[quote="mihail911"]wouldn't it be 128..because the distance between m and n is 7 and the distance between n and p is 9 so the distance between m and p is 16....so 16 must be the length of the diagonal...so now you have 16=x*sqrt 2...solve and you get the length of the side is 8*sqrt 2...square it and you get 128...im not sure[/quote]
No, no one is telling you that the points are collinear, so the diagonal isn't 16.

[quote="CircleSquared"]In my diagram, the 4 triangles are congruent.  Each one is a right triangle with legs 7 and 9.  So the area of the square is $(\sqrt{7^{2}+9^{2}})^{2}=130$, I think.[/quote]
that's correct :)

wait...but how do you know that they are congruent and that each has legs of side length 7 and 9?

You can't see it in the diagram, but I purposely tried to draw it so that as many angles were right as possible to fit the solution.  So all 4 triangles are right.
That means the acute angles of each triangle are complementary.
Since the angles of the square are each 90, the triangle angles that make up the angles of the square are complementary, so each corresponding pair of acute angles in the triangles is congruent (it's hard to explain, as I seem to have forgotten to label the diagram).
Also, I used the fact that all the sides of a square are congruent.   Then by AAS congruence, the 4 triangles are congruent.

The lengths 7 and 9 are given in the problem.  I just used the definition of distance between 2 lines.


geometry
3D geometry
LaTeX
geometry unsolved

$ ABCDEFGH$ is a cube of side $ 2$ with $ A, B, C, D$ above $ F, G, H, E$, respectively.

(a) Find the area of the quadrilateral $ AMHN$ where $ M$ is the midpoint of $ BC$ and $ N$ is the midpoint of $ EF$.

(b) Let $ P$ be the midpoint of $ AB$, and $ Q$ the midpoint of $ HE$. Let $ AM$ meet $ CP$ at $ X$, and $ HN$ meet $ FQ$ at $ Y$. Find the length of $ XY$.

hi
(a)AMHN is a rhomb , so it's area is 2*6^0.5
(b)   (2*11^0.5)/3
i'm sorry that i can't write in LaTeX


Does anyone know how to solve this exercise? I have no idea...thank you
[img]http://img143.imageshack.us/img143/2049/checkift1eo.jpg[/img]

If you let a, b, and c equal 1, then the expression would equal 1/4.
If you let a, b, and c equal anything greater than 1, the denominator will grow faster than the numerator, therefore making the expression smaller.
I think any decimals would still be the same. I don't know how to prove it, though.  :blush:

[quote="perfectnumber628"]well, $\frac{a}{2a}=\frac{1}{2}$, so then adding more stuff (that's postive) to the denominator will just make the fraction even smaller.  so, is the expression always less than or equal to $\frac{1}{2}$?[/quote]

Hes got it right, the expression < 1/2 because they cant = 0 and a/2a is 1/2.

edit: and whats the sum there for? its not going to anything.

When you put in the sigma, did you mean that we have to add up all the possible expressions resulting when you plug in variables for a, b, and c?

If not, you can use small fractions and get below 1/2 (I think)
if:
a=1/8, b=1/16, and c=1/32
then $\frac{1/8}{2(1/8)+1/16+1/32}=4/11=.363636....$

[quote="swirly3"]When you put in the sigma, did you mean that we have to add up all the possible expressions resulting when you plug in variables for a, b, and c?[/quote]
Yeah, I believe that's what he means.
In other words, we need to prove/disprove that $\frac{a}{2a+b+c}+ \frac{b}{a+2b+c} +\frac{c}{a+b+2c} \le \frac34$ for all positive reals $a$, $b$, $c$.

This problem is definitely beyond the GS level and needs to be moved to Intermediate/Pre-Olympiad forum.

yea, this is way above GS level.

[hide]
Make the substitutions $x=\frac{a}{a+b+c},\ y=\frac{b}{a+b+c},\ z=\frac{c}{a+b+c}$ so that $x+y+z=1$ and the original inequality is equivalent to $\sum \frac{x}{2x+y+z} \leq \frac 34$ (just divide numerator and denominator by $a+b+c$).

$\sum \frac{x}{2x+y+z}=\sum \frac{x}{x+1}=3-\sum \frac{1}{x+1}$.

By AM-HM, $\sum \frac{1}{x+1}\geq \frac{9}{\sum (x+1)}=\frac 94$. So $3-\sum \frac{1}{x+1}\leq \frac 34$ which proves the inequality.
[/hide]

[quote="VenDiddy"]From what to what does the $\sum$ go to?[/quote]

Just to explain, the $\sum$ is really a "cyclic sum."

$\sum \frac{a}{2a + b + c}$ really means

$\frac{a}{2a + b + c} + \frac{b}{2b + c + a} + \frac{c}{2c + a + b}$.


Prove that if $ a,b,c>0$ and $ \frac {1} {a^2}\plus{}\frac {1} {b^2}\plus{}\frac {1} {c^2} \equal{}\frac {1} {2}$ then:

$ a^4\plus{}b^4\plus{}c^4\plus{}a^2b^2c^2 \ge 3abc(a\plus{}b\plus{}c)$

[quote="moldovan"]Prove that if $ a,b,c > 0$ and $ \frac {1} {a^2} \plus{} \frac {1} {b^2} \plus{} \frac {1} {c^2} \equal{} \frac {1} {2}$ then:

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2c^2 \ge 3abc(a \plus{} b \plus{} c)$[/quote]
It's obvious because $ a^4\plus{}b^4\plus{}c^4\geq abc(a\plus{}b\plus{}c)$ and $ a^2b^2c^2\geq2abc(a\plus{}b\plus{}c).$

Hello arqady...
Could you please explain how $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \geq abc(a \plus{} b \plus{} c)$ under the given condition?? :maybe:

EDIT: I get now...

$ \left(\frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b}\right)(a\plus{}b\plus{}c) \geq (ab\plus{}bc\plus{}ca)^2 \geq (a\plus{}b\plus{}c)^2$

I think....is it right?? :maybe:

$ a^2b^2c^2 \equal{} 2\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\right)a^2b^2c^2 \equal{} 2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\ge 2abc(a\plus{}b\plus{}c)$ by Muirhead (or weighted AM-GM). The first inequality is also only Muirhead.

$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\ge a\plus{}b\plus{}c$ is wrong for $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt{6}}$.

[quote="Sunkern_sunflora"]Hello arqady...
Could you please explain how $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \geq abc(a \plus{} b \plus{} c)$ under the given condition?? :maybe:

[/quote]
Even  $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \geq 6\sqrt6abc(a \plus{} b \plus{} c).$ :wink:

Actually $ (ab)^2\plus{}(bc)^2\plus{}(ca)^2\ge ab\cdot bc\plus{}bc\cdot ca\plus{}ca\cdot ab\equal{}abc(a\plus{}b\plus{}c)$.

[quote="arqady"]Even  $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \geq 6\sqrt6abc(a \plus{} b \plus{} c).$ :wink:[/quote]

But what if $ a\equal{}b\equal{}c$? :maybe:

[quote="nayel"]Actually $ (ab)^2 \plus{} (bc)^2 \plus{} (ca)^2\ge ab\cdot bc \plus{} bc\cdot ca \plus{} ca\cdot ab \equal{} abc(a \plus{} b \plus{} c)$.

[quote="arqady"]Even  $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \geq 6\sqrt6abc(a \plus{} b \plus{} c).$ :wink:[/quote]

But what if $ a \equal{} b \equal{} c$? :maybe:[/quote]
Oh... I meant to prove that $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \geq 6\sqrt6(a \plus{} b \plus{} c)$ with the same conditions.
The trace of this problem see in the bertram's post here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1259510#1259510

Oh OK. :)

First we have $ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\ge abc(a\plus{}b\plus{}c)$.

Then $ abc\equal{}\sqrt{2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}\ge\sqrt 6\sqrt[3]{a^2b^2c^2}\implies abc\ge 6\sqrt 6$. :)


number theory
greatest common divisor
search
number theory unsolved

Let $ a, b, c$ be three natural numbers such that $ a < b < c$ and $ gcd (c \minus{} a, c \minus{} b) \equal{} 1$. Suppose there exists an integer $ d$ such that $ a \plus{} d, b \plus{} d, c \plus{} d$ form the sides of a right-angled triangle. Prove that there exist integers, $ l,m$ such that $ c \plus{} d \equal{} l^{2} \plus{} m^{2} .$
[b][Weightage 17/100][/b]

we obviously have ($ k,r,s$ are integers with $ (r,s)\equal{}1$ and $ r \not \equiv s \mod{2}$):
$ (a\plus{}d,b\plus{}d,c\plus{}d)\equal{}(k(r^2\minus{}s^2),k(2rs),k(r^2\plus{}s^2))$ or $ (a\plus{}d,b\plus{}d,c\plus{}d)\equal{}(k(2rs),k(r^2\minus{}s^2),k(r^2\plus{}s^2))$.
by computing $ c\minus{}b$ and $ c\minus{}a$ we get $ k\equal{}1$.so we have $ c\plus{}d\equal{}r^2\plus{}s^2$.as we wanted.

Since $a<b<c$ and gcd$(c-a, c-b) = 1$. There exist $x$, $y$ such that $(c-a)x + (c-b)y = 1 \implies c(x+y) + a (-x) + b (-y) = 1.$
Therefore gcd$(a,b,.c)$ = 1. Thus for any $d$, we have gcd$(a+d, b+d, c+d) = 1$ and that means $(a+d, b+d, c+d) $ is a primitive Pythagorean triplet. Therefore there exist $l,m$ $(l>m)$ with opposite parity and gcd$(l,m)$ = 1 such that $a+d = l^2 - m^2, b+d = 2lm$ and $c+d = l^2 + m^2.$

[quote=SigmaG]You have to prove that $c+d=l^2+m^2$ and not $(c+d)^2=l^2+m^2$. Note the difference.[/quote]

OMG, thanks...
Maybe I was drunk! :(

Sorry for the spams above..So to compensate I am giving my solution:-
Observe that $gcd(a+d,b+d,c+d)=1$ as $gcd(c-a,c-b)=1$(given).
Thus $(a+d,b+d,c+d)$ is a primitive pythagorean triplet..
Thus there exist $u,v,w$ such that $c+d=u^2 +v^2,b+d-2uv,a+d=u^2 - v^2$.
DONE...
QED!


What is the sum of the proper divisors of 256?

The proper divisors are all the divisors except the number itself, so the answer is $ 1\plus{}2\plus{}4\plus{}\cdots\plus{}128\equal{}255$.

There is a formula here: http://216.235.203.102/NETCOMMUNITY/Document.Doc?id=120 it should be towards
the very end of the last page. It gives a more general formula to find the sum of the factors of any number. In this case however it is quite useless.

for any number that is equal to 2^x=n, where x is a positive integer, the sum of the proper factors is equal to n - 1.


For an triangle ABC.
  zH=zA+zB+zC-2ZO.   This is the relation of Sylvester???


Let $a,b,c$ positive reals, such that $abc=1$

Prove that $\frac{a}{b}+\frac{b}{c}+\frac{1}{a}\geq a+b+1$

I have universal inequality:Prove that:$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq x+y+z$
where $x;y;z$ are positive  real numbers satisfying $abc \geq 1$


calculus
derivative
calculus computations

Remark that $ \frac{d}{dx}(xy)\equal{}y\plus{}xy',\ \frac{d}{dx}y^2\equal{}2yy'$,

Differentiating both sides with respect to $ x$ gives the answer (e).


i dont think thats possible is it? 29 is prime.

Just use the quadratic formula  :wink: .

[quote="7h3.D3m0n.117"]Just use the quadratic formula  :wink: .[/quote]

Which is this:

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

We use this when the equation is set up like this:

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

So in your case, $ a,b,c$ would equal $ 2,6,\minus{}29$ respectively.

So plug those values to get your answer.

$ \star$ It should get you [u]2[/u] answers, positive value and negative value.

Not necessarily a positive and negative value.

Consider the quadratic $ x^{2}\minus{}2x\plus{}1$

applying the quadratic formula gives us

$ \frac{2 \pm \sqrt{4 \minus{} 4}}{2}$

Both solutions are positive in this case.

[quote="SnipedYou"]Not necessarily a positive and negative value.

Consider the quadratic $ x^2\minus{}2x\plus{}1$

applying the quadratic formula gives us

$ \frac{2\pm{\sqrt{4\minus{}4}}}{2}$

Both solutions are positive in this case.
[/quote]

[quote="t0rajir0u"][quote="DonkeyKong"]$ \star$ It should get you [u]2[/u] answers, positive value and negative value.[/quote]

True when $ ac$ is negative, but not true when $ ac$ is positive.

In any case, we are looking to factor, not to solve, so the last step is to write the quadratic equation as

$ ax^2 \plus{} bx \plus{} c \equal{} a \left( x \minus{} \frac { \minus{} b \plus{} \sqrt {b^2 \minus{} 4ac}}{2a} \right) \left( x \minus{} \frac { \minus{} b \minus{} \sqrt {b^2 \minus{} 4ac}}{2a} \right)$[/quote]

Oh.. Yeah.. I am sorry.. You guys are right.. Good catch..

[quote="7h3.D3m0n.117"]Just use the quadratic formula  :wink: .[/quote]
that wont help you factor

[quote="t0rajir0u"]Actually, as I've already demonstrated, the quadratic formula is in fact a foolproof method to find a factorization.  It's just tedious.[/quote]
Foolproof?? You mean Fullproof??
 :?:
What...

http://dictionary.reference.com/browse/foolproof

The quadratic formula always gives the correct roots, and since the roots give a factorization, the quadratic formula always leads to a factorization.  In many cases it is simply much more work than it's worth.

use quadratic formula, but consider 2 before factoring  :lol:


[quote]There are three piles of objects. The first contains 3 objects, the second contains 5, and the third contains 7. Two players alternate picking any number of objects (at least one) from one of the piles. The player taking the last object loses.

(1) Determine which player has a winning strategy.

(2) Find this strategy.[/quote]
I was able to find the answer using a very tedious recursive method. But when I try to generalize it to any 3-pile object, the pattern seems unclear. Any ideas?

[hide="Answer"]The first player has a winning strategy. The actual strategy is too complicated to type up.
[/hide]

If I'm not misunderstanding, you're asking about [url=http://en.wikipedia.org/wiki/Nim]Nim[/url].

The theory of [url=http://en.wikipedia.org/wiki/Nimber]Nimbers[/url] turns out to have applications far beyond the game of Nim itself due to the [url=http://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem]Sprague-Grundy theorem[/url].


function
algebra
polynomial
inequalities
real analysis
real analysis unsolved

Show that the set of even functions on $ C[\minus{}1,1]$ form a proper, closed subalgebra of $ C[\minus{}1,1]$.

How would I show that $ (fg)h\equal{}f(gh)$, $ f(g\plus{}h)\equal{}fg\plus{}fh$, and $ (\alpha f)g\equal{}\alpha (fg)$?  That the even functions form a proper subset is because there are also odd functions, and functions that are neither even nor odd.

If multiplication is pointwise, verifying the axioms is trivial (and you also need to verify that the sum/product of two even functions is even); the only hard part is proving closure, i.e. the limit of even functions is even.  (Hint:  for any sequence $ f_n$ of even functions, consider the sequence $ f_n(x) \minus{} f_n(\minus{}x)$.)

How would I prove the sum/product of two even functions is even?  If they're polynomials it's easy but how about just $ f(x)g(x)\equal{}f(\minus{}x)g(\minus{}x)$ or $ f(x)\plus{}g(x)\equal{}f(\minus{}x)\plus{}g(\minus{}x)$?  Why are those necessarily even?

And I guess we're trying $ f_{n}(x)\rightarrow f(x)$, where $ f$ is even?

[quote="JRav"]How would I prove the sum/product of two even functions is even?  If they're polynomials it's easy but how about just $ f(x)g(x) \equal{} f( \minus{} x)g( \minus{} x)$ or $ f(x) \plus{} g(x) \equal{} f( \minus{} x) \plus{} g( \minus{} x)$?  Why are those necessarily even?[/quote]
It's because of exactly what you've written down.  If $ f(x) \equal{} f(\minus{}x)$ and $ g(x) \equal{} g(\minus{}x)$ then $ f(x) g(x) \equal{} f(\minus{}x) g(\minus{}x)$ and $ f(x) \plus{} g(x) \equal{} f(\minus{}x) \plus{} g(\minus{}x)$.  It doesn't get any plainer than this.

If you don't see that $ f(x)g(x)\equal{}f(\minus{}x)g(\minus{}x)$ is even, write $ h(x)\equal{}f(x)g(x)$. Then you have $ h(x)\equal{}h(\minus{}x)$  :D

[quote="t0rajir0u"](Hint: for any sequence $ f_{n}$ of even functions, consider the sequence $ f_{n}(x)\minus{}f_{n}(\minus{}x)$.)[/quote]
Would this work?
Let $ (f_{n})$ be a sequence of even functions with $ f_{n}\rightarrow f$.  Then
$ \left|f_{n}(x)\minus{}f_{n}(\minus{}x)\right|\leq\left|f_{n}(x)\minus{}f(x)\right|\plus{}\left|f(x)\minus{}f_{n}(\minus{}x)\right|\equal{}$
$ \left|f_{n}(x)\minus{}f(x)\right|\plus{}\left|f(\minus{}x)\minus{}f_{n}(\minus{}x)\right|<\frac{\varepsilon}{2}\plus{}\frac{\varepsilon}{2}\equal{}\varepsilon$
Then $ f_{n}(x)$ converges to $ f_{n}(\minus{}x)$ so it is even.

Did I mix up some of the functions or inequalities?

In fact $ f_n(x) \minus{} f_n( \minus{} x) \equal{} 0$ for all $ n$ and $ x$, and you are [i]given[/i] this!  You want to show that if the $ f_n$'s are even and $ f_n\to f$, then $ f(x) \minus{} f( \minus{} x) \equal{} 0$ (do what you did with the roles of the $ f_n$'s and $ f$ reversed)

Okay, that's what I was suspecting.  Thanks.


Any AoPSers going to mathcamp from the NY metro area?  What flight are you going to be on?

I have two options that cost the same amount, and I'm wondering which ones you guys are going to be on.

1) Take Jet Blue from JFK to Portland 7:55 PM to 10:55 PM on Jul 3.  Take Jet Blue from Portland to JFK 12:00AM to 8:15 AM on Aug 7.
Pro: Direct flight.  Con: Inconvienent times

2) Take American Airlines from La Guardia to Portland via Chicago on Jul 3.  Take American Airlines from Portland to La Guardia via Chicago on Aug 7.
Pro: Convienent times.  Con: Connecting flight.

I'm down in central Jersey, but fly out of Newark as there are practically NO airports in Jersey... ironic that it's the most densely populated state, but a lot of things about Jersey work out that way.

Anyway, if you're willing to fly out of Newark, there's a Continental flight EWR-PDX out at 8am on July 3, and 7am August 7... I believe.  Both flights are nonstop, AND the hours are pretty good.  If you want details/flight numbers, let me know.

I had the same problem; this was the only direct flight I could find in either Newark of Phili.  Everything else had me connecting through Chicago or Cincinnati... Or something really out of the way like Dallas.

Hope it all works out

well, I've decided to take JetBlue because I don't want to connect.

Ya, I was considering that EWR flight, but frankly, I do not want to wake up at 4 in the morning and such.  I'm a night person, not a morning person.

Hmm...anyone else flying from Houston?

Continental...


calculus
integration
logarithms
limit
real analysis
real analysis unsolved

I think the answer is not infinity , but I cannot figure it out .

$\int_{1}^{\infty}\sum_{p}\frac{1}{p^{x}}\ dx \ , \ p\in\text{Primes}$

We can interchange summation and integration because all the terms are positive (Tonelli's theorem). Since $\int_{1}^{\infty}\frac{1}{p^{x}}dx=\frac{1}{p\log p}$, the question reduces to the convergence of $\sum_{n=1}^{\infty}\frac{1}{p_{n}\log p_{n}}$, where $p_{n}$ is the $n$th prime. By the Prime Number theorem $\lim_{n\to\infty}\frac{p_{n}}{n\log n}=1$. Apply the Limit Comparison Test.


If $ x\plus{}y|xy$ show that $ x\equal{}y$.

sorry, the problem was to find whether $ x\plus{}y|2xy$ (or also $ x\plus{}y|x^2 \plus{} y^2$)

well... GREAT!!!!

what I really want is to characterize all solutions.

Okay then, let $ g \equal{} \gcd(x,y)$ and say $ x \equal{} gx'$ and $ y \equal{} gy'$ where $ \gcd(x',y') \equal{} 1$. Then we need

$ x\plus{}y|2xy$
$ gx' \plus{} gy'|2g^2x'y'$
$ x' \plus{} y'|2gx'y'$

But $ \gcd(x',y') \equal{} 1$ so $ \gcd(x',x' \plus{} y') \equal{} \gcd(y',x' \plus{} y') \equal{} 1$, so this is equivalent to

$ x' \plus{} y'|2g$

so the characterization of all solutions is $ (x,y) \equal{} (gx',gy')$ where $ x' \plus{} y'|2g$.


inequalities
calculus
logarithms
inequalities proposed

Let $k\ge\frac{\sqrt{5}-1}2$. For positive reals $x_{1},...,x_{n}$ with product $1$, prove
\[\sum_{i=1}^{n}\frac{k+1}{k+x_{i}}\le \sum_{i=1}^{n}x_{i}.\]
I have a beautiful proof. :)

[hide="more on $k$..."]Much more is true. The inequality holds in fact for $k\ge k_{min}\approx .16836895$ which is asymptotically ($n\to\infty$) best possible. But my proof of that (starting with n-1EV theorem, followed by some calculus) is a bit ugly.
In fact, I can show that $k_{min}$ is the unique positive solution $x$ of the equation

 – allow a moment for taking a deep breath –

$2x(2+x)\ln\left(\frac{1-2x-2x^{2}-\sqrt{1-4x(1+x)^{2}}}{2(1+x)}\right)$ $+(1+2x)\sqrt{1-4x(1+x)^{2}}+1+2x^{2}+2x^{3}=0$
:D[/hide] :P
Are there nice direct solutions for cases like $k=\frac12$ ? Even a proof of that only for $n=3$, i.e. proving
$\sum\frac3{2x+1}\le x+y+z$ for $xyz=1$ would be welcome!

before dealing with the problem let me remind you that your last inequality is equaivalent to $abc=1$
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sum_{cyc}\frac{a}{a+2c}$
and Darij posted somewhere in the forum the stronger inequality to which he had a solution with computer algebra and was searching a nice one.(don't know if he found it)
$abc=1$
$a+b+c\geq 3\sum_{cyc}\frac{a}{a+2c}$
note that darij's inequality directly proves your's (but is surely very hard to prove...)

[quote="Albanian Eagle"]and Darij posted somewhere in the forum the stronger inequality to which he had a solution with computer algebra and was searching a nice one.(don't know if he found it)...[/quote]I'm afraid he hasn't  :(  one more reason for featuring the original problem.
[b]Please try it guys, you will enjoy it! [/b]It uses only AM-GM  :lol: [hide="hint"]
Start with writing it as $RHS-LHS\ge0$ and adding  $n(k+1+\frac1k)$  on both sides.[/hide]


Some time ago I tried to nest hides (see my post on the Test Forum) and it didn't work as expected. I wonder whether it is just a well-known bug that is low on the priority list (I know that there are many more important things to take care of) or I stumbled upon something new...

Best wishes and thanks for maintaining this wonderful site,
Fedja

Fedja,

Just couriosity, what is the bug you are talking about. I saw your post, and the tests done there for other people, and it seems, for me, that it works very well.

Best,

There is a bug. Look carefully at the test-the-hide-bug topic in the test forum (I don't remember which one is it exactly) ;)


I dont know if this a right place for this problem, but anyway, here is it.
Prove that $\tan{20}\tan{40}\tan{60}\tan{80}=3.$

hello, since $ tan(60^{\circ})\equal{}\sqrt{3}$ it remains to show that
$ \tan(\frac{\pi}{9})\tan(\frac{2\pi}{9})\tan(\frac{4\pi}{9})\equal{}\sqrt{3}$
and we have
$ \tan(\frac{\pi}{9})\equal{}\frac{(\minus{}1\minus{}i\sqrt{3})^{4/3}\minus{}(\minus{}1\plus{}i\sqrt{3})^{4/3}}{(\minus{}1\minus{}i\sqrt{3})^{1/3}(\minus{}i\plus{}\sqrt{3})\minus{}(\minus{}1\plus{}i\sqrt{3})^{1/3}(i\plus{}\sqrt{3})}$
Sonnhard.

My CTRL key has gone again, so I can't write in LaTeX.

Writing tan as sin/cos, we can simplify numerator and denominator separately. For the numerator, using the difference of two cosines is twice the sine of half the sum by the sine of half the difference, cos two theta is one minus sine squared theta and cos 60 = 1/2, we get 3/16. For the denominator, using the sum of two cosines is twice the cos of half the sum by the cos of half the difference, cos two theta is two cos squared theta minus one and cos 60 = 1/2, we get 1/16. Therefore the given expression equals 3/16 divided by 1/16, that is 3.

I hope this makes sense.

[quote="Raz"]I dont know if this a right place for this problem, but anyway, here is it.
Prove that $ \tan{20}\tan{40}\tan{60}\tan{80} \equal{} 3.$[/quote]

my solution :
 :thumbup:
[img]http://www.mathmontada.net/vb/uploaded/17_1257516062.gif[/img]
with my best wishes
mrmohamed

Use $ \tan A \tan (60^{\circ}\plus{}A) \tan (60^{\circ}\minus{}A)\equal{}tan3A$


exponential
composite number
number theory

Look here (and juse search function next time please): http://www.mathlinks.ro/Forum/viewtopic.php?highlight=5%5E100&t=56542


How do you find a formula for the magic number for an NxN magic square?

That website only confuses me, though I don't know why. Think about it this way: the sum of all the elements is $\frac{n^2(n^2+1)}{2}$ and since there are n rows each with equal sum, divide by n to get the sum of each row: $\frac{n(n^2+1)}{2}$. This is the same result that the website got.

[b]n(n :^2: + 1)/2 .[/b]

Sum of all the numbers = n :^2: (n :^2: + 1) / 2
divided by the number of rows, n.

*points up*

that's kinda off topic but CS is a first person shooter if i think that's what G-UNIT is referring to  :D

Ah yes. I played that once. But it's too much graphics and memory and expensive. Now if you want to play something simple, free, and just as fun, check out [url=http://lieroxtreme.thegaminguniverse.com]Liero[/url].

Anyway, thanks for the input everyone. I was wondering if there was an algorithm for filling in a grid of any size. I think if N is odd, then the center square is filled with the center term.

there are three methods actually, one is for odd ordered magic squares, singly even (a magic square that has an order of a multiple of 4), and doubly even (a magic square that has an order which is equal to a multiple of 4 plus 2). A link that describes how to solve them is: [url]http://mathworld.wolfram.com/MagicSquare.html[/url]


Let $ P(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$, $ Q(x) \equal{} x^2 \plus{} x \plus{} 2007$, $ P,Q\in \mathbb {R}[x]$, such that, $ P(x)$ has exactly $ 3$ roots ($ 3$ roots are all distinct), and $ P(Q(x))$ has no root. Prove that: $ P(2007) > \frac {1}{4^3}$.

You are wrong. Only $ P(2007)>\frac{1}{4^3}$ is true.

[quote="mathVNpro"]Let $ P(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$, $ Q(x) \equal{} x^2 \plus{} x \plus{} 2007$, $ P,Q\in \mathbb {R}[x]$, such that, $ P(x)$ has exactly $ 3$ roots ($ 3$ roots are all distinct), and $ P(Q(x))$ has no root. Prove that: $ P(2007) > \frac {1}{4}$.[/quote]

Let $ r_1$, $ r_2$ and $ r_3$ the three distinct real roots of $ P(x)$.

$ P(Q(x))$ has no root $ \iff$  $ Q(x)\equal{}r_i$ has no root $ \iff$  so $ x^2 \plus{} x \plus{} 2007 \minus{} r_i \equal{} 0$ has no root $ \iff$ $ 2007\minus{}r_i>\frac{1}{4}$

So $ P(2007)\equal{}\prod (2007 \minus{} r_i) > \frac{1}{4^3}$, as Rust previously said.

As counter-example of your statement :
$ P(x)\equal{} (x \minus{} 2007 \plus{} \frac{4}{8}) (x \minus{} 2007 \plus{} \frac{5}{8})(x \minus{} 2007 \plus{} \frac{6}{8})$

$ P(Q(x))\equal{}0$ has no roots and $ P(2007)\equal{}\frac{15}{64} < \frac{1}{4}$


How can you solve for $x$ in terms of $y$ for the following expression in order to find the ranges?

$y=\frac{2x^2}{x^2-4}$

Remember the plus/minus :)

But I got a different answer.

[hide]
$yx^2 - 4y = 2x^2$
$(y-2)x^2=4y$
$x=\pm 2\sqrt\frac{y}{y-2}$
[/hide]

why your title is "Ranges"? I thought we are supposed to find the range of $y$...

[quote="NoSoupForYou"][hide="My solution"]$y=\frac{2x^2}{x^2-4}\Rightarrow y=\frac{1}{2-\frac{1}{2}x^2}\Rightarrow 2-\frac{1}{2}x^2=\frac{1}{y}\Rightarrow \frac{1}{2}x^2=2-\frac{1}{y}\Rightarrow x=\sqrt{4-\frac{2}{y}}$[/hide][/quote]
i found your mistake. it is this one:
$y=\frac{2x^2}{x^2-4}\Rightarrow y=\frac{1}{2-\frac{1}{2}x^2}$ wrong
$y=\frac{2x^2}{x^2-4}\Rightarrow y=\frac{1}{2-\frac{1}{2}x^{-2}}$ correct

better viewing of Iversonfan2005's steps:
$y=\frac{2x^2}{(x^2-4)}$ (i put in $x^2$ to numerator)

$(x^2-4)y=2x^2$

$x^2y-4y=2x^2$

$x^2(y-2)=4y$

$x^2=\frac{4y}{(y-2)}$

$x=\pm 2\frac{\sqrt{y}}{\sqrt{y-2}}$ (i put in $\pm$ for you)


In how many ways can $ 3$ Americans, $ 4$ Germans, $ 2$ Frenchmen, and $ 3$ Russians sit around a circular table if those of the same country sit together?

There are $ 3!$ ways to arrange the countries around the table and $ 3!4!2!3!$ ways to arrange the people after that.  The answer is thus $ 3!3!4!2!3!\equal{}6 \cdot 6 \cdot 24 \cdot 2 \cdot 6\equal{}\boxed{10368}$.

[quote="AIME15"]There are $ 3!$ ways to arrange the countries around the table and $ 3!4!2!3!$ ways to arrange the people after that.  The answer is thus $ 3!3!4!2!3! \equal{} 6 \cdot 6 \cdot 24 \cdot 2 \cdot 6 \equal{} \boxed{10368}$.[/quote]

What is it $ 3!$ instead of $ 4!$ ways to arrange the countries?

When you arrange $ n$ objects around a circle, there are $ (n \minus{} 1)!$ ways to do so. See if you can prove it as an exercise.

When arranging objects in a circle, you must take into account that it is possible to rotate the table.  Thus, there are $ \frac{n!}{n}\equal{}(n\minus{}1)!$ ways to arrange the objects in a circle.

Or, fix one of the objects in a certain spot.  There are then $ (n\minus{}1)$ objects left to permute, for a total of $ (n\minus{}1)!$ ways.


[quote="GameBot"]The cube shown has edges of length 2 cm. What is the area of triangle $ BCD$? Express your answer in simplest radical form.

[asy]import three;

draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)--cycle3,linewidth(2));

filldraw((0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)--cycle3,gray,linewidth(2));

draw((1,0,1)--(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle3,linewidth(2));

draw((1,0,1)--(1,1,0)--(0,1,1)--(1,0,1)--cycle3,linewidth(2));

label("B",(1,1,0),SE);
label("C",(1,0,1),NW);
label("D",(0,1,1),NE);[/asy][/quote]

Equilateral triangle, side length $ 2\sqrt{2}$. Plugging in gives

$ 2\sqrt{3}$.


algebra
polynomial
linear algebra
matrix

Let $ A$ and $ B$ be $ n\times n$ matrices over a field $ F$ such that $ A^{2} \equal{} A$ and $ B^{2} \equal{} B$. Suppose that the matrices $ A$ and $ B$ have the same rank. Prove that $ A$ and $ B$ are similar.

Key word: idempotent.

$ A$ is diagonalizable over $ F$ iff its minimal polynomial splits into linear factors and has no repeated factors. If $ A^2\equal{}A,$ then the minimal polynomial divides $ x(x\minus{}1)$ and thus $ A$ is diagonalizable. [As an aside, its minimal polynomial is $ x(x\minus{}1)$ unless it is either the identity matrix or the zero matrix.]

Since $ A$ and $ B$ are diagonalizable and each one has only $ 0$ and/or $ 1$ as eigenvalues, each is similar to a diagonal matrix with some $ 1$'s and some $ 0$'s on the main diagonal. We can choose to arrange that with the $ 1$'s first and the $ 0$'s later. These are the same precisely when the number of $ 1$'s is the same, and that number is the rank.

Note also that for such a matrix, $ \text{rank}(A)\equal{}\text{trace}(A).$ Hence we can say that $ A$ and $ B$ are similar iff $ \text{trace}(A)\equal{}\text{trace}(B).$ Saying it this way saves us the computational trouble of row eliminating to find the rank.

[quote="Kent Merryfield"]Note also that for such a matrix, $ \text{rank}(A) \equal{} \text{trace}(A).$ Hence we can say that $ A$ and $ B$ are similar iff $ \text{trace}(A) \equal{} \text{trace}(B).$[/quote]

Only over a field of characteristic $ 0$.

  darij

Oh, yeah - I was thinking about real matrices and I forgot that point. You do have to be a little more careful in finite characteristic.


topology
advanced fields
advanced fields unsolved

How can I find the closure and boundary in $ \mathbb{R}$ of the following set:
$ A \equal{} \{cos((\pi/4)n) \plus{} n/(n \plus{} 1): n\in\mathbb{N}\}$ ?

thanks for the help

Being countable, the set has no interior. So its closure and its boundary are the same thing. It's easier to think of the closure: think of all convergent subsequences of the given sequence and add their limits to the set.


hey everybody!
  I looked all over online but could not get a satisfactory answer (maybe it's the exact words I googled) to this question: [b]how does one calculate the trigonomic ratios (i.e. sine, cosine, and tangent - cosecant, secant, and cotangent are trivial) of a given angle? (Step by step, please) [/b]

Challenge Question:
How do calculators do it?

[url=http://en.wikipedia.org/wiki/Taylor_series]Taylor series[/url].  Those for the sine and cosine converge quite quickly, and there is a nice theorem that tells you exactly how many terms you need to reach a desired accuracy.

COOL!!!!  :lol: Thanks for the super speedy response!

A couple of questions:
Is it possible to calculate the trigonomic ratios (without a calculator) without using calculus?

Is the Taylor series also how calculators calculate it so fast?

[b]And what is that theorem????[/b]

[quote="theax"]Is it possible to calculate the trigonomic ratios (without a calculator) without using calculus? [/quote]
For an [i]arbitrary[/i] angle?  No.  For arbitrary angles, the trigonometric ratios aren't rational or even algebraic, so there's really nothing else you could possibly do.  (This is also how calculators compute exponentials and logarithms.)

Edit:  Well, okay, this is sort of a lie.  You could write the angle (after dividing by $ \pi$) in binary and repeatedly apply the angle-addition and half-angle formulas, but this is definitely more work than using the Taylor series.

[quote="theax"]Is the Taylor series also how calculators calculate it so fast?[/quote]
Yes.  That was intended as the answer to both questions.

[quote="theax"][b]And what is that theorem????[/b][/quote]
Predictably enough, it's called [url=http://en.wikipedia.org/wiki/Taylor%27s_theorem]Taylor's theorem[/url].


Given four positive integers a,b,c,d such that a>b>c>d>0 and $ ac\plus{}bd\equal{}(b\plus{}d\plus{}a\minus{}c)(b\plus{}d\minus{}a\plus{}c)$. Prove that ad+bc is not a prime.

[quote="mr.danh"]Given four positive integers a,b,c,d such that a>b>c>d>0 and $ ac \plus{} bd \equal{} (b \plus{} d \plus{} a \minus{} c)(b \plus{} d \minus{} a \plus{} c)$. Prove that ad+bc is not a prime.[/quote]
A nice proof in http://www.mathlinks.ro/viewtopic.php?t=150525
it is same your n_3,mr danh


prove that $\displaystyle \sum_{n=3}^\infty( (\log_{n-1} n) -1)$ diverges

[hide="Possible Hint"] By using the change of base property, you can find the terms are going to be $(\frac{\log 4} {\log 3}-1, \frac{\log5} {\log 4}-1, \frac{\log 6} {\log 5}-1, ...)$, all of which are positive, but each term is smaller than the last so that alone won't do. I'm not sure if this helps  :? [/hide]

[quote]$\ln (1+\frac1{n-1})>\frac1{n-1}$ [/quote]
are you sure? :huh:  ln (3/2)<1/2

edit:You're on the right track though; it turns out 3 is the only n that it doesnt work for

Actually, I'm wrong... Too hasty.

Thanks.

Try using $\log(1+x)>x-x^2/2$ instead.  This allows you to establish a result analogous to the one Farenhajt proved earlier, which is enough to conclude.

Alternatively, that sum of positive reals will converge only if the product, $\Pi_{n=3}^{\infty}\log_{n-1}n$ converges.  Using change of base, this product is $\Pi_{n=3}^{\infty}\frac{\log(n)}{\log(n-1)}$ which certainly diverges.


Single-voice Oriental-sounding melodies are super-easy to compose. Here's one I just made up (see below for notation)

4/4
$ (35)|5\cdot3_{1/2}(65)(35)|5_{3} (35) |1^{\cdot} \cdot 1_{1/2}^{\cdot} (2^{\cdot} 1^{\cdot} ) (6 1^{\cdot}) |5_{3}$ ... and so on

[hide="notation"]
(This is the best LaTeX I could come up with to simulate the notation for music you see in Chinese folk song books...)
Below is all for 4/4 time - easily adjusted to other time sigs.
[b]A number indicates pitch.[/b]
$ 1$ = tonic (a C in C major)
$ 2$ = 2nd degree / 1 tone above tonic / "re" (a D in C Major)
$ 3$ = 3rd degree / 2 tones above tonic / "mi" (an E in C Major)
$ 4$ = 4rd degree / 5 semitones above tonic / "fa" (an F in C Major)
... etc.
[b]Superscripted and Subscripted dots indicate octave.[/b]
In C major:
$ 1$ = Middle C (or any specific C you'd like)
$ 1^{\cdot}$ = C above middle C
$ 1_{\cdot}$ = C below middle C
[b]Solo numbers or numbers within parentheses have a total value of 1 beat.[/b]
$ 5$ = a "so" note for a duration of 1 beat
$ (35)$ = a pair of beamed eighth notes
[b]Subscripted numbers indicate number of beats.[/b]
$ 2_{1/2}$ = an eighth note
$ 5_{3}$ = a dotted half note
[b]A dot after a number is a dotted quarter note.[/b]
$ 5\cdot$ = a "so" note for a duration of 3/2 beats
[b]Vertical Bars divide measures.[/b]

[/hide]

OK, I realize this topic will have a rather limited audience here on AoPS, but I really want this to happen...

Each poster gets to add two measures to the song. Hopefully we can get a nice song  :)

wow, this is very interesting.  I play the guzheng, and am used to reading this type of music, but am having trouble understanding this notation.  Is there software to compose this type of music?  I know there is for 5 lined...

Edit: I think I'm getting it now, are we trying to stay pentatonic or not?

Yeah, let's stay pentatonic. (although they can be used well and keep with the style, 7's and 4's often make it sound bad... )

yeah, adding in 4's and 7's requires a skilled composer, but using them often adds a special flavor to the piece, for example, qin sang qu has many 4's and 7's


function
limit
integration
calculus
absolute value
real analysis
real analysis unsolved

Let $ f\in L^1[0,\infty)$. Show there is a positive strictly increasing measurable function such that $ \lim_{x\to\infty}a(x)\equal{}\infty$ and that $ \int_0^\infty a(x)f(x)dx<\infty$.

See http://www.mathlinks.ro/viewtopic.php?t=202699

You did not say $ f$ is positive, but look at the absolute value and write the integral as an infinite sum of integrals over smaller intervals and apply the above


AMC USA(J)MO USAMO ARML trigonometry inequalities geometry

Let's start out with an old classic:

Find all positive integers $n$ for which $2^{n}+1$ is divisible by $n^{2}$

Hint 1:
[hide]Let $p$ be the smallest prime dividing $n$.  What must $p$ be?[/hide]

Hint 2:
[hide]How many times may $p$ occur in the prime factorization of $n$?[/hide]

Hint 3:
[hide]What other primes can divide $n$?[/hide]

Well, since $2^{n}+1$ must be odd for $n>0$, $n^{2}$ is odd $\Rightarrow n$ is odd.

So, $2^{n}+1$ is divisible by 3...

(and $n=3$ is a solution)

but I couldn't get any further.

Look at the hints:
Let $p$ be the smallest prime dividing $n$.
We know that $2^{2n}=1$ mod p as does $2^{p-1}$.   Why do these two facts together, along with the definition of $p$, imply that $p=3$?  Then, ask yourself, if $n$ has $k$ factors of $3$ how many factors of $3$ does $2^{n}+1$ have?

Don't worry if this seems hard even with the hints... it is an old IMO #3, but a well-known classic.

Jeremy, the problems you're posting are way too hard for this forum. Please tone down the difficulty. But I don't mean to drive you away as has apparently been done with Arnav.

I'll leave this problem for someone else anyway, since I've seen it.

Hm ok...
Yes, in hindsight the problem I posted in the ARML forum was too difficult.  I came up with it myself, and so found it hard to judge to the difficulty  :( .  On the other hand, I think it's fair enough to post some nice Olympiad problems here.  We certainly have enough people in North Carolina who are capable of solving them.  I don't mean this to be a place for people to post Olympiad problems they can't solve, but rather ones that they found particularly elegant or generally enlightening.  With hints, I think any Olympiad problem can be made quite approachable.  Anyways, even if you have seen  my problem and so do not wish to give a solution, feel free to post your own problem.   It's probably a good thing to have more than one problem open at a time.

Yay! I can finally do IMO geo problems! Here's a pretty easy one for you NC USAMOers:
[hide="1."]Points A, B, C, and D are on a line. Draw circles with diameters AC and BD, and label their intersections X and Y. Now choose a point O on line XY. Let BO and CO intersect the circles on which B and C lie at points F and E, respectively. Prove that AE, DF, and XY are concurrent.[/hide]
And on the same test, something harder and much cooler:
[hide="2."]In hexagon ABCDEF, AB=BC=CD=x and DE=EF=FA=y. Angles BCD=EFA=60. Choose points G and H inside the hexagon so that AGB=EHD=120. Then prove that $GA+GB+HD+HE+GH\ge CF$.
[/hide]
[hide="hint"]Cyclic quads.[/hide]
And I love this blue MOSP hw problem!
[hide="MOSP06-2.25"]Let p be an odd prime, and let $S=${$n_{1},n_{2},...,n_{k}$} be a set of perfect squares relatively prime to p. Minimize k so that there is necessarily a subset of S with the product of its elements one more than a multiple of p.[/hide]
[hide="hint"]Primitive roots.[/hide]

[hide="The second problem"]Whenever there are rigid 60 degree angles in a problem the solution generally involves either trigonometry or Ptolemy.

As this is an inequality, we should be especially attracted to Ptolemy.

After noticing that DBAE is a kite, the solution effectively pops out.

Reflect C over line BE to a point C'

Then BC'A is equilateral and G lies on its circumcircle.

It follows that BG+GA=C'G

Similarly, let F' be the image of F under reflection.

Then DH+HE=HF'

It follows that DH+HE+BG+GA+GH=F'H+HG+GC'

By the triangle inequality, this is at least C'F', which is the same as CF.[/hide]

[hide="Problem 3"]

I couldn't solve this last year but now it looks trivial  :)  :)

Let $\omega$ be a primitive root of $\mathbb{Z}_{p}-\{0\}$.

Then $\omega^{2}$ is a generator for the group of all $\frac{p-1}{2}$ quadratic residues.

If we take our set S to be $\frac{p-1}{2}-1$ perfect squares all congruent to $\omega^{2}$, we see that the minimum $k$ is at least $\frac{p-1}{2}$.

We claim in fact that $k$ is exactly $\frac{p-1}{2}$

In fact, we prove the following slightly more general fact:

Let $G$ be the cyclic group of order $n$.  Then given a collection of any $n$ (not necessarily distinct) elements of $G$, there exists a subcollection of those $n$ elements with product $1$.

Proof:

Let the collection consist of elements $x_{1},x_{2},...,x_{n}$

Consider the $n$ subcollections

$x_{1}$
$x_{1},x_{2}$
$x_{1},x_{2},x_{3}$
...
$x_{1},x_{2},...,x_{n}$

If any of these subcollections has product $1$, we are done.  Otherwise, some two of them must have equal product by the pigeonhole principle.  Let these two subcollections be

$x_{1},...,x_{i}$ and $x_{1},...,x_{j}$, with $j>i$

Then we may take $x_{i+1},x_{i+2},...,x_{j}$ as the desired subcollection with product 1.[/hide]

I'll leave the easy geometry problem for someone else.  In the meanwhile, here is one of my recent favorites:

On an infinite (in both directions) strip of squares, indexed by the integers, are placed several stones (more than one may be placed on a single square).  We perform a sequence of moves of one of the following types:

a) Remove one stone from each of the squares $n-1$ and $n$ and place one stone on square $n+1$

b) Remove two stones from square $n$ and place one stone on each of the squares $n+1$ and $n-2$

Prove that any sequence of such moves will lead to a position in which no further moves can be made, and moreover that this position is independent of the sequence of moves.

This is equivalent to saying all integers have a unique representation base φ with digits 0 and 1 and the further restraint that there are no two adjacent 1's.

That's not a solution, just a really cool way to restate the problem.

in the first geometry problem are the points necessarily in that order on the line?  i think they have to be...

Yes, they are. Otherwise, it doesn't really even make sense.

Another fun geometry problem (IMO 1960/3)

A right-angled triangle $ABC$ is given for which the hypotenuse $BC$ has length $a$ and is divided into $n$ equal segments, where $n$ is odd.  Let $\alpha$ be the angle with which the point $A$ sees the segment containing the midpoint of the hypotenuse.  Prove that

$\tan{\alpha}= \frac{4nh}{\left(n^{2}-1\right)a}$

where $h$ is the height of the triangle (more specifically the length of the perpendicular from $A$ to $BC$)

[hide="Hint"]
haha this isn't really a hint.  but anyway, think about where you can get tangent from.  and if you say sine and cosine - no.  not unless you have a deathwish.[/hide]

[quote="n^4+4"]On an infinite (in both directions) strip of squares, indexed by the integers, are placed several stones (more than one may be placed on a single square).  We perform a sequence of moves of one of the following types:

a) Remove one stone from each of the squares $n-1$ and $n$ and place one stone on square $n+1$

b) Remove two stones from square $n$ and place one stone on each of the squares $n+1$ and $n-2$

Prove that any sequence of such moves will lead to a position in which no further moves can be made, and moreover that this position is independent of the sequence of moves.[/quote]

This is one of those very few Olympiad problems that is hard to write-up, so my proof is a bit non-rigorous. Or is this okay?
[hide="proof outline"]Let's assume that the number of stones is minimal in some new sequence. Then no two stones are adjacent and there are no more than 2 squares in one spot (or else you can apply (b) and then change n and n+1 to n+2 for a smaller number). Among these possible positions, choose the one that has the fewest 2's (if there are none, we are done). But if there is a 2 at n, apply (b). Then the new stone at n+1 must not pile onto another stone, since no stones are adjacent. But the stone at n-2 could do so. If there are 3 stones there, we can reduce the number of stones for a contradiction. If there are just 2 stones there, we can iterate until we reach the stone at the left-most place (for the set must be bounded), and then we have reduced by 1 the number of 2's. If there is 1 stone there, we are done because we have reduced the number of 2's by one. So there may be no 2's, and we are done.[/hide]

As for the new problem... what's wrong with tan=sin/cos? just use Stuart's Theorem to find the edges of the triangle, and then use two area formulas and solve for the sin of that angle, and then convert to tan... :rotfl:

Oh man JOHN BERMAN uhhh yeah that is pretty true it was on last year's APMO and it took me quite a while to do the writeup (which ended up being like 2 pages) but eh here I will try it again:

OK first of all move a can only occur a finite number of times, obviously. Hence if this can go on forever, we have to keep doing move b. However, if we consider this thing in base 2, we get a monovariant and hence can not go on forever. Well actually blurgle yes it can but you can show it's bounded on the left side by considering base 1/2.

OK now that we've established that it reaches some ending position, let's suppose there are two ending positions. Suppose k is smallest number such that the positions match for each t > k, and there's a stone on k for only one of the positions. Well the other position is not allowed to have two consecutive stones, so clearly the arrangement which optimizes the value in base phi is when there are stones on k-1, k-3, k-5, etc.

However, we have

$\sum_{i=1}^\infty \phi^{k-2i+1}= \frac{\phi^{k-1}}{1-\frac{1}{\phi^{2}}}$
$= \frac{\phi^{k+1}}{\phi^{2}-1}= \phi^{k}\cdot \frac{\phi}{\phi^{2}-1}= \phi^{k}$.

So basically since the position without a stone on k has a smaller value in base phi than the position with a stone on k. However, the base phi value is invariant under the operations, so this is impossible.

zhai - no offense meant, but two things.

1. hide your solutions for politeness' sake.

2. move to nc or leave our forum.  i mean, the occasional comment is fine, but dude - it's like [b]NORTH CAROLINA[/b] usamo practice.  the key words there are in bold.  i guess that it's okay, but the first problem ticked me off a little.  nice solution, though.

There is nothing wrong with him posting. The only thing I thought was odd was talking about his score in a NC score thread... You're just sore he isn't moving here...

But please refrain from putting full names in caps.

@Teki-Teki:  Let me quote my first post.

[quote="I wrote"]Hello everyone!

Anyone can use this forum, but mostly for anyone from North Carolina. Any competition that takes place in North Carolina can be discussed, as well as preparation for national competitions such as ARML. I hope more North Carolinians get to know each other through this forum, so please post who you are and something about yourself. You do not have to say your name if you do not wish to.

I moderate this forum, and if you have any questions, you can contact me by pm or email at soccerboy1992@gmail.com.

I have been to ARML once (this year), I have participated in mathcounts for the last two years, and will participate again next year, and I am going into 8th grade and help other students at my school in North Carolina for our math club.[/quote]

The point is for NC related things to be discussed, but anyone can post, if they want.  If we wanted to limit it, then this would be a private forum.

Anyway, back on topic (I'll try to solve the easy geometry problem now)...

okay, sorry to zhai.  yes, i am sore about him not moving here.  i did NOT pick up on your sarcasm, jb.  ok?  drop it.

whee, jeremy, easy mod question.  now to find the right base?

well - jb, i guess there's nothing really wrong with that other than it's like brute force all the way, and you don't really know the lengths of the sides of the big right triangle and stewart's theorem would get really ugly super fast.

yeah.  not a bad idea though.  (actually, stewart's theorem wouldn't get that ugly if you see a quick trick, but after that i don't know)

Alright guys let's see a solution posted to one of the three open problems.

If nobody answers by the end of the day I will post solutions to the easy geo and the weird trig problems.

Also, who is The Herald of Doom and who is Stephen?

[quote="n^4+4"]Here's another quick problem for beginners to think about:

Solve $x^{2}=y^{3}+7$ in integers[/quote]

Dude what the hell how is this a problem for beginners...

y must be odd or else $x^{2}\equiv 3 \mod 4$, then

$x^{2}+1 = (y+2)((y-1)^{2}+3)$ so that $(y-1)^{2}+3 \equiv 3 \mod 4$. Then there is a prime p with $p | (y-1)^{2}+3, p \equiv 3 \mod 4$. However -1 is not a quadratic residue of p, contradiction.

yay i have been allowed to post until USAMO in only this thread

guys i have found that i [b]SUCK AT INEQUALITIES[/b] so can you help?

Inequalities are a strange subject at the moment, what with all the Muirheading and Schuring.  There might be a relatively nonstandard inequality on the USAMO this year.  On the other hand, attempts to predict this sort of thing in the past have been very wrong.

A good intro to the subject can be found at the bottom of http://www.unl.edu/amc/a-activities/a4-for-students/s-index.shtml

On the Resources page of this site there is a slightly more advanced article by Thomas Mildorf.

yes i have seen mildorf's
we got a bunch from AMP too (including a slightly simplified one from Mildorf), but I can't do them :blush:

URGH

Berman tried to teach em to me but I was like "huh? what? dude that's cheap" or something like that because I suck and he was like "this is easy"

Then here's a fun problem for you combining two common weaknesses: geometry and inequalities.  That's right - it's a [b]GEOMETRIC INEQUALITY[/b]!

Anyway, here we go.  Consider a triangle $ABC$.  Let the interior angle bisector of $\angle A$ meet side $BC$ at $D$, and define $E,F$ accordingly opposite $B,C$ respectively.   $AD, BE, CF$ meet at $I$, the incenter of $ABC$.  Show that

$\frac{1}{4}< \frac{IA \cdot IB \cdot IC}{DA \cdot EB \cdot FC}\le \frac{8}{27}$.

Find when equality holds on the right, with proof.  Also find the limiting case on the left. (i.e., find with proof when the bound on the left is almost an equality.)

[hide="Hint"]
The geometry behind this problem is not very hard, and neither is the inequality.  (both can be 'brute-forced' by expanding once you get a nicer form.)  don't consider the fraction as a whole.  remember your area formulas.  when does equality occur in a lot of traditional inequalities?  what geometric inequalities are always strict?

ok, so that's like practically a guide to solving the problem if you pick up on the uber-obvious hints.

hey look, i can put YouTube stuff on here now
[/hide]

[hide="EDIT Hidden for Teki-Tekis sake"]

um well i have equality on the right which is trivial, an equilateral

however, i should actually try to do the whole thing... :ninja:

EDIT: well the expression is equal to $\frac{r^{3}}{h_{a}h_{b}h_{c}}= \frac{abc}{a+b+c}$ if i'm not mistaken

EDIT AGAIN: well I'm wrong... it should be $\frac{(h_{a}-r)(h_{b}-r)(h_{c}-r)}{h_{a}h_{b}h_{c}}$
[/hide]

Since Damien has not bothered to hide his hint, which makes the inequality easier, I will now post a solution.

[hide]
First note by similar triangles that $\frac{IA}{DA}= 1-\frac{r}{h_{a}}$, where $r$ is the inradius and $h_{a}$ is the length of the altitude from $A$ to $BC$.  Now, note that $r = \frac{2K}{a+b+c}$, where $K$ is the area of $ABC$, and $h_{a}= \frac{2K}{a}$.  Hence our inequality becomes

$\frac{1}{4}< \frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+c\right)^{3}}\le \frac{8}{27}$

Now to prove the upper bound.  Consider $x,y,z$ such that $y+z = a, z+x = b, x+y = c$.  It is obvious that $x,y,z$ are the lengths of the tangents from the incircle to $A,B,C$ respectively and hence are positive.  We want to show that

$\frac{\left(2x+y+z\right)\left(x+2y+z\right)\left(x+y+2z\right)}{\left(2x+2y+2z\right)^{3}}\le \frac{8}{27}$

or equivalently,

$\frac{4\left(x+y+z\right)}{3}\ge \sqrt[3]{\left(2x+y+z\right)\left(x+2y+z\right)\left(x+y+2z\right)}$ which follows from separating terms in the left-hand side and applying AM-GM.

Good.  We have one bound.  Now for the other.

We're going to expand on this one to get our desired inequality as

$\sum_{cyc}a^{3}+3\sum_{sym}a^{2}b+6abc < 4\sum_{sym}a^{2}b+8abc$.

Canceling terms this becomes:

$\sum_{cyc}a^{3}< \sum_{sym}a^{2}b+2abc$

We drop the $2abc$ term and the resultant inequality follows from summing triangle inequalities - $a^{3}< a^{2}b+a^{2}c$

Hence we establish our desired inequality.  Note that we can also prove the upper bound by expanding and canceling terms as well, and then using AM-GM.

Now note that equality above occurs when $x = y = z$ and thus $a = b = c$, so $ABC$ is equilateral.  Equality below is approached when the triangle inequalities become more strict, as $ABC$ becomes degenerate.  These are typical results.[/hide]

why thank you i believe most of us need it

why thank you jeremy i'm going to steal Damien's.

umm...since i can't edit my old post anymore everybody please note that Damien is being a responsible AoPSer and hiding all hints/partial solutions/full solutions (yeah right what full solutions?).  let's give Damien a round of applause. :clap2:

okay i should stop being mean to damien good luck damien hopefully we will see each other at MOP (you at red, me at blue)

good luck to everyone else too (unless you don't need it, then give it to someone else).

Can I have Arnav's?

Good luck all!

Stop being mean to Damien/Tianyi.

well thanks for being nice everyone  :wink:

I
SHALL
NOT
FAIL
(hopefully)

ugh


In the addition problem below, there are three two-digit numbers in which different letters represent different digits. What digits do A, B, and C represent?
            A A
            B B
         + C C
_______________
         B A C

[hide="Solution"]We have that $ A\plus{}B \equal{} 10$, and that $ B\plus{}C \equal{} 9$. If $ B \equal{} 2$, then $ A \equal{} 8$ and $ C \equal{} 7$, but this is invalid. If $ B \equal{} 1$, then $ A \equal{} 9$ and $ C \equal{} 8$. This is invalid. We know $ 1 \le B \le 2$. So, $ \boxed{A\equal{}9}$, $ \boxed{B\equal{}1}$, and $ \boxed{C\equal{}8}$.

[/hide]


calculus
integration
derivative
function
limit
calculus computations

Hi,,,
I need to find the value of [b]B[/b] in the following integration problem:

a<inf, b>0 , -inf<x<inf,

$ \int \frac{B}{1\plus{}(\frac{x\minus{}a}{b})^2}  dx \equal{} 1$

Do you know what derivative of $ \arctan x$ is?

yes...
but the result is going to be real value or constant value in terms of a,b and B ?
moreover what is the value of arctan(inf) ?

You treat $ a$, $ b$ and $ B$ as given constants when calculating the integral. The value of the integral will depend on them. So what you want to do is to compose arctan with a linear function in order to get the right derivative corresponding to your integrand. Or, which is the same thing, you make a linear substitution $ t \equal{} \frac{x\minus{}a}{b}$

$ \lim_{x \to \infty} \arctan x \equal{} \frac {\pi}{2}$. Draw a picture of arctan.

Thanx
Onemore thing waht is the integration of this one:

$ \int  B x^2 e^{\frac{\minus{}x^2}{a^2}} dx$ for x>0 and a>0

Are you sure you wrote this correctly? This one is fundamentally more difficult than the previous problem.

First of all you must know the value of $ \int_0^\infty e^{\minus{}x^2} \, dx$ which can't be done with fundamental substitutions/integration by parts methods as far as I know. After that I would integrate $ \int_0^\infty 1\cdot e^{\minus{}\frac{x^2}{a^2}}$ by parts, upon which your integral will appear in the answer.

[quote="0714446459923"]wouldn't it be much simpler to use the substitution
$ u \equal{} \frac {x^2}{a^2}$

then you'd be left with something like $ a^3\Gamma(1.5)$.[/quote]

What is gamma $ \Gamma(1.5)$ ?

This has been discussed recently- see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=182542]here[/url].

[quote="Kalle"][quote="0714446459923"]wouldn't it be much simpler to use the substitution
$ u \equal{} \frac {x^2}{a^2}$

then you'd be left with something like $ a^3\Gamma(1.5)$.[/quote]
How is this at all useful? You have to do essentially the same computation to calculate $ \Gamma(1.5)$[/quote]

it's great if you happen to have $ \Gamma(.5)$ memorized. I suppose otherwise, it wouldn't work out so well


Go to profile and then go to blog options, and then click create a blog :wink:

create one in [url=http://blogspot.com]blogspot[/url].
[img]http://disneyano.altervista.org/immagini/koda8.jpg[/img]

[quote="neelnal"]How to make one? I want to join AoPs blogs, but it asks for a http:// website blog and I don't have a blog already on someother website or anything. Can someone basically tell me how to make a blog?[/quote]

More likely than not, you clicked on the Blogs link on the Community tab off to the left of the site. That is for adding an outside blog to the AoPS ring. Go to "AoPS Blogs" instead to have your blog hosted on the AoPS server.

:)

or just make a xanga, myspace, facebook, whatever.


Google has created a revolutionary free broadband service. Simply flush a length of fiber optic cable down the toilet, and it will be automatically connected to one of their TiSP nodes.

:roll:

Of course, it's just another one of Google's April Fools jokes, like Google Gulp, Google Romance, etc. What do you think about it?

this product is awesome!!!!! I'm getting it!!!!!!!

By the way, here is a link to it:

[url]http://www.google.com/tisp/[/url]

It's a fairly average joke, but the execution is just flawless.

Wait till I join the google team.
Then like you won't even need the fibre optic.
Flushing your you-know-what each morning will be enough. :wink:


limit
algebra
polynomial
function
linear algebra

In AoPS volume 2, when proving Binet's formula, it starts out writing the Fibonacci sequence as the sum of two geometric sequences. Where is the motivation for this? Are most recursive sequences able to be written as the sum of geometric sequences?

the motivation is that as you take more and more terms the fibonacci sequence behaves more and more like a geometric sequence. more precisely, $ \lim_{n\to \infty} \frac{F_{n\plus{}1}}{F_n}\equal{}\frac{\sqrt{5}\plus{}1}{2}$

[quote="NuncChaos"]Are most recursive sequences able to be written as the sum of geometric sequences?[/quote]  Almost (if by "recursive sequences" we mean those where each term is given by a sum of multiplies of a fixed set of preceding terms): try to find a counter-example by looking at the equation $ a_{n \plus{} 1} \equal{} 4a_{n} \minus{} 4a_{n \minus{} 1}$ (or even more simply $ a_{n \plus{} 1} \equal{} 2a_n \minus{} a_{n \minus{} 1}$) with various initial values.  But the "bad behavior" from these examples (extra polynomial factors) is the only kind of bad behavior that can occur.  Doubtless the proof of this has been given on the forum, probably repeatedly -- try searching for things like "linear recurrence" and/or "generating function" or the like.  It can also be found in the first chapter of Wilf's [i]Generatingfunctionology[/i], which is freely available online.  The generating-functional method gives more motivation.  It is possible to prove that the solutions have the particular form that they do without resorting to that particular piece of technology, but instead you need some linear algebra: show that "things of the right form" form a space of a particular dimension, show that the solution space has the same dimension, and thus conclude that "things of the right form" fill up the entire solution space.

Ok, thanks!

off to work through generatingfunctionology..


I've been wanting to do this for awhile, although I'm a little apprehensive about doing so. My school has a few problems that should be corrected, however I'm concerned about how they will react to my list of improvements. Here's some of the points I have thought of thus far:

$\bullet$ As of now, classes presented after the base requirements for the school are labelled as Honors classes. For example, if a student has taken 2 years of science, and decide to take a science class in Junior year, it will be Honors. I especially do not like this rule as the chemistry class I am a part of now is taught at a sub-subpar level (which I will address again later), yet it is labelled as an Honors class for all juniors. Also, for Seniors, every class (besides English, which requires all four years, so it still has a Regular and Honors separation) is labelled as an Honors class. In my opinion, it should not be like this, and Juniors and Seniors should receive Regular grading for Regular classes. [Addendum: Honors classes award the student 5 points extra on the overall grade for that class, to be factored in Class Rank and Honor Roll]

$\bullet$ True Honors classes for Mathematics. As it is now, students that take mathematical Honors courses are put into a math class one year ahead of the rest of their peers. That is, eight graders that take Honors math have a separate class, Algebra, while the rest of their peers take Pre-Algebra; then, during ninth grade Honors math students take geometry, and the others take Algebra, and so forth. Honors students do not receive their own class, they are placed with students of the next grade level. However, even though these students are a grade higher, does not mean they are on par with the Honors students, and that's where complications arise. Since the teacher must cater to all students, he or she still must teach at that "Regular" level, and no true Honors mathematical education is received. What I would like is for the Honors students to have their own class, where they can be pushed like a true Honors class should be.

$\bullet$ Honors classes should be challenging and rewarding. This ties in a lot with my first point and second point. The chemistry class I am part of now has covered a meager four chapters over the course of the entire semester. This, in my opinion, is horrid. We have not covered anything sufficiently hard, only basic things such as the quantum numbers, some very basic history, and as of now we're covering the periodic table. On Friday our assignment was to read a one page worksheet on the lab we'll be doing on Monday. He spent roughly 3 minutes talking in class, the other 43 minutes was spent socializing (or for me, writing some of these points down) and doing nothing of importence. I understand these things need to be covered, but it's such basic information that it should be covered within a few weeks at most. The entire book (16 chapters) is covered in the course of two years. I think this is ridiculous. On the other end of the spectrum, however, my Anatomy/Physiology class has covered 10 chapters of college level material. I knew what a periodic table was in ninth grade, but do you think I could have told you what happens after an action potential reaches a neuromuscular junction? My other "true" honors class, English, is also a pretty rigorous and rewarding class.

$\bullet$ Honors classes should be taught by competent teachers. I seem to be harping on my Chemistry teacher, but he is the epitome of slack teachers, and his role as the Honors Chemistry teacher needs to be changed. This is an improvement dedicated solely to him. There needs to be another teacher that teaches Honors Chemistry I, Chemistry II, and Physics. I've pretty much summed up his lack of true, inspired teaching in latter paragraphs, so I won't go into them again. Thinking back, he is the only truly uninspired Honors teacher I've had so far in High School.

$\bullet$ Adoption of the Advanced Placement program. Depending on the completion of the previous suggestions I believe the ultimate thing to do is to adopt the Advanced Placement program. Not only are AP programs extremely beneficial on transcripts, but they are beneficial in the fact that they reward students that truly want to learn in an accelerated way. Unfortunately I am not sure of the cost of this and restrictions etc., or if it could even be done.

My school prides itself on the education it presents (often flaunting the fact that we score higher on the ACT than the top schools in Lincoln [img]http://img27.exs.cx/img27/5379/rolleyes2hu.gif[/img]) I would think that my school would take these suggestions to heart and hopefully implement them, but I'm not entirely sure how they would react to my suggestions for the Chemistry teacher.

If I could receive some input from you guys as to how exactly I should present this or if at all, that would be great, thanks. [img]http://img86.exs.cx/img86/9913/smile6qv.gif[/img]

sounds like my own school...

my school (junior high ) didn't like either mine or my parents suggestions and the high school was completly unresposive to other suggestions.

So, I admire your thoughts, but I think you're in for a difficult time.  First, you're probably going to be going up against a lot of things, the most significant being institutional inertia -- an institution at rest remains that way unless a massive force is applied to it.  There are probably going to be a lot of sub-reasons that the school will name (if they respond to your complaints) that will fall under this category -- no flexibility to schedule, limited availability of teachers (perhaps due to contract), not enough students to form honors classes, "We already have honors classes -- what are you complaining about?", not enough money in the budgent, and similar things.  There are only some of these that you will be able to actually alter -- for example, if a reason is difficulty in scheduling, that's relatively easy to get around as compared to budgetary or contractural constraints.  So, you should just be aware that what you want to do will probably not be easy.

Second, you need to find a good way to frame your argument.  You're going to have to show that this is for the better of some significant group of people, not just you, so you're probably going to want more students than just you advocating for this.  Most schools simply are not going to change for 1 or 5 or even more students.  It would also help if you could get student's parents involved, since they are more likely to have sway as they pay taxes and presumably vote for the school board and are generally more convincing advocates than a bunch of middle or high schoolers.  (No offense intended by that last part -- just a fact of life that adults prefer to listen to other adults.)  So, you will probably need to get a number of students and their parents involved.  Then, you're going to have to present your case in a non-antagonistic way.  Avoid insulting people, including your fellow students, your teachers and especially the administrators.  It's easier to do things when people don't get angry whenever you open your mouth.  One thing you can try to do is make life easier for administrators -- a good way to do this is to try and recruit teachers willing to run Honors classes for you.  That way, the administrator has an easier life, since he already knows who the teacher could be.

I'm stopping for the night, but this is of course not a comprehensive guide to getting things done and I'm sure you or other members of the forum will have good thoughts on how to make this work.  Good luck!

the whole honors class of mine well almost came together with our parents and the administrators flat said no.  It has worked the way it is for a long time and we see no reason to change it.  Kids in honors classes are the only ones that do not fail state standardized tests and we need to put them into classes where they will help other students to get to their level before they go surging ahead to the point where we can not keep up with them.

It seems to me that my school system is afraid of us getting to be too much for the teachers.  Our honors classes especially in math gets bigger every year.  3 years ago it was 15 kids and now in my math class its 54 and we have to have 2 classes at the junior high level, in high school it will be as said above.  we will be in classes with sophmores and so on.  in my other classes we now number 31 for the different advanced/honors english, history, science, and such.  We have no advanced reading we mearly test out which seems bad for many who test out one year fail the readng portion of our standardized test if it was only a fluke that they got out like 75% of them were this year.  MY school says they are afraid of this happening if they do separate advanced class and allow more skipping or accelarated in any class.  That is my schools main argument against it.

Any suggestions?

"Confronting" is probably a waste of your time.  "Circumventing" may be
a more realistic/efficient possibility, as in signing up for
external AP, ACH, GRE, etc  exams to document progress by you
and any similarly inclined schoolmates, or taking external or correspondence
courses.  If you can convince the school to either support you in the effort
(e.g. count self-study as a course) or give credit after the fact
you may free some time on your school schedule.  I don't imagine
most schools are in the business of acting on the purportedly helpful
administrative suggestions of their inmates/students.


geometry
Gauss
perpendicular bisector
geometry unsolved

It's true for any point $ H$ inside triangle $ ABC$ such that $ \angle HBA\equal{}\angle HCA$.

Let $ E$, $ F$ the projections of $ H$ on $ AB$, $ AC$, respectively. Hexagon $ AEPHFQ$ is cyclic, therefore, we can easily prove that $ \angle PEF\equal{}\angle PFE$ and $ \angle QEF\equal{}\angle QFE$. Hence, we conclude that $ P$, $ Q$ lie on the perpendicular bisector of $ EF$ $ (1)$

On the other hand, denote by $ K$, $ L$ the middle points of $ BH$, $ CH$, respectively. We have $ \triangle MKE\equal{}\triangle FLM$, i.e. $ ME\equal{}MF$ $ (2)$

Combining $ (1)$ and $ (2)$, the result follows.

[quote="jayme"]Dear jbmorgan and Mathlinkers,
the center of the Euler's circle of ABC is also on this line.
Nice generalization of April
Sincerely
Jean-Louis[/quote]

I don't think so. If $ \overline{PQM}$  passes through center $ E$ of Euler's circle of $ \triangle ABC$ then for every point $ H$ satisfies $ \widehat{HBA}\equal{}\widehat{HCA}$ we must have $ H$ lies on $ ME$, thus the locus of $ H$ is a line (or even a part of a line. But it is wrong, the locus of $ H$ is a part of a hypebol curve. :blush:

Dear Mashimaru and Mathlinkers,
this result comes from Droz-Farny...
Please, see again your figure and toys with it...
Sincerely
Jean-Louis

[size=134][color=red][b]Very nice[/b][/color] [color=darkblue]the [u][b]April[/b]'s extension and its proof[/u] ![/color][/size]

[color=darkblue]I remained without words. Thank you ! The [u][b]April[/b]'s proof[/u] is a combination of the following lemmas :[/color]

[quote][color=darkred][b][u]Lemma 1[/u].[/b] Let $ AQHP$ be a rectangle inscribed in the circle $ w\equal{}C(O)$ . Consider the points $ \{E,F\}\subset w$ so that

the line $ AP$ separates $ E$ , $ F$ and $ \widehat {PAE}\equiv\widehat {PAF}$ . Then $ Q$ , $ P$ belong to the bisector of the segment $ [EF]$ .[/color]

[hide="Proof."]
[color=darkblue]Construct  the tangent $ t\equal{}PP$ in the point $ P$ to the circle $ w$ . Observe that $ \widehat {PAE}\equiv\widehat {PAF}\ \implies\ t\parallel EF$

and $ O\in PQ\implies \overline {POQ}\perp t$ and $ QE\equal{}QF$ , i.e. $ Q$ , $ P$ belong to the bisector of the segment $ [EF]$ .[/color][/hide][/quote]

[quote][color=darkred][b][u]Lemma 2 (well-known)[/u].[/b] Let $ PBC$ be a triangle. Denote the midpoint $ M$ of the side $ [BC]$ . Construct outside of $ \triangle\ PBC$

two $ P$-right triangles $ PBE$ , $ PCF$ so that $ \widehat {PBE}\equiv\widehat {PCF}$ . Then the point $ M$ belongs to the bisector of the segment $ [EF]$ .[/color][/quote]

Another “smart proof” for extension:
CH intersects AB at R, BH – AC at S. The midpoints of diagonals AH, RS and BC of the complete quadrilateral ARHSBC are collinear (Gauss line), but BCSR is cyclic ($ \angle ABH \equal{} \angle ACH$), hence the line connecting the midpoints of 2 opposite sides is the perpendicular bisector of the line connecting the feet of the perpendiculars from the intersection of its diagonals (H) to the other opposite sides (EF); from above see that PQ is the perpendicular bisector of EF.

Best regards,
sunken rock

Consider the homothety centered at $H$ sending the nine-point circle of $\triangle ABC$ to the circle $\omega=\odot(ABC)$. Then it is well-known that $X$ and $M$ ($X$ is the midpoint of $HA$) map to $A$ and $A'$, resp., where $O$, the center of $\omega$ is on $AA'$. Thus, $AA'||XM$. It is also known that $AX$ and $AA'$ are isogonals w.r.t. $\angle A$, and thus if $Z=XM\cap AP$, then $\angle XAZ=\angle ZAA'=\angle AZX$, whence $XA=XZ$. It follows from Thales that $\angle AZH=90^\circ$, so $Z=P$. Also, $APHQ$ is a rectangle, and line $XM$ passes through its center and vertex $P$, so this line also passes through $Q$. $\blacksquare$

[quote=jbmorgan]Let $ ABC$ be a triangle with orthocenter $ H$. Let the projections of $ H$ on the internal and external bisectors of $ A$ be $ P$ and $ Q$ respectively. Let $ M$ be the midpoint of $ BC$.

Prove that $ P,Q,M$ are collinear.[/quote]

Also see [url=http://www.artofproblemsolving.com/community/c6h1215129]Passes through midpoint[/url] and [url=http://www.artofproblemsolving.com/community/c6h1203203]MN bisects AC[/url].


1. Julie baked cupcakes for her family at home and for a party at school. She iced 4 cupcakes with red frosting, 2 cupcakes with orange frosting, 2 with yellow, 2 with green, 3 with blue and the last 3 with violet frosting. Each cupcake is iced with exactly one color of frosting. Julie plans to take 10 cupcakes to her party, and will take either all of the cupcakes of a particular color or none of the cupcakes of that color. How many different combinations of cupcakes could she take to her party?

2. What is the greatest number of whole 4-inch by 6-inch notecards that can be cut from a 25-inch by 27-inch sheet of poster board if every cut must be made parallel to a side of the poster board? NO tapinf or gluing is allowed.

3. A data set for a class of 25 sixth graders has their ages listed as the integer values of either 10 or 11 years. The median age in the data is 0.36 years greater than the mean. How many 10-year-olds are in the class?

(from 2007 state round)

I didn't find any nice neat ways to solve these problems.  See attached for solutions.


algebra
polynomial
number theory
least common multiple
algebra open

For a polynomial $ f\in\mathbb Z[x]$ with leading coefficient 1, suppose that all its zeros (in $ \mathbb C$) have even multiplicities. Prove that there is $ g\in\mathbb Z[x]$ such that $ f\equal{}g^2$ (otherwise stated, $ \sqrt{f}$ has also integer coefficients).
A similar result is true for any $ k\in\mathbb N$ instead of 2.

My proof is straightforward but ugly. Has anyone a nicer solution?

First off, the algebraic numbers are a perfect field, so irreducible polynomials are separable.

Now, given any root $ \alpha$ of $ f(x)$, consider $ f_{\alpha}(x)$ its minimal polynomial over $ \mathbb{Z}$.  Then $ f_{\alpha}(x) | f(x)$, and moreover each root of $ f_{\alpha}(x)$ (the conjugates of $ \alpha$) is distinct and must have the same multiplicity as $ \alpha$.  Let the multiplicity of $ \alpha$ in $ f(x)$ be $ 2m_{\alpha}$.  Then

$ g(x) \equal{} \prod_{f(\alpha) \equal{} 0} (x \minus{} \alpha)^{m_{\alpha}} \equal{} \prod f_{\alpha}(x)^{m_{\alpha}} \equal{} \sqrt{f(x)}$

is a product of polynomials with integer coefficients.  :)

Thank you very much.

In your term $ \prod f_{\alpha}(x)^{m_{\alpha}}$, I guess you mean the $ \prod$ only over certain $ {\alpha}$'s that represent the 'irreducible components' of $ g$, well-defined because $ f_{\alpha}$ and $ m_{\alpha}$ are the same for all roots $ {\alpha}$ of a same component?

Is the following argument also a valid proof?
Define $ h$ to be the simultaneous minimal polynomial for all the roots (which should be the same as the $ lcm$ of all the $ f_{\alpha}$'s, isn't it?). $ h$ may be reducible, but all roots of $ h$ must be distinct, and considering each component of $ h$, we are done as $ h|g$. :)

[quote="t0rajir0u"]Yes, I mean all distinct $ \alpha$ up to conjugation.

The argument is troublesome; it's not obvious why $ h$ should have degree less than $ g$ (at least, not to me) without a discussion of the individual minimal polynomials.[/quote]
But we should have $ h(x) \equal{} \prod_{f(\alpha) \equal{} 0} (x \minus{} \alpha)$ as all roots of $ h$ are different. (They are different for each $ f_{\alpha}$ and thus for their lcm too, aren't they?), and as we have obviously $ h|f$, so $ g$ must already contain $ h$. Still a flaw in that?  :?:

Sorry; I ran out of time typing that response.  The real issue is that $ h | g$ over $ \mathbb{C}[x]$ does not necessarily mean that $ h | g$ over $ \mathbb{Z}[x]$.


Let $A$ is a plane. $m,n$ are postive integers.
In each coordinat $P(m,n)$ we write any positive integer.
No natural number repeats itself more than $2005$ times in that plane!
Prove that there are $m,n$ such that $P(m,n)>mn$

[quote="Farenhajt"]Huh? Write $1$ in all points, then you have no claim.[/quote]
No, I meant that for all positive integers $m,n$ you write a positive integer number in coordinat $(m,n)$!

[quote="Farenhajt"]Yes, and I say: In every $(m,n)$ write number $1$ - and your claim doesn't hold.

Or your wording is a bit odd - you mean that no natural number repeats itself more than $2006$ times in that grid?[/quote]
Yes, that's what I meant.

Denote firstly $T(m,n)=\{P(k,l) \in [1,mn] \}$ and $f(n)$ as number of $P(k,l)$ in $T(n,n)$ with repeated values where every repeated value is counted in (example: so $P(k_1,l_1)=P(k_2,l_2)=...=P(k_r,l_r)$ is counted in $f$, $r-1$ times). Suppose that our hypothesis is false, then it implies that $f(n) \leq 2004n^2$ for all $n$ (otherwise we would have contradiction to assumption that every integer appears no more than 2005 times). Now note that $|T(n,n)| \geq \sum_{k \leq n^2} d(t)$ where $d(t)$ is number of divisors of $t$ (because for each $t \leq n^2$ we can choose $P(a,b)$ with $ab=t$ in exactly $d(t)$ ways). Now we bound $\sum_{k \leq n^2} d(k)$ from below:

$\sum_{k \leq n^2} d(k) = \sum_{k \leq n^2} \sum_{a,b | ab=k} 1 = \sum_{ab \leq n^2} 1 = \sum_{a \leq n} \sum_{b \leq n^2/a } 1 +\sum_{b \leq n} \sum_{a \leq n^2/b } 1 - \sum_{a \leq n} \sum_{b \leq n } 1 = 2 \sum_{a \leq n} [\frac{n^2}{a}] - n= 2n^2 \sum_{a \leq n} \frac{1}{a} -n^2+ \Omega(n) = 2n^2 log (n+1) - n^2 + \Omega(n)$

And from these it is easy to get:

$2004n^2 \geq f(n) \geq |T(n,n)| - n^2 = 2n^2 log (n+1) - 2n^2 + \Omega(n)$ which gives a contradiction because for large enough $n$ right side is larger than left side.

Note: This beautiful estimation of $\sum_{n \leq x} d(n)$ does not belong to me - it can be found  in Narkiewicz's "Teoria Liczb" - stronger one of course  :)

Writting $> f ( x ) + O ( g ( x ) )$ is [b]non sense[/b]. C.f Narkiewicz p.1 ``Oznaczenia'' or Me, p. this post "In what follow":

Recall the definition of big-oh notation. $f ( x ) = O ( g ( x ) )$ means that $f ( x ) \leqslant c \cdot g ( x )$ for some real constant $c$ as $x \geqslant x_0$
(in the most common case) for some $x_0 \in R$. Thus you can write $f ( x ) = O ( g ( x ) )$ or even $f ( x ) < O ( g ( x ) )$[b] but not [/b]$f ( x ) > O ( g ( x )$

Apart that[b] [color=red]nasty[/color] [/b]remark (which homever lead me too often to mistake) I didn't dwell into the proof :)
(BTW the estimate is correct if only you change $O$ to $\Omega$, or put $- n$ in place of $O ( n )$)

Thanks [b]{x}[/b] for reminding  :) - now it should be okay

(also I believe I just could write: $2n^2 \sum_{a \leq n} \frac{1}{a} -n^2+ O(n) = 2n^2( log (n+1)+O(1)) - n^2 + O(n)$ couldn't I?)

[quote="Megus"]

(also I believe I just could write: $2n^2 \sum_{a \leq n} \frac{1}{a} -n^2+ O(n) = 2n^2( log (n+1)+O(1)) - n^2 + O(n)$ couldn't I?)[/quote]
You can write this if you want an upper estimate. Any statement with $O$ is actually an upper estimate. ;)
Homever you want an lower estimate, so you need to change all $O$ to $\Omega$. (otherwise  if we assume $k,f,g,h,b$ takes only positive values then  $k(x)=f(x)+O(g(x))>h(x)+\Omega(b(x))$, is equivalent to $k(x)<f(x)+cg(x)>h(x)+cb(x)$ which is absurd)
What you can write is:
$2 \sum_{a \leq n} [\frac{n^2}{a}] - n^2= 2n^2 \sum_{a \leq n} \frac{1}{a} -n^2 + \Omega(n) = 2n^2 log (n+1) - n^2 + \Omega(n)$

The formal definition is:
$f ( n ) = O ( g ( n ) )$ (with $n \rightarrow \infty$)
If and only if , there is some constant $C$ and $n_0$ such that
$|f ( n ) | \leqslant C|g ( n ) |$ for all $n \geq n_0$.
$f ( n ) = \Omega ( g ( n ) ) \Longleftrightarrow |f ( n ) | \geqslant C|g ( n) |$ for all $n \geq k$ where $k$ and $C > 0$ are constant.

Take a look at: http://www.math.uvic.ca/faculty/gmacgill/guide/big-O.pdf
A good reference for all this notation (far better than I) is "Matematyka Konkretna", Graham, Knuth, Patashnik, PWN. (perhaps you have it ?)

Thanks again  - now I guess everything is fixed and correct - also I suppose I won't make such a mistake next time  :)

P.S. I don't have "Matematyka konkretna" though I've had it few times in my hands  and I didn't like it very much :D

[quote="Megus"]... and I didn't like it very much :D[/quote]

Impossible, it is THE book :)
Anyway, take care when using $\Omega$. It's quite tricky, and ... personally I never fell sure when using it :(

Here is my solution.

Let $S_k=\{(m,n)|mn\le k\} \forall k\ge 2$ .
and $T_k=\{P(mn)|(m,n)\in S_k\}$

We have $|S_k|=\sum\limits_{i=1}^k[\frac{k}{i}]$

Suppose $P(m,n)\le mn \forall m,n$. Thus, $P(m,n)\le k \forall (m,n)\in S_k$. (*)

$(*)$ follows that $|T_k|\le k$. (**)

Because no natural number repeats itself more than $2005$  times, hence $|T_k|\ge \frac{S_k}{2005}$ $(***)$.

From $(**)$ and $(***)$ we have $2005k\ge \sum\limits_{i=1}^k[\frac{k}{i}]>k\sum\limits_{i=1}^k\frac{1}{i} -k$.

Thus, $2006>\sum\limits_{i=1}^k\frac{1}{i} \forall k \ge 2$. Which is a contraction.

Therefore, there are $m, n$ such that $P(m,n)>mn$


[quote="JBL"][quote="M4RI0"]it's not true for $n=1$ and $n=2$.[/quote]

Or $n = 3$.  Mod 5, the expression is $1+2^{1987}+(-2)^{1987}\equiv 1$, so it's not divisble by 5.

Or $n = 4$, since mod 6 $3^{k}= 3$, $4^{k}= 4$, $1^{k}= 1$ and $2^{2k-1}= 2$ for $k\geq 1$ so the whole expression is $\equiv 1+2+3+4\equiv 4 \not\equiv0 \pmod6$.

Or for $n = 2k$ with $k$ odd, because the expression will be odd (exactly $k$ odd terms in the sum) while $n+2$ will be even.[/quote]
I had difficulty proving this, so there isn't a proof? :huh:

Maube there's an adittional condition or you made a typo.

no, no additional condition or typo. Maybe we can make our own additional condition. :D

[quote="FieryHydra"][quote="Inspired By Nature"]Prove $(1^{1987}+2^{1987}+...+n^{1987})$ is divisible by $(n+2)$.[/quote]
Here is the generalisation: Prove $(1^{k}+2^{k}+...+n^{k})$ is divisible by $\frac{n(n+1)}{2}$ where $k$ is an odd positive integer.

I am pretty sure  [b]Inspired By Nature [/b]got the problem from a source that didn't have the correct Russian to English translation.[/quote]
yeah, probably.


1. [hide]$3a-2ab$[/hide]
2[hide]$c^4d^5ad^-1a^9c^3$
$c^(-180c \times a^2 \times d^2)$
i simply multiplied the different powers[/hide]
4.[hide]$24ab-8ac$[/hide]

[quote="Mike33"]Simplify:
a-5ab+2a+ab+2ab

c^4d^5ad^-1a^9c^3

Find the greatest common factor of:

8c^dw^5-10c^6d^2w^10

indicated Operation:
-3^2+(-2)^2

Distributive property;
8a(3b-c)[/quote]

Someone should put that in LaTeX language.

Click for the answer.

1.[hide="$a-5ab+2a+ab+2ab$"]$3a-2ab$[/hide]

2.[hide="$c^4d^5ad^{-1}a^9c^3$"]$c^7d^4a^{10}$[/hide]

4.[hide="$-3^2+(-2)^2$"]$-5$[/hide]

5.[hide="$8a(3b-c)$"]$24ab-8ac$[/hide]

The one about the GCF is confusing because it wasn't written in LaTeX. I think it might be [hide="this"]$2cdw^5$[/hide], but you have to take off the ^ from before the d.


I got few questions for AMCs...

First, is AMC competition entered as a school, or can an individual may enter?
Second, how hard is the difficult level between AMC10 and AMC12?
Third, how can an individual prepare for AMC? Is there a website other than AMC official website? And is there a transcript for AMC discussion in MathJams?

[quote="Silverfalcon"]I got few questions for AMCs...[/quote]

See the evolving [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693]AMC FAQ[/url] for many of the answers to your questions, and follow the links and read them for more information.

Hope this helps!


$\prod \left(\frac{x}{x+y+z} + \frac{x}{y+z}\right)$
$=xyz \prod \left(\frac{1}{x+y+z} + \frac{1}{y+z}\right)$
$=\prod \sqrt{yz}  \left(\frac{1}{x+y+z} + \frac{1}{y+z}\right)$
$=\prod  \left(\frac{\sqrt{yz}}{x+y+z} + \frac{\sqrt{yz}}{y+z}\right)$
$\leq \prod  \left(\frac{y+z}{2(x+y+z)} + \frac{1}{2}\right)$
$\leq \left[\frac {\sum (\frac{y+z}{2(x+y+z)} + \frac{1}{2})}{3}\right]^3$
$= \left[\frac{1 + \frac{3}{2}}{3}\right]^3$
$=\frac{125}{216}$

The two inequalities follow from AM-GM

thanks,in fact,this inequality is my inequality for n=3.i used the same method,so i want to know another solution


solve in nonegative integers the equation       $ \sqrt{x}\plus{} \sqrt{y}\plus{} \sqrt{z}\equal{} \sqrt{2009}$    :spider:

[quote="Pirkuliyev Rovsen"]solve in nonegative integers the equation       $ \sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z} \equal{} \sqrt {2009}$    :spider:[/quote]

$ (0,0,2009)$ or $ (0,2009,0)$ or $ (2009,0,0)$

There is another  solution may be??????  :lol:  :rotfl:

There are other solutions as well greenhand. In fact this works:

$ x \equal{} 41a^2$
$ y \equal{} 41b^2$
$ z \equal{} 41c^2$

For $ a,b,c$ non-negative integers satisfying $ a \plus{} b \plus{} c \equal{} 7$

I think this gives all solutions, can't be too sure though.


A newly designed kite is the union of a semicircular region and an isosceles triangle. The base of the triangle coincides with the diameter of the semicircle. The semicircumference is $ 10\pi$ inches and the legs of the isosceles triangle are 40 inches. Find the total number of inches of tubing, to the nearest tenth, needed for crossbars $ \overline{AB}$ and $ \overline{CD}$.

[asy]pair D=(0,0), C=(4,0), A=(3,1), B=(3,-1);
draw(D--A..C..B--D);
draw(D--C);
draw(A--B);
label("A",A,N);
label("B",B,S);
label("D",D,NW);
label("C",C,E);[/asy]

The semicircumference is $ 10\pi$ so $ AB\equal{}20$.  Let the intersection of $ AB$ and $ CD$ be $ E$.  Then, by the Pythagorean theorem, \[ DE\equal{}\sqrt{40^2\minus{}10^2}\equal{}10\sqrt{15}\] Clearly, $ EC\equal{}10$.  Thus, \[ AB\plus{}CD\equal{}30\plus{}10\sqrt{15} \approx \boxed{68.7}\]


Find the smallest real number a for which there exists an infinite sequence [tex]a_1;a_2;...[/tex]of positive intergers satisfying the following properties:
1) For each [tex]n\in N[/tex] then [tex] n<a_n<2005n [/tex]
2) For every [tex]n\ge 2[/tex] then [tex]a_n^a\ge 4 max\{a_ia_j|i+j=n\}[/tex]

$\forall n\geq 2,(2005n)^a\geq {a_{2n}}^a\geq 4a_n^2>n^2$,so $a\geq 2$.
For $a\geq 2$,put $a_n=2n,\forall n$


if $ a_1, a_2, ... a_n \in Z$ prove that
$ \prod_{1 \leq k < l \leq n} (l \minus{} k)$    $ |$ $ \prod_{1 \leq k < l \leq n} (a_l \minus{} a_k)$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Here is a very nice problem. I don't know if it is classical, but I've found it this morning:
   Let $G$ an infinite multiplicative subgroup of $GL_n(C)$ in which any element has finite order. Prove that the set of traces of elements of $G$ is infinite. Deduce that if $G$ is a subgroup of $GL_n(Z)$ in which every element has finite order, then $ |G|$ is at most $ (2n+1)^{n^2}$.

1/ Here is a proof(I think..):
Let $T=\{tr(M) | M \in G\}$ be the set of the traces of elements of $G$.
If $T$ were finite, we can construct an injection of $G$ into $T^r$ as follows,
where $r \leq n^2$.
Consider $V=vect\{g \in G\}=vect\{g_1,..,g_r\}$, with $g_i \in G$.
Consider $f:M->T^r$, $f(M)=(Tr(Mg_1),..,tr(Mg_r))$. Then $f$ is an injection
because if $f(A)=f(B)$ then $Tr((A-B)g)=0$ for any $g \in G$.
So $Tr((A-B)A^{-1})=f(I_n-BA^{-1})=0$. So $Tr(BA^{-1})=n$. But all the eigenvalues
of $BA^{-1}$ have modulus $1$ and $BA^{-1}$ is diagonalisable. So $A=B$.
$G$ is finite since $T^r$ is finite.

2/
Let $G$ be a finite subgroup of $Gl_n(Z)$ in which every element has finite order.
${-n} \geq Tr(g) \geq {n}$, so $G$ is finite, and $|T^r| \leq (2n+1)^{r}$.
So $|G| \leq (2n+1)^{n^2}$.


quadratics
number theory unsolved
number theory

Prove that the sum of the squares of $ 1984$ consecutive positive integers cannot be a perfect square.

someone may wanna doubblecheck this as i was calculating this manually.
consider the problem modulo $ 64$

-note $ 1984 \equal{} 64\cdot 31$
-of any $ 64$ consecutive integers there are $ 8$ of them that square to $ 0,16,4$or$ 36$ and $ 4$ which square to $ 1,9,25,49,17,57,41$ and $ 33$ by modulo $ 64$
-sum of squares of any $ 64$ consecutive integers is therefore congurent to $ 4\cdot (1 \plus{} 9 \plus{} 25 \plus{} 49 \plus{} 17 \plus{} 57 \plus{} 41 \plus{} 33) \plus{} 8\cdot (0 \plus{} 4 \plus{} 16 \plus{} 64) \equal{} 1376 \equal{} 32$ modulo $ 64$
-therefore sum of any $ 1984$ consecutive squers equals $ 32\cdot 31 \equal{} 32$ modulo $ 64$ and consequently can't be a perfect square because all quadratic residues are $ 0,1,4,9,16,17,25,33,36,41,49,57$

Let $ S\equal{} k \plus{} ...\plus{} (k\plus{}1983)\equal{}1984k\plus{}1983*992$ .The highest power of 2 that divides S is 5.So S can't be a perfect square.


[hide]After rewriting it as $x^2+2x+1$, we see that it's follows the form $(a+b)^2=a^2+2ab+b^2$. So the solution to the equation is -1. [/hide]

[hide]
turn the equation into $x^2+2x+1$
factor and it becomes

$(x+1)^2=0\\\\x+1=0\\\\x=-1$[/hide]

[quote="236factorial"][quote="anirudh"]is this classroom level? looks more like mathcounts.  :)[/quote]

Easy quadratics are of classroom level  ;) Harder ones belong ni the basics.[/quote]
Because this requires even a smige of thought, this is going to the basics.

You son't even need the formula for this question.
[hide]

Since the solution is negative, $x$ is a negative number. It has to be odd, because the solution is $-1$. So the answer has to be $x = \framebox{-1}$
Factoring the equation to $(x + 1)(x + a) = 0$ If you solve this equation, you get that $ a = 1$, which means that there is only 1 solution.  [/hide]


$ABC$ is a right triangle, $AE$ bisect $\angle CAB$ and $CD$ bisect $\angle ACB$. If $AE=9$ and $CD=8 \sqrt{2}$, find the length of the hypotenuse $AC$.

Let a = BC, b = CA, c = AB.

$CD^2 = l_c^2 = \frac{2a^2b}{b + a},\ AE^2 = l_a^2 = \frac{2c^2b}{b + c}$

$\frac{l_c^2}{l_a^2} = \frac{a^2(b + c)}{c^2(b + a)} = \frac{\sin^2 A}{\cos^2 A} \cdot \frac{1 + \cos A}{1 + \sin A} = \frac{128}{81}$

Denote $x = \cos A$.

$\frac{3^4(1 - x^2)(1 + x)}{2^7x^2} - 1 = \sqrt{1 - x^2}$

$3^8(1 - x^2)^2(1 + x)^2 - 2^8 \cdot 3^4 (1 - x^2)(1 + x) + 2^{14} x^6 = 0$

$(2^{14} + 3^8)x^6 + ... + (3^8 - 2^8 \cdot 3^4) = 0$

$5 \cdot 13 \cdot 353 x^6 + ... - 81 \cdot 25 \cdot 7 = 0$

Looking for a rational root $x = \frac p q,\ 0 < x < 1$, p divides the absolute term, q divides the coefficient at $x^6$. We try $\frac 1 5, \frac 3 5, ...$, each time calculating $\sqrt{1 - x^2}$ and substituting to the original equation for x. $x = \cos A = \frac 3 5$ and $\sin A = \sqrt{1 - x^2} = \frac 4 5$ satisfies the equation, so that $a : b : c = 4 : 5 : 3$. In a triangle $\triangle A'B'C'$ with a' = 4, b' = 5, c' = 3, we get $l_a'^2 = \frac{45}{4},\ l_a' = \frac 3 2 \sqrt 5$. The similarity coefficient of $\triangle ABC \sim \triangle A'B'C'$ is $k = \frac{l_a}{l_a'} = \frac{6}{\sqrt 5}$, therefore $b = kb' = 6 \sqrt 5$.


linear algebra
matrix
linear algebra unsolved

Define a matrix $ A_t = \left( \begin{array}{ccc} 2 & - 1 & 1 \\
- 1 & t & 1 \\
1 & 1 & 2 \end{array} \right)$.

(1) Find the dimension of real linear space $ V_t = \{\text{x}\in{\mathbb{R}^3|A_t\text{x} = 0\}\ (t\in{\mathbb{R}})}$.

(2) Find $ t\in{\mathbb{C}}$ such that the characteristic equation of $ A_t: \text{det}(A_t - xI) = 0$ has double root and judge for those $ t$ whether $ A_t$ can be diagonalizable or not.

As it happens, the last question can be answered independently of all others. $ A_t$ is diagonalizable for all $ t$ because it is real symmetric.

Sorry, I have just edited partly.

kunny


if $x,y,z\geq 0$ and $x+y+z=1$ then prove:
$x^3+y^3+z^3+6xyz \geq \frac{1}{4}$

[quote="silouan"]homogenize to get

$x^3+y^3+z^3+6xyz\ge \frac{(x+y+z)^3}{4}$

or equivalent

$x^3+y^3+z^3+8xyz\geq (x+y)(y+z)(z+x)$

Recall Schur's ineq $x^3+y^3+z^3+5xyz\geq (x+y)(y+z)(z+x)$

now it is obvious![/quote]
i am happy to see you again. :D

here is my solution:
$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)+3xyz= (x+y+z)^2-3(xy+yz+xz)+3xyz=1-3(xy+yz+zx)+3xyz$
now we should prove this:(?)
$\frac{1}{4}\geq (xy+yz+zx)-3xyz$
and here is the proof:
$xy+yz+zx-3xyz=xy(1-z)+yz(1-x)+xz(1-y)=xy(x+y)+yz(y+z)+ xz(x+z)=x^2(y+z)+y^2(x+z)+z^2(x+y)=x^2(1-x)+y^2(1-y)+ z^2(1-z)\leq \frac{x+y+z}{4}=\frac{1}{4}$
and we R done. :D

[quote="silouan"]homogenize to get

or equivalent

$x^3+y^3+z^3+8xyz\geq (x+y)(y+z)(z+x)$

Recall Schur's ineq $x^3+y^3+z^3+5xyz\geq (x+y)(y+z)(z+x)$

[/quote]

i don't understant
can you explain it for you?


Let $a,b,c$ be reals such that

$\cos{a}+\cos{b}+\cos{c} = \sin{a}+\sin{b}+\sin{c} = 0$.

Prove that

$\cos{(a+b+c)} = \frac{1}{3}(\cos{3a}+\cos{3b}+\cos{3c})$
$\sin{(a+b+c)} = \frac{1}{3}(\sin{3a}+\sin{3b}+\sin{3c})$.

[hide="hint"]
complex numbers.
[hide="further hint"]
$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-yx)$[/hide][/hide]

I shouldn't have read your hints before doing this, but oh well:

[hide]The first expression is equivalent to

$e^{ia} + e^{ib} + e^{ic} = 0$.

If we multiply both sides by

$(e^{2ia} + e^{2ib} + e^{2ic} - e^{ia} e^{ib} - e^{ia} e^{ic} - e^{ib} e^{ic})$

and expand, we get

$e^{3ia} + e^{3ib} + e^{3ic} - 3e^{ia} e^{ib} e^{ic} = 0$.

Re-arranging and applying some basic exponent rules, we get

$e^{i(a + b + c)} = \frac{1}{3} \left(e^{3ia} + e^{3ib} + e^{3ic}\right)$.

Equating real and imaginary parts then gives the desired result.[/hide]


geometry
circumcircle
LaTeX
geometry unsolved

$ ABCD$ is a square. $ X$ is a point of $ AB$, $ Y$ a point of $ CD$ and $ Z$ a point of segment $ XY$. $ P \ne Z$ is the intersection between the circumference circumscribed to triangle $ AXZ$ and the circumference circumscribed to triangle $ CYZ$. Find the locus described from $ P$ to varying of $ X$, $ Y$, $ Z$. What happens if the point $ Y$ is taken on side $ BC$?

Are you sure that the problem is correct?

Hello :) !
Thats correct. for first case(y on CD), the locus will be AC.

CYPZ is cyclic then CPZ=CYZ
AB || CD  then  AXZ=CYZ
then AXZ=CPZ
AXZP is cyclic then  APZ+AXZ=180
then  APZ+CPZ=180     then P is on AC.

for second case (Y on BC), I think the locus will be circumcircle of ABCD.
How can I write like LaTeX? Please help me!

[quote="Kulianov"][color=darkred]For a fixed square $ ABCD$ denote the mobile points $ X\in (AB)$, $ Y\in (CD)$, $ Z\in (XY)$.
Let $ P$ be the second intersection $ P$ between the circumcircles of the triangles $ AXZ$ and $ CYZ$.
Find the geometrical locus of the point $ P$. What happens if the point $ Y$ is taken on side $ BC$?[/color][/quote]
[color=darkblue][b]Proof.[/b] $ m(\widehat{APC})=m(\widehat{APZ})+m(\widehat{CPZ})=\{\begin{array}{cccc}1\blacktriangleright Y\in (CD)\ : \ & m(\widehat{AXY})+m(\widehat{DYX})=180^{\circ}& \implies & P\in AC\\\\ 2\blacktriangleright Y\in (BC)\ : \ & m(\widehat{BXY})+m(\widehat{BYX})=90^{\circ}& \implies & PA\perp PC\end{array}$.[/color]


Given triangle ABC.D moves in AB.E is the point of  intersection of second-common touch of two incircles of triangle ACD,BCD and [CD].Determine the locus of E when D varies..

See http://www.mathlinks.ro/Forum/viewtopic.php?t=28329 .

  darij

Let A1, A2, A3, A4 be the touch points of the incircle of ACD with CD, common tangent, CA and AD respectively. Let B1,B2,B3,B4 be the touch points of the incircle of BCD with CD, common tangent, CB and BD respectively.

It results that CE = CA1-EA1=CA3-EA2. Similarly CE = CB1-EB1 = CB3-EB2.

Since A2B2=A4B4, 2CE = CA3+CB3-A4B4 = CA3+A3A-AA4-A4B4-B4B+BB3 = CA-AB+BC.


Harold, Tanya, and Ulysses paint a very long picket fence.
Harold starts with the first picket and paints every $ h$th picket;
Tanya starts with the second picket and paints everth $ t$th picket; and
Ulysses starts with the third picket and paints every $ u$th picket.
Call the positive integer $ 100h \plus{} 10t \plus{} u$ $ \textit{paintable}$ when the triple $ (h,t,u)$ of positive integers results in every picket being painted exaclty once.  Find the sum of all the paintable integers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63397]Old AIME problem[/url]. (do a search next time; any part of your post would have worked)