Please look at this picture to get what I'm trying to solve.
http://img352.imageshack.us/my.php?image=exampleyz4.jpg

Lets say the Red dot is 748, 298 (Middle of the Orb around us).
Lets say the Aqua dot is 630, 320.
The circle around these dots is our In-Game Mini Map.

Now, when we move to the Aqua dot, our X,Y will now be (748, 298) instead of (630, 320) since where ever we move, we're always in the center of the orb around us.  So, my problem is, what is my formula to figure out the X,Y of the Red dot after moving to the Aqua dot?

From what I understand, there are three entities here, and they have these initial coordinates.
You: $ (748,298)$
Red: $ (748,298)$
Blue: $ (630,320)$
(main thing that was confusing is that Red and You are not explicitly separated in your post)
The coordinates of You are always considered to be $ (748,298)$, and everything moves relative to it.

Think of it this way. If you move to the Blue dot, its coordinates will need to become $ (748,298)$. The vector from the point $ (630,320)$ to the point $ (748,298)$ is $ \langle 118,\minus{}32 \rangle$. Move the red dot by this same vector to get its new coordinates: $ (748,298)$ added to $ \langle 118,\minus{}32 \rangle$ is $ (866,266)$.

This is not a direct formula, but that should be enough to determine it.

Okay so, if there was a Black dot at (650, 260) and I'm at (748, 298).  I now move to the Black dot, which is now (748, 298) since I'm now on top of the Black dot. Now, there is a Blue dot at (620, 280).  I move to the Blue dot which is now (748, 298) since I'm now on top of it.  Now, I want to move back to the Black dot from the Blue Dot. I would do:

748x - 650x = 98
298y - 260y = 30

748x + 98 = 846
298 + 30 = 328

So my X,Y to get back to the black dot would be (846, 328) ??

Here is the actual Mini Map I'm referring to:

http://img209.imageshack.us/my.php?image=examplexw9.png

The Green Dot is where I started (Coordinates right now arn't important).
The Blue Dot was my first move which was (723, 345) which is in the middle of the Blue Dot.
I'm the little white square that I outlined with red, which is (748, 298).
From where I am now, the Blue Dot is (807, 303) (807, 303 is not excat but close).

Using the given formula, it just doesn't quite add up.  Is there anything else we can try?

Let me get this straight. While You were at Green, you went to the (723, 345) where (748, 298) = Green. This (723, 345) is the point marked Blue.

Now, as You are at Blue, you move to the current position.

With this current position, Blue = (807, 303)

then Green -> Blue = <-25, 47>
Blue -> Green = <25, -47>
Current -> Blue = <59, 5>
Blue -> Current = <-59, -5>

Green -> Current = <-84, 42>
Current -> Green = <84, -42>
so from your current position, Green is at (832, 256).

From Blue, current position = (689, 293)
and Green = (773, 251)

From Green, current position = (664, 340)

(Sorry if I made any arithmetical mistakes...)

Is that what you're asking? And how come I don't remember anything in Runescape about pixel hunting?

I've been trying to use that formula for awhile and it doesn't really work.  Look at it like this, when I listed the (807, 303) which is the Blue dot while I'm at my current position, we arn't supposed to know that position.  So, what I want to figure out is how am I to get (807, 303) from using my current position (748, 298) and the Blue dot (723, 345). I should of said this before I posted but, I just listed (807, 303) as to check our answers.  Thanks for your reply.

I'm sorry, I cannot understand your posts at all. As far as I can see this is about relative motion and should have the simple solution of translating all points by the same amount, but you claim it's not working. Unless there's some complication you're not making clear, it should.

Here is my last try. Let's suppose that in an initial position we have the following coordinates for "You" and some other objects

"You": $ (x_0,y_0)$
O1: $ (x_1,y_1)$
O2: $ (x_2,y_2)$
O3: $ (x_3,y_3)$

"You" want to move to O1. Because of the way the map works, the position of "You" needs to stay $ (x_0,y_0)$. So what we do instead is shift all of the other objects so that O1 and "You" coincide. Equivalently, O1 will have coordinates $ (x_0,y_0)$ after the shift.

In order to get $ (x_1,y_1)$ to $ (x_0,y_0)$, we have to shift by the vector
\[ [\text{destination}] \minus{} [\text{start}] \equal{} (x_0,y_0) \minus{} (x_1,y_1) \equal{} \boxed{(x_0\minus{}x_1,y_0\minus{}y_1)}\]
Apply this transformation to every object except "You" - that is, O1, O2, and O3. Our new coordinates will then be

"You": $ (x_0,y_0)$
O1: $ (x_1,y_1) \plus{} (x_0\minus{}x_1,y_0\minus{}y_1) \equal{} (x_0,y_0)$
O2: $ (x_2,y_2) \plus{} (x_0\minus{}x_1,y_0\minus{}y_1) \equal{} (x_2\plus{}x_0\minus{}x_1,y_2\plus{}y_0\minus{}y_1)$
O3: $ (x_3,y_3) \plus{} (x_0\minus{}x_1,y_0\minus{}y_1) \equal{} (x_3\plus{}x_0\minus{}x_1,y_3\plus{}y_0\minus{}y_1)$

Being general this should work for any example you might have, unless as said above you are not making something clear.

MellowMelon, I just re-did your first formula and now it works. I guess I was adding something else to get my false results.  So:

(728, 298) - Position I'm going to move to + (748, 298).

So again, it was my fault and you gave me a perfectly working formula.  Thanks again MellowMelon and to math_explorer for your response. What a relief... :p


Να λυθει το συστημα στους πραγματικους για $ n\geq3$:

$ x_n^3\equal{}x_1\plus{}x_2\plus{}1$

$ x_1^3\equal{}x_2\plus{}x_3\plus{}1$
.
.
.
$ x_{n\minus{}1}^3\equal{}x_n\plus{}x_1\plus{}1$

Spera Alexandre. Poli oraia kai dyskoli askisi. Vazw mia lisi pou mou irthe alla epidi den exo poli xrono kai den tin exw koitaksei arketa den pairnw kai orko gia to an xanei kapou i oxi (sorry gia ta greeklish kai gia to kako latex alla ta exei paiksei afto to pc kai exei xalasei kai to laptop mou)

An kanoume afaireseis kata melh me oles tis diadoxikes eksisoseis exoume to systima:

$ {x_i}^3 - {x_{i + 1}}^3 = x_{i + 1} - x_{i + 3}$ i isodinama
$ (x_{i + 1} - x_{i + 3})({x_i} - {x_{i + 1}) = ({x_i} - {x_{i + 1})^2(x_ix_{i + 1} + {x_i}^2 + {x_{i + 1}}^2) > = 0, x_{n+i} = x_i.}}$

Etsi prokeiptoun oi anisotikes sxeseis:

$ x_1 > = x_2 > = x_4$ i anapoda
$ x_2 > = x_3 > = x_5$ i anapoda
.
.
.
$ x_n > = x_1 > = x_3$ i anapoda

An $ x_{i + 1} > = x_{i + 3}$ kai $ x_{i + 1} = < x_{i + 2}$ tote tha einai $ x_{i + 2} > = x_{i + 3}$
Akoma an $ x_{i + 3} > = x_{i + 1}$ kai $ x_{i + 1} > = x_{i + 2}$ tote tha einai $ x_{i + 3} > = x_{i + 2}$

Ara oi parapanw triples anisotites tha exoun eite oles tin idia fora, eite i fora tous tha pigainei enalaks. Kai stis dio periptoseis odigoumaste sto $ x_1 = x_2 = ... = x_n = x$ (giati?). Opote $ x^3 = 2x + 1$ kai $ x_i = - 1$ i $ x_i = (1 + / - riza5)/2$ gia kathe  $ 1 = < i = < n$.


geometry
rhombus
search
geometry proposed

If ABCD is a rhombus, and the circumradii of ABD, ACD are 12.5, 25. What is the area of the rhombus?


Draw a grid on the diagram for 25 little squares, each 1 sq unit. 7 of these.

Also, you can take the somewhat large rectangle in the bottom left corner. It is 4*4=16, and subtractind the 3*3=9 square out, you get the l shape as 7 units^2.


Let $\{a_{n}\}_{n \ge 1}$ be a sequence of positive numbers such that \[a_{n+1}^{2}= a_{n}+1, \;\; n \in \mathbb{N}.\] Show that the sequence contains an irrational number.

[i]Reductio ad absurdum..[/i]
So, let us assume that all those $a_n$ are rational, so we may write, for some coprime $p_n, q_n, \ \ a_n = \frac {p_n}{q_n}, n \geq 1.$ Hence, using the condition we get that $q_{n+1}^2 = q_n,$ so that from some point $a_n$ will be an integer.
Now, let's assume that we've reached this point, then clearly $a_n$ cannot equal $1,$ whence $a_{n+1} < a_n.$
Then, from this point, the sequence is a strictly decreasing sequence of positive, which is a contradiction, so we're done.

or simply observe that by the condition each $ a_n \plus{} 1$ must be a perfect square, and $ (a_n \plus{} 1)*(a_{n \plus{} 1} \plus{} 1) \equal{} (a_n*a_{n \plus{} 1})^2$ will get $ a_n \equal{} 1$, which is a contradiction since 2 is not a perfect square.

p.s.  i don't and can't believe this is a putnam problem.

It's definitely not on the 1995 Putnam, at least. Check your sources again.


If $ a,b,c\in [1,2]$ and $ a\plus{}b\plus{}c\equal{}4$ , then :

$ a\sqrt{b\plus{}c}\plus{}b\sqrt{c\plus{}a}\plus{}c\sqrt{a\plus{}b}\le 4\sqrt{3}$

[quote="Pain rinnegan"]If $ a,b,c\in [1,2]$ and $ a \plus{} b \plus{} c \equal{} 4$ , then :

$ a\sqrt {b \plus{} c} \plus{} b\sqrt {c \plus{} a} \plus{} c\sqrt {a \plus{} b}\le 4\sqrt {3}$[/quote]

[hide]Dividing by $ 4$ ,we have that $ \frac {a}{4} \plus{} \frac {b}{4} \plus{} \frac {c}{4} \equal{} 1$

Define $ f(x) \equal{} \sqrt {x}$ ,which is concave since $ f''(x) < 0$

By Jensen's inequality we obtain

$ LHS \leq f\left(\frac {\frac {a}{4}(b \plus{} c) \plus{} \frac {b}{4}(c \plus{} a) \plus{} \frac {c}{4}(a \plus{} b)}{\frac {a}{4} \plus{} \frac {b}{4} \plus{} \frac {c}{4}}\right)\leq \sqrt 3 \iff a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \leq 12 \iff a(4 \minus{} a) \plus{} b(4 \minus{} b) \plus{} c(4 \minus{} c) \leq 12 \iff a^2 \plus{} b^2 \plus{} c^2 \geq 4$

Since $ 2(a^2 \plus{} b^2 \plus{} c^2 ) \geq 2(ab \plus{} bc \plus{} ca) \implies a^2 \plus{} b^2 \plus{} c^2 \geq \frac {16}{3} > 4$[/hide]

I'm having again troubles with equality case.If I'm wrong again ,can you post the correct solution ?
Thank you great math

No,enndb0x, your solution is right until the substitution changes it sign. It must be

$ a(b\plus{}c)\equal{}a(4\minus{}a)$ instead of $ a(b\plus{}c)\equal{}a(a\minus{}4)$

AM-GM or Holder are less messy than jensen: $ (\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^3 \le 3(a\plus{}b\plus{}c)$

assuming $ a\ge b \ge c$, by Chebyshev

$ LHS \le \frac{1}{3}(a\plus{}b\plus{}c)\left(\sum \sqrt{a\plus{}b}\right) \le \frac{1}{3}\cdot 4 \sqrt[3]{3\cdot 2(a\plus{}b\plus{}c)} \equal{} \frac{4}{3}\sqrt[3]{24} < 4\sqrt{3}$

and there's no equality

Cauchy gives $ a\sqrt {b \plus{} c} \plus{} b\sqrt {c \plus{} a} \plus{} c\sqrt {a \plus{} b}\le \sqrt {2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$ , like [hide="this"]$ LHS\equal{}\sum{\sqrt{a}\sqrt{a(b\plus{}c)}}\le\sqrt{(a\plus{}b\plus{}c)[a(b\plus{}c)\plus{}b(c\plus{}a)\plus{}c(a\plus{}b)]}\equal{}RHS$[/hide] .

and with $ a,b,c\in [1,2]$ we can prove : $ 3(a \plus{} b \plus{} c)\ge 6 \plus{} ab \plus{} bc \plus{} ca$ ([hide="Try it!"]$ 3(a \plus{} b \plus{} c)\ge 6 \plus{} ab \plus{} bc \plus{} ca\Leftrightarrow (a \minus{} 1)(2 \minus{} b) \plus{} (b \minus{} 1)(2 \minus{} c) \plus{} (c \minus{} 1)(2 \minus{} a)\ge 0$[/hide])

Still no equality case .

Perhaps it helps that $ a$, $ b$, and $ c$ are sides of a triangle...

Can we find a better constant than $ 4\sqrt{3}$?

You are right, enndb0x!

To Altheman, the solution of ocha has the best constant since the equality case is $ a\equal{}b\equal{}c$ and the fact that we do not need to use Chebyshev in this case because AM-GM works well


IMO Shortlist
Set systems
Extremal combinatorics
combinatorics

Let $ N \equal{} \{1,2 \ldots, n\}, n \geq 2.$ A collection $ F \equal{} \{A_1, \ldots, A_t\}$ of subsets $ A_i \subseteq N,$ $ i \equal{} 1, \ldots, t,$ is said to be separating, if for every pair $ \{x,y\} \subseteq N,$ there is a set $ A_i \in F$ so that $ A_i \cap \{x,y\}$ contains just one element. $ F$ is said to be covering, if every element of $ N$ is contained in at least one set $ A_i \in F.$ What is the smallest value $ f(n)$ of $ t,$ so there is a set $ F \equal{} \{A_1, \ldots, A_t\}$ which is simultaneously separating and covering?

I believe the answer is $f(n)=\lceil\log[2]{(n+1)}\rceil\ (*)$.

For each $i\in\overline{1,n}$, let $v_i$ be the binary vector of length $f(n)$ whose $j$’th component is $1$ iff $i$ belongs to the $j$’th set $A_j$. The hypothesis tells us that these vectors are non-zero and no two are equal, so a necessary and sufficient condition for their existence is $2^{f(n)}-1\ge n$, which gives $(*)$.

[quote="grobber"]I believe the answer is $f(n)=\lceil\log[2]{(n+1)}\rceil\ (*)$.[/quote]

I am hereby to confirm that your claim is correct, as I am possessing the book [i]'An Olympiad Down Under'[/i] which contains detailed solutions of all the long-listed problems on the 29th IMO.

The solution appeared in the book writes:
$f(n)= \lceil \log_2 {n} \rceil +1$ if $n=2^r$ and $f(n)= \lceil \log_2 {n} \rceil $ if $n\neq2^r$
Indeed, I have checked that it's the same as yours.

Let $s(n)$ denote the smallest number of seperating subsets of $\{1, \cdots, n\}$. $cs(n)$ denote the smallest number of both seperating and covering subset of $\{1, \cdots, n\}$. We easily get the recursion:
a. $s(n) = 1 + min\{max(s(2), s(n-2)), max(s(3), s(n-3)), \cdots \}$
b. $s(n)+1 = cs(n)$.

From $a$, we easily get $s(n) = \lceil log_2 n \rceil$, so $cs(n) = s(n) + 1 = \lceil log_2 n \rceil + 1$


The cells of the square $8 \times 8$ are colored in $N$ colors. Columns $A$ and $B$ of the square are said to be [i]antiequal[/i] if there exist colors $C_A$ and $C_B$ such that:
(i) $C_A$ occurs in $A$ but doesn't occur in $B$;
(ii) $C_B$ occurs in $B$ but doesn't occur in $A$.
Similarly define antiequal rows. Find the minimal number $N$ for which there exist a coloring such that every two columns are antiequal and every two rows are antiequal.

It can be done with 5 colors, even an 10x10 board. Take the 10 three-color subsets $S_1,...S_{10}$. Note that $S_i\cap S_j\neq \{\}$ for all $i,j$. Color cell $(i,j)$ with any color $\in S_i\cap S_j$. Done.
It is easy to show that it's not possible with 4 colors. For that, we would need 8 subsets $S_1,...S_8$ (corresponding to the unions of the colors of each, say, row) such that  $S_i\not\subset S_j$ for all $i,j$. So at least 4 of those can only have two colors, as there are only 4 three-color subsets. But if there is a three-color subset $\{A,B,C\}$, this already excludes 3 of the 6 two-color subsets. Which means it isn't possible.
For a smaller board, those subsets may exist but it turns out that even for a 5x5 one, a coloring with 4 colors is not possible.


I think these are good problems but fit a little better at Intermediate level.

I dont' know. I help coach at our middle school Mathcounts team, and each member of the team can understand how to do those types of problems. I think if these are pretty essential to MC, that they should belong in the Beginner's Forum. And our team isn't that strong either - in practice sets they rarely get over 10 problems on chapters.  :?

[quote="white_horse_king88"]1) The average of a, b, s, and t is 6 and the average of s and t is 3. WHat is the average of a and b?

2) The vertices of a square are at (-3,4), (3,4), (3,-2), and (-3,-2). WHat is the area of the square?

3) In a stack of six cards, each card is labeled with a different integer 0 through 5. If two cards are selected at random without replacement, what is the probability that their sum will be 3?

4) What is the difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd cuonting numbers?

5) What is the probability that a randomly drawn positive factor of 48 is less than 11?

6) What is the units digit of $133^{1987}$? [hide="Hint"] Try 3^{1987} and find a pattern. [/hide]

Answer in spoiler please.[/quote]
[hide="The Answers"]
1)
a + b + s + t = 24
s + t = 6
a + b = 18
9

2)
Each side has a length of 6, so the answer's 36.

3)
There are four ways out of thirtysix total combinations, so...
1/9

4)
Each even counting number is one more than its odd counting number counterpart, so (2003)(1), or 2003, is the answer.

5)
The postive factors of 48:
1
2
3
4
6
8
12
16
24
48
6 of these 10 are less than 11, so the answer's 3/5.

6)
3^1 ends in 3
3^2 ends in 9
3^3 ends in 7
3^4 ends in 1
3^5 ends in 3
...
3^1987 ends in 7
7
[/hide]

[quote]There are four ways out of thirtysix total combinations, so...
[/quote]
Would there be 30 combinations since there is no replacement?

Good they are all right except 3. That's a trivial mistake - I'm sure you'll see it upon review.


A man rows his boat ashore, and immediately gets out and goes into a restaurant. He sees on the menu "pelican sandwich", which happens to be a food he likes a lot. He takes one bite of the sandwich, and runs out of the restaurant and throws himself into the ocean. Explain.

Isn't this like the albatross problem? I stilll don't understand the suicide though.

I don't know what the albatross problem is, but there definitely exists a reason for the suicide in the problem.

This problem has something to do with eating a person right?

-edit-
I meant abalone problem... lol :rotfl:

But how does that fit into "pelican sandwich"?  Did the people at the restaurant lie?

[quote="me@home"]I've heard this before, and I think a lot of people have.
[hide="spoiler"]
yes it has to do with cannibalism and the man knew he was eating one of his friends... that had been stranded with him or something like that[/hide][/quote]

yeah me too, but why does that make the person suicide? That's stupid

how does the person realize that he's eating the people?

[hide="spoiler 2"]
the pelican sandwich tastes different from the "pelican" (which was really a human) that whoever he was stranded with had given him to eat
[/hide]

Since everyone seems to be spoiling it, I'll just fully explain it.

[hide]The man and two of his friends were stranded on an island. One of his friends died, and the other suggested that they eat him. He said no, and so the other friend made him a "pelican sandwich", which was actually made of his dead friend. The man ate the pelican sandwich, and was then rescued. When he got back, he ate a real pelican sandwich, realized that it tasted totally different, and that he had eaten his friend. Unable to live with himself, he commited suicide.[/hide]

why can't he still live? I still don't get that part

Actually, he was stranded with his wife and a friend. He ate his wife, and then killed himself once he realized what he had done.

Sorry for starting the "everybody's spoiling" trend, but most people do already know this one (as evinced by the fact that everyone is in fact spoiling it). Thanks for filling in the rest of the story... I had forgotten it.

can you give me a link where you found this problem (if it's on the internet of course)


function
calculus
derivative
complex numbers
complex analysis

Are the following function analytical?
1) f(z)=e^k   (where z=a+bi, k=a-bi)
2) f(z)=z|z|²
3) f(z)=z|z|

$ f(z)\equal{}f(x\plus{}iy)\equal{}e^{x\plus{}iy}\equal{}e^{x}\left(\cos(y)\plus{}i\sin(y)\right)$. So saying that $ f(z)\equal{}u\plus{}iv$, then $ u\equal{}e^x\cos(y)$ and $ v\equal{}e^x\sin(y)$.

Are the equations $ \frac{du}{dx}\equal{}\frac{dv}{dy}$ and $ \frac{du}{dy}\equal{}\minus{}\frac{dv}{dx}$ satisfied for all points?

Similarly break those functions into real and imaginary parts and use the Cauchy-Riemann Equations.

Additionaly, the real and imaginary parts of f, the functions u(x,y) and v(x,y), must be and have continuous partial derivatives.

One characterization of the Cauchy-Riemann equations: $ f$ is analytic if $ \frac{\partial f}{\partial\overline{z}}\equal{}0$, where $ \frac{\partial f}{\partial\overline{z}}\equal{}\frac12\left(\frac{\partial f}{\partial x}\plus{}i\frac{\partial f}{\partial y}\right)$. This also comes with the operator $ \frac{\partial}{\partial z}\equal{}\frac12\left(\frac{\partial}{\partial x}\minus{}i\frac{\partial}{\partial y}\right)$, which is the complex derivative when that exists.

These operators commute, and $ \frac{\partial}{\partial z}$ is zero on conjugates of analytic functions. They obey the usual chain and product rules, so calculation is easy:

$ f_1(z)\equal{}e^{\overline{z}}$:
$ \frac{\partial f_1}{\partial\overline{z}}\equal{}\frac{\partial \overline{z}}{\partial\overline{z}}\cdot e^{\overline{z}}\equal{}e^{\overline{z}}$

$ f_2(z)\equal{}z^2\cdot \overline{z}$:
$ \frac{\partial f_2}{\partial\overline{z}}\equal{}0\cdot\overline{z}\plus{}z^2\cdot 1\equal{}z^2$

$ f_3(z)\equal{}z\sqrt{z\overline{z}}$:
$ \frac{\partial f_3}{\partial\overline{z}}\equal{}0\cdot |z|\plus{}z\frac{0\cdot\overline{z}\plus{}z\cdot 1}{2\sqrt{z\overline{z}}}\equal{} \frac{z^2}{2|z|}$

None of these are zero, so none of these are analytic. The second and third are at least complex differentiable at the origin, with derivative zero.


[quote="mr. math"]Whats the most efficient way to solve this:

  AB
[u]x C[/u]
  CD

What is the largest possible value of the four digit number ABCD?[/quote]

(10A+B)C=10C+D
10AC+BC=10C+D
C(10A-10+B)=D

If A=2, then C(10+B)=D, and D must be one digit. This doesn't work, so A=1:

BC=D.

The maximum is 9*1=9, so 1919.

I'm not sure this is right.


geometry trigonometry AMC USA(J)MO USAMO search probability

I was going to get a geometry book for one of my friends' birthday.  Checkin through the lists of books under Resources, I thought Challenging Problems in Geometry or Geometry Revisted would be best.  Has anyone read these books?  The kid has taken high school geometry, and is currently taking trigonometry and pre-calculus.  (geometry is his favorite type of math).  Which book do you think would be best for him?  What did you think of these books?  Can you think of any others that might be better?  thanks

I've had Challenging Problems in Geometry for quite some time and I just checked out Geom. Revisited.  They are both excellent books.  Challenging Problems, well, has mostly problems whereas Geom revisited tries to teach the material more.  The material in Challenging Problems isn't as advanced as Geometry Revisited, I would say.   It's really hard to say which he would like more.  I don't think you could go wrong either way though ;)

It depends on exactly how much your friend likes geometry and how good he/she is at solving problems. I would say Geometry Revisited (for anyone not looking into USAMO problems) would just be another book on the shelf. It's much too difficult for light reading.

I have Geometry Revisited. Excellent book for perusing through and refreshing your memories of USAMO-type geometry... (although I would never call it a perusing, I do mean like 5-hour study for a unit, because I study 10hours on math in extreme :D )

So if you feel like your friend is at that level... well it would be a good gift.

As for the other book, i do not know anything about it.

Questions of this sort are best suited to "Other Problem Solving Topics."  In addition, I'm sure that this has been discussed here before and you might do a search for other opinions on those books.

Ok, so I [i]thought[/i] his favorite subject was geometry, but when books came up in a conversation I snatched the opportunity to ask him some questions on what kind of books he might like.  Quoted exactly:

"I like numbers.... and probability? and and... fractals... and ..... 0 and i and and.... umm.... pi and and.... euler stuff and... and.... polygons and stars and
yeah...
am I missing anything other than subjects?
yeah
maybe I'll settle with numbers...
rational and irrational"

So now I'm confused, I'm thinking of getting two if they're not very expensive but does anyone have suggestions for this book based on what he said?  (and btw, i think I've chosen Challenging Problems in Geometry over Geometry Revisited, but I'll probably buy them both and figure it out after looking at them.. then keep the other  :lol: )

Both Challenging Problems in Geometry and Geometry revisited are excellent books... Geometry Revisited have more "lectures" or "lessons", but it gives out a LOT of CONCEPTS(theorems/formulas etc.) As i said before, G Revisited is good for USAMO prep, or maybe IMO (not sure about that :D ).

Challenging Problems in Geometry is more problem-oriented, althoguh ive heard that it also does some explaining (i dont own it). It was cheap too so doesnt hurt you to get both books. Only if you dont study then it will be a waste. But I certainly recommend both books.  Great books! :lol:

Ah yes, I am going to get both, but when I said I would get two, I meant I'd get two for him, so I guess technically three.  So what I'm asking is for recommendations of other books that I haven't mentioned here that people think he might like, based on what he said.

uggh, but he is mentioning TOO MANY topics. I do not know a book that covers all those topics... maybe AoPS???

Perhaps [i]What is Mathematics?[/i] by Courant and Robbins is the sort of book you're looking for.


It is now OK to discuss to 2009 AMC 10B/12B.

Reminder: we will be having a Math Jam on Thursday (tonight!) to discuss the problems from the AMC 10B/12B.  The Math Jam will start at 7 PM Eastern / 4 PM Pacific (note earlier-that-usual start time for this Math Jam).  Click on the "Math Jams" link at the top of the page for more information about Math Jams.

The [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=250]transcript[/url] from last night's Math Jam is now available.


probability
search
combinatorics unsolved
combinatorics

Given 2 very large numbers, what is the probability that they are co primes?

Here's a very old one:[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=358097764&t=1089[/url]


topology
function
trigonometry
real analysis
real analysis theorems

As I recently started taking a course in introductory topology, I am met for the hundreth time in my education of some type of discussion trying to describe the real numbers as quickly as possible.

Most books I've had start with the least upper bound property and then show Bolzano-Weierstrass for real sequences and/or show that Cauchy sequences converge. Always these proofs seem to be unintuitive, tricky and hard to understand. The way I would like to do both proofs is to look at the $\liminf$ of the sequence. It is easy to show that there is a subsequence converging to the $\liminf$, and it is easy to show that the Cauchy sequence converges to the $\liminf$. Furthermore, doing the proof this way is very natural.

However, I have not seen this done. Rather, one extracts a monotone sequence or something similar. Why is this? Is there some type of fallacy to just looking at the $\liminf$?

When I first saw the argument you speak of in I don't know what book, I wondered the same thing.
Furthermore, I think that you can also use the "bisection method" and then Cantor's Lemma (nested intervals), but that's a little more complicated.

I don't see what's unintuitive about it. I started from the reals are the unique Archimedian field with the property that every nested sequence of closed intervals has a common member. The derivation of Cauchy's criterion from there is delicate and depends on Bolzano-Weierstrass but I think breaking it into steps is just as good an approach.

Hard to understand at this level mostly has to do with 'maturity'. And either way one is going to have to be familiar with the multiple approaches at one point or another :ninja:

I feel that your post is a little condescending towards me. First of all, I don't know if we're discussing the same type of proof. I have never seen a formal construction of the real numbers. What I am talking about is how texts simply list the defining properties and then give some very specialized argument to prove completeness that does not provide any particular insight to the real numbers. By "hard to understand" I just mean hard to put into a bigger picture, not hard to follow as "proofs". On the other hand, the $\liminf$ concept is natural and we know from the start that the limit should be the $\liminf$ so why not just use it?

Could you say what textbooks you are talking about?

If you want a formal construction of the real numbers, there's one in Rudin's "Principles of Mathematical Analysis".

Some of my ideas:

There is nothing "real" about the real numbers, They are a construction, a figment of human mathematical imagination. I have a colleague (of a combinatorial/discete bent) who says, "Show me a real number - anywhere" and he has a point.

But once you do accept the construction (either Dedekind cuts or Weierstrassian equivalence classes of Cauchy sequences), then what is the essence of what "complete" means? It means that there are enough real numbers for there to be a number everywhere you think there ought to be a number. The true power of the concept is that it allows you to assert the existence of limits without [i]a priori[/i] naming the limit. It allows you do discuss numbers and functions whose names you do not know. I'm talking about numbers like "the unique solution of $\cos x=x$" or functions like the antiderivative of $e^{x^{4}}.$

The following notions are either logically equivalent to the completeness of the real numbers, or this notion together with a separate assertion of the Archimedean property is equivalent to completeness.

1. A nonempty set with an upper bound has a least upper bound.

2. If we partition $\mathbb{R}$ into sets $A$ and $B$ where every element of $A$ is less than every element of $B,$ then either $A$ has a greatest element or $B$ has a least element.

3. Monotone bounded sequences converge.

4. Cauchy sequences converge.

5. Bounded sequences have convergent subsequences (Bolzano-Weierstrass).

6. A nested decreasing sequence of closed intervals has a nonempty intersection.

7. A decimal expansion represents a real number.

Note that in all seven cases the essential assertion is that [b]there exists[/b] a real number with the desired property - whether you know the name of that real number or not. In particular, #3, 4, and 5, have as conclusion that a sequence (or subsequence) converges, but that's really the assertion that there is a real number out there that can be the limit.

An intersting exercise is to start with any one of these seven notions (separately assuming Archimedean, if necessary) as an axiom and finding a pathway for proving the other six as theorems.

To some extent, Kalle is talking about alternate pathways through the seven properties that I gave, perhaps also including  a #3.5: bounded sequences have limsups and liminfs. I'll just say that there are alternate pathways; my own might well be $1\implies 3\implies 6\implies 5\implies 4.$ but you can also (and I think this is Kalle's suggestion) use $1\implies 3\implies 3.5\implies 4.$

Try $3\implies 6\implies 2\implies 1$ as a loop-closing pathway.

§outh§star alluded to the uniqueness result - that up to field isomorphism, there is only one complete ordered field. This result is intimately tied to constructions of the reals and to these equivalent notions of completeness.

("Archimedean" is the notion that you can "empty the ocean with a teaspoon." That is, if $x,y>0,$ then there exists a natural number $n$ such that $nx>y.$)


blah. can we just keep this as open field for engel questions? fyty has posted a few and i would like to. blah must go now will post some in evening.

Some of Engel's solutions have a few...holes for beginners.

A lot of Engel solutions aren't anywhere near complete.

Yeah, often he just jots down a random point or hint.

[quote="MysticTerminator"]tiling problems from Engel: find recursions for a_n, where a_n is the number of ways you can tile a ...

1. 2 by n rectangle with 2 by 1 dominoes.
2. 2 by n rectangle with 1 by 1 squares and L-trominoes.
3. 2 by n rectangle with 2 by 2 squares and L-trominoes.
4. 3 by n rectangle with 2 by 1 dominoes.
5. 4 by n rectangle with 3 by 1 dominoes.
6. 2 by n rectangle with 1 by 1 and 2 by 1 tiles.
7. 4 by n rectangle with 2 by 1 dominoes.
8. 2 by 2 by n box with 1 by 1 by 2 bricks.[/quote]
1. Make a table of the first few values.
[tab] n | a_n
[tab]---------
[tab] 1 | 1
[tab] 2 | 2
[tab] 3 | 3
[tab] 4 | 5
[tab] 5 | 8
At this point, I claim a_0=1 a_1 = 1, a_n=a_(n-1)+a_(n-2) for n>1[tab]

..which is easily proven by seeing that (assuming the n side is horizontal) if you start with a vertical domino, theres an n-1 x 2 rectangle left, and if you start horizontal, there must be a horizontal under that leaving a n-2 x 2 rectangle.

[quote="Rep123max"]whats engel?[/quote]

http://www.amazon.com/exec/obidos/ASIN/0387982191/artofproblems-20

MysticTerminator wrote:tiling problems from Engel: find recursions for a_n, where a_n is the number of ways you can tile a ...

2. 2 by n rectangle with 1 by 1 squares and L-trominoes.
3. 2 by n rectangle with 2 by 2 squares and L-trominoes.
4. 3 by n rectangle with 2 by 1 dominoes.
5. 4 by n rectangle with 3 by 1 dominoes.
6. 2 by n rectangle with 1 by 1 and 2 by 1 tiles.
7. 4 by n rectangle with 2 by 1 dominoes.
8. 2 by 2 by n box with 1 by 1 by 2 bricks.

2. [hide]If we put 2 squares on the left-most edge, we get an - 1.  We can fill up the leftmost two rows in 4 ways with a tetromino and a square.  So that gives us 4an - 2.  And we can fill up the three leftmost rows in 2 ways with tetrominoes.  So

an = an - 1 + 4an - 2 + 2an - 3[/hide]

Does everyone understand how to go about the rest of these problems?  They are definitely more doable then the first set of ten he gave.  I would recommend that everyone try them and that whoever does the splitting leaves them in Intermediate, hopefully moving the harder ones to Advanced.


Hic co ket qua vong 2 lau qua roi ma may thay van chua chiu tap trung doi tuyen de hoc nua. :ninja:  :ninja: . Nam ngoai hinh nhu 22 la tap trung rui.
 Sao toi cai nam cua minh cai gi no cung con rua bo. Vong 1 cho lau lac, vong 2 cung vay. Bo 5 ban doi tuyen con lai ai co thong tin gi lam on noi cho tui biet voi. Sot ruot qua. :?


calculus
integration
function
real analysis
real analysis unsolved

What is solution $ \int_{}^{} e^{ \minus{}x^{2} }dx \equal{} ?$

It is what it is.  (There is no elementary form for this integral.  Up to a constant multiple and constant of integration, it's the error function erf, but this is just another way of saying my first sentence ;).)


topology
geometry
vector
function
calculus
integration
advanced fields

Let $ U\equal{}(\minus{}1,1) \times (\minus{}1,1) \subset \mathbb{R}^2$ and $ X , Y$ are two smooth vector fields on $ U$ s.t. in any point of $ U$ there are linearly independent. Prove that, there exist non-zero smooth functions $ f, g : U \rightarrow \mathbb{R}$ s.t. $ [fX, gY] \equal{}0$. (where $ [,]$ is the Lie bracket and $ (fX)(x)\equal{}f(x)X(x)$.)

The main idea is based on investigation how the flow of $ X$, $ \phi$ changes when it's multiplied by $ f$ and the answer is: it changes by the integral of $ f$, that is, $ \phi (\int f)$ and the rest is to show that (locally) for what kind of suitable functions the flows commute. (note that, the problem is locally correct.)


inequalities
real analysis
real analysis unsolved

We have $x_{2k}=4(\frac 1{2k-1}-\frac 1{2k-3}+\ldots)+(-1)^kx_0$ and $x_{2k+1}=4(\frac 1{2k}-\frac 1{2k-2}+\ldots)+(-1)^kx_1$. Because the sequence is decreasing, it must have a limit, and this alone determines $x_0,x_1$. I think $x_0=\pi, x_1=\frac{\pi}2$. However, the inequalities $x_{n+1}<x_n$. I'm too lazy to do that, but it must be Ok.. :)


function
calculus
trigonometry
limit
analytic geometry
graphing lines
slope

Q. Approximate tan(61degrees)
This is how my book solved this problem.
f(pi/3 + pi/180) = tan(pi/3) +(sec(pi/3))^2 * pi/180 = 1.802

Here' how I solved it.

f(60 + 1) = tan(60degrees) + (sec60)^2 * 1

And obviously I got the wrong answer since I used degrees. But why can I not use degrees in this problem?  Shouldn't I get the same answer?

The problem here is $ 1\neq 1^\circ$. :)

hmmm... then what about pi/180 and pi/180 radian? Are they not equal?

Never, ever use degrees in calculus.  All the calculus trig formulas you know (starting with the most basic one, $ \lim_{x \to 0} \frac {\sin x}{x} \equal{} 1$, from which all others follow) are true [i]under the assumption that the arguments of trig functions are taken in radians[/i].  To convince yourself of this, it might help to look at the graph of $ \sin x$ when $ x$ is plotted in degrees -- does the slope of the graph really look like $ 1$ at the origin?

Of course, translating from degrees is a very simple change of variables $ t \equal{} \frac {\pi}{180} \cdot x$ (or the other way around), and so you could restate all your trig formulas in terms of degrees.  The result would be a proliferation of factors of $ \frac {\pi}{180}$, coming from this change of variables and the chain rule.  In particular, if the variable $ x$ is in degrees then it is [b][i]not[/i][/b] true that $ \frac {d}{dx} \tan x \equal{} \sec^2 x$.

You're already using a fair amount of [[LaTeX]] code -- why not go all the way and actually use it for real?  See that link and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

[quote="JBL"]Never, ever use degrees in calculus. [/quote]
About 20 years ago, just before graphing calculators became popular, I had an exercise I used as homework in a sophomore-level differential equations class for engineers. I took a linear damped oscillator problem, and gave it physical values in MKS units: such and such a mass, spring strength, damping force. I gave three versions of the problem, with two of the constants the same and one of them (I think I used the damping constant) given as three moderately differing values. That gave me three versions of the problem which were, of course, overdamped, critically damped, and underdamped. I gave each the same initial conditions: zero position, same initial velocity. I then asked for careful graphs of these solutions over a certain interval, and strongly suggested the use of graph paper. (As I said, before graphing calculators and before every student had a computer.)

My point was supposed to be that despite the three different algebraic forms of the answers, the graphs were quite similar. They both rose to a maximum in the same general area - the underdamped case a little higher, the overdamped case a little lower - then went back down to near zero and stayed there. (The fact that the mildly underdamped case passed below zero was barely visible at the resolution of the graph.)

But I'd get several papers in which the overdamped and critically damped cases looked as they were supposed to look - but the underdamped case was smashed flat against the horizontal axis as a very different graph. What?

I finally figured out what was wrong: those students had their (scientific) calculators set on degrees instead of radians.

When JBL says "calculus," he is most emphatically including differential equations within that word.

Thank you for the reply, JBL. I understand the point you're making in the second paragraph. But I don't really understand this: "To convince yourself of this, it might help to look at the graph of sin x when x is plotted in degrees -- does the slope of the graph really look like 1 at the origin?" To me, the graph of sinx looks the same whether the domain is in degree or in radian.  And the slope at the origin almost looks like 1 to me...

When I plot $ y \equal{} \sin x$ for $ x \geq 0$, it reaches its first local maximum at $ x \equal{} \frac {\pi}{2}$ -- it has an average rate of change of $ \frac {1}{\frac {\pi}{2}} \equal{} \frac {2}{\pi} \approx 0.6366\ldots$ between the origin and this first maximum.  When I plot $ y \equal{} \sin x^\circ$ for $ x \geq 0$, it reaches its first local maximum at $ x \equal{} 90$ -- it has an average rate of change of $ \frac {1}{90} \approx 0.0111\ldots$ between the origin and this first maximum.  Perhaps you are forgetting that to judge slope from a graph, you need one unit on your $ x$-axis to be the same length as one unit on your $ y$-axis?  (My computer refers to this as "aspect ratio 1.")  Of course, the two graphs will look the same if you stretch the $ x$-axis (and [i]not[/i] the $ y$-axis) by a factor of $ \frac{\pi}{180}$ -- but this changes all the slopes (in a way that is exactly equivalent to what I said about changing variables).

[quote]When I plot $ y \equal{} \sin x^\circ$ for $ x \geq 0,$ it reaches its first local maximum at ...[/quote]
... at somewhere that's way off the right edge of the paper. Hence the problem with my ODE graphing problem.

I see. I understand your point now. Thank you!


Show that cos(x)-siny>1/8
where x=sqrt(3)/2 and y=1/2

It is easy to show that x > sin x for all positive x. (Take derivative.)

So it is sufficient to prove cos x - y > 1/8

<=> cos x > 5/8

The algebraic form of cos x = 1 - x^2 / 2! + x^4 / 4! ...

It is easy to see that x^p / p! > x^(p + 2) / (p + 2)! for x < 1 and p integer.

So cos x > 1 - x^2 / 2! = 1 - (3/4)/2 = 1 - 3/8 = 5/8, as desired.


Are there plugins for programs such as google talk or aim so that we can use latex in them? Im looking for a program similar to the math jams run on AoPS.

[url=http://sourceforge.net/projects/pidgin-latex]Pidgin-LaTeX[/url] [i]A plugin for Pidgin which translates LaTeX code into images in your IM conversation[/i].
[url=http://linux.softpedia.com/get/Communications/Chat/Gaim-LaTeX-3544.shtml]Gaim-LaTeX[/url] [i]Gaim-LaTeX is a plugin which translate LaTeX code into image in your IM conversation[/i].
[url=http://www.informatik.uni-freiburg.de/~mechnich/sim/index.html]LaTeX patch for Simple Instant Messenger[/url] [i]This patch gives SIM the ability to display latex formulas by calling the external application tex2im.[/i]

Oh wow thanks.  I found that skype had an extremely useful ability to have public chats where I can moderate the chat and run everything quite smoothly.  Does anyone know any plugins for Skype to view latex?  Or just to change font to subscipt and superscript and stuff.  Thanks

I can't help with Skype but there is [url=http://thewe.net/tex/]TEX THE WORLD[/url] [quote]TEX THE WORLD converts TeX formulas anywhere, on any website, into rendered images. All you have to do is put them between [; and ;] (for example: [;e^{\pi i} + 1 = 0;])
When using interactive websites, like chats, TEX THE WORLD runs every 3 seconds, so chats will be converted as well.
It requires Firefox and Greasemonkey.[/quote]The website shows its use in Google Chat.


Harvard
college
geometry
linear algebra
Science Olympiad
number theory

Hello and welcome to the new Chemistry forum at AoPS!

I've volunteered to moderate this forum, and I'd like to briefly introduce myself.  I have my Ph.D. in physical chemistry and am an associate professor of chemistry at a small state university.  I've been teaching and doing computational/theoretical chemistry research for the past 15 years or so.   I don't know much about how chemistry is taught in high school, or about chemistry competitions, but I'd like to learn.   So I'll be following along on the various discussions and contributing whenever there might be a content question.    And be sure to let me know if I can help in any other way.

Other introductions are welcome here also - let us know what you're studying or what interests you that's related to chemistry.

David

I'm a South Carolina high school senior planning to major in Chemistry or Chemical Engineering in College.

I go to the same school as the person above and i'm taking ap chem next year. so until then, you won't see me browsing the chemistry forum that much as the honors chem knowledge diffused out of my head. :lol:

i am a grade 10 boy in Beijing, China. I am taking both Chemistry and Physics competition course in my school and i find both of them are interesting and difficult. I am not very good at Chem but i'd like to know interesting things in this subject and i will post problems and funny discovers in my daily life!

[quote="Kent Merryfield"]I'm a professor of mathematics and I'm here for the mathematics. But:

My father was a chemical engineer.
My mother was a chemist.
My wife is a biochemist, and a fellow faculty member.
My own bachelors degree was a double major in math and chemistry, although I went on to specialize in mathematics in graduate school.

I've used Michaelis-Menten enzyme kinetics for homework in a differential equations class, and messy equation-balancing in a linear algebra class.[/quote]
Wow, gifted family. :)

I'm not taking chemistry right now, but:
I did AP Chem last year as a sophmore and taught myself honors chem the summer before. I studies polymers and organic chemistry quite a bit for science olympiad last year. I'm thinking about a double major in chemistry and mathematics.

I am a junior in highschool, I take AP Chemistry and I enjoy chemistry. I am not sure whether I will major in chemistry or anything related, but I always try and keep my options open, and just take courses. Also, I have yet to take serious physics (my school is somewhat strange), so I do not have a complete view about sciences.

However, I am doing my best this year in linear algebra and chemistry, so  :lol:

hmm... I'm a sophomore in high school and I took AP Chemistry last year as well as the USNCO, which I got high honors in.  I took USNCO again this year and am praying that I made the summer camp.  I also took the US Physics Team exam and miraculously (and I seriously mean it) was a semifinalist.  When I took the qualifiying round for that... i had only learned mechanics... and I guessed on I think 16 out of 20 multiple choice questions and did only 3 of the 4 free response questions.  Yet somehow I still made it... O_o.  And my major is: not chemistry or physics (no idea what it will be though).

I am in the 11th grade and interested in cell/molecular biology, ecology and evolution, and mathematics. I do, however, have a passion for chemistry also!!!!!

I'm the son of a chemistry major who became an industrial engineer. My son now is taking an "honors chemistry" course at seventh-grade age, taught from the Russo AP/college introductory textbook. He will probably learn and review some more chemistry next year and then take the AP chemistry test in May 2006.

i'm a senior in AP chemistry. at this point, i'm not 100% of what i might want to do for a living, but i like chemistry alot and will definately take chem classes in college

I'm Jello whoa.  I just typed Jello.  I think I must be hungry...

Anyhow, I'm Jason LaRue; I'm a sophomore at Palmetto High School in Miami.  I took AP Chem this year, and managed to qualify for USNCO study camp.  Of course, I can't go, so one of you will have to tell me all about it.  I plan to go to Harvard and double major in Math and Chem, but we'll see how that goes.

(And yes, I really did type 'Jello.'  I wasn't just trying to funny.  Cut me a break, it's past midnight now.)

Hi, I'm a sohphmore in high school in Massachusetts, and just took chemistry for the first time this year. I really enjoyed it and took the local UNSCO, though I didn't make it past that round. Next year I'm taking AP Chemistry.

In September I will be sophmore in high school, taking an AP Chemistry class.
Chemistry is some-what new to me, but I am looking foward understanding some of the basic (and not so basic) concepts :D

Hey, I am 1234567890, and I'm not that at all great at chemistry, but I am in AP Chemistry this year as a 11th grader. I did a practice USNCO Exam and only got a 40,  but hopefully, I'll improve. I might come here once in a while to ask about USNCO and ask math-related chemistry questions.

While I am here, could you tell me about the USNCO process and how we register if our teachers don't know how?

Hi guys,

We have gathered together some of the best cheat and
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[quote="amirhtlusa"]ok, hello my name is Amir Talebi, and i line in WA,
i came to US like two years ago, i was born in Iran
i started chemistry about 5 years ago, when i was in 6 or 7 th grade probably
when i was in Iran, i took the National Chemistry Olypmiad,
i was ranked 12th in the country, and with hopefully i'm going to take the USACO next year...(frankly i think CO is very easy...)
now i'm a junior in HS, i took the AP chem Class this year. it was an easy class.
this forum in one of my favorites... :lol:[/quote]

[quote="xushuzhen"]Hi guys,

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[/quote]

Is that chemistry?  :roll:

Anyway:

I'm an 8th grader who is trying to learn chemistry by myself but not doing excessively well. I did read chemistry back in 4th grade but I never really learned anything there :D

My father is .........:|  (well, nothing to say about him)

Anyway, my mother is working in bank as senior officer.

 i devote myself to math and science, I just love it.

Chemistry and number theory are two things that i love most.

Philosophy is kinda freaky yet mesmerizing to me.

That's all
 :P

Hi!
I'm a year 13 chemistry student (doing my A levels). I also am doing an A -level in Physics and maths, along with another AS in Further maths!

I would like to become a chemical engineer in the future.

I hope that I would be a good addition to this forum!

Hi new to this forum been studying chemistry for a while but always have problem with working out calulations one in particular is atomic mass unit can anyone ofter any help

yeah, just post the problem that you have on the forum.
this thread is for introductions. :)

Allo Allo...

I'm a rather "mature" student in that I'm already into my thirties, but I'm studying in order to return to university for autumn of `07.  The programs I'm aiming for require biology and chemistry so I'm starting with the grade eleven courses since I never took those subjects the first time I went through school.  The biggest problem with doing my courses online is the lack of anyone to ask for help when I have trouble understanding something...and it *has* been a while since I last had to study.

So expect to see me on here pretty often, usually scratching my head, but I'll try to help answer anything I see that I may have already grasped, as well.   :D

I am a non traditional student (that means old) and i love sciencies and history. I have a degree in Anthropology, and have decided to go back to school. I took my last math class 14 years ago, and i have never had a chemistry class. I have taught myself (to a point) algebra this summer. Im now working on chemistry,  Im starting school again this fall with hopes of earning a computer/electrical engineering degree. You will see me around here on the recieving end of getting help.

Hey guys, I'm a rising junior in NY. I took honors chem in my uber tiny school, and found it easy under a great teacher. This year, I want to start a chem olympiad team in my school; I only know one other kid in my grade who loves chemistry as much as I do though, so it's gonna be tough to get a clique. I'm self-studying AP Chem this coming year in hopes of a)Getting far in nationals/making IChO and b)maybe winning the Siemens AP scholar award  :maybe:

I plan to major in something involving extensive biochemistry; I'm stuck between molecular parasitology/virology and myology/kinesiology(the latter area because I'm a HUGE sports/athletics fan). I hope to seem on-par with the rest of you after studying Brown's book.  :lol:

I'm a Canadian junior who is taking her final year of Chemistry [I fast-tracked]. I will be looking for all the help as I am preparing for various Chemistry-focused competitions/Olympiads.

Chemistry is by far, my favourite branch of science. =)

Hello.  I'm a theoretical physicist, which is essentially an applied mathematician.  I coach mostly math teams, but I also have a physics club.  My B.S. is in chemistry from UCLA, with concentration in theoretical chemical physics.  My masters is in physics and my Ph.D. in Chemical Physics, both from Harvard.

I greatly enjoyed doing my doctoral thesis and somehow managed to spend five years and write three hundred pages of equations that describe how lithium flouride breaks apart if you zap it with a laser.  It has a title so memorable that even I, the author, can't recall it off the top of my head, though it is something catchy like "Multichannel Formulation of the Quantum Scattering Theory of Photodissiciation."

Edit: Hah!  The title is on the web.  "Multi-channel theory of molecular photodissociation"  That wasn't quite as excruciating as I remembered.  The actual thesis is buried somewhere in a box, from when we moved.  :-)

My father is the chemist who discovered the iridium anomalies in the Cretaceous-Tertiary boundary that led the team of Alvarez, Alvarez, Asaro, and Michel to pospose an asteroid or comet hit the earth and caused mass extinction 65 million years ago, including the demise of the dinosaurs.

I have two students doing the USNCO this year (both of them AoPSers :-)), so I came over here to look around.  Looks like a great forum.

Best,
Catherine Asaro


Prove that $\left(\forall\right)\ n\in N^*,\ \left(\exists\right)!\ \ p\in N^*$ so that $\ \frac{1}{2p+1-\sqrt{8n+1}}\ge \frac 12.$

[hide="Solution:"]
[color=blue][b]
Prove that:
$\left(\forall\right)\ n\in N^*,\ \left(\exists\right)!\ \ p\in N^*$ so that $\ \frac{1}{2p+1-\sqrt{8n+1}}\ge \frac 12.$
[/b][/color]

[u][b]Solution:[/b][/u]

$\frac{1}{{2p + 1 - \sqrt{8n + 1} }} \ge \frac{1}{2} \Leftrightarrow$

$\Leftrightarrow \left( {2p + 1 - \sqrt{1 + 8n} > 0} \right) \wedge \left( {2 \ge 2p + 1 - \sqrt{1 + 8n} } \right) \Leftrightarrow$

$\Leftrightarrow \frac{{ - 1 + \sqrt{1 + 8n} }}{2} < p \le \frac{{1 + \sqrt{1 + 8n} }}{2}\mathop \Leftrightarrow \limits_{\mathop {p \in \mathbb{N}^* }\limits^ \uparrow }$

$\Leftrightarrow p = \left[ {\frac{{1 + \sqrt{1 + 8n} }}{2}} \right]$
[/hide]


This was an interesting article, and you might recognize some of the names, too:

[url]http://www.sciencenews.org/20020713/mathtrek.asp[/url]

I think the sciencenews server must be down, because that is its website and I just visited it by a different route and couldn't get access.  (http://www.sciserv.org is the website of Science Service, who publishes Science News.)

The link works now.

Wow, interesting...I am at a loss of words

Aha. Jeff Lagarias. I met him at a conference once. He said some very interesting things there. about mathematical motivation that people might find interesting. First, he said ``If you can't solve a problem, give up.'' But that isn't very good advice, so he changed it to ``If you are working on a hard problem, keep working on it, since it might actually be an easy problem.''

Hey, I know Jeff.  We once wrote a paper together.

Hey that's what my mom says :lol: "If you can't solve a problem after you chew on it a while, give up and look for the answer" It's actually not a half bad advice since rarely will the answer pop up after thinking about it for a couple of years :)


How do you solve number thirteen from this years AIME 1?  I got it down to
x^36-x^19-x^17+1=0
but couldn't get any further.

For anyone who doesn't know the problem, when the solutions to that equation are put into cos x + i*sin x form, you have to find the sum of the 5 smallest x's between 0 and 360 (right?  It was degrees, not radians I think).

x^36-x^19-x^17+1=0

group and factor

(x^17+1)(x^19-1)=0

X^17 = -1
x^19 = 1

easy from there with cis / demoivre's , whatever u wanna call it

actually, it should be (x^17-1)(x^19-1)=0

Also, the question was find the sums of the smallest p such that
cis(2( :pi: )(p)) is a root. Most stuff involving rcis :theta:  or e^(i :theta: ) will be in radians.


Find the rightmost nonzero digit of $ \frac{100!}{5^{20}}$ (here $ n!\equal{}1\cdot2\cdot3\cdot\ldots\cdot
n$).

There are 24 powers of 5 in $ 100!$. Thus the last digit is determined by $ 5^4 \times a^{10}$, where $ a \equal{} 1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8 \times 9$.

Note that the last three digits of $ a$ are $ 576$ and that in determining the rightmost nonzero digit of $ a^{10}$ upon multiplication by 4, only the last 3 digits are important. Considering the last three digits, some trial and error and reasoning confirms that the last three digits of $ a^{10}$ are $ 376$. Multiplied by $ 5^4$, we get $ \boxed {5}$ as the rightmost nonzero digit.

Uhh, don't all the powers of 5 cancel out with powers of 2 to make 0's?

Straightforward parity
[hide]$ \frac {100!}{5^{20}} \equal{} 20!(1*2*3*4)(6*7*8*9*11)(12....$
Since this whole term includes at least 1 $ 5$, we need to see if the product of all the other numbers are even, which it's true,  so the last digit is $ \boxed{0}$[/hide]

Err, we are looking for the rightmost [b]nonzero[/b] digit, we already know the last digit is a $ 0$  :P

[hide="I think..."]Take the 4 remaining powers of 5 and multiply by 16 to form 4 0's at the end. Then consider the rest of the numbers, without factors of 5 or the number 16 mod 10. This is equivalent to $ 2^{10} \cdot 3^{10} \cdot 4^{10} \cdot 6^{9} \cdot 7^{10} \cdot 8^{10} \cdot 9^{10} \pmod {10}$ Do whatever you want to simplify it, but it comes out to $ \boxed6$.[/hide]

Sorry, you're right, the answer is actually $ 6$.  :blush:  The last four digits matter, not the last three. The last four digits can be reasoned to be $ 1376$ as follows:

We know that for $ 0 \leq x \leq 9$, $ \mod 1000$, $ (10n \plus{} 1)(10n \plus{} 2)(10n \plus{} 3)(10n \plus{} 4) \equiv 24$, and similarly for $ (10n \plus{} 6)(10n \plus{} 7)(10n \plus{} 8)(10n \plus{} 9)$. An expansion will easily convince anyone. So we get $ (10n \plus{} 1)(10n \plus{} 2)(10n \plus{} 3)(10n \plus{} 4)(10n \plus{} 6)(10n \plus{} 7)(10n \plus{} 8)(10n \plus{} 9) \equiv 576 (\mod 1000)$.

So we want to know what $ 576^{10} (\mod 10000)$ is. We get $ 576^2 \equiv 776 (\mod 1000)$ and $ 576^4 \equiv 176 (\mod 1000)$. And also, $ 576^6 \equiv 576 (\mod 1000)$. Thus $ 576^{10} \equiv 576 \cdot 176 \equiv 1376 (\mod 10000)$.

What's your solution? I'm pretty sure it is 6.

I did not get $ 6$ :S

Dividing by $ 5^{20}$ (i.e. 1 power of 5 from EVERY multiple of 5 from 1 to 100),
we get the following multiplication:

$ 1\times2\times3\times4\times6\times7\times8\times9\linebreak \times\linebreak 11\times12\times13\times14\times16\times17\times18\times19\linebreak \times...\times\linebreak 91\times92\times93\times94\times96\times97\times98\times99\linebreak \linebreak \times1\times2\times3\times...\times18\times19\times20\linebreak$

(The last row was when i divided all the powers of 5)
Each of the first rows produces a rightmost digit of $ 6$.
Only the LAST ROW could produce anymore 0's.
Deleting these 0's, the last row produces a rightmost digit of $ 4$.

Thus, our desired result is the rightmost digit of

$ 6^{9}\times4$

which is obviously

$ 4$.

PS sorry for the bad LATEXing im not very good at it

Yeah you're right :), after recalculating it, it occured to me that I miscalculated the last step

oopss..
didn't see ur hidden solution, which is 99% similar to mine  :)
Sorry

Ah, I have gone horribly wrong again..  :blush:

Altheman, where did you calculate that? Lol..

Excellent explanation of the solution, iniquitus. :D


Prove that :

$ \mathrm{arctg}^3\ x\plus{}\mathrm{arcctg}^3\ x\plus{}\frac{3\pi}{2}\mathrm{arctg}\ x\cdot \mathrm{arcctg}\ x\equal{}\frac{\pi^3}{8}\ ,\ (\forall)x\in \mathbb{R}$

[quote="alex2008"]Prove that :

$ \mathrm{arctg}^3\ x \plus{} \mathrm{arcctg}^3\ x \plus{} \frac {3\pi}{2}\mathrm{arctg}\ x\cdot \mathrm{arcctg}\ x \equal{} \frac {\pi^3}{8}\ ,\ (\forall)x\in \mathbb{R}$[/quote]

Let $ tan p\equal{}x$, $ cot q\equal{}x$.
Then $ cot q\equal{} tan (\frac{\pi}{2}\minus{}q)\equal{}tan p$

So the question becomes $ (\frac{\pi}{2}\minus{}q)^3\plus{}q^3\plus{}\frac{3 \pi}{2}(\frac{\pi}{2}\minus{}q)q\equal{}\frac{\pi^3}{8}$

which is true after simplification.


geometry
3D geometry
tetrahedron
Triangle
IMO Shortlist

Let $ ABCD$ be a regular tetrahedron and $ M,N$ distinct points in the planes $ ABC$ and $ ADC$ respectively. Show that the segments $ MN,BN,MD$ are the sides of a triangle.

WLOG $AB=1$. Then the height is $h=\sqrt{\frac2 3}$ and $BN+MD>2h>1>MN$. Now it remains to show that $|BN-MD|<MN$. Let the points $M', N'$ be symmetric to $M, N$ respectively wrt the bisector plane of dihedron $AC$. If $M'=N$ then $BN=MD$ and everything is obvious. Otherwise define $O=MN\cap M'N'$. We have $|BN-MD|=|DN'-MD|<MN'<MO+ON'=MN$ as desired. The result follows.


To get Karma, you have to rate the quality of the problem. But, when you move your cusor over any other star than the worst, it still displays the worst. for example, if i move my cursor over ^^|^|^^ it still says learned nothing or quality is very poor. Is it just me?

This should be fixed now. Sorry for the inconvenience!


Hey...Quick Question here...

I am currently going through Vol. 1 and a lot of it is repititon of what I have already done. Most of the basics things I know, but I may not get some "advanced" topics or problems right away.

So, how should I attack the book? Should I just go through everything I already know and just go to parts and problems that are a bit more advanced?

Also, is it okay to do some problems from AoPS Vol 2 without having gone through Vol 1?

Also, how long do you think it should take to go through AoPS Vol 1 IF you have about 1 or 2 hours per day to spare?

How long it takes to get through the book depends on if you do every single example and every single problem.

If you feel comfortable with the concepts in book 1, do the problems. If you don't understand one, look back at the chapter and read more about it.

You [i]could[/i] do problems in Vl. II without having done Vol I. Some sections will be more difficult to you than others but you'll do fine.

well, eventhough i knew most of the problems in the first book on geometry, i did all of them, repetition helps you get better, i also would suggest that you take your time, things sink in better if you work that way

Go here:

http://www.artofproblemsolving.com/Books/AoPS_B_About.php

You certainly do not have to necessarily go through V. 1 to do V. 2.  In fact, it helps me to jump into problems right away and then get advice for how to do them later.  You may not realize it, but pretty soon you'll be noticing things you didn't notice before due to all the ideas that have gone through your head from the various problems.  Yeah well just go at your own pace, and you mostly don't have to do things in any order... just whatever you feel like doing.  Math should be a fun thing and not so much hardcore studying :-P.

Yeah I find problem-solving math very fun, but I really want to do well on the AMC and AIME....I don't know i will just see what happens.

Whoa! Those are expensive.  Anywhere else I can get those or aother materials to study for the AMC 10/12?
Thanks

Well the Art and craft of problem Solving is pretty expensive considering its length.

Any other responses?

The books are worth the price.  I've done / am doing them kind of randomly and fragmented, but as I read a chapter, if most of the stuff I have seen before I'll just jump right into the problems and if I come across something really foreign will go back into the text and see if I can find where they talk about it.

You'll find that while you know a lot of stuff they talk about, they definitely go more in depth and introduce more concepts that are worthwhile to look at.


Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$
Prove that:
$ \sum\frac{ab\plus{}1}{a^{2}\plus{}b^{2}}\geq\frac{9}{a\plus{}b\plus{}c}$

I'm not too good with inequalities, but does:

$ \sum{\frac{1\plus{}c}{a/b\plus{}b/a}}\geq 9/(a\plus{}b\plus{}c)$ followed by a substitution in the denominator work?  We could even multiply through and denote f(x)=x+1/x so we would get $ \sum\frac{f(a)\plus{}f(b)\plus{}c\plus{}c^{2}}{f(a/b)}\geq 9$.

[quote="chien than"]Let $ a;b;c$ be positive real numbers satisfying $ abc \equal{} 1$
Prove that:
$ \sum\frac{ab\plus{}1}{a^{2}\plus{}b^{2}}\geq\frac{9}{a\plus{}b\plus{}c}$[/quote]
I think it's wrong. Try $ a \equal{} b \equal{} 2$ and $ c \equal{}\frac{1}{4}.$ :wink:

I know almost nothing about ineqs, can someone help me further :)

$ \frac{9}{a\plus{}b\plus{}c}\leq\frac{9}{3}\equal{} 3$, thus if we can prove that it is $ \geq3$ then we're done.

$ \sum\frac{ab\plus{}1}{a^{2}\plus{}b^{2}}\equal{}\sum\frac{ab(c\plus{}1)}{a^{2}\plus{}b^{2}}\geq 2\sum\frac{\sqrt{ab}}{a^{2}\plus{}b^{2}}\geq3$

And now i am stuck.

arqady's right; this doesn't work.

Assume b=1 and assume a>1 so c<1=1/a.

The function then turns into $ \frac{(a+1)^{2}}{a^{2}+1}+\frac{2}{(a^{2}+1/a^{2})}\geq\frac{9}{(1+a+1/a)}$
After more simplification, $ (1+\frac{a^{2}+1}{a})(1+\frac{2a}{a^{2}+1}+\frac{2}{a^{2}+1/a^{2}})\geq 9$ or $ \frac{2a}{a^{2}+1}+\frac{2}{a^{2}+1/a^{2}}+\frac{a^{2}+1}{a}(1+\frac{2}{a^{2}+1/a^{2}})\geq 6$ or ${ \frac{2}{a+1/a}+\frac{2}{a^{2}+1/a^{2}}+\frac{a^{2}+1}{a}(\frac{(a^{2}+1)^{2}}{a^{4}+1}})\geq 6$
Of course, the left hand side must eventually exceed the right and at a=1, they are equal, but until some point between 4 and 5, the left hand side is smaller than the right hand side.


Euler and Guass are at a meeting:

Euler "How old is your single son?"
Guass "I am 5 times his age."
Euler "That does not give me the answer."
Guass "In 5 years, I will be 4 times his age."
Euler "Aha.  Now I understand."
Guass "How old are your 3 sons?"
Euler "The product of their ages is 36."
Guass "That does not give me the answer."
Euler "The sum of their ages is today's date."
Guass "Sorry, but that doesn't tell me their ages."
Euler "Oh - I forgot to tell you that my youngest son has red hair."
Guass "Aha.  Now I understand."
Euler "And if you take your age, subtract my age, and subtract your son's age, you get the same result you would if you subtracted your son's age from my age twice."
Guass "From youngest to oldest, your family, if I am correct, goes from youngest to oldest [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i].
Euler "And yours [i]e[/i], [i]f[/i]."

Compute the list {[i]a, b, c, d, e, f[/i]}

What's the date? if it's June 6, I guess you use 6, but that cannot lead to their product to be 36.

[hide]From Gauss' first few pieces of the conversation, we have $x=5y$ and $x+5=4(y+5)$ where $x$ is Gauss' current age and $y$ is his son's current age. Solving this system gives $x=75$ and $y=15$.

When Euler tells Gauss that the product of his sons' ages is 36, Gauss can deduce that there are 8 sets of possible ages. The fact that Guass cannot narrow down the ages to one set after being told that their ages sum to the date indicates that the sum of the ages in two of the sets is the same, and thus Gauss cannot decide between the two. The two sets with the same sum are $(2,2,9)$ and $(1,6,6)$, so the date is the 13th (finding the date is another question that would have made this problem interesting).

Then Euler confides that his youngest son has red hair. Most people agree that red hair has little to do with age (since hair color usually doesn't change) but the fact that there is a youngest child tells us that the second set of ages must be the correct choice. Thus the ages of Euler's children are $1$, $6$, and $6$.

The final piece of information, if not written incorrectly, shows that Euler's age is $\frac{75}{2}$ so ${a,b,c,d,e,f}={1,6,6,\frac{75}{2},15,75}$.[/hide]

[quote="miyomiyo"]Euler and Guass are at a meeting:

Guass "How old are your 3 sons?"
Euler "The product of their ages is 36."
Guass "That does not give me the answer."
Euler "The sum of their ages is today's date."
Guass "Sorry, but that doesn't tell me their ages."
[/quote]

Guass's last statement should give you the date.


Alcumus FTW AMC USA(J)MO USAMO MATHCOUNTS

Lets say I am in 7th grade.

My Level of smartness:
This year I got a 44/46 in Chapter and a 40/46 in state, and sadly didnt make nats.
I am pretty good at countdown, maybe a an average 1650 rating..

The Resources I have:
The whole Introductory Set
Alcumus
FTW
AOPS
idk anything else

AND A MONTH LEFT TO STUDY UNTIL NATS

What would be the best method of studying for me?
Also how many hours per day?

Don't count your hours. Just work on your weak spots. Get books and practice old tests
Good Luck

What do you mean a month left to study until nats? And which state do you live in?

How to study depends on your weaknesses. If you mostly make sill mistakes, then you should study differently then if you're bad at, say, probability.

Work through the entire Intro set, which, if you work at a decent pace, you can finish by the end of summer.  Buy AoPS Volume 1 and work through that afterwards as a good review and to learn a bit more on topics (as well as a lot of practice problems).

[quote="KingJamesMvP"]Just wondering for those who have been to a math summer camp. How much does it really help?[/quote]

Well, going to mop = very very very very helpful for me. But I'm not sure if that counts.

Try "play for the win"..it improves your skills altought it must taken not too much time..

[quote="The Pink Panther"]How do you go to mop?[/quote]

I think that you have to be in the top few places on the USAMO to be invited to MOP.

Does MOP stand for the Mathematical Olympiad Program?

yes, and it is also called MOSP (Math Olympiad Summer Program)

Then why is it in this forum? Obvioiusly, none of us have been to the program and aren't going to in the near future.

[quote="pythag011"]Well, going to mop = very very very very helpful for me. But I'm not sure if that counts.[/quote]

Ahem. There's your counterexample.

Well, this is a middle school forum. I'm just saying this MOP should be moved to the high school forum.

[quote="The Pink Panther"]How do you go to mop?[/quote]

You have to be insanely out of your mind beastly smart. Other than that, its pretty easy :P

Also, which state do u live in, because depending on which state you live in, your practice time should vary (if MC is your only goal). For example, a 40 at states for the MI team would have easily gotten you on the team (I think you would have been 3rd written), and MI didnt actually do that bad at nats (15th)

[quote="Math Champion"]So, I'm assuming you live in MI, Thunder?[/quote]

Yeah lol :lol:

[quote="maybach"]Does anyone know what the top score was for Indiana?[/quote]

Can't you just check online at the official website?

This is getting really really off-topic.

I'm not sure what you mean by a month left to study until nats, but if you really only have a month, I would recommend going through Volume 1: The basics. Everything you need for mathcounts you can find in that book. If you have more time, which you should, I'd do the introduction series, as they go into the material in Volume 1 in more depth.


limit
real analysis
real analysis unsolved

if $ f \in W^1_p(U)$, how to show that: $ f^ \plus{} ,f^ \minus{} \in W^1_p(U)$ and

$ D(f^ \plus{} ) \equal{} Df$ a.e on $ \{f > 0\}$, and $ D(f^ \plus{} ) \equal{} 0$ a.e on $ \{f < \equal{} 0\}$
$ D(f^ \minus{} ) \equal{} \minus{} Df$ a.e on $ \{f < 0\}$, and $ D(f^ \minus{} ) \equal{} 0$ a.e on $ \{f > \equal{} 0\}$.

I was given a hint: $ f^ \plus{} \equal{} \lim_{\epsilon \minus{} > 0} F_{\epsilon}(f)$, where
$ F_{\epsilon}(z) \equal{} (z^2 \plus{} {\epsilon}^2)^{1/2} \minus{} {\epsilon}$ if $ z > \equal{} 0$ and $ F_{\epsilon}(z) \equal{} 0$ if $ z < 0$.

Compute $ \lim_{\varepsilon\to 0^\plus{}} DF_\varepsilon(f)$ using the chain rule and show that
\[ \lim_{\varepsilon\to 0^\plus{}} DF_\varepsilon(f) \equal{} D\left(\lim_{\varepsilon\to 0^\plus{}}F_\varepsilon(f)\right) \equal{} Df^\plus{}
\]

[quote="blahblahblah"]Compute $ \lim_{\varepsilon\to 0^ \plus{} } DF_\varepsilon(f)$ using the chain rule and show that
\[ \lim_{\varepsilon\to 0^ \plus{} } DF_\varepsilon(f) \equal{} D\left(\lim_{\varepsilon\to 0^ \plus{} }F_\varepsilon(f)\right) \equal{} Df^ \plus{}
\]
[/quote]
can you give more detail? how to show:

$ \lim_{\varepsilon\to 0^ \plus{} } DF_\varepsilon(f) \equal{} D\left(\lim_{\varepsilon\to 0^ \plus{} }F_\varepsilon(f)\right)$?

The difference between $ DF_{\epsilon}$ and $ DF^\plus{}$ is $ \left(\frac{F}{\sqrt{F^2\plus{}\epsilon^2}}\minus{}1\right)DF$ on the set where $ F>0$. This is at most $ |DF|$ in all cases but is much less (because of cancellation) when $ F$ is much larger than $ \epsilon$. So you should estimate the $ L^p$ norm differently on two different sets, say $ 0<F<\epsilon^{1/2}$ and $ F\ge \epsilon^{1/2}$.


From the University of Maryland Mathematics Competition, Part 1, 2001:

Suppose that there are some blabs, some blibs, and some blubs. Suppose also that all blabs are blibs and some blubs are blabs. On the basis of this information, which of the statements X, Y, Z must be true? X: All blabs are blubs; Y: Some blabs are not blubs; Z: Some blibs are blubs.
a. X only   b. Y only   c. Z only   d. X and Y only   e. Y and Z only

[hide]Consider the statement "all blabs are blubs." THis doesn't have to be true, since just because some blubs are blabs, it doesn't mean that all blabs have to be blubs. Statement Y doesn't have to be true because while it isn't necessary, there is a possibility that all blabs are blubs. (although X and Y can't be both true or false at the same time). Statement Z must be true; since all blabs are blibs and some blubs are blabs, some blubs are blibs, and the reverse has to be true (there's more than 0 blubs). Thus c. is the right choice. [/hide]

Why is Y incorrect? I thought "some" implies not all or am I mistaken?

I agree with breez

[quote="236factorial"]onsider the statement "all blabs are blubs." THis doesn't have to be true, since just because some blubs are blabs, it doesn't mean that all blabs have to be blubs.[/quote]

The last thing you said, "it doesn't mean that all blabs have to be blubs," you contradicted what the statement was.  All blabs [i]are[/i] blubs.  However, the converse is not true.  Only [i]some[/i] blubs are blabs, but any blabs will always be blubs.  And carrying out some more reasoning, I believe [b]E[/b] is the correct choice.

[quote="E^(pi*i)=-1"]The correct answer is C.  "Some" does not imply "not all".[/quote]

I dont' think I could've been wrong especially with the answer keys  :D


The question reads:
Without dividing, determine the remainder when [i]f[/i](x)=X^3 + 5X^2 - x - 21 is divided by X^2 + 3X - 10.

I got this on a test last year and I saw it today and tried to do it and it's the only question I was unable to do. I was able to factor the top to (x+3)(X^2 +2X-7) and the bottom to (X+5)(X-2) but i tried a few things and haven't been able to get an answer. What do i do?

And if you do post an answer, could you explain how you got there?

Thanks,
Boggy

[quote="FantasyLover"]without dividing??

dont you mean like 'literally dividing'??

synthetic division is kinda division...  :D[/quote]

I don't know you may be correct. I remember whacking myself when he told us what to do. Is there a way to do it with neither synthetic or real division?

hmm idk...

but even if theres such way, i think synthetic division will be more useful when dividing a polynomial.

I thought synthetic division only applied to division by binomials...

hm cant it divided by polynomials

like u were supposed to divide polynomial $ 4x^3\plus{}4x^2\plus{}2x\plus{}1$

by $ x^2\plus{}2x\plus{}1$, since the root of $ x^2\plus{}2x\plus{}1\equal{}0$ is -1, the synthetic division would be

-1 | 4 4 2 1
    |_______

no synthetic division is only by binomials, however say you want to divide P(x) by the polynomial (x-1)(x-2), you could use synthetic to first divide by 1 and then by 2.

the kth symmetric sum is$ \minus{} 1^k a_{n \minus{} k}/a_n$
now I understand this fantasy lover lol

Here is the answer to the problem:
$ (x\plus{}3)(x^2\minus{}2x\plus{}7) \equal{} (x\plus{}5)(x\minus{}2)Q(x) \plus{} ax \plus{} b$
Plugging in the roots of the equation $ (2,\minus{}5)$ allow us to get a system and solve for $ a,b.$
There, no division  :P

it can be done without literally dividing or synthetic division ..

[quote="mathemonster"]no synthetic division is only by binomials, however say you want to divide P(x) by the polynomial (x-1)(x-2), you could use synthetic to first divide by 1 and then by 2.

the kth symmetric sum is$ \minus{} 1^k a_{n \minus{} k}/a_n$
now I understand this fantasy lover lol

Here is the answer to the problem:
$ (x \plus{} 3)(x^2 \minus{} 2x \plus{} 7) \equal{} (x \plus{} 5)(x \minus{} 2)Q(x) \plus{} ax \plus{} b$
Plugging in the roots of the equation $ (2, \minus{} 5)$ allow us to get a system and solve for $ a,b.$
There, no division  :P[/quote]

Could you elaborate on how to continue to get the answer. I understand the equation but not how to continue.

I tried something, tell me if this is correct.

I took X^2 + 3X - 10. and thought what would be the closest way to get to X^3 + 5X^2 - x - 21 with multiplication. I found that It should be (X+2) since it yields X^3 + 5X^2 then other stuff.

So i took that and did the full expansion
X^3 + 5X^2 - 4x - 20 and then subtracted this from the original equation to get 3x-1 so is (3x-1)/(X^2 + 3X - 10) the answer?

[quote="mathemonster"]Plug in what I said to and see what two equations you will get then solve. Your method is not rigorous it is more like guesswork but it is correct.[/quote]

I was a bit confused by what you wanted me to do.

My method isn't guess work. You simply find out what numbers you have to use for the first few numbers to be correct (in this case 2), then do the multiplication and subtract to find the remainder. And it makes sense because a remainder is in essence the difference between what would divide perfectly and what you got.

[quote="mathemonster"]ok maybe thats what your teacher wants you to do then  :wink:
plugging in we get
2a + b = 35
-5a + b = -84
7a = 35 + 84 a = 17  b = 1
so your answer is wrong, the remainder is 17x + 1
however i did this in my head so I might have calculated incorrectly[/quote]

I got my answer when i did division, could you check by doing division in case i did the division wrong.

I know what I do may seem like guessing but it's not, there is only one combination that was satisfy what i did and it's mathematically found. But off of that tangent...

Well check to make sure you got the right one using division. If so then show your division cause then i'm screwing mine up.

Another method is to use the method of congruences.

Let $ g(x) \equal{} x^2\plus{}3x\minus{}10$

$ x^2 \equiv 10\minus{}3x \bmod{g(x)}$

Then $ x^3 \equiv x.x^2 \equiv x(10\minus{}3x) \equiv 10x\minus{}3x^2 \equiv 10x\minus{}3(10\minus{}3x) \equal{} 19x\minus{}30 \bmod{g(x)}$

Thus $ f(x) \equiv 19x\minus{}30\plus{}5(10\minus{}3x)\minus{}x\minus{}21 \equal{} 3x\minus{}1 \bmod{f(x)}$

You can do it one by one. First factor the divisor into $ (x \minus{} 2)(x \plus{} 5)$. Then do synthetic division. Use either the first or the second factor as the divisor of the polynomial function. Of course you will have a sum of a  depressed polynomial and a rational expression. By remainder theorem, substitute the remaining factor of the divisor to the depressed polynomial (the factor must be in the form $ x \minus{} r$. Substitute r in the depressed polynomial) then add the remainder to the rational expression obtained from the previous

Either the remainder is a linear or a constant.

Let me try it by long division and by my proposed method to confirm my answer. I'm not even sure about this  :D

[quote="hsbhatt"]
$ x^2 \equiv 10 \minus{} 3x \bmod{g(x)}$

Then $ x^3 \equiv x.x^2 \equiv x(10 \minus{} 3x) \equiv 10x \minus{} 3x^2 \equiv 10x \minus{} 3(10 \minus{} 3x) \equal{} 19x \minus{} 30 \bmod{g(x)}$

Thus $ f(x) \equiv 19x \minus{} 30 \plus{} 5(10 \minus{} 3x) \minus{} x \minus{} 21 \equal{} 3x \minus{} 1 \bmod{f(x)}$[/quote]

Mod is remainder right so like 5 mod 5 = 0, I have no recollection of ever learning mod so i'm not sure, why is there a mod on the first line i quoted? I understand that if the first line has it, so does the second and why the last one has it but not the first one.


Prove that the equation $t^{2}+x^{2}+y^{2}=2007^{2007}$ has no solution in integers.

$n^{2}\equiv 0, 1, 4 (\text{mod 8})$
so, $a^{2}+b^{2}+c^{2}\not \equiv 7 (\text{mod 8})$

Therefore there is more no solution than $2007^{2007}\equiv 7 (\text{mod 8})$


Three congruent circles G1, G2, G3 have a common point P.
G1 and G2 intersect again at C.
G2 and G3 intersect again at A.
G3 and G1 intersect again at B.
(a) Prove that P is the orthocenter of ABC.
(b) Prove that the circumcircle of ABC is congruent to the given circles.

a) [hide]When you make the circles, their overlappings form three congruent football shapes with one of their corners joing at P.  Their second corners of each football is A, B, and C and ABC is an equilateral triangle.  That proves that P is the orthocenter of the ABC. [/hide]

b) [hide]If you put a point say D horizontally opposite of say point C on say circle G1, you will form another equilateral triangle, BCD and BCD :cong: ABC.  That proves that the circle with points A, B, and C on the circle is congruent to G1, G2, and G3.[/hide]

What is the name of a shape that is like a football?

frodo wrote:a) [hide]When you make the circles, their overlappings form three congruent football shapes...[/hide]

This isn't true. The problem doesn't say that the circles are necessarily evenly spaced around P.

You can find a soln here [url]
http://www.mathlinks.ro/phpBB/viewtopic.php?t=2033 [/url]

Here's another one:

(a) Let R be the radius of Gi. Let Oi be the center of the circle Gi. It's obvious that O2AO3P, O3BO1P, O1CO2P are rhombi. This means that O2C || PO1 || O3B, and since O2C=O3B=R, we get O2CBO3 to be a parallelogram, which means O2O3 ||BC. O2O3 is perpendicular to AP, so BC is also perpendicular to AP. By the same way: BP perp. AC, CP perp. AB => P is the orthocenter of ABC.

(b) You can see from the proof above that O2O3 = BC and the other analogous ones also hold, so ABC = O1O2O3 (equal triangles), so their circumradii are also equal, but the circumradius of O1O2O3 is R, so the circumradius of ABC must also be R.

[quote="frodo"]
What is the name of a shape that is like a football?[/quote]

It took me a while to think about it, but I guess it would be called a Lens shaped region. I don't think there's any more specific name than that.

By the way, nice solutions grobber!

a lot of people just call it a football . . .


Just to clear up some simple stuff about logs, is log a+log b+ log c = log a*b*c?

Just to clear up on something.. your not really in 4th grade

i just realized this forum is not a place for jokes or kidding around (not even once is allowed!)


calculus
integration
logarithms
calculus computations

I'll do $\int \ln x dx$, since it only differs by a constant from both of your integrals there.

We'll do it by parts.  Put $u=\ln x$, $dv=dx$, $v=x$, $du=1/x$

Then

$\int \ln x dx = uv-\int vdu=x\ln x-\int \frac{x}{x}=x\ln x-\int 1 = x\ln x - x$.

[quote="pkothari13"]I am unsure as to where to post this problem.  Obviously, this problem is probably trivial to those who frequent this part of the forum, but I am just learning basic Calculus, and I figured this would be the place to post my question.

How would one go about finding the following:

$\displaystyle\int \log_a x\,dx$  What about cases like $\displaystyle\int \log_a kx\,dx$ where $k$ is some constant?

Thanks for the help in advance.[/quote]
For the other part,\[ \int \log_a kx\,dx=\int(\log_ax+\log_ak)\,dx=\int \log_a x\,dx+\int \log_a k\,dx,  \] where the second term is just the integral of a constant.

you can simlify $\int\log_ax{ }dx$ to $\int\frac{lnx}{lna}dx$ and $lna$ is just a constant so it becomes $\frac{1}{lna}\int{lnx}dx=\left(\frac{1}{lna}\right)(xlnx-x)+C$


blogs
Support
geometry
3D geometry
sphere

No, not likely.  Looks to me the Russians are just trying to assert their influence over the region encompassing the old Soviet republics.  The Russians are neither stupid nor crazy, and neither are the Americans; neither powers have any interest in armed conflict.  A Cold War II, on the other hand, seems to already have been underway for some time.  I think Georgia is more a symptom of increasing tensions over the past few years than it is a cause.

I agree with Joe.

Bush and others have said that this is a regional conflict (saying that we won't fight, just talk in defense of Georgia).

One spokesman (Gates, I think) said that we didn't avoid fighting each other through the cold war just to end up fighting now.

The question for us is how much can we punish Russia, without eliminating the ability to punish them again or hurting ourselves too much.  Most folks, including Russia, don't understand how much we have been supporting their economy, trying to avoid it falling into anarchy and have the potential for a dangerous radical government there.  We now have a slightly dangerous (to little neighbors) autocratic government.

I'd not bet that we (Western world, not just  US) let them into the Group of 8, or the World Trade Organization anytime soon.  The intelligence sharing and efforts to suppress their regions seeking to be free might just fall from a level 8 to a level 5 or 6.  (Sorry Ivan, the information is just a bit too old to allow it to be useful.)

The payment for their space launches might just get a bit more bogged down in paperwork and the US efforts to compete might just be a bit faster and more capable.

In 5 years, I believe that this will be viewed by Russia and others as a major error.

I understand that their fighting was very poor and perhaps undermined the fear others had of their military potential.  It certainly didn't seem to slow down Poland for signing a joint defense agreement with the US, in exchange for hosting a missile defense system that Russia can't match and which decreases the political value of their ICMBs.

I note that the dollar seems to be coming back up.  Almost as if it were down in value at the will of the US government.  Those Russian oil revenues will be a bit less this month and by the end of the year, there could be a sizable loss of Russian income.  (Oil is largely valued in dollars even if sold in other currency, so exchange rates affect oil revenues a great deal.)

Bottom line, if we end up in WWIII, Georgia will be like Chamberlain ceding little counties to Hitler, not the main precipitating event, like the assassination of the Archduke of Serbia.

This is a silly topic: why would the United States go to war to defend Georgia?  (A more interesting topic is, how did Saakashvili come to believe that we would?)

I read a [url=http://lefarkins.blogspot.com/search/label/confrontation%20in%20the%20caucasus]blog[/url] that has had very good coverage of this entire war -- if you click that link, you'll see the posts on the subject, newest on top.  Scroll down to the bottom and click "older posts" if you want to start from the beginning.

[quote="wizard2378"]One spokesman (Gates, I think) said that we didn't avoid fighting each other through the cold war just to end up fighting now.[/quote]  If you mean Robert Gates, he's the Secretary of Defense.  This is rather different than being a spokesperson.

[quote="wizard2378"]Most folks, including Russia, don't understand how much we have been supporting their economy, trying to avoid it falling into anarchy and have the potential for a dangerous radical government there.  We now have a slightly dangerous (to little neighbors) autocratic government.

I'd not bet that we (Western world, not just  US) let them into the Group of 8, or the World Trade Organization anytime soon.[/quote]  Where do you get this stuff?  Russia still has a nuclear arsenal sufficient to destroy the world several times over, and they blew a Georgian army trained and supplied by the US and Israel out of the water without a problem.  Their military and economic strength have been steadily on the rise since the mid-1990s, in large part thanks to high oil and natural gas prices.  They already belong to the G8.  This stuff (especially the last bit) isn't so hard to learn about ....

[quote="wizard2378"]The question for us is how much can we punish Russia, without eliminating the ability to punish them again or hurting ourselves too much.  Most folks, including Russia, don't understand how much we have been supporting their economy, trying to avoid it falling into anarchy and have the potential for a dangerous radical government there.  We now have a slightly dangerous (to little neighbors) autocratic government.[/quote]
In the past several years Russia received less than 1 billion annually from the US; and of that, nearly all was spent on accounting for their nuclear stockpile, which is clearly more in the interests of the US than of Russia.  Russian GDP growth exceeds that of the US by multiples.  Russia has relatively minor federal and national debt.  As a matter of fact, until recently Russia held about 100 billion dollars of [i]our[/i] debt via Freddie and Fannie.  Russia does not have an autocratic government; it is a federation with a constitution, a legislative body, a judicial branch with federal and local courts, judicial review of legislative and executive acts, and universal suffrage (with average voter turnout higher than in the US).  Fundamentally the government is structured similarly to our own.  As for its insignificant danger, JBL already pointed out the country's nuclear capability.  In short, each of your claims and insinuations is ridiculous.
Source: [url=https://www.cia.gov/library/publications/the-world-factbook/geos/rs.html]CIA - The World Factbook - Russia[/url]

In my personal opinion, if Russia adopts a 'dangerous radical' position towards the United States, it would be in no small part because too many in our country think, speak, and push policy based on the ludicrous world-view reflected in your inflammatory post.

[quote="wizard2378"]I understand that their fighting was very poor and perhaps undermined the fear others had of their military potential. It certainly didn't seem to slow down Poland for signing a joint defense agreement with the US, in exchange for hosting a missile defense system that Russia can't match and which decreases the political value of their ICMBs.[/quote]Poland's actions indicate precisely the opposite of what you assert.  And Russia has stated it has no desire to match missile defense (Russian strategic doctrine is retaliatory, not first strike), but rather will simply modernize its ICBMs to circumvent missile defense.

[quote="wizard2378"]I note that the dollar seems to be coming back up. Almost as if it were down in value at the will of the US government. Those Russian oil revenues will be a bit less this month and by the end of the year, there could be a sizable loss of Russian income. (Oil is largely valued in dollars even if sold in other currency, so exchange rates affect oil revenues a great deal.)[/quote]  You are suggesting my government's fiscal policy is dictated by how best to mischievously destabilize the commies' [u]nominal[/u] profits?

[quote="JBL"]They already belong to both the G8 and the WTO. This stuff (especially the last bit) isn't so hard to learn about[/quote]
I may be wrong, but I don't believe Russia is in the WTO.  It was due to join within a year or so, but I imagine recent events may delay that.

Anyway, I'm too far off topic.  The US isn't going to war with Russia any time soon.  Well, unless maybe there's a critical mass of people in our government who come to realize we need to punish that harmless autocratic government (that has kept anarchy at bay only by thanklessly feeding off of our generous aid, mind you) in order to prevent it from being taken over by dangerous radicals that hate us.

My apologies to wizard -- you are correct about the WTO.  Lessons learned:

1) Just because someone is wrong doesn't necessarily mean they're entirely wrong.

2) Lectures about fact-checking should be more carefully fact-checked than usual.  :)

[quote="crazychinesemaniac"][quote="(^_^)"][quote]it is a federation with a constitution, a legislative body, a judicial branch with federal and local courts, judicial review of legislative and executive acts, and universal suffrage (with average voter turnout higher than in the US)[/quote]

that means nothing, and the current russian government is not known for protecting its citizens' freedom and human rights.[/quote]

you forgot to say: at all costs... :|[/quote]

Hmm... I wonder where did you take that information. You exaggerated things a lot. Some things in Russia are not perfect yet but not as far as you put it. No one limits somebody's freedom.  :D By the way, the current Russian government is the one who finally started to resurge Russia from the Eltsin's epoch. Neocon's US media make a fool of you. :)

[hide="Off-topic"]@wizard: the archduke wasn't Serbian: he was the Austro-Hungarian crown prince Franz Ferdinand (assassinated by a Serb in Sarajevo).

[/hide]

Now, back to the topic. War is out of question: the Georgian conflict was Saakashvili's major mistake, some weird miscalculation, counting on US support in the campaign (but still, it wasn't very likely that US would do much more than verbal support-so real reasons for Saakashvili's blunder still remain blurred).

The Western diplomacy indeed does some heating up of the situation, but there's no real threat. For instance:

[quote]

Russia could have its sights set on other countries after recognising the independence of Georgian breakaway regions South Ossetia and Abkhzazia, French Foreign Minister Bernard Kouchner said Wednesday.

Likely objectives could be "the Crimea, Ukraine and Moldova," Kouchner said, emphasising that this was "very dangerous."
[/quote]

Objectively speaking, this is very unlikely (Moldova, Ukraine? There can only be some kind of political/diplomatic pressure, no military intervention whatsoever), but sounds like an efficient mind-game... On the other hand, we could have heard similar mind-game comments from the "No-independent-Kosovo" block in February 2008.

[quote="crazychinesemaniac"]Does anyone other than me think that if this problem with Georgia, The US and Russia isn't resolved soon, we could go into World War 3?[/quote]
I hope not.I am not "news-watching" lover,but I am convinced that the conflict will be settled soon,IMHO:furthermore,at currently moment unleashing Third World War is not good neither for US,or for Russia(it is bad for Georgia in any case).
I see only one way to settle this conflict,US must not interfere in this conflict(I am not saying that UN shouldn't interfere,assuming that US and UN are not the same yet),of course,Russia must recognize the independence of Osetia...

[quote="hsiljak"][hide="Off-topic"]@wizard: the archduke wasn't Serbian: he was the Austro-Hungarian crown prince Franz Ferdinand (assassinated by a Serb in Sarajevo).[/hide][/quote] I actually sometimes wonder if wizard isn't some bizarre satire/performance art piece.  I mean, the comment you responded to and the stuff about currency?  What is there to say about such nonsense?  I think that was one of those posts that one should just rate as spam and ignore.

[quote="hsiljak"]Objectively speaking, this is very unlikely (Moldova, Ukraine? There can only be some kind of political/diplomatic pressure, no military intervention whatsoever), but sounds like an efficient mind-game... On the other hand, we could have heard similar mind-game comments from the "No-independent-Kosovo" block in February 2008.[/quote]  Well, Moldova is pretty solidly in the Russian sphere, isn't it?  For example, Russian troops are present in Transnistria, a region claimed by Moldova.  (Similar to South Ossetia or Abkhazia in Georgia.)  And it seems certainly the case that the Russians believe that Ukraine should be in their sphere of influence, although I imagine they wouldn't try to enforce that militarily.

Yes, there are Russian troops in Transnistria, but Moldova, squeezed between NATO-wannabe Ukraine (which will probably never enter the Pact, for the sake of world peace) and NATO member Romania could hardly be a terrain for campaign like in Georgia. And the Molodovan government is more careful in its actions against the Transnistrians: not like the latent state of war in Georgia for tle last few years. On the other hand, Russians may put some pressure on the Ukraine, but without any real threat-it would be too risky. BTW, Ukraine has been trying to get closer to NATO in the last few years, but Brussels doesn't want to push it, as Russia is a big player in the world security game: so the status quo is satisfying for both sides at the moment.

I basically agree, except to note that a "[url=http://www.npr.org/templates/story/story.php?storyId=5070407&ft=1&f=3]real threat[/url]" need not be [url=http://www.iht.com/articles/ap/2006/12/26/business/EU_FIN_Belarus_Russia_Gas_War.php]military[/url] in nature.

Remember the gas crisis in 2007?

http://en.wikipedia.org/wiki/Russia-Belarus_energy_dispute

When it comes to profit and power, Russia can blackmail even the greatest allies (or pets :maybe: ?): i.e. Belarus. I agree on the statement that "the real threat" doesn't have to be military-with an addition to this gas story - shutting of the Russia-Ukraine/Belarus gas pipes threatens Western European industry.

EDIT: now I saw your second link, sorry for doubling ;) I definitely agree: modern politics have a wide spectrum of threatening and "convincing".


[quote="jagdish"][b](1) find the value of  sin25*sin35*sin85
here angle are in degree. and * means multiplication.[/b][/quote]

Let $ x\equal{}\frac{\pi}{36}$ (5 degrees)

We are looking for $ A\equal{}\sin(\frac{\pi}{6}\minus{}x)\sin(\frac{\pi}{6}\plus{}x)\sin(\frac{\pi}{2}\minus{}x)$

$ A\equal{}(\frac{1}{2}\cos(x) \minus{} \frac{\sqrt 3}{2}\sin(x)) (\frac{1}{2}\cos(x) \plus{} \frac{\sqrt 3}{2}\sin(x)) \cos(x)$

$ A\equal{}\frac{1}{4}(\cos^2(x) \minus{} 3\sin^2(x))  \cos(x)$

$ A\equal{}\frac{1}{4}(4\cos^3(x) \minus{} 3 \cos(x))$

$ A\equal{}\frac{1}{4}\cos(3x)$

$ A\equal{}\frac{1}{4}\cos(\frac{\pi}{12})$

And, since $ \cos(\frac{\pi}{6})\equal{}2\cos^2(\frac{\pi}{12}) \minus{} 1$ and $ \cos(\frac{\pi}{12})>0$, we get $ \cos(\frac{\pi}{12})\equal{}\sqrt{\frac{2\plus{}\sqrt 3}{4}}$

And so $ A\equal{}\frac{\sqrt{2\plus{}\sqrt{3}}}{8}$

And with $ \sqrt{2\plus{}\sqrt{3}}\equal{}{1\plus{}\sqrt{3}\over\sqrt{2}}$ this simplifies further into $ A\equal{}{\sqrt{2}\plus{}\sqrt{6}\over 16}$


geometry
3D geometry
sphere
inequalities unsolved
inequalities

Let $x-4=a,\ y-5=b,\ z-6=c$, your problem is that if $a,\ b,\ c\geq 0,$ and $(a+4)^{2}+(b+5)^{2}+(c+6)^{2}\geq 90$, then prove $a+b+c\geq 1$.

Proof: Consider 3-D $abc$ plane, we are to find the condition for which the plane $a+b+c=k$ has at least one intersection points with the exterior part or the surface of the sphere with the center $(-4,\-5,\-6)$ and radius $3\sqrt{10}$. Thus using the fomula of the distance between a point and a plane, the condition gives the distance between the center and the plane is greater than or equal to the radius $\Longleftrightarrow \frac{|-4-5-6-k|}{\sqrt{1^{2}+1^{2}+1^{2}}}\geq 3\sqrt{10}$, since $k\geq 0$, yielding $k\geq 3\sqrt{5}(\sqrt{6}-\sqrt{5})\geq 1$.