nora.91"]umm.... Coming back to the topic of discussion ...

I support [b]ROGER FEDERER[/b] !  :winner_first:[/quote]

BOOYAH!fav. perfect. all technique and skill.

Of course, jonathan, a little bit of luck is also involved.

Yes, he demolished Soderling. 6-1, 7-6, 6-4  :o

Seeing however that no one really cares. Lets continue the conversation on Zuming h4x0r skills in Starcraft.

[quote="pinkmuskrat"]Hmm. Let's see...

Men's Singles- 2009 Roland Garros

QF-
Max Schindler(WC) def. Rafael Nadal(1), 6-3 6-2 6-7(5) 7-5
Victor Wang(Q) def. Andy Murray(3), 6-7(3) 6-4 6-4 6-7(14) 7-5
Maria Sharapova(I) def. Novak Djokovic(4), 6-3 2-6 6-1 7-6(2)
Runpeng Liu(LL) def. Roger Federer(2), 6-4 7-6(9) 7-6(8)

SF-
Max Schindler(WC) def. Victor Wang(Q), 6-3 6-7(1) 6-7(0) 6-4 6-3
Runpeng Liu(LL) def. Maria Sharapova(I), 7-5 6-1 6-2

F-
Max Schindler(WC) def. Runpeng Liu(LL), 6-7(5) 6-7(12) 7-6(8) 7-6(33) 6-4

I'm confident I'll win the Fantasy league on ATP this week.[/quote]

Awww... I did okay in the fantasy.... I guess
The guy who won made my quarterfinals, atleast.....

One of d most interesting moments of the match was [b]Federer almost being attacked by that weird guy [/b] :rotfl:

OH MY GOSH! WHAT AN UPSET! MAX JUST BEAT FEDERER 6-0 6-0 6-(-1)!!!!

[quote="Mewto55555"]You also had Nadal getting defeated pretty darn early.[/quote]

You're right. I didn't think of that. Thanks for making me feel better. :)


Prove that, in a quadrilateral, the sum of any three sides is greater than the fourth side.

[quote="darkdieuguerre"][hide="Argument"](This being in HSB, I'm not going to use Cauchy-Schwarz; the proof is practically the same as that of the triangle inequality.)

Let the sides be of length $ a$, $ b$, $ c$, and $ d$, with $ a \le b \le c \le d$. It is sufficient to prove $ a \plus{} b \plus{} c > d$. Assume this is not the case. If $ a \plus{} b \plus{} c \equal{} d$, then the only way a quadrilateral can be constructed is if it is degenerate. If $ a \plus{} b \plus{} c < d$, then no quadrilateral can be constructed. Quadrilaterals cannot be degenerate, so therefore the desired claim is proven.

[/hide][/quote]

OK, but don't you need to prove that in the first case, the triangle is degenerate, and in the latter case, that the quadrilateral can't be constructed.

Another way would be to draw a diagonal as andrewtom said.
[hide]Let the four sides of the quadrilateral be $ a,b,c,$ and $ d.$ We draw a diagonal and let the diagonal's length be $ e,$ so that the two triangles we form have side lengths $ a,b,e$ and $ c,d,e.$

From the triangle inequality on the first triangle, we have $ a\plus{}b > e,$ and from the triangle inequality on the second triangle, we have $ c\plus{}e > d.$ Adding these two inequalities, we have $ a\plus{}b\plus{}c\plus{}e > e\plus{}d,$ the $ e's$ cancel, leaving us with $ a\plus{}b\plus{}c > d,$ as desired.   [/hide]


calculus
integration
function
calculus computations

Expand to gamma function ;

$ \int_{0}^{\infty}y^{n}e^{-my}dy-\int_{0}^{1}y^{n}e^{-my}dy.$ First term is obvious (using adequate substitution) , second term, use partial integration n times, and I get :

$ \frac{n!}{m^{n+1}}+\frac{e^{-m}}{m}+\frac{n e^{-m}}{m^{2}}+\dots+\frac{n! e^{-m}}{m^{n+1}}-\frac{n!}{m^{n+1}}$ ,and
$ \frac{n!}{m^{n+1}}(e^{-m}\sum_{k=0}^{n}\frac{m^{k}}{k!})$,

Edit: Misan, your result is correct, I forgot to include the last term $ -\frac{n!}{m^{n+1}}$ and the sum actually starts from 0, then we get your result. I'll correct it.

Nice integral though. :wink:


superior algebra
superior algebra unsolved

Prove that there is no a such that (a-inverse)*(123)*a=(13)(578).

Does this have anything to do with the fact that (123) has order 3, so then [(a-inverse)*(123)*a]^3=e?

Not only does it have something to do with it, but that constitutes a proof.

Why would it be the proof though? Isn't the question asking for some permutation a...and in the above case we are only looking at the conjugation expression cubed..?

[b]rawkerrxx[/b] I write here the full solution:
$ (a^{\minus{}1}(123)a)^{3} \equal{} a^{\minus{}1}(123)aa^{\minus{}1}(123)aa^{\minus{}1}(123)a \equal{} a^{\minus{}1}(123)(123)(123)a \equal{} a^{\minus{}1}a \equal{} e$.
 In the other hand, by  a simple computation you can see that $ ((13)(578))^{3} \neq e$. So the equality you gave can't hold, that is, there is no such $ a$.


Let $S$ be the set of all points with coordinates $(x,y,z)$, where $x$, $y$, and $z$ are chosen from the set ${0,1,2}$. How many equilateral triangles have all their vertices in $S$?

Not sure but here's what i got

[hide="Solution"]
Notice that the whole set of possible points forms a $3$x$3$ unit cube. 8 equilateral triangles may be formed from one cube, no matter what the size. Therefore, the amount of cubes and equilateral triangles there are in a $3$x$3$ cube are
\[27(8)+8(8)+1(8)=288 \]
[/hide]

There are more ways to make equilateral triangles, I believe, but I'm not sure. Is it possible to make one with like sidelength $\sqrt{2}$ or $\sqrt{3}$?

oh oops i think i can account for more now.


By the method in my post of http://www.mathlinks.ro/viewtopic.php?t=288699 , we can draw a circle orthogonal to the given circle, tangent to one of given lines and its center lies on the bisector of angle which is formed by two given lines. This circle is just the wanted circle in the problem of this topic. See at the picture 3.

[hide=" To jrrbc"][b]Problem:[/b]
Given a circle $ (O)$ and two lines $ d'$ and $ l$. Draw a second circle orthogonal to the circle $ (O)$ and tangent to both lines $ d'$, $ l$.
[b]Full construction:[/b]
Draw the line $ d$ be the bisector of angle formed by two lines $ d'$ and $ l$ ($ d'$ and $ l$ are symmetric w.r.t the line $ d$)
Let H be the orthogonal projection of B on d'. The circle of diameter BH cuts the circle (B;BA) at F. Let E be the orthogonal projection of F on BH. Denote C be the midpoint of BE. Draw the circle (C) of diameter BE, it meets the segment AC at D. Draw the circle (A;AD). Construct two tangent lines from B to the circle (A), they cut the line d' at two points P,Q. Then, the incircle of triangle BPQ is just the wanted circle.
See at the picture 3 in my above post.[/hide]


geometry
MATHCOUNTS
LaTeX
function
rhombus
perimeter
AMC 8

I am doing Math Counts and AMC 8 next year...and I am just wondering if there are any formulas I should memorize for those. Especially for geometry. I am very bad at geometry.

Geometry:

Pick's Thereom and the Shoelace Thereom are Useful.

Other's you NEED to know:

diagnoals in a polygon:  n(n-3)/2 where n equals number of sides
degrees in a polygon 180(n-2) n=number of sides
degrees of the measure of an interior angle in a regular polygon:  180(n-2)/n, where n=number of sides

Also, I find subbing in numbers extremely helpful when you have a problem where you have to find ratios.

Pythagorean Theorem. Definitely. Also, you should know basic area/volume formulas.

The formulas for triangular, square (obviously), pentagonal, and hexagonal numbers are useful too.

Triangular:
$ T_n\equal{}\frac{n(n\plus{}1)}{2}$
Square (I'm pretty sure everyone knows this xD):
$ S_n\equal{}n^2$
Pentagonal:
$ P_n\equal{}\frac{n(3n\minus{}1)}{2}$
Hexagonal:
$ H_n\equal{}n(2n\minus{}1)$

Given a right angled triangle $ ABC$ with hypotenuse $ BC$, draw the perpendicular $ AH$ from $ A$ to $ BC$,

$ \boxed{AB^2\equal{}BH*BC,\ AC^2\equal{}CH*BC,\ AH^2\equal{}BH*CH}$, which is very useful.

@kunny: I don't think that will come in useful very often in MC and AMC8. There is already a thread about this, found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=292950]here[/url].

For what age do you take a test in $ MC,\ AMC 8?$

usually around middle school age so... 11-14 I guess...

Thank you! I should have examined in advance, now I could make a confirmation the level in [url=http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/2008-8-Brochure-Q.pdf]this[/url], and I have read Teachers Manual.

Oh, I have just noticed (MATHCOUNTS, AMC 8, MOEMS, AGES 10-13), which is convinient to us.

[quote="randomtext"]The formulas for triangular, square (obviously), pentagonal, and hexagonal numbers are useful too.
[/quote]

wait, do we [i]really[/i] need to know triangular, pentagonal, and hexagonal numbers?

What are those useful for, anyway?(I'm referring to triangular, pentagonal, and hexagonal numbers)

For the MATHCOUNTS problems that involve them. :wink:

I mean, an example of a problem that actually uses them.

"What is the 5th pentagonal number?"

That's one problem I remember, and you can find some others by looking at old tests.

Yea, I only suggested them in the first place because I remember seeing them (to be specific, I think they were asking for a huge hexagonal number) on an old team round. I didn't know the formula back then so I did a lot of calculating (and got the wrong answer anyway .-. I tend to make dumb mistakes)

Wait does that me H base n?
[hide="fail"]
I finally figured out izzy's siggy.[/hide]

No $ H_{n} \equal{} n(2n \minus{} 1)$ means the $ n^{th}$ hexagonal number is $ n(2n \minus{} 1)$

[hide="Note"]
It would have been easier to figure out her signature if she had written it in latex like
[code]${{i}\choose{u}}$[/code] instead of [code]$\binom{i}{u}$[/code][/hide]

Triangle numbers

1/2n^2+1/2n

Square

1n^2+0n

Pentagonal

3/2n^2-1/2n

Hexagon

2n^2-1n

Bonus points for seeing a pattern

You will need to know

-Areas
-Volumes
-Surface Area
-Pythagorean
-Pick's Theorem
-Heron's Formula
-Diagonal of a Convex Polygon
-Degrees
-All the properties of circles. Very useful.
-Linear and Parabolic functions.

Anything I didn't cover?

i'm not getting parabolic equations.
checked on wiki

[quote="jonathanchou711"]Parabolic functions.[/quote]At the high school level, this is commonly known as 'quadratic functions'. With the same name, 'parabolic' functions is another class of functions studied mainly in Partial Differential Equations.

Ooops. Not equations. Functions. That slipped. sorry.

I am no expert but I would master area formulas for most shapes.  Also, try to memorize those all the formulas everyone told you and PRACTICE.  I found that just memorizing alone is not very helpful.  DO PROBLEMS.

I missed a state competition problem for not knowing the formula for the area of a rhombus and that kicked me out of the "Top 50 Highest Scoring Students" :(

Remember:  It's 1/2(d1 * d2) where d1 and d2 are the diagonals!

Heron's formula helps with figuring out area of irregular triangles. S is the semi-perimeter (half perimeter) of the triangle you're trying to find the area of. Let A, B, C just be the lengths.
The area of the triange is the square root of ((((Sx (S-A) (S-B) (S-C).

This formula helps with the sum of consecutive odd numbers. It only works if you start with 1, 3, 5 etc.
Let the amount of numbers in the sequence be N. The sum is N squared.
Example: 1+3+5+7+=???
There are four numbers in the sequence. So the answer is four squared. Ans is 16. Check it yourself. :oops:

This next formula is just like the one before it, except it finds the sum of consecutive EVEN numbers starting from 2, 4, 6, etc. Let the amount of numbers in the sequence be N. The sum is N(N+1). Once again you can try it yourself.  :)

There are 2 special types of right triangles. The 30-60-90 degree triangle and the 45-45-90 degree triangle. Mathcount problem makers love using these two types, no matter how subtle. A special rule about the the 45-45-90 triangle is that the two shorter legs are of equal length. The 30-60-90 triangle has appeared MORE often for more on problems so this is even more vital than the 45-45-90. The shorter leg of this triangle is half of the hypotenuse. The longer leg (but not the hypotenuse) is the hypotenuse divided by the square root of 3. I hope this helps and good luck to you all Mathcount contestants. Hopefully I'll see you in the Nationals!  :rotfl:

[quote="flamex12"]The longer leg (but not the hypotenuse) is the hypotenuse divided by the square root of 3. [/quote]

Its actually the hypoteneuse divided by $ \frac{2}{\sqrt3}$ I find it easier to remember as multiplying the short leg by $ \sqrt3$


geometry
3D geometry
linear algebra
absolute value

Hi,
could anyone help me with this problem?:

If [tex] T:R^n \rightarrow R^n [/tex] is a linear transformation with [tex]||T(x)|| = ||x||[/tex], show that for any n-cube c in [tex]R^n[/tex], [tex]vol(c) = vol(T \circ c) [/tex]

How would I go about finding the volume of c and T composite c?

This is linear algebra, not calculus. Your transformation is an isometry, and the image is congruent to the original. In particular, the image of a box is a congruent box with equal volume.

For any linear transformation, volume is scaled by the absolute value of the determinant.


Find all $f:(0,\infty)\rightarrow(0,\infty) \mbox{ such that } f(f(x)+y)=xf(1+xy) \mbox{ for all }x,y>0$

Yes, grobber, it's the function f for which f(1+xf(y))=yf(x+y). But I wouldn't say this is the same problem.

Well, I'm not saying that either; one reason is that I couldn't solve that one, but I could solve this one :).


[b]1.[/b] Se considera numarul complex $ z$, astfel incat sa aiba loc simultan relatiile:
\[ |z\plus{}1| \le 1\] \[ |z^2 \plus{}1| \le 1\] \[ |z^3 \plus{} 1| \le 1\]
Sa se arate ca $ z\equal{}0$.

[b]2.[/b] Fie $ x, y, z \in (0, \infty)$. Sa se arate ca:
a) $ \arctan(x) \plus{} \arctan(y) < \frac{\pi}{2}$ daca si numai daca $ xy<1$
b) $ \arctan(x) \plus{} \arctan(y) \plus{} \arctan(z) < \pi$ daca si numai daca $ xyz < x\plus{}y\plus{}z$

[b]3.[/b] Pentru $ a \in \mathbb{R}$ se defineste functia $ f_a : \mathbb{R} \minus{}> \mathbb{R}^2$ cu $ f_a (t) \equal{} (\sin t , \cos at)$ pentru orice $ t \in \mathbb{R}$
a) Sa se arate ca functia $ f_{\pi}$ nu este periodica
b) Sa se determine valorile reale ale parametrului $ a$ pentru care functia $ f_a$ este periodica

Pe 4 imi este lene sa-l scriu. Ma duc sa mananc. :P

Pentru $ 1$ vezi http://www.mathlinks.ro/viewtopic.php?t=184077


geometry
geometric transformation
dilation

[quote="Cyber-Shreyas"]Imagine you are in total zero gravity space and inside a car. You have nothing around you, just a void, nothing at all. Now this car is capable of moving at constant velocity.... and since there are no distorting forces, it doesn't need any force or acce.

Once it has attained const velocity. Will it be possible for you to detect whether you are moving or not?? Ofc without using a speedometer or other gadgets. :D[/quote]

Without looking out the windows, no. Since $ a\equal{}0$ the resultant forces on your body are $ 0$ so you wouldn't be able to notice it

even if you are near the speed of light, you still can't tell, by einstein's principle of relativity right? the car+you=inertial frame, and you can't tell if an inertial frame is at rest or not.

i wrote "as long as u r [b]not[/b]near the speed of light"

[quote="rituraj007"]as long as u r not moving at speed near that of light--u can apply relativity
and use a light clock(i m supposing that u have 1)[/quote]

Nice try, but I said you aren't allowed to use any gadgets...

i got an idea--y not apply bio?????????
body systems can tell u a lot
u will feel some sensation in stomach if ur moving
but ofc everything is determined by quantum interactions betw our body quarks and others

@chronoz--refer to second last page of last chapter in HCVerma(2)((if u have one)).

its clearly written that heart beat and b.p. will be affected--and that wud confirm movement. :D  :D

The real question is ... moving relative to what?  :P You can not define movement without defining a reference point.

Even if you are moving near the speed of light, time dilation would occur, but it would only be observed for a reference frame to which you are moving near the speed of light to.  As bowei correctly said, you are in an inertial reference frame, and it would be impossible to tell whether you are moving.  Think of it this way, you could be lying perfectly still somewhere on the Earth, but the Earth is rotating at some speed, and is also revolving around the sun at some speed, which is revolving around the milky way galaxy at some speed, so you see, everything is moving relative to everything else, yet you would not be able to detect this by simply sitting somewhere on Earth.


complex numbers
superior algebra
superior algebra unsolved

Hi, i need to show that $ f(z): C \minus{}> R$ is a homomorphism under addition

$ z \equal{} x \plus{} y*i$

I can do it complex numbers to complex numbers, but not sure how to map complex to real numbers.

That's what I have done so far:

$ f(z\plus{}m) \equal{} f(z) \plus{} f(m) \equal{} (x \plus{} y*i) \plus{} (x' \plus{} y'*i) \equal{} (x \plus{}x') \plus{} i*(y \plus{} y')$

need to show that

$ f(x \plus{} y*i) \equal{} x \plus{} y$

not sure what do do next...

thanx!

[quote="Antonio_26"]Hi, i need to show that $ f(z): C \minus{} > R$ is a homomorphism under addition

$ z \equal{} x \plus{} y*i$

I can do it complex numbers to complex numbers, but not sure how to map complex to real numbers.

That's what I have done so far:

$ f(z \plus{} m) \equal{} f(z) \plus{} f(m) \equal{} (x \plus{} y*i) \plus{} (x' \plus{} y'*i) \equal{} (x \plus{} x') \plus{} i*(y \plus{} y')$

need to show that

$ f(x \plus{} y*i) \equal{} x \plus{} y$

not sure what do do next...

thanx![/quote]You'll need to show the addition for $ \mathbb{C}$ is carried over to $ \mathbb{R}$ through the mapping $ f(z \plus{} _{\mathbb{C}}z') \equal{} f(z) \plus{} _{\mathbb{R}}f(z')$ (definition of homomorphism).

So if your mapping is $ f(z) \equal{} f(x \plus{} yi) \equal{} x \plus{} y$, let $ z \equal{} x \plus{} yi,z' \equal{} x' \plus{} y'i$,

$ \begin{aligned} f(z \plus{} z') & \equal{} f((x \plus{} yi) \plus{} (x' \plus{} y'i)) \equal{} f((x \plus{} x') \plus{} (y \plus{} y')i) \\
& \equal{} (x \plus{} x') \plus{} (y \plus{} y') \equal{} (x \plus{} y) \plus{} (x' \plus{} y') \equal{} f(z) \plus{} f(z'). \end{aligned}$

Thus, $ f: \mathbb{C}\longrightarrow\mathbb{R}$ is a homomorphism.

Thank you.

I understand that $ ker(f) \equal{} x \minus{} x*i$

i.e.: $ f(z) \equal{} 0$ iff $ x \plus{} y \equal{} 0$  i.e. $ x \equal{} \minus{} y$ so $ ker(f) \equal{} x \minus{} x*i$.

and $ im(f) \equal{} f / ker(f)$

$ im(f) \equal{} (x \plus{} y*i) / (x \minus{} x*i)$

$ f$ is injective if $ ker(f) \equal{} 0$ but i have $ ker(f) \equal{} x \minus{} x*i$. Not sure if $ ker(f)$ is correct.

also not sure how to show that $ f$ is surjective using $ im(f)$, and therefore, isomorphic.

any help would be appreciated.


AMC AIME probability logarithms trigonometry email inequalities \/closed

Hi,

I'm hoping to make AIME next year (I didn't because I crushed on AMC-10 for probability) and I'm hoping to take AMC-12 next year since it's easier to qualify.

So, should I take AMC-10 Problem Series since there are chances that I might take AMC-10 next year or take AIME Problem Series? I do have all AMC-10 Questions from 00 to 05.

If you're goal is to qualify for the AIME, the AMC 10 Problem Series is a better class.  It focuses on problems that would be similar to problem 10 - 25 on the AMC 10 and 10 - 20 on the AMC 12 (minus selected curriculum such as logarithms and trigonometry that do not appear on the AMC 10).

The AMC 12 Problem Series is best for students who are already very confident of making the AIME, but want to improve their scores.

Beyond the problem series, I recommend that any student working to qualify for the AIME take any Introductory subject class that covers the subject in which they consider themself weakest.  Turning that subject into a strength could add 10 or more points to an AMC score.

Mr. Crawford, i'm in the EXACT same situation as Silverfalcon. i too bombed the AMC 10 and fell short and i'm also going to take the AMC 12 next year to get a better chance at making the AIME. What would be your recommendations for an Independent Study class?

If you have not yet done well on the AMC exams, one of the Introductory Independent Study courses is almost surely the right way to go.  If you email me and tell me which chapters in AoPS Volume 1 you are most comfortable with and what your weakest subject(s) are, then I can give a more definite recommendation.

What problem series should someone who got a 4 on the AIME, but a 10 w/out dumb mistakes (as many people made) go for? AMC 12 or AIME? or AIME or Intermediate Algebra?

On a side note, to what depth does Intermediate Algebra cover?I can answer 50% of the problems in the pre-test but would like to take it if it has Olympiad level questions.

Regarding the AIME series, what does one do each day in the classroom? How many problems? Any HW?

[quote="1234567890"]What problem series should someone who got a 4 on the AIME, but a 10 w/out dumb mistakes (as many people made) go for? AMC 12 or AIME? or AIME or Intermediate Algebra?[/quote]

Either would be helpful, the AIME Problem Series would probably be more helpful.  Also, it's best to go into that class not thinking that you should already get 10 without careless errors, but that there are probably some methods that help eliminate errors or give you cleaner, more efficient solutions that allow you to find your own errors more easily.

[quote="1234567890"]
On a side note, to what depth does Intermediate Algebra cover?I can answer 50% of the problems in the pre-test but would like to take it if it has Olympiad level questions.[/quote]

If you have the time, I recommend taking Intermediate level subject classes.  Those focus in detail on each topic and method that would help on the AIME.  The Intermediate Algebra class does bring you to the doorstep of Olympiad Algebra and Inequalities problems, though there is not a lot of material that would be considered Olympiad level on its own.  The best thing that class does in regards to Olympiad problems is that many of the Message Board problems are phrased such that you write proofs.

[quote="1234567890"]Regarding the AIME series, what does one do each day in the classroom? How many problems? Any HW?[/quote]

Each day of the AIME Problem Series involves class discussions of around 6-8 problems in which we discuss how you see through to clean methods on your own -- not simply a solution.  Each week the Message Board is populated with 10 or more practice problems of the variety discussed in class.

Many students have completed that class and gone on to qualify for the USAMO.  You might seek opinions from students in that class as to how it helped them prepare.

[quote="dts"][quote="1234567890"]Well, actually I could answer more like 6 or 7 on the AIME in a regular year. This was pretty easier.

So in the AIME Problem Series, we discuss past AIME problems, or are they made up?[/quote]

Mostly, past AIME problems.[/quote]

The AIME B Problem Series -- the one we're most likely going to teach this summer -- pulls about 1/4 of its problems from other sources simply because we ran out of AIME problems for some important topics.


geometry rectangle trapezoid parallelogram

Draw parallelogram BZDW.
Setting BW as x, we can show the area in two different terms.
$ 12 * x \plus{} 36 * 2 \rightarrow 12x \plus{} 72$
and
$ 120 * 2 \rightarrow 240$

Setting them equal to another, $ 12x \plus{} 72 \equal{} 240$, so $ x \equal{} 14$
$ \frac {14 * 6}{2} \equal{} 7 * 6 \equal{} \boxed{42}$.


modular arithmetic
combinatorics unsolved
combinatorics

In conference there $n>2$ mathematicians. Every two mathematicians communicate in one of the $n$ offical languages of the conference. For any three different offical languages the exists three mathematicians who communicate with each other in these three languages. Find all $n$ such that this is possible.

For $n$ odd, we give a construction: let $n = 2k+1$ and let both the mathematicians and the languages be numbered $0, 1, \ldots, 2k$.  Have mathematicians $i$ and $j$ speak language $(i+j)(k+1) \pmod n$.  Then any triple of distinct languages $a, b, c$ is spoken by the mathematicians numbered $a+b-c, a-b+c$ and $-a+b+c$.

Now, each language must be spoken by the same number of pairs of mathematicians (just sum over all triples), so it is clearly a requirement that $\frac1n{n \choose 2}$ be an integer, so $n$ odd is necessary.  Thus, the answer is exactly the odd integers larger than 2.

[quote="JBL"]For $n$ odd, we give a construction: let $n = 2k+1$ and let both the mathematicians and the languages be numbered $0, 1, \ldots, 2k$.  Have mathematicians $i$ and $j$ speak language $(i+j)(k+1) \pmod n$.  Then any triple of distinct languages $a, b, c$ is spoken by the mathematicians numbered $a+b-c, a-b+c$ and $-a+b+c$.[/quote]

Ok!

[quote]Now, each language must be spoken by the same number of pairs of mathematicians (just sum over all triples), so it is clearly a requirement that $\frac1n{n \choose 2}$ be an integer[/quote]

Why? (because English of mine is not good) Can you concrete?

[quote="N.T.TUAN"][quote]Now, each language must be spoken by the same number of pairs of mathematicians (just sum over all triples), so it is clearly a requirement that $\frac1n{n \choose 2}$ be an integer[/quote]

Why? (because English of mine is not good) Can you concrete?[/quote]

Let us count the number of pairs of mathematicians who communicate in language $0$.  There are $n \choose 3$ triples of languages and $n \choose 3$ triples of mathematicians, so each triple of languages is represented by exactly one triple of mathematicians.  There are $n-1 \choose 2$ triples of languages which include language $0$, and so language $0$ is spoken in exactly $n-1\choose 2$ triples of mathematicians.  Given a pair of mathematicians $(x, y)$ who speak language $0$, they are members of $n-2$ triples, so the number of such pairs times $n-2$ must be $n-1 \choose 2$.  Thus the number of such triples is $\frac{1}{n-2}{n-1 \choose 2}= \frac{n-1}{2}$ which is equal to the value I gave, and which is an integer only when $n$ is odd.

[quote="JBL"] Given a pair of mathematicians $(x, y)$ who speak language $0$, they are members of $n-2$ triples, so the number of such pairs times $n-2$ must be $n-1 \choose 2$.  [/quote]

Will you  count number of triples $\{x,y,z\}$ (not $(x,y,z)$, i think) mathematicians speak language $0$ by two way? :maybe:

An equivalent formulation of this problem appeared in China MO 2009. See [url=https://artofproblemsolving.com/community/c6h250260p1371303]here[/url].


سلام
میشه یکی لطف کنه و منابع المپیاد سال سوم رو بگه؟ و اگه میشه بگه چه کتاب هایی رو بخونیم بهتر هست.

پ.ن: چرا انقدر فروم فارسی کم کار هست؟

manabe' sale sevvom ba sale 2vvom va avval farghi nemikone!!,tarkibiate  alipour[entesharate fatemi],nazarie adade mirzakhani,beheshti[entesharate fatemi],hendese ahmadipoor[entesharate khoshkhan],jabr mehdi safa[entesharate khoshkhan].


A 2-inch elastic band is fastened to the wall at one end, and there'a bug at the other end. Every minute (beginning at time 0), the band is instantaneously and uniformly stretched by 1 inch, and then the bug walks 1 inch toward the fastened end, Will the bug ever reach the wall?

[hide]Yes, the bug will reach the wall.

Place the x-axis along the rubber band. With the attached end coinciding with the origin and the rubber band stretching out in the positive x direction.

Then we can label the position of the bug immediately before the $n$th stretch by $x_{n}$. Since the bug is at the end of the 2-inch rubber band before the 1st stretch, $x_{1}=2$. Then when the rubber band is stretched (uniformly), the bug is at $\frac{2+1}{2}x_{1}=3$. From there the bug moves 1-inch to the left, so we have $x_{2}=2$.

Using the same type of reasoning, it is not hard to see that $x_{n}=\frac{n+1}{n}x_{n-1}-1$. Using induction, we can prove that this leads to $x_{n}=(n+1)(\frac{1}{k+1}x_{k}-(\frac{1}{n}+\frac{1}{n-1}+\ldots+\frac{1}{k+2}))-1$ for $n\geq3$ and $k<n$. Knowing that $x_{1}=2$, set $k=1$ to get $x_{n}=(n+1)(1-(\frac{1}{n}+\frac{1}{n-1}+\ldots+\frac{1}{3}))-1$, which holds for $n\geq3$. If we want the bug to reach the wall, we need $x_{n}\leq0$ for some $n$. Let's use the equation to see if there are any possible values of $n$.

$x_{n}=0\geq (n+1)(1-(\frac{1}{n}+\frac{1}{n-1}+\ldots+\frac{1}{3}))-1$. So $\frac{n}{n+1}\leq (\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n})$. As $n$ approaches infinity, the right hand side tends to infinity whereas the left hand side tends to $1$. Thus the inequality will eventually be satisfied, i.e., the bug will eventually reach the wall.[/hide]

I just noticed that my first sentence was a contradiction. The math is fine (i think). I meant to say, [hide]yes the bug will reach the wall.[/hide]

[quote="Treething"]If I read it correctly, it's basically a harmonic series, meaning the bug will reach the wall.[/quote]

right


Hi folks, I am new here and not really good at math, just need a help if anybody can. - I am neither sure if it is a real math problem nor I am writing in appropriate place.

Given any number(real number) "n" and if it is odd, then "3n+1" becomes "n",(3n+1-->n) otherwise "n" will become "n/2"(n/2-->n). I am wondering - is possible to prove that n, followed by this sequence will eventually become "1" at the end always? thanks.

So... If $n$ is even, the next term would be $\frac{n}2$, and if $n$ is odd, the next term would be $3n+1$?

If the original author means what i-like-pie wrote, this is an open question known as Collatz conjecture.

aaww.... conjecture. Try putting this in the olympiad forum. They might have some ideas, but, if n is odd, 3n+1 is always even. If n is congruent to 1 mod 4, 3n+1 will be congruent to 0 mod 4. If n is congruent to 3 mod 4, 3n+1 will be congruent to 2 mod 4, thus being divisible by 2, but not any other power of 2.

[quote="nobody89"][quote="JBL"]This is the famous (if unexcitingly named) [url=http://mathworld.wolfram.com/CollatzProblem.html]3n + 1 problem[/url] (also known as the Collatz Problem, as puuhikki noted).  It's unsolved, and probably very hard.[/quote]
WOW, thanks for letting me know this - I didn't know that. Actually it's a "programming problem" but while trying to solve it I encountered this question(will it reach 1?) and thought to address it here, hmm, it turns out...

thanks for everybody!

PS, sorry for misexplanation of the problem, the original problem is like "i_like_pie" said.[/quote]

Yeah, for whatever reason it's pretty frequently referenced in programming problems. I've encountered it at least twice in IOI problems (it's one of the first problems in "Programming Challenges" IIRC).


The score was out this morning
I got a 790 on the Math Level II Test.  :( (It was my first attempt.. and I didn't study hard enough)
Should I retake the test  :?:  I think that 790 is not good enough for getting into MIT..

Thanks for any advices!

[quote="chilly_snow"][quote="Geehoon"]in my opinion, 790 and 800 are almost the same..
I understand how you feel, but I don't think it's worthy to retake it
(person who got into MIT with 760 physics are saying this..)[/quote]

Thanks for replying!

One more question.
Does the university care about when you take the SATs?[/quote]

I don't think so. Unless you take something multiple times then they might look at more than one test date. And MS scores don't count but I don't think that's a problem here.

Yay! SAT I scores in:

Critical Reading 780
Math 800
Writing 790 (MC 80, Essay 8)

Very happy with my scores, though I thought my essay was better than that. Oh well, I can't complain :D.

congrats paladin on great score
as to the question about timing, when you took an exam is irrelevant as long as you did take it and did well on it. they are arranged chronologically on the score report, so that gives a sense of relative timing but that's about it.
also, i don't recommend taking any exam more than once unless you really bombed it (and no, 790 is not considered bombing!) :D

my SAT score was actually in that vicinity (800 math, 660 verbal and 710 on SAT II writing)... so i dont' see why not, unless you are a sophomore or younger or feel that you can really pull up some of your scores to a total around 2300  which would be great. i'm not agaist retaking the SAT I, i was mainly talking about subject tests in my previous post

A combined SAT of 2150 should put you near the 25-40th percentile at the most elite schools - if you are not a minority, an international student, or don't have some really good reason for admission, then you would probably be in a bit of trouble.

2250 is a whole different game.

You're also a USAMO qualifier, which puts you under the 'really good reason for admission' category.

Anyways, now that I know that your question isn't hypothetical, I would say yes, you should retake it.  If you could raise it another hundred or so points, I don't see how you could be rejected, short of insulting your interviewer and sending bomb threats to the admissions office.

Thats really not the greatest thing to say in my opinion.  It could incite an ego, and also might make for a heartbreaking situation should he be rejected.  Also, its a topic no one can speak with absolute certainty on.

(underrepresented) Minority admissions are easier to determine with certainty than, say, admissions for some chinese kid.  There are preciously few minorities who can compete on an equal playing field (in terms of statistics and (non-soft) accomplishments) with the rest of the applicant body - with (multiple?) USAMO qualifications and a SAT comfortably in the 50th percentile of applicants, he would only need a solid essay and moderate extracurriculars to be almost ensured acceptance.

The AMC doesn't release ethnic breakdowns for the USAMO, as far as I know (nor should they) but the number of underrepresented minorities qualifying for the USAMO most years must be at most 5 - probably less if you consider kids who qualify the normal way (e.g, are not the highest scorer in a state without meeting the cutoff score)

[url=http://opr.princeton.edu/faculty/tje/espenshadessqptii.pdf]This study[/url] estimates that blacks get the equivalent of +230 pts added to their SAT score.  Just thought I would throw that out.  I found that article while reading wikipedia: http://en.wikipedia.org/wiki/Affirmative_Action

[quote="aznphatso"]Sorry Paladin, but I believe I three-upped you :-P
Writing 800 (MC 78 Essay 12)
CR 800
Math 800

The adrenaline high from getting those results at 3 AM (on West coast) caused me to pretty much bomb an English test that day, and a Bio test the day after, but WOOOOO because it's now Winter Break. woot

But congrats nonetheless 2370 is a great score![/quote]

thats pretty impressive, did you do a lot of practice tests, or are you just a talented test taker?

[quote="aznphatso"]Sorry Paladin, but I believe I three-upped you :-P
Writing 800 (MC 78 Essay 12)
CR 800
Math 800

The adrenaline high from getting those results at 3 AM (on West coast) caused me to pretty much bomb an English test that day, and a Bio test the day after, but WOOOOO because it's now Winter Break. woot

But congrats nonetheless 2370 is a great score![/quote]

Haha nice. I bow down to your complete domination of this standardized test  :omighty:

^^well, I wouldn't say complete domination, considering he got a 78 out of 80 on the Writing MC :P , j.k. that score is unbelievable, you're one of the 100 or so that get a perfect score on the SAT

[quote="blahblahblah"]Of more interest, perhaps, is the fact that there were almost no (e.g, ~200) black students scoring over 1500 on the (old) SAT 1 in a given year (we discussed this before; look in the archives of this forum).  If we call 2250 the 'new' 1500, that is why I was saying that 2250 was such an important score.[/quote]
I remember reading an old post by Warpedklown, which showed that he got 1540 on the old SAT (I assume that he was either in his freshman or sophomore year when he took the old SAT). Therefore, he is capable of getting over 2300 on the new SAT. He should retake the test and be a little bit more serious about it.

Practice a LOT. don't memorize word lists because that's just a waste of time. Learn idiomatic expressions such as "different _____ (From or than?)" Reading a lot of stuff like good literature helps.

[quote="tekno10m"]Hey Paladin and aznphatso, and anybody else, do you guys have any advice on how to prepare for the SAT?  Just do practice tests?[/quote]

That's what I did... there is a list of basically every type of grammar problem to appear on the SAT Writing MC, hence the reason I didn't miss any. I look at a sentence from the section now and it's like OH GOD IT'S SO MESSED UP *BUBBLES IN* next question. Yeah, Writing is easy to get good at with practice and a little guidance.

As for Critical Reading, I have no idea how I got 780. On all the practice tests I took I ranged from low 700s to like 760 at most. The curve may have been generous or I got lucky, too. But, as has been said many times on this forum, read a lot. As long as you read, you'll get both vocabulary and reading comprehension practice so that's just cool. Yeah, practice tests are quite useful as well, just to get you familiar with the question types and the general themes behind their questions.

is there a book that has all the grammer errors on the test?

[quote="WarpedKlown1335"]Hehehe I think that 1540 I mentioned before was on a practice SAT.   :blush:

Pretty close to predicting my SAT I score though, without writing I would've had a 1530.[/quote]

You did better on reading than writing :o , I think writing is much easier (the MC part) than reading because reading actually requires you to think :P

[quote="paladin8"]That's what I did... there is a list of basically every type of grammar problem to appear on the SAT Writing MC, hence the reason I didn't miss any. I look at a sentence from the section now and it's like OH GOD IT'S SO MESSED UP *BUBBLES IN* next question. Yeah, Writing is easy to get good at with practice and a little guidance.
[/quote]

Hey paladin, where did you get this "list"?

Now I'm interested. :lol:

it seems like you would be better off if you don't put your ethnicity if you're asian or even caucasian...

then why would anyone put their ethnicity/race if they're not minorities?

It still helps if you're black or hispanic though.

[quote="tokenadult"]This is a bit of topic shift, but the race/ethnicity question on college applications is OPTIONAL. A lot of applicants (and a growing number of applicants, year by year) don't answer it, as a matter of principle, and that applies to applicants from all the various ethnic groups presupposed by the questions.[/quote]

i'm sorry about not staying on topic.. but wouldn't people in admission committee know anyways since your last name probably shows your ethnicity?

In some cases, it  can be obvious like if you're last name is Gonzalez or Hernandez or other last names that are dead give aways to your ethnicity.  Yet,  in many others its not so.

I doubt you could tell if someone was black though from looking at names alone.  I mean, if you solely knew my name, you'd think I was European perhaps.  0_o.

yannik is gonna pown all other black kids applying to same college as him   :P


function
analytic geometry
algebra unsolved
algebra

For which values of $ k$ and $ d$ has the system $ x^3\plus{}y^3\equal{}2$ and $ y\equal{}kx\plus{}d$ no real solutions $ (x,y)$?

we have $ x^3\plus{}(kx\plus{}d)^3\equal{}2$ so $ (k^3\plus{}1)x^3\plus{}3kd^2x\plus{}3k^2dx^2\plus{}d^3\minus{}2\equal{}0$ so  we must have $ k^3\plus{}1\equal{}0$ (since every equation in odd degree has a real solution). and so $ k\equal{}\minus{}1$.
now we have : $ \minus{}3d^2x\plus{}3dx^2\plus{}d^3\minus{}2\equal{}0 (*)$ , if $ d\equal{}0$ then$ \minus{}2\equal{}0$ which is contradiction ,
$ (*)$ don't have solution iff $ 9d^4\minus{}12d(d^3\minus{}2)<0$ so $ d(8\minus{}d^3)<0$so $ d(2\sqrt{2}\minus{}d)<0$ .
so the system has non real solution if  and only if  $ k\equal{}\minus{}1$ and ($ d(2\sqrt{2}\minus{}d)<0$).

There's also a nice solution if you draw the function $ y\equal{}f(x)\equal{}\sqrt[3]{2\minus{}x^3}$ in a coordinate system and then look for all lines $ y\equal{}kx\plus{}d$ that don't intersect the function $ f(x)$.

yes but we must draw a good figure to this function , and there are many cases to discuss.

Good figure yes, but there aren't many cases, I mean, if you have a good figure, then you can show that the functions lies between the two lines $ y\equal{}\minus{}x\plus{}2$ and $ y\equal{}\minus{}x$.


calculus
derivative
Pythagorean Theorem
geometry
calculus computations

A 20 foot ladder slides down a wall at 5ft/sec. At what speed is the bottom sliding out when the top is 10 feet from the floor (in ft/sec).

Taking the derivatives of the pythagorean theorem,
$ x\left(\frac{dx}{dt}\right)\plus{}y\left(\frac{dy}{dt}\right)\equal{}z\left(\frac{dz}{dt}\right)$
But $ \left(\frac{dz}{dt}\right)\equal{}0$ so
$ x\left(\frac{dx}{dt}\right)\equal{}\minus{}y\left(\frac{dy}{dt}\right)$
Then from pythagoras, $ x^{2}\plus{}y^{2}\equal{}z^{2}$.  We are looking for $ x$ at $ y\equal{}10$ so $ x\equal{}\sqrt{400\minus{}100}\equal{}\sqrt{300}$.
Then we have
$ \sqrt{300}\left(\frac{dx}{dt}\right)\equal{}\minus{}10(5)\Rightarrow\left(\frac{dx}{dt}\right)\equal{}\frac{50}{\sqrt{300}}\equal{}\frac{5}{\sqrt{3}}$
The negative sign just gives direction so it isn't really important.

I won't solve the problem for you (is this your homework?) but I'll help you start.

Let x be the height of the top ladder off the ground, and y be the distance of the bottom of the latter from the wall. By the Pythagorean Theorem, we know that $ x^2 \plus{} y^2\equal{}20^2$

Why don't you try implicit differentiation, solve for dy/dt, and see where that gets you.

its not homework. im practicing for the AP Calculus exam coming up in about a month I think.

[quote]I won't solve the problem for you.[/quote]
In case you didn't notice, I already did. :lol: