analytic geometry
algebra
polynomial
limit
ARML
ratio
linear algebra

If $P_1 = (2, 0)$, $P_2 = (0, 4)$, and $P_3 = (0, 0)$ and the points follow the recurrence(s) $P_{2n-1} = \frac{4}{5} P_{2n-4} + \frac{1}{5} P_{2n-3}$ and $P_{2n} = \frac{1}{5} P_{2n-3} + \frac{4}{5} P_{2n-2}$. Find $P_{\infty}$.

P.S. You could've split it up into the sequences ${x_i}$ and ${y_i}$ if you wanted, but I chose not to.

P.S.S. I don't really know what level this is at, but I couldn't do it and it's really frustrating. :oops:. And as an added bonus :lol: I will reveal what inspired it and so doing open up that question to be solved in an easier way, thus making me look like a fool :D.

Edit: I edited the points because $P_1$ wass just a starter but didn't actually make the recursion work. Oops.

You can solve this without any "advanced" methods (linear algebra) by noting that this is an
averaging operation, therefore affine (sum of coefficients = 1, or sum of coefficients = 0 when all terms are brought to one side of the equation), so that one can get a simpler linear transformation on the differences $Q_i$ = $P_i - P_{i-1}$.    If it helps, you can look at two steps of the operation as a single transformation (maybe on just the even or just the odd $P_i$).

This kind of problem can be done reasonably systematically.

Let's focus on the $x$-coordinates, so $x_1 = 2$, $x_2 = 0$, $x_3 = 0$, and
\begin{eqnarray}
x_{2n - 1} &=& \frac{1}{5} x_{2n - 3} + \frac{4}{5} x_{2n - 4}, \\
x_{2n} &=& \frac{4}{5} x_{2n - 2} + \frac{1}{5} x_{2n - 3}.
\end{eqnarray}

From (2), we get
\[x_{2n - 3} = 5x_{2n} - 4x_{2n - 2}. \quad (3)\]
Shifting, we also get
\[x_{2n - 1} = 5x_{2n + 2} - 4x_{2n}. \quad (4)\]

Substituting (3) and (4) into (1), we get
\[5x_{2n + 2} - 4x_{2n} = \frac{1}{5} (5x_{2n} - 4x_{2n - 2}) + \frac{4}{5} x_{2n - 4},\]
or
\[25x_{2n + 2} = 25x_{2n} - 4x_{2n - 2} + 4x_{2n - 4}.\]

Now, let $a_k = x_{2k}$, so
\[25a_k = 25a_{k - 1} - 4a_{k - 2} + 4a_{k - 3}.\]
The characteristic polynomial of this recurrence is $25t^3 - 25t^2 + 4t - 4 = (t - 1)(25t^2 + 4)$, so the roots are 1 and $\pm 2/(5i)$.  We also calculate that
\[a_1 = x_2 = 0, \quad a_2 = x_4 = \frac{2}{5}, \quad a_3 = \frac{8}{25}.\]

Solving, we get
\[a_n = \frac{8}{29} + \left( \frac{-8 - 9i}{58} \right) \left( \frac{2}{5} i \right)^n + \left( \frac{-8 + 9i}{58} \right) \left( -\frac{2}{5} i \right)^n.\]

Now, it is evident that
\[\lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} a_n = \frac{8}{29},\]
and the same calculations can be done for the $y$-coordinates.

Thanks a bunch jhaussmann5. This post has two purposes. Explain a little bit of what he said better for anyone who wants to see and then stating the original problem.
[quote="jhaussmann5"]....Now, let $a_k = x_{2k}$, so
\[25a_k = 25a_{k - 1} - 4a_{k - 2} + 4a_{k - 3}.\]
The characteristic polynomial of this recurrence is $25t^3 - 25t^2 + 4t - 4 = (t - 1)(25t^2 + 4)$, so the roots are 1 and $\pm 2/(5i)$...[/quote]

Taking $25a_k = 25a_{k - 1} - 4a_{k - 2} + 4a_{k - 3}$ and bringing everything over, we get $25a_k - 25a_{k - 1} + 4a_{k - 2} - 4a_{k - 3} = 0$. It is well-known and provable enough**(maybe not by me, though I can do it in a case-by-case fashion at this point) that this recursion, since it is defined by the three terms before it is of the form $a_k = px^k + qy^k + rz^k$. For any recursion such as this where we have $\alpha a_k + \beta a_{k - 1} + \gamma a_{k - 2} + \epsilon a_{k - 3} = 0$, we know that $x,y,z$ are the solutions of the polynomial $\alpha\omega^3 + \beta\omega^2 + \gamma\omega + \epsilon = 0$. So, for our given recursion, jhaussman did that and found that $x = 1$, $y = -2/(5i) = \frac{2}{5}i$, and $z = 2/(5i) = -\frac{2}{5}i$.

[quote="jhaussmann5"]...We also calculate that
\[a_1 = x_2 = 0, \quad a_2 = x_4 = \frac{2}{5}, \quad a_3 = \frac{8}{25}.\]

Solving, we get
\[a_n = \frac{8}{29} + \left(\frac{-8 - 9i}{58} \right) \left( \frac{2}{5} i \right)^n + \left( \frac{-8 + 9i}{58} \right) \left( -\frac{2}{5} i \right)^n.\]...[/quote]

This solving jhaussmann5 speaks of is this system:

$0 = a_1 = px + qy + rz = p + q\left(\frac{2}{5}i\right) + r\left(-\frac{2}{5}i\right)$
$\frac{2}{5} = a_2 = px^2 + qy^2 + rz^2 = p - q\left(\frac{4}{25}\right) - r\left(\frac{4}{25}\right)$
$\frac{8}{25} = a_3 = px^3 + qy^3 + rz^3 = p - q\left(\frac{8}{125}i\right) + r\left(\frac{8}{125}i\right)$

And, actually, the recurrence he gives:
$a_n = \frac{8}{29} + \left( \frac{-8 - 9i}{58} \right) \left( \frac{2}{5} i \right)^n + \left( \frac{-8 + 9i}{58} \right) \left( -\frac{2}{5} i \right)^n$

Gives $a_0 = 0$, $a_1 = \frac{2}{5}$, and $a_2 = \frac{8}{25}$, but it still satisfies the same relation.

[quote="jhaussmann5"]...Now, it is evident that
\[\lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} a_n = \frac{8}{29},\]...[/quote]

$\lim_{n \rightarrow \infty} a_n = \frac{8}{29} \Rightarrow \lim_{n \rightarrow \infty} \left( \frac{-8 - 9i}{58} \right) \left( \frac{2}{5} i \right)^n + \left( \frac{-8 + 9i}{58} \right) \left( -\frac{2}{5} i \right)^n = 0$ because $\left(\frac{2}{5}\right)^n$ approaches $0$ as $n$ approaches $\infty$.

[quote="jhaussmann5"]...and the same calculations can be done for the $y$-coordinates...[/quote]
We will do the same for the y-coordinates. We have $y_1 = 0, y_2 = 4$, and $y_3 = 0$. By the given recurrences:

$y_4 = \frac{1}{5} y_{1} + \frac{4}{5} y_{2} = 0 + \frac{4\cdot 4}{5} = \frac{16}{5}$

$y_6 = \frac{1}{5} y_{3} + \frac{4}{5} y_{4} = 0 + \frac{4 \cdot 16}{5 \cdot 5} = \frac{64}{25}$

So, if we say $y_{2k} = b_k$, $b_i$ satisfies the recurrence, and these tell us $b_1 = y_2 = 4, b_2 = y_4 = \frac{16}{5}$, and $b_3 = y_6 = \frac{64}{25}$.

So, we now have the system:
$4 = b_1 = px + qy + rz = p + q\left(\frac{2}{5}i\right) + r\left(-\frac{2}{5}i\right)$
$\frac{16}{5} = b_2 = px^2 + qy^2 + rz^2 = p - q\left(\frac{4}{25}\right) - r\left(\frac{4}{25}\right)$
$\frac{64}{25} = b_3 = px^3 + qy^3 + rz^3 = p - q\left(\frac{8}{125}i\right) + r\left(\frac{8}{125}i\right)$

Which has solutions ${p,q,r} = {\frac{80}{29},\frac{-40-45i}{29},\frac{-40+45i}{29}}$.

Thus, $\lim_{n \rightarrow \infty} y_n = \lim_{n \rightarrow \infty} b_n = \frac{80}{29}$.

And finally, $\lim_{n \rightarrow \infty} P_n = \left(\frac{8}{29},\frac{80}{29}\right)$.

Now, for the real problem. Before I start I add that I think there is an error somewhere because when I did it before and cheated with a calculator I got $\lim_{n \rightarrow \infty} P_n = \left(\frac{8}{29},\frac{20}{29}\right)$.

So, here goes, and I credit this problem to the Chicago ARML coaches, a list can be found [url=http://www.chicagoarml.org]here.[/url]

While up late at night in Iowa City, making sure her team members were getting enough sleep to compete, an ARML coach noticed a spider on the ceiling behaving quite strangely. She watched the spider start at one corner of the 2' x 4' ceiling tiles, walk the shorter side of the tile (call this segment 1), walk diagonally across the middle of the tile (segment 2), then return to its starting point along the side of the tile (segment 3). Then things became even stranger: the spider went across the tile to the closest point on segment 2 (forming segment 4), and continued in this fashion, forming segment $n$ by moving to the closest point on segment $n-2$. If the spider continued this behavior "indefinitely," the coach mused, where would the spider end up in relation to the original corner?

I said $P_0 = (0,0)$ was the origin and segment $k$ connects point $P_k$ and $P_{k-1}$, so, from this, I got the givens from above -- $P_1 = (2,0), P_2 = (0,4)$ and $P_3 = (0,0)$ -- by orienting the tile in the corner of Quadrant I and setting the shorter side on the x-axis. After this, we notice the spider goes perpendicular and divides the given side into a 4:1 ratio and the rest is trivial to get to my recurrence.

My challenge: Find $P_{\infty}$ in a geometric way. (If not possible, a different way.)

**From the post above, the double asterisk refers to this question:

If anyone would like to offer a proof of this fact:

Given the recurrence $t_k$:

$c_k t_k + c_{k-1} t_{k-1} + c_{k-2} t_{k-2} + \cdots + c_{k-(n-1)} t_{k-(n-1)} + c_{k-n} t_{k-n} = 0$.

Where the $t_i$ are the terms of the recurrence, the $c_i$ are the coefficients, $n$ is the number of terms needed to find the next one in succession, and $k$ is any arbitrarily chosen number.

(1)Prove that the closed form of $t_k$ can be written as:

$a_1(b_1)^k + a_2(b_2)^k + a_3(b_3)^k + \cdots + a_{n-1}(b_{n-1})^k + a_n(b_n)^k$

(2) And that $b_1,b_2,\ldots$ are the solutions of the polynomial:

$c_k x^n + c_{k-1} x^{n-1} + c_{n-2} x^{k-2} + \cdots + c_{k-(n-1)} x^{n-(n-1)} + c_{k-n} x^{n-n} = 0$

please do so :lol:.
[hide="Hint"]Make a geometric series.[/hide]

[quote="Pork_Chop8"]...Which has solutions ${p,q,r} = {\frac{80}{29},\frac{-40-45i}{29},\frac{-40+45i}{29}}$.

Thus, $\lim_{n \rightarrow \infty} y_n = \lim_{n \rightarrow \infty} b_n = \frac{80}{29}$.

And finally, $\lim_{n \rightarrow \infty} P_n = \left(\frac{8}{29},\frac{80}{29}\right)$.

Now, for the real problem. Before I start I add that I think there is an error somewhere because when I did it before and cheated with a calculator I got $\lim_{n \rightarrow \infty} P_n = \left(\frac{8}{29},\frac{20}{29}\right)$...[/quote]

There was a discrepancy between the real problem and my recursion. My original recursion was for $P_1 = (0,0), P_2 = (2,0), P_3 = (0,4)$ and $P_4 = (0,0)$. But, $P_1$ didn't contribute to the recurrence the way it should, so I trashed it, but in doing that I forgot to change the indices because all odds became evens and all evens became odds so it screwed up my problem.
So, instead, I re-did it and had:

$x_1 = 0, x_2 = 2, x_3 = 0, x_4 = 0, x_6 = 0$

$y_1 = 0, y_2 = 0, y_3 = 4, y_4 = 0, y_6 = \frac{4}{5}$

So, I get:

$x_{2k} = a_k = p(1)^k + q(\frac{2}{5}i)^k + r(-\frac{2}{5}i)^k$

Plugging in $k = 1,2,3$ we get a system and solving we get $\left(p,q,r\right) = \left(\frac{8}{29},\frac{50-125i}{58},\frac{50+125i}{58}\right)$

and

$y_{2k} = b_k = l(1)^k + m(\frac{2}{5}i)^k + n(-\frac{2}{5}i)^k$

Plugging in $k = 1,2,3$ we get a system and solving we get $\left(l,m,n\right) = \left(\frac{20}{29},\frac{125+50i}{58},\frac{125-50i}{58}\right)$

So, finally, $\lim_{n \rightarrow \infty} P_n = \left(\frac{8}{29},\frac{20}{29}\right)$, as desired.

P.S. sorry for the infinity of posts in my own thread, but I've been pondering this problem since May 10th or so, so you can imagine my great joy and enthusiasm for it's solving.

P.S.S. I would really appreciate it if some great geometer [b](AHEM YETTI)[/b] could find a different way to do this and maybe even explain what significance the point $\left(\frac{8}{29},\frac{20}{29}\right)$ holds for the triangle with vertices $(0,0), (2,0)$ and $(0,4)$.

If your recursion or something like it (i.e., linear) is correct, then there is a linear transformation $L$ of the plane that sends triangle $P_{n-1} P_{n} P_{n+1}$ to $P_{n} P_{n+1} P_{n+2}$ and you are looking for its unique fixed point.  In particular, we know the coordinates of $P_{\infty}$ are rational if the first few $P_i$ are, because it is found by solving a system of linear equations with rational coefficients (namely $L(v) = v$).  There should be a geometric construction as well:  given a linear map that sends triangle $T_1$ to triangle $T_2$, construct its fixed point. I think it can be done with a ruler only.

fleeting guest's idea of the fixed point is a good one, but one has to be careful about which triangles one chooses.

Note that triangles $P_1 P_2 P_3$ and $P_8 P_6 P_7$ are homothetic (just draw a diagram).  Then this fixed point lies on $P_1 P_8$ and $P_2 P_6$.  We have that
\[P_1 = (0,0), \quad P_2 = (2,0), \quad P_6 = \left( 0, \frac{4}{5} \right), \quad P_8 = \left( \frac{8}{25}, \frac{4}{5} \right).\]

Then the line $P_1 P_8$ is given by
\[y = \frac{5}{2} x,\]
and the line $P_2 P_6$ is given by
\[y = -\frac{2}{5} x + \frac{4}{5}.\]
Solving, we get
\[(x,y) = \left( \frac{8}{29}, \frac{20}{29} \right).\]

Once you exhibit a linear transformation that takes triangle $n$ to triangle $n+C$, for some constant $C$ and arbitrary $n$, it will be contracting, and one can solve for the (unique) fixed point.

By "care is needed", do you mean that $C = 1$ is impossible (that may be true, I didn't check)?  I'm fairly sure that $C = 2$ is possible, and $C=4$ is almost trivial, because:

if the segments traversed are numbered 1,2,... as in the problem statement, then
[i]segment $n$ is parallel to segment $n+4$[/i], so that triangles $i$ and $i+4$ are homothetic (their sides are parallel).
Drawing lines between the corresponding vertices locates the center of similarity.

But a slightly more complicated construction (with ruler only, I think) would find the fixed point for triangles related by a more complicated (and in this case, angle-preserving) linear transformation.


calculus
integration
complex analysis
complex analysis unsolved

Use the definition to compute the following integral:

$ \int_C e^z$ where $ C$ is the contour from $ 1\minus{}i$ to $ 1\plus{}i$.

I do not see how to do this problem. I know the definition is $ \int_C f(z)dz \equal{} \int_C f(z(t))z'(t)dt$. However, I don't see how to use it in this problem. Thanks.

Yes, that is the right formula.  But in order to use it, you need a parametrization.  Fortunately, this is a really easy path to parametrize.  The path connecting $ z_{1}\equal{}1\minus{}i$ and $ z_{2}\equal{}1\plus{}i$ is just a straight line.  The equation given two points is $ z(t)\equal{}z_{1}\plus{}t(z_{2}\minus{}z_{1})\equal{}1\minus{}i\plus{}2ti$, so just use this.


trigonometry
inequalities
geometry
inradius
circumcircle
inequalities proposed

Let$ABC$is triangle.
Prove that:$\frac{(cosA+cosB+cosC-1)^2}{2}\ge cosAcosBcosC$.

From the wellknown identity $\cos A+\cos B+\cos C=1+\frac rR$

results that the proposed inequality is equivalently with the wellknown inequality

$\overline {\underline {\left| \ \cos A\cos B\cos C\le \frac 12\left(\frac rR\right)^2\ \right| }}\le \frac 18.$

oh yeah! Prove that:
$cosAcosBcosC\le \frac{r^2}{2R^2}$.

[b]You ask Zhaobin[/b]. [i]You don't want, but you wish a nice solution[/i] ! From the Zhaobin's reply,

$P\equiv\prod \cos A\le \prod (1-\cos A)\Longleftrightarrow P\le 8\prod \sin ^2\frac A2\Longleftrightarrow$

$P\le 8\left[\frac{(s-a)(s-b)(s-c)}{abc}\right]^2\Longleftrightarrow P\le 8\left(\frac{sr^2}{4Rsr}\right)^2\Longleftrightarrow P\le \frac 12\left(\frac rR\right)^2.$

Thank.I have a stupid question. :blush:  :blush:

No [i]Conga[/i], you have not a stupid problem ! Here is a solution of the remarkable inequality:
\[ \overline {\underline {\left| \ \prod \cos A\le \frac 12 \left(\frac rR\right)^2\ \right| }}  \] I suppose w.l.o.g. that the triangle $ABC$ is acute. Prove easily that

$(a^2+b^2-c^2)(a^2+c^2-b^2)\le (a+b-c)^2(a+c-b)^2\Longleftrightarrow (b-c)^2(b^2+c^2-a^2)\ge 0$.

Therefore, the following inequality is true: $\prod (b^2+c^2-a^2)\le \prod (b+c-a)^2\ \ (1).$

The inequality $(1)$ is equivalently with the inequality:

$a^2b^2c^2\cos A\cos B\cos C\le8(s-a)^2(s-b)^2(s-c)^2\Longleftrightarrow$

$P\equiv \cos A\cos B\cos C\le \frac{8(s-a)^2(s-b)^2(s-c)^2}{(4Rsr)^2}\Longleftrightarrow P\le \frac 12\left(\frac rR\right)^2.$

[u]Remarks.[/u] I note $2s=a+b+c$, the inradius $r$, the circumradius $R$

and I used the relations $S=sr=\sqrt{s(s-a)(s-b)(s-c)}$, $4RS=abc=4Rsr$,

$(s-a)(s-b)(s-c)=sr^2$, $b^2+c^2-a^2=2bc\cos A$.

The inequality
\[ \cos A\cos B\cos C\leq\frac{r^2}{2R^2}  \]
was proposed in a Vietnamese problem solving journal this December. For more problems from Vietnam, please visit

math4u.de

Pham Van Thuan

And, this is one problem :For$a,b,c>0;r>1$.
Prove that: $P=\sum (a^2+\frac{1}{ab})^r\ge 3.2^r$.
Hee.It's easy. :rotfl:  :D

I think your solution is difficult.I can't think that:
$(a^2+b^2-c^2)(a^2+c^2-b^2)\le (a+b-c)^2(a+c-b)^2$.

[quote="zhaobin"]I think it is equal to:
$\left(1-\cos A \right) \left(1-\cos B \right) \left(1-\cos C \right) \ge \cos A \cos B \cos C$
which is well-known.you  can search. :)[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=518862&p=3396140#p3396140


The sum of three different numbers is 67. The two larger numbers differ by 7 and the two smaller numbers differ by 3. What is the value of the largest number?


Let k be a real constant and a,b be fixed complex numbers. Describe the locus of points z that satisfy:
$ arg(\frac{z\minus{}a}{z\minus{}b})\equal{}k$
I tried very hard to do this but still can't get any good results. Can someone help me on it? I think it has a nice way to do than simply just bash everything out.

[quote="Altheman"]The equation essentially says that the angle formed by a,z,b is constant...[/quote]
But it asks to describe the locus of points z? So I tried to find the equation that can relate a,z,b to k but failed, that is to transform $ \frac {z \minus{} a}{z \minus{} b}$ into $ x \plus{} yi$. Now I have just read the book further and it says the formula that I think can help me solve this one: $ tan(argz)$=$ \frac {y}{x}$ when $ z \equal{} x \plus{} yi$. Do you think it works Altheman? If not can you show me how to do it? That idea I mentioned above must be the final things I haven't thought of when solving this problem. If it won't work, then I really really need help!

You can do it by the algebraic method. But if a, b and k are specified its very easy to use geometry. You can see that a and b are the end-points of a chord of a circle (obviuosly passing thru a and b). We now have to find its centre and radius

If you will read this post http://www.goiit.com/posts/list/algebra-complex-number-97942.htm#442807 it may be useful


inequalities
LaTeX
inequalities unsolved

[quote="Lee Sang Hoon"]a,b,c>=0. prove that 1/(4a^2+bc)^(1/2) + 1/(4b^2+ca)^(1/2) + 1/(4c^2+ab)^(1/2)>=4/(a+b+c)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146809

$ a,b,c\geq0$ prove that $ \sqrt{\frac{1}{4a^2\plus{}bc}}\plus{}\sqrt{\frac{1}{4b^2\plus{}ca}}\plus{}\sqrt{\frac{1}{4c^2\plus{}ab}}\geq\frac{4}{a\plus{}b\plus{}c}$

I translated because I want to train my latex :D !


http://www.artofproblemsolving.com/Wiki/index.php/Main_Page

Firefox 3, Mathlinks theme (but aops theme on wiki), Windows XP.

Oops forgot to declare some global vars. Fixed now :D


induction
number theory proposed
number theory

Prove that for every positive integer $ n$ there exists a positive integer with the following properties:
(i) It has exactly $ n$ digits.
(ii) None of the digits is equal to zero.
(ii) It is divisable by the sum of its digits.

Here's an approach that should work for all $ n$ large enough, say $ n \ge 11$.

Fact: For every positive integer $ k$, there is a unique $ k$-digit positive integer whose digits are all 4's and 5's that is divisible by $ 2^k$.  (Proof: Induction on $ k$.)

Let $ n$ be a positive integer.  Let $ k$ be a positive integer such that $ 3n \le 2^k < 6n$.  We will choose our positive integer to have a digit sum of $ 2^k$.  Its rightmost $ k$ digits will be the number given by the fact above.  The first $ n \minus{} k$ digits can have any digit sum between $ n \minus{} k$ and $ 9(n \minus{}k)$.  So we have plenty of slack to make the entire digit sum equal to $ 2^k$.


Four consecutive positive integers have a product of 840.
What is the largest of the four integers?

We could solve a fourth degree equation, but its much simpler to realize that our four integers are all small, and that there's a factor of $ 5$ and $ 7$ in $ 840$. From this, it is easily found that $ 4(5)(6)(7)\equal{}840$, so the answer is $ \boxed{7}$.

hello, ok if we solve the equation $ n(n\plus{}1)(n\plus{}2)(n\plus{}3)\equal{}840$ in positive integers we get the solution $ n\equal{}4$.
Sonnhard.

[quote="gaussintraining"]We could solve a fourth degree equation[/quote]
Come to think of it, this has a nice solution (just somewhat long for FTW).
Let our first integer be $ n$. Then, $ n(n\plus{}1)(n\plus{}2)(n\plus{}3)\equal{}840$.
Rearranging,

$ (n)(n\plus{}3)(n\plus{}1)(n\plus{}2)\equal{}840$

$ \implies (n^2\plus{}3n)(n^2\plus{}3n\plus{}2)\equal{}840$.

Letting $ y\equal{}n^2\plus{}3n$, we have $ y(y\plus{}2)\equal{}840 \implies (y\plus{}1)^2\equal{}841$.

Clearly, $ y$ must be positive, so $ y\plus{}1\equal{}29 \implies n^2\plus{}3n\plus{}1\equal{}29 \implies n^2\plus{}3n\minus{}28$.

This factors as $ (n\plus{}7)(n\minus{}4)$, so $ n\equal{}4$, and $ n\plus{}3\equal{}7$.

Yeah, but you can't really do that in 45 seconds and expect to be faster than your opponent...

Simple guesswork is all that is required.

Make a guess that the integers are 3, 4, 5, 6, and 3 * 4 * 5 * 6 = 360, which is too low. But you know that 360 = 3 *120 and 840 = 7 * 120, so swapping the 3 with a 7 solves the problem.

or just know that 7!=5040, so you could spare a 6 then gg.. so 4*5*6*7


Minor problem:

When I try to click "Download PDF Version" in the AIME section of the Resources page, it points me to the forum index (http://www.artofproblemsolving.com/Forum/) instead.

Is this intentional? :maybe:

I think that's because there aren't pdf's to all of the problems.

No. Its some PDF generator which is disabled because no one can fix it :(


How would you do this?

Find the eccentricity of the ellipse with the equation 9x^2 +3y^2 - 36x - 6y + 12 = 0

And how do you find the foci of an ellipse?

Ugh...it's been a while since I've done conic sections:
[hide="Solution"]We can rearrange the original equation to get $\frac{(x-2)^{2}}{3}+\frac{(y-1)^{2}}{9}=1$.
So the eccentricity is $\sqrt{1-\frac{3}{9}}=\frac{\sqrt{6}}{3}$
Well, the foci of this ellipse are $\sqrt{6}\text{ (which is }\sqrt{9-3})$ from the center along the major axis, so they're $(2, 1+\sqrt{6}), (2, 1-\sqrt{6})$[/hide]
I think I did it right. Feel free to correct me.


let $ p(x)$ be a polynome with integer coefficients and wish satisfy $ p(p(...p(x)...))\equal{}x$ where $ p(x)$ occure $ n$-times(for n greater than 1).
prove that $ p(p(x))\equal{}x$.

Obviously the polynom will be linear.
Let's say $ ax\plus{}b$
then we know that
$ a^{n}\equal{}1$ and $ b\sum_{i\equal{}1}^{n\minus{}1}a^{i}\equal{}0$  so either $ a\equal{}1$ and $ b\equal{}0$ ( in this case $ P(x)\equal{}x$)
or $ a\equal{}\minus{}1$ n is even
in the last case $ P(P(x))\equal{}a^{2}x\plus{}b(1\plus{}a)\equal{}x$  :)

Maybe I'm wrong but I guess that he hasn't meant the polynomial equality but the normal equality for given integer $ x$? In this case, for example look here: http://www.mathlinks.ro/Forum/viewtopic.php?p=572821#p572821.

Yes maybe he also meant to ask that any integer fixed point of $ P_{k}$ is a fixed point of $ P_{2}$ but shouldn't he have mentioned that $ x$ is an integer at least :D ?


modular arithmetic
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I have searched through the forums and searched on the internet for a long time and i still do not understand what the term "mod" means.  I would appreciate it greatly if someone could explain it to me and what its uses are.  There are several solutions in this forum that i dont understand because of this "mod."  It is inhibiting my learning!

Thank you very much,
jeremy

[quote="xxreddevilzxx"]modulus means this:

if $a \equiv b(mod c)$
that means that when a is divided by c, it leaves a remainder of b.[/quote]

so that would mean

$2005 \equiv 1(mod 2004)$

so we dont know the product, only the dividend, divisor, and remainder

What is it used for?

so if the number on the left (a) is less than the "mod" (m) by 1 or 2, the remainder is -1 and -2, respectively?

[quote="xxreddevilzxx"]if $a \equiv b(mod c)$
that means that when a is divided by c, it leaves a remainder of b.[/quote]That's not quite right.  For example, $19 \equiv 9 \pmod 5$ is true, even though it doesn't fit your definition.

There is also a great article about mods, remainders, and congruences in the old magazine Quantum.

[quote="Jeremy"][quote="Rep123max"]

So here is an example of a problem:

What is the remainder when $2^{1000}$ is divided by 3?

To solve this, $2\equiv -1\pmod 3$, so $2^{1000}\equiv (-1)^{1000}\equiv 1\pmod 3$, so it leaves a remainder of 3.

[/quote]

Can someone prove this?  i understand things better if i see it in proof.

If you can't, that is fine.  I think i understand the basic concept of mods.  i just would like to see [i]why[/i] that statement is true.

Thanks a bunch
jeremy[/quote]

To answer the question "What is the remainder when $a^{n}$ is divided by m?" you can use this reasoning:

Let the remainder be "r" s.t. $a^{n}\equiv r\pmod m$  Letting:

$a\equiv b\pmod m$

we can use the property of multiplying mods (as described by Rep123max), to rewrite:

$a\cdot a\cdot a\cdot \ldots$ n times as congruent to $b\cdot b\cdot b\cdot \ldots \pmod m$ n times, which gives you:

$a^{n} \equiv b^{n} \pmod m \equiv r \pmod m$

and then just plug in a=2, n=1000, and m=3.  But I think Rep123max made a typo cause the remainder should be 1, not 3.  Does that make sense?

Mod stands for Modulus. It's like a 12 hour clock. When you look at the time you cast out the 12's and keep the remainder. So 13 o'clock is 1 o'clock. 17 o'clock is 5, and so on.

In Modular Arithmatic the modulus can be any number > 1.

On a '7-hour clock' you would only have the times 0 to 6 o'clock. The only difference is 7 o'clock would be 0 o'clock because 7 has no remainder, or remainder zero, when divided by 7.

Binary code is a kind of modular arithematic with a modulus of 2 so the only numbers that exist in binary are 0 and 1 because when any number is divided by 2 it leaves a remainder of 0 or 1.

[quote="Rep123max"]Hmmm. For the best information, consult AoPS Volume 1, some chapter that I can't remember. But anyways, $a\equiv b\pmod m$ if $\frac{a-b}{m}$. Another way to think of that is that if the previous equivalence is true, then the two numbers leave the same remainder when divided by what ever the mod base is.

You can add mods, like $2\pmod 5+4\pmod 5=6\pmod 5=1\pmod 5$, you can subtract mods like $2\pmod 5-4\pmod 5=-2\pmod 5=3\pmod 5$. You can multiply mods like $2\pmod 5\cdot 4\pmod 5=8\pmod 5=3\pmod 5$. BUT, you cannot always divided congruences! (Note, that addition, subtraction, and multiplication assumes that it is the same base.)

..[/quote]

Can you give the proof of the operations of mod?
I have only heard of mod before....

[quote="Rep123max"]

So here is an example of a problem:

What is the remainder when $2^{1000}$ is divided by 3?

To solve this, $2\equiv -1\pmod 3$, so $2^{1000}\equiv (-1)^{1000}\equiv 1\pmod 3$, so it leaves a remainder of 3.

Hopefully this helped give you an introduction to mods.[/quote]

As far as I know, mods cant be negative, so 2=2(mod3) because 2/3 is 0 R 2, isnt it?
And also, when anything is divided by 3 theres no way the remainder can be 3 because then you just add one to the quotient and the remainder is 0.

[quote="shobber"]
Can you give the proof of the operations of mod?
I have only heard of mod before....[/quote]

Here's my proof for the addition:
$a \equiv b \pmod m$  and   $c \equiv d \pmod m$
means
$m|(a-b)$ and $m|(c-d)$
so $m$ also divides $(a-b)+(c-d)$, that is $(a+c)-(b+d)$, so, by the definition of mod,
$a+c \equiv b+d \pmod m$

I think proofs for the others operations are quite similar to this (hoping this is correct!).

[quote="Jeremy"]I have searched through the forums and searched on the internet for a long time and i still do not understand what the term "mod" means.  I would appreciate it greatly if someone could explain it to me and what its uses are.  There are several solutions in this forum that i dont understand because of this "mod."  It is inhibiting my learning![/quote]There are several intuitive ways to think of what "mod" is that I know of from my limited exposure to modular arithmetic.

Suppose you divide a natural number $n$ by another natural number $m$ (natural numbers for simplicity) and you get $n=q\cdot{m}+r, 0\le{r}<n$. Then you can say $n\equiv{r}\pmod{m}$, or "$n$ is congruent to $r$, mod $m$." So you could say that any number that gives the remainder $r$ upon division by $n$ is congruent to $r$, mod $n$.

Another way to think about it is the definition that $a\equiv{b}\pmod{c}$ iff $c$ divides $a-b$, or $\frac{a-b}{c}$ is an integer (that is, two numbers are congruent in a modulus if that modulus divides their difference, or their difference is a multiple of the modulus). In the above example, $n=q\cdot{m}+r\Longrightarrow\frac{n-r}{m}=q$, so $\frac{n-r}{m}$ is an integer, so $n\equiv{r}\pmod{m}$. A special case would be when $r=0$, or $n=qm$. In this case $n\equiv{0}\pmod{m}$.

The last way I can think of is to look at the integers in a certain modulus as a ring of numbers (that's probably where the word ring in higher algebra comes from, but I don't even know what it means). The first row shows the natural numbers written in order (could you say this is in $\text{mod}\infty$ or is that just silly?), and then each row shows the natural numbers written in the next greatest modulus, starting with one.

1,2,3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,...
1: 0,0,0,0,0,0,0,0,0,...
2: 1,0,1,0,1,0,1,0,1,0,1,0,...
3: 1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,...
4: 1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0,...
5: 1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,...
.............................................................
n: 1,2,3,...,n-1,0,1,2,3,...,n-1,...

When you count in a certain modulus, you count from $1$ to $n-1$ as you normally would, but then $n$ becomes $0$, since $n$ divides $n$, and $n+1$ becomes $1$, and so on. It's easy to see if you write it out as an actual ring of numbers from, for example, $0$ to $5$ (for $\text{mod }6$).

This was meant to just give a basic intuitive understanding of what "mod" means. The arithmetic of mods, their applications to problem-solving, and the math that uses them will come later. I don't know much about these myself.

From what I remember about number theory, it's ok to have negative mods.   For n>0, $-n\equiv (m-n)\pmod m$

[quote="AntonioMainenti"]
The last way I can think of is to look at the integers in a certain modulus as a ring of numbers (that's probably where the word ring in higher algebra comes from, but I don't even know what it means). [/quote]

Actually, the integers $\bmod n$ do form a ring, but what is of particular interest is that the integers $\bmod p$, where $p$ is a prime, form a far more powerful construction, called a field.  The main thing separating a ring and a field, in this case, is that if $p$ is prime, for every $a\in\mathbb{Z}_p\setminus\{0\}$, there exists $b\in\mathbb{Z}_p\setminus\{0\}$ such that $ab=1\bmod p$.

Can you prove that this is true if and only if $p$ is prime?

This is quite easy to prove, but you have to know something about the Extended Euclidean algoritm first.
It says: gcd(a,b)=1 if and only if there exist $x,y \in N$ so that $xa + yb=1.$

Theorem:
For $a,n \in N$ and $n>0$ you have:
The class of a in the ring $Z/nZ$ only has a inverse if gcd(a,n)=1

Proof
(i) if $gcd(a,n)=1$ then $xa+yn=1$, thus $(xa + yn) Mod n=(xa+ y0) Mod n = xa Mod n = 1 Mod n$
     thus x is the inverse of a
(ii) if $xa = 1 mod n$ then $xa - 1 = yn \Rightarrow xa+ yn=1 \Rightarrow gcd(a,n)=1$

Obviously, only primes are relatively prime with every integer, so only in prime rings has every element an inverse.

[quote="Meester"]This is quite easy to prove, [/quote]

Well, its only interesting pedagogically if you don't know the proof already.

[quote="Rep123max"].

So here is an example of a problem:

What is the remainder when $2^{1000}$ is divided by 3?

To solve this, $2\equiv -1\pmod 3$, so $2^{1000}\equiv (-1)^{1000}\equiv 1\pmod 3$, so it leaves a remainder of 3.

Hopefully this helped give you an introduction to mods. [/quote]
you have made mistake (i'm pretty sure it's a tryp, because the reamainder is $1$ not $3$(because $1^{1000}=1\not=3$) :lol:

[quote="kops723"].As far as I know, mods cant be negative, so 2=2(mod3) because 2/3 is 0 R 2, isnt it? [/quote]
yes mods [b]can[/b] be negative

where can I find a lot of practice problems with mods?


With so little information it's impossible to diagnose the problem. You don't say what system or editor you are using. Compile this minimal document and then show us what the log says:
[code]\documentclass{article}
\begin{document}
Hello World!
\end{document}[/code]

Make sure you close the output file on Adobe before writing a new one onto it. I've gotten the fatal error message before by forgetting to do that.

That shouldn't be necessary unless you are using version 8.0 of Adobe Reader which has a bug in it, but was corrected in later versions.

[quote="pytheagle"]My $ \text{\LaTeX}$ used to work fine, but now it's screwed up. I can't output anything.

I keep getting EMERGENCY STOP and FATAL ERROR OCCURED, NO OUTPUT PDF FILE PRODUCED.

Adobe Reader says it cannot find the document.[/quote]

Another possible problem is that you're not choosing the right application... It happened to me before......

I figured out why; I had a "%" in saved file's name. For some reason, it won't work if I have a percent symbol in its name.

If you have symbols like % or \$, the code will not work. Also, certain words will not work in the file name either. (I don't happen to recall which words, but there are a handful that are not allowed in the name of a tex document.)


I think that this one is stronger than problem S 63. from mathematical reflections.
Problem S 63. states that, if positive reals $ a$, $ b$ and $ c$ satisfy $ ab+bc+ca\geq 3$ then:

$ \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{3}{\sqrt{2}}$

[quote="delegat"]
Problem S 63. states that, if positive reals $ a$, $ b$ and $ c$ satisfy $ ab + bc + ca\geq 3$ then:

$ \frac {a}{\sqrt {a + b}} + \frac {b}{\sqrt {b + c}} + \frac {c}{\sqrt {c + a}}\geq\frac {3}{\sqrt{2}}$[/quote]
Your problem is very easy:
$ \sum_{cyc}\frac{a}{\sqrt{a+b}}\geq\frac{\sqrt a+\sqrt b+\sqrt c}{\sqrt2}\geq\frac{\sqrt[4]{16(ab+ac+bc)}}{\sqrt2}\geq\sqrt[4]{12}>\frac{3}{2}.$ :wink:

Can you detail please,Arqady:
$ \sum_{cyc}\frac{a}{\sqrt{a\plus{}b}}\equal{}\sum_{cyc} \frac{{\sqrt a}^2}{ \sqrt{a\plus{}b}} \geq \frac{ (\sqrt a\plus{}\sqrt b\plus{}\sqrt c)^2}{\sqrt{a\plus{}b}\plus{}\sqrt{b\plus{}c}\plus{}\sqrt{b\plus{}c}}$ from Engel form of Cauchy,
[b]but why this:[/b]
$ \frac{ (\sqrt a\plus{}\sqrt b\plus{}\sqrt c)^2}{\sqrt{a\plus{}b}\plus{}\sqrt{b\plus{}c}\plus{}\sqrt{b\plus{}c}}\geq \frac{\sqrt a\plus{}\sqrt b\plus{}\sqrt c}{\sqrt2}$
[b]and why :[/b]
$ \frac{\sqrt a\plus{}\sqrt b\plus{}\sqrt c}{\sqrt2}\geq\frac{\sqrt[4]{16(ab\plus{}ac\plus{}bc)}}{\sqrt2}$

Thanx !

[quote="soruz"]Can you detail please,Arqady:
$ \sum_{cyc}\frac {a}{\sqrt {a \plus{} b}} \equal{} \sum_{cyc} \frac {{\sqrt a}^2}{\sqrt {a \plus{} b}} \geq \frac { (\sqrt a \plus{} \sqrt b \plus{} \sqrt c)^2}{\sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {b \plus{} c}}$ from Engel form of Cauchy,
[b]but why this:[/b]
$ \frac { (\sqrt a \plus{} \sqrt b \plus{} \sqrt c)^2}{\sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {b \plus{} c}}\geq \frac {\sqrt a \plus{} \sqrt b \plus{} \sqrt c}{\sqrt2}$[/quote]

No, not so!
$ \sum_{cyc}\frac {a}{\sqrt {a \plus{} b}}\geq\frac {\sqrt a \plus{} \sqrt b \plus{} \sqrt c}{\sqrt2}.$
The proof see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56071
[quote="soruz"]
[b]and why :[/b]
$ \frac {\sqrt a \plus{} \sqrt b \plus{} \sqrt c}{\sqrt2}\geq\frac {\sqrt [4]{16(ab \plus{} ac \plus{} bc)}}{\sqrt2}$
[/quote]
Let $ a\equal{}x^2,$ $ b\equal{}y^2$ and $ c\equal{}z^2,$ where $ x,$ $ y$ and $ z$ are non-negative. Then
$ \frac {\sqrt a \plus{} \sqrt b \plus{} \sqrt c}{\sqrt2}\geq\frac {\sqrt [4]{16(ab \plus{} ac \plus{} bc)}}{\sqrt2}\Leftrightarrow(x\plus{}y\plus{}z)^4\geq16(x^2y^2\plus{}x^2z^2\plus{}y^2z^2)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^4\plus{}4x^3y\plus{}4x^3z\minus{}10x^2y^2\plus{}12x^2yz)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^4\minus{}x^3y\minus{}x^3z\plus{}x^2yz)\plus{}5\sum_{cyc}(x^3y\plus{}x^3z\minus{}2x^2y^2)\plus{}11\sum_{cyc}x^2yz\geq0.$ :wink:

[quote="arqady"]If $ x,y,z$ be positive reals, then
\[ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\geq\sqrt [4]{\frac {27(yz \plus{} zx \plus{} xy)}{4}},
\]
with equality if and only if $ x \equal{} y \equal{} z$.[/quote]
$ \sum_{cyc}x\sqrt [4]{\frac {4}{27(x \plus{} y)^2(yz \plus{} zx \plus{} xy)}}\geq \sum_{cyc}\frac {x(x \plus{} y \plus{} 2z)}{x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 3(yz \plus{} zx \plus{} xy)} \equal{} 1$.
[hide="Proof."]$ 4[x^2 \plus{} y^2 \plus{} z^2 \plus{} 3(yz \plus{} zx \plus{} xy)]^4 \minus{} 27(x \plus{} y)^2(yz \plus{} zx \plus{} xy)(x \plus{} y \plus{} 2z)^4$

$ \equal{} (x^2 \plus{} y^2 \minus{} 2z^2)^2[z^2 \plus{} 3z(x \plus{} y) \plus{} x^2 \plus{} 3xy \plus{} y^2][z^2 \plus{} 9z(x \plus{} y) \plus{} 4x^2 \plus{} 9xy \plus{} 4y^2]$

$ \plus{} 27(x \plus{} y)^2(x \plus{} y \plus{} 2z)^2(z^2 \minus{} xy)^2\geq0$.[/hide]

[url=http://forum.cnool.net/topic_show.jsp?id=3955823&oldpage=1&thesisid=494]Thanks for remembering me.[/url]

[quote="arqady"]If $ x,y,z$ be positive reals, then
\[ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\geq\sqrt [4]{\frac {27(yz \plus{} zx \plus{} xy)}{4}}
\]
[/quote]

I think I have solved it.

From Holder's Inequality we have

$ \left(\sum\frac {x}{\sqrt {x \plus{} y}}\right)\left(\sum\frac {x}{\sqrt {x \plus{} y}}\right)\left(\sum x(x \plus{} y)\right)\ge (x \plus{} y \plus{} z)^3$

$ \iff \sum\frac {x}{\sqrt {x \plus{} y}}\ge \sqrt {\frac {(x \plus{} y \plus{} z)^3}{x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx}}$.

So it remains to show that

$ \left(\frac {(x \plus{} y \plus{} z)^3}{x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx}\right)^2\ge \frac {27}{4}(yz \plus{} zx \plus{} xy)$

Let $ x \plus{} y \plus{} z \equal{} 1$ and $ xy \plus{} yz \plus{} zx \equal{} t$. Thus we need to prove that

$ \left(\frac {1}{1 \minus{} t}\right)^2\ge \frac {27t}{4}$

$ \iff t(1 \minus{} t)^2\le\frac {4}{27}$

$ \iff t^2(t \minus{} 2) \plus{} t\le\frac {4}{27}$

Which is obvious using $ t\le \frac 13$.

Hope I'm right.

[quote="nayel"]$ \left(\frac {1}{1 \minus{} t}\right)^2\ge \frac {27t}{4}$
Which is obvious using $ t\le \frac 13$.
Hope I'm right.[/quote]
$ 4 \minus{} 27t(1 \minus{} t)^2 \equal{} (4 \minus{} 3t)(1 \minus{} 3t)^2\ge 0$.

Thank you very much Ji Chen and nayel! :lol:
For $ x,y,z > 0$ and $ xy \plus{} xz \plus{} yz \equal{} 3$ the nayel's way gives that the following inequality holds:
\[ \frac {x}{\sqrt {4x \plus{} 5y}} \plus{} \frac {y}{\sqrt {4y \plus{} 5z}} \plus{} \frac {z}{\sqrt {4z \plus{} 5x}}\geq1
\]
But it doesn't help for the following inequality:
\[ \frac {x}{\sqrt {5x \plus{} 4y}} \plus{} \frac {y}{\sqrt {5y \plus{} 4z}} \plus{} \frac {z}{\sqrt {5z \plus{} 4x}}\geq1
\]

Nice solution, nayel. I am wondering if
\[ \frac x{\sqrt{x\plus{}y}}\plus{}\frac y{\sqrt{y\plus{}z}}\plus{}\frac z{\sqrt{z\plus{}x}}\leq3\sqrt[4]{\frac{x^2\plus{}y^2\plus{}z^2}{12}}.\]

Thank you everyone. :blush:

For Mr pvthuan's problem, I wonder if the following inequality holds:
\[ \left(\sum_{cyc}(x\plus{}y)\right)\left(\sum_{cyc}\frac{x}{x\plus{}y}\right)\le 3\sqrt{3(x^2\plus{}y^2\plus{}z^2)}\]
It's clearly true when $ x\le y\le z$, but I still can't prove it generally. :maybe:

No need for that anymore, I think I have another solution that only uses Cauchy-Schwarz inequality.

[quote="pvthuan"]I am wondering if
\[ \frac x{\sqrt {x + y}} + \frac y{\sqrt {y + z}} + \frac z{\sqrt {z + x}}\leq3\sqrt [4]{\frac {x^2 + y^2 + z^2}{12}}.
\]
[/quote]

We need to prove that
\[ \sum_{cyc}\frac {x}{\sqrt {x + y}}\le\sqrt [4]{\frac {27}{4}(x^2 + y^2 + z^2)}
\]
We have
\begin{eqnarray*}\sum_{cyc}\frac {x}{\sqrt {x + y}} & = & \sum_{cyc}\left(\sqrt [4]{\frac {x}{x + y}}\right)\left(\sqrt [4]{\frac {x^3}{x + y}}\right) \\
& \le & \sqrt {\left(\sum_{cyc}\sqrt {\frac {x}{x + y}}\right)\left(\sum_{cyc}\sqrt {\frac {x^3}{x + y}}\right)} \\
& \le & \sqrt {\frac {3}{\sqrt 2}\sum_{cyc}x\sqrt {\frac {x}{x + y}}}\end{eqnarray*}
Thus we need to show that
\[ \left(\frac {3}{\sqrt 2}\sum_{cyc}x\sqrt {\frac {x}{x + y}}\right)^2\le\frac {27}{4}(x^2 + y^2 + z^2)
\]
Or
\[ \left(\sum_{cyc}x\sqrt {\frac {x}{x + y}}\right)^2\le \frac 32(x^2 + y^2 + z^2)
\]
But we have
\begin{eqnarray*}\left(\sum_{cyc}x\sqrt {\frac {x}{x + y}}\right)^2 & = & \left(\frac {\sum x\sqrt x\sqrt {y + z}\sqrt {z + x}}{\sqrt {(x + y)(y + z)(z + x)}}\right)^2 \\
& = & \frac {\left(\sum(x\sqrt {y + z})(\sqrt x\sqrt {z + x})\right)^2}{(x + y)(y + z)(z + x)} \\
& \le & \frac {\left(\sum x^2(y + z)\right)\left(\sum x(z + x)\right)}{(x + y)(y + z)(z + x)}\end{eqnarray*}
Let $ p = x + y + z, q = xy + yz + zx, r = xyz$. Then it remains to show that
\begin{eqnarray*} & & \frac {(pq - 3r)(p^2 - q)}{pq - r}\le\frac 32(p^2 - 2q) \\
\ & \iff & 2(p^3q - pq^2 - 3p^2r + 3qr)\le 3(p^3q - p^2r - 2pq^2 + 2qr) \\
\ & \iff & p^3q + 3p^2r\ge 4pq^2 \\
\ & \iff & p^2q + 3pr\ge 4q^2 \\
\ & \iff & (x + y + z)^2(xy + yz + zx) + 3xyz(x + y + z)\ge 4(xy + yz + zx)^2 \\
\ & \iff & (x^2 + y^2 + z^2)(xy + yz + zx) + 2(xy + yz + zx)^2 + 3xyz(x + y + z)\ge 4(xy + yz + zx)^2 \\
\ & \iff & (x^2 + y^2 + z^2)(xy + yz + zx) + 3xyz(x + y + z)\ge 2(xy + yz + zx)^2 \\
\ & \iff & (x^2 + y^2 + z^2)(xy + yz + zx)\ge 2(x^2y^2 + y^2z^2 + z^2x^2) + xyz(x + y + z) \\
\ & \iff & \sum_{sym}x^3y\ge 2\sum_{sym} x^2y^2 \\
\ & \iff & \sum_{sym}xy(x - y)^2\ge 0\end{eqnarray*}
Which is obviously true.

This is a correct solution. Congratulations, nayel.

[quote="arqady"] For $ x,y,z > 0$ and $ xy \plus{} xz \plus{} yz \equal{} 3$, the following inequality holds:
\[ \frac {x}{\sqrt {5x \plus{} 4y}} \plus{} \frac {y}{\sqrt {5y \plus{} 4z}} \plus{} \frac {z}{\sqrt {5z \plus{} 4x}}\geq1\][/quote]
see:[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=75967]http://www.mathlinks.ro/Forum/viewtopic.php?t=75967[/url]


articles Support geometry calculus integration

Wikipedia is the source of all our knowledge, amen.

And since I'm an editor  you better speak nice about our beloved Wiki.

Now seriously, the fact that anyone can edit wikipedia I'd say in general makes it more reliable and not less, I certainly don't see why somebody is going to spend time editing some ornitology article if he doesn't know the subject. And there are the discussion boards, and with respect to scientfic articles I think most of them are quite well done, there are others were people have complained and have been corrected.
The fact you mention can be read in the information about wikipedia in the site, also it says that  wikipedia do falls behind in comprehenssion. The E. Britannica articles are larger. However Wikipedia beats all encyclopedias in the amount of articles which is kind'a obvious.

D.

[quote="dhernandez"]Wikipedia is the source of all our knowledge, amen.

And since I'm an editor  you better speak nice about our beloved Wiki.[/quote]

Facetious or not, that's not a great way to actually get someone to listen to you.

[quote]Now seriously, the fact that anyone can edit wikipedia I'd say in general makes it more reliable and not less, I certainly don't see why somebody is going to spend time editing some ornitology article if he doesn't know the subject.[/quote]

Um, because, it's fun to mess with people?  Did you not hear about the Steven Colbert incident where he got all of his fans to support his edit that was blatantly false?  Additionally, people can have something wrong (mistakenly) and put it into the article thinking it's correct with all good intentions.  It won't always get caught.

[quote]
The fact you mention can be read in the information about wikipedia in the site, also it says that  wikipedia do falls behind in comprehenssion. The E. Britannica articles are larger. However Wikipedia beats all encyclopedias in the amount of articles which is kind'a obvious.[/quote]

Traditional encycylopedias are still just second hand sources.  Depending on the scope of one's article one should go to the primary sources.

I suggest not citing wikipedia as a source.  Use it to get a general overview of the topic you're researching and then if it has links to other sources, go use them and cite them.  If the wikipedia article doesn't cite its sources then it's a bad article and shouldn't be cited.  There isn't any original work which is published on wikipedia.  They get their information from other sources (those sources which are often times secondary themselves).  It's best to stay as close to the primary source as reasonably possible.

[quote="joml88"]Facetious or not, that's not a great way to actually get someone to listen to you.[/quote]

can't see why are you going to stop listening at my first sentence which does not mean to be any argument and since the real stuff comes inmediately after. Facetious or not you took it way too seriously I'm afraid.

[quote="joml88"]Um, because, it's fun to mess with people?  Did you not hear about the Steven Colbert incident where he got all of his fans to support his edit that was blatantly false?[/quote]

Yeah, and have you heard of this to be the rule or the exception? I'm not saying you should trust wikipedia in anyway close to absolutely, well I did say something like this in my facetious first phrase, but let me put it in a much more harmless way. The advantages of wikipedia are overwhelming when compared to these deficiencies. Granted, articles specially about people or political issues can be wrong and you should be careful. But what I state is that the controversial articles are but a few, I strongly doubt your power to "mess with people" when editing that [i]ornitology[/i] article I told you about.

[quote="joml88"]Additionally, people can have something wrong (mistakenly) and put it into the article thinking it's correct with all good intentions.  It won't always get caught.[/quote]

It do gets caught at the moment someone with more knowledge of the matter sees your article. And since there seem to be a lot of people with the desire to edit, that is, the desire to visit articles on subjects they are familiar with I would say that it will get caught most times.

[quote="joml88"]I suggest not citing wikipedia as a source.  Use it to get a general overview of the topic you're researching and then if it has links to other sources, go use them and cite them.  If the wikipedia article doesn't cite its sources then it's a bad article and shouldn't be cited.  There isn't any original work which is published on wikipedia.
  They get their information from other sources (those sources which are often times secondary themselves).  It's best to stay as close to the primary source as reasonably possible.[/quote]

I'm not so sure about that. I can't prove it to you but I do have edited wikipedia a few times and well, what do you mean by original work? There aren't much things strictly original in scientific books either, most original stuff is published in magazines and then when it gets well known and understood and if it is sufficiently general someone puts it in his book. If the original source is recent you put references, if it's not you don't in general. You mean we should get always to the first articles and read Riemann's work if we want to know about differential geometry? I think not, I think knowledge is filtered through time and people goes on understanding what is vital and what is secondary.
In that sense, things in wikipedia clearly are not original but I think people many times do write their own articles, not copy from others. If I'm writing about the Eiffel Tower, what would be an original article? And wikipedia as a secondary source [i]should be[/i] excellent in principle since everyone can edit the references or ask for them as opposed to E. Britannica

I agree though that if you want detailed information on one subject you should go elsewhere. But in such a case you are not looking for an article of an encyclopedia. To conclude what II mean is that if you are not looking for specifics but say you heard about Peter Unknown in some Trivial Pursuit question and want to know something more, wikipedia is excellent. If you study the Unknown family and their rather irrelevant contributions to the Bermuda Triangle history don't use it

D

I must admit that Wikipedia contains lots of information, but I know that my teachers say that we can't use it. I recommend just asking.

I use wikipedia to look for facts, and then I find those facts in other sources and cite those.

Encyclopedias in general don't count as sources for us. However, wikipedia is still tremendously useful for uncited homeworks, such as history IDs.

I believe research has shown that Wikipedia's rate of errors is somewhat higher than Britannica's - 35%ish more I think.

But definitely do not use Wikipedia as a citation for any significant research papers. In fact, like joml said, if you're looking for very credible sources, it's best to go back to the primary ones. Wikipedia is reliable because there are so many people who monitor it, and it has features like automatic edit notifications to deter people who go in and make random edits to screw others over.

I'd recommend you use it as an overview, a jumping off point. And then from there, look for other sources to either confirm or refute the information you find.

[quote="MithsApprentice"]But definitely do not use Wikipedia as a citation for any significant research papers.[/quote]

I don't use it for things that really matter, especially when I am citing my sources.  It is a nice point to start, to get everything together, then work out from there.

As for the error rate and people destroying articles, it is rather obvious ususally - I was using an article in history today and there was a sentance that went the likes of "{insert name here} has recognised as being the hero of the {insert battle here}."  It was quite obviousally fake, but the school computers have been banned from editing (reoccuring vandals :rotfl:) so I couldn't fix it...  Then someone had fixed it my the time I had gotten home.  So that speaks well for it.

[quote="MithsApprentice"]But definitely do not use Wikipedia as a citation for any significant research papers.[/quote]
I agree. The only teachers I have really used Wikipedia for were really laid back and they didn't seem to care a whole lot. :wink:

It doesn't really matter what the error rate is. Maybe 23.9 hours out of 24 every article is absolutely perfect, but the problem is that you have no guarantee that that is the case at the particular moment you choose to visit, and this makes it unreliable.

Our teachers allow us to use any website. They dont care. :lol:

I'm not going to go back and address every one of your points.  I don't have that kind of time (sorry).  But I will say this:  I'm sure the people editing wikipedia are smart.  Perhaps some of them even qualify as experts in certain fields.  But the fact is that you cannot be certain that the article you are reading has been checked by an expert.  On the other hand, there are other resources that you can read and BE SURE they have been checked by articles.  For example, scientific journals consist of a bunch of articles submitted by scientists.  The good journals have the articles peer reviewed where, confidentially, the paper is sent to a bunch of qualified professionals in that particular field to make sure that the article is legitimate and suggest revisions that should be made before the article should be published.

[quote="joml88"]The good journals have the articles peer reviewed where, confidentially, the paper is sent to a bunch of qualified professionals in that particular field to make sure that the article is legitimate and suggest revisions that should be made before the article should be published.[/quote]
I'm pretty sure there was an incident not too long ago (a few years maybe) where some guy sent an essay to a scholarly journal which was an elaborate joke, but the journal published it. It may have been some kind of philosophical thing, like about postmodernism maybe.
Anyway the point is that nothing is infallible. Judging by the fields in which I have sufficient knowledge to make judgments about the accuracy, I think Wikipedia is reliable enough that it is useful.

Again an important point is that different subject areas have different probabilities of being (a) edited by a well-meaning non-expert who makes mistakes and (b) intentionally vandalized. Politically-charged and pop culture/famous person - related articles are most vulnerable to (b), while highly technical material is extremely unlikely to be subject to it. Also, my (unsupported) belief is that people in category (a) are much more likely to harm intermediate-level articles, that is, ones that are widely known, but still only by people with decent education. Nobody's going to accidentally put wrong stuff in the page about the Radon-Nikodym Theorem or something, but someone might put wrong or misleading stuff in the "integral" page just because he or she doesn't know the theory of integration in great detail (not realizing that there is such a thing and thinking it's just an elaborate application of algebra, or whatever).

[quote="pieterminate"]Found this interesting thing on some article...
[quote]"The primary trade-off is not really 'speed for accuracy' as some might suggest, but rather 'speed for guarantee,'" says Friedman. "If the information absolutely must not be wrong for each and every person who views it, then collaborative content is not going to work very well."[/quote][/quote]

That's basically one of my main points.  If you have the choice between an article written by people doing it for free in their free time and a book/article/etc written by an expert or experts in the field who are being paid and are reviewed by the peers you should probably choose the latter.  It's a matter of good scholarship.  If you've ever played that game in elementary school where the first person whispers to the person behind her a sentence and then he whispers what he heard to the person behind him and so on down the line who do you think is the best person to ask what the first student originally said.  That original student, right?  Not the 3rd or 4th student.

At the very least, if a teacher is going to allow students to cite wikipedia she should make sure the class knows the importance of citing sources that are both quality and close to the primary sources.  I place the importance of making students cite their work in middle/high school mostly in that it will potentially prepare them for what they might be doing in the future.

Once again, I will say that I think wikipedia is an awesome resource.  It ranks 2nd behind google on my list of useful internet resources.

There are a ton of journals out there, so you still have to be careful even when you're citing a primary source.  But still, if you cite a primary source, and that article is wrong you can blame them ( :D ) wheres blaming wikipedia for falsely reporting something that you could have looked up and checked yourself still leaves you with some of the blame.

[quote="byrdgyrl"]Do your teachers allow you to use Wikipedia as a source on school research papers, projects, etc?  Why or why not?

I have read somewhere that it has the same error rate as Britannica, but it still disturbs some teachers that theoretically anyone can edit it.

Personally, my teachers are fine with me sourcing it just as long as it is not my primary source - common knowledge tells us that you should not rely on a single source ever, no matter what it is.  Just wanted to hear your opinions.[/quote]

my history teacher doesn't because of inaccurate info. Some cubing thing in there I think is messed, so yeah there is something wrong. Even though it is not allowed, I still use it sometimes if I can't find any other good sources

My teachers could care less about what we use, as long as we get the right information.

Wickipedia can be messed with very easily, so I would HIGHLY RECCOMEND NOT USING WIKIPEDIA AS A SOURCE!! Looking up stuff on Wikipedia is fine, but be very careful about the content.

Yeah wikipedia can easily be wrong! The very principle that makes it so useful allows for this snag - its open-sourcedness.

[quote="dhernandez"][quote="Koda"]Yeah wikipedia can easily be wrong! [/quote]

Statistics don't say so

D[/quote]

He's saying that it's easier for wrong information to be 'published' in wikipedia than it is for wrong information to be published in a renonwned journal or something of that sort.

And what statistics are you talking about?  Where are they?  Who compiled them?

Alot of people abuse wikipedia. For example, last year, I was writing an essay on Susan B. Anthony. I found the wikipedia article, and it said she was black in more than one place. Also, someone apprarently edited it and added at the bottom 'SUSAN B ANTHONY WAS A MOTHERB---ING B---- WHO -----S LOTS OF ------.'

I no longer trust it.

wikipedia is better than you think, 13375P34K43V312.

It's also productive when you need basic or simple details, or don't know and can't find a tiny thing.

So don't attack it like a pig.

Wikipedia is the Number 1 source I go to for research but I hate it how normal people can just screw around in it.

For example, a person in my class wanted to mod the American Revolution part to say "IT NEVER HAPPENED" in big bold letters but Wikipedia blocked all edits from our school IP address  :rotfl:  :rotfl: .

[quote="byrdgyrl"]I have read somewhere that it has the same error rate as Britannica, but it still disturbs some teachers that theoretically anyone can edit it.[/quote]

That study was published in Nature and quoted by CNN and other news outlets.

My Physics teacher is dogmatically opposed to Wikipedia as a source for the research experiment we have to do...

I usually use Wikipedia unless it's been obviously vandalized (like "Education in the United States" was a few days ago) but I check with other sources just to be sure.


1. A book weighing 3N is resting on a table 1 meter above the floor. How much force is the table exerting on the book?

2. How much work is the table doing on the book?

1. -3N since the book is in equilibrium the down force (3N) must equal the the normal (up) force but in opposite direction.

2. No work is being done- no displacement = no work W= force * displacement


linear algebra
matrix
SFFT
special factorizations

Let $ A,B \in M_n(\mathbb{C})$ s.t. $ 2A(B\minus{}A)\equal{}A\plus{}B$. Prove that $ AB\equal{}BA$.

$ 2AB\minus{}2A^2\equal{}A\plus{}B$ =>$ (2A\minus{}I_{n})(\minus{}A\plus{}B\minus{}I_{n})\equal{}I_{n}$ =>
$ (\minus{}A\plus{}B\minus{}I_{n})(2A\minus{}I_{n})\equal{}I_{n}$  => $ 2BA\minus{}2A^2\equal{}A\plus{}B$ => $ AB\equal{}BA$

There is another problem which is similar:

Let A, B be $ n\times n$ matrices satisfying A + B = AB. Prove that A and B commute.

Simon's favorite factoring trick?

$ AB\minus{}A\minus{}B\equal{}0$

$ AB\minus{}A\minus{}B\plus{}I\equal{}I$

$ (A\minus{}I)(B\minus{}I)\equal{}I$

But a matrix commutes with its inverse, so

$ (B\minus{}I)(A\minus{}I)\equal{}I$ which leads to

$ BA\equal{}A\plus{}B.$

Since $ A\plus{}B$ also equals $ AB,$ we have $ AB\equal{}BA.$

This proof should work in any ring with identity.

[quote="Kent Merryfield"]This proof should work in any ring with identity.[/quote]
no. ab=1 does not imply ba=1 in general, but in matrix algebras it's true.


algebra function domain AMC USA(J)MO USAMO logarithms

I was doing this question for a review section on logs.

     Write $ln (x^2 - 9)$

     Looks simple right? Its just $ln (x+3) + ln (x-3)$

     But the problem is that the book im using doesn't state that the domain of the function is changed when you do this expansion of the logs. The book doesn't include this in the answer and my teacher didn't include it in his solution of the problem. Should you just state the new domain of the expanded function (I think so), don't state it, or not expand it?

I'm not sure about what convention you use, but I just thought of something interesting. You could also write $x^2=|x^2|=|x|^2$ so $\ln(x^2-9)=\ln(|x|^2-9)=\ln(|x|+3)+\ln(|x|-3)$. Then you wouldn't need to change the domain.

well...it depends on the setting of where you are...in school, they most likely will not bother with those domain things, however, if you are say taking the usamo, you definitely want to pay attention to that restriction, but it is just easier to always put the restriction anyway because your teacher cannot say that you are wrong

$ab>0\Longrightarrow \ln ab=\ln |a|+\ln |b|\ , \ln\frac ab=\ln |a|-\ln |b|\ .$
$a\ne 0\ ,\ n\in N^*\ ,\ 2|n\Longrightarrow \ln a^n=n\cdot \ln |a|\ .$

[b]Example.[/b] $|x|>3\Longleftrightarrow x\in (-\infty ,-3)\cup (3,\infty )\Longrightarrow\boxed {\ \ln (x^2-9)=\ln |x-3|+\ln |x+3|\ }\ .$
$(\ln |x|)'=\frac 1x$ for any $x\ne 0\ \ ;\ \ \int \frac 1x\ dx=\ln |x|+ C$ for any interval $I\subset R^*\ .$

Use the absolute value signs. For example:

$\ln((-4)^2 - 9) = \ln 7 \neq \ln(-4-3) + \ln (-4+3) = \ln(-7)+\ln(-1)$

So you really should use the absolute value signs rather than changing the domain. $\ln{(n^2-9)}$ is defined for all $n\neq\pm3$ but $\ln(n+3)+\ln(n-3)$ is not.

Note: this only considers the logarithms in the real number system.

[quote="Virgil Nicula"][quote="Hamster1800"]$\ln{(n^2-9)}$ is defined for all $n\neq\pm3$[/quote]It is falsely. $\ln (n^2-9)$ (over $R$) is defined for any $|n|>3$, i.e. $n<-3$ or $n>3\ .$[/quote]

Yes yes...I deviated from using $\mathbb{R}$ only for a bit.


geometry parallelogram trigonometry circumcircle

Draw two intersecting circles $\omega_{1}$ and $\omega_{2}$, with the radius of $\omega_{1}$ smaller than that of $\omega_{1}$, and label their centers $O_{1}$ and $O_{2}$ respectively. Let their common chord be $BC$, and pick a point $A$ outside of both circles such that $\triangle ABC$ is acute and $AO_{2}< AO_{1}$. Draw points $D=\overline{AC}\cap \omega_{1}$,  $\ E =\overline{AB}\cap \omega_{1}$,  $\ M =\overline{AC}\cap \omega_{2}$,  and $\ N =\overline{AB}\cap \omega_{2}$. A diagram is attached.

Find a method of determining the distance $O_{1}O_{2}$ with the [u]least[/u] possible number of straight line measurements on triangle $ABC$.

the circles are defined with 4 lengths...i ono if there is a better way that does not define the diagram, but the length is fixed. EB, BC, CE, NE

Construct the parallelogram $O_{1}O_{2}MF$.

Then it's easy to prove that $FM=FD$
[hide="Indeed"]If $\overrightarrow{P_{1}P_{2}}$ is the projection of $\overrightarrow{O_{1}O_{2}}$ onto $MC$ then $\overrightarrow{P_{1}P_{2}}=\overrightarrow{CP_{2}}-\overrightarrow{CP_{1}}= \frac{\overrightarrow{CM}-\overrightarrow{CD}}{2}=\frac{\overrightarrow{DM}}{2}$

But $\overrightarrow{FM}=\overrightarrow{O_{1}O_{2}}\Rightarrow$

The projection of $\overrightarrow{FM}$ onto $CM$ equals $\frac{\overrightarrow{DM}}{2}$

$\Rightarrow FM=FD$
$\hline$[/hide]

So $O_{1}O_{2}= FM=FD.$

Let $H$ be the symmetric point of $M$ w.r.t. $F$.

Then from the right triangle $MHD$ we have $MD=2MF\cdot\sin(\angle MHD)$

But $\angle MHD = \angle ACB$ (the two angles have perpendicular sides)

Hence $MD=2MF\sin C = 2O_{1}O_{2}\sin C \Rightarrow \boxed{O_{1}O_{2}= \frac{MD}{2\sin C}}\ \ (1)$

One segment that we should measure is the segment $MD$

[u]Can we bring lines on[/u] $ABC$ [u]or we should only measure the distances?[/u]

[color=blue][b]1) [/b]If we are allowed to bring the altitude from $A$ in $\triangle ABC$ then we need 2 lengths for the calculation of $\sin C$.
[u][b]3 total lengths[/b][/u].[/color]

[color=brown][b]2) [/b]If we are not allowed to bring lines on $\triangle ABC$ then we will calculate $\sin C$ from the formula $\sin C = \frac{AB}{2R}$,
where $R$ is the circumradius of $ABC$.

Then $(1)$ becomes $O_{1}O_{2}=\frac{MD}{AB}R$

In order to calculate $R$ we will need all the sides of the triangle $ABC$,
[u][b]4 total  lengths[/b][/u][/color]


a country has 100 towns ,any two towns are connected with one-direction road,after verifying we have known that we cant go from any town to others according to the rule, prove that we can choose one town and switch the direction of its road ,such that we can travel from each town to any other town.

I have to admit that I don't quite understand what you mean. Are you saying that at the beginning there is no town from which we can go to every other town, and then we must choose a town and switch the directions of all the roads having an endpoint in that town?

no! maybe there exists just one town that we cant travel from that to every other town.

I think the statement is that the corresponding graph is not connected.

Pierre.


a,b,c positive reals

prove

a+b+c/2+b/c+a +c/a+b

Hi akashram;

Can you post the rest of that problem, there seems to be something missing.

[quote="akashram"]I am sorry  :blush:

the problem is a+b+c/2+b/c+a  +c/a+b> or equal to 5/2

sorry I have problems with latex[/quote]

What did you mean ?

$ a\plus{}b\plus{}\frac{c}{2}\plus{}\frac{b}{c}\plus{}a  \plus{}\frac{c}{a}\plus{}b\ge \frac{5}{2}$
or
$ \frac{a\plus{}b\plus{}c}{2}\plus{}\frac{b}{c\plus{}a}  \plus{}\frac{c}{a\plus{}b}\ge \frac{5}{2}$ ?

I'm pretty sure he means the latter case, although I'll let him verify before trying to post a solution.

[quote="Shanku"][quote="akashram"]I am sorry  :blush:

the problem is a+b+c/2+b/c+a  +c/a+b> or equal to 5/2

sorry I have problems with latex[/quote]

What did you mean ?

$ a \plus{} b \plus{} \frac {c}{2} \plus{} \frac {b}{c} \plus{} a \plus{} \frac {c}{a} \plus{} b\ge \frac {5}{2}$
or
$ \frac {a \plus{} b \plus{} c}{2} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge \frac {5}{2}$ ?[/quote]

if it´s the latter case, just put a=o.8, b=0.1=c. the inequality does not hold.

[quote="Shanku"][quote="akashram"]I am sorry  :blush:

the problem is a+b+c/2+b/c+a  +c/a+b> or equal to 5/2

sorry I have problems with latex[/quote]

What did you mean ?

$ a \plus{} b \plus{} \frac {c}{2} \plus{} \frac {b}{c} \plus{} a \plus{} \frac {c}{a} \plus{} b\ge \frac {5}{2}$
or
$ \frac {a \plus{} b \plus{} c}{2} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge \frac {5}{2}$ ?[/quote]

If it is the former case, it does not hold.

Let $ a \equal{} b \equal{} c \equal{} \frac{1}{10}$

The left-hand side becomes:

$ \frac{1}{10} \plus{} \frac{1}{10} \plus{} \frac{1}{20} \plus{} 1 \plus{} \frac{1}{10} \plus{} 1 \plus{} \frac{1}{10} \equal{}$

$ 2 \plus{}\frac{4}{10} \plus{} \frac{1}{20} \equal{}$

$ 2 \plus{} \frac{9}{20} < 2 \plus{} \frac{10}{20} \equal{} \frac52$

I meant the latter case and the inequality does hold

Hi;

For the first interpretation:

$ a \plus{} b \plus{} \frac {c}{2} \plus{} \frac {b}{c} \plus{} a \plus{} \frac {c}{a} \plus{} b\ge\frac {5}{2}$

fails for a=1, b = 1/64, c = 1/4  this yields 79 / 32 = 2.46

The second one also fails:

$ \frac {a \plus{} b \plus{} c}{2} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge\frac {5}{2}$

a = 3/4, b = 1/2, c = 1 yields 619 / 280 = 2.21


The MathLinks clock seems to have gone a bit bonkers after the upgrade...I checked my profile and it's still set to +0800 GMT, but the time shown is 1 hour ahead of time.  I don't think both my computer clock and my other clock are both 1 hour behind time :D.

P.S. I know my screen resolution is small, only 800*600, but is it possible to alter the settings such that the page doesn't get cut off on the right?  I think it's because of the "Watched Topics" and "Preferences" at the top...(Sorry for being so picky.)

even my clock runs an hour ahead even though i have set it to GMT + 5.5 HOURS

okay let's see if ->THIS<- helped :D hmm

well, i think the clock of the forum is one hour ahead of time..

I mentioned this a long time ago, but no one seemed to care then :D:D

A long time ago?  Are you referring to the Daylight Savings period?  Or did I just miss a post somewhere recently (which is probably more likely)?

errr :) I will fix the damn clock don't worry :D some time now :)


hello all,
This is my first post here on the site. I'm not sure whether this topic belongs to this particular section of the forum.

I just graduated from high-school, and I'm looking for a way to cut college costs through spending less time there. Lots of people told me to concider taking APs, or CLEP exams. APs here in my country seem to be no choice, at least not at this time of the year. So the CLEP sounded like the thing i had to do.

The problem with the CLEP exam, calculus in particular, is that it has no good study guides or references. I would very much appreciate the help from anyone who has already taken/wants to take the test, and has some idea about study guides.

Any suggestions for taking other tests too?
Thanks to all in advance  :)

[quote="aboodz"]
 I would very much appreciate the help from anyone who has already taken/wants to take the test, and has some idea about study guides.

Any suggestions for taking other tests too?
Thanks to all in advance  :)[/quote]

I have neither taken the test nor do I want to take it, this should be appreciated nevertheless.

Single Variable Calculus:Early Transcendentals is a pretty good start. Just at the level that is required (I'm assuming - since it buiilds on principles from high school)

[url=http://www.amazon.com/gp/product/0534496784/701-1174975-5081966?v=glance&n=916520&s=gateway&v=glance]Check it out[/url]

What country are you in? Does the college have its own tests for testing out of some course requirements?

well i live in Jordan, but im looking to attend a university in the USA, most likely in UTAustin, or any other good universities in NY or Texas.

I have taken some calculus in high-school, and im really good at it (at least the part we took). Math consisted of 6 units; Limits, Diferentiation, Applications of Differentiations, Integration, Applications of Integration, Probabilities. So I actually have a good idea about calculus. But what book could help me with the kind of questions that come in the test (the CLEP), prolly the only training i need is the way questions are. Most reviews on the amazon say that the official 'College Board' study guide isnt that good, and other books include no calculus exam.

So,could anyone help?

Wow, I live in Austin and have been to UT-Austin many times.  If you have questions feel free to PM me and I'll [i]try[/i] to help.  (Sorry, I'm not that knowledgeable on CLEP tests.)


Can someone explain this for me in a little more detail?

In how many 0's does 58! end when expressed in base 12?

I understand that you have to find how many powers of 12 go into 58!, since each of these would be one more zero at the end of the number...but I don't understand why dividing 58 repeatedly by 2 and adding up the quotients(truncated if not an integer) each time, and then dividing it by 3 and adding up the quotients(truncated if not an integer), can give you the number of 12's. Basically, I'm asking what these quotients mean. I know the Number Theory has this exact example, but I don't understand what the quotients signify. If it's unclear, there's a latex version...

[hide="written out"]
$ \frac{58}{2}\equal{}29,\frac{29}{2}\equal{}14.5,\frac{14}{2}\equal{}7,\frac{7}{2}\equal{}3.5,\frac{3}{2}\equal{}1.5$

$ 29\plus{}14\plus{}7\plus{}3\plus{}1\equal{}54$

$ \frac{58}{3}\equal{}19\frac{1}{3},\frac{19}{3}\equal{}6\frac{1}{3},\frac{6}{3}\equal{}2$
$ 19\plus{}6\plus{}2\equal{}27$

What is the significance of the quotients? What do they mean? And because of that, what do the sums of the quotients mean?

The answer to the problem is that there are 27 0's...[/hide]

It shows the number of times those factors appear in that number.

[quote="Icy"]OH! So...$ 3^{27}$ and $ 2^{54}$ are the greatest multiples of 2 and 3 that are factors of 54!, so $ 12^{27}$ is the largest power of 12 in 58! ?[/quote]

Exactly.


function
calculus
integration
complex analysis
complex analysis unsolved

Let $f(z)$ be analytic function in the open unit disk $|z|<1$,$f(0)=0$.Prove that the series $\sum_{k=1}^{\infty}f(z^k)$ converges for $|z|<1$ and represents an analytic function

$f(z^n)$ is $z^n$ multiplied with some function of $z$ which is at most $M$ for some $M>0$ for all $|z|\le r<1$, and this proves the convergence of $\sum f(z^n)$ in the disk $D(0,r)$. We do this for $r\to 1$, and we have the convergence.

For the second part, we can see that the integral of $\tilde f(z)=\sum f(z^n)$ is zero on every closed contour in $D(0,1)$, and we can use Morera's Theorem, stating that in such a case, the function is analytic (the function $\tilde f$ is clearly continuous, which is required as a hypothesis in Morera's Theorem, so we can apply it).


Show that one integer of the form $ 4^{a}(8b+7)$, with $ a, b$ non-negative integers, can't be written as a sum of three squares.


Factor the expression $ ab\plus{}5b\plus{}2a\plus{}10$.

From Simon's favorite factoring trick, $ ab\plus{}5b\plus{}2a\plus{}10\equal{}(a\plus{}5)(b\plus{}2)$

There's not much more I can say about this...

Simon's Favorite Factoring Trick:

$ mn \plus{} mx \plus{} ny \plus{} xy \equal{} (m \plus{} y)(n \plus{} x)$


Triangle ABC has medians BD and CE.  H is the midpoint of BE, and J is the midpoint of CD.  Is HJ trisected by BD and CE?

It 's trivival.
$ HJ \cap BD = T; HJ \cap EC=V$
$ \Longrightarrow HT=VJ=\frac{1}{2}ED;TV=\frac{BC-ED}{2}=\frac{1}{2}ED$
$ \Longrightarrow HT=TV=VJ=\frac{1}{2}ED$


Find the least number $n$ such that $1999$ can be expressed as the sum of $n$ fourth powers of positive integrers numbers.


If $n \geq 3$,
 then prove that $2^n$ can be represented
in the form $ 2^n = 7x^2 + y^2 $ with $x, y $ odd positive integers.

very famous problem :)

Arthur Engel gives an elementary solution (by Erik Muller) in his Problem-Solving Strategies on page 126. Engel claims it to be one of the most difficult olympiad-problems ever.


Let $ p,q$ are distinct primes. For $ n > 2$, find all value of integer $ c$ such that the polynomial $ f(x) \equal{} x^n \plus{} cx^{n \minus{} 1} \plus{} pq$ can be factorized into the product of non constant 2 polynomials with integer coefficient.

It's just division.

Let one of the factor polynomial $ g(x)$ is of degree $ k \le n\minus{}2$ , say $ a_k x^k\plus{}a_{k\minus{}1} x^{k\minus{}1}\plus{}\dots \plus{}a_1 x\plus{}a_0$. Obviously $ a_k\equal{}1$
[b]Case 1[/b] $ f(x)$ has a foctor polynomial $ g(x)$ of the form WLOG $ a_k x^k\plus{}a_{k\minus{}1} x^{k\minus{}1}\plus{}\dots \plus{}a_1 x \pm p$ (as $ p, q$ can interchange).Then constant term of $ \frac{f(x) \mp q g(x)}{x}$ is divisible by $ \pm p$ , i.e. $ p|a_1$ as $ p \nmid q$ , by similar division we get that for all $ i \le k$  if $ p|a_1,a_2,\dots ,a_{i\minus{}1}$ , then $ p|a_i$ , so $ p|a_k$, which is impossible, so $ g(x)$ is impossible to be of this form.And if $ f(x)$ has a factor polynomial of the form  $ a_{n\minus{}1} x^{n\minus{}1}\plus{}a_{n\minus{}2} x^{n\minus{}2}\plus{}\dots \plus{}a_1 x \pm p$, then it has a factor polynomial of the form $ x \mp q$, which is impossible.
[b]Case 2[/b] $ f(x)$ has a factor polynomial $ g(x)$ of the form $ a_k x^k\plus{}a_{k\minus{}1} x^{k\minus{}1}\plus{}\dots \plus{}a_1 x \pm pq$. By similar claim we get $ pq|a_k$ which is impossible again. So now only [b]Case 3[/b] is possible.
[b]Case 3[/b] $ f(x)\equal{}(x \pm 1)(x^{n\minus{}1}\plus{}a_{n\minus{}2} x^{n\minus{}2}\plus{}\dots \plus{}a_1 x \pm pq)$
$ \Leftrightarrow (\mp 1)^n \plus{} c(\mp 1)^{n\minus{}1}\plus{}pq\equal{}0$
So the possible values of $ c$ are :
when $ n \;\; even$, $ c\equal{}pq\plus{}1$ or $ \minus{}pq\minus{}1$
when $ n \;\; odd$, $ c\equal{}\minus{}pq\plus{}1$  or  $ \minus{}pq\minus{}1$


number theory
greatest common divisor
number theory unsolved

Find the greatest common divisor of all number in the set $( a^{41} - a | a \in \mathbb{N} and \geq 2 )$ . What is your guess if 41 is replaced by a natural number $n$

You proved that 13530 is a common divisor of all numbers of this form...
But how would you prove that it is really the [b]greatest[/b] common divisor? :D

By Labiliau's remark, or a generalization there of, it follows that  the given primes $p$ that satisfy $p - 1|40$ have their multiplicity in the desired gcd is exactly 1. Now we shall show that no other primes occur in the gcd. For each prime $q$ that does not abide with $q - 1|40$, consider a primitive root $r_{q}$ of that prime (which exists). It follows that firstly $gcd(r_{q},q) = 1$, and as well that $r_{q}^{40} \neq 1$ (mod  $q$), and this proves the desired result. Thus the presented number is indeed the greatest gcd.


number theory
least common multiple
number theory unsolved

Determine the number of integers $ n$ with $ 1 \le n \le N\equal{}1990^{1990}$ such that $ n^2\minus{}1$ and $ N$ are coprime.

Note that $ 1990^{1990} \equal{} (2\times5\times199)^{1990} \equal{} 2^{1990}\times5^{1990}\times199^{1990}$.

Since $ 2$, $ 5$ and $ 199$ are the only prime factors of $ 1990^{1990}$ and $ \mbox{lcm}(2,5,199) \equal{} 1990$, we just have to find the values $ \bmod1990$ for which $ 2\nmid(n^2 \minus{} 1)$, $ 5\nmid(n^2 \minus{} 1)$ and $ 199\nmid(n^2 \minus{} 1)$.

We want to find the values $ \bmod1990$ for which $ n^2 \minus{} 1$ is odd, what happens if and only if $ n$ is even.

Therefore, $ n$ must be even, what happens with $ \frac {1}{2}$ of the values $ \bmod1990$.

If $ 5|n^2 \minus{} 1$ we have $ n^2 \minus{} 1\equiv_50\Leftrightarrow(n \minus{} 1)(n \plus{} 1)\equiv_50$, what happens if and only if $ n\equiv_51$ or $ n\equiv_54$, what happens with $ \frac {2}{5}$ of the values $ \bmod1990$.

Therefore, the opposite case happens with $ \frac {3}{5}$ of the values $ \bmod1990$.

If $ 199|n^2 \minus{} 1$ we have $ n^2 \minus{} 1\equiv_{199}0\Leftrightarrow(n \minus{} 1)(n \plus{} 1)\equiv_{199}0$, what happens if and only if $ n\equiv_{199}1$ or $ n\equiv_{199}198$, what happens with $ \frac {2}{199}$ of the values $ \bmod1990$.

Therefore, the opposite case happens with $ \frac {197}{199}$ of the values $ \bmod1990$.

So we have $ 1990\times\frac {1}{2}\times\frac {3}{5}\times\frac {197}{199} \equal{} 591$ values $ \bmod1990$ in the asked conditions.

So we have $ 591\times\frac {1990^{1990}}{1990} \equal{} 591\times1990^{1989}$ numbers in the asked conditions.


topology
group theory
advanced fields
advanced fields unsolved

Prove that if a finite group $G$ acts freely on $\mathbb{R}^{n}$ for $n \geq 1$, then $G$ is trivial.
* How about the case of cyclic group $G$.

i'm trying the "easy" thing: take the traslation action of $\mathbb{Z}$ on $\mathbb{R}^{n}$..
this seems to be free...

about the principal question, you should ask for the action to be continuous, i think..
otherwise, you can just consider $\mathfrak{c}$ copies of $G$ with the obvious action of $G$, and take a bijection of this set with $\mathbb{R}^{n}$...

Yes, that's free- but $\mathbb{Z}$ isn't finite.

This is obviously meant as a problem in algebraic topology, and that means that $G$ acts continuously. In fact, each element acts as a homeomorphism.

An equivalent statement:
Suppose $\phi$ is a homeomorphism from $\mathbb{R}^{n}$ to itself of finite order $k>1$. Then $\phi^{j}$ has a fixed point for some $j<k$.

You may obviously assume that $G$ is cyclic.

A different approach: the quotient space is a manifold, hence has no cohomology in dimensions $>n$, but its cohomology is also the group cohomology of $G$ which is $\mathbb Z/\#G$ for cylic $G$ and even dimensions. owk

Hi.

A few thoughts about the problem for $n = |G| = 2$:

Suppose $\varphi:{\mathbb R}^{2}\to{\mathbb R}^{2}$ is a homeomorphism with $\varphi^{2}= \text{id}$ and $\varphi(x)\neq x$ for all $x\in{\mathbb R}^{2}$. Choose an arbitrary $x\in{\mathbb R}^{2}$ and let $\gamma: [0,1]\to{\mathbb R}^{2}$ be a path connecting $x$ and $\varphi(x)$. Since $\varphi$ has no fixed point, the map $\tilde{f}_{\gamma}: [0,1]\to{\mathbb S}^{1}$ given by $\tilde{f}_{\gamma}(t) : = \frac{(\varphi\circ\gamma)(t)}{\|(\varphi\circ\gamma)(t)\|}$ is well-defined and since $\varphi^{2}= \text{id}$ we have $\tilde{f}_{\gamma}(1) =-\tilde{f}_{\gamma}(0)$. Now think of $[0,1]$ as the upper half of the unit circle ${\mathbb S}^{1}$ and extend $\tilde{f}_{\gamma}$ to a map $f_{\gamma}:{\mathbb S}^{1}\to{\mathbb S}^{1}$ via $f_{\gamma}(-x) =-f_{\gamma}(x)$. Since any two paths connecting $x$ and $\varphi$ are homotopic we see that the homotopy-class of $f_{\gamma}$ does not depend on the chosen $\gamma$. Now, consider $\eta: [0,1]\to{\mathbb R}^{2}, \eta : = \overline{\varphi\circ\gamma}$. Then $f_{\eta}= \overline{f_{\gamma}}$, but since $[\overline{f_{\gamma}}]$ is the inverse of $[f_{\gamma}]$ in $\pi_{1}({\mathbb S}^{1})$, this means that $-[f_{\gamma}] = [\overline{f_{\gamma}}] = [f_{\eta}] = [f_{\gamma}]$ and $f_{\gamma}$ is nullhomotopic, which is not possible since $f(-x) =-f(x)$ and by Borsuk-Ulam $f$ has odd degree.

Since Borsuk-Ulam is also true in higher dimensions $n$, this seems to work also for arbitrary $n$ if one constructs the map $f:{\mathbb S}^{n-1}\to{\mathbb S}^{n-1}$ inductively.

Hanno

Hello.

As owk said, one may look at the quotient space ${\mathbb R}^{n}/ G$ having ${\mathbb R}^{n}$ as its universal covering. Hence, $\pi_{1}({\mathbb R}^{n}/ G) \cong G, \pi_{k}({\mathbb R}^{n}/ G) = 0, k\geq 2$. Is there any classification of such spaces?

Hanno


If $ u_n/v_n$ converges to $ 1$ then we write $ u_n\sim{v_n}$.
Here $ a_n$ tends to $ \infty$ then $ a_{n\plus{}1}\sim{a}_n^2$.
Then (good exercise) there exists $ \lambda>1$ s.t. $ a_n\sim\lambda^{2^n}$.
$ ln(a_n)\sim{2}^nln(\lambda)$. $ ln(u_n)\equal{}1/2^n\sum_{k\equal{}0}^n{l}n(a_k)\sim{l}n(\lambda)/2^n\sum_{k\equal{}0}^n{2^k}\sim{2}ln(\lambda)$ and $ u_n$ converges to $ \lambda^2$.
It remains to find explicitly (if it's feasible) $ \lambda(\approx{2.81722})$.


Triangle ABC is acute-angled

Prove Sqrt{1+8[(CosB)^2]}/SinA + Sqrt{1+8[(CosC)^2]}/SinB +
+Sqrt{1+8[(CosA)^2]}/SinC >=6

This is problem E2119

See http://www.mathlinks.ro/Forum/viewtopic.php?t=18568 .

  Darij


Can someone help me with this problem from the AMC 12 2004:

A plane contains points A and B with AB = 1.  Let S be the union of all disks of radius 1 in the plane that coves AP.  What is the area of S?

[hide]

picture included .

due to my inability to use paint, the pictures looks horrible.  Arcs CE and DF are each actually (1/3 of) a circular arc.

this of it this way:

you have a wooden meter stick and a hula hoop of radius 1 meter, lay both flat on the ground, place the hula hoop around the stick.  let one end of the stick be A, and the other be B.

the area is area is divided up into 5 parts:

the two parts that are each 1/3 of a circle of radius 2, bounded by BC and BE or AD and AF

the two part counted twice by the previous two arcs

the two parts are not accounted for by the previous two arcs

Find the two arcs(TwoBigArcs):

affix one point on your hula hoop to A and move it around.  you only have limited movement due to the other end of the ruler that's standing in it's way.  doing this is essentially the same same as just having the diameter of the hula hoop and moving it around, except there are limits to how far the diameter can move.

the limits are the two cases where the hula hoop is in contact with both ends (A and B) of the stick.

these cases are important, so draw them out.  the diameter must go through the center of the circle, so we have BC and BE as bounds to of the two arcs, as it is impossible to move the hoop such that the longest distance between some point ont he hoop is 2meters from point A.

Now notice that AB=1(given) and that OB=OA=1 (radii of a circle of radius 1)

so triangle OAB is equilateral.  the same argument works for triangle PAB.  so angle OBP is 120, and thus the arc CE is of measure 120, making the area of Sector BCE 1/3 of the area of a circle of radius 2.

now do the same thing for the arc DF.

Finding the areas of the two sectors that the two arcs do not enclose (TwoSmallArcs):

If you drew a nice diagram, it should be apparent that these two sectors are sectors DOC and FPE.  each is 1/6 of a circle of radius 1, due to angles DOC and FPE being equal to 60.

Finding the area counted twice(CountedTwice):

notice how rhombus BOAP is enclosed by both sectors BCE and ADF.

the area is simply the sum of the areas of the two  equilateral triangles BOA and BAP.

Calculating:

Total area=TwoBigArcs+TwoSmallArcs-CountedTwice

=2x(120/360 x :pi:2^2)+2(60/360 x :pi:1^2)-2(1 x :rt3:/2)/2

so the answer should be 3:pi:-:rt3:/2

*note: i am not held responsible for calculation errors.  it is kinda late   [/hide]

Another way to do it:
[hide]
The left most circle and right most circle that cover $AB$ are the ones with $AB$ for a radius.  These two circles intersect at two points, $C$ and $D$.  The diameter through $B$ in the right one intersects the circle at $E$ and the diameter through $A$ in the right one intersects the circle at $F$.

The highest point in $S$ is the circle that has $AB$ as a chord.  The radii drawn to the endpoints of this chord form an equilateral triangle with side 1, so we know that this ceneter is $C$.  The lowest point, naturally, will have center $D$.

Now, we can rotate point $F$ about $A$ onto a point on circle C, diameterically opposite $A$.  If we let this point be $F'$, then we can draw a sixth of a circle centered at $C$ through $F$ and $F'$.  This sector, of radius 2, will contain points in $S$, since we can rotate Circle B as we did Point F onto Circle C, creating circles that are contained in $S$ while in motion.  We can also rotate Circle C downwards onto Circle D, so we have a $120^\circ$ sector with radius 2 on the right.

We can do the same thing on the left, creating two sectors with area $\frac{4\pi}{3}$.  We also have two sixths of circles C and D that we haven't counted.

These each have area $\frac{\pi}{6}$.

The two $120^\circ$ sectors overlap, so we must subtract that overlap, which comes in the form of Rhombus $ACBD$.  This has twice the area of equilateral $ABC$.  Therefore, we must subtract $\frac{\sqrt{3}}{2}$.

The area of $S$ is $2\left(\frac{4\pi}{3}\right)+2\left(\frac{\pi}{6}\right)-\frac{\sqrt{3}}{2}=\boxed{3\pi-\frac{\sqrt{3}}{2}}$; $\boxed{\mathbb C}$.
[/hide]


search
number theory
relatively prime
number theory unsolved

find all integer solutions to the equation:
$m^{2}-1=n^{3}$

Quite easy? As (m+1, m-1) | 2...

Edit--actually, it is not quite as easy as I thought. So I will try a new approach:

$m^{2}= n^{3}+1$
$m^{2}= (n+1)(n^{2}-n+1)$

The two RHS factors are relatively prime except for some easy cases like n = 2. So now we have that $n^{2}-n+1$ is a square, which is contradiction by bounding... so maybe it is easy after all?

[quote="TomciO"]I don't know what is easy for you but I would like you to finish your solution...[/quote]

He is done...$n=2$ is the only solution...as he proved

[quote="probability1.01"]
The two RHS factors are relatively prime except for some easy cases like n = 2[/quote]

$n=5$?
$n=9999998$?

Probability1.01 is false. gcd(n^2-n+1,n+1)=1 or 3 so we must solve and the system n^2-n+1=3a^2, and n+1=3b^2 (a,b)=1.it have many solutions.

Yeah, you're right :(. I suppose you might be able to pound it out with some pell equations, but I'll work on it more later.

To be honest, it's really famous equation and I think that solution which doesn't use elliptic curves is not known. Actually, Euler's solved it without elliptic curves but it can be proved that his solution is essentialy the same as the elliptic curves one.

this is mordell equation for k=1, and its a special case of Catalan's Conjecture.

there are only 3 solutions, (-1,0), (0,1), (2,3)

n^3+1>=0 so n>=-1^
 if n=-1 there is one solution
 if n=0 there isn't any solution
so we can take n to be positive
write as n^2=(m+1)(m^2-m+1)
 if gcd((m+1),(m^2-m+1))=1
if m=1 there is one solution if not
m^2-m+1=x^2 for some integer x>1
but (m-1)^2<m^2-m+1<m^2
so it can't be a square number
 if gcd((m+1),(m^2-m+1))=3
m must be in the form of 3k-1 for positive integer k
so (9k^2-9k+3)/3=3k^2-3k+1 and k must be square numbers
but i cant get this part i think it is going to go like first case but it doesnt
i will work hard on it but if someone has a solution please write
and i look at the link at mathwolfram there isnt anything they didnt write anything about the solution so can i think about that this solution was very hard?

Take a look at p=2, q=3 in the PDF http://www.math.leidenuniv.nl/~jdaems/scriptie/Catalan.pdf

It's a very nice PDF by the way.

Here's an interesting one (not so hard)...

Find all solutions to \[p^{n}-q^{m}=1\] where $p, q$ are primes.

ya I posted while the pdf was loading...

Wait, are you sure?  I didn't find anything in the pdf on this one, and it's anyways a much simpler one which requires only very elementary number theory (I could solve it)...
Maybe I just didn't understand their terminology...

Well your problem is trivial for specific p,q so now I just need to look over the proofs for stuff like 2,3 2,5 3,5 3,7 etc that i've done before and generalize
i''ll post after i eat dinner...

Well clearly, one number is 2...that's the first step  :)

Is it true that $n|\phi (2^{n}\pm 1)$? I tried it for low cases and it seems true...
that kind of crushes the problem .. lol

For the $-$ yes, for the $+$ I don't know.

It"s true for general $2$ and $1$ and posted by Darij some time ago (3 months or so).

Okay I'll search for it... by general 2 and 1 do you mean any m and n work?


geometry
parallelogram
analytic geometry

Parallelogram $ ABCD$ has $ A(0, 1)$, $ B(2, 1)$ and $ C(4, 0)$ as three of its vertices. What is the sum of the coordinates of point $ D$?

This isn't possible as both (6,0) and (2,0) are possible points for D.

[hide="Solution"]--(2,1)--(4,0)--(2,0)--cycle); label ("$ A$",(0,1),NW);label ("$ B$",(2,1),NE); label ("$ C$",(4,0),SE); label ("$ D$",(2,0),SW); [/asy]

We know that the $ x$ coordinate of $ A$ is 0, and the $ x$ coordinate of $ B$ is 2, thus the difference between the $ x$ coordinate of $ C$ and $ D$ must be 2 also. Thus, the $ x$ coordinate of $ D$ is 2.

Since both lines are parallel, the $ y$ coordinates of $ C$ and $ D$ must be the same; 0.

From this we get $ D \equal{} (2,0)$. Thus, the answer is $ \boxed{2}$.

Although, I think it can be 6 also...[/hide]

Thus, the answer is either 6 or 2?

Note, however, that it is parallelogram $ ABCD$, meaning that the points taken in clockwise or counterclockwise order (one of the two) must be in order, so the only possibility is $ D\equal{}(2,0)\implies2\plus{}0\equal{}\boxed2$.


Let $ f: R-->N$ be a function satisfying:  $ f(x+1/f(y))=f(y+1/f(x))$ for every $ x,y \in R$.
   Prove that there exist a positive integers number $ n$ such that $ f(x)$ is not equal $ n$ for every $ x \in R$

As the title, this is one of the shortlisted problems in IMO 2008.

[Moderator edit: http://www.mathlinks.ro/viewtopic.php?t=287857 ]


there is a progression, which does not contain square numbers .

2 3 5 6 7 8 10 11 12 13 14 15 17 .....

then show that the general term is [n+ sqrt{n} + 1/2]  ([x] is gauss' notation)

Let $ f(n)\equal{}\left\lfloor n\plus{}\sqrt{n}\plus{}\frac{1}{2}\right\rfloor$. Then consider $ f(a^2\plus{}a)$ and $ f(a^2\plus{}a\plus{}1)$ for nonnegative integers $ a$. Since $ \left( a\plus{}\frac{1}{2}\right)^2\equal{}a^2\plus{}a\plus{}\frac{1}{4}$,
\[ a^2\plus{}2a\plus{}\frac{1}{2}\leq a^2\plus{}a\plus{}\sqrt{a^2\plus{}a}\plus{}\frac{1}{2}<a^2\plus{}2a\plus{}1\]
\[ \Rightarrow f(a^2\plus{}a)\equal{}a^2\plus{}2a\equal{}(a\plus{}1)^2\minus{}1\]
and
\[ a^2\plus{}2a\plus{}2<a^2\plus{}a\plus{}1\plus{}\sqrt{a^2\plus{}a\plus{}1}\plus{}\frac{1}{2}\leq a^2\plus{}2a\plus{}\frac{5}{2}\]
\[ \Rightarrow f(a^2\plus{}a\plus{}1)\equal{}a^2\plus{}2a\plus{}2\equal{}(a\plus{}1)^2\plus{}1\]
Since $ f$ is increasing, $ f(n)$ cannot be a square for any positive integer $ n$, for if $ f(n)\equal{}(b\plus{}1)^2$, we would need $ b^2\plus{}b<n<b^2\plus{}b\plus{}1$.

On the other hand, $ n\plus{}1\plus{}\sqrt{n\plus{}1}\plus{}\frac{1}{2}$ and $ n\plus{}\sqrt{n}\plus{}\frac{1}{2}$ differ by more than 1, so $ f(n\plus{}1)\geq f(n)\plus{}1$. This shows that for $ m\neq n$, $ f(m)\neq f(n)$.

So for every positive integer $ a$, the set $ S_a\equal{}\{ f(1),f(2),\cdots ,f(a^2\plus{}a)\}$ consists of $ a^2\plus{}a$ positive integers, each at most $ a^2\plus{}2a$, with none of them equal to one of $ 1,4,\cdots a^2$ (all less than $ a^2\plus{}2a$, and there are $ a\equal{}(a^2\plus{}2a)\minus{}(a^2\plus{}a)$ of them). The only possibility is that $ S_a$ contains all of the nonsquares that are at most $ a^2\plus{}2a$. This says every nonsquare is in the range of $ f$.


number theory
prime numbers
number theory unsolved

Given an integer 'n',find out the minimalest  integer 'm' that it has
n divisors.
For example:
n=4,m=6;
n=5,m=16;
n=18,m=180;
Who can tell me how to get 'm'?

It  seems like it is like this. For $ n \equal{} \pi_1^{\alpha_1}\pi_2^{\alpha_2}\cdots \pi_s^{\alpha_s}$, where $ \pi_1 > \pi_2 > \cdots > \pi_s$ are the primes dividing $ n$, and denoting $ (p_k)_{k\geq 1}$ the sequence $ 2,3,5,7,\ldots$ of the prime numbers, then $ m \equal{} \left ( p_1^{\pi_1\minus{}1}p_2^{\pi_1\minus{}1}\cdots p_{\alpha_1}^{\pi_1\minus{}1} \right ) \left ( p_{\alpha_1\plus{}1}^{\pi_2\minus{}1}p_{\alpha_1\plus{}2}^{\pi_2\minus{}1}\cdots p_{\alpha_1 \plus{}\alpha_2}^{\pi_2\minus{}1} \right ) \cdots \left ( p_{\alpha_1\plus{}\alpha_2 \plus{} \cdots \plus{}\alpha_{s\minus{}1}\plus{}1}^{\pi_s\minus{}1}p_{\alpha_1\plus{}\alpha_2 \plus{} \cdots \plus{} \alpha_{s\minus{}1}\plus{}2}^{\pi_s\minus{}1}\cdots p_{\alpha_1\plus{}\alpha_2 \plus{} \cdots \plus{} \alpha_{s\minus{}1}\plus{}\alpha_s}^{\pi_s\minus{}1} \right )$.

Woo!How did you get it?
I tried some 'n'and'm',they are all rignt.
Can you prove it?And can you give me a detailed answer.thanks.

These are the highly composite numbers for certain values of m. See [url]http://en.wikipedia.org/wiki/Highly_composite_number[/url]. There is no general way to find very large highly composite numbers without a computer. I don't think your problem has an easy solution.

I have seen the link.I can programme to solve it.JUST for fun!
Thank you all.

I programmed using $ m \equal{}\left ( p_{1}^{\pi_{1}\minus{}1}p_{2}^{\pi_{1}\minus{}1}\cdots p_{\alpha_{1}}^{\pi_{1}\minus{}1}\right )\left ( p_{\alpha_{1}\plus{}1}^{\pi_{2}\minus{}1}p_{\alpha_{1}\plus{}2}^{\pi_{2}\minus{}1}\cdots p_{\alpha_{1}\plus{}\alpha_{2}}^{\pi_{2}\minus{}1}\right )\cdots\left ( p_{\alpha_{1}\plus{}\alpha_{2}\plus{}\cdots\plus{}\alpha_{s\minus{}1}\plus{}1}^{\pi_{s}\minus{}1}p_{\alpha_{1}\plus{}\alpha_{2}\plus{}\cdots\plus{}\alpha_{s\minus{}1}\plus{}2}^{\pi_{s}\minus{}1}\cdots p_{\alpha_{1}\plus{}\alpha_{2}\plus{}\cdots\plus{}\alpha_{s\minus{}1}\plus{}\alpha_{s}}^{\pi_{s}\minus{}1}\right )$,but when n=8 ,I got m=30,the right one is 24.
WHY this formula do not work when n=2^k,k=(1,3,5,7....).

WHY this formula do not work when n=2^k,k=(1,3,5,7....).when k=(2,4,6,8..) it is wrong either.


Let $ a,b,c,k>0$ and $ ab\plus{}bc\plus{}ca\equal{}1$.Find max P with $ P\equal{}\frac{k(1\minus{}a^2)}{1\plus{}a^2}\plus{}\frac{1\minus{}b^2}{1\plus{}b^2}\plus{}\frac{1\minus{}c^2}{1\plus{}c^2}$

[quote="vo thanh van"]Let $ a,b,c,k > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$.Find max P with $ P \equal{} \frac {k(1 \minus{} a^2)}{1 \plus{} a^2} \plus{} \frac {1 \minus{} b^2}{1 \plus{} b^2} \plus{} \frac {1 \minus{} c^2}{1 \plus{} c^2}$[/quote]
Since$ a,b,c,>0$ and $ ab\plus{}bc\plus{}ca\equal{}1$,We put:$ a\equal{}tan\frac{A}{2},b\equal{}tan\frac{B}{2},c\equal{}tan\frac{C}{2} ,A,B,C$ to be angles of a certain triangle
Exercise become:Find max P with $ P\equal{}k cosA\plus{}cosB\plus{}cosC$
It very easy :)

[quote="quykhtn-qa1"][quote="vo thanh van"]Let $ a,b,c,k > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$.Find max P with $ P \equal{} \frac {k(1 \minus{} a^2)}{1 \plus{} a^2} \plus{} \frac {1 \minus{} b^2}{1 \plus{} b^2} \plus{} \frac {1 \minus{} c^2}{1 \plus{} c^2}$[/quote]
Since$ a,b,c, > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$,We put:$ a \equal{} tan\frac {A}{2},b \equal{} tan\frac {B}{2},c \equal{} tan\frac {C}{2} ,A,B,C$ to be angles of a certain triangle
Exercise become:Find max P with $ P \equal{} k cosA \plus{} cosB \plus{} cosC$
It very easy :)[/quote]
It's the same my solution but it's not nice.I want to know a nice and short solution for this problem :)

The condition is equivalent to $ 1\equal{}\frac{a^2b^2c^2}{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}$. Thus, homogenizing
$ \sum_{cyc}\frac{(a^5\plus{}b^5)a^2b^2c^2}{a^2b^2(a^3\plus{}b^3)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)} \ge 1$
$ \Leftrightarrow \sum_{cyc} c^2 \cdot \frac{a^5\plus{}b^5}{a^3\plus{}b^3} \ge \sum_{cyc} c^2 \cdot \frac{a^2\plus{}b^2}{2}$
$ \Leftrightarrow \sum_{cyc} c^2\cdot \frac{(a\minus{}b)^2(a\plus{}b)(a^2\plus{}ab\plus{}b^2)}{2(a^3\plus{}b^3)} \ge 0,$
which is true.

It is just simple enough!
$ 2 (a^5+b^5) \ge (a^2+b^2)(a^3+b^3)$
We are done!

I have a nice solution with k>$ \frac{1}{2}$ :D

[quote="tuandokim"]I have a nice solution with k>$ \frac {1}{2}$ :D[/quote]
Post your solution,please :)