calculus geometry trigonometry function HMMT AMC AIME \/closed

I have a two questions regarding the calculus class

1. when will the next one be offered?

2. I took the "are you ready for this" test and was able to solve almost all of the problems quickly, but because its been emphasized so much that one shouldn't take calculus without plenty of algebra preparation, I'm wondering how much algebra that should be. What are the main concepts one should be familiar with (looking at the interm algebra textbook TOC, I know it pretty well except for the last 3 chapters)? What level of AMC/AIME/USAMO questions should we be comfortable solving?

In my opinion, AMC/AIME learning is very separate from calculus learning.  For the AMC/AIME, you need to be very comfortable with a variety of problem-solving strategies and little tricks.  That is, you usually need to be able to go outside the box a little bit.  With calculus, I didn't get that feeling.  Very rarely does a calculus class try to see how far outside the box you can think.  You will get a question, and that question will somehow depend on some concept you learned.  For example, knowing how to solve a related rates problem or average value problem depends on your knowledge of the derivative.  Seeing a volume or area problem should immediately trigger integral.  I think calculus problems are far more straightforward than AMC/AIME problems.

As for preparation, I would make sure that you're very comfortable with trig, in particular, trig identities and values of trig functions in radians.  Know how to complete the square, partial fraction decomposition, and know how to work with and re-arrange fractions.  If you are comfortable with these, then you should be ready for calculus.

JRav is pretty much correct about the typical calculus class being much easier than AMC, just as all the rest of the typical high school curriculum is easier than the AMC.  That said, our calculus class is not the typical calculus class, just as our Intro Algebra class is a wholly different beast than a typical Algebra I class.  (JRav, I believe, is not referring to our calculus class in these comments.)  We do include problem solving in the calculus class, and pull problems from contests like HMMT and the Putnam.  The AP BC test is not the purpose of our course, so our course is more challenging than that test (which is pretty easy).

JRav's assessment of needed algebra skills is accurate.  I'd add being comfortable with functions and their graphs.

I do think students should be able to pass the AMC before taking calculus.  Of course, few students do so because high school calculus classes are designed with a much more recipe-like approach.  There are many 5's on the AP who fail the AMC.  There are few people who pass the AMC who will not eventually earn a 5 on the AP when they take it.  The main reason I suggest passing the AMC before calculus is simply that if you haven't, then you haven't really learned how to solve problems.  You've raced through the curriculum and are in what I call the "Calculus Trap" -- the belief that the goal of a math education in high school is calculus, when it should be the ability to solve problems you haven't seen before.  This is much more important than the machinery of calculus.  (I might even make my restriction tighter and suggest that a student get 3-5 AIME problems, too.  Mainly, I would want to create incentives that make students realize that AMC-style problem solving should be the goal of a middle & high school education, not mere fact memorization and procedure performance.)

And finally, to answer your first question last -- we expect to offer the class again in fall, 2009.

Will you ever create a Calculus textbook? Just wondering- no pressure. :D

[quote="AIME15"]They are already writing it - I read that DPatrick is supposed to have it done by the end of this year.[/quote]

This is correct.

I look forward to this!  In the mean time, if you want a really rigorous calculus book that completely and rigorously develops the theory, there is no better book that Spivak.  Many people will agree with me, and many people will prefer Apostol or something else.  But Spivak is really good.  He starts you off with some  numerical examples to try, and then he moves on to more difficult proofs, and finally, he finishes with proofs that are absolutely insane by his own admission.  He doesn't really teach problem-solving per se, but he teaches you calculus.


Is there a way that i can find the Latex source of a PDF documnet that has been created by texnic center?

I tried to open the PDF document by Texnic center again
, It opened it , But it shows some Weird Symbols :D

I'm afraid it can't be done*. The latex code is not in the pdf and any equations you have there behave like pictures rather than text. This is why it's more important to keep backup copies of the original tex file than the pdf which can always be regenerated.

The weird symbols may be due to peculiar fonts that haven't been included in the pdf and the document properties may give some clues.

*It is possible to include latex code in pdfs (by using pop-ups for example) but that is rather unlikely.

If you deleted the tex file, you can always revive it from the Recycle Bin...at least on a pc


linear algebra
matrix
vector
inequalities

Hi, I wonder if anyone who could help me here a bit, thanks.

Let $Q \in R^{n \times m} (n > m)$ be an isometry with the columns $q_1, q_2, ..., q_m$

$1.$ Show that $QQ^Tv=0$ is $v$ is orthogonal to $q_1, q_2, ..., q_m$

$2.$ Show that $QQ^Tq_i = q_i, i=1,...m$

$3.$ Show that $(QQ^T)^2= QQ^T$ . Thus $QQ^T$ is a projector

hi

I don't know how much algebra you have had yet, but I will give some hints

first of all, with the dimension you give that cannot be an isometry I think, isometries have invertible matrices
i think you meant distance preserving map, injective, but not surjective

consider a few things like these

when $v\in \mathbb{R}^n$  , work out for yourself that

${Q^T v=\begin{bmatrix}<q_{1},v>\\ \vdots \\{ <q_{n},v>}\end{bmatrix}$

the $<a,b>$ stands for the standard inproduct in $\mathbb{R}^n$

now try to understand that $<q_{i},q_{j}> \ \ =0$ when $i\neq j$  and $<q_{i},q_{j}>\ =1$ when $i= j$

these things immediately give you the first part

the second part of the exercise will be seen immediately when you work out $Q Q^T q_{i}=Q (Q^T q_{i} )$

the third part  : i will try to give you a very intuitive proof
 if it doesn't work, come back :
when you apply $Q Q^T$ to a vector $v$ you will get a vector in $\mathbb{R}^n$ that is a linear combination of the $q_{i}$
because of the second part of the question those $q_{i}$ are not bothered by applying $Q Q^T$ again
and thus our $Q Q^T v$ seen a linear combination of the $q_{i}$ won't be bothered either

Thanks for your help.
I think I got the first one as you said which is about the pairwise orthogonal.
But  I still dont understand what u mean for the part 3. Could u explain more? Thx in advance!! :blush:

[quote="fredbel6"]when you apply $Q Q^T$ to a vector $v$ you will get a vector in $\mathbb{R}^n$ that is a linear combination of the $q_{i}$
because of the second part of the question those $q_{i}$ are not bothered by applying $Q Q^T$ again
and thus our $Q Q^T v$ seen a linear combination of the $q_{i}$ won't be bothered either[/quote]

Please tell me what exactly of my explanation you don't get

On a side note : a different more matrix like way to work would be to prove that $Q^T Q=1$ with $1$ the $m\times m$ identity matrix
then you get $(Q Q^T)(Q Q^T)=Q (Q^T Q) Q^T=Q Q^T$

Hi,

I think I know how to know that $QQ^T=I$, but I dont know if I am in the right track.
I guess you were saying that if I could prove $QQ^T=I$ then I could plug into $(Q Q^T)(Q Q^T)=Q (Q^T Q) Q^T=Q Q^T$ then provided the inequality of part $3$, I think it could show the inequality of part $2$, $QQ^Tq_i= q_i$, right?

no oh no
first of all i think you mean equality

the first two parts have been done and i haven't come back to them
i just gave two different ways solve the third

Hi, I guess I am not following right...

I mean I know how to prove the $QQ^T=I$, then my question is can I use it to prove the part $2$ and part $3$?

Please help, Thx in advance.


just out of pure curiosity, wat do you all do when ur really stuck on a problem?

depends what circumstances. like if you're talking about amc's on something else.

it matters a lot
-jorian

There should be multiple choices.

In the end I answered "AOPS" because that's the final step.

1) Skip it
2) Come back to it
3) Go over it for a while
4) Post on AOPS

[quote="i_like_pie"]What if nobody here knows how to solve it? :ninja:[/quote]

Then the problem writer's have not created a solvable problem.  We have torajir0u and darij.  I don't think there are any "unsolvables" for them.

Or you're posting in the wrong forum. If it's a contest problem, then if there is no way to solve it (not no good way, just plain no way), or there is no answer, then you shouldn't be doing that contest.

well, for cases like mathcounts thats definately not true
-jorian

Why post it here if you aren't ALLOWED to? :wink:  :wink:  :wink:


problem 4
Find all functions f from R to R and f (x) f (y) – f (xy) = x + y

Very easy; since no one posted a solution, here's mine:

Let $P(x,y)$ be the assertion $f(x)f(y)-f(xy)=x+y$.

$P(0,0)\implies f(0)^2-f(0)=0\implies f(0)=1$ or $f(0)=0$.

If $f(0)=1$ then $P(x,0)\implies f(x)=x+1\ \forall x\in \mathbb{R}$, which is indeed a solution.

Else, $f(0)=0$ and we get $P(x,0)\implies x=0\ \forall x\in \mathbb{R}$, a contradiction.

It follows $\boxed{f(x)=x+1}\ \forall x\in \mathbb{R}$ is the only solution. $\blacksquare$

Arkan


Let $n$ be a positive integer.
1. Write down all ordered pairs of integers $(a, b)$, where:

a) $1 \le a < b \le n$;
b) $a+b > n$;
c) $a$ and $b$ are relatively prime

2. For each ordered pair $a, b$, compute the reciprocal of the product of the two numbers, in other words, compute $\frac{1}{ab}$.

3. Add up all of these fractions.

Prove that this must equal $\frac{1}{2}$.

Note: The last sentence is my own conjecture, the problem asks you to experiment and conjecture something.

[hide="Solution"] We'll prove by induction.  It's obvious for $n = 2$.

Given that it's true for $n = k$, consider how the summation changes as we move to $n = k+1$.  We lose all the pairs $(a, b)$ such that $a+b = k+1$, and we gain all the pairs $(a, b)$ such that $b = k+1$.

The sum of the pairs we lose is

$\frac{1}{2}\sum \frac{1}{i(k+1-i)}= \frac{1}{2(k+1)}\sum \frac{1}{i}+\frac{1}{k+1-i}= \frac{1}{k+1}\sum \frac{1}{i}$

where $i$ is taken over the units $\bmod k+1$ (we divide by $2$ because we count each pair twice; recall that if $i$ is a unit then $k+1-i$ is also a unit), and the sum of the pairs we gain is

$\sum \frac{1}{i(k+1)}= \frac{1}{k+1}\sum \frac{1}{i}$

Where $i$ is again taken over the units $\bmod k+1$.  Clearly the two are equal, so the summation is constant as a function of $n$ by induction.  QED. [/hide]

[quote="t0rajir0u"][hide="Solution"]

where $i$ is taken over the units $\bmod k+1$ [/quote]

Can anybody please explain, the meaning of the above line ?


Let $S$ be a set with $n$ elements. In how many different ways can one select two not necessarily distinct or disjoint subsets of $S$ so that the union of the two subsets is $S$? The order of selection does NOT matter.

[hide] For each element, we have 3 possibilities: in subset A, in subset B, or in both. Therefore, we have $3^{n}$ possibilities. However, this choice takes order into account. Therefore, we must divide by $2$ for all subsets except the pair $S,S$, whose reverse is itself. The answer is $\frac{3^{n}+1}{2}$. [/hide]

This was just like an AIME II 2002 Problem except 10 elements.

[quote="t0rajir0u"]Well, it couldn't have been just like this problem, since $n = 10$ gives a number far too large to have been an AIME answer.[/quote]

Then they asked us to find the last three digits of the number.

[quote="Phelpedo"]The problem was quite different but incorporated a similar idea for the solution.[/quote]

Oh, I just remembered, the solution that is all.


Prove that there exist infinitely many collections of positive integers $ (a,b,c,d,e,f)$ such that $ a < b < c$ and the equalities $ ab \minus{} c \equal{} de,$ $ bc \minus{} a \equal{} ef$ and $ ac \minus{} b \equal{} df$ hold.

Sistem equations equavalent to :
$(a-1)(b+c)=d(e+f), \ (b-1)(a+c)=e(d+f), \ (c-1)(a+b)=f(e+d).$
Let $p=a+b+c-1, \ q=e+f+g, \ then \ p^2-x^2=q^2-y^2, \ (x,y)=(b+c+1-a,e+f-d),(a+c+1-b,d+f-e),(a+b+1-c,d+e-f)$.
It given solutions $a=b,f=0,a^2-1=de$. For example:
$a=b,c=1,f=0,d=a+1,e=a-1$ give solution for any $a$.
I am sorry, I forget conditions.

[quote="Rust"]$a=d+1,b=e+1,c=f+1,f=e+d,$
give infinyty solutions, where  d,e are  any natural.[/quote]

This choice of $a,b,c,d,e,f$ does not work: $ac-b=df+2d, bc-a=ef+2e,$ but it is easy to repair this solution...


Bright costs 60% more than Fresh, so for the same volume it costs one dollar and sixty cents. Glow costs 25% less than Bright for the same volume, so it costs one dollar and twenty cents per unit of volume.

I think I messed up, though, since I didn't use the volumes, but since it's asking for the amount per unit of volume...

im pretty sure this is right but im not sure

[hide] ok u want to convert everything into fresh which is F(F=fresh,B=Bright,G=Glow)
so u know that B=1.6F (price)so you take u take .75 of that and find that G=1.2F(price)
Since you dont care about Brights volume you know that G=(4/3)F(volume)
so u can make a table:
                 B                F               G
Price-   (8/5)F               F            (6/5)F

Volume-  dont care       F                (4/3)F

so now u divide and get 9/10 price per volume so the answer is [b]90[/b] cents[/hide]

[hide]
Let $B$ be bright, $F$ be Fresh and $G$ be glow
$B_c = 1.6F_c$
$B_v = .75G_v$
$G_c = .75B_c$
$G_v = \frac 43F_v$

in order of given

If Bright costs 1.6 that of Fresh, and Glow costs .75 that of Bright, Glow costs $.75*1.6 = 1.2$
The total cost of Glow will be 1.2. We must now find the volume. It is given that Fresh is 1 dollar per unit of volume. If Glow is 4/3 of that, the price per units of volume will be $ \frac {1.2}{\frac 43} $ = $9/10$ or 90 cents
[/hide]

[quote="236factorial"]A consumer report revealed the following information about three tubes of toothpaste. Bright is 60% more expensive than Fresh and has 25% less volume than Glow. Glow is 25% less expensive than Bright and has 33 1/3 % more volume than Fresh. Fresh costs 1 dollar per unit of volume. what is the number of cents per unit of volume of Glow?

I took about 5 minutes and got something like 72 which I don't even know if it's anywhere close.[/quote]I think it's 90. javascript:emoticon(';)')
Wink

Sorry everyone, I thought that 1.6 was 5/3... :blush:
I got it now. THanks everyone (it really wasnt' that hard)

[quote="JesusFreak197"]Oh, right, that's almost as easy to do as I thought. It's one dollar and twenty cents for the same volume as Fresh, but it has $\frac{1}{3}$ more volume than Fresh, so it's 90 cents. ;)[/quote]

I'm surprised that it's the last problem on a National sprint. I heard that on this year's sprint at nationals, only one person got the last problem correctly.

[quote="noneoftheabove"]yeah that guy was on my team the problem wasnt that hard though i just forgot that it was a space diagonal[/quote]

Space diagonal? What are you talking about?

He's referring to this year's #30, but I don't know what it was.

This year, I think most people just didn't have time to do #30, because there were a lot of time consuming questions before #30.

[quote="bbb"] javascript:emoticon(';)')
[/quote]
what's up with the javascript: emoticon thingy???


trigonometry
LaTeX
email
geometry
trig identities
Law of Sines

How do you work these problems i cant figure it out the equation is a2=b2+c2-2bc Cos A i dunno what goes where

$a^2 = b^2+c^2-2bc\cos{A}$

a=side facing angle A.
b and c=the other two sides...
A= angle A, which faces side a.

you can change around to make these:

$b^2 = a^2+c^2-2ac\cos{B}$
$c^2 = a^2+b^2-2ab\cos{C}$

so can anyone give me a step by step or explain or something

Umm...what exactly is the problem that you need help with?

[quote="brock"]i know that but i mean like if you have a triange points J K and H   the measure of angle J is 55 and h=6.3 and k=6.7 how do you solve[/quote]Are h and k the lengths of the sides opposite point H and point K, respectively?

[quote="brock"]i know that but i mean like if you have a triange points J K and H   the measure of angle J is 55 and h=6.3 and k=6.7 how do you solve[/quote]

applie the method...

$j^2 = 6.3^2 + 6.7^2 - 2 \cdot 6.3 \cdot 6.7 \cdot \cos{55}$

then from there, since you have all the three sides, use Law of Sines to finish off the angles.

Just a note, you use Law of Cosines to solve triangles when you know two sides and their included angle OR all three sides.

[quote="brock"]ya i get a little bit confused on that to just the order so you have all 3 sides h=6.3 k=6.7 and you might wanna check it but i got j=36.16 and where do i go from there on the sines formula[/quote]

notice it says $j^2$?... u didnt square root... as i used my calc...

$j = \sqrt{36.15867724} = 6.01 =6.0$ which sounds right...

about finishing off... now we have enough info:
Law of Sines

$\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}$

o i see you square both sides duhh lol i feel stupid ok and how do you put the sybols on there like sq. rt. and squared

and having j=6.1 h=6.3 and k=6.7 on Sin A / a and Sin B / b what goes where say to find angle J

You can use $\LaTeX$.  There is a little tutorial thing somewhere on this site.

i do not understand your question... what?

[quote="brock"]ok the sides of the triangle are h=6.3 k=6.7 and j=6.0 to i have angle J=55 with the law of sines  equation how do you find the other 2 angles what s the process[/quote]

$\displaystyle \frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}$

the important thing here is matching the sides with the corresponding angles...

$\displaystyle \frac{j}{\sin{J}} = \frac{h}{\sin{H}} = \frac{k}{\sin{K}}$
$\displaystyle \frac{6.0}{\sin{55}} = \frac{6.7}{\sin{K}} = \frac{6.3}{\sin{H}}$

from here, separate the 3-way equation into two two-way equations:
$\displaystyle \frac{6.0}{\sin{55}} = \frac{6.7}{\sin{K}}$
$\displaystyle \frac{6.0}{\sin{55}} = \frac{6.3}{\sin{H}}$

now solve for each...

hope that cleared up, brock.

for more question, just PM me, or aim me at es7abl1sh3d

or email: es7abl1sh3d@gmail.com

[quote]

and with that 6.0/sin 55 = 6.7/sin h do you cross multiply and divide 6.7 x sin 55 /6.0 [/quote]

assuming that what is in the quote in what you actually asked, then yes. Cross multiply...

or you can simply get the value $\frac{6.0}{\sin{55}}$... then multiply the do other stuff to get the answer.


modular arithmetic
quadratics
search
algebra
quadratic formula
special factorizations

For $n\leqslant 100$, find all $n$ such that $n^{4}-n^{2}+57$ is divisible by $73$.

I was just about to post this exact problem.  :o

[hide="Tricks"] The quadratic formula applies in prime moduli :o  (Once you understand modular division in prime moduli, you'll see that every step in completing the square is completely valid.)

$n^{2}\equiv \frac{1 \pm \sqrt{1-4 \cdot 57}}{2}\bmod 73$

After a lot of rather annoying searching, we find that $\sqrt{1-4 \cdot 57}\equiv \pm 24 \bmod 73$ (there are ways to speed this search up, but I don't remember what they are.)  Hence

$n^{2}\equiv \frac{1 \pm 24}{2}\equiv \frac{50}{2}, \frac{25}{2}\equiv 25, 49 \bmod 73$

It's very easy to take the root of these;

$n \equiv \boxed{ 5, 68, 7, 66 }\bmod 73$ [/hide]

If anyone's interested, I might post a little something about modular division and modular $n^{th}$-roots on my blog.

[quote="boxedexe"][quote="junggi"]How did you come up with the first step?[/quote]
Add $16(73)$.[/quote]
I mean how did you think of doing that?
It seems kinda random to me....or was it just adding 73 until you got a square?

[quote="junggi"]or was it just adding 73 until you got a square?[/quote]

Yep, it probably was.  (Though if you want to solve the problem along those routes, I think my solution is a little more straightforward and familiar :) )

Using Legendre Symbol, you obtain that $57\equiv-16\pmod{73}$ is a quadratic residue modulo $73$.  Then you want to find $x\mod 73$ such that $x^{2}\equiv-16\pmod{73}\implies x\equiv \pm\sqrt{-16}\pmod{73}$.  There are exactly two solutions to this congruency since $73$ is prime.  Using primitive roots, you can solve this to get $x\equiv \pm 35\pmod{73}$.

Say it with me now:  Google, Wikipedia, Mathworld.

http://mathworld.wolfram.com/LegendreSymbol.html
http://en.wikipedia.org/wiki/Legendre_symbol

[hide]
\[n^{4}-n^{2}+57\equiv n^{4}-74n^{2}+57\equiv (n^{2}-37)^{2}+57-37^{2}\pmod{73}\]
Since $37^{2}\equiv (2^{-1})^{2}\equiv 4^{-1}\equiv-18\pmod{73}$, the above reduces to
\[(n^{2}-37)^{2}\equiv-2\pmod{73}\]
By inspection, we find $73=1+2\cdot 6^{2}$ giving
\[-2\equiv (6^{-1})^{2}\equiv (-12)^{2}\pmod{73}\]
(Often, a nice way to find a square root of some small number $s$ modulo a prime $p$ is to find a multiple of $p$ of the form $a^{2}-sb^{2}$.) So our quartic factors into $(n^{2}-25)(n^{2}-49)$ which conveniently factors further:
\[(n+5)(n-5)(n+7)(n-7)\equiv 0\pmod{73}\]
[/hide]


symmetry
trigonometry
inequalities
algebra unsolved
algebra

Solve inequation with variable $x$: $\sqrt{1+x}+\sqrt{1-x}\le 2-\frac{x^{2}}{4}$

Sketching the graphs of $y=\sqrt{1+x}+\sqrt{1-x}$ and $y=2-\frac{x^{2}}{4}$ for $-1\leq x\leq 1,$ we can find the answer immediately,
$-1\leq x\leq 1.$ Especially the graph of $y=\sqrt{1+x}+\sqrt{1-x}$ can be obtained by composing two graphs $y=\sqrt{1+x}$ and $y=\sqrt{1-x},$ and has the symmetry with respect to the $y$ axis and endpoints $(-1,\ \sqrt{2}),\ (1,\sqrt{2}).$

Comment:
Most people will solve by the squaring both sides of the given inequality.

$\sqrt{1+x}+\sqrt{1-x}\leq 2-\frac{x^{2}}{4}$

$\Longleftrightarrow (\sqrt{1+x}+\sqrt{1-x})^{2}\leq \left(2-\frac{x^{2}}{4}\right)^{2}$ under the conditions  $1+x\geq 0$ and $1-x\geq 0$ and $2-\frac{x^{2}}{4}\geq 0.$

I will leave it you from now on.

I find a method can use to solve it as:
Take $x=\ cos 2t$ with $0\le t\le \frac{\pi}{2}$, the inequation reduce:
$\sqrt{2}(\sin t+\cos t)\le 2-\frac{cos^{2}2t}{4}$

But now i have no idea to continue

Let $x=2\sin \theta,$ we can rewite the given inequality as $|\sin \theta+\cos \theta|+|\sin \theta-\cos \theta|\leq 2-\sin^{2}\theta,$ where $-\frac{1}{2}\leq \sin \theta \leq \frac{1}{2}.$

Yes, a nice substitution. Thanks a lot. I really have solved ít!

Another approach:
condition $-1\le x\le 1$
Put $t=\sqrt{1-x}+\sqrt{1-x}$ then:$t^{2}=2+2\sqrt{1-x^{2}}\Rightarrow x^{2}=t^{2}-\frac{1}{4}t^{4}$
The equation is equavalent with: $t=2-\frac{1}{4}\left(t^{2}-\frac{t^{4}}{4}\right)\Leftrightarrow (t-2)^{2}(t^{2}+4t+8)=0\Leftrightarrow t=2$
Hence the equ has a root $x=0$


The number $ 5^{2009}$ is calculated. The sum of this huge number's digits is then obtained, then the sum of the digits of this result is found and so on until we get only a single digit. What is this digit?

What about  for $ 4^{2008}$

[hide="hint"]
Maybe this can help.
[url]http://www.mathlinks.ro/viewtopic.php?t=273755[/url]
[/hide]

$ 5^{2009}$ gives $ 2$.
$ 4^{2008}$ gives $ 4$

[hide="note"]this is pretty much the same as the 'divisibility test' problem[/hide]

Dude, just take it $ \mod 9$.

[hide="Solution"]
$ \varphi(9)\equal{}6$
$ 2009\equiv 5\bmod 6$.
$ 5^5\equiv 5\cdot 25^2\equiv 5\cdot 49\equiv 20\equiv \boxed{2}\bmod 9$
[/hide]


topology
advanced fields
advanced fields unsolved

Consider the well-known bijection $ f: (0,1)\to (0,1)\times (0,1)$ given by
$ f(0.a_1a_2a_3a_4\ldots) \equal{} (0.a_1a_3\ldots,0.a_2a_4\ldots)$

($ 0.a_1a_2a_3a_4\ldots$ is the decimal representation of some $ 0 < x < 1$)

(i) Is $ f$ continuous?
(ii) Is $ f$ a homeomorphism?

A bijection?  What's the preimage of $ (0.\bar{9}, 0.\bar{1})$?

I'm sorry. It should be $ f: [0,1]\to [0,1]\times[0,1]$, with $ 0$ and $ 1$ written as $ 0.00\ldots$ and $ 0.99\ldots$ respectively

(i): it seems to be continuous, since $ |x\minus{}y|<10^{2n}$ implies $ |f(x)\minus{}f(y)|<2\cdot 10^n$ (if i'm not completely gone).
(ii): no, it can't be continuous, since any point (different from 0,1) disconnects the first set, but leaves the second connected.

[quote="ma_go"](i): it seems to be continuous,
(ii): no, it can't be continuous, [/quote]
???

I think he means that $ f$ is continuous but $ f^{\minus{}1}$ is not.

I still didn't get the second part, Could you provide an elementary proof of $ f^{\minus{}1}$ is not continuous over $ [0,1]\times [0,1]$?

I think ma_go means (ii) no, it can't be homeomorphic...since that is what the statement is proving.

It is a fairly common technique. If you remove a point from two homeomorphic sets, it shouldn't make the two topologies different somehow.

That would be a way to show $ (0,1)$ and $ [0,1]$ are not homeomorphic. You always disconnect $ (0,1)$ when removing a point, but you can remove a point from $ [0,1]$ and keep it connected.

Same idea applies. You never disconnect $ [0,1]\times [0,1]$ when removing a point, but you always disconnect $ [0,1]$ when removing a point (except for $ 0$ and $ 1$ itsef).


Prove that
$\large\liminf_{n\to\infty}\mbox{ } n|\sin n|=0$  where   $\large n\in\mathbf{N}$

[quote="Rust"]It is easy to prove, that $\large\liminf_{n\to\infty}\mbox{ } n^2|\sin n|<\frac{1}{\sqrt 5}.$[/quote]
Without knowledge of Hurwitz' theorem ( http://mathworld.wolfram.com/HurwitzsIrrationalNumberTheorem.html ) this seems not so easy (and I'm also not sure if you really get that bound with it).
Also a pleasure for other topics: please give proofes instead of such answers, I don't think they really help the author of the topic. Or at least give a short reasoning (but proving is better than reasoning/sketching).

I am sorry. I thik, this problem is not solved.
Let $x_0=\frac{1}{\pi},a_0=0,P_0=Q_{-1}=0,Q_0=1,P_1=1$
$a_{k+1}=[\frac{1}{x_k}],x_{k+1}=\frac{1}{x_k}-a_{k+1},P_{k+1}=a_{k+1}P_k+P_{k-1} \ (k>0),$
$Q_{k+1}=a_{k+1}Q_k+Q_{k-1}$ (Defined continiusly fraction).
We have $Q_k<n<Q_{k+1} \Longrightarrow |\sin n|<|\sin {Q_k}|$. Therefore
$\large\liminf_{n\to\infty}\mbox{ } n|\sin n|=\large\liminf_{k\to\infty}\mbox{} Q_k|\sin {Q_k}|.$
We have $\sin {\frac{\pi}{2Q_{k+1}}}<|\sin{Q_k}|<\sin {\frac{\pi}{Q_{k+1}}}$. Therefore this problem equavalent to
$\large\limsup_{k\to\infty}\mbox{ } a_k=\infty.$


Okay, I have a question -

Aren't SBC and AT&T the same thing now?  So how can someone like SBC more than AT&T, or vice versa?

Uhh
Yeah
I guess

I've never heard anyone say they like SBC more than AT&T or vice versa though

[quote="akalra1 (me)"]aargh ATT

SBC... all better[/quote]
anyways, chek the "just a question forum" (its locked)
anyways, SBC owns ATT, but SBC calls it ATT and its weird
anyways, its AT&T

$3$

At least I've beaten shobhit and arnav to it.

thats not a godd thing, its gonna get deleted


algebra
polynomial
number theory
greatest common divisor
algebra unsolved

Find all $ m,n\in\mathbb{N}$ such that
$ 1\plus{}x\plus{}x^2\plus{}...\plus{}x^m|1\plus{}x^n\plus{}x^{2n}\plus{}...\plus{}x^{mn}$

See, if you write is as $ \frac{(x^{n(m\plus{}1)}\minus{}1) (x\minus{}1)}{(x^{m\plus{}1}\minus{}1)(x^n \minus{} 1)}$

then you will see that both $ (x^{m\plus{}1}\minus{}1)$ and $ (x^n \minus{} 1)$ are factors of $ (x^{n(m\plus{}1)}\minus{}1)$. x=1 is not a repeated root for any of these three polynomials. So, it is enough if $ x^{m\plus{}1}\minus{}1$ and $ x^n \minus{}1$ have only x-1 as the common factor. Which means we must have gcd (m+1,n)  = 1.

You can also argue using complex roots of unity, but this solution is straightforward enough I guess


Are these really the official results? I can see some strange results there.

Yes, I am pretty sure these are the official results.

Why do you say that some of these results are strange?

Probably I did not read correctly, but some guys that were fantastically good (and they still are, surely) have some small marks.

Like Kappa (Gabriel Kreindler)? I think he got first prize last year, this year he got third prize.

(But oh well, he got 40 points at IMO :D)

We can see some of the former (and current) Romanian IMO gold medalists there: Andrei Negut, Ana Caraiani, Dragos Micnea (11th grade this year!) and ...  :blush: Kappa. I suspect he didn't care too much about the contest itself :)


If Bob can eat $50 inch^3$ of ice cream in a half hour, and his ice cream melts at a rate of $60 inch^3$ an hour, how much ice cream would he eat in a two scoop bowl, each scoop having a radius of $6$ in. (assuming that melted ice cream can't be consumed)?
[hide="answer"]$360pi$ if I'm not mistaken[/hide]

I got [hide] $\pi180$. (assuming that each scoop is half a sphere with $r=6$, and the bowl contains $\pi288 inch^3$ ice cream[/hide]

So are we assuming that it is half a scoop or a whole scoop, because then the problem would change, and then your answers would be correct?

I thought it was pretty obvious it was a whole scoop  :? . Can you go to a store and order half a scoop of ice cream?

[quote="k8reindeer"]well but since when are scoops perfect? they're not dainty little spheres are they? it's hard to classify your typical ice cream scoop as sphere or half sphere. I'd say something in between, but then I wouldn't know how to find the volume...[/quote]
In math, everything is perfect :D. like the board problem at nats on the sprint back page.

Yeah, sorry. It is $\pi360$ :blush:
[hide]$0.625(\pi576)=\pi360$[/hide]
That is what happens when you do problems at 1:00 am :wallbash_red:


bad inequalitie
for all ai >0
prove that
(a1/(a1+a2))+..........+(an/(an+a1)) >(n/4)

I  think  he ment  this  ineq :  [tex]\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+...+\frac{a_n}{a_1+a_2}>\frac{n}{4}[/tex]  which is  an old  problem (ASU , 1969)


A standard die is rolled six times. What is the probability that the result of each roll is odd? Express your answer as a common fraction.

Answer:  there is 1/2 chance for odd, so (1/2)^6=1/64

[quote="theGoodGuy"]to clear that up...the only way to get an odd product is to have all of the numbers that you are multiplying be odd. For that to happen, there is a 1/2 chance per roll. so (1/2)^6=1/64[/quote]
Ummmm, I'm not sure why you added [quote="theGoodGuy"]the only way to get an odd product is to have all of the numbers that you are multiplying be odd[/quote]  The problem asked nothing of the product of the rolled numbers.  Either way, the answer is correct.  :)

$\frac{1}{2}$ of the numbers on a standard die are odd and vice versa.

Since you roll the die $6$ times and all the results must be odd, $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot = \frac{1}{2}^6=\boxed{\frac{1}{64}}$.

[quote=Shnarble]$\frac{1}{2}$ of the numbers on a standard die are odd and vice versa.

Since you roll the die $6$ times and all the results must be odd, $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot = (\frac{1}{2})^6=\boxed{\frac{1}{64}}$.[/quote]

Remember parenthesis  :-D


probability
parameterization
expected value
calculus
calculus computations

In analyzing the risk of a catastrophic even, and insure uses the exponential distribution with mean $\gamma$ as the distribution of the time until the even occurs. The insure had $n$ independent catastrophe policies of this type. find the expected time until the insure will have the first catastrophe claim.

isnt this just a gamma distribution with parameters $\alpha=n$ and $\beta=\frac{1}{\gamma}$? so the mean would be $\frac{\alpha}{\beta}=n\gamma$?

The sum $S = X_{1}+X_{2}+\cdots+X_{n}$ of $n$ independent and identically distributed exponential random variables $X_{1}, X_{2}, \ldots, X_{n}$ is gamma distributed.  Specifically, if $X_{k}$ are exponential with mean $\theta$, then $S$ is gamma with mean $n\theta$ and variance $n\theta^{2}$.

It's noteworthy that the sum $S$ is a random variable that describes the time until the $n$-th claim if claims followed a Poisson process (whose [b]interarrival[/b] times are exponentially distributed).

However, according to the model you specified, the events are not interarrival times, but rather they run concurrently.  So the time until the $k$-th event is NOT gamma; rather, it is the $k$-th order statistic $Y_{k}$.  Fortunately, the first such order statistic $Y_{1}$ is exponential, which we show by recalling that $Y_{k}$ has PDF

$f_{Y_{k}}(x) = \frac{n!}{(k-1)!(n-k)!}F_{X}(x)^{k-1}(1-F_{X}(x))^{n-k}f_{X}(x).$

If $k = 1$, we immediately obtain

$f_{Y_{1}}(x) = n(e^{-x/\theta})^{n-1}\frac{1}{\theta}e^{-x/\theta}= \frac{n}{\theta}e^{-x(n/\theta)},$

which is exponential with mean $\theta/n$.  Note that this answer makes sense because as the number of policies $n$ held by the insured increases, the expected waiting time until the first claim [b]decreases[/b].  This would not be the case if one and only one policy at a time were in force at all times until the last policy claim--such a scenario would correspond to the gamma distribution previously mentioned.

To check your understanding, here are some questions:

1.  What is the PDF of the second order statistic $Y_{2}$, and what does it represent?
2.  Which of the $Y_{k}$ belong to the gamma or exponential family of distributions?
3.  Prove that ${\rm E}[Y_{k}] = \theta \sum_{j=1}^{k}\frac{1}{n-k+j}$.

-Actuaries do it with frequency and severity.

i dont have too much time right now so ill only be able to do #1.
$Y_{2}=\frac{n!}{(n-2)!}\left[1-e^{\frac{-t}{\gamma}}\right]\left[e^{\frac{-t}{\gamma}}\right]^{n-2}\frac{1}{\gamma}e^{\frac{-1}{\gamma}}$

This is the distribution for the time until the 2nd claim is filed.

o and 1 question. what do you mean by "a Poisson process (whose interarrival times are exponentially distributed). "?

Correct.

So a Poisson process is a situation where, for example, we have a particular "rare" event for which the number of occurrences in a given interval of time is Poisson distributed, and are independent if the intervals are not overlapping.  For example, we might model the arrival of subway trains as a Poisson process--suppose that we are waiting for a train, and we know that they arrive at a Poisson rate of $\lambda = 4$ per hour.  Then the probability that we see $k$ trains in 1 hour is

$\Pr[N(1) = k] = e^{-\lambda}\frac{\lambda^{k}}{k!}= e^{-4}\frac{4^{k}}{k!}.$

Furthermore, it is not hard to see that the probability that we see $k$ trains in $t$ hours is

$\Pr[N(t) = k] = e^{-\lambda t}\frac{(\lambda t)^{k}}{k!}.$

Besides considering the counting random variable $N(t)$, we can consider waiting time random variables $T_{n}$.  The interarrival waiting time is the time we would wait between events, and it so happens that this is exponentially distributed.  Therefore the Poisson process is also modeled by the sum of IID exponential waiting times.  The total waiting time $T_{n}$ until the $n$-th event is gamma with parameters $\alpha = n$ and $\theta = 1/\lambda$, where $\lambda$ is the Poisson rate parameter.

For more about Poisson processes, see the textbook [u]Probability Models[/u] by Sheldon Ross.  It is one of the topics featured on SOA Exam MLC, and CAS Exam 3.

The Poisson process is the model which underlies most uses of the Poisson distribution. The key to it all is that there's really only one parameter, and that is a [i]rate.[/i] You only told me one thing about the subway trains - that they come at a rate of 4 per hour. So if you want to count how many trains will come in a one minute, or in seven hours, you know exactly how to do that: multiply the rate by the time interval to get the expected value, and the number will be given by a Poisson distribution with that expected value.

By the way, subway trains make a bad example, and they almost certainly do not follow a Poisson process. Suppose you observe 5 trains in a single 10-minute period. What are the chances that you'll see another one in the next 10 minutes? I'd have to say that the chances of that are much lower than usual, because you've pretty much used up the local supply of trains available. But if it were truly a Poisson process, the fact that you've just observed 5 trains in 10 minutes [i]makes absolutely no difference![/i] The distribution over the next time period is independent of the distribution in any other time period.

Sometimes we uses the Poisson distribution or Poisson process model not so much because we know that it is correct, but because we truly only have one piece of information to go on - the rate - and this is the only model that uses that little information.

Now for an observation to disturb your sleep: suppose that these are mystical unreal subway trains and they really do follow a Poisson process after all. Pick a random time to enter the station.

How long will it be until the next train arrives? That has an expected value of 15 minutes.

How long has it been since the last train arrived? That also has an expected value of 15 minutes.

Now, how long is the train-free interval we have entered? That is, what is the length of time between the last arrival and the next one? That has an expected value of ... (drumroll, please) ... 15 minutes.

Oh, and we've been waiting for 30 minutes already and haven't seen a train yet. How much longer are we going to have to wait? The expected remaining time is ... 15 minutes.


group theory
abstract algebra
superior algebra
superior algebra unsolved

1) Let Inn(G) = the set (fx : x in G), where fx(y) = xyx^(-1).
Show that Inn(G) is normal in Aut(G).

2) Suppose that N is normal in G and P is a p-Sylow subgroup of G.
Show that the intersection of N and P is a p-Sylow subgroup of N.

3) Prove that, if G is a group, then G/Z(G) is cyclic iff Z(G) = G (iff G is abelian).

1) Take $g \in Aut(G)$, and $f_{x}$ in $Inn(G)$. Then $gfg^{-1}(y)=g(xg^{-1}(y)x^{-1})=g(x)yg(x)^{-1}$, meaning $gfg^{-1} \in Inn(G)$.

2) Take $x \in N \cap P$, and consider the subgroup $<x>$ of $P$. Such a subgroup (as a $p$-group of $P$) is contained in a $p$-Sylow $S_{P}$ of $P$, $S_{P}$ being contained in a $p$-Sylow $S_{G}$ of $G$, which is conjugated to $P$. Thus we can write $S_{P}=aPa^{-1}$ for some $a \in G$, and we have $S_{G} \cap N = S_{P}$ (because $S_{P}$ is maximal). But $S_{G} \cap N = a(P \cap N)a^{-1}$ (since $N$ is normal) which has same cardinal than $P \cap N$. Whence $P \cap N = S_{P}$, and we are done.

3) $G/Z(G)$ is cyclic means there is a generator $aZ(G)$ of this group. Now let $x,y \in G$, we can write $x=a^{k}z_{x}$ and $y=a^{m}z_{y}$ and $xy=a^{k}a^{m}z_{x}z_{y}=a^{m}a^{k}z_{y}z_{x}=yx$, whence $G$ is abelian, and $Z(G)=G$. The other way is obvious.


trigonometry
vector
calculus
calculus computations

Hey everyone:

I have one more problem that I'm having trouble with.

Here it is:

A projectile is fired with an initial speed of 160 ft/s at a target 600 ft away, and there is a hill 300 ft high midway between the starting point and the target. At what initial angle of elevation should the projectile be fired?

[img]http://i113.photobucket.com/albums/n225/ccsballplaya30/projectileproblem.jpg[/img]

Again, any help is greatly appreciated
Thanks in advance

:lol:

Actually, there's a maximum height formula that is derived from the kinematic equations:

$h_{max}= \frac{v_{0}^{2}\sin^{2}\theta}{2g}$

Just use that.

[hide="To derive it"]

Resolve the initial velocity vector into components. The vertical component is $v_{0}\sin \theta$, a quantity which we shall call $v_{0y}$. Since the vertical component of the velocity vector will be $0$ when the object reaches its maximum height, we wish to determine the height of the object at that point.

Since $v_{y}^{2}= v_{0y}^{2}-2gy$ for a projectile fired from the ground (a kinematic equation (if acceleration is constant) which you can derive from some calculus and then algebraic manipulation of $a(t) = c$ where $c$ is constant), we see that:

$0 = (v_{0}\sin \theta)^{2}-2gy$

Rearranging, $y = h_{max}= \frac{v_{0}^{2}\sin^{2}\theta}{2g}$.
[/hide]


geometry
rectangle
algebra
polynomial
geometric transformation
rotation

I have been considering ways to count the number of ways to place polyominoes on an n by n chess board (and more generally a m by n chess board). For small cases such as squares and rectangles it is fairly easy, but at the moment I am trying to consider the more general problem of

What is the total number of ways to place polyominoes of any size on a chess board of n by n?

At the moment I seem to be fairly stuck, using a mixture of computation to bash out some small values and further more guessing and proving a formula.

Polyominoes of any size? Some polyominoes don't even tile rectangular boards, even if the square count doesn't contradict.

EDIT: polynomials --> polyominoes  :oops:

Well then they wouldn't add to the total.

Perhaps a sum would be better, I'm trying to work out a general formula for T

$ T \equal{} \sum_{k\equal{}1}^\infty \sum_{l \equal{} \forall \text{polyominoes size k}} f(l)$

Where $ f(x)$ is the number of ways to place a polyomino x on a n by n chess board

The problem is equivalent to finding the total number of ways of partitioning an $ n\times n$ grid into polyominoes, where rotation of the grid are distinguishable.


[quote="Litlle 1000t"]$ a,b,c,d$ are positive reals.Prove that $ {a^{4}+b^{4}+c^{4}+d^{4}+2abcd}\ge{{a^{2}}{b^{2}}+{a^{2}}{c^{2}}+{a^{2}}{d^{2}}+{b^{2}}{c^{2}}+{d^{2}}{b^{2}}+{c^{2}}{d^{2}}}$[/quote]
It's Turkevic's ( not Turkevici's :wink: ) inequality.
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6397


Hi,
For the NC Math Contest State Level, which university is better to go to, or which one are you guys going to: UNCG or Wesleyan College(the one in Rocky Mount)?
Thanks!

My naive guess is that you will have less competition at Wesleyan, which is presumably good.

It doesn't matter; the results are complied from all three sites:

well, you can get regional awards too, though top 10 in state are compiled from all 3 sites. But UNCG is almost definitely harder because Triangle/Triad area people all go there

That's your decision. I say it's good, but I bet you disagree.

[quote="xiaoyan"]Actually, I do agree.  If it's harder, than winning something there will bring more glory. lol.

But then, I'm disagreeing with the fact that you think i'll disagree.....
so am i disagreeing or agreeing?
hm....[/quote]

flip a coin

Lol.  I did and it landed on the edge and just stood there. I thought this was rare so I didn't knock it down.  lol.  Just kidding. But it would be interesting if that really happened.
lol.

PS.  Nice google chrome avatar.

Your trophies probably are a lot better. I went to the Algebra 1 contest last year and my trophy could have been bought at a thrift store for 50 cents.

I didn't get anything...we had the Florida Keys Field trip at the time, plus if i make state this year i can't participate again because of the DC field trip!  :rotfl:  :rotfl:  :rotfl:

[quote="b-flat"]Also, the trophies are much better at the non-UNCG regions.  That is the only thing that I remember when I won the state Algebra II contest; the other two regions got much better trophies than I did.

Hence, I would go to the Rocky Mount one.[/quote]

hmm my trophy wasn't bad

except 2nd through 10th had unfortunate trophies (they were all little tiny things that were exactly the same size? poor 2nd place)

I thought calculators were banned everywhere (as well they should be :P)?

What the state contest is heavily reliant on calculators, and my local contest is even more reliant on them.

Well there's a new rule banning calculators at state (see website), so it seems like the regions would be wise to follow suit?

do yall have any explained problems from algebra 2 contests at either regional or state level?

[quote="xiaoyan"]Well, regionals are diff because the problems are actually made by the University.  Thus, if some really difficult professor were to make the qualifier problem set, the least they could do is to allow calc.  Moreover, it isn't unfair to the other regions cuz there isn't a overall rating so there isn't any peer pressure from the other qualifier sites.[/quote]

Au contraire, if a really good professor made the test, they definitely would not be allowing calcs.

[quote="xiaoyan"]OH NOOOOO
I just checked the Weslayan banned calcs this year. Greensboro hasn't been updated since last year, so I don't know. But I REALLY REALLY hope they didn't ban calcs as well. I don't totally depend on my calc, but it helps ALOT.[/quote]

Umm right. :roll:

lol.  The whole problem is that there really is a lack of good professors.
Which, in turn, is good for test takers.

ok so pretty much

bring a calculator, but don't expect to use it. Using your brain is a much better idea.

lol.
That's actually what I was thinkin.
Thanks!

PS.  I looked at the other websites and noticed that UNCG is the only one that starts at 9:30 (or some time close).  Rocky Mount and Asheville start almost 1 -2 hours later.  I wonder why....

Wow, haven't checked this forum in a WHILE.  Wesleyan trophies aren't much better, watermellon.  lol. I guess we have the economy to blame for that...

Btw. Is it me or was the questions harder than those of previous years? I mean I never saw sigma and hyperbolas on the Algebra II's of previous years (not that i didn't know them, just surprised)... And the average scores were lower too..

[quote="Watermelon99"]About UNCG. The trophies are not that great, but the campus is fun. Its also a lot harder cuz LIGON GT MAGNET MIDDLE DOMINATES THE COMPETITION THERE. Hi Xiaoyan I'm in your online class :).[/quote]

yeah, ligon does dominate there. i would know since i went there last year
central region almost always dominates too. thats an advantage of going there, if you win central you almost always win state too, so you dont have to sit around and worry about that

[quote="xiaoyan"]@ligonppl: How is ur school so good? My teach said u guys were a math magnet school, but still... dang...[/quote]

we have really good teachers, and we get like most of the asians from the cary area (at least we used to, reassignment made them go to carnage)

@xiaoyan: what school do you go to?

Jack Britt High School.  :blush:  Yep, we aren't that famous.

i go to enloe now. pretty good school (only NCSSM is better). but ligon is going to get owned in the near future, because i heard that carnage has like 20 6th graders in algebra 1

Sighz... I feel so inferior on AoPs... Like all of you guys go to top notch schools.... I just got mediocrely good at math by doing alot of research...

Yeah, since all the Cary kids are going to Carnage.

And then the School Board is just going to switch the school back to Ligon because the people at Ligon won't be good in the future!  I wish they could just make up their mind.


Due to the fact that I really can not copy directly from m/word in the forum pages, I attached pdf file with the equations clearly understood,

Please help me Iam not that good in maths but Iam trying my best,,,

Thank you very much,,,

1. $b = 8(4)-2(12)$
2. $3(4)^{2}+2(12) = 2b$
3. $(2(12)^{2}-(4)^{3})^{2}= \frac{3b}{4}$
4. $\frac{2}{b}= 3(4)^{2}-12$

Just plug in the answer you got for $b$ back into the equations and see if it checks out. If not, look over your work again.  :ninja:

Thank you so much Imerial effect,,, if you can desribe me the whole process of solving you are the best,,,  :roll: ,,,, and could you please tell me how do you copy the equations in the forum. I copy paste from microsoft word but the tezt appear without some mmathematical signs if you just know,,,

Thank you waiting for your reply about the solutions of these equations

I just want someone to tell me wether or not my results are correct , and try to solve the last two equations contained in the pdf file ,,,

Please help me

make sure when you see $3x^{2}$ you think $3(x^{2})$ not $(3x)^{2}$ you made that mistake at least twice and got a couple problems wrong because of it.

For the last two questions:

$\frac{6x+7}{2}-x-\frac{11}{2}=0$
$6x+7-2x-11=0$ (multiply up by 2)
$4x-4=0$ (collecting like terms)
$x=1$

$\frac{x+3}{x-2}=1+\frac{x+1}{2(x-3)}$
$2(x+3)(x-3)=2(x-2)(x-3)+(x+1)(x-2)$ (multiplying up by $2(x-3)(x-2)$)
$2x^{2}-18=2x^{2}-10x+12+x^{2}-x-2$ (multiplying out)
$x^{2}-11x+28=0$ (collecting like terms on one side)
$(x-7)(x-4)=0$
$x=7$ or $4$.

(By the way, to type equations use $LaTeX$ (like \frac{x+3}{x-2}=1+\frac{x+1}{2(x-3)} and put dollar signs around your code)


calculus
integration
function
vector
algebra
polynomial
real analysis

I made up this example since I'm having trouble with the concept of bounded when applied to linear functional. To be honest I just have never dealt with functional before and I would like to think about a lot of examples until the concept becomes less unfamiliar. (Was this something I should have picked up in a previous course? I'm taking Real Analysis using Royden.)

Here is my first attempt at making up a functional:
Let $ F: L^1 \rightarrow \mathbb{R}$. be defined by $ F(f) \equal{} Sup \{ f(x) | 0 \leq x \leq 1 \}$.

Now is this even linear?
for $ a \geq 0$
$ F(af) \equal{} Sup \{ af(x) | 0 \leq x \leq 1 \} \equal{} a Sup \{ f(x) | 0 \leq x \leq 1 \} \equal{} a F(f)$

for $ a \leq 0$
$ F(af) \equal{} Sup \{ af(x) | 0 \leq x \leq 1 \} \equal{} a Inf \{ f(x) | 0 \leq x \leq 1 \} \neq a F(f)$

OK. So, much for that example. It's not even linear. I'll try to make up another one, though I'm open to suggestions. I know about the standard examples like integration... And I have a few examples to work with in $ l^p$ spaces... but I want to think about more examples for $ L^P$ spaces... though I guess, in the end they will all end up being integration anyway, that's what the Rieze representation theorem says... But it's good to try things and see where they fail to be linear-- or fail to be bounded.  That's what I'm trying to do.

evaluation map $ e_c(f) \equal{} f(c)$ (c a constant) is linear but not bounded.

Strictly speaking, you cannot evaluate an $ L^p$-function at a point ($ L^p$ consists of classes of functions differing on a set of measure $ 0$ rather than of functions themselves). Still, you can view it as an example of a discontinuous linear functional on the space of, say, continuous functions with $ L^p$-norm. I'm not sure if one can construct a discontinuous linear functional defined on the entire $ L^p[0,1]$ without some form of transfinite induction.

[quote="fedja"]Strictly speaking, you cannot evaluate an $ L^p$-function at a point ($ L^p$ consists of classes of functions differing on a set of measure $ 0$ rather than of functions themselves).[/quote]

Then are you saying $ F(f) \equal{} Sup\{ f(x) | 0\leq x\leq 1\}$ won't make sense for $ f \in L^p$? But won't this value be the same for any of the equiv. class of functions in $ L^p$? because a difference between functions on a set of measure 0 won't change the Sup?

[quote="fedja"]Still, you can view it as an example of a discontinuous linear functional on the space of, say, continuous functions with $ L^p$-norm. [/quote]

I don't quite know what you are saying here-- $ F(f) \equal{} Sup\{ f(x) | 0\leq x\leq 1\}$ isn't even linear. ??

1) $ \sup f$ won't make sense but $ \text{ess sup\,}f$ will and people are just too lazy to write $ \text{ess}$ most of the time.

2) By default, you answer to the previous post here, not to the original one. My comment related to Soarer's claim, not to your claim about $ \sup$.

[quote="fedja"]1) $ \sup f$ won't make sense but $ \text{ess sup\,}f$ will and people are just too lazy to write $ \text{ess}$ most of the time.[/quote]

I was not being lazy, I really do need to know the difference between these things-- so thanks for pointing that out!

[quote="fedja"]2) By default, you answer to the previous post here, not to the original one. My comment related to Soarer's claim, not to your claim about $ \sup$.[/quote]

Soarer's example is in the wikipedia... but for the vector space of real-valued polynomials not one of the Lp spaces... maybe he/she was thinking of it on the vector space of real-valued polynomials ... ?

[quote="fedja"]Strictly speaking, you cannot evaluate an $ L^p$-function at a point ($ L^p$ consists of classes of functions differing on a set of measure $ 0$ rather than of functions themselves). [/quote]
oops, what's wrong with me.. sorry.

[quote]I'm not sure if one can construct a discontinuous linear functional defined on the entire $ L^p[0,1]$ without some form of transfinite induction.[/quote]
May I know where one suchconstruction can be found?

[quote="thefuturebird"]
I was not being lazy, I really do need to know the difference between these things-- so thanks for pointing that out!
[/quote]
It is not about your individual laziness at all but about general sloppyness of mathematical writing, which is completely tolerable because (almost) everybody understands what is meant, but which can be somewhat confusing sometimes when you meet it for the first time.

As to constructing a discontinuous functional using Axiom of Choice, take any normed vector space $ X$ containing a dense subspace $ Y\ne X$, ($ X\equal{}L^p, Y\equal{}C$ work just fine) take any linear functional $ \psi$ on $ X/Y$ (here is where AC seems to come into play: without it, how do you know that there exists at least one?) and define $ \varphi(x)\equal{}\psi(x\plus{}Y)$.

[quote="fedja"]
As to constructing a discontinuous functional using Axiom of Choice, take any normed vector space $ X$ containing a dense subspace $ Y\ne X$, ($ X \equal{} L^p, Y \equal{} C$ work just fine) take any linear functional $ \psi$ on $ X/Y$ (here is where AC seems to come into play: without it, how do you know that there exists at least one?) and define $ \varphi(x) \equal{} \psi(x \plus{} Y)$.[/quote]

Thanks!


calculus
integration
number theory proposed
number theory

Let $ a, b, c, m, n$ be positive integers. Consider the trinomial $ f (x) = ax^{2}+bx+c$. Show that there exist $ n$ consecutive natural numbers $ a_{1}, a_{2}, . . . , a_{n}$ such that each of the numbers $ f (a_{1}), f (a_{2}), . . . , f (a_{n})$ has at least $ m$ different prime factors.

Let $ f$ be any integral polynomial. Then by some other known problem there are $ mn$ different primes $ p_{i,j}$, $ 1 \leq i \leq n , 1 \leq j \leq m$, and integers $ n_{i,j}$ such that $ p_{i,j}| f(n_{i,j})$.
Then simply take $ a_{0}$ such that $ a_{0}\equiv n_{i,j}-i \mod p_{i,j}$ by the chinese remainder theorem and take $ a_{k}= a_{0}+k$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=168465
More we can prove that each of them has at least $ p$ prime divisor.


Given a function $ f: R*\rightarrow R$, such that $ f(xy)\equal{}f(x)\plus{}f(y)$, and $ \lim_{x\rightarrow 0}\frac{f(x)}{x}\equal{}l \in R$. Prove that $ l\equal{}0$.

[b]My proof[/b]:
$ f(xy)\equal{}f(x)\plus{}f(y) \rightarrow f(x^2)\equal{}2f(x) \rightarrow \frac{f(x^2)}{x^2}\equal{}\frac{2f(x)}{x^2} \rightarrow \lim_{x \rightarrow 0\plus{}} \frac{f(x^2)}{x^2}\equal{}\lim_{x \rightarrow 0\plus{}}\frac{2f(x)}{x^2} \rightarrow l\equal{}l(\plus{}\infty)$, impossible if $ l\neq0$. So l=0.

[quote="Spribo"]Given a function $ f: R*\rightarrow R$, such that $ f(xy) \equal{} f(x) \plus{} f(y)$, and $ \lim_{x\rightarrow 0}\frac {f(x)}{x} \equal{} l \in R$. Prove that $ l \equal{} 0$.

[b]My proof[/b]:
$ f(xy) \equal{} f(x) \plus{} f(y) \rightarrow f(x^2) \equal{} 2f(x) \rightarrow \frac {f(x^2)}{x^2} \equal{} \frac {2f(x)}{x^2} \rightarrow \lim_{x \rightarrow 0 \plus{} } \frac {f(x^2)}{x^2} \equal{} \lim_{x \rightarrow 0 \plus{} }\frac {2f(x)}{x^2} \rightarrow l \equal{} l( \plus{} \infty)$, impossible if $ l\neq0$. So l=0.[/quote]

Yes, that's ok. May be you could write (a little bit simpler, avoiding $ \infty$) : $ f(x^2) \equal{} 2f(x)$ $ \implies$ $ x\frac{f(x^2)}{x^2}\equal{}2\frac{f(x)}x$ and then, setting $ x\to 0$ : $ 0\equal{}2l$


geometry
power of a point
radical axis
perpendicular bisector
geometry solved

Here is the very classical construction due to Napoleon :
Take any point A lying on r; draw any circle r' with center A; this circle intersects r at B and C.
The circles going through A and centered respectively at B and C intersect again at D.
The circle centered at D going through A intersects r' at E and F.
The circles going through A and centered respectively at E and F intersect again at the center of r.
Kind regards.

[quote="ssupermath"]Let $r$ be a given circle. The center of $r$ is not given. Find the centre of $r$ using only the compass.[/quote]

[quote="jpe"]Here is the very classical construction due to Napoleon :
Take any point A lying on r; draw any circle r' with center A; this circle intersects r at B and C.
The circles going through A and centered respectively at B and C intersect again at D.
The circle centered at D going through A intersects r' at E and F.
The circles going through A and centered respectively at E and F intersect again at the center of r.[/quote]

Just for the sake of completeness, here is a proof of this construction:

Let O be the center of the circle r. Since the points A and B both lie on the circle r, we have OA = OB, so that the triangle AOB is isosceles.

Since the points B and C are the two common points of the circles r and r', they are symmetric to each other with respect to the line joining the centers O and A of these circles, i. e. with respect to the line OA. Hence, $OA\perp BC$.

Now consider the two circles through the point A and centered at B and C. These two circles have the points A and D in common; thus, the points A and D are symmetric to each other with respect to the line joining the centers B and C of these circles, i. e. with respect to the line BC. Hence, $AD\perp BC$.

From $OA\perp BC$ and $AD\perp BC$, it follows that the points O, A and D all lie on one and the same line perpendicular to the line BC. Now, since the points A and D are symmetric to each other with respect to the line BC, we have AB = DB, so that the triangle ABD is isosceles.

Since the two isosceles triangles AOB and ABD have a common base angle (< OAB = < BAD), they are similar. Thus, AB : AO = AD : AB. In other words, $AB^{2}=AO\cdot AD$.

Let M be the midpoint of the segment OA. Then, using directed segments on the line OA, we have AM = MO, so that

$AB^{2}-AD^{2}=AO\cdot AD-AD^{2}=AD\cdot\left(AO-AD\right)=AD\cdot DO$
$=\left(AM+MD\right)\cdot\left(MO-MD\right)=\left(AM+MD\right)\cdot\left(AM-MD\right)=AM^{2}-MD^{2}$.

Thus, $MD^{2}-AD^{2}=AM^{2}-AB^{2}$. But $MD^{2}-AD^{2}$ is the power of the point M with respect to the circle centered at D and passing through A (since the center of this circle is D and the radius of this circle is clearly AD), and $AM^{2}-AB^{2}$ is the power of the point M with respect to the circle r' (since the center of this circle is A and its radius is AB, as it passes through the point B). Hence, the point M has equal powers with respect to the circle centered at D and passing through A and to the circle r'.

Since the circle centered at D and passing through A and the circle r' intersect at the points E and F, their radical axis is the line EF. On the other hand, this radical axis must be perpendicular to the line OA (since the line OA is the line joining the centers D and A of these two circles), and, of course, it must pass through the point M, since this point M has equal power with respect to the two circles. Thus, the line EF is perpendicular to the line OA and passes through the point M, which is the midpoint of the segment OA. In other words, the line EF is the perpendicular bisector of the segment OA. Consequently, EO = EA and FO = FA. Hence, it becomes clear that the circles going through A and centered respectively at E and F intersect again at the point O, i. e. at the center of the circle r.

And this proves Napoleon's construction of the center of r.

A different construction can be obtained from the problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=57006 .

  Darij


algebra
polynomial
superior algebra
superior algebra unsolved

Theorem 33.3: Let E be a field of p^n elements contained in the algebraic closure $ \bar{\mathbb{Z}}_p$ of $ \mathbb{Z}_p$. The elements fo E are precisely the zeros in $ \bar{\mathbb{Z}}_p$ of the polynomial $ x^{p^n} \minus{} x$ in $ \mathbb{Z}_p[x]$.

Theorem 33.12: Let p be a prime and let $ n \in \mathbb{Z}^ \plus{}$. If E and E' are fields of order p^n, then E is isomorphic to E'.

The proof of Theorem 33.12 is six lines long but I think it follows directly from theorem 33.3. Am I wrong?


summer program
Mathcamp
algebra unsolved
algebra

Given is a sequence of numbers $a_1,a_2,...$ with the property:

$a_1=2, a_2=3$ and $a_{n+1}=2a_{n-1}$ or $a_{n+1}=3a_n-2a_{n-1}$ for all $n>=2$.

Prove that no number between $1600$ and $2000$ can be an element of the sequence.

This was actually one of the problems in the Mathcamp team problem solving contest.

One very ugly solution i saw included listing out all paths until the numbers topped 2000  :P

[quote="harazi"]Fine, fine, but of course one must observe (very natural observation, you will probably agree with me) that each term is the sum of 2 powers of 2. Try to prove this.[/quote]
Actually, it is easier to prove inductively a stronger statement (this is not uncommon), namely $a_n=2^{x_n}+2^{y_n}$, with $x_n,y_n\in \mathbb{N}$ and $x_n\le x_{n+1},y_n\le y_{n+1}, x_n+y_n+1=x_{n+1}+y_{n+1}.$

[quote="fourierseries"]This was actually one of the problems in the Mathcamp team problem solving contest.

[/quote]
The problem comes from the Dutch Olympiad, 1994 or 1995.


[quote="stancioiu sorin"]Let $\ ABC$ be a triangle and we take $\ D$ a point so that $\ m(<ADB)=90$, $\ D$ lies on $\ BC$.Also, let's take $\ M$ the midpoint of $\ BC$.If we know that $\ m(<BAD)=m(<CAM)$, find $\ m(<BAC).$[/quote]

Are you sure the problem is complete?, because if $M=D$, $\measuredangle BAC$ can take any values

Sorry, I wanted to say that $\ AD\bot BC$, $\ D$ lies on $\ (BC)$.

[quote="stancioiu sorin"]Also, we cannot have $\ M=D$, because always $\ AM>AD$. ;)[/quote]

Acording to what I get, the triangle would always be isosceles and the $\measuredangle BAC$ can take any value.

Any other condition you forgot?

[quote="stancioiu sorin"]No, the problem is $\ correct$. ;)[/quote]

Ok. Then here's my solution:

Let: $\measuredangle BAM=\measuredangle DAC=\alpha$, $\measuredangle MAD=\theta$ and $P$($P \in AC$) such that $BP \perp AC$

Then obviously: $BM=MC=MP$ and $\measuredangle PBM = \measuredangle BPM = \alpha$  :arrow:  $\measuredangle PMC=2 \alpha$, also we observe BAPM is ciclyc. So $\measuredangle BAP= \measuredangle PMC=2 \alpha$. Then we have: $\theta = 0^\circ$

If $P$ is outside $ABC$ the proceed is similar.

Can't see what's wrong. :huh:

Let the circumcentre of $\triangle{ABC}$ be $O$. Then $\angle{CAO}=\angle{BAD}$. Thus $\angle{CAO}=\angle{CAM}$. Then discuss.


There is a parallelogram ABCD. Its diagonales cut in a point O . There is a line which pass into the point O and cuts the side AB in a point  K and the side CD in a point L . Prove that point K and L are symetric with each other in conjuction with O.

(Please do this proof without congruence of triangle)

Hint: it suffices to show that $ AB$ and $ CD$ are equidistant from $ O$ to show that $ K$ and $ L$ are symmetric about $ O$. And that's trivial to prove.


give a,b,c in real and for n in natural we have:
             $[n*a]+[n*b]=[n*c]$
prove that one of a,b is integer

I'll sketch the proof:
First you prove that $a+b=c$.
Next you will consider only the case when $a,b<1,c=a+b<1$
Next you will prove it for rationals.
Then you will consider it for irrationals(the most difficult part.)
If you will not be able to  solve the problem I'll give a strict proof.

sure you have the nice solution and it's long to write all the solution because with your thinking i try to have the solution and i successful .But the solution too long and make me sleepy when i write it in the forum :D  :D

proof that $a+b=c$

first, write $a=a_1+a', b=b_1+b', c=c_1+c', a_1,b_1,c_1 \in \mathbb{Z}, 0 \leq a',b',c'<1$

we have $a_1+b_1=[a]+[b]=[c]=c_1$

next take $n=10,10^2,10^3,...$

from $[na']+[nb']=[nc']$, we have that the $n$-th decimal digit of $c'$ is the sum of the the $n$-th decimal digit of $a'$ and the $n$-th decimal digit of $b'$ for $n=1,2,3,...$

this implies that $a'+b'=c'$

thus $a+b=(a_1+b_1)+(a'+b')=c_1+c'=c$

proof that $a+b=c$

suppose that $a+b>c$, so $a+b-\delta=c$
let $[n\delta]>1$
$[na]+[nb]=[nc]=[na+nb-n\delta] \leq [na+nb]-[n\delta] < [na+nb]-1$
which is contradiction because $[na+nb] \leq [na]+[nb]+1$

suppose that $a+b<c$, so $a+b+\delta=c$
let $[n\delta]>0$
$[na]+[nb]=[nc]=[na+nb+n\delta] \geq [na+nb]+[n\delta] > [na+nb]$
which is contradiction because $[na+nb] \geq [na]+[nb]$


[b]If $ a,b,c , A,B,C \in \mathbb{R} $ s.t. $ aC - 2 b B + c A = 0 $ and $ ac-b^2 > 0 $ , Prove that $ AC - B^2 <0 $[/b]

I had previously posted this in the Calculus forum coz I found his problem in an Indian Calculus book - probably a mistake of the author - that happens in India!! So, any ideas????

The obvious idea is to consider the two quadratics $ax^2 + 2bx + c = 0$ and $Ax^2 + 2Bx + C= 0$ and find their discriminants... However, couldn't figure anything beyond this!!!


In $\triangle ABC$, $AB = 8, BC = 7, CA = 6$ and point $P$ is chosen on the extension of side $BC$ such that $\triangle PAB$ is similar to the $\triangle PCA$. The length of $PC$ is

$\text{(A)} \ 7 \qquad \text{(B)} \ 8 \qquad \text{(C)} \ 9 \qquad \text{(D)} \ 10 \qquad \text{(E)} \ 11$

[hide]
P is on the opposite of AC as B
$PC = x, PA = y\\ \frac{x}{6} = \frac{y}{8}, \frac{y}{x} = \frac{x+7}{y}$ by similar triangles
$\frac{y}{x} = \frac{4}{3}\\ \frac{4}{3} = \frac{3}{4} + \frac{7}{y}\\ y = 12\\ x = 9 = PC$
[/hide]


[quote="236factorial"]IF the sum of the medians in a triangle is 9, what is the perimeter of the triangle?[/quote]
[hide]
If we assume that this triangle is equilateral...

then each side would be (in terms of median $m$)

$\frac{2m\sqrt{3}}{3}$

That's one median and since in an equilateral triangle, all medians are equal, then we can subsitute $3$ for $m$.

$\frac{6\sqrt{3}}{3}$

$2\sqrt{3}$

Multiply by 3.

$6\sqrt{3}$

I just hope that holds for all triangles.[/hide]

[quote="pianoforte"][quote="Drunken_Math"][quote="236factorial"]IF the sum of the medians in a triangle is 9, what is the perimeter of the triangle?[/quote]
[hide]
If we assume that this triangle is equilateral...

then each side would be (in terms of median $m$)

$\frac{2m\sqrt{3}}{3}$

That's one median and since in an equilateral triangle, all medians are equal, then we can subsitute $3$ for $m$.

$\frac{6\sqrt{3}}{3}$

$2\sqrt{3}$

I just hope that holds for all triangles.[/hide][/quote]

[hide]wouldn't it be $6\sqrt{3}$ then?[/hide][/quote]

Yes it would!  :blush:

I fixed it.

hey drunken_math u just proved for equilateral triangles , how do you know it would be applicable to other triangles too??? :huh:  :huh:  :huh:

Because it doesn't specify what kind of triangle, so if it works for one, it should work for all.

Reword the question to say, if the sum of the squares of the medians in any triangle is 9, what is the sum of the squares of the three sides? (competition errata).


Let d be any divisor of an integer of the form n^2 + 1. Prove that d − 3 is not divisible by 4.

Let in contrary $ d \equiv-1 \pmod{4}$. Then $ d$ has at least one prime factor of the kind $ p \equiv-1 \pmod{4}\Rightarrow p=4k-1$
It follows from $ n^{2}+1\equiv 0 \pmod{p}$ that -1 is quadratic residue and hence $ 1=(-1/p)=(-1)^{\frac{p-1}{2}}=(-1)^{2k-1}=-1$


For some time now, I have been bugged by finding a solution to this

f(x)->b  as x->a, g(y)->c  as y->b, but g(f(x))!->c as x->a

For some reason, this eludes me, even though it should be simple enough. Naturally f or g or both must be discontinuous.

Since you've assumed the limits, the hole is at the point itself.
Take $f(x)=b$, $g(y)=c$ if $y\neq b$ and $g(b)=d$ for some $d\neq c$.


Let $f: \mathbb{S}^{n}\to \mathbb{S}^{n}$ be an involution. Prove that if $f$ has a fixed point, then $f$ has at least $2$ fixed point.

It's equivalent to another problem you posted, namely [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62062]this one[/url].

In that other problem we're concerned with involutions on $\mathbb R^{n}$; we want to prove that such an involution $f$ has a fixed point. Assuming the contrary, it's clear that $f$ goes to infinity at infinity, so we can extend $f$ by continuity to the one-point compactification of $\mathbb R^{n}$ by making it fix the point at infinity. Then, we get a contradiction from the problem in this topic. Conversely, we can convert this problem to that other one through stereographic projection from what we assume to be the unique fixed point.

When you posted that other problem I went along the lines of the arguments presented on sci.math, making $\mathbb Z/2$ act on $\mathbb R^{n}$, but at the time I had no idea what Eilenberg-MacLane spaces were, so I couldn't finish it :).


Let $ f: \mathbf{R}^n\rightarrow \mathbf{R}^k$ be $ C^r$ and let $ S \equal{} \{x \in \mathbf{R}^n : f (x) \equal{} c\}$ be a level set of $ f$. S is called a [b]regular level set[/b] if at every point $ x_0 \in S$ the linear map $ Df (x_0 ) : \mathbf{R}^n \rightarrow \mathbf{R}^k$ has rank $ k$. Show that regular level sets are $ C^r$ submanifolds of $ \mathbf{R}^n$ with dimension $ n\minus{}k$.

This is a standard theorem. In my textbook it's the "Constant-Rank Level Set Theorem", at least if we include the extra word "embedded", make everything $ C^{\infty}$, and put it in a generic smooth manifold. The proof is very short, invoking a variant of the inverse function theorem (the rank theorem) to produce slice charts.


http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1995

On question #2, I got up to the two equations as follows:

$ x^{2\plus{}\sqrt{2}}\equal{} 1995$
$ x^{2\minus{}\sqrt{2}}\equal{} 1995$

I proceeded to multiply the equations together and I got

$ x^{4}\equal{} 1995^{2}$.

Then I took the square root of each side to get $ x^{2}\equal{} 1995$.  But that leaves only one positive root, and the correct answer is $ 1995$.  I got the correct answer by changing the base to $ 1995$ instead of $ x$ earlier in the problem, but how come it doesn't work both ways?

Check your math for when you multiplied the two equations together. ;) The method using the $ x$ does work.

Dang I really don't see where I went wrong.  :o

Ergh. I apparently made that post at 4:00 in the morning (I don't even remember getting on AOPS that late last night). No wonder it was gibberish.

Yeah, the multiplying the equations together part is where you went wrong. Notice that you concluded $ x^{2}\equal{} 1995$ from that. That clearly contradicts your two given equations.

im pretty sure the exponents should be $ \frac{2\pm\sqrt{2}}{2}$

Your main problem is that your equations contradict [b]each other[/b].  ;)

[quote="Kyron"]I [b][i]believe[/i][/b] that there is some rule that when multiplying equations together both must be equal to zero.[/quote]

No, multiplying equations is perfectly valid because equality is transitive.

Ahh i see...but why in the other method did the equation contradiction not occur.

NOOOO

Could you give us the original equations?

He did. See the link in the first post and how he said "question #2"?

[quote="brady00n"]why in the other method did the equation contradiction not occur.[/quote]

You shouldn't expect contradictions to occur.  I think what you're asking is why the contradiction occurred in the first method you tried, and to know that we would have to know what that method was.  So... how exactly did you get there?


Can someone help me understand sine, cosine, and tan in an easy way? I need to know this for summer, as I am going to take a geometry course.

thank you..

since it's your first year, nothing in detail. In Geometry, I believe you don't get too in-depth on trigonometry. Just always remember
$SOH$ $CAH$ and $TOA$ This will guide through you through many trigonometry problems.

Basically it means
$sin=\frac{Opposite}{Hypotenuse}$
$cos=\frac{Adjacent}{Hypotenuse}$
$tan=\frac{Opposite}{Adjacent}$

Now I could go over this, but I think it's too big of a subject to put into in a single post.

Just remember $SOH$ $CAH$ and $TOA$ and you will probably be ahead of everyone in ur geometry class who will probably repeat these phrases to memorize it continuously, when you already know it!

You pronounce it "SO KA TOE-AH" by the way.

Here's a good neumonic:

$S$ome $O$ld $H$ippy $C$aught $A$nother $H$ippy $T$ripping $O$n $A$cid

You'll never forget that  :blush:

i've never really needed to remember sohcahtoa...
also, just remember, if you need to remember whether $sin{60}$ is $\frac{1}{2}$ or $\frac{\sqrt{3}}{2}$ or something else having to do with a 30-60-90, just imagine the triangle in your head and use sohcahtoa

Yes, I don't believe you need much more than that for a Geometry course. Now, if you want to learn more, there's always lots of fun stuff... :D
Oh, and SOHCAHTOA only applies for right triangles, in case you didn't notice.

Well, when you use it enough you don't eally have to think

I remember that cosine is the sine of a compement angle and tan is the point where the ray intersects the tangent of unit circle going though the point $(0,1)$. Therefore I need only to remember what is sine.

soh cah toa is so 2000

Old Hippies Are High On Acid
oh - ah - oa


superior algebra
superior algebra unsolved

Find an extension $ L/K$ such that there is no intermediate field $ M$ with $ L/M$ purely transcendental and $ M/K$ algebraic.

Take $ K$ any algebraic closed field. Then take $ L$ to be any nontrivial extension of $ K(X)$.

The only algebraic extension of $ K$ is $ K$ itself as it is algebraically closed.
Thus if such $ M$ exists, then $ M\equal{}K$. Now $ M(X)$ is surely not algebraically closed, so e.g. if we take $ L$ to be the algebraic closure of $ K(X)$, then we are done.


Hello everyone,

I am having some trouble with the following two kinematic problems and I would appreciate any help or hints.

Thank you!

---

1. Two rugby players are running towards each other. They are 37 m apart. If one is accelerating from rest at $ 0.5 m/s^2$ and the other was already moving at 3.1 m/s and maintains her speed,

a) how long before they crunch together?
b) how fast was the accelerating player going?
[u]
My Work :[/u]

I am unsure of this, but I think I would need to set two equations equal to each other: each equation representative of each of the two rugby players. However, I do not know which equations I would need to set equal to each other.

2. A police car stopped at a set of lights has a speeder pass it at 27.78 m/s. If the police car can accelerate at $ 3.6 m/s^2$,

a) how long does it take to catch the speeder?

[u]My Work :[/u]

I think that the concept is the same; I would need to set one equation for the police car and one for the speeder equal to each other.

1.(a) In time $ t$, the first player travels a distance of $ \frac{1}{2}\,(0.5)\,t^2$ (using $ s\equal{}ut\plus{}\frac{1}{2}\,at^2$) and the other a distance of $ 3.1t$. You have $ \frac{1}{2}\,(0.5)\,t^2\plus{}3.1t\equal{}37$.

(b) Use $ v\equal{}u\plus{}at$.

2. In time $ t$, the police car travels a distance of $ \frac{1}{2}\,(3.6)\,t^2$ (by $ s\equal{}ut\plus{}\frac{1}{2}\,at^2$) while the speeding car travels a distance of $ 27.78t$. Assuming the police car starts moving as soon as the speeder passes it, you want $ \frac{1}{2}\,(3.6)\,t^2\equal{}27.78t$ for when the police car just catches the speeder.

Thank you for your reply and help, Jane_Bennet. I got the answers!


AMC
AMC 12
geometry
conics
ellipse
rotation
rhombus

A plane contains points $A$ and $B$ with $AB=1$.  Let $S$ be the union of all disks of radius 1 in the plane that cover $AB$.  What is the area of $S$?

[hide="not the most elegant..."]
Drawing such an AB, and making circles with center A and B, we find the limits at those sides.  Then make two more circles centered at the intersection points of the first two.  Drawing more circles that fit, you can kinda conjecture the locus of all circles will be one large circle centered at the midpoint of AB with radius $\frac 32$... so the area is $\frac{9 \pi} 4$
hmm... probably wrong.
[/hide]

Sort of a conjecture here...

[hide="Answer"]We will assume that this gives an ellipse. When the segment $AB$ is a radius of the given circle, we see that we must have a major axis of length $3$ here, so $a=\frac{3}{2}$. When the center of the circle is either directly above or below the midpoint of the segment, notice that we form an equilateral triangle between $A$, $B$, and the center of the circle. Therefore, $b=\frac{\sqrt{3}}{2}$. From this, we have that the area of the ellipse is $\boxed{\frac{3\pi \sqrt{3}}{4}}$.[/hide]

Right... ellipse makes more sense, but I think
[hide]
that the other axis would be $\frac{\sqrt 3}{2} + 1$ instead.  Since you add the radius to the height of the equilateral triangle to get the total length.  Then the area is... well, really nasty.
[/hide]

Well, the axis is $\sqrt{3}$, since it's $2b$. Where did you get $\frac{\sqrt{3}}{2}+1$ from? From the direction you're looking at at that point, $AB$ is flat, so it gives no distance. Basically, as I said, the center of the circle is either directly above or below the midpoint of the segment. The circles in this case will have both $A$ and $B$ on the circle, and when you draw the radii, they're also length $1$, and therefore you get equilateral triangles with height $\frac{\sqrt{3}}{2}$. They have the same base, so you just add the heights.

[b]EDIT:[/b] Nevermind, you're right. I just figured out what you were saying. Yeah, the area should be [hide="Answer"]$\pi \left(\frac{3}{2}\right)\left(\frac{\sqrt{3}}{2}+1\right)=\boxed{\frac{3\pi \sqrt{3}+6\pi}{4}}$.[/hide]

Well, none of those are possible answers, here are the choices:

$2\pi +\sqrt{3}$

$\frac{8\pi}{3}$

$3\pi -\frac{\sqrt{3}}{2}$

$\frac{10\pi}{3}-\sqrt{3}$

$4\pi-2\sqrt{3}$

[hide="solution"]say the center of the circle is $O$ for $AB$ to be contained in the circle, it is necessary and sufficient that 0<AO<1, and 0<BO<1

say we have $(0,0)$, and $(1,0)$, $O$ is an arbitrary point $(x,y)$, thus we have

$0<x^2+y^2<1$, and $0<(x-1)^2+(y)^2<1$

notice this is the intersection of 2 circles, this is clearly 2 segments that subtend a 120 degree angle, this is the locus of the centers of the circles

now we have to consider the extension of the radii, giving the whole circle, not just the center,

we have to find the area of this extended out 1 unit as far as possible

take 1 of the circle segments, and have a concentric circle radius of 2, Say the segment is AB, extend OA and OB to touch the larger circle, this gives that 1 unit extension for any point on the arc of the smaller circle, this is because any point on the smaller circle's segment has a minimum distance to the outer circle of 1, but also we are missing the part where we have a circle centered at A or B, there is 60 degrees of circle that are not accounted for, and this is a circle radius of 1, finding the areas, we get

$2*(\frac{2^2\pi}{3}+\frac{\pi}{3}-\frac{\sqrt{3}}{2})=\frac{10\pi}{3}-\sqrt{3}\implies \boxed{D}$

sorry for the horrid explaination...i have to go to bed soon, also a diagram would help[/hide]

[quote="t0rajir0u"][hide="Thoughts"]
Consider the bounds on the center of the circle.  It cannot be more than $1$ away from either point.  Draw a circle of radius $1$ around $A$ and another circle around $B$.  The center of the circle in question must be within the area formed by the overlap - hence, this is not an ellipse.

But I don't see a way to calculate this area  :(
[/hide][/quote]

You are right it is not an ellipse.

It is two intersecting arcs of 120 degrees, so you have to find a way of calculating the intersection.

the area is $\frac{2}{3}* \pi*4-\sqrt\frac{3}{2}+{1}{3}*\pi = 3\pi-\sqrt\frac{3}{2}$. (C). Four $60$ degree arcs ($\frac{2}{3}$ of radius two circle), minus the overlap of the wedges (two equilateral triangles), and + the portions of the radius one circle that filled in the gaps at the top and bottom $2*\frac{1}{6}= \frac{1}{3}$ of circle with radius one.)

i think this is pretty difficult for an AMC problem

[quote="8parks11"]the area is...

i think this is pretty difficult for an AMC problem[/quote]Look closely at that explanation.  Draw it out, do whatever you have to, but remember, it's not that difficult for an--well a problem anyway.

If you know that:

Because this shows up on the AMC, it's not that difficult (as math on those competitions goes).  Once you know that, it's just a matter of open thinking--really open.  That, for my part is one of the most difficult things about geometry, but not unconquerable...  Sketch lightly, do it neatly, but remember that most of what solves a problem is the conceptual stuff done in your head.  I've solved these things by 15 minutes of drawing Everything (every possible line, etc.), but this isn't the way.  A clue like the Tip on the slice created by our two circles, [b]A[/b]U[b]B[/b], represents a discontinuity.  As the new--or sliding--circle drawn along that path (the outline of this slice) makes its way, the Farthest point created by each instance of it always comes where our slice hits the Center in an exactly parallel pass to that farthest point on this circle.  Connect it back to the starting circle ([b]A[/b] or [b]B[/b]) if you need to; otherwise just picture a stack with the circles and three points of interest like a circus display.  It's as if on a high wire, you hit the end; that makes the circles move free~~kinda Grounded to the radius 1 thing.

Funny illusions like that help me at least.  Adding little areas of what's left is the easy thing.  You only have to get yourself to scream at 'oh maybe an ellipse' --rule-breaking things--if without an explanation.  Lines Points Arcs --and lots of overlaps.

Cheers with this year's maths, you guys.

The answer is $\boxed{3\pi-\frac{\sqrt{3}}{2}}$:
[hide="Explanation"]
The left most circle and right most circle that cover $AB$ are the ones with $AB$ for a radius.  These two circles intersect at two points, $C$ and $D$.  The diameter through $B$ in the right one intersects the circle at $E$ and the diameter through $A$ in the right one intersects the circle at $F$.

The highest point in $S$ is the circle that has $AB$ as a chord.  The radii drawn to the endpoints of this chord form an equilateral triangle with side 1, so we know that this ceneter is $C$.  The lowest point, naturally, will have center $D$.

Now, we can rotate point $F$ about $A$ onto a point on circle C, diameterically opposite $A$.  If we let this point be $F'$, then we can draw a sixth of a circle centered at $C$ through $F$ and $F'$.  This sector, of radius 2, will contain points in $S$, since we can rotate Circle B as we did Point F onto Circle C, creating circles that are contained in $S$ while in motion.  We can also rotate Circle C downwards onto Circle D, so we have a $120^\circ$ sector with radius 2 on the right.

We can do the same thing on the left, creating two sectors with area $\frac{4\pi}{3}$.  We also have two sixths of circles C and D that we haven't counted.

These each have area $\frac{\pi}{6}$.

The two $120^\circ$ sectors overlap, so we must subtract that overlap, which comes in the form of Rhombus $ACBD$.  This has twice the area of equilateral $ABC$.  Therefore, we must subtract $\frac{\sqrt{3}}{2}$.

The area of $S$ is $2\left(\frac{4\pi}{3}\right)+2\left(\frac{\pi}{6}\right)-\frac{\sqrt{3}}{2}=\boxed{3\pi-\frac{\sqrt{3}}{2}}$; $\boxed{\mathbb C}$.
[/hide]


calculus
integration
blogs
number theory
prime factorization
relatively prime

Find the least positive integral value of $z^{3}$ for which there exist defferent positive integers $x,y,$ and $z$ that satisfy $x^{2}+y^{2}=z^{3}$.

Is it [hide="this?"]35,120,25[/hide]
I'm too lazy to explain solution.

[hide="the correct answer is"] $125$[/hide]

There has to be a better solution than brute force...

well...like i just looked at units digits until i found pairs that worked

oh wow come on...are they serious...?

ok so (1+2i)^3=-(11+2i)

thats pretty slick...i do have to say...

i have to remember the random stuff gaussian integer stuff that i learned from you...

Quite interesting... from that method we may obtain the identity

$(a^{2}+b^{2})^{3}= (a^{3}-3ab^{2})^{2}+(b^{3}-3ba^{2})^{2}$

from the expansion

$(a+bi)^{3}= (a^{3}-3ab^{2})+i(3a^{2}b-b^{3})$,

or even a more general one that will probably end up looking ugly from the expansion $(a+bi)^{n}$.

[quote="t0rajir0u"][hide="Hint"] In the best case, the prime factors of a number representible as a sum of two squares must be primes congruent to $1 \bmod 4$.  What is the smallest such prime? [/hide][hide="In other words..."] $z^{3}= (x+yi)(x-yi)$ [/hide][/quote]

I don't quite get your hint >.<

[hide="In simple terms"] Every odd prime representible as a sum of two squares is congruent to $1 \bmod 4$.  This is not hard to see, since a sum of two squares can only be congruent to $0, 1, 2 \bmod 4$.

What is hard to see is that the converse is true - every prime congruent to $1 \bmod 4$ is representible as a sum of two squares!

Now, we also have the identity

$(a^{2}+b^{2})(c^{2}+d^{2}) = (ac \pm bd)^{2}+(ad \mp bc)^{2}$

So the product of a sum of squares is also a sum of squares.  In other words, a product of primes congruent to $1 \bmod 4$ is representible as a sum of two squares.

The converse is [i]almost[/i] true - every number representible as a sum of two squares must have the property that the primes $\equiv 3 \bmod 4$ in its prime factorization have even exponents.  For example,

$45 = 3^{2}\cdot 5 = 6^{2}+3^{2}$ [/hide]
[hide="How it relates to the problem"] Now, those primes $\equiv 3 \bmod 4$ only occur in the prime factorization of a sum of two squares when the squares are both divisible by that prime.  If we have the additional constraint that the squares are relatively prime, then we only have prime factors $\equiv 1 \bmod 4$ (and possibly some $2$s).

If $z^{3}$ can only have prime factors $\equiv 1 \bmod 4$, it follows that $z$ can only have prime factors $\equiv 1 \bmod 4$.  $5$ is the smallest such prime, so $z = 5$ is the smallest case.   ($z = 2$ produces $x = y$.)  [/hide]
The proof of these results requires some interesting number theory.  I believe I wrote an early blog post or two about it.

oh, okay, I think I get it now, but it's a little late. I'll digest this tomorrow :P.

One thing to note: this is interesting math; the application of gaussian integers. However, it takes almost no time to just bust out the casework. Like I computed this in under 2 mins, and I am slow as --- when it comes to computation.


Let $S \in R^{nxn}$ be a matrix such that $S^{t}=-S$, that is, $S$ is a skew-symmetric matrix. $(,)$ is a standard inner product

$(1)$ Show that $(Sv, w)=-(v,Sw)$ for all $v, w \in R^{n}$

$(2)$ Show that $\sigma(S) \subset{0}$

$(3)$ Show that $I+S$ and $I-S$ are invertible matrices.

$(4)$ Define $A=(I+S)(I-S)^{-1}$ Show that $A^{t}A=I$

$(5)$ Show that if $A$ is as in part $(4)$ then $-1 \notin \sigma(A)$

For (1), let matrix notation help you. Consider vectors to be column matrices and write $(u,v)=u^{T}v.$ Then,

$(Sv,w)=(Sv)^{T}w=v^{T}S^{T}w=-v^{T}Sw=-v^{T}(Sw)=-(v,Sw).$

For (2), suppose $\lambda\in\mathbb{R},$ $v\ne 0,$ and $Sv=\lambda v.$

Then $\lambda\|v\|^{2}=v^{T}(\lambda v)=v^{T}Sv=-v^{T}S^{T}v=-(Sv)^{T}v=-\lambda\|v\|^{2}.$

We conclude that $\lambda=0.$

If we repeat this calculation using the complex inner product $u^{*}v,$ allowing $\lambda$ to be complex, we get that $\lambda=-\overline{\lambda},$ or $\lambda$ is pure imaginary.

(3) We have shown in (2) that $1$ and $-1$ are not eigenvalues of $S.$

First a lemma: for $v,w\in\mathbb{R}^{n},$

(i) $w^{T}Sv=-v^{T}Sw$

(ii) in particular, $w^{T}Sw=0.$

Proof: note that the transpose of a $1\times 1$ matrix is itself. Hence $w^{T}Sv=(w^{T}Sv)^{T}=v^{T}S^{T}w=-v^{T}Sw.$

For part (ii), just let $w=v$ in this expression.

Now, for part (4):

Let $v$ be arbitrary and let $w=Av.$ Then

$(I-S)w=(I+S)v$

Left-multiply that twice; first by $w^{T},$ then by $v^{T}.$

$w^{T}w-w^{T}Sw=w^{T}v+w^{T}Sv$ or $\|w\|^{2}=w^{T}v+w^{T}Sv$

$v^{T}w-v^{T}Sw=v^{T}v+v^{T}Sv$ or $v^{T}w-v^{T}Sw=\|v\|^{2}$

But by the lemma, $w^{T}v+w^{T}Sv=v^{T}w-v^{T}Sw.$ Hence we have proved that $\|w\|^{2}=\|v\|^{2}.$

$A$ has the property that $\|Av\|^{2}=\|v\|^{2}$ for all $v.$ That makes $A$ an orthogonal matrix: $A^{T}A=I.$

(5) Suppose $Av=-v.$ Then,

$-(I-S)v=(I+S)v$

$-v+Sv=v+Sv$

$-v=v,$ or $v=0.$

Hence $-1$ is not an eigenvalue of $A.$

---

Let $n=2.$ Then any such $S$ can be written as

$S=\begin{pmatrix}0&-t\\t&0\end{pmatrix}$ for some $t\in\mathbb{R}.$

Given that, we have

$A=\frac{1}{1+t^{2}}\begin{pmatrix}1-t^{2}&-2t \\2t&1-t^{2}\end{pmatrix}.$

The eigenvalues of $A$ are $\frac{(1-t^{2})\pm 2ti}{1+t^{2}}=e^{2i\theta}$ where $\theta=\pm\arctan(t).$ The possible values of the spectrum are every point on the unit circle except for $-1.$

I just realized that my alleged proof of (4) above actually shows that $B=(I-S)^{-1}(I+S)$ is orthogonal, rather than $A=(I+S)(I-S)^{-1}.$

So here's the correct proof:

Suppose $w=Av=(I+S)(I-S)^{-1}v.$

Let $u=(I-S)^{-1}v.$ Then:

$v=(I-S)u$ and

$w=(I+S)u.$

$\|v\|^{2}=v^{T}v=u^{T}(I-S)^{T}(I-S)u=u^{T}(I+S)(I-S)u=u^{T}(I-S^{2})u$

$\|w\|^{2}=w^{T}w=u^{T}(I+S)^{T}(I+S)u=u^{T}(I-S)(I+S)u=u^{T}(I-S^{2})u$

So $\|v\|^{2}=\|w\|^{2}$ or $\|Av\|^{2}=\|v\|^{2}$ for all $v.$ As before, this is sufficient to show that $A$ is orthogonal.

:blush: I have just taken part in a linear algebra course  so many concepts are very strange with me :( .
If you can,please explain to me  some concepts which was mentioned above:[b]standard inner product[/b] and what is $\sigma (S)$.
thanks and I apologize if I disturb you :(

The standard inner product is what you would have called the dot product in your vector calculus or physics class. The standard inner product of vectors $v$ and $w$ is $(v,w)=\sum_{j=1}^{n}v_{j}w_{j}.$

You'll note that I used the alternative notation $v^{T}w,$ considering the vectors $v$ and $w$ to be column vectors. I find that to be a very useful notation - it lets matrix mutiplication, applied in natural ways, carry a substantial portion of the burden of a proof like this.

$\sigma(S)$ is also called the [i]spectrum[/i] of $S.$ In a finite-dimensional case, which is what we have here, this is simply the set of all of the eigenvalues of $S.$

(But note that when Sandie18 asked for "$\sigma(S)\subset \{0\},$ she was asking for something that only makes sense if you're only talking about that portion of the spectrum which lies in the real numbers. More generally, the spectrum of a skew symmetric real matrix - or a skew-hermitian complex matrix - lies within the imaginary axis in the complex plane.)


If $ a \bigtriangleup b \equal{} (a\plus{}b) \plus{} ab\plus{} b$, what is $ (5\bigtriangleup 7)\bigtriangleup 3$?

First, figure out $ 5\bigtriangleup 7$. This is $ (5\plus{}7)\plus{}5\cdot7\plus{}7\equal{}54$. Then plug $ 54$ back in with $ 3$ to find $ 54\bigtriangleup 3$: $ (54\plus{}3)\plus{}54\cdot3\plus{}3\equal{}\fbox{222}$.


probability
inequalities
triangle inequality

Here are to questions.....

$Question 1$ I have a straight stick of finite length (perhaps a chop stick !!) and I break it into $3$ parts at random.
$a$. Whats the probability that these $3$ parts can be used to form a triangle?
$b$. What about equilateral triangle?

$Question 2$. $a$ and $b$ are rational numbers...Solve $a^b = b^a$

I am sure $Question 2$ has been posted here but I tried searching, I could not find it... so, if possible, you can send the link where it has been posted earlier.

I have a doubt on question 1b: the straight stick is like the line of all real numbers, cause it's a continuum, so couldn't I cut it in an $\infty$ number of way? If I can, I have just 1 possible case to have an equilateral triangle, so if the probability is defined as good cases/possible cases, shoulodn't the probability to have an equilateral triangle $\rightarrow 0$?

Nice link about $x^y=y^x$: http://www.qc.edu.hk/math/Advanced%20Level/y%5Ex_x%5Ey.htm

[quote="srulikbd"][hide]1.b. I think there is only one case satisfying it. 1.a. I just saw the solution yesterday in my riddles book-hint- u need to know that if u have an equilateral triangle, if u draw altitude to the 3 sides, there sum is constant and equal to the height of the triangle( how do I prove it?).[/hide][/quote]

you can just use the triangle inequality, can't you?

what is it?
http://en.wikipedia.org/wiki/Triangle_inequality this?
I didn't understand it...

[quote="236factorial"][quote="srulikbd"]what is it?
http://en.wikipedia.org/wiki/Triangle_inequality this?
[/quote]

yes

Simply stated, the sum of the lengths of two sides of a triangle must be greater than the length of the third side of the triangle.[/quote]

Thats true, but still hopw do you calculate the probability?

#2...take a=b, that is an infinite number of solutions there, also $(2,4)$, works, $(-4,-2)$

[quote="236factorial"][quote="srulikbd"]what is it?
http://en.wikipedia.org/wiki/Triangle_inequality this?
[/quote]

yes

Simply stated, the sum of the lengths of two sides of a triangle must be greater than the length of the third side of the triangle.[/quote]
the theorem stating that for any triangle, the measure of a given side must be less than the sum of the other two sides but greater than the difference between the two sides.
I knew the first part (for any triangle, the measure of a given side must be less than the sum of the other two sides ) but the second I didn't know  :)  (but greater than the difference between the two sides.
) is it result from the first part, like if we have that $a+b>c,a+c>b,c+b>a$ we can conclude that $a>c-b$ etc?

[quote="JBL"]Try searching the forum -- both of these have appeared numerous times before.[/quote]

That's the strange thing: I tried 3 times and it didn't appear any result....

An idea for Question 1:
[hide]
Here is our chopstick:
|-------------|
I want to split it in three parts: 2 cuts.
The first cut is made and there are two possibilities:
|--------|----| More than half way
|----|--------| Less than half way
(No exactly half way; that has probability 0 if it is truly random.)
If it is less than halfway, flip it over so that it is more than half way.
The second cut is made:
|------*--|----|  If the second cut is after the half way mark (*), the sum of the rightmost pieces is less than the leftmost piece.  Otherwise, a triangle can be made.

Does that mean the answer is one half?[/hide]


Asymptote
limit
trigonometry
calculus
calculus computations

Show that the polar curve $ 4 \plus{} 2\sec(\theta)$ has a vertical asymptote at $ x\equal{}2$ by showing that the $ \lim_{r \rightarrow \pm \infty} x \equal{} 2$.

This seems like a simple enough question, but I'm not sure if I'm doing this right.

$ \lim_{\theta \rightarrow \frac{\pi}{2}} 4 \plus{} 2\sec(\theta) \equal{} 4 \plus{} \lim_{\theta \rightarrow \frac{\pi}{2}} 2\sec(\theta) \equal{} \pm \infty$

I can't, however, show that the above statement is equivalent to what's being asked.

Assuming it's $ r$ that's equal to $ 4 \plus{} 2\sec \theta$: You need to isolate $ x$ in terms of $ r$ and then send $ |r|$ to $ \infty$. [hide]Since $ x \equal{} r\cos \theta$, we get $ \sec \theta \equal{} \frac {r}{x}$, then
\[ r \equal{} 4 \plus{} \frac {2r}{x}\Rightarrow x \equal{} \frac {2r}{r \minus{} 4} \equal{} \frac {2}{1 \minus{} \frac {4}{r}}
\]
which indeed tends to 2 as $ |r|$ goes to $ \infty$.[/hide]
Your progress shows that $ |r|\rightarrow \infty$ as $ \theta \rightarrow \frac{\pi}{2}$, but you'll have to show how that relates to $ x$.

How about the case where $ r \equal{} \tan(\theta)\sin(\theta)$? I simply got $ x \equal{} \sin^{2}(\theta)$, which makes it easy to show that $ 0 \leq x < 1$ but not that $ x \equal{} 1$ is an asymptote. The reasoning I have is that for $ \sin^{2}(\theta) \equal{} 1$ we need $ \theta \equal{} \frac {\pi}{2}$ but plugging into our original equation gives us an indeterminate form (division by 0), a contradiction. Is that our proof, or is there a neater way?


Let be $ a,b,c\in \mathbb{R}_\plus{}$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$ . Prove that :

$ 6abc\le a\plus{}b\plus{}c\minus{}a^3\minus{}b^3\minus{}c^3$

We have $ a\plus{}b\plus{}c\minus{}a^3\minus{}b^3\minus{}c^3\equal{}a(1\minus{}a^2)\plus{}b(1\minus{}b^2)\plus{}c(1\minus{}c^2)$
$ \equal{}a(b^2\plus{}c^2)\plus{}b(a^2\plus{}c^2)\plus{}c(a^2\plus{}b^2)$
, which $ \ge 6abc$ by AM-GM.

[quote="alex2008"]Let be $ a,b,c\in \mathbb{R}_ \plus{}$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ . Prove that :

$ 6abc\le a \plus{} b \plus{} c \minus{} a^3 \minus{} b^3 \minus{} c^3$[/quote]
your inequality is equivalent as $ \sum a(b\minus{}c)^2 \geq 0$ wich is true.


Hi everybody,

Im at the about 95 level on the AMC 12 and I really want to qualify for AIME. Would it be useful to do the problem series class? Also, will AoPS V2 help me at all if i do the class or will just the class be fine?

Thanks very much

For passing the AMC 12, you might want to spend some time with AoPS Volume 1.  The AMC 12 Problem Series will also be helpful.  Volume 2 will be more appropriate for the hardest AMC 12 problems, and for the AIME.