AMC
AMC 12
AIME
AMC 12 A
USA(J)MO
USAMO
AMC 12 B

Please vote (best AMC12 score)+10*(AIME score). (Note that you should wait till you take the AIME II to vote, if you're taking the AIME II :P)

224 here. (114 AMC12A, 11 AIME I)
I had a 172 last year, haha.

My index with AMC 12 is 205.5, with AMC 10 is 240, do I have a chance of making USAMO?

150+14*10=290
10 points away from the goal :wink:

Haha none of the above.
139.5+6(10)=199.5 :P

Yes, we are indeed Lemonadeless  :(

Did you show this to Lemonade?

:rotfl:

I wonder if there is an [i]unofficial[/i] Lemonade?  :maybe:

138+60=198.  Same AIME score as last year.  I am really mad.

[quote="solafidefarms"]Aww. Did anyone get a 300 this year? If I remember right, last year David Benjamin and Sam Elder and someone else perfect-300ed.[/quote]

That was Haitao Mao.

epic epic epically failed....204 :( so much for making usamo...oh well there's next year

at least I beat my last year's score .. 133.5/10 = 233.5

Eh...with amc12b it would be 132+8=212, with amc10a 144+8=224. I'm pretty unhappy about my AIME score, but at least I probably made usamo and I beat my last year's score (which was also horrible).

126+80=206

50 points better than last year but still

I hope I make USAMO... :(

150+(9~11)*10=240~260. (Our school doesn't give us back the AIME I until the AIME II... don't know why, so I don't know whether #8 or #14 I got correct or not).

AIME goal: 10. Possible...

A note to the pollsters, the index having a lot of choices between 190-220, but one choice for above 220 causes a lot of people to concentrate on that choice (after all this is AOPS).

Wow I'm pissed. I left the test fairly happy, thinking I got about 9 right. I guess I didn't get any of the details right, since I was fairly confident I knew how to do 1-8, 11, and 14, and even had plenty of time to check my answer. Turns out I only got 4 right. WTF? How did I miss 1, 2, 4, 6, and 7??? Whatever, guess that blows my hopes for USAMO, even though it would have been close anyways. I doubt I did anything wrong except computation, too. Damn.

Guess I should start working on those Putnams for next year. ;)

Index: 154?

129 + 120 = 249.

I'm almost happy.  :huh:

Scores over time: 8th/9th/10th

AMC 12: 120/141/130.5 (bubbling error gragh)
AIME: 7/13/10
USAMO: 6/13/???

So by my current pattern of 10th = (8th+9th)/2, I will get 9.5 on the USAMO...that would be quite disappointing.

@james: I suppose my poll, unlike the other, is primarily concerned with borderline-USAMO indexes.

Wow, you guys have definately humbled my 192.5 (112.5/8), which is pretty bad for a 12th grader, although I'd be an 11th grader if I hadn't skipped a year.

I really need to study up on geometry. My school only offers a basic geometry course, so any question that starts with "Consider triangle ABC" is bound to stump me. Maybe that's why no one else in my school qualified for the AIME :(

Right now I'm studying mass point geometry, which I've heard will be very useful. Any other suggestions on what to know would be greatly appreciated.

[quote="Dudemcbob"]...
Right now I'm studying mass point geometry, which I've heard will be very useful. Any other suggestions on what to know would be greatly appreciated.[/quote]

Number 5 can be solved with mass points...

I got $ 144\plus{}10*10\equal{}244$, with the 144 from the AMC 12B. I was really disappointed in my AIME score, however (not [i]really[/i], but disappointed enough).

Somewhere between 250-260 bleh, whatever, should have been at least 270 (2 arithmetic errors ftw)

230.

150 (on AMC 10) + 8*10.

I missed #1 and #9. And got lazy with 14 and was off. Doh.

192.5. 112.5 AMC 12B and 8 AIME 1.  (#2-#8, #14).  Got 807 for #1, and realized that it was 840 like an hour after taking it...DOH! Overcounted on #9, and somehow thought that (x1-x2) had to be 29 on #11.

How do you convert an AMC10 score for the equation?

Hm I thought you had an 8 on the AIME. Ah whatever.

Argh you all should stop saying "fail." It makes my self-esteem even lower.  :(

I failed, like plenty of others.
A chronicle of my performance on the AMCs & AIME over the past few years:
[hide]2007: Took the AMC 12 as a freshman (because Karthik persuaded me to). 99(fail)+60 = 159.
2008: Took the AMC 10. 138+80 = 218. Made it to the USAMO. Got a single-digit score.
2009: Took the AMC 12. 136.5+70(fail) = 206.5. [/hide]
All I can do now is hope (or take the AIME II, whichever comes to mind on that day).

mine's 189 :(
i ABSOLUTELY phailed the AIME I today with a 6.... :(

[quote="georgiamathdude"]Would a score between 205 ~210 offer at least a somewhat good chance of making usamo? (like 50-50  :blush: ?)

If not... I will infiltrate amc headquarters and change my score.  :ninja:[/quote]

Depends. I thought this year's test was easier than normal (only slightly) and the AMCs, well I can't accurately judge those.

So the index should be around 205~210, giving you 50-50 chances, at least.

WHy so many people think the cutoff is over 220? Does it ever happen before?
Also, when is the cut off announced? I guess after AIME 2?

AMC10+10*AIME
139.5+8*10=219.5
I guess i'll probably make it


function
number theory
prime factorization
number theory unsolved

Define $ \mu: N$->$ Z$ by $ \mu(n) \equal{} ( \minus{} 1)^k$ if $ n$ is a product of $ k$ distinct primes (including the case $ k \equal{} 0$) and $ \mu(n) \equal{} 0$ otherwise. Prove that $ \sum_{d|n}\;\mu(d) \equal{} 0$ unless $ n \equal{} 1$.

Mobius already defined function $ \mu$ (in fact I think it has been named like this based on the Greek initial letter of his name)!
[hide]The relation is the simplest application of Mobius' inversion formula: take $ f(1) \equal{} 1$ and $ f(n) \equal{} 0$ otherwise, and $ F(n) \equal{} \sum_{d \mid n} f(d) \equal{} 1$. Then $ f(n) \equal{} \sum_{d \mid n} \mu(d)F\Big ( \frac {n} {d} \Big ) \equal{} \sum_{d \mid n} \mu(d)$.[/hide]

Suppose that $ n \equal{} p_{1}^{a_{1}}\cdots p_{k}^{a_{k}}.$ Then, it is not difficult to see that $ n$ has exactly $ \binom{k}{r}$ divisors whose prime factorization contains exactly $ r$ primes, all of them with exponent $ 1$. Since those are the only factors of $ n$ for which $ \mu(d)$ doesn't vanish, we conclude that

$ \sum_{d|n} \mu(d) \equal{} 1 \plus{} \binom{k}{1}( \minus{} 1)^{1} \plus{} \binom{k}{2}( \minus{} 1)^{2} \plus{} \ldots \plus{} \binom{k}{k} ( \minus{} 1)^{k} \equal{} (1 \minus{} 1)^{k} \equal{} 0.$

Let $ n \equal{} p_{1}^{a_{1}}\cdots p_{k}^{a_{k}}.$ The only terms that do not vanish in the sum are the mobius' of the product of primes and 1

Let $ d$ be such a number

Case 1: $ k$ is odd.  In this case consider the pairing $ (d,n/d)$
Case 2: $ k$ is even.  If $ p_{1}$ divides $ d$, take $ (d,d/p_{1})$. If $ p_{1}$ doesnt divide d, take $ (d,dp_{1})$

In the sum, consider the above pairings.


floor function
ceiling function
function
algebra
polynomial
quadratics
combinatorics unsolved

To the problem :

How many different pairs of natural numbers add to make one-thousand?

The solution is:

By writing the sums as: 1 + 999, 2 + 998, 3 + 997, ... , 499 + 501, 500 + 500.
It is clear that the number of different pairs is five hundred.

Now, the problem is :

To find the number of different triplets of natural numbers when added make one-thousand.
That is, find all (a,b,c) such that a+b+c = 1000

Let $m$ and $n$ be positive inntegers. There are $\binom{n-1}{m-1}$ ordered $m-$ tuples $(x_{1},x_{2},...,x_{m})\in\mathbb{N}^{m}$ such that $x_{1}+x_{2}+...+x_{m}=n$.

The question appears to be asking for the number of [i]unordered[/i] solutions.  The sequence (shifted slightly) is [url=http://www.research.att.com/~njas/sequences/A001399]A001399[/url].

Let  $f_{2}(n)$ be the number of different ways to write $n$ as a sum of two natural numbers.
(E.g. $f_{2}(1000) = 500$.) Similarly let  $f_{3}(n)$ be the number of different ways to write $n$ as a sum of three natural numbers.

I suppose it is not so difficult to give explicit formulae for $f_{2}$ and $f_{3}$.

For the first it is easy to see that $f_{2}(n) = \lfloor n/2 \rfloor$.

For the second we do cases for the largest summand $l$ and the smallest summand
$s$. Every admissible choice of $l$ and $s$ corresponds to exactly one partition
of $n$ into a sum. First observe that
\[\lceil n/3 \rceil \le l \le n-2. \]
If in fact $\lfloor n/2 \rfloor \le l \le n-2$ then the possible values for $s$ are exactly
\[s \in \{1 ,\ldots, \lfloor (n-l)/2 \rfloor \}. \]
(Then indeed, for the medium summand $m : = n-s-l$ we have s $s \le m \le l$.)

If however $\lceil n/3 \rceil \le l \le \lfloor n/2 \rfloor-1$ then the possible values
for $s$ are exactly
\[s \in \{n-2l, \ldots, \lfloor (n-l)/2 \rfloor \}. \]
(Then indeed, for the medium summand $m : = n-s-l$ we have s $s \le m \le l$.)

So we have
\[f_{3}(n) = \sum_{l= \lceil n/3 \rceil }^{\lfloor n/2 \rfloor-1}(\lfloor (n-l)/2 \rfloor-(n-2l)+1)+\sum_{\lfloor n/2 \rfloor}^{n-2}\lfloor (n-l)/2 \rfloor \]
i.e.
\[f_{3}(n) = \sum_{l= \lceil n/3 \rceil }^{n-2}\lfloor (n-l)/2 \rfloor-\sum_{l= \lceil n/3 \rceil }^{\lfloor n/2 \rfloor-1}(n-2l-1). \]
Of course, this can be simplified, but I am too lazy to do that. Distinguishing the cases
n mod 6 one can get rid of the floor-functions, plug in the formulas for the sums and
thus gets a polynomial expression in $n$. :)

I hope this is correct. Doing the calculation by hand I thus got
$f_{3}(1000) = 333^{2}-166^{2}= 83333$ but
it is very likely that I made a mistake somewhere, so that is not reliable.

[quote="Jean-Louis"]JBL,

I'm having trouble with the link you posted.[/quote]

I don't know why -- it works fine for me.  It's the Online Encyclopaedia of Integer Sequences, sequence number A001399.  solyaris' answer is correct; the cases mod 6 can be summarized as "the nearest integer to $\frac{n^{2}}{12}$."  (6 is not a quadratic residue mod 12, so there are never any ties.)

An equivalent question (the number of solutions to $x+2y+3z = n$) was posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134892

hi.... i think we can use burnside's lemma, which is the following:

no. of way= (999 x 998 / 2 + 3 x 499 + 2 x 0) / 6 = 83333, which is the same as solyaris. For those who dont know abstract algebra, my solution is just a more sophisticated way to undo the "double counting". I really wonder why such a question will occur in any olympiad since abstract algebra should be uni maths (right? :wink: )

what is wrong with this argument:[hide]to find the number of solutions to $a+b+c=n$, we let $c=1,2,\dots, n-2$, and count the number of solutions to the equations $a+b=n-1,a+b=n-2,\dots, a+b=2$ and find the desired number $\lfloor\frac{n-1}{2}\rfloor+\lfloor\frac{n-2}{2}\rfloor+\dots+\lfloor\frac{2}{2}\rfloor$[/hide]

Ok, Let me clarify myself. For the UNORDERED solutions of a + b + c = 1000, they can only be either of these two:

(a, b, c) where a, b, c are distinct and (a, b, b)

So, the first term (999*998)/2 actually count the first triplet 6 times and the second triplet 3 times. To smooth things out, I add 3 times of the no. of second triplet, which is 499 since they range from ( 998, 1, 1)..... all the way until (2, 499,499). then i divide by six to get the number of distinct unordered solutions.

Let's try the following generalisation:

Find the number of UNORDERED solutions for this:

a+b+c+d=10000

Good Luck!

Now I see! This solution obviously is a lot more elegant than mine,
and also the generalization to 4 or more summands is much easier
than in my approach. Very nice!

Thank you, solyaris! Your solution is very nice also! But for the conditions of 5 or more summands we will have a hard time when we use burnside's lemma since we have to consider permutational group of 5 or more elements, which will involve a lot of cases (unlike in the case of three summands, in which the elements of the permutational group of three elements are only e, (12), (23), (31), (123) and (132).)

But I think the case of four summands is still manageable... so please give a try...

[quote="JBL"]The first thing wrong with it is that to use it requires you to add 997 terms.  The second thing that's wrong with it is that you don't have unordered solutions: you've counted $(1, 1, 998)$ and $(1,998,1)$ separately.[/quote]

well i'm not so good at counting. :blush: but can my argument be made correct somehow? :maybe:


1.Can you arrange the numbers 1, 2,..., 9 along a circle such that the sum of two neighbors is never divisible by 3, 5, or 7?

2.Each Karabas is familiar with 4 Karabases and with 7 Barabases, while
each Barabas is familiar with 7 Barabases and with 9 Karabases. Which
group is larger, Karabases or Barabases?

3. There are 20 children in a kindergarden. Any two children have the
same grandfather. Prove that one of the grandfathers has at least 14
grandchildren attending the kindergarden.

4.At four corners of a 3X3 chess board there are 2 white Knights (in the
bottom row) and 2 black Knights ( in the top row). Is it possible to
place them at the corners in alternating colours by moving the Knights
according to chess rules?

5. Three chips are placed at three vertices of a regular pentagon. It is
allowed to move any chip to any free vertex along a diagonal of the
pentagon. Is it possible to reach the situation when one chip returns to
its original position while two others trade their positions by applying
this operation?

6.Metro Map is a connected graph. Prove that one of the stations can
be closed (a vertex and all exiting edges are removed), so that one can
always reach any station from any other station.

[quote="game.guy"]3. PIE

4. No. The kings can't move far enough without being in checkmate.[/quote]3. PIE is not the solution. Try again.

4. The problem has nothing to do with being on checkmate and the pieces are [u]knights[/u], [b]not[/b] [u]kings[/u]. Try again. Hint: find an invariant.

Are you sure about number 3? That problem is in my Intermediate Counting and Probability book, and PIE is used to solve it.


Hello,

I need to put some table into my latex text. I know to do this, but some my tables shouldn't put nothing at the bottom of table. The problem is when first I put a table that it need to have the description "[i]Table 1.1: my_description[/i]" , the I put some table that it doesn't need to have the description and finally I put a table that it need to have description. When I do it, the tabla's number is increasing to 1 for all my table that I've put. For this reason, my document has for example [b]Table 1.1[/b] , [b]Table 1.6[/b] and [b]Table 1.8[/b] on the bottom of tables. The table 1.2,1.3,1.4,1.5 and 1.7 shouldn't have description and for it, I don't put anything on the \caption. My question is what should I do to have the sequence [b]Table 1.1[/b] , [b]Table 1.2[/b] and [b]Table 1.3[/b], i.e, I want have consecutive order and for this reason I should skip the tables [b]Table 1.2[/b] , [b]Table 1.3/b] and [b]Table 1.4[/b], [b]Table 1.5[/b] , [b]Table 1.7[/b].?

Thanks in advance.

[quote="stevem"]For those tables you don't want captioned don't use floats. Thus for those tables with captions use
[code]\begin{table}
\begin{tabular}{ccc}
...
\end{tabular}
\caption{this is a table}
\end{table}[/code]but for those without captions use
[code]\begin{tabular}{ccc}
...
\end{tabular}[/code]and they won't affect the numbering of the other tables.

If, however, for some reason you have to have floats for all tables, then the tables use the table counter. So after using table without a caption, use \addtocounter{table}{-1} to reduce the counter so that the next table uses the value that you want.[/quote]

Thank you very much.


where can in find the information on business oppertunities in pakistan?

The solution is http://www.iin.com.pk. Industrial Information Network http://www.iin.com.pk is the first Pakistani B2B and information portal designed specifically to cater the trade information needs of businesses for various industrial sectors. The project is initially focused on facilitation of Textile and Leather sectors with more sectors to be added later.
The portal provides services like b2b, yellow pages and event calendar. In addition it also contains considerable information on government rules and regulations, financing in Pakistan, business startup guides, sectoral analysis reports about Pakistan, pre-feasibility reports, cost of doing business in Pakistan, latest business news about Pakistan and many more.
IIN is designed to become a one-stop-shop for businesses for all their Information and eCommerce needs.

Visit http://www.iin.com.pk for all of your information needs about Pakistan


Altitudes PP',QQ',RR' and medians PX,QY,RZ are drawn in triangle PQR. Prove that the length of the broken line P'YR'XQ'ZP' is equal to the perimeter of triangle PQR.

Circle (X) with diameter QR cuts RP, PQ at Q', R' => XQ' + R'X = QR.
Circle (Y) with diameter RP cuts PQ, QR, at R', P' => YR' + P'Y = RP.
Circle (Z) with diameter PQ cuts QR, RP at P', Q' => ZP' + Q'Z = PQ.


If a and b are integers such that a+ b< 1000 and a/b=0.625 , what is the greatest possible value of b?

well $\frac a b=0.625$ so  $a=0.625b$ and because $a+b<1000=>1.625b<1000$ so $b<\frac {8000} {13}$and b is integer so $b = [\frac{8000}{13}]=615$
You might try posting more difficult problems from now on :P  :lol:

ok ..I didn't post this on because I didn't how to do ... but I posted it on because I got the same answer like yours but they marked me wrong and gave out the answer was 608 .. so I am kinda lost.. this was a math problem in the Official SAT guide... thankx anyways

You are wrong.

$0.625=\frac {5}{8}\implies 8a=5b$.  So $b$ is a multiple of $8$.

We see $\frac {13}{8}b<1000$, which implies $b\leq 615$.  The smallest multiple of $8$ which is $\leq 615$ is $608$.

yes sorry i should have found $ a$ too and see if it's integer :oops:  :blush:


There is no nontrivial integer solution for the following equation
x^2+y^2=u^2+v^2
xy=3uv

Let $s=x^2+y^2=u^2+v^2$ and $p=xy=3uv$. Then both $s^2 - 4p^2 = (x^2-y^2)^2$ and $9s^2 - 4p^2=9(u^2-v^2)^2$ are squares. The problem then becomes similar to that of [url=http://maths.paisley.ac.uk/allanm/ECRNT/dickdood/conc.htm]concordant numbers[/url] and is easily treated with elliptic curves.

This problem is solved by 4 number theorem.You can find this kind of problems and their solutions in  Mathlinks Contest.


superior algebra
superior algebra unsolved

Let G=Sym(n) .Let C be a cyclic group. Suppose that $ \theta \in Hom(C, G).$ then how to show that the generators for $ \theta (C)$ in G all have the same cycle type and so are conjugate in G


limit
trigonometry
logarithms
calculus
calculus computations

Find the limit: $ \lim_{n \to \infty} \cos^{n} \left(\frac {1}{\sqrt {n}}\right)$

Let $ a_n \equal{} \cos^n(\frac{1}{\sqrt n})$.

We use the identity

$ 1 \minus{}\frac{x^2}{2} \le \cos x \le 1 \minus{} \frac{x^2}{2} \plus{} \frac{x^4}{24}$

From this we get that

$ (1\minus{}\frac{1}{2n})^n \le a_n \le (1\minus{}\frac{1}{2n} \plus{} \frac{1}{24n^2})^n$

LHS converges to $ e^{\minus{}1/2}$. We show that RHS converges to the same value.

For any $ M > 0$, for sufficiently large $ n$ we have that

$ (1\minus{}\frac{1}{2n})^n < (1\minus{}\frac{1}{2n} \plus{} \frac{1}{24n^2})^n < (1\minus{}\frac{1}{2n} \plus{} \frac{1}{Mn})^n \equal{} (1 \plus{} \frac{2\minus{}M}{2Mn})^n$

RHS converges to $ e^{\frac{2\minus{}M}{2M}} \equal{} e^{\frac{1}{M}\minus{}\frac{1}{2}}$. Since this is true for every large M, the result follows.

Alternatively:
We have,
$ \lim_{x \to \infty} x \log \cos \left(\frac {1}{\sqrt {x}} \right) = \lim_{t \to 0} \left(\frac {\log \cos t}{t^2}\right) = - \frac {1}{2}\lim_{t \to 0}\left(\frac {1}{\cos t} \cdot \frac {\sin t}{t} \right) = - \frac {1}{2}$  (I have used L'Hopital)

Taking the exponential and noting that $ \cos^n\left(\frac{1}{\sqrt{n}} \right) = \exp\left({n \log \cos \left(\frac{1}{\sqrt{n}} \right) \right)}$ we get the answer.

I thought you said beginners  :roll:  :D

Or to avoid mentioning L'Hopital:
Let's note that $ \cos t \approx 1 \minus{} \frac {t^2}{2}$ and $ \log \left(1 \plus{} t\right) \approx t$ for small t.

We get: $ \lim_{x\to\infty}x\log\cos\left(\frac {1}{\sqrt {x}}\right) \equal{} \lim_{t\to 0}\left(\frac {\log\cos t}{t^{2}}\right) \equal{} \minus{} \frac {1}{2}$.

Now taking the exponential gives the result.


$ x,y,z\geq 0$,prove that
\[ \frac{x^3\plus{}y^3\plus{}z^3}{3}\geq xyz\plus{}\sqrt{9\plus{}6\sqrt{3}}\displaystyle|(y\minus{}z)(z\minus{}x)(x\minus{}y)\displaystyle|\]

[quote="Tourish"]$ x,y,z\geq 0$,prove that
\[ \frac {x^3 \plus{} y^3 \plus{} z^3}{3}\geq xyz \plus{} \sqrt {9 \plus{} 6\sqrt {3}}\displaystyle|(y \minus{} z)(z \minus{} x)(x \minus{} y)\displaystyle|\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101916


$ ABCD$ is a quadrilateral, points $ M, P$ are chosen on $ AB$ and $ AC$ such that $ \frac{AM}{AB} \equal{} \frac{CP}{CD}$. Find all locus of midpoints $ I$ of $ MP$ as $ M, P$ vary on $ AB, AC$.

[quote="outback"]$ ABCD$ is a quadrilateral, points $ M, P$ are chosen on $ AB$ and $ CD$ such that $ \frac {AM}{AB} \equal{} \frac {CP}{CD}$. Find all locus of midpoints $ I$ of $ MP$ as $ M, P$ vary on $ AB, AC$.[/quote]
The locus of the midpoind $ I$ of the segment $ MP,$ is the line connecting the midpoints of the side-segments $ BC,\ AD,$ of the given quadrilateral $ ABCD,$ as a direct application of the [b][size=100]ERIQ theorem[/size][/b] (= [b][size=100]E[/size][/b]qual [b][size=100]R[/size][/b]atios [b][size=100]I[/size][/b]n [b][size=100]Q[/size][/b]uadrilateral ).

See for two elementary proofs of this powerful theorem, in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Collinearity of orthocenters[/url], of shobber.

Kostas Vittas.

I think if the problem is $ \frac{AM}{AB}\equal{}\frac{DP}{DC}$ then yes, but I don't think it is so when $ \frac{AM}{AB}\equal{}\frac{CP}{CD}$...

[quote="outback"]I think if the problem is $ \frac {AM}{AB} \equal{} \frac {DP}{DC}$ then yes, but I don't think it is so when $ \frac {AM}{AB} \equal{} \frac {CP}{CD}$...[/quote]
You are right dear [b][size=100]outback[/size][/b] and sorry for such that my inattentiveness in the formulation of the proposed problem.

But, because of the [b][size=100]ERIQ theorem[/size][/b] is also true, for non-convex quadrilateral ( and for 3D space quadrilateral too ), we can consider here, the non-convex quadrilateral $ ABDC$ and then, the locus of the midpoint $ I$ of the segment $ MP,$ is the line connecting the midpoints $ X,\ Y$ of the segments $ AC,\ BD$ respectively, such that $ \frac {XI}{XY} \equal{} \frac {AM}{AB} \equal{} \frac {CP}{CD}.$

Sorry again, Kostas Vittas.

[quote="mr.danh"]Dear Kostas,
I think you had better read this problem in details. outback wrote P are chosen on AC, not CD.  :)
On the diagonal AC, let K such that CK=CD. And point I will move on the line connecting the midpoints O,H of AC,BK. But the locus of I is the segment OH iff CA\ge CD. When CA<CD, this locus is a subset of OH.[/quote]
Thank you dear [b][size=100]mr.danh[/size][/b] for your remark and pointing out my blindness.

I saw that [b][size=100]outback[/size][/b] wrote that the point $ P$ is on $ AC,$ but because of he gave the relation of $ \frac {AM}{AB} \equal{} \frac {CP}{CD},$ I thought that there was a typo in the wording of the problem.

Any way, I hope that this time I am not mistaken, but I think that the author had in mind the [b][size=100]ERIQ theorem[/size][/b], when he inspired this problem ( probably no ).

$ \bullet$ So, let $ D'$ be the point on $ AC,$ such that $ CD' \equal{} CD$ and then, we have to find the locus of $ I,$ when $ P$ is on $ AC$ such that $ \frac {AM}{AB} \equal{} \frac {CP}{CD'}.$

We consider the degenerated non-convex quadrilateral $ ABD'C$ and based on the [b][size=100]ERIQ theorem[/size][/b], we conclude that the locus of the midpoint $ I$ of the segment $ MP,$ is the line connecting the midpoints $ X,\ Y$ of the segments $ AC,\ BD'$respectively, such that $ \frac {XI}{XY} \equal{} \frac {AM}{AB} \equal{} \frac {CP}{CD}.$

Best regards, Kostas Vittas.


modular arithmetic
Diophantine equation
number theory unsolved
number theory

Prove that for every natural number $ n$,there exist natural numbers $ a,b$,such that $ n \equal{} \frac {a^2 \plus{} b}{a \plus{} b^2}$

At first glance, I thought of proving that the set $ \mathcal{C} \equal{} \left\{(x,y): \frac {x^2 \plus{} y}{x \plus{} y^2} < 2n\right\}$ contains at least a point $ (a,b)\in\mathbb{N}^2$ such that $ \frac {a^2 \plus{} b}{a \plus{} b^2}\equiv 0\pmod n$.  Minkowski's Theorem may be of some help here, but I have not tried it yet.  You may try to find $ a,b$ in terms of $ n$ to obtain $ n\equal{}\frac{a^2\plus{}b}{a\plus{}b^2}$.  I'll work on this after I write my exams.

If $ n \equal{} 1$, take $ a \equal{} b \equal{} 1$.
If $ n \equal{} (2s)^2$, take $ a \equal{} 2s^2 \plus{} 8s^5, b \equal{} 4s^4$.
If $ n \equal{} m^2 > 1, m \equal{} 2s \plus{} 1, s > 0$, take $ b \equal{} \frac {m^3 \minus{} m^2 \plus{} m \minus{} 1}{4} \equal{} 2s^3 \plus{} 2s^2 \plus{} s$ and $ a \equal{} \frac {m^2 \plus{} 2mb \plus{} 1}{2} \equal{} 4s^4 \plus{} 6s^3 \plus{} 6s^2 \plus{} 3s \plus{} 1$.
Direct checking gives that these are solutions. So we may assume that $ n > 1$ isn't a square.

Consider $ c^2 \minus{} nd^2 \equal{} 1 \minus{} n^3$. One solution is $ c \equal{} \minus{} 1, d \equal{} n$.
Let $ x,y > 0$ be a solution of the Pell-equation $ x^2 \minus{} ny^2 \equal{} 1$. By the reasoning of http://www.mathlinks.ro/Forum/viewtopic.php?t=77121 with $ p \equal{} 2n$, we may assume that $ x \equiv 1 \mod 2n$, $ y \equiv 0 \mod 2n$.

Now look at $ e \equal{} cx \plus{} ndy, f \equal{} dx \plus{} cy$:
- It is $ e^2 \minus{} nf^2 \equal{} 1 \minus{} n^3$ by straightforward calculation.
- We have $ e \equiv 1 \cdot ( \minus{} 1) \plus{} n \cdot 0 \cdot d \equiv \minus{} 1 \mod 2n$.
- Additionally:
\[ e \equal{} \minus{} x \plus{} n^2y > 0 \\
\iff n^2y > x \\
\iff n^4y^2 > x^2 \equal{} 1 \plus{} ny^2 \\
\iff (n^4 \minus{} n)y^2 > 1
\]
As $ n > 1,y > 0$, this is surely true, so $ e > 0$.

Now we can take $ b \equal{} \frac {e \plus{} 1}{2n}$. By what we did, $ b$ is a positive integer. Now straightorward calculation shows that $ f^2 \equal{} n^2 \plus{} 4nb^2 \minus{} 4b$, so $ a \equal{} \frac {n \plus{} \sqrt {n^2 \plus{} 4nb^2 \minus{} 4b}}{2} \equal{} \frac {n \plus{}f}{2}$ is a positive integer and that $ \frac {a^2 \plus{} b}{a \plus{} b^2} \equal{} n$.

[quote="Lei_Lei"]Erken, May I know the source of the problem?
[/quote]
I don't know,my teacher gave it to me. :)
Thanks for help.

[quote="ZetaX"]Please ask him if there is a shorter and nicer solution  :wink:[/quote]
OK.Thanks for help.


1)If we dont have a PDF uploader, can we print and fax solutions in?

2)The rules say you are allowed calculators. Does that mean when writing solutions we can say that we are using a calculator to compute something. For example: if a and b are relatively hard to compute(assume we know a and b), and we want to show $ a>b$, can we just say a calculator says that a=2.39323124 and b=2.14047592, so $ a>b$?(hypothetical case. :))

1) No. there is a pdf uploader on usamts.org.

[quote="abacadaea"]1)If we dont have a PDF uploader, can we print and fax solutions in?[/quote]No -- we are not accepting any solutions by fax.  I don't know what you mean by "If we don't have a PDF uploader"....the upload page in on the USAMTS website.  Please read the submission instructions carefully:
[url]http://www.usamts.org/Rules/U_RulesSubmitting.php[/url]

[quote]2)The rules say you are allowed calculators. Does that mean when writing solutions we can say that we are using a calculator to compute something. For example: if a and b are relatively hard to compute(assume we know a and b), and we want to show $ a > b$, can we just say a calculator says that a=2.39323124 and b=2.14047592, so $ a > b$?(hypothetical case. :))[/quote]Hypothetical answer: for something like the situation you describe, we would [i]probably[/i] allow it.  It's unlikely that would be the point of the problem, though.  To clarify: you are always allowed to [i]use[/i] a calculator, but if you use a calculator to [i]prove[/i] something, that's where there's a higher level of scrutiny.


Find all functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[f\left(xf(x)+f(y)\right)=\left(f(f(x))\right)^{2}+y \quad \forall x,y \in \mathbb R.\]

[quote="Kondr"]Find all functions $f: \mathbb{R}\rightarrow\mathbb{R}$:

$f\left(xf(x)+f(y)\right)=\left(f(f(x))\right)^{2}+y$[/quote]

$E1(x,y)$ : $f\left(xf(x)+f(y)\right)=\left(f(f(x))\right)^{2}+y$
Let $f(0)=a$ and $f(a)=b$

$E1(0,x)$ : $f(f(x))=x+b^{2}$ $\implies$ $E2(x)$ : $f(x+b^{2})=f(x)+b^{2}$

$E1(0,0)$ : $b=b^{2}$ $\implies$ $b=0$ or $b=1$

1) If $b=1$, we have $f(f(x))=x+1$ and $f(x+1)=f(x)+1$ and so $E1(x,y)$ becomes $f(xf(x)+f(y))=(x+1)^{2}+y$.
Then $f(0)=a$ implies $f(-1)=a-1$
Then $E1(-1,0)$ gives $f(a-f(-1))=0$ so $f(1)=0$ and $f(f(1))=a$ but $f(f(1))=2$ and so $a=2$ and, since $f(0)=a=2$, we have $f(n)=n+2$ for any integer n.
But then $E1(1,0)$ is wrong

2) So $b=0$, and then  $E1(0,x)$ gives $f(f(x))=x$ and then $E1(x,0)$ gives $f(xf(x))=x^{2}$ and $E1(f(x),0)$ gives $f(f(x)x)=f(x)^{2}$. So $f(x)^{2}=x^{2}$

So either $f(x)=x$, either $f(x)=-x$. The question then is to check if it is possible to have $f(x)=x$ for some $x$ and $f(x)=-x$ for some other $x$ :
Let $x$ such that $f(x)=x$ and $y$ such that $f(y)=-y$
Then $E1(x,y)$ gives $f(x^{2}-y)=x^{2}+y$ and so either $x^{2}-y=x^{2}+y$, either $-x^{2}+y=x^{2}+y$, and so either $x=0$, either $y=0$

So either $f(x)=x$ for any $x$, either $f(x)=-x$ for any $x$

It is easy to verify that these two solutions are available.

So the two solutions are $f(x)=x$ and $f(x)=-x$


The sum of the reciprocals of the roots of the equation $ x^2 \plus{} px \plus{} q \equal{} 0$ is:

$ \textbf{(A)}\ \minus{}\frac{p}{q} \qquad
\textbf{(B)}\ \frac{q}{p}\qquad
\textbf{(C)}\ \frac{p}{q}\qquad
\textbf{(D)}\ \minus{}\frac{q}{p}\qquad
\textbf{(E)}\ pq$

[hide="Idea"]
Let [i]r[/i] and [i]s[/i] be the roots of the equation.

$ \frac 1r \plus{} \frac 1s \equal{} \frac {s \plus{} r}{rs}$

Somewhat common knowledge: $ s \plus{} r \equal{} \minus{} p$ and $ rs \equal{} q$
So: $ \frac {s \plus{} r}{rs} \equal{} \frac { \minus{} p}q$

Which is ([b]A[/b]).
[/hide]
Geesh I'm bad at LaTex. . . Took 5+ min.


4x-5y-z-7=0 is plane equation. M(-1,2,0) is point of space. P is point of the plane and MN is perpendicular at this plane and MP=NP. Then find the coordinate of point N

Hi anuonhu. I'm not sure that I understand the question. I think it means that $ MN$ is perpendicular to the plane and cuts it at $ P$, in which case, I get the following solution.

[hide="Click to reveal hidden solution"]

The vector equation of the line $ MPN$ is $ r\equal{} \minus{}i \plus{} 2j \plus{} \lambda(4i \minus{}5j\minus{}k)$ so $ P$ is where $ 4(\minus{}1 \plus{}4\lambda) \minus{}5(2\minus{}5\lambda) \minus{}(\minus{}\lambda) \equal{} 7$, that is where $ \lambda \equal{} \frac{1}{2}$ and so $ P$ is ($ 1, \frac{\minus{}1}{2}, \frac{\minus{}1}{2})$ and so $ N$ is $ (3, \minus{}3, \minus{}1)$. We can check this by finding the lengths of the line-segments $ MP$ and $ PN$ which are both $ \frac{\sqrt{42}}{2}$.

[/hide]

Dear AndrewTom thanks for your solution! and please try this problem! http://www.mathlinks.ro/viewtopic.php?t=303974

Sorry... this is a dumb question but what are vectors?


find all functions $ f : \mathbb R \rightarrow \mathbb R$  such that for all $ x\notin \{\minus{}1,1\}$ :
$ f\left( {\frac{{x \minus{} 3}}{{x \plus{} 1}}} \right) \plus{} f\left( {\frac{{3 \plus{} x}}{{1 \minus{} x}}} \right) \equal{} x$

take $ g(x)\equal{}\frac{x\minus{}3}{x\plus{}1}$, we have $ gog(x)\equal{}\frac{3\plus{}x}{1\minus{}x}$ and $ gogog(x)\equal{}x$ so:
$ (i)$ $ fog(x)\plus{}fogog(x)\equal{}x$
$ (ii)$ $ fogog(x)\plus{}f(x)\equal{}g(x)$
$ (iii)$ $ f(x)\plus{}fog(x)\equal{}gog(x)$

$ (iii)\plus{}(ii)\minus{}(i)$ gives $ f(x)\equal{}\frac{g(x)\plus{}gog(x)\minus{}x}{2}$


calculus
integration
probability
real analysis
real analysis unsolved

I don't know how to prove this proposition.

$ \mathcal{M}$=$ \{$all the probability measure on $ ([0,1], \mathcal{B}[0,1])\}$

given $ \mu\in \mathcal{M}$, we say $ \nu\in\mathcal{M}$ is a best reponse to $ \mu$ iff $ \nu$  maximize the following integral:

$ \int_{x > y} y d\mu d\nu + \int_{x < y} x d\mu d\nu + R(\mu \times \nu (x < y) - \mu \times \nu(x > y)) + \int_{x = y} x d\mu d\nu$

where $ R$ is a given constant $ R>0$

and given $ \nu\in \mathscr{M}$, we say $ \mu\in\mathscr{M}$ is a best reponse to $ \nu$ iff $ \mu$  maximize the following integral:

$ \int_{x > y} y d\mu d\nu + \int_{x < y} x d\mu d\nu - R(\mu \times \nu (x < y) - \mu \times \nu(x > y)) + \int_{x = y} x d\mu d\nu$

My question is how to prove that if $ \mu^*$ and$ \nu^*$ are best reponse to each other, show that $ \mu\{0\} = 1$ and $ \nu\{0\} = 1$.

Thanks


hello

i am having a problem with this question please i need STEP BY STEP SOLUTION

A) How many solutions are there to the equation
x1 + x2 + x3 = 12,
where x1 , x2 , x3 and x4 are integers satisfying
x1 ≥ 0, x2 > 2 and x3 ≥ 1?

**** THE OTHER QUESTION SIMILAR TO IT
B) How many solutions are there to the equation
x1 + x2 + x3 = 12,
where x1 , x2 , x3 are NON NEGATIVE INTEGERS satisfying
X3 >=4 ?

for b)  think of this:
[hide]
How big could $x_{3}$ get before $x_{1}$ and $x_{2}$ are no longer positive integers?
[/hide]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133548

Please refer to this.


For a positive integer $ n$, let $ \theta(n)$ denote the number of integers $ 0 \leq x < 2010$ such that $ x^2 \minus{} n$ is divisible by $ 2010$. Determine the remainder when $ \displaystyle \sum_{n \equal{} 0}^{2009} n \cdot \theta(n)$ is divided by $ 2010$.

[hide="Solution"]
We re-phrase the sum as follows: for each integer $ 0\le x < 2010$, find the sum of the residues $ x^2 \pmod{2010}$. Indeed, if we look at this sum, every integer $ n$ is represented $ \theta(n)$ times amongst the squares.

Since the answer requests the remainder when the sum is divided by $ 2010$, we can simply consider the sum of all of the squares $ 0 \le x < 2010$ (using the sum of consecutive squares formula) and find the resulting remainder. Thus,
\begin{align*}\sum_{n = 0}^{2009}\, n \cdot\, \theta(n) & = \sum_{x = 0}^{2009} x^2 = \frac {2009(2009 + 1)(2\cdot 2009 + 1)}{6} \\
& \equiv \frac {( - 1)(2010)( - 1)}{6} \equiv \boxed{335} \pmod{2010}.\end{align*}
[/hide]


ratio
trigonometry
Pythagorean Theorem
geometry
AMC 10

A circle of radius $ 1$ is surrounded by $ 4$ circles of radius $ r$ as shown. What is $ r$?
[asy]defaultpen(linewidth(.9pt));

real r = 1 + sqrt(2);
pair A = dir(45)*(r + 1);
pair B = dir(135)*(r + 1);
pair C = dir(-135)*(r + 1);
pair D = dir(-45)*(r + 1);

draw(Circle(origin,1));
draw(Circle(A,r));draw(Circle(B,r));draw(Circle(C,r));draw(Circle(D,r));

draw(A--(dir(45)*r + A));
draw(B--(dir(45)*r + B));
draw(C--(dir(45)*r + C));
draw(D--(dir(45)*r + D));
draw(origin--(dir(25)));

label("$r$",midpoint(A--(dir(45)*r + A)), SE);
label("$r$",midpoint(B--(dir(45)*r + B)), SE);
label("$r$",midpoint(C--(dir(45)*r + C)), SE);
label("$r$",midpoint(D--(dir(45)*r + D)), SE);
label("$1$",origin,W);[/asy]$ \textbf{(A)}\ \sqrt {2}\qquad \textbf{(B)}\ 1 \plus{} \sqrt {2}\qquad \textbf{(C)}\ \sqrt {6}\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 2 \plus{} \sqrt {2}$

[hide="Click for solution"]
Draw the square that has vertices at the centers of each of the circles with radius $ r$.  Then the side length is $ 2r$ and the diagonal length is $ 2+2r$.  Thus, we have:

\begin{align*}
(2r)^2+(2r)^2&=(2+2r)^2
\\ 8r^2&=4r^2+8r+4
\\ 4r^2-8r-4&=0
\\ r^2-2r-1&=0
\\ r&=\frac{2\pm\sqrt{4+4}}{2}
\\ &=\frac{2\pm 2\sqrt{2}}{2}
\end{align*}

We want the positive value, so we have $ 1 +\sqrt{2}=r$, or $ \boxed{\textbf{(B)}}$.
[/hide]

[hide="Faster"]
Draw a 45-45-90 triangle with sides $ 2r$, $ 2r$, and $ 2r\plus{}2$.
By the 45-45-90 ratio (or by noting that $ \sin 45^\circ\equal{}\frac{\sqrt{2}}{2}$), $ 2r\plus{}2\equal{}2r\sqrt{2}\implies 2\equal{}2r\sqrt{2}\minus{}2r\equal{}2r(\sqrt{2}\minus{}1)\implies r\equal{}\frac{1}{\sqrt{2}\minus{}1}$.
$ \frac{1}{\sqrt{2}\minus{}1}\cdot\frac{\sqrt{2}\plus{}1}{\sqrt{2}\plus{}1}\equal{}\boxed{\textbf{(B)}\ 1\plus{}\sqrt{2}}$.

If you'd rather not use trigonometry, or you don't know triangle ratios, you can use the Pythagorean theorem to get $ (2r)^2\plus{}(2r)^2\equal{}2\cdot(2r)^2\equal{}(2r\plus{}2)^2$, and then take the square root of both sides to arrive at the same thing.
[/hide]

Solution to prob. 18 on AMC 10 Contest B 2007

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

I used this solution when I solved this problem, but it is a little bit more complicated:
Draw a square that circumscribed about 4 circles,
Connect the diagonals
Then 4sqrt2 r = 2sqrt2r + 2r + 2
solve the equation you get r= 1 + sqrt 2

[asy]
defaultpen(linewidth(.9pt));
real r = 1 + sqrt(2);
pair A = dir(45)*(r + 1);
pair B = dir(135)*(r + 1);
pair C = dir(-135)*(r + 1);
pair D = dir(-45)*(r + 1);
draw(Circle(origin,1));
draw(Circle(A,r));draw(Circle(B,r));draw(Circle(C,r));draw(Circle(D,r));
draw(A--(dir(45)*r + A));
draw(B--(dir(45)*r + B));
draw(C--(dir(45)*r + C));
draw(D--(dir(45)*r + D));
draw(origin--(dir(0)));
D(A--origin--IP(CR(A,r),CR(D,r))--cycle);
label("$r$",midpoint(A--(dir(45)*r + A)), SE);
label("$r$",midpoint(B--(dir(45)*r + B)), SE);
label("$r$",midpoint(C--(dir(45)*r + C)), SE);
label("$r$",midpoint(D--(dir(45)*r + D)), SE);
label("$1$",origin,W);[/asy]

We have a right triangle with sides $r$, $r$ and $r+1$. So $r\sqrt{2}=r+1\implies r\sqrt{2}-r=1\implies r=1+\sqrt{2}$.


geometry
circumcircle
geometric transformation
reflection
cyclic quadrilateral
geometry unsolved

Consider a generic convex cyclic quadrilateral q = ABCD inscribed in the circle k(K, r) and having finite intersection points F,G of opposite sides.Line e = FG is the polar of the intersection point E of the diagonals AC,BD. The circle c with diameter FG is orthogonal to k.Also, the midpoints X, Y of the diagonals and the center H of c are collinear.
I managed to prove that FG is the polar,but can't prove orthogonality,collinearity. May I get your nice help ? Please ~

The midpoints $ X, Y, H$ of the 3 diagonals $ AC, BD, FG$ of a complete quadrilateral formed by the lines $ (AB, CD, BC, DA)$ are always collinear, regardless of $ ABCD$ being cyclic, $ XYH$ is its Newton line. According to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=176253]complete quadrilateral[/url], at least 30 different proofs have been proposed. Here is one:

Assume $ F \equiv AB \cap CD,$ $ G \equiv BC \cap DA.$ $ (X), (Y), (H)$ are 3 circles with diameters $ AC, BD, FG.$ Pick 3 diameter edpoints not on one line (4 different possibilities) and consider the resulting triangle, say the $ \triangle ABG.$ $ X$ is on midline parallel to $ BG,$ $ Y$ is on midline parallel to $ GA,$ $ H$ is on midline parallel to $ AB.$ If a circle is centered on a triangle midline and passes through the corresponding vertex, it also passes through the corresponding altitude foot. So, $ (X), (Y), (H)$ pass through the feet $ H_a, H_b, H_g$ of the A-, B-, G-altitudes of the $ \triangle ABG.$ Let the altitudes cut the circumcircle $ (O_1)$ of this triangle at $ K_a, K_b, K_g$ and let $ H_1$ be its orthocenter. $ K_a, K_b, K_g$ are reflections of $ H_1$ in $ H_a, H_b, H_g.$ Power of the orthocenter $ H_1$ to the circumcircle $ (O_1)$ is

$ p(H_1, (O_1)) \equal{} \overline{H_1A} \cdot \overline{H_1K_a} \equal{} \overline{H_1B} \cdot \overline{H_1K_b} \equal{} \overline{H_1G} \cdot \overline{H_1K_g}.$

Power of $ H_1$ to the circle $ (X)$ is

$ p(H_1, (X)) \equal{} \overline{H_1A} \cdot \overline{H_1H_a} \equal{} \frac {_1}{^2} \overline{H_1A} \cdot \overline{H_1K_a} \equal{} \frac {_1}{^2} p(H_1, (O_1))$

and similarly, powers of $ H_1$ to $ (Y), (H)$ are equal to $ \frac {_1}{^2} p(H_1, (O_1)).$ So the powers of $ H_1$ to $ (X), (Y), (H)$ are equal. Now, pick some other triangle, say the $ \triangle DAF$ with the circumcircle $ (O_2)$ and orthocenter $ H_2.$ In exactly the same way, you can show that powers of $ H_2$ to $ (X), (Y), (H)$ are all equal to half the power of $ H_2$ to $ (O_2).$ So the powers of $ H_2$ to $ (X), (Y), (H)$ are also equal. But if 3 circles have more than one point with the same powers to all 3, they are coaxal and their centers $ X, Y, H$ are collinear.

If you managed to show that $ FG$ is a polar of $ E,$ you should also manage to show that $ GE$ is a polar of $ F$ and $ EF$ is a polar of $ G.$ Lines connecting poles $ E, F, G$ to the circumcenter $ (O)$ of $ ABCD$ are perpendicular to their polars, so that $ EO, FO, GO$ are altitudes of the $ \triangle EFG$ and $ O$ its orthocenter. Inversion in $ (O)$ takes the vertices $ E, F, G$ into its altitude feet $ E', F', G'$ on their polars. Since the angles $ \angle FF'G, \angle GG'F$ are right and $ FG$ is a diameter of $ (H),$ it follows that $ F', G' \in (H).$ So, the inversion in $ (O)$ takes two points $ F, G$ on $ (H),$ which are not on $ (O)$, into another two points $ F', G'$ on $ (H).$ This is possible only if the inversion in $ (O)$ takes the entire circle $ (H)$ into itself; even one point not on $ (O)$ would be enough for this. But this means that $ (H)$ is perpendicular to the inversion circle $ (O).$ Similarly, the circles with diameters $ GE, EF$ are also perpendicular to $ (O).$


Give me solutions!

Beginning with the second day of a seven-day week, Potter ate 4 more dog biscuits each day than he had eaten the previous day. He ate a total of 105 biscuits during the week. How many biscuits did he eat on the fifth day?

In the diagram both circles have center 0 , $ \overline{AC}$ and $ \overline{BD}$ are diameters of the larger circle, $ \overline{AC}\perp \overline{BD}$, A0 = 2, and OE = 1.  Find the area of the shaded region. Express your answer as a common fraction in terms of x.
See attachment prob1.

A fair, 8-sided die has faces numbered 1, 2, 3, 4, 5, 6, 7, and 8.  If the die is rolled twice, what is the probability that the sum of the two numbers will be 9? Express your answer as a common fraction.

An 8 inch by 10 inch rectangular picture is surrounded by a border that is 2 inches wide on all sides. Find the ratio of the area of the border to the area of the picture. Express your answer as a common fraction.

If :$ \frac{1}{x} \plus{} \frac{1}{y} \equal{} 3$ and $ \frac{1}{x} \minus{} \frac{1}{y} \equal{} \minus{}7$ what is the value of x + y ? Express your answer as a common fraction.

Do you know what a common fraction is?  Look it up in the material I gave you!

[hide="First one"]Set up an equation.
Let $ x$ be how many he ate on Tuesday.
The equation should be$ x \plus{} (x \plus{} 4) \plus{} (x \plus{} 8) \plus{} (x \plus{} 12) \plus{} (x \plus{} 16) \plus{} (x \plus{} 20) \plus{} (x \plus{} 24) \equal{} 105$
Solve.
$ 7x \plus{} 84 \equal{} 105$
$ 7x \equal{} 21$
$ x \equal{} 3$

Then you plug it in on the fifth sequence, which is $ (x \plus{} 16)$ to get $ \boxed{19}$biscuits. [/hide]

[hide="Second one"]Get the area of a square by $ \frac{d_1\times d_2}2$
So area of a square is $ 8$.
I would get only $ \frac34$ of a square, so $ 8\times\frac34 \equal{} 6$ for now.
Then you get $ \frac34$ area of a smaller circle.
So that gives us $ 1\pi\times\frac34 \equal{} \frac {3\pi}{4}$

The area of a shaded area is $ 6 \minus{} \frac {3\pi}4\equal{} \boxed{\frac{24\minus{}3\pi}{4}}$  [/hide]

Ah... I made a stupid mistake on this one... got the correct answer, but put down something else...
[hide="Third one"]$ 8\times8 \equal{} 64$total possible.

We want two numbers sum to be equal to 9, so $ (1,8),(2,7),(3,6),(4,5)$, but the another one can also have this so $ 4\times2 \equal{} 8$

So $ \frac8{64} \equal{} \boxed{\frac18}$

Or you can do $ \frac4{32}$ also..then simplify.[/hide]

I don't know how I missed this one..I'm embrassed..
[hide="Fourth one"]Area of a picture=$ 80$
Area of a frame= $ (10 \plus{} 4)\times(8 \plus{} 4) \minus{} 80 \equal{} 88$

So $ \frac {88}{80} \equal{} \boxed{\frac {11}{10}}$[/hide]

[hide="Fifth one"]Click this link.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173099[/url]
It's #5[/hide]

#2  Check the area of your square.  (it is now corrected)

The rest are right!

[quote="danb80"]#2  Check the area of your square.

The rest are right![/quote]

Oh.. I edited it


algebra
polynomial
inequalities
algebra unsolved

Does there exist a polynomial with integer coefficient that has root ((2008)^(1/2008) + (2008)^(-1/2008))?  Prove or disprove.

[quote="t0rajir0u"]A sum of two algebraic numbers is algebraic.  This is well-known.[/quote]

How do you prove that?  And how do you prove these two numbers are algebraic?
Thanks a lot!

$ 2008^{\frac {1}{2008}}$ is a root of the polynomial $ x^{2008} \minus{} 2008 \equal{} 0$, and $ 2008^{ \minus{} \frac {1}{2008} }$ is a root of the polynomial $ 2008x^{2008} \minus{} 1 \equal{} 0$.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202382&search_id=1574381402]this thread[/url].

Alternately, here we can explicitly construct this polynomial:  start with the [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomial[/url] $ T_{2008}(x)$.  The polynomial $ P(x) \equal{} 2 T_{2008} \left( \frac{x}{2} \right)$ satisfies

$ P \left( x \plus{} \frac{1}{x} \right) \equal{} x^{2008} \plus{} \frac{1}{x^{2008}} \implies$
$ P(\alpha) \equal{} 4016$

where $ \alpha$ is the root we desire, so $ P(x) \minus{} 4016$ is the polynomial we seek.


Ascertain for  $ n\in\mathbb N^*$ the real numbers  $ \{x,y,z\}\subset[0,1]$ so that

$ \boxed {\ x^n\plus{}(1\minus{}y)^n\equal{}y^n\plus{}(1\minus{}z)^n\equal{}z^n\plus{}(1\minus{}x)^n\equal{}\frac {1}{2^{n\minus{}1}}\ }$ .

[hide]By PM-AM, for $ a\in[0,1]$, we have $ \sqrt [n]{{a^n \plus{} (1 \minus{} a)^n\over 2}}\geqslant {a \plus{} 1 \minus{} a\over 2} \equal{} {1\over 2}\implies a^n \plus{} (1 \minus{} a)^n\geqslant {1\over 2^{n \minus{} 1}}$. Adding up the equations, we obviously get the last inequality triplicated, hence $ x \equal{} 1 \minus{} x \equal{} y \equal{} 1 \minus{} y \equal{} z \equal{} 1 \minus{} z \equal{} {1\over 2}$ is the only solution.[/hide]

[hide="EDIT - Appendix"]The case $ n\equal{}1$ must be discussed separately, but then we simply get $ x\equal{}y\equal{}z$. So, the solutions are

$ (t,t,t)$ for $ n\equal{}1$ and $ \left({1\over 2},{1\over 2},{1\over 2}\right)$ for $ n\geqslant 2$[/hide]


A right triangle's hypotenuse has projections of length 4 and 9. Find the area of the triangle.
What do you do in this problem? I have no idea where to start.

[hide="i think :maybe: "]I think they mean that when the hypotenus is projected onto the x-y axes, you get a length of 4 and 9.  The area of the triangle is 18.[/hide]


function
integration
calculus
calculus computations

We know that 0!=1.Can anyone give a proof that using Gamma function?

This is a definition. We define $0! := 1$ so there is nothing to prove.

I think he means: prove that $0!=\Gamma(1)=1$. It is easy because $\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}dt$ so
$\Gamma(1)=\int_0^{\infty}t^0e^{-t}dt=0-(-e^{-0})=1$.

Misha

BTW: please choose a proper topic name next time.


Consider function $ f: R\to R$ which satisfies the conditions for any mutually distinct real numbers $ a,b,c,d$ satisfying $ \frac {a \minus{} b}{b \minus{} c} \plus{} \frac {a \minus{} d}{d \minus{} c} \equal{} 0$, $ f(a),f(b),f(c),f(d)$ are mutully different and $ \frac {f(a) \minus{} f(b)}{f(b) \minus{} f(c)} \plus{} \frac {f(a) \minus{} f(d)}{f(d) \minus{} f(c)} \equal{} 0.$ Prove that function $ f$ is linear

This is the solution I came up with during the test:

w.l.o.g we can assume f(0)=0 (or consider the function f(x)-f(0)) and we can also multiply a nonzero constant to f in order to make f(1)=1

First we take $ a\equal{}0$ in the condition , we get that if distinct nonzero real numbers b,c,d satisfying $ \frac {1}{b} \plus{} \frac {1}{d} \equal{} \frac {2}{c}$, then $ \frac {1}{f(b)} \plus{} \frac {1}{f(d)} \equal{} \frac {2}{f(c)}$. So it reminds me to consider a new function g(x) defined as $ g(x)\equal{}\frac {1}{f(\frac {1}{x})}$ ($ g: R/{0}\to R/{0}$).
It is easy to check that g is also satisfying the original condition and since if distinct nonzero real numbers b,c,d satisfying $ \frac {1}{b} \plus{} \frac {1}{d} \equal{} \frac {2}{c}$, then $ \frac {1}{f(b)} \plus{} \frac {1}{f(d)} \equal{} \frac {2}{f(c)}$, we get that for distinct nonzero real numbers x,y,z, if $ x\plus{}y\equal{}2z$, then $ g(x)\plus{}g(y)\equal{}2g(z)$.
i.e. for distinct nonzero real numbers x and y, if $ x\plus{}y \not \equal{} 0$, then $ g(x)\plus{}g(y)\equal{}2g(\frac {x\plus{}y}{2})$
since f(1)=1, we have g(1)=1.using $ g(x)\plus{}g(y)\equal{}2g(\frac {x\plus{}y}{2})$  we can easily get for any nonzero rational number x, $ g(x)\equal{}x$. Then it suffices to prove g is monotone.

my method is a little tricky, I think.
we prove a lemma: for nonzero real numbers x>y>z, if $ \frac {1}{x} \plus{} \frac {1}{y} \not \equal{} \frac {1}{z}$, then $ g(x)\minus{}g(z)$ and $ g(y)\minus{}g(z)$ are both positive or negative.
lemma's proof: in the condition we take $ a\equal{}x$, $ c\equal{}y$, $ b\equal{}z\plus{}\sqrt {(x\minus{}z)(y\minus{}z)}$, $ d\equal{}z\minus{}\sqrt {(x\minus{}z)(y\minus{}z)}$. According to $ x>y>z$, we have $ a>b>c>d$.It is obvious that a and c is not 0.if b or d is 0, then $ z^2 \equal{} (x\minus{}z)(y\minus{}z)$, which leads to $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{z}$, a contradiction.So a,b,c,d are distinct nonzero real number,i.e. they satisfies the condition in th original problem. Thus we get $ \frac {f(a) \minus{} f(b)}{f(b) \minus{} f(c)} \plus{} \frac {f(a) \minus{} f(d)}{f(d) \minus{} f(c)} \equal{} 0$
It equals that $ \frac {g(a) \minus{} g(b)}{g(b) \minus{} g(c)} \plus{} \frac {g(a) \minus{} g(d)}{g(d) \minus{} g(c)} \equal{} 0$
i.e. $ {[2g(a)\minus{}g(b)\minus{}g(d)]}{[2g(c)\minus{}g(b)\minus{}g(d)]} \equal{} {{g(b)\minus{}g(d)}^2}$
since $ {g(b)\plus{}g(d)}\equal{}2g(\frac {b\plus{}d}{2})$ we get $ {[g(a)\minus{}g(\frac {b\plus{}d}{2})]}{[g(c)\minus{}g(\frac {b\plus{}d}{2})]} \equal{} \frac {[g(b)\minus{}g(d)]^2}{4}$
So $ [g(x)\minus{}g(z)][g(y)\minus{}g(z)] \equal{} \frac {[g(b)\minus{}g(d)]^2}{4} \ge 0$, the lemma is proved

To finish the answer, we assume there exist a real number $ x$ such that $ g(x) \not \equal{} x$
case 1: $ g(x)>x$, then we can find a rational number y satisfying $ x<y<g(x)$ and a rational number z satisfying $ g(z)>g(x)$ and $ \frac {1}{z} \plus{} \frac {1}{y} \not \equal{} \frac {1}{x}$. So g(z)=z and g(y)=y. we have z>y>x but $ g(z)\minus{}g(x)$ is positive while $ g(y)\minus{}g(x)$ is negative, a contradiction.
case 2: $ g(x)<x$ is similar to get a contradiction.

so the hypothesis of the existence of such an x is wrong.i.e. g(x)=x for all nonzero x.Accoding to the definition of g we get f(x)=x for nonzero real number x. Also f(0)=0, so f(x)=x for all real x, and the original f must be a linear function.
Q.E.D.

[quote="linboll"]
we have g(1)=1.using $ g(x) \plus{} g(y) \equal{} 2g(\frac {x \plus{} y}{2})$  we can easily get for any nonzero rational number x, $ g(x) \equal{} x$.

[/quote]
Sorry :blush: .I don't know how to prove $ g(x)\equal{}x \forall x \in Q$ :?: .We can not substitute $ x\equal{}y$.Could anyone explain please? :maybe:

[quote="KDS"][quote="linboll"]
we have g(1)=1.using $ g(x) \plus{} g(y) \equal{} 2g(\frac {x \plus{} y}{2})$  we can easily get for any nonzero rational number x, $ g(x) \equal{} x$.

[/quote]
Sorry.I don't know how to prove $ g(x)\equal{}x \forall x \in Q$ :?: .We can not substitute $ x\equal{}y$.Could anyone explain please?[/quote]

I think this is wrong. In particular $g(x)$ can be any function of a kind $g(x)=kx+b$ where $k+b = 1 (x \not = 0)$

It seems, it's possible to prove that $g(x)$ should be only of that kind.
In this version, we are to check (?) if functions of a kind $\frac{1}{k \frac{1}{x} +b}$ may do. Put it into original :) so we may check that only linear functions are suitable.

Does anyone know how to solve it with less efforts? :)

[quote="linboll"]we can easily get for any nonzero rational number x, $ g(x)\equal{}x$. Then it suffices to prove g is monotone.[/quote]
We can tweak linboll's solution slightly to get a correct solution.

Indeed, if we have $g(x)+g(y)=2g(\frac{x+y}{2})$ then starting with 2 known values of $g$ we can get a dense subset $D\subset\mathbb{R}^*$ in which $g$ is a linear function, where $\mathbb{R}*=\mathbb{R}\backslash \{0\}$.

My recommendation is to take $g(\sqrt{2}+1)=a$, and work from there (with $g(1)=1$), since the subset constructed will exclude 0 (because $\sqrt{2}$ is irrational).

Following which, we can finish with his lemma that $g$ is increasing, swapping out $\mathbb{Q}$ for $D$. Conclude first that $g$ is linear, then reason that  if $g(x)\neq x$ then we can pick a value of $x$ such that $g(x)=0$, which shouldn't be possible. Thus $g(x)=x$ for all $x$, and we are done.

The problem is same as problem N. 138. of the KöMaL magazine (April 1997.)
http://db.komal.hu/KomalHU/feladat.phtml?id=56447

[quote=linboll]Thus we get $ \frac {f(a) \minus{} f(b)}{f(b) \minus{} f(c)} \plus{} \frac {f(a) \minus{} f(d)}{f(d) \minus{} f(c)} \equal{} 0$
It equals that $ \frac {g(a) \minus{} g(b)}{g(b) \minus{} g(c)} \plus{} \frac {g(a) \minus{} g(d)}{g(d) \minus{} g(c)} \equal{} 0$
[/quote]

how could you get that?

It is clear that $f $ must be injective.
Postcomposing  with Möbius transformation $\implies f(0)\to 0,f(1)\to 1,f (\infty)\to\infty. $
Assume that $f(0)=0,f(1)=1,f(\infty)=\infty. $
[u][b][color=#f00]Claim-[/color][/b][/u]
$f $ is a ring homomorphism.
[u][b][color=#00f]Proof-[/color][/b][/u]
Note that $\left [0,2x: x,\infty\right] $ is a harmonic range, so $f(2x)=2f(x).$ We know that $\left[a,b:\frac {(a+b)}{2},\infty\right]$ is a harmonic range. $f\left(\frac {(a+b)}{2}\right)=\frac{f(a)+f(b)}{2}\implies f(a+b)=2f\left (\frac{(a+b)}{2}\right)=f(a)+f(b). $ So $f $ is additive.We can prove that $f(t)=t\,\forall t\,\text {in the prime subfield}$ either $\mathbb Q $ or $\mathbb F_p. $ Using this we prove  $f(-x)=-f(x)\,\forall x$
We know $\left[0,-b:\frac {(b^2-b)}{2},\frac {(1-b)}{2}\right]$  is a harmonic range, as $\left [0,b:\frac {(b^2+b)}{2},\frac {(1+b)}{2}\right]. $
Since, $f\left(\frac {(1-b)}{2}\right)=\frac {(1-f(b))}{2},f\left(\frac{(b^2-b)}{2}\right)=\frac {\left(f(b)^2+f(b)\right)}{2}. $
Adding the properties yields  $f(b^2)=b^2. $ It is not known if any of these are $=, $ that is possible if $b=0,\pm 1. $
So $f(x^2)=f(x)^2\,\forall x. $ Thus, $f ((a+b)^2)=f(a^2)+2f(ab)+f(b^2)=(f(a)+f(b))^2=f(a)^2+2f(a)(b)+f(b)^2\implies f(ab)=f(a)f(b)\,\forall a,b. $
Hence, $f $ is additive and multiplicative. $f $ is a ring homomorphism.
[u][b][color=#f00]Note-[/color][/b][/u]
Homogeneous coordinates for $P^1 (k) $ should be in the form $[a,b] $ for $a,b\in K. $ Since, we don't know notation for harmonic range.
Let $[x,y: z,w] $ be the 4 points of the harmonic range. Cause  $P^1 (k)=k\cup\infty. $


calculus
integration
trigonometry
calculus computations

I hope you can work out by more ways

$ \int_{0}^{1}\frac {x^{4} \plus{} 1}{x^{6} \plus{} 1}dx$

Note that $ (x^6\plus{}1)\equal{}(x^2\plus{}1)(x^4\minus{}x^2\plus{}1)$

Subtract and add $ x^2$ from the numerator of the integral.

Thus our integral is

$ \int_{0}^{1}{\frac{1}{1\plus{}x^2}} \plus{} \int_{0}^{1}{\frac{x^2}{x^6\plus{}1}}$

For the second, the subsitution $ x^3\equal{}t$ should do the trick.

I think the final answer is $ \frac{\pi}{3}$.

In a book it tells
${ \int \frac{1+x^{4}}{1+x^{6}}}dx$
$ =-\frac{1}{3}arctan\frac{3x(x^{2}-1)}{x^{4}-4x^{2}+1}$

Could you work it out ?

$ \int \frac{1}{1\plus{}x^{2}}dx\plus{}\int \frac{x^{2}}{1\plus{}x^{6}}dx$

$ \equal{}\tan ^{\minus{}1}x\plus{}\frac{1}{3}\tan ^{\minus{}1}x^{3}$

$ \equal{}\frac{1}{3}\tan ^{\minus{}1}\frac{x^{3}\minus{}3x}{3x^{2}\minus{}1}\plus{}\frac{1}{3}\tan ^{\minus{}1}x^{3}$

$ \equal{}\frac{1}{3}\tan ^{\minus{}1}\frac{\frac{x^{3}\minus{}3x}{3x^{2}\minus{}1}\plus{}x^{3}}{1\minus{}\left( \frac{x^{3}\minus{}3x}{3x^{2}\minus{}1}\right) \left( x^{3}\right) }$

$ \equal{}\frac{1}{3}\tan ^{\minus{}1}\frac{3x\left( 1\minus{}x^{2}\right) }{x^{4}\minus{}4x^{2}\plus{}1}$

$ \equal{}\minus{}\frac{1}{3}\tan ^{\minus{}1}\frac{3x\left( x^{2}\minus{}1\right) }{x^{4}\minus{}4x^{2}\plus{}1}$

However, I do not understand why that book requires the reader to arrive at such a cumbersome representation.

You know in $ \int_{0}^{1}\frac{x^{4}\plus{}1}{x^{6}\plus{}1}dx$ if we sub $ x\equal{}1$ and $ x\equal{}0$ we will get wrong answer


Yes, I know there have been many threads about the insanely hard online riddle [url=http://deathball.net/notpron/]Notpron[/url], but they're all too old to be revived.

I'm currently stuck on level 16. (Oh, and no spoilers please... or at least put them under hide tags... thanks.)

I'm now on level 23 and I can't find the thing I'm supposed to be looking for in the picture.

[quote="142857"]I'm stuck on lv12 now. Don't know what to do with the dragon.[/quote]

Are you sure you got ALL the clues? What would JayPack do? [hide="Semi-spoiler"]The dragon actually has little to do with anything.[/hide]

[quote="sapphyre571"]I think it's kinda funny that pron actually shows up on notpron on about level 28.[/quote]

Actually, I took a wrong turn at level 21 and ran into a rather unpleasant picture, too.


Kenton watched 2000 adults board a cruise ship. Half of the adults were women. If 20$ \%$ of the women and 9$ \%$ of the men were wearing sunglasses, what was the total number of men and women wearing sunglasses?

So there are $ 1000$ women and $ 1000$ men.  $ 20\%$ of $ 1000$ is $ 200$.  $ 9\%$ of $ 1000$ is $ 90$.  Thus, we simply add for a grand total of $ \boxed{290}$.


[quote="Poincare"]nope. Not at all. My mom sneezes when see sees a triangle and my dad screams when he sees a function.[/quote]

The poll says "anybody that is not on aops."

So I assume you don't mean former IMO gold medalists?

hmm i don't get help but i do take math classes outside of school... it's not really help though... it's just cuz i'm so bored in school... so i put yes... but it's like 1/2 and 1/2...

I am completely self-teaching outside of AoPS.

I take a pretty good math class (more like school) that almost every good (and bad) math student in the area goes to.


trigonometry
geometry
vector
geometric transformation
rotation
rhombus
complex numbers

I'm trying to solve these without too much algebra bashing, so using pictures would be cool

1. Let $ z$ lie on the unit circle. Show that $ |1\minus{}z| \equal{} 2\sin \equal{}\left(\dfrac{\arg z}{2}\right)$.

2. Determine the locus $ z$ for:

a) $ Re \left(\dfrac{z\minus{}1\minus{}i}{z\plus{}1\plus{}i}\right)\equal{}0$

b) $ Im \left(\dfrac{z\minus{}1\minus{}i}{z\plus{}1\plus{}i}\right)\equal{}0$.

3. Show that if $ |z|\equal{}1$, then $ Im\left(\dfrac{z}{(z\plus{}1)^2}\right)\equal{}0$.

4. Let $ k$ be a constant and let $ a,b$ be fixed complex numbers. Describe the locus of points $ z$ that satisfy $ \arg\left(\dfrac{z\minus{}a}{z\minus{}b}\right)\equal{}k$.

--

also I'm having trouble with geometry with complex numbers as well. What happens to $ z$ if I have $ |1\minus{}z|$? What about $ |z\minus{}1|$?

$ |z \minus{} w|$ describes the distance from $ z$ to $ w$, so $ |z \minus{} 1| \equal{} |1 \minus{} z|$ is the distance from $ z$ to $ (1, 0)$.  $ |z| \equal{} 1$ describes the unit circle.  Similarly, expressions like $ z \minus{} (1 \plus{} i)$ describes the vector from $ (1, 1)$ to $ z$.  Quotients describe a rotation and a scaling.  The argument of a quotient is the angle (of rotation) between the two corresponding vectors, and the length of a quotient describes the scaling.

2 (a) z lies on a circle with 1+i and -1-i as diameters
(b) z lies on the straighline joining 1+i and -1-i

[quote="AstroPhys"]For geometric representations of such problems, I like to look at complex numbers in exponential form.  What does raising a complex number to a power do? etc.[/quote]
See Formulation 2 [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=216343]here[/url].

A geometric solution of question 3.

Let w=z+1 then Iw-1I = IzI =1 . So w lies on  the fixed  circle of unit radius whose centre is at the point 1 on the real line.
 Denote the variable points z and w by P and Q respectively.  Let O be centre of  the circle  IzI =1, and C the centre of the circle  Iw-1I =1.
Since Q is obtained from translating P directly one unit to the right, PQ is parallel to the real line and the figure OPQC is a rhombus with $ \angle POC$  =2$ \angle QOC$ for all positions of P.

From the diagram we see that argz=$ \angle POC$ and argw =$ \angle QOC$ giving argz =2argw

Now arg$ \frac{z}{(z+1)^2}$ =argz - 2arg(z+1) =0 from which it follows that Im$ \frac{z}{(z+1)^2}$ must be zero.

  For question 1, if  put w= 1-z, then w will lie on the unit circle translated one unit to the right and we  can construct a rhombic figure as in question 3.It is not clear in your post what we have to prove.

Question4 Theory
In the complex plane let P(z) be variable and $ P_1$(a) $ P_2$(b) be fixed.Let P vary so that the $ \angle P_1PP_2$ =k a real constant.

The set of points z such that arg$ \frac{z-a}_{z-b}$ =k is

1.k=0 ,the whole line passing through $ P_1$ and $ P_2$ with the segment  $ P_1P_2$ removed.

2.k=$ \frac{\pi}{2}$,a semicircle with points $ P_1$ and $ P_2$ at opposite ends of the diameter.

3.0<k<$ \pi$ , an arc of a circle terminating at $ P_1$ and $ P_2$.

4.k=$ \pi$ ,the line segment $ P_1P_2$.

If a given value of k gives a certain arc the rest of the circle to which the arc belongs(i.e. the supplementary arc) corresponds to the value $ k\pm\pi$


MATHCOUNTS
ARML
AMC
AMC 10
AIME
summer program
MathPath

Haha, might as well.

This I really have no clue on, however, AOPSers might be able to give their input.
Which team do you think will win?

I'll ask a mod to maybe add a poll later..

Why don't you start a poll instead. It's so much easier to look at.

Because I want to hear opinions and reasons first.

Don't count on Illinois--we've got no returning people--we were all 8th graders

I wouldn't count on Texas either.
Though we have the likely #1 written, our team won't be too stellar.
I think there was one 7th grader this year with like 38 or 37, and if you wanted to do it scorewise, we're not that bad, since we had 5 or 6 7th graders in the top 20, which took about a score of... 32 or 33 or so.

Still, we had a pretty good 2 year run. But not too much talent to go on now, similar to Indiana.

[quote="jclarkemathL314159"]I mean this seriously this time....

Massachusetts.
We have three returning people, and a bunch of promising people coming up as well. There were 2 6th graders and at least 15 7th graders who scored above 30 at states this year.[/quote]
There are probably 100 6th or 7th graders in California who could score over 30 at states.  :P
I have a feeling Maine's not gonna win next year. And our state coach (from my school) for the last 5 years is retiring.

If you're looking at people with a shot of helping a team to win, you should be looking at 7th graders who scored 35+ this year, there were tons of 7th graders in CT too who scored above 30.
And you can take CT out of the list of places that might win, two returning people 90th (a good day) and 188th (a bad day) but our really good people are gone from middle school. (Yifan Cao, Jason Hoch)

With 3 returning people, MA might do good considering their rank this year, but I still think WA has a better chance.

[quote="ehehheehee"]Does this matter?  And usually, when we try to guess this, we don't get it right...[/quote]

What do you mean by that? No one really thought MI was going to win.

Michigan doesn't have a chance, even though I still want them to win and do better than 13th Place.  I also hope that Mrs. Leybourn is the coach for the 4th straight year.

We only have one good person, Allen Yuan, and he is not that great.  No offense, kyyuanmathcount. :lol:

[quote="Ignite168"]Don't listen to Zenith Xan, VA has no chance next year. Their only returning vet scored 22 on this year's test.

I think WA has a chance, not IN though.[/quote]

Uh huh. And our only two returning vets from last year only had 24s on the 2004-2005 test ;)

Anyway, I still don't think that VA will win next year. I predict that Guam will win next year...Not really. Truly I'd agree that Washington has a good chance at winning but I believe that (since no top 3s repeated this year) they'll get somewhere from 4th to 10th. California has the best odds, but Texas and Illinois will probably be powerhouses too. All in all, I say it'll end up:

1) Illinois
2) Texas
3) California

[quote="pianoforte"]My predictions:

1. Kansas
2. Guam
3. Virgin Islands
4. Canada[/quote]

No...Obviously Taiwan will win. I mean, have you SEEN their AMC scores?

Besides, Canada could easily beat Guam and the Virgin Islands, probably Kansas too.

George Yu and George Silvis both got the same score at States and Nationals. :o

37 and 38 respectively, I think.

No, crap I didn't realize that -_-. I did copy and paste with three [hide]'s and then added my slashes manually at the end.

Neither will Michigan, in fact we shouldn't be put on any poles (for obvious reasons).

The only person who has a chance in making Countdown is our 70th Place student. (Not revealing names...)

[quote="Ignite168"][quote="samath"][quote="Ignite168"]I think WA has a chance, not IN though.[/quote]

I see we get respect AFTER the competition  :roll:

I'm going to try to be an assistant coach for Washington next year.  Hurray!  I get to see Washington win![/quote]

No it's just because you got that 15th place girl in 7th grade. I think after next year, WA won't place as high for a while.[/quote]

I got 19th in seventh grade, and there was another seventh grader then.  But we didn't get the respect until after we got second :P.  That's my point.

Well, I hope that MN will at least get 25th place at Nats. I made something like 17th at state, just missed the countdown. That's what I love.

I also love getting 14th Place at National Mathcounts just missing the Countdown Round by 2 places.

[quote="Phelpedo"][quote="biffanddoc"][/quote]
Which backs up my claim even more.... If you get more at nats then state, especially with a rather easy state round this year and hard nationals, then that is some serious improvement.[/quote]

Just because he improved from state to nationals doesn't mean that he's going to keep that improvement curve all the way until next year. The fact is that he got 108th at nationals, which is considerably less impressive than many other 7th graders. As a side note, has anyone ever gone from a triple-digit rank in 7th grade to making countdown in 8th?[/quote]

Well I did do better in 6th than in 7th =p ...

Treething, I was just wondering, where is your trophy right now being placed exactly?

I don't think I can combine any of my 11 trophies to equal the area of yours.

I didn't make it to Nationals as a 6th Grader because I wasn't on the team and didn't do well as an individual.  But, in 2004, I remember that they stopped doing this for some reason, but they gave a trophy to the best 6th Grader, 7th Grader, and 8th Grader.  Yay, I was best 6th Grader.

[quote="Treething"][quote="Phelpedo"][quote="biffanddoc"]
Which backs up my claim even more.... If you get more at nats then state, especially with a rather easy state round this year and hard nationals, then that is some serious improvement.[/quote]

Just because he improved from state to nationals doesn't mean that he's going to keep that improvement curve all the way until next year. The fact is that he got 108th at nationals, which is considerably less impressive than many other 7th graders. As a side note, has anyone ever gone from a triple-digit rank in 7th grade to making countdown in 8th?[/quote]

Well I did do better in 6th than in 7th =p ...[/quote]

Well that's just weird. Connie Wu did the same: 29th, 56th, 33rd in 6th, 7th, and 8th grade.

[quote="Phelpedo"][/quote][quote="Treething"][quote="Phelpedo"][quote="biffanddoc"]
Which backs up my claim even more.... If you get more at nats then state, especially with a rather easy state round this year and hard nationals, then that is some serious improvement.[/quote]

Just because he improved from state to nationals doesn't mean that he's going to keep that improvement curve all the way until next year. The fact is that he got 108th at nationals, which is considerably less impressive than many other 7th graders. As a side note, has anyone ever gone from a triple-digit rank in 7th grade to making countdown in 8th?[/quote]

Well I did do better in 6th than in 7th =p ...[/quote][quote="Phelpedo"]

Well that's just weird. Connie Wu did the same: 29th, 56th, 33rd in 6th, 7th, and 8th grade.[/quote]

Respectively

[quote="piggyinc"][quote][b] Ignite 168 wrote:[/b]
No it's just because you got that 15th place girl in 7th grade. I think after next year, WA won't place as high for a while.
[/quote]

Not really: we have a bunch of sixies and supersmart fifth grader (Darryl Wu) coming in as well, and they'll be in Mathcounts by next year.

-Oink, Oink, Oink[/quote]
Don't awaken old topics. This thread has the history of being discussion without information, it should stay that way I think.


algebra
polynomial
vector
linear algebra
combinatorics unsolved
combinatorics

we say a family $F$ of k-element subsets of $[n]$ is a k-forest if for every $f\in F$ there is a partition $[n]=V_{1,f}\cup \dots \cup V_{k,f}$,such that $f$ is the only member of $F$ which intersects every $V_{i,F}$.prove that $|F|\leq {n-1\choose k-1}$

Thus we have $k|n$ right?
I'm asking this because I think it's not necessarily true, and maybe there is a typo ==)

not necessarily;the sets $V_{i,F}$ need not be of the same size.for instance if $n=7,k=3$ we could take the family $\mathcal{F}$ to be the collection $\mathcal{F}=\{A | 1 \in A\}$.then for each set $A$ we take the partition $V_{i,A}$ to be $V_{1,A}=\{1\} \cup ([n]\setminus A);V_{i,A} =\{i^{th}$ element of $A\}$ for other $i$.

I have an idea, but it needs to be completed.

By induction.
Suppose a contrary, i.e. there are at least $C_{n-1}^{k-1}+1$ such $k$-tuples. Consider two sets:
1) $A$ is the set of $k$-tuples containing $n$;
2) $B$ is the set of $k$-tuples not containing $n$.

[*] I want now that $A$ forms a $k-1$-forest on $[n-1]$ and $B$ forms $k$-forest on $[n-1]$.

Then we have exactly one of two possibilities:
\[|A|\geq C_{n-2}^{k-2}+1\mbox{ or }|B|\geq C_{n-2}^{k-1}+1.\]
Both of them leads to contradiction.

But I don't know how to justify [*].

Sorry. It was nonsense. We can't guarantee that $A$ is $k-1$-forest.

I saw this problem as an exercise for a course regarding algebraic methods in combinatorics. There was no solution, but there was a hint, which, unfortunately, I looked at... :(

We may assume WLOG that for all partitions as the ones described, $1\in V_{1,f}$. Now consider the polynomials $P_f({\bf x})=\prod_{i=2}^k\sum_{j\in V_{i,f}}x_j$ (over $\mathbb Z_2$), where ${\bf x}=(x_2,x_3,\ldots,x_n)$ are characteristic vectors of subsets of $[n]\setminus \{1\}$. Let $m=|\mathcal F|$. It's easy to see that for $i,j\in\overline{1,m}$, we have $P_{f_i}(f_j)=\delta_{ij}$ (the Kronecker symbol), meaning that these polynomials are linearly independent in the vector space of multilinear homogenous polynomials of degree $k-1$ in $n-1$ unknowns over $\mathbb Z_2$. This space has dimension $\binom{n-1}{k-1}$, so that's pretty much it, unless there are huge mistakes in there (it was just a hint; I still had to do some work, and I might have made mistakes :)).

[quote="grobber"]I saw this problem as an exercise for a course regarding algebraic methods in combinatorics. [/quote]
Do you have an electronic version of this course? Or some links?

It's [url=http://www.math.cmu.edu/~pikhurko/]here[/url]. It's the one called "Algebraic Methods in Combinatorics, Cambridge 2002" (among the last teaching links on the page).

Thanks Grobber!
How do you find all these stuffs? :)  :)  :)

Pierre.

I look for them. It's surprising, really: they're all over the place :). That course looks like gold :). Unfortunately, there are other things I have to read.. :(

Wow, I must also thank you Grobber. And really how did you find it? Because I tried looking for such materials by typing "algebraic methods in combinatorics" and there were so many results that I gave up. Also if you have any other links to articles regarding linear algebra in combinatorics I'd greatly appreciate  :D

sorry for my stupid problem but i don't know where is that course at this page, can someone tell me how should I find it??


The minute hand of a watch is 8mm long, and the hour hand is 4 mm long.  How fast is the distance between the tips changing when it is 1:00?

I haven't learned integration, or even L'hospital's rule, so don't use any of that plz.

This is in section "related rates"

Thanks

This is actually an kinematics problem, but it has some trig involved.
It asks, essentially, for the relative velocity modulus of the hands tips.
The minute hand has an angular velocity of $ 2\pi$ rad/h, that multiplied by his length gives his tip velocity. Same thing for the small hour hand with angular velocity of $ {{\pi}\over 6}$ rad/h.
Vectorial subtraction gives the answer. ( at this time the velocities are making an angle of 30º, so you can use cosine law).

so, if you draw the triangle, you can cut it into two right triangles

the first one goes partially up the minute hand up to $ 4\cos \theta$
and goes right length $ 4\sin \theta$

and the second one has the same base $ 4\sin \theta$
with height $ 8\minus{}4\cos \theta$

so the distance in question $ s^2\equal{}(8\minus{}4\cos \theta)^2\plus{}(4\sin \theta )^2$
$ \left(\frac{s}{4}\right)^2\equal{}(2\minus{}\cos \theta)^2\plus{}\sin^2 \theta\equal{}5\minus{}4\cos \theta$
taking $ \frac{d}{dt}$ we get

$ 2\frac{s}{4} \frac{s'}{4}\equal{}4\theta ' \sin \theta$

so $ s'\equal{}\frac{32\theta' \sin \theta}{s}$

so $ \theta' \equal{} \frac{\pi}{6}\minus{}2\pi\equal{}\minus{}\frac{11\pi}{6}$ and $ \theta \equal{} \frac{\pi}{6}$ at 1pm,
and $ s\equal{}\sqrt{(8\minus{}2\sqrt 3)^2\plus{} 2^2}\equal{}4\sqrt{5\minus{}\sqrt 3}$

giving $ s'\equal{}\minus{}\frac{88 \pi}{3\sqrt{5\minus{}\sqrt 3}}$


modular arithmetic
number theory
relatively prime

Given that the $ gcd(a,b)\equal{}1$, find all ordered pairs $ (m,n)$ such that $ a^{m}\plus{}b^{m}$ divides into $ a^{n}\plus{}b^{n}$.

[hide="What I have so far"]$ \frac{a^{n}\plus{}b^{n}}{a^{m}\plus{}b^{m}}\equal{}k$ where $ k$ is an integer. This fraction turns into $ \frac{(a\minus{}b)(a^{n\minus{}1}\plus{}a^{n\minus{}2}b\plus{}...\plus{}b^{n\minus{}1})}{(a\minus{}b)(a^{m\minus{}1}\plus{}a^{m\minus{}2}b\plus{}...\plus{}b^{m\minus{}1})}\equal{}k$. From here since $ a\neq b$ we can divide $ a\minus{}b$ to have $ \frac{a^{n\minus{}1}\plus{}a^{n\minus{}2}b\plus{}...\plus{}b^{n\minus{}1}}{a^{m\minus{}1}\plus{}a^{m\minus{}2}b\plus{}...\plus{}b^{m\minus{}1}}\equal{}k$. From here is where I am stuck...any help?[/hide]

You need to show that $ n$ or $ m$ is odd, and the factorization does not work if there is an even exponent. Also, the factorization would only work for $ a^n \minus{} b^n$, for evens. (it's fine if the power is odd).

One obvious solution is $ (m,n) \equal{} \boxed{(k,k)}$, for some $ k$.

Are there any restrictions about the actual numbers $ a,b,m,n$ themselves?

The only other restrictions that I forgot to mention were that $ a$, and $ b$ are positive integers.

I have proceeded with this problem as follows:
Since $ a^m\plus{}b^m|a^n\plus{}b^n$ hence $ m\leq n\implies m\plus{}k\equal{}n$
Plugging this value of $ n$ into the given relation we have $ a^m\plus{}b^m | a^m \cdot a^k \plus{}b^m\cdot b^k$.
So now we let $ a^m \cdot a^k \plus{}b^m\cdot b^k\equal{}la^m\plus{}lb^m\implies a^m(a^k\minus{}l)\plus{}b^m(b^k\minus{}l)\equal{}0$
$ \implies a^m(a^k\minus{}l)\equal{}b^m(l\minus{}b^k)$
Now since $ \gcd (a,b)\equal{}1$; hence this equation is only possible when $ a^k\minus{}l\equal{}l\minus{}b^k\equal{}0$
$ \implies a^k\equal{}b^k\implies \boxed{k\equal{}0}$.
Hence $ m\equal{}n$ is forced; $ \implies$ all possible pairs of $ m,n$ are
$ \boxed{(m,n)\equal{}(t,t)}$ where $ t$ is any positive integer.

[quote="Potla"]Now since $ \gcd (a,b) \equal{} 1$; hence this equation is only possible when $ a^k \minus{} l \equal{} l \minus{} b^k \equal{} 0$
$ \implies a^k \equal{} b^k\implies \boxed{k \equal{} 0}$.[/quote]

This is not necessarily true, since all m,n odd with m|n also work.
Consider m=1, k=2, l=a^2-ab+b^2, for example.

[hide="hopefully correct"]
As above, let $ n\equal{}m\plus{}k$. Since $ m\equal{}n$ obviously works, let k be positive.
Then we want $ a^ka^m\plus{}b^kb^m\equiv 0 \pmod{a^m\plus{}b^m}$
$ a^ka^m\plus{}b^kb^m\minus{}(a^k\plus{}b^k)(a^m\plus{}b^m)\equiv 0 \pmod{a^m\plus{}b^m}$
$ a^kb^m\plus{}b^ka^m\equiv 0 \pmod{a^m\plus{}b^m}$
If $ k<m$ then we have
$ a^kb^k(a^{m\minus{}k}\plus{}b^{m\minus{}k})\equiv 0\pmod{a^m\plus{}b^m}$.
Because $ gcd(a,b)\equal{}1$, both $ a^k$ and $ b^k$ are relatively prime to $ a^m\plus{}b^m$. This means we need
$ a^{m\minus{}k}\plus{}b^{m\minus{}k}\equiv 0\pmod{a^m\plus{}b^m}$.
Which cannot be true because the LHS is greater than 0 and less than $ a^m\plus{}b^m$.
This means we need $ k\ge m$, in which case
$ a^mb^m(a^{k\minus{}m}\plus{}b^{k\minus{}m})\equiv 0\pmod{a^m\plus{}b^m}$, or
$ a^{k\minus{}m}\plus{}b^{k\minus{}m}\equiv 0\pmod{a^m\plus{}b^m}$.
Now we need $ k\minus{}m\ge m$, so this is the same situation as the beginning, except $ n$ is replaced by $ k\minus{}m$. We have the solution $ k\minus{}m\equal{}m$, or $ k\equal{}2m$, and then we can repeat the exact same process with $ k\minus{}m\equal{}m\plus{}l$ where $ l$ is positive, to get $ l\minus{}m\ge m$. The solution $ l\minus{}m\equal{}m \implies l\equal{}2m\implies k\minus{}2m\equal{}2m\implies k\equal{}4m$ works, and then we can let $ l\minus{}m\equal{}m\plus{}a$, where a is another positive number, etc. Therefore, the only possible solutions are $ k\equal{}0, 2m, 4m, 6m,\cdots$, and they also all work. This implies $ n\equal{}m, 3m, 5m, \cdots$ so the final answer is any ordered pair of the form $ (m, (2c\minus{}1)m)$, where $ m$ and $ c$ are positive integers. [/hide]
This could probably be stated in a simpler and shorter way, but oh well.

[quote="Potla"]$ \implies a^m(a^k \minus{} l) \equal{} b^m(l \minus{} b^k)$
Now since $ \gcd (a,b) \equal{} 1$; hence this equation is only possible when $ a^k \minus{} l \equal{} l \minus{} b^k \equal{} 0$[/quote]
I think my previous step is incorrect .... :( Because of anderson's counter example: [quote="andersonw"][quote="Potla"]Now since $ \gcd (a,b) \equal{} 1$; hence this equation is only possible when $ a^k \minus{} l \equal{} l \minus{} b^k \equal{} 0$
$ \implies a^k \equal{} b^k\implies \boxed{k \equal{} 0}$.[/quote]
This is not necessarily true, since all m,n odd with m|n also work.
Consider $ m\equal{}1, k\equal{}2, l\equal{}a^2\minus{}ab\plus{}b^2,$ for example.[/quote]

[quote="JBL"][quote="Potla"]So now we let $ a^m \cdot a^k \plus{} b^m\cdot b^k \equal{} la^m \plus{} lb^m$[/quote]  You might want to reconsider this step  :wink:[/quote]
Can you explain JBL, why I can't assume $ a^m \cdot a^k \plus{} b^m\cdot b^k \equal{} l(a^m \plus{} b^m)$ where $ l\in\mathbb N$ ? I think it is correct here since $ (a^m \plus{} b^m)|(a^m \cdot a^k \plus{} b^m\cdot b^k)$....

@ all: sorry for that stupid mistake :(

@Potla: my apologies, I was reading quickly and misunderstood what you wrote.


consider $a;b$ satisfies the equation:$ax^3-x^2+bx-1=0$ has three positive roots;
Find the minimum of $P=\frac{5a^2-3ab+2}{a^2(b-a)}$

Minimum which I got is $a \sqrt{5}$ hope it is correct but i could easily miscalculated at some computations because there were a lot of them, so I just point out idea of my solution.

1) Make use of Viete's Formulas to have $a,b$ described by $x,y,z$ [roots] only
2) Substitute and simplify $P$ as much as possible
3) Use AM-GM to get the minimum


quadratics
function
integration
geometry
3D geometry
calculus
calculus computations

Say that you have a tent-like figure, given by the equations:

$y=\frac{81}{2}-x^2=\frac{81}{2}-z^2$

The problem is to find the volume under these in $\left[-\frac{9}{\sqrt{2}},\frac{9}{\sqrt{2}}\right]$x$\left[-\frac{9}{\sqrt{2}},\frac{9}{\sqrt{2}}\right]$.

I don't know if this makes any sense so far, but try thinking of it as a tent, with the quadratic functions as supports.

Anyways, I did horizontal slicing and came up with:

$\int_{0}^{\frac{81}{2}}(81-2y)dy=1640.25$

I noticed that this is exactly half of the cube with length 9, width 9, and height 81/2.

Does this look correct or did I do something wrong?

Thanks.

The statement of the problem isn't very clean. Here's what I think you mean:

Find the volume of the region bounded by the $x$-$y$ plane and the surfaces $z=\frac{81}2-x^2$ and $z=\frac{81}2-y^2.$

Then the horizontal slices at height $z$ are squares with corners at $\left(\pm\sqrt{\frac{81}2-z},\pm\sqrt{\frac{81}2-z}\,\right).$ Such a square has area $162-4z^2.$ Hence the volume is

$\int_0^{\frac{81}2}162-4z^2\,dz=\frac{81^2}2=3280.5.$

This is half of the box - I wouldn't call it a cube - of dimensions $9\sqrt{2}\times 9\sqrt{2}\times \frac{81}2$ that the figure fits into.

In summary: you were nearly correct, just some issues about factors of two. And "half the box" is correct.


Is given a right triangle ABC  inscribed in a circle (The triangle is in the circle). The catetes are respectively 13 and 24 .
The bisector of the right angle intersect the circle in a point D . Find the area of the tetrahedron ABCD.

A right angle inscribed in a circle cuts off a 180 degree arc, making the hypotenuse of triangle ABC a diameter. Its length can be calculated (Pythagoras) from the lengths of the legs of the right triangle. The area of right triangle ABC can also be calculated, taking one leg as the base and the other as the altitude.
To connect the ends of the hypotenuse of ABC to point D, we draw two chords, making another triangle. We must find out more about this triangle.
The (also inscribed) angle at D also cuts off a 180 degree arc, so it is also a right angle.
The two halves of the original right angle are 45 degree angles, so they subtend two 90 degree arcs. Since the arcs are congruent, so are the corresponding chords (connecting D to the ends of the diameter/hypotenuse of ABC).
In conclusion, the triangle formed by that diameter and the chords connecting it to D is a right isosceles triangle.
Since we know the length of its hypotenuse (diameter) we can calculate its area (and its legs if we cared).
The quadrilateral ABCD is made up of the two right triangles. Its area is the sum of the areas.

[quote="mewto55555"]Whats a catete? A leg?[/quote]I assumed so. It may be from British English or from another language. In Spanish the word is "cateto".
In other countries right triangles do not have legs, and clocks do not have hands.


Prove that if f is a polynomial with integer coefficients, and there exists an integer k such
that none of the integers f(1), f(2), . . ., f(k) is divisible by k, then f(x) has no integer root.

If $ f(r)=0$, then $ f(r)\equiv 0\mod k$. But this is a contradiction since all residues mod $ k$ produce a nonzero residue when they are put in the function.


integration
inequalities
real analysis
real analysis solved

Suppose $1<p<\infty$. Define
$\mathcal{F} = \left\{ f\in L([0,1]): \int_0^1 |f(x)|dx = 1, \int_0^1 |f(x)|^p dx = 2 \right\},$
Prove that $\forall 0<\varepsilon<1$, $\exists \delta>0$, such that
$m(\{ x\in[0,1]: |f(x)|>\varepsilon \}) \geq \delta, \quad f\in \mathcal{F}.$

Let $E=\{x\mid |f(x)|>\epsilon\}$, $f\in \mathcal{F}$. Suppose WLOG $f\geq 0$.
We have $\int_{E}f(x)dx>1-\epsilon(1-m(E))$. Moreover, form Holder inequality \[\int_{E}f(x)dx=\int_{E}1\cdot f(x)dx \leq m(E)^{1/q}||f||_p\leq 2m(E)^{1/q},\] where $1/p+1/q=1$.
Define $m=m(E)$. Then we have obtained $1-\epsilon(1-m)<2m^{1/q}$. It follows $1-\epsilon<2m^{1/q}$, i.e. $m>\left(\frac{1-\epsilon}{2}\right)^q$ for all $f\in\mathcal F$.


p is prime with =1(mod 4). prove geometircally that there is a integer a,b such p=a^2+b^2

Are you serious or kidding :?: If the former the proof must be amazing

@Leonhard Euler: If you are from Vietnam, read namdung's post in forum diendantoanhoc.net, please  :wink:


geometry
3D geometry
sphere
Functional Analysis
real analysis
real analysis unsolved

if we take a banach space and as usual define the norm for elements in the dual to be ||L||= sup {L(x): ||x||=1} then is it true that this sup is achieved? If so give a proof and if not give me a condition that would ensure that it is.

I might need this for a problem I am working on so accurate complete answers please.

Let $X$ be a Banach space. The following are equivalent:
(i) every bounded linear functional on $X$ attains its norm on the unit sphere;
(ii) $X$ is reflexive.

(ii)=>(i) follows from the Banach-Alaoglu theorem. (i)=>(ii) is a hard theorem due to R.C.James.

I think I found Banach-Alaoglu theorem online so if i find the other one I understand.

Key words - james, theorem, reflexive - will get you a bunch of links.

what is an example of it being false in a non-relfexive space?

Let $c_{0}$ be the space of sequences converging to $0$. It is known that $c_{0}^{*}=\ell_{1}$, but $(\ell_{1})^{*}=\ell_{\infty}$, so $c_{0}$ is not reflexive. Pick a sequence $x\in\ell_{1}$ with infinitely many nonzero terms, e.g. $x_{n}=1/n^{2}$. Summation against this sequence is a linear functional on $c_{0}$ whose norm is $\parallel x\parallel_{1}$. (Precisely, $\varphi(y)=\sum_{n}x_{n}y_{n}$ for all $y\in c_{0}$). However, the norm is not attained. Indeed, if $\varphi(y)=\parallel x\parallel_{1}$ and $\sup_{n}|y_{n}|=1$, then $y_{n}=\mathrm{sign}(x_{n})$ for all $n$, which means $y\notin c_{0}$.

Thank you very much mlok, now i can continue my research.


inequalities
function
real analysis
real analysis theorems

Hello, I recently decided to learn Analysis and the biggest problem I am running into is without an instructor/teacher it is hard to tell if my answers are correct. Do I assume things that shouldn't be assumed without proof, do I state things that are too obvious, etc. I would appreciate it if someone could look over this solution to a basic problem from my book (Principles of Mathematical Analysis by Rudin).

Question: "Fix $ b\in\mathbb{R} > 1$. Let $ x\in\mathbb{R}$, define $ B(x)$ to the set of all numbers $ b^t$, where $ t\in\mathbb{Q}$ and $ t\leq{x}$. Prove that :

                                                                 $ b^x = \sup{B(x)}$"

Answer: First note that $ x\in\mathbb{R}\implies\exists{t\leq{x}}\backepsilon{t}\in\mathbb{Q}\quad\therefore{B(x)\ne{\emptyset}\,\forall{x}\in\mathbb{R}}$. Next note that due to the restictions on $ b$, $ b^x\in\uparrow$, so the inequality $ t\leq{x}\implies{b^t\leq{b^x}}$ holds. So since $ \forall{t}\in{B(x)}t\leq{x}$ we may state that $ B(x)$ is bounded above by $ b^x$. Therefore $ B(x)$ meets the requirements for its supremum to exist. So finally noting as stated earleir $ t < x\implies{b^t}<{b^x}$ we can state that if $ t\<x$ then $ b^t$ is not an upper bound of $ B(x)$. So we may finally conclude that $ \sup{B(x)} = b^x$

Please any constructive criticism is greatly appreciated.

That's not the correct problem statement.  The point of this problem is to construct the function $ b^x$, and this is the step that does it.  The correct problem statement is that $ x$ is rational.

One general strategy for doing problems of this nature is to prove the inequalities $ b^x \le \sup B(x)$ and $ b^x \ge \sup B(x)$.  The first is clear since $ b^x \in B(x)$.  The second you've got in your second line, but then you restate it with strict inequalities (which doesn't get you the result you want), so you can clean this proof up a bit.  It might also be a good exercise for you to show that $ b^x$ is in fact a strictly increasing function.

[quote="Mathstud28"]No, that is the exact quote from the book.[/quote]

I opened my copy of Rudin just to make sure.  You are wrong and t0r is right.

The point is that $ \sup B(x)$ makes sense for all $ x\in\mathbb{R}$ even when $ b^x$ (at least at this stage) doesn't, so you can take it as a definition.  But it has to at least hold for $ r\in\mathbb{Q}$ to be a good definition!

The exact quote from the third edition is as follows:

"Fix $ b > 1$.
...
c)  If $ x$ is real, define $ B(x)$ to be the set of all numbers $ b^t$, where $ t$ is rational and $ t \le x$.  Prove that

$ b^r \equal{} \sup B(r)$

when $ r$ is [b]rational[/b] (emphasis added).  Hence it makes sense to [b]define[/b]

$ b^x \equal{} \sup B(x)$

for every real $ x$."

Oh, you are correct. I am very sorry I should have verified after you said something. Thank you again.


quadratics conics parabola symmetry function

A projectile is fired at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 100 feet per second.  The height of the projectile is given by
h(x) = (-32x^2)/(100)^2  + x, where x is the horizontal distance of the projectile from the firing point.

(a) How far from the firing point is the height of the projectile a maximum?

(b) Find the maximum height of the projectile.

(c) How far from the firing point will the projectile strike the ground?

this is confusing.......
is the "+ x" part of the denominator or outside of the fraction?

The "+ x" lies OUTSIDE of the fraction.

It is NOT part of the fraction.

Can you do it now?

Here it is:

A projectile is fired at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 100 feet per second. The height of the projectile is given by
h(x) = [(-32x^2)/(100)^2] + (x), where x is the horizontal distance of the projectile from the firing point.

(a) How far from the firing point is the height of the projectile a maximum?

(b) Find the maximum height of the projectile.

(c) How far from the firing point will the projectile strike the ground?

[hide]
this is a quadratic so results in a parabola.
since the x^2 term is multiplied by a negative constant, the parabola opens downward. that means the maximum height of the parabola is its vertex, or whatever you call it. to find the x value of the vertex, use the equation -b/2a where "b" is the constant of the x term and "a" the constant of the x^2 term
plug this in to find the x value to be 625/4, the answer to #2.
plus this value back into the quadratic to get the y value to be 1875/8 which is the answer to #1.
Since this parabola has a vertical line of symmetry, because its a function, the part of the parabola where the projectile is on the ground is on the opposite side of the part where the projectile has landed after its flight, and the part where the projectile is at its highest is at the exact middle, or the vertex. thus the difference of the x value of the very beginning where the projectile is at the ground and the x value of the the maximum height is equal to the difference of the x value from the highest to the x value of when the projectile lands. Since we know the least value of x is 0, the difference to the max height is 625/4 so add another 625/4 to get 625/2, the distance it has traveled before it lands.

I probably made some really dumb mistakes in my reasoning/ calculations, so please verify this.[/hide]

I like all replies, especially the last reply.


[quote="cgyao15"]$ 3\sqrt8 -\sqrt{18}$=?[/quote]

Fixed your $ \text{\LaTeX}$. :)


if  y =  2x +3 and x < 2 which of the following represents all the possible values for y ?

a) y < 7   b)  y > 7  c)  y < 5  d) y > 5 e) 5 < y  < 7

[hide=" click to reveal hidden content"]

(a)

[/hide]


Let a,b,c   >= 0 and a 2 <=b 2 +c 2 , b 2  <= c 2 +a 2 , c 2  <= a 2 +b 2
Prove that :
(a+b+c)((a 2 +b 2 +c 2)(a 3+b 3+c 3) >= 4(a 6+b 6+ c6)

When does the inequality hold?

[quote="RaduSorici"]Is it that hard?[/quote]It doesn't matter. People at not machines to instantly answer your questions ;) Leave them for a few days, and then ask if it is hard or not :P

By the way I solved it before and I got that result that you got and I was surprised too so that is why I posted it and that is why I wrote what I wrote because I thought I solved it incorrectly.  :D

Radu you have to write $4000-5000$ posts and maybe you will reach grobber !
It was time when grobber was a new member ;)

[quote="prowler"]We can try to prove that $ (\sum a^2)^3\geq 4(\sum a^6)$
Knowing that $a^2, b^2, c^2$ are sides of the triangle we get :

$ (\sum a)^3\geq 4(\sum a^3)$  where $ a,b,c $new sides of a triangle.

$ (\sum a)^3\geq 4(\sum a^3 - 3abc)+12abc$

Using $ (\sum a^2) = 2p^2 - 2r(4R+r)$, $ (\sum ab) = p^2 + r(4R+r)$

$8p^3\geq 4*2p*(p^2 - 3r(4R+r)) + 12*4pRr$ or

$24pr(4R+r) \geq 48pRr$ or $4R+r\geq 2R$ c.t.d.

So there is no case of equality.
It is possible that I made a mistake, but where I can’t see now.[/quote]

We can prove that $(a^2+b^2+c^2)^3\geq4(a^6+b^6+c^6)$ in quite a different way.
Using that $(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)$ it remains to prove that:
$(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq(a^6+b^6+c^6)$. And after opening the paranthesis:
$2a^2b^2c^2+a^4(b^2+c^2)+ b^4(a^2+c^2)+ c^4(a^2+b^2)\geq(a^6+b^6+c^6)$,which follows from the given condition!

By the way equality holds when a=b=c=0

[quote="Sailor"]
By the way equality holds when a=b=c=0[/quote]
What about $a=0, b=c$ ?

the case where $a=0,b=c$ or any other means that the triangle is degenerated so $R=r=0$

[quote="Armo"]Does that change anything? ;)[/quote]
No it doesn't,
you are right about the equality case, but I just wanted to show that equality occurs!


Prove that from \[ \begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & 7 & 8\\ \hline 9 & 10 & 11 & 12\\ \hline 13 & 15 & 14 & \\ \hline\end{tabular} \] we cannot arrive at  \[ \begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & 7 & 8\\ \hline 9 & 10 & 11 & 12\\ \hline 13 & 14 & 15 & \\ \hline\end{tabular} \].

This is an old game.. where you can move a number in that space, just it's neighbour as you can see here:

you can move from: $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & 7 & 8\\ \hline 9 & 10 & 11 & 12\\ \hline 13 & 15 & 14 & \\ \hline\end{tabular}$ and get to: $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & 7 & 8\\ \hline 9 & 10 & 11 & \\ \hline 13 & 15 & 14 & 12\\ \hline\end{tabular}$ or to $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & 7 & 8\\ \hline 9 & 10 & 11 & 12\\ \hline 13 & 15 & & 14\\ \hline\end{tabular}$ but not to: $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & 7 & 8\\ \hline 9 & 10 & & 12\\ \hline 13 & 15 & 14 & 11\\ \hline\end{tabular}$.

or another example: from $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & & 8\\ \hline 9 & 10 & 11 & 12\\ \hline 13 & 15 & 14 & 7 \\ \hline\end{tabular}$ you can only make 4 moves:

1) $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & & 4\\ \hline 5 & 6 & 3 & 8\\ \hline 9 & 10 & 11 & 12\\ \hline 13 & 15 & 14 & 7 \\ \hline\end{tabular}$
or

2) $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & 11 & 8\\ \hline 9 & 10 & & 12\\ \hline 13 & 15 & 14 & 7 \\ \hline\end{tabular}$

or

3) $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & 6 & 8 & \\ \hline 9 & 10 & 11 & 12\\ \hline 13 & 15 & 14 & 7 \\ \hline\end{tabular}$

or
4) $\begin{tabular}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 5 & & 6 & 8\\ \hline 9 & 10 & 11 & 12\\ \hline 13 & 15 & 14 & 7 \\ \hline\end{tabular}$.

other moves are not allowed. I hope I was clear enough!

This is famous Sam Lloyd's Fifteen puzzle.

To prove that one position can or cannot get into the other, one needs to associate to each position a permutation and show that every legitimate move corresponds to multiplication by an even permutation. Then all positions are split into two classes depending on the oddness of the corresponding permutation. And it is impossible to tansform position from one class into position from the other class with legitimate moves.

See http://mathworld.wolfram.com/15Puzzle.html

On that page you posted it says that:

///   Stated more simply, if the permutation symbol $(-1)^{i(p)}$  of the list is $+1$ , the position is possible, whereas is the signature is -1, it is not. This can be formally proved using alternating groups. ///

How do you prove it using alternating groups?

Help? Does anybody have a mathmatical proof to how I cannot get from that position to the other?

thank you very much!  :blush:  :blush:

If you want the open square to end up back where it started, an even number of moves is required. Each move is a transposition of the empty square and another square (an odd permutation) so the full sequence is an even permutation. The target configuration is one transposition away- that's odd, so it's impossible.

In the standard configuration, every even permutation (fixing the blank square) is possible; that's harder to prove.


We want to colour all the squares of an $ nxn$ board of red or black. The colorations should be such that any subsquare of $ 2x2$ of the board have exactly two squares of each color. If $ n\geq 2$  how many such colorations are possible?

[quote="trejos"]We want to colour all the squares of an $ n\times n$ board of red or black. The colorations should be such that any subsquare of $ 2\times 2$ of the board have exactly two squares of each color. If $ n\geq 2$  how many such colorations are possible?[/quote]
Just notice that there is a bijection :) : [url]http://www.mathlinks.ro/viewtopic.php?p=877703#877703[/url]

[code]And it should be $n\times n$ not $nxn$.[/code]


There is one exception: If $a_{0}=-1/2$, then $a_{n}=-1/2$ for all $n$.

[quote="Bazooka"]Let $-1<a_{0}<0$ and $a_{n+1}=2a_{n}^{2}-1$ .Prove that  $\exists n$ such that: $a_{n}>0$[/quote]

I know a solution and answer is $a_{0}=-\frac{1}{2}$
Hint :you can prove $|a_{0}+1/2|\leq K_{n}$ and $lim K_{n}=0$