trigonometry
function
conics
hyperbola
ellipse
complex analysis
complex numbers

This is from a problem sheet from my course in complex analysis, and I don't think it's meant to be very difficult, but I can't figure out how to do it.

"Consider the conformal map given by $\cos z$. What are the images of lines parallell to the real and imaginary axes? What is the image of the strip $-\pi < \Re z < \pi$"

Are the curves of the lines meant to be possible to describe in some kind of simple form? I can't figure it out :(

$\cos z=(\exp(iz)+1/\exp(iz))/2$, the composition of multiplication by $i$, exponential function, $z+1/z$, and division by $2$.

WW$\cos z$D?  :D

Is this correct?

1) A line parallel to the imaginary axis is mapped to a branch of an hyperbola (except for degenerate cases)

2) A line parallel to the real axis is mapped to an ellipse (except the real axis itself).

3) $-\pi < \Re z < \pi$ is mapped to $\mathbb{C}\backslash(-\infty,-1]$

For 1) and 2) I just really wrote down the function $\cos z$. For 3) I used your hint, mlok, but I had to figure out what $\frac{1}{2}(z+\frac{1}{z})$ did to $\mathbb{C}\backslash(-\infty, 0]$. This was quite a tedious computation! Is there a smarter way? I guess you could use 1) and 2) to figure it out

This is often called Joukowski function ('J' read as 'Zh'). This [url=http://www.lerc.nasa.gov/WWW/K-12/airplane/map.html]NASA webpage[/url] explains why it is useful.

Write $z$ in polar form $re^{i\theta}$: the image has real part $(r+1/r)\cos\theta$ and imaginary part $(r-1/r)\sin\theta$. Thus, circles are mapped into confocal ellipses. Lines passing through the origin are mapped into confocal hyperbolas (the family of curves orthogonal to the ellipses). The unit circle is an exception: it is collapsed into interval $[-1,1]$  (degenerate ellipse).

Ah, that's another way to get 1) and 2) then. Do you know if my answer to 3) is correct? I feel unsure.

I used that $\frac12 (z+1/z) = G^{-1}(G(z)^{2})$, where $G$ is the Möbius transform $G(z) = \frac{z-1}{z+1}$ to find the answer to 3).

Yes, 3) is correct. The $\mathbb C\setminus (-\infty,0]$ is obtained by removing the point $-r$ from each circle $|z|=r$. This circle is transformed (bijectively) into the ellipse with semi-axes $(r+1/r)/2$ and $|r-1/r|/2$ (with an appropriate interpretation when $r=1$). Take the point $-(r+1/r)/2$ out of each ellipse, and you end up with $\mathbb{C}\setminus (-\infty,-1]$.


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inequalities
Cauchy Inequality
inequalities open

This problem has been posted by keira_khtn,but no one solves it.I hope that there is a genius to solve it.Thanks.
a,b,c,d are four positive reals such that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/aff83c32798b199ebdc07217113187d1.gif[/img]
prove that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/b3ee436a7bf95052149c30987ae3b2dd.gif[/img] :)

Why no one helps me?What can I do is only to wait.

But...But...I have waited for weeks. :(  :( Please do it,OK?

You can cry as long as you want, but I have solved your problem at least a month ago. I also posted a generalization in $n$ variables. So, try to search on the forum.

I have also solve it,use mixing variables.It is similar to siuhochung method herehttp://www.mathlinks.ro/Forum/topic-32527.html:.I will post full solution soon

But ...but,I have been waiting so long. :(

Then ok, I wil try to give you the generalization: if $ x_1+...+x_n=1$ then the minimal value of $ (1-x_1)...(1-x_n)(\frac{1}{x_1}+...+\frac{1}{x_n})$ is taken when all variables are equal.
   The nice elementary solution I've found ( note that Myth found a proof using Lagrange multipliers) is with the help of the difficult inequality from Mathlinks contest: $(\frac{x_2^2+...+x_n^2}{n-1})...(\frac{x_1^2+...+x_{n-1}^2}{n-1})(\frac{1}{x_1}+...+\frac{1}{x_n})^2\geq n^2 (\frac{x_1^2+...+x_n^2}{n})^{n-1}$. For a proof of this inequality look in the topic 'an intriguing inequality', which is on the first page of inequalities forum. Apply Cauchy inequality for the last inequality and you'll have the result I stated in the beginning and also your problem.

Can you complete your solution for the generalization,harazi?? ;)

But there is nothing to complete. Definitely I won't prove again the inequality from Mathlinks contest. If you want I can prove how it implies the generalization. Using Cauchy we have
   $\prod {\frac{x_2^2+...+x_n^2}{n-1}}\cdot n\cdot (\frac{1}{x_1^2}+...+\frac{1}{x_n^2})\geq n^2 (\frac{x_1^2+...+x_n^2}{n})^{n-1}$. Now, put $ x_1^2=a_1,...,x_n^2=a_n$ and use the fact that $a_1+...+a_n=1$.

[quote="harazi"]   The nice elementary solution I've found ( note that Myth found a proof using Lagrange multipliers) is with the help of the difficult inequality from Mathlinks contest: $(\frac{x_2^2+...+x_n^2}{n-1})...(\frac{x_1^2+...+x_{n-1}^2}{n-1})(\frac{1}{x_1}+...+\frac{1}{x_n})^2\geq n^2 (\frac{x_1^2+...+x_n^2}{n})^{n-1}$. the beginning and .[/quote]It's very strong :blush: .Can you give me the link?

I know it is very strong. The topic where you can find it is " An intriguing inequality" by manlio. I think it is on the first 3 pages of the unsolved section inequalities forum. It is easy to find.


Since　　$ 2\cos80^{\circ} \cos40^{\circ} \equal{}\cos120^{\circ} \plus{}\cos40^{\circ} \equal{}\minus{}\frac{1}{2}\plus{}\cos40^{\circ}$, we get

given　$ \equal{}\frac{4\cos80^{\circ} \cos40^{\circ}\plus{}1 }{\cos40^{\circ} }$

$ \equal{}\frac{\minus{}1\plus{}2\cos40^{\circ} \plus{}1}{\cos40^{\circ} }\equal{}2$

[quote="Dr Sonnhard Graubner"]hello, using $ 4\cos(80^{\circ}) = 4\left(\frac {1}{2}( - 1 - i\sqrt {3})(\frac {1}{2}( - 1 + i\sqrt {3}))^{\frac {1}{3}} + \frac {1}{2}( - 1 + i\sqrt {3})(\frac {1}{2}( - 1 - i\sqrt {3}))^{\frac {1}{3}}\right)$ and
$ \cos(40^{\circ}) = \frac {1}{2}\left((\frac {1}{2}( - 1 - i\sqrt {3}))^{\frac {1}{3}} + (\frac {1}{2}( - 1 + i\sqrt {3}))^{\frac {1}{3}}\right)$ on can show that $ 4\cos(80^{\circ}) + \frac {1}{\cos(40^{\circ})} = 2$
Sonnhard.[/quote]

Do you use the CAS(CAS is short for computer algebra system. Matematica, Maple and Maxima...)?  :maybe:
The posting of the answer is equal with [b]spam[/b] if the answer that it solved with a CAS!! :spam:

Yeah. I also posted a trigonometric problem and you know what he had posted? An answer correct to at least 19 places! Smart guy, huh?

In HSB (or HS), the solution that is aided by computer (or CAS) is not so welcome, I think...
For example, my solution is as follows.

\begin{eqnarray*}
4 \cos 80^\circ + \frac{1}{\cos 40^\circ}
& = & 4(2 \cos^2 40^\circ - 1) + \frac{1}{\cos 40^\circ} \\
& = & \frac{8 \cos^3 40^\circ - 4 \cos 40^\circ + 1}{\cos 40^\circ} \\
& = & \frac{2(4 \cos^3 40^\circ - 3 \cos 40^\circ) + 2 \cos 40^\circ + 1}{\cos 40^\circ} \\
& = & \frac{2 \cos 120^\circ + 2 \cos 40^\circ + 1}{\cos 40^\circ} \\
& = & \frac{-1 + 2 \cos 40^\circ + 1}{\cos 40^\circ} \\
& = & \frac{2 \cos 40^\circ}{\cos 40^\circ} \\
& = & 2
\end{eqnarray*}


function
integration
trigonometry
calculus
calculus computations

Suppose we have a continuous function $ f$ on $ [0,\pi]$ such that its Fourier series converges pointwise to $ f$, and $ \int_0^{\pi}f(t)dt\equal{}0 \ (\star)$. We can write $ f(x)\equal{}\frac{a_0}{2}\plus{}\sum_1^{\infty}a_k \cos(k x)$. Because of  ($ \star$), $ a_0\equal{}0$ so if we let $ x\equal{}\frac{\pi}{2}$ we get $ f(\frac{\pi}{2})\equal{}0$?? This looks weird..


[quote="billzhao"]I realized that on the search results page, the source fields of the topics are not displayed. Could this be fixed? Or at least be added as an option?[/quote]Will be done. Ideally all the features present on the ex-board will be present here.


This problem wasted so much time in an Olympiad I took... ugh  :mad:

$1^2, 2^2, 3^2 ... 13^2$ are divided into 3 groups of equal sum. What are the 3 groups?
For example, one "group" could consist of $1^2, 5^2, 7^2, 13^2$

$1^2+2^2+...13^2 = \frac{13(14)(27)}{6} = 819$

Each group will be $819/3 = 273.$

I started with the largest square, $169.$ This means you have $104$ left over which you can easily recognize as $100 + 4$.  You have one group.

I took the next largest square, $121$, thus leaving me $129.$  From here, I somewhat did trial and error by subracting first $100$, then $81$, and then $64.$  When subtracting $64$, my remainder was $65$ which I recognized to be $16+49.$
So my second group was now $144+64+49+16.$

Whatever was left over obviously has to add up to $273$ because we already partitioned off $2*273 = 546$, and $819-546=273.$

In the end, your groups are:

$1.)$ $13^2, 10^2, 2^2$

$2.)$ $12^2, 8^2, 7^2, 4^2$

$3.)$ $11^2, 9^2, 6^2, 5^2, 3^2, 1^2$

[quote="pkothari13"]This means you have $104$ left over which you can easily recognize as $100 + 4$.  You have one group.
[/quote]

What made you decide one 100 + 4 instead of say, 36+64+4? Did you just get lucky?

$100=64+36$
$10^2=6^2+8^2$
that is a popular pythagorian triple,no need luck there.


We still haven't got ours back yet! Grrr...oh well, I'll get back to you :)

Oh, and welcome back! Wow, tons of people are returning, I'll name this..."the homecoming era" :lol:, see if you can count up all the people who came back to AoPS recently.

[quote="cats..."]?  you said that you did not take it?[/quote]
see above :lol:

[quote="cats..."]?  you said that you did not take it?[/quote]

Oh, I originally thought I couldn't take it because we did not have school that day, but we got to take it anyways.

[quote="TheSouthpaw"]BAM! TheSouthpaw makes his long awaited return.

So, MAML results are in!!! What'd everyone get?

120 here, superior to the surprisingly low qualification score of 103.[/quote]

hm, 103 is suprisingly high i think, based on previous years... i think its higher this year becuase they threw out questions #25 and 27 and marked everyone with a right answer for those 2 problems.

Btw i got 124.


Είμαι καινούργιος στο  forum και θα ήθελα βοήθεια στην παρακάτω άσκηση.

Να βρεθούν όλες οι υποομάδες της $ \mathbb{Z}_{12}^*\equal{}\{1,5,7,11\}$


algebra
polynomial
search
LaTeX
Galois Theory
superior algebra
superior algebra unsolved

Suppose I have an element (in general some kind of expression containing irrational roots) and I want to find the minimal polynomial over $ \mathbf{Q}[x]$. Is there any good ways of doing this? For instance, suppose I wish to find the minimal polynomial of, say, $ \frac{1}{2^{1/5}\minus{} \sqrt{3}}$. I guess I could set $ x \equal{} \frac{1}{2^{1/5}\minus{} \sqrt{3}}$ and then try to multiply/square the different sides until I get rid of the roots, and then check if the polynomial is irreducible. Does this work? Is there any better methods?

Try the technique outlined by jmerry in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=405354180&t=202382]this thread.[/url]  (Note that all you need to do is calculate the minimal polynomial of $ \sqrt[5]{2} \minus{} \sqrt{3}$ and reverse the coefficients.)

Thanks, great! Nice solution.

I dont think, however, this problem was meant to be solved that way (in my case). It would work to set $ x \plus{}\sqrt{3} \equal{} ^5\sqrt{5}$, raise both sides to the power of 5, then move every term containing $ \sqrt{5}$ to one side and square both sides, right?

Btw, how do you write the fifth root nice in Latex (I used ^5\sqrt.. but it isnt very nice).

I guess so; jmerry's technique is best suited to proof, especially when $ \alpha, \beta$ are not specified and may themselves have complicated minimal polynomials.  In this case, both polynomials are simpler, so the form of the conjugates of both are simple to determine and we can in fact explicitly write the minimal polynomial in factored form (using $ \omega$ to denote a primitive fifth root of unity) as

$ \prod_{0 \le i \le 4, 0 \le j \le 1} \left( x \minus{} \sqrt [5]{2} \omega^i \plus{} \sqrt {3} ( \minus{} 1)^j \right)$.

This polynomial is fixed by every member of the Galois group of $ \mathbb{Q}[ \sqrt [5]{2}, \sqrt {3} ]$, so it has rational (in fact integer) coefficients, and no subset of its factors has the same property.  It also has the expected degree of $ 10$.  Of course, it's not that easy to calculate the coefficients this way.

Isn't the degree $ 40$? You need to attach $ \zeta_5$ for the extension to be Galois.

I don't think it's true that the minimal polynomial of $ \sqrt[5]{2} \minus{} \sqrt{3}$ is the minimal polynomial of $ \mathbb{Q}[ \sqrt[5]{2}, \sqrt{3}, e^{ \frac{2 \pi i}{5} } ]$, but I could be wrong.

What's the minimal polynomial of a field extension¿

The minimal polynomial of $ \sqrt[5]{2}\minus{}\sqrt{3}$ is of degree $ 10$, no doubt, since it has $ 5\times 2\equal{}10$ conjugates over $ \mathbb{Q}$. But how can we talk about the Galois group of $ \mathbb{Q}(\sqrt[5]{2},\sqrt{3})$? I meant that the deg. of the smallest Galois extension that contains it is $ 40$.


geometry
parallelogram
circumcircle
geometry proposed

Let $ ABCD$ be a parallelogram with $ \angle BAD \equal{} 60$ and denote by $ E$ the intersection of its diagonals. The circumcircle of triangle $ ACD$ meets the line $ BA$ at $ K \ne A$, the line $ BD$ at $ P \ne D$ and the line $ BC$ at $ L\ne C$. The line $ EP$ intersects the circumcircle of triangle $ CEL$ at points $ E$ and $ M$. Prove that triangles $ KLM$ and $ CAP$ are congruent.

$ \angle LPM\equal{}\angle LAD\equal{}\angle ABC\equal{}\angle LBK, \angle LKB\equal{}\angle LCA\equal{}\angle LME$
Therefore $ \angle BLK\equal{}\angle PLM \Rightarrow \angle KLM\equal{}\angle CLP\equal{}\angle PAC$
But LBKM is cyclic then $ \angle LKM\equal{}\angle LBM\equal{}\angle ADB\equal{}\angle ACP$
On the other hand, $ \angle ABL\equal{}\angle ALB\equal{}60^o$ so $ AB\equal{}LB, BK\equal{}BC$
$ \Rightarrow AK\equal{}LC \Rightarrow AC\equal{}LK$
We are done.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.MidEuro2009Problem9

Vo Duc Dien


[quote="ElChapin"]How many distinct $(a_{1},a_{2},...,a_{n})$ satisfy the equation:

$a_{1}^{n+1}+a_{2}^{n+1}+...+a_{n}^{n+1}-a_{1}a_{2}...a_{n}=na_{n}a_{n-1}...a_{1}-1$

 where $n$ is the product of primes greater than 2

 :)   :lol:  :!:[/quote]
It is equivalent to
$\sum_{k=1}^{n}a_{k}^{n+1}+1=(n+1)a_{1}a_{2}...a_{n}$
But by AM-GM we have
$\frac{1}{n+1}\sum_{k=1}^{n}a_{k}^{n+1}+1\geq a_{1}a_{2}...a_{n}$ with equality only if $a_{1}=a_{2}=...=a_{n}$ but they are distinct so there is no solution to the given equation.

[quote="Svejk"]
with equality only if $a_{1}=a_{2}=...=a_{n}$ but they are distinct so there is no solution to the given equation.[/quote]

Are you sure?

Or may be I wasn't clear enough:

for example if $n=3$:

How many distinct terns $(a_{1},a_{2},a_{3})$ satisfy the equation. In this case look that $(-1,1,-1)$ is a solution

[quote="Svejk"]But by AM-GM we have
$\frac{1}{n+1}\sum_{k=1}^{n}a_{k}^{n+1}+1\geq a_{1}a_{2}...a_{n}$ with equality only if $a_{1}=a_{2}=...=a_{n}$ but they are distinct so there is no solution to the given equation.[/quote]
you're wrong  :wink:  , $a_{i}$ can be negative

come on this is a nice easy problem
why such a disinterst?
I'm really sure this is pre-olympiad or less level so I can't believe nobody has a solution yet.
:(   :wink:   :)

So we are looking for all ordered pairs, not solutions with distinct $a_{i}$? Then the answer is just building off of Svejk: since n is a product of odd primes, n+1 is even, so we can apply AM-GM. We get that $|a_{1}|=|a_{2}|=...=|a_{n}|=1$ and an even number of them can be negative so the sum is $\binom{n}{0}+\binom{n}{2}+...+\binom{n}{n-1}=2^{n-1}$.


Find all prime numbers that divide $ 50!$.


[hide]
Since it is an odd number, none of the factors will be even.
11*11=121, so both of the numbers will not be greater than 11, since there integers.
1*123 are the first ones that come to mind.
1+2+3=6, which is a multiple of 3,
3*41 is another pair of factors.
There are no more pairs, so the answer is $1+3+41+123=\framebox{168}$[/hide]

[quote="236factorial"]What is the sum of all the positive integer factors of 123?[/quote]

[hide]Using a calculator program.  :)

The factors of 123 are 1, 3, 41, and 123

The Sum of those is $1+3+41+123=168$[/hide]

[quote="236factorial"]What is the sum of all the positive integer factors of 123?[/quote]
[hide]
123=3x41=123x1

the sum is 3+41+123+1=124+44=168
[/hide]

[quote="nvidura"]41 + 3 = 44[/quote]

That's not correct (you forgot the 1 and 123), and next times, please hide your answer (I understand you're new so it's okay).


[hide]There are no such integers N. Look at the cubes of the resides mod 9: $\{0, 1, 2, 3, 4, 5, 6, 7, 8\}^3 \equiv \{0, 1, 8, 0, 1, 8, 0, 1, 8\} \mod 9$. So its not possible for the cube of an integer to leave a remainder of 2 when divided by 9.[/hide]

[quote="white_horse_king88"]Find all integers $N$ such that $N^3 = 9k + 2$, where $k$ is an integer.[/quote]

[hide]
Rewrite as:
$n^3 \equiv 2 \pmod{9}$

$1^3 \equiv 1 \pmod{9}$
$2^3 \equiv 8 \pmod{9}$
$3^3 \equiv 0 \pmod{9}$
$4^3 \equiv 1 \pmod{9}$
$5^3 \equiv 8 \pmod{9}$
$6^3 \equiv 0\pmod{9}$
$7^3 \equiv 1\pmod{9}$
$8^3 \equiv 8\pmod{9}$
and so on, so there are no solutions.[/hide]


algebra
function
domain
Ring Theory
number theory theorems
number theory

How do you define coprime numbers in $Z[i]$ and why is
$(y+i,y-i)=1$ when $y$ is even (i read this in my book but that was in a solution of one problem and there was no prove for that or a definition of coprime numbers)
And one more question, can i find somewhere on the net some nice articles about $Z[i]$ or Euclidian rings at all

Thank you

In a (commutative with unity) ring $R$, element $a$ divides element $b$, written $a \mid b$, if there is an element $c \in R$ such that $ac=b$.  A unit of the ring is a divisor of 1, equivalently a multiplicatively invertible element.  Two elements are coprime if their only common divisors are units.

In $Z[i]$, suppose $x$ divides $y+i$ and $y-i$.  Then $x$ divides their difference, $2i$ and their product, $y^2+1$.  If $y$ is even, then $y^2+1$ is odd, say $2k+1$.  So $x$ divides $2k+1$ and $2i.(-ik) = 2k$.  Hence $x$ divides their difference, namely $1$.  So $y \pm i$ are  coprime.

For links, try the article [url=http://en.wikipedia.org/wiki/Gaussian_integer]Gaussian integer[/url] in [url=http://en.wikipedia.org/wiki/Main_Page]Wikipedia[/url] or [url=http://mathworld.wolfram.com/EuclideanDomain.html]Euclidean Domain[/url] in [url=http://mathworld.wolfram.com/]MathWorld[/url].


modular arithmetic
number theory
relatively prime

Let $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{z}$ and $ x \perp y$, where $ x,y,z \in N$. Prove that $ x\plus{}y\equal{}k^2$ where $ k \in N$.

[hide="No Solutions?"]Multiply through by the denominators to get $ yz\plus{}xz\equal{}xy$.  Consider this equation $ \pmod{x}$.  We get $ yz\equiv 0\pmod {x}$.  Since $ x$ and $ y$ are relatively prime, we have $ z\equiv 0\pmod{x}$, so $ x|z$.  Since the given equation is symmetric in $ x$ and $ y$, similarly we have $ y|z$.  But $ x$ and $ y$ are relatively prime, so $ xy|z$.  Let $ z\equal{}nxy$.

Substituting, we have $ y\cdot nxy\plus{}x\cdot nxy\equal{}xy$.  Dividing by $ xy$, we get $ yn\plus{}xn\equal{}1$.  But $ x,y,n$ are positive integers, so there are no solutions.[/hide]

Did I mess something up?  :(

Does $ x \perp y$ mean they're relatively prime? I think it's supposed to say $ x|y$

If it's $ x|y$, then let $ y\equal{}xn$.  We then have
$ \frac{n\plus{}1}{n}\equal{}\frac{x}{z}$.  Then $ x\equal{}k(n\plus{}1), z\equal{}kn$, for some $ k$.  It's not necessary that $ x\plus{}y\equal{}x(n\plus{}1)\equal{}k(n\plus{}1)^2$ is a perfect square.  Take, for example, $ x\equal{}6, z\equal{}4, y\equal{}12$.

[quote="matt276eagles"]No Solutions?[/quote]
Right - no solutions with given assumptions. Let ignore them and decide in $ \mathbb{Z}$

Let $ u \equal{} gcd(x,y)$ then $ x \equal{} ux',\ y \equal{} uy'$ with coprimes $ x'$ and $ y'$. Now the equation becomes

$ (x' \plus{} y')z \equal{} ux'y'$

Since $ x' \plus{} y'$ is coprime with $ x'$ and $ y'$ we have from this that both $ x'$ and $ y'$ devides $ z$ and so $ x'y'$ devides $ z$:

$ z \equal{} vx'y'$

Substitution to the above gives $ u \equal{} (x' \plus{} y')v$ and so

$ x \equal{} (x' \plus{} y')x'v$
$ y \equal{} (x' \plus{} y')y'v$
$ z \equal{} x'y'v$

Here we are done because this satisfy the equation $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{z}$ for arbitrary $ x'\ne 0 , y'\ne 0, x' \plus{} y'\ne 0, v \ne 0$

From this is evident that $ x \plus{} y$ would be perfect square only if $ v$ is perfect square but this has no connection with restriction $ x \perp y$ exept trivial cases $ x \plus{} y \equal{} 1$ which are impossible in $ \mathbb{N}$


inequalities
function
inequalities unsolved

Given positive reals such that $ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ac$. Prove that the following inequality holds: \[ a\plus{}b\plus{}c\plus{}1\geq4abc.\]

Let $ a \plus{} b \plus{} c \equal{} 3u$, $ ab \plus{} bc \plus{} ca \equal{} 3v^2$, $ abc \equal{} w^3$. Fix $ u$ and $ v^2$. Then we have to prove $ f(w^3) \equal{} \minus{} Aw^3 \plus{} B \ge 0$, where $ A \ge 0$. Hence the function assumes its minimum when $ w^3$ assumes its maximum. This is equivalent to $ b \equal{} c \equal{} t$. Hence we have to prove $ a \plus{} 2t \plus{} 1 \ge 4at^2$ with the condition $ a \plus{} 2t \equal{} 2at \plus{} t^2$ (*). Using this, the inequality is equivalent to $ (t \minus{} 1)^2(4t^2 \minus{} 1) \ge 0$, which is true, as (*) implies $ 2 \ge t \ge \frac {1}{2}$.

[quote="Lukashenko"]Given positive reals such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ac$. Prove that the following inequality holds:
\[ a \plus{} b \plus{} c \plus{} 1\geq4abc.\]
[/quote]
Let $ x \equal{} \frac{1}{a}, y \equal{} \frac{1}{b}, z \equal{} \frac{1}{z}$. Then we have $ xy\plus{}yz\plus{}zx\equal{}x\plus{}y\plus{}z$ and should prove $ xy\plus{}yz\plus{}zx\plus{}xyz \ge 4$.
Let $ f(x,y,z) \equal{} xy\plus{}yz\plus{}zx\plus{}xyz\minus{}4$. If there exists $ x,y,z$ such that $ g(x,y,z) \equal{} xy\plus{}yz\plus{}zx\minus{}(x\plus{}y\plus{}z)$ and $ f(x,y,z) < 0$, then let $ k$ be the number such that $ f(x\plus{}k,y\plus{}k,z\plus{}k) \equal{} 0$. Then $ g(x\plus{}k,y\plus{}k,z\plus{}k) \equal{} g(x,y,z) \plus{}3k^2\minus{}3k\plus{}2k(x\plus{}y\plus{}z) > g(x,y,z) \equal{} 0$. Since we get $ x\plus{}y\plus{}z \ge 3$ from $ x\plus{}y\plus{}z \equal{} xy\plus{}yz\plus{}zx \le \frac{(x\plus{}y\plus{}z)^2}{3}$.  So it suffices to prove that $ g(x,y,z) \le 0$ whenever $ f(x,y,z) \equal{} 0$. That is:
$ xy\plus{}yz\plus{}zx \le x\plus{}y\plus{}z$ when $ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$.
From $ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$ we find that there exists $ n,m,p>0$ such that $ x \equal{} \frac{2n}{m\plus{}p}, y \equal{} \frac{2m}{n\plus{}p}, z \equal{} \frac{2p}{n\plus{}m}$. Inserting and expanding gives:
$ n^3\plus{}m^3\plus{}p^3\plus{}3nmp \ge nm(n\plus{}m) \plus{} mp(m\plus{}p) \plus{} pn(p\plus{}n)$ which is schur.

[quote="Lukashenko"]Given positive reals such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ac$. Prove that the following inequality holds:
\[ a \plus{} b \plus{} c \plus{} 1\geq4abc.\]
[/quote]
If $ a\plus{}b < 1$ then $ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ca\equal{}c(a\plus{}b)\plus{}ab<c\plus{}ab < c\plus{}a < c\plus{}a\plus{}b$ and we obtain a contradiction. So $ a\plus{}b,b\plus{}c,c\plus{}a \ge 1$
Let $ a \equal{} x \plus{} \frac{1}{2}, b \equal{} y \plus{} \frac{1}{2}, c \equal{} z \plus{} \frac{1}{2}$. Then the condition is:
$ xy\plus{}yz\plus{}zx \equal{} \frac{3}{4}$ and $ a\plus{}b,b\plus{}c,c\plus{}a \ge 1 \iff x\plus{}y,y\plus{}z,z\plus{}x \ge 0$. So at most one of $ x,y,z$ is negative.
Inserting gives:
$ 8xyz \le 1$
If one of $ x,y,z$ is negative it is obvious. If not, then $ \frac{1}{4} \equal{} \frac{xy\plus{}yz\plus{}zx}{3} \ge \sqrt[3]{x^2y^2z^2} \iff 8xyz \le 1$ and we are done.

Extraordinary solutions. Thank you all. Shouldn't there be a solution based on classical well-known theorems-inequalities?

[quote="Lukashenko"]Given positive reals such that $ a + b + c = ab + bc + ac$. Prove that the following inequality holds:
\[ a + b + c + 1\geq4abc.\]
[/quote]
Let $ a = \frac {1}{x}$, $ b = \frac {1}{y}$ and $ c = \frac {1}{z}$. The condition remains the same and we have to prove $ x + y + z + xyz \ge 4$. We have $ z = \frac {x + y - xy}{x + y - 1}$ ($ x + y - 1$ is positive, as one easily sees). Hence the inequality is equivalent to
\[ (x + y)(x + y - 1) + (x + y - xy) + (x + y - xy)xy \&\ge 4(x + y - 1) \\
\Leftrightarrow x^2 + y^2 + xy + x^2y + xy^2 - x^2y^2 - 4x - 4y + 4 \&\ge 0.\]
But this is equivalent to
\[ (x - 1)(y - 1)(4 - xy) + (x - y)^2 \ge 0\]
We can clearly choose $ \alpha$, $ \beta \in \{x,y,z\}$, such that $ (\alpha - 1)(\beta - 1) \ge 0$. Moreover, if $ xy \ge 4$, we are done, as $ xy + yz + zx + xyz > xy = 4$.


linear algebra
matrix
function
real analysis
linear algebra unsolved

Let $ A(\cdot)$ be a smooth function from $ [\minus{}1,1]$ to the set of invertible symmetric real $ n \times n$ matrices such that $ A''(t) \equal{} A'(t)A^{\minus{}1}(t)A'(t)$, with $ A(\minus{}1) \equal{} \left( \begin{matrix} 4 & 3 \\ 3 & 2 \end{matrix} \right)$  and $ A(0) \equal{} \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right)$. Find $ A(1)$.

Since $ A: [\minus{}1,1]\rightarrow Sym(2,\mathbb{R})\cap GL(2,\mathbb{R})\,,\,t\mapsto A(t)$ is smooth and $ [\minus{}1,1]$ is a connected interval, no sign change can occur in the function $ \det A(t)$. However, the determinants of the two matrices $ A(0),A(\minus{}1)$ have opposite signs, hence the function $ A$ [b]does not exist[/b].

OK hjbrasch is right.
Assume now that the 2 conditions are $ A(0)\equal{}I$ and $ A(u)\equal{}C$ where $ u\not\equal{}0$ and $ C$ is a real invertible matrix (we leave the useless hypothesis: symmetric).
Then $ A(t)\equal{}e^{Bt}$,  where $ B$ is some real matrix, satisfies the required conditions except perhaps the last one: $ A(u)\equal{}C$ ie. $ e^{Bu}\equal{}C$;  thus $ C$ must be the exponential of a real matrix  or equivalently the square of a real matrix.
   Obviously it isn't the case here.

[quote="hjbrasch"]... hence the function $ A$ [b]does not exist[/b].[/quote]You're right, my mistake  :oops:

The answer I was looking for is $ A(1) \equal{} A( \minus{} 1)^{ \minus{} 1}$. Loup blanc has given the reason why. This is actually correct if $ A(t)$ is allowed to be complex and symmetric ([i]not hermitean[/i]) and $ \det(A(t))$ is allowed to trace out a curve in the complex plane (a portion of the unit circle in this particular example).

Anyway thanks for a concise argument.


algebra
function
domain
algebra unsolved

Find all f(x) with the following properties;
(1) f : R -> R
(2) {f(x)+f(y)}{f(a)+f(b)} = f(xa-yb)+f(xb+ya)

The answer is kx^2, but I don't know why

[quote="gus135791"]Find all f(x) with the following properties;
(1) f(xy) = f(x)f(y)
(2) f(x) + f(y) = {f(x+y)+f(x-y)}/2

The answer is kx^2, but I don't know why[/quote]

domain and range?!

[quote="Bugi"]It's IMO 2002/5. Can be found in Resources.[/quote]

THanks a lot!!!!!!!!!


$A,B,C$ are three angles of the triangle $ABC$.

Show that if $\sin^{2}{A}+\sin^{2}{B}=\sin{C}$ and $|A-B|<\frac{\pi}{2}$ then ${C}=\frac{\pi}{2}$.

see http://www.mathlinks.ro/Forum/viewtopic.php?t=78819

First of all, the condition $ |A-B|<\frac{\pi}{2}$ will help because it yields $ \cos(A-B)>0$
$ \hline$

$ C = \pi-(A+B)$

$ \sin C = \sin (A+B)$

$ \sin C = \sin A \cos B+\sin B \cos A$

$ \sin^{2}A+\sin^{2}B = \sin A \cos B+\sin B \cos A$

$ \boxed{\sin A (\sin A-\cos B)+\sin B(\sin B-\cos A) = 0}$

$ \boxed{\sin A \cdot X+\sin B \cdot Y = 0}\ \ (1)$

$ \hline$

It is $ \sin A, \sin B>0$. We will see that $ X\cdot Y\geq 0$

We have:
$ X\cdot Y = (\sin A-\cos B)(\sin B-\cos A)=$

$ \sin A \sin B+\cos A \cos B-\sin A \cos A-\sin B\cos B=$

$ \cos(A-B)-\frac{\sin 2A+\sin 2B}{2}=$

$ \cos(A-B)-\sin(A+B)\cos(A-B)=$

$ \cos(A-B)[1-\sin(A+B)]$

We have $ \cos(A-B)>0$
Also the factor $ 1-\sin (A+B)$ is in the interval $ [0,1)$

So $ X\cdot Y\geq 0$. This means that either $ X,Y\geq 0$ or $ X,Y\leq 0$

At any case, the only solution of $ (1)$ is $ X=Y=0$

Then $ \boxed{\sin A = \cos B\ \ \wedge\ \ \sin B = \cos A}\ \ (2)$

$ \sin A, \sin B>0 \Rightarrow \cos A, \cos B>0\Rightarrow A,B\in\left(0,\frac{\pi}{2}\right)$

Then from $ (2)$ we get $ A+B=\frac{\pi}{2}\Rightarrow C = \frac{\pi}{2}$

[quote="pontios"][color=darkred]Show that in any $ \triangle ABC$ exists the implication $ \|\begin{array}{c}|A-B|<\frac{\pi}{2}\\\\ \sin^{2}{A}+\sin^{2}{B}=\sin{C}\end{array}\|\implies{C}=\frac{\pi}{2}$.[/color][/quote]
[color=darkblue]$ \|\begin{array}{c}\sin^{2}A+\sin^{2}B=\sin C\ge \sin^{2}C\implies a^{2}+b^{2}\ge c^{2}\implies\boxed{\cos C\ge 0}\\\\ |A-B|<\frac{\pi}{2}\implies \boxed{\cos |A-B|>0}\\\\ \sin^{2}A+\sin^{2}B=\sin C\Longleftrightarrow\cos 2A+\cos 2B=2\cdot (1-\sin C)\Longleftrightarrow-2\cdot\boxed{\cos |A-B|}\cdot\boxed{\cos C}=2(1-\sin C)\end{array}\|$ $ \implies$ $ \sin C\ge 1$ $ \implies$ $ C=\frac{\pi}{2}$.

[b]Remark.[/b] I used only the relations $ \{\begin{array}{c}2\sin^{2}x=1-\cos 2x\\\ x+y=\pi\implies \cos x+\cos y=0\end{array}$.[/color]


Hi there... i had a grade 8 student ask me this question and i as stopped in my tracks :S i'm a grade 12 student just to add insult to injury :blush: !

Anyways... here goes the question... i hope you guys can help me out haha...

There is a bag of quarters, nickels, dimes and pennies, they average out to 17 cents... Someone takes away a penny and now the average is 18 cents. How many nickels are there?

[hide="Solution"]There are a total of 17 coins to start, with a value of 289 cents ($17\times17$).  Take a penny away, and the value is 288 cents, divided by 16 gives you the average value of 18 cents.  The only combination of coins that can total the original value is 11 quarters, 2 nickels, and 4 pennies.  So the answer is 2.[/hide]

THANK YOU!!!! I'd kiss you but that's queer so I won't  ;)
but how did you start off with the solution?

IE: How did you get the 200+ cents and 17 coins as a start?

The product of a number one less and a number one more than a specific number is one less than the square of that specific number.  For example, $24\times26 = 624$ and $25^2=625$.

In this case, you could recognize that $16\times18 = 17^2 - 1$.  Or you could simply use guess and check.

but how did you get 288 cents from 17 coins?

Edit: nm i got it! thanks again


Consider a set of 30 elements $ \{a_1,a_2,\ldots ,a_{30}\}$. For the first round, pick $ a_1$ and then a different element at random. For the second round, pick $ a_2$ and another element randomly. For the third round, pick $ a_3$ and another element at random. Etc. What is the probability that after $ n$ rounds, each of the 30 elements is picked at least once?

So the smallest possible $ n$ is 15 (example: $ a_1,a_{30}$ then $ a_2,a_{29}$, ..., and finally $ a_{15},a_{16}$), and the largest $ n$ that matters is 30.

[hide="not sure  :blush: "]
for $ 15 \le n \le 30$

the required prob = $ \frac{(30\minus{}n)! C^n_{2n\minus{}30}}{30^{30\minus{}n}}$

[/hide]

That looks about right I think. Do you have to multiply by $ (\frac{29}{30})^{2n\minus{}30}$, since for example you can't pick $ a_1,a_1$?

chrischris' formula is wrong.  Consider the case $ n \equal{} 29$, for example: in this case, we win if one of our 29 random choices gets us 30, and lose otherwise.  This "otherwise" has probability $ \left(\frac {29}{30}\right)^{29}$, so the answer should be $ 1 \minus{} \left(\frac {29}{30}\right)^{29}$.  There's no fix to a formula of that sort that will make this answer come out.

This is some variation of the coupon collector problem.  To write down the answer, you're probably going to want to use inclusion-exclusion.

JBL, I see what you mean. But for the case $ n \equal{} 29$, should it be $ 1 \minus{} (\frac {28}{29})^{29}$ because we are certain we will not pick the same element in a given round? For $ n \equal{} 28$ would it be $ 1 \minus{} 2(\frac {28}{29})^{28}$? But for $ n\equal{}27$, $ 1\minus{}3(\frac{28}{29})^{27}$ gives a negative number.

I'm not sure how to come up with a general formula.

[quote="tempo10050"]JBL, I see what you mean. But for the case $ n \equal{} 29$, should it be $ 1 \minus{} (\frac {28}{29})^{29}$ because we are certain we will not pick the same element in a given round?[/quote] Yes, you're right.

[quote="tempo10050"]For $ n \equal{} 28$ would it be $ 1 \minus{} 2(\frac {28}{29})^{28}$?[/quote]  No, it wouldn't.  For $ n \equal{} 28$, we can fail by never picking 29 (probability $ \left(\frac {28}{29}\right)^{28}$) or by never picking 30 (probability $ \left(\frac {28}{29}\right)^{28}$), but we've double-counted the cases in which we never pick either 29 or 30 (probability $ \left(\frac {27}{29}\right)^{28}$), and so the probability that we want is $ 1 \minus{} 2\left(\frac {28}{29}\right)^{28} \plus{} \left(\frac {27}{29}\right)^{28}$.  As I said in my previous post, you can write down the general formula using the principle of inclusion-exclusion.  (You'll get some big sum.  It might look a little nicer if you factor a $ 29^{\minus{}n}$ out of every term.)

Ok I understand now. But are you sure the inclusion exclusion formula is right for the case $ n\equal{}15$ since it should be $ \frac{15!}{29^{15}}$?

Yes -- in that case the formula does give the value you've written, though it's not obvious just from looking at it that it does so.

I wonder if I did something wrong but I went ahead and computed the sum for $ n\equal{}15$ and got 1.68*10^-10 or so whereas 15!/29^15 is about 1.52*10^-10.

Either you implemented it wrong (probably not because the numbers are so close) or you have calculator round-off error (likely) or both -- you might try comparing just the numerator $ \sum_{a \equal{} 0}^{15} ( \minus{} 1)^a \binom{15}{a} (29 \minus{} a)^{15}$ to $ 15!$, as possibly this will avoid some of that error, but I really do promise that they give the same value :)


Show that for all real numbers $x,y,z$ we have
\[ 11(x+y+z)^3\leq 54xyz+27(x+y+z)(x^2+y^2+z^2)  \]

Just expand it !
  The ineq is equivalent to
 $8\sum_{sym}x^3-6\sum_{sym}x^2y-12xyz \geq 0$ , which is true by Muirhead (or AM-GM)

this is the same that $\sum{8x^3 - 6x^2y - 2xyz} \geq 0$, that holds by Muirhead.  :)

Aren't the last two posts exactly the same??

Bomb

Well, This inequality is quite strange for me.
In fact, I can solve it by this way:
Because the inequality is homogeneous then we put $x+y+z=3$
And we have to prove $f(x,y,z)=3(x^2+y^2+z^2)+2xyz-11 \geq 0$
Put $x=3-y-z$ then we get
$f(y,z)=(6-2z)(y+\frac z2 -\frac32)^2+\frac12 (z+5)(z-1)^2$
I wonder if $f(y,z)$ is really greater than 0 because $y,z$ is real number.And after a few minute calculation Well... let take $x=y=-1$ $z=-2$ then it is wrong!
If $y,z$ are positive number than this equality is trivia!

Not true for all real . Counter example : x=y=0 , z=-1

LHS = -11 , RHS = -27


greatest common divisor
Ring Theory
number theory
number theory unsolved

[quote="fredfox"]Let x,y and z be integers such that X^3-2=Y^2.
Prove that x=3,y=5.
(x>0 and y>0)[/quote]
[hide]
$ x^{3}\equal{}(y\plus{}i\sqrt{2})(y\minus{}i\sqrt{2})$
GCF of the two terms on the RHS is at most 2, but if 2|x then the LHS would be -2 mod 8, which isn't a residue.
Since $ i\sqrt{2}$ is a UFD, we have that both $ y \pm i\sqrt{2}$ are cubes.
$ (a\plus{}bi\sqrt{2})^{3}\equal{}a^{3}\minus{}6ab^{2}\plus{}i(a^{2}b\sqrt{2}\minus{}2\sqrt{2}b^{3})$
$ a^{2}b\minus{}2b^{3}\equal{}1$
b|1, b=1 or -1.
If b=1, then we have that $ a\equal{}\sqrt{3}$ which doesn't work.
If b=-1, then we have that $ a\equal{}\pm 1$. a=1, b=-1 means that $ y\plus{}i\sqrt{2}\equal{}\minus{}5\minus{}i\sqrt{2}$.
So $ y\equal{}\pm 5$, but since y>0 we have that y=5. x=3 follows.[/hide]

$ \text{HINT: }\\{x^3 \minus{} 2 \equal{} y^2}\Longleftrightarrow{x^3 \minus{} 27 \equal{} y^2 \minus{} 25}\Longleftrightarrow{(x \minus{} 3).(x^2 \plus{} 3x \plus{} 9) \equal{} (y \minus{} 5).(y \plus{} 5)}\Longleftrightarrow$
$ \Longleftrightarrow{x \minus{} 3 \equal{} \frac {y \minus{} 5}{k}}\text{ and }x^2 \plus{} 3x \plus{} 9 \equal{} k.(y \plus{} 5);k\in\mathbb{Z}^*.$

[quote="diophantient"][quote="fredfox"]Let x,y and z be integers such that X^3-2=Y^2.
Prove that x=3,y=5.
(x>0 and y>0)[/quote]
[hide]
$ x^{3} \equal{} (y \plus{} i\sqrt {2})(y \minus{} i\sqrt {2})$
GCF of the two terms on the RHS is at most 2, but if 2|x then the LHS would be -2 mod 8, which isn't a residue.
Since $ i\sqrt {2}$ is a UFD, we have that both $ y \pm i\sqrt {2}$ are cubes.
$ (a \plus{} bi\sqrt {2})^{3} \equal{} a^{3} \minus{} 6ab^{2} \plus{} i(a^{2}b\sqrt {2} \minus{} 2\sqrt {2}b^{3})$
$ a^{2}b \minus{} 2b^{3} \equal{} 1$
b|1, b=1 or -1.
If b=1, then we have that $ a \equal{} \sqrt {3}$ which doesn't work.
If b=-1, then we have that $ a \equal{} \pm 1$. a=1, b=-1 means that $ y \plus{} i\sqrt {2} \equal{} \minus{} 5 \minus{} i\sqrt {2}$.
So $ y \equal{} \pm 5$, but since y>0 we have that y=5. x=3 follows.[/hide][/quote]

how do you know that the gcd is at most 2, in general how do you determine the gcd of two factors in a ring?

$ \gcd(y\plus{}\sqrt{\minus{}2} , y\minus{}\sqrt{\minus{}2}) \equal{}\gcd(y\plus{}\sqrt{\minus{}2}, 2 \sqrt{\minus{}2} )$ which divides $ 2\sqrt{\minus{}2}$. As it is an integer, it divides at most $ 2$ (in fact, we don't need to exclude $ 2 \sqrt{\minus{}2}$ as $ \gcd$ and can still argue as before).


:rotfl:  :pilot:
please prove this inequality

You have some kind of typo:
try $ x\equal{}y\equal{}\frac{\sqrt{2}}{2}$


Let's post our favorite mathematical memories from HCSSiM!  Any year of HCSSiM is fine.

Here are some of mine:

Working on a hard problem in problem session and being really disgruntled when we ended early to go have ice cream.  I really liked the ice cream, but I couldn't talk to anyone because I was concentrating on the problem.  I went to bed thinking about it, woke up thinking about it, and wrote a really long explanation the next afternoon.

Discovering Dyck's Theorem and being frightened that all the ideas about surfaces we'd been using were going to crash in on themselves and turn worthless.  Fortunately, we needed to make only a small adjustment to how we'd been thinking about things.

Trying to work on a proof with the alumn mini mini when we were all talking at the same time (we all had so many ideas) and having trouble thinking and listening at the same time because it all went so fast.  We were yelling and joking and running in and out of sarah-marie's classroom asking s-m and Emily for advice, and eventually we figured out something, and we went through the classrooms shouting the proof outline at high speed to whoever would sit still.

Okay, so Kelly's PTT about the probabilistic method was excellent.   That is my prototypical "sexy proof."  That was...a moment where I suddenly considered a whole new way of looking at something.

The second problem set of workshop last year was kind of weird.  I was in this kind of state of odd focus, where I could take a solid whack at anything and could think really fast.  It was really kind of disconcerting, but very, very fun.

I'll think of some more in a minute.

I remembered another one!

I'd gotten nothing done in problem session, and then seventeen minutes before the end of problem session I realized that if you draw a Hamilton cycle on a polyhedron and then cut along it, you get two pieces, and if you take the dual of each piece in a certain way, it's a tree.  I wrote it up in the seventeen minutes remaining with apologies that it was rushed, but Tom understood what I meant and thought it was a cool idea.

Mathematical ehh? So toga parties and no-pants sundays dont count?

Lets see, Kelly's probabilistic method primetime was pretty sweet, that definitely made me want to skip dinner and do math right then.  Also, Zeb's primetime was like that too A- it was Zeb B- that whole any string of digits is the start of a power of two was awsome.

Or last year in the algebraic topology mini, Wing was telling us how he originally wanted to name the mini: "infinity ways to lynch a torus" right after we classified the homotopy group of the torus.

Or Cary and Zeb dressing as pirates to discuss complex numbers, a + b(aye!) or (arrrr)cis(theta).  Only to be topped by s-m and Emily a few days later, when they dressed as a pirate and anti-victor for a nice excursion into graph theory.

Infinite proofs of infinite primes was also classic, my favorite is lemma: the twin prime conjecture, proof Duh! And of course the finitely many proofs of finitely many primes was excellent aswell the last night.

Well I should probably stop there, but I could definitely go on.

I have to go soon so I'll only give  a couple...
The last day of Wing's mini where he sat in the back and told us what to write on the board... Ravi's Prime Time about the sum of inverse squares... the "Louis vs. Dan" problem set problem in s-m's class... Cale's crazy polyhedral Mathematica-animated buckyballs... YP singalongs... folding stuff in Tom's maxi [a regular heptagon!]....  the mind-blowing prime time on lines, circles, and planes...  I miss HCSSiM.

OK, I have to go.  Bye!


[hide="solution"]
Straightforward...

$ (\minus{}1)^k*\frac{n}{k!} \plus{} (\minus{}1)^{k\minus{}1}\frac{1}{(k\minus{}1)!}$

$ \equal{} (\minus{}1)^k\left(\frac{n}{k!} \minus{} \frac1{(k\minus{}1)!}\right)$

$ \equal{} (\minus{}1)^k*\left(\frac{n\minus{}k}{k!}\right)$

[/hide]


I haven't studied much about the incompleteness theorem, but i would like to ask whether it is possible for some famous conjectures such as Goldbach conjecture turns out to be unprovable. Can we prove that Goldbach conjecture is provable or unprovable? If this statement is unprovable, then mathematicians will work on it for centuries but still getting nothing.

However, if we can prove that Goldbach conjecture is provable, then it is only a matter of time that we find out the proof. Since the world is deterministic, the prove is already here, hidden in somewhere in the world. Why can't we claim that we have already got the proof? Also we can use machines to simulate human mind and find out the proof. The algorithm will find out the proof after finite number of calculations. This is just the same as proving Goldbach conjecture for even numbers greater than $ 10^{10^{10}}$ would mean that Goldbach conjecture is proved. Can we claim that we have proved everything if we construct an algorithm that simulate human thinking process? Does such algorithm exists? Or is this an unprovable statement as we human construct something that resembles ourselves, which causes a recursion?

It's all about journey, not destination. Even if the destination is unreachable, we'll probably find interesting things along the way.

The mere knowledge that Goldbach's conjecture is true (or false) is pretty much worthless unless it expands our understanding of arithmetics. If an oracle, or a supercomputer, finds that the conjecture is certainly true, but does not explain why it is true, why should anybody care?

[quote="modeler"]I haven't studied much about the incompleteness theorem, but i would like to ask whether it is possible for some famous conjectures such as Goldbach conjecture turns out to be unprovable. Can we prove that Goldbach conjecture is provable or unprovable? [/quote]

The Incompleteness Theorem merely states that certain statements [i]exist[/i] that are true but unprovable.  These statements are constructed in a recursive way, and do not correspond to the statements that we usually talk about.  It's just that a sufficiently powerful system is [i]capable[/i] of expressing true but unprovable statements, not that many of the normal statements we make in that system are actually unprovable.

The Goldbach Conjecture seems quite provable to me, anyway.  Some mathematicians suspect that the Collatz Conjecture might not be provable, but personally, I doubt it.

[quote="modeler"]Also we can use machines to simulate human mind and find out the proof.[/quote]

If that were true, there would be no need for mathematicians.  I'd like to see your program that does this :)

If we would have arbitrary fast supercomputers, we could check all "logical expressions" up to very big lengths. If we assume there is a "reasonable" proof for something, such a sequence proving it exists. Then it could theoretically be found by a computer using brute force (or a bit trickier ideas).
But untill know, mathematicians are very much more efficient than any computer if it commes down to proving.

[quote="modeler"]I haven't studied much about the incompleteness theorem, but i would like to ask whether it is possible for some famous conjectures such as Goldbach conjecture turns out to be unprovable. Can we prove that Goldbach conjecture is provable or unprovable? If this statement is unprovable, then mathematicians will work on it for centuries but still getting nothing.
[/quote]

I'm not sure about this, but I believe that while there are ways one can show that a statement is unprovable in a system (for example showing that both the statement and its negation are consistent in some model of the system), the only way to [i]prove[/i] that a statement is provable is with an actual proof of the statement.

So if Goldbach is provable (which I think almost everyone believes that it is), the only way to prove it would be with a proof!

If Goldbach is unprovable, then the only way to know is with a consistency proof - which may be just as difficult to find (if it exists) as an actual proof of the statement (if it exists).


[quote="cincodemayo5590"](something i just made up)
Define a "good word" as one who has no two identical letters adjacent to one another. How many "good words" can be created from the letters in $MATHEMATICS$? How many "good words" are there are of lengths 1-11?

(Extra points) How many of these "good words" are actually english words?

 :D[/quote]

[hide]Um...
I just did $\frac{11!}{2!2!2!}-3\displaystyle\left(\frac{9!}{2!2!}\right)+\frac{6!}{2!2!}=4717620$[/hide]


[quote="polya.szego"]Hi folks,

Is it true that if the sum of some integers can be expressed directly as a power of 2, e.g., 2^p, where p is in the naturals, then the common divisor of such integers must also be able to be expressible in terms of 2?

Thanks all.[/quote]

Sure : if $ \sum a_k\equal{}2^p$, then $ \gcd(a_k)$ divides each $ a_k$, so divides their sum, so divides $ 2^p$, so is $ 2^q$ for some $ q\in[0,p]$

What about the sum of the first $ n$ odd numbers?
$ 1 \plus{} 3 \plus{} 5 \plus{} 7 \equal{} 16$....

Edit: Oops, my mistake, I thought we're excluding the trivial $ 2^0$ case as the $ \gcd$.


superior algebra
superior algebra unsolved

Show that
c=sqrt(1+sqrt(2)+sqrt(3))

is a constructible number by finding a sequence of square root extension fields extending from the field Q of rational numbers to a square root extension field containing the number c.

[quote="ZetaX"]Adjoin $ \sqrt 2$ then $ \sqrt 3$ then the whole thing.[/quote]

Can you be more specific, please, because I have really hard time in understanding these constructible numbers and fields. Thanks

That's not all that different from a set of instructions for actually constructing $ c$ using compass and straightedge as in Euclid.

Given a fixed length which will will call $ 1:$

First construct the geometric mean of $ 1$ and $ 2,$ hence $ \sqrt{2}.$

Then, construct the geometric mean of $ 1$ and $ 3,$ hence $ \sqrt{3}.$

(Or alternatively, construct the diagonal of an isosceles right triangle to get $ \sqrt{2},$ then use that as a the leg of a right triangle with other leg $ 1$ to get $ \sqrt{3}$ as the hypotenuse.)

Lay out $ 1,\sqrt{2},$ and $ \sqrt{3}$ end to end to make $ 1\plus{}\sqrt{2}\plus{}\sqrt{3},$ then use the geometric mean construction to make $ \sqrt{1\plus{}\sqrt{2}\plus{}\sqrt{3}}.$


let $a\ge2$ and $n$ integers such that :$a^{n-1}\equiv1(modn)$ and $a^{p}\not\equiv1(modn)$ for all prime factor $p$ of $n-1$

proof that$n$ is a prime.

It's wrong: $n=91, a=3$.

But when you change the condition $a^{p}\not\equiv 1\mod n$ into  $a^\frac{n-1}p \not\equiv 1\mod n$, then it's correct (and easy since $a$ is a primitive root then).

sorry i have made a mestake ! and i 'll correct it

let $a\ge2$ and $n$ integers such that :$a^{n-1}\equiv1(modn)$ and $a^{m}\not\equiv1(modn)$ for all divisor $m$ of $n-1$

proof that$n$ is a prime.


Next time do it here: http://www.mathlinks.ro/index.php?f=224


Well, for me, tomorrow is state.

At chapter I missed #6 and #11 on the sprint, so my score was a 44.

I was the only one in my chapter to get #29, the one about the # of isosceles triangles in a 3x3 grid. That problem is a very slightly easier version of an old national target.

EDIT: THe poll got deleted, because...? I'm not sure if it was an accident or not. Since no mod commented, I'll re-instate it. If a mod gives a comment, which would let me know for sure it was deleted rather than an error on my part, then take the poll down again. I'm leaving for the hotel for state soon.

Edit #2: The poll disappeared... I edited it to re-add the options, and wow the poll pops back up... with new options. and it disappears again.

I did horrible! got a 26, 6th written, 1st cd

I expect ~ 34 (i was undercounting i thought) after i took the test

-jorian

I got a 38.

I thought this year's was harder than last years.

i don't know my score, but i got 5th written, 6th after CD.
I know for sure I got 2 sprint problems wrong, but that's it.

I got around a 29 >_<

That was HARD IMO. Besides the fact that I was nervous. :(

I had like 17 on the sprint and 6 on the target.

I still managed 13th written though. *whew*

I better pull my act together for the state.

I got 30... 22 on sprint and 4 on target. I suck. :(

I wasn't nervous, I just SUCK.  I can't believe I made it to state with a 30!

I was worried, I'd been there last year...I was shocked when my D+ math test got me 2nd written, 3rd CD.  I was in a "sudden victory" 3-2 loss though (pretty sure they called it sudden death last year?) on that spinners, product probability odd/even question.  I need to get my act together for state too.

Got 19 or 20 sprint, 6 (12) on target.

I think that 70% in MathCounts should be an A+.
 :D

I got a 41 (25 on sprint and 8 on target)

22 Sprint and 12 on Target for a total of 34. 10th written.

23 on sprint and 7 target.  1st written.

Ugh, 29/14. I was a complete idiot on 8 on target, I thought I could just take the average of the speeds...

I don't know I was never told, and when I asked, my coach sayed she didn't know. Probably 1...

[quote="jhredsox"]Lol, there were idiots saying that in the room i was in as well.

What sprint did you get wrong?

-jorian[/quote]

Can we stop with the "there were idiots" and "your method sucks" comments.  If someone is an idiot because they did that target wrong then you are saying Perfect628 is an idiot.  And if I'm not mistaken we could have gotten all Targets wrong and scored higher than I believe you said you did.

Can we stick to facts and not opinions before this forum degenerates any further??

18 on sprint and 6 targets gives me a score of 30, which was also 3rd written. After cd I was fourth.

What I meant was, there were like 30 people in the room I was in (there were many rooms) and everyone was like, isn't 54 the answer? I just wanted to get it across that I was amused at this and used the word "idiots" to make perfect hopefully feel less of one as he wasn't the only one to make that mistake.

Sorry about that.

About the team round, I guess I should've said I personally think there's a better way to go.

Sorry again,
Jorian

I didn't ACTUALLY think I was an idiot, I meant I made a dumb mistake. Actually, I realized it was wrong, but with about 15 seconds left, so I guessed 52.7. kinda close...

I got 24/14=38, 1st written.

I was mad at myself because I missed the target question with the quadrilatteral because I didn't add the areas of the two triangles I divided it into.... :ewpu:

22/16. 5th.    I screwed up the counting/probability problems in the sprint

I love C/P problems, but I think I may have miscounted the chocolate box thing...not really C/P, but w/e...

i think i got 20 on the chocolate box problem. what was the right answer?

I don't remember my answer-I think I put 18-ur right.

god i wish i was in a easy chapter
even though my chapter was pretty easy too
I got 39 and got 3rd
23/16
idk what happened on sprint
i thought i did pretty well
look at my other posts for more detail

for me, i really didin't THAT much about my score at chapters(22/16) since to me, all that matters is making it to State and give myself at least a chance, no matter how small, at making Nats.


Just wondering, are there any USAPhO takers or Physics Olympiad team members here browsing the AOPS forums?

not me... i'm still in eighth grade

but I did read the bios of the team... some REALLY strong math people


parameterization geometry calculus MIT college

I'm guessing it's okay to talk about the semifinal round now, since it was around this date last year, too.

What did you guys think? I epic failed on number 4 because I forgot that sound has a finite speed -_-

I also said for number 3 that the core has to have an infinitely high temperature because I modeled the planet as a bunch of thin shells; I said that since it was in a steady-state, if the outer shell was losing heat with power P, it must be gaining heat with power P and so on, so every shell was transmitting P. That was wrong. lol

annnd... yeah, didn't do too well on number B1, but got most of B2.

Overall the topics seemed more diverse this year, more calculation heavy (number 4....), and slightly more difficult than last year's semi-final exam but not as bad as 2006. Thoughts?

Well I only know mechanics and electrostatics, but here were my attempts to guess how to do problems:

A1 - didn't have any idea, don't really know voltage and air

A2 - I don't really know gases, but I remember ideal gas law from chemistry, and I think I might have gotten most of it

A3 - Don't know thermodynamics

A4 - Since they gave you the equation, I think you could figure this out without knowing waves - I think I got like parts a and b and maybe c

B1 - solved completely

B2 - I got a little bit of the first couple parts, but that's it

How well do you normally have to do to get into physics camp? I didn't think I had a chance, but I did sort of well for not knowing thermo, electomagnestism, etc. It seems like it would be pretty easy if you actually knew all these areas.

From looking at threads from past years, it seems like the cutoff for the camp depends on whether or not you are a senior.

I feel like I got most of A2 and A4, some of A1.  My A3 was blank.

I don't remember which one of the Bs was capacitors and which one was mechanics, but I got most of the mechanics one and a bit of the other one.

I wrote up solutions to all of the questions, but the thermodynamic ones I kind of BS-ed (my knowledge of thermodynamics is rather rudimentary).  For A3, I also modeled the planet as consisting of thin shells.  A4 did involve a lot of computation, but for the most part, I thought it was quite easy (it didn't really require any prior knowledge).

A1 and B2 were solvable if you knew the formula for the capacitance of a parallel-plate capacitor.  I had to rederive it using Gauss's law because I didn't remember it.

The optional part C was more of a math problem than a physics problem! (It didn't require any prior knowledge, either.)  But that's cool.

What was the optional part? I didn't do that.

The optional part was an optics problem, but it didn't require you to know any formulas beforehand.  Basically, the problem was about a new type of lens that's being researched right now, involving a glob of liquid sitting on a metal surface whose shape can be adjusted by varying the voltage across the surface or something.  The question asked you to find some parameters in terms of others and to calculate the volume of the glob in terms of the parameters.  It was just a bit of geometry and maybe simple calculus.

You should try the optional part! (if you take the test again next year)

I'm a senior, and I doubt I'll make it, but...

I got A1, A2, most of A3 and all but (d) for A4.

I got all of B1 and all but one letter on B2.

I hope this is good enough =/. I heard results come out next week.

That's really good.  If you feel confident about your answers, then I really do think you have a chance at making the camp.  But this is my first year doing the physics olympiad, so I wouldn't know.

Do you need calculus to participate in USAPHO?

[quote="genericme"]Generally, yeah (at least in the later rounds).[/quote]

genericme, what grade are you in?

Hmmm I probably knew how to do all of the problems but did not necessarily have enough time for them. Seems like a lot of it is speed and familiarity.

Dude you're obviously going to make the camp... because you have a reputation... and everyone knows how great you are.

anyone know when results are supposed to come out?

According to Artzilla, they should be out this week!

Yeah, AAPT called my school...

and asked for the contact information of...

THE OTHER TWO PEOPLE WHO TOOK THE EXAM FROM MY SCHOOL!!!!

*feels sad*

...anyway, they said that they will be releasing at the latest this upcoming Monday. So I first said this week, now I say at the latest this upcoming Monday.

Good luck everyone =/.
(also, post if u make the team!)

That stinks.

I know you said you were a senior earlier in the thread.  What were the grades of the other two from your school?  Also, how did they do compared to you on the test?  I'm just trying to get an idea of what the cutoff may be.

The other people didn't think they did THAT well, but at this point it doent look like it matters so much. I'm at MIT now, but I'll be home in a few hours. But...my mom said I didn't get anything in the mail.


Γεια σε ολους ! να βρειτε τον τυπο της ακολουθιας των ακεραιων αριθμων που δεν ειναι τελεια τετραγωνα.στη συνεχεια προσπαθηστε αν γινεται να γενικευσετε το αποτελεσμα(για καθε δυναμη >2)

Λοιπόν για να μη βάλλω ολόκληρη τη λύση μου και χαλάσω την προσπάθεια των άλλων απλά βάζω το εξής .Η ακολουθία πουμας δίνει τέλεια τετράγωνα είναι
$a_{n}=a_{n-1}+2n-1$  kai $a_{1}=1$  $a_{2}=4$        :wink:

(χωρις αποδειξη)(ειναι ευκολη γιατι αποδεικνυεται οτι το f(n) βρισκεται μεταξυ δυο διαδοχικων τετραγωνων,απλως εχει αρκετες πραξεις)
Η συναρτηση f(n)=[n+(n)^(1/2)+1/2] ,oπου [χ] ειναι το ακεραιο μερος του χ,
δινει ολους τους αριθμους,εκτος απο τα τελεια τετραγωνα.

ΡΑΛΛΗΣ :D

Μπραβο Ραλλη.ομως οι πρωτες αγκυλες δεν χρειαζονται.δηλ η παρασταση δεν ειναι ολη στο ακεραιο μερος.........θα ηθελα να δω την προσπαθεια του σιλουανου. και το σημαντικοτερο μιας και το ελυσες..υπαρχει τυπος που δινει τους ακεραιους που δεν ειναι κυβοι ?δηλ γενικευεται?

Δεν εχω βρει τιποτα για αυτο με τους κυβους που λες,αλλα θα το ψαξω με το ιδιο σκεπτικο(να περιεχεται μεταξυ δυο διαδοχικων κυβων)
Εχω βρει επισης την ακολουθια που δινει ολους τους αριθμους εκτος απο τους τριγωνικους αριθμους   ( αυτους που ειναι της μορφης ν(ν+1)/2  ) και ειναι
f(n)=n+[(2n)^(1/2)]+1/2  οπου [χ]=ακεραιο μερος του χ.

ΡΑΛΛΗΣ :D

λοιπον ενας φιλος, μου προτεινε τη γενικευση του προβληματος. δηλαδη με δεδομενη την ακολουθια Α ειναι δυνατο  να κατασκευαστει μια ακολουθια Β ωστε καθε φυσικος να ανηκει ή στη μια ακολουθια ή στην αλλη .και η ενωση τους να ειναι το συνολο των (θετικων) ακεραιων. ????

να δωσω και την πηγη που ισως να το λυνει.εχει να κανει με ενα ενδιαφερον θεωρημα των lambek & Moser: εστω f(n) μια ακολουθια φυσικων .οριζουμε την f* ακολουθια ως εξης:αν f*(n)=k tote f(k) <n<=f(k+1) ,το f*(n) δειχνει ουσιαστικα τον αριθμο των ορων της ακολουθιας f που ειναι μικροτεροι ή ισoi του n. οι f,f* λεγονται αντιστροφες ακολουθιες.το θεωρημα λεει οτι δυο ακολουθιες f,f*ειναι αντιστροφες(κατα των πανω ορισμο) αν και μονο αν οι F(n)=f(n)+n και οι G(n)=f*(n)+n ειναι συμληρωματικες ακολουθιες ,δηλ η F και η G (η ενωση τους ) μας δινει το συνολο Ν των φυσικων.(αρκει οι f,f*  να μην ειναι φθινουσες ακολουθιες φυσικων ή αλλιως οι F,G να ειναι γνησια αυξουσες).  :)


Let $A_1,...,A_n$ matrices $nxn$ with elements in a certain field $K$. Suppose they are nilpotent and they commute. Then prove that $ A_1A_2...A_n=0$. Try not the give the classical solution using endomorphisms, but rather something purely algebraic.

They are simultaneously trigonalizable (in $M_n(\tilde K)$, where $\tilde K$ is an algebraic closure of $K$, so that we don't have any problems), and it's easy to prove that the product of $k$ upper-triangular nilpotent matrices has $0$ on the first $k-1$ diagonals above the main diagonal. The result follows from here.

Is it ok?

[quote="harazi"]... something purely algebraic.[/quote]

It seems you gave a formulae with sum and product long time ago on the forum,
I couln't find the post


search
trigonometry
real analysis
real analysis unsolved

Some of the musings I posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=864823984&t=204380]here[/url] led to the following question:  does the series

$ \sum_{p \text{ prime}} \frac {\exp(2 \pi i/p)}{p}$

converge?  This might be very easy.

$ \sum_p \frac{1}{p}$ diverges isn't it ? and $ \cos(\frac{2i \pi}{p}) \to 1$  :maybe:

I thought so - ah, I had the wrong mental picture of what the partial sums looked like.  My apologies.


analytic geometry
graphing lines
slope
USAMTS
Euler
projective geometry
geometry

this is a general question.
what are the methods commonly used to show that a series of points are collinear? like if you're given 3 points, how would you show that they are collinear?

thankS!

Several ways:

If you have coordinates, then show that the slopes between pairs of points are equal.

If you have triangles, or other things, then try [url=http://www.artofproblemsolving.com/Wiki/index.php/Menelaus%27_Theorem]Menelaus's Theorem[/url]

Comment on the first approach
If the slope of $ \overline{ab}$ is the same as the slope of $ \overline{cd}$ does not imply that $ \overline{abcd}$ is a line
Though if the slope of $ \overline{ab}$ is equal to the slope of $ \overline{bc}$ then that implies $ \overline{abc}$ is a line.

or you can be really cool and apply a pole-polar transformation and prove that the polars of the points with respect to some circle are concurrent, and the concurrency problem may be slightly easier than directly proving collinearity.

thanks guys, that helps a lot!

@1337...
im not that good. :blush:

This isn't because you saw one of those USAMTS problems (number two, most recent one?), is it?

well...sort of...i saw that problem from USAMTS but this question has been nagging me ever since i finished aops vol 1...they never mentioned anything about that in the book, and there is a lot of problems that involve proving collinearity.

Yea, it's a good thing to know in general.

The slopes way is usually the way to go if you're doing analytic geo, whereas menelaus's thm. is good for euclidian geo.

Alternatively you could look for bijections to well-known collinearities, such as the Euler Line or the Simson Line. Then Pappus and Pascal can also be helpful.

Also homothecy helps, if you have ratios of segments...

You could also apply various transformations that preserve collinearity and see if that simplifies it a little.


Các anh cho em hỏi có sách PH hàm nào hay không ạ?(Trừ cuốn của Nguyễn Văn Mạu+Nguyễn Trọng Tuấn)
Nếu có file nào upload len cho em thì càng tốt!
Thanks!

Em vào blog của anh, có một cuốn trong đó đấy!

Chỗ nào anh nhỉ,dẫn link cho em được không ạ?
Hay upload len mathlinks cung duoc!

Chỗ mục download trên cùng đấy em!  :wink:

Chienthan vào đó theo link :

http://trungtuan.wordpress.com/download

Dạo này blog anh Tuân cứ gọi là đắt sô nha.
 [hide="hỏi admin tí"] có khi nào bị nghẽn mạng do quá nhiều người xem chưa anh  :D [/hide]

[quote="thaithuan_GC"][quote="quangchuc"]Ai co ebook cuon thi quoc gia 1990-2006 bang tieng anh thi post len cho em voi!! :lol:
Thank you very much!! :)[/quote]
Ủa, vậy bạn có bản Tiếng Việt cuốn thi 1990-2006 sao? Có thể giới thiệu với mình đc ko?[/quote]
Có sách TV rồi mà anh,cuốn của toán tuổi trẻ

[quote="N.T.TUAN"][quote="quangchuc"]Ai co ebook cuon thi quoc gia 1990-2006 bang tieng anh thi post len cho em voi!! :lol:
Thank you very much!! :)[/quote]
http://mathscope.org/forum/index.php[/quote]
Anh có thể đưa link rõ hơn được không ạ.


pigeonhole principle
ratio
geometric series
geometric sequence

Is it possible for the numbers $1,2,...100$ to be the terms of 12 geometrical progressions?

EDIT: the seems to be a rather weak bound (as in it can't be done with more than 12 progressions). I'm not sure though.

[hide] By the pigeonhole principle, there must be at least one progression with at least 9 terms. We can, WLOG, set the first term of this progression to be $a$. The other terms will be $ar, ar^{2}, ...ar^{8}$. Let $r=\frac{p}{q}$. Then $ar^{8}= a\frac{p^{8}}{q^{8}}$, an integer. However, no eighth power of any integer divides a number $1$ through $100$. Therefore, $q=1$. But this is also impossible, as $p^{8}> 100$ for all positive integers $p$. [/hide]

It dosnt have to be the first 8 terms in the series to be the integers. It can be any terms at all. Pigenhole was the right track though.

Since were dealing with a geometric series, ie multiplication involved, handy numbers to look at are of course:

[hide="Dont cheat"]Primes[/hide]

I didn't phrase it very well. You're right, the first 8 terms don't have to be integers, but if they aren't, then no terms afterwords can be integers. It is impossible to have a geometric series that goes integer, ...non-integer, integer, for the reason outlined (the power of primes in the denominator).

[quote="Phelpedo"]I didn't phrase it very well. You're right, the first 8 terms don't have to be integers, but if they aren't, then no terms afterwords can be integers. [b]It is impossible to have a geometric series that goes integer, ...non-integer, integer[/b], for the reason outlined (the power of primes in the denominator).[/quote]

$1$, $\sqrt{2}$, $2$, $\sqrt{8}$, $4$,...

Good point. So what's the best way to write it up? I think the primes argument runs into similar complications.

I haven't looked at the problem, but maybe when you said "integer, noninteger, integer", you meant "integer, noninteger rational, integer".

That is what I meant, but an irrational common ratio is certainly possible, so there's still a problem :(

[hide="The solution"]Consider a geometric series that contains 3 distinct primes, say $p<q<s$
Wlog these can be written as:

$p=a$    $q=ar^{n}$    $s=ar^{m}$ (Note r dosnt have to be rational) with $n,m\in \mathbb{Z}$

Then $q^{m}=a^{m}r^{mn}\rightarrow r^{mn}=\frac{q^{m}}{a^{m}}$ and $s^{n}=a^{n}r^{mn}=a^{n}\frac{q^{m}}{a^{m}}$

And thus $a^{m}s^{n}=a^{n}q^{m}$

And so $p^{m}s^{n}=p^{n}q^{m}$

Now the fact that these numbers are prime comes into play. Since they are distinct, we see that $m=n$ looking at the p'th power. Thus $q=s$ contradiction. Therefore no geometric progression can contain 3 primes, even with irrational ratio. Since there are 25 primes less than 100, if we could split these numbers into 12 geo-series, by pigeon-hole 3 primes must be in one of then. Contradiction.[/hide]


Solve in positive integers the equation:

(10^a) + (2^b) - (3^c) =1997.

Oh my friend I think that this is too difficult problem (according to solution that I saw).

Namely this problem was given by [b]S.Shikh[/b] in one of  Belarusian IMO 1997 team trainings tests (test 3 problem 3).

I do not have time to post solution now. Unique solution is $ (a,b,c)\equal{}(3,10,3)$

For example, in solution $ \mod 193$, $ \mod 97$, $ \mod 257$ were considered.


calculus
integration
function
real analysis
real analysis unsolved

let [tex] f: A   -> R [/tex] be an integrable function, where A is a rectangle.  If g = f at all but a finite number of points, show that g is integrable and [tex] \int_{A}f = \int_{A}g [/tex]

We prove that $\int f dxdy = 0$ if $f\neq 0$ on finite points $\{(x_i,y_i)\}_1^n$ in $A$.

According to the definition (Riemann Integral), the Riemann sum
$\sum f(x,y) \Delta \sigma \leq \sum_{i=1}^n |f(x_i,y_i)| \Delta \sigma$
But $\Delta \sigma$ there can be arbitrary small,
hence
$\sum f(x,y) \Delta \sigma \to 0$
and
$\int f dxdy = 0$.


1. Given $ a_1 < a_2 < a_3 < ...$ is a sequence of positive integers satisfy :$ a_n \leq a_{n \plus{} 1} \plus{} 30$ for every $ n \geq 1 (n \in N*)$ If we denote $ x \equal{}\frac{a_1a_2a_3... a_n}{10^n} .$ then prove that x is'nt $ \in Q.$

[quote="Allnames"]Sorry, But I think $ a_1a_2..a_n$ belongs to $ \mathbb N$. So that $ x$ must belongs to $ \mathbb Q$. AM I mad?[/quote]
....>"< I mean we must prove that the sequence of a_{n} is not a circulation sequence to conclude the L.E.D  ...>"< but I don't know how to say , because I'm bad at English :D ...The sequence is an infinitive sequence of positive integers :D

P/s: ai ng viet trong nay giup e voi :| ngu tieng Anh... :|

Do you mean to say $ \sum_i\frac {a_i}{10^i}$? Because $ a_i$ are not digits for high enough $ i$.  :lol:

EDIT: Oh, I see you must mean the concatenation of the $ a_i$, preceded by $ 0$ and the decimal point.

By the way, the proof sketch is to notice that as $ i \to \infty$ we can get arbitrarily long sequences of zeroes.

[quote="Kirill"]Maybe, long14893 means that $ x \equal{} \lim_{n \minus{} > \infty} \frac {a_{1}a_{2}a_{3}... a_{n}}{10^{n}}$[/quote]
How about $ a_1 \equal{} 10;a_2 \equal{} 20;...;a_n \equal{} 10n$?. Now $ lim_{n\to \infty }x \equal{} \infty$.
And long14893 assumed that $ a_1 < a_2 < ... < a_n$ then the condition $ a_n\le a_{n \plus{} 1} \plus{} 30$ will be always true?.Am I mad?

Indeed, I think he must mean the concatenation that I mentioned.

The reason for the condition is that you could choose e.g. $ a_1 \equal{} 1$, $ a_2 \equal{} 11$, $ a_3 \equal{} 111$, $ a_4 \equal{} 1111$, ... and the concatenation of that is $ 0.\bar{1} \equal{} \frac {1}{9}$ which is rational.
But with the condition, you can get that as $ i \to \infty$ we can get arbitrarily long sequences of zeroes (followed by other digits) in the decimal, implying that the decimal is irrational. :wink:

EDIT: I said that the concatenation is $ 0.\bar{9}$ when it is clearly $ 0.\bar{1}$...

[quote="HTA"][quote="HTA"]Kirill is right .

In other words , What the author does mean is that the question asks for $ \bar{0,a_1a_2a_3...a_n.....}$ is not in Q .[/quote]

yes , what i mean is also concatenation . It is not a product but it's like you write a_1 , a_2 ,.. next to each other . Im sorry because I can not make the "bar" above 0,a_1a_2 ,... any longer . btw , i think this is the only way that could make the problem make sense .[/quote]
exactly .................................................................
sr because Im very bad at English , and I haven't used Latex for a long time , so I could'nt remember how to write the "bar" above $ 0,a_1a_2....$ ......=.='
[hide]Ai người Việt dịch hộ em cái ak:|em còn 1 lỗi sai nhỏ ở bài đầu là thay $ a_n$ thành $ a_{n\plus{}1}$ nhưng k edit dc:|[/hide]


I've got a problem here that i just cant get my head around. But i'm sure that there is some of you out there that can  do it blindfolded.

" A cancer patient is to be laid inside an elliptical lead coated tank with the tumour exactly at the focal point. The radioactive isotope that generates the radiation is being produced at the rate of 8mg per second, but has a half-life of only 15 seconds.
A tumour 2cm in diameter needs '110mg of radiation' to kill it. If the bombardment starts at the first production of the isotope, find the length (time) of the exposure required."

Wouldn't we need to know the size of the tmor?

this sounds more like a science question than a math question. you might want to try the science forums instead... or the mods can move it there

How long will it take to produce 110mg of isotope at 8mg/sec rate?

Best,
Borek


inequalities
LaTeX
inequalities proposed

Prove that n( (n+1)^(2/n) - 1) < ∑1^n (2i+1)/i^2 < n(1 - 1/n^(2/(n-1)) ) + 4

I know it may not make sense, but the actual inequality is so overwhelming that it's impossible for me to type in LaTeX. :D

:lol:  :P  $It\ was\ my\ pleasure\ !$  :oops:  :blush:


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function
quadratics
probability
number theory
prime numbers

1)  Show that there exist an infinite number of primes.

2)  Show that there exist an infinite number of primes of the form $4n+3$.

3)  Show that there exist an infinite number of primes of the form $6n+5$.

4)  Show that there exist an infinite number of primes of the form $4n+1$.

A Solution to number 1. Originally proposed by Euclid I think.

[hide] Assume that we have a finite number of prime numbers.

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

Consider the following number:

$P_{1}\times P_{2}\times P_{3}\times............ \times P_{n}+1$ This new number cannot possibly be a prime and so it must be composite. However, if it is composite then one of the prime numbers must be able to divide it but we see that none can. A contradiction and thus there exists an infinite number of primes.[/hide]

[hide="3"] This is similar to the proof of 1. above, but a little more complex. We basically assume there are a finite number of primes of the form $6n+5$, then construct a number which cannot have any prime factors.
Let all the primes of the form $6n+5$ be $p_{1}, p_{2}, ..., p_{m}$. Now construct the number:
$N = 6(p_{1}p_{2}... p_{m})-1 = 6(p_{1}p_{2}... p_{m}-1)+5$
We can see from the first representation that none of the p's can divide N (because $\bmod p_{anything}$, N is always equivalent to $-1$). Now we shall show that no prime of the form $6n+1$ can divide N.
The product of any two numbers $6a+1$ and $6b+1$ is
$(6a+1)(6b+1) = 36ab+6a+6b+1 = 6(6ab+a+b)+1$,
also expressible as $6n+1$. But N is expressible as $6n+5$ (from the second representation) and so cannot be the product of just $6n+1$'s.
Remember now that all primes can be expressed as either $6n+1$ or $6n+5$, and any integer $>1$ is expressible as the product of prime(s). We showed above that N cannot be have only $6n+1$ numbers as factors, but cannot have any $6n+5$ numbers as factors. This is a contradiction. Hence, our initial assumption that there are a finite number of primes of the form $6n+5$ is false. So there must be an infinite number of them.
This same reasoning can be extended to question 2, and in fact can be used to prove that there are an infinite number of primes of the form $bn-1$ if $\varphi(b) = 2$.
Btw t0rajir0u, this problem was an exercise in the book "What Is Mathematics?". Is that where you got it from?
[/hide]

[quote="boostage"]This same reasoning can be extended to question 2, and in fact can be used to prove that there are an infinite number of primes of the form $bn-1$ if $\varphi(b) = 2$. [/quote]
Interesting. Could you link me to a proof (Google is rather nasty with $\varphi$ and the Search function on ML doesn't seem to work...) or post a proof here?  :)

EDIT: Hmm, I think I got it  :P  Didn't really pay attention to the minus in $bn-1$. Thanks anyway for pointing this nice fact out!

Question 4
[hide]Assume that we have a list of primes $p_{1}, p_{2}, p_{3}, ..., p_{r}$ all of which are congruent to $1$ modulo $4$.

Consider the number
	$A = (2p_{1}p_{2}p_{3}...p_{r})^{2}+1$

Now $A$ can be factored into primes as:
	$A = q_{1}q_{2}q_{3}...q_{s}$

It is obvious that $q_{1}, q_{2}, q_{3}, ..., q_{s}$ are not in original list, since none of the $p_{i}$'s divide $A$. So if we can show that atleast $q_{i}$'s is congruent to $1$ modulo $4$, we are done.

It should be noted that $A$ is odd, thus all $q_{i}$'s are odd.
Also, each $q_{i}$ divides $A$, so
	$(2p_{1}p_{2}p_{3}...p_{r})^{2}+1 = A \equiv 0 (mod q_{i})$

This implies that $x = 2p_{1}p_{2}...p_{r}$ os a solution to congruence
	$x^{2}\equiv-1 (mod q_{i})$

so, $-1$ is a quadratic residue modulo $q_{i}$. From Quadratic Reciprocity we know that
	$q_{i}\equiv 1 (mod 4)$

QED.[/hide]

[quote="boostage"]Btw t0rajir0u, this problem was an exercise in the book "What Is Mathematics?". Is that where you got it from?
[/quote]

Actually, it showed up on a problem set at PROMYS.  Nice solution :)

[url=http://www-users.cs.york.ac.uk/~susan/cyc/p/primeprf.htm]one is well known[/url]

The  most "wise guy" proof for 1, I have seen ..

Obviously the probablity that two numbers are coprime to eachother =
 $\frac 6{\pi^{2}}= (1-1/2^{2})(1-1/3^{2})(1-1/5^{2}) \dots (1-1/p_{n}^{2}) \dots$
Obviously if number of primes are finite, then the RHS is rational, but we know $\pi^{2}$ is irrational. QED. :)

[quote="Gyan"]Obviously[/quote]

 :rotfl:

I understand that using Dirichlet L-series it is easy to apply a similar process to proving Dirichlet's Theorem?   :P


Let H be the set of all positive integers that leave remainder $1$ on division by $4$. Say that a number in H, other than $1$, is H-prime if it cannot be expressed as the product of two smaller numbers belonging to H. Thus $5$ and $9$ are the first two H-primes.

Find four H-primes $p_{1}, p_{2}, p_{3}, p_{4}$, no two equal, such that $p_{1}p_{2}= p_{3}p_{4}$.

[hide="Method I assume i_like_pie used"]
Consider the product of two primes that are $\equiv 3\pmod 4$.  $3^{2}\equiv 1\pmod 4$, but this number cannot be written as a product of two smaller numbers $\equiv 1\pmod 4$.  So just find four primes $\equiv 3\pmod 4$.  i_like_pie used $3,7,11,11$.
[/hide]


rotation
geometry
geometric transformation
reflection

Starting with 4 colors, red, white, blue, and green, hwo many ways can the corners(vertices) of an equilateral triangle be colored?  Note a vertex is dimensionless so that it can't actually be "colored".  it is more proper to say we are assigning colors to the vertices.  Assume you can look at the triangle from front or back so that some back views are equivalent to some front views, and visa versa.  Similarly assume rotated views are equivalent.  We say, in other words that reflections and rotations are allowed.

this question is so long that it's confusing, is it talking about 2D tri or 3d?

Since an equilateral triangle is given, I would have to say that is 2D.

I don't know if the triangle being equilateral would have any bearing on the solution.

The total number of ways you can assign a color to each corner would be: $_4P_3$ or $24$ ways

[quote="deej21"]Actually it would be $4^3=64$ ways, since each of the 3 corners can be one of 4 colors.[/quote]

Yeah, $4^3$ is the right one.

this post is for deej21 and jose:

Are you including cases where a color can appear more than once on the triangle?

I only considered the cases where you had three different colors.

If you are then it WOULD be $4^3$.

[quote="breez"]But rotations and flips are allowed.[/quote]
Yeah, you guys are overcounting. :P

Sorry, the triangle didn´t appear as I wished.

But suppose A, B and C are the vertex. Then, look my post before and see.

One of my posts didn't appear.

So, here´s my explanation:

[hide="See here."]Suppose A, B and C are the three vertex. Then, you can colour A with 4 colours, B with 4 colours, and C with the same 4. Then, The result is $4*4*4=4^3$.[/hide]

The equation is $\binom{4}{3}$, as order doesn't matter. This gives an answer of $24.$


The area of the shaded region is 71 square meters. What is the number of square meters in the area of the larger of the two inner squares given that all dimensions are positive integers?

[asy]path A = (0,0)--(12,0)--(12,8)--(0,8)--cycle;
path B= (1,7)--(5,7)--(5,3)--(1,3)--cycle;
path C = (10,2)--(10,5)--(7,5)--(7,2)--cycle;
fill(A,grey); fill(B,white); fill(C,white);
label("12m",(0,0)--(12,0),S); label("8m",(12,0)--(12,8),E);[/asy]

The area of the entire thing without holes is 96. Thus the holes have area of 96-71=25. In other words we want to find positive a,b such that a^2+b^2=25. This happens as a 3-4-5 triangle => larger hole has area of [b]16[/b].


Let $p$ be prime number and $n$ be non negative integers.

(1) Let $m$ be integer such that $0\leq m\leq n.$ How many numbers are there among integers 1 through $p^{n+1}$ which can be divided by $p^{m}$ not but by $p^{m+1}?$

(2) For two integers $x,\ y$ 1 through $p^{n+1},$ how many pairs of $(x, y)$ such that the product $xy$ can be divided by $p^{n+1}?$

[quote="kunny"]Let $ p$ be prime number and $ n$ be non negative integers.

(1) Let $ m$ be integer such that $ 0\leq m\leq n.$ How many numbers are there among integers 1 through $ p^{n \plus{} 1}$ which can be divided by $ p^{m}$ not but by $ p^{m \plus{} 1}?$

(2) For two integers $ x,\ y$ 1 through $ p^{n \plus{} 1},$ how many pairs of $ (x, y)$ such that the product $ xy$ can be divided by $ p^{n \plus{} 1}?$[/quote]
1) call x is a integer $ x\in[1,p^{n\plus{}1}]$ such that $ p^m\parallel{}x$
Then $ x\equal{}k.p^m  \Longrightarrow k\in [1,p^{n\minus{}m\plus{}1}]$  and $ (k,p)\equal{}1$
There are exactly $ \varphi (p^{n\plus{}m\minus{}1})$ k  satisfy condition .
So the number is $ \varphi (p^{n\minus{}m\plus{}1})\equal{}p^{n\minus{}m}(p\minus{}1)$
2)Let $ v_p(x)\equal{}k \Longrightarrow v_p (y)\geq n\plus{}1\minus{}k,\forall k\equal{}0,..,n\plus{}1$
The number of $ y\in [1,p^{n\plus{}1}]$ such that $ v_{p}(y)\geq n\plus{}1\minus{}k$ is $ p^k$
The number of $ x\in [1,p^{n\plus{}1}]$ such that $ v_p (x)\equal{}k$ is $ \varphi (p^{n\minus{}k\plus{}1})$
So the number  of $ (x,y)$ satisfy is
$ |S|\equal{}\sum_{k\equal{}0}^{n\plus{}1}p^k.\varphi (p^{n\plus{}1\minus{}k})$


A 48-foot-wide building has a roof that rises 4 feet for each 12-foot horizontal change. If the roof has a 2-foot overhang, what is the distance from the peak, P, to the edge of the overhang, O?

[asy]pair A,B,C,D,E;
A = (0,0); B = (3,1); C = (6,0);
D = (3*A+B)/4; E = (3*C+B)/4;
draw(A--B--C);
draw(D--E);
label("$O$",A,S);
label("$P$",B,N);
label("2 ft", A--D, NW);
label("48 ft", D--E ,S);[/asy]

24 feet from the edge to the peak, which is 2 horizontal changes of 12 feet.  Then it rises by 8 feet total.  So now we have a right triangle with sides 12, 8, and $ \sqrt{144\plus{}64}\equal{}\sqrt{210}$.  So the total length is $ \sqrt{210}\plus{}2$ feet.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Does there exist a subgroup of a product of groups such that it is not a product of the subgroups?

In other words, if you take $ G_i$ as groups (maybe not necessarily countable?), is it true that every subgroup $ H\leq \times G_i$ is of the form $ H\equal{}\times H_i$ where each $ H_i\leq G_i$?

No, consider the subgroup $ \{(0,0),(1,1)\}$ of $ (\mathbb Z/2 \mathbb Z)^2$.
Or do you want it to be isomorphic to such a product¿

Yeah, up to isomorphism. I would consider that $ \mathbb{Z}_2\leq\mathbb{Z}_2\times\mathbb{Z}_2$ which can be done by taking $ \mathbb{Z}_2\times \{1\}$.

It's easy to give infinite counterexamples like $ \mathbb Z$ as subgroup of $ \prod \mathbb Z / n \mathbb Z$.
Im pretty sure that there are finite counterexamples as well.

Hmm...this will sound really dumb probably. Why is $ \mathbb{Z}$ a subgroup of that product? It can't be viewed as "the last factor", or else it could be written as $ \prod \{1\}\times\mathbb{Z}$ which is a product of subgroups.

We embed the element $ m$ as the tuple $ \{m\mod n\mathbb{Z}\}$. These are different elements for each $ m$, so it works.


Hi,

There is a square ABCD, with 3 parallel lines, p1, p2 and p3 passing through vertexs A, B and C respectively. The minimum distance between p1 and p2 is 6cm and the minimum distance between p2 and p3 is 7cm. Calculate the area of this square...

tlhoo

ummm wouldn't that just make the diagonal of the square [hide]6cm + 7cm = 13cm?[/hide]

[hide]let P be intersection of AC and p2.  let U,V on p2 such that CU, AV perpendicular to p2.

let side length of square be x

then AP:PC=6:7

AP+PC=sqrt(2)x

so AP=6sqrt(2)x/13

PC=7sqrt(2)x/13

Let BP=y

then x^2(PC+AP)=(sqrt(2)x)(y^2+84x^2/169)

x^2=y^2+84x^2/169

85x^2/169=y^2

then y*13=x^2

hence y^2=x^4/169

x^4=85x^2

x^2=85[/hide]

Use spoiler.

Nukular -- that's only true if the lines are in one particular orientation (parallel to a diagonal), in which case the distances between the two pairs of lines would have to be the same.

[quote="JBL"]Use spoiler.[/quote]

forgot... thx

let p2 be the x-axis and let one vertex of the square be the origin.
alpha is the angle between the x-axis and the square side a going to p1
beta is the angle between the x-axis and the square side a going to p3

so we have

  sin(alpha) = 6/a ; sin(beta) = 7/a

now consider the equation

  alpha + beta = 90

taking the cosine on both sides using the addition law and trigonometric pythagoras gives

  sqr(1-(6/a)^2) * sqr(1-(7/a)^2) - 42/a^2 =0

  <==> sqr(a^2-36) * sqr(a^2-49) = 42

  <==> (a^2-36) * (a^2-49) = 42^2

  <==> a^4 - 85a^2 = 0

  ==> a = sqr(85)

  ==> Area = 85.

Draw 2 lines perpendicular to p1 and p3 that passes through vertices b and d.  these lines along with the 2 parallel lines passing through a and c form a larger square, each with side 13.  Subtract out the area of the four small triangles (169-84) and get 85.

[hide]Draw the perpendiculars from A and C to p2. The resulting triangles (altitude-side-partofp2) are congruent and right. The legs of the triangles are therefore 6 and 7, so the hypotenuse is sqrt(85). That's the side of the square, so the area is 85.[/hide]


Find the best value $ k$ to this ineq is true for all $ a,b,c\in R$

$ k\left\vert\ (a\plus{}b\plus{}c) [a^2b^2(a\minus{}b)\plus{}b^2c^2(b\minus{}c)\plus{}c^2a^2(c\minus{}a)]\right\vert\ \le (a^3\plus{}b^3\plus{}c^3)^2$

[quote="Allnames"]Find the best value $ k$ to this ineq is true for all $ a,b,c\in R$

$ k\left\vert\ (a \plus{} b \plus{} c) [a^2b^2(a \minus{} b) \plus{} b^2c^2(b \minus{} c) \plus{} c^2a^2(c \minus{} a)]\right\vert\ \le (a^3 \plus{} b^3 \plus{} c^3)^2$[/quote]
Let $ c\equal{}0$, then the inequality becomes
$ k \left\vert\ a^2b^2(a\plus{}b)(a\minus{}b)\right\vert\ \le (a^3\plus{}b^3)^2$
$ \Leftrightarrow k \left\vert\ a^2b^2(a\minus{}b)\right\vert\ \le \left\vert\ (a\plus{}b)\right\vert\ (a^2\minus{}ab\plus{}b^2)^2$
Now, let $ a\equal{}\minus{}1,b\equal{}1$, then we get $ k \le 0$.
Moreover, it is clear that the inequality holds for the case $ k\equal{}0$.
Thus, we get the conclusion
$ k_{\max} \equal{}0.$
We have done.
:)


Ring Theory
algebra
polynomial
superior algebra
superior algebra unsolved

let F be a field that is a subring of some larger field K. Let t be any fixed element in K, and suppose that t is a root of some nontrivial polynomial with coefficients in the smaller field F. Prove that the smallest such polynomial (i.e., the one of lowest degree) must be irreducible in F[X].

suppose that $ t\in K$ is a root of $ p(x)\in F[x]$, where $ p(x)$ is of minimal degree among such polynomials. assume that $ p(x)$ is reducible in $ F[x]$. then we can write $ p(x) \equal{} u(x)v(x)$, where $ u(x),v(x)\in F[x]$ and each has lower degree than $ p(x)$. letting $ x \equal{} t$ we have $ 0 \equal{} p(t) \equal{} u(t)v(t)$, which means, as $ K$ is a field, that one of $ u(t),v(t)$ must equal $ 0$. but then $ t$ is a root of a polynomial of smaller degree than $ p$--contradiction.


AMC
AIME
email
USAMTS
blogs
analytic geometry
symmetry

[quote="mathemonster"]when can we start talking about the problems btw? deadline was mail by today at 12 pm.[/quote]Don't discuss the problems until we say it's OK.  This will likely not be until after the solutions are posted later this week.

[quote="mathemonster"]why not tommorow? then it will be too late for someone who sees it to alter their solns[/quote]Because we maintain a 1-2 day grace period for technical issues (like we discover that an uploaded file got corrupted or something), or on a limited case-by-case basis we might allow a student to submit a day late with a really compelling excuse (usually involving the words "emergency room").

By the way, the AMC does the same sort of thing: they usually ask us to hold discuss for a few days after the AMC or AIME so that they can clear up technical issues before releasing the solutions.

[quote="not_trig"]Is it due at midnight (for internet submission)?[/quote]Yes, midnight Pacific time.  If you wait until the last possible moment, though, and it doesn't work because of some fault with your computer, we will not be sympathetic.  (In other words, this is not a "compelling excuse" as I described it in the last post.)

[quote="timwu"]Uh...I forgot to send the Entry form and my fax machine doesn't work in long distance...will it be ok to e-mail the scanned copy or send it tomorrow?[/quote]Do not email an Entry form.  Just mail it as soon as possible.

[quote="DPatrick"]Don't discuss the problems until we say it's OK.  This will likely not be until after the solutions are posted later this week.[/quote]
 :)

[quote="DPatrick"][quote="BOGTRO"]Don't know if this should go here but when will the solutions be graded?[/quote]We are hoping to release Round 1 scores and comments sometime during the week of Nov 3-7 if all goes well.[/quote]

when you say graded, do you mean that you grade them and give us our scores on the MY USAMTS page, or you post them on the official results page, or do they both happen at the same time?

graded=scores on MY USAMTS

What official results page do you mean?  Top scorers page?  That should go up at the same time

Everything (your individual scores, your comments, and the leaderboard and statistics) will go up at the same time.

[quote="cognos599"][quote="not_trig"]#1 - brute force
#2 - write them as $ (n \minus{} 1)/(blah)$, then bound and use SFFT
#3 - triangle inequality gogo
#4 - umm law of cosines twice?
#5 - induction + factoring stuff (except I decided to not remember to prove it was integral)

this was not a hard round, except I waited too long to do it ><
I think #1 was unfortunately the hardest problem - to USAMTS people: please less brute force problems?[/quote]

wait a minute. you had to used SFFT on #2?[/quote]

kind of, I ended up with something like 2(a+b) > ab

Number one...yes brute force

Unfortunately I was one off..

I got #1 correct (it was very boring  :mad: , why must there be brute force?), but I want to know if there is a non-counting (nice) proof of it. I tried to make a rotation equation, but that went nowhere at all.

Well, using Burnside's Lemma really reduces the amount of actual brute force required. :wink:

For #4, here's an outline of a coordinates bash that isn't [i]that[/i] bad.

Set the side length to be $ 2s$, then place the tetrahedron at $ A\equal{}(0,s,0)$, $ B\equal{}(0,\minus{}s,0)$, $ C\equal{}(s\sqrt{3},0,0)$, and $ D\equal{}(\frac{s\sqrt{3}}{3}, 0, \frac{2s\sqrt{6}}{3})$
(I think this placement is probably one of the most "optimal", but I could be wrong)

Then, let point P be at $ (x, y, z)$. We can immediately deduce that $ y\equal{}0$ from $ PA\equal{}PB$, and solving for $ x$, $ z$, and $ s$ turns out not to be too tedious (in the end, you get a quartic that's like the one given in the official solutions).

[quote="pythag011"][quote="andersonw"]For #4, here's an outline of a coordinates bash that isn't [i]that[/i] bad.

Set the side length to be $ 2s$, then place the tetrahedron at $ A \equal{} (0,s,0)$, $ B \equal{} (0, \minus{} s,0)$, $ C \equal{} (s\sqrt {3},0,0)$, and $ D \equal{} (\frac {s\sqrt {3}}{3}, 0, \frac {2s\sqrt {6}}{3})$
(I think this placement is probably one of the most "optimal", but I could be wrong)

Then, let point P be at $ (x, y, z)$. We can immediately deduce that $ y \equal{} 0$ from $ PA \equal{} PB$, and solving for $ x$, $ z$, and $ s$ turns out not to be too tedious (in the end, you get a quartic that's like the one given in the official solutions).[/quote]

Hmm...there is a method which I believe uses less work (also, you end up with a quadratic, not a quartic.)

OK, so take a homethety so the side length is 1. then, use your placement. Look at the set of points ( a plane) for which PA = PB. Also look at the set of points (another plane) for which PC = PD. Consider their intersection (a line). We know that (PA/PC)^2 = 11/17, so we can use distance formula and multiply by PC^2 to get a quadratic.

The advantage of this method is that we get a quadratic, not a quartic.l[/quote]

Well, the quartic you get is $ ax^4\plus{}bx^2\plus{}c$ for some a, b, c, so it's pretty much a quadratic. Sorry, I should have been more clear about that.

Your way is also nice, though.  :lol:

@FantasyLover: I actually tried those coordinates (or something very similar) first, the algebra was too messy for me.

[quote="Yongyi781"]How did the graders let you get by without proving that the given coordinates actually give a tetrahedron?

I inscribed it in a cube, so its coordinates were
(0, 0, 0)
(a, a, 0)
(a, 0, a)
(0, a, a).[/quote]

That's very nice, I hadn't thought of that.

For the first question: I already proved that the coordinates gave a regular tetrahedron in my solution.

Yes, I was actually surprised (and pleased) to see that I had NO computational errors on #4 despite my last-minute-ness.

Let's see,
#1 - no time to list the cases
#2 - made a linguistic error (or not, depending on the circumstances)
#3 - I accidentally used the bounds $ \minus{}100<x<100$ instead of $ \minus{}1<x<1$ which confused my grader and unfortunately earned me a 2
#4 - perfect! :lol:
#5 - Forgot to prove that $ 4|(x^{r}\plus{}x^{\minus{}r})^2$ implies that $ x^r$ is groovy... had no time, it was 11:50 PM.

On number 4, I oriented the tetrahedron so that the point was on the $ z$-axis, $ AB$ was parallel to the $ x$-axis, and $ CD$ was parallel to the $ y$-axis - this vastly simplified the coordinate-bashing.

I put my vertices at $ (x,0,0) \implies (\minus{}x,0,0)$ then used the fact it was equalateral to get the coordinates of D and C. Unfortunately I put the wrong coordinate for one point slighlty of into my solver so I got $ 5.81$


inequalities
geometry
circumcircle
inequalities proposed

Consider the triangle $ ABC$ with side-lenghts equal to $ a,b,c$. Let $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$, $ R$-the radius of circumcircle of the triangle $ ABC$, $ r$-the radius of the incircle of the triangle $ ABC$ and let $ l_a,l_b,l_c$ be the lenghts of bisectors drawn from $ A,B$ and $ C$, respectively, in the triangle $ ABC$. Prove that:
$ l_al_b\plus{}l_bl_c\plus{}l_cl_a\leq p\sqrt{3r^2\plus{}12Rr}$

[i]Proposer[/i]: [b]Baltag Valeriu[/b]

[quote="DreamTeam"]Consider the triangle $ ABC$ with side-lenghts equal to $ a,b,c$. Let $ p \equal{} \frac {a \plus{} b \plus{} c}{2}$, $ R$-the radius of circumcircle of the triangle $ ABC$, $ r$-the radius of the incircle of the triangle $ ABC$ and let $ l_a,l_b,l_c$ be the lenghts of bisectors drawn from $ A,B$ and $ C$, respectively, in the triangle $ ABC$. Prove that:
$ l_al_b \plus{} l_bl_c \plus{} l_cl_a\leq p\sqrt {3r^2 \plus{} 12Rr}$

[i]Proposer[/i]: [b]Baltag Valeriu[/b][/quote]
$ l_a \le \sqrt{p(p\minus{}a)},l_b \le \sqrt{p(p\minus{}b)},l_c \le \sqrt{p(p\minus{}c)}$;

$ l_al_b \plus{} l_bl_c \plus{} l_cl_a \equal{} p(\sqrt{(p\minus{}a)(p\minus{}b)}\plus{}\sqrt{(p\minus{}b)(p\minus{}c)}\plus{}\sqrt{(p\minus{}c)(p\minus{}a)}) \le p\sqrt{3((p\minus{}a)(p\minus{}b)\plus{}(p\minus{}b)(p\minus{}c)\plus{}(p\minus{}c)(p\minus{}a))}\equal{} \sqrt{3(ab\plus{}bc\plus{}ca\minus{}p^2)}\equal{}p\sqrt {3r^2 \plus{} 12Rr}$; :)


For a,b,c,d,e,f real positive numbers define

A=(a+b+c)
B=(d+e+f)
C=(ad+be+cf)
D=(a^2+b^2+c^2)
E=(d^2+e^2+f^2)

Prove that

  2ABC+3DE \geq 3CC+EAA+BBD

All you need is...courage:
   Believe me or not, the inequality reduces to (ae-af+cd-ce+bf-bd)^2>=0.

Excellent perceived by intuition, Harazi.
Now I would like to check if it is valid for n variables.


i attachment equation from second degree in this topic and i need some help to get values of X, Y, and a