The concept is really quite similar to solving equations: What you do to one side, you must do to the other and leave the sign as it is. The only difference is that when you multiply or divide by negatives, you have to flip the sign.

[quote="linzk3"]can you solve for v?

-5v-8<3v+32

pls && thx[/quote]

[hide="with simple algebra..."]
$\\-5v-8<3v+32\\-40<8v\\-5<v$[/hide]

[quote="laughinghead505"]...when you multiply or divide by negatives, you have to flip the sign.[/quote]

$-3v>6$
$v<-2$

or

$-\frac{1}{2}v>4$
$v<-8$

The most important thing is the flipping sign, but it is easy . :lol:
[hide]$-5v-8<3v+32\\-5v<3v+40\\-8v<40$
Now the flip

$v>-5$[/hide]

You won't ever fear inequalities if you keep in mind a few simple facts.

[b]Definition:[/b]  $a > b \Leftrightarrow a-b > 0$

[b]Definition:[/b]  $a < b \Leftrightarrow b > a$

[b]Axiom:[/b]  $a > 0, b > 0 \implies ab > 0$

In our case, we make the following simple algebraic transformations.

$-5v-8 < 3v+32$
$3v+32 >-5v-8$ (definition)
$(3v+32)-(-5v-8) > 0$ (definition)
$8v+40 > 0$ (simplification)
$v+5 > 0$ (multiplication by $\frac{1}{8}> 0$)
$v >-5$ (definition)


superior algebra
superior algebra unsolved

soit $ P$ un espace fỉbré principal analytipue complexe, de base $ X$, et de groupe structural G le groupe linéeire complexe $ GL_r(\mathbb{C})$. Prenons pour fibré type $ F$ l'escape $ \mathbb{C}^r$ sur lequel $ G$ opère de facon évidente, et soit $ V = P \times_GF$ l'escape fibré associé à $ P$ et de fibré type $ F$ (rappelons que $ V$ est l'escape quotient de $ P\times F$ par la relation d'équivalence $ (p.g,f)\equiv(p,gf)$ pour $ p\in P, g\in G, f\in F$). Je ne sais pas ce concept " et soit $ V = P \times_GF$ l'escape fibré associé à $ P$ et de fibré type $ F$ (rappelons que $ V$ est l'escape quotient de $ P\times F$ par la relation d'équivalence $ (p.g,f)\equiv(p,gf)$ pour $ p\in P, g\in G, f\in F$". Aide-moi

Write in English or restrict your posts to the local communities.

I read this concept in artile "Un theorème de dualité" of Jean - Pierre Serre. But I do not understand this concept. Help me, please


linear algebra
matrix
advanced fields
advanced fields unsolved

Find A=(a(i,j)) \in M_4(R) skew-symetric  tA = -A
(where tA is the transpose of A) such that the differential form
w =  \sum _{i,j}a(i,j)dx_i/\dx_j  verify
w/\w = dx_1/\dx_2/\dx_3/\dx_4

[quote="Moubinool"]Find $A=(a_{i,j}) \in M_4(\mathbb{R})$ skew-symetric  $^t A = -A$
(where $^t A$ is the transpose of $A$) such that the differential form
$w =  \sum _{i,j}a_{i,j} dx_i \wedge dx_j$  verifies
$w \wedge w = dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4$[/quote]

Just take $A$ defined by $a_{1,2} = a_{3,4}=a_{3,1}=a_{2,4}=
a_{4,1}=a_{3,2}= \frac{\sqrt{6}}{12}$ and  give $0$ to all the other coefficients that
are not given by the skew-symetry.

Fabrice


in one of the examples in chapter 4 of AoPS dealing with proportions, there is a problem.

It says since [i]a[/i] is directly proportional to [i]b[/i] and also directly proportional to [i]c[/i] it must be directly proportional to [i]bc[/i] can anyone explain why this is clearly?? Thanks.

When something is directly proportional to something else, we say this:
$A = kB$
This means that A is directly proportional to B with constant of propotionality k.
Inverse proportionality is similar:
$A = k \cdot \frac{1}{B}$
This means that A is inversely proportional to B with constant of proportionality k.

I'll use a real life example.
The shadow of an object is directly proportional to it's height. We can write:
$S = kH$
Where H is height, little k is constant of proportionality, and S is shadow length.
The shadow of an object is also directly proportional to the ange it makes with the plane. We can write this as
$S = kA$
Where S is shadow length, k is C.P., and A is the angle it makes with the plane.
Because both of these are true at the same time, we can see that
$S = kAH$
And thus, two things can be directly proportional to a third simultaneously.

I hope that made sense.


group theory
abstract algebra
invariant
superior algebra
superior algebra theorems

If $ G$ is a group, then the $ i^{th}$ derived subgroup $ G^{\right(i\left)}$ is a fully unvariant subgroup, whence $ G^{\right(i\left)}$  is normal.

It's a characteristic (=invariant under all automorphisms) subgroup by definition (it's generators are bijectively mapped to such generators).


geometry
perimeter
rectangle
3D geometry
prism

please answer the question with a diagram if possible.

1)The inner perimeter of a rectangle room is 250m.Its height is 6m.

What is the amount to be spent to paint the walls of the room at the

the rate of Rs.12/-per square metre.

Our math teacher told us that this
 problem is wrong but i don't know. :maybe:

Can you please clarify this question? I'm not sure what Rs.12/-per square metre. is supposed to mean.

Basically 12 Rupees(money) per square meter.

yeah u r right but i am not getting anything to figure out.

It is money required to paint one square meter.

 Rupees is the currency of India.

Oh!

Don't tell me to prove this but:

Since the area of the walls don't matter on the different lengths of the walls (1x250, 50x50 etc.) we can easily place any values as the lengths.

Let's set the length and width at 50. Since the walls are 6 meters high, the area of each of the four faces are $ 50\cdot 6\equal{}300$. The total area is then $ 300\cdot 6\equal{}1800$.

So we know that it costs one "Rupee" for 12 square meters. $ 1800/12\equal{}150$

The answer is $ \boxed{150\text{ rupees}}$

can anybody explain with a diagram in a better manner.

@dojo You Are wrong  50x50=2500 not 250.

Dojo, since the inner perimeter and the height are given, we assume that the problem means "lateral surface area" when is says "walls". Further, the inner perimeter of a room with wall lengths 50 is 200, not 250. Thus, we need to take $ 250 \cdot 6$ to find the area of the walls. Dividing this by 12, we get $ \$125$.

well, since the perimeter of the figure is 250, then the sum of the sides is 250/2, or 125. The height is 6, so 125-6 is 119, or the width. Multiplying, you get the area to be 714 sq. meters. Multiplying by the currency gets you 8568 rupees.

The solution would have been changed if I knew what the "inner perimeter" was. Please verify or change your post. I just substituted "perimeter" for "inner perimeter"

wat r u all talking about? u all r taking own values and explaining.

I got a different answer.

First, to avoid complicating things, I'll pretend the room has a square base.

$ \frac{250}{4} \equal{} 62.5$. This gives us the length (in meters) of the sides of the square.

$ 62.5 \cdot 6 \equal{} 375$. 6 is the height, so multiplying length by width gives us area. This is only for a single wall.

As there are four walls, the total number of square meters $ 375 \cdot 4 \equal{} 1500$ square meters.

Now we make sure we answer the problem. Since the rate is 12 Rupees per square meter, the number of Rupees is $ 1500 \cdot 12 \equal{} \boxed{18000}$.

I'm sure I made some stupid mistake somewhere.

I agree with googology, I got $ \boxed{18000}$ rupees, too.

I mean, the lengths don't matter, the total area will always be $ Ph$ where $ P\equal{}$ perimeter and $ h\equal{}$ height.

And the units are given in meters, too. \[ 250\cdot6\cdot12\equal{}\boxed{18000}\].

Thank you  akhil and googology101.now I understood it better.

Lateral Surface Area of any prism = Perimeter of the base x height.

We know that, Cuboid is a rectangular prism

Lateral Surface area of room = perimeter x height
                                       = 250 x 6
                                       = 1500 sq. mt.

Area of the four walls of room = 1500 sq. mt

Total cost of painting the walls @Rs.12 per sq mt = 1500 x 12
                                                                   = Rs. 18000

basically, you can only solve it if your smart. :rotfl:


search
function
MATHCOUNTS
FTW
geometry
3D geometry
AMC

i need help.
how can i practice for getting to the nats?
i thought i was good...
then i got 40th in state (my team got 12th). :(
also, is there any way i can practice to get faster for the countdown round?
 :starwars:   :bruce:  :ddr:
sorry. i was bored.

Countdown Round: One word.  FTW.

You should practice by taking previous States, I find this helpful.

And also, I am currently working on a Mock MC State, for those who want to practice on new, fresh problems.

technically, thats three words.
anyway, can anybody else help me?

Hint: Why does AoPS have a search function?

hint: thanks for making me feel stupid.

edit: a search turned up nothing i wanted.

What did you search? :wink:

Try "Mathcounts tips".

What state are you from?
If you find most problems to be difficult and you don't have an idea of how to start or approach them, you should get Volume 1. Most Mathcounts problems can be solved by using methods similar to many of the ones used in Volume 1.

On the other hand, if you're finding it hard to finish in time or just want more practice, you should do many, many problems. This can also help with finding how to start problems, as many problems will be similar to old ones. Also, doing lots of problems familiarizes you with techniques and you will be able to solve many more problems. That being said, if you're preparing specifically for Mathcounts, I recommend doing old Mathcounts problems :P  If you don't have many, ask your coach, go on FTW, or go on [url=http://mathcounts.saab.org/mc.cgi]this website[/url] and do a lot of problems. If you end up finishing all the problems, which is hard to do considering the many problems there are on Saab, you can do problems from other competitions. The AMC competitions are great, except for the fact that they're multiple choice.

Basically, if you live in a weak to decent state, doing the above *well* should be quite helpful. If you live in a strong state, such as Texas, then I don't really have much advice :P [/hide]

For AMC's, just try doing the problems without looking at the choices.

whats the point of countdown round i thought it was just for fun..

In some states it counts...some it's just for fun.

Anyone know if it counts in california?

Mathcounts....

I didn't know that word until 6th grade. My general math team coach told all of us to come after school one day to take a test. COOL I said, MT rocks. I got there on test day, and read the front page of the test... Mathcounts 2004 Chapter Sprint Round.

WTF?!?!?!? What is this???

I took the test and ended up tying an 8th grader for first. The top 12 of our math team moved on to 'the next round'.... basically I went to extra practices after school. We took a lot more tests, and she announced the top 4 and the next 4. I got 3rd in the top 4, but I STILL didn't understand. Coach explained to me that I was on the official school team and the format of Mathcounts. Chapter Competition is a Saturday a few months from now, in our cafeteria.

I don't want to! There aren't any other 6th graders going! I'm scared.

We practiced Countdown and I found out I was good. I LIKE this, I said.  Chapter day: I tie for 10th, but lose. No CD for me :( My team got 3rd, thus making it to state. DUDE I want to make NATIONALS. But I didn't want to prepare, so I didn't past what was mandatory. Went to state, 10 minutes from my house. Our team gets third, I got 22nd. :( No nationals for me.

7th grade: I was READY and I made the school team again. This time, I practiced my *** off, taking around two test home a day and doing them. My coach says I've probably taken every test she has. Chapter: 5th place! AND I won CD! I felt good. I joined AoPS, took EVEN MORE TESTS, and prepared for state.

State: On the opposite of the state, I have to stay in a hotel which I hate, I don't eat a good breakfast, and I have no charms or anything. Great start to state day. **testing... waiting.... more waiting....**

Calling out names for CD, 10 names down, 2 left.... I haven't been called.... DYNAMO729 **phew** its all over. Uh oh, i still have CD left! Won that, and then found out I was in the top 4! My HOURS upon HOURS had paid off! I was going to Nationals! (I did not my best at Nationals, but hey I still made it.)

------------------------------------------------------------------------------------------------------

WHAT is the MORAL of my OVERLY LONG story?!?!?!? If you really want to make Nats, do this:

1. Practice, practice, PRACTICE. Even if you have the most natural ability of any other competitor, if you don't practice, you don't win.
2. Don't think good or bad of your ability. Simply try to improve it.
3. Don't give up.
4. Don't invest time in getting lucky, because you have no control over that. Practice at getting GOOD.

Sorry for the longness, I just have a different way of getting to the point :)

Anyways, good luck to you, and I hope to see you at Nationals this year.

[quote="1=2"]I'm pretty sure it counts in California.

It doesn't count in NY, though.[/quote]
Considering that California has two competitions, one in the north and one in the south, countdown round being official wouldn't really be possible. Hence, it's unofficial and probably held once in the north and once in the south.

ok....
so ill practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice and practice....
and then i'll be seeing you all in the nats! (i don't see myself as overly confident. i see myself as optimistic.)
dynamo729, you were probably the most helpful. i KNOW i will be seeing you there.

[quote="asterday1"](i don't see myself as overly confident. i see myself as optimistic.)
dynamo729, you were probably the most helpful. i KNOW i will be seeing you there.[/quote]

Great! I think thats a lot of the reason most people aren't optimistic, is because it might be takes as swagger or bigheadedness.

Glad to help. Hah, only if I make it too :P

The MC book of "The Greatest MC problems" is amazing.
It helped me tremendously!

what i would also do is take 3 warmups (from the school handbook), and work on them as a sprint round. if you feel as though you are not that comfortable with the problems, try earlier ones like w1, w2, and w3 as a test, as you get better, you can start doing higher ones, or you can mix it up.

Here are my suggestions:

1. Practice!

I know this has been said a hundred million and fifty times, but that's because it's true. You will never succeed if you don't practice. On the Minnesota MathCounts site you can find a bunch of chapter, state, and school tests. Also, do FTW - it's help with speed and is a great resource for problems. Get someone to trade tests with. And why stop there? Write up your own test, trade with a friend, and solve those. Even solving your own problems can be a nice  challenge. You're on AoPS, so that's a good start.

2. Read, read, read!

AoPS books are fantastic. I know they are very expensive, but see if you can borrow from someone. I borrow all my AoPS books from one of my friends. Make sure you do ALL the problems, and reread a million times if necessary.

3. Understand, don't memorize

It's good to know formulas, but make sure you actually understand them. Know how to derive them as well as use them. For example, you probably know the sum of the first $ n$ integers is $ \frac{n(n\plus{}1)}{2}$. But do you know how to derive it? How to prove it? Make sure you don't blindly memorize things.

4. Memorize a certain amount

Know your squares from $ 1^2 \sim  30^2$ and cubes from $ 1^3 \sim  15^3$. Make yourself familiar with useful formulas(but see #3!). Use the search function - there was once an excellent thread with some 30 pages on formulas, along with fantastic links. See if you can find it.

5. Google is your friend

You can find an amazing amount of fantastic resources with any search engine. Try "MathCounts", "MathCounts tips", "MathCounts tests", and anything else you can think of. I found a lot of great stuff.

6. Use the search function on AoPS

As math154 said, try "MathCounts tips" and other phrases - I found lots and lots of things you're looking for.

Oh, and good luck! Hope to see you at Nationals(if I make it!)

[quote="isabella2296"]Here are my suggestions:

1. Practice!

I know this has been said a hundred million and fifty times, but that's because it's true. You will never succeed if you don't practice. On the Minnesota MathCounts site you can find a bunch of chapter, state, and school tests. Also, do FTW - it's help with speed and is a great resource for problems. Get someone to trade tests with. And why stop there? Write up your own test, trade with a friend, and solve those. Even solving your own problems can be a nice  challenge. You're on AoPS, so that's a good start.

2. Read, read, read!

AoPS books are fantastic. I know they are very expensive, but see if you can borrow from someone. I borrow all my AoPS books from one of my friends. Make sure you do ALL the problems, and reread a million times if necessary.

3. Understand, don't memorize

It's good to know formulas, but make sure you actually understand them. Know how to derive them as well as use them. For example, you probably know the sum of the first $ n$ integers is $ \frac {n(n \plus{} 1)}{2}$. But do you know how to derive it? How to prove it? Make sure you don't blindly memorize things.

4. Memorize a certain amount

Know your squares from $ 1^2 \sim 30^2$ and cubes from $ 1^3 \sim 15^3$. Make yourself familiar with useful formulas(but see #3!). Use the search function - there was once an excellent thread with some 30 pages on formulas, along with fantastic links. See if you can find it.

5. Google is your friend

You can find an amazing amount of fantastic resources with any search engine. Try "MathCounts", "MathCounts tips", "MathCounts tests", and anything else you can think of. I found a lot of great stuff.

6. Use the search function on AoPS

As math154 said, try "MathCounts tips" and other phrases - I found lots and lots of things you're looking for.

Oh, and good luck! Hope to see you at Nationals(if I make it!)[/quote]
*GASP* Does it really help to memorize THAT MANY squares and cubes????? :o and izzy is that what makes you so beastly at FTW?

Hmmm... from what I heard, I think you only need to mem cubes up to $ 12$.  But $ 15$ does help.

Yeah, she has just played many games. [i](currently 4050?)[/i][/i]

[quote="isabella2296"]I've seen a problem where a knowledge of what $ 14^3$ is was very helpful. Just to be safe, memorize a few cubes/squares than you deem necessary. :)

Oh, and I'm not beastly at FTW. I'm actually pretty bad.

EDIT: Actually 4026. :)[/quote]
I just think you are because you have a rating >1600 and im only at 1276 :(

would you mind telling me a couple formulas I should program in my calculator. (besides volume, area, Pythagorean, hero's, and a decimal to binary convertor.) Are there also any specific apps i should put on it? ( I have a TI-84 plus)
 :juggle:  :play_ball:  :starwars:

And could someone tell me how to even make a program?  or download one?  would i have to install linux on my calc?

[quote="ernie"]No... you wouldn't.

Actually, it depends.  What type of calculator?

You can only program on Graphing Calculators.

You do not need an operating system, the operating system on your calc is what you need.

Oh, and I would suggest to make a triangle solver.  :P[/quote]

What do you mean by triangle solver?

If you need help programming ur calcs ASK [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231004]here[/url]

ask on that link that AIME posted

we have started a calc prog thread...

okay well it helps a lot if you change the way you study.

instead of just doing contests year by year, try studying subject by subject.

you get much better fundamentals, and good fundamentals means more consistent scores. then after you reach say, a consistent 35-40 on state tests then you should start fine tuning and start concentrating on minimizing your mistakes

and btw, calculator programs are often more negative effect than positive.   and i'm not just talking about the "learning' aspect either.

once you use programs,  on target, instead of looking for the best way to do a problem, you will start trying to find ways in which you can use the programs. often times, this new "advantageous" (because you can make use of your super-cool-amazing-jaw-dropping program) way is faulty and you will instead get the wrong answer.

you might not believe me, but we math people think in weird ways.

[quote="ajain"][quote="asterday1"]i will ask again.
What programs should i make for MC
and
what apps should i download?[/quote]

YOu should probably get "cabamap".  THe program's digit limit is really high, so you could do 423479847 squared without having an E sign.  To download it, you could probably google it.[/quote]

also mirage OS, especially if you have to take school math classes.

These are useful (TI-83 family).

[url=http://www.ticalc.org/archives/files/fileinfo/323/32326.html]Cabamap[/url]--works with large integers.
[url=http://www.ticalc.org/archives/files/fileinfo/400/40005.html]Noshell[/url]--once activated, you can access archived programs without unarchiving them.
[url=http://www.ticalc.org/archives/files/fileinfo/226/22626.html]Omnicalc[/url]--lots of little things that can help. Make sure to start this before starting Noshell.
[url=http://education.ti.com/educationportal/sites/US/productDetail/us_poly_83_84.html]Polynomial Root Finder and Simultaneous Equation Solver[/url]--goes up to 10th degree/10 variables.

EDIT:
As mewto said, get MirageOS or CrnchyOS or both (I usually prefer CrnchyOS because it does not Garbage Collect).
It seems they have made [url=http://www.ticalc.org/archives/files/fileinfo/148/14876.html]Phoenix 4.0[/url]. :)

Holy guacamole now I need to find my link cable. You know what they say about the Texas team on the Team Round right?

They calc bash?  IDK.  what do they say bout them?

[quote="mewto55555"]They spend the first 18 minutes playing games then start on the problems. They always get 10s.[/quote]

What they really should do is to start on the problems, then spend the 18 minutes playing games.

[quote="CA Math"]The one thing that helps the most is doing a lot of problems.
Experience and familiarity with　the problems helps you solve them with some more ease.[/quote]

Yes, like doing old mathcounts problems.


inequalities function calculus derivative LaTeX

Prove that $\displaystyle x^4 > x -\frac 12$ for all real numbers $x$.

isn't this a little too easy? :?

the minimum of the function x^4 - x is \sqrt (1/4)

and for this value it's still true... :?

Let me rephrase it: ... WITHOUT using derivatives :)

It was a problem in the Russian olympiad for the youngest participants,
and they certainly do not know derivatives.

well just want to test some more latex if you don't mind :D:D

\[ x^4 -x + \frac 12 > 0 \Leftrightarrow x^4-x^2 + \frac 14 + x^2-x+\frac 14 > 0 \Leftrightarrow \]
\[ \left(x^2- \frac 12\right)^2 + \left(x - \frac 12\right)^2 >0 \] which is true as $x$ cannot be simultaneously equal with $\displaystyle \frac 12$ and $\displaystyle \pm \frac 1{ \sqrt 2}$.

[quote="Arne"]Let me rephrase it: ... WITHOUT using derivatives :)[/quote]

Now THAT makes it an olympiad problem ;-)

?

I see no "a" in the question...

can you please explain your solution?

yep, indeed that's also a solution.

You'll even get a better approach for its minimal value (same result as derivates approach), but that is not really the question here I think :)


Triangle $ ABC$ is an isosceles right triangle with the measure of
angle $ A$ equal to 90 degrees. The length of segment $ AC$ is 6 cm.
What is the area of triangle $ ABC$, in square centimeters?

We know that $ c \equal{} a\sqrt2$, where $ c$ is the hypotenuse and $ a$ is the side.  Thus, $ a \equal{} \frac{6}{\sqrt2}$ or $ 3\sqrt2$.  The area would simply be $ \frac{9\sqrt4}{2} \implies \boxed{9}$.

[quote="ernie"]We know that $ c \equal{} a\sqrt2$, where $ c$ is the hypotenuse and $ a$ is the side.  Thus, $ a \equal{} \frac {6}{\sqrt2}$ or $ 3\sqrt2$.  The area would simply be $ \frac {9\sqrt4}{2} \implies \boxed{9}$.[/quote]

Given that the problem says that A is where the right angle is, wouldn't that make AC the leg of 6 and not the hypotenuse? (Knowing one leg of an isoceles right triangle is the same as the other, we have both the altitude and base of the triangle as 6.) By 1/2(b)(h), we would have (1/2)(6)(6)= [b]18 square units.
[/b]
An image of the triangle I refer to can be found here:
[img]http://www.testdesigner.com/imgs/Triangles/200x200/right_abc_rotation01.gif[/img]


The sequence $\{a_{n}\}$ is defined as $a_{1}=1$ and $a_{n}=\frac{n}{a_{n-1}}$.
Find $\lim_{n\to\infty}\frac{a_{n}}{\sqrt{n}}$
1) for even $n$;
2) for odd $n$.

I'm not sure that there is a limit, but they appear to be approaching values that are reciprocals of each other.

Well, by some observation $a_{n}=\frac{n(n-2)(n-4)\ldots}{(n-1)(n-3)\ldots}$.  I don't know  what that converges to, of it it even converges, but from that we divide by $\sqrt{n}$ to get $\frac{\sqrt{n}}{n-1}\frac{(n-2)(n-4)\ldots}{(n-3)(n-5)\ldots}= \frac{\sqrt{n}}{n-1}a_{n-2}$

So if the individual $a_{n}$ converge to some real number, the entire thing will approach $0$.  But then again, the $a_{n}$ might be diverging, which would change the problem a bit.

[hide]
To simplify things, we can set $b_{n}=\frac{a_{n}}{\sqrt{n}}$, then
\[b_{n}b_{n-1}=\sqrt{\frac{n}{n-1}}\]
That explains the part that $b_{2k}$ and $b_{2k+1}$ approach reciprocals of each other as $k$ gets arbitrarily large.

Now we only need to consider the limit for odd $n$. If $n=2m+1$, we have
\begin{eqnarray*}a_{2m+1}&=&\frac{3}{2}\cdot \frac{5}{4}\cdot \frac{7}{6}\cdot \cdots \cdot \frac{2m+1}{2m}\\ a_{2m+1}^{2}&=&\frac{3}{2}\cdot \frac{3}{2}\cdot \frac{5}{4}\cdot \frac{5}{4}\cdot \cdots \cdot \frac{2m+1}{2m}\cdot \frac{2m+1}{2m}\\ &\geq &\frac{3}{2}\cdot \frac{4}{3}\cdot \frac{5}{4}\cdot \frac{7}{6}\cdot \cdots \cdot \frac{2m+1}{2m}\cdot \frac{2m+2}{2m+1}\\ &=&m+1\\ b_{2m+1}&\geq &\sqrt{\frac{m+1}{2m+1}}\\ &>&\frac{1}{2}\\ \end{eqnarray*}
But also, from our formula for $b_{n}b_{n-1}$, we can get
\[b_{2m+1}=\frac{\sqrt{(2m+1)(2m-1)}}{2m}b_{2m-1}<b_{2m-1}\]
So $b_{2m+1}$ is decreasing and bounded below, so it converges to [b]something[/b].

As to the value to which it converges...  :maybe:
[/hide]


logarithms
function
calculus
calculus computations

How should i do this equation?

$x^2(e^x-6)=0$

The equation isn't made easier with the $e^x$ and the $x^2$ in the same equation.

Thx in advance.

Indeed. It's easier when it's factored like that. Either $x^2=0$ or $e^x-6=0$. Solve each separately.

Thx for your help on that one. With this solution as a startup i didn't find many problems in the next handful of simular equations.

But how would you prove this equation doesn't intersect with the X-axis?

$e^{2x}+2e^x+1=0$

I can easily see it but i don't know how to prove it mathematically.

Simply put $y=e^x$ clearly $y>0$ the you get $(y+1)^2=0$  which doesn`t have postive roots.

OK, great. I have a problem with getting two results in this one though:

$e^{2x}<3e^x-2$

$e^{2x}-3e^x+2<0$

$2x-3x+ln(2)<0$

$-x+ln(2)<0$

$x>ln(2)$

... but i know there is another result: $x<0$. Again i have a problem showing it mathematically.

It pains me to see the way you manipulated logaritms .Are you sure you know how to logaritmate? I don`t think so  :P  Just a few reminders $\ln ab=\ln a+\ln b$... Plus the fact the function $\ln$ is not defined in the point $0$! Try not to use stuff you are not sure of.
Your inequation is one of the second degree: take $y=e^x$ you`ll get  $y^2<3y-2$ which is equivalent with $\left ( y-\frac{3}{2} \right ) ^2 <\frac{1}{4}$ from where $(y-2)(y-1)<0$ so $1<y<2$ equivalent to $1<e^x<2$ now you can use logaritms an you get $0<x<\ln 2$.

I have said it before, but ok, you didn't heard that, so let me tell you. My textbook is very lacky if you ask me :(

Quite hard to follow you there i have to say, but when i transformed $e^x$ into $y$ i could easily do a $y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

OK, this should be the last question for today.

$(ln(x))^2-3ln(x)=0$

$-x^3+x^2-1=0$

In case this is right, and i finally understood something :lol:, how would you find the intersection with the x-axis? My guess, from looking at the graph on my calculator, is that it is an irrational number. Have tried putting in a few numbers and guess where it intersects.


function
complex analysis
algebra
domain
complex analysis unsolved

Let $u: C\rightarrow R$, prove that $u$ is harmonic (subharmonic) function iff for all holomorphic function $f: C\rightarrow C$ then $u_{o}f$ is harmonic function(respectively, subharmonic function)

Let D and D' be domains in $\mathbb{C}$.
Let $f: D'\to D$ be horlomorphic and $u%Error. "inHA" is a bad command.
(D)$
We consider the point $z'\in D'$ and the imagine is $f'(z')$.
Choose $\varepsilon >0$ enough small such that:
$f[B(z',\varepsilon)]\subset B\left(f(z'),R)\right)\subset D$.
Because of $u$ is harmonic on D then $u$ is harmonic on $B\left(f(z'),R)\right)$. Hence there exist the function $g$ is horlomorphic on $B\left(f(z'),R)\right)$ such that $u=Reg$. In that on $B(z',\varepsilon)$ we have
$u_{o}f=Re(g_{o}f)$ and therefor $u_{o}f$ is harmonic on this one.


limit
real analysis
real analysis unsolved

Let $A,B$ be subsets of the positive integers such that $B\subset A$ and
\[\sum_{a\in A}a^{-1}= \infty \]

Moreover impose the condition that $B$ has positive relative density in $A$, that is, for some $t>0$,
\[\lim_{n\to\infty}\frac{|B\cap\{1,\ldots,n\}|}{|A\cap\{1,\ldots,n\}|}= t \]

Is it then true that
\[\sum_{b\in B}b^{-1}= \infty \]
?

Let $a_{n}=\#\{a\in A\colon 2^{n}\le a<2^{n+1}\}$, and similarly for $b_{n}$. Observe that $\sum_{a\in A}a^{-1}=\infty$ iff $\sum_{n}2^{-n}a_{n}=\infty$. The density assumption implies $\lim_{n\to\infty}\frac{b_{n}}{a_{n}}=t$. And it looks like I never used $B\subset A$.  :maybe:

But what if we impose a weaker density assumption
\[\liminf_{n\to\infty}\frac{|B\cap\{1,\ldots,n\}|}{|A\cap\{1,\ldots,n\}|}>0\]
instead?

[quote="mlok"]And it looks like I never used $B\subset A$.[/quote]

It isn't necessary in my proof either (which uses the Cauchy condensation test and is more or less the same as yours) but it's natural to impose the condition that B is 'inside' A if you're looking at B as having positive relative density to A.

I'm not sure as to what the answer to your problem is.

[quote="mlok"]But what if we impose a weaker density assumption
\[\liminf_{n\to\infty}\frac{|B\cap\{1,\ldots,n\}|}{|A\cap\{1,\ldots,n\}|}>0 \]
instead?[/quote]
Note that $\sum_{a\in A}a^{-1}=\sum_{n\ge 1}\frac 1{n(n+1)}|A\cap\{1,\ldots,n\}|$. Should I say more?

That's a neat solution. :)  There must be a counterexample with $\limsup_{n\to\infty}\frac{|B\cap\{1,\ldots,n\}|}{|A\cap\{1,\ldots,n\}|}>0$, but probably not a simple one.

[quote="mlok"]... a counterexample with $\limsup_{n\to\infty}\frac{|B\cap\{1,\ldots,n\}|}{|A\cap\{1,\ldots,n\}|}>0$, ...[/quote]

$A_{k}= N \cap [k 2^{k},(k+1) 2^{k})$
$A = \bigcup_{k=1}^{\infty}A_{k}$
$B = \bigcup_{k=1}^{\infty}A_{k^{2}}$


If (2x)^3 = .√128 and x = 2^(1/n), what is the value of n?

 :o

[quote="FullMetalNaruto"]If $ (2x)^3 \equal{} \sqrt {128}$ and $ x \equal{} 2^{\frac 1n}$, what is the value of $ n$?[/quote]

[hide="Solution"]
$ 128\equal{}2^{7}$

$ 2^3x^3\equal{}2^{\frac 72}$

$ 3\lg x\equal{}\frac 72\minus{}3\equal{}\frac 3n\implies \boxed{n\equal{}6}$
[/hide]

EDIT: Darn, beaten.


induction
pigeonhole principle
combinatorics unsolved
combinatorics

Let $n$ be positive integer, set $M = \{ 1, 2, \ldots, 2n \}$. Find the minimum positive integer $k$ such that for any subset $A$ (with $k$ elements) of set $M$, there exist four pairwise distinct elements in $A$ whose sum is $4n + 1$.

There exist the generlized problem in the imo shotlists (1995-2001, it is in a book so i dont know which year :) ) . and it is this :

Let $n,m,k$ be natural numbers such that $ \ 1< n \le m-1 \le k$) .find the maximum of the numbers of the subset $S$ form the set $\{1,2,3,...,k\}$ which sum of the each $n$ elements of $S$ dont be equal to $m$.

And i am going to check for the answer of it  :)

:read: Well we can easily show that the answer is $\le k - \alpha$ where $\alpha $ is the biggest natural number such that is $\le \frac{m}{n}-\frac{n-1}{2}$ and by my supprise the answer is exactly $k - \alpha$ and the proof is by the induction :what?:  :what?:  :what?:  :!: on the $n$ .
and the proof is highly complicated for me at least :wacko:

$k=n+3$ will do.
 to start with note that the set $\{n-1,n,n+1,n+2,\ldots,2n\}$ is a set of $n+2$ elements and has a minimum sum (with $4$ distinct elements) of $n-1+n+n+1+n+2=4n+2$, so $k\geq n+3$.also for a set of size atleast $n+2$ we can form pairs of distinct elements that sum to $2n$ and pigeonhole $\Rightarrow$ there exists a pair of distinct elements with sum $2n$.once we remove such a pair from the $n+3$-set and we are still left with an $n+1$ set from which another pair,this time with sum $2n+1$ (again pigeonhole) can be extracted and so we are through.

a minor adjustment (your proof is correct):

After you have chosen a pair that adds up to 2n. Now you need to subtract 4, not 2, elements from the k. (because say a+b=2n, not only will we not be able to use a and b in forming a pair with sum 2n+1, we can't use 2n+1-a and 2n+1-b either.) But now we have effectively narrowed now the number of valid pairs that add up to 2n to just n-2, which means another pigeonhole indeed gets us home. (So you see, it's not because we have 2n+1 elements and n possible pairs, but cuz we have 2n-1 elements and n-1 possible pairs.)  :D

well any subset of $\{1,2,\ldots,2n\}$ of size $n+1$ contains a pair of distinct elements which sum up to $2n+1$.this is independent of what i do before getting to that $n+1$-set. the more important thing is to get the sum to $2n$ first because choosing $n+1$ elements from $\{1,2,\ldots,2n\}$ will not guarantee $\textbf{two}$ distinct elements to add up to $2n$, for eg:$\{1,2,\ldots,n-1,n,2n\}$.


geometry
trapezoid
trigonometry
angle bisector

Not too bad(once you see it), but I like it. :)

In triangle ABC, the internal angle bisector of angle B and the external angle bisector of angle C intersect at E.  A line is drawn through E parralel to BC interscting the triangle at points F and G, with F on AC and G on AB.  If FC= a and GB = b, find the length of FG in terms of a and b.

Drop normals from the points $E, F, G$ to the line $BC$ and let $E', F', G'$ be the feet of these normals. Since the lines $EF \equiv EG \parallel BC$ are parallel, $EE' = FF' = GG' = h$ and $F'G'  = FG$. The angle $\angle EBE' = \frac{\angle B}{2}$, because $BE$ is the internal bisector of the angle $\angle B$, and the angle $\angle ECE' = 180^o - \frac{\angle C}{2} - 90^o = 90^o - \frac{\angle C}{2}$, because $CE$ is the external bisector of the angle $\angle C$. Then

$h = FF' = FC \sin{\widehat C} = a \sin{\widehat C},\ \ h = GG' = GB \sin{\widehat B} = b \sin{\widehat B}$

$h = EE' = E'B \tan{\frac{\widehat B}{2}} = E'C \tan{\left(90^o - \frac{\widehat C}{2}\right)} = \frac{E'C}{\tan{\frac{\widehat C}{2}}}$

WLOG, assume that $\angle B < \angle C$. Then $b > a$ and the points $B, C, E'$ follow on the line $BC$ in this order.

$BC = E'B - E'C = \frac{h}{\tan{\frac{\widehat B}{2}}} - h \tan{\frac{\widehat C}{2}} =$

$= \frac{h \cos{\frac{\widehat B}{2}}}{\sin{\frac{\widehat B}{2}}} - \frac{h \sin{\frac{\widehat C}{2}}}{\cos{\frac{\widehat C}{2}}} = \frac{2h \cos^2{\frac{\widehat B}{2}}}{\sin{\widehat B}} - \frac{2h \sin^2{\frac{\widehat C}{2}}}{\sin{\widehat C}} =$

$b (1 + \cos{\widehat B}) - a (1 - \cos{\widehat C}) =  b - a + b \cos{\widehat B} + a \cos{\widehat C} =$

$= b - a + BG' + CF' = BG' + F'G' + CF'$

As a result, $FG = F'G' = b - a$. If we assumed $\angle B > \angle C$, it would be $a > b$, The points $C, B, E'$ would follow on the line $BC$ in this order and we would similarly obtain $FG = a - b$. Combining these 2 results yields $FG = |a - b|$.

Very nice!  I think all of that works, and that is what I got, although I am pretty sure there is a bit easier method.

[hide="Hint"]
Look for isosceles triangles...
[/hide]

WLOG, assume that $\angle B < \angle C$, hence $a < b$

The angles $\angle GEB = \angle EBC = \angle EBG$, the 1st pair because the transversal $EB$ cuts the parallel lines $EG \parallel BC$, the 2nd pair because $BE$ bisects the angle $\angle GBC \equiv \angle ABC$. Hence, the triangle $\triangle EGB$ is isosceles with $GE = GB = b$.

The angles $\angle FEC = 180^o - \angle BCF = \angle ECF$, the 1st pair because the transversal $EC$ cuts the parallel lines $EF \parallel BC$, the 2nd pair because $CE$ bisects the angle $180^o - \angle BCF \equiv 180^o - \angle BCA$. Hence, the triangle $\triangle EFC$ is isosceles with $FE = FC = a$.

$FG = GE - FE = b - a$

Similarly, if $\angle B > \angle C$ and $a > b$, we get $GF = FE - GE = a - b$. As a result, $FG = |a - b|$. Nice.

Good solution, yetti.  Indeed it is nice, one of my favorites.  :)


inequalities
function
inequalities unsolved

Let $ a,b,c,d\geq0$ s.t. $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$.Prove that:
$ a^3\plus{}b^3\plus{}c^3\plus{}d^3\leq8$

It suffices to prove that
$ \left(\sum a^3\right)^{\frac {1}{3}}\leqslant \left(\sum a^{2}\right)^{\frac {1}{2}}$.
Let $ x \equal{} a^{2}$ etc. Theinequality is equivalent to
$ \sum x^{\frac {3}{2}} \leqslant \left(\sum x\right)^{\frac {3}{2}}$
which follows from the Karamata inequality for the convex function $ f\left(u\right) \equal{} u^{\frac{3}{2}}$ and $ \left(x,y,z,t\right)\prec\left(x \plus{} y \plus{} z \plus{} t,0,0,0\right)$.

I just have seen that this inequality is too easy. :)
Squaring the both sides weneed to prove that:
$ (a^3\plus{}b^3\plus{}c^3\plus{}d^3)^2\leq (a^2\plus{}b^2\plus{}c^2\plus{}d^2)^3$.
Using $ CBS$ we infer that:
$ (a^3\plus{}b^3\plus{}c^3\plus{}d^3)^2\leq (a^2\plus{}b^2\plus{}c^2\plus{}d^2)(a^4\plus{}b^4\plus{}b^4\plus{}d^4)$.So we need only to prove that:
$ a^4\plus{}b^4\plus{}c^4\plus{}d^4\leq(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2$, which is easy.

[quote="Inequalities Master"]Let $ a,b,c,d\geq0$ s.t. $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$.Prove that:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3\leq8$[/quote]
We have :
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$
$ 0\leq a,b,c,d \leq 2$
$ 0 \leq a^3 \leq 2a^2 , 0 \leq b^3 \leq 2b^2 , 0 \leq c^3 \leq 2c^2 ,0 \leq d^3 \leq 2d^2$
$ 0\leq a^3\plus{}b^3\plus{}c^3\plus{}d^3 \leq 2(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\equal{}8$

[quote="conan_naruto236"][quote="Inequalities Master"]Let $ a,b,c,d\geq0$ s.t. $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$.Prove that:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3\leq8$[/quote]
We have :
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$
$ 0\leq a,b,c,d \leq 2$
$ 0 \leq a^3 \leq 2a^2 , 0 \leq b^3 \leq 2b^2 , 0 \leq c^3 \leq 2c^2 ,0 \leq d^3 \leq 2d^2$
$ 0\leq a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \leq 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \equal{} 8$[/quote]

 Nice solution :)


It completely depends on the person. Someone else with more specific info should comment, but I'm sure math competitions inspire thousands, and many continue with mathematics or related fields in their careers. Also remember real-world applications of math are often quite different than, say proving two triangles are similar in 5 minutes. Perhaps a better question to ask regards IMO, or other competitions of higher level than AMC. The 8, 10, and 12 are... well not super easy, it depends on the person, but probably not the most "prominent" students necessarily, even with perfect scores.

As certain as mathematics competitions inspired people to continue in anything related in it, there have been people who were disgusted by math because of competitions.

[quote="CEM_College_Champ"]I wonder what happened to prominent AMC8/10/12 participants... did some of them become mathematicians themselves, like some IMO medalists?

Or their fates are like that of some other IMO medalists, burning out after either middle or high school without getting anywhere useful?

Or their lives reverted back to normalcy once in college?[/quote]

Not so much AMC prominent, but MathCounts prominent here - so I guess similar to AMC8.

I feel like if you graphed my math ability vs. time from age 0 - 14, extrapolated one curve for continued progress, and another for burnout, I'd fall directly in-between. $ \frac{f_0(x)\plus{}f_1(x)}{2}$, if you will. From what I know about other MathCounts medalists, most of them end up like that; often, they end up focusing on something other than math.

I'm assuming that a similar phenomenon occurs for AMC10/12 participants.


Is it true that $ e^{x^2}\plus{}x\geq e^{x}$ for all $ x$?

I know that; I was posting it as an exercise :)

Trying to keep the explanation at the level of a first calculus course ...

Let $ f(x)=e^{x^2}+x-e^x.$ Then $ f'(x)=2xe^{x^2}+1-e^x$ and ${ f''(x)=(2+4x^2)e^{x^2}-e^x}.$

Note that $ f(0)=f'(0)=0.$  We need to claim that $ f''(x)>0$ for all $ x.$

$ f''(x)>0$ iff $ (2+4x^2)e^{x^2-x}-1>0.$

$ e^{x^2-x}\ge e^{-\frac14}>1-\frac14=\frac34$ and $ (2+4x^2)\ge 2.$

Hence $ (2+4x^2)e^{x^2-x}\ge \frac34\cdot2>1.$

Since $ f''(x)>0$ for all $ x,$ it follows that $ f'(x)>f'(0)=0$ for $ x>0$ and $ f'(x)<f'(0)=0$ for $ x<0.$ The function $ f$ is decreasing on $ (-\infty,0]$ and increasing on $ [0,\infty)$, hence has an absolute minimum at $ 0,$ where its value is zero. Hence $ f(x)\ge 0$ for all $ x.$

I never did find the actual minimum value of $ f''(x).$ Who needs to be that exact? We got what we needed.


Where can I find romanian math olympiad year 1950-1990?
 Does anyone have it?
If there is,May I request to post in the forum?
 I really want to do it ,but I could't find anywhere.
thank you


geometry
analytic geometry
graphing lines
slope
MATHCOUNTS
FTW

What is the positive value of [i]m[/i] such that the triangle bounded by the lines
y=0
x=12
y=[i]m[/i]x
and has an area of 1872 sq units?
*the [i]m[/i] is the [i]m[/i] in slope intercept form, or the slope of the line of y=[i]m[/i]x
EDIT: Here's the answer. [hide]Answer:26[/hide]
I still don't understand how to get the answer, though.  :(

[hide]
312???

Just use $ A\equal{}\frac{bh}{2}$.
[/hide]

You're trying to find the slope of the hypotenuse, not the height...

[hide]
So you have a right triangle with vertices (0,0), (12,0), and (12, x)
Since the area is 1872, $ 1872 \equal{} (x*12)/2$, so $ x \equal{} 312$.
Now, the coordinates of the third vertex is (12,312), so the slope of the third line must be $ \frac{312}{12} \equal{} \boxed{26}$.[/hide]

where did u come across this problem?

I need help

Verizon, many MathCounts resources can be found on the yeshar.biz site, and the direct link to them can be found in the thread titled: "MathCounts Resources" (or something similar). I suggest you do Chapter/State sprints.

To make my post non-offtopic, alkjash's method is the easiest method I could think of, although there are more than one.

I also found it on FTW, and put 312 :roll:  I should start reading as well.

[quote="verizonwires"]so thats were andrewliu got these problems right?[/quote]
yeah- that's where I got them.... unfortunately, it only goes up to 2005.....

If the triangle is going to have an area of 1872 sq units, the height has to be 312, because the formula is b*h/2. If the height is 312, then the slope has to be 312/12, or the answer, [b]26[/b].

P.S. Please tell me if I am wrong.

[quote="alkjash"][quote="returnofkiladi"]So...like when ur teacher gives u a mathcounts test for class, u just memorize all the answers and do that.
Ur math coach wudnt be happy...
 :o  :(  :mad:[/quote]

I seriously doubt you could memorize a few thousand problems and their answers...[/quote]

Well, I'm not sure about "andrewliu"'s school, but I have MathCounts once a week on thursdays, and thus I can study like a couple of Sprints and Targets every day and I should have most of the questions in the bag. I mean, the coordinator of MC, might give a few hints about what he will give, like a previous school or chapter round. I mean, it's when they give u the new tests when you figure out if the scores were well deserved.

[quote="returnofkiladi"]
Well, I'm not sure about "andrewliu"'s school, but I have MathCounts once a week on thursdays, and thus I can study like a couple of Sprints and Targets every day and I should have most of the questions in the bag. I mean, the coordinator of MC, might give a few hints about what he will give, like a previous school or chapter round. I mean, it's when they give u the new tests when you figure out if the scores were well deserved.[/quote]

Never has anyone been able to just memorize answers to most of the previous mathcounts tests and gotten into the top 10 on the new one.  I think.

Duh.... Of course. There are all these losers who think they can get away with memorizing answers. It happens to be my pet-peeve!
 :mad:  :mad:  :mad:

[quote="returnofkiladi"]I think that andrewliu is an unethical person. Who's with me???[/quote]

I think you should take a nice, deep breath. Currently: You mood seems to be strangely similar to that of an angry mod--armed with a nice set of torches, pitchforks, and various other shiny objects. I'm fairly sure that andrewliu is not a wholly unethical person. Chillax.

seriously- sure, if you memorize it could increase your score on past tests- but, when you take an unseen test..... (your "respect level" probably goes down 20 points.........) I highly doubt memorizing is worth it.....


There is a rectangular sheet of mass $ M$ and length $ a$ and width $ b$.it is hung from a cieling using two identical inextensible,massless threads of length $ L$ .They are attached symmetrically to the sheet and are $ d$ distance apart.Now the sheet is moved in it's plane(the  threads, sheet and the point of attachment to cieling lie in one plane) a slight distance by keeping the threads tight.Find the time period of the small oscillations

Am I correctly assuming that the strings are attached to a single point on the ceiling? or is the arrangement something like the ballistic pendulum?

no they are attached at different points

Your question is confusing. Perhaps you should include a picture, if it's possible.

let me try for a picture.Can u tell me which part u have doubt in?

here is the picture probably will make things clearer.the pendulum is made to oscillate in it's plane

According to my calculations it is:
$ T \equal{} 2 \pi \sqrt{\frac{L}{g}}$

yes that's right [b]@petko[/b] Can u post ur solution

Ok, here it is:
[hide="solution"]
Rectangle's center of mass is moving on a circle of radius $ L$ (but it is not rotating!). Let the $ \varphi$ be the angle between thread and the perpendicular.
Kinetic energy of rectangle is then
$ E_k \equal{} \frac{M (L \dot \varphi)^2}{2}$
And the potential energy
$ E_p \equal{} \minus{} M g L \cos \varphi \plus{} C$
Because oscillations are small we can assume that $ \cos \varphi \approx 1 \minus{} \varphi^2/2$ hence
$ E_p \equal{} \frac{1}{2}MgL \varphi^2 \plus{} C'$
$ C,C'$ are some constants.
Because energy is conserved, therefore
$ \frac{\mbox{d}(E_p \plus{} E_k)}{\mbox{d}t} \equal{} 0 \iff M g L \varphi \dot \varphi \plus{} M L^2 \dot \varphi \ddot \varphi \equal{} 0 \iff \ddot \varphi \plus{} \frac{g}{L} \varphi \equal{} 0$
After solving this equations we get the answer - $ T \equal{} 2 \pi \sqrt{L/g}$
[/hide]


Perhaps you were at WSMC in Blaine this year.
And it's possible you wondered why there were a few teams labeled 'Northlake.'

That would be because a certain school in the North-of-Lake-Washington-area (hence the name) did/could not send a "faculty member" to Blaine.  This meant that the certain school in question could not be liable for the actions done by the students.  Since using the school name at Blaine would imply that the students were chaperoned by school faculty member(s), and this was not the case, the students had to:

a) Choose a new school name
OR
b) bring a school staff member.

Since b) would not happen (approximately 1% of the school faculty knows that a math team exists), a) was chosen.  Blaine co-coordinator Gail Wildermuth tagged the teams from the school in question as "Northlake," and that's the name the students went under for the competition.

Hmmm...I've never had that problem there...

Do you mind posting which individuals and teams won? The information doesn't seem to be online yet.

(I'm from Seattle, but I'm at a boarding school on the east coast right now, though I tend to be curious about what's going on back there.)

9-10 Team results (all I have):
1. Garfield 1 : 142
2. Lakeside 4 : 136
3. Newport 1 : 133.5
4. NWCS : 131.5
5. Interlake 1 : 128.5
6. North Lake : 127
7. Interlake 2 : 122
8. Jefferson 1 : 119.5
9. Newport 4 : 111
10. Lakeside 2 :108

And umm from the back of my memory (where all unmemorable events go :P ):
Individuals:
1. Andy Morgan : 39
2. Darryl Wu : 39

the Blaine website has been updated, information is [url=http://www.blainesd.org/mathchamps/high_school/HS09.html]here[/url]


lim         (sin x * sin y) / (x^2 + y^2)  = ?
  (x,y)->(0,0)

    Find the limit if exists or prove that it does not exist.[/list]


Let $ x\geq0$. Prove that

$ \sqrt{x}\plus{}\sqrt[3]{x}\plus{}\sqrt[6]{x}\leq2\plus{}x$

 :maybe:

Let $ \sqrt[6]x\equal{}a$. The inequality is equivalen to \[ (a\minus{}1)^2(a^4\plus{}2a^3\plus{}3a^2\plus{}3a\plus{}2)\ge0\] which is obviously true.

Since $ (1,0,0)$ majorises $ (\frac{1}{2},\frac{1}{3},\frac{1}{6})$
By Murihead inequality we obtain:

$ \sum_{sym}x^1.1^0.1^0\geq \sum_{sym}x^{\frac{1}{2}}.1^{\frac{1}{3}}.1^{\frac{1}{6}}$

$ \Leftrightarrow 2\plus{}x\geq\sqrt{x}\plus{}\sqrt[3]{x}\plus{}\sqrt[6]{x}$

 :D  :D  !!


If the value of Tan5X is known find the number of distinct values of CosX

The period of $\cos x$ is $2\pi$, while the period of $\tan 5x$ is $\frac \pi 5$. In addition, $\tan 5x$ is one-to-one function of x in the intervals $\left[\frac{k\pi}{5}, \frac{(k + 1)\pi}{5}\right)$, except for the midpoints of these intervals $\frac{(2k + 1)\pi}{10}$, where $\tan 5x$ is undefined. Thus in the interval $[-\pi, +\pi)$, $\tan 5\pi$ has 10 periods and it attains the same value y for 10 different arguments of

$-\pi \le x_{-5} < x_{-4} < x_{-3} < x_{-2} < x_{-1} < 0 \le x_0 < x_1 < x_2 < x_3 < x_4 < +\pi$

Since $\cos x$ is monotonically increasing in the interval $[-\pi, 0)$, the values of $\cos x_i,\ i = -5, -4, -3, -2, -1$ are different from each other. Likewise, since $\cos x$ is monotonically decreasing in the interval $[0, +\pi)$, the values of $\cos x_i,\ i = 0, 1, 2, 3, 4$ are different from each other. Assume that there exist 2 arguments $x_m, x_n$ (i.e., $-5 \le m < 0,\ 0 \le n < 5$), such that

$\tan 5x_m = \tan 5x_n\ \ \text{\&}\ \ \cos x_m = \cos x_n$

From the 1st equation,

$x_n = x_m + \frac{(n - m)\pi}{5} = x_m + \frac{k\pi}{5},\ 0 < k = n - m < 10$

Substituting this to the 2nd equation.

$\cos x_m =  \cos\left(x_m + \frac{k\pi}{5}\right) = \cos x_m \cos \frac{k\pi}{5} - \sin x_m \sin \frac{k\pi}{5}$

$\tan x_m = \frac{\cos \frac{k\pi}{5} - 1}{\sin \frac{k\pi}{5}}$

$\cos x_m = \pm\frac{1}{\sqrt{1 + \tan^2 \frac{k\pi}{5}}} = \pm \frac{\sin \frac{k\pi}{5}}{\sqrt{\sin^2 \frac{k\pi}{5} + \cos^2 \frac{k\pi}{5} - 2 \cos \frac{k\pi}{5} + 1}} = \pm \frac{\sin \frac{k\pi}{5}}{\sqrt{2\left(1 - \cos \frac{k\pi}{5}\right)}} =$

${= \pm \frac{\sin \frac{k\pi}{5}}{\sqrt{4 \sin^2 \frac{k\pi}{10}}} = \pm \frac{2 \sin \frac{k\pi}{10} \cos \frac{k\pi}{10}}{2 \sin \frac{k\pi}{10}}} = \pm \cos \frac{k\pi}{10}$

Since $x_m \in [-\pi, 0)$, this implies that our assumption can be true only if $x_m = -\frac{p\pi}{10},\ p = 1, 2, ..., 10$. But since $\tan 5x$ is undefined for $x = \frac{(2k + 1)\pi}{10}$, i.e., for an odd p, the possibilities are only $x_m = -\frac{p\pi}{10},\ p = 2, 4, 6, 8, 10$ or $x_m = -\frac{q\pi}{5},\ q = 1, 2, 3, 4, 5$. This implies that $\tan 5x_m = \tan q\pi = 0$. In conclusion, either the values of $\cos x_i$ for the 10 arguments $x_i \in [-\pi, +\pi),\ i = -5, -4, ..., +4$ satisfying the problem conditions are all different from each other, or the 10 arguments are equal to $x_i = \frac{i\pi}{5},\ i = -5, -4, ..., +4$, in which case we have only 6 distinct values of $\cos x_i$ in the interval $[-\pi, +\pi)$, because $\cos x = \cos (-x)$ is an even function of x, namely

$\cos x_{-5} = \cos \left(\frac{-5\pi}{5}\right) = \cos (-\pi) = -1$

$\cos x_0 = \cos \left(\frac{0\pi}{5}\right) = \cos 0 = 1$

$\cos x_i = \cos \frac{i\pi}{5} = \cos\left(-\frac{i\pi}{5}\right) = \cos x_{-i},\ \ i = 1, 2, 3, 4$


[b]When you get stuck on a problem you're working on with no time limits, how can you motivate yourself again?
How do you get over periods of time where you just don't feel like doing math?[/b]
[u]Related Questions[/u]
[hide]
How do you know when to stop trying a question?
How do you prevent yourself from stressing over a problem you're stuck on?
[/hide]
[u]3 Reasons Why People Give Up on Problems[/u]
[hide]
"To  quote Art and Craft of Problem Solving, inexperienced problem solvers give up because of the following:
They do not know where to start.
They make progress using one approach but then get stuck.
They try several methods to attack the problem, which does not work, so they give up."

- co-captain of the math team at my high school
[/hide]

1) Usually my motivations for working on problems aren't vulnerable to how difficult the problem is. What I mean is that if I spend a lot of time on a problem and I don't get anywhere, I assume it is because the problem is hard, but that doesn't mean my original reasons for trying to do the problem are no longer valid. So I can keep working on it. (But see below for ways to deal with hard problems.)

2) If I don't feel like doing math, I usually either find other work to do or go to sleep. If I don't feel like working on a particular problem, I'll work on a different problem. It's good to have 2 or 3 things available to work on. (By "things" I mean math homework assignments, reading, one or two hard problems, etc.) Then, if one is giving you trouble, you can switch to something else and come back to the troublesome thing later. This is especially good when you're getting frustrated and upset, because it gives you a chance to relax and you'll be much more efficient when you come back to the original problem. (This may be a good strategy on difficult math tests, too, like AIME or USAMO.)

3) Sometimes I do decide a problem is too hard and the time necessary for solving it is not worth the benefit of doing the problem. Usually it is difficult to know how hard a problem is before you know the solution, though. You can sometimes judge this by the source of the problem, or by asking someone who knows how to do it (assuming it's not an open problem). If you know you can usually do a similar problem in $ n$ hours, and you've spent $ n\minus{}1$ hours on this one with no progress, why give up already? Maybe it will take you $ n$ more hours, but you shouldn't expect it to take (say) $ 50n^2$ hours. (I'm assuming $ n$ is not too small.) However, if it's an unsolved problem, or something much harder than the problems you usually do, you might not know whether or not you have any chance of solving it. Then you might want to give up, or at least take a LONG break.

4) That's hard. As I said above, usually if I'm starting to get stressed I just take a break. Maybe that means working on another problem, or maybe it means just daydreaming, sleeping, eating, playing a game, etc. Just get the problem off my mind, and once that's done I come back to it. I do that even on exams. If there's only one problem left, or all of the remaining ones are stressful, I'll just sit there and daydream for 10 minutes or something, and then go back to work.

To some extent, I agree with the quote you put in there. The only thing I'd add is that if you are faced with a difficult problem, it's not good to just try a bunch of different tricks on it. I've heard that some people train for the IMO that way, and maybe it will work, because the style of IMO problems is to be vulnerable to a certain set of tricks, which (while large) is certainly finite. However, in my opinion, the best way to approach a difficult problem is to try to develop a strategy for solving it. (This is especially important for the more complicated kinds of problems that don't appear on timed contests.) Break the problem into simpler steps. Try to isolate the difficulty in solving the problem. Once you have a strategy, everything else will be much easier (assuming your strategy works...). I almost never start working right away when I see a problem. I always think about strategies for solving it, and once I have one that seems promising, I try to implement it. This seems to be much more effective than the "try all my tricks" method I used in high school.

Sorry, I'm not as experienced as all the other AoPSer's like Xevarion and the others, but here goes nothing!

Probably a bad habit from me, but I find that I like to focus on other problems first (but still keeping in mind the original problem)  :D . Like once I was thinking about a problem for days, then I started thinking about other problems. And then in the shower I finally noticed the solution, and like rushed to the computer to type in the solution. It was kinda' like the Archimedes and Gold Crown story, except I had a towel and some other clothers  :D

This certainly has happened to me before (when I got LoZ TP). So it was the summer, and I decided to start playing video games. And then not before long, I began getting crazy and playing 3 hours a day. When you don't feel like doing math, either

1. Try to sit down again and do it.

2. If you still don't feel like doing math, try taking away whatever is stopping you from doing math (for example, get your parents to control your random internet browsing time, or simply get more sleep).

3. Hmm... I'm not sure

For me, I'm not sure when to stop. But I would say it would be when the problem is really getting on your nerves and annoying you. Math is supposed to be fun. Enjoy it!

Hmm, not sure. Maybe just get more sleep (good for relieving stress in anything, really), and in your clumps of free time do some math (there isn't any time limit, so no real need to get nervous). Or maybe, after you've tried a long time, find someone who can guide you or get you started a bit.


integration
calculus
logarithms
calculus computations

$I=\int\frac{dx}{1+\sqrt{x+1}+\sqrt{x}}=\int\frac{1+\sqrt{x+1}-\sqrt{x}}{2(1+\sqrt{x+1})}dx$

$=\frac{x}{2}-\frac{1}{2}\int\frac{\sqrt{x}}{{1+\sqrt{x+1}}}dx$

$=\frac{x}{2}+\frac{1}{2}\int\frac{1-\sqrt{x+1}}{\sqrt{x}}dx=\frac{x}{2}+\sqrt{x}-\frac{1}{2}\int\sqrt{\frac{x+1}{x}}dx$.

For calculating the last integral we use the substitution $x=t^{2}$. We get that:

$\frac{1}{2}\int\sqrt{\frac{x+1}{x}}dx=\int\sqrt{1+t^{2}}dt=\frac{1}{2}\ln(t+\sqrt{1+t^{2}})+\frac{t\sqrt{1+t^{2}}}{2}+C$

$=\frac{1}{2}\ln (\sqrt{x}+\sqrt{1+x})+\frac{\sqrt{x(1+x)}}{2}+C$.


If the six solutions of $x^6=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, then the product of those solutions with $a>0$ is

$\text{(A)} \ -2 \qquad \text{(B)} \ 0 \qquad \text{(C)} \ 2i \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 16$

[hide]Write $-64$ as $64e^{i\pi}$ (Euler's Formula). When we take the 6th roots, we only want those with $e$ exponents between $\pi/2$ and $-\pi/2$.

The only two roots that satisfy this are $2e^{\pi/6}$ and $2e^{-\pi/6}$. Their product is $4$.[/hide]

how did you get the roots from 64e^ipi?
why the r of all the roots is all 2?


Only four integers between 100 and 1000 equal the sum of the cubes of their digits. Three of these are 153, 370, and 407. What is the fourth?

[hide="Not much of a strategy"]

Okay, first I look at the numbers you gave us and noticed that 370 ended in 0. Exploring this in my mind, I found that if you add 1 to get 371, you would be adding one to the sum of the cubes of 370 because $ 0^{3}=0$ and $ 1^{3}=1$, so you just add 1 to each side.

Therefore, the answer is $ \boxed{371}$

[/hide]


algorithm calculus integration trigonometry

Helo!
I bet that every programmer has struggled with this, in java i am trying to calculate pi using the algorithm, atan(-1) * 4
using integral calculus i know that that = pi but all i can get is a few decimal digits of pi because i can't go more than a double sooo, i was wondering how i could fix that.

As I have recently discovered while writing a program that uses trigonometry, Java is not very precise. In it's documentation, it specifically says that doubles and floats are not to be used for accurate calculations; there is the [url=http://java.sun.com/javase/6/docs/api/java/math/BigDecimal.html]BigDecimal[/url] class for that.


In problem GG–55, we focused on linear inequalities. How can we use those ideas with non–linear inequalities? In this problem, we will focus on the quadratic inequality given below:

a) To find our dividing point(s), solve the equality x2 – 3x – 18 = 0.

b) In the case of this parabola, we seem to have two dividing points! Plot these dividing points on a number line.

c) How many regions (sections or "pieces ") do two different points create on the number line? Identify each region.

d) For what x–values is x2 – 3x – 18  0? Test a number in each region. Be sure to consider whether the dividing points themselves are part of the solution.

e) Suppose we change the direction of the inequality symbol.

[quote="jbf13"]a) To find our dividing point(s), solve the equality x2 – 3x – 18 = 0.

b) In the case of this parabola, we seem to have two dividing points! Plot these dividing points on a number line.

c) How many regions (sections or "pieces ") do two different points create on the number line? Identify each region.

d) For what x–values is x2 – 3x – 18  0? Test a number in each region. Be sure to consider whether the dividing points themselves are part of the solution.

e) Suppose we change the direction of the inequality symbol.[/quote]

[hide="edited in"]a) $x^{2}-3x-18=0$ is $(x-6)(x+3)=0$, so $x=6,-3$ are our dividing points
b) Plot that
c) 3 regions: $(-\infty,-3),[-3,6],(6,\infty)$ or else. It depends on what type of inequality it is: $\le$ and $\ge$ or $<$ and $>$.
d) $x^{2}-3x-18 \le 0$ then test $x=-5,0,7$ and find the solution $[-3,6]$.
e) $x^{2}-3x-18>0$ then test $x=-5,0,7$ and find the solution $(-\infty,-3) \cup (6,\infty)$.
.[/hide]
[quote]In problem GG–55, we focused on linear inequalities.[/quote]
What is GG-55 anyway?

[quote="kohjhsd"][quote="jbf13"]a) To find our dividing point(s), solve the equality x2 – 3x – 18 = 0.

b) In the case of this parabola, we seem to have two dividing points! Plot these dividing points on a number line.

c) How many regions (sections or "pieces ") do two different points create on the number line? Identify each region.

d) For what x–values is x2 – 3x – 18  0? Test a number in each region. Be sure to consider whether the dividing points themselves are part of the solution.

e) Suppose we change the direction of the inequality symbol.[/quote]

[hide]a) $x^{2}-3x-18=0$ is $(x-6)(x+3)=0$, so $x=6,-3$ are our dividing points
b) Plot that
c) 3 regions: $(-\infty,-3),(-3,6),(6,\infty)$. This does not include $-3$ and $6$ though.[/hide]
d) Where did the inequality go?
e) Cannot be done without d).

[quote]In problem GG–55, we focused on linear inequalities.[/quote]
What is GG-55 anyway?[/quote]

A problem in my book.

[quote="kohjhsd"][quote="minsoens"][quote="kohjhsd"]d) Where did the inequality go?[/quote]hmm.. i think the inequality is $x^{2}-3x-18\le 0$. see [url=http://www.google.com/search?hl=en&rls=GGLG%2CGGLG%3A2005-48%2CGGLG%3Aen&q=%22In+the+case+of+this+parabola%2C+we+seem+to+have+two+dividing+points%22+%22For+what+x%E2%80%93values+is%22]here[/url][/quote]

I guess that's the book.
I edited things in anyway :([/quote]

Thanks man!


Let $ (O) $ is the circumcircle of the triangle $ ABC $ . A point $ M $ moves on segment $ AC $ . $ BM $ and $ (O) $ intersect at $ P $ . A line through $ A $ and perpendicular to $ AB $ interscet a line through $ P $ and perpendicular to $ PC $ at $ Q $ . Prove that $ O , M , Q $ are in a straight line

With my usual arrogance, I am saying that the problem is trivial once seen from the right viewpoint.

If B' and C' are the points diametrically opposite to the points B and C on the circle (O), then the segments BB' and CC' are diameters of the circle (O) and hence pass through the center O of this circle. In other words, the point of intersection of the lines BB' and CC' is the point O.

Since the segment BB' is a diameter of (O), we have < BAB' = 90, so that the point B' lies on the perpendicular to the line AB through the point A. The point Q also lies on this perpendicular; hence, we can say that the point Q lies on the line B'A.

Since the segment CC' is a diameter of (O), we have < CPC' = 90, and hence the point C' lies on the perpendicular to the line PC through the point P. The point Q also lies on this perpendicular; therefore, the point Q lies on the line C'P.

Hence the point Q is the point of intersection of the lines B'A and C'P.

Finally, the point M is the point of intersection of the lines AC and PB.

Now, after the Pascal theorem, applied to the cyclic hexagon BB'ACC'P, we obtain that the point of intersection of the lines BB' and CC', the point of intersection of the lines B'A and C'P, and the point of intersection of the lines AC and PB lie on one line. In other words, the points O, Q and M lie on one line.

  Darij

Yes, yes, I know it's overkill, but I still like doing this :):

In general, let $X'$ be the antipode of $X$. For reasons which are easy to see (congruent angles), we have $(A',B';C,P)=(P',C';B,A)$, so the mapping $AA'\leftrightarrow PP',AC\leftrightarrow PB,AB'\leftrightarrow PC',AP\leftrightarrow PA$ is a projective transformation between the pencil through $A$ and the one through $P$, mapping $AP$ to itself, so the locus of the intersection of two corresponding lines is a line, and thus $O,M,Q$ are collinear.

I think I have a shorter proof.
Let $AB$ and $CP$ meet at $K$ and $KQ$ and $BC$ meet at $R$.
One can show in twwo ways that $KR$ is altitude in $KBC$. One is
the inversion of pole $K$ and power $KA\times KB$ transforms $(KAQP)$ into $BC$
So $AQRB$ is cyclic.
The other way is to note $AQ \bigcap KC=\{M\}, PQ \bigcap KB=\{N\}$ and to show that $Q$ is orthocenter of $MNK$ and $MN||BC$.
Now to prove the problem we must show
$\frac{d(O,BP)}{d(O,AC}=\frac{d(Q,BP)}{d(Q,AC}$ which is now obvious since we know the angles
$AKQ$ and $PQK$

Well done Iura, but I think Grobber's solutions is much more ''scholl-like'' if you know what I mean.


Whenever I open a Geogebra image, it flashes.  Also, whenever I try to make a geogebra image, it shows up like this:

I am on OS X 10.5 Leopard, with the latest version of FF and java.

Works just fine for me, on all Windows browser. I just tested on a Mac, and indeed it doesn't work on Firefox, but it does work on Safari ...


vector
inequalities
trigonometry
geometry
3D geometry
inequalities proposed

for $x,y,z,\alpha,\beta,\gamma>0$ and $\alpha + \beta +\gamma = \pi$
find the maximum of
$f(\alpha,\beta,\gamma)= xsin \alpha + ysin \beta +zsin \gamma$
 :D

I haven't done it yet,but I think Cauchy and vector method will work here.  :maybe:

Using Cauchy-Schwarz :

f(u,v,w) <=  :Sigma: x * :Sigma: sin(u)

:Sigma: sin(u) = 2  +  cos(u)cos(v)cos(w) using basic trigonometry.

:Sigma: sin(u) is maximum is prod cos(u) > 0 so if u,v,w in [0..Pi/2].

cos is convave in [0..Pi/2] so Jensen + AM-GM proves :Sigma: sin(u) <= 9/4

Max of f(u,v,w) is then 3/4*sqrt (:Sigma: x)

Correct ?

[quote="belenos"]Max of f(u,v,w) is then 3/4*sqrt (:Sigma: x)

Correct ?[/quote]

When does f attain  the maximum? It seems to me the "=" signs at two places (Cauchy and Jensen) don't hold at the same time.

Oops, you're right I only found an upper bound ...

I have done a closely related problem before. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13402[/url] for a version with $\alpha + \beta +\gamma = 2\pi$.

There is no nice answer in general- the problem requires solving a cubic as an intermediate step. By Lagrange multipliers, $x \cos \alpha = y \cos \beta = z \cos \gamma$ at the maximizing point.
Let $x \cos \alpha =Ay \cos \beta = z \cos \gamma = A$ at the maximizing point. This gives $1 - (\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})A^2 - \frac{2}{xyz}A^3 = 0$. Solve for the unique positive value of $A$; the value of $x \sin \alpha + y \sin \beta + z \sin \gamma$ is $\sqrt{x^2-A^2} + \sqrt{y^2-A^2} + \sqrt{z^2-A^2}$. An explicit form would require cube roots.


Support AMC AIME USA(J)MO USAMO function geometry

How in depth does the camp go, and to what type of audience does it support (i.e. AMC, AIME, preparing for USAMO, or higher, IMO, etc.)?

What kind of topics are lectured there? Thanks!

[quote="hockeyadi23"]I went to AMP last summer, and I must say it was definately one of the most fun, and difficult camps I've ever been to. The staff there is amazing. The teachers, including Zumzum, Titu, Josh, John Kane, and of course Dorin were all geniouses, and fantastic teachers. Seriously like I was in the middle ("Red") group and we were doing problems at the IMO level or higher, every day. If nothing else desribes AMP, it's hard, because there's no lack of challenging problems in the upper five groups. And at first, I was just getting owned by the problems. But after a few days and weeks, I solved an IMO problem with no help from anyone... it was huge for me.

AMP has five "groups" essentially. At the bottom is White, which does Mathcounts/AMC stuff, and then there's Yellow, Red, Blue, and Black. These five groups cover more or less the same material, but the Black group is all IMO-ready people, Blue is people close to making MOsP, Red is ppl who are USAMO qualifiers but aren't yet ready for MOP (blue and red are essentially the same), and Yellow is the largest group, with effectively everyone else. My entire dorm was in red, but this was only by chance (you have to take this six-hour placement test on the first weekend).

I know they're making a lot of changes to AMP, but the camp life there was top-notch. The counselors/assistants were not only smart, but also chill. The likes of Mildorf, David Vincent, and my counselor--Khoa--were seriously amazing, and made the three weeks fly by. If you have any questions just IM me (hockeyadi21) or something b/c I don't check this board often, but like the topic we did ranged from the Lagrange Interpolation, to Generating Functions, to IMO-level algebraic identities/inequalities, to Zuming-level geometry, to Titu's proofwriting class which just was cool. Word.[/quote]
The five group thing isn't there anymore I heard

[quote="hockeyadi23"]I went to AMP last summer, and I must say it was definately one of the most fun, and difficult camps I've ever been to. The staff there is amazing. The teachers, including Zumzum, Titu, Josh, John Kane, and of course Dorin were all geniouses, and fantastic teachers. Seriously like I was in the middle ("Red") group and we were doing problems at the IMO level or higher, every day. If nothing else desribes AMP, it's hard, because there's no lack of challenging problems in the upper five groups. And at first, I was just getting owned by the problems. But after a few days and weeks, I solved an IMO problem with no help from anyone... it was huge for me.

AMP has five "groups" essentially. At the bottom is White, which does Mathcounts/AMC stuff, and then there's Yellow, Red, Blue, and Black. These five groups cover more or less the same material, but the Black group is all IMO-ready people, Blue is people close to making MOsP, Red is ppl who are USAMO qualifiers but aren't yet ready for MOP (blue and red are essentially the same), and Yellow is the largest group, with effectively everyone else. My entire dorm was in red, but this was only by chance (you have to take this six-hour placement test on the first weekend).

I know they're making a lot of changes to AMP, but the camp life there was top-notch. The counselors/assistants were not only smart, but also chill. The likes of Mildorf, David Vincent, and my counselor--Khoa--were seriously amazing, and made the three weeks fly by. If you have any questions just IM me (hockeyadi21) or something b/c I don't check this board often, but like the topic we did ranged from the Lagrange Interpolation, to Generating Functions, to IMO-level algebraic identities/inequalities, to Zuming-level geometry, to Titu's proofwriting class which just was cool. Word.[/quote]

I was also at the camp last year. I have to say, some of the levels regarding difficulty in this post are definitely exaggerated

I concur. Though it'll probably be suitable for many people up to and including people who have made Red MOP.

The counselors are amazing. The food on the other hand, ran into problems often (though they got it sorted out by the last week, so next year should be fine. Experience: A single apartment unit will go through 20+ waffles per day)


I'm learning about a hyperbolic function and I want to introduce a problem that baffles me a lot.

How would you write the function y=  Ae^x + Be^x, where A >0 and B>0 in the form of y= K cosh(x-c), where K and c are constants? In other words, what would K and c be equivalent of using the constants A and B in the preceding equation.

I feel pretty nescient today and not having a great time due to the bodily discomfort, but I hope to learn a lot more about a hyperbolic function as it has ample application in engineering fields.

This works in a similar way to how it does with the usual trigonometric functions sin(x) and cos(x).

I'm assuming there was a small typo in your original post, with the problem being $ A\mathrm{e}^{x} \plus{} B \mathrm{e}^{\minus{}x}$ expressed in terms of the $ \cosh$ function.

We need to know the definitions of $ \cosh$ and $ \sinh$ in terms of the exponential function:
$ \cosh(x) \equal{} \frac{\mathrm{e}^{x}\plus{}\mathrm{e}^{\minus{}x}}{2}$ and $ \sinh(x) \equal{} \frac{\mathrm{e}^{x}\minus{}\mathrm{e}^{\minus{}x}}{2}$.

From these we find that:
$ \mathrm{e}^{x} \equal{} \cosh(x) \plus{} \sinh(x)$ and $ \mathrm{e}^{\minus{}x} \equal{} \cosh(x) \minus{} \sinh(x)$
and there is the analogue of the usual addition formula for $ \cos(x)$ as:
$ \cosh(x\plus{}y) \equal{} \cosh(x) \cosh(y) \plus{} \sinh(x) \sinh(y)$.

So, your question about $ A \mathrm{e}^{x} \plus{} B \mathrm{e}^{\minus{}x}$ can be rewritten as:
$ A ( \cosh(x) \plus{} \sinh(x) ) \plus{} B (\cosh(x) \minus{} \sinh(x) )$ = $ C \cosh(x) \plus{} D \sinh(x)$ with $ C \equal{} A\plus{}B$ and $ D \equal{} A\minus{}B$.

Now you can consider $ K \cosh (x\plus{}d)$, expand the $ \cosh$ bit by using the addition formula, and then you equate the coefficients of $ \cosh$ and $ \sinh$ in what you get with the definitions of $ C$ and $ D$. You should be able to get $ K$ and $ d$ in terms of $ C$ and $ D$.


Can somebody send me problem from CMO 2003 and 2004


geometry
calculus
derivative
function
trigonometry
linear algebra
matrix

So I am working on a research project with one of my professors to model the movement of cells. This is basically the common cellular slime mold example. with a couple of small changes.

I was hoping  that anyone who is interested my look at my work and let me know if they see any problems or have any suggestions.

[b]This is the problem:[/b] an electrode is implanted in the brain which causes a release of chemical "c", microglia a specific cell in the brain responds to chemical "c" and degrades chemical "c". I am trying to solve for the time where time where the release of the chemical form the electrode is in equilibrium with the microglia.

So this is what I have.

m(x,t)= density of microglia per unit area
c(x,t)= concentration of chemical c per unit area

$\frac{dm}{dt}=\frac{-d}{dx}(-\mu\frac{dm}{dx}+\chi*m\frac{dc}{dx})$
$\frac{dc}{dt}=\frac{-d}{dx}(-D\frac{dc}{dx})+f-km$

Where
$\mu$= microglia motility
$\chi$= chemotactic coefficient
D= diffusion rate
f= rate of chemical secretion caused by the electrode
k= rate of microglia's chemical degradation

Because of my inability to solve this PDE we are going to make it an ODE. By assuming after some amount of time it becomes homogeneous.
Where
$m(x,t)=\overline{m}$
 $c(x,t)=\overline{c}$
 and
$\frac{d\overline{m}}{dt}=\frac{d\overline{c}}{dt}= 0$
$\frac{d\overline{m}}{dx}=\frac{d\overline{c}}{dx}= 0$

This means when substituting into the first equations that $f=k\overline{c}$

Then in order to determine the stability of this system I am adding the derivative and looking to see whether this perturbation amplifies or attenuates

$m(x,t)=\overline{m}+m^\prime(x,t)$
 $c(x,t)=\overline{c}+c^\prime(x,t)$

This means that
$\frac{dm^\prime}{dt}=(\mu\frac{d^{2}m^\prime}{dx^{2}}+\chi*\overline{m}\frac{d^{2}c^\prime}{dx^{2}})$

$\frac{dc^\prime}{dt}=D\frac{d^{2}c^\prime}{dx^{2}}=f^\prime-kc^\prime$

Because these are liner functions we can approximate them as such

$m^\prime(x,t)=M\exp^{\sigma*t}\cos{qx}$
$c^\prime(x,t)=C\exp^{\sigma*t}\cos{qx}$
$f^\prime(x,t)=F\exp^{\sigma*t}\cos{qx}$

After substituting these into the previous equations we get

$M(\sigma+\mu*q^{2})-C(\chi*\overline{a}q^{2})=0$
$-F+C(\sigma+Dq^{2}+k)=0$

This is really where I am getting stuck I would like to find a solution where M, C and F are not equal to 0, to do this I know that I must find the determinant of the equations then I believe I can solve for $\sigma$ but I don’t have a square matrix to solve for a det.

q in this instance is a wavenumber defined as $q=\frac{n\pi}{L}$
$L=x$ and $n=(0,1,2,3,...)$

So if anyone has any advice on what I can do to solve this problem let me know. Or if you see any huge mistakes which I know there are, but I am not sure how to fix them.

Thanks all for the help

[quote="newmexiconate"]So I am working on a research project with one of my professors to model the movement of cells. This is basically the common cellular slime mold example. with a couple of small changes.

ooo

m(x,t)= density of microglia per unit area
c(x,t)= concentration of chemical c per unit area

$ \frac {dm}{dt} \equal{} \frac { \minus{} d}{dx}( \minus{} \mu\frac {dm}{dx} \plus{} \chi*m\frac {dc}{dx})$
$ \frac {dc}{dt} \equal{} \frac { \minus{} d}{dx}( \minus{} D\frac {dc}{dx}) \plus{} f \minus{} km$

Where
$ \mu$= microglia motility
$ \chi$= chemotactic coefficient
D= diffusion rate
f= rate of chemical secretion caused by the electrode
k= rate of microglia's chemical degradation

ooo

Because of my inability to solve this PDE we are going to make it an ODE.
[/quote]

Are you still interested in solving this PDE? If so, please let me know.


Another example would be

'The quick, brown fox jumped over the lazy dog'

[hide]Does it have every letter of the alphabet in it?[/hide]

its the shortest sentence in the english language that has every letter

It's not the shortest, the shortest is: "Brick quiz whangs jumpy veldt fox."  BTW, these are called pangrams

If you allow abbreviations:

Mr. Jock, TV quiz PhD, bags few lynx.

[quote="Ihatepie"]What is so unusual about the sentence below?
(Other than the fact that it doesn't make much sense.)

"Jackdaws love my big sphinx of quartz."[/quote]

[hide="Hint"] Look at each letter. [/hide]


function
integration
calculus
real analysis
real analysis solved

Let $ f_n$ be a sequence of nonnegative continuous functions defined on [0,1] such that $\sum_{n\geq 0} {\int_{0}^{1} {f_n(x)dx}}< infinity $.
  a) Find such a sequence with the property that the set of those numbers x such that $\sum_{n\geq 0} {f_n(x)}=infinity$ is dense in [0,1].
  b) Prove that the set of those x such that $\sum_{n\geq 0}{f_n(x)}<infinity$ is always dense in [0,1].

This doesn't solve any part of the problem, but I think it helps a lot with the first part. It's an example of such a family of functions none of which is continuous:

Let $(r_n)_{n\ge 0}$ be an enumeration of the rationals in $[0,1]$. We take $f_n(x)=\frac 1{n+1}$ if $x=r_i$ for some $i\in\overline{0,n}$ and $0$ everywhere else. Since the number of points of discontinuity is finite for each of the functions, it means they're integrable (Riemann), and their integrals are all $0$. We can then adjust the $f_n$'s to make them continuous, because we can find continuous functions which approximate those we have used as well as we want, so their integrals will be as close to $0$ as we want.

Sorry if I had to give a specific example.. I'll think about that.

Here's an example:

For any real $x$, let $d(x)$ be the distance from $x$ to the nearest integer. Define $f_n(x)=\max(1-2^nd(2^nx),0)$. The graph of $f_n$ on the interval $[0,1]$ consists (in the net) of $2^n$ spikes, each of height 1 and width $2\cdot2^{-2n}$. Then each $f_n$ is continuous (being the maximum of two continuous functions), nonnegative, and
\[\int_0^1f_n(x)dx=2^{-n}.\]
Thus, $\sum_{n=0}^{\infty}\int_0^1f_n=\sum_{n=0}^{\infty}2^{-n}=2<\infty.$

On the other hand, at any dyadic rational (any number $x$ of the form $\frac j{2^k}$ for integers $j$ and $k$), the sum $\sum_nf_n(x)$ is eventually a sum of all 1's, and hence equals infinity. The dyadic rationals are dense.

The second part of this would be at home in a Lebesgue integral course. By the Monotone Convergence theorem,
\[\sum_{n=0}^{\infty}\int_0^1 f_n(x)dx=\int_0^1\sum_{n=0}^{\infty}f_n(x)dx<\infty\]
But since the sum of the $f_n$ is integrable, it must be finite almost everywhere. "Almost everywhere" certainly implies dense, since you can't leave out any interval.


Consider all possible sets of $ n$ weights whose masses in grammes are distinct integers not exceeding $ 21$. What is the smallest $ n$ such that in every such set of weights there are two pairs whose total masses are equal?

$ a\plus{}b\equal{}c\plus{}d \Longrightarrow a\minus{}c\equal{}d\minus{}b$, so the problem is equivalent to finding a sequence whose pairwise positive differences are unique. The set of differences can only be $ \{1,2,3,\ldots 20\}$. There are $ {n\choose 2}$ pairs of numbers, but it is possible for a difference to be achieved twice if one number is common to both pairs, eg $ a\minus{}b\equal{}b\minus{}c \Longrightarrow 2b \equal{} a\plus{}c$. Since it is not possible for two different pairs of numbers to have the same average, there can be at most $ n$ such pairs sharing a common number. $ {n \choose 2} \le 20\plus{}n \Longrightarrow n \le 7$.

This is achievable with the first $ 7$ Fibonacci numbers, $ 1,2,3,5,8,13,21$ (consider the Fibonacci base).


For which primes $p >3$ is $6$ a square modulo $p$?

[hide="Solution"] $( \frac{6}{p} \right) = ( \frac{2}{p}) ( \frac{3}{p})$

It's well known that $( \frac{2}{p}) = (-1)^{ \frac{p^{2}-1}{8}}$; in other words, it is $1$ when $p \equiv 1, 7 \bmod 8$ and $-1$ when $p \equiv 3, 5 \bmod 8$.

$( \frac{3}{p}) = (-1)^{ \frac{p-1}{2}}( \frac{p}{3})$

Well, $( \frac{p}{3}) = 1$ when $p \equiv 1 \bmod 6$ and it equals $-1$ when $p \equiv 5 \bmod 6$.  Then $( \frac{3}{p}) = 1$ when $p \equiv 1, 11 \bmod 12$ and it equals $-1$ when $p \equiv 5, 7 \bmod 12$.

Using CRT we have our result $\bmod 24$:

- If $p \equiv \boxed{ 1, 5, 19, 23 \bmod 24 }$ then $( \frac{6}{p}) = 1$.
- If $p \equiv 7, 11, 13, 17 \bmod 24$ then $( \frac{6}{p}) =-1$.  [/hide]


题目1，有问题。

[quote="feizhenpeng"][size=150]国家集训队考试试题（一）
2006年3月19日，星期日
1、设$ABCD$是一个梯形并且$AB$//$CD$，$ABCD$内部有两个圆$w_1$和$w_2$满足$w_1$与三边$DA$，$AB$，$BC$相切，圆$w_2$与三边$BC$，$CD$，$DA$相切．令$l_1$是过点$A$的异于直线$AD$的圆$w_2$的另一条切线，$l_2$是过点$C$的异于直线$CB$的圆$w_1$的另一条切线．证明：$l_1$//$l_2$．
2、求所有的正整数对$(a,n)$使得$\frac{(a+1)^n-a^n}{n}$是整数．
3、设$a_1,a_2,$…,$a_n$是给定的$n$个实数$(n\geq 1)$．求证：存在实数$b_1,b_2,$…,$b_n$满足下列条件：
（a）对任何的$1\leq i\leq n$，$a_i-b_i$是正整数；
（b）$\sum_{1\leq i<j\leq n}{(b_i-b_j)^2} \leq \frac{n^2-1}{12}$．[/size][/quote]


function
real analysis
real analysis unsolved

It is writen in this book: "It is given a function f: I->R who has primitive and it is bounded. Let $ x_n $ be the sequence of  the points in which f is not continuous." Why is the set of the points in which f is not continuous numerable? Sorry for mistakes in tehnic words. :blush: You can edit it if iou realize what i want to say :blush:

In 1881, Volterra gave an example of such a function: It is bounded and has a primitive on a closed interval, but it is not Riemann-integrable.

Thank you four your reply, but my question was not refering to the existence of this funtions, i was wondering why that afirmation is true.

[quote="grobber"]But liyi indirectly answered your question :):

If a function is not Riemann integrable, then its set of discontinuities does not have null Lebesgue measure, so it certainly cannot be countable. This means that the example constructed by Volterra is bounded, has a primitive, but does not have a countable set of discontinuities.

So, you see, that property [b]is not[/b] true, actually :).[/quote]
Yes, that is exactly what I want to imply.


Circle $c_1,c_2$ intersect at points $A$ and $B$ ,and the circle $c_3$ tangents them from outside at the points $C$ and $D$ respectively. The line $AB$ intersects the circle $c_3$ in the points $P$ and $Q$. Prove that $\angle PKC=\angle QKC$.

I'm sorry this question is wrong :o  :mad:

i saw the right form of it that the circle C3 is tangent to them from in side.but i think it doesnt mather wether its tsngent to them from inside or out side.so plz tell the right problem ;)


integration
calculus
calculus computations

find all solutions for the following:

$ x''\plus{}5x'\plus{}4\equal{}0$

$ x''\plus{}5x'\plus{}4 \equal{} 0\Longleftrightarrow x''\equal{}\minus{}5\left(x'\plus{}\frac{4}{5}\right)$

$ \therefore x'\equal{}Ae^{\minus{}5t}\minus{}\frac{4}{5}$

$ \therefore x\equal{}C_{1}e^{\minus{}5t}\minus{}\frac{4}{5}t\plus{}C_{2}$.

Another way.

Set x' = p, and so x'' = p'. Now substitute in your differential equation to get p' + 5p = -4, which is a first order linear d.eq. whose general solution is

$ p \equal{} e^{\minus{}\int5\,dt}\left(\int(\minus{}4)e^{\int5\,dt}\plus{}C\right) \equal{} e^{\minus{}5t}\left(C\minus{}\frac{4}{5}e^{5t}\right)$

$ \Rightarrow  p \equal{} Ce^{\minus{}5t}\minus{}\frac{4}{5}$

and so

$ x' \equal{} Ce^{\minus{}5t}\minus{}\frac{4}{5}\Rightarrow  x \equal{} C_{1}e^{\minus{}5t}\minus{}\frac{4}{5}t\plus{}C_{2}$.


$ \{\begin{array}{ccc} x+y+z & = & 1 \\
x^4+y^4+z^4 & = & 1 . \end{array}$

Thanks a lot :D

Someone helps me to do this problem plz  :maybe:

hungnsl, actually I don't know whether $ x,y,z$ in your question can be negative. If $ x,y,z$ can be negative , there will be infinitely many solutions.

I think $ (x,y,z)>0$ is not true!  we can have $ min{(x,y,z)}\prec0$ so $ min{(x,y,z)}\succ0$ must be mentioned :wink:

[quote="mathematikos"]Hi!! this is my solution! :)
Let $ (x,y,z)\neq(0,1,0);(0,0,1)and(1,0,0)$:
$ \{\begin{array}{ccc} x + y + z & = & 1 \\
x^4 + y^4 + z^4 & = & 1 . \end{array}$
$ \Longleftrightarrow$
$ |x|\prec1$
$ |y|\prec1$
$ |z|\prec1$
and:
$ (x + y) > 0$
$ (x + z) > 0$
$ (y + z) > 0$
then we can prove that $ (x,y,z) > 0$ so:
$ x + y + z$>$ x^{4} + y^{4} + z^{4}$
$ \Longrightarrow$1>1 it a contradiction!! so we conclude:
$ S = (0,1,0);(0,0,1)(1,0,0)$[/quote]

I think your roots is right but your solution isn't OK

if x+y, y+z, z+x > 0 it's not enough to show that x,y,z >0  :wink:

yes,it's what a i try to say in my second message :wink:


The download section for LaTeX no longer has instructions on how to convert your .dvi files to .pdf could someone please tell me how to do this or point me to a site with instructions. thanks.

The reason is is that it is built into MiKTeX as the default. In TeXnicCenter choose LaTeX=>PDF when you compile your .tex file.

[quote="stevem"]The reason is is that it is built into MiKTeX as the default. In TeXnicCenter choose[color=red] LaTeX=>PDF [/color]when you compile your .tex file.[/quote]
Sorry, But I can't find it? Where it?  :maybe: I Download and install MiKTeX and TeXnicCenter as in http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

Help me!

wow i never noticed that before. thanks a lot.  :)

[quote="N.T.TUAN"]
Sorry, But I can't find it? Where it?  :maybe: I Download and install MiKTeX and TeXnicCenter as in http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php
[/quote]

When you set it up did you tell TeXnicCenter where to find Adobe Reader and MiKTeX? Is it in Build, Define Output Profiles? If so, put the path to Adobe Reader there. Otherwise run the Wizard.
Finally, if still no joy, in the same place import the pdf.tco file below (unzip it first). IMPORTANT you must change [b]all[/b] the paths on [b]all[/b] the tabs to agree with your computer.

But it looks as if you don't even have the LaTeX=>DVI option so you haven't run the Wizard which pops up when you first run TexNicCenter.

At this time, I don't understand what you said. But I will try. See you again. Thank you very much  :oops:

I can't  :(  Now, I only have basic-miktex-2.5.2580 and TXCSetup_1Beta7_01. Please hint to me, step.1, step.2,...,etc. Thank you very much.

1. Open TeXnicCenter
2. Go to Build, Define Output Profiles
3. Click the Wizard button in the bottom left corner
4. This is from http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php :
[quote]The program will then walk you through the configuration wizard:
When it askes you for the "full path of the directory where the executables" are located, type C:\Program Files\MiKTeX 2.5\miktex\bin (Note: if you changed the default location when installing MiKTeX, then you'll need to replace "C:\Program Files\MiKTeX 2.5" with the directory to which you installed MiKTeX.)
If it asks you to pick a PostScript viewer, you may just leave everything blank and just click "Next". Similarly if it asks you to pick a DVI viewer, just leave everything blank and click "Next". (It may or may not ask you these things, depending on how your computer is configured.)[/quote]
4. Then OK and TeXnicCenter should be set up ready to use

[quote="stevem"]
2. Go to Build, Define Output Profiles[/quote]
And I have... But How I can contiuous?

Ok, What is here? Thank you, sorry for my English  :)

I have already answered that in step 4 of my post; I will repeat it again.

4. This is from http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php :
[quote]The program will then walk you through the configuration wizard:
When it askes you for the "full path of the directory where the executables" are located, type C:\Program Files\MiKTeX 2.5\miktex\bin (Note: if you changed the default location when installing MiKTeX, then you'll need to replace "C:\Program Files\MiKTeX 2.5" with the directory to which you installed MiKTeX.)
If it asks you to pick a PostScript viewer, you may just leave everything blank and just click "Next". Similarly if it asks you to pick a DVI viewer, just leave everything blank and click "Next". (It may or may not ask you these things, depending on how your computer is configured.)[/quote]

Wow, this is  my first LaTeX document, please see  :)
Thank again.


function
algebra
functional equation
algebra unsolved

Find all function f: $ Q \to Q$ such that $ f(2f(x)\plus{}f(y))\equal{}2x\plus{}y$ with all $ x;y \in Q$

[quote="boy own me"]Find all function f: $ Q \to Q$ such that $ f(2f(x) \plus{} f(y)) \equal{} 2x \plus{} y$ with all $ x;y \in Q$[/quote]

It's easy to see that the function is bijective.

Now let $ f(k)\equal{}0$ , then

$ f(2f(k)\plus{}f(k))\equal{}3k$ gives $ f(0)\equal{}3k$

$ f(2f(0)\plus{}f(0))\equal{}0$ gives $ 9k\equal{}k \Rightarrow k\equal{}0$ as the function is bijective.

Now put $ x\equal{}0$ into the original equation we have $ f(f(y))\equal{}y...............(1)$
put $ y\equal{}0$ into the original equation we have $ f(2f(x))\equal{}2x$ , which gives $ 2f(x)\equal{}f(2x)............(2)$ after using $ (1)$

Now $ f(f(2f(x)\plus{}f(y)))\equal{}f(2x\plus{}y)$ gives $ f(2x\plus{}y)\equal{}2f(x)\plus{}f(y)\equal{}f(2x)\plus{}f(y)$ after using $ (1)$ and then $ (2)$.

So we have $ f(x)\plus{}f(y)\equal{}f(x\plus{}y) \Rightarrow kf(x)\equal{}f(kx)$ for any rational number $ x$ and integer $ k$.

Now let $ f(x)\equal{}\frac{q}{p}$ , where $ p,q$ are integers , and $ f(1)\equal{}m$ ,
we have $ pf(\frac{1}{p})\equal{}f(1)\equal{}m \Rightarrow f(\frac{1}{p})\equal{}\frac{m}{p}$
$ f(x)\equal{}f(\frac{q}{p})\equal{}qf(\frac{1}{p})\equal{}\frac{q}{p}m\equal{}mx$

Now plug in $ f(x)\equal{}mx$ and $ f(y)\equal{}my$ into the oringinal equation we know the only possible $ m$s are $ 1$ and $ \minus{}1$

So $ f(x)\equal{}x$ and $ f(x)\equal{}\minus{}x$ are the only possible functions.

Let $ a$ satisfy $ f(a) \equal{} 0$, then the functional equation implies $ f(0) \equal{} 3a$, so if $ f$ has a zero it is unique.  Since $ f(3f(0)) \equal{} 0$ it has exactly one zero, and moreover we must have $ a \equal{} 3f(0) \equal{} 9a$, so $ a \equal{} 0$.  Let $ x \equal{} 0$, then $ f(f(y)) \equal{} y$ for all $ y\in \mathbb{Q}$.  Now apply $ f$ to both sides of the original functional equation, then $ 2f(x) \plus{} f(y) \equal{} f(2x \plus{} y)$ for all $ x,y\in \mathbb{Q}$.  But this implies $ f(2x) \equal{} 2f(x)$, and hence $ f(2x) \plus{} f(y) \equal{} f(2x \plus{} y)\Leftrightarrow f(x) \plus{} f(y) \equal{} f(x \plus{} y)$ for all $ x,y\in \mathbb{Q}$.

Therefore $ f$ satisfies Cauchy's equation, and we proceed as usual (see, for example, [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]here[/url]) to prove that all solutions are of the form $ f(x) \equal{} ax$ for some $ a$, and verify that $ a \equal{} \pm 1$ are solutions.