AMC
AIME
geometry
USA(J)MO
USAMO
geometric transformation
reflection

Is it just me or AIME II was harder AIME I?

i was careless on AIME 1 but i thought both were close.  They were both alot hareder than previous years though.

I would argue that AIME II is not necessarily HARDER mathematically. However, the way it is so cleverly designed gives loads and loads of room for counting errors. (What a bummer :-( )

In any case, I opened up the thing and it was like..."When did I sign up for the verbal SAT?"

it was a little wordy and i wouldve liked more pictures and geometry.  if i remember correctly i dont think the first one had too many geometry problems either.   :(

[quote="SirErnest"]it was a little wordy and i wouldve liked more pictures and geometry. [/quote]

A "little" wordy huh? #5 and #6 were each about the size of a paragraph in a history term paper.

[quote] if i remember correctly i dont think the first one had too many geometry problems either.   :([/quote]

You remember incorrectly--it was practically all geometry. ;)

Since when is more geometry better? I for one am unable to think in terms of similar triangles. Had to solve the paper folding problem with a couple of pythagoras equations. They should have more of those questions everybody hates, like #15 on the AIME1 and AIME2. Those are easier =P.

I thought AIME II was harder, but the two tests were roughly comparable in difficulty. However, I only took the II, so that opinion probably has a heavy tilt to it.

AIME 1 was easier, in my opinion...but i dunno, not by much, both were not that hard mathematically, the stupid errors definitely took their toll

AIME1 had 8 geometry questions i believe...AIME2 only had like 3...

[quote="**********"][quote="MithsApprentice"][quote="**********"]
AIME1 had 8 geometry questions i believe...AIME2 only had like 3...[/quote]

let's recount shall we...

1, 3, 7, 11, 12, 13

6 geometry problems... :-)[/quote]

i never claimed to be good at counting, hehe...3 isnt geometry...so ur down to 5...regardless, its still less than the aime1[/quote]

3 is too geometry!...it's 3-D geometry...but still geometry...

I thought the II was a lot harder than the I.

[quote="tokenadult"]A given test taker can only take one AIME per year, so where is everyone getting the impression that one test is harder than the other this year? How would anyone know?[/quote]

What do you mean? The problems for both tests are all available in the two appropriate math jam transcripts.

[quote="tokenadult"]A given test taker can only take one AIME per year, so where is everyone getting the impression that one test is harder than the other this year? How would anyone know?[/quote]

We took one officially and the other unofficially for practice. :-)

Test questions are released 24 hours after the test, meaning that you can post and discuss them and stuff.

[quote="Fierytycoon"]What do you mean? The problems for both tests are all available in the two appropriate math jam transcripts.[/quote]

What I mean is that the experience of taking the test "for keeps" and discussing the other test in a math jam is different. Taking the real test that counts on the real day is always a different experience from taking a practice test. I noticed when people were discussing the AMC 10/12 A and B tests the same phenomenon seemed to occur: most people thought the test that they took was the harder test, if they only took one. In that instance, people could take both the A and B versions of the same test in the same year, as a few AoPSers reported that they did, and I think the consensus is that the overall statistical result matches people's perceptions that the A tests were harder (especially for the 10), but no one is allowed to take both AIMEs for real in one year, so how could anyone match results between the two tests?

[quote="MithsApprentice"]What I did was take the AIME A test just like I would be taking the actual thing. (My official score is the B score) Comparing that score with my actual score, I feel that the B is more difficult. I realize it is probably different taking it as a mock test (probably doing less checking and recalculations) than the actual one. However, I still believe that an almost-fair comparison can be taken that way.[/quote]

I got a dismal 3 on the AIME B (several stupid mistakes, most notably on 14), and I believe that I would have gotten around 8 on the AIME A.  However, I'm Canadian, so this matters little, as I qualified for (and wrote) the CMO earlier this year.  For that matter, I royally messed up the CMO, too. ::shrugs::

I took the AIME1 and skimmed the AIME2, and it definately looked harder.  And I don't think I'm unique in that belief, so if the people who took the 1 and the people who took the 2 thought it was harder, it likely was.  Granted, I'm lazy and hate reading long math problems.  I put off question 5 of the AIME1 (or whatever that one was about the 2 day math competition) until there were <20 minutes at the very end because I didn't want to read the whole thing.

Well, I didn't take either officially, but I looked at most of the problems on each test, and I thought that the second AIME was significantly easier. There weren't too many questions on the second AIME that were just messy computations (more or less) the way there were on the first one. At least, I tend to think of the geometry problems around #10 on the first AIME as being messy computations that I mess up far too often. On the second AIME, only #12 was a messy computation. At least, of the ones I have seen so far, only #12 fits that description.

[quote="tokenadult"]A given test taker can only take one AIME per year, so where is everyone getting the impression that one test is harder than the other this year? How would anyone know?[/quote]

A lot of the people who officially took the II practiced with the I and have some idea of what their score would have been. Also, kids who have been doing this forever have established something of a pecking order which tends to be consistant. If kids A, B, and C consistantly do better than kids X, Y, Z, but on this particular occasion X,Y,Z take the first test and do better than A,B,C taking the second test, it suggests a difference in the tests.

I think describing which AIME was "harder" is a bit misleading.  If you like number theory, the second AIME was easier.

I personally felt like the "middle" problems were a bit harder on the second AIME while the "hard" problems were easier on the second AIME (mostly because I thought problem 15 from the first AIME was enormously time consuming).  It wouldn't surprise me if students who are the kind who attempt 7 or 8 problems thought the second was harder while those who are the kind who attempt 10-15 thought the second was easier.

I personally made several little errors as I worked through the first 2/3 of the second AIME.  I misread at least two problems and made a couple of small algebra mistakes all of which I had to go back and fix.  (This is a lesson to organize your work.  I was not being very organized at all.  From what I can see, organization makes even more of a difference now than back when I was taking the AIME.)

Overall, I think the committee did a great job of balancing the difficulty for the kinds of students who would be fighting for 7 or 8 correct (where the AIME floor is likely to be most years).

I only took the AIME2 as a real test, but...

I think it was easier to mess up on it.
I solved 14 problems but made 4 stupid mistakes - most notably on #5, where i made the same stupid mistake a lot of people made...:(
And #14, where i fluctuated between 100 and 109 only to realize 2 minutes after test was over that it was 108... However, #15 was easier than AIME 1.

On AIME1, problems 1-14 were all about the same difficulty, while 15 was a lot harder.

Ok, well that's not true.  There was a big difference in difficulty among the first 14 problems, though compared to the 15, I suppose they seem relatively equal.

Well, compared to the 2003 test, for example, I think there is no doubt that #1 on 2004 was much harder and #14 was much easier, so the difference between 1 and 14 was significantly less than it has been in previous years.

[quote="MithsApprentice"]What I did was take the AIME A test just like I would be taking the actual thing. (My official score is the B score) [/quote]

I'm at least as confused as tokenadult about this.  I thought you were *required* to take the AIME-I, unless there was an unavoidable conflict, such as your school being closed.  So how could anybody take the AIME-I under real test conditions, but then take the AIME-II for their official score?

[quote="MithsApprentice"][quote="JBL"]I assume by "real test conditions" he means that he put himself in a room for three hours with a pencil, paper, and the test, and acted as if he was taking the exam officially.[/quote]

correct..everything except the actual answer sheet...

It was spring break week...[/quote]

But you're homeschooled, so you can take Spring Break whenever you want.  You took Spring Break during AIME I week on purpose, so that you'd have more time to prepare for AIME II?

[quote="rcv"] I'm at least as confused as tokenadult about this.  I thought you were *required* to take the AIME-I, unless there was an unavoidable conflict, such as your school being closed. [/quote]

Well, you didn't have to provide "proof" that the school was on spring break. The difference in price was an incentive to take the 1st date if you could. Also, planning to do the 1st date gives you a fall-back if someone gets sick or has an emergency.

I had all 5 of my group's qualifiers take the 2nd date because just one of them had an AMC-sanctioned conflict and I didn't want to deal with exam arrangements twice. I feel a little bad about that now, because I think the II date was harder. I had 2 kids who had qualified for the USAMO last year who just missed this year after taking the II. One of them had qualified for each of the past 3 years. I can't help but think they would be in if I had had everyone take the I. Overall, I think it's better to do the first date if you can. Any benefit from extra studying time is trivial.

In last two years  the AIME II is also harder than AIME I.  I think they made the AIME II harder than AIME I.  So taking AIME I is always a good choice.

I thought last year's AIME II was easier than the I. I havent taken a look at this year's II, but I don't think that its ever just "easier" or "harder." Depends on what kind of problems you are better at.

[quote="JP"]The AIME in general is getting easier compare to 90s.  It is harder than those in 80s.  However, there happened to be fewer pictures and more words.  I misread #6 on AIME I and did not understand #14 at all.

Also, the easy problems are getting harder, the hard problems are getting easier.[/quote]

I partially agree with this.  The easier questions are slightly harder now and most of the problems past number 10 are a shade easier.  The geometry seems easier to me, but that might be because I wasn't big on geometry and have learned more since.

The tests in the 90s varied a lot.  If you look at 1994 or 1995 you might think they were hard.  If you look at some of the others, they might look a little easier than of late.

I think that there are now more long word problems that are easier to stumble over and more comp sci algorithmic problems.  I think this reflects the times and makes the AIME a little different, though perhaps not particularly easier or harder.

By looking at the [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2004-aa/04aimestats.html]score distribution[/url], I would conclude that the AIME II was definitely harder. I only took the AIME I.

My theory about that was that since most people will take the AIME1, you'd probably have most of the super-geniuses take the first one and you might have a higher average test-taking ability on the first one, so that could skew the average score somewhat, but looking at the average AMC scores of the people taking AIME (which was [i]higher[/i] on the AIME2, with a significantly lower average AIME score), it does look like the second one was harder. Of course, even with statistics, this is still subjective, because people are better in different specializations. I got a 9 taking the AIME1 in practice (with a 12-13 after adding in questions I solved minus a stupid mistake) and got a 10 on the AIME2, with a 14 after disregarding stupid mistakes.

That's very interesting.  I had just about the same thing.  I didn't make any mistakes on the AIME I test which I took informally, though, and I would've then gotten a 13 if it were official :-(.  But I definitely can't say I'm not satisfied with 11.


Hey, i have a simple calc question for an implicit differentiation problem.

Find the second derivative of y^2=x^3.

The book says the answer is 3x/4y, but I keep getting a more complicated result. I thought this problem would be just another easy one, i guess i must just be making an algebra error of something.

Can someone do the problem showing steps, thanks.

What answer did you get?  What was your method?

i simply differentiated both sides with respect to x to solve for the first derivative and for this i got dy/dx=3x^2/2y.

then i tried to do the derivative of this with the quotient rule, but i got a long answer:

d^2y/dx^2=((6x)(2y)-(3x^2)(2(dy/dx)))(2y)^2

After simplifying this it didnt go down to 3x/4y. I am baffled bc/ implicit differentiation isn't supposed to be hard.

please help

Think about what the second derivative is: the derivative of the first derivative.

[tex]y^2=x^3[/tex]
[tex]\frac{d}{dx}(y^2)=\frac{d}{dx}(x^3)[/tex]
[tex]\frac{d}{dx}(y^2)=?[/tex]
let [tex]g(x)=y[/tex] and [tex]f(x)=x^2[/tex].  So, [tex]\frac{d}{dx}f(g(x))=f'(g(x)) \cdot g'(x)=2y \cdot \frac{dy}{dx}[/tex]

So now you have: [tex]2y \cdot \frac{dy}{dx} = 3x^2[/tex]

Solve for [tex]\frac{dy}{dx}[/tex] to get: [tex]\frac{dy}{dx}= \frac{3x^2}{2y}[/tex]

Now take the derivative of both sides, taking the derivative of the first derivative which is dy/dx:

[tex]\frac{d}{dx}(\frac{dy}{dx})=\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{3x^2}{2y})[/tex]

What is [tex]\frac{d}{dx}(\frac{3x^2}{2y})[/tex]?

Apply the quotient rule, let [tex]f(x)=3x^2[/tex] and [tex]g(x)=2y[/tex], since:

[tex]\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}[/tex]

You get:

[tex]f'(x)=6x[/tex] and [tex]g'(x)=2 \cdot \frac{dy}{dx}= 2 \cdot (\frac{3x^2}{2y})[/tex].

Thus:
[tex]\displaystyle\frac{d^2y}{dx^2}=\frac{(6x)(2y)-(3x^2)(\frac{3x^2}{y})}{4y^2}[/tex]

Simplify the RHS to:

[tex]\displaystyle\frac{d^2y}{dx^2}=\frac{12xy^2-9x^4}{4y^3} [/tex]

Hmm, so either I am wrong or your textbook is, and I think we both know where the error most likely lies =).

PS It's summer, why are you learning calculus?  I ask b/c I am doing the same thing.

ya i noticed quite a few, i dont wanna say anything until i give u a chance to fix it tho

So basically all we differ on is the exponent of y on the bottom?

Okay, I realized I added wrong.  I agree w/what you said.

ya, i got cubed because u multiply by y/y to get rid of the fraction in the numerator, which should make 4y^3, not 4y^5 in the denominator.

thanks for the confirmation of my answer tho, im takin calc in a crappy community college over the summer, and i have im leavin before the class ends so it might as well be an independent study, and i guess in just not so confident of my calc skills yet. thanks

Hm, let me give it a try. Keep in mind that I hate calculus. ;)

[tex]\displaystyle 2y{{dy}\over{dx}}=3x^2[/tex], so
[tex]\displaystyle{{dy}\over{dx}}={{3x^2}\over{2y}}={{3}\over{2}}x^2y^{-1}[/tex]
[tex]\displaystyle{{d^2y}\over{dx^2}}={{3}\over{2}}\left(2xy^{-1}+x^2(-1)y^{-2}{{dy}\over{dx}}\right)[/tex]; substitute in dy/dx to get
[tex]\displaystyle ={{3}\over{2}}\left({{2x}\over{y}}-{{x^2}\over{y^2}}\left({{3x^2}\over{2y}}\right)}\right)[/tex]; substitute in y2 to get
[tex]\displaystyle ={{3}\over{2}}\left({{2x}\over{y}}-{{x^2}\over{x^3}}\left({{3x^2}\over{2y}}\right)}\right)[/tex]
[tex]\displaystyle ={{3}\over{2}}\left({{2x}\over{y}}-{{3x}\over{2y}}\right)}\right)[/tex]
[tex]\displaystyle ={{3x}\over{4y}}~.[/tex]

[quote="brian216"]i got ((12xy^2)-(9x^4))/4y^3[/quote]

Why don't you sub in y2 = x3? That changes the expression into 3x/4y as required...

[tex]\frac{d^2y}{dx^2}=\frac{12xy^2-9x^4}{4y^3}[/tex]
[tex]\frac{d^2y}{dx^2}=\frac{12x(y^2)-9x^4}{4y(y^2)}=\frac{12x^4-9x^4}{4x^3y}=\frac{3x}{4y}[/tex]

Why do you get rid of some y's but not others?  Both answers are right...

Brian I am taking calculus over the summer too, at University of Rhode Island...  You class might not be that great since we did implicit differentiation on second day.  Either that or my class is just super fast.  Anyway, good luck!  If you are trying to skip out of HS calc I would study a little on your own.

Well, I can change the original equation into y = x3/2, but then it wouldn't be implicit differentiation any more. So I just change the even powers of y in hope that it yields an expression that looks nice. Comparatively, 3x/4y is obviously simpler than (12xy2 - 9x4) / 4y3.

Of course I can change the remaining y too, but then I get a fraction in my exponent.

ya, i knew i was just being stuid algebra-wise. i forgot about substituting from the original expression.

thanks guys

You can do this without the quotient rule, as well:

[tex]y^2 = x^3[/tex]
[tex]2y\frac{dy}{dx}=3x^2[/tex]
Then we can differentiate that again, to get
[tex]2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2}=6x[/tex]

Then we can substitute in for [tex]\frac{dy}{dx}[/tex] and solve that.  It makes for the limiting of fractions, which usually makes life easier.


What's the best way to incorporate Geometer Sketchpad images into latex files? I can't import .gs pictures directly, so is there some gs-->.png converter?

See Valentin's answer at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=339812#339812


linear algebra
matrix
function
algebra
rational function

use cramers rule to solve the following problem

2x-y=4
3x+y-z=10
y+z=3

\[ \\2x-y+0z=4\\3x+y-z=10\\0x+y+z=3  \]
$x=\frac{\left|\begin{matrix}4&-1&0\\10&1&-1\\3&1&1\end{matrix}\right|}{\Delta}$  $y=\frac{\left|\begin{matrix}2&4&0\\3&10&-1\\0&3&1\end{matrix}\right|}{\Delta}$  $z=\frac{\left|\begin{matrix}2&-1&4\\3&1&10\\0&1&3\end{matrix}\right|}{\Delta}$  $\Delta=\left|\begin{matrix}2&-1&0\\3&1&-1\\0&1&1\end{matrix}\right|$

And the result follows.

Masoud Zargar

In practice, it is essentially always faster to use Gaussian elimination. What is Cramer's rule actually good for? Theoretical applications: the solution is a rational function of the various coefficients, which we could write down if we needed to.

Cramer's rule is handy to have when the size of the matrix is very small but the coefficients are not simple - I'm thinking in particular of cases in which the coefficients are algebraic expressions rather than numbers. But I think my definition of "very small" stops at $3\times 3$ unless the matrix is quite sparse.


A nice one:
Fp={M s.t. M^p=In} where M are n*n real matrices.
Find the accumulation points of Fp.

In $M_n(C)$ think matrices with eigenvalues of modulus 1

That's pretty clear, but we have to determine which ones. I'm not really sure (this is sort of new to me), but I think we can even say that the eigenvalues of the matrices in the adherence are roots of $x^p-1$. Is this true?

He did mention they were "real matrices", so it's $M_n(\mathbb R)$.


first of all, there are four parts that are overlapped by two strips.  Each of them are 1 by 1 squares.  so, we have some area, which is going to be $1+1+1+1 = 4$.

Now, since we have taken out the overlapped areas, each of the strips are going to decrease its length by $2$, since $1$ has been taken out from each ends.  therefore, the final area is $4((10-2)(1)) + 4 = 32 + 4 = 36$.

thus, the answer is $A$.

[hide]
$A$, since they overlap perpendicularly, the area lost is $1+1+1+1=4$
Since each of the strips area is 10, $4(10)-4=40-4=36$.[/hide]


number theory
greatest common divisor
number theory proposed

aind all triple $(a,b,c)$ such that:
$a^2b,b^2c,c^2a|a^3+b^3+c^3$

I'm wondering if $a,b,c$ are positive integers or not ?

Note that $(a, b, c)$ works iff $(ka, kb, kc)$ works, so consider only triples $(a, b, c)$ with their $GCD$ equal to $1$. We'll show that any such triple not only has $(a, b, c)=1$ but also $(a, b)=(b, c)=(a, c)=1$. Indeed, suppose two of them are divisible by some prime $p$, say $a, b$. Then $p \mid a^2b \mid a^3+b^3+c^3$, and since $p \mid a, b$ we have $p \mid c^3 \Rightarrow p \mid c$ contradicting $(a, b, c)=1$.

Then $a, b, c$ are pairwise coprime. The conditions given imply $a^2, b^2, c^2 \mid a^3+b^3+c^3$ which in turn implies $(abc)^2 \mid a^3+b^3+c^3$ (since $a, b, c$ are pairwise coprime), and if this is satisfied then the original condition is too, so the problem becomes symmetric in $a, b, c$. Then WLOG assume $a \leq b \leq c$. Then $a^2b^2c^2 \leq a^3+b^3+c^3 \leq ac^2+bc^2+c^3$, so $a^2b^2 \leq a+b+c$. Thus $c \geq a^2b^2-a-b$.

Now, remember that $c^2 \mid a^3+b^3$.

Suppose $a \leq 2$. Then $c \geq 4b^2-2b \geq 2b^2$, so a number that is at least $4b^4$ divides $a^3+b^3$ which is at most $2b^3$, absurd. Then $a=1$. We now have $c \geq b^2-b-1, c^2 \mid b^3+1$. However, for $b \geq 4$, it's easy to see that $c^2>b^3+1$. Then $b \leq 3$.

If $b=1$ we have $c^2 \mid c^3+2$, so $c^2 \mid 2, c=1$, and $(1, 1, 1)$ works. If $b=2$ we have $4c^2 \mid c^3+9$. Then $c^2 \mid c^3+9$, so $c^2 \mid 9$. Since $a \leq b \leq c$ we have $c \leq 2$ so $c=3$, and $(1, 2, 3)$ works. If $b=3$ we have $9c^2 \mid c^3+28$, so $c^2 \mid c^3+28, c^2 \mid 28$. The largest square dividing $28$ is $4$ so $c \leq 2<b$, absurd.

Thus all solutions are the permutations of $(n, n, n)$ and $(n, 2n, 3n)$ for all $n$.


Find a relativley simple formula for

$\cos(2x)\cos(3x)\cos(4x)\dots\cos(nx)$ for $n\in\mathbb{N}$.

hello, Mathematica gives the following answer
$ \prod_{k\equal{}0}^n\cos(ix)\equal{}2^{\minus{}n\minus{}1} e^{\frac{1}{2} i n (n\plus{}1) x} \sec (x) \left(\minus{}1;e^{\minus{}2 i
   x}\right){}_{n\plus{}1}$
Sonnhard.


integration
linear algebra
matrix
function

[b] Let [/b] $ A \in M_{n} \mathbb{(R)}$ , $ x \in R$ , [b] x >0 [/b] ;
[b]Prove that : [/b]
 $ \int_{0}^{\infty} {(A+xI)^{n}.e^{-x}.dx} = n! (I+\sum_{k=1}^{n} \frac {A^{k}}{k!})$
[b]I is identity nxn matrix !!![/b]
[b]"..Do math , love math to love native place , love country.."[/b]

$ \int_{0}^{\plus{}\infty} {x^{n}e^{\minus{}x}}\equal{}n!$
[b]I am sorry because I do'nt write gamma function[/b]


[quote="Ligouras"]Let $ a, b, c$ positive reals such that $ 0 \leq a, b, c \leq 1$. Prove that

$ (a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2)\geq (a^2 \plus{} b^2 \plus{} c^2 \minus{} 1)^2$[/quote]
I am sorry, but your inequality is not true for $ a\equal{}b\equal{}c \rightarrow 0.$

By the way, HAPPY NEW YEAR!


$AH$ is a height of $\triangle ABC$ and $AD$ is a bisector of $\angle BAC$. Let $E$ be a projection of $D$ onto $AC$. $AB$ and $EH$ intersect at point $N$. Let $M$ be the midpoint of $BC$. $MN$ is perpendicular to $AD$. Given $\angle BAC=\alpha$ find $\angle ABC$.

The answer is (without a proof):
[hide]
$\angle ABC = 90^{\circ} - \frac{\alpha} {4}$
or
$\angle ABC = \angle BAC = 45^{\circ}$
[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose G is a group of order $ 5203$ which equals $ 11^2*43$.

Need to find values of $ n_{11}$ and $ n_{43}$ and find out if G is simple.

The factors of $ 5203$ are: $ 1, 11, 43, 121$ and $ 5203$.

By Sylow's theorems: $ n_p$ divides $ 5203$ and $ n_p \equal{} 1$ (mod p), i.e. $ n_{11} \equal{} 1$ (mod 11) and $ n_{43}\equal{}1$ (mod 43)

Then $ n_{11}$ and $ n_{43}$ both equal $ 1$.

Is this correct or have I missed anything?

Thank you for your help!

[quote="SD5225272"]All good.  All groups of this order are abelian.

Steve[/quote]

Thank you!


function
geometry
geometric transformation
reflection
geometry proposed

Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Find all positive integers $n$ such that $PA^{n}+PB^{n}+PC^{n}$ does not depend upon $P$.

When $P = A, B, C$ the function $f(n) = PA^{n}+PB^{n}+PC^{n}$ takes on the value $2s^{n}$ where $s$ is the side length of the triangle.

On the other hand, when $P$ is midway between any two of $A, B, C$ then $f(n)$ takes the value $(2+2^{n}) r^{n}$ where $r$ is the circumradius.

Clearly if we are to have $2s^{n}= (2+2^{n})r^{n}$ then $\frac{s}{r}= \sqrt{3}= \sqrt[n]{1+2^{n-1}}$ which holds for $n = 2, 4$ and nothing else (I think).

t0rajir0u,

I know $n=4$ also holds, value of $PA^{4}+PB^{4}+PC^{4}$ is $18R^{4}$. I will check your solution later :)

Surely Math Relec =Mathematical reflection? you should write clearly your source! :D This is a problem in Mathematical reflection

This problem is proposed by Oleg Mushkarov.

I have known the reverse of this problem from Japanese university entrance exam as follows, so I have studied the very Our problem before.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=754146#754146]Here is a Link.[/url]

[quote="N.T.TUAN"]t0rajir0u,

I know $n=4$ also holds[/quote]

Sorry, I made an error.  It's been corrected.   :oops:

[quote="gemath"]This is a problem in Mathematical reflection[/quote]
Yes, my good friend :D

Hint
If $f(P)=PA^{n}+PB^{n}+PC^{n}$ does not depend $P$ forall $P\in (O)$ then $f(A)=f(M)\Rightarrow n=2$ or $n=4$ with $M$ is midpoint of small arc $BC$
If $n=2 ,n=4$ then really $f(P)=const$  :)


factorial
search
modular arithmetic
number theory unsolved
number theory

It has been posted recently. Search for it :wink:

[quote="sax"]2^n = a ! + b ! + c ![/quote]

Wlog say $ a\geq b\geq c$ Then $ c! |2^n$ and so either $ c \equal{} 1$, either $ c \equal{} 2$

1) $ c \equal{} 1$
$ n > 0$ and so $ a! \plus{} b!$ must be odd and so $ b \equal{} 1$ and we want $ 2^n \equal{} a! \plus{} 2$
If $ a\geq 4$, RHS$ \equal{} 2\pmod 4$ and so $ n \equal{} 1$ which is impossible.
So $ a < 4$ and the only solutions $ 8 \equal{} 3! \plus{} 1! \plus{} 1!$ and $ 4 \equal{} 2! \plus{} 1! \plus{} 1!$

2) $ c \equal{} 2$
So $ 2^{n \minus{} 1} \equal{} \frac {a!}{2} \plus{} \frac {b!}{2} \plus{} 1$
If $ b\geq 4$, RHS$ \equal{} 2\pmod 4$ and so $ n\leq 2$ which is impossible. So either $ b \equal{} 2$, either $ b \equal{} 3$

2.1) $ b \equal{} 2$
$ 2^{n \minus{} 1} \equal{} \frac {a!}{2} \plus{} 2$
If $ a\geq 6$,  RHS$ \equal{} 2\pmod 4$ and so $ n\leq 2$ which is impossible. So $ a\in\{2,3,4,5\}$  and no solution

2.2) $ b \equal{} 3$
$ 2^{n \minus{} 1} \equal{} \frac {a!}{2} \plus{} 4$
If $ a\geq 6$, RHS$ \equal{} 4\pmod 8$ and so $ n\leq 3$ which is impossible. So $ a\in\{3,4,5\}$ and two solutions : $ a \equal{} 4$ and $ a \equal{} 5$

3) Synthesis of solutions :
$ 4 \equal{} 2! \plus{} 1! \plus{} 1!$
$ 8 \equal{} 3! \plus{} 1! \plus{} 1!$
$ 32 \equal{} 4! \plus{} 3! \plus{} 2!$
$ 128 \equal{} 5! \plus{} 3! \plus{} 2!$

[quote="sax"]thanks, but why c! divides 2^n?[/quote]

uhhh, this is the first line of my demo. I hope you'll not encounter problems at each line  :wink:

Since $ a\geq b\geq c$, we know that $ c! | a!$ and $ c! | b!$ and so $ c! | a!\plus{}b!\plus{}c!$ and so $ c!|2^n$


analytic geometry
graphing lines
slope
trigonometry

[quote="lomos_lupin"]Two slopes with the Name of $M$ and $m$ (With masses $M$ ane $m$)are on each other . There is No friction at all , Anywhere. ( Look at the Picture)

What is the accerelation of $M$ at the time zero ?

PS. Please note that the usual approach of decomposing the gravity force doesnt work.[/quote]

Hmm.. why not? Isn't the force just $mg \cos t \sin t$ where $t$ is the angle of the slope M, which means the acceleration is $\frac{m}{M}\sin 2t$?

Not exactly, object $m$ is accerelating itselt, the force it exerts on object $M$ doesnt decompose like that.

Thats would have worked if we had friction.

this problem is unsolvable because the center of mass of m needs to be known because it certain positions torque would be caused but at others it wouldnt.  If there is no torque, the force on $m$ is $mg\sin(\theta)$ and the force on $M$ is $\frac{mgsin(2\theta)}{2}$ This can only be done if one knows the angle of the incline of $M$

the answer
$a_{M}=\frac{mgcosa sina}{M(1-\frac{m}{M}sin^{2}a})$
no tongue. :huh:


limit
logarithms
real analysis
real analysis unsolved

Here is a limit, for those interested in challenges! :)
$\lim_{n\to\infty}\sqrt[n^{2}]{\binom{n+1}{1}\binom{n+2}{2}\binom{n+3}{3}... \binom{2n}{n}}$.


LaTeX
inequalities
inequalities unsolved

if any one can rewrite this exercice with LaTEX
THNKS

[color=red]Note by Megus: I rewrited it, as you asked but next time you won't be so lucky -try to learn LaTeX - it's surely not hard[/color]

It is just AM-GM: above inequality is equivalent to following:

$\sqrt[2n+1]{(n+1)^{n+1}(n-1)^n} < \frac {n(2n+1) +1}{2n+1} < n$

As $n$ is integer, we have : $1-\frac{1}{n^2}>=0$ and  $(1+\frac{1}{n})>=0$.

We can thus use the AM-GM :

$\sqrt [n+1] {(1-\frac{1}{n^2})^n  (1+\frac{1}{n})}=< \frac {n-n*\frac{1}{n^2}+1+\frac{1}{n}}{n+1}=1$

Then $(1-\frac{1}{n^2})^n  (1+\frac{1}{n})=< 1$

But we have the equality only if $1-\frac{1}{n^2}=1+\frac{1}{n}$ only if $n=-1$ and we check that this solution is false. Thus :

$(1-\frac{1}{n^2})^n  (1+\frac{1}{n})< 1$


Consider $ 9$ positive integer numbers which are divisors of $ 30^{2009}$ .
Prove that one can find two of them such that their product a square .

$ 30^{2009}\equal{}2^{2009}3^{2009}5^{2009}$. Pick a set of 9 divisors of this number. They will each be of the form $ 2^a3^b5^c$. By the Pigeonhole Principle, there will be a subset of 5 numbers such that, for every element, a has the same parity. Any two of these numbers multiplied will be of the form $ 2^{2x}3^y5^z$. Similarly, within this set, there will be a subset of 3 numbers such that, for every element, b has the same parity. Any two of these numbers multiplied will be of the form $ 2^{2x}3^{2y}5^z$. Finally, within this set, there must be a pair of numbers such that, for both numbers, c has the same parity. These two numbers multiplied will be of the form $ 2^{2x}3^{2y}5^{2z}$, which is a perfect square, as desired.

[hide="Slightly different but same concept"]
There are 2 cases for the parity of a, for the parity of b, and for c, for a total of $ 2^3\equal{}8$ so out of 9, two must be the same by the Pigeonhole Principle[/hide]


in this game there are 3 rows one with 1 stick, one with 3, and the other with 5.
[hide="for you visual people it looks like this"]

|
|||
|||||
[/hide]
you can remove as many sticks from one row as you would like during your turn. the point is to avoid removing the last stick.
describe a winning stategy if you go first.

[hide="here is what i have"]
if you remove three stick from the row with 5 then it is always possible to win. I dont know how to prove this with math or anything i just figured it out with trial and error[/hide]

Two players.  And this game is more commonly called Nim.

Yes I have heard of NIM too.
I think it is also possible to win removing one stick from row 2. Again I did not do mathematical calculations, but checked the actual possibilities.

If he picks up the other two, you pickup the entire row 3 and win.
If he picks up one of them, you pick up four from row3 and win again.
If he picks up three from row3, you pickup row1 and win.
If he picks up four from row3, you pick up one from row2 and win.
If he picks up the entire row3, you pick up the rest of row2 and win.
If he picks up row1, you pick up three from row3 and win.

and so on.

No no this is not done. It was wrong. Sorry!  :blush:

we could play this game on aim or sometin

ill teach it to mi friends

it sounds fun

[quote="pascal_1623"]I know a game like this. You count up to 100, whoever gets to 100 first wins. But, you can only count 4 numbers at a time (e.g. 1, 2, 3, 4).[/quote]

[hide="And the strategy..."]
If the other person says x, you say 5-x. Since 100 is divisible by 5, the second player always wins.
[/hide]

first player wins if you are allowed to count max n at a time where n + 1 is not divisible by 100.

[url]http://www.transience.com.au/pearl3.html[/url]

Try this game, it's similar.

Ouch, is there the limit of the levels?, I reach level 11 (100%) and it really hurts to make a lot of LP (Losing position) (Four piles, I wonder if the highest level consist of 7 piles  :maybe: )

Probably not, computers can generate stuff like that.

However, I'm pretty sure there is some modeleing thing you can do.

If you have two piles left, you just have to make them equal, and you win.


function
limit
real analysis
real analysis solved

Let $f$ be a continious function from $\mathbb{R}\to\mathbb{R}$ such that $|f(x)-f(y)|\geq|x-y|$ for all $x,y$. Prove that the range of $f$ is all of $\mathbb{R}$

$f$ is injective, so assume WLOG it's strictly increasing. if it has a real limit $\ell$ when $x$ goes to $\infty$, then for some $\varepsilon>0,c>0$ we have $|f(x)-f(y)|<\varepsilon$ whenever $x,y>c$, which is a contradiction. This means that $\lim_{x\to\infty}f(x)=\infty$. In the same way we show that $\lim_{x\to-\infty}f(x)=-\infty$.

[quote="alekk"]what about the following pb:
let $K$ be a metric compact set.
Let $f$ be a continuous function from $K$ to $K$ such that $d(f(x);f(y)) \geq d(x,y)$ for all $x,y$ in $K$. Is $f$ a bijection ?[/quote]

Yes, I think it does follow that $f$ is surjective (and, of course, it's also injective).

Let $K_n=f^{(n)}(K)$. $(K_n)_n$ is a descending (i.e. $K_{n+1}\subseteq K_n$) sequence of compact sets, so $\cap_{n\ge 0}K_n$ is non-void. Call that intersection $F$. $F$ is a compact subset of $K$ satisfying $f(F)=F$. Assume now that $F\ne K$. There is a point $x\in K\setminus F$ s.t. $d(x,F)=t>0$. We have $d(f^{(n)}(x),f^{(n)}(F))=d(f^{(n)}(x),F)\ge t,\ \forall n\ge 0$. However, $f^{(n)}(x)\in K_n,\ \forall n\ge 0$. Let $F'$ be the set of points which have distance $\ge t$ from $F$. $F'$ is a compact subset of $K$. From what we have above we see that $K_n\cap F'\ne\emptyset,\ \forall n\ge 0$. This means that $F\cap F'\ne\emptyset\ (*)$, which is absurd ($(*)$ follows from $F\cap F'=\bigcap_{n\ge 0}(K_n\cap F')$).
Is it Ok? :?

Now, to finish this topic, prove that $f$ is actually an isometry, i.e., $d(f(x),f(y))=d(x,y)$ for all $x,y\in K$. :)

There is a proof of this in post number $12$, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=isometry&t=18269]here[/url]. Is your solution different, fedja?


Let $ a,b,c$ be positive real numbers *edited  :wink:  *.

Prove that

$ \sqrt {\frac {\left(\sum \frac 1 a \right)\cdot \left(\sum a\right)}{2}}\geq \sum \sqrt [4]{\frac {a^2}{(a \plus{} b)(a \plus{} c)}}$

[quote="Georg-A."]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$.

Prove that

$ \sqrt {\frac {\left(\sum \frac 1 a \right)\cdot \left(\sum a\right)}{2}}\geq \sum \sqrt [4]{\frac {a^2}{(a \plus{} b)(a \plus{} c)}}$[/quote]
The condition $ abc\equal{}1$ is irrelevant..
$ LHS \ge \sqrt{\frac{9}{2}}$ by cauchy-schwarz.
And using $ \sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z} \ge \sqrt{3}\sqrt{x\plus{}y\plus{}z}$ on the $ RHS$ we only have to prove:
$ \sqrt{\frac{9}{2}} \ge \sqrt { 3\sum_{cyc} \sqrt{\frac{a^2}{(a\plus{}b)(a\plus{}c)}}}$
But $ \sum_{cyc} \sqrt{\frac{a^2}{(a\plus{}b)(a\plus{}c)}} \equal{} \sum_{cyc} \sqrt{\frac{a}{a\plus{}b} \cdot \frac{a}{a\plus{}c}} \le \sum_{cyc} \frac{\frac{a}{a\plus{}b}\plus{}\frac{a}{a\plus{}c}}{2} \equal{} \frac{3}{2}$, and we are done :)

[quote="Mathias_DK"]
The condition $ abc \equal{} 1$ is irrelevant..
[/quote]

Of course you`re right - i had changed the inequality not noticing that the condition becomes unnecessary :blush:

Anyway, thanks for your solution :)


there are N points in the plane such that the total number of pairwise distances of these N points is at most n. Prove that $ N <\equal{} (n \plus{} 1)^2$.

We shall have maximum number of pairwise distances if all the points on the plane are distinct. Hence we shall have $ N\choose2$ distances. By the condition given, $ N\choose 2$=n
So we have to prove  $ N<\equal{}$ $ N\choose 2$ $ N\choose 2$ + 1 + 2$ N\choose 2$ and this I feel is trivial.

I don't come here too often, so I haven't been able to keep an eye. I hope you didn't post this before the deadline, mnrc_2.

Assume $ N > (n\plus{}1)^2$. Suppose $ N \equal{} (n\plus{}1)^2 \plus{} 1$.

Consider a point P out of those given, which is a "boundary" point, i.e. a line through P has all the other $ (n\plus{}1)^2$ points on one side of it (some may be on the line itself).

Apply the pigeon hole principle to prove that at least n + 3 of these remaining points must lie at the same distance from P. Obviously, from the given and the assumed conditions, these points all lie on a semicircle with center P.

It is now easy to show that these last n + 3 points define at least n + 2 distinct distances among them, which is a contradiction.

The same reasoning works for $ n^2\plus{}n$ instead of $ (n\plus{}1)^2$.

Note that a boundary point is just a vertex of the convex hull.

Pierre.

Yup, it is. Yes, and I forgot to mention the rather obvious fact that whenever you prove it for $ N \equal{} (n\plus{}1)^2 \plus{}1$, you can obviously prove it for any N greater than that. That's VERY trivial.

But, as far as I know, $ n^2\plus{}n$ is not greater than $ (n\plus{}1)^2$ :wink:

Pierre.

Well, lol, I was referring to MY earlier solution, not yours  :wink:

Because I had forgotten to extend my proof to cover all possible N's given in the problem, so ...

Yes, of course the reasoning holds true for any N greater than $ n^2 \plus{} n$.

Rik

i don't understand any of the 2 proofs. please elaborate a bit further?

Ah, that sweet Erdos distance problem again! :). Alex Iosevich put a nice expository paper about it online: [url]http://www.math.missouri.edu/%7Eiosevich/erdoshs.pdf[/url]


A circle is inscribed in a triangle. If the area of the triangle is 15 units squared and the area of the circle is 25pi, what is the area of the part of the triangle that is not blocked by the circle?
(I wish I had a diagram. :noo:)

note: there is a few typos,

[quote="007math"]How can you inscribe a circle with an area of $ 25\pi$ into a triangle with an area of $ 15$?   :spam: <-Exactly.[/quote]

Oops, typo. It's supposed to be triangle inscribed in circle!
And what you're caculating is the part of the circle that is showing from behind the triangle. :noo: I hate typos

In that case, wouldn't it just be $ 25 \pi \minus{}15$?


The equiangular convex hexagon $ ABCDEF$ has $ AB \equal{} 1$, $ BC \equal{} 4$, $ CD \equal{} 2$, and $ DE \equal{} 4$. The area of the hexagon is

$ \textbf{(A)}\ \frac{15}{2}\sqrt{3}\qquad
\textbf{(B)}\ 9\sqrt{3}\qquad
\textbf{(C)}\ 16\qquad
\textbf{(D)}\ \frac{39}{4}\sqrt{3}\qquad
\textbf{(E)}\ \frac{43}{4}\sqrt{3}$

[hide="Click for solution"]
We extend $ \overline{FA}$ and $ \overline{CB}$ to meet at $ X$, $ \overline{BC}$ and $ \overline{ED}$ to meet at $ Y$, and $ \overline{DE}$ and $ \overline{AF}$ to meet at $ Z$.  The interior angles of the hexagon are $ 120^\circ$, so the triangles $ \triangle XYZ$, $ \triangle ABX$, $ \triangle CDY$, and $ \triangle EFZ$ are equilateral.  Since $ AB=1$, $ BX=1$.  Since $ CD=2$, $ CY=2$.  Thus, $ XY=7$ and $ YZ=7$.  Since $ YD=2$ and $ DE=4$, $ EZ=1$.  The area of the hexagon can thus be found by subtracting the areas of the three small triangles from the area of the large triangle:

$ 7^2 \left (\frac{\sqrt{3}}{4} \right ) -1^2 \left (\frac{\sqrt{3}}{4} \right )-2^2 \left ( \frac{\sqrt{3}}{4} \right )-1^2 \left (\frac{\sqrt{3}}{4} \right )=\frac{43\sqrt{3}}{4}$, or answer choice $ \boxed{\textbf{(E)}}$.
[/hide]

Or, you can inscribe the hexagon in a rectangle whose sides contain AB and DE, and the other two sides contain C and F. Let the corners of the rectangle be G,H,J, and K, where GEF,DHC,CJB, and AKF are 30-60-90 right triangles. Let GE=x, AK=y. Since GH=JK, we have
$ x \plus{} 2 \equal{} y$
Also, $ GF \equal{} \sqrt {3} \cdot x$ and $ FK \equal{} \sqrt {3} \cdot y$, so since GK=HJ, we have
$ x \plus{} y \equal{} 3$
Solving, x=$ \frac{1}{2}$ and AK=$ \frac {5}{2}$. The area of the hexagon is the area of the rectangle minus the area of the four right triangles, which is $ \frac {43\sqrt {3}}{4}$


I just looked at this forum, and I know I posted the 20,000th user topic, where in the world is it?????????? Why did 10,000th user replace it?

I haven't seen it but maybe it's because 10000th User's was posted a while ago? Does it even matter? Also, is it really necessary to make a topic over something stupid like this?

There is already a thread discussing the 20,000th user. hello, please address future questions like this to me via PM.


inequalities
logarithms
inequalities solved

Oops. :blush:  I can't use Muirhead now.

Well, I think we can use the most basic method...
Assume $x_1 \leq x_2 \leq \cdots \leq x_n$. Then tranform the inequality into: $ (\frac{x_i}{x_{i+1}})^{(x_i-x_{i+1})}$ or sth like that....
Can we prove it like that?

I am pretty sure we can, yes.

Anyway, the Chebyshef is also very nice... suppose WLOG that the given numbers are in decreasing order, then by Chebyshef, since

\[\ln{x_1} \geq \ln{x_2} \geq ... \geq \ln{x_n}\] we have

\[x_1 \ln{x_1} + x_2 \ln{x_2} + ... + x_n \ln{x_n} \geq \frac{1}{n}\left(x_1 + x_2 + ... + x_n\right)\left(\ln{x_1} + \ln{x_2} + ... + \ln{x_n}\right)\]

and hence \[e^{\textrm{RHS}} \leq e^{\textrm{LHS}}\] from which the result follows.

Or Jensen. By the way this is a well-known inequality for sure. ;)


MATHCOUNTS
percent
number theory
prime numbers

What percent of two-digit positive integer is prime?

This is just to get you thinking about memorizing all the prime numbers.

It's even handier to know which numbers ``look prime'' but aren't: the two most important of those are probably 91 and 301.

Yup yup, correct, just to get you thinking...just watch out for those divisible by 3 or 7 and you should be fine for all prime under 100. 3, 7, 11 and 13 if under 200 :)


how many $\alpha \in [0,\frac{\pi}{2}]$ satisfied the equation $\cos \alpha \cos 2\alpha \cos 4\alpha=\frac18$?

Take the graph .I think  there are three roots of this equation.

The double-angle identity $\sin 2x = 2 \sin x \cos x$ can help.  Multiplying both sides of the original equation by $8 \sin \alpha$ gives \[\sin \alpha (2 \cos \alpha) (2 \cos 2\alpha) (2 \cos 4 \alpha) = \sin \alpha \, .\] By applying the double-angle identity three times, the left-hand side is $\sin 8 \alpha$.  So we get the equation \[\sin 8 \alpha = \sin \alpha \, .\] From here, it should be smooth sailing.  We have to remember that $\alpha = 0$ is an extraneous solution.

[quote="Ravi B"] \[\sin 8 \alpha = \sin \alpha \, .\] [/quote]
Give $7\alpha =2\pi k, k\not=0$ or $9\alpha =\pi (2k+1)$. Therefore 5 solutions.

Basically, after you get sin(8a) = sin(a), you need to find the values in the interval such that 8a = a + 2n(pi)

[quote="Ravi B"]Don't forget about the supplementary case $8 \alpha = \pi-\alpha+2n\pi$.  That's why Rust wrote two equations.[/quote]
Thank you all for the nice proof. :)


integration
real analysis
real analysis unsolved

[quote="joao1"]Consider:

3(3 + x^2)(dx/dt) = 2((1 + x^2)^2)e^(-(t^2)).

If x(0) <= 1 show that there is a M > 0 st |x(t)| < M for t >=0.[/quote]

Solved the differential equation we got:
$3arctgx+\frac{3x}{2(1+x^{2})}=\int_{0}^{t}{e^{-y^{2}}}dy+K$ so
couse $x(0)\le 1$ we have $3arctgx+\frac{3x}{2(1+x^{2})}\le \int_{0}^{t}{e^{-y^{2}}}dy+\frac{3(\pi+1)}{4}\le \frac{3\pi+2\sqrt{\pi}+3}{4}$ where from we got that $|x(t)|<M$


algebra
polynomial
superior algebra
superior algebra unsolved

For any prime $ p$, find an irreducible polynomial in $ \mathbb{Q}[x]$ of degree $ p$ that has exactly $ q$ real roots, where $ q \in \{0, ..., p \}$.

This is not true if $ q\equal{}0$ (and $ p\not \equal{} 2$)  :D  :wink:

Let $ r$ be the number of distinct real and $ s$ be the number of distinct pairs of purely complex roots of a polynomial of degree $ n$.
As an irreducible polynomial over $ \mathbb Q$ has no double roots, we have $ n \equal{} r \plus{} 2s$. But this requires us to have $ n \equiv r \mod 2$.

So we necessarily need $ q \equiv p \mod 2$.


Consider an $n\times m$ grid. A paintbrush whose width is the same as the width of a grid square is used to paint a line from one corner of the grid to the opposite corner. How many squares will have paint in their interiors.

You can assume that the length of the side of square is $ 1$. Now put the grid into $ xy$-plane where the corners of the grid are $ (0,0),(0,m),(n,0),(n,m)$. Then the equation of the diagonal is $ y \equal{} \frac {m}{n}x$. The boundary of the painted line is $ y \equal{} \frac {m}{n}x \plus{} c$ for some $ c\in\mathbb{R}$. We know that its distance to the origin is $ 1$, so $ \frac {|c|}{\sqrt {n^2 \plus{} m^2}} \equal{} 1$. Therefore $ c \equal{} \pm \sqrt {n^2 \plus{} m^2}$.

Next we compute how many integer points there are between the diagonal and the line $ y \equal{} \frac {m}{n}x \minus{} \sqrt {n^2 \plus{} m^2}$. The result is $ \sum_{x \equal{} 0}^{n \minus{} 1}\left (\left \lfloor \frac {m}{n}x\right \rfloor \minus{} \left \lfloor \frac {m}{n}x \minus{} \sqrt {n^2 \plus{} m^2}\right \rfloor \minus{} \gcd(n,m)\right ) \plus{} \gcd(n,m)$. We have to take the sum twice as the diagonal splits the rectangle into two congruent triangle. But now we have computed the points in a diagonal twice so we have to add $ \gcd(n,m)$ to the sum. Therefore the result is
\[ 2 \sum_{x \equal{} 0}^{n \minus{} 1}\left (\left \lfloor \frac {m}{n}x\right \rfloor \minus{} \left \lfloor \frac {m}{n}x \minus{} \sqrt {n^2 \plus{} m^2}\right \rfloor \minus{} \gcd(n,m)\right ) \plus{} \gcd (n,m) \equal{} 2 \sum_{x \equal{} 0}^{n \minus{} 1} \left (\left \lfloor \frac {m}{n}x\right \rfloor \minus{} \left \lfloor \frac {m}{n}x \minus{} \sqrt {n^2 \plus{} m^2}\right \rfloor\right ) \minus{} \gcd(n,m).
\]


geometry
3D geometry
rectangle
algebra
system of equations

I know this forum isn't very active, but does anyone have the questions and/or the winners from the Manhattan Mathematical Olympiad, which was scheduled for Saturday, April 14?

Somewhat late, but if it's still of interest, I have the winners:

Grades 5-6:
1st Kaitlin Wichmann
2nd John Edward Seaton
3rd Albert Cochrane

Grades 7-8:
1st Leiming Tang
(no other competitor solved a single problem fully)

Grades 9-12:
1st Stephen Rong (me  :D )
2nd Ingrid Hwang
3rd Li Zeng

Grand Prize Winner: Kaitlyn Wichmann
Best Solution Winner: Stephen Rong

Grades 5-6

1. There are 64 cities in the country Moonland. Prove that there will be at least three of them which have the same number of rainy days in September 2007.

2. Matches from a bow are placed on the table in such a way that they form a (wrong) equality in Roman numbers (each segment on the picture below is a single match). Change a position of exactly one match (without removing or breaking it) and get a correct equality in Roman numbers:

VI-IV=XI

3. A craftsman has 4 ox f paint in order to paint all faces of a cube with the edge equal to 1 in. He cuts the cube into 27 smaller identical cubes. How much more paint does he need in order to paint completely faces of all smaller cubes?

4. Arrange the whole numbers 1 through 15 in a row so that the sum of any two adjacent numbers is is a perfect square. In how many ways this can be done? [Count backwards and forwards as one]

Grades 7-8

1. Suppose that we know that for all integer numbers $x$ the value $ax^{2}+bx+c$ is also integer. Can we conclude that the numbers $a,b,c$ are integers?

2. Let $x,y$ be integer numbers such that $3x+7y$ is divisible by 19. Prove that $43x+75y$ is also divisible by 19.

3. What is the biggest power of 2 which divides the product of 2007 consecutive numbers $2008\cdot2009\cdot2010\cdots4014$?

4. It is easy to show that the sum of five acute angles of a regular star is equal to 180 degrees. Prove that the sum of five angles of an irregular star is also 180 degrees.

[Diagram]

Grades 9-12

1. One throws randomly 120 squares of the size 1x1 in a 20x25 rectangle. Prove that one can still place in the rectangle a circle of the diameter equal to 1 in such a way that it does not have common points with any of the squares.

2. How many digits has the number $2^{70}$ (product of 70 factors of 2)?

3. Prove that the equation $x^{2}-2y^{2}=1$ has infinitely many integer solutions (i.e. there are infinitely many integers $x$ and $y$ which satisfy this equation).

4. Find all $x,y,z$ which satisfy the following system of equations:

$\frac{xy}{x+y}=\frac{8}{3}$
$\frac{zy}{z+y}=\frac{12}{5}$
$\frac{zx}{z+x}=\frac{24}{7}$.


[quote="qweretyq"]I know exactly how to prepare to try and make USAMO, because the material is taught in school and all you have to do is go over AMCs AIMEs etc...

but how do you prepare for the USAPhO?
I couldn't even get past to becoming a semifinalist because I couldn't even comprehend some of the problems on the test. It was different from previous years. Where can I learn this material, and what is the fastest way to go about doing it?

I would really like to at least make semifinals next year. Thanks.[/quote]

I'd say just read the Tipler book (the [i]full[/i] one, with Electricity and Magnetism as well). For Mechanics, I suggest doing all of Dr. Dell's first tier problems on physicstree: http://www.physicstree.net/vTreeTest.swf. For Electricity and Magnetism, do all of the Tipler problems with 3 dots. It would also be good to know Thermodynamics, Optics, and Quantum Mechanics. Thermo I believe is in the Tipler book as well, and you can also just go ahead and find some past Olympiad thermo problems to do. Optics and Quantum Mechanics you can learn by getting a book on them, or taking the course if your school offers it.

Doing past olympiad problems helps here, just like in the AMC/AIME. The main emphasis in the prelims is on the exclusively mechanics free response section. I believe I got something like 6 right on the MC section, but answered all the Mechanics problems and ended up with a 96 overall on the prelims. For the semi-finals, of course, you basically need to know all of physics.

[quote="qweretyq"][url=http://www.amazon.com/gp/product/0716743892/sr=8-1/qid=1144639374/ref=pd_bbs_1/102-2921698-9849724?%5Fencoding=UTF8]This One?[/url][/quote]

Well the cover's different from the one I have, but the title is the same, perhaps it's a newer version. It looks good, I'd go ahead and buy it.


In a group of 30 high school students, 8 take French, 12 take Spanish and 3 take both languages.  How many students of the group take neither French nor Spanish?

[hide]there are 3 repeats, so take them away from a language like Spanish. So now there are 17 students taking languages, and 13 who don't. [/hide]


What books can I buy to get a lot of number theory problems? As I worked through some old AMC 10s, I realized that I need to improve my understanding of number theory if I want to get the last few problems correct. In order to get a better understanding, I need to solve a lot of problems. So far, all I have is the Introduction to Number Theory and 104 Number Theory Problems. I also have AoPS Vol. 1 and 2 in which I haven't yet done the number theory sections, but will do so in the future. Thanks in advance.

Introduction to Number Theory will help you solve most of those tricky AMC 10 NT problems.  Did you go through it yet?  The last few chapters are especially important (congruences, bases, etc.)

Well I kind of "skimmed" the last few chapters. I better do the problems when I get the time.

By the way, how would I prepare for AIME Geometry problems?

Well, ultimately, you have to learn the concepts needed, and how to apply them. You can either learn the concepts first, then do lots of problems to practice, or look at problems and when you don't understand how to do them, learn the related concepts.


Of the 100 students at Lakeland Middle School, 56 take French, 60 take Spanish, and 20 take Latin. All students take at least one of these classes. What is the minimum number of students who take more than one of these subjects?

(56+60+20)-100=36.

If someone takes 3 languages they are overcounted twice, so 18 are taking 3 languages.

18 all languages
2 Latin
42 Spanish
38 French
--------
100 people

To obtain minimum, we assume that everybody either takes one language or two, not all three.  Since $(56+60+20)-100=36,$ and the people taking two languages are overcounted once, the answer is $\dfrac{36}{2}=\boxed{18\text{ people}}$.


MIT
college
linear algebra
linear algebra theorems

hi math's lover,
i need help to find a good course on linera algebra especially on reduction.for instance course wich is delivred by prestigious university(Princeton,MIT,...).

first of all thank you for the links.secandely how can i read the cours?
I downlowd something which is on the page related to course but it doesn't contain lessons.

Yes, it does contain lessons. It's something like an offline web material, so you can simply browse it as a webpage in your browser. Unzip it someplace and run Welcome.htm. Using the links on your left hand side you can access all course materials (both off and online).

I'm assuming you downloaded the course:

http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/DownloadthisCourse/index.htm

In case you're still experiencing difficulties:

http://ocw.mit.edu/OcwWeb/web/help/faq4/index.htm#t2


ratio quadratics inequalities trigonometry search geometry

Hello. I want to improve my proofs. If you wouldn't mind could you post some tips for looking at a problem involving a proof. Also, speaking of proofs I thought that this would be a good opportunity to post some problems to see how people would solve them.

[b]Problem 1[/b]
You are given a pair of triangles for which
$ (a)$ two sides of one triangle are equal in length to two sides of the second triangle, and
$ (b)$ the triangles are similar, but not necessarily congruent.
Prove that the ratio of the sides that correspond under the similarity is $ \frac{1}{2}\cdot(\sqrt{5}\minus{}1)$.

I don't think such ratio is unique. For example, we can have the sides $ 1,x,x^2$ and $ x,x^2,x^3$ with the ratio $ x>1$.
Your answer looks like the triangle should be right, so $ 1\plus{}x^2\equal{}x^4\implies (x^2)^2\minus{}x^2\minus{}1\equal{}0\implies x^2\equal{}\frac{1\plus{}\sqrt5}2$, but the square of your answer is $ \frac{3\minus{}\sqrt5}2$, which isn't exactly the same...

From the conditions of the problem, the 2 equal sides are smallest in one, and biggest in other triangle, call them a and b, and call the other sides c and d.

First: c,a,b
Second: a,b,d    sorted by length

$ \frac bd\equal{}\frac ab\equal{}\frac ca$

$ d\equal{}\frac{b^2}{a},c\equal{}\frac{a^2}{b}$

The ratio is $ \frac ba$, and it needn't to be your fraction.

I know I've seen this problem somewhere, but I can't seem to remember where...

The problem should be 'Prove that the ratio of the sides that correspond lies in between $ \frac {\sqrt {5} \minus{} 1}{2}$' and $ \frac {\sqrt {5} \plus{} 1}{2}$.

Well, let's try it now. If a=b, it's trivial.
c+a>b
$ \frac{a^2}{b}\plus{}a>b$
$ a^2\plus{}ab>b^2$

Similarly $ a\plus{}b>\frac{b^2}{a}$

$ a^2\plus{}ab>b^2$, same as above

Now solve the quadratics for equality case, and it immediately follows that interval of the solutions is between the solutions of the quadratics.

It's from Canada, 2007:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229080[/url]

Sorry, I forgot to copy the second part.

I don't think one single inequality will suffice. It can be shown through case work that after normalizing the smaller triangle to side length of 1 for the smallest side, the triangles must be $ 1,x,x^2$ and $ x,x^2,x^3$ for $ x > 1$.
Then through the inequalities $ 1 \plus{} x > x^2,x^2 \plus{} x > 1,x^2 \plus{} 1 > x$ we have
$ x^2 \minus{} x \minus{} 1 < 0\implies x < \frac {1 \plus{} \sqrt5}2$
$ x^2 \plus{} x \plus{} 1 > 0\implies x > \frac { \minus{} 1 \plus{} \sqrt5}2$
$ x^2 \minus{} x \plus{} 1 > x^2 \minus{} 2x \plus{} 1 \equal{} (x \minus{} 1)^2 > 0$ which is always true.

I posted the following in another thread (it's a trig "marathon") but no one responded

Here's a quick trig identity I made up to verify

$ (\sin x \plus{} \sin y)(\sin x \minus{} \sin y) \equal{} \sin(x \plus{} y)\sin(x \minus{} y)$

By the way this problem was already solved, but someone might feel free to find a synthetic geometric proof for it:

Squares $ ABMN$ and $ BCPQ$ are constructed externally on the sides $ AB$ and $ BC$ of the triangle $ ABC$. If $ BD$ is a median in $ \triangle ABC$, prove that its extension is a height in $ \triangle BMQ$

The sine problem:

Let $ \sin x\equal{}a,\cos x\equal{}b,\sin y\equal{}c,\cos y\equal{}d$

$ \text{LHS}\equal{}a^2\minus{}c^2$

$ \text{RHS}\equal{}(ad\plus{}bc)(ad\minus{}bc)\equal{}a^2d^2\minus{}b^2c^2\equal{}a^2(1\minus{}c^2)\minus{}(1\minus{}a^2)c^2\equal{}a^2\minus{}c^2$

Now the 2 sides are equal.

The second is well known, search for it.

Hehe I made up that trig identity just to trick the newbies into thinking $ \sin(x\plus{}y)\equal{}\sin x\plus{}\sin y$ :P
By the way I tried to search using the keywords "square side median construct externally triangle" but the search didn't return anything... so Bugi could you just post your solution (using synthetic geometry)?
Also feel free to post the next problem :!:


inequalities
geometry
3D geometry
calculus
derivative
symmetry
algebra

What about this pb?
a,b,c positive reals less than 1.
prove: a2+b2+c2 <= a2b+b2c+c2a+1
i have a nice sol (::)

is the inequality like so:
a^2+b^2+c^2<=a^2b+b^2c+c^2a+1?

or you can notice that
f(a,b,c)=a^2 (1-b)+ b^2(1-c) +c^2(1-a)
is convex in each variable and we are done

I found a nice solution by considering a, b, c segments on the sides of a unit cube, so, for example, a^2(1-b) is a volume, but it would be really hard for me to explain it, because I have no pictures.

Define f(a,b,c) = a^2(1-b) + b^2(1-c) + c^2(1-a)
The problem is equivalent to asking us to prove that f(a,b,c) <= 1
for positive reals all less than 1. I'll prove the slightly
stronger case where a,b,c <= 1.
Taking partial derivatives:
df/da = 2a(1-b) - c2
d2f/da2 = 2(1-b) > 0 (except when b = 1, when it equals 0)
By cyclic symmetry, we see that d2f/db2 > 0 and d2f/dc2 >0 all
except possibly at edge points in the domain
Let a1,b1,c1, be points where f achieves a max.
Suppose one of these three points occurs at a value other than 0 or 1.
Suppose it's a1
Then d2f/da2(a1,b1,c1) > 0
That means that either (of both) of f(0,b1,c1) and f(1,b1,c1) is
greater than f(a1,b1,c1) Contradiction. Analogous contradctions
will be had if the point not = 0 or 1 is b1 or c1.
Therefore, a1 = 0 or 1, b1= 0 or 1, and c1 = 0 or 1
So there are eight possibilities for (a1,b1,c1):
(0,0,0)
(0,0,1)
(0,1,0)
(0,1,1)
(1,0,0)
(1,0,1)
(1,1,0)
(1,1,1)
Plug each of them into the formula for f, and you'll see that the
max of f is indeed <= 1.

cheers!! :D


Prove that : If $ x,y,z,t \geq 0$ then :
1) $ 3(x^2\plus{}y^2\plus{}z^2\plus{}t^2)\plus{}4\sqrt{xyzt} \geq (x\plus{}y\plus{}z\plus{}t)^2$
2) $ \frac{x^3}{(x\plus{}y)^3}\plus{}\frac{y^3}{(y\plus{}z)^3}\plus{}\frac{z^3}{(z\plus{}x)^3} \geq \frac{3}{8}$

[quote="QuyBac"]Prove that : If $ x,y,z,t \geq 0$ then :
1) $ 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \plus{} 4\sqrt {xyzt} \geq (x \plus{} y \plus{} z \plus{} t)^2$
2) $ \frac {x^3}{(x \plus{} y)^3} \plus{} \frac {y^3}{(y \plus{} z)^3} \plus{} \frac {z^3}{(z \plus{} x)^3} \geq \frac {3}{8}$[/quote]

$ 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \plus{} 4\sqrt {xyzt} \geq (x \plus{} y \plus{} z \plus{} t)^2$

substitution $ x\equal{}x^2,y\equal{}y^2,z\equal{}z^2,t\equal{}t^2$

we only prove

$ 3(x^4 \plus{} y^4 \plus{} z^4 \plus{} t^4) \plus{} 4{xyzt} \minus{}(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2)^2\geq 0....(**)$

assume $ x\geq y \geq z \geq t$

substitution $ x\equal{}a\plus{}b\plus{}c\plus{}t,y\equal{}b\plus{}c\plus{}t,z\equal{}c\plus{}t$ $ (a,b,c\geq 0)$

$ (**) \Longleftrightarrow (8b^2\plus{}6c^2\plus{}8ab\plus{}4ac\plus{}8bc\plus{}6a^2)t^2\plus{}(4c^3\plus{}16a^2c$ $ \plus{}8b^3\plus{}4ac^2\plus{}8a^3\plus{}20abc\plus{}12ab^2\plus{}20b^2c\plus{}20a^2b\plus{}8bc^2)t$

$ \plus{}8a^2c^2\plus{}8a^3b\plus{}2b^4\plus{}20a^2bc\plus{}8b^3c\plus{}4ab^3\plus{}8b^2c^2\plus{}2a^4$ $ \plus{}12ab^2c\plus{}8abc^2\plus{}8a^3c\plus{}10a^2b^2\geq 0$

2) $ \frac {x^3}{(x \plus{} y)^3} \plus{} \frac {y^3}{(y \plus{} z)^3} \plus{} \frac {z^3}{(z \plus{} x)^3} \equal{} \sum \frac {x^4}{x(x \plus{} y)^3}$ $ \geq \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{x(x \plus{} y)^3 \plus{} y(y \plus{} z)^3 \plus{} z(z \plus{} x)^3}$

we only prove

$ \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{x(x \plus{} y)^3 \plus{} y(y \plus{} z)^3 \plus{} z(z \plus{} x)^3}\geq \frac {3}{8}......(*)$

assume $ x \equal{} max{(x,y,z)}$

$ (*)\Longleftrightarrow (5x^2 \plus{} ( \minus{} 4y \plus{} 2z)x \plus{} 3y^2 \plus{} 9z^2 \minus{} 7yz)(x \minus{} y)(x \minus{} z) \plus{}$ $ (5y^2 \minus{} 2yz \plus{} 5z^2)(y \minus{} z)^2\geq 0$

easy prove

$ 5x^2 \plus{} ( \minus{} 4y \plus{} 2z)x \plus{} 3y^2 \plus{} 9z^2 \minus{} 7yz\geq 0$

$ 5y^2 \minus{} 2yz \plus{} 5z^2\geq 0$

[quote="QuyBac"]Prove that : If $ x,y,z,t \geq 0$ then :
1) $ 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \plus{} 4\sqrt {xyzt} \geq (x \plus{} y \plus{} z \plus{} t)^2$
[/quote]

We assume that $ x \ge y \ge z \ge t$
$ f(x;y;z;t): \equal{} 3({x^2} \plus{} {y^2} \plus{} {z^2} \plus{} {t^2}) \plus{} 4\sqrt {xyzt}  \minus{} {(x \plus{} y \plus{} z \plus{} t)^2}$
So
$ f(x;y;z;t) \minus{} f\left( {\sqrt {xy} ;\sqrt {xy} ;z;t} \right) \equal{} 2{\left( {\sqrt x  \minus{} \sqrt y } \right)^2}\left( {x \plus{} y \minus{} z \minus{} t \plus{} 2\sqrt {xy} } \right) \ge 0$
$ \Rightarrow f(x;y;z;t) \ge f\left( {\sqrt {xy} ;\sqrt {xy} ;z;t} \right)$
By S.M.V theorem, we must to prove that
$ f(a;a;a;t) \ge 0$
$ \Leftrightarrow 3\left( {9{a^2} \plus{} {t^2}} \right) \plus{} 4\sqrt {{a^3}t}  \ge {\left( {3a \plus{} t} \right)^2}$
Easy to prove

[quote]2) $ \frac {x^3}{(x \plus{} y)^3} \plus{} \frac {y^3}{(y \plus{} z)^3} \plus{} \frac {z^3}{(z \plus{} x)^3} \geq \frac {3}{8}$[/quote]
This ineq can be proved by Cauchy-Schwarz and Schur  :lol:  :lol:

[quote="bigbang195"][quote="QuyBac"]Prove that : If $ x,y,z,t \geq 0$ then :
1) $ 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \plus{} 4\sqrt {xyzt} \geq (x \plus{} y \plus{} z \plus{} t)^2$
2) $ \frac {x^3}{(x \plus{} y)^3} \plus{} \frac {y^3}{(y \plus{} z)^3} \plus{} \frac {z^3}{(z \plus{} x)^3} \geq \frac {3}{8}$[/quote]

$ \sum \frac {a^3}{(c \plus{} b)^3} \ge \frac {(\sum \frac {a}{b \plus{} c})^3}{9} \ge^{Nessbit}\frac {(\frac {3}{2})^3}{9} \equal{} \frac {3}{8}$[/quote]
Sorry but your solution is wrong, since you proved $ \sum_{cyc}\frac{a^3}{(b\plus{}c)^3}\geq \frac 38$ which is weaker than the original inequality.

However, I think that the second can also be proved using Mixing Variables method, but I don't have a proof yet.

[quote="Potla"][quote="bigbang195"][quote="QuyBac"]Prove that : If $ x,y,z,t \geq 0$ then :
1) $ 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \plus{} 4\sqrt {xyzt} \geq (x \plus{} y \plus{} z \plus{} t)^2$
2) $ \frac {x^3}{(x \plus{} y)^3} \plus{} \frac {y^3}{(y \plus{} z)^3} \plus{} \frac {z^3}{(z \plus{} x)^3} \geq \frac {3}{8}$[/quote]

$ \sum \frac {a^3}{(c \plus{} b)^3} \ge \frac {(\sum \frac {a}{b \plus{} c})^3}{9} \ge^{Nessbit}\frac {(\frac {3}{2})^3}{9} \equal{} \frac {3}{8}$[/quote]
Sorry but your solution is wrong, since you proved $ \sum_{cyc}\frac {a^3}{(b \plus{} c)^3}\geq \frac 38$ which is weaker than the original inequality.

However, I think that the second can also be proved using Mixing Variables method, but I don't have a proof yet.[/quote]

$ \sum \frac {a}{b \plus{} c} \ge \frac{3}{2}$ this inequality is right !

[quote="QuyBac"]Prove that : If $ x,y,z,t \geq 0$ then :
1) $ 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \plus{} 4\sqrt {xyzt} \geq (x \plus{} y \plus{} z \plus{} t)^2$
2) $ \frac {x^3}{(x \plus{} y)^3} \plus{} \frac {y^3}{(y \plus{} z)^3} \plus{} \frac {z^3}{(z \plus{} x)^3} \geq \frac {3}{8}$[/quote]

For (1), it is Turkevici's Inequality. And it is [b]Problem 6.34 - page 301[/b] in new book (where it is presented with three nice and simple proofs):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.


You want to measure every possible weight from 1 to 100 with 5 different weights using a scale. If you can place weights on either side, what weights should you choose?

[hide="Solution"] Balanced ternary notation dictates the choices $1, 3, 9, 27, 81$.  For more information:

http://en.wikipedia.org/wiki/Balanced_ternary [/hide]

[hide]1, 3, 9, 27, 81
Reason: Any number from 1 to 100 can be expressed as a five-digit base-3 number with each digit as a -1, 0, or 1.[/hide]

This reminds me of the ARML power round...

That's exactly what I was thinking - ARML power

This is such a well-known problem that I'm surprised it was even on the ARML this year. However, you should still prove that powers of 3 work, and nothing else works (is balanced terniary a citable fact?).

How do you prove all numbers can be expressed in balanced ternary? How many balanced-trits does it take to represent a number n?

[quote="JavaMan"]How do you prove all numbers can be expressed in balanced ternary?[/quote]

Induction is a good way to go.

[quote="boxedexe"]The question is equivalent to the following: How would you convert an unbalanced ternary number into a balanced one?  This is quite easy to answer.

Masoud Zargar[/quote]

Well this is easy.  you can find unique representation if the division theorem holds (so you just repeatedly apply it and you get something unique, and with some induction on the way).

So everything boils down to showing that division theorem holds for balance ternary.  But note that the only difference between balance ternary from the regular one is that instead of writing:

$a = 3b+2$

you write:

$a = 3(b+1)-1$

So there is a bijection for cases of division theorem between regular and balanced ternary.  So since we have unique $a$ and $b$ in the regular version, we must also have unique $a$ and $b$ in the balanced version.  Therefore every integer is representable and unique in the balanced ternary system.

[quote="boxedexe"]The question is equivalent to the following: How would you convert an unbalanced ternary number into a balanced one?  This is quite easy to answer.[/quote]

As I wrote in the Power Round, this has a nice interpretation in terms of generating functions.  The generating function for the positive integers in base three is

$(1+x+x^{2})(1+x^{3}+x^{6})(1+x^{9}+x^{18})... = 1+x+x^{2}+x^{3}+...$

Which illustrates that all naturals have a unique ternary representation (this expansion is easily proven by multiplying by $1-x$).  Dividing each factor by its middle term,

$(x^{-1}+1+x^{1})(x^{-3}+1+x^{3})(x^{-9}+1+x^{9})... = ...+x^{-2}+x^{-1}+1+x+x^{2}+x^{3}+...$

Which illustrates that all integers have a unique balanced ternary representation. ;)


inequalities
integration
inequalities unsolved

Let $ a,b,c>0$ and $ \sum \frac{1}{a}\equal{}1$. Prove that
\[ \sum \frac{a}{b(b\plus{}1)^2}\geq \frac{243}{(a\plus{}b\plus{}c)(a\plus{}b\plus{}c\plus{}3)^2}.\]

by Minkowski's ineq we get:
$ (a \plus{} b \plus{} c)(a \plus{} b \plus{} c \plus{} 3)^2 \ge (\sum_{}^{} \sqrt [3]{a(a \plus{} 1)^2})^3$
so it's enough to prove:
$ (\sum_{}^{} \frac {a}{b(b \plus{} 1)^2}) (\sum_{}^{} \sqrt [3]{b(b \plus{} 1)^2})^3 \ge 3^5$
or
$ (\sum_{}^{} \frac {a}{b(b \plus{} 1)^2})^ \frac {1}{4} (\sum_{}^{} \sqrt [3]{b(b \plus{} 1)^2})^\frac {3}{4} \ge 3^\frac {5}{4}$
using Holder and inequality between power means we have:
$ L \ge \sum_{}^{} a^ \frac {1}{4} \ge 3 ( \frac {3}{\sum_{}^{} \frac {1}{a} })^ \frac {1}{4} \equal{} 3^ \frac {5}{4}$
done  :lol:

Another way is to notice that LHS is
\[ 2\int_{0}^1 (\sum \frac {ax}{(x \plus{} b)^3})dx.\]


There's a cool forum up that's solely dedicated to college and graduate school. You can ask your questions about the SAT, ACT, AP tests, GRE, etc. If you have any questions about this stuff, then you should join today!

[url]http://s215234424.onlinehome.us/index.php[/url]

There's also CollegeConfidential, though that place is unbelievably neurotic, and I tend to avoid it at all costs.

Well, this forum is relatively new (I think) and there's not very many people. And the administrator is very friendly and will help you and answer any of your questions.


function
real analysis
real analysis unsolved

$d: K \to  \mathbb{R}^+$, the distance from a point $x \in K$ to $D$, is a continuous function on the compact $K$. So it reaches its minimum to a point $x_0 \in K$, and of course $dist(x_0,D)>0$ because $D$ is closed and $x_0$ does not belong to $D$.

[quote="Kalle"]This isn't really a problem from a book or anything, just something that I can't figure out. Suppose we have two disjoint compact sets D and K in R^n. How can I show that the distance between D and K, that is the infimum of $|x-y|$ taken over all x in D and all y in K, is strictly larger than zero?[/quote]One simple way could be like this. Form the product $K \times D \subseteq \mathbb{R}^n \times \mathbb{R}^n$. The product space is compact, and the map $K \times D \ni (x,y) \mapsto |x-y| \in \mathbb{R^+}$ is continuous. Hence the image of that map is compact and in particular closed. So if the infimum of $|x-y|$ were zero, there'd exists $x' \in K$, $y' \in D$ w. $|x' - y'| = 0$, contradicting $K \cap D = \emptyset$.


Let $ x_1,x_2,...,x_n$ such that
$ 0 < x_1\leq x_2\leq x_3\leq...\leq x_n$
 and $ P \equal{} x_2^k(x_1 \minus{} x_3) \plus{} x_3^k(x_2 \minus{} x_4) \plus{} ..... \plus{} x_1^k(x_n \minus{} x_2)$
Prove that
$ P\geq 0$ if $ k > 1$
$ P\leq 0$ if $ 0 < k < 1$

Rewrite the expression as:
\[ (x_{n\minus{}1}\minus{}x_1)x_{n\minus{}1}^k \plus{} (x_n\minus{}x_2)x_n^k \minus{} \sum_{i\equal{}1}^{n\minus{}2} {(x_{i\plus{}2}\minus{}x_i)x_{i\plus{}1}^k}\]
\[ \equal{} (\sum_{i\equal{}1}^{n\minus{}2} (x_{i\plus{}1}\minus{}x_i))x_{n\minus{}1}^k \plus{} (\sum_{i\equal{}2}^{n\minus{}1} (x_{i\plus{}1}\minus{}x_i))x_n^k \minus{} \sum_{i\equal{}1}^{n\minus{}2} {(x_{i\plus{}2}\minus{}x_{i\plus{}1}\plus{}x_{i\plus{}1}\minus{}x_i)x_{i\plus{}1}^k}\]
\[ \equal{}(\sum_{i\equal{}1}^{n\minus{}2} (x_{i\plus{}1}\minus{}x_i))(x_{n\minus{}1}^k \minus{} x_{i\plus{}1}^k) \plus{} (\sum_{i\equal{}2}^{n\minus{}1} (x_{i\plus{}1}\minus{}x_i))(x_{n}^k \minus{} x_i^k)\]


abstract algebra
group theory
superior algebra
superior algebra open

Find a group of finite presentation that has $(\mathbb{Q},+)$ as a subgroup.
Such a group exists if you were doubting it.

A presentation for a group si $G=<S|R>$ where $S$ is a set (the generators of $G$) and $R$ is a set of finite length words (they are relations between the generators) in the elements (letters) of $S$.

A finite presentation means $S$ and $R$ are finite.
Finitely generated means just that $S$ is finite.

For example:
$\mathbb{F}_2=<a,b>$ is the free group on two generators with no realtions.
$\mathbb{Z}^n=<a_1,\ldots,a_n|a_ia_j=a_ja_i\forall i,j=\overline{1,n}>$ is the free abelian group on $n$ generators.

A presentation for $(\mathbb{Q},+)$ is $<a_i,i\geq 2|a_2^2=1,\ a_{n+1}^{n+1}=a_n (\forall n\geq 2)>$. The correspondence is $\frac 1{n!}=a_n$.
Notice that for this presentation $S$ and $R$ are both infinite.
There is no finite presentation for $(\mathbb{Q},+)$ because it is not finitely generated. But this is not what I'm asking for, is it?

[quote="perfect_radio"]If you say that such a group exists, then why are you putting it in the [i]Open Questions[/i] subforum?[/quote]
The proof of the existence is not constructive. It is an open problem to give an example, but it is known that it must exist.


inequalities
linear algebra
matrix
induction
algebra
polynomial
complex numbers

Show that for any square matrix A exist an unitary matrix C such that $CA{C^{-1}}$ is upper triangular

use induction
pick a random eigenvalue of the matrix
now take an eigenvector of it, and extend it to a basis
the first column will be completely zero in your new matrix, except for the very first element, which will be the eigenvalue
now you can use induction

[quote="alekk"]or use Gram-Schmidt process ..[/quote]

1. How to pronounce Schmidt? I totally don't know how to pronounce a word with the string Schm

2. Do we need the property that the characteristic polynomial of A splits?

[quote="fredbel6"]Definitely , because two unitary equivalent (or simply equivalent for that matter) have the same characteristic polynomial, and an upper or lower triangle matrix obviously has a splitting char poly because you can just read the eigenvalues of the diagonal

But why do you ask?  I didn't know unitary matrices was defined in any other field other than the complex numbers?  Could you explain?[/quote]

oh, sorry I thought in R you can call it unitary/orthogonal.

Take an eigenvector $X_1$ coressponding to an eigenvalue of the matrix $A$,called $a_1$.Then construct an orthonormate basis $X_1,...,X_n$(here you use the schmidt diag process).The matrix $A$ written in this base has $a_1$ in the first position of the first column,and 0 elsewhere on that column.It rests to see why the basis transformation matrix it is an unitary matrix(i.e ...$Matrix_BX=$unitary,where $B$ is a certain basis of $C^n$ and $X$ is a orthonormate basis of $C^n$).After that you can continue the process inductively choosing the $(n-1)\cdot(n-1)$sub-matrix of $A$,and so on...
I hope it's ok.

[quote="Soarer"]
oh, sorry I thought in R you can call it unitary/orthogonal.[/quote]

In $\mathbb{R}$ I believe we call it orthogonal, and in $\mathbb{C}$ we call it unitary.
Well, I understand your question now.

In $\mathbb{C}$ every polynomial splits, but in $\mathbb{R}$ that is not true, and then Schur's theorem should be corrected : every matrix with a splitting characteristic polynomial over the reals is orthogonally equivalent to an upper diagonal matrix


algorithm
combinatorics proposed
combinatorics

Let $n>2$ be an integer. In a country there are $n$ cities and every two of them are connected by a direct road. Each road is assigned an integer from the set $\{1, 2,\ldots ,m\}$ (different roads may be assigned the same number). The [i]priority[/i] of a city is the sum of the numbers assigned to roads which lead to it. Find the smallest $m$ for which it is possible that all cities have a different priority.

We claim that the answer is 3. If we reduce each of the numbers assigned on the roads by 1, then the priority of each city is reduced by $ n\minus{}1$, so all city still have distinct priorities. Now, each road is assigned an integer from $ \{0,1,2,\ldots,m\minus{}1\}$. If $ m\minus{}1\equal{}0$, all roads have the same priority. If $ m\minus{}1\equal{}1$, each road is assigned either 0 or 1. The greatest possible priority of a city is $ n\minus{}1$ and the smallest possible priority is $ 0$. So the priorities must be $ 0,1,2,\ldots,n\minus{}1$. All roads from the city of priority $ 0$ must be assigned 0, while all roads from the city of priority $ n\minus{}1$ must be assigned $ 1$. Considering the road connecting these two cities, we get a contradiction. Thus $ m\minus{}1\ge2$. We prove that $ m\minus{}1\equal{}2$ is possible.

We will find an algorithm to assign each road an integer from $ \{1,2,3\}$. We divide into two cases:

(i) $ n$ is even ($ n\equal{}2k$)
Let the cities be $ C_1,C_2,\ldots,C_n$ such that the priority of $ C_i$ is less than that of $ C_j$ for all $ i<j$. We assign the roads from $ C_1,C_2,C_3,\ldots,C_n$ consecutively as follows: (see figure 1)

The value of $ X$ is clearly greater than $ 2k\minus{}4$. In each assignment, the priority of each city does not increase by more than 2.

(ii) $ n$ is odd ($ n\equal{}2k\plus{}1$)
Similar to case (i), only different table (see figure 2).

easier solution:
first consider 3 points or country its easy to prove that m>=3  becouse if m<=2 thus there are at least tow road that has a same number and ...
now we prove by apriori that if m=3 its possible to put number on roads suppose that for n=k it is possible now consider another country with n roads  put the smallest priority of n city in priority of new city and suppose that its k(k>n). now k=x1+x2+...+xn(x i=1,2,3)and the biggest number of road connect to city which has a biggest priority and ....if k=n-1 so we cant write k=x1+...+xn .now consider k=3n.it will be possible.if its not tru help me

Obviously for $m=2$ our priority can have just $n-1$ different value we have $n$ vertices =>    contradiction

Now we prove $m=3$ is our suitable solution. for this we should just build this graph for any $n$.

[b]obvious claim [/b]: about each graph at least one of the states is true:
[u]state 1[/u]: the minimum of priorities isn't equal to $n-1$
[u]state 2[/u]: the maximum of priorities isn't equal to $3(n-1)$

[b]proof[/b]: if about one graph none of the states is true so the priority of one of the vertex for example $A$ is $n-1$ and the priority of $B$ is $3(n-1)$ so all of the contact edges of $A$ are $1$ and all of the edges of $B$ are $3$. Consider the edge between $A$ and $B$=>contradiction

Now just we should use induction. Our conditions are true for $n=3,4,...,k-1$ We want to prove for $n=k$
so cosider the graph when $n=k-1$ .
 if the [u]state 1[/u] is true for this graph we just add one vertex in this graph and put the number of the edges between this new vertex and the others $1$
if the [u]state 2[/u] is true for this graph add one vertex and put new edges $n-1$
So we are done  :agent:


Team Penguin can only do it at 6:30 PM EST Thursday. Is that time O.K. for SagginWolley?

NO, ZENITH XAN GETS HOME AT 22:30PM EST ON THURSDAYS.

WEDNESDAY 17:30-18:30 EST OR SUNDAY FROM 12:00 TO 20:00 EST OR 21:00-23:59 EST

TUESDAY AND MONDAY TIMES MIGHT BE AVAILABLE BUT WILL BE HIGHLY LIMITED.

wait, we can do the tournament on sunday?

If that's the case, I will be home from HMMT and we are going to be home all day...

Any time on Sunday should be good, let me check with abacadaea and zolojetto...

[quote="abacadaea"]we will try every hour i guess because my parents are always planning something. 12:00 is good. 13:00-15:00 we are usually doing things. erm, post which hours between 12:00 and 20:00 on sunday are good for you. i will probably be able to work around most things.[/quote]
zolojetto has told me that any time on Sunday will be good for him. Any time on Sunday is good for me too.

OK. @SagginWolley: How is Sunday 4 O'clock? That is the best time for Penguin.


geometry trigonometry MATHCOUNTS AMC AIME ratio

I know everyone is involved into chapter problems, but here's a problem from the handbook of 2003-2004 (the only problem I haven't solved!):

There are two non congruent triangles. Each have one side with 10 cm, another side 12cm, and both have the angle facing the 10cm side measuring 45 degrees. In square centimeters, what is the sum of the areas of the two triangles?

I drew a picture-

[hide]I got 72 as the answer.  First I realised that though the two triangles are not congruent, they still have the same area. Next what i did was draw a perpendicular line and that made a right icoceles triangle. Then I did 2x 2 =144. So x= :sqrt: 72. The area of the triangle then is 36. The area of both triangles is 72.[/hide]

[quote="GoBraves"][hide]... the two triangles ... have the same area. ...[/hide][/quote]
I am intrigued by your solution, even though I don't think the assertion quoted above is correct

Here's what I figure. Since the figures are noncongruent, there are only two possible ways that you can come up with the figure, with the 12 being the longer side (creating an obtuse triangle I think), and the twelve being the shorter side (like in gobraves' drawing...  I'm still working on a solution tho...

hockey's on the right track, after i did it, I found the problem from that handbook(on my hard drive)

Nice picture Braves, but I think that's only one of the triangles

30(sqrt 3) + 50. I'm guessing by making two triangles that are not congruent. Please correct me if I'm wrong. Is anybody even listening?

jason, I'm listening.  That's not the right answer.  The answer can be found with using so little math it is quite elegant!!  The only math I used was probably that of an average 5th or 6th grade class.  It's seeing how to use it that makes it special :)

[quote="frost13"]The answer can be found with using so little math it is quite elegant!![/quote]
Yes!  I am sure frost13 and I are thinking of the same solution.

This was a request for help.  If somebody doesn't solve this in the next day or two, frost13 or I will have to post our elegant solutions, thus depriving you guys of finding it for yourselves.

Hopefully rcv will mean the "next day" part, not the "or two" part

[quote="GoBraves"][quote="kidwithshirt"][quote="rcv"]dasherm:  I believe your answer is incorrect.

Everybody:  This is MATHCOUNTS.  Please solve this without using trigonometry.

There is a wonderful solution.  Anybody who finds it on their own will feel a great satisfaction.  Even though this is a request for help, please hide your solutions.[/quote]

So trig is not permited in mathcounts?[/quote]

Its permitted, any method of solving is permitted (except cheating), but all of the problems can be solved without trig.[/quote]

...and no one wants to do something like sine of 42 in the middle of the sprint round... Just like the AIMEs are designed for if you have Pre-Calc knowledge, the Mathcounts problems are designed for anyone who has taken Algebra 1 and has a bit more knowledge than that..

Also, using trigonometry ratios when solving a problem relating to 30-60-90 or 45-45-90 triangles is much harder than just memorizing the special triangles relationships.

Yup... My dad tried a mathcounts problem with a 30:60:90 problem and went into all this trig... and it was so much more extensive than knowing the ratios and avoiding trig.

Too bad they can't have a trigonometry table at the competitions for those who don't know their special triangles  :lol: ...

[quote="hockeyadi23"][quote="smiley"][hide="Here is my answer."]
I drew out both triangles, ontop of each other.  Then, I moved the obtuse triangle to the other side and flipped it.  Since we knew that the angle opposite the side length 10, we have a 45-45-90 right triangle, with base and height 12.  THerefore, the area is:
$\frac{12\cdot 12}{2}=72$
My answer is 72.
[/hide][/quote]

I don't exactly get your solution... where did you put in the ten? Can you label out your pic?[/quote]

Smiley's got it, here is a similar picture with one less line and with lables.  The two original triangles have been combined on their common edge of 10 to form one isoceles triangle with sides 12 and two 45 degree angles making in a right isoceles.

that's a really nice solution... I would've never thought of that.


[quote="Treething"]Mary and Jerry have three kids: Harry, Larry, and Steve. If three family members are picked at random, what is the probability that they are all male? Express your answer as a percentage.[/quote]

[hide]$\frac{\binom{4}{3}}{\binom{5}{3}}=\frac{4}{10}=\boxed{40}$percent[/hide]

SplashD, your answer is "[Unparseable or potentially dangerous latex formula. Error 6]" ;)

[hide]
The answer is $\frac{{4 \choose 3}}{{5 \choose 3}}=\frac{2}{5}=40 \%$
[/hide]

Oh, and why can't I quote SplashD?

[size=75][color=darkred]Fixed your LaTeX
-nebula42[/color][/size]

The quote option is hidden under one of the hide tags in his signature.

[hide="answer"]There are $\binom43$ ways to choose 3 boys and $\binom53$ total ways to choose them, making the answer $\frac{\binom43}{\binom53}=\frac{4}{10}=\frac25=40\%$[/hide]

mathnerd, you need to have \frac{{4\choose3}{{5\choose3}} or what I have  :)

[hide="Assuming mary is female and jerry, harry larry, and steve are male"]$\frac{\binom43}{\binom53}=\frac25=40\%$[/hide]Treething said [b]%[/b]

[quote="mathnerd314"]SplashD, your answer is "[Unparseable or potentially dangerous latex formula. Error 6]" ;)
[/quote]

Yeah, I put % in $LaTeX$ instead of \%.

[quote="mathnerd314"]SplashD, your answer is "[Unparseable or potentially dangerous latex formula. Error 6]" ;)

[hide]
The answer is $\frac{4 \choose 3}{5 \choose 3}=\frac{2}{5}$
[/hide]

Oh, and why can't I quote SplashD?[/quote]

And your answer has an error.

You can quote SplashD, just click in his signature, "Click here to reveal hidden content", or something like that, and then the quote will come.


function
complex analysis
complex analysis unsolved

Consider an entire function $ F$ that is real valued on the segment $ [0,1]$. Prove that it is real-valued on the whole real axis.

Its Taylor expansion at $ z\equal{}1/2$ has real coefficients.


we have a equation:$x^8+9y^8=10z^4$.How many solutions does it have satisfise:
$x,y,z\in Z^+$ and $gcd(x,y,z)=1$?
 :( $\leftarrow$ I unsolve it ,help me.

[quote="Rust"]It is equavalent $x^8-y^8=10(z^4-y^8)$. It had onlu solutution x=y=z=1.[/quote]
I don't understand your solution. :(


For how many real values of x is
square root (120 − square root (x) an integer?

A. 3 B. 6 C. 9 D. 10 E. 11

hello, is your term $ \sqrt{120\minus{}\sqrt{x}}$?
Sonnhard.

Assuming he meant $ \sqrt{120\minus{}\sqrt{x}}$:

We pretty much want $ 120 \minus{} \sqrt{x}$ to equal a square, so when square rooted we would achieve an integer.  The squares of $ 0\minus{}10$ are below $ 120$, so therefore we have $ \boxed{11}$ possible values of $ x$ where $ x$ is a real and $ \sqrt{120\minus{}\sqrt{x}}$ is an integer.


Maximize the following expressions

$a^{2}+b^{2}+c^{2}+d^{2}-ab-ac-ad-bc-bd-cd \qquad a,b,c,d\in [0,1]$

and

$\sum_{sym}a^{2}b \qquad a,b,c >0, a+b+c=1$

[quote="venatrix"]Maximize the following expressions

$a^{2}+b^{2}+c^{2}+d^{2}-ab-ac-ad-bc-bd-cd \qquad a,b,c,d\in [0,1]$
[/quote]
Let $A(a,b,c,d)=a^{2}+b^{2}+c^{2}+d^{2}-ab-ac-ad-bc-bd-cd.$
Since, $\frac{\partial^{2}A}{\partial a^{2}}=\frac{\partial^{2}A}{\partial b^{2}}=\frac{\partial^{2}A}{\partial c^{2}}=\frac{\partial^{2}A}{\partial d^{2}}=2>0$ then
$\max_{\{a,b,c,d\}\subset[0,1]}A(a,b,c,d)=\max\{A(1,1,1,1),A(1,1,1,0),A(1,1,0,0),A(1,0,0,0),A(0,0,0,0)\}=$
$=\max\{-2,0,1,1,0\}=1.$
[quote="venatrix"]
$\sum_{sym}a^{2}b \qquad a,b,c >0, a+b+c=1$[/quote]
See the Sung-yoon Kim's post for $\sum_{cyc}a^{2}b$ here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960
For $\sum_{sym}a^{2}b$ it's a trivial question: $\sum_{sym}a^{2}b\leq\frac{1}{4}.$ :wink:


IMO
Brocard
angles
geometry
Triangle
geometric inequality
imo 1991

Let $ \,ABC\,$ be a triangle and $ \,P\,$ an interior point of  $ \,ABC\,$. Show that at least one of the angles $ \,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $ 30^{\circ }$.

A solution was given in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=16171#p16171]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273 post #6[/url] by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=694]Johann Peter Dirichlet[/url]:

"Let ABC be a triangle and X an interior point of ABC. Show that at least one of the angles XAB, XBC, XCA is less than or equal to 30 degrees."

The solution is obvious with the Brocard point! The anterior messages contains all necessary theorems. P is the Brocard poin of the triangle ABC. If X=P then aren't nothing to prove. If X<>P then X is in the interior of one of the triangles PAB,PBC,PCA (wlog PAB).Then the problem is obvious (connect the point X to point A and verify the angle).

Here's a solution that I found right now.  It is very late at my place, so I will skip the obvious parts.

[i]Solution.[/i] Assume to the contrary that all the angles are greater than $ 30^{\circ}$.  Let $ PA\equal{}x,PB\equal{}y,PC\equal{}z$.  Then it follows that $ x^2\equal{}z^2\plus{}b^2\minus{}2bz\cos\angle PCA>z^2\plus{}b^2\minus{}bz\sqrt{3}$.  Similarly, we find anologous inequalities.  Adding them, we obtain $ a^2\plus{}b^2\plus{}c^2<\sqrt{3}(bz\plus{}ay\plus{}cx)$.  But $ [ABC]\equal{}\frac{1}{2}\sum cx\sin\angle PAB>\frac{ay\plus{}bz\plus{}cx}{4}$.  Thus, $ a^2\plus{}b^2\plus{}c^2<4\sqrt{3}[ABC]$.  This is a contradiction to a well-known inequality (an old IMO problem that can be easily proven using Heron's formula along with AM-GM). $ \Box$

Let $ d(P,AB)\equal{}x$ , $ d(P,BC)\equal{}y$ , $ d(P,CA)\equal{}z$
lemma. $ PA \plus{} PB \plus{} PC \geq 2(x \plus{} y \plus{} z)$

At least one of $ PA\geq 2x$ , $ PB\geq 2y$ , $ PC\geq 2z$ holds by lemma.
if we assume $ PA\geq 2x$ , $ \frac {x}{PA}\leq\frac {1}{2}$ and $ \angle{PAB}\leq 30^{\circ}$

[hide="Solution"]Let $ \angle PAC \equal{} \gamma$, $ \angle PCB \equal{} \angle \alpha$, $ \angle PBA \equal{} \angle \beta$, $ \angle PAB \equal{} \alpha'$, $ \angle PBC \equal{} \beta'$, and $ \angle PCA \equal{} \gamma'$.

We will use contradiction. Suppose that $ \alpha, \beta, \gamma > 30^{\circ}$. Now, observe that $ \sin \alpha, \sin \beta, \sin \beta > 1/2$, since $ 30^{\circ} < \alpha, \beta, \gamma < 150^{\circ}$. Now, observe that $ 1/8 < \sin \alpha \sin \beta \sin \gamma \equal{} \sin \alpha' \sin \beta' \sin \gamma'$ by Ceva's. Squaring both sides, $ 1/64 < (\sin \alpha' \sin \beta' \sin \gamma')^2 \equal{} \sin \alpha \sin \alpha' \sin \beta \sin \beta' \sin \gamma \sin \gamma'$. Now, observe that $ \minus{}\ln \sin x$ is convex, so by Jensen's,

$ \minus{}(\ln(\sin \alpha \sin \alpha' \sin \beta \sin \beta' \sin \gamma \sin \gamma'))/6 \geq \minus{}\ln(\sin(\frac{\alpha \plus{} \alpha' \plus{} \beta \plus{} \beta' \plus{} \gamma \plus{} \gamma'}{6}))$

which is equal to $ \minus{}\ln(1/2)$. Rearranging, we get that $ \sin \alpha \sin \alpha' \sin \beta \sin \beta' \sin \gamma \sin \gamma' \leq 1/64$, which is a contradiction. [/hide]

erdos-mordell inequality (when converted to trig form0 gives an immediate contradiction when we assume that all marked angles are at least 30.

Here's a very simple solution that I found for this problem:

If we draw a Triangle and interior point P, we label the vertices A, B, C anyway we want WLOG. Since the labeling of the vertices of the triangle can be interchanged, it suffices to show that at least one of the 6 angles $ \,\angle PAB,\;\angle PBC,\;\angle PCA,\;\angle PBA ,\;\angle PCB ,\;\angle PAC\,$ is less than or equal to 30. We prove this statement by contradiction. First, we assume that all six of these angles are greater than 30. Then, the sum of these angles must be greater than 180, which is a contradiction since the sum of these six angles must be 180. Thus, at least one of these 6 angles must be less than or equal to 30. We can label the vertices A, B, and C so that either $ \,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is one of these six angles.

Can someone look through and see if I made any mistakes in this proof?

Assume that $\,\angle MAB, \,\angle MBC, \,\angle MCA$ are all greater than $30^{\circ}$.

By $\textbf{sine Ceva},$

$\sin(MAC)\sin(MBA)\sin(MCB)$
= $\sin(MAB)\sin(MBC)\sin(MCA) > sin^3(30^{\circ})=\frac{1}{8}$.

Also, from other side, $\,\angle MAC + \,\angle MBA + \,\angle MCB < 180^{\circ} -3*30^{\circ} = 90^{\circ}$.

Note that the function $f(x)=\log(\sin(x))$ is concave on $[0,\pi]$. Applying Jensen, we get a contradiction here.

[i]Trivialized by Jensen's[/i]

But more trivialized by Erdos-Mordell inequality:

Let the projections of $M$ onto sides $BC,AC,$ and $AB$ be $X,Y,$ and $S$, respectively. By Erdos-Mordell, $$MA+MB+MC \geq 2(MX+MY+MS).$$ This implies that at least one of $MA \geq 2MS$, $MB \geq 2MX$, or $MC \geq 2MY$ must be true. WLOG, let $MA \geq 2MS$. Since in right $\triangle SAM$ we have $\sin SAM \leq \frac{MS}{MA}=\frac{1}{2}$, which implies $\angle SAM \leq 30^\circ$ and we are done. $\mathbb{S.A.M.!}$

Same as everyone else.

[hide=Solution]Denote the feet of the perpendiculars from $P$ to $BC, AC, AB$ by $X,Y,Z$ respectively. Then Erdos-Mordell gives $PA+PB+PC \geq 2(PX+PY+PZ)$, so at least one of the inequalities $PA \geq 2MZ, PB \geq 2PX, PC \geq 2PY$ holds true. This means at least one of the inequalities $\sin \angle PAB \leq \frac{1}{2}, \sin \angle PBC \leq \frac{1}{2}, \sin \angle PCA \leq \frac{1}{2}$ holds true, so at least one of the angles is less than or equal to $30$ as desired.[/hide]


Given triangle ABC. Prove that $ 8\cos A\cos B\cos C\le\cos (A\minus{}B)\cos (B\minus{}C)\cos (C\minus{}A)$

This was posted not long ago
Isn't It from an on-going competition!!

T/6 379

N.N.Trung, are you Vietnamese? If so, you should have common sense.

Should be locked!

Japanese Communities Moderator

kunny


group theory
abstract algebra
Support
superior algebra
superior algebra unsolved

Suppose we have a $ k$-cycle in the group $ S_{n}$ where $ k\leqslant n$.  Is there any reasonable way to calculate the number of distinct subgroups of $ S_{n}$ generated by all $ k$-cycles?

For example, if we take all $ \left(abcde\right)\in S_{12}$, how many of the subgroups $ \left<\left(abcde\right)\right>$ are distinct?  I know that there are $ \binom{12}{4}$ such $ 5$-cycles, but they don't all generate distinct subgroups.

Let's say that the support of a cycle is the set of its non-fixed elements. Any two $ k$ cycles with distinct supports generate distinct subgroups.

Two $ k$ cycles with the same support generate the same subgroup iff one of them is the $ p$th power of the other, where $ p\in \mathbb Z_k^*$. There are $ \varphi(k)$ such values $ p$. Thus we get $ \frac{(k\minus{}1)!}{\varphi(k)}$ distinct subgroups from $ k$-cycles with fixed support.

Since there are $ \binom{n}{k}$ possible supports, there are $ \frac{(k\minus{}1)!}{\varphi(k)}\binom{n}{k}$ distinct subgroups generated by $ k$-cycles. Right?  :maybe:


Anyone play Halo 2 on Xbox Live in here?  It would be pretty cool to play a game with some AoPSers.  If anyone wants to add me to their friend list my Gamertag is cyclotomic.

Man, I wish I had an XBOX. Do you have HALO 1 for the PC?

Halo 2 + Xbox Live = Awesome

My handle is Pitcany.

Want it, except I'd feel guilty about having it and wasting the money because I hardly ever get to play Halo2 as it is with school and everything.

Oh well, at least there's still LAN parties.

i have halo 2, but not with x-box live. u haf 2 pay extra for that cause its connected to the world wide web. but i never play it anyways....my brother does. :lol:

halo 2 is awesome but i can't get xbox live cuz i have stupid dial up and my xbox is not compatible anymore as i had some modifications done to it. o well. i'll just go to a friends house to play online.

[b][i][u]i have halo 2 but i dont have live.. i wish i did..[/u][/i][/b]

Yo EFUZZY.
I got Halo 1, and i'm willing to play. Add Divine eggnog to your aim list, and im me when im online. My username is [hide="click text"]sh!tbr!ck[/hide]. I stopped cussing a long time ago, but that is my best file[/hide]

how do you get Halo 1 demo online version for PC ?

I really want to try it...........

never played halo before except one time when I went over my friend's house....

well, it was halo 1 though.....

sheepwarrior , you sound like my chinese friend Raymond...

well, I'll try...it never worked forme though...

could you give me the site address?bittorrent...


Putnam
function
number theory
relatively prime
college contests

Hey,
[quote="sexynsmartjenny"]
As for the solution to this problem, it used height functions, and it says that for every next composed function, the height of the solutions are at least greater than 2 plus the height of the previous composed function.  But how does that mean that the intersection is empty?  I just don't see it. And what is a height function?  Please help me.  Please?   :blush:
[/quote]
It would be helpful to know what the original problem was,
Putnam 2001 B-4
Let $ {S}$ denote the set of rational numbers different from $ {{\{}{{ - 1, 0, 1}}{\}}}$. Define $ {{f: S}{\to}{S}}$ by $ {{f(x)} = x-\frac1x}$. Prove or disprove that
$ {\bigcap_{n = 1}^{\infty}}{{f}^{(n)}}{(S)} = {\emptyset}$
where $ {{f}^{(n)}}$ denotes $ {f}$ composed with itself $ {n}$ times.

As for what a height function is, I'm not all that sure myself. In the [i]Art and Craft of Problem Solving[/i], 2nd Edition, Paul Zeitz touches on the concept but doesn't elaborate. In anycase, I found these websites pretty useful for some background information,

http://www.math.brown.edu/~jhs/HtSurveyMSRIJan06.pdf
http://planetmath.org/encyclopedia/HeightFunction.html

Thanks,

-APSStudent

I doubt that any of those links contain information relevant to this problem. As for the solution: since $ x\minus{}1$ and $ x$ are relatively prime, a rational number in $ S$ must have an infinite number of prime factors, which is absurd. (Maybe this is where the "height" comes in: the number of prime factors?)

Keyword missing: the set of numbers [i]different[\i] from $ \{ \minus{} 1,0,1\}$.

The idea: start with all rational numbers except those three- $ 1\mapsto 0\mapsto \infty$ and $ \minus{} 1\mapsto 0\mapsto\infty$. Apply the function to get $ f(S)$, which is contained in $ S$. Apply it again, getting $ f^2(S)\subset f(S)$. Repeat. No $ f^k(S)$ is empty, but each $ q$ appears only in a finite subset of them.

The height idea in the proof: show that $ f(r)$ is always more complicated than $ r$ in some sense for rational $ r$ other than the initial excluded list.

Is the function well-defined?

Isn't $ f(1/2) \equal{} \frac {1/2 \minus{} 1}{1/2} \equal{} \frac { \minus{} 1/2}{1/2} \equal{} \minus{} 1$?

Edit: Internet tells me that the correct $ f$ is $ f(x) \equal{} x \minus{} 1/x$.

Edit 2: Really neat idea, jmerry.

Edit 3: Jenny, the "height function" here must be the following $ H$ or something similar: If $ a/b$ is fully reduced (i.e. $ \gcd(a,b)\equal{}1$), then $ H(a,b)\equal{}\max(a,b)$.

I also recall seeing the height function $ H(a, b) \equal{} a \plus{} b$ in Tate-Silverman (unless I'm remembering incorrectly).