A better reason is that $\frac{a}{b} = c \iff a = b \times c$. But if $b = 0$ and a does not, then there is no value c that fits this second equation. And if a does equal zero, [i]any[/i] value of c fits the second equation. Therefore division by zero is not well defined.

how would you define infinity?
Would infinity +1 = infinity?
I think that is why dividing by zero or having infinite solutions to something is actually
"undefined"

Infinity is grouped into infinitly many subsets: $\aleph{0}$, $\aleph {1}$, $\aleph {2}$, etc..., where $\aleph{0}$ is the number of points in a line, $\aleph {1}$ is the number of lines in space, and $\aleph {2}$ is the number of curves in space.

$\aleph{n}+a=\aleph{n}$, where $a$ is a finite number.

[quote="4everwise"]Can someone justify if this is the proof to why you cannot divide by 0?

I believe it's somewhere along the lines of the fact that if you divide by a number less than zero, the answer you get is larger than the original number.

For example, if you divided 3 by .00000000000000000000009, it would be an extremely large number. Since 0 is actually 0.00000000 and etc, the answer would be an infinately large number.[/quote]

The proof that you cannot divide by zero goes as follows:

Division by $a$ exists if and only if there is a number $b$ such that $ab = 1$, but when $a=0,ab =0$ and therefore $0=1$, a logical contradiction.

[quote="A+MATH"]You could also try to do long division when your dividing by zero. As you can see you get no where pretty fast.[/quote]

 :lol:

[quote="elementsofmathkid"]You can't divide by zero because it's the 11th commandment.

Thou shall not divide by zero.

My teacher said so....  :lol:  :P[/quote]
that was funny! :lol:  :lol:  :lol:

[quote="NoSoupForYou"][quote="elementsofmathkid"]Thou shall not divide by zero.[/quote]

Wouldn't you say, $\text{\"Thou shalt not divide by zero.\"}$ ? You know, with like a [i]t[/i]?[/quote]

I don't know.... why do you think it would be shalt? I have no idea. You might be right.  :)

That's on one of the posters my teacher has

dividing by 0 is impossible because dividing by 0 would be infinity, and infinity is an idea.

[color=indigo]one way to look at it is 0*x = 0, no matter what x is. so it wouldn't make sense to divide anything by zero. this doesn't really work because 0/0, but it still kinda works[/color]

In my opinion, i'd say that $\infty \cdot \infty = \infty$ I understand that $0\cdot\infty$ is indeterminant but I think that $0\cdot \infty = \infty$ because even though 0x = 0 for any and all x, infinity is not a number, and therefore (in my opinion) doesn't follow the rules like that one that normal numbers do.

lmao@elementsofmathkid 10th commendement eh? :rotfl:

[quote="dareonion"]just pointing out another reason...

you notice that theoretically speaking, the answer to 1/0 could be either positive infinity or negative infinity.  this contradicts the fact that division always results in ONE quotient.[/quote]

I'll say it again for those who have not followed the discussion to date.

$\frac{1}{0}$ [b]is not[/b] $\infty$; it is not $+\infty$, it is not $-\infty$.  $\frac{1}{0}$ is not a number. It is undefined.

well we could say that by long division we'd have:

$\frac 10 = \infty + \frac 10$

and by substitution we'd find that:

$\frac 10 = \infty \cdot \infty$ which has been declared indeterminant and thus $\frac 10$ is undefined

Sure, its good to be thinking of advanced topics in earlier years, but I would suggest waiting until you start learning basic calculus to attempt working with infinity very much.  Don't get in the habit of using $\infty$ like any number.

For example, does $\infty-\infty=0$?
Let $S_1$ be the series $1+2+3+4+5+6+...$ continuing forever.  Clearly $S_1$ is approaching $\infty$ (but note that I'm not saying $S_1 = \infty$, because again, you can't treat it like a number)
Let $S_2$ be the series $2+3+4+5+6+...$ which also clearly approaches $\infty$.
So write out $S_1-S_2$ like so--  $1+2-2+3-3+4-4+5-5+6-6+... = 1+0+0+0+0+0+0+... = 1$

Similar arguements can be made to make "$\infty-\infty$" equal anything.  Just don't try to use or manipulate equations with $\infty$ in them, because really they make no sense.

So how about earlier when someone said $\frac{1}{\infty}=0$?  Is this not ture?  The concept  makes sense, but again it can't be written like that, so limits were invented.  The real way to say this is $\displaystyle \lim_{n \to \infty} \frac{1}{n} = 0$, meaning that as n gets closer and closer to infinity, 1/n gets closer and closer to 0.

For more on this, I would seek teachers or books, because fundamentals of calculus aren't the kinds of thinks that most people will come up with on their own  ;)
Moral of story: avoid using $\infty$ in expressions

So getting back to the original question, I believe all of this has been said before, but here it is all in one place:

For any $k \neq 0$ we attempt to solve the equation $\frac{k}{0}=x$.
Multiplying through by 0 we get $k=0\times x$ which means $k=0$.  But we defined $k \neq 0$ so there is contradiction and it is thus $undefined$.

So now we try for $k=0$. to solve $\frac{k}{0}=x$.  After multiplying this out, there is no contradiction, we get $0=0$, but this is true for all $x$.  So $\frac{0}{0}$ is not undefined, but you can't determine a specific value for it, so it is called $indeterminant$.

[quote]as n gets closer and closer to infinity, 1/n gets closer and closer to 0[/quote]
Saying this implies that you consider infinity as a number : how could you a number approach infinity if infinity was not a number ? However, there is a problem : [tex]1[/tex] is as close to [tex]\infty[/tex] as [tex]9999999999^{99999999^{99999999}}[/tex], so you can't say a number approaches infinity.

Also, as I tried to explain before, if [tex]\frac{a}{b}[/tex] [tex]=[/tex] [tex]c[/tex], [tex]a[/tex] does not "always" equal [tex]b \times c[/tex]. But OK, my explanations were lousy and not clear at all, so I'll try a second time  :D

I've seen several times people saying that, if [tex]\frac{1}{\infty}[/tex] [tex]=[/tex] [tex]0[/tex], [tex]1[/tex] [tex]=[/tex] [tex]0 \times \infty[/tex], and that [tex]0 \times \infty[/tex] equals 0 and not 1, therefore you can't divide by infinity. Their reasoning is false. [tex]0 \times \infty[/tex] [u]can[/u] equal [tex]1[/tex], but you can't tell this from the equation [tex]\frac{1}{\infty}[/tex] [tex]=[/tex] [tex]0[/tex]. Why ? Because if you do, you are using yourself (but you don't realize it) in your reasoning the fact that [tex]0 \times \infty[/tex] [tex]=[/tex] [tex]1[/tex].
You don't believe me ? What have you done to each side of the equation [tex]\frac{1}{\infty}[/tex] [tex]=[/tex] [tex]0[/tex] to reach [tex]1[/tex] [tex]=[/tex] [tex]0 \times \infty[/tex] ? You've multiplied by infinity. The right side is all right, you have [tex]0[/tex], you multiply it by [tex]\infty[/tex] and you get [tex]0 \times \infty[/tex]. OK. What about the left side now ? You had [tex]\frac{1}{\infty}[/tex], i.e. [tex]0[/tex] as the first equation [tex]\frac{1}{\infty}[/tex] [tex]=[/tex] [tex]0[/tex] says. You multiply it by [tex]\infty[/tex] and.....what the heck ???? You get [tex]1[/tex] ? [tex]\frac{1}{\infty} \times \infty[/tex] [tex]=[/tex] [tex]1[/tex] ? [tex]0 \times \infty[/tex] [tex]=[/tex] [tex]1[/tex] ? You see that in your reasoning you had used [tex]0 \times \infty[/tex] [tex]=[/tex] [tex]1[/tex]. What you had just proved is that if [tex]\frac{1}{\infty}[/tex] [tex]=[/tex] [tex]0[/tex] [b]AND[/b] [tex]0 \times \infty[/tex] [tex]=[/tex] [tex]1[/tex], [tex]0 \times \infty[/tex] [tex]=[/tex] [tex]1[/tex]. Nice one !

The problem is that it would be correct only if [tex]\frac{\infty}{\infty}[/tex] "always" equalled [tex]1[/tex]. OK, a number divided by itself is 1, so [tex]\frac{\infty}{\infty}[/tex] [tex]=[/tex] [tex]1[/tex]. But [tex]\infty[/tex] [tex]=[/tex] [tex]2 \times \infty[/tex], so [tex]\frac{\infty}{\infty}[/tex] [tex]=[/tex] [tex]\frac{2 \times \infty}{\infty}[/tex] [tex]=[/tex] [tex]2 \times \frac{\infty}{\infty}[/tex] [tex]=[/tex] [tex]2 \times 1[/tex] [tex]=[/tex] [tex]2[/tex]. You can replace the 2 by anything, [tex]\frac{\infty}{\infty}[/tex] can equal anything. That's the weakness of the reasoning.

But let's come back to the goal of my post. You can do the same thing for 0, [tex]\frac{0}{0}[/tex] can equal anything. Therefore, after multiplying by zero [tex]\frac{k}{0}[/tex] [tex]=[/tex] [tex]x[/tex], you don't necessarily get [tex]k[/tex] [tex]=[/tex] [tex]0 \times x[/tex] as I've seen in the previous post, because you can't say [tex]\frac{0}{0}[/tex] "always" equals [tex]1[/tex].

By the way, I don't see why [tex]\infty[/tex] wouldn't be a number. You can compare several infinities (as Georg Cantor did) and tell if one infinity is bigger than the other. Infinity can also be an amount of things : it's the amount of counting numbers, even/odd numbers, the amount of points in a segment, etc. Since it can be a quantity, why wouldn't it be a number ?

If you say "As $x$ approaches $\infty$...", that means that as $x$ gets bigger and bigger. Lots of people say it. It's a perfectly sound statement, mathematics-wise.  ;)

It seems you haven't understood me. I never said noone says it, I just said it's wrong to say it, and showing me that everyone in universe except me says "x approaches infinity" wouldn't change my mind. You just can't approach infinity.

It seems you haven't understood me. $x$ approaches infinity is a completely valid statement. There's nothing wrong about it. It's not wrong to say it. Prestigious mathematicians have defined a limit as "the mathematical value toward which a function goes as the independent variable approaches infinity."  ;) Mods, can you explain better?

I'm confused at who your post was targetted towards, unless it was a lot further back in the discussion, where I did not check.

If you don't like the wording as n approaches infinity you can say "as n increases without bounds" meaning as n gets bigger and bigger but never reaching a bound, or a limiting value.

When I say don't treat it as a number, I mean you can't use it in most arithmetic or algebraic manipulations.  For all real numbers $k$, you could say that if $2+k = x+k$,  then $2=x$, but saying that $2+\infty = x+ \infty$ implies $2=x$, then that is nonsense.  The same goes for saying for all numbers $k$, $0 \times k = 0$.  True... assuming infinity isn't one of these "numbers."  It doesn't have the same properties as numbers.

If you have [tex]2 + \infty = x + \infty[/tex] it doesn't mean that [tex]2 = x[/tex]. It's the same thing as what I explained in a previous post (#42), you can't substract [tex]\infty[/tex] to each side and say [tex]\infty - \infty[/tex] is 0. You (djshowdown2) said it yourself (#41), [tex]\infty - \infty[/tex] can equal anything.

However you're right : infinity is special and doesn't obey to rules. But why wouldn't it be a number ? Zero is special too, zero times (nearly) any number is zero, [tex]a + 0 = a[/tex], [tex]a - 0 = a[/tex], [tex]0 - a = -a[/tex], [tex]\frac{1}{0} = [/tex]...oops, I'm going too far ;)


hope this belongs here
Quest:
show that i (imaginary) is a zero of f (x)= x^3+ (3-i)x^2 - 4ix -1

just plug in $i$...and the expression is equal to 0, so $i$ is a root


summer program
MathPath
geometry
induction

Hi everyone! I'm at MathPath and having a good time!

Post here if you are at MathPath!

Also post what you think of it and what is going on here!

I'll start:

MathPath so far is great.

We're doing a Chess tournament (WHICH ISN'T WORKING SO WELL)
We do Soccer/Basketball/Volleyball/Pickleball/Salsa dancing (lol) on most nights.
The classes are fun. We had Professor John Conway do talks, Mr. M talks about writing math,
Mr. L does Problem of the day (Yes... with prizes)
Mr. V does Breakouts, etc.

Normal Day:

6:45-7:15: Wake Up/ Get ready
Wake up, get ready
7:15-8:15: Breakfast
About a 10 min walk to the cafeteria
8:30-9:30: History of Mathematics
Glen discusses, well, the history of Mathematics
9:45-10:35: Breakout session
Several sessions run at once, with you choosing which ones you want. Examples are: Anyl Geometry, Nats MC, Reg MC, Induction, etc.
10:45-11:45: Plenery Lecture
A select person speaks about a certain topic, which varies.
12:00-1:00: Lunch
See Breakfast
1:15-2:15: Break
There's Ping Pong, Pool, Chess, Piano, Fusball, etc.
2:30-2:50: Writing Mathematics
Qualifying test papers are used to illustrate good & not-so-good examples of writing math.
3:00-4:20: Breakout session
See above.
4:30-5:00: Annoucments, Problem of the day.
Staff make nessacary annoucments and the Problem of the day is discussed.
5:10-6:15: Dinner
See Breakfast
6:20-6:55: Break
6:55-8:30: Evening Activities
Soccer, Volleyball, Basketball, etc.
9:00- ~9:15: Counselor meeting
Counselors make any Announcments needed.
9:30-10:00: Quiet Time
Get ready for Bed
10:00 Lights Out!

-BOGTRO.
- If you are here, see my name tag. It has paperclips on it.

err not at mathpath... but just a random comment that someone will probably tell the mod to edit =D.

Mr. M is the name of a guy in Rurouni Kenshin!! Sano makes fun of him cuz he's a fat sumo wrestler =D... okies end randomness.

Slightly on topic, but not really
Wait what only like 2 hours of break in total o.o?

??? Delete the above and this post.

Clarification: Mr. M is NOT Rurouni Kenshin. His name is Steven Maurer.

Mr. M is NOT a sumo wrestler. lol

And we have break from 6:10-10:00.

Well yay. Post something useful. Such as what you think of it and what we do here (fr people not here)

Im at mathpath, and its cool. hey BOGTRO whos ur favorite teacher/counselor? Oh, I know, it's Charles! You really should take salsa dancing. and oh yeah, are u forgetting something? that should give you a good enough pic of who i am. :P

I LOSE!!!

Too bad MP will be over in 2 days  :(

Time passes really fast doesn't it?
Just a few more days to normal school life again... :(


We had to do this park project thing in class where we were supposed to design a park with an area of six acres. My group made a shape and labeled it with dimensions using acres, but someone pointed out that acres are 2 dimensional. Any ideas?

Acres are two dimensional, but note that they are only the area of [i]land[/i] used. How about you make a 2 "acre" x 3 "acre" land, and then add what you will to the landscape? If you're designing a park, you can add trees or a pond or benches, something like that.

Hope this helps.

That's a nice idea. :thumbup: Normally, I wouldn't put so much effort (lazy as I am) into something like this, but it's worth half our grade.


Stop asking this!

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=144432

[quote="arqady"]Somebody can explain to me, why it is forbidden to post the solution of this problem? Thanks.[/quote]
Why is it forbidden?
If you do include the solution though, please surround it with hide tags.

[quote="toan hoc"]Let $a;b;c$ be non-negative realnumbers satisfying $a+b+c=1$.prove that:
$\sqrt{a+(b-c)^{2}}+\sqrt{b+(c-a)^{2}}+\sqrt{c+(a-b)^{2}}\geq \sqrt{3}$[/quote]

Just out of interest, since I am just starting to learn the SMV-Theorem posted in the Inequality Forum, can someone tell me if this can be solved using SMV-Theorem? If it can't, then why not.

Thanks.

[quote="arqady"][quote="toan hoc"]Nobody can prove it!
[/quote]
OK ( it's not hard ) but explain, why was closed this topic:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=144432[/quote]
i think this topic was closed because this problem is in the newest magazine in Vietnam and its not expire ....


A pretty simple logarithm problem for beginners.

For some positive real $ A$ and $ B$:

\[ \log A \plus{} \log B \equal{} 7\]

\[ \log A \minus{} \log B \equal{} 8\]

Find $ A$ and $ B$.

[hide="Solution"]
Add the equations vertically, for $ 2\log A \equal{} 15$, so $ \log A^2 \equal{} 15$, so $ A^2 \equal{} 10^{15}$, so $ A \equal{} 10^{7.5}$
Similar procedure for B.
[/hide]

Yup.

Now, another approach is to directly apply the logarithm property on each equation separately. Like...

$ \log(AB) \equal{} 7$ and $ \log(A/B) \equal{} 8$, does that help us with anything?

Yes. Now we can find that

$ AB\equal{}10^7$
$ \frac{A}{B}\equal{}10^8$.
Multiplying the two equations we get

$ A^2\equal{}10^{15}$

and we're done.

Bye
GM


topology
real analysis
real analysis unsolved

Give me an example of a nowhere dense set with positive measure.

Thanks.

This is trivial if you do not say which space you are working with, and with what measure. I could perfectly take $\mathbb{R}$ with the usual topology and the counting measure; in this case, the set $\{1\}$ will be nowhere dense in $\mathbb{R}$, and will have positive measure $1$.

If now you consider $\mathbb{R}$ with the euclidean topology and the Lebesgue measure, you can do as follows:

Consider a sequence $(r_{n})_{n \geq 1}$ of all the rational numbers in $[0,1]$, and the set $E=[0,1]-\cup_{n \geq 1}{B(r_{n},2^{-n-1})}$, where $B(x,r)$ denotes the open ball with center $x$ and radius $r>0$. Clearly $E$ is nowhere dense (since it is closed with empty interior), and $\lambda(E)=\frac{1}{2}$.


superior algebra
superior algebra unsolved

Let $ G$ be a group and $ a,b\in G$ .
If $ a^{ \minus{} 1}*b^2*a \equal{} b^3$ and $ b^{ \minus{} 1}*a^2*b \equal{} a^3$. Show that $ a \equal{} b \equal{} e$ where $ e$ is the identity element of $ G$


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x^2}{(1\plus{}x\tan x)(x\minus{}\tan x)\cos ^ 2 x}\ dx.$

That's correct.

I should have posted the problem as $ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x^2}{(x\sin x\plus{}\cos x)(x\cos x\minus{}\sin x)}\ dx.$ :wink:

In that case, my approach would be rather different. :lol:

$ \frac{x^{2}}{(x \sin x \plus{} \cos x)(x\cos x \minus{} \sin x)}$

$ \equal{} \frac{x^{2}\cos ^{2} x \minus{} x\sin x\cos x\plus{} x^{2}\sin^{2} x \plus{} x\sin x\cos x}{(x\sin x \plus{} \cos x)(x\cos x\minus{}\sin x)}$

$ \equal{} \frac{x\cos x}{x\sin x \plus{} \cos x} \plus{} \frac{x \sin x}{x\cos x \minus{} \sin x}$

$ \equal{} \frac{x\cos x}{x\sin x\plus{} \cos x} \minus{} \frac{\minus{}x\sin x}{x\cos x\minus{}\sin x}$

$ \Rightarrow$ integral $ \equal{} \ln(x\sin x \plus{}\cos x) \minus{} \ln(x\cos x\minus{}\sin x) \plus{} c$

Putting in the limits, I get

$ \ln \frac{(\pi\sqrt{3}\plus{}3)(\pi \minus{} 4)}{(\pi\minus{}3\sqrt{3})(\pi\plus{}4)}$

I have no word to add it more than that. :lol:

[quote]Indochina wrote ;

=$ \int_{\pi /4}^{\pi /3}\frac{x^{2}\sec ^{2}x}{\left( 1\plus{}x\tan x\right) \left( x\minus{}\tan x\right) }dx$

=$ \allowbreak \int_{\pi /4}^{\pi /3}\left( \frac{\tan x\plus{}x\sec ^{2}x}{1\plus{}x\tan x}\minus{}\frac{1\minus{}\sec ^{2}x}{x\minus{}\tan x}\right) dx$

[/quote]

How can we easily split the integral so easily??

[quote="IndoChina"]$ \int_{\pi /4}^{\pi /3}\frac{x^{2}}{\left( 1\plus{}x\tan x\right) \left( x\minus{}\tan x\right) \cos ^{2}x}dx$

$ \equal{}\int_{\pi /4}^{\pi /3}\frac{x^{2}\sec ^{2}x}{\left( 1\plus{}x\tan x\right) \left( x\minus{}\tan x\right) }dx$

$ \equal{}\allowbreak \int_{\pi /4}^{\pi /3}\left( \frac{\tan x\plus{}x\sec ^{2}x}{1\plus{}x\tan x}\minus{}\frac{1\minus{}\sec ^{2}x}{x\minus{}\tan x}\right) dx$

$ \equal{}\left[ \allowbreak \ln \left( \frac{1\plus{}x\tan x}{x\minus{}\tan x}\right) \right] _{\pi /4}^{\pi /3}$

$ \equal{}\allowbreak \ln \frac{\left( 3\plus{}\pi \sqrt{3}\right) \left( 4\minus{}\pi \right) }{\left( 3\sqrt{3}\minus{}\pi \right) \left( 4\plus{}\pi \right) }$[/quote]
Sir here is problem how this step come directly .or how can we think about such a step
$ \equal{}\allowbreak \int_{\pi /4}^{\pi /3}\left( \frac{\tan x\plus{}x\sec ^{2}x}{1\plus{}x\tan x}\minus{}\frac{1\minus{}\sec ^{2}x}{x\minus{}\tan x}\right) dx$

[b]See[/b] the [u]exercise[/u] [b]1.6[/b] from [url=http://www.artofproblemsolving.com/blog/38414][color=darkred][b]here[/b][/color][/url], i.e. $1.\blacktriangleright\ \mathrm{(sixth\ ex.\ in\ first\ row)}\ \int\frac{x^2}{(x^2-4)\sin x+4x\cos x}\ \mathrm{dx}$

which by the substitution $x:=2x$ obtain immediately the [u][b]Kunny[/b]'s proposed exercise[/u]. There is another  its proof.

[b][u]Remark[/u].[/b] Indeed, $\frac{x^{2}}{(x \sin x \plus{} \cos x)(x\cos x \minus{} \sin x)}=\frac {2x^2}{\left(x^2-1\right)\sin 2x+2x\cos 2x}$ .


modular arithmetic
induction
number theory unsolved
number theory

[quote="moldovan"]An integer is said to be digitally divisible if:

$ (i)$ none of its digits is zero, and
$ (ii)$ it is divisible by the sum of its digits (e.g. $ 322$ is digitally divisible).

Prove that there are infinitely many digitally divisible integers.[/quote]

$ f(n)\equal{}\frac{10^{3^n}\minus{}1}{9}$ for all $ n$ are digitally divisible integers.

[b]Proof:[/b] Sum of its digit $ \equal{}3^n$. Clearly its true for $ n\equal{}1$.
Assume that $ 3^k|f(k)$.
$ \frac{f(k\plus{}1)}{f(k)}\equal{}10^{2(3^k)}\plus{}10^{3^k}\plus{}1 \equiv 1\plus{}1\plus{}1\equal{}0 \pmod{3} \Rightarrow$ the statement is true for $ n\equal{}k\plus{}1$ also. So the statement is true by principle of mathematical induction, done.

Suppose there are finitely many such numbers - and let $ a$ be the biggest "digitally divisible" number. Also suppose that $ a$ consists of $ l$ digits. And let the sum of its digits be $ k$. According to the statement, $ k$ divides $ a$.

Designate the number $ (10^{2l} \plus{} 10^{l} \plus{} 1)\cdot a$ as $ x$. Clearly the sum of digits of $ x$ is $ 3k$. None of its digits is zero, and $ x$ is divisible by $ 3k$,  as $ 10^{2l} \plus{} 10^{l} \plus{} 1$ is a multiple of $ 3$. Clearly, $ x>a$, so we obtain a contradiction.


If  $ a,b,c,d$  are  positive numbers, then
$ \frac{c}{a\plus{}2b}\plus{}\frac{d}{b\plus{}2c}\plus{}\frac{a}{c\plus{}2d}\plus{}\frac{b}{d\plus{}2a} \ge \frac{4}{3}$.

[quote="soruz"]If  $ a,b,c,d$  are  positive numbers, then
$ \frac {c}{a \plus{} 2b} \plus{} \frac {d}{b \plus{} 2c} \plus{} \frac {a}{c \plus{} 2d} \plus{} \frac {b}{d \plus{} 2a} \ge \frac {4}{3}$.[/quote]

I think, $ \frac {c}{a \plus{} 3b} \plus{} \frac {d}{b \plus{} 3c} \plus{} \frac {a}{c \plus{} 3d} \plus{} \frac {b}{d \plus{} 3a} \ge 1$ nicer. :wink:


You have n people, a1 a2 a3....an , each having a probability 'p' of dying in a certain year. If that year is fixed for all , find the probability of one who dies the 1st.
My solution is below but i wanted to know if there is another method.

[hide]
the usual is if we assume , a1 to die 1st we have p
then if a1 doesnt die and a2 dies we have p + (1-p)*p
so on and so forth for the 1st person to die we get the some as
1-(1-p)^n

The some of a G.P.
Is there any other method?[/hide]

[quote="ocha"]Is $ p$ the same for all $ a_i$?
What do you mean when you say "that year is fixed for all"; everybody will die in that year??[/quote]
Yes. all of them will be dying since they are of the same age<forgot to state this sorry>
You have to find the probability that one of them dies and dies the 1st.

I'm not quite sure what you meant by your solution (there are places in your argument whose meaning I failed to understand; it doesn't mean they are incorrect, just unclear), but here is
[hide="my argument"]
Let $ A_j$ be the event that the $ j$-th person dies and dies first. Note that the events $ A_j$ are disjoint, have equal probability (by symmetry), and their union is the event that there will be at least one death in the given year. Thus, $ P(A_j)\equal{}\frac 1n[1\minus{}(1\minus{}p)^n]$
[/hide]


I've mentioned this before, and I wondering if there's any development on that.

Due to the huge number of problems on the forum, it's becoming increasingly difficult to look for a particular one. Thus, it would a good idea to allow the viewing of all the first post of all the threads in a forum (the first post usually contains the problem).

Alternately, there could be an option in the search menu that allows the result to display only the first post of all the threads that were hit.

Speaking of searching, the "display results as posts" option does not allow the display of LaTeX, and that's quite a nuisance.

There is a method to view only first posts in threads with the source field. Just search for "*a* *e* *i* *u* *o*" with the option only in source fields. This will basically cover all source fields possible, and if you also select the option display as posts, you will obtain only the first posts in each thread that has a source ;)

About the other issues, they will be solved in the near future.

Can you do that in artofproblemsolving style?

You must know that all the problems with sources are from Advanced and College sections, so the Getting Started, Intermediate and most of PreOlympiad problems cannot be found in this matter. I just checked, for now, that method cannot be applied on the AoPS skin. We'll make it available there soon.


geometry
trigonometry
trig identities
Law of Sines
similar triangles

In a baseball game, a batter hits a ground ball in the general direction of the shortstop, but not directly at the shortstop.  The shortstop knows the direction of the ball at the moment the batter makes contact, and both the ball and the shortstop travel at constant but possibly different speeds, starting the moment the batter makes contact.  The shortstop chooses a direction to run in, and given any choice of direction, there is a maximum ball speed $ v$ such that the shortstop can reach the ball.  What direction should the shortstop choose, with respect to the batter and the path of the ball, to maximize $ v$?

The other consideration is, of course, that if you take too long getting to the ball you aren't going to make the throw to first.

[quote="MellowMelon"][hide="Solution"]Suppose the short stop is at $ A$ and the ball is hit at $ O$. We want to find a point $ B$ such that $ \angle AOB$ is some fixed value and $ AB/OB$ is minimal. By law of sines on $ \triangle AOB$ this occurs when $ \angle OAB$ is a right angle. So the shortstop should always run at an angle perpendicular to the point from where the ball is hit, no matter which direction the ball is going.[/hide][/quote]

Ah, nice.  My solution used similar triangles and basically required you to guess the answer:

[hide="Solution"]Define $ A,O,B$ as MellowMelon did.  Note that $ A$ and $ O$ are fixed, and $ B$ is on a fixed line $ l$ through $ O$.  Draw the perpendicular to line $ AO$ through $ O$, and let this meet $ l$ at $ B'$.  Drop a perpendicular from $ B$ to line $ AO$, meeting $ AO$ at $ P$.  Since $ OB\ge PB$ with equality when $ O\equal{}P$, we have $ AB/OB\le AB/PB\equal{}AB'/OB'$, so $ B'$ is the unique point that minimizes $ AB/OB$.  Therefore, the shortstop should run perpendicular to $ AO$.[/hide]

worthawholebean: True.  The shortstop should only run in this direction if the ball is actually travelling at that maximum speed.  But I'm not sure if a more accurate reformulation of the problem would be interesting to solve.


Jane's quiz scores were 98, 97, 92, 85 and 93. What was her mean score?

Respectively, each of 98, 97, 92, 85 and 93 are 2, 3, 8, 15, and 7 away from 100. 2, 3, 8, 15, and 7 add up to 35, so the average is 35/5=7 less than 100, which is 93


In a circle we enscribe a regular $2001$-gon and inside it a regular $667$-gon with shared vertices.

Prove that the surface in the $2001$-gon but not in the $667$-gon is of the form $k.sin^3\left(\frac{\pi}{2001}\right).cos^3\left(\frac{\pi}{2001}\right)$ with $k$ a positive integer. Find $k$.

First we have the following: the area of a regular $ n$-gon, inscribed in a circle of radius $ r$, is $ \frac 12nr^2\sin\frac {2\pi}{n}$. (It can be easily verified from summing the areas of the isosceles triangles two of the sides of which are of length $ r$ and the angle between them is $ \frac {2\pi}{n}$.)

Hence the area of the surface stated equals
\begin{align*} & \frac 12r^2\left(2001\sin\frac {2\pi}{2001} - 667\sin\frac {2\pi}{667}\right) \\
= & \frac {667}{2}r^2(3\sin 2\theta - \sin 6\theta) \\
= & 4\times\frac{667}{2}r^2\sin^32\theta \\
= & 2\times 4\times\frac{667}{2}r^2\sin^3\theta\cos^3\theta\end{align*}
(here $ \theta = \frac {\pi}{2001}$ and $ r$ is the radius of the given circle)

Hence $ k = 2\times 4\times\frac{667}{2}r^2=2668r^2$. (This is the best we can do since nothing about the lengths of the polygon is given)

Yeah, I guess the original problem will have stated a unit circle. :)


http://www.math.swt.edu/~haz/prob_sets/notes/node11.html - with proof
http://mathworld.wolfram.com/DiophantineEquation.html - with extra stuff

Is there a simpler way?

This is the team round, where you can use a calculator.  Here's the slacker way to do it with the calculator:

37x+73y=2016
rewrite as:
y = (-37x + 2016)/73

Plug into Ti-83+ as Y_1 = (-37x + 2016)/73
Look at table.  Scroll down until you see an integral value for Y.

When x=17, y=19.  There's your answer.


Solve the equation: $2sin(17x)+\sqrt{3}cos(5x)+sin(5x)=0$

divide by 2, and we get

$\sin 17x+\sin 60\cos5x+\cos 60\sin 5x=0$

$\sin 17x=-\sin (5x+60)$

the rest is trivial


linear algebra
matrix
algebra
polynomial

Consider the following matrix that I'd like to know how to show it has a full rank.

$\begin{bmatrix}\phi_1(t_1)&\phi_1(t_1)&...&\phi_m(t_1)\\\phi_1(t_2)&\phi_2(t_2)&...&\phi_m(t_2)\\\phi_1(t_3)&\phi_2(t_3)&...&\phi_m(t_3)\\...\\...\\ \phi_1(t_n)&\phi_2(t_n)&...&\phi_m(t_n)\end{bmatrix} \begin{bmatrix} x_1\\x_2\\.\\.\\x_m \end{bmatrix}=\begin{bmatrix} y_1\\y_2\\y_3\\.\\.\\y_n \end{bmatrix}$

I know that every nonzero polynomial of degree $\leq m-1$ is zero at most $m-1$ destinct points. But how would I use this fact to show it has a full rank. Please help, thanks in advance.

It is equivalent to showing that a $mXm$ matrix carved from the original matrix has rank $m$ and denote it by $P_i(t_j)$. For that we consider the first $m$ lines . We can write : $\phi_i=a_{i1}+a_{i2}X+...+a_{im}X^{m-1}$, denote by $P_{ij}$ the matrix of the coeficients of the polynomials $\phi_i$. . Noticing the fact $P_i(t_j)=P_{ik}(t_j^{k-1})$ we can say that $\det(P_i(t_j))=\det(P)\cdot V(t_1,t_2,...,t_m)$. Where $V$ stands for Vandermonde. I hope that I understood correctly what you are asking

Hi, Thanks for your helping, but I dont think I have learned that far to be able to understand your explanation which is " Noticing the fact $P_i(t_j)=P_{ik}(t_j^{k-1})$ we can say that $\det(P_i(t_j))=\det(P)\cdot V(t_1,t_2,...,t_m)$. Where $V$ stands for Vandermonde. " Is that the only to show it has a full rank or any other simple way? Sorry for not having that much further background to understand.

What I did was to compute the value of the determinant of a matrix like this one :
$P_i(x_j)=\left (\begin{array}{ccccc} P_1(x_1)&P_1(x_2)&P_1(x_3)&...&P_1(x_n)\\P_2(x_1)&P_2(x_2)&P_2(x_3)&...&P_2(x_n)\\...&...&...&...&...\\P_n(x_1)&P_n(x_2)&P_n(x_3)&...&P_n(x_n) \end{array} \right )$
where $P_i=a_{i1}+a_{i2}X+...+a_{in}X^{n-1}$ . And I got $\det(P_i(x_j))=\det(P)\cdot Vandermonde(x_1,x_2,..x_n)$ where $P$ is the matrix obtained from the coefficients of $P_i$.

   My idea was that if that big matrix should have full rank then it should have it`s rank equal to $\min(m,n)$ and this would mean that a determinant of the type above should be different from zero.


geometry
3D geometry
limit
irrational number

Here is something I think is true but can't prove.

If a and b are two numbers and a/b is irrational, then ab is also irrational.

And vice versa.

I've started (in spoiler), if you can give a hint without actually proving it that would be nice.

[hide]Assume ab is rational, that is, ab=n/m for integers n, m.  Then m*a=n*(1/b).  Since m and n are integers, it seems to me that a and 1/b will have to have a common, irrational factor that makes a/b rational, contradicting the given, but I can't seem to show it.[/hide]

[b]Not true:[/b]  let   a = 3^(1/3)  (that is cube root of 3) and b = 9^(1/3)  (= Cube root of 9)

a/b  is irrational
ab is rational = 3

Thanks.  It is true if you limit your numbers to square roots, though, since if ab=r then a/b=r/b^2 and since b^2 would be rational, r/b^2 would be rational, which is a contradiction.  I guess from that I thought they would all work.

Reminds me of an interesting problem: Prove that is possible to  have
[b]a^b [/b]as rational while both [b] a[/b]  and [b]b[/b] are irratonal

  (Or give an example where this is true)

(I will give one  "well known" solution which is fun  - if no one else has given it  in a few days..)

<kill -joy > If you are thinking of   e^(ln 2)  or some thing like that .. you may have to give a simple proof that e and (ln 2) both are  irrational  . </kill joy>

[quote="nr1337"]More examples:

\[
e,\frac{1}{e}
\]
\[
\pi,\frac{1}{\pi}
\]
\[
\phi,\frac{1}{\phi}
\]

etc...[/quote]

What does the phi mean?

It refers to the golden section, or $\frac{1+\sqrt(5)}{2}$

[quote="Gyan"]
(Or give an example where this is true)

(I will give one  "well known" solution which is fun  - if no one else has given it  in a few days..)
[/quote]

And the coolest thing about that solution (which I'm sure is the one that I'm thinking of) is that it does *not* give an example where it is true :) Which is probably why you had that in brackets.

[quote="pakagawa"]How can we calculate e again?

I know it's euler's "number" and theorem.

And by the way, can any root work for that certain equation?[/quote]

\[e=\lim_{n\to \infty}{\left(\frac{n+1}{n}\right)^n}\]

You could just put big numbers in there to get a good approximation or you could use the alternate form of the above to approximate it:

\[e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots\]

I can't quite read Gyan's mind, but the "well known" example that immediately popped into my mind is [hide]e^(i*pi)=-1 ...then again, that involves an unreal number....[/hide]

I'm not psychic either, but theres a more well-known example I think Gyan is thinking of. What number do you think of when you think "irrational number"?

James wrote:Here's the well-known solution (I assume it is obvious that the square root of two is irrational):

[hide]Call sqrt(2)^sqrt(2) x. Then x^sqrt(2) is rational. The great thing about this proof is that it is not necessary to show that x is irrational. If it is rational, then sqrt(2)^sqrt(2) is rational.[/hide]

Yeah, thats the idea - although you can state it slightly more clearly:

[hide]

If x = sqrt(2)^sqrt(2) is rational, then that is an example. If it is irrational, then x^sqrt(2) = 2 is rational. Therefore either sqrt(2)^sqrt(2) is rational, or it is irrational in which case (sqrt(2)^sqrt(2))^sqrt(2) is rational. The proof doesn't actually say which of those two examples is the correct one.

[/hide]


http://mathworld.wolfram.com/SkolemSequence.html

Show that a necessary condition for a Skolem Sequence of order n to exist is n=4k or n=4k+1

btw since this forum allows editing of headers then may i make a suggestion to add something like "RESOLVED: " in the header after a question have been solved

I can't prove that, but I can prove 10c.

Let (a1,a2,.....,an) be the indices of the first occurences of the integers from 1 to n.

Then there is a set (a1+1,a2+2,.....,an+n) which represents the indices of the second occurences.

Clearly, there are 2n indices, so the sum of the indices is equal to:

2n(2n+1)/2

This must be equal to 2(a1,a2,......,an)+(1+2+....+n)

So 2(a1,a2,......,an)+(1+2+....+n)=2n(2n+1)/2

So (a1,a2, ..... , an) = n(3n+1)/4

Clearly, n(3n+1)  :equiv: 0 mod 4

So n :equiv: 0 mod 4 or n :equiv: 1 mod 4.

The proof is not mine, but I cleaned up the explanation and notation.


A string in the shape of a 2 X 8 inch rectangle can be shaped into a square whose side is

A) 4 inches B) 5 inches C) 16 inches D) 20 inches

i think ill choose b

just a perimeter problem

Since the perimeter is $ 4 \plus{} 16 \equal{} 20$, the side length is $ 20 \div 4 \equal{} \boxed{5}$, or B.


Find all functions $ f: \mathbb N^2 \to \mathbb R$ satisfying
\[f(x,x)\equal{}x, \quad  f(x,y)\equal{}f(y,x),\quad \text{and} \quad(x\plus{}y)f(x,y)\equal{}yf(x,x\plus{}y)\]
for all positive integers $x$ and $y$.

[quote="math10"]Find all function $ f$,defined onthe sets of ordered pairs of positive integers,satisfying properties:
 $ f(x,x) \equal{} x; f(x,y) \equal{} f(y,x)$ and $ (x \plus{} y)f(x,y) \equal{} yf(x,x \plus{} y)$[/quote]

Must we understand that $ f(x)$ is from $ \mathbb N^2\to\mathbb R$ ? (my question is about image)

[quote="math10"]Find all function $ f$,defined onthe sets of ordered pairs of positive integers,satisfying properties:
 $ f(x,x) \equal{} x; f(x,y) \equal{} f(y,x)$ and $ (x \plus{} y)f(x,y) \equal{} yf(x,x \plus{} y)$[/quote]

Let $ P(x,y)$ be the assertion $ (x \plus{} y)f(x,y) \equal{} yf(x,x \plus{} y)$

1) There is at most one solution
If $ x\equal{}y$, $ f(x,x)\equal{}x$
$ P(1,x)$ $ \implies$ $ (1\plus{}x)f(1,x) \equal{} xf(1,1\plus{}x)$ $ \implies$ $ f(1,x\plus{}1)\equal{}\frac{x\plus{}1}xf(1,x)$ $ \implies$ $ f(1,x)\equal{}x$
If $ x\neq y$, Wlog say $ x<y$, then $ yf(x,y\minus{}x) \equal{} (y\minus{}x)f(x,y)$ and so $ f(x,y)\equal{}\frac{y}{y\minus{}x}f(x,y\minus{}x)$
And so we have a finite descent method with and ending situiation either $ f(1,x)$, either $ f(x,x)$
Q.E.D

2) $ f(x,y)\equal{}lcm(x,y)\equal{}\frac{xy}{\gcd(x,y)}$ fits the equation
$ lcm(x,x)\equal{}x$
$ lcm(x,y)\equal{}lcm(y,x)$
$ \gcd(x,x\plus{}y)\equal{}\gcd(x,y)$ $ \implies$ $ lcm(x,x\plus{}y)\equal{}\frac{x(x\plus{}y)}{\gcd(x,x\plus{}y)}$ $ \equal{}\frac{x(x\plus{}y)}{\gcd(x,y)}$ $ \equal{}\frac{x\plus{}y}{y}\frac{xy}{\gcd(x,y)}$ $ \equal{}\frac{x\plus{}y}{y}lcm(x,y)$
Q.E.D.

3) Hence the only solution is $ f(x,y)\equal{}lcm(x,y)\equal{}\frac{xy}{\gcd(x,y)}$

Yes,you had right.
This is nice problem,nice solution :lol:


calculus
integration
SFFT
special factorizations

Find all possible values for $9a+9b = ab$

right...forgot about the negative ones.  :oops:

btw, the method of factoring is called simon's favorite factoring trick, or SFFT


I think $N=2^{n-1}$ . Anyway,  a more precise interval can be founded..
Let $P_n,S_n$ be  $P_n = a_1...a_n  , \ S_n = \frac  1{a_1} + ... +\frac  1{a_n}$
I obtain, by using induction : $a_{n+1} = P_n+1 ,\ S_n = 1 - \frac 1{P_n}$
If $u,v$ are positive integers and  $u \leq P_n < v$ then $u^2 < P_{n+1} < v^2$

Finally : $1-\frac 1{P_k^{N_n}} \leq  S_{n+k-1} <  1-\frac 1{(1+P_k)^{N_n}}$, where $N_n=2^{n-1}$
For ex. , with $k=1 \Longrightarrow 1-\frac 1{2^{N_n}} \leq  S_n <  1-\frac 1{3^{N_n}}$.
 :cool:


Five friends compete in a dart-throwing contest. Each one has two darts to
throw at the same circular target, and each individual's score is the sum of the
scores in the target regions that are hit. The scores for the target regions are the
whole numbers 1 through 10. Each throw hits the target in a region with a
different value (e.g., two people can't both hit target value 3). The scores are: Alice 16 points, Ben 4
points, Cindy 7 points, Dave 11 points, and Ellen 17 points. Who hits the
region worth 6 points?

Since you know that Ben had to have got a $ 1$ and $ 3$ because you can't make four points with any other combination.

Then the only combination available for Cindy is $ 2$ and $ 5$.

Then the only combination available for Dave is $ 4$ and $ 7$.

Which makes the only combination available to Alice $ 6$ and $ 10$.

So therefore the answer is $ Alice$.


Inca o chestie din programa, pe care n-o gasesc niciunde: Ce e teorema de completare la o baza?(la spatii vectoriale) V-as ruga sa imi raspundeti in maxim 1 saptamana :)

Daca $M$ este o multime de vectori liniar independenti, atunci exista o baza $B$ a spatiului vectorial astfel ca $M\subseteq B.$

(related:) Lema substitutiei, Teorema lui Steinitz.

Din cate am vazut, nu s-a prea dat in nici un an problema cu spatii vectoriale, asa ca nu prea insist pe ele.  Vreau doar sa acopar programa, pt. ca in caz ca totusi se da ceva, sa nu ma zgarai pe ochi dupa aceea ca nu am citit inainte.


ratio
limit
real analysis
real analysis unsolved

If $ 0 < p \le |a_n| \le q$ for all $ n \in N$, find the radius of convergence of $ \sum a_n x^n$.

Observe that $ \left|\frac{a_{n\plus{}1}}{a_n} \right| \leq \frac{q}{p}$. So, for $ |x| < \frac{p}{q}$ we have  $ \left|\frac{a_{n\plus{}1}x^{n\plus{}1}}{a_nx^n}\right| \leq \frac{q}{p} |x| < 1$.

The ratio test is inferior to the root test, and the radius of convergence is actually $ 1,$ not $ \frac pq.$

$ \lim \sqrt[n]{p}\le \limsup\sqrt[n]{|a_n|}\le \lim\sqrt[n]{q},$ so $ \limsup\sqrt[n]{|a_n|}\equal{}1.$

Furthermore, the series does not converge when $ |x|\equal{}1,$ as the terms do not tend to zero.


[quote="Rep123max"]Hello.

I'm reading [i]Free to Choose[/i] and am on the section of the Federal Reserve. There's one paragraph that's just not clicking with me.

Initially, it was expected that the Federal Reserve Banks would operate mostly by direct loans to banks, on the security of the banks' own assets, in particular, the promissory notes corresponding to loans by banks to businesses. In many such loans, the banks "discounted" the notes--that is, paid out less than the face amount, the discount representing the interest charged by the banks. The Federal Reserve in turn "rediscounted" the promissory notes, thereby charging the banks interest on the loans.

I'm missing the connection between the Fed and the promissory note in the last sentence. And while we're at it, can somebody explain what exactly a promissory note is to me? Thanks[/quote]
A promissory note is in effect a check. (A contract made by one person to pay money to another.) The last sentence basically says that the Fed re-discounted the notes, agreed to accept less in repayment of the notes than what they were originally worth, and this effectively tinkered with the interest rate that banks were charged in repaying what they had barowed from the Federal Reserve.

[quote="Rep123max"]I'm sorry but it's still not clicking. So do banks issue promissory notes? And how does the Fed come into contact with the promissory note?

Thanks[/quote]

As I understand it, the Federal Reserve is the [i]maker[/i] of the promissory note, and a given bank is the [i]payee[/i]. The Fed issues promissory notes with the expectation that they will be repaid with interest. When the Fed discounts a promissory note, it in effect reduces the interest that the bank holding the note must repay.


function
superior algebra
superior algebra unsolved

Find a bijection between the real line R and the unbounded interval (0, inf) = {x in R: x>0}.

Can the bijection be f: x-->x+1?

No. That maps the real line to the whole real line, not just the positive part.

That's not injective, and not surjective onto the positive numbers either.

How is it not surjective?  Isn't there a pre-image x for every image |x|+1?
Maybe I'm missing something...any hints?

It's not surjective because you can't get $ \frac12$, or anything else less than $ 1$.

The easiest examples to think of are continuous. A continuous injective function on the real line is either increasing or decreasing. You want something that increases to $ \infty$ on one end and only decreases to zero on the other.

By the way, this is definitely the wrong forum, and you shouldn't post multiple copies of threads.


Did I do this one correctly?

If a = f - 2 and b = 4f + 5, what is an expression for b in terms of a?

It's multiple choice:

4a
4a - 4
4a + 2
4a + 7
4a + 13

a = f - 2 and b = 4f + 5

add 2 to each side ----> a = f and b = 4f + 7

my answer is 4a + 7

Here are our equations:

$ a \equal{} f \minus{} 2$

$ b \equal{} 4f \plus{}5$

Let's find $ f$ in terms of $ b$ and then substitute it in our first equation.

$ b \equal{} 4f\plus{}5$

$ b \minus{} 5 \equal{} 4f$

$ \frac {b \minus{} 5}{4} \equal{} f$

$ a \equal{} f \minus{} 2$

$ a \equal{} \frac {b \minus{} 5}{4} \minus{} 2$

Now we just clean up this equation.

$ 4(a) \equal{} 4(\frac {b \minus{} 5}{4}) \minus{} 4(2)$

$ 4a \equal{} b \minus{} 5 \minus{} 8$

$ 4a \equal{} b \minus{} 13$

Since we want $ b$ in terms of $ a$...

$ b \equal{} 4a \plus{}13$

This is my first instinct on solving it. Your solution was much easier, you just made a mechanical error.

What you did:

$ a \equal{} f \minus{} 2$

$ a \plus{} 2 \equal{} f$

$ b \equal{} 4(a \plus{} 2) \plus{} 5$

$ b \equal{} 4a \plus{} 8 \plus{} 5$

$ b \equal{} 4a \plus{}13$

I think you did $ 8 \plus{} 5 \equal{} 7$ by mistake.

PS: It's okay -- I do stuff like that ALL the time.

Duh-my brain is fried. Thank you for showing me where my error is. It really helps.

No. You added 2 incorrectly. You don't add two to both equations,

but rather, both sides.

Equation (1): $ a\equal{}f\minus{}2 \implies a\plus{}2\equal{}f$.

Equation (2): $ b\equal{}4f\plus{}5 \implies b\equal{}4(a\plus{}2)\plus{}5$. The $ 4(a\plus{}2)$

comes from $ f\equal{}a\plus{}2$ and substituting.

So, $ b\equal{}4(a\plus{}2)\plus{}5\equal{}4a\plus{}8\plus{}5\equal{}\boxed{4a\plus{}13}$.


function
integration
real analysis
real analysis unsolved

Prove that if $ f: \mathbb{R}\to{\mathbb{R}}$ is a function for which exists $ f^{(4)}(x)$,  and $ f(x)\ge{0}$, $ f^{4}(x) < \equal{} 0$ for every $ x\in{\mathbb{R}}$, then exist $ a,b,c\in{R}$ so that $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$.

$ f(x)\equal{}\int_u^x(\int_u^y(\int_u^z(f^{(3)}_{(t)}\minus{}f^{(3)}_{(u)})dt)dz)dy$$ \plus{}f^{(3)}_{(u)}(x\minus{}u)^3 \plus{}f^{(2)}_{(u)}(x\minus{}u)^2 \plus{}f^{(1)}_{(u)}(x\minus{}u) \plus{}f(u)$
  $ f^{(3)}$ is decreasing because of $ f^{(4)}_{(t)} \leq 0$
$ u \leq x \Longrightarrow u\leq t \leq z \leq y \leq x \Longrightarrow f^{(3)}_{(t)}\minus{}f^{(3)}_{(u)}\leq 0$ $ \Longrightarrow \int_u^x(\int_u^y(\int_u^z(f^{(3)}_{(t)}\minus{}f^{(3)}_{(u)})dt)dz)dy \leq 0$

$ u \geq x \Longrightarrow u\geq t \geq z \geq y \geq x \Longrightarrow f^{(3)}_{(t)}\minus{}f^{(3)}_{(u)}\geq 0$ $ \Longrightarrow \int_u^x(\int_u^y(\int_u^z(f^{(3)}_{(t)}\minus{}f^{(3)}_{(u)})dt)dz)dy \equal{}\minus{}\int_x^u(\int_y^u(\int_z^u(f^{(3)}_{(t)}\minus{}f^{(3)}_{(u)})dt)dz)dy \leq 0$

So $ \forall_x,\ 0\leq f(x) \leq f^{(3)}_{(u)}(x\minus{}u)^3 \plus{}f^{(2)}_{(u)}(x\minus{}u)^2 \plus{}f^{(1)}_{(u)}(x\minus{}u) \plus{}f(u)$, so we have to have  $ f^{(3)}_{(u)}\equal{}0$  for all $ u$.
Finally $ f(x) \equal{} ax^2 \plus{}bx \plus{}c$, with $ (a> 0  ,\ 4ac \geq  b^2)$  or $ (a\equal{}b\equal{}0,  c\geq 0)$
 :cool:


[size=100]Every day, Pat goes to wok in the city. She gets to the city on a train. Every afternoon, she comes back at 5 PM, and her sister Sal is waiting there in her car to pick up Pat (Sal leaves home before 5).

One day, Pat gets a monster headache  :( , and decides to come home early. She gets on a different train, and arrives at 4 PM. She decides to get some fresh air, so she walks along the same path Sal drives to take Pat home. Eventually, Pat and Sal intercept, Pat gets in Sal's car, and they go home.

She got home 10 minutes earlier, the car drove at a constant speed, Pat walked at a constant speed and the interception took no time at all. How many minutes did Pat walk?[/size]

[hide="Huge Hint"]Draw a picture![/hide]

Are you sure enough information is provided to solve the problem?

Yes. I already solved it.
[hide="Hint"]Draw a time-distance graph. [/hide]

P.S. Can someone please tell me how to put images on this forum (not as an attachment)?

upload it to photobucket or imageshack or putfile and copy/paste the URL

ask somebody else who doesn't enjoy using terms that may or may not confuse people who are technically challenged

Hmm, I'm pretty sure I've got the answer, I just have no idea how to word it, but I'll try...[hide]
Ok first of all, I know this is usually unacceptable but due to lack of any other method I will use all distances in minutes of the car's rate. By making the diagram, if 10 minutes are saved, this means that Sal drives 5 minutes less than usual both ways. So if the total distance from house to train station is $x$ then Sal drives $x-5$ minutes. Regardless of what $x$ value is entered you will still get the time of intersection at $4: 55$, since $5$ minutes are saved in getting to the station and since she reaches the station at $5: 00$. So $4: 55$ $-$ $4: 00$ $=$ $0: 55$.
 [b]ANSWER[/b]: $55$ [b]MINUTES[/b][/hide]

You got the right answer(atleast that's the one I got :P) , but these kind of problems can be solved very easily by using a distance-time graph. Can I see you make one?

Ok, maybe I don't understand the question properly.  But she must be walking over an hour because she started walking at 4 pm.  She usually gets picked up at 5 pm, and arrives home at some time clocktime_home=5pm + t_travel, where t_travel=usual travel time from train station to home.  If she got home 10 minutes earlier than usual, then she arrives at clocktime_new=5pm+(t_travel-10min). But she started walking at 4 pm, which means her total travel time is: t_new=1hr+(t_travel-10min)=t_travel+50min.  In that time, she travels the same distance as usual, but part of the way at velocity v_walking, and the remaining time at v_car, which is the usual speed for the car.  Thus, since the distances are equal:

vcar*t_travel = v_walking*t_walking + v_car*t_car  (1)

where t_new=t_walking+t_car (her time walking plus her time in the car) was given above.  Thus, t_car=t_new-t_walking=t_travel+50min-t_walking.  Substituting this into (1) and solving for t_walking gives:

t_walking = (50 min.)*v_car/(vcar-v_walking)

So, if this is correct, then the velocity of the car, and the velocity walking need to be known before we get the answer. Since t_walking has to be greater than 1 hr., v_walking>v_car/6.

Any comments?

Could you use $\LaTeX$?
To express clocktime_home, just put clocktime_{home} around dollar signs $\$$.
You get something like this: $clocktime_{home}$.

This is the first problem given in the section after the famous monk
problem in ACoPS, and the hint given is to draw a distance time graph, if that helps. The solution above seems to be a description of that solution...


Does anyone know the exact date and location of the 2004 junior Balkan Math Olympiad ?

It seems it should be set by now.

Thanks !
Steve

Sorry Steve, we (Romanian committee) have no idea (yet) when it will be. We are still waiting for the official faxes.


If $ x^2 \minus{} 2 \times x \equal{} 1$, then what's the numerical value of:
$ {(x \plus{} 1)} \times {(x^7 \plus{} \frac{1}{x^9})} \times {(x \minus{} 1)}$

[hide="WRONG Solution"]
$ x^{2} \minus{} 2x \minus{} 1 \equal{} 0 \rightarrow \left( x \minus{} 1 \right) ^{2} \equal{} 0 \rightarrow x \equal{} 1$
The expression is equal to $ 0$.
[/hide]

EDIT Knew there was something wrong with it. Thanks.

[quote="AstroPhys"][hide="Solution"]
$ x^{2} \minus{} 2x \minus{} 1 \equal{} 0 \rightarrow \left( x \minus{} 1 \right) ^{2} \equal{} 0 \rightarrow x \equal{} 1$
The expression is equal to $ 0$.
[/hide][/quote]

$ x^{2} \minus{} 2x \plus{} 1 \equal{} 0 \rightarrow \left( x \minus{} 1 \right) ^{2} \equal{} 0$, not as mentioned by you:
$ x^{2} \minus{} 2x \minus{} 1 \equal{} 0 \rightarrow \left( x \minus{} 1 \right) ^{2} \equal{} 0$

[hide="My attempt"]the given equation can be written as $ x^2\minus{}2x\minus{}1\equal{}0$

the solutions are $ 1\pm\sqrt2$

then put values in the given equation

but it yields a scary situation while putting value of $ x$ in given equation

probably there is a better way out[/hide]

$ (x\plus{}1)(x\minus{}1) \equal{} x^2\minus{}1 \equal{} 2x$

Hence, $ (x\minus{}1)(x\plus{}1) \left(x^7 \plus{} \frac{1}{x^9} \right) \equal{} 2 \left(x^8 \plus{} \frac{1}{x^8} \right)$

Now, $ x \minus{} \frac{1}{x} \equal{} 2 \Rightarrow x^2 \plus{} \frac{1}{x^2} \equal{} \left(x \minus{} \frac{1}{x} \right)^2 \minus{} 2 \equal{} 6$

This yields $ x^4 \plus{} \frac{1}{x^4}  \equal{} 6^2 \minus{} 2 \equal{} 34$ and so $ x^8 \plus{} \frac{1}{x^8} \equal{} 34^2 \minus{} 2 \equal{} 1154$

Hence the expression equals  $ \boxed{2308}$


induction
strong induction
combinatorics proposed
combinatorics

I found this problem and I can't solve this...
For every integer $ n > 1$ compute maximal numbers of subsets $ (1,2,...,n)$ such that any two of this subsets are disjoint or one of them contain in the another one.

Let the answer for $ n$ be $ a_n$.

Given an optimal configuration for $ n$, which is going to include $ S \equal{} \{ 1,2, \ldots n \}$, pick a nonempty subset $ T$ such that there is no subset $ U$ in the configuration with $ T \subset U \subset S$. Now every subset in the configuration except $ S$ and the empty set must be either a subset of $ T$, or a subset of $ T^c$. If $ |T| \equal{} k$, then $ a_k \plus{} a_{n\minus{}k}$ counts the maximum number of subsets for $ n$, with the empty set counted twice and $ S$ counted zero times (these cancel). So $ a_n \equal{} a_k \plus{} a_{n\minus{}k}$.

We get the recurrence $ a_n \equal{} \max_{1 \leq k \leq n\minus{}1} ( a_k \plus{} a_{n\minus{}k} )$. Now we can show by strong induction that $ a_n \equal{} 2n$.

i think the answer is $ 2n\minus{}1$. atleast it works for the smaller numbers :)

Are you counting the empty set and the original set? For $ n \equal{} 1, 2$, the power set works as a configuration.

EDIT: By a request, a full explanation of the strong induction mentioned in my post.

For $ n \equal{} 1$, the $ 2$ subsets $ \{ \}, \{ 1 \}$ form a valid configuration, and there are no more to include. So $ a_1 \equal{} 2$. Assume for a fixed $ n$ that $ a_k \equal{} 2k$ for all $ 1 \leq k < n$. We have $ a_n \equal{} \max_{1 \leq k \leq n\minus{}1} ( a_k \plus{} a_{n\minus{}k} )$. By the inductive hypothesis, $ a_n \equal{} \max_{1 \leq k \leq n\minus{}1} ( 2k \plus{} 2(n\minus{}k) ) \equal{} \max_{1 \leq k \leq n\minus{}1} 2n \equal{} 2n$. This completes the induction. It also shows that it doesn't matter what the $ T$ mentioned in my above post is.


calculus
integration
quadratics
number theory solved
number theory

Well, the notation you use is not used to me, but i think you want rings for that the integral closure of $\mathbb{Z}$ in a (quadratic ¿) extension of $\mathbb{Q}$ is euclidean. (then they are euclidean for the discriminants $-11,-8,-7,-4,-3,5,8,12,13,17,21,24,28,29,33,37,41,44,57,73,76$).

[quote="chuckzeduck"]Can someone give me a proof that there are only 21 Euclidean fields?

Thanks.

Charlie[/quote]

There are some articles by H. Lenstra in old issues of Mathematical Intelligencer, that discuss his and others' work on the problem.   I believe that all these classifications are conditional on some conjectures such as General Riemann Hypothesis and the analytic continuation of Artin L-functions.  It is a long time since I saw the articles, but I think that at the time the list was complete up to one possible exception, as in the list of 10 quadratic fields with class number 1.   The class number one problem was eventually settled completely but I don't know if it is so for Euclidean rings.

As much as I know, only all imaginary-quadratic extensions with class number $1$ (there are $9$ of them) are known. But for real-quadratic fields it's still open if there are infinetely many of them, so even for all extensions. I don't know if all euclidean rings (of that type) are known, but the quadratic ones are (see my post above).

Exactly ZetaX. Sorry about the notation. I though it might be more helpful to refer to these rings as euclidean quadratic fields.

I know that it was proved in the 50's that all imaginary-quadratic extensions with class number $1$ were found. (2 seperate proofs). One finds a lower bound for $\delta_0$ as -200. I think the other finds a lower bound around $-5*10^6$. There are $9$ of these.

Thank you

The list of imaginary quadratic fields of class number 1 was not known to be complete until the late 1960's or early 1970's.   Heegner's proof from the 1950's was completed (and shown to be essentially correct, though that was far from clear when it was published) by Stark, and Baker had a different proof using Diophantine approximation.   The "effective" version of the problem was done by Goldfeld, Gross, and Zagier in the 1980's.

I really suggest looking at Lenstra's article to see whether the list of quadratic Euclidean rings was actually complete, or just presumed to be complete as with the class number 1 problem.


quadratics
geometry
3D geometry
Harvard
college
calculus
AMC

[quote="236factorial"]I'm just wondering, does any paticular study show that asian kids are better in subjects like math than other culture's kids? I seem to hear a lot about how asians are smart, etc, etc.[/quote]

You dont need a study to prove that. Thats just obvious.

[quote="236factorial"]I'm just wondering, does any paticular study show that asian kids are better in subjects like math than other culture's kids? I seem to hear a lot about how asians are smart, etc, etc.[/quote]

  well other than the fact that subjecs like math and music are highly prmoted in asian society and families, i cant say ive seen any such study....

Asians hold subjects such as math and science in high esteem. Most Maericans are highly enthusiastic at sports. Where I'm hinting towards is that whatever peoples' cultures value more is what the majority of the people of those ethnicities will be good at.

you're right... i didn't see much chinese people swimming in the Olympics last year (although Japan was really good).
I swim, and I'm chinese. Looks like that doesn't work out too well.

but sometimes it is annoying to hear

"oh you're good in math $just$ $because$ you're asian!"

not that I am great in math, but I'd much prefer to hear
"oh you're good in math because $you$ $love$ $it$"

lol about chinese not swimming at the olympic....

FYI: Women backstroke champion is chinese....in 2004 olympic

the problem is that chinese ppl don't have too many pools as we have here....

little kids were not expose to such sports at a young age....the few that r got good and swim to win a gold....

same as much of the other sports in china....if it can be promoted to everybody...with 1.3 billion ppl there is a high chance there will be a champion...

look at table tennis..... :D  :D  :D ...this year....china sweep all three world cups and this is the 6th or 7th time they done this....it cause there are table tennis tables every where in China....

I used to live in Korea, then I immigrated to Canada three years ago. I got 7 years of education in Korea and 3 years in Canada(7+3=10, but I'm finishing grade 11 next month. Where did the one year go? :? ). There are huge differences between their education.

1) Korea seems to pressure students hugely, while Canada does not push students at all. If you are not paying attention in Canadian class, your teacher will not bother you as long as you are not distracting other students. It's completely different in Korea.
2) Koreans tend to learn things earlier than Canadians. For example, Koreans learn to solve quadratic equation in early grade 9, Canadians learn it in high school.
3) Meaning of 'application problem' is different in both countries. In Korea, application problem is a problem with higher difficulty, which requires perfect understanding of the unit that you were learning. In Canada, application problem is a problem with real life setting, but exactly same problem that you did thousands of times in class.

I'm not saying that Korean education is better. Both has pros and cons(I personally think both education is horrible). However, this differences in education might explain the question.

my communications professor is Taiwanese and she doesnt like the bias against asians being good in math, she told a story that she said to others trhat she was prof and they thought math, which got her disappointed. LOL. I also, think that the Chinese are very strong at mathematics, but not all. Also, Russians are, they have strong mathemtical education in Russia. GO RUSSIAAAAAAAAAA!!!. I was actually born in St Petersburg, hehe.

so about Russians also being "good" at math... after knowing me for about 3 months, a friend of mine started speaking to me in russian because "i'm good in math, and i don't look asian, so i have to be russian" i really don't like such views!
one of my good friends in school- who is chinese- is having alot of trouble in math class. she's really good in humanities though... she gets mad when people assume she's good just because she's asian and one time actually yelled at someone for saying that

Interesting. All the less smart people in our grade are chinese (but i'm in shanghai so what can you expect) but all the smarter people are also chinese, and the non-asians are inthe middle somewhere...

see, the thing i dont think people understand is that the people who are asian and pressured to do well in math/science... aren't they already higher income families (in those asian countries)? i mean, a third of the world lives in asia. the ones that we come in contact with are generally those who are upper class and many of them have immigrated to the U.S. there have to also be asians that are non-mathematical. i mean, aren't a bunch of chinese people, in rural areas... aren't they farmers? while they may be gifted, i don't think THOSE people are being pushed as much as the ones WE come in contact with, if that makes any sense; whereas the asians that come to america come into contact with ALL americans: the good, the bad, and the ugly. maybe that is the difference? what we are exposed to is not actually the general trend...

I may be mistaken, but I don't think most first generation immigrants are upper-class.  I think they're working class.

I think it's simply that Asians place a high value on education--and in math in general.  Not that I think Asian culture is better than American culture--they're just different.  For one thing, I think many Asian parents put waaay too much pressure on their kids to succeed academically, but I also think that Asian parents happen to encourage their children in math more--which I think is a good thing.  I'm exposed to both cultures because I'm a third-generation immigrant from Korea, so I'm raised 100% American, but my mother tells me stories of how she was raised by Korean immigrants.

I don't think that biology plays a significant factor.  Asian people are better at math because their culture emphasizes it more.

[quote="star99"]In competitive swimming, about 3/5 of the kids are asian.[/quote]

REally? About 1 in every 20 in my league are asian.

[quote]but the schools in my area have one very good thing, they try to meet everyones needs. there are different levels and you can be one higher than your grade level, if you learn faster. then there is also a privte school in which there is different levels for each subject. you can be in 10th grade math, 7th grade science, and 6th grade history, and you could be 14.[/quote]

wow, you're very lucky! :)

We have honor classes, where you can take two years of meaterial in just one year of education. this is only offered for one year. You'll be ahead by one year, but that's it. Coming up next is AP.

that happens at the high school i'm going to as well. in 9th grade, you can take geometry and algebra 2 in the same year.

It's very encouraging to see so many thoughtful views on this topic in this AoPS community.  I have decided to share my own ideas here.

Great mathematicians have lived in every culture that was at its time "modern", which might simply be defined by the fact that cultural technology was at least near the higher levels of what was going on in the world at the time.  There have been great mathematicians in Africa, India, East Asia, Europe, Russia, and more recently, in the United States.

The genetic similarity of the entirety of [i]homo sapiens[/i] suggests [to me] that mathematical achievement (or general intelligence of any kind) is far far more culturally based than it is based in the genetic differences between what we call races.  In fact, genetically speaking, it is difficult to find many consistent genetic markers that seperate the races.

Beyond that idea that cultures that value mathematics produce more top math students, I think that in the United States that we see first and second generation Asian and European immigrants performing better at mathematics that third, fourth, fifth, etc.  This suggests that there is either a diminishing cultural value tied to mathematics the longer families are in the U.S., that there is a selection bias in immigration, or more likely, both.

These are not simply beliefs of my own, but in fact cuts to the heart of our goal of making AoPS a culturally influential community where students who learn to do great things can be more overtly aware of the value of their efforts.  Indeed, we see many students who come here to the forums, buy the AoPS books, or take classes here who quickly skyrocket in ability -- including DPopov, himself Russian and I believe partially schooled in Russia (correct me if I am wrong about that).  While a small few students might do something like improve from being able to score a 1 on the AIME to a 13 in a single year in Russia, he was able to do that here as well.  Certainly it took an interest in mathematics, but an immersion in the cultural of problem solving certainly helps.

Hopefully, this and other communities will help generate a greater cultural enthusiasm for education so that U.S. students remain aware of the power of problem solving.

i agree, problem solving, and mathematics in general, is becoming more popular in the world today. maybe its because problem solving resources are becoming more available, or, maybe the quality of math education has risen.

by the way, i think DPopov is Bulgarian.

I'm a native Chinese, and I've never been abroad, so I don't know the situations in foreign nations such as USA.

We know, Indians(I mean native Americans) look the same as the asians in far east asia(black hairs, light yellow skin), so I wonder are they as good at math as asians?

I also notice that sometimes different books from different countries (problem solving ones) have quite different content and problem types within them. doing all types of problems from different places can get you more knowledge of all the types of problems, and expand your problem solving techniques.

In our town, however, there are no challenging problems. No one has ever came close to nationals in Mathcounts here. I don't think I will be the first one in so many years in my entire area (including some 25 school districts). The students in my honors classes don't even get the concepts well enough, so we're always going at a slow pace. Our final exam was so easy; we didn't have to show work, and the problems were as basic as the first problems in the textbook (concept practice parts). They had no challenge in them. It was much much easier than the review sheets we got, besides that they were easy enough.

I have no idea of the situation in China, but I've heard that the problems on their exams are sometimes on the level of Olympiads. Whether that is or isn't the case, I still think that the problems in china are still harder than problems given in most schools here in the US.

My school doesn't have the slightest clue about the AMC. Not one person from NY last year got a perfect score on the AMC 8, and that's not a hard test. that's sort of bad with 18 million people (of course only a small fraction actually took the test). i don't think i'll get a perfect score, but I think I'll score well. But nooo, my school doesn't know what AMC stands for  :roll:

[quote="qxy"]
We know, Indians(I mean native Americans) look the same as the asians in far east asia(black hairs, light yellow skin), so I wonder are they as good at math as asians?[/quote]

We'll know when they're free of their current political nightmare.

[quote="h78reg"]my school doesn't offer AMC either. :( which is weird because few people would even think of taking it and its a pretty cheap test anyway.......[/quote]

You better believe its cheap. I think its 10 dollars for 10 tests.

[quote="star99"]my school takes it as a sort of fun thing.  the 6th 7th and 8th graders all take the AMC 8.  one kid even got a perfect.  i guess thats because the teachers are very motivational (mostly).  they evenm have a math club[/quote]

We have a math club too (mathcounts  :roll: ), but it barely ever meets.

we have no math club, no math anything for that matter..... :(


yummy, I will log in to runescape now, I got the feeling that I need to play it now

^^

jk

er... AoPS is addicting too, I think   :lol:

you know, people spend hours here and grows levels up  (like, from "new member" to "poincare conjecture", "navier-stroke equation" and levels l tending to infinity)

I'm newbie yet   :(

I was just kidding, I got your point man

I was like level 80 in runescape and did every quest... then I quit cause it got boring. Math is a lot funner.

[quote="now a ranger"]http://spacers.lowtech.org/runespacers/rs_addiction.html
http://www.progressiveu.org/162238-runescape-addiction

same for warcraft

http://us.i1.yimg.com/videogames.yahoo.com/ongoingfeature?eid=505289&page=0

Moral: Games can ruin your life is gone out of control.[/quote]

the second one wasnt about warcraft...it was about WoW...
warcraft is the game for battlenet where you can chat, and hold clan channels, and op and kick, and you cuold play maps and in other words, its an RTS. WoW is jus an MMORPG
they are very different, because with warcraft you get better by having real skills. with WoW, you get better by being online for a longer time.

anyways i quit runescape over a month ago...the 4 consecutive deaths i got promoted me to quit (i was members with money skills and friends).
if i hadnt quit then then i mite've never had the chance to quit

To be honest, I think the moral of the these addiction stories is bs. Sure, these games are fun, but to simply dismiss them as pointless life-ruining things based on these examples makes absolutely no sense. If every single person who played these games got addicted, I wouldn't be saying this. But the truth is, there are tons of non-addicted people. And the addicted people are always the ones who have no lives and don't care for anything(which, I'm sure, none of us are).

Why would you not play them cause certain people are dumbasses and overuse them?

That's like saying because that girl died from drinking gallons of water, you shouldn't play water.

back to chopping yew trees and fletching them

[quote="now a ranger"]aops is an oppurtunity to learn math and discuss problems as well as competitions.

WC and Runescape are just useless games with no point.[/quote]

WoW and Runescape don't have any less point than AoPS does; it's all about perspective.

wonder wut yew log prices and yew long prices went to...i'm too lazy to check the rs forums...


Prove that the number $ \underbrace{ 222...2}_{n}\underbrace{111...1}_{n}$ is a triangular number.(a triangular number $ t_n$ is defined by $ t_n\equal{}\frac{n(n\plus{}1)}{2}$)

[quote="balan razvan"]Prove that the number $ \underbrace{ 222...2}_{n}\underbrace{111...1}_{n}$ is a triangular number.(a triangular number $ t_n$ is defined by $ t_n \equal{} \frac {n(n \plus{} 1)}{2}$)[/quote]

$ \underbrace{111...1}_{n}\equal{}a$  ---> $ \underbrace{999...9}_{n}\equal{}9a$ --> $ 10^{n}\equal{}9a\plus{}1$ ----->

$ \underbrace{ 222...2}_{n}\underbrace{111...1}_{n}\equal{}10^{n}*2a\plus{}a\equal{}(9a\plus{}1)2a\plus{}a\equal{}18a^{2}\plus{}2a\plus{}a\equal{}18a^{2}\plus{}3a\equal{}3a(6a\plus{}1)\equal{}\frac{6a(6a\plus{}1)}{2}$


The positive integer 4 can be represented as a sum of 1's or 2's in five ways, such as $ 1\plus{}2\plus{}1$ and $ 1\plus{}1\plus{}2$. In how many ways can the positive integer 5 be represented as a sum of 1's or 2's?

[i]Case 1: Two 2's are used.[/i]

$ 2 \plus{} 2 \plus{} 1$ (3 ways to arrange)

[i]Case 2: One 2 is used.[/i]

$ 2 \plus{} 1 \plus{} 1 \plus{} 1$ (4 ways to arrange)

[i]Case 3: No 2's are used.[/i]

$ 1\plus{}1\plus{}1\plus{}1\plus{}1$ (1 way to arrange)

Thus, the answer is $ \boxed{8}$.

Also, you can take the 5th distinct Fibonacci number, which is also $ \boxed{8}$.

----------

Can someone please explain the last line?

Let $ f(n)$ be the number of ways to get $ n$ using 1's and 2's.  Clearly $ f(1)\equal{}1$ and $ f(2)\equal{}2$.  Note that $ f(n)\equal{}F_{n\plus{}1}$

We can either start the sum with a $ 1$ or a $ 2$.  If $ 1$, then we just need the number of ways to express $ f(n\minus{}1)$ as the sum of 1's and 2's.  If we start with a $ 2$, then it's the number of ways to express $ f(n\minus{}2)$ as a sum.

So \[ f(n)\equal{}f(n\minus{}1)\plus{}f(n\minus{}2)\]


If x, y, z are pozitive real numbers, than
1/(x2 + xy + y2)+1/(y2 + yz + z2)+1/(z2 + zx + x2) \geq 9/(x + y + z)2.

Here's a nice approach for this (although not a complete solution):

Let  \varepsilon be cos(2P1/3)+isin(2Pi/3).
We can see that x^2+xy+y^2=(x- \varepsilon y)(x- \varepsilon^2 y) = |x- \varepsilon y|^2. Our inequality becomes
\sum {cyl}1/|x-  \varepsilony|^2 >= 9/(x+y+z)^2.
This means we can interpret it like this: we may assume x+y+z=1. Then let ABC be an equilateral triangle with height 1. Let P be a variable point inside ABC. Let x=d(P,BC); y=d(P,CA); z=d(P,AB). D, E, F are the projections of P on BC, CA, AB respectively. Our inequality becomes 1/DE^2+1/EF^2+1/FD^2 >= 9/(x+y+z)^2=9.

I haven't worked on this yet (maybe it's not even simple  :D ).

wlog, lets assume that x+y+z=1 and that x>=y>=z.
1/(x^2+xy+y^2) =1/((x+y)^2- (\sqrt xy)^2)) = 1/(2(x+y))(1/x+y- \sqrt xy) + 1/(x+y+ \sqrt xy)
but x+y>x+z>y+z and x+y- \sqrt xy)>x+z-root xz>y+z-root yz
hence sum 1/x+y )(1/x+y-rootxy) >=1/3 sum(1/x+y)(sum 1/x+y-root xy)
sum 1/x+y>= 9/2 and sum 1/x+y-rootxy>=9
hence sum 1/x+y* x+y-rootxy  >=27/2
similarly sum 1/x+y. x+y+rootxy >=1/3.(9/2)(9/3)=9/2
hence the total sum >= 1/2(27/2+9/2)=9 = 9/(x+y+z)^2
well, its done! :D

Well, here's an application of my previous idea (even though that idea didn't yield the solution  :D ): Let ABC be an equilateral triangle of side l. Let P be a variable point inside ABC. Prove that 1/PA^2+1/PB^2+1/PC^2>=9/l^2. This needs the inequality Gil posted, and since ali posted a solution for that one, this inequality can now be quickly solved. However, I think it would be pretty tough otherwise.

Ali, I don't understand  how can you prove that

   x+y-root xy>x+z-root xz>y+z-root yz

and

   sum 1/(x+y-rootxy)>=9.

Can you please explain it ?

Thank you.

you are absolutely right!
x+y- \sqrt xy > x+z- \sqrt xz
is not true if  \sqrt y+ \sqrt z> \sqrt x which is possible. ( I am sorry for this stupid mistake)
so the solution is incomplete.

I found this problem very challenging. I will try to write a solution. First, we can suppose that x+y+z=1. Then, we observe that x^2+y^2+xy=1-(xy+yz+zx)-z. Let us denote t=1-(xy+yz+zx). We have to prove that 1/(t-x)+1/(t-y)+1/(t-z)>=9. Clearing denominators, we arrive at the inequality
9t^3-21t^2+12t-1-9xyz<=0, which can be also written as
  (3+9(xy+yz+zx))(1-xy-yz-zx)(xy+yz+zx)<=9xyz+1. Since (3xy+3yz+3zx-1)^2>=0, we must have (3+9(xy+yz+zx))(1-xy-yz-zx)<=4 and it remains to show that 4(xy+yz+zx)<=9xyz+1, which can be written (x+y-z)(y+z-x)(z+x-y)<=xyz, well known.


How many triangles of any size are in the picture?

[asy]size(75);
for(int i = 0; i<5; ++i)
for(int j = i+1; j<5; ++j)
draw(dir(i*72+18)--dir(j*72+18));[/asy]

There are 10 points...unless I miscounted o.o jk
anyway
$ \dbinom{10}{3}$ is the number of possible selections of 3 points. However, the inner pentagon has not been accounted for (we don't have lines there). Fortunately, we can easily correct for this. I don't want to explain; we have counted 40 invalid triangles; therefore we must subtract this later. But even accounting for this, we still are including degenerate triangles (aka 3 points on the same line). Thus we subtract these; there are 5 lines each with 4 points and $ \dbinom{4}{3} \equal{} 4$ overcounted degenerate triangles. Hence our final answer is $ 120 \minus{} 40 \minus{} 5\cdot4 \equal{} \boxed{60}$.

Sorry if I missed anything else; I am too lazy to check.

@sinozvufenaeg,
Most or all of these problem are on FTW. Also, look at math154 solution it's very helpful. I don't know how to do this.

[quote=sinozvufenaeg]The answer  on FTW is apparently 35...  Can anyone please explain this problem?[/quote]

hes right; ftw ans is indeed 35

\BUMPING REAL HARD

I'm guessing that non-degenerate triangles do not count.

[hide = Solution]
There are a couple of distinct types of triangles here.

Type 1: No points from the inner pentagon.

This type is simple: just choose $3$ points from the outer pentagon, making $\binom{5}{3} = 10$ cases.

Type 2: One point from the inner pentagon.

Look at any single point from the inner pentagon. We can count that it makes $4$ triangles with points from the outer pentagon. Since there are $5$ points in the inner pentagon, there are $4 \cdot 5 = 20$ possible cases in this type.

Type 3: Two points from the inner pentagon.

These two points must be adjacent to each other, or they cannot be connected. From here, we see that there is only one point from the outer pentagon that makes a triangle with two adjacent points on the inner pentagon. This gives $5$ cases.

In total, that is $35$ cases.
[/hide]

@math154 Your idea is nice, but sadly, it does not work since there are some cases that you have not subtracted. For example, consider the bottom left point of the outer pentagon and the top two points of the inner pentagon. These do not make a valid triangle, but your method counts it. It may work with a bit of refining.

I think it is the 3rd triangular number squared -1 for the pentagon in the middle... But I'm not sure if it works for irregular shapes like this...

This pentagon is very interesting. you can see 7 different triangles (flips will be considered different), and there will be 5 of each (this is a very nice shape). :-D

With casework:

Case 1: Picking the single, small triangles on the perimeter of the big pentagon: $\textbf{10}$
Case 2: Picking two adjacent small triangles on the perimeter of the big pentagon to make a slightly bigger triangle: $\textbf{10}$
Case 3: Picking the triangle comprised of three adjacent vertices on the big pentagon: $\textbf{5}$
Case 4: Picking the triangle from one vertex of the big pentagon to the opposite side of the big pentagon: $\textbf{5}$
Case 5: Picking the triangle on a vertex of the big pentagon, another vertex of the big pentagon with a vertex between the previous one, and a vertex of the small pentagon: $\textbf{5}$

10 + 10 + 5 + 5 + 5 = $\boxed{35}$


[hide="hint"]Make a table to show the amount of money they earn each year for each of the jobs.[/hide]

[quote="dwx314"]So, you've just graduated college and are looking for a job. You've selected two jobs you really want to do, but can't decide between them. So you decide to compare salaries.

Job 1: Starting with an annual salary of 100,000 dollars, and a 20,000 increase
every year.

Job 2: Starting with a semiannual salary of 50,000, and an increase of 5,000
every 6 months

In all other respects, the two jobs are exactly alike.

Which is the better offer? (After the first year.)[/quote]

[hide]
It would have to be job 2 because after half a year, job 2 has 50,000 AND the 5,000 increase. Then after a year, Job 1 has 100,000 while Job 2 is 105,000.[/hide]

im not getting what you are asking, are you asking which one is a better offer for next year? or forever?
wouldnt' it be one because the because the salary taht they start with is the same but the first one's increase is bigger?

Yeah, I don't completely understand the question either. It says which job would you be better off with after the first year? Why does it have to be after the first year? I think you could know at the end of the first year, unless you have some weird way of figuring the salary...

Easy problem, made so confusing....

It's asking which job will give you more money after the first year.

so does the increase count in the first year? if so, than job one is better with 120000 and job 2 at 110000, but if not than job two is better at 105000 and job 1 at 100000

[hide]The thing is it says semi-annual salary.

YEAR 1-job1
100,000

first 6 months-job2
50,000
last 6 months-job2
55,000

job2 is better.[/hide]


In a space city there are $ 100$ space stations. Every pair is connected by a tunnel. In $ 100$ of the tunnels there is two-way traffic and in the rest of them there is one-way traffic. $ 4$ space stations are said to be closely linked if it is possible to reach every one of them from any one of the other $ 3$ along tunnels. What is the maximum possible number of sets of four closely linked stations?

http://web.archive.org/web/20040318185049/www.kalva.demon.co.uk/china/chi99.html

check A3 solution on this page :) (the big S button)

Wow, such a cool reference. You must know a lot of problems, greentreeroad  :)


algebra
polynomial
inequalities
function
number theory
least common multiple
abstract algebra

Let $ q_{0}, q_{1}, \cdots$ be a sequence of integers such that

a) for any $ m > n$, $ m \minus{} n$ is a factor of $ q_{m} \minus{} q_{n}$,

b) item $ |q_n| \le n^{10}$ for all integers $ n \ge 0$.

Show that there exists a polynomial $ Q(x)$ satisfying $ q_{n} \equal{} Q(n)$ for all $ n$.

Still working on this.  A related question:  are there non-polynomial sequences that satisfy condition a) alone?

[quote="Peter"]How about $ q_n \equal{} n!$?[/quote]
$ 5!\minus{}2!$ must be divisible by $ 3$,but it is not true.

Indeed, I'm sorry. Interesting question, I can't find a counterexample either.

$ q_n\equal{}P_n(n), P_n(x)\equal{}a_0\plus{}a_1x\plus{}a_2x(x\minus{}1)\plus{}...\plus{}a_nx(x\minus{}1)...(x\plus{}1\minus{}n)$, were $ a_i\in Z$.
For any integer sequence $ a_i$ we get sequence $ q_n\equal{}P_n(n)$, suth that $ m\minus{}n|q_m\minus{}q_n$.

Rust's counter example is false, as his sequence is not bounded by any polynomial

I did see solution of this problem once, it 's proof used Lagrange interpolation and an easy to show inequality on "Least common multiple"

Obviously, Rust answered all that posts before (except the starting post)  :wink:

Nice (and I was thinking along the same lines).  It is clear that we can "mod out" by any sequence of the form Rust gives, but I don't see a way to readily turn this into a solution (at least, I cannot pin down the intuitive notion that the growth rate of $ q_k$ mod polynomials of the form Rust gives is either better-than-polynomial or zero).

Idea:  consider the generating function $ P(x) \equal{} \sum_{k \equal{} 0}^{\infty} \frac {q_k}{k!} x^k$.  If $ q_k$ is a polynomial sequence then $ P(x)$ is given by a product of a polynomial and $ e^x$ (or something along those lines); in particular, $ P(x)$ satisfies a linear homogeneous differential equation.  $ q_k$ being bounded by a polynomial implies convergence for all $ x$.  The first condition says that the coefficients of $ P(x) \minus{} P^{(m \minus{} n)}(x)$ have numerators divisible by $ m \minus{} n$, or something along those lines.

[quote="Rust"][quote="mto"]Rust's counter example is false, as his sequence is not bounded by any polynomial
[/quote]
Were is false?
If $ m \minus{} n|q_m \minus{} q_n \ \forall m > n$, then exist integer sequence $ a_i$, suth that $ q_n \equal{} P_n(n)$.
If $ a_i$ integer sequence, then it define $ q_n \equal{} P_n(n)$, suth that $ m \minus{} n|P_m(m) \minus{} P_n(n) \equal{} P_m(m) \minus{} P_m(n)$.
We can define polinom P(x):
\[ P(x) \equal{} \sum_{k \equal{} 0}^{\infty}a_k\binom{x}{k}k!,
\]
were $ \binom{x}{0}0! \equal{} 1,\binom{x}{k}k! \equal{} x(x \minus{} 1)...(x \plus{} 1 \minus{} k)$ and $ degP(x) \equal{} \infty$ if there are infinetely many $ a_i\not \equal{} 0$.
Then $ q_n \equal{} P(n) \equal{} P_n(n)$ give solution.[/quote]

But we don't have  |qn| <n^10 for all n  .... :mad:

 I don't remember solution of my teacher for this one ( it was a long time). But it 's based on Lagrange interpolation and inequality on LCM
of consecutive integers

[quote="mto]
But we don't have  |qn| <n^10 for all n  .... :mad:

 I don't remember solution of my teacher for this one ( it was a long time). But it 's based on Lagrange interpolation and inequality on LCM
of consecutive integers[/quote]
I give all solutions. If you want $ |q_n|<n^{10}$ you can take $ a_i\equal{}0, \ \forall i\not \equal{}10, \ a_{10}\equal{}1.$

[quote="mto"]But we don't have  |qn| <n^10 for all n  .... :mad:[/quote]
Rust was addressing my question, not the original one.

[quote="mto"]I don't remember solution of my teacher for this one ( it was a long time). But it 's based on Lagrange interpolation and inequality on LCM
of consecutive integers[/quote]
Here's what I've got along those lines:

Write a sum of Newton polynomials $ Q(x)$ of degree $ 10$ such that $ q_k \equal{} Q(k), k \equal{} 0, 1, ... 10$.  Then

$ p_k \equal{} 10! (q_k \minus{} Q(k))$

satisfies condition a) (because $ 10! Q(k)$ is an integer polynomial).  Moreover, $ p_k \equal{} 0, k \equal{} 0, 1, ... 10$.  It follows that

$ m \minus{} k | p_m, k \equal{} 0, 1, ... 10 \implies$
$ \text{lcm}(m, m\minus{}1, ... m\minus{}10) | p_m$.

What I'm having trouble with is the inequality that would show that $ \text{lcm}(m, m\minus{}1, ... m\minus{}10)$ grows faster than $ m^{10}$.

[quote="t0rajir0u"][quote="mto"]But we don't have  |qn| <n^10 for all n  .... :mad:[/quote]
Rust was addressing my question, not the original one.

[quote="mto"]I don't remember solution of my teacher for this one ( it was a long time). But it 's based on Lagrange interpolation and inequality on LCM
of consecutive integers[/quote]
Here's what I've got along those lines:

[b]Write a sum of Newton polynomials $ Q(x)$ of degree $ 10$ such that $ q_k \equal{} Q(k), k \equal{} 0, 1, ... 10$.[/b]  Then

$ p_k \equal{} 10! (q_k \minus{} Q(k))$

satisfies condition a) (because $ 10! Q(k)$ is an integer polynomial).  Moreover, $ p_k \equal{} 0, k \equal{} 0, 1, ... 10$.  It follows that

$ m \minus{} k | p_m, k \equal{} 0, 1, ... 10 \implies$
$ \text{lcm}(m, m \minus{} 1, ... m \minus{} 10) | p_m$.

What I'm having trouble with is the inequality that would show that $ \text{lcm}(m, m \minus{} 1, ... m \minus{} 10)$ grows faster than $ m^{10}$.[/quote]

U could replace 10 by any integer K here, then what u have to show is the lcm(m,m-1,..,m-K) grows faster than m^10

Remark that the common divisor of any two numbers in m,m-1,...,m-K does not exceed K. From this, we could show that

LCM(m,m-1,..,m-K) > m.(m-1)...(m-K) / [ K^M]  for certain constant M

I came to the same conclusion.  :)

Edit:  The set of sequences satisfying condition a) is a $ \mathbb{Z}$-module $ \mathcal{A}$, a submodule of which is the set of sequences Rust gives $ \mathcal{P}$ (a submodule of which in turn is the set of integer polynomials).  The quotient $ \mathcal{A} / \mathcal{P}$ consists of (equivalence classes of) sequences $ \{ q_k \}$ such that $ q_k$ is defined $ \bmod k!$.  Hence we can pick a representative such that $ 0 \le q_k < k!$.

It follows that $ q_0 \equal{} q_1 \equal{} q_2 \equal{} q_3 \equal{} 0$.  (Edit 2: )  We then have $ q_4 \equal{} 0, 12$ and $ q_5 \equal{} 0, 60$.  For Rust's claim to be true (that $ \mathcal{A} \equal{} \mathcal{P}$ so the quotient is trivial) we need to show that $ q_4, q_5 \neq 0$ gives a contradiction.  But in fact I think it's possible that it doesn't.

One more observation.  If $ q_m$ satisfies condition a), then so does the forward difference $ \Delta q_m \equal{} q_{m\plus{}1} \minus{} q_m$.  The connection to Newton polynomials is clear.  In particular, following Rust's lead we can write [i]any[/i] integer sequence $ \{ q_k \}$ in the form

$ q_n \equal{} \sum_{i\equal{}0} a_i {n \choose i}$

where $ a_0 \equal{} q_0, a_1 \equal{} q_1 \minus{} q_0$, and in general $ a_n \equal{} (\Delta^n q)_0$.  We then have

$ \Delta^k q_n \equal{} \sum_{i\equal{}0} a_{i\plus{}k} {n \choose i}$.

The finite differences eventually becoming the zero sequence corresponds to $ a_n$ eventually becoming zero and $ q_n$ becoming a polynomial.  So perhaps we should consider $ \Delta^{11} q_n$.  (Unfortunately, the current bounded condition on $ q_n$ does not easily translate into bounds on $ \Delta q_n$ because of the possibility that $ q_n$ alternates between positive and negative values, so the bounds must be taken together with the divisibility condition.)

When $ a_n$ is eventually zero, it seems not too hard to prove that the divisibility condition is satisfied if and only if $ k! | a_k$.  Otherwise, I'm not so sure.

In fact,Taiwan 1996 is a special case of USA MO 1995.http://www.artofproblemsolving.com/Forum/viewtopic.php?p=353086&sid=985c19f15cb1fc80ca3958605bdec44e#p353086


[hide="It is easy..."]Since you paddle $45$ minutes upstream, and the water is against you, your speed is $1\frac{1}{4}$ mph, and so you go $\frac{15}{16}$ miles. Now, the hat went downstream for $45$ minutes. (Let's hope that the hat doesn't know how to swim :D ). So it went $\frac{9}{4}$ miles downstream. Now, for every hour ($h$), the hat moves another $3$ miles downstream. So it's rate is $\frac{9}{4}+3h$.

Since you go downstream after $45$ minutes, your speed is $7\frac{1}{4}$ mph. So your rate is $\frac{29}{4}h-\frac{15}{16}$ (since you lost $\frac{15}{16}$ miles going upstream).

Now it's a simple algebra problem. All you do is solve for $h$:
$\frac{9}{4}+3h=\frac{29}{4}h-\frac{15}{16}$

$\frac{51}{16}= \frac{17}{4}h$

$51 = 68h$

$h = \frac{3}{4}$

The answer is $\frac{3}{4}$ hours, or $45$ minutes.[/hide]

Yeah, try to find a trick that makes this problem really easy.

[quote="pascal_1623"]You are rowing a boat up stream.
The river flows at 3 miles per hour; your speed against the current is $4\frac{1}{4}$ miles per hour.
You lose your hat in the water, then you paddle for 45 minutes up stream without noticing you lost it.
How long will it take before you get your hat back?[/quote]

[hide]$\boxed{45 \text{ minutes}}$[/hide]


geometry
algebra
polynomial
analytic geometry
topology
advanced fields
advanced fields unsolved

I have 3 questions:

1. If $ V$ is a variety over an algebraically closed field $ k$, what is the associated scheme $ t(V)$?

2. In the proof of the Prop 2.6, why is $ \alpha : X \to t(X)$ a continuous map? Why dose $ \alpha$ induce

a bijection between the set of open subsets of $ X$ and the set of open subsets of $ t(X)$?

3. Why is the structure sheaf of the scheme $ t(V)$ $ \alpha_{*}(O_V)$?

Thanks!

[quote="foggy"]1. If $ V$ is a variety over an algebraically closed field $ k$, what is the associated scheme $ t(V)$?[/quote]
Here the subtle difference between classical varieties and varieties in the sense of algebraic geometry comes in: If $ V$ is a (classical) affine variety then it is the set of zeros of (finitely many) polynomials in $ k^n$, but a variety in algebraic geometry is a scheme. (Here you should look up the definition which becomes important in the sequel. ;) ) To be more precise: If $ A$ is the coordinate ring of $ V$, then $ t(V) \equal{} Spec(A)$. You can recover $ V$ by taking the closed points of the topological space of $ t(V)$. Of course this is meant up to homeomorphism. ... but this will be dealt with in more detail below.

NB: A point $ x$ is called closed in a topological space $ X$ iff $ \{x\}$ is a closed subset in $ X$. Note that is important to say in which topological space the point is meant to be closed: $ x$ can be closed in $ X$, but not in $ X'$ containing $ X$!
[quote]2. In the proof of the Prop 2.6, why is $ \alpha : X \to t(X)$ a continuous map?[/quote]This is true for every topological space. Just check using the definition of the topology on $ t(V)$.
[quote]Why dose $ \alpha$ induce a bijection between the set of open subsets of $ X$ and the set of open subsets of $ t(X)$?[/quote]This is the only algebraic ingredient here. What you definitely have to use is the Hilbert Nullstellensatz. Write up what you have found out and where you get stuck ... and we'll see. ;)
[quote]3. Why is the structure sheaf of the scheme $ t(V)$ $ \alpha_{*}(O_V)$?[/quote]If you know from which $ k$-algebra the affine scheme $ t(V)$ should stem, then there is a good candidate of morphism (of locally ring spaces) which should become an isomorphism. But it is given in the proof in Hartshorne's book. Use that $ k$ is algebraically closed !!!

I am still not clear about the 3rd question. Would you like give a more specific explanation?

I think the question
[quote="foggy"]3. Why is the structure sheaf of the scheme $ t(V)$ $ \alpha_{*}(O_V)$?[/quote]
should be rephrased as

        [i]"Why does the sheaf $ \alpha_{*}(O_V)$ make the topological space $ t(V)$ a scheme?"[/i]

For this purpose you need to show that $ (t(V),\alpha_{*}(O_V))$ is a scheme; in fact it is an affine scheme if $ V$ is affine. In addition the statement requires a morphism of locally ringed spaces $ t_V: V\to t(V)$.

I think that the structure sheaf of the scheme $ t(V)$ is $ O_{t(V)}$,

but actually it's  $ \alpha_{*}(O_V)$ in the book. This is my question. :?:

In the proposition you construct a sheaf on $ t(V)$ - this is just an abstract topological space constructed on the basis of $ V$. A sheaf on $ t(V)$  does not exist from the very beginning of the proposition. But it claims that there [b]is[/b] a sheaf $ \mathcal F$ on $ t(V)$ such that $ (t(V),\mathcal F)$ [b]is a scheme[/b]. The proposition says that this works if you define $ \mathcal F: \equal{} \alpha_*\mathcal O_V$. Thus you have to show that $ (t(V),\alpha_*\mathcal O_V)$ [b]is a scheme[/b].

Note that there are situations that a topological space $ X$ (with additional assumptions, of course) can be equipped with a couple of sheaves $ \{\mathcal F_i\}_{i\in I}$ such that each of the pairs $ (X,\mathcal F_i)$ is a scheme. In other words the structure sheaf of a scheme is [b]not[/b] uniquely determined by the topological space as such.


What's source of it? I don't want to solve non-real problem. Please give the begining problem not problem after expanding?

It's not the expanding of another problem, it is just a new problem I think but I have no solution.Can anyone help me ?

[quote="spanferkel"]Are you sure it is not $\sum(2a^3b^3(a-b)+2\boxed{b^3c^2}(a-b)^2+a^3c^2(a-b)^2)\geq 0$? Otherwise the expression is not homogenous  :?:[/quote]

You are right sorry I made a typo now it is corrected


Please use the poll only if you know your score for certain.  If you have a guess, just add it as a post if you like.

I really screwed up on my sprints. Our team came second (8 for a team score), I came fourth in the written (23!! sprint, 8 target) and first in countdown...

I cannot believe I got a 23 on the sprints! Maybe I wasn't warm or something, but the sad thing is I solved all but two during the targets b/c I had extra time (the target round was relatively easy this time around)

O well I hope I do better on states... what killed me was my:
1) Bad time management (I got confused abt how much time was left)
2) STUPID ERRORS!!! (I've been working on that)

I would have gotten a 39 if i had remembered that if 2x=3y, (x,y) does not equal (2,3)
*hits himself*

I am pretty sure I got 29 sprints and 8 targets
I did not get the reversed-age problem in time.

oh yea... that reversed age question was really hard! i tried to do it algebriacally but it didn't work out for me... the sixth grader on our team got it right because he did guess and check... but he said it was a really lucky guess and check, because i specnt forever doing guess and check when i coldn't do it algebraically :(

Same with me, it was definately much easier.

The fact that I improved 5 points from last year kind of raised a flag.

The target was MUCH easier.. the sprint involved 4 stupid errors, two in which I didn't see anything, and the age problem in which I found the answer while I was doing the Target ;)

i got the age problem right, but i got question #1 on sprint wrong which is pathetic, i dont know how i read the chart wrong, there wasnt even any math involved. oh well, its probably because CT has their chapter competitions early, i can hardly wait to wake up early again for states.

[color=indigo]I got only got 29 but still managed to get 9th place and this is the same chapter that my brother got 44 in three years ago and got 2nd so no clue how that worked... my teammate only got 38 and he won the whole thing! I think everyone (including me because  I was doing much better on practice tests) must have had no concentration that day or something... [/color]

Our chapter was pretty sad, with only six people doing the countdown round. With a score of 27, I was seventh in the competition. The highest score was something like 32, then 31, 29, 29, 29, 28, and then me, 27. Our team (which was the favorite to win) got a 35.5 and finished second, mainly because one of our teammates got a 14!!!

How do you know your scores?
I know they have the answers up on the mathcounts site but there's know way that I remember what I put down for 38 problems.

Qualified for states anyway.

Usually a team's coach will show his or her students the sheet that says what each person got.

[quote="yif man12"]I did some of the sprints in my free time. I started from 30 and went backwards. They didn't seem as hard as they did in past years, but maybe it's because I know more. Just my opinion.[/quote]

What was your score? Personally, I think it is better to start with the first one, because the beginning is easy. It's like countdown round questions--you can finish the whole page in less than a minute. If you start from the end, you might not get to the first ones.

Heh, I didn't compete. I proctored one of the competitions and did some just to see how it would be. I wouldn't advise doing what I did. :)

OK, got scores. Me:24/8. Team members, 16/3, 10/2,10/2.

Wow...  We had everyone over 35 in our school team

The weekend before Chapters, I was sick. So I got all the previous chapters and states, then had a Mathcounts marathon. For the tests, I got 4th place in chapter. For the Countdown round, I went up to first.  State was the next weekend, so I did the same thing, but this time with my team. In state, I got fifth place after the tests, then I moved up to second in the Countdown round. BTW, last year I didn't get in the top 10 in chapter. :what?: