i tried using algebra...didn't work...so i used common sense :lol:

anyways, note that both x and y must be less than or equal to 1...since $x^{m}\le x^{n}$, where $m \ge n$...same logic with y.

Therefore $x^{5}+y^{5}\le 2$

i know it's not too formal...but it makes sense, right?

It doesn't work because you could have $x > 1$ and $y < 1$ (or vice versa).

[quote="falconwing64"]what?...x and y must be positive...
can you give me an example?[/quote]

ie, x=1.01, y=3/4, or something like that...

[quote="arqady"]$x$ and $y$ are positive numbers such that $x^{2}+y^{3}\ge x^{3}+y^{4}.$
Prove that $x^{5}+y^{5}\le 2.$[/quote]

0 [u]>[/u]x^2(x-1) + y^3 (y-1)
0 [u]>[/u](x^2(x-1) + y^3 (y-1))^2
0 [u]>[/u]x^4(x^2-2x+1) + 2x^2y^3(x-1)(y-1)+y^6(y^2-2y-1)

When you expand all of this and go on, you will be able to enstrangle a x^5 + y^5 from it and then all of the stuf on the other side will have to be less than 2, and hence, by logic, x^5 + y^5 are most definitely less than 2.

I don't know where the "equal" comes into play tho.

[quote="neelnal"]
0 [u]>[/u](x^2(x-1) + y^3 (y-1))^2[/quote]
I'm pretty sure it's the other way around.

Does this help?[I doubt it,but heh,who knows]
I got that $ x^{5}+y^{5}\leq\frac{x^{2}+x^{7}+y^{3}+y^{6}}{2}$

[quote="andyciup"]Does this help?[I doubt it,but heh,who knows]
I got that $x^{5}+y^{5}\leq\frac{x^{2}+x^{7}+y^{3}+y^{6}}{2}$[/quote]

That is not correct.

EDIT: My bad. It is correct. Forgot the less than sign.

[quote="silouan"]I also proved that $x^{3}+y^{3}\leq x^{2}+y^{2}\leq x+y\leq 2$
Could this help ?[/quote]
Yes!  :lol:

$ x^3 \plus{} y^4 \leq x^2 \plus{} y^3$
$ x^2(x\minus{}1) \leq y^3(1\minus{}y)$

Equality stands for $ x \equal{} y \equal{} 1$ therefore $ x^5 \plus{} y^5 \equal{} 2$

For $ x>1$ then $ (x \minus{} 1)>0$, therefore $ (1 \minus{} y)>0 \Leftrightarrow y<1$ then $ x^2(x \minus{} 1)>1$ and $ y^3(1 \minus{} y)<1$ therefore
$ x^2(x \minus{} 1)>y^3(1 \minus{} y)$ which is contradiction to inequality

For $ x<1$ then $ (x \minus{} 1)<0$ so inequality stands for every $ y\leq1$ because $ x^2(x \minus{} 1)<0$ and $ y^3(1\minus{}y)\geq0$
$ x<1$ and $ y<1 \Rightarrow x^5 \plus{} y^5<2$

Let's see what happens if $ x<1$ and $ y>1$
Then $ x^2(x\minus{}1)>\minus{}1$ and $ y^3(1 \minus{} y)<\minus{}1$ which is again contradicting the inequality.

So, inequality stands only for $ x\leq1$ and $ y\leq1$ and for those numbers $ x^5 \plus{} y^5\leq2$

That method has really big problems. $ x>1$ means $ x^2(x\minus{}1)$ is greater than 0, not necessarily 1. We can have $ x>1$ or $ y>1$ (but not both); see K18o7's post.

This problem requires a lot more work than that. I solved it in some other post by showing $ x^3\plus{}y^3\leq x^2\plus{}y^2$ which reduces the problem to factoring a polynomial of degree 7 (!), but you're free to find a nicer solution if you can.

[quote="NapoleonXIV"]
For $ x > 1$ then $ (x \minus{} 1) > 0$, therefore $ (1 \minus{} y) > 0 \Leftrightarrow y < 1$ then $ x^2(x \minus{} 1) > 1$ [/quote]
Why $ x^2(x \minus{} 1) > 1$ $ ?$ :wink:

Ups, my bad.  :oops:
Have to be more careful about those thinks.
Thanks for pointing it out.

[quote=arqady]$x$ and $y$ are positive numbers such that $x^{2}+y^{3}\ge x^{3}+y^{4}.$
Prove that $x^{5}+y^{5}\le 2.$[/quote]
Let $x$ and $y$ are positive numbers such that $x^{5}+y^{5}\ge 2.$  Prove that $$x^{2}+y^{3}\le x^{3}+y^{4}.$$

[quote=Ji Chen][quote="arqady"]$ x$ and $ y$ are positive numbers such that $ x^{2} \plus{} y^{3}\ge x^{3} \plus{} y^{4}.$ Prove that $ x^{5} \plus{} y^{5}\le 2.$[/quote]$ x^5 \plus{} y^5$

$ \equal{} 2 \minus{} \frac {5(x^2 \plus{} y)(x^2 \plus{} y^3 \minus{} x^3 \minus{} y^4)}{2} \minus{} \frac {5x^2y^4 \plus{} 3x^5 \minus{} 5x^2y^3 \plus{} 3y^5 \minus{} 5x^4 \plus{} 5x^3y \minus{} 5y^4 \minus{} 5x^2y \plus{} 4}{2}$

$ \leq 2,$

where $ 5x^2y^4 \plus{} 3x^5 \minus{} 5x^2y^3 \plus{} 3y^5 \minus{} 5x^4 \plus{} 5x^3y \minus{} 5y^4 \minus{} 5x^2y \plus{} 4\geq0$ holds for all $ x,y\geq0.$

See : http://www.mathlinks.ro/viewtopic.php?t=227889[/quote]

This   $ 5x^2y^4 + 3x^5 - 5x^2y^3 + 3y^5 - 5x^4 + 5x^3y - 5y^4 - 5x^2y + 4\geq0$ holds for all $ x,y\geq0.$
 it is not easy to prove.


function
factorial
integration
calculus
complex numbers
puzzles

out of curiousity, i googled "i !" and it came up with an answer 0.498015 - 0.154949 i
how is i! defined?

in terms of the gamma function
the gamma function is an extension of the factorial to the complex plane
it is defined (if you really want to know)

$ \int^{\infty}_0 t^{x\minus{}1} e^{\minus{}t} dt$

(note that: for integers, Gamma[n]=(n-1)!)
[url]http://mathworld.wolfram.com/GammaFunction.html[/url]

That definition is only unambiguously valid if the real part of $ x$ is greater than $ 0$. For $ x$ on the imaginary axis, you're taking a limit from the right; the integral converges as an improper integral, but it doesn't converge absolutely. For negative real parts, you need the notion of analytic continuation, or a different definition.

Well, he wants $ i!$, thus $ \Gamma(1\plus{}i)$, so the definition is valid.


function
real analysis
real analysis solved

Prove that there is no function f:R->R such that |f(x)-f(y)|>1, for all x distinct from y.

Assume there is such a function. For each $t=f(x)$, there are no points from $f(\mathbb R)$ in the interval $(t-1,t+1)$. This means that the intervals of length $1$ centered around the points of $f(\mathbb R)$ are disjoint, so $f(\mathbb R)$ is countable. However, $f$ is obviously an injection, so $f$ is an injection from an uncountable set to a countable set, which is absurd.

This solution i know too.I'm wondering if there's another one.


logarithms
inequalities
function
real analysis
real analysis solved

What is the limit of the sequence:

a(n)=prod((2^k-1)/2^k) with k from 1 to n.

Misha

I wanted to solve another problem of this section and I came to this one as the last part of my solution. I have some ideas for this problem but not a complete solution. Perhaps you, Myth, or somebody else can help me.

Misha

Among the things that can be said:

It converges (and in the language of infinite products, "converges" includes that the limit is not zero.)

$\prod_{k=1}^{\infty} \frac{2^k-1}{2^k} < e^{-1}.$

Rounded to 6 decimal places, the limit is .288788.

A general principle: if $a_k$ is eventually constant in sign, $\prod_{k=1}^{\infty} (1+a_k)$ converges if and only if $\sum_{k=1}^{\infty} a_k$ converges. This comes from taking the logarithm of the product and noting that $\ln(1+a_k)$ is comparable to $a_k$.

[quote="Kent Merryfield"]It converges (and in the language of infinite products, "converges" includes that the limit is not zero.)
$\prod_{k=1}^{\infty} \frac{2^k-1}{2^k} < e^{-1}.$
Rounded to 6 decimal places, the limit is .288788.
A general principle: if $a_k$ is eventually constant in sign, $\prod_{k=1}^{\infty} (1+a_k)$ converges if and only if $\sum_{k=1}^{\infty} a_k$ converges. This comes from taking the logarithm of the product and noting that $\ln(1+a_k)$ is comparable to $a_k$.[/quote]
I am sure that everyone who read this topic knows these simple facts from calculus.

Misha! You better give us link to the source problem.

[quote="Kent Merryfield"]Among the things that can be said:
$\prod_{k=1}^{\infty} \frac{2^k-1}{2^k} < e^{-1}.$
[/quote]

How did you show this? I have the following proof:

$\prod_{k=1}^{\infty} \frac{2^k-1}{2^k} \leq e^{-1}.$

because of:

$ln(\prod_{k=1}^{n} \frac{2^k-1}{2^k})\leq nln(\frac{\sum_{k=1}^{n}\frac{2^k-1}{2^k}}{n})=nln(\frac{n-1+\frac{1}{2^n}}{n})$

The last term converges to 1/e (you can show it with L'Hopital).

Misha

But it is only an inequality for the limit. How can you prove what the limit is (an not only that the limit exists)?

Misha

[quote="Misha123"][quote="Kent Merryfield"]Among the things that can be said:
$\prod_{k=1}^{\infty} \frac{2^k-1}{2^k} < e^{-1}.$
[/quote]

How did you show this? I have the following proof:

$\prod_{k=1}^{\infty} \frac{2^k-1}{2^k} \leq e^{-1}.$

because of:

$ln(\prod_{k=1}^{n} \frac{2^k-1}{2^k})\leq nln(\frac{\sum_{k=1}^{n}\frac{2^k-1}{2^k}}{n})=nln(\frac{n-1+\frac{1}{2^n}}{n})$

The last term converges to 1/e (you can show it with L'Hopital).
[/quote]
I believe it is too complicated proof ;)
Classical proof is based on well known inequality $1+x<e^x$ for all $x\neq 0$.

[quote="Myth"]Classical proof is based on well known inequality $ 1+x<e^x $ for all $ x\neq 0 $.[/quote]
That's almost what I had in mind. The inequality I'm actually using is $\ln(1+x)<x$ for all $x\ne 0$.
$\ln\left(\prod_{k=1}^{\infty}(1-2^{-k})\right)=\sum_{k=1}^{\infty}\ln(1-2^{-k})< -\sum_{k=1}^{\infty}2^{-k}=-1$.

But then you could just use a calculator or spreadsheet to multiply out the first 20 or so terms of the product - that's where that 6-digit estimate is coming from.

[quote="Kent Merryfield"]But then you could just use a calculator or spreadsheet to multiply out the first 20 or so terms of the product - that's where that 6-digit estimate is coming from.[/quote]
I did the same thing using MathCad yesterday ;)
In my opinion, it is impossible to obtain nice answer in this problem.
Misha! Where is link to initial problem? :?

for those of you who like very mad equations: we have
$\prod_{k=1}^{\infty} \frac{2^k-1}{2^k} = 2^{-\frac 38}3^{\frac 18}5^{\frac {1}{12}} \left( 20\left(\frac{1}{45}+\frac{16}{3}\sum_{k=1}^{\infty} \frac{k^3}{2^k-1}\right)^3-\frac{4}{18225}\left( 1 - 504\sum_{k=1}^{\infty} \frac{k^5}{2^k-1}\right)^2\right)^{\frac{1}{24}}$.

i hope there are no typos or miscalculations in it, but as far as i checked it, it seems to be right.
this doesn't actually help us since both sums are very hard to evaluate. i derived that formula on my own, but please don't ask for a proof, my proof is VERY VERY long and requires some advanced analysis. maybe you find an easier one but i doubt.

Peter

hmm...there's seems to be an error somewhere in the calculation though i can't find it...or maybe the approximations are too bad but i doubt they are...

Peter

*walkingwithheadagainstthewall*

yeah, the calculations are actually correct. however, my formula is computationally nonsense(the first three factors give about 1, thus the term under the 24th root must be about $10^{-10}$, but it is hard to calculate both addends two that precision), resulting in an enourmous error. the problem was that i wanted the left side to be about 2(i don't know why), so i wondered why the term under the 24th root isn't about $10^{10}$...
anyway, i must admit that there's a small mistake: a factor of $2^{\frac{1}{12}}$ is missing on the right.

Peter

this problem was posted before
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=1958&highlight=sequence[/url]

Seems to be really a hard problem.

Misha

hmm...harazi's equation at http://www.mathlinks.ro/Forum/viewtopic.php?t=1958 seems interesting combined with my one...in fact one can probably hope to express the appearing sums in terms of the other and thus get some weird equations the limit satisfies...anyway i don't really believe in it.

Peter

arghlb...

harazi's equation is just a very simple application of an idea i used as well in my proof for my equation...but i'm too stupid to see that one can do it that simple...

Peter


Is it true that for any 2^n, where n is a natural number greater than or equal to 2,
there exist two primes p and q, where 2^(n-1)<p<2^n, 2^n<q<2^(n+1),
such that p, 2^n, q form an A.P. ?

I have test it for n=2,3...,7,
the results are as follow:
n=2, p=3, q=5
n=3, p=5, q=11
n=4, p=13, q=19
n=5, p=23, q=41
n=6, p=61, q=67
n=7, p=107, q=149

This problem looks like  [u][i][b][color=red]Goldbach's conjecture[/color][/b][/i][/u].

$ p\plus{}q\equal{}2^{n\plus{}1}$


How many times a day (24h)the pointer of a clock make 90 degrees?

I would say 22 for each time the hour hand goes around, and thus, 44.  That is, if it's a 12-hour clock.  For every interval in which the hour hand is (out of the 12 hour intervals), the minute hand creates a 90 degree angle with it twice.  Except for the two intervals between 2 and 4: in those, it only makes 3.  Then, if you look at the interval between 8 and 10, the same applies.


function
algebra
polynomial
number theory proposed
number theory

assume that $n\geq 2 $ is integer and let $x_1,x_2,...,x_n$ be real numbers.consider $N={n\choose 2}$ sums $x_i+x_j,1\leq i<j\leq n$and denote them $y_1,y_2,...,y_N$for which n are the $x_1,x_2,...,x_n$ uniquely determined by $y_1,...,y_N$

This is a splendid problem, but known long time ago. Anyway, if you don't know that, I don't know what chances you have to solve it. The answer is that those numbes are uniquely defined by those sums if and only if n is not a power of 2.

it seems that I solve it incomplete ( because  i know you're right) so enlighten us :D

I might be wrong, I don't know. But I know there is a theorem of Selfridge dealing with this thing. Anyway, here is what I think:
   Suppose we have n numbers $ x_1,...,x_n $ and another n numbers $ y_1,...,y_n $ such that the collections of pairwise sums of two numbers from the two groups are the same. Define the functions $ f(x)=x^{x_1}+...+x^{x_n} , g(x)=x^{b_1}+...+x^{b_n} $. Then we have $ f^2(x)-f(x^2)=g^2(x)-g(x^2) $ and from here we find that $ (f(x)+g(x))(f(x)-g(x))=f(x^2)-g(x^2) $ . Note that using rational aproximation for these numbers ( Dirichlet theorem in general form) we may assume they are natural ( I did not understand well this argument, but one of my teachers said it's correct!) and thus f and g are polynomials. Now, since $ f(1)=g(1)=1 $ we can find a natural r and a polynomial h such that $ h(1)<>0 $ and $ f(x)-g(x)=(x-1)^rh(x) $. Substitute this in the relation above and then make x=1. You will find that $ n=2^{r-1} $. Thus, n is a power of 2. I think this can be done directly for the reals if we work with analytic functions, but I don't know sure. Now, I prove that for n a power of 2 the numbers aren't uniquely determined. Let $ A_1={0,3}, B_1={1,2} $ and $ A_{n+1}=A_n union {2^{n+1}+B_n} , B_{n+1}=B_n union {2^{n+1}+A_n} $. It's not hard to show that we can take the elements of $A_n$ to be the n numbers $ x_i$ and the elements of $ B_n $ to be $ y_i$.


i  kind of need some help in permutations and combinations
can anyone suggest something?

Could you be a bit more specific in your query? :)
I see in particular that this is your first post. So is this a more homework-related question, or a math-competition oriented one?

Look all over this forum. There's many many problems.
You could also use [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]this.[/url]
And [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]this.[/url]
And [url=http://www.amazon.com/exec/obidos/ASIN/0883856158/artofproblems-20]this.[/url]
And [url=http://www.amazon.com/exec/obidos/ASIN/0817643176/artofproblems-20]this.[/url]
And [url=http://www.amazon.com/Path-Combinatorics-Undergraduates-Counting-Strategies/dp/0817642889/ref=sr_1_2?ie=UTF8&s=books&qid=1219586040&sr=1-2]this.[/url]
And [url=http://www.math.upenn.edu/~wilf/DownldGF.html]this.[/url]
And while you're at it look all over this forum again.

i want it not for any competition in general but to just improve in olympiads

1=2 why have you givn him Genreating functions? That is that too much for some one who is beginning.
Geniusbliss, you have sufficient knowledge in permutations and combinations as i personally know you. So u need not have posted this topic at all.


[quote="CheeseIsGood"][hide]
1.X=30
   Y=10
30+10=40
30-10=20
So, X^2-Y^2 is 900-100, which is 800
2.
Y=10
X=2
Z=8
10+2=12
2+8=10
10+8=18

So, X+Y+X=20.
[/hide]
I think there probably are faster ways of doing this than [hide=" my method"]guess and check.[/hide][/quote]

Faster way for both:

[hide]
1. Multiply the equations by each other, you get x^2-y^2=800
2. Add the equations together, you get 2x+2y+2z = 40. Divide by 2.

[/hide]

[hide]
number 1, multiply
number 2, add and divide by 2[/hide]

[quote="Zenox"]Try and do them in 30 seconds or less:

1. X+Y=40
    X-Y=20

What is X^2-Y^2?

2. X+Y = 12
    X+Z = 10
    Y+Z = 18

What is X+Y+Z?

Feel free to post some more of this type: They are handy for Sprint + Countdown  :)[/quote]

[hide="answers"]
1. X^2-Y^2 = (X+Y)(X-Y)
Therefore it is 40*20 and the answer is hence 800.

2. 2X+2Y+2z=40
X+Y+Z=20

[/hide]

Anyone got any other problems similar to these?

[quote="Zenox"][hide]

First one I divided the first equation by two and then subtracted the second from the first, which gives y = 0, so x = 8. And that means X + Y^2 also = 8

One the second one I'm not quite sure, but I had X^2 + 2xy + Y^2 = 49, so 49-2xy = z^2.

Then I used trial and error to get z = 5.

[/hide][/quote]
The first solution is correct, but I don't think the second one is.  The second one should be infinitely many solutions since you are dealing with 3 variables and 2 equations. You need one more equation to find z.

[quote="Zenox"][hide]

First one I divided the first equation by two and then subtracted the second from the first, which gives y = 0, so x = 8. And that means X + Y^2 also = 8

One the second one I'm not quite sure, but I had X^2 + 2xy + Y^2 = 49, so 49-2xy = z^2.

Then I used trial and error to get z = 5.

[/hide][/quote]

Both of them are correct.

[hide="dont read unless you want to no the solution."]The second one is 4^2+3^2=5^2[/hide]

Also, Ill edit the second question.


AMC
AIME
MATHCOUNTS
USA(J)MO
USAMO
AMC 10
AMC 10 A

What do you guys think the floor score will be?

I think it will be a 7, too bad I got a 3 or 4.

I heard that Saiem said there was a 150/6, meaning the floor will very likely be a 6.

If you seem to think that the index will be >210...

- [u]More[/u] of an expansion
- Harder AMC grading
- Harder AMC
- Harder AIME

... like a 4 ...

this AIME was a beast

as far as I can tell, beasted everyone at my school (that includes 2 MOPpers?)

I'm absolutely positive that there was a 144/6, so probably at most 6 or 7.

I'm hoping it's a 5 but from polls, it seems that it will be around 6 or 7.

I know what it feels like to hope for a floor equal to your score, and then have it be 1 higher, so I hope it works out for everyone...good luck!

*prays that it's a 6* *prays that it's a 6*

when do they usually announce the floor? This is my first time doing amc's

Beginning of April, AMCDirector said.
No chance for me. I got a 5. Though it would be cool if it was.  :lol:

Wow, this thread relieves me...I got a 7 because of my really blonde log mistake, but this gives me reassurance.

Hopefully it's 6 though, in case I made a bubbling error.

[quote="mysmartmouth"]It seems like the way they determine floor, it would be a 6 this year.  Except that would result in an abnormally high number of people, even with the expansion, so I'm gonna guess 7.[/quote]

They'd rather have more than go under what they promised.

i think a LOT of people got 6, so the floor's probably gonna be 7.

I don't think it was that much harder this year. I've gotten the same score on the AIME since 2005.

I don't think the tiebreaker is fair, but I'm confident that with the floor at 7, there will be around 500 USAMO qualifiers. I suppose it all depends on AIME II.

[quote="randomdragoon"][quote="xxazurewrathxx"]I don't think it's going to be a 6. I'm pretty sure it'll be a 7 as the floor. If you look at last year, 432 people made the USAMO with an 8 as the floor. This year's AIME isn't hard enough to make it drop 2 points, and the expansion only carries to 500.[/quote]
Ahem.

1. Two point drop for me.
2. There's a 150/6; not everyone has to drop by 2 points, only someone 20 points above the border does.[/quote]

Uh, 1. You are one person.
2. It'd be upsetting, but a 150/6 might not make it. The index, I think is going to be above 210. Low 210s, most likely, so a 7 is the Aime cutoff.

Makes sense based on last year statistics. Tons of people are reporting 7s and 8s. Even with stupid mistakes.

[quote="white_horse_king88"]
Uh, 1. You are one person.
2. It'd be upsetting, but a 150/6 might not make it. The index, I think is going to be above 210. Low 210s, most likely, so a 7 is the Aime cutoff.

Makes sense based on last year statistics. Tons of people are reporting 7s and 8s. Even with stupid mistakes.[/quote]

Yep same as my prediction: index 212, floor 7.

[quote="white_horse_king88"]So someone got a 147/6?[/quote]
147 isn't possible, is it?

Why not? Oh wait... you're right  :blush: .

[quote="white_horse_king88"]So someone got a 147/6?[/quote]

He's expecting a 117/9 to be the lowest of the top 330 indicies.

[quote="Treething"]I can't see how the floor would be [b][size=150]6[/size][/b] with [b][size=150]9[/size][/b] fairly easy problems.[/quote]
I can't see how the floor would be [b][size=150]6[/size][/b] with [b][size=150]15[/size][/b] fairly easy problems.  :roll:

BUT THAT RUINS THE NATURAL DIVINITY OF THE NUMBER SIX AND NINE

oh wait i'm supposed to be subtle

man #9, 10, 12, 13, 14, and 15 were hard

[quote="Treething"]Hmmmm.

I was pretty uninspired when I took the AIME (unofficially of course, through the questions posted on AoPS), and I managed to brute force my way to nine correct answers through basically algebra and pattern finding.

Questions 1 to 5 were honestly very easy if you knew how to not make stupid mistakes. For number 6 I made a giant list of sorts, and numbers 7 and 8 were also fairly simple.

Question 11 was the most retarded assfest ever (especially if you didn't know the formula for a sum of squares), but it was pretty simple as well, too easy for a #11 in my opinion.

I can't see how the floor would be [b][size=150]6[/size][/b] with [b][size=150]9[/size][/b] fairly easy problems.[/quote]

Because the floor score does not revolve around what you think the difficulty of the AIME is.

A general consensus would tell you that most people don't have a high opinion of the test's difficulty. Only its brute force nastiness. So that says nothing about who managed to not make stupid mistakes. So some people who usually get 7s on practices (this may not be enough on a typical AIME) get 7 on this AIME. So, the fact that some really great Olympiad performers got low AIME scores says nothing along the lines of a deflated floor score. So 7 sounds about right.

[quote="Treething"]Hmmmm.

I was pretty uninspired when I took the AIME (unofficially of course, through the questions posted on AoPS), and I managed to brute force my way to nine correct answers through basically algebra and pattern finding.

Questions 1 to 5 were honestly very easy if you knew how to not make stupid mistakes. For number 6 I made a giant list of sorts, and numbers 7 and 8 were also fairly simple.

Question 11 was the most retarded assfest ever (especially if you didn't know the formula for a sum of squares), but it was pretty simple as well, too easy for a #11 in my opinion.

I can't see how the floor would be [b][size=150]6[/size][/b] with [b][size=150]9[/size][/b] fairly easy problems.[/quote]
Nor does the floor score revolve around your obsession with a certain number


Two sides of scalene $ \bigtriangleup ABC$ measure $ 3$ centimeters and $ 5$ centimeters. How many different whole centimeter lengths are possible for the third side?

The general formula for this is $ 2s\minus{}1$ where $ s$ is the smaller side, thus $ 2\cdot3\minus{}1\equal{}5$

The triangle must be scalene...
The formula above gives the correct answer for any triangle, but we need to account for when the third side is 3 or 5, so $ 5\minus{}2\boxed{3}$
If you don't know the formula, go prove it now
[hide="hint"]
The triangle inequality states that for a triangle with sides $ a$, $ b$, and $ c$, $ a\plus{}b>c$
[/hide]


geometry
geometric transformation
homothety
angle bisector
geometry proposed

A circle $\cal C$ is given. In the plane, there are also two lines, say p and q, that do not intersect or touch the circle $\cal C$. For every circle $\cal K$, which is tangent to both the lines p and q, denote by $X$ the intersection of the two common (external) tangents  of the circles $\cal C$ and $\cal K$. Find the locus of $X$.

The locus of the point X, is a line passes through the intersection point of the given lines p and q.

     This problem is a direct application of the follows second theorem of concurrent segments ( see for
definition and first theorem, the post ''Hexahedron’s concurrensy'' ).

     THEOREM  – Let AA', BB', CC' be three concurrent segments at a point S and taken per
two of them ( as opposite sides ), form three quadrilaterals. Prove that:

    a) – The intersection points of diagonals of two quadrilaterals and the intersection point of
the opposite sides, the other than S, of the third one, are collinear.

    b) – The intersection points of the opposite sides, the other than S, of all the quadtilaterals, are collinear.

   This theorem is also true, for every configuration of the given segments ant their edges, in the plane,
or in 3D space and there are many interesting, it’s applications.

    A simple proof of this theorem, has been posted in Hyacinthos forum.

    http://groups.yahoo.com/group/Hyacinthos/message/13326

    I will post here next time, the solution of Eric problem, based on the above theorem.

    Kostas Vittas.

Let S be, the intersection point of the given lines p and q. The center of every circle (K), lies on the line s, angle bisector between p, q.

     The point X is the external homothety center of the circles (C), (K).

     So, if K1, is an arbitrary point on the line s, and P1, the orthogonal projection of K1 on the line p ( or on the line q ), then we define the constant point M on the circle (C), such that the segment line CM, where C is the center of the circle (C), to be perpendicular to the line p
( parallel to the line K1P1, to the same direction ).

     Hence, the line connecting the points P1, M, intersects the line CK1, at a point so be it X1, external homothety center of the circles (C), (K1) and also one point of the locus as we want.

     Similarly from another arbitrary point K2, on the line s, we can define the points P2, X2, on the lines p, CK2, respectively.

     We have now, the configuration of three parallel segments ( concurrent at infinity ) CM, K1P1, K2P2 and based on the second theorem of concurrent segments, we conclude that the points S, X1, X2, are collinear.

     Hence the locus of X, is a line passes through the intersection point S, of the given lines p, q and the solution is completed.

     NOTES.

a) – By the same way we can prove that the locus of the internal homothety center of the circles (C), (K), is another line, also passes through the point S.

b) – It is clear that doesn’t matter for the place of the circle (C), with respect to the given lines p and q.

     Kostas Vittas.


The math teachers visiting the Indianapolis 500 Speedway timed the car that was practicing on the track and found it took 57 seconds for the car to complete the 2.5-mile lap. To the nearest whole number in miles per hour, what was the average speed of the car?

\[ \frac{2.5\text{ miles}}{57\text{ seconds}}\cdot \frac{3600\text{ seconds}}{1\text{ hour}} \approx \boxed{158 \text{ mph}}\]

EDIT: beaten


Does anyone here have any Orkut invites?
It would be great if i can hav one.

Thx.
tlhoo

Orkut is actually like Friendster....some community place to meet friends....but it is from Google....therefore, like GMail....one has to be invited to be a member of Orkut.


Hello and welcome to the AoPS instute of technology, the proud sponsor and host of the 2008 series, AoPS mini-Olympics.

The cost is only $ \$$5/event.

Only open to AoPS city citizens

[quote="1=2"]I will join. I want to get away from the factory.[/quote]

This is not the revamped one.
Are you in the orginal one?

Hey AoPS city people!!! Monetary prizes for winners.

[quote="zapi2007"]I'm just curious.
How are these miniolympics carried out anyway?[/quote]

Just sign up
Youll figure out

(Lucky for many of you AoPSers its nothing physical)

aw, i was hoping it would be physical
does it have anything to do with who can type the fastest or something  :lol:

I'll join.

Im actually considering two divisions:
Physical and non-Pyhsical...

How about:

Track and Field, Ball Sports, Aquatic Sports, Racing Sports (e.g. biking, equestrian), Accuracy Sports (e.g. archery, rifling), Strength Sports (e.g. weightlifting, wrestling), and Miscellaneous Sports (e.g. fencing)

Dont lose interest you guys!!
We have to wait till Day 11... :(

VERY old thread.

Please lock!

unless Dojo does come back and revive this with a good plan.

I join (if this isn't revival)

5 real dollars?

i'll join.

but i hope this don't die like all the other gamez i've been on...

I shall explain very carefully why we should let this topic die.

First of all, this topic is extremely old, and it has less than 40 posts. Plus there's a span of more than one year (Friday, December 5, 2008 to Yesterday) where nobody posted. Therefore the user named eagle2 must have gone out of his way to actually find some topic like this, and then post in it. However, chances are if it hasn't been replied to for more than a year, people have abandoned it. Also eagle2 has been on AoPS for more than a month, which is more than enough time to read the rules of this forum, and he clearly hasn't. This topic has been revived just for the sake of reviving it, and it should either be successfully revived by the topic creator (Dojo), or be locked forever to prevent more posts.

I leave it up to Dojo to decide the fate of this topic.

Or the mods. They can lock this topic if they want.

Wow I just realized that...

Mods, please lock.

I actually thought that this was a new thread...

Yes, Please lock.

Eagle2, please know what you are doing before reviving a long-dead thread.

oh, ok.... forgot to check validity...  :(

I didn't notice it guys.. You guys are soooomean :(

[quote="A PM"]Lock this before 10 people make posts on it accusing the person of reviving the topic. [/quote]

Oops. Too late.


pigeonhole principle
combinatorics
IMO
IMO 1972

Prove that from a set of ten distinct two-digit numbers, it is always possible to find two disjoint subsets whose members have the same sum.

note that if we get two sets $A,B$ (not necessarily disjoint) with this property, then the same is also true for the sets $A\setminus (A\cap B),B\setminus(A\cap B)$.

to get two distinct subsets with the same sum, note that there are $2^{10}-1$ different non-empty subsets. from amongst the integers $10,\ldots,99$ any subset of size atmost $10$ can have a maximum sum of atmost $91+92+\cdots+99=945$ and so necessarily the sums of various subsets from among all the subsets of the chosen set of $10$ elements can be atmost $945<1023$ and so the problem is solved.

[quote="TheLivingShadow"]I like your solution but I have a question, does the fact that there are 1023 subsets but at most the sum of one of these subsets is 945 mean that there are at least 78 subsets with equal sum? If not, then why?[/quote]

No it doesn't mean that , because $78$ set can have the same sum.

@seshadri: You have a typo, not $ 91+92+\cdots+99=945 $ but $ 90+91+92+\cdots+99=945 $.

Sorry everybody for bringing back an old topic.

[quote="limac"]wait...say we have the two disjoint subsets A and B each with sum k. So the sum of all the 10 elements is 2k, an even number. But the problem does allow us to pick the 10 numbers 10, 20, 30, ..., 80, 90, 91 for instance, with an odd sum. Am I missing something?[/quote]
Yes nothing wrong! But the subsets may not have 5 elements each, so I  do not see any problem.

yeah, but the sum has to be divided evenly between the two subsets, although we don't require both to have 5 elements each. So I just let that common sum be k for both of the subsets.

The subsets need not form a partition of the ten-element set.  In your example, I might take {30} and {10, 20}, for example.

Hi everybody!
I have only a question: subset1 $\cup$ subset = $A$ ?

Better late than never :D
Number of possible subsets = $2^{10} - 1 = 1023$
The subset with the maximum sum of elements is $\{90, 91 , \cdots , 99 \}$ (sum $= 945$)
Subset with the minimum sum of elements is $\{10\}$
Let $S_1 , S_2 , \cdots S_{1023}$ be the sum of elements of all the possible subsets.
There are 1023 'sums' but the number of 'possible sums' is less than $1023$ (because of the maximum bound) hence by PHP we can conclude there exist at-least 2 sums $S_i$ and $S_j$ ($i \neq j$) such that $S_i = S_j$.
Recall that the original question asked us to prove that there exists disjoint subsets, now since we know that there exists at-least $2$ subsets with the same sum of elements, if they are not disjoint then remove the common element from the $2$ subsets, this strategy will keep the sum of elements of the $2$ subsets same but then they will be disjoint, and we are done.

[hide=Solution]Note that the total amount of subsets of a set with $10$ elements, excluding the empty set, is $2^{10}-1=1023$. However, we cannot have the set itself as a subset, because there are no subsets disjoint to this. That gives a new total of $1022$. The largest possible sum is $91+92+93+\ldots+99=855$ (remember that we cannot have all 10 elements) and the smallest possible sum is just using only $10$ for a sum of $10$. This gives a total of $846$. Clearly $846<1022$, so at least two different subsets indeed have the same sum. To make sure they are disjoint, take out any elements that are in both subsets. The resulting subsets will still have the same sum, as the same thing was subtracted, and we're done.[/hide]

[quote=OlympusHero][hide=Solution]The total amount of subsets, excluding the empty set, is $2^{10}-1=1023$. The maximum possible sum is $90+91+92+\ldots+99=945$, but this is smaller than $1023$. By the Pigeonhole Principle, we hence have that there are two disjoint subsets whose members have the same sum.[/hide][/quote]

You have to prove that there are $945$ possible sums.

Wait am I missing something or is it clear that every sum before 945 can be made except for those less than $10+11+12+13+\ldots+19=145$ right?

Bumping. Please don't get mad at me because I bumped 24 minutes before 24 hours. The question is in the above post.

[quote=jasperE3]You have to prove that there are $945$ possible sums.[/quote]You need to show that there are [i]at most[/i] that many distinct sums.  The main missing bit from OlympusHero's post was the note about disjointness from #2.

Okay, for anyone curious; I completely misread the problem - I'll fix it now. Thanks @jasperE3 and @oolite.

To begin, we find that the total number of subsets of a set with 10 elements, excluding the empty set and the entire set (since no other set can be disjoint to it) is $2^{10}-2=1022$.

We now need to find the amount of possible sums of these subsets. The smallest sum occurs in the subset $\{10\}$, whose sum of 10. The largest sum occurs in the subset $\{91,92,...,98,99\}$, which has a sum of
$$91+92+...+98+99=\frac{9*(91+99)}{2}=9*95= 855$$
Since $855<1022$, there must be two subsets with the same sum by the Pigeonhole Principle.

For sets with the same sum that are not disjoint, we can remove all elements that are common in both sets, preserving the fact that both sets have the same sum.

Therefore, we have proved that in a set of 10 distinct two-digit numbers, it is always possible to find two disjoint subsets whose members have the same sum. $\square$

First notice that the total number of possible subsets is $2^{10}-2=1022$.
Next, notice that the smallest possible sum is $10,$ and the largest possible sum is $$91+92+...+98+99=\frac{9\cdot(91+99)}{2}=9\cdot95= 855.$$ So, by the Pigeonhole Principle, there must be $2,$ subsets that have the same sum. To make the set disjoint, we simply just remove the intersections of the $2$ sets and hence we are done. $\blacksquare$

The number of possible subsets is $\binom{10}{1}+\binom{10}{2}+\dots+\binom{10}{9}=1022.$ The possible sums range from $10$ to $91+92+\dots+99=855,$ so there are $846$ sums. Hence, by the Pigeonhole Principle, there must exist two distinct subsets $A$ and $B$ with the same sums. Notice that $A-A\cap B$ and $B-A\cap B$ satisfies the problem. $\square$

The number of total subsets is just $2^{10}-2=1022$. Now the maximum possible sum is $91+92+\ldots+99=855<1022$. Thus, at least two subsets must have the same sum. To make them disjoint, just remove their intersection.


trigonometry
inequalities
inequalities unsolved

Prove that for angles $ A$, $ B$, $ C$ in a triangle, then

$ \sqrt{3}(\sin{\frac{A}{2}}\sin{\frac{B}{2}}\plus{}\sin{\frac{B}{2}}\sin{\frac{C}{2}}\plus{}\sin{\frac{C}{2}}\sin{\frac{A}{2}}) \le 2 \cos{\frac{A}{2}} \cos{\frac{B}{2}} \cos{\frac{C}{2}}$

Thanks

i don't like this solution, but anyway here it goes... perform the substitution $ \sin\frac{A}{2}\equal{}\sqrt{\frac{yz}{(x\plus{}y)(x\plus{}z)}}, \cos\frac{A}{2}\equal{}\sqrt{\frac{x(x\plus{}y\plus{}z)}{(x\plus{}y)(x\plus{}z)}}$, for some positive reals $ x,y,z$.

after some little manipulation it follows that the inequality is equivalent to

$ \sqrt3\sum\sqrt{z(y\plus{}z)(x\plus{}z)}\leq 2(x\plus{}y\plus{}z)\sqrt{x\plus{}y\plus{}z}$, or equivalently,

$ \sum\sqrt{\bigl(4z(x\plus{}y\plus{}z)\bigl)\bigl(3(y\plus{}z)(x\plus{}z)\bigl)}\leq 4(x\plus{}y\plus{}z)^2$

from am-gm we have that
$ \sum\sqrt{\bigl(4z(x\plus{}y\plus{}z)\bigl)\bigl(3(y\plus{}z)(x\plus{}z)\bigl)}\leq\sum\dfrac{7z^2\plus{}7xz\plus{}7yz\plus{}3xy}{2}$

we conclude the proof using the well-known $ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}yz\plus{}zx$, and we're done  :D

Can you explain to me how we are allowed to make those substitutions?

I am confused on how to show that it is okay to make those as substitutions.

if we let $ a,b,c$ to be the corresponding sidelengths, we have that $ \sin\frac{A}{2}\equal{}\sqrt{\dfrac{(s\minus{}b)(s\minus{}c)}{bc}}, \cos\frac{A}{2}\equal{}\sqrt{\dfrac{s(s\minus{}a)}{bc}}$, where $ s$ is the semiperimeter... then we take $ (s\minus{}a,s\minus{}b,s\minus{}c)\equal{}(x,y,z)$ and that's it... this substitution is very useful in proving trig inequalities, although it "kills" its sense  :ninja:

Thanks.

Another question, how did you know how to group the terms?  Like why did you put the 4 with one term and the 3 with the other, and not the other way around?

when you use am-gm like $ 2\sqrt{ab}\leq a\plus{}b$ you have equality iff $ a\equal{}b$... in this case it's easy to note that we have equality if $ x\equal{}y\equal{}z$, so it would be a bad idea to write make something like $ a\equal{}3x(x\plus{}y\plus{}z)$ and $ b\equal{}4(x\plus{}y)(x\plus{}z)$...


linear algebra
matrix
linear algebra unsolved

Let $ n \in \mathbb{N}, n \ge 2$ and $ A \in M_{n,2}(\mathbb{C})$, $ B \in M_{2,n}(\mathbb{C})$ with th property that there exists $ k \in \mathbb{N}$ such that $ (AB)^k\equal{}O_n$. Prove that:

$ a)$ $ (AB)^3\equal{}O_n$

$ b)$ If rank $ (AB) \not\equal{} 2$, then $ (AB)^2\equal{}O_n$

$ rank(A)\leq{2}$ then $ rank(AB)\leq{2}$. Moreover $ AB$ is nilpotent then $ AB$ is similar to $ diag(\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},0_{n\minus{}3})$ or $ diag(\begin{pmatrix}0&1\\0&0\end{pmatrix},0_{n\minus{}2})$ or $ diag(\begin{pmatrix}0&1\\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix},0_{n\minus{}4})$ or $ AB\equal{}0$.
After it's easy.

I understand [color=blue]loup blanc [/color]!
You can use Theorem Cayley Hamilton !
1)The fist : $ B(AB)^kA \equal{} (BA)^{k \plus{} 1} \equal{} O$ ,$ BA \in M_2(\mathbb{C})$ have eigenvalues are $ O$
Then use Theorem Cayley Hamilton $ (BA)^2 \equal{} O$ then $ (AB)^3 \equal{} A(BA)^2B \equal{} O$
2) We have  $ rank(AB) \leq rank(A) \leq 2$ then $ rank(AB) \neq 2$ Implies $ rank(AB) \equal{} O$ or $ 1$
 Implies with $ rank(AB) \equal{} O$
Now if $ rank(AB) \equal{} 1$ then $ (AB)^2 \equal{} Tr(AB)AB$ from (1) easy, too ....


inequalities
algebra
n-variable inequality
IMO Shortlist
IMO Longlist

Prove the inequality

[b]a.)[/b]  $
\left( a_{1}+a_{2}+...+a_{k}\right) ^{2}\leq k\left(
a_{1}^{2}+a_{2}^{2}+...+a_{k}^{2}\right) ,  $

where $k\geq 1$ is a natural number and $a_{1},$ $a_{2},$ $...,$ $a_{k}$ are arbitrary real numbers.

[b]b.)[/b] Using the inequality (1), show that if the real numbers $a_{1},$ $a_{2},$ $...,$ $a_{n}$ satisfy the inequality

\[
a_{1}+a_{2}+...+a_{n}\geq \sqrt{\left( n-1\right) \left(
a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\right) },
\]

then all of these numbers $a_{1},$ $a_{2},$ $\ldots,$ $a_{n}$ are non-negative.

In that case, assume WLOG that $a_n<0$. Then $a_1+\ldots+a_{n-1}+a_n<a_1+\ldots+a_{n-1}\le \sqrt{(n-1)(a_1^2+\ldots+a_{n-1}^2)}\le \sqrt{(n-1)(a_1^2+\dots+a_{n-1}^2+a_n^2)}$, which contradicts the hypothesis of b).

So, for b) :

Assume that for such numbers, we have $a_n < 0$. Then :
$a_1 + \cdots + a_{n-1} > a_1 + \cdots a_{n-1} + a_n \geq \sqrt {(n-1)(a_1^2 + \cdots +a_n^2)} > 0$.
It follows from (1) that
$(n-1)(a_1^2 + \cdots + a_{n-1}^2) \geq (a_1 + \cdots + a_{n-1})^2 > (n-1)(a_1^2 + \cdots +a_n^2)$
so that $(n-1)a_n^2 < 0$, a contradiction.

Pierre.

Part A is extremely obvious using any of these theorems:
QM-AM, Jensen's inequality, Chebycheff's inequality, Engel's inequality.
I think It can also be done by Induction but that's much more than these 3 line solutions :)


calculus
derivative
function
real analysis
real analysis theorems

prove that  the existence of a partial derivative at a point does not necessarily imply the continuity of z=f(x,y) at that point.

The classic example:

$ f(x,y)\equal{}\begin{cases}\frac{xy}{x^2\plus{}y^2},&(x,y)\ne(0,0)\\0,&(x,y)\equal{}(0,0)\end{cases}$

Both partial derivatives exist (and are zero) at the origin, but the function is discontinuous there.

$ \chi(x, y) \equal{} \begin{cases} 1, & xy \equal{} 0 \\ 0, & xy \neq 0\end{cases}$ also works nicely at the origin.


Find all prime numbers like $p$ and $q$ such that :
$p^2-2q^2=1$

[hide]If $q=3$, $p^2=19$ which has no solution.
If $q\neq 3$, $q^2\equiv 1\pmod{3}\Rightarrow 3|p^2\Rightarrow p=3\Rightarrow q=2$
making the only solution $(3,2)$[/hide]


I look for the math teem or  the math circle in Lexington KY.

okay. Good luck. By the way, I need more posts.

Proffeser Jelly Bean.  If you want a higher post rating and higher posts, then post relevant posts not I need posts or this or that.  Post answers to questions or something.  Also plz stop spamming.   :spam:  :spam:  :spam: That's all,
Thanks

[quote="Butler"]What kind of math team small_unicorn? There are several different ones.[/quote]
I am in 6 grade. I look for any unschooler team or the  open  team. I  look for  math circle as  Boston  math circle or russian  math circle.

[quote="Butler"]What kind of math team small_unicorn? There are several different ones.[/quote]
I  like open contest  Turnir Gorodov . I do not like doing many easy problems .
I want learning non school math  more then a contest.


Suppse that $n\geq 3$ be an inetegr number and $x\in \mathbb{R}$ s.t. $x,x^2$ and $x^n$ have a same decimal representation. Prove that $x$ is an integer number.

I'm sorry, but have taken these problems from? :?

I think this appeared on the Irish MO in 1998 as well.

Here it is:
http://www.kalva.demon.co.uk/irish/soln/sol985.html


When is the list of USAMO qualifiers coming out? D:

Not until AT LEAST AIME2 is administered (and then some more time after that - about a week)

When can I receive my AIME 1 score? Will it happen before or at the same time as the result of USAMO qualifiers is announced?


Fifteen bands of diameter $2.25$ inches are put together to form the 'triangle' shown. What is the minimum length in inches, of the metal band required to enclose the balls?

[img]7205[/img]

[i]Note: Picture is not drawn to scale :P [/i]

[hide]Split the band into 2 parts: the rounded corners and the straight parts.

The combined length of the three rounded corners is the circumference of one of the circles, or $2.25\pi$.

The length of each straight part is $4\cdot2.25$. Their combined length is $27$.

$2.25\pi+27\approx\boxed{34.07\text{ inches}}$[/hide]

[quote="#H34N1"]How do you know that the three rounded corners are each $\frac{1}{3}$ of the band.[/quote]

Draw an equilateral triangle, by connecting the centers of all the circles. Now draw perpendiculars from the center of the corner circles to the band. You have two 90-degree angles and one 60-degree angle, the last one has to be 120- degrees which is 1/3 of the circumference of the circle.


graph theory
combinatorics proposed
combinatorics

A country has $ 1998$ airports connected by some direct flights. For any three airports, some two are not connected by a direct flight. What is the maximum number of direct flights that can be offered?

I'm assuming that the graph is undirected, i.e., if there is a flight from $ X$ to $ Y$, then there is also a flight form $ Y$ to $ X$.  Then the maximum number of possible edges is $ 999^{2}$.  This is possible by constructing a bipartite graph between two distinct sets of $ 999$ points, that is, partition the set of vertices $ V$ into two disjoints sets $ V_{1}$ and $ V_{2}$ so that they each are of size $ 999$, and for any $ v_{1}\in V_{1}$ and $ v_{2}\in V_{2}$, let $ (v_{1},v_{2})\in E$.

Now suppose a graph $ G$ has $ 2n$ vertices.  Then if $ G$ has no triangle, I will show the number of edges $ e$ cannot be greater than $ n^{2}$.  Suppose $ e\ge n^{2}+1$.  Then any two adjacent vertices $ x$ and $ y$ cannot share a common point, so $ d(x)+d(y)\le 2n$.  If we let the set of all edges be $ E$, then we have

$ \sum_{x\in V}d(x)^{2}=\sum_{(x,y)\in E}(d(x)+d(y))\le 2en$.

But by Cauchy-Schwartz, we have

$ \sum_{x\in V}d(x)^{2}\ge \frac{(\sum_{x\in V}d(x))^{2}}{|V|}=\frac{2e^{2}}{n}$.

This leads to the fact that $ \frac{2e^{2}}{n}\le 2en$, implying that $ e\le n^{2}$, a contradiction.

Note that this problem can be immediately killed by Turan's Theorem, which states:

If a graph $ G=(V,E)$ on $ n$ vertices does not have a $ (k+1)$-clique for some $ k\ge 2$, then $ |E|\le \left (1-\frac{1}{k}\right )\frac{n^{2}}{2}$.

Quite simple. Take the vertices as the cities and connect edges for those that are connected. It is easy to see that this a 3-free graph. Hence, using Turan's, we will find the maximum number of directed flights, given the constraint of a 3 free graph. The answer is clearly $1/4 \cdot 1998^2=998,001$

For the construction, consider a bipartite graph(split the 1998 vertices into 2 groups) and let them be say, A and B. Then for each vertex in A, connect this vertex to 999 vertices in B. Do this for all the vertices in A. This gives the desired graph.


$P$ and $Q$ are two different point of circle $x^2+y^2=4x$ . The tangent lines at $P$ and $Q$ touch another circle $x^2+y^2=1$ at $R$ , $S$ respectively . Prove that the equation of circle passing through $P,Q,R,S$ is $x^2+y^2=2x+2$.

uh... welll the circle going through $x^2 + y^2 = 2x+2$ would intersect $x^2 + y^2 =1$ and $x^2+y^2 = 4x$ in two places each... giving us 4 pts of intersection

if $P$ and $Q$ lie on $x^2 + y^2 = 4x$ and $x^2 + y^2 = 2x+2$, and $R$ and $S$ lie on $x^2+y^2 = 1$ and $x^2 + y^2 = 2x + 2$, then that means $P,Q,R,S$ must be those 4 pts of intersection

but those 4 pts of intersection are constant... so does this mean that the points $P,Q,R,S$ are fixed? which i dont think is true....

"Touch" means "is tangent to," so in fact $P$ and $Q$ are fixed.  (Drawing a picture helps here.)

Yes , you are right JBL . $P$,$Q$,$R$,$S$ are all fixed since the two tangent lines at both the circle will give us exactly 4 different point . And " touch " means tangent .


Hello!
What is the code for putting a box around your answer?
Thanks!

If you want a bigger box you can also use [code]$\boxed{\ 100\ }$[/code]

Which gives:
$ \boxed{\ 100\ }$ instead of $ \boxed{100}$

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196392

I wish people would read...

[quote="Temperal"][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196392[/url]

I wish people would read...[/quote]
I wish people would pay attention to dates and times. :dry:

That topic was created [i]after[/i] this one.

What I said is still true, then, but the other way around.

[quote=hunter117]I want to put a really long equation into a box and it seems like the equation cant really fit in one line. How do I make the box vertically taller so the equation can fit?[/quote]

Well, I don't know how to do that, but you can create two boxes, or you can just create one box and leave the rest that won't fit draping down onto the next. My really long conclusion sentence took up one whole line and there really was no need for me to create another box.

Also, you can do  \boxed{text\{your sentence or equation}} (with dollar signs) to give something like $\boxed{\text{Hi!}}$

[quote=LynnSaratoga]Also, you can do $ \ b o x e d { t e x t \ { y o u r  s e n t e n c e   o r   e q u a t i o n } } $ to give something like $\boxed{\text{Hi!}}$[/quote]

Do you mean
[code]$\boxed{\text{your sentence or equation}}$[/code]?
(that gives $\boxed{\text{your sentence or equation}}$)

[quote=nkouevda][b]\boxed{<expression>}[/b]

[b]\boxed{e^{i\pi}+1=0}[/b] produces: $ \boxed{e^{i\pi} \plus{} 1 \equal{} 0}$[/quote]

but you have to put $'s on them

sorry to revive a dead thread, but you can also use \fbox{text} . Example: $\fbox{testing}$

it's probably find because the thread was from 2008

it's a million years old anyway, people keep reviving it :)


I know that there is a brute force way to do this but im looking for a formal way.  I also note that can be achieved in more than one way.

How many integers can be expressed as the sum of two or more different members of the set $ \{0, 1, 2, 4, 8, 16, 31\}$?

[hide="hint"]It's screaming for base 2.[/hide]
[hide="solution"]In base 2 it seems to be $ 0,1,10,100,1000,10000 and 11111$. Note that just by adding, you can get all numbers from 1 to 111110, which, in base 10, is all the numbers from 1 to 63.
$ \boxed{62}$.[/hide]

EDIT-I was being an idiot.

Just my method
[hide]
Rewrite the numbers in binary form:

$ 0, 1, 10, 100, 1000, 10000, 11111$

There're two kinds of numbers
(1) Without $ 11111$
(2) With $ 11111$

Because the sum of first five numbers are $ 11111$, (1) and (2) are equal.

Since every number has an unique binary expression, there're

$ \binom{5}{1}\plus{}\binom{5}{2} \plus{} \binom{5}{3} \plus{} \binom{5}{4} \plus{} \binom{5}{5} \equal{} (1 \plus{} 1)^5 \minus{} \binom{5}{0} \equal{} 31$

different sums in each kind, and total is $ 31*2 \equal{} 62$

[/hide]

To xpmath: it's a good method, but it should be $ 1$ throught $ 111110$.

[quote="timwu"]Just my method
[hide]
Rewrite the numbers in binary form:

$ 0, 1, 10, 100, 1000, 10000, 11111$

There're two kinds of numbers
(1) Without $ 11111$
(2) With $ 11111$

Because the sum of first five numbers are $ 11111$, (1) and (2) are equal.

Since every number has an unique binary expression, there're

$ \binom{5}{1} \plus{} \binom{5}{2} \plus{} \binom{5}{3} \plus{} \binom{5}{4} \plus{} \binom{5}{5} \equal{} (1 \plus{} 1)^5 \minus{} \binom{5}{0} \equal{} 31$

different sums in each kind, and total is $ 31*2 \equal{} 62$

[/hide]

To xpmath: it's a good method, but it should be $ 1$ throught $ 111110$.[/quote]

I agree with you, and have a different [hide="solution."]With the first 5 powers of 2(2^0 through 2^4), we can get any sum less than 32 except the powers of 2. With the 0, we can get those. And with the 31 around, we can get 31 more numbers, getting 62.[/hide]


quadratics
geometry
calculus
trigonometry
algebra
quadratic formula
special factorizations

I've noticed that all too many algebra 1 textbooks teach the quadratic formula before the students have learned the methods necessary to derive it. Being a hardcore AoPSer, I've learned all of my formulas after I could derive them on my own.

Also, my public school math class isn't challenging at all! What should I do?

ask the teacher to give you some more challenging things to do during class
you can ask your teacher to give you some sites to look at after school and do some extra hard problems for homework instead of what they usually give
basically, team up with your teacher and ask him to help you find some harder problems and let him teach your harder concepts

Personally, I think you should turn to independent study. If you're above average, don't waste your time bugging your teacher for challenges. You're in AoPS, so that's a good start. Do FTW. Take some classes. Buy/borrow the books. That should keep you busy. :)

A reason this happens is because both are thought of as methods used to solve a quadratic. But the problem is that by the time the students figure out that they could use the completing the square method to derive the quadratic formula, it's probably taken too long. So I guess they just explain it in the book. Since either way, they would probably show the derivation, it might not matter too much...though a student might be able to follow along without actually understanding the reasoning behind the steps they used.

In other words, someone might understand the solution, but they wouldn't know why the given steps were used, or how the solver knew they would lead to the solution.

Yeah, I have AoPS books...

"Your way is faster and better and clearer, but you have to do it my way because the other students don't know your way." -- my math teacher

*sobs*

[quote="nick42"]ha, we never learned how to complete the square. i had to figure it out by myself.[/quote]

Yeah, same here.

What is completing the square (I've already taken Algebra I)?

Completing the square is an algebraic maneuver.

For example, take the problem $ 4x^2 \minus{} 2x \minus{} 5 \equal{} 0$

First, move the number without an $ x$ coefficient to the right side.

$ 4x^2 \minus{} 2x \equal{} 5$

Then, divide both sides by the $ x^2$ coefficient.

$ x^2 \minus{} \frac {1}{2} x \equal{} \frac {5}{4}$

Then, take the coefficient of the $ x$ term and halve it.  Then, square this and add the result to both sides.

$ x^2 \minus{} \frac {1}{2} x \plus{} \frac {1}{16} \equal{} \frac {21}{16}$

Now, you can factor this as a binomial square.

$ (x \minus{} \frac {1}{4})^2 \equal{} \frac {21}{16}$

Take the square root of both sides.

$ x \minus{} \frac {1}{4} \equal{} \plus{} \frac {\sqrt {21}}{4}$ or $ x \minus{} \frac14 \equal{} \minus{} \frac {\sqrt {21}}{4}$

Finally, solve for $ x$.

$ x \equal{} \frac {1}{4} \plus{} \frac {\sqrt {21}}{4}$ or $ x \equal{} \frac {1}{4} \minus{} \frac {\sqrt {21}}{4}$

How to prove quadratic formula:

$ ax^2\plus{}bx\plus{}c\equal{}0$

$ x^2\plus{}\frac{b}{a}x\plus{}\frac{c}{a}\equal{}0$

$ \left(x\plus{}\frac{b}{2a}\right)^2\equal{}\frac{b^2}{4a^2}\minus{}\frac{c}{a}$

$ \left(x\plus{}\frac{b}{2a}\right)^2\equal{}\frac{b^2\minus{}4ac}{4a^2}$

$ x\plus{}\frac{b}{2a}\equal{}\frac{\pm\sqrt{b^2\minus{}4ac}}{2a}$

$ x\equal{}\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$

Some do it this way (it comes out to be the same thing):

$ ax^2\plus{}bx\plus{}c\equal{}0$

$ ax^2\plus{}bx\equal{}\minus{}c$

$ 4a(ax^2\plus{}bx)\equal{}4a(\minus{}c)$

$ 4a^2x^2\plus{}4abx\equal{}\minus{}4ac$

$ 4a^2x^2\plus{}4abx\plus{}b^2\equal{}b^2\minus{}4ac$

$ (2ax\plus{}b)^2\equal{}b^2\minus{}4ac$

$ 2ax\plus{}b\equal{}\pm\sqrt{b^2\minus{}4ac}$

$ 2ax\equal{}\minus{}b\pm\sqrt{b^2\minus{}4ac}$

$ x\equal{}\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$

And there's an alternative (rarely used), if $ c\neq0$

$ x\equal{}\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$

$ x\equal{}\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}\cdot\frac{\minus{}b\minus{}\sqrt{b^2\minus{}4ac}}{\minus{}b\minus{}\sqrt{b^2\minus{}4ac}}$

$ x\equal{}\frac{4ac}{2a(\minus{}b\minus{}\sqrt{b^2\minus{}4ac})}$

$ x\equal{}\frac{2c}{\minus{}b\minus{}\sqrt{b^2\minus{}4ac}}$

My school probably wouldn't care. Ha we probably don't learn in it in my school until like 9th grade. :maybe:

Probably. The quadratic formula would be Alg I ( :maybe: ), which is what the "average" student takes in 9th grade.

At our schools, geometry is 9th grade, and algebra 1 is 7th grade, 8th grade is like a huge review calld "advanced math" aka easier than AMC 8 level...

For me:

6th/7th grade: Pre-Algebra
8th grade: Algebra I
9th grade: Geometry
10th grade: Algebra II

this is not spam. I am testing to see if forum is working now

6th Grade: Math 6(learn how to add fractions!)

7th Grade: Math 7(review how to add fractions!)/Pre-Algebra(yes, this is supposed to be an advanced course)

8th Grade: Pre-Algebra/Algebra 1

here is what you do if you are really bored. Go tell your school you want to skip a grade. Go to some summer thing and then skip it. That way you get a challenge. I'm in 8th grade taking algebra 2. lol right now we are learning how to graph linear equations xD

How it is normally done in my school:

8th grade: Pre-Algebra
9: Alg I
10: Alg II
11: Goemetry
12: Pre-calc
(But a lot of people are advanced anyways, so only about half the people actually follow this. Most are advanced 1 year.)

How I'm doing it:

6: Alg I honors
7:Alg II honors
8: Geometry/Pre-calc
9: Calc BC
10: College math

My old school was -

6th - Pre-Algebra
7th - Algebra 1
8th - Geometry

My current school -

6th - ?
7th - Pre-Algebra
8th - Algebra 1/Beginning of Geometry

What I've been Doing -

6th - Pre-Algebra - Geometry
7th - Alebra 1/Geometry

I forgot to say what I did:

5th Grade: Pre-Algebra(learning Algebra I on my own)

6th Grade: Algebra 1(learning Geometry and Pre-Calculus on my own)

7th Grade: Geometry(learning Calculus on my own)

8th Grade: Algebra II(not really sure what I'm gonna learn on my own xD )

9th Grade: Pre-Calc?

10th Grade: Calc?

11th Grade: Beats me

The norm here:

5th ----------------5th grade math
6th-----------------6th grade math
7th-----------------7th grade math
8th-----------------Pre-Algebra
9th-----------------Algebra I
10th----------------Geometry
11th----------------Algebra II
12th----------------Pre-Calculus
Freshman college----Calculus

me

5th ----------------6th grade math
6th-----------------7th grade math
7th-----------------Algebra I
8th-----------------Teaching myself Algebra II although the teacher gives me tests (The only way I could have taken Geometry was by going to the local high school, which my mom would have to take me to and from every day aka no way)
9th-----------------Geometry
10th----------------Algebra II
11th----------------Pre-Calculus
12th----------------Calculus
Freshman college----Who knows?

I would like to learn Trig, because it is one of my biggest weaknesses, when I am taking the AMC 10.

What happens if I'm already halfway through vol. 2 and I'm still in 7th grade???

Norm at my school:
7: Pre-Algebra
8: Algebra
 (my middle school doesn't offer 6th. I went to 6th in elementary school)

Me:
7: Advanced Algebra
8: Geometry

What I should be doing:
7: Second half of Algebra 2
8: AoPS Classes
9: More AoPS Classes

But of course you can do it on your own. :)


calculus
integration
topology
logarithms
derivative
Ring Theory
absolute value

As I learned [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=698615]here[/url] (with some guidance from mlok), both the size and additive structure of a set $A \subset \mathbb{R}$ influence the size of the set $A+A$. So, in order to get the kind of result I want about a Minkowski sum, I need to figure out what effect the additive structure has on the size of the sum set.

So, what about multiplicative structure? Will I face a similar obstacle if I want to consider the set $A\cdot A = \{a_{1}\cdot a_{2}| a_{1}, a_{2}\in A\}$? In the other thread mlok suggested that I should look at the Fourier transform of a measure on $A$ (for addition). Does anyone have any ideas about how I could investigate the multiplicative structure of a set with this goal of determining size in mind? Is there some appropriate integral transform?

Multiplication is just like addition: take a logarithm. Negative numbers are slightly different, but nothing too complicated.

But what relation is there between the measure of the set $\log(A) = \{\log(a) | a \in A\}$ and the original set $A$?

I feel like multiplicative structure is stronger in some sense, because if $\text{dim}_\mathcal{H}(A) = s$ and it contains a number larger that 1 in absolute value, then $\mathcal{H}_{s}(A) = 0$ or $\infty$.

But maybe we can in fact conclude something similar when we have additive structure and it just requires some cleverness with exponential or something....

The Hausdorff measure of $\{\log a \colon a\in A\}$ is within a multiplicative constant of the measure of $A$ provided that $A$ is bounded away from $0$ and $\infty$. This is because $\log$ satisfies a bi-Lipschitz estimate there (simply put, its derivative is bounded away from $0$ and $\infty$). Considering the exhaustion of $A$ by $A\cap (1/n,n)$, you'll find that passing to logarithms does not change the Hausdorff dimension.

Of course, multiplication [b]within integers[/b] is quite different from addition.

By the way, this discussion reminded me of a [url=http://arxiv.org/abs/math.CA/0607233]recent paper[/url] which contains the following amusing passage.
[quote]the inspiration for this paper is a preprint ... in which the authors assert that a set having Hausdorff dimension $d/2$ but distance set of dimension $1/2$ must have a lot of additive structure. This is obtained from a Fourier analytic estimate. We wrote this paper because we felt that the authors understated the case. Not only does the set have additive properties, but it also has multiplicative properties. Thus, a fortiori... the set does not exist.[/quote]

Wow, that's the second time I've seen Nets Katz's name this week! If I keep being interested in GMT maybe I should go there and see if he'll be my PhD advisor.

By the way thanks very much for those references; they seem to be related to what I'm trying to do, and maybe if I read some actual papers about this stuff, I will stop bumbling around like an idiot filling notebook pages and forum threads without any actual results...

Um, that particular paper is not an introductory reading. Although the references could be of some use.

I still want to know if $\mathcal H^{s}(A)>0$ implies $\mathcal H^{s}(A+A)=\infty$ for fractional values of $s$.  :maybe:

[quote="mlok"]Um, that particular paper is not an introductory reading. Although the references could be of some use. [/quote]
Yeah I am a little more interested in some of the references. Would you mind telling me whether the first one, Bourgain, "On the Erdos-Volkmann and Katz-Tao ring conjectures," contains a proof of the Erdos-Volkmann ring conjecture? I don't want to look because that's what I'm trying to prove...

[quote="mlok"]I still want to know if $\mathcal H^{s}(A)>0$ implies $\mathcal H^{s}(A+A)=\infty$ for fractional values of $s$.  :maybe: [/quote]
I am interested too. I'll keep thinking about that one; my winter break starts next week, and then I can REALLY start working on GMT. ;)

[quote="mlok"]It does, but you can read that paper without any danger of understanding the proof.  :lol:

A [b]much[/b] simpler proof was found independently by Edgar and Miller ("Borel subrings of the reals").[/quote]
I see. My professor intends to give us Edgar and Miller's proof on Tuesday, and he challenged us to try to prove the (slightly) easier version where you assume the subring is an $F_\sigma$ set (instead of just Borel) before he does it for us. I don't think he actually thinks we'll do it, but I decided I would try anyway.


vector
function
real analysis
real analysis unsolved

How is vector-valued continuity defined? (I know that isn't very specific, but it should be clearer with my example). Today in class we were told that (something essentially the same as this) $\vec{r}(t)=t\vec{i}$ for $t\in[-1,1]$ is not continuous at $t=0$. This doesn't seem right because the limit from the right and left both exist and go to zero. I was told as an explaination that since for any $\varepsilon > 0$ the function on $[0-\varepsilon, 0+\varepsilon]$ you have vectors pointing to both the left and the right and thus it is not continuous.

That's a continuous function by the usual definition. Your class is obviously talking about something else.

OK, so I simplified it since the only reason I was given for it not being continuous was that it changed signs at zero. The real problem was:

$\vec{x}(t)=t\vec{e}_{1}+\frac{1}{3}t^{3}\vec{e}_{2}$, so the point was to show why the usual definition of curvature doesn't work. We have $\vec{k}=\frac{-2t}{(1+t^{4})^{2}}(t^{2}\vec{e}_{2}-\vec{e}_{1})$. So there is a sign change at the point of inflection which is at $0$. But $|k|$ goes to $0$ from both the left and right, and not only did my prof say that it was discontinuous for the given reason, this is an example from the book illustrating the different definition of curvature because we want it to be continuous at inflection points.

It looks continuous to me. I have no idea what the problem is.
It's true that there's no well-defined normal to the curve at the inflection point, but zero curvature shouldn't cause trouble.


[hide="Hint"]Write $ \tan(x/2)$ as $ \frac{\sin(x/2)}{\cos(x/2)}$ and multiply the numerator and denominator by $ \cos^2(x/2)$.[/hide]

i'm stuck.

[hide]$ z\equal{}tan(\frac{x}{2})$
$ z\equal{}\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})}$
$ z\equal{}\frac{sin(\frac{x}{2})cos^2(\frac{x}{2})}{cos(\frac{x}{2})cos^2(\frac{x}{2})}$
$ z\equal{}\frac{sin(\frac{x}{2})cos^2(\frac{x}{2})}{cos(\frac{x}{2})(\frac{1\plus{}cos(x)}{2})}$
$ z\equal{}\frac{\frac{sin(x)}{2}}{\frac{1\plus{}cos(x)}{2}}$
$ z\equal{}\frac{sin(x)}{1\plus{}cos(x)}$[/hide]

So much for $ z$.  What about $ \frac {1 \minus{} z^2}{1 \plus{} z^2}$?

[hide="Complex-number interpretation"] $ (1 \plus{} zi)^2 \equal{} (1 \minus{} z^2) \plus{} 2zi$.

Now, $ 1 \plus{} zi$ is a vector with argument $ \arctan z \equal{} \frac {x}{2}$, so $ (1 \plus{} zi)^2$ has argument $ x$ and we conclude that

$ \cos x \equal{} \frac {1 \minus{} z^2}{1 \plus{} z^2}$
$ \sin x \equal{} \frac {2z}{1 \plus{} z^2}$.

In fact, when $ z$ is rational this gives all the possible pairs of rational values of $ \sin x, \cos x$, which leads to an immediate proof of the Pythagorean triple generating formulae.  :)[/hide]


Prove:
If $ a+b+c = \frac{\pi}{2}$ then
1. $ sin^2a+sin^2b+sin^2c=1-2\sinh\sinh\sinh$
2 $ \tanh \tanh+\tanh\tanh+\tanh\tanh=1$

[quote="serialk11r"]Prove:
If $ a + b + c = \frac {\pi}{2}$ then
1. $ sin^2a + sin^2b + sin^2c = 1 - 2\sinh\sinh\sinh$
2 $ \tanh \tanh + \tanh\tanh + \tanh\tanh = 1$[/quote]
Fixing your code...
Prove:
If $ a + b + c = \frac {\pi}{2}$ then
1. $ \sin^2a + \sin^2b + \sin^2c = 1 - 2\sin a\sin b\sin c$
2 $ \tan a \tan b + \tan b\tan c + \tan a\tan c = 1$


geometry
geometric transformation
rotation
trigonometry

9. A uniform ladder in static equilibrium leans against a frictionless wall.  The ladder has length L and mass M.  Given
that the angle between the ladder and floor is $ 30^\circ$ degrees, what would be the value of the coefficient of static friction ($ \mu_s$) between the ladder and the ground?

(Image attached)

A) $ \mu_s\ge\frac{1}{\sqrt3}$
B) $ \mu_s\ge\frac{2}{\sqrt3}$
C) $ \mu_s\ge\frac{\sqrt3}{2}$
D) $ \mu_s\ge\frac{1}{2}$
E) $ \mu_s\ge\sqrt3$

It seems that this should be easy... but for some reason I'm not getting it!

Here're the forces:

1) $ mg$ acting down on the center

2) $ mg$ acting upwards on the botton edge

3) $ F_{f}$ acting to the right on the botton edge

4) $ F_{N}$ acting to the left on the top edge (exerted by the wall)

You can arbitrarily pick any axis of rotation; I picked the bottom edge because $ mg$ and $ F_{f}$ do not go into the torque equation if the axis of rotation is located there.

[hide="solution"]So, with forces, $ F_g \equal{} N_{bottom}$ and $ f_s\ge N$ (I'm just gonna call the normal force of the wall $ N$).
With torque, placing the axis at the bottom edge, $ N\sin\theta L \equal{} F_g \cos\theta \frac {L}{2}$.  Simplifying, one gets that $ N \equal{} \frac {Mg\sqrt3}{2}$.

Then, substituting back into forces, $ \mu_s Mg \equal{} \frac {Mg\sqrt3}{2}$ so then $ \mu_s \ge \frac {\sqrt3}{2}$. [/hide]

Edit: Found my mistake!


1. a plane is flying at a speed and direction which in still air would be 500 mph due north. there is a wind of 100 mph blowing toward 30 degrees east of north. give the resultant ground speed of the plane in mph.

[hide="answer?"]588.7[/hide]

2. the ratio of the areas of two similar hexagons is 4:3. the sides of the larger hexagon are in the ratio 6:5:5:4:4:3. what is the length of the longest side of the smaller hexagon if the perimeter of the larger hexagon is 54?

[hide="answer?"]$\frac{\sqrt{3}}{18}$[/hide]


what are Fermat points  :o

I know just one Fermat point.

There are two Fermat points: You get one by constructing the equilateral triangles on the outside, and the other by constructing the equilateral triangles on the inside:
[url=http://mathworld.wolfram.com/FermatPoints.html]mathworld.wolfram.com/FermatPoints.html[/url]
[url=http://mathworld.wolfram.com/FirstFermatPoint.html]mathworld.wolfram.com/FirstFermatPoint.html[/url]
[url=http://mathworld.wolfram.com/SecondFermatPoint.html]mathworld.wolfram.com/SecondFermatPoint.html[/url]

[quote="epitomy01"]Prove the midpoints of the two fermat points F_1 and F_2 lies on the nine-point circle, and the line F_1 F_2 also passes through the midpoint of GH (G= centroid, H = orthocentre) and the symmedian point. Prove also that the isogonal conjugates of F_1 and F_2 lie on all three Apollonius circles (i.e. the isodynamic points and the fermat points are isogonal conjugates), and that the two isogonal conjugates of F_1 and F_2 lie on a line through the circumcentre O. Finally prove F_1 F_2 N O is concyclic, where N is the nine-point center and further that this circle is orthogonal to the circle having HG as diameter.[/quote]

The last proposition is the Lester Theorem I have never seen any proof for this theorem  :?

Dear all,
an answer to this situation, is given on my website in an article intitled "La fascinante figure de Cundy" (volume 2 (2008))
http://perso.orange.fr/jl.ayme
Sincerely
Jean-Louis.


geometry
rectangle
calculus
integration
function
combinatorics proposed
combinatorics

[quote="Valentin Vornicu"]How many ways are there to divide a $2\times n$ rectangle into rectangles having integral sides, where $n$ is a positive integer?[/quote]

This seems interesting, however, are two divisions considered different if they differ only in the placement of the different subrectangles?

Daniel

to dhernandez: I think it should be considered different.
and based on this ,I got the number is $\frac{4+\sqrt{2}}{14}*(3+\sqrt{2})^n+\frac{4-\sqrt{2}}{14}*(3-\sqrt{2})^n$

Hawk Tiger could you please post your solution

You can use the fact that every rectangle placed across(parallel to the side of length $2$) will create two $2Xk$ rectangles, and then you could find a recursion relation. :)

Well i solved this problem, but i took a lot of time calculating the final formula. I found the recurrence $r_{n}= 6r_{n-1}-7 r_{n-2}$ this was easy with $r_{1}= 2$ and $r_{2}=8$. I know how to solve this recurrence relations in two ways, gerenating functions or solving the caracteristic equation. Is there a faster way of finding the general term? I don't think there exists a way to count this without using recurrences, is there? I hope someone can help.

[quote="jasamper88"]Well i solved this problem, but i took a lot of time calculating the final formula. I found the recurrence $ r_{n} \equal{} 6r_{n \minus{} 1} \minus{} 7 r_{n \minus{} 2}$ this was easy with $ r_{1} \equal{} 2$ and $ r_{2} \equal{} 8$. I know how to solve this recurrence relations in two ways, gerenating functions or solving the caracteristic equation. Is there a faster way of finding the general term? I don't think there exists a way to count this without using recurrences, is there? I hope someone can help.[/quote]
Please explain how did you get that recursion.


[i]One class which include $ n$ members participates in the Mathematical Contest . The contest has four  rounds .

   In each round , the head teacher will choose some students to establish a team .

   Let $ E_1 \ ; \ E_2 ; \ \ E_3 ; E_4$ be the teams that takes the exam in the frist , the second , the third and the fourth round respectively .

   How many ways the teacher can establish the four teams $ E_1 \ ; \ E_2 ; \ \ E_3 ; E_4$ such that :

    $ E_1 \ \bigcap \ E_2 \ \ne \ \emptyset \ ; \ E_2 \ \bigcap \ E_3 \ \ne \ \emptyset ; E_3 \ \bigcap \ E_4 \ \ne \ \emptyset$[/i]

Please give your posts more informative titles: whether this problem is "nice" or "hard" is debatable, and in any event there are thousands of nice, hard problems on the forum; however, there is no doubt that this is a problem about "subsets with non-empty intersections," and giving such a title would make it much easier to locate the problem.

This is a natural-looking situation for inclusion-exclusion:
The number of ways to choose four subsets $ (E_1, E_2, E_3, E_4)$ of an $ n$-set is $ (2^n)^4 = 16^n$.
The number of ways to do this so that $ E_1 \cap E_2 = \varnothing$ (likewise $ E_2, E_3$ or $ E_3, E_4$) is $ 3^n \cdot (2^n)^2 = 12^n$: every element can go in $ E_1$ or in $ E_2$ or in neither, and then independently either in or not in $ E_3$ and $ E_4$.
Similarly, the number of ways to do it so that $ E_1 \cap E_2 = E_3 \cap E_4 = \varnothing$ is $ (3^n)^2 = 9^n$.
The number of ways to do it so that $ E_1 \cap E_2 = E_2 \cap E_3 = \varnothing$ (likewise $ E_2, E_3, E_4$) is $ 5^n \cdot 2^n = 10^n$: for each element, we may choose to put it into $ E_2$, into $ E_1$ and $ E_3$, into just $ E_1$, into just $ E_3$, or into none of these, and then independently we may choose whether or not to put it into $ E_4$.
Finally, the number of ways to do it so that $ E_1 \cap E_2 = E_2 \cap E_3 = E_3 \cap E_4 = \varnothing$ is $ 8^n$: every element can go into $ E_1$ and $ E_3$, $ E_1$ and $ E_4$, $ E_2$ and $ E_4$, into a single one of the $ E_i$, or into none of them.  (Or, we could just remember that the number of subsets of a 4-set having no two consecutive elements is a certain Fibonacci number.)

Now, for the inclusion-exclusion part is straightforward and gives us the final answer of $ 16^n - 3 \cdot 12^n + 2\cdot 10^n + 9^n - 8^n$.


is it possible to trace someone's IP address from their AIM screenname?

I have a suspicion that this kid has 2 screenamees, but he won't admit it. is there any way I could blow his cover?

Yes, but you need to intiate a file transfer or a direct connection.  There are websites that show you how.

Can't you just use nbtstat? I'm not an advanced user of it, but I think it can be done that way.

I have another question

how does your sn just appear on someone else's buddylist? i get a lot of random IMs asking me who I am, the people say my sn just appeared on their buddylist

[quote="iamagenius"]how does your sn just appear on someone else's buddylist? i get a lot of random IMs asking me who I am, the people say my sn just appeared on their buddylist[/quote]

lmao he's probably an internet pred. some person did that to me and i talked to him for a minute before he started asking personal questions (what grade you in, where do you live, age, etc.) and then i blocked him.

i use AIM express which allows you to label your buddy contacts. sometimes when im at my friend's house using AIM 6.0 i forget who some people are sometimes.

so there's no way my sn could have randomly appeared on someone else's buddy list? i guess it could be a prank.........


If $ \sum_{r\equal{}0}^{21}f(\frac{r}{11}\plus{}2x)\equal{}constant$ for all real x and f(x) is periodic ,then what could be period of f(x)

[quote="pankajsinha"]If $ \sum_{r \equal{} 0}^{21}f(\frac {r}{11} \plus{} 2x) \equal{} constant$ for all real x and f(x) is periodic ,then what could be period of f(x)[/quote]

I think the question is not very clear.
Are you speaking about "what could be least period ?" (when it exists) ?
Or are you speaking about "what is the set $ P$ of reals "$ a$" such that it exists a solution to the equation such that $ f(x \plus{} a) \equal{} f(x)$ $ \forall x$ ?

Replacing $ x$ by $ x \plus{} \frac {1}{22}$ immediatly gives $ f(x \plus{} 1) \equal{} f(x)$
So, $ N\subseteq P$
But maybe some functions with period $ \frac 1n$ may exist.
But, if the question is "what could be least period ?", then, for example, $ \frac 1{11}$ cant be accepted.

So, could you be more precise in your question, please ?

[quote="pankajsinha"]This is how the question was given to me.If possible find the least period.

On replacing $ x$ by $ x \plus{} \frac {1}{22}$ we immediately get $ f(2x) \equal{} f(2 \plus{} 2x)$ and hence $ f(x) \equal{} f(2 \plus{} x)$.

So one possible solution can be that period could be multiple of 2.[/quote]

Oops  :oops: you're right, $ f(2x\plus{}2)\equal{}f(2x)$, and not $ f(x\plus{}1)\equal{}f(x)$  :oops:

But you're wrong, the least period cant be a multiple of $ 2$. Since $ 2$ is "a" period, the least one can only be $ \frac 2n$ for some $ n$. So the question is "what are the "n" for which it exists a solution.

It's easy to construct a solution for which the least period is $ 2$
It's sure we cant construct a solution for which the least period is $ \frac 2{22}$
To be continued ...


Check this my fellow mathematicians and babylon tower mates:

You are driving a car,the car gets out of your control and you will hit 2 upcoming cars.The only thing you can do is choose which car you will hit,you cant do anything to avoid the car.In one car,Mother Teresa is sitting and in the other 5 teenagers.Which car do you choose and why?

Edit:we assume that Mother Teresa still lives

[quote="zscool"]considering that mother teresa has been dead since 1997, i think its an easy choice[/quote] lol. I still would have chosen 5 teenagers anyway, cuz there were 5 young lives to save vs 1 life (cruel I know  :().

anyway try not to think that you are driving that bad :D

when I first got my car, right in the first week I managed to trash it completely hiting a policeman :( ... fortunately for me that I had insurance covering the repairs ... :)

I know what you mean,after you get your drivers license,you are like a kamikaze on the road (I got my license 6 months ago hehe).
[quote]considering that mother teresa has been dead since 1997, i think its an easy choice[/quote]
We assume that you are driving the car before she died or we assume that she still lives. :D

I believe each life is special in its own way...so since it's 5 to 1, I'd try to avoid the car with more people.

It has been decided,mommie will die again

Considering how great of a person Mother Teresa was, she would probably prefer that you would end her life than to end the life of 5 teenagers who are still pretty young.

zscool,i disagree with you.Take a look at :

[url]http://slate.msn.com/id/2090083[/url]
[url]http://www.secularhumanism.org/library/fi/hitchens_16_4.html[/url]

This is the oldest open thread in Fun Factory.

Lemme add to it.

[quote=gauss202]This is a variation of a classic problem from Moral Philosophy called the Trolly problem. It has many variations, but in it's standard form it goes as so:

Suppose that you are the controller of a train switch. Coming towards your switch is a fast moving train. Down the path that the train is currently set to travel, there are four man working on the tracks. They will not have time to get out of the way. Along the other path, there is a single man working on the tracks. He also will not have time to get out of the way if the train comes down his path.

The question is in two parts. Are you morally permitted to switch the direction of the tracks so that the train will avoid the four men in it's current path, and kill the man working on the other track? If the answer is yes to the first question, are you morally obligated to switch?

But once we think we understand those questions, a paradox arrises when we consider a seemingly similar situation. Suppose now that you are a doctor with four sick patients - each in critical need of different organs, and you have one healthy patient who is an organ match for each of them. Are you morally permitted to kill this one patient to save the other four? Are you morally obligated to do it?

What makes these two situations different?

Then you can add all kinds of variations to each situation - like what if the one person was someone you knew or a loved one? What if the one person was a woman or a child? What if you knew that the one person was a bad person, and the four others were good, or vice versa? What if it was a billion people instead of four?

This leads ultimately to the question of, is morality really rational? And lots of questions of how we rationalize killing in war.[/quote]

Ethics and morality, are things that we humans made up. Thus, there is no "right" answer. HA end of discussion.

[quote=DuoDuoling0]Do you know g1zq on AoPS? he/she joined in 2nd grade lol I joined in 3rd but im in 4th now[/quote]

no...i don't


integration
function
algebra
domain
topology
advanced fields
advanced fields solved

I am stuck on this, can someone help?

Suppose f, g are analytic on the open unit disc. And suppose that
|f|^2 + |g|^2 =1 for all z in the unit disc.

Prove that both f,g are constants.

We can use the following: if $f$ is analytic in a disc of radius $>r$, and
$\displaystyle f(z) = \sum_{n=0}^\infty a_nz^n$, then
$\displaystyle \frac1{2\pi} \int_0^{2\pi} |f(re^{i\theta})|^2 d\theta = \sum_{n=0}^\infty |a_n|^2r^{2n}$.
Do this for both $f$ and for $\displaystyle g(z) = \sum_{n=0}^\infty b_nz^n$ and we get that
$\displaystyle \forall r\ge 0, \sum_{n=0}^\infty (|a_n|^2+|b_n|^2)r^{2n} = 1.$
If any $a_n$ or $b_n$ is nonzero for any $n>0$, then this sum is a strictly increasing function of $r$. The only way it can be constant in $r$ is for $a_n=b_n=0  \forall n>0$. Thus, $f$ and $g$ are constant.

Ah, that was unnecessarily complicated. Let's try it again, with "subharmonic" as the key notion.

Use $\nabla^2$ to stand for the Laplace operator, $\displaystyle \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$. One can show that if $f$ is analytic, then $\nabla^2(|f|^2) = 4|f^\prime|^2$.

Now suppose that $f$ and $g$ are each analytic on connected domains and that on some open set common to both domains, $|f|^2+|g|^2 = $ a constant. Take the Laplacian of this expression (and divide by 4) to get $|f^\prime|^2+|g^\prime|^2 = 0$ on that open set. But then, each of $f^\prime$ and $g^\prime$ must be zero, which means that each of $f$ and $g$ are constant on that open set, hence by analytic continuation constant on their respective domains.

[quote="Kent Merryfield"]Use $\nabla^2$ to stand for the Laplace operator[/quote]
the standard notation is $\Delta$?  :maybe:

Wow, both solutions are excellent. Thanks a lot.


For all x,y,z>0 prove that :
$\sqrt[3]{xyz}+\frac{|x-y|+|y-z|+|z-x|}{3}\geq$$\frac{x+y+z}{3}$

I think you need only a hint : $assume that a \ge b \ge c$ .

[quote="pavel kozlov"]We consider, that $x\leq y\leq z$.
Then $\frac{x+y+z-((y-x)+(z-x)+(z-y))}{3}=\frac{x+y+z-(2z-2x)}{3}=\frac{3x+y-z}{3}\leq x\leq \sqrt[3]{xyz}$
 Q-E-D[/quote]
thank you


Two users have pretty big post counts (see Memberlist). I recall this happening before, and they had similar post counts. If I recall correctly, last time, during the period of big post counts, everyone else's didn't change, but I haven't checked if that's the case now as well.

This is like the (6th?) thread.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193396
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191973


If the Euler line of\[ \Delta ABC\] be parallel to BC,
show that \[ tanBtanC \equal{} 3\]

[hide="hint"]note that $ tanBtanBCH \equal{} 1$ H is the orthocenter [/hide]


search
MATHCOUNTS
geometry
area of a triangle
Heron\u0027s formula

Does any1 have some helpful formulas that could really help in Mathcoutnts? I don't want ones like pythagorean and stuff like that

Don't memorize formulas, see how they work.....  :wink:

this is repetitive

there was a formula topic started by goldendomer that was like 12 pages long

search it up

You shouldn't memorize formulae... LEARN HOW TO DERIVE THEM! (Or be stuck on proof problems :D )

[quote="Stargirl240"]MathCounts problems normally don't require you to know many formulas, so just practice problems instead of memorizing formulas! Well, the Pythagorean theorm is one that you should know already, but otherwise, nothing else is required.[/quote]

Actually, I think remembering certain formulas really helps.  Especially for Nationals tests, it can be a real time saver.  Here's the ones i recommend:

Heron's Formula (area of a triangle) : $\sqrt{s(s-a)(s-b)(s-c)}$
Pythagorean Theroem: You should know this
Choose Numbers: $nCk=\dbinom{n}{k}=\frac{n!}{k!(n-k)!}$

You just have to know how to use fundemental math :)

Can you use search for the formula thread?  I'll say its OK to revive that dinosaur so we can avoid making future threads like this.  Thanks.


geometry
probability
Putnam
college contests

A dart, thrown at random, hits a square target. Assuming that any two parts of the target of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge.


Description:  A parent of a [url=http://www.mathpath.org]Mathpath[/url] alumnus teams up with Mathpath Executive Director Dr. George Thomas to answer general questions from prospective applicants.

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here for details and the complete Math Jam schedule.[/url]

Most of us know that AoPS members are well represented among National MATHCOUNTS competitors.  But did you know that Mathpath alumni also have done very well at Nationals?

[color=red]Two girls[/color] made it to the 2005 National Countdown Round.  Both of them were Mathpath alumni!  Both of them were semifinalists (top 4) in the competition.

Overall, [color=red]six of the twelve[/color] 2005 National Countdown Round competitors were Mathpath alumni!

If you want to learn more about Mathpath, attend the Math Jam on March 3 at 7:30 PM ET (4:30 PM PT).


What is the average of the first 1000 positive odd numbers?

[hide]1+2001=2002/2=1001 multiply that by 1000 to get 1001000 :idea: our answer[/hide]

[hide]
using a formula i get that the sum is:
1001000
and divide that by 100 and the answer is:
1001[/hide]

[quote="236factorial"]you found the sum  :D[/quote]

You said: find the average of the first 1000 [b]odd[/b] numbers, not [b]even[/b]. What you should do is find the sum using $n^2=1000^2=1000000$, not $n(n+1)=1000(1001)=1001000$.  Then what you do is $\frac{1000000}{1000}=1000$.  The answer is not $1001$, but $1000$. :D

Masoud Zargar

Thank you.  Since many people didn't answer it correctly, I answered it; otehr than that, I'm not suppsoed to answer this.

Masoud Zargar


A sequence of numbers is called of Fibonacci-type if each term, after the first two, is the sum of the previous two. Prove that the set of positive integers can be partitioned into the disjoint union of infinite Fibonacci-type sequences.

[i]Kmal, problem A. 244, September, 2000[/i]

this one seems not hard for me.
we know that an integer N can be written as the sum of non-consecutive Fibonacci numbers (this form is unique)

N = f_{n1} + f_{n2} + ... + f_{nk}

n1+1<n2, n2+1<n3, ... n(k-1)+1<nk

Once we have such a decomposition, we make the next number
N' = f_{n1+1} + f_{n2+1} + ... + f_{nk+1}

in this way, we get many Fibonacci-type sequences..
the left thing is to prove that they are disjoint. i believe this is easy.

This problem appeared on an other forum some times ago. Here is how I solved it :

The idea is to be able to build Fib sequences by giving the first element and an injective (and simple !) function which gives the second element.

i.e. u(1)  IN, u(n+1) = f(u(n)). We must have f^2(n) = f(n) + n which was a good start to think of the following :

f(n) = floor (n * (1 + sqrt(5))/2 + 1/2).   ((1 + sqrt(5))/2 being the golden number)

I claim we have  f^2(n) = f(n) + n.
f is also strictly increasing so injective.

Let a  IN, let's define fib(a,*) this way : fib(a,1) = a,  fib(a,n+1) = f(fib(a,n)).

For ex. si a = 1 :
1 2 3 5 8 13 21 34 55 89 144 ...

If a = 4 (the first one not appearing in fib(1,*)) :
4 6 10 16 26 42 68 110 178 288 ...

Si a = 7 (the first one not appearing in fib([12],*) :
7 11 18 29 47 76 123 199 322 521 ...

That way we can define a sequence a(n) as the smallest integer not appearing in the previous fib(a(i),*), i<n.

It is clear that, by construction, all integers are reached.

Let m and m the smallest such as If f^n(a) = f^m(b) then by injectivity f^n-1(a) = f^m-1(b) which contradict "smallestness" (cas m=1 or n=1 easy).

So the fib(a(n),*) sequences works and gives and explicit way to build the sequences :D  :D

With a small computer program, I computed some special values :

100 = fib(62,2)
1000 = fib(90,6)
2003 = fib(473,4)  (it was in 2003  :)
11111= fib(619,7)


solve this system of e's in R

$x+y=1$
$x^4+y^4=1$

If you multiply the first equation with $x$ you get: $x^4+xy^3=x$ and $x^4=x(1-y^3)$. From this and the second equation it follows: $x(1-y^3)=1-y^4$ Because of the symmetry: $y(1-x^3)=1-x^4$. If $x\neq1$ and $y\neq1$ we get:

$x=y+\frac{1}{y^2+y+1}$ and $y=x+\frac{1}{x^2+x+1}$.

It is a contradiction because we have $x>y$ and $y>x$. So we must have $x=1$ or $y=1$. Now we get the both solutions: $(1,0)$ and $(0,1)$.

Misha


graph theory
combinatorics unsolved
combinatorics

There are six points in the plane. Any three of them are vertices of a triangle whose sides are of different length. Prove that there exists a triangle whose smallest side is the largest side of another triangle.

[hide]
It is well known that any 2 colouring of the complete graph on 6 vertices contains a monchromatic triangle.

Initially colour all edges Blue. Now consider each triangle and colour its largest side Red.

Since there must be a monochromatic triangle, and since each triangle has at least one Red edge, there must be a monchromatic Red triangle. Thus there is some triangle, whose smallest side is the largest side of some (other) triangle.
[/hide]


algebra
polynomial
number theory
prime numbers
number theory unsolved

Let $ m$ be an arbitrary natural number.Prove that there exist infinitely many prime numbers $ p$,such that $ m\plus{}p^3$ are composite.

[quote="TTsphn"]A good problem . Here is my solution .
Lemma :Consider $ q$ is a prime number .Then exist infinite prime in the form
$ kp \plus{} 1$
To prove this lemma only need consider the number $ \frac {a^p \minus{} 1}{a \minus{} 1}$
Now apply this to your problem .
Call $ q$ is a prime divisor of $ m \plus{} 1$ then : exist infinite $ p\in P$ such that
$ q|p \minus{} 1$ so $ q|p^3 \minus{} 1$
It mean that $ m \plus{} p^3\equiv m \plus{} 1 \equiv 0 (\mod q )$
Because $ m \plus{} p^3 > q$ so $ m^3 \plus{} p$ is a composite .[/quote]
Actually this is the official solution, so i'm looking for other proof.And you made some typos,for example you need to prove that there exist infinitely many prime numbers of the form $ kq\plus{}1$, where $ q$ is the prime divisor of $ m\plus{}1$.

The lemma was posted often enough before, he touched the idea (consider primes dividing  $ \frac {a^p \minus{} 1}{a \minus{} 1}$).

This problem generalizes directly to: The only polynomials $ f(x)$ such that $ f(p)$ is prime for all primes $ p$ are the trivial ones (constant or $ f(x)\equal{}x$).
The only ones that map primes to prime-powers are that of type $ f(x)\equal{}x^n$ (and constant ones).

Both have beenposted and proven before (also using Dirichlet's theorem).

[quote="ZetaX"]The lemma was posted often enough before, he touched the idea (consider primes dividing  $ \frac {a^p \minus{} 1}{a \minus{} 1}$).

This problem generalizes directly to: The only polynomials $ f(x)$ such that $ f(p)$ is prime for all primes $ p$ are the trivial ones (constant or $ f(x) \equal{} x$).
The only ones that map primes to prime-powers are that of type $ f(x) \equal{} x^n$ (and constant ones).

Both have beenposted and proven before (also using Dirichlet's theorem).[/quote]
Yes and the first one was posted by me :)
But i believe that the original problem has another solution,without using Dirichlet's theorem and its special cases,even if they can be proved easily.


algebra
function
domain
analytic geometry
calculus
integration
trigonometry

Calculate the volume of this domain:
$ A \equal{} \{ (x,y,z) \in R^3 , x^2 \plus{} y^2 \minus{} ax \leq 0 , x^2\plus{}y^2\plus{}z^2 \leq a^2 \}, a > 0$

My approach:

I've tried to use cylindrical coordinates and spherical coordinates but they didn't work.
So, I tried to "brute force" it. We have $ x^2 \minus{} ax \leq \minus{}y^2 \leq 0$ then $ 0 \leq x \leq a$. (1)
Also, $ y^2 \leq ax \minus{} x^2$ leads to $ \minus{}\sqrt{ax\minus{}x^2} \leq y \leq \sqrt{ax\minus{}x^2}$ (2)
And, $ \minus{}\sqrt{a^2\minus{}x^2\minus{}y^2} \leq z \leq \sqrt{a^2\minus{}x^2\minus{}y^2}$ (3).

From (1),(2),(3), we have the integral : Volume = $ \int^2_0 \int^{\sqrt{ax\minus{}x^2}}_{\minus{}\sqrt{ax\minus{}x^2}} \int^{\sqrt{a^2\minus{}x^2\minus{}y^2}}_{\minus{}\sqrt{a^2\minus{}x^2\minus{}y^2}}dzdydx$ which is very difficult to calculate.

I have no idea what to do, could you please give me a hint ?

I think you wrote off cylindrical coordinates too quickly.

$ x^2\plus{}y^2\minus{}ax\le0$ is a cylinder, just not centered on an axis. In polar (that is, cylindrical) coordinates, it is

$ r^2\minus{}ar\cos\theta\le0,$ which rearranges to

$ r\le a\cos\theta.$

Draw what this looks like in the $ xy$ plane. It's a circle. The inequality only makes sense for $ \cos\theta>0,$ so we'll restrict $ \theta$ to $ \minus{}\frac{\pi}2\le\theta\le\frac{\pi}2.$

Then the sphere can be written as $ r^2\plus{}z^2\le a^2.$ Try setting up the integral this way:

$ \int_{\minus{}\frac{\pi}2}^{\frac{\pi}2}\int_0^{a\cos\theta} \int_{\minus{}\sqrt{a^2\minus{}r^2}}^{\sqrt{a^2\minus{}r^2}}r\,dz\,dr\,d\theta$

$ \equal{}\int_{\minus{}\frac{\pi}2}^{\frac{\pi}2}\int_0^{a\cos\theta} 2r\sqrt{a^2\minus{}r^2}\,dr\,d\theta$

You can finish that.

Hi,

Yes thank you. I've thought of that but I didn't have the courage to integrate $ 2r\sqrt{a^2\minus{}r^2}$ and put $ a\cos{\theta}$ to it.

Also, I think the restriction for $ \theta$ is $ 0 \leq \theta \leq \frac{\pi}{2}$, because we only want $ x$ to be in [0,a], so we can choose this condition. (If we choose $ \minus{}\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ , there will be |sin(x)| in the outer integral, which is difficult to integrate)

With $ 0\le \theta\le \pi/2$ you get half of the solid. Double the result to get the full volume.

[quote]... there will be |sin(x)| in the outer integral, which is difficult to integrate)
[/quote]
That is [i]not[/i] difficult to integrate. Absolute values can be easy if you only remember to break the definition up into pieces. For instance:

$ \int_{\minus{}\frac{\pi}2}^{\frac{\pi}2}|\sin x|\,dx\equal{} \int_{\minus{}\frac{\pi}2}^0\minus{}\sin x\,dx\plus{} \int_0^{\frac{\pi}2}\sin x\,dx\equal{}1\plus{}1\equal{}2.$

(Or, in this case, use symmetry, with $ |\sin x|$ an even function. Which is approximately what mlok said.)

Yes , you guys are right. We can't use $ 0 \leq x \leq \frac {\pi}{2}$.
Thank you so much, this will help me much for the next exams.
Salut!


LaTeX
algebra
system of equations
algebra unsolved

hi
Find all non-zero real numbers x, y, z which satisfy the system of equations:
(x^2 + xy + y^2)(y^2 + yz + z^2)(z^2 + zx + x^2) = xyz
(x^4 + x^2y^2 + y^4)(y^4 + y^2z^2 + z^4)(z^4 + z^2x^2 + x^4) = x^3y^3z^3
at first please wirte it with Latex and then please solve it.

It has already posted   :wink:  see here : http://www.mathlinks.ro/viewtopic.php?p=1746106#1746106


The edges of an octahedron are weighted with the integers [tex]1,2,...,11, n[/tex] (with [tex]n>11[/tex]). Each vertex is weighted with the sum of the weights of the four edges passing through this vertex.
For what integers [tex]n[/tex] can all vertices have equal weight?

[hide="A partial solution"]There are six vertices.  Each of them must have equal weight, so the total weight is of the form $6m = 2(1 + 2 + ... + 11 + n)$ and $n = 3m - 66$ for some positive integer m.  In addition, the vertex weights must be at least n + 11 because the two endpoints of the edge of value n must each have weight at least this great.  (This is beacause the smallest sum of three members of {1, 2, ..., 11} you can make in two distinct ways is 11 = 1 + 3 + 7 = 2 + 4 + 5.)  Thus $m \geq n + 11$ and so $n \geq 3n + 33 - 66$.  Thus $n \leq 16$.  We also know that 3 divides n, so the only available options are n = 12, n = 15.  I have constructed a solution for n = 12: if the octahedron has edges AB, AC, AD, AE, BC, BE, BF, CD, CF, DE, DF, EF, then they have values
AB = 12
AC = 3
AD = 6
AE = 5
BC = 11
BE = 2
BF = 1
CD = 4
CF = 8
DE = 9
DF = 7
EF = 10.  I imagine n = 15 is solvable as well, but I haven't done any more work to determine whether that is in fact true.[/hide]


trigonometry
geometry
analytic geometry
graphing lines
slope
ratio
algebra

simplify: $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

ABCD is a square and M and N are the midpoints of BC and CD respectfully. Then what is the sine of angle MAN?

I remember seeing that radical problem somewhere, and generally speaking,

$\frac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+\sqrt{x+y}} = \sqrt{x} + \sqrt{y} - \sqrt{x+y}$  It ends up being just a difference of squares.

[hide="2"]Dunno if it's a good idea to resort to analytical geometry but that's what I did. Let B be the origin and A, D, and C be (0,2), (2,2), and (2,0), respectively. (It doesn't matter what the side lengths are) Then M is (1,0) and N is (2,1). Finding the equation of the line through (0,2) and (2,1), we get $y=-\frac{1}{2}x+2$. A line perpendicular to this will have slope 2, and such a line that passes through M would be y=2x-2. Solving the system of equations, we get that the line perpendicular to AN passing through M intersects AN at $\left(\frac 85,\frac 65\right)$. The distance between this intersection point and M is then $\frac{3\sqrt{5}}{5}$. We can also find the length of AM as $\sqrt{5}$, and since the sine can be expressed as the ratio of the two, our answer is $\frac 35$.[/hide]
That was probably not the smartest solution, if it's even right, but oh well.

You are right, LynnelleYe. :)

Here is my solution.

Let $A(0,0)$, we can set $M(1,-2),N(2,-1)$, we have $\triangle AMN=\frac{1}{2}|2\cdot (-2)-(-1)\cdot 1|=\frac{3}{2}$.

On the other hand, $\triangle AMN=\frac{1}{2}AM\cdot AN\sin \angle MAN=\frac{1}{2}\sqrt{5}\cdot \sqrt{5}\sin\angle MAN$.

Equating these two equations we  get $\sin \angle MAN=\frac{3}{5}$.

kunny

[quote="shobber"][hide="2"]
$\sin{x}=\cos{2\angle{BAM}}=1-2\sin^2{\angle{BAM}}=1-\frac{2}{5}=\frac{3}{5}$[/hide][/quote]

Very simple! ;)  :)