Didn't know where to put this, but this seems as good a place as any...
The question goes,
An extensible string of length a is hung horizontally at points A and B a distance a apart.
Given that the string has mass that amounts to a Non-Negligible weight, find the maximum vertical displacement of the string from the horizontal line connecting A and B.

So I guess the weight of the string makes it hang like a parabola, the tension in the string must vary from point to point, oughtn't it?
And needn't it be assumed that the string is uniform?

This belongs in the physics forum.  Presumably, you want a string of fixed length longer than distance $AB$, and of uniform density (in the sense of mass per unit length).  Then you have a classical result of physics -- and it is not, in fact, a parabola.  (I actually don't remember what the name of the shape is, but I suspect someone in the physics forum can tell you.)

The shape is called a catenary. Its equation involves a hyperbolic cosine.

Am I to move this then? I suppose the parabola guess was an honest mistake, try this link
http://whistleralley.com/hanging/hanging.htm

great explanation, the differential equation is pretty hefty.
Thanks!


quadratics
FTW
algebra
quadratic formula

Okay then... AoPS just logged me out, and deleted my post... -Tries again-

 I need help with the Quadratic Formula.

 Find the value of $ b^2\minus{}4ac$ for each equation

 3. $ x^2\plus{}10x\plus{}13\equal{}0$

 Help?

 And I have to go, so don't expect a fast reply.

[hide]Quadratic equations can be written in the form $ ax^2\plus{}bx\plus{}c\equal{}0$.

Your equation is already in this form, so simply plug in $ a$, $ b$, and $ c$ to find $ b^2\minus{}4ac$.[/hide]

[quote="JRav"]$ x \equal{} \frac { \minus{} 10\pm\sqrt {100 \minus{} 4(13)}}{2} \equal{} \frac { \minus{} 10\pm\sqrt {48}}{2} \equal{} \minus{} 5\pm 2\sqrt {3}$[/quote]
The question isn't asking for $ x$. :roll:

[quote="JRav"]I thought I'd just do the whole thing. :lol:[/quote]0 points, answered something that was not asked  :D

Was only problem #3 posted, or is something wrong with my computer?

It's really sad that after about 10 posts, no one has the RIGHT answer.
[color=white]Just kidding no one is trying. WHITE TEXT FTW
[/color]
$ 10^2\minus{}4*1*13$
$ \equal{} 100\minus{}52$
$ \equal{} 48$

[quote="moogra"]It's really sad that after about 10 posts, no one has the RIGHT answer.[/quote]
In this case, the point is to explain [i]how[/i] to solve it, not necessarily to give the answer. :wink:

lol I had white text for a reason. Oh wells. I always thought people quoted the whole post.

Anyways solution:
A quadratic equation is USUALLY written as $ ax^2\plus{}bx\plus{}c$. We want to find $ b^2\minus{}4ac$. So we take the $ b$ value, the $ a$ value and the $ c$ value. Then we square the $ b$ value and subtract 4*the a value* the c value.

I sure hope somebody is on at this time... I have fourteen problems that I need to get done and get help with fast...

12. $ r^2\plus{}10r\plus{}9\equal{}0$
13. $ \minus{}a^2\plus{}5a\minus{}6\equal{}0$
14. $ \minus{}b^2\plus{}6b\plus{}7\equal{}0$
15. $ 9d^2\minus{}4d\plus{}3\equal{}0$
16. $ 3v^2\plus{}5v\minus{}8\equal{}0$
17. $ x^2\minus{}8x\equal{}0$
18. $ \minus{}p^2\plus{}6p\minus{}5\equal{}0$
19. $ w^2\plus{}w\minus{}7\equal{}0$
20. $ z^2\plus{}9z\minus{}2\equal{}0$
21. $ 4g^2\plus{}8g\minus{}3\equal{}0$
22. $ \minus{}2t^2\plus{}4t\equal{}3$
23. $ 2c^2\plus{}14c\equal{}10$
24. $ 5z^2\plus{}12\equal{}\minus{}6z$
25. $ 8k^2\minus{}5k\equal{}7$

 Sorry if this is way too much, but I /have/ to get this done FAST, and doing it with Dad's help is near-impossible, and you guys are REALLY helpful... ^^;

We aren't here to do your homework. Use the solutions that were posted to figure out how to plug numbers into a formula already given to you.

EDIT: I like your dinosaurs.

EDIT 2: (Since I like your dinosaurs):

The quadratic formula is: $ \frac{\minus{}b^2\minus{}\sqrt{b^2\minus{}4ac}}{2a}$.

Any quadratic equation is in the form $ ax^2\plus{}bx\plus{}c$

Now, for example, in number $ 12$, the equation is:

$ r^2\plus{}10r\plus{}9\equal{}0$ (you can factor this)

Now, don't freak out just because the variables in the quadratic equation aren't in this one. Notice that if you have:

$ a\equal{}1$
$ x\equal{}r$
$ b\equal{}10$
$ c\equal{}9$

$ ax^2\plus{}bx\plus{}c\equal{}r^2\plus{}10r\plus{}9$

By changing the "letter" of $ x$, we don't nullify the quadratic formula, we just use different "letters." Also, $ a, b, c$ are just coefficients and they are meant to stand for integers (so in this case $ b$ stands for $ 10$). If there is no "$ b$" that means $ b\equal{}1$ because $ 1\cdot\text{any number}\equal{}\text{that number}$

[quote="happyme"]By changing the "letter" of $ x$, we don't nullify the quadratic formula, we just use different "letters." Also, $ a,b,c$ are just coefficients and they are meant to stand for integers (so in this case $ b$ stands for $ 10$). If there is no "$ b$" that means $ b\equal{}1$ because $ 1\cdot\text{any number}\equal{}\text{that number}$.[/quote]

To clarify this:  If there is no coefficient in front of the $ b$ term (like $ 5x^2\plus{}x\minus{}8$), then $ b\equal{}1$.  If there is no $ b$ term at all (like $ 5x^2\plus{}8$), then $ b\equal{}0$.

(Hehehe, thank you! Although, they are dragons, not dinasaurs... ^^)

 Okay, let me try some like that...

I thought
[hide="that the quadratic formula was"]
Given the equation $ ax^2\plus{}bx\plus{}c\equal{}0$, $ x\equal{}\frac{\minus{}b\pm \sqrt{b^2\minus{}4ac}}{2a}$
[/hide]

It is, but you are utilizing an important part of the Quadratic Formula, the discriminant, in the original problem.

[quote="happyme"
The quadratic formula is: $ \frac { \minus{} b^2 \minus{} \sqrt {b^2 \minus{} 4ac}}{2a}$.
[/quote]

Errrrr......  :roll: $ \frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$

The general layout of a quadratic equation is ax^2+bx+c.

So, for x^2+10x+13,

b^2-4ac will be 10^2-(4*1*13) = 100-52 = 48

Hey, the original problem has been answered like 3 times, and only about once with a complete solution, so can we please move on?

@Verridith: Are these actually all the ones that you need help with? Because you're doing the same thing in each one: using the given coefficients to find the discriminant. IF you put any effort into this at all, I think that you would get all of these done pretty quickly.

Maybe post the specific ones you don't get and [i]why[/i] you don't get them. Post your solution as well, so we can see that you're actually trying. This is not for homework help. Many people have already given you ways to do this problem. If you still can't get it, then you're not trying.

P.S. Where's 16?


linear algebra
matrix
inequalities
logarithms
linear algebra unsolved

Let $T\in\mathcal M_{n}(\mathbb R)$ be a matrix that : \[\parallel I-T\parallel < 1.\] a) Prove that $T$ is invertible.
b) Prove that there is a matrix $S\in\mathcal M_{n}(\mathbb R)$ that $e^{S}=T$.
(For a real $n\times n$ matrix $X$, we define $\parallel X\parallel =\text{trace}(X^{T}X)$)

This should have been posted in the analysis forum, I think. Is there any algebraic solution?

Yes, you are right. But I wonder if anybody has an algebraic solution.

I have seen this problem before( Of course I couldn't solve it). But there is a mistake in question you have submitted. The norm is in fact the [b][i]operator norm[/i][/b].Not the [b][i]trace norm[/i].[/b]
operator norm is defined as follows:

$\parallel .\parallel =inf${$C\geq 0: |T(v)|\leq C|v|,v\in R^{n}$}

Of course any two norms on a finite-dimention space are equivalent!!

Of course, the inverse is $I+T+T^{2}+T^{3}+\cdots$; the sum converges by the norm inequality and the fact that $\|T^{n}\|\le\|T\|^{n}$.
The trace of $X^{T}X$ isn't really a norm. You need to take a square root, and even then it has the undesirable property that $\|I\|>1$.
If we use an operator norm instead, we can also use the fact that $|Tx|\le \|T\| |x|<|x|$ when $x\neq0$, and thus $(I-T)x\neq0$. To apply this here, note that $\|X\|_{2}\le\sqrt{\text{tr}(X^{T}X)}$, with equality only if $X$ has rank at most 1.

For (b), $\log T=-(1-T)-\frac12(1-T)^{2}+\frac13(1-T)^{3}+\cdots$; the power series converges for the same reasons the earlier one did.


Να και μια ωραία από συνδυαστική.Εστω ένα (2ν+1)-γωνο τέτοιο ώστε κάθε του πλευρά να έχει με δοσμένο κύκλο ένα κοινό σημείο.Να δείξετε ότι μια τουλάχιστον πλευρά του εφάπτεται στον κύκλο

Μακαρι τετοια προβληματα να εμπαιναν σε διαγωνισμους :(  (αν και ομολογουμενως αυτο ειναι σχετικα ευκολο). Τελος παντων να μια λυση...

[hide]Εστω οτι δεν υπαρχει τετοια πλευρα. Τοτε η πρωτη πλευρα θα εχει ενα σημειο μεσα και ενα σημειο εξω απο τον κυκλο, ομοιως και η δευτερη και η τριτη κτλ. Αλλα ευκολα βλεπουμε πως 2η-3η πλευρα συναντονται εξω απο τον κυκλο το ιδιο 4η-5η, 6η-7 ... (2ν)η-(2ν+1)η. Ολες οι αλλες συναντονται μεσα στον κυκλο. Αρα η (2ν+1)η-1η συναντουντε μεσα στον κυκλο. Αυτο ομως ειναι ατοπο γιατι τοτε ειτε η 1η θα ειχε 2 σημεια στο εσωτερικο του κυκλου(δεν θα τον ετεμνε) ειτε δεν θα υπηρχε.[/hide]


Anyone want to help organize a large game (20+ players) sometime? It would be rather interesting. If you want to join just post to here and when I get 20 replies we can start talking about a date, time, number of questions, etc.

You can't have more than 20 players unless you have THECHAMP and he hacks.

yeah the maximum is 20 players. Pacman also set up a thread like this during the summer. Didn't work out. I'll see how this thread turns out.


What is the probability that Lea and Dan select diagonals of the same length from a regular hexagon if they select different diagonals? Express your answer as a common fraction.


Find the sum of the roots of $iz^2-(2+2i)z+2-i=0$

[quote="riddler"][hide="answer"]
$2-2i$?
[/hide][/quote]

I think so  :)


AMC
AIME
quadratics
modular arithmetic
rotation
induction
number theory

Ah, FLT:

[hide]....tells us that $7^{12}\equiv 1\bmod 13$

Next, $7^{99}\equiv 7^3\bmod 13$

But since $7^2\equiv 10\bmod 13$, we get that:

$7^{99}\equiv 70\equiv 5\bmod 13$

Therefore $5$ is the answer.[/hide]

I'm pretty sure that FLT is is Fermat's Last Theorem, but I don't know how it would apply to here.  According to  [url=http://en.wikipedia.org/wiki/FLT]this website[/url], it's [quote]There are no positive integers a, b, and c such that $a^n + b^n = c^n $ in which n is a natural number greater than 2.[/quote]  Is that right?

Yeah, I would be appreciated if I can learn this.

I know this kind of problem appears a lot on AIME :D

He meant Fermat's Little Theorem, not Fermat's Last Theorem.  You can find it in most books on number theory.  Or look in MathWorld.

little thereom wouldn't work unless u use quadratic residues at least i don't think

Fermat's Little Theorem does work.  Blah X 3 showed how to use it.

oh wow, i'm an idiot $a^{p-1} \equiv 1 \mod p$ I see how it is used and I didn't see that he showed it before. Sorry!

How did blah get this:
$7^{99}\equiv 7^3\bmod 13$?

FLT says that $7^{12}\equiv 1\pmod{13}$.  So we can think of $7^{99}$ as:

\[7^{99}=(7^{12})^8\cdot(7^3)\]

And because $7^{12}\equiv 1\pmod{13}$ we can plug 1 into that above equal to get:

\[7^{99}\equiv 1^8\cdot7^3\equiv 7^3\pmod{13}\]

Yeah, but it's not really basic.  We'll work in $\mathbb{Z}_p$ where $p$ is a prime.  Then all of the elements $\{1,2,\ldots, p-1\}$ are just a permutation of the elements $\{a,2a,\ldots a(p-1)\}$.

Multiplying all of them, we get $a^{p-1}\cdot \prod a_i\equiv \prod a_i\bmod p$.  But since $\prod a_i$ is non-zero, it is also invertible, so multiplying by $(\prod a_i)^{-1}$, we get $a^{p-1}\equiv 1\bmod p$

Assuming you understand the notation I'm working in, see if you can 'fill in the gaps'.  (an element $a$ in $\mathbb{Z}_p$ is invertible if there exists $b$ such that $ab\equiv 1\bmod p$.

1) Prove that $\{a,2a,\ldots a(p-1)\}$ is a set of $p-1$ distinct elements.
2) Prove that $\{1,2,\ldots, p-1\}$ and $\{a,2a,\ldots a(p-1)\}$ are actually the same set.
2) Prove that $\prod a_i$ is invertible

Heres an interesting proof.

[hide="Counting proof"]

If we have a circle of p points, how many ways are there to colour the points if we have a colours but don't want to colour them all the same colour?

Well, if we ignore the fact that rotating a circle can give the same circle, there are p choices for each point, making $a^p$, but we subtract off the a possibilities of them all being the same colour, making $a^p - a$. On the other hand, when counting like this we have repeated every option p times, since for any circle we can rotate it by 0, 1, 2, 3, .., p-1. So the number of ways is $\frac{a^p-a}{p}$, and that must be an integer since its the number of ways.

(The reason it doesn't work for p not a prime is because if p had any factors then it would be possible to rotate the circle by one of its factors and end up with the same arrangement).

Example: p=3, n=2: if the colours are B and W we have:
{BBW, BWB, WBB}, {BWW, WBW, WWB}, and we have repeated each circle three times.
[/hide]

I may not have written that very well, but you get the point. I hope.

Here's proof by induction:
[hide]
Base Case:  $a=1$

We have $1^p\equiv 1 \pmod p.$

That  works.

Assume that $a^p\equiv a \pmod p.$

Now we want to prove that $(a+1)^{p}\equiv a+1\pmod p.$

We can finish up by using the fact that $(a+b)^p\equiv a^p + b^p \mod p$ where $b=1$ and our assumption.

You can verify this last step by using the binomial theorem on $(a+b)^p$ and then by showing that $\binom{p}{k}$ is divisible by prime $p$ for all $k$ such that $0<k<p$. [/hide]

[quote="blahblahblah"]Anyone who has Engel can just look in the book anyways  ;)[/quote]

I didn't actually get it from Engel, but if it appears in there, cool! I worked that out from my abstract algebra course :) You can do real cool things by counting stuff.


i need a recommendation for a piano piece to learn. i'm having a really hard time finding good-sounding ones. by that i mean melodies that are pleasing to the ear.

i hope i can get some suggestions. preferably medium to fast in speed.

PS an example of a piece i think is "good-sounding" is fantasie impromptu, which i know. any others?

http://gh.ffshrine.org/sheet-music.php

Try the Final Fantasy piano collections.  That's a pretty nerdy and fanboyish answer, I know, but it's good melodic music.

Etincelles(=sparks) op36
Caprice Espagnol op37 -both by Moritz Moskowski

Song without Words in F by Tchaikovsky

Die Forelle by Schubert transcribed for piano solo by Liszt

Chopin Waltzes -the fast ones

op18, op34No3, op42 op70No1.I think op18 will best meet your needs

Medtner Fairy Tale Op. 20 No. 1

This one is of a medium difficulty. There are a few things that make this one in particular tricky. There is unbounded 2 against 3 but that's ok for you because you've done the Chopin's Fantasy Impromptu. The other major problem is touches, you need to have a very good portato to do this one otherwise it's too blended (legato) or too choppy (staccato). Be read to voice like crazy as well. You'll need a lot of voicing the top of octaves and chords and you'll sometimes need to voice tricky situations to bring the original theme out from a myriad of other notes.

Rachmaninoff Sonata No. 2

No information needed. Good luck.

Liszt Hungarian Rhapsody

Pick one... Any one. It will probably meet your description.

The list can go on forever, but seriously consider the Medtner. It's a beautiful piece.


quadratics
algebra
quadratic formula
superior algebra
superior algebra unsolved

find the splitting field of x^3+x+2 over Q

i hv found the solution in a book but cant understand it properly.if someone give a detail solution then it will be of a grt help for me...
thank you

Note $ x^{3}\plus{}x\plus{}2\equal{}(x\plus{}1)(x^2\minus{}x\plus{}2)$. Find roots $ \alpha,\beta$ of $ x^{2}\minus{}x\plus{}2$ via the quadratic formula, and the splitting field will be by definition $ \mathbb{Q}(\minus{}1,\alpha,\beta)$. You may then want to show that this is equal to a simple extension, which by my guess is $ \mathbb{Q}(\sqrt{\minus{}7})$.

Any questions or corrections, please let me know.


Uri buys two burgers and a soda for $ \$2.10$, and Gen buys a burger and two sodas for $ \$2.40$. How many cents does a soda cost?

We have \[ 2b\plus{}s\equal{}210\] \[ b\plus{}2s\equal{}240\] Adding the two equations, we get \[ 3b\plus{}3s\equal{}450\Rightarrow b\plus{}s\equal{}150\] Subtracting this from the second equation gives us $ s\equal{}\boxed{90}$


The letters A,B,C, and D represent four distinct integers from 0 through 9. When A is added to B the result is C. When B is subtracted from A the result is D. How many possible ways are there to assign values to A,B,C and D?

[hide="quick guess"] 25? I'll post a solution if I'm right. [/hide]

[quote="The QuattoMaster 6000"]Listing all of the possibilties out won't take too long.
[hide] $a$ Cannot be 0, 1, or 2.  So let's start with 3.  When $a=3$, b can be 2 or 1.  When $a$ is 4, b can be 3 or 1.  You can keep on doing this.  Shown below is the list:
a=1, b=no solution
a=2, b=no solution
a=3, b=1,2
a=4, b=1,3
a=5, b=1,2,3,4
a=6, b=1,2
a=7, b=1,2
a=8, b=1
a=9, b=no solution
a=0, b=no solution
Counting all of the possibities, we get $13$ soutions. [/hide][/quote]

Yup thats correct.  :)

Is there a logical way (not listing the possibilities) to do this?

Not really because fo the cases, but it shouldn't take you long to list them out.

If the numbers were larger then I'd agree there might be an easier way.


trigonometry
analytic geometry
algebra
system of equations

[hide="HINT"]Multiply the first equation by 5 and the second equation by 8. Then subtract the two equations and then you will have the y variable cancelled out and then you can easily solve for x. After solving for x plug your x value into the original equation and then solve for y. Hope that helped :D [/hide]

What does "£" mean?

@wiseidiot I think Paulineal means that they are two different problems since Paulineal put find the value of x and y twice. In that case there would be an infinite amount of answers. If it was one problem take then your hint is good.

[quote="shentang"]What does "£" mean? [/quote]

British pound currancy symbol; analgous to our "$ \$$" sign.

These are no ordinary equations; in fact, these are Diophantine. Should be moved to Intermediate? In anycase, because this his homework, full solutions should not be posted, only hints.

[quote="7h3.D3m0n.117"]These are no ordinary equations; in fact, these are Diophantine.[/quote]I would assume that it's a system of equations?
If they were Diophantine, then there would be no solutions...

@ paulineal:

Are you sure these were not 2 halves of the same problem, because, as you can see, they would be impossible to solve, as has been stated many times.

Why aren't there solutions if they were 2 halves of a problem (I haven't tried yet but it seems possible)
Just DIOPHANTINE is impossible because that assumes integer values. I'm sure apples (just as an exaple) can be 1.32 dollars etc.

yes but i think i must be really thick because i didn't understand his answer :rotfl:

[quote="Paulineal"]Fred buys a box of 7 hardback books and 8 softback books for a price of £8.37
Diane buys a box of 12 hardback books and 5 softback books for a price of £10.95
Given that all hardback books are sold for one price (x), and all softback books are sold for one price (Y), form equations and solve them to find the price of each book.[/quote]

Yeah, this would be a system of equations.

$ 7x\plus{}8y\equal{}8.37$
$ 12x\plus{}5y\equal{}10.95$

You are basically trying to find a value of $ x$ and $ y$ that would satisfy [i]both[/i] equations. [url=http://www.artofproblemsolving.com/Wiki/index.php/System_of_equations]Here[/url] are some methods to solve these types of problems.

[quote="fishythefish"]That means that these are two halves of the same problem and the solution starts as what wiseidiot said. If these were two separate problems, then each would have infinite solutions.[/quote]

not necessarily.  If we used cents, thus having whole numbers, then there are not an infinite number of solutions, and in some cases just 1

[quote="Paulineal"]I'm helping my daughter with some homework and it would seem i have totally forgotten anything i ever learned about algebra.
She has to solve some what should be quite easy equasions and i can't remember how to do it.
Any help would be appreciated....thankyou.

7x  +  8y = £8.37.........find the value of x and y

12x  +  5y = £10.95.........find the value of x and y

I know this should be easy but what can i say, lol, i'm getting old[/quote]

[hide]
So what you have to do is make it so one equation has the same amount of x's.

so Multiply the first equation by 12 and the second equation by 7.

12(7x+8y)=12(£8.37)

and 7(12x+5y)=7(£10.95)

so 84x+96y= 100.44 and
    84x+35y= 76.65

so just minus the second equation from the first and solve for y. Then insert y into one of the equations and solve for x.

If you need the full solution, just post.[/hide]

Ahhhhhh, finally i get it.
Thank you so much.
Sometimes it just takes me a while  :lol:

@alanchou:

I agree, if the amount in cents were prime, and we could only use whole number prices. But that was never indicated. For all we know, there could be a sale like 3 for $ \$$13 (unlikely for a textbook), but my point is there could be any number of prices like this.

[quote="shaun ong"]why cant we just use 7x+8y=8.37 and 12x+5y=10.95 and jz put it in the value of cents?[/quote]

You mean Brute Force it until you find a solution? This method works, but remember, we're finding possible solutions for $ x$ and $ y$ that fit [i]both[/i] equations. This will be extremely tedious and might take several hours if not days to generate a single answer.

New method would be using a graphing calculator and finding the intersecting point.

It teaches you that algebra and graphing are closely related?

Which is why teachers generally don't give calcs unless you absolutely need it, like to find the exact value of trig functions. This is easy, even by hand.

[quote="hunter34"]But does that actually teach you something?[/quote]

It teaches you how to use the calc button on the graphing calculator. Calc->intersect is used.

You should definitely learn to make the most out of your calculator, but don't become too dependent on it.

Besides, if the intersection has irrational coordinates, It's not going to help you much.

just saying, substitute x for y.  You have 2 equations!!! solve by substitution!!
(calculators work too...)

[quote="Paulineal"]Hi,
Thanks every one for your reply's, i'll try and elaborate a bit more.
It isn't homework as such, just a worksheet that could be done at home if they wanted to by choice......no marks or anything for getting things right....just practice sheets really.
Maybe i've read the question wrong so i'll write it out in full.

Fred buys a box of 7 hardback books and 8 softback books for a price of £8.37
Diane buys a box of 12 hardback books and 5 softback books for a price of £10.95
Given that all hardback books are sold for one price (x), and all softback books are sold for one price (Y), form equations and solve them to find the price of each book.

My daughter gave up and decided not to do the work sheet but i thought i'd try it and now it's driving me mad....maybe i got the equations wrong and thats how i can't solve it, i'm just really curious now but wouldn't dare admit to the teacher that i couldn't do my 11 yr olds worksheet...lol
Thanks again to every one who posted.
Pauline.[/quote]
Use substitution to solve... (right?)

To solve systems of equations, there are two methods we can use. They are called "substitution" and "elimination". Since this is a homework problem, I will give examples of other problems using first substitution and then elimination.

[b]SUBSTITUTION[/b]

Look at the following problem:

$ 4x \plus{} y \equal{} 22$

$ 3x \plus{} 8y \equal{} 31$

We often use substitution when in one of our equations there is a variable with a coefficient of 1 -- in other words, when the variable is "by itself", such as the $ y$ in our first equation.

We want to find $ y$ in terms of $ x$, so:

$ y \equal{} 22 \minus{} 4x$

Now, we [i]substitute[/i] $ 22 \minus{} 4x$ into our second equation.

$ 3x \plus{} 8(22 \minus{} 4x) \equal{} 31$

$ 3x \plus{} 176 \minus{} 32x \equal{} 31$

$ \minus{}29x \equal{} \minus{}145$

$ x \equal{} 5$

Okay, so now we know what $ x$ is equal to, $ 5$. We can substitute this into the equation we made of $ y$ in terms of $ x$.

$ y \equal{} 22 \minus{} 4x$

$ y \equal{} 22 \minus{} 4(5)$

$ y \equal{} 2$

Aha! Now we finally know the values of both $ x$ and $ y$. $ x$ is 5, and $ y$ is 2.

[b]ELIMINATION[/b]

However, in systems of equations, there isn't always going to be a variable with a coefficient of 1. Look at the following two equations.

$ 2x \plus{} 4y \equal{} 26$

$ 5x \plus{} 2y \equal{} 41$

All the variables have coefficients that aren't 1. So we use a method called elimination. This involves eliminating a variable of our choice. Let's get rid of the $ y$ variable. We will soon see why the $ y$ variable is perfect for elimination.

What we want to do is make sure the $ y$ terms have the same coefficient and that one is negative and the other is positive. Since one is a 4 and the other is a 2, we have some multiplying to do. Let's multiply the second equation by $ \minus{}2$.

Just remember: when we multiply an equation by a number, we have to make sure we multiply each term on each side of the equation, like this:

$ \minus{}2(5x \plus{} 2y) \equal{} 4(\minus{}2)$

$ \minus{}10x \minus{} 4y \equal{} \minus{}82$

Okay, we multiplied our equation. Now comes the fun part. We can add our equations together!

$ 2x \plus{} 4y \equal{} 26$

$ \minus{}10x \minus{} 4y \equal{} \minus{}82$

When we add the $ y$ terms, they drop out, because $ 4\plus{}(\minus{}4) \equal{} 0$. This is true for any number $ n$ and $ \minus{}n$.

$ \minus{}8x \equal{} \minus{}56$

$ x \equal{} 7$

Now we have a step just like in substitution: we substitute our $ x$ value into an equation. It doesn't really matter which equation we do it in, but let's choose one that's nice and easy. I like our first equation.

$ 2x \plus{} 4y \equal{} 26$

$ 2(7) \plus{} 4y \equal{} 26$

$ 14 \plus{} 4y \equal{} 26$

$ 4y \equal{} 12$

$ y \equal{} 3$

Great! Now we have both values of $ x$ and $ y$. $ x$ is 7, and $ y$ is 3.

Those are the two ways to solve systems of equations. You can choose either one to solve the problem on the worksheet. If you have any questions regarding solving systems of equations, please, feel free to ask.

Those are the two most common ways to solve systems of equations.  However, for certain systems, it's better to try other methods.

Ex:
$ 5x\plus{}3y\equal{}17$
$ 8x\plus{}10y\equal{}22$

Before using the process of substitution or elimination, if we add the equations, we get
$ 13x\plus{}13y\equal{}39$
So $ x\plus{}y\equal{}3$, so we can quickly use this to find $ x\equal{}4$ and $ y\equal{}\minus{}1$.  Don't rely on trying to find something like this too much, but when something looks ugly, try to find an easier way, and note what you are looking for.

Yes, this is contributing to the revival of an old topic, just thought I'd like to add something.

And I'd like to add something as well. In my last post on this topic, I gave a little explanation on substitution and elimination. But I realized I neglected to explain about certain equations.

Let's take a look at this equation:

$ \minus{}3x \plus{} y \equal{} 5$

$ \minus{}6x \plus{} 2y \equal{} 8$

We can multiply the top equation by $ \minus{}2$, so that we can use elimination. You could use substitution, but in this case, elimination is a lot quicker.

$ \minus{}2(\minus{}3x \plus{} y) \equal{} \minus{}2(5)$

$ 6x \minus{} 2y \equal{} \minus{}10$

Do you notice anything about this equation? It's almost the spitting image of our second one! What happens when we add our equations?

$ 0 \equal{} \minus{}2$

This can't be right. All our variables dropped out, not just the $ x$ or the $ y$! And since when is $ 0$ equal to $ \minus{}2$?

So what did we do wrong? Nothing! These two equations simply have no solutions. You will never find two numbers that work in both equations.

If we were to graph these two equations(remember the old $ y \equal{} mx \plus{} b$?) we would find that the two lines are parallel! They never intersect and never will. If we graphed a typical systems of equations, we would find that the lines intersected in one point, and one point only. The coordinates of that point would be our solution.

Now look at this:

$ 4x \plus{} 4y \equal{} 8$

$ 12x \plus{} 12y \equal{} 24$

You might notice something rather odd, but let's just start trying to solve. All the variables have coefficients that are not equal to $ 1$, so we must use elimination.

$ \minus{}3(4x \plus{} 4y) \equal{} \minus{}3(8)$

$ \minus{}12x \minus{} 12y \equal{} \minus{}24$

This strangely resembles our second equation. Let's add.

$ 0 \equal{} 0$

 What happened here? Everything just disappeared! You may have noticed the reason why. Our two equations were the one and same! Take the first equation and multiply by $ 3$. You will get the second equation.

What happens if we graph? We find that they are the same line! We call this infinitely many solutions.

[quote="isabella2296"]To solve systems of equations, there are two methods we can use. They are called "substitution" and "elimination". Since this is a homework problem, I will give examples of other problems using first substitution and then elimination.

[b]SUBSTITUTION[/b]

Look at the following problem:

$ 4x \plus{} y \equal{} 22$

$ 3x \plus{} 8y \equal{} 31$

We often use substitution when in one of our equations there is a variable with a coefficient of 1 -- in other words, when the variable is "by itself", such as the $ y$ in our first equation.

We want to find $ y$ in terms of $ x$, so:

$ y \equal{} 22 \minus{} 4x$

Now, we [i]substitute[/i] $ 22 \minus{} 4x$ into our second equation.

$ 3x \plus{} 8(22 \minus{} 4x) \equal{} 31$

$ 3x \plus{} 176 \minus{} 32x \equal{} 31$

$ \minus{} 29x \equal{} \minus{} 145$

$ x \equal{} 5$

Okay, so now we know what $ x$ is equal to, $ 5$. We can substitute this into the equation we made of $ y$ in terms of $ x$.

$ y \equal{} 22 \minus{} 4x$

$ y \equal{} 22 \minus{} 4(5)$

$ y \equal{} 2$

Aha! Now we finally know the values of both $ x$ and $ y$. $ x$ is 5, and $ y$ is 2.

[b]ELIMINATION[/b]

However, in systems of equations, there isn't always going to be a variable with a coefficient of 1. Look at the following two equations.

$ 2x \plus{} 4y \equal{} 26$

$ 5x \plus{} 2y \equal{} 41$

All the variables have coefficients that aren't 1. So we use a method called elimination. This involves eliminating a variable of our choice. Let's get rid of the $ y$ variable. We will soon see why the $ y$ variable is perfect for elimination.

What we want to do is make sure the $ y$ terms have the same coefficient and that one is negative and the other is positive. Since one is a 4 and the other is a 2, we have some multiplying to do. Let's multiply the second equation by $ \minus{} 2$.

Just remember: when we multiply an equation by a number, we have to make sure we multiply each term on each side of the equation, like this:

$ \minus{} 2(5x \plus{} 2y) \equal{} 4( \minus{} 2)$

$ \minus{} 10x \minus{} 4y \equal{} \minus{} 82$

Okay, we multiplied our equation. Now comes the fun part. We can add our equations together!

$ 2x \plus{} 4y \equal{} 26$

$ \minus{} 10x \minus{} 4y \equal{} \minus{} 82$

When we add the $ y$ terms, they drop out, because $ 4 \plus{} ( \minus{} 4) \equal{} 0$. This is true for any number $ n$ and $ \minus{} n$.

$ \minus{} 8x \equal{} \minus{} 56$

$ x \equal{} 7$

Now we have a step just like in substitution: we substitute our $ x$ value into an equation. It doesn't really matter which equation we do it in, but let's choose one that's nice and easy. I like our first equation.

$ 2x \plus{} 4y \equal{} 26$

$ 2(7) \plus{} 4y \equal{} 26$

$ 14 \plus{} 4y \equal{} 26$

$ 4y \equal{} 12$

$ y \equal{} 3$

Great! Now we have both values of $ x$ and $ y$. $ x$ is 7, and $ y$ is 3.

Those are the two ways to solve systems of equations. You can choose either one to solve the problem on the worksheet. If you have any questions regarding solving systems of equations, please, feel free to ask.[/quote]

[quote="isabella2296"]And I'd like to add something as well. In my last post on this topic, I gave a little explanation on substitution and elimination. But I realized I neglected to explain about certain equations.

Let's take a look at this equation:

$ \minus{} 3x \plus{} y \equal{} 5$

$ \minus{} 6x \plus{} 2y \equal{} 8$

We can multiply the top equation by $ \minus{} 2$, so that we can use elimination. You could use substitution, but in this case, elimination is a lot quicker.

$ \minus{} 2( \minus{} 3x \plus{} y) \equal{} \minus{} 2(5)$

$ 6x \minus{} 2y \equal{} \minus{} 10$

Do you notice anything about this equation? It's almost the spitting image of our second one! What happens when we add our equations?

$ 0 \equal{} \minus{} 2$

This can't be right. All our variables dropped out, not just the $ x$ or the $ y$! And since when is $ 0$ equal to $ \minus{} 2$?

So what did we do wrong? Nothing! These two equations simply have no solutions. You will never find two numbers that work in both equations.

If we were to graph these two equations(remember the old $ y \equal{} mx \plus{} b$?) we would find that the two lines are parallel! They never intersect and never will. If we graphed a typical systems of equations, we would find that the lines intersected in one point, and one point only. The coordinates of that point would be our solution.

Now look at this:

$ 4x \plus{} 4y \equal{} 8$

$ 12x \plus{} 12y \equal{} 24$

You might notice something rather odd, but let's just start trying to solve. All the variables have coefficients that are not equal to $ 1$, so we must use elimination.

$ \minus{} 3(4x \plus{} 4y) \equal{} \minus{} 3(8)$

$ \minus{} 12x \minus{} 12y \equal{} \minus{} 24$

This strangely resembles our second equation. Let's add.

$ 0 \equal{} 0$

 What happened here? Everything just disappeared! You may have noticed the reason why. Our two equations were the one and same! Take the first equation and multiply by $ 3$. You will get the second equation.

What happens if we graph? We find that they are the same line! We call this infinitely many solutions.[/quote]

Sorry to revive a month-old topic, but I would just like to commend isabella2296 on taking time to write one of the best guides to systems of equations I've seen!


This is [b]Singapore 2000[/b]

See [url=http://www.mathlinks.ro/viewtopic.php?t=119243]here[/url]

Maybe this one is harder, with the same condition, prove that

$ (a^2\plus{}b^2)^6\ge ab[(\sqrt{a^3c}\plus{}\sqrt{b^3d})(\sqrt{bc}\plus{}\sqrt{ad})]^2$

@alex2008: Do you have any other Singapore Olympiad's problems?? thx

[quote="IW@IT"]Maybe this one is harder, with the same condition, prove that

$ (a^2 \plus{} b^2)^6\ge ab[(\sqrt {a^3c} \plus{} \sqrt {b^3d})(\sqrt {bc} \plus{} \sqrt {ad})]^2$[/quote]
We have: $ (\sqrt{a^3c}\plus{}\sqrt{b^3d})^2 \le (a^2\plus{}b^2)(ac\plus{}bd) \le (a^2\plus{}b^2)\sqrt{(a^2\plus{}b^2)(c^2\plus{}d^2)}\equal{}(a^2\plus{}b^2)^3$
And $ (\sqrt{bc}\plus{}\sqrt{ad})^2 \le (a\plus{}b)(c\plus{}d) \le\sqrt{2(a^2\plus{}b^2)2(c^2\plus{}d^2)} \equal{} 2(a^2\plus{}b^2)^2$;
  $ ab \le \frac{a^2\plus{}b^2}{2}$
Then,we have Q.E.D. :)


What are the last three digits of $ 7^{1051}$?

I know you set it mod 1000, but i'm stuck...

In general, $ a^{\phi(m)}\equiv 1 \pmod m$ if $ \gcd(a,m) \equal{} 1$.
Since $ 1000 \equal{} 2^3\cdot5^3$, $ \phi(1000) \equal{} \frac12\cdot\frac45\cdot1000 \equal{} 400$. Thus, $ 7^{1051}\equiv7^{251}\pmod{1000}$. There's no point in finishing this, but the answer is $ \boxed{743}$. :wink:

I don't quite understand the totient function...

Also, how'd you get from $ 7^{1051}\equiv7^{251}\pmod{1000}$ to $ 743$?

The last step was using a calculator app (Cabamap), but for the first part: by Euler's Totient Theorem, $ 7^{\phi(1000)}\equiv7^{400}\equiv1\pmod{10}$. That's what the Totient Theorem says.

http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_theorem
http://www.artofproblemsolving.com/Wiki/index.php/Totient_function

Can someone show me how to solve the last part, then?

Thanks math154!

$ \lambda (1000) \equal{} 100 \implies 7^{100} \equiv 1 \pmod {1000} \quad {\color{blue} [\star]}$

$ \therefore 7^{1051} \equiv 7^{51} \pmod {1000}$

$ {\color{blue} [\star]}$ : $ \lambda$ is the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url].

Now I'll be flamed for posting a noob question...(sigh)

How'd you simplify $ 7^{51}$?

I don't think you can without a calculator. :maybe:

If you evaluate $ 401^2$, you get $ 160801$, and the remainder when divided by $ 1000$ is $ 801$.


geometry
3D geometry
tetrahedron
inequalities
inequalities proposed

Let $ T$ be a tetrahedron with vertices $ (0,0,0),(a,0,0),(0,b,0),(0,0,c)$ for some $ a,b,c > 0$. Let $ V$ be the volume of $ T$, and let $ l$ be the sum of the lenghts of the six edges of $ T$. Prove that:

$ V\leq\frac{l^{3}}{6\left(3\plus{}3\sqrt{2}\right)^{3}}$


I 've a bilinear form f: V x V --> Reals, written as a sum of a symmetric and skew-symmetric form:

f=g+h, with v,w belongs to V;

g(v,w)= 1/2(f(v,w)+f(w,v)) (symmetric)
h(v,w)= 1/2(f(v,w)-f(w,v)) (skew-symmetric)

I want to prove that f is unique. So, I assume f=g'+h' also suffices. This give me:

0=g+h-(g'+h')=>g+h=g'+h'

Hence, f=g+h is unique. Does this prove make sense, is there a more elegant proof?

thnx

Your question doesn't really make sense... are you trying to show that $ g$ and $ h$ are uniquely determined from $ f$? That statement is true. Right now it looks like you are asking whether or not $ f$ is uniquely determined from $ f$...

I was trying to show that f can be uniquely written as g+h. So the proof is ok?

No, all you did was show that g+h = g'+h', which is just f = f. You need to show g = g' and h = h'.

Oh, I suggest this is the right statement for the proof, isn't it?

Assume g'(v,w)= 1/2(f(v,w)+f(w,v)) and h'(v,w)= 1/2(f(v,w)-f(w,v)). Then

g-g'=0 and h-h'=0

So, b is uniquely written as a sum of g and h.

I believe the problem is to show $ f$ can be uniquely written $ f \equal{} g \plus{} h$, where $ g$ is symmetric and $ h$ is skew-symmetric; in other words, you have to show that formula is the only formula that works.

Yes, that is my goal. So, assume f can also be written as

f=g'+h'

with the properties:

g'(v,w)=g'(w,v) for all v,w that belongs to V
h'(v,w)=-h'(w,v) for all v,w that belongs to V

Could you give me hand to proceed?


trigonometry
geometry
angle bisector
geometry proposed

Prove that any triangle having two equal internal angle bisectors (each measured from a vertex to the opposite side) is isosceles.

Length of an angle bisector for sides a and b:

$ l_a \equal{} \cos\frac {A}{2} \frac {2bc}{b \plus{} c}, l_b \equal{} \cos\frac {B}{2} \frac {2ac}{a \plus{} c}$

Divide: $ \frac {\cos\frac {A}{2}}{\cos\frac {B}{2}} \equal{} \frac {ab \plus{} ac}{ab \plus{} bc}$

$ \cos\frac {A}{2} \equal{} \sqrt {\frac{\cos A \plus{} 1}{2}}$

$ \frac {\cos A \plus{} 1}{\cos B \plus{} 1} \equal{} \frac {a^2b^2 \plus{} a^2c^2 \plus{} 2a^2bc}{b^2a^2 \plus{} b^2c^2 \plus{} 2ab^2c}$

Substitute the formulas for cosines, and expand and verify. I don't like Brutal solutions...

We have a simple geometrical theorem that ...........
If  angleB > angleA then l(angle bisector of angleA) > l(angle bisector of angle B) which can be proved using sine rule ........
Hence the result follows ........

Steiner-Lehmus Theorem !
[url]http://en.wikipedia.org/wiki/Steiner%E2%80%93Lehmus_theorem[/url]

[hide=so here we go!]WLOG angle bisector emanating from vertex $A$ and vertex $B$ are equal for $\triangle{ABC}$\\
so we have $l_{a}=l_{b} \implies l_{a}^2=l_{b}^2$\\
gives , $b-a=ab\cdot(\frac{a}{(b+c)^2}-\frac{b}{(c+a)^2})\implies b-a=-\frac{ab(b-a)(a^2+ab+2ac+b^2+2bc+c^2)}{(a+c)^2(b+c)^2} \implies b=a$ can only be possible other wise product of squares will be negative which is not possible , hence   which proves the triangle is isosceles $\blacksquare$[/hide]

If only modern olympiad problems were like this... Sigh

[quote=djmathman][quote=lifeismathematics]WLOG median emanating from vertex $A$ and vertex $B$ are equal for $\triangle{ABC}$[/quote]
These aren't medians :)[/quote]

Oops! I read it miserably wrong, thanks i fixed it


$ a,b,c>0$ ; $ a\plus{}b\plus{}c\equal{}3$

Prove that : $ a^{2}(a\plus{}b)\plus{}b^{2}(b\plus{}c)\plus{}c^{2}(c\plus{}a) \geq 6$

It's equivalent to
\begin{align*}&\sum a^3+\sum a^2b\ge\frac{2(a+b+c)^3}{9}\\
&\iff 7\sum a^3+9\sum a^2b\ge 6(a+b)(b+c)(c+a)\\
&\iff 7\sum a^3+3\sum a^2b\ge 6\sum ab^2+12abc\end{align*}
which is obvious since $ 6\sum a^3\ge 6\sum ab^2$ and $ \sum a^3+3\sum a^2b\ge 12abc$.


lolz 38.

That was way back when I still had 1400ish.

Lol all I ever get now is 2 or 3 points for beating someone with 1600+... phailure

Lol all I ever get now is 1 point for beating someone 1600ish...

If I remember correctly, I've gained 43 points in a game, awhile back, when I was ~1300 :P

[b][i][size=125]480[/size][/i][/b]

see points lost thread for reason

34.
The message is too small. Please make the message longer before submitting.

Hmm not too long ago i saw Knockout gain 75 in one game.. i lost nine, i guess r15 lost like 64...

23. This number will go Higher as I get better, hopefully. In like 15 min. I got about 75 or so.

I beat mousy twice in normal games and Jongao in a countdown

It boosted me +92 rating in just 15 minutes!

Now if that isn't luck, then I don't know.....

For me, about 17, I know splatyango got about 25 because she was yelling at me at school about how she did so wonderful...  :P

[quote="babybear"]lol i got 31 by beating mousy in a cd :D[/quote]

The title said games, not cd. In a cd, it is generally easier to get rating than in games.

Mine is probably 20 max...

ok @dragon, well i beat both mousy, pinky, Mchow and zomgzord in the same game and got +30 yesterday :P

dDavidd2 just beat mousy like 2 hours ago. He pwned 4-0 lol :)

[quote="batteredbutnotdefeated"]w00t 54 points!

unfortunately i'm still dwindling in the 1900's....[/quote]

1900's!?

Yes, the nice thing about having a low rating is that you go up really easily. When you get to about 1550, gaining rating becomes a pain.

heh it happens later, about 1750.
Really annoying how you get +0, +1, (if you're really lucky) +2 when you win, and how you get -10, -15, (if you lose to a person with ~1400) -20 when you lose.

[quote="RoFlLoLcOpT"]
Really annoying how you get +0, +1, (if you're really lucky) +2 when you win, and how you get -10, -15, (if you lose to a person with ~1400) -20 when you lose.[/quote]

But that's because as an owner of an inflat - er, high rating, you're expected to beat everyone below you.

i gained 159 in one day! epic failz :P
29 when i beat yaofan and jxl in a normal, then 28 more when i beat jxl in a cd... lolz :P

Woot! I got 32 from beating Jongao, mousy, and other guys in the 1500s in a normal game!

 :P sorry mousy and Jongao I know u love ur ratings.

Note: Why don't people leave even if you lose internet and come back and say "Please leave, I lost internent.."..... are people that concerned about gaining free rating?!

People get suspicious sometimes; I've been lied to so many times where certain users claim their Internet has died/phone is ringing/flashies, etc. Also, I generally don't leave if I would have won anyway, like if I was up 120-40 in a normal of 3-0 in a countdown.

[quote="HiDN428"]went from 1063 to 1455 in 2.5 daysish.

on a cd my highest is +29 for beating r15 when i was in the 1000s. lol

on a normal game. idk. like +17[/quote]

 :rotfl:  :rotfl:  :rotfl:

bleh i gained 50 in a game........i was a really low ratin and i think the person i was playin fail memmed 5 times...so...

Beat Eli (AIME15) in countdown 4-2 when the new problems came out. Twice. :D

Good Job Mewto55555!  You are so far winning in most points gained in a game!


I'm sorry if this problem was posted before.
Find all primes $ p$, that satisfies to the following conditions:there are $ m,n\in Z$, such that $ p=m^{2}+n^{2}$ and $ p|m^{3}+n^{3}-4$.

Let $ m=sn, s^{2}=-1(mod \ p)$. From $ 16\equiv (m^{3}+n^{3})^{2}\equiv 2m^{3}n^{3}\mod p$ we get $ m^{3}=2(1-s),n^{3}=2(1+s)$.
Therefore $ mn=2t^{-1},t^{3}\equiv 1\mod p$. Because $ 4\equiv m^{3}+n^{3}=-mn(m+n)$ we get $ m+n=-2t$.
It is easy to check, that p=2 and p=5 solution.
If (p>5) and $ 3\not |p-1$, then $ t=1$ we get $ n^{2}+(p-2-n)^{2}\ge \frac{(p-2)^{2}+1}{2}>p$ contradition.
If $ 12|(p-1)$, then $ t^{-1}\equiv t^{2}$, therefore $ m,n=-t(1\pm s)$. It give $ m=-t(1+s),n=-t(1-s)$.
In these case we have t,s,  suth that
$ \frac{p-\sqrt{2p}}{2}<t,ts<\frac{p-\sqrt p }{2}, \ t^{3}=1\mod p, s^{2}=-1\mod p.$
I don't found p, satisfyed these condition.


Euler
geometry
circumcircle
geometry unsolved

Let a triangle ABC and T a point inside it s.t angles ATB,BTC,CTA are equal.Prove that the Euler lines of triangles of triangles ATB,BTC,CTA are concurent.

Since $ \angle ATB=\angle BTC=\angle CTA$, $ T$ is fermat point. Let $ D$ be a point outside of $ \triangle ABC$ such $ \triangle ADB$ is equilateral and $ M$ be midpoint of $ AB$. Since $ T$ is fermat point, $ D,T,C$ are collinear. Let $ G,O,G^{'}$ be centroid of $ \triangle ABC$, circumcenter of $ \triangle ATB$, centroid of $ \triangle ATB$ respectively. Then $ \frac{GC}{MG}=\frac{G^'T}{MG^'}=\frac{OD}{MO}=2$. Hence $ G,O,G^'$ are collinear. Hence Euler line of triangle $ ATB,BTC,CTA$ pass $ G$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=6188

  darij


in english, i have to write a persuasive essay (about anything) and I can't think of anything to write about.

I would really appreciate it if you could share some ideas with me. I have google searched ideas and haven't found much luck.  :(

Sorry if this is in the wrong forum.

hm... here are just some controversial issues http://www.sac.edu/students/library/nealley/websites/controversial.htm

Yeah, it's the wrong forum, just like this old gem:[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139375&search_id=1632586280[/url] Put these topics in The Round Table.

P.S. I'm sorry but I had to bring it up.

Heh... no problem

Good luck with that essay

Plastic Soup?

It's a huge body of trash floating in the Pacific directly between Japan and Hawaii, and it's twice the land mass of the whole U.S.

[quote="Cyrazeno"]
Good luck with that essay[/quote]

thanks, this is supposed to be a big part of my grade. yikes...

Lawl
1=2, did you get that off of Digg?
It said that may be false information. I believe it, though...

Aanjohri, don't waste your time in the games and fun factory forum! Work on that essay... unless, of course, you [i]want[/i] to get a bad grade.  :P (highly doubt that)  :wink:

What are you guys talking about???
Everyone knows writing about the advantages of canting keels on modern ocean racing sailboats is a surefire way to an A!
 :P

Dude, ya gotta write about the ever consuming, assimilating powers of consumerism, capitalism and the perpetual power of the corrupting media.


In any triangle prove
$ \sqrt{2(b^2\plus{}c^2)\minus{}a^2}  \le \sqrt{2(a^2\plus{}c^2)\minus{}b^2}\plus{}\sqrt{2(a^2\plus{}b^2)\minus{}c^2}$.

[quote="soruz"]In any triangle prove
$ \sqrt {2(b^2 \plus{} c^2) \minus{} a^2} \le \sqrt {2(a^2 \plus{} c^2) \minus{} b^2} \plus{} \sqrt {2(a^2 \plus{} b^2) \minus{} c^2}$.[/quote]
It equivalent to:
$ m_{a} \le m_{b}\plus{}m_{c}$
It is true because $ m_a,m_b,m_c$ are the length sides of a triangle.


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Prove that in a Noetherian ring (commutative, with unit) there are only finitely many minimal prime ideals.

a quick way to prove it is this:

the ideal $(0)$ has a primary ideal decomposition $(0)=\cap_{i=1}^nq_i$.
If $p$ is a minimal prime ideal of $A$ (our noetherian ring), then $(0)\subset p\Rightarrow \cap_{i=1}^nq_i\subseteq p\Rightarrow q_j\subseteq p$ for some $j$. By passing to radicals we have $\sqrt{q_j}\subseteq p$. Since $\sqrt{q_j}$ is a prime ideal and $p$ is a minimal prime ideal, we get $q_j=p$.
So any minimal prime ideal of $A$ is one of the prime ideals $\{\sqrt{q_i}|i=\overline{1,n}\}$. The converse is not true.  Every $\sqrt{q_i}$ is prime, but not necessarely a minimal prime ideal of $A$.

Another beautiful way to prove it (I think it's due to Emmy Noether):
Let's show that the number of minimal primes over any ideal $I$ is finite. Take an ideal $J$ maximal with respect to the property of having an infinite number of minimal primes lying over it. Since $J$ can't be prime, we can find $r,s\not\in J$ with $rs\in J$. So $rs$ is contained in every minimal prime of $J$. Thus any minimal prime contains $r$ or $s$. But by maximality of $J$,  there are only finitely many prime ideals minimal over $J+Rr$ and $J+Rs$. So $J$ has a finite number of minimal primes lying over it. Contradiction.

yet an alternative: every noetherian space is a finite union of irreducible closed subsets (otherwise choose a minimal closed counterexample), and the irreducible components of $ Spec(A)$ correspond to minimal prime ideals of $ A$. finally, $ A$ noetherian implies $ Spec(A)$ noetherian, so we're done.

by the way, does the converse "$ Spec(A)$ noetherian $ \Rightarrow A$ noetherian" also hold? certainly it's not true, but I can't find a counterexample.

[quote="-oo-"]by the way, does the converse "$ Spec(A)$ noetherian $ \Rightarrow A$ noetherian" also hold? certainly it's not true, but I can't find a counterexample.[/quote]
Try $ A\equal{}k[x_1,x_2,x_3,\ldots]/(x_1^2,x_2^2,x_3^2,\ldots)$.


Round 4 scores are now available on the [url=http://www.usamts.org]USAMTS website[/url].

We will be making an announcement about prizes (prize eligibility, the prize selection process, and the prize distribution timetable) sometime in early-to-mid April.  Please note that you must have submitted a Year 20 (2008-09) Entry Form and must have a current US mailing address in your profile in order to receive a prize.  [b]Please do not inquire about prizes until we make this announcement.[/b]

When will Round 2 scores be avaliable for the ones that did not turn one back in?  With the prize announcements?

Thanks

EDIT: nvm.  I read the other thread and it said that they will not be considering this until the first week of April.


When a class of math students lined up for a class picture on Picture Day, they found that when they stood in rows of four there was one person left over. When they formed rows of five students, there were two extra students, and when they lined up in rows of seven, there were three students left over. What is the fewest number of students possible in this class?

Treat this like three different division problems, with the people left over as the remainder. Thus:

Find the smallest number such that it has a remainder of $ 1$ when divided by $ 4$, a remainder of $ 2$ when divided by $ 5$, and a remainder of $ 3$ when divided by $ 7$. The smallest such number is $ \fbox{17}$.

WE have that

$ x \equiv 1 \mod 4$

$ x \equiv 2 \mod 5$

$ x \equiv 3 \mod 7$

Loking at the first two we can change them to:

$ x \equiv \minus{}3 \mod 4$

$ x \equiv \minus{}3 \mod 5$

By the Chinese Remainder Theorem we have $ x \equiv \minus{}3 \equiv 17 \mod 20$

We do the same thing to the 3mod7 and have that $ x \equiv 17 \mod 7$

Applying the Chinese Remainder Theorem again we have $ x \equiv 17 \mod 140$

The smallesst positive $ x$ that satisfies this is $ x\equal{}17$

eerrmm... what is the "Chinese Remainder Theorum"? It seems like you know what you're doing, but a little more explanation would be good. Thanks.


A straight tube with mass M and length L is placed on the edge of a frictionless table with half of its length outside the table as shown in the figure. A small fly with mass m lands on the left end of the tube inside the table and walks to the other end outside the table. The tube does not overturn.

(i) Where is the center of mass of the system when the fly is at the end of the tube outside the table?

(ii) Suppose a second fly with mass m’ lands smoothly on top of the first fly. What is the largest possible value of m’ where the tube remains stable?

To solve this problem one needs to know that the center of gravity doesn't move when an object moves on a rigid object on a frictionless surface.  In other words as the fly moves off the table, the tube moves back a sufficient amount that the center of gravity of the tube and fly together doesn't move.

Once one knows where the center of the tube/fly system is and where the fly is located, it is a trivial exercize to calculate how large the second fly can be to move the center of the tube/fly/fly sustem to the edge of the frictionless table.


geometry
perimeter
analytic geometry
graphing lines
slope
trigonometry
LaTeX

I got the perimeter in terms of one variable y and numbers, and used tangents sum forumula on both the triangle and square which I solved the system for y. Then I plugged y in to find AB and AC.
Anyone else?

I used Pythagorean to prove that $BM+CN=MN+BC=18$, drew the altitude $AZ$, found that $SZ=2$ (where $S$ is the point where circle $O$ is tangent to $BC$), then used that to relate $BR$ and $CT$, $R$ and $T$ being the points where $O$ is tangent to $BM$ and $CN$, respectively, and also used Pythagorean to prove that $AR=AT=12$. Then, I used the relation between $BR$ and $CT$ to find the measure of each one, and added those to $AR$ and $AT$ to get $AB=19$ and $AC=17$. :)

Wow, that was rather convoluted. :P

I let the two sides be x and n, and used the fact that two tangents from one point to a circle are congruent as well as the fact that an angle bisector in a triangle divides the opposite side into segments proportional to the side lengths to get two equations in x and n, which I then solved.

i used infinite geometric series, the pythagorean theorem, and the formula for the radius of an inscribed circle.

I had two variables b and c, which were the tangents of the angles at B and C. I then used the equations of the circles and the fact that the slope of a perpendicular is -1/m (perpendicular relating to tangency). I then wrote two equations in b and c, one based on the fact that XY=1, the other based on the fact that the lower circle is tangent to all four sides of MNCB. After solving for b and c, I used trig to find AB and AC.

I used the fact that two tangents from one point to a circle are equivalent extensively, and I also used the fact that the length from A to either of the two points of tangency on the circle inscribed in MNBC is equal to the semiperimeter of the triangle AMN, which, with the fact that MX = YN.  Since XY = 1, and MN = 6, it follows that MX = 2.5 and MY = 3.5.  Then I got the length from A to the two points of tangency on the circle inscribed in AMN by using the fact that M and N are midpoints to get some equations from which the solution immediately follows.

Yeah, I used the fact that tangents were congruent and I had a whole bunch of variables but solving for them, I limited it to one variable, then I used the fact that the smaller triangle was similar to the bigger triangle, and by solving I found the answers to be $17,19$.

First, I proved the following as lemmas, even I probably could have invoked all of them without proof:

Lemma 1: if a quadrilateral can be circumscribed around a circle, the sum of one pair of opposite sides equals the sum of the other pair of opposite sides (as applied to the problem, this means that $MB+MC=BC+MN$).

Lemma 2: If the incircle of triangle PQR touches side PQ at S, then $\displaystyle PS=\frac{PQ+PR-QR}{2}$ (as applied to the problem, this means $\displaystyle MY=\frac{MN+MA-AN}{2}$.

Lemma 3: If the excircle of triangle PQR (the one tangent to PQ) touches PQ at T, then $\displaystyle PT=\frac{PQ+QR-PR}{2}$ (as applied to the problem, this means $\displaystyle MX=\frac{MN+AN-MA}{2}$)

These lemmas, plus some trivial stuff involving midpoint properties and the fact that $XY=1$, quickly give us a pair of simple simultaneous equations for $AB$ and $AC$.

To the administrators: Great job with the new feature that detects an odd number of dollar signs! :thumbup:   Not that I'd ever make that mistake when using $\LaTeX$... :whistling:


integration
calculus
conics
ellipse
function
calculus computations

Let $ x(t) \equal{} \frac{e^t \plus{} e^{\minus{}t}}{2}$ and $ y(t) \equal{} \frac{e^t \minus{} e^{\minus{}t}}{2}$ generate a curve from $ t \equal{} \minus{}1$ to $ t \equal{} 1$. What is the length of the arc?

I can find $ \frac{dy}{dx}$ just fine, but I'm not sure how to convert that into a "nice" form.

[quote="An0maly"]Let $ x(t) \equal{} \frac {e^t \plus{} e^{ \minus{} t}}{2}$ and $ y(t) \equal{} \frac {e^t \minus{} e^{ \minus{} t}}{2}$ generate a curve from $ t \equal{} \minus{} 1$ to $ t \equal{} 1$. What is the length of the arc?

I can find $ \frac {dy}{dx}$ just fine, but I'm not sure how to convert that into a "nice" form.[/quote]

The length of the arc $ L \equal{} \int_{ \minus{} 1}^{1}\sqrt {x^{'2} \plus{} y^{'2}}dt$ The result of  integral is expressed in the form of ellipse function


This is kind of hard because I don't know how to draw the figure with latex or something. :blush:

A right prism has quadrilateral base $ EFGH$ and opposite face, $ ABCD$. $ A$ aligns with $ E$, $ B$ aligns with $ F$, $ C$ aligns with $ G$ and $ D$ aligns with $ H$. The base $ EFGH$ has right angles at $ E$ and $ G$. The height $ AE$ is $ 32$.
$ AD\equal{}20$
$ AB\equal{}15$
$ BC\equal{}15$
$ HG\equal{}20$

The distance from $ A$ to $ G$ is:

(A)41
(B)40
(C)34
(D)36
(E)44

[hide="hint"]All you need to do to keep making right triangles. Don't be a hero and try to solve this without a diagram.[/hide]

13375P34K43V312, you're on the right track.

[b]To everyone: draw a diagram. I haven't because I can't draw with a computer, but you can with paper and pencil or pen or whatever.[/b]

Im going to say $ 40$ just because i dont want to do the problem, and i noticed that $ 40^{2}\minus{}32^{2}\equal{} 24^{2}$, and $ 24$ is an integer.

and $ 24$ is another cool number so work with because $ 24$ divided by $ 2$ is $ 12$, which happens to be $ 15^{2}\minus{}9^{2}\equal{} 20^{2}\minus{}16^{2}$

hehe problaby wrong but who cares, i only spent 1 sec doing it..

what ever happened to the diagnol formula $ \sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}\equal{}d$ or can we not use it in this problem?


Show that there are no positive integers $m,n$s.t.$(2^{n}-1)(3^{n}-1)=m^{2}$.

It equivalent to $6^{n}-2^{n}-3^{n}=m^{2}-1$.
Firstly, show that $m$ and $n$ are divisible by 6 (it's not diffucult, viewing remainders of division by 4 and 3). Then $m^{2}-1$ is divisible by 5, but $6^{n}-2^{n}-3^{n}$ is not divisible by 5, because since $n$ divisible by 6, $6^{n}\equiv 1 \mod 5$, $2^{n}\equiv\pm1\mod 5$ and $3^{n}\equiv 4 \mathrm{\enskip or\enskip}1\mod 5$.

[quote="TomciO"]I think that's wrong. The fact that $6|m$ doesn't imply that $m^{2}-1$ is divisible by $5$... (for example $m=12$).[/quote]Yes, you are right, here I've mistaked. :(

It is obviously in following cases.
For $2\not |n$ the order of two is one.
For $2|n$, $3\not |n$ the order of three is one.

The case $6|n$ is true if abc hypothesis is true.

See http://mathworld.wolfram.com/abcConjecture.html
and in russian http://ega-math.narod.ru/Lang/HypotABC.htm

If you say that exists the simple solution, we will be finding. :)

Are you sure this problem can be done with elementary techniques? I think it is an extremely difficult problem. Even though it is Chinese TST...

my opinion is the same of harazi.

i try to solve this with elementary ideas for hours!

first i prove that 12|m, and later my idea is find, with hensel's lema, one prime p such that the sum of number of factors p in $2^{n}-1$ and $3^{n}-1$ is odd! (but, if $p \leq 61$, i'm don't have nice results)

I just put that comment because I perfectly remember having seen this problem in the introduction of an article on powers in recurence sequences and they said it was proved quite recently. This is what makes me think this is damn difficult. Unfortunately, I don't remember any reference about that article.  :blush:

We can check if $n=12k$ deliver solutions to $(2^{n}-1)(3^{n}-1)=m^{2}$ modulo prime $p$ (I tested primes below $10^{6}$), and discover that:
$p=17$ gives $2^{3}|n$;
$p=31$ gives $5|n$;
$p=43$ gives $7|n$.

Therefore, $2^{3}\cdot 3\cdot 5\cdot 7=840|n.$

Similarly, we can check if $n=840k$ deliver solutions to $(2^{n}-1)(3^{n}-1)=m^{2}$ modulo prime $p$, and discover that:
$p=193$ gives $2^{4}|n$;
$p=311$ gives $31|n$;
$p=379$ gives $3^{3}|n$;
$p=491$ gives $7^{2}|n$;
$p=601$ gives $5^{2}|n$;
$p=1171$ gives $13|n$;
$p=2311$ gives $11|n$;
$p=55987$ gives $31\cdot 43|n$;
$p=58171$ gives $277|n$;
$p=71191$ gives $113|n$;
$p=585131$ gives $643|n$.

Therefore, $n$ must be divisible by
\[2^{4}\cdot 3^{3}\cdot 5^{2}\cdot 7^{2}\cdot 11\cdot 13\cdot 31\cdot 43\cdot 113\cdot 277\cdot 643 = 2030276593861916400. \]
Similarly we can proceed further to grow up the divisor of $n.$ But I have no proof that the divisor can be grown infinitely. Such proof would mean that there no $n$ satisfying $(2^{n}-1)(3^{n}-1)=m^{2}.$

Very simple and completely elementary proof can be found here
[url]http://www.sav.sk/index.php?lang=sk&charset=&doc=publish-journal&part=list_articles&journal_issue_no=1145[/url]
See "On Diophantine equations of the form $(x^{n}-1)(y^{m}-1)=z^{2}$" (P. G. Walsh)

i have a nice solution use pell"s equation.


In triangle ABC points D and E lie on sides AB and BC, respectively, so that AD/DB=1/2. Segments CD and AE intersect at point F. What is the value of CF/FD? Express your answer as a common fraction

Um, we need a restriction on $ E$ to do the problem.

Edit: If I'm not mistaken, the original problem is:

[quote="2008-2009 MATHCOUNTS Handbook Volume II: Triangles ULTRA Stretch"]
In triangle ABC points D and E lie on sides AB and BC, respectively, so that $ \frac {AD}{DB} \equal{} \frac35$ and $ \frac {CE}{EB} \equal{} \frac12$. Segments CD and AE intersect at point F. What is the value of $ \frac {CF}{FD}$? Express your answer as a common fraction.
[/quote]

[hide="Solution"]
Without loss of generality, let $ [ABC] \equal{} 1$. Then $ [BDE] \equal{} \frac {BD}{BA}\cdot\frac {BE}{BC}\cdot[ABC] \equal{} \frac58\cdot\frac23 \equal{} \frac5{12}$, and

$ [ADEC] \equal{} 1 \minus{} \frac5{12} \equal{} \frac7{12}$. Also, $ \frac {[ADE]}{[BED]} \equal{} \frac {AD}{DB} \equal{} \frac35\implies[ADE] \equal{} \frac35\cdot\frac5{12} \equal{} \frac14$, and similarly,

$ \frac {[CED]}{[BED]} \equal{} \frac {CE}{EB} \equal{} \frac12\implies[CED] \equal{} \frac12\cdot\frac5{12} \equal{} \frac5{24}$. Now, let $ x \equal{} [DEF]$ so that

$ [ADF] \equal{} [ADE] \minus{} x \equal{} \frac14 \minus{} x$, $ [CEF] \equal{} [CED] \minus{} x \equal{} \frac5{24} \minus{} x$, and

$ [AFC] \equal{} [ADEC] \minus{} \left(\frac14 \minus{} x \plus{} x \plus{} \frac5{24} \minus{} x\right) \equal{} \frac7{12} \minus{} \left(\frac14 \minus{} x \plus{} x \plus{} \frac5{24} \minus{} x\right)$

$ \equal{} \frac18 \plus{} x$. Finally,

$ \frac {CF}{FD} \equal{} \frac {\frac18 \plus{} x}{\frac14 \minus{} x} \equal{} \frac {\frac5{24} \minus{} x}x \equal{} \frac {\frac18 \plus{} \frac5{24}}{\frac14} \equal{} \boxed{\frac43}$, where we have used the fact that $ \frac {p_1}{q_1} \equal{} \frac {p_2}{q_2} \equal{} \frac {p_1 \plus{} p_2}{q_1 \plus{} q_2}$.
[/hide]

I think you can do it a bit quicker using mass point geometry.

[geogebra]58aef56b2855aca6e8d3e79de925d5426bdc1cc7[/geogebra]

Hence CF/FD=8/6=4/3.

Are mass points MC level...??? :huh: Why does that work?

First, we are given that AD/DB=3/5. Assuming that the fulcrum is D, and since we need to balance it we can assign assign weights for the points, namely A=5 and B=3, hence D=5+3=8. Also, since CE/EB=1/2, and B=3, C=3*2=6, and E=3+6=9. Now we know that for exactly one point on CD, the weight is 6+8=14. Similarly, for exactly one point on AE, the weight is 9+5=14. So those two points are the same, namely F, so F=14. Hence CF/FD=8/6=4/3.

Um, sorry if I'm being stupid, but why does adding the two endpoints of the "lever" give the "fulcrum?"

That is true by definition. The following link gives a good introduction to mass point geometry:
http://mathcircle.berkeley.edu/mpgeo.pdf


integration
trigonometry
logarithms
function
calculus
real analysis
real analysis unsolved

[color=darkred][b]$ \mathrm I\ \blacktriangleright$ Here, for $ \int\frac {f(x)}{g(x)}\ \mathrm {dx}$ find $ h(x)$ and $ k$ so that $ f(x) \equal{} h(x)\cdot g(x) \plus{} k\cdot g'(x)\ :$[/b]

$ \mathrm {I.1}\ \ \int\frac {\sin x \plus{} 5\cos x}{3\sin x \plus{} 2\cos x}\ \mathrm {dx}\ .$

$ \mathrm {I.2}\ \ \int\frac {\sin x}{e^x \minus{} \sin x \minus{} \cos x}\ \mathrm {dx}\ .$

$ \mathrm {I.3}\ \ \int\frac {3e^x \plus{} 5\sin x \plus{} 10\cos x}{e^x \plus{} 4\sin x \plus{} 3\cos x}\ \mathrm {dx}\ .$

$ \mathrm {I.4}\ \ \int\frac {2\minus{}x\arctan x}{(x^2\plus{}1)(x\plus{}\arctan x)}\ \mathrm {dx}\ .$

===========================================

$ \mathrm {II}\blacktriangleright$ [b]Here ascertain $ S\equiv A \plus{} B$ and $ D\equiv A \minus{} B$ :[/b]

$ \mathrm {II.1}\ \ A\equiv\int\frac {\sin 2x\cos x}{1 \plus{} \cos^4x}\ \mathrm {dx}$ and $ B\equiv\int\frac {\cos 2x\sin x}{1 \plus{} \cos^4x}\ \mathrm {dx}\ .$

$ \mathrm {II.2}\ \ A\equiv\int\frac {\sin^2x\cos x}{\sin x \plus{} \cos x}\ \mathrm {dx}$ and $ B\equiv\int\frac {\cos^2x\sin x}{\sin x \plus{} \cos x}\ .$

$ \mathrm {II.3}\ \ A\equiv\int\frac {\sin x}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx}$ and $ B\equiv\int\frac {\cos x}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx}\ .$

$ \mathrm {II.4}\ \ A\equiv\int\frac {2\sin^2x \plus{} 3\cos^2x}{\sin x \minus{} \cos x}\ \mathrm {dx}$ and $ B\equiv\int\frac {2\cos^2x \plus{} 3\sin^2x}{\sin x \minus{} \cos x}\ \mathrm {dx}\ .$[/color]

[quote="Virgil Nicula"]$ \mathrm {I.2}\ \ \int\frac {\sin x}{e^x - \sin x - \cos x}\ \mathrm {dx}\ .$[/quote]
\begin{eqnarray*}\int\frac {\sin x}{e^x - \sin x - \cos x}\,dx & = & \frac12\int\frac { - 2\sin x}{\sin x + \cos x - e^x}\,dx \\
& = & \frac12\int\frac {\cos x - \sin x - e^x - (\sin x + \cos x - e^x)}{\sin x + \cos x - e^x}\,dx \\
& = & \frac12\left(\int\frac {(\sin x + \cos x - e^x)'}{\sin x + \cos x - e^x}\,dx - \int dx\right) \\
& = & \frac12\left(\ln\left|\sin x + \cos x - e^x\right| - x\right) + k\end{eqnarray*}

[quote="Virgil Nicula"]$ \mathrm {I.4}\ \ \int\frac {2 - x\arctan x}{(x^2 + 1)(x + \arctan x)}\ \mathrm {dx}\ .$[/quote]
\begin{eqnarray*}\int\frac{2-x\arctan x}{(x^2+1)(x+\arctan x)}\,dx&=&\int\frac1{x^2+1}\left(\frac{2+x^2-x^2-x\arctan x}{x+\arctan x}\right)\,dx\\&=&\int\frac1{x^2+1}\left(\frac{2+x^2}{x+\arctan x}-x\right)\,dx\\&=&\int\left(\frac{2+x^2}{(x^2+1)(x+\arctan x)}-\frac x{x^2+1}\right)\,dx\\&=&\int\frac{(x+\arctan x)'}{x+\arctan x}\,dx-\frac12\int\frac{(x^2+1)'}{(x^2+1)}\,dx\\&=&\ln|x+\arctan x|-\frac12\ln(x^2+1)+k\end{eqnarray*}

[color=darkblue][b]Another exercises :[/b][/color]
[quote="Virgil Nicula"][color=darkred]$ \mathrm {II.5}\ A \equal{} \int\frac {x^2}{x^4 \plus{} ax^2 \plus{} b}\ \mathrm {dx}\ \ ;\ \ B \equal{} \int\frac {1}{x^4 \plus{} ax^2 \plus{} b}\ \mathrm {dx}$, [b]where[/b] $ b > 0\ .$

$ \mathrm {II.6}\ A \equal{} \int\frac {\sin^2x\plus{}\sin x}{1\plus{}\sin x\plus{}\cos x}\ \mathrm {dx}\ \ ;\ \ B \equal{} \int\frac {\cos^2x\plus{}\cos x}{1\plus{}\sin x\plus{}\cos x}\ \mathrm {dx}\ .$[/color][/quote]

[quote="Virgil Nicula"]$ \mathrm {I.1}\ \ K \equal{} \int\frac {\sin x \plus{} 5\cos x}{3\sin x \plus{} 2\cos x}\ \mathrm {dx}\ .$[/quote]

[color=darkblue][b]Proof.[/b] For the denominator-function $ f(x) \equal{} 3\sin x \plus{} 2\cos x$ I"ll ascertain

the constants $ a,b$ so that the numerator $ \sin x \plus{} 5\cos x \equal{} af(x) \plus{} bf'(x)$, i.e.

$ \sin x \plus{} 5\cos x \equal{} a(3\sin x \plus{} 2\cos x) \plus{} b(3\cos x \minus{} 2\sin x)$ $ \implies$ $ a \equal{} b \equal{} 1\ .$

Therefore, $ K \equal{} \int\frac {af(x) \plus{} bf'(x)}{f(x)}\ \mathrm {dx} \equal{}$ $ \int\left(a \plus{} b\cdot\frac {f'(x)}{f(x)}\right)\ \mathrm {dx}$ $ \implies$

$ K \equal{} x \plus{} \ln |3\sin x \plus{} 2\cos x| \plus{} \mathcal C\ .$[/color]

[quote="Virgil Nicula"]$ \mathrm {I.3}\ \ L \equal{} \int\frac {3e^x \plus{} 5\sin x \plus{} 10\cos x}{e^x \plus{} 4\sin x \plus{} 3\cos x}\ \mathrm {dx}\ .$[/quote]
[b]Proof.[/b] Denote $ f(x) \equal{} e^x \plus{} 4\sin x \plus{} 3\cos x$. I"ll ascertain the constants $ a,b$

so that  the numerator is equally to $ af(x) \plus{} bf'(x)$, i.e. $ 3e^x \plus{} 5\sin x \plus{} 10\cos x \equal{}$

$ a(e^x \plus{} 4\sin x \plus{} 3\cos x) \plus{} b(e^x \plus{} 4\cos x \minus{} 3\sin x)$ $ \implies$ $ a \equal{} 2$ and $ b \equal{} 1\ .$

In conclusion, $ L \equal{} 2x \plus{} \ln\left|e^x \plus{} 4\sin x \plus{} 3\cos x\right| \plus{} \mathcal R\ .$

[color=darkred][b]Remark.[/b] Remain only the set II of exercises ![/color]

I like this integrals. First part is very interesting. I had no time to solve them. It's very interesting if we don't know this trick how we will sove these integrals ;).
I'm sending a solution to II-1. In part two I think we may use also different techniques (universal substitution as well as trig formulae).

P.S. I modified my topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=168846 - so you may take a look at the complete solutions. You also may try:
http://www.mathlinks.ro/Forum/viewtopic.php?t=169021.

[color=darkred][u][b]Attention borislav_mirchev ![/b][/u]

The common denominator of $ A$ and $ B$ is $ 1 + \cos^4x$ and not $ 1 + \cos ^2x\ .$

I didn't speak well ! In fact, here isn't the method $ A + B\ ,\ A - B$

but $ A + mB\ ,\ A + nB$, where $ \{m,n\}\subset \mathcal R\ .$[/color]

[quote="Virgil Nicula"]$ \mathrm {II.1}\ \ A\equiv\int\frac {\sin 2x\cos x}{1 + \cos^4x}\ \mathrm {dx}\ \ ;\ \ B\equiv\int\frac {\cos 2x\sin x}{1 + \cos^4x}\ \mathrm {dx}\ .$[/quote]
[color=darkblue]$ \blacktriangleright\sin 2x\cos x - 2\cos 2x\sin x = 2\sin x(\cos^2x - \cos 2x) = 2\sin x(1 - \cos^2x)\ .$

$ \boxed {A - 2B} = 2\cdot\int\frac {1 - \cos^2x}{1 + \cos^4x}\cdot\sin x\ \mathrm {dx} =$ $ 2\cdot F(\cos x) + \mathcal C\ ,\ F = \int\frac {t^2 - 1}{t^4 + 1}\ \mathrm {dt} =$

$ \int\frac {1 - \frac {1}{t^2}}{t^2 + \frac {1}{t^2}}\ \mathrm {dt} =$ $ \int\frac {(t + \frac 1t)'}{(t + \frac 1t)^2 - 2}\ \mathrm {dt} =$ $ \frac {1}{2\sqrt 2}\cdot\ln\frac {t^2 - \sqrt 2\cdot t + 1}{t^2 + \sqrt 2\cdot t + 1} + \mathcal C\ .$

In conclusion, $ \boxed {\ A - 2B = \frac {\sqrt 2}{2}\cdot \ln|\frac {\cos^2x - \sqrt 2\cdot\cos x + 1}{\cos^2x + \sqrt 2\cdot\cos x + 1}| + \mathcal C\ }\ .$

$ \blacktriangleright 3\sin 2x\cos x - 2\cos 2x\sin x = 2\sin x(3\cos^2x - \cos 2x) = 2\sin x(1 + \cos^2x)\ .$

$ \boxed {3A - 2B} = 2\cdot\int\frac {\cos^2x + 1}{\cos^4x + 1}\cdot\sin x\ \mathrm {dx} =$ $ - 2\cdot G(\cos x) + \mathcal C\ ,\ G = \int\frac {t^2 + 1}{t^4 + 1}\ \mathrm {dt} =$

$ \int\frac {1 + \frac {1}{t^2}}{t^2 + \frac {1}{t^2}}\ \mathrm {dt} =$ $ \int\frac {(t - \frac 1t)'}{(t - \frac 1t)^2 + 2}\ \mathrm {dt} =$ $ \frac {\sqrt 2}{2}\cdot\arctan\frac {t^2 - 1}{t\sqrt 2} + \mathcal C\ .$

In conclusion, $ \boxed {\ 3A - 2B = \sqrt 2\cdot\arctan\frac {\sin^2x}{\sqrt 2\cos x} + \mathcal C\ }\ .$[/color]

I'm sorry! I don't know from where I decided number is 2 instead 4. But now we know even with number 2 it may be solved with method A+B A-B.

[quote="borislav_mirchev"]$ \mathrm {II.1}\ \ A\equiv\int\frac {\sin 2x\cos x}{1 \plus{} \cos^2x}\ \mathrm {dx}\ \ ;\ \ B\equiv\int\frac {\cos 2x\sin x}{1 \plus{} \cos^2x}\ \mathrm {dx}\ .$[/quote]
$ A \minus{} B \equal{} \int\frac {\sin x}{1 \plus{} \cos^2x}\ \mathrm {dx}$ $ \implies$ $ \boxed {A \minus{} B \equal{} \minus{} \arctan (\cos x) \plus{} \mathcal C}\ .$

$ 3\sin 2x\cos x \minus{} 2\cos 2x\sin x \equal{} 2\sin x(1 \plus{} \cos^2x)$ $ \implies$

$ \boxed {3A \minus{} 2B \equal{} \minus{} 2\cdot \cos x \plus{} \mathcal C}\ .$ In conclusion,

$ \left\{\begin{array}{c} A \minus{} B \equal{} \minus{} \arctan (\cos x) \plus{} \mathcal C\ . \\
 \\
3A \minus{} 2B \equal{} \minus{} 2\cdot \cos x \plus{} \mathcal C\ .\end{array}\right\|$ $ \implies$ $ \boxed {\begin{array}{c} A \equal{} 2\arctan (\cos x) \minus{} 2\cos x \plus{} \mathcal C\ . \\
 \\
B \equal{} 3\arctan (\cos x) \minus{} 2\cos x \plus{} \mathcal C\ .\end{array}}$

Is it correct solution to II-2?

P.S. Edited - for final answer - "-" should be instead of "+" before ln|1+sin2x|.

[quote="Virgil Nicula"]$ \mathrm {II.2}\ \ A\equiv\int\frac {\sin^2x\cos x}{\sin x \plus{} \cos x}\ \mathrm {dx}\ \ ,\ \ B\equiv\int\frac {\cos^2x\sin x}{\sin x \plus{} \cos x}\ .$[/quote]
$ A\plus{}B\equal{}\int\sin x\cos x\ \mathrm {dx}\equal{}\int \sin x(\sin x)'\ \mathrm {dx}$ $ \implies$ $ \boxed {A\plus{}B\equal{}\frac 12\cdot\sin^2x\plus{}\mathcal C}\ .$

$ B\minus{}A\equal{}\int \frac {\sin x\cos x(\cos x\minus{}\sin x)}{\sin x\plus{}\cos x}\ \mathrm {dx}\equal{}$

$ \frac 12\cdot\int\frac {(\sin x\plus{}\cos x)^2\minus{}1}{\sin x\plus{}\cos x}\cdot (\sin x\plus{}\cos x)'\ \mathrm {dx}\equal{}$

$ \frac 12\cdot\left[\int(\sin x\plus{}\cos x)(\sin x\plus{}\cos x)'\ \mathrm {dx}\minus{}\int\frac {(\sin x\plus{}\cos x)'}{\sin x\plus{}\cos x}\ \mathrm {dx}\right]\implies$

$ \boxed {B\minus{}A\equal{}\frac 14\cdot(\sin x\plus{}\cos x)^2\minus{}\frac 12\cdot\ln|\sin x\plus{}\cos x|\plus{}\mathcal C}$ [b]a.s.o.[/b]

[quote="Virgil Nicula"]$ \mathrm {II.3}\ \ A\equiv\int\frac {\sin x}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx}\ \ ,\ \ B\equiv\int\frac {\cos x}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx}\ .$[/quote]
$ \blacktriangleright B \minus{} A \equal{}$ $ \int\frac {\cos x \minus{} \sin x}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx} \equal{}$ $ \int\frac {(1 \plus{} \sin x \plus{} \cos x)'}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx}$ $ \implies$

$ \boxed {B \minus{} A \equal{} \ln|1 \plus{} \sin x \plus{} \cos x| \plus{} \mathcal C}\ .$

$ \blacktriangleright A \plus{} B \equal{} \int\frac {\sin x \plus{} \cos x}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx} \equal{}$ $ \int\left(1 \minus{} \frac {1}{1 \plus{} \sin x \plus{} \cos x}\right)\ \mathrm {dx}$ $ \implies$

$ A \plus{} B \equal{} x \minus{} C\ \ ,\ \ C \equal{} \int\frac {1}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx}$ $ \implies$

$ C \equal{}\int\frac {1}{2\cos^2\frac x2\plus{}2\sin\frac x2\cos\frac x2}\ \mathrm {dx}\equal{}$ $ \frac 12\cdot \int\frac {\sin^2\frac x2\plus{}\cos^2\frac x2}{\cos\frac x2\left(\cos\frac x2\plus{}\sin\frac x2\right)}\ \mathrm {dx}\equal{}$

$ \int\frac {1 \plus{} \tan^2\frac x2}{2\left(1 \plus{} \tan\frac x2\right)}\ \mathrm {dx}\implies$ $ C \equal{} \int\frac {\left(1 \plus{} \tan\frac x2\right)'}{\left(1 \plus{} \tan\frac x2\right)}\ \mathrm {dx} \equal{}$ $ \ln\left|1 \plus{} \tan\frac x2\right| \plus{} \mathcal C\implies$

$ \boxed {A \plus{} B \equal{} x \minus{} \ln\left|1 \plus{} \tan\frac x2\right| \plus{} \mathcal C}\ .$

[quote="Virgil Nicula"]$ \mathrm {II.4}\ \ A\equiv\int\frac {2\sin^2x \plus{} 3\cos^2x}{\sin x \minus{} \cos x}\ \mathrm {dx}\ \ ,\ \ B\equiv\int\frac {2\cos^2x \plus{} 3\sin^2x}{\sin x \minus{} \cos x}\ \mathrm {dx}\ .$[/quote]
$ B \minus{} A \equal{} \int\frac {\sin^2x \minus{} \cos^2x}{\sin x \minus{} \cos x}\ \mathrm {dx} \equal{}$ $ \int (\sin x \plus{} \cos x)\ \mathrm {dx}$ $ \implies$

$ \boxed {B \minus{} A \equal{} \sin x \minus{} \cos x \plus{} \mathcal C}\ .$ Other side, $ A \plus{} B \equal{} 5\cdot C$, where

$ C \equal{} \int\frac {1}{\sin x \minus{} \cos x}\ \mathrm {dx} \equal{}$ $ \int\frac {\sin x \minus{} \cos x}{(\sin x \minus{} \cos x)^2}\ \mathrm {dx} \equal{}$ $ \int\frac {(\sin x \plus{} \cos x)'}{(\sin x \plus{} \cos x)^2 \minus{} 2}\ \mathrm {dx}\ .$

In conclusion, $ \boxed {A \plus{} B \equal{} \frac {5}{2\sqrt 2}\cdot \ln\left|\frac {\sin x \plus{} \cos x \minus{} \sqrt 2}{\sin x \plus{} \cos x \plus{} \sqrt 2}\right| \plus{} \mathcal C}\ .$

[color=darkred][b]We have still two exercises :[/b][/color]

[quote="Virgil Nicula"][color=darkblue][b]Another exercises :[/b][/color]
[color=darkred]$ \mathrm {II.5}\ A \equal{} \int\frac {x^2}{x^4 \plus{} ax^2 \plus{} b}\ \mathrm {dx}\ \ ;\ \ B \equal{} \int\frac {1}{x^4 \plus{} ax^2 \plus{} b}\ \mathrm {dx}$, [b]where[/b] $ b > 0\ .$

$ \mathrm {II.6}\ A \equal{} \int\frac {\sin^2x \plus{} \sin x}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx}\ \ ;\ \ B \equal{} \int\frac {\cos^2x \plus{} \cos x}{1 \plus{} \sin x \plus{} \cos x}\ \mathrm {dx}\ .$[/color][/quote]

$ A \equal{} \int \frac {sin^{2}x \plus{} sinx}{1 \plus{} sinx \plus{} cosx} \,dx$
$ B \equal{} \int \frac {cos^{2}x \plus{} cosx}{1 \plus{} sinx \plus{} cosx} \,dx$
$ A \plus{} B \equal{} \int \frac {sin^{2}x \plus{} cos^{2}x \plus{} sinx \plus{} cosx}{1 \plus{} sinx \plus{} cosx} \,dx$
Using: $ sin^{2}x \plus{} cos^{2}x \equal{} 1$ follows
$ A \plus{} B \equal{} \int \frac {1 \plus{} sinx \plus{} cosx}{1 \plus{} sinx \plus{} cosx} \,dx$
$ A \plus{} B \equal{} x \plus{} C$
$ A \minus{} B \equal{} \int \frac {sin^{2}x \minus{} cos^{2}x \plus{} sinx \minus{} cosx}{1 \plus{} sinx \plus{} cosx} \,dx$
$ A \minus{} B \equal{} \int \frac {(sinx \minus{} cosx)(1 \plus{} sinx \plus{} cosx)}{1 \plus{} sinx \plus{} cosx} \,dx$
$ A \minus{} B \equal{} \minus{} sinx \minus{} cosx \plus{} C$
$ A \equal{} \frac {1}{2}(x \minus{} sinx \minus{} cosx) \plus{} C$
$ B \equal{} \frac {1}{2}(x \plus{} sinx \plus{} cosx) \plus{} C$

It is not difficult. But what about II - 5? I cannot invent short way (at first look) to solve it. I can find A and B as integrals one by one but it is not short and I don't use your method. What is your idea?

[quote="Virgil Nicula"]$ \mathrm {II.5}\ A \equal{} \int\frac {x^2}{x^4 \plus{} ax^2 \plus{} b}\ \mathrm {dx}\ \ ,\ \ B \equal{} \int\frac {1}{x^4 \plus{} ax^2 \plus{} b}\ \mathrm {dx}$, [b]where[/b] $ b > 0\ .$[/quote]
$ A\pm B\sqrt b \equal{} \int\frac {x^2\pm \sqrt b}{x^4 \plus{} ax^2 \plus{} b}\ \mathrm {dx} \equal{}$ $ \int\frac {1\pm \frac {\sqrt b}{x^2}}{x^2 \plus{} \frac {b}{x^2} \plus{} a}\ \mathrm {dx}$ $ \implies$

$ A\pm B\sqrt b\equal{}\int\frac {\left(x\mp \frac {\sqrt b}{x}\right)'}{\left(x\mp \frac {\sqrt b}{x}\right)^2 \plus{} a\pm 2\sqrt b}\ \mathrm {dx}\ .$

$ \boxed {\begin{array}{cccc} A \plus{} B\sqrt b \equal{} F\left(x \minus{} \frac {\sqrt b}{x}\right) \plus{} \mathcal C & \mathrm {where} & F(t) \equal{} \int\frac {1}{t^2 \plus{} a \plus{} \sqrt b}\ \mathrm {dt}\ . \\
 \\
A \minus{} B\sqrt b \equal{} G\left(x \plus{} \frac {\sqrt b}{x}\right) \plus{} \mathcal C & \mathrm {where} & G(t) \equal{} \int\frac {1}{t^2 \plus{} a \minus{} \sqrt b}\ \mathrm {dt}\ .\end{array}}$

It was very interesting and beautiful!
Did you take a look at the links above?

Yes ! I saw all. They are nice and correctly. I hope that such you will try solve any integral first time without using the "integration techniques". Success !

Remarkable substitutions in integration calculus ...

[quote="borislav_mirchev"]What do you mean when you say "integral technics"?[/quote]
It should be "integration techniques"

Well, english is not my mother language but one may sometimes understand "strange dialects".


Introduce yourself, your team, etc.

I'm going with the Alabama team, hopefully we can do well now that we're forced into A Division.  ;)

alabama is closer to las vegas than pennsylvania?

WHOA THIS ROCKS MY WORLD

Haha, go figure, I guess our team wants to go chill in Vegas instead

In airline terms, Las Vegas is easier to get to from a lot of places than State College, PA. When ARML West was in Las Vegas before, up until 2000 or 2001, Alabama sometimes came to Las Vegas.

Anyway, I'll be there - as a coach. And we're close enough that we're coming by bus. Expect three Southern California teams again, and expect to see a number of people whose AoPS usernames you probably know: Teddy, litoxe27182818, JAM, pkerichang, isaacchao, ambierona, kkelove7, Takeide, and so on - as well as a lot of people who don't post here or seldom post here.

Southern California A will be in A division; Southern California B and C will be in B division.

If I can get the details worked out, I might be going with a Utah Team. I also know that there will be a Nevada team there.

ARML goes to Las Vegas?  That's really weird.  I think my parents said that once there was an MAA conference (might have been AMS, though; I'm not sure) at Las Vegas, but afterwards it was voted never to hold the conference there again, on the grounds that it was too difficult to concentrate.

ARML first set up a western site around 1995, and it was in Las Vegas, on the UNLV campus. A few years ago, they switched to San Jose and the San Jose State campus. The competition is moving back to Las Vegas this year, and there will probably be more teams there than either of the last two years.
As for that "too distracting" issue, it shouldn't be a problem for high school students. It's a bit surreal to see slot machines in the airport terminal and the monuments of the Strip looming nearby, but the legal gambling age is 21. There isn't anything special about Las Vegas as a tourist destination if you're a 17-year old math nerd.

ehm,,, ehm...

WA is going there, and just as a side note, everyone else might want to start fight for second place, cause we are taking over the first place  :P

we do have a really strong team though!

wooo...we got competitions...

As Mr. Merrifield posted above, I will be on the So. Cal. ARML team coming HOPEFULLY on team A (a.k.a. division A).  We're deciding the teams TOMORROW!!!!!

Eric - I probably won't finalize the teams for another couple of weeks. Tommorow is not the last word.

I'm going to las vegas with the SD Math Circle team. :)

[quote="Teddy"][quote="amirhtlusa"]everyone else might want to start fight for second place[/quote]yeah, behind a team from California.[/quote]

Yup! our plan is to throw Teddy all the hard problems (number 9 and 10 in team round)...He'll be the Adam Hesterberg from So. Cal. :rotfl: (quote by JAM)

Number of USAMO qualifiers by ARML team territory, amended:

San Diego: $2\ \longrightarrow\ 4$
Southern California: $10\ \longrightarrow\ 12$
SFBA: $35\ \longrightarrow\ 39$
Northern California: $2\ \longrightarrow\ 2$
Washington: $14\longrightarrow\ 15.$

The two new ones in Southern California (from Santa Monica and Claremont) are not students I've ever had contact with or heard of before.

Also, that 39/2 split between SFBA and Northern California is based on the assumption that the boundary between them is somewhere near Interstate 5. If the boundary is further west - say, the crest of the Oakland hills - then you could reassign one or two people from places like Danville.

If SFBA brought all those people to Las Vegas.....Hope we still have a chance to beat them...(just imagine, a region that sends three teams with 9 USAMO qualifiers on the C team...)

Going from WA, and I still think we have a chance to give SFBA a run for their money :D

Representing: (drumroll) the San Diego Math League.

Representing Colorado: currently tied for 4th in Power Contest  :first:

Teams represented by posts on this topic:

Washington
Southern California
San Diego (new team)
Colorado (new team)
Utah, with one Utah student saying that there will be a Nevada team. (Is that two different teams? Last year it was one Utah-Nevada team)
Alabama

Not heard from yet:

SFBA (Come on - there must be SFBA AoPS-ers!)
Northern California

Anyone else? New Mexico? Oregon? Arizona?

I'm from SFBA. I've never done this before, but I figure I'll try this year.
And yes, we could definitely field several teams filled with all USAMO qualifiers ^^.

Do the various teams normally associate outside of the competition, or do you just stay with your own team (or is there even any time to do anything outside of the competition)?

At the Penn State site we mostly stayed with our own team, but you could find other teams around campus and in the dorms. It'd probably be easy to start up ultimate frisbee or something with another team. We also played cards with some New York kids in the dorm lobby. As for extra time, it depends on when you get there. The day of competition keeps very busy until the afternoon.

[quote="pavolas"]NorCal is here![/quote]
Who's coaching you this year - Dr. Sallee?

No, but Dr. Sallee will be going with us to the competition.
Our coach is Spyridon Michalakis.

Cicatriz - the timing is a little different at the western site. We do the Team Round and the Power Round Friday evening, and then the Individual Round and the Relay Round on Saturday morning. For us, everything is over shortly after noon on Saturday, and we mostly tend to leave right after that.

We're going to get in probably about 1:00 on Friday, if anyone is looking for any of us. I haven't seen the dorm setup in Las Vegas, so I don't know what the common areas will look like, or the lawns outside. (But, this being Las Vegas, it will probably be 100° out there.)

In the first year I went, 2002, we came back to the dorm after the Friday evening competition to find a basketball game on the TV in the common area: NBA playoffs, Lakers versus Kings. With the Northern California (meaning Sacrmento/Davis) team and the Southern California team sharing the same dorm, that was a very loud room.

Haha, I can imagine.

Thanks for the info.

San Diego! We're gonna get smashed... we have like 5 eighth graders on the A team, including me

I'll be there as a coach for SFBA.

In response to your question about associating with other teams, 101101101, it really depends on if you want to.  When I was on the Minnesota team, we kept to ourselves to some degree, but we also hung out with the Chicago teams a bit; we'd play them in frisbee every year, and last year we also started a bridge game against them.

I don't really know what to expect as this is my first time with SFBA, but I'm sure it will be different just because of the logistical differences.  MN would usually get to Iowa around mid-afternoon Friday, go play some frisbee, go to dinner, and then meet up downstairs in the dorm for games.  We're in a much different position, since on Friday we will have come off of the all-night (most likely) drive to Vegas, and we start the competition Friday night.  There's also the issue of finding a frisbee field to play on that would complicate any games of Ultimate.

If you want to get to know people from other teams, you'll have no problem doing so.

[quote="moralityjustice"][quote="mrosett"]Let me think... How many USAMO qualifier does Utah have?
...
I'll give you a hint.... if x equals the number of qualifiers Utah has, x satisfies x=1.
Somehow I don't think we'll out muscle SFBA[/quote]

Haha. I like your bubbly wit, Max.[/quote]

Bubbly? That is a way my wit has never been described. And believe me, my wit has been described in all sorts of ways: boring, grating, repetitive, unoriginal, obnoxious.... Bubbly is probably a step up
As it turns out, the Utah team will have two Arizona members, one of whom was a USAMO qualifier. So, my bubbly wit aside, the Utah team will 2 USAMO qualifiers. On the other hand, we're going to be an up-and-coming team. We only have one senior. Believe it or not, no matter how hard I try I can't talk any seniors into skipping their high school graduation to come to a math contest.
Oh- about the Nevada thing. Last year we were a joint team with them. By that I mean that I went along as the sole Utah representative on the Nevada team. This year we are separate teams (although we were planning to be a joint team until about February), and I am about 90% sure that they are going to have a team. Among other things, the contest is in Las Vegas, and they managed to get a team when it was in CA. I'm not sure though- I've kind of fallen out of touch with the girl who is organizing it.

[quote="mrosett"]Among other things, the contest is in Las Vegas, and they managed to get a team when it was in CA. I'm not sure though- I've kind of fallen out of touch with the girl who is organizing it.[/quote]
She has an AoPS username: angelanana, but she seldom posts. Her most recent post was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61474]this,[/url] last November in the Nevada forum.

[quote="samath"]Hurray for Washgington![/quote]
ditto.


group theory
abstract algebra
search
superior algebra
superior algebra unsolved

Prove that every subgroup of index 2 is normal subgroup

Check [url=http://planetmath.org/encyclopedia/ASubgroupOfIndex2IsNormal.html]this[/url] page or search the forum. :)


real analysis
advanced fields
advanced fields unsolved

Prove that a connected and locally connected complete metric space is path connected.

maybe (the beginning of) an idea:

Take an arbitrary point $x \in X$, where $X$ is connected, locally connected complete metric space. One wants to show that there is a path in $X$ between $x$ and any other point $y \in X$.

If $r>0$ is an arbitary real number, one will say that $x$ and $y$ are $r$-connected if there exists a finite sequence of open sets $B_{0}, B_{1}, B_{2}, \ldots, B_{n}$ of diameter at most $r$ such that:
1/ $B_{i}\cap B_{i+1}$ is non-empty
2/ if $|i-j|>1$ then $B_{i}\cap B_{j}$ is empty
3/ $x \in B_{0}$ and $y \in B_{n}$
4/ every $B_{i}$ are connected subset of $X$

First, remark that $x$ is $r$-connected to any other point $y \in X$. To show that, let us note $E$ the set of points that are $r$-connected to $x$. $E$ is non-empty since $x \in E$, and it suffices to show that $E$ is open and closed in $X$.($X$ is connected)
1/ $E$ is clearly open since if $B_{0}, B_{1}, B_{2}, \ldots, B_{n}$ are open subsets that connect $x$ and $y$, then $B_{n}\subset E$
2/ Suppose that $(y_{n})_{n \geq 0}\in E$ converges towards $y \in X$. Take $B$ a neigborhood containing $y$, of diameter at most $r$, that is connected ($X$ is locally connected) and $y_{n}$ such that $d(y,y_{n})<r$. Then one easily show that $y \in E$, by constructing a sequence of subsets that connect $x$ and $y$, extracted from the subsets that connect $x$ and $y_{n}$ etc..

Now, I don't really know how to properly conlude, but here is the idea. Take $x,y \in X$. Then $x$ and $y$ are $\frac{1}{2^{1}}$ connected by subsets $B_{0}^{1}, B_{1}^{1}, \ldots, B_{n_{1}}^{1}$. Take some points $x_{i}^{1}\in B_{i}^{1}\cap B_{i+1}^{1}$. Then, because each $B_{i}^{1}$ is also connected, locally connected, one can do the same reasonning to show that $x_{i}^{1}$ and $x_{i+1}^{1}$ are $\frac{1}{2^{2}}$-connected by open subsets included in $B_{i+1}^{1}$. All these new subsets, $B_{0}^{2}, B_{1}^{2}, \ldots, B_{n_{2}}^{2}$, of diameter at most $\frac{1}{2^{2}}$, are included in  $\bigcup_{1}^{n_{1}}B_{i}^{1}$, and deleting some of them, one can suppose that they connect properly (in the sense above) $x$ and $y$.

If one call $P_{k}= \bigcup_{1}^{n_{k}}B_{i}^{k}$, then $P_{k+1}\subset P_{k}$, and because of the completeness of $X$, I *think* one can show that $\cap_{k \geq 1}P_{k}$ is a path connecting $x$ and $y$  :)  Has it any chance to work ?

I think you can finish like this: choose a uniform partition of $[0,1]$ by $n_{1}$ points $t_{i}^{1}$ (including endpoints) and set $\gamma(t_{i}^{1})=x_{i}^{1}$. On the next step, add a uniform partition of each subinterval $[t_{i-1}^{1},t_{i}^{1}]$ by $n_{2}$ points $t_{ij}^{2}$ and set $\gamma(t_{ij}^{2})=x_{ij}^{2}$... You will get a uniformly continuous map $\gamma$ from a dense subset of $[0,1]$ into $X$, which can be extended to $[0,1]$.


geometry
3D geometry
perpendicular bisector
advanced fields
advanced fields unsolved

I'm at IMO training camp right now, and our leader has asked us a question that has really frustrated some of my teammates. It goes as follows:

Can one put uncountablity many topological "Y" shapes in the plane such that no two intersect.

[quote="grobber"][url=http://www.mathlinks.ro/viewtopic.php?t=6352]Here[/url]'s something you might find interesting.

I know that problem deals with $T$'s and yours deals with topological $T$'s (it's the same as topological $Y$'s), but I think ilyasu'sargument can be modified to give the answer "no" for this problem as well.[/quote]

The other thread dealt with non-topological Ts. So this problem is much different and more difficult. Though we know that the answer is indeed "no".

[quote="ilyasu1"]Associate $3$ points in $Q^2$ to a T - so that they would be extremely close to the "3-point-boundry" of T.
There are countably many such triples, and two disjoint T's have a different set of triples.
Result follows.[/quote]

The problem is how does one define "extremely close". It is not trivial at all.

Yeah, I had a wrong argument and that's why I said it worked..

Here's something else that might work (but I think it becomes sort of nasty): We could partition the $Y$'s in a countable set of classes $S_{m,n,p,q},\ m,n,p,q\in \mathbb N^*$ where their significance is the following: the "arms" of a $Y$ in such a class are in the intervals $[\frac 1m,\frac 1{m-1}),\ [\frac 1n,\frac 1{n-1}),\ [\frac 1p,\frac 1{p-1})$ and $\liminf_{\mbox{U,V,T are points on the Y in this order}}\angle {UVT}\in [\frac 1q,\frac 1{q-1})$. Then we could try to show that only countably many $Y$'s from a certain class could be drawn on the plane. Maybe this is easier because the conditions give a certain "rigidity" to the $Y$'s of each class.

yep i think ilyusa's argument is not right...one must say a few more words to show that. for example, what would you do if you had a big T and at one vertex of the T is a a point in whose $\epsilon$-neighbourhood there are infinitely many small T's.

When I thought about the problem (the older one, regarding $T$'s), I wanted to do something like in my previous post on this topic. In that case, we only needed to consider classes according to the length of the arms of the $T$'s, and it was easier to show that each class can have only countably many representatives in the plane.

My supervisor told me that initial problem (about T's) was invented in Moscow State University many-many years ago. And proof was similar to ilusha's one:
we must choose triangle with rational vertices which sides intersect at least one of the sprouts of Y and which contains "center" of Y.

i already recognized that my counterexample to ilyusas argument is nonsense(not the points itself but the triples are different). myth's proof is just beautiful, but i've got a proof with a completely other idea:

first i show the following

LEMMA
given an uncountability S of points in $R^n$, there is a point P in S such that in every open neighbourhood of P there are uncountable many points of S.

Proof: by dividing the $R^n$ into unit (hyper-)cubes(there are countable many of them), one cube has to contain uncountable many points of S, thus we may assume that S is bounded. now assume that every point P of S has a open neighborhood $N(P)$ of P s.t. there are only countably many points of S in $N(P)$. now, the union N of all $N(P)$ is open. now, let all the points in N with distance smaller or equal to $\frac{1}{n}$ to the boundary of N be in $M(n)$, thus $M(n)$ is closed. since the the $N(P)$ cover $M(n)$, there is a finite subcover of the $N(P)$ covering $M(n)$, thus $M(n)$ contains only countably many points of S. but this means that N contains only countably many points of S, but all points of S are included in N, contradiction.

now take S as a part of $R^8$ to be the set of all ordered quadruples of points of our Y's. the lemma shows that there is "the" Y such that there are uncountably many Y's with vertices as close as we want to the vertices of "the" Y(it's not unique, but i have to distinct it from the other Y's). the three "arms" of "the" Y dissect the neighborhood of the center of "the" Y into three parts, one of them must contain uncountably many centers of other Y's(call it the "upper" part, so the vertex on the other side is the "lower" vertex). now take such a Y. to get to the (neighborhood) of the "lower" vertex, we must somehow get around one of the other two vertices and uncountably many go over one vertex, call it the "right" vertex. now, draw the perpendicular bisector to the line through the center and the "right" vertex of "the" Y. then two "arms" of our other Y must cross that perpendicular bisector. but the area contained in this perpendicular bisector and the two arms of the other Y isn't used by any other "arm" of an Y anymore(at least those in close neighborhood to "the" Y(such that the centers of them do not lie in that triangle), thus uncountably many). thus continuing, we always get a positive area not intersecting the others and the sum of these areas finite, and are left with uncountably many Y's, thus get a sum of uncountably positive numbers which is finite, contradiction.

that proof is far not so nice as myth's one, but at least it is my own :)
i hope there is no mistake in it(or at least one can repair it).

Peter

I think, such solutions are unreadable without appropriate figures and text formatting.