search
calculus
trigonometry
derivative
calculus computations

An aircraft is flying northward searching for wreckage on the strip of ocean between 100m
and 300m to the east. How high should the aircraft fly to maximize the apparent (angular)
size of the search strip?

This is the same as the "picture on the gallery wall" problem. The answer is that the plane's altitude should be the geometric mean of 100 m and 300 m, namely $ 100\sqrt{3}\approxeq 173.$

A non-calculus solution:

Draw a picture with the horizontal axis representing the ocean surface and the vertical axis representing the set of possible locations for the plane (with the flight path perpendicular to the plane we're drawing). Let $ O$ be the origin and $ A$ and $ B$ be the points on the horizontal axis such that $ OA\equal{}100$ and $ OB\equal{}300$ - thus $ A$ and $ B$ bound the search strip.

Construct the circle that passes through $ A$ and $ B$ and is tangent to the vertical axis. Call the point of tangency $ C.$

If $ D$ is any point other than $ C$ on the vertical axis, then $ \angle ADB<\angle ACB.$ Why? If $ P$ is any other point on the circle, then $ \angle APB\equal{}\angle ACB,$ but $ D$ lies outside the circle, making that angle smaller.

Hence $ C$ is our optimal location for the airplane's flight. By the power of the point, $ OC^2\equal{}OA\cdot OB.$ Hence the answer above involving the geometric mean.

How do you solve the problem in using calculus?

Let $ a\equal{}100$ and $ b\equal{}300.$ Let $ y$ be the altitude. Find positive $ y$ that maximizes \[ \theta\equal{}\tan^{\minus{}1}\left(\frac by\right)\minus{}\tan^{\minus{}1}\left(\frac ay\right).\]
Take the derivative and if you're clever enough with the algebra, you should be able to show without too much pain that there's a critical point where $ y^2\equal{}ab.$

Let $ P(0,\ y)$ be the points of the aircraft, $ A(a,\ 0),\ B(b,\ 0)\ (0 < a < b)$ with $ \alpha \equal{} \angle{POA},\ \beta \equal{} \angle{POB}$, for $ \theta \equal{} \beta \minus{} \alpha \ (0 < \theta < \frac {\pi}{2})$, we are to maximize $ \tan \theta$ since $ \tan \theta$ increases monotonously for $ 0 < \theta < \frac {\pi}{2}.$

$ \tan \theta \equal{} \tan (\beta \minus{} \alpha) \equal{} \frac {\tan \beta \minus{} \tan \alpha}{1 \plus{} \tan \beta \tan \alpha}$

$ \equal{} \frac {\frac {b}{y} \minus{} \frac {a}{y}}{1 \plus{} \frac {b}{y}\cdot \frac {a}{y}} \equal{} \frac {b \minus{} a}{y \plus{} \frac {ab}{y}}\leq \frac {b \minus{} a}{2\sqrt {y\cdot \frac {ab}{y}}}$, the equality holds $ y \equal{} \frac {ab}{y} > 0$, yielding the desired altitude is $ \sqrt {ab}.$


Since this is a forum for collections of high-school problems, I thought this would be the right place to announce that the new PEN is there. The document is available on the supporting site http://www.problem-solving.be/pen.

wow, thank you very much!  :wink:  The site on the book  :)


If $ a\geq b\geq c\geq d\geq e$, then
$ (a\plus{}b\plus{}c\plus{}d\plus{}e)^2\geq 8(ac\plus{}bd\plus{}ce)$.
For $ e\geq 0$, determine when equality occurs.

[quote="Inequalities Master"]If $ a\geq b\geq c\geq d\geq e$, then
$ (a \plus{} b \plus{} c \plus{} d \plus{} e)^2\geq 8(ac \plus{} bd \plus{} ce)$.
For $ e\geq 0$, determine when equality occurs.[/quote]
Let $ a\equal{}e\plus{}x\plus{}y\plus{}z\plus{}t,$ $ b\equal{}e\plus{}x\plus{}y\plus{}z,$ $ c\equal{}e\plus{}x\plus{}y$ and $ d\equal{}e\plus{}x,$
where $ x,$ $ y,$ $ z$ and $ t$ are non-negative numbers.
Then your inequality is equivalent to the following:
$ (e\minus{}y\plus{}t)^2\plus{}4z(e\plus{}y\plus{}z\plus{}t)\geq0,$ which obvious.
The equality occurs when $ e\equal{}y\minus{}t$ and $ e\plus{}z\plus{}y\plus{}t\equal{}0,$ which gives $ e\equal{}0$ and $ a\equal{}b\equal{}c\equal{}d$
or when $ e\equal{}y\minus{}t$ and $ z\equal{}0,$ which gives $ \{(x\plus{}2y,x\plus{}2y\minus{}t,x\plus{}2y\minus{}t,x\plus{}y\minus{}t,y\minus{}t)|x\geq0,y\geq t\geq0\}.$


geometry
3D geometry
tetrahedron
geometry unsolved

A regular tetrahedron ABCD with edge b is inscribed in a cone with
a vertex angle 90° in such way that AB is on a generator of the cone.
Find the distance from the vertex of the cone to the straight line CD

I don't know why but I got that it is impossible to inscribe the regular tetrahedron in that cone. :(
Apparently, I am wrong... :?


probability
analytic geometry
geometry
inequalities

Two mathematicians take a morning coffee break each day.  They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly $m$ mintues.  The probability that either one arrives while the other is in the cafeteria is $40 \%,$ and $m=a-b\sqrt{c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime.  Find $a+b+c.$

[hide="Hint"]Interpret this problem geometrically.[/hide]

[hide="Solution"]
Let the two mathematicians be $M_1$ and $M_2$.  Consider plotting the times that they are on break on a coordinate plane and shading in the places where they would be there at the same time as such:

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=641[/img]

We can count the area that we don't want in terms of $m$ and solve:

\begin{eqnarray*} \frac{(60-m)^2}{60^2} &=& .6\\ (60-m)^2 &=& 36\cdot 60\\ 60 - m &=& 12\sqrt{15}\\ \Rightarrow m &=& 60-12\sqrt{15} \end{eqnarray*}

So the answer is $60+12+15=\boxed{087}$.[/hide]

sorry for reviving
can somebody explain how you draw the graph (joml, you're image wasn't attached).
i saw it on Kalva but don't quite get how the image was constructed.

[quote="joml88"][hide="Solution"]
Let the two mathematicians be $ M_1$ and $ M_2$.  Consider plotting the times that they are on break on a coordinate plane and shading in the places where they would be there at the same time as such:

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=641[/img]

We can count the area that we don't want in terms of $ m$ and solve:
\begin{eqnarray*} \frac {(60 - m)^2}{60^2} & = & .6 \\
(60 - m)^2 & = & 36\cdot 60 \\
60 - m & = & 12\sqrt {15} \\
\Rightarrow m & = & 60 - 12\sqrt {15} \end{eqnarray*}
So the answer is $ 60 + 12 + 15 = \boxed{087}$.[/hide][/quote]

Anyone please attach a graph for this problem.

[quote="lingomaniac88"][hide="Graph"]
[asy]pair a,b,c,d,w,x,y,z;
real m = 60-12*sqrt(15);
a=(0,0);
b=(60,0);
c=(60,60);
d=(0,60);
w=(m,0);
x=(60,60-m);
y=(0,m);
z=(60-m,60);
draw(a--b--c--d--a,linewidth(1));
draw(w--x,linewidth(1));
draw(y--z,linewidth(1));
fill(a--w--x--c--z--y--cycle,gray(0.7));[/asy]
[/hide][/quote]

I'm sorry, but can you elaborate the graph a little bit?

We let the time for one of the mathematician to arrive be $ x$. Then the time $ y$ for the other mathematician must satisfy $ x\minus{}m<y<x\plus{}m$. Graph the inequality with $ y\equal{}y(x)$.

Or you can just reason it out.
If the first mathematician comes at $ 9$, then the next mathematician can come from $ 9$ to $ 9: m$.  Only when the first mathematician comes at $ 9: m$ does the graph start lifting itself off of the $ x$ axis.  This continues until you get to $ 10: (60\minus{}m)$, in which the graph starts hitting the top horizontal line, and you get the same graph as you did near the origin.

Draw a diagram. The undesirable area is just two 45-45-90 right triangles with leg length $\frac{60-m}{60}$, so they have total area $\left(\frac{60-m}{60}\right)^2$. We hence have $1-\left(\frac{60-m}{60}\right)^2 = \frac{2}{5}$. The only reasonable solution is $m=60-12\sqrt{15} \implies 60+12+15=\boxed{87}$.


What is the largest integer which cannot be expressed as $ 2008x\plus{}2009y\plus{}2010z$ for some positive integers $ x,y$, and $ z$?

[hide="Solution"]On a side note, I really liked this problem because I actually solved it correctly during PUMaC, and I'll be presenting my solution as I solved it during the competition, even though it probably isn't the best solution.

Consider the possible values of $ 2008x\plus{}2009y\plus{}2010z$ when $ x\plus{}y\plus{}z\equal{}1$. The possible values are $ 2008$ through $ 2010$. In general, when $ x\plus{}y\plus{}z\equal{}a$, $ 2008x\plus{}2009y\plus{}2010z$ takes on values from $ 2008a$ through $ 2010a$. Therefore, we desire the value of $ a$ such that $ 2010a < 2008(a\plus{}1)$, but $ 2010(a\plus{}1) \ge 2008(a\plus{}2)$. $ 2010a < 2008(a\plus{}1)$ implies that $ a < 1004$. We note that $ 2010 \times 1004 \equal{} 2008 \times 1005$. Therefore, the two ranges that do not overlap are $ 2008\times 1003$ to $ 2010\times 1003$ and $ 2008 \times 1004$ through $ 2010 \times 1004$. The maximum number in the first range is $ 2016030$. The minimum number in the second range is $ 2016032$. Therefore, the answer is $ \boxed{2016031}$.

[/hide]

[quote="darkdieuguerre"][hide="Solution"]On a side note, I really liked this problem because I actually solved it correctly during PUMaC, and I'll be presenting my solution as I solved it during the competition, even though it probably isn't the best solution.

Consider the possible values of $ 2008x \plus{} 2009y \plus{} 2010z$ when $ x \plus{} y \plus{} z \equal{} 1$. The possible values are $ 2008$ through $ 2010$. In general, when $ x \plus{} y \plus{} z \equal{} a$, $ 2008x \plus{} 2009y \plus{} 2010z$ takes on values from $ 2008a$ through $ 2010a$. Therefore, we desire the value of $ a$ such that $ 2010a < 2008(a \plus{} 1)$, but $ 2010(a \plus{} 1) \ge 2008(a \plus{} 2)$. $ 2010a < 2008(a \plus{} 1)$ implies that $ a < 1004$. We note that $ 2010 \times 1004 \equal{} 2008 \times 1005$. Therefore, the two ranges that do not overlap are $ 2008\times 1003$ to $ 2010\times 1003$ and $ 2008 \times 1004$ through $ 2010 \times 1004$. The maximum number in the first range is $ 2016030$. The minimum number in the second range is $ 2016032$. Therefore, the answer is $ \boxed{2016031}$.

[/hide][/quote]
Since x,y, and z are positive integers $ x\plus{}y\plus{}z$ can't be 1 right?

True; the lower bound is then $ x\plus{}y\plus{}z\equal{}3$.


Find max. numbers $A$ wich is true ineq.:

$\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{x^{2}+z^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}\geq A$

$x,y,z$ are positve reals numberes! :wink:

[b]Romania Team Selection Test 1993:[/b]
Find the greatest real numbers $A$ such that $\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{x^{2}+z^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}> A$  is true  for all positve reals numberes $x,y,z$ .
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=town&t=24768]Osh town olympiad:[/url]
For $a,b$ and $c$ positive reals prove that\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=293319#p293319
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=479664]Macedonia,1995[/url]

For $a,b$ and $c$ positive reals prove that\[\sqrt{\frac{a}{b+c}}+2\sqrt{\frac{b}{c+a}}+4\sqrt{\frac{c}{a+b}}\leq \sqrt{7\left(\frac{a}{b+c}+\frac{2b}{c+a}+\frac{4c}{a+b}\right)} .\]
\[\sqrt{\frac{a}{b}}+2\sqrt{\frac{b}{c}}+3\sqrt{\frac{c}{a}}\leq \sqrt{6\left(\frac{a}{b}+\frac{2b}{c}+\frac{3c}{a}\right)} .\]
( Daniel Sitaru)

[quote=ytChen][quote=sqing]Let $a_1,a_2,\cdots,a_n $ $(n\ge 2)$ be positive real numbers. Prove that$$\sqrt{\frac{a_1}{a_2+a_3}}+2\sqrt{\frac{a_2}{a_3+a_4}}+\cdots+2^{n-2}\sqrt{\frac{a_{n-1}}{a_n+a_1}}+2^{n-1}\sqrt{\frac{a_n}{a_1+a_2}}\leq \sqrt{(2^n-1)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)}.$$$$\sqrt{\frac{a_1}{a_2}}+2\sqrt{\frac{a_2}{a_3}}+\cdots+(n-1)\sqrt{\frac{a_{n-1}}{a_n}}+n\sqrt{\frac{a_n}{a_1}}\leq \sqrt{\frac{n(n+1)}{2}\left(\frac{a_1}{a_2}+\frac{2a_2}{a_3}+\cdots+\frac{(n-1)a_{n-1}}{a_n}+\frac{na_n}{a_1}\right)}.$$[/quote]

[b][i]Solution[/i][/b]. Since that $a_1,a_2,\cdots,a_n $ $(n\ge 2)$ are positive real numbers, the Cauchy-Schwarz's inequality gives that
\begin{align*}&(2^n-1)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)\\
=&\left(1+2+\cdots+2^{n-1}\right)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)\\
\ge&\left(\sqrt{\frac{a_1}{a_2+a_3}}+2\sqrt{\frac{a_2}{a_3+a_4}}+\cdots+2^{n-2}\sqrt{\frac{a_{n-1}}{a_n+a_1}}+2^{n-1}\sqrt{\frac{a_n}{a_1+a_2}}\right)^2,
\end{align*}which yields the desired inequality immediately. The second inequality can be shown likewise. $\blacksquare$[/quote]
Thank you very much.


What's the difference between nSpire and Ti-84 Silver? I just got one from ARML.

If it comes with an interchangeable keypad, then the short answer is that TI-NSpire = TI-84 + other fancy stuff.

NSpire is more like TI-89 + fancy stuff, because it can do symbolic computations (which the TI-84 cannot). That is, it can solve equations involving multiple variables, solve in terms of variables and other symbols, give exact answers like $ \sqrt{2}$ instead of $ 1.414...$, etc.

Only the CAS can do symbolic computations, so a normal N-spire keyboard is pretty much an 84 except with a different layout and a few more features.

Oh, it's not an NSpire CAS. Well, I feel much better about not winning one then :D


Could you solve this "simple" equation for me?
$ 2^{\frac{1}{x^2}}\plus{}2^x\equal{}3$


Let $ X_n$ be a sequence of real numbers such that the sequences $ |x_n \plus{} 1|$ and $ |x_n \minus{} 1|$ have limits.Prove that $ |x_n|$ has a limit.

We will use $ a_n\rightarrow L$ to denote $ lim_{n\rightarrow \plus{} \infty}a_n \equal{} L$.

Let's define $ L$ with $ |x_n \plus{} 1|\rightarrow L$ and $ K$ with $ |x_n \minus{} 1|\rightarrow K$.

First, we will prove that $ x_n \plus{} 1$ can have at most two limit points: $ L$ and $ \minus{} L$. Suppose it has a limit point other than $ L$ or $ \minus{} L$. We will denote this point with $ \alpha$. Then we have a subsequence $ X_{p_n}$ such that $ x_{p_n} \plus{} 1\rightarrow \alpha$. Then, clearly $ |x_{p_n} \plus{} 1|\rightarrow |\alpha|$. However, $ |x_{p_n} \plus{} 1|\rightarrow L$ and, as sequences cannot have two limits, $ L \equal{} |\alpha|$, which is in obvious contradiction with $ \alpha\neq L \wedge \alpha\neq \minus{} L$. By analogy, $ x_n \minus{} 1$ can have at most two limit points: $ K$ and $ \minus{} K$.

Now, we can easily conclude that $ x_n$ can have at most two limit points: $ L \minus{} 1$, $ \minus{} L \minus{} 1$ (from $ x_n \plus{} 1$). It is obvious that, if it has only one limit point, it is indeed convergent. Let us suppose it has two limit points. Since from $ x_n \minus{} 1$ we concluded that $ K \plus{} 1$ and $ \minus{} K \plus{} 1$ are those limit points of $ x_n$, it is clear that one limit point involving $ K$ most be the same as one of the limit points involving $ L$ and that the other limit point involving $ K$ is the same as other limit point involving $ L$. We therefore have two cases.

First, let us examine $ L \minus{} 1 \equal{} K \plus{} 1$, $ \minus{} L \minus{} 1 \equal{} \minus{} K \plus{} 1$. We get $ L \minus{} K \equal{} 2$ and $ L \minus{} K \equal{} \minus{} 2$, which is clearly a contradiction. This leaves us with our second case: $ L \minus{} 1 \equal{} \minus{} K \plus{} 1$, $ \minus{} L \minus{} 1 \equal{} K \plus{} 1$. This leads to $ L \plus{} K \equal{} 2$ and $ L \plus{} K \equal{} \minus{} 2$, also an obvious contradiction.

So, $ x_n$ has indeed only one limit point (as it must have at least one limit point). This leads to final result - $ x_n$ is convergent. Obviously, this means $ |x_n|$ is also convergent.

[size=75]This proof lacks some rigor and it can indeed contain mistakes, so, please, read it carefully and make sure everything is correct. I'm also somewhat surprised with the final result - it not only seems to solve the problem, it gives a stronger property for $ x_n$.[/size]
[size=59]Edit: just adding my surprise at the result.[/size]
[size=59]Edit no.2: Spelling. D'oh![/size]

What about the following -
$ |x_n \plus{} 1| \to L_1, |x_n \minus{} 1| \to L_2 \implies x_n^2 \plus{} 2x_n \plus{} 1 \to L_1^2, x_n^2 \minus{} 2x_n \plus{} 1 \to L_2^2$
$ \implies (x_n^2 \plus{} 2x_n \plus{} 1) \plus{} (x_n^2 \minus{} 2x_n \plus{} 1) \equal{} 2(x_n^2 \plus{} 1) \to L_1^2 \plus{} L_2^2$
$ \implies x_n^2 \to \frac {L_1^2 \plus{} L_2^2 \minus{} 2}{2}$
$ \implies |x_n| \to \sqrt {\frac {L_1^2 \plus{} L_2^2 \minus{} 2}{2}}$

Actually, the convergence $ x_n$ follows even more easily:
$ (x_n^2 \plus{} 2x_n \plus{} 1) \minus{} (x_n^2 \minus{} 2x_n \plus{} 1) \equal{} 4x_n \to L_1^2 \minus{} L_2^2$
$ \implies x_n \to \frac {L_1^2 \minus{} L_2^2}{4}$


Find the number of permutations of {1,2,...,n} with no consecutive integers. e.g. both ...,1,2,... and ...,2,1,... are invalid.
i think i saw this problem on this forum sometime ago, but can't find it now, can anyone help?

I believe this is the sequence you are looking for:
http://www.research.att.com/~njas/sequences/A078603
The encyclopaedia does not have a general formula that I see.

For the half-sequence, [url=http://www.research.att.com/~njas/sequences/A002816]A002816[/url], the encclopaedia gives a recursion, though.

no they are different, the one u gave considers that 1 is adjacent to n, but mine doesn't. the first 5 terms for urs are 1,0,0,2,6 and for mine are 1,0,0,2,14.

In that case, [url=http://www.research.att.com/~njas/sequences/A002464]this (A002464)[/url] is the sequence you're looking for, with recursion and generating function.

Right, that's the same link maxal gave, thanx a lot anyway! But i'm still wondering how to get the recursion in that link. Does anyone know a combinatorical interpretation of that formula?

Hi, the general formula is:

$\sum_{r=0}^{n-1} \sum_{c=0}^{r}{(-1)^r 2^c (n-r)! \binom{n-r}{c} \binom{r-1}{c-1}}$

This formula use the inclusion/exclusion principle.


calculus
integration
arithmetic sequence

The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length

$\text{(A)}\ 22 \qquad \text{(B)}\ 58 \qquad \text{(C)}\ 81 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 361$

[hide]A Pythagorean triple that forms an arithmetic sequence is the 3-4-5 triangle.  So, we need a length that is divisible by either 3, 4, or 5.  The only one out of the five is 81.  $\boxed{C}$.

[/hide]

[quote="math92"][hide]
the answer is D because 91 is a multiple of 7 and there is a triple that is 7,24,25[/hide][/quote]

But the triple 7, 24, 25 isn't an arithmetic progression.

[hide]
Assume that the smallest leg is x, the other leg is x+d, and the hypotenuse is x+2d. Then,
x^2+(x+d)^2 = (x+2d)^2
2x^2+2dx+d^2 = x^2+4dx+4d^2
To make the left side a perfect square, we subtract (x^2+4dx) from both sides:
x^2-2dx+d^2 = 4d^2 or (x-d)^2 = 4d^2
Taking the square root of both sides, we get,
x-d = 2d or x = 3d
Since all sides have integer lengths, d must be an integer. This means that the smallest leg (x) must be divisible by 3. The only number in the list divisible by 3 is 81, and thus it is the answer. [/hide]


Let AB be a diameter of a circle and C, D two points in the circle such that CD is not a diameter. AC and BD intersect in L and the tangents to the circle at C and D meet at N. Prove that LN is perpendicular to AB.

The proof is very easy using polar relationship:

Let k be the circle with diameter AB, and O be the center of this circle, i. e. the midpoint of the segment AB. Also, let P be the point of intersection of the lines AB and CD. In the following, "pole" and "polar" will always mean pole and polar with respect to the circle k.

We will prove that the line LN is the polar of the point P. Then, it will follow that $LN \perp OP$ (since O is the center of the circle k), or, in other words, $LN \perp AB$, and the problem will be solved.

In order to show the line LN is the polar of the point P, we must prove two things:

[b]a)[/b] The polar of the point P passes through the point L.
[b]b)[/b] The polar of the point P passes through the point N.

Well, since the point N is the point where the tangents to the circle k at the points C and D meet, it follows that the points C and D are the points where the tangents from N to the circle k meet this circle k. And hence, the line CD is the polar of the point N. In other words, the point N is the pole of the line CD. Now, since the point P lies on the line CD, the polar of the point P passes through the pole of the line CD, i. e. through the point N. This proves assertion [b]b)[/b].

It is well-known that for any four points A, B, C, D on a circle, the point of intersection of the lines AC and BD lies on the polar of the point of intersection of the lines AB and CD with respect to that circle. Applying this to the four points A, B, C, D from our problem, we conclude that the point of intersection L of the lines AC and BD lies on the polar of the point of intersection P of the lines AB and CD. Thus, the polar of the point P passes through the point L. This proves assertion [b]a)[/b], and thus, the solution is complete.

  Darij

Using the Pascal 's theorem for 4 points.
  If AB is not a diameter, then LN will pass the intersection of the two tangents of (O) at A and B. Fix A and let B approach the point such that AB is a diameter. Therefore we will obtain LN will be parallel to the two tangent at A and B of (O) which implies LN is perpendicular to AB.
       This proof in fact is something like polar too since Pascal 's theorem for 4 points can be proved by polar.

We assume that $C$ is closer to $A$ than $D$ is to $B$. We consider two cases-

(1) $C$ and $D$ are on two different sides of $AB$.
Like the above post apply Pascal's Theorem for the degenerate cyclic hexagon $ACCBDD$. We have that $T=BC\cap LN$ is collinear with the points $C$ and $D$. As $AB$ is the diameter we have that $\angle LCB=\angle ADB=90^{\circ}$. Thus $A$ is the orthocenter of $LTB$.
Hence $AB\perp LN$.

(2) $C$ and $D$ are on the same side of $AB$.
Again applying Pascal's theorem on the same hexagon we have that $U=AD\cap BC$ is collinear with $L$ and $N$. As $AB$ is the diameter we get that $\angle ACB=\angle ADB=90^{\circ}$. Hence $U$ is the orthocenter of $ALB$.
Thus $LN\perp AB$.

Dear Mathlinkers,
just for history, the case of Pascal's theorem with a degenerated hexagon with two tangents comes from MacLaurin.
Sincerely
Jean-Louis


algebra
polynomial
superior algebra
superior algebra unsolved

Show that for every nonprime integer $n \geq 4$, there is a polynomial $P$ of degree $n$ such that $P$ is irreducible in $\mathbb{Z}[X]$ but $P$ is reducible in $\mathbb{Z}/p\mathbb{Z}[X]$ for every prime $p$.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=64273 for the beginnings ;)


function
algebra
polynomial
trigonometry
rational function
complex numbers
algebra unsolved

I don't know if someone has created this problem before.

Does there exist a period function f: N->N with odd period,which includes addition, subtraction, multiplication, division or power only? (Please don't use complex number)

Do you mean does there exist a periodic polynomial or rational function? If so - No. If you meant some other question, then please clarify your point.

Any function with only addition, subtraction, multiplication, division or power, such as f(x)=(-1)^x .

Interesting question. $f(x)=(-1)^x$ is periodic, but its period is 2, which is even. We're looking for a function with an odd period.

[quote="Rust"]All periodic function $f: N\to N$ defined function $g: (Z/TZ)\to N$, and all function $g: (Z/TZ)\to N$ give periodic function $f(n)=g(n(mod \ T))$.[/quote]
I'd say using "$mod$" is not allowed.

Let's clarify the rules:
It should be a function with only addition, subtraction, multiplication, division or power. Is it allowed to use infinite sums or multiplications?
The expression is not allowed to contain complex constants (would it help that much?). Is the function allowed to contain real constants, or just natural numbers?

[quote="ZetaX"]I just want to say that complex ones help: $\zeta_n^k + \zeta_n^{-k}$ as function of $k$ has period $n$ ($\zeta_n$ being a primitive $n$-th root of unity).[/quote]
But then it isn't a $f: \mathbb{N}\to\mathbb{N}$, not because it contains complex numbers, but because its result isn't a natural number.

On the other side, the Fibonacci sequence can be written $fib(n)=\frac{a^n-b^n}{a-b}$ where $a,b\in\mathbb{R}$ are the two real roots of $x^2=x+1$, and
I know the Fibonacci sequence isn't periodic, but it's a good example to show that we can use non-natural real numbers in a expression which result is always a natural number.
I should have written:
The function $f: \mathbb{N}\to\mathbb{N}$ is not allowed to contain complex constants (as modeler stated, but would it help that much?, it would be difficult for them to give only natural numbers as a result). Is the function $f: \mathbb{N}\to\mathbb{N}$ allowed to contain real constants, or just natural numbers?

My approach works (to restrict the image to integers should be nonsense, too)...
It's totally neglectible if the result is an integer or not, just kill the irrational (algebraic) part by some more terms; this can e.g. be done by taking norms (and $n=2,3,4$ give integers already).
This is the same effect as your Fiboniacci numbers, with the difference that image set is finite.

Proof that we have just to disprove the existence of one special given function:
Let $f: \mathbb{N} \to \mathbb{R}$ be periodic of the desired type with minimal period $n$.
Claim: for any $a_1 , a_2 , ... , a_n \in \mathbb{R}$ (or complex when allowed to be) there is a function $h$ of the desired type with $h(k) = a_k$ and period $n$.
Proof:
[hide="Not to be viewed by young children or the faint at heart"]
Take $h(x) = \frac{1}{g(0)} \cdot \sum_{k=1}^{n} a_k g(x+n-k)$ with:
\[ g(x) = \prod_{k=1}^n \prod_{\substack{ y\in \{ 1,2,...,n\} \\ f(y) \neq f(k)}} \left( f(x+k) - f(y) \right) \]
This clearly works since $g$ is periodic with period $n$ and fulfills $g(0) \neq 0$ and $g(x)=0$ for $x=1,2,...,n-1$.
The first one more or less by definition, the second one since other wise we would have $f(x+k) \neq f(y)$ for all $k$ and $f(y) \neq f(k)$, thus $f(x+k)=f(k)$ for all $k$. But the last one is not possible since that means that $x<n$ is a period of $f$.
[/hide]
So we can for example assume that the sequence is given by $\zeta^n + \zeta^{-n}$.

Thank you for your wonderful proof, although I don't really understand it.
By the way, I would like to ask how you reject the case that the function includes some trigo functions, it seems that your proof works for all functions.

What part is not understandable¿ :)
But till know we are left to show that a term of type $\zeta_n^{k}+\zeta_n^{-k}$ can't be expressed with the given conditions (they can expressed with e.g. $\cos$ !).


Find the number of ways of arranging the digits [b]1,2,...,n[/b]

such that: no pair of successive integers are adjacent.

Let $ \{c_1,\dots,c_n\}$ be a permutation of $ \{1,\dots,n\}$ with $ |c_i\minus{}c_{i\plus{}1}|>1 \quad \forall 1\leq i\leq n\minus{}1$. Let $ f(n)$ be number of all such permutations and $ r_k$ denote the number of such permutations in which $ r$ appears in $ k$th place.

$ f(n)\equal{}n_1\plus{}\dots\plus{}n_n$

We may construct the recursion,

$ (n\plus{}1)_1\equal{}f(n)\minus{}n_1$
$ (n\plus{}1)_2\equal{}f(n)\minus{}(n_1\plus{}n_2)$
$ \dots$
$ (n\plus{}1)_k\equal{}f(n)\minus{}(n_{k\minus{}1}\plus{}n_k)$
$ \dots$
$ (n\plus{}1)_{n\plus{}1}\equal{}f(n)\minus{}n_n$

$ \Rightarrow  f(n\plus{}1)\equal{}(n\minus{}1)f(n); \quad f(4)\equal{}3$

Srikanth, are you sure that $ (n \plus{} 1)_k \equal{} f(n) \minus{} (n_{k \minus{} 1} \plus{} n_k)$ ? What about a permutation like $ \left(...,3,n \plus{} 1,2,...\right)$ (with $ n \plus{} 1$ on the $ k$-th place)? It certainly occurs on the left hand side, but it does not on the right, because $ \left(...,3,2,...\right)$ is not a valid permutation (since $ 3$ and $ 2$ are adjacent).

[b]EDIT:[/b] For the problem, see also http://www.mathlinks.ro/Forum/viewtopic.php?t=65890 .

  darij

@Darij: As you have pointed out we miss some possibilities on right side. So my previous post only gives a lower bound for $ f(n)$.

It is sequence [url=http://www.research.att.com/~njas/sequences/A002464]A002464[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] -- there's a not-very-nice recursion and an even-less-nice summation.

Sequence [url=http://www.research.att.com/~njas/sequences/A001100]A001100[/url] gives the number of permutations with exactly $ k$ pairs of successive adjacent pairs.

Other related questions:
1) What if we consider $ 1$ and $ n$ to be successive?
2) What if we consider the first and last members of our permutation to be adjacent?
3) What if we do both (1) and (2)?

Here is a system of recursive equations developing from my previous post. The recursions look quite entangled, hope somebody can help me and we may prove the recursion stated on the oeis page (as Joel pointed out).

Let $ g(n,k)$ be the number of permutations of $ \{1,\dots,n\}$ with exactly $ k$ pairs such that $ c_i\minus{}c_{i\plus{}1}\equal{}1$

$ h(n,k)$ be the number of permutations of $ \{1,\dots,n\}$ with exactly $ k$ pairs such that $ c_i\minus{}c_{i\plus{}1}\equal{}1$ and the unordered pair $ (n\minus{}1,n)$ does  appear.

We have the recursion,

$ f(n)\equal{}(n\minus{}2)f(n\minus{}1)\plus{}g(n\minus{}1,1)\minus{}h(n\minus{}1,1)$

$ (n\minus{}2)f(n\minus{}1)$ is derived from my previous post. $ g(n\minus{}1,1)\minus{}h(n\minus{}1,1)$ accounts to to possibilities I missed as pointed out by Darij. We are going to insert $ n$ ( into a permutation with exactly one pair) between the pair. This gives $ g(n\minus{}1,1)$. We remove the possibilities when the unordered pair was $ (n\minus{}2,n\minus{}1)$ as this would give rise to the unordered pair $ (n\minus{}1,n)$ inevitably.

We can find recursions for $ g$ as follows,

$ g(n\minus{}1,1)\equal{}2g(n\minus{}2,2)\minus{}h(n\minus{}1,2)$
$ g(n\minus{}2,2)\equal{}3g(n\minus{}3,3)\minus{}h(n\minus{}2,3)$
$ \dots$
$ g(n\minus{}k,k)\equal{}(k\plus{}1)g(n\minus{}k\minus{}1,k\plus{}1)\minus{}h(n\minus{}k,k\plus{}1)$
$ \dots$
until $ g(r,r\minus{}1)\equal{}2$ appears.

Let me explain the first step and the rest analogously follow. We insert $ n\minus{}1$ ( into a permutation counted by $ g(n\minus{}2,2)$) between one pair destroying exacly one pair among the two. We subtract $ h(n\minus{}1,2)$ to remove the possibilities when the unordered pair $ (n\minus{}2,n\minus{}1)$ is created during the process.

We can find the recursions of $ h$ as follows,

$ h(n,1)\equal{}2f(n\minus{}1)$
$ h(n,2)\equal{}h(n\minus{}1,2)\plus{}h(n\minus{}1,1)\plus{}2g(n\minus{}1,1)$
$ \dots$
until $ h(r,r\minus{}1)\equal{}2$ appears.

First step is trivial, inserting $ n$ on either sides of $ n\minus{}1$. Second step is obtained as follows. We insert $ n$ between the unordered pair $ (n\minus{}2,n\minus{}1)$ (in a permutation counted by $ h(n\minus{}1,2)$) creating the unordered pair $ (n\minus{}1,n)$. $ h(n\minus{}1,1)$ is added to take care of possibilities when we have both unordered pairs $ (n\minus{}2,n\minus{}1)$ and $ (n\minus{}1,n)$.$ 2g(n\minus{}1,1)$ counts the possibility of creating the unordered pair $ (n\minus{}1,n)$ without destroying a unordered pair.

I donot know how to go ahead from here.


function
algebra
polynomial
complex analysis
binomial theorem

Does $ z \cdot e^{1/z} \cdot e^{\minus{}1/z^2}$ have essential singularity at 0? Why or why not.

A function has an essential singularity at $ a$, iff it's Laurent expansion around $ a$ has arbitrary small nonzero coefficients (so infinitely many coefficients of negative index). Easy calculation shows that this is true here.

What can be said about the general expression:

$ h(z) \equal{} f(z)e^{g(z)}$?

I don't know.  I suspect if $ g(z)$ is meromorphic, then $ h(z)$ will have an essential singularity at each singularity of $ g(z)$.  I think maybe this is also the case for any non-polynomial entire function, that is:

$ h(z) \equal{} E(g(z))$ has an essential singularity if $ g(z)$ is meromorphic and $ E(z)$ is a non-polynomial entire function.  Not sure though.

I suspect further there is no way to eliminate or even qualitatively affect the essential singularity of $ h(z)$ by multiplication with another function $ f(z)$ unless it too has an essential singularity at the same spot as $ h(z)$ and so I think the $ z$ term had no effect on the singular nature of the function above.  Kindly correct if any of my suspicions are wrong.

[edit]
Also:

If $ f(z) \equal{} ze^{1/z}e^{ \minus{} 1/z^2}$ then:

$ f(z) \equal{} \mathop\sum\limits_{n \equal{} 0}^{\infty}\frac {1}{n!}\mathop\sum\limits_{k \equal{} 0}^{n}\binom{n}{k}\frac {z^{n \minus{} k}}{z^{2n \minus{} 1}}( \minus{} 1)^k \equal{} 1 \plus{} z \plus{} \mathop\sum\limits_{m \equal{} 1}^{\infty}\frac {a_m}{z^m}$
(use expansion for Exp function then Binomial Theorem)
where:

$ a_m \equal{} \mathop\sum\limits_{k \equal{} 0}^{F(m)}\frac {( \minus{} 1)^k}{k!(m \plus{} 1 \minus{} 2k)!}$
(will need to work with some of the terms like m=10,11 to figure $ a_m$ out) with $ F(m) \equal{} m/2$ if $ m$ is even and $ F(m) \equal{} (m \plus{} 1)/2$ if odd.  This shows explicitly the coefficients are finite for all Laurent terms.  Note in particular:

$ \mathop\text{Res}\limits_{z \equal{} 0} f(z) \equal{} \mathop\sum\limits_{k \equal{} 0}^{1}\frac {( \minus{} 1)^k}{k!(m \plus{} 1 \minus{} 2k)!} \equal{} \minus{} 1/2$

It is much easier to recognize essential singularities by the absence of the limit (finite or infinite) of the function at the point. For the example in the original problem, just look at what happens when $ z$ approaches $ 0$ along the real and the imaginary axes.


geometry
incenter
inequalities unsolved
inequalities

Prove that for all acute triangles ABC with incenter I

$\frac{1}{IA}+\frac{1}{IB}+\frac{1}{IC} \geq \frac{1}{3R}+\frac{4}{3r}$

We have :
  $ IA = \frac{r}{sin\frac{A}{2}} $
  $ IB = \frac{r}{sin\frac{B}{2}} $
  $ IC = \frac{r}{sin\frac{C}{2}} $
  $ r = 4R sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} $
 The ineq is equivalent to
  $ sin\frac{A}{2} + sin\frac{B}{2} + sin\frac{C}{2} \geq \frac{4}{3}(1+sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}) $ , which is [b]Jack Garfulkel[/b] 's ineq


AMC
AIME
rotation
analytic geometry
graphing lines
slope
geometry

L, L' are the lines through the origin that pass through the first quadrant (x, y > 0) and make angles π/70 and π/54 respectively with the x-axis. Given any line M, the line R(M) is obtained by reflecting M first in L and then in L'. R^n(M) is obtained by applying R n times. If M is the line y = 19x/92, find the smallest n such that R^n(M) = M.

kalva says that for each R() operation we rotate M by π/70 + π/54 counterclockwise, but since the slope of M is so much greater than that of L and L', wouldn't we be rotating by π/54 - π/70 counterclockwise?

Well this is how I did the problem, and you seem to be right, the angle of rotation was not  :pi: /54+ :pi: /70 but rather 2(:pi: /54 - :pi: /70 )

This is what I did, Id imagine you did the same thing judging from your comments

[hide]

This is a bit hard to explain, (and very easy to see if drawn out), but lets call the original angle between M and L - A.  Reflecting over L, now the angle formed between L' and L is clearly A + (:pi: /54 - :pi: /70 ), then reflecting over L', the angle is still the same, but with respect to L, it is A + (:pi: /54 - :pi: /70 ) from L', which is another (:pi: /54 - :pi: /70 ) from L.  So it has rotated 2(:pi: /54 - :pi: /70 ) counter clockwise.

Solving

n*2(:pi: /54 - :pi: /70 )=k* :pi:

n*2*(16/(54*70))=k* :pi:

n*2(4)/(27*35)=k :pi:

the smallest value of n=27*35, and the line will rotate 8 :pi:  degrees,

n=945.

[/hide]

The thing I am wondering about, using (:pi: /54 - :pi: /70 ), I'd imagine you got the correct answer, so the solution I posted is more for me to me to understand the problem, rather than help. Still good practice for tomorrow = ).  And a question... i sorta feel my solution is wrong because it doesn't required the slope of the line, yet I think thats the right answer, do i have any problems here?

Yes, your answer is correct. Its what I got too. kalva seems to have gotten it too, except perhaps in the wrong way (or maybe I just didn't understand its solution).

[hide]Quote:now the angle formed between L' and L is clearly A + (  :pi: /54 -   :pi: /70 ), [/hide]

I think this is just a typo, since the angle between L' and L is constant, but no big deal.

hah, yeah, L should probably be the new line... L... line, same thing? I should stop writing "clearly"...

Without having gone and read it, there's probably just a typo on the Kalva site.


If $ a,b,c>0$ with $ a\plus{}b\plus{}c\equal{}1$ then prove that :
 \[ \frac{c\minus{}a}{\sqrt{a^2\plus{}3b}}\plus{}\frac{a\minus{}b}{\sqrt{b^2\plus{}3c}}\plus{}\frac{b\minus{}c}{\sqrt{c^2\plus{}3a}}\geq 0\]

[i]I dont have a solution for it or a counter example.[/i]

[quote="Svejk"]If $ a,b,c > 0$ with $ a \plus{} b \plus{} c \equal{} 1$ then prove that :
\[ \frac {c \minus{} a}{\sqrt {a^2 \plus{} 3b}} \plus{} \frac {a \minus{} b}{\sqrt {b^2 \plus{} 3c}} \plus{} \frac {b \minus{} c}{\sqrt {c^2 \plus{} 3a}}\geq 0
\]
[i]I dont have a solution for it or a counter example.[/i][/quote]

try a=0.4, b=0.6, c=0 :wink:


[hide]a-b+c-d+e-f+g-h+...=0.

a^2-b^2+c^2-d^2...=

(a+b)(a-b)+(c+d)(c-d)...

since a+b is always the same thing as c+d or e+f(everyone played equal amounts of games), then replace all c+d, e+f... with a+b.

(a+b)(a-b+c-d+e-f...)

a-b+c-d+e-f = 0.

so (a+b)(0) = 0[/hide]

[quote="JBL"]Wow, that definitely needs more explanation.  Like, for example, what do all of your letters stand for?  And why do you have apparantly endless sums?[/quote]

I have endless sums becuase you didnt specify how many teams there were.  team 1: a is wins b is losses.  team 2:c is wins d is losses. team 3:e is wins f is losses.

A good way to use notation here:

w1 = team 1's wins
l1 = team 1's losses
...
wn = team n's wins
ln = team n's losses

And then your sums can have an end, too.

Ah, and now that I understand what you wrote -- good job!

[hide]let T be total number of games

its equilvalent in proving

Sum (w_i^2-l_i^2)=0

LHS=Sum((w_i+l_i)(w_i-l_i))=T*Sum(w_i-l_i)

but as JBL said its trivial that Sum(w_i-l_i)=Sum(w_i)-Sum(l_i)=0 so LHS=0 and we are done

[/hide]

Not for say, a 1960's Putnam, which is when it would be from.  1964, maybe?  Not sure.  Long time ago.  Early question.  Not hard.

Are all the questions in the Putnam free response?

[quote="JBL"]This might be from an old Putnam, I just don't remember.  I may even have put it on here once before.  Anyhow:

Given a round robin tournament, in which each team plays every other exactly once.  We have trivially that the sum of the number of wins of each team equals the sum of the number of losses of each team.  Prove that the sum of the squares of the number of wins of each team equals the sum of the squares of the number of losses of each team.

For example, in a 4 team tournament in which finishing records are 3-0, 1-2, 1-2 and 1-2, we have
3 :^2: + 1 :^2: + 1 :^2: + 1 :^2: = 0 :^2: + 2 :^2: + 2 :^2: + 2 :^2:[/quote]

Yes. And each (of the twelve) questions is scored from 0 to 10.

The [url=kalva.demon.co.uk]kalva[/url] website has all of the Putnam exams (give or take a question here and there) if you're interested in taking a look.


Mr.Jensen's running shoes had 1 inch of tread when new. He has run $ 440$ miels in the shoes and has $ \frac{3}{8}$ inches of tread left. Assuming the tread wears at a constant rate, how many more miles can Mr. Jensen run before his shoes have no tread left?

Since he has $ \frac{3}{8}$ left, he used $ 1\minus{}\frac{3}{8}\equal{}\frac{5}{8}$ to run the $ 440$ miles. Calling $ x$ the number of miles he can run, $ \frac{5}{8}x\equal{}440 \implies x\equal{}704$. So, the answer is $ 704\minus{}440\equal{}\boxed{264}$.


geometry
circumcircle
power of a point
radical axis
geometry unsolved

An acute angled triangle ABC has orthocentre H.
The circle passing through H with centre the midpoint of BC intersects the line BC at A1 and A2. Similarily, the circle passing through CA intersects the line at B1 and B2 and the same with line AB at points C1 and C2.
Prove that the six points, A1 A2 B1 B2 C1 C2 lie on a circle.

Someone randomly gave me this problem and I dunno if this is ambiguously worded, or if it's a well-known problem.

The generalization of this fact is called the Droz-Farny theorem.
It states:
Let $ P$ and $ Q$ be isogonal points in triangle $ ABC$. $ P_a$ is the foot of perpendicular from $ P$ to $ BC$. $ A_1$ and $ A_2$ are the points of intersection of $ BC$ and the circle with the center $ P_a$ and of radius $ P_aQ$. Analogously we define $ B_1,B_2,C_1,C_2$. Then the points $ A_1,A_2,B_1,B_2,C_1,C_2$ lie on the same circle. This theorem is well-known but in case of $ P\equal{}O$, and $ Q\equal{}H$ it takes the simplest form :)
I know a short prove, and if you are intrested in I could post it.

A sketch to prove the initial problem:
See that $AH$ is the radical axis of the circles centered on $AB$, $AC$, etc.

Best regards,
sunken rock

Let $D,E$ be midpoints of $BC,AC$, then $DE\perp CH$ so $CD^2-DH^2=CE^2-EH^2$. Let $O$ be the circumcentre with circumradius $R$. We shall show that $O$ is the circumcentre of $A_1A_2B_1B_2C_1C_2$. Clearly $OA_1=OA_2$. Now we show $OA_1=OB_1$ and we are done. Note that $OA_1^2=R^2-DC^2+DA_1^2=R^2+DH^2-DC^2=R^2+EH^2-CE^2=OB_1^2$.


$ x,y,z$ are positive real numbers and $ xyz\equal{}1$. Prove that
     $ \sqrt{\frac{x}{x\plus{}8}}\plus{}\sqrt{\frac{y}{y\plus{}8}}\plus{}\sqrt{\frac{z}{z\plus{}8}}>\equal{}1$.

[quote="lasha"]$ x,y,z$ are positive real numbers and $ xyz \equal{} 1$. Prove that
     $ \sqrt {\frac {x}{x \plus{} 8}} \plus{} \sqrt {\frac {y}{y \plus{} 8}} \plus{} \sqrt {\frac {z}{z \plus{} 8}} > \equal{} 1$.[/quote]

I think this problem is old and not hard  :P
Put $ a\equal{}\frac1{x},b\equal{}\frac1{y},c\equal{}\frac1{z}$,we have  $ abc\equal{}1$

ineq <=>$ \frac1{\sqrt{1\plus{}8a}}\plus{}\frac1{\sqrt{1\plus{}8b}}\plus{}\frac1{\sqrt{1\plus{}8c}}\ge 1$

<=>$ \sum\sqrt{(1\plus{}8a)(1\plus{}8b)}\ge \sqrt{(1\plus{}8a)(1\plus{}8b)(1\plus{}8c)}$

<=>$ (\sum\sqrt{(1\plus{}8a)(1\plus{}8b)})^2\ge (1\plus{}8a)(1\plus{}8b)(1\plus{}8c)$

<=>$ 4(a\plus{}b\plus{}c)\plus{}\sqrt{(1\plus{}8a)(1\plus{}8b)(1\plus{}8c)}(\sqrt{1\plus{}8a}\plus{}\sqrt{1\plus{}8b}\plus{}\sqrt{1\plus{}8c}) \ge 255$

By AM-GM we easy prove:

$ a\plus{}b\plus{}c\ge3$
$ (1\plus{}8a)(1\plus{}8b)(1\plus{}8c)\ge 729$  and
$ \sqrt{1\plus{}8a}\plus{}\sqrt{1\plus{}8b}\plus{}\sqrt{1\plus{}8c}\ge 3\sqrt[6]{729}\equal{}9$
so we are done! :wink:

Very famous inequality :P

Just substitute $x=\frac{a^{2}}{bc} ; y=\frac{b^{2}}{ca} ; z=\frac{c^{2}}{ab}$

So we have to show that $\sum \frac{a}{\sqrt{a^{2}+8bc}}\geq1$

Now see here: https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_2


Απιστευτο και ομως αληθινο.Η παρακατω ασκηση μας δοθηκε στο σχολειο και η εκφωνηση  της δειχνει απλη,αλλα τα φαινομενα απατουν.

      Αν σε ενα τριγωνο τα μηκη δυο εσωτερικων διχοτομων ειναι ισα δειξτε οτι ειναι ισοσκελες.

     Ο  Κ. Βηττας θεωρει οτι ειναι ενα πολυ δυσκολο προβλημα και προτεινε να το βαλω για ανταλλαγη σκεψεων.

Aσχετος ειμαι στη γεωμετρια αλλα θυμαμαι περσυ ενας μαθηματικος μου ειχε αναφερει το προβλημα και μου ειχε πει οτι ειναι πολυ δυσκολο. Το προσπαθησα χωρις επιτυχια, και τελικα δεν τον ρωτησα τη λυση, οποτε θα με ενδιεφερε και εμενα να την δω. Απο' οτι καταλαβα απο γνωσεις θελει μονο μεχρι και ισοτητα τριγωνων... :maybe:

ΤΟ πρόβλημα πράγματι θεωρείται δύσκολο για σχολική άσκηση, αλλά δεν θα έλεγα ότι είναι πάνω από επίπεδο ευκλείδη (και πολύ λέω...). Μάλιστα απ' ότι θυμάμαι έχει πέσει και σε διαγωνισμό της ΕΜΕ!
Για τη λύση αν $ ABC$ το τρίγωνο και $ BD$,$ CE$ οι ίσες διχοτόμοι έχουμε:
Εστω ότι $ \angle{B}>\angle{C}$ τότε αν πάρουμε σημείο $ X$ στην $ EC$ ώστε $ \angle{DBX}\equal{}C/2<B/2$ το Χ θα είναι ενδιάμεσα στα E και C και $ BCDX$ εγγράψιμο. Oμως $ \angle{XBC}>BCD$ άρα κι οι αντίστοιχες χορδές θα είναι $ BD<CX<CE$ που είναι άτοπο από υπόθεση :wink:

Δεν την είχα ξαναδεί τη λύση αυτή mim. Υπάρχουν καμιά 50αρια λύσεις αλλά όλες σχεδόν ψιλομετρικές με Stewart και τριγωνομετρία πολλές από αυτές, :wink:  αλλά καμιά τόσο elementary.

ontos to provlhma einai gnosto kai theoreitai arketa duskolo gia sxoliki askisi.

poios sas kanei o geometria sto sxoleio? (duo periptoseis uparxoun...o kathigitis einai polu kalos kai kanei polu sovarh douleia kai dinei prokliseis sta paidia h o kathigitis einai psilo kobleksikos kai thelei na kanei epideiksi isxuos me apotelesma oi mathites anti na misisoun mono auton na katalhgoun na misoun kai ti geometria...tespa elpizo na mhn einai to deutero!)

Δεν ηξερα οτι η ασκηση ειναι γνωστη και απο παλιο διαγωνισμο της ΕΜΕ.Πολυ ωραια η λυση του bilot (Δημητρης ?).Εχω μια λυση επισης με
 εις ατοπο απαγωγη και μια με τριγωνομετρια.Θα τις ανεβασω αυριο μηπως και εμφανιστουν και αλλες ιδεες.

nick οι ΒΔ,ΓΕ ειναι διχοτομοι.Ετσι η ΙΕ δεν ειναι καθετη στην ΑΒ.Μην συγχεεις το ιχνος της διχοτομου με το σημειο επαφης της πλευρας ΑΒ με τον εγγ.κυκλο.

[quote="nick.cha"]Δεν ειμαι και πολυ σιγουρος για αυτη αλλα τεσπα.
Εστω ΑΒΓ το τριγωνο, ΒΔ και ΓΕ οι διχοτομοι και Ι το εγγεντρο.
Αρχικα θελω νδο τα τριγωνα ΑΒΙ και ΑΓΙ ειναι ισα. Εχω τις γωνιες Α1=Α2 και ΑΙ κοινη. Αρα αρκει νδο ΒΙ=ΓΙ.
Ειναι ΙΕ καθετο ΑΒ, ΙΔ καθετο ΑΓ(αφου Ι εγγεντρο).
Ευκολα παιρνω οτι τα ορθογωνια τριγωνα ΑΙΕ=ΑΔΙ, αρα ΙΕ=ΙΔ.
Αφου ΒΔ=ΓΕ(υποθεση), ΔΙ=ΕΙ με αφαιρεση ΒΙ=ΓΙ.
Αν ειναι σωστη νομιζω πως ειναι η πιο απλη λυση που μπορει να δωθει με ισοτητες τριγωνων.
Ελπιζω να βοηθησα!!!
 :wink:[/quote]
ΟΚ ακυρο αυτο. Για αυτο κατι δεν μου πηγαινε καλα απο την αρχη.   :oops:

Δημητρη πολυ ωραια λυση!
Εγω την ελυσα ετσι,
πηρα,$ t \equal{} \frac {a \plus{} b \plus{} c}{2}$
και
$ d_a \equal{} d_b$
 $ \leftrightarrow \frac {2bc}{b\plus{}c} \sqrt {\frac {t(t \minus{} a)}{bc} } \equal{} \frac {2ac}{a \plus{} c} \sqrt {\frac {t(t \minus{} b)}{ca} }$
$ \leftrightarrow(a \minus{} b)( ac^2 \plus{} bc^2 \plus{} 3abc \plus{} c^3 \plus{} ab^2 \plus{} a^2b) \equal{} 0$
$ \leftrightarrow a \equal{} b$

ΑΠΟΔΕΙΞΗ οτι $ d_a \equal{} \frac {2bc}{b \plus{} c} \sqrt {\frac {t(t \minus{} a)}{bc} }$
[hide]- Ισχυει
$ 2cos^2{\frac {A}{2}} \minus{} 1 \equal{} cos{A} \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc}$
$ \leftrightarrow cos^2{\frac {A}{2}} \equal{} \frac {t(t \minus{} a)}{bc}$(1)
και,
Αν Δ το ιχνος της διχοτομου,
$ (ABC) \equal{} ( ABD) \plus{} (ADC)$
$ \leftrightarrow \frac {1}{2}bc sinA \equal{} \frac {1}{2}cd_a sin{\frac {A}{2}} \plus{} \frac {1}{2}bd_a sin{\frac {A}{2}}$
$ \leftrightarrow d_a \equal{} \frac {2bc}{b \plus{} c} cos{\frac {A}{2}}$(2)
απο (1),(2) το ζητουμενο.[/hide]

Ηλία αυτή τη λύση ήξερα κι εγώ... :wink:

Νομίζω ότι παλιά ο Πανάκης είχε βγάλει ένα βιβλιαράκι ΤΟ ΙΣΟΣΚΕΛΕΣ ΤΡΙΓΩΝΟ που είχε καμμιά δεκαπενταριά τρόπους στο διάσημο αυτό πρόβλημα

Να μια γεωμετρικη λυση βασισμενη σε elementary γνωσεις.

       Εστω οι ισες διχοτομοι $ CE = BD$. Εστω οτι WLOG $ \angle{B} > \angle{C}$. Τοτε $ BE < DC$. Απο το $ E$ φερνω παραλληλη προς την $ BD$ και οριζω σημειο $ K$ ωστε $ EK = BD$. Τοτε το  BEKD //μο. Αρα $ EB = KD < DC$. Επειδη $ BD = EC$ ισχυει$ EC = EK$. Αρα EKC ισοσκελες αρα $ \angle{EKC} = \angle{ECK}$.

Επειδη $ KD < DC$ ισχυει $ \angle{DCK} < \angle{DKC}$. Ομως $ \angle{EKD} = \angle{EBD} = \angle{\frac {B}{2}} > \angle{\frac {C}{2} = \angle{ECD}}$.
Αρα συνολικα $ \angle{EKC} > \angle{ECK}$ πραγμα ατοπο αφου $ EKC$: ισοσκελες.

http://www.mathematica.gr/forum/viewtopic.php?f=28&t=71&start=10
ΠΡΟΚΕΙΤΑΙ ΓΙΑ ΤΟ ΘΕΩΡΗΜΑ LEHMUS STEINER ΚΑΙ ΕΧΕΙ ΣΥΖΗΤΗΘΕΙ ΚΑΙ ΕΔΩ


calculus
derivative
logarithms
calculus computations

Let y = lnx^2.
a. Find the inverse of y = lnx^2 ie find x = f(y).
b. Find the first derivative of the inverse. Show that x' = 1/y'.

I worked out the following:
[b]Inverse[/b]
x=√(e^y)

I am not sure if it is correct and I am having trouble finding the derivative of the inverse.

If anyone can show me some clear workings it would be great.

hello, from $ y \equal{} \ln(x^2)$ follows $ |x| \equal{} \sqrt{e^y}$ and you have to distinguish two
cases $ x > 0$ or $ x < 0$.
Sonnhard.

a.

hm i got $ e^{\sqrt {y}}$

$ y \equal{} \ln x^2 \implies \sqrt {y} \equal{} \log_e x \implies x \equal{} e^{\sqrt {y}}$

EDIT:

meh, i thought it was $ (\ln x)^2$

hello, i think he meant $ y\equal{}\ln(x^2)$ and from here we have
$ e^y\equal{}x^2 \Leftrightarrow |x|\equal{}\sqrt{e^y}$.
Sonnhard.


[quote="phamduyhiep"]Let $a,b,c$ be the positive real numbers such that : $abc=a+b+c+2$.
Prove that :
$3abc-\frac{16}{(a-1)(b-1)(c-1)}\leq 8$
when does the equality hold ?[/quote]
Try $a=\frac{1}{2}$ and $b=c=5.$ :wink:

It is the same after making the substiutions $a=\frac{y+z}{x}$ and the similar ones,with an older post of mine as a solution to it.( with the condition that x,y,z are sides of a triangle) :lol: http://www.mathlinks.ro/Forum/viewtopic.php?t=103274

Sorry because I posted carelessly, please try it with $1<a,b,c<2$ :blush:
About that links  :roll: That is an problem in Mathematical&Youngth Magazine 07/2006 .


induction
combinatorics proposed
combinatorics

Suppose that $ \{a_{1},a_{2},...,a_{n}\}, \{b_{1},b_{2},...,b_{k}\}$ are two sets of positive integers with the same sum of elements, and that sum is less than $ nk$. Show that it is possible to omit some, but not all, summands from both sides of the equality $ a_{1}+a_{2}+...+a_{n}=b_{1}+b_{2}+...+b_{k}$ is such a way that the equality remains valid.

By induction on $ n+k.$
One can easily see that $ n, k > 1,$ hence the base case when $ n+k = 4$ is clear.
Now, assume the problem's statement holds for some $ i = n+k \geq 4.$
Let's take a look at $ a_{1}+\ldots+a_{n}= b_{1}+\ldots+b_{k}< nk,$ with $ n+k = i+1.$
WLOG, $ a_{1}= \max_{j}a_{j}, \ \ b_{1}= \max_{j}b_{j}$ and $ a_{1}> b_{1}.$
Then we may rewrite the equality as $ a_{1}-b_{1}+\ldots+a_{n}= b_{2}+\ldots b_{k},$ where $ b_{2}+\ldots+b_{k}< n(k-1).$
Finally, since $ n+k-1 = i,$ the induction hypothesis yields the desired conclusion.


Let $p > 5$ be a prime. Consider  the set $X = \{p - n^2 \mid n \in \mathbb N\}$. Prove that there exist two distinct elements $x, y \in X$ such that $x\neq 1$ and $x \mid y$.

[i]Albania[/i]


[quote="r_boris"]Μπόρης Ροδόλφος. Ετών >50. Ναυπηγός ΕΜΠ + Μαθηματικός. Γοητευμένος από το περιεχόμενο του forum. Μόλις βρω χρόνο να ασχοληθώ με τα LATEX θα προτείνω κι' εγώ κάποιες ασκήσεις, ίσως και παραμύθια! Καλή συνέχεια[/quote]

Ροδόλφε,
 καλώς ήρθες. Όταν έχεις χρόνο ,γράψε και στείλλε μας ό,τι θεωρείς καλό και ενδιαφέρον. Η κοινότητα σου σφίγγει το χέρι και σου εύχεται καλές ... επνεύσεις !
  [b]Τα μαθηματικά μας εννώνουν ! [/b]

Μπάμπη
Συγχαρητήρια για την πολύ καλή δουλειά στο forum και για τα βιβλία σου επίσης. Πράγματι μας ενώνουν και όχι μόνον. Θα προσπαθήσω να επικοινωνήσω μαζί σου τηλεφωνικά σύντομα
R B

Να καλωσορίσω κι εγώ με τη σειρά μου το νέο μέλος. :) Κύριε Μπόρη εύχομαι καλή αρχή και καλή δουλειά... :wink:

Ροδόλφε, καλώς ήρθες κι' από μενα, επίσης ετών >50.  Καλή συνέχεια και γόνιμη συμμετοχή στα δρώμενα του ολοένα και πιο δυνατού, ελληνικού ''Μathlinks forum''.

Κώστας Βήττας.


calculus
integration
function
trigonometry
logarithms
calculus computations

You will need:-

$ \int \tan ^{n}\theta d\theta \equal{}\frac{1}{n\minus{}1}\tan ^{n\minus{}1}\theta \minus{}\int \tan ^{n\minus{}2}\theta d\theta$

$ \int \tan \theta d\theta \equal{}\allowbreak \ln |\sec \theta |$

hello, after the formula above we get $ \frac{1}{6}\tan(x)^6\minus{}\frac{1}{4}\tan(x)^4\plus{}\frac{1}{2}\tan(x)^2\minus{}\frac{1}{2}\ln(1\plus{}\tan(x)^2)\plus{}C$.
Sonnhard.

[quote="IndoChina"]You will need:-

$ \int \tan ^{n}\theta d\theta \equal{} \frac {1}{n \minus{} 1}\tan ^{n \minus{} 1}\theta \minus{} \int \tan ^{n \minus{} 2}\theta d\theta$

[/quote]

how  do u get  that?  is that a standard reduction formula?

$ \tan ^ n \theta \equal{}\tan ^{n\minus{}2} \theta \cdot \tan ^ 2 \theta$.

$ \int \tan ^{n}xdx$

$ \equal{}\int \tan ^{n\minus{}2}x\sec ^{2}xdx\minus{}\int \tan ^{n\minus{}2}xdx$

$ \equal{}\frac{1}{n\minus{}1}\tan ^{n\minus{}1}x\minus{}\int \tan ^{n\minus{}2}xdx$

[quote="kunny"]$ \tan ^ n \theta \equal{} \tan ^{n \minus{} 2} \theta \cdot \tan ^ 2 \theta$.[/quote]

yup  got it now.  thanks.


This was a marathon problem posted by superwizard a few days ago, and I didn't know how to solve these types of problems quickly, and couldn't understand the solutions given, so can someone give a very detailed solution along with how to do all the rest of these problems? Thanks.

Problem 302

A number leaves a remainder of 1 when divided by 2, a remainder of 1 when divided by 3, a remainder of 4 when divided by 5, a remainder of 5 when divided by 8, and a remainder of 3 when divided by 7. What is the smallest number this could be?

The answer was:
[hide] [/hide]

[hide]from the congruences, we have:

n=1 mod6
n=4 mod5
n=5 mod8
n=3 mod7

combining the first two, we have 5n=5 mod 30
                                             6n=24 mod 30.

Subtracting the second from the first, we get n=19 mod 30

now combine n= 5 mod 8 and n= 3 mod 7

we have: 8n=24 mod 56
and         7n=35 mod 56
              n=-11 mod 56= 45 mod 56.

now we have  n=19 mod 30 and
                   n=45 mod 56

28n=532 mod 840
15n=675 mod 840.

2*(equation#2)-equation 1=

2n=818 mod 840

because 2, 818, and 840 all have a common factor 2, we have

[b]n=409 mod 420

so n=409[/b].[/hide]

wait...409 is 1 mod 8, what did i do wrong?

my solution (dunno how fast it is, but it's all i can think of) is...
[hide]Well you can start with just the first two statements.
The first number that works is 7, then 13, ...
In this set, the first number satisfying the third statement is 1 less than a multiple of 5, so we have 19, then 49, 79 ...
For the fourth statement, we want 3 less than a multiple of 8, and the first one to work for that is 109, then 229, 349 ...
Finally, we need the smallest one that is 3 more than a multiple of 7, so we have $\fbox{829}$ (829-3=826, which is divisible by 7) [/hide]

@indianamath: What did you do in your first step (after writing congruences) ? If you multiplied the first congruence by 5 (looks like you might have, but I'm not sure) then you should have gotten $5n \equiv 5 \mod 6$ and it looks like you did similar things throughout your solution. As far as I know, the part after the mod usually doesn't change, but the other parts do change.

[hide] actually, im not very familiar with modular stuff, but i noticed that 829 = 409 mod 420, though not the least value of it, and 829 is the correct answer (sry, i forgot to post the answer where i left a space for it on the question)[/hide]

umm... can someone elaborate a little more on the modular arithmetic thing?

[url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]Modular Arithmetic from the AoPS Wiki[/url]

umm... maybe im just being stupid here, but I still don't get how to solve this question using modulos.

I found something in the Test Forum
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137325
In case it gets deleted:

[color=red]<snip>[/color]

[size=75][color=red]<Sorry, vamathletes.  The Chinese Remainder Theorem is beyond the scope of MATHCOUNTS.  You may include a link, but you may not include a discussion of the method, even if you think the target topic will get deleted.  -- moderator rcv>[/color][/size]


geometry
parallelogram
combinatorics unsolved
combinatorics

Let $x,y$ be reals. Prove that there exist integers $m,n$ such that:
               $\left| \sqrt{( x-m )^{2}+( y-n )^{2}}-100 \right| < \frac{1}{14}$
( I'm trying to use the famous theorem that says : For every real number $\alpha$ and every positive integer $m$, there exists an integer $p$ and a positive integer $q$ ( $q \leq m$ ) such that:     $\left| \alpha-\frac{p}{q}\right| < \frac{1}{qm}$ )
[b]Can this help?!!![/b]

If the property holds for $(x,y)$ it will also be true for $(x+k,y+l)$, $\forall k,l\in \mathbb Z$ since we just need to make $m'=m+k,\;\;n'=n+l$ to get another appropiate pair. This means we can reduce the problem to $x,y\in [0,1)$. In addition, if the property holds for $(x,y)$ it will also hold for $(-x,y)$ (just make $m'=-m$), so we can also assume $x\in [\frac{1}{2},1)$.
Let $k\in \mathbb{N}$ and $f(k,y)=\sqrt{ 100^{2}-(k+y)^{2}}-99$ for $k\in \mathbb{N}^{*},\;\;y\in [0,1)$. Note that $f$ is decreasing for both variables, $f(k)<1$ and
$0\leq k<10 \Leftarrow k^{2}<\frac{399}{4}\Leftarrow (k+y)^{2}<\frac{4\times 100^{2}-199^{2}}{4}\Leftarrow$
$\Leftarrow \sqrt{100^{2}-(k+y)^{2}}>\frac{199}{2}\Leftarrow f(k,y)>\frac{1}{2}$.
Also $k<10\Rightarrow f(k,y)-f(k+1,y)=\sqrt{ 100^{2}-(k+y)^{2}}-\sqrt{ 100^{2}-(k+y+1)^{2}}=$

$=\frac{(k+y+1)^{2}-(k+y)^{2}}{\sqrt{ 100^{2}-(k+y)^{2}}+\sqrt{ 100^{2}-(k+1+y)^{2}}}=$

$=\frac{2(k+y)+1}{\sqrt{ 100^{2}-(k+y)^{2}}+\sqrt{ 100^{2}-(k+1+y)^{2}}}<\frac{2(k+y)+1}{\sqrt{ 100^{2}-10^{2}}+\sqrt{ 100^{2}-11^{2}}}=$
$=\frac{4k+2}{399 }<\frac{42}{399}<\frac{2}{14}$.
Now consider the set $B(k)=\{ (x,y)\in [\frac{1}{2},1]\times [0,1)/ |\sqrt{(x+99)^{2}+(y+k)^{2}}-100|<\frac{1}{14}\}$; it is a circular band around the arc $A(k)=\{ (x,y)\in [\frac{1}{2},1)\times [0,1)/ \sqrt{(x+99)^{2}+(y+k)^{2}}=100^{2}\}=$
$=\{ (x,y)\in [\frac{1}{2},1]\times [0,1)/ x=\sqrt{100^{2}-(y+k)^{2}}-99 \}$
which has $(f(k,0),0)$ and $(f(k,1),1)=(f(k+1,0),1)$ as endpoints and center $O(k)=(-99,-k)$. Given an arbitray point $P\in A(k)$ and a line $l$, $P\in l$, if we call $QR$ the segment $l\cap B(k)$ is clear that $PQ\geq \frac{1}{14}$, $PR\geq \frac{1}{14}$, as can be easily seen by noticing the circle of center $P$ and radius $\frac{1}{14}$ is tangent to the two circles which delimit $B(k)$. In particular, if we consider a line parallel to the $x$-axis we get that  $B'(k)=\bigcup_{|\delta|\leq \frac{1}{14}}(A(k)+\delta) \subset B_{k}$.
On the other hand $f(0,1)>1-\frac{1}{14}$ and $\forall k,y\; \;0\leq k\leq 10\;\;\;\frac{f(k,y)-f(k+1,y)}{2}<\frac{1}{14}$.
So $[\frac{1}{2},1)\times [0,1) \subset \bigcup_{k=0}^{10}B'(k) \subset \bigcup_{k=0}^{10}B(k)$. And we are done, since for any $(x,y)\in B(k)$, $m=-99$, $n=k$ satisfy the conditions and if the property holds for $[\frac{1}{2},1)\times [0,1)$ holds also for all $\mathbb{R}\times \mathbb{R}$ as we said on top.

Even if what I have written is a complete nonsense, master Yoda's wisdom may still enlight us:
[quote]Enough tough ain't, try if you . [/quote]
(I'm upset, I have cronical tea overdosis.)

Excusse me Mimoide, but what do you mean by :
Even if what I have written is a complete nonsense, master Yoda's wisdom may still enlight us:
Quote:
Enough tough ain't, try if you .  :huh:
and your solution is reaeaeaeaeally tough Dude, can you explain more?

Imagine two circles centered at the origin with radius $100+\frac{1}{14}$ and  $100-\frac{1}{14}$. Consider the set of points, $B$ between them (what I call a band, though there must exist an appropriate name). The problem reduces to show that for any $x,y\in \mathbb{R}$ there exist $(x',y')\in B$ such that  $\{ x' \}=\{x\}$, $\{y' \}=\{ y \}$, which reduces the problem to the interval $[0,1)$. Now, if you look at how the band behaves near to  point $(100;0)$ and $(-100,0)$ you notice that it is almost vertical, so the fractional part of the points in their proximity, which is just a point in $[0,1)\times[0,1)$ must appear similar to the union of successive parallelograms almost $\frac{1}{14}$ wide crossing that square bottom-top, each one slightly (approximately $\frac{1}{7}$) to the left (right) of the previous one as the distance from $(100,0)$, ($(-100,0)$) grows. The solution is just a formalization of this.

Sorry; not only bad solution, but bad explanation to the bad solution.

[quote] Even if what I have written is a complete nonsense, master Yoda's wisdom may still enlight us:
Quote:
Enough tough ain't, try if you .

(I'm upset, I have cronical tea overdosis.)
[/quote]
This is just a mental excrement to deviate people's attention from the  previous and noticeably bigger one.    :P

you mentioned circles. that was exactly what the real problem was...
you know lattice points ( points $(x,y)$ where $x,y$ are integers) The problem said we draw circles with center of all lattice points with radius $\frac{1}{14}$, prove that each circle with radius 100 in plane will intersect atleast one of these small circles....
now can you give an easiyer solution...  :?:


Let $ a,b,c$ be nonnegative real numbers such that

$ \frac 1{2a^2 \plus{} ab \plus{} 3bc} \plus{} \frac 1{2b^2 \plus{} bc \plus{} 3ca} \plus{} \frac 1{2c^2 \plus{} ca \plus{} 3ab} \equal{} \frac 9{2(a \plus{} b \plus{} c)^2}$.

Prove that

$ \frac {a(b \minus{} c)}{a \plus{} b\minus{}c} \plus{} \frac {b(c \minus{} a)}{b \plus{} c\minus{}a} \plus{} \frac {c(a \minus{} b)}{c\plus{}a\minus{}b} \equal{} 0$.

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac 1{a^2 \plus{} 2ab \plus{} 3bc} \plus{} \frac 1{b^2 \plus{} 2bc \plus{} 3ca} \plus{} \frac 1{c^2 \plus{} 2ca \plus{} 3ab} \ge \frac 1{2}$.[/quote]

$ LHS \ge \frac{9}{(a\plus{}b\plus{}c)^2\plus{}3(ab\plus{}bc\plus{}ca)} \ge \frac{9}{2(a\plus{}b\plus{}c)^2}$

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac 1{a^2 \plus{} 2ab \plus{} 3bc} \plus{} \frac 1{b^2 \plus{} 2bc \plus{} 3ca} \plus{} \frac 1{c^2 \plus{} 2ca \plus{} 3ab} \ge \frac 1{2}$.[/quote]

From Cauchy we have that

 $ \sum\frac{1}{a^{2}\plus{}2ab\plus{}3bc}\geq \frac{9}{\sum a^{2}\plus{}2\sum ab\plus{}3\sum ab}\equal{}\frac{9}{\left(\sum a\right)^{2}\plus{}3\sum ab}\equal{}\frac{9}{9\plus{}3\sum ab}\equal{}\frac{3}{3\plus{}\sum ab}$

So, now we have to prove that $ \frac{3}{3\plus{}\sum ab}\geq\frac{1}{2}\Longleftrightarrow 3\geq \sum ab\Longleftrightarrow 3\cdot 3\geq 3\sum ab\Longleftrightarrow \left( \sum a\right)^{2}\geq 3\sum ab$ which is obvious true.

[quote="quykhtn-qa1"][quote="Γιώργος"][quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac 1{a^2 \plus{} 2ab \plus{} 3bc} \plus{} \frac 1{b^2 \plus{} 2bc \plus{} 3ca} \plus{} \frac 1{c^2 \plus{} 2ca \plus{} 3ab} \ge \frac 1{2}$.[/quote]

From Cauchy we have that

 $ \sum\frac {1}{a^{2} \plus{} 2ab \plus{} 3bc}\geq \frac {9}{\sum a^{2} \plus{} 2\sum ab \plus{} 3\sum ab} \equal{} \frac {9}{\left(\sum a\right)^{2} \plus{} 3\sum ab} \equal{} \frac {9}{9 \plus{} 3\sum ab} \equal{} \frac {3}{3 \plus{} \sum ab}$

So, now we have to prove that $ \frac {3}{3 \plus{} \sum ab}\geq\frac {1}{2}\Longleftrightarrow 3\geq \sum ab\Longleftrightarrow 3\cdot 3\geq 3\sum ab\Longleftrightarrow \left( \sum a\right)^{2}\geq 3\sum ab$ which is obvious true.[/quote]
it similar sergic primazon :wink:[/quote]

Yeah, i know. I didnt see his solution, for i was posting the above solution that moment :)

I'm sorry since I made a mistake.  :oops:  I reedited the initial in(equality).

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that

$ \frac 1{2a^2 \plus{} ab \plus{} 3bc} \plus{} \frac 1{2b^2 \plus{} bc \plus{} 3ca} \plus{} \frac 1{2c^2 \plus{} ca \plus{} 3ab} \equal{} \frac 9{2(a \plus{} b \plus{} c)^2}$.

Prove that

$ \frac {a(b \minus{} c)}{a \minus{} b} \plus{} \frac {b(c \minus{} a)}{b \minus{} c} \plus{} \frac {c(a \minus{} b)}{c \minus{} a} \equal{} 0$.[/quote]
I think we must prove
$ \frac {a(b \minus{} c)}{a \minus{} b} \plus{} \frac {b(c \minus{} a)}{b \minus{} c} \plus{} \frac {c(a \minus{} b)}{c \minus{} a} \plus{}2(a\plus{}b\plus{}c) \equal{}0.$
As from the given hypothesise, we have $ 2a^2\plus{}ab\plus{}3bc\equal{}2b^2\plus{}bc\plus{}3ca\equal{}2c^2\plus{}ca\plus{}3ab$, then
$ 2a^2\plus{}ab\plus{}3bc\minus{}(2b^2\plus{}bc\plus{}3ca)\equal{}0$, or $ (a\minus{}b)(2a\plus{}2b\minus{}3c) \equal{}b(c\minus{}a)$, then $ \frac{b(c\minus{}a)}{a\minus{}b} \equal{}2a\plus{}2b\minus{}3c.$
So
$ \sum \frac{a(b\minus{}c)}{a\minus{}b} \plus{}2(a\plus{}b\plus{}c) \equal{}\sum \frac{b(a\minus{}c)}{a\minus{}b} \plus{}a\plus{}b\plus{}c \equal{}\minus{}\sum (2a\plus{}2b\minus{}3c)\plus{}a\plus{}b\plus{}c \equal{}0.$
Am I right, VasC?  :blush:  :maybe:

[quote="can_hang2007"]I think we must prove
$ \frac {a(b \minus{} c)}{a \minus{} b} \plus{} \frac {b(c \minus{} a)}{b \minus{} c} \plus{} \frac {c(a \minus{} b)}{c \minus{} a} \plus{} 2(a \plus{} b \plus{} c) \equal{} 0.$
Am I right, VasC?  :blush:  :maybe:[/quote]
Yes, and no.  :wink:
No, because there is other variants, too. For instant
$ \frac {a(b \minus{} c)}{a \plus{}b\minus{}c} \plus{} \frac {b(c \minus{} a)}{b\plus{}c\minus{}a} \plus{} \frac {c(a \minus{} b)}{c\plus{}a\minus{}b} \equal{} 0.$ :P

[quote="Vasc"][quote="can_hang2007"]I think we must prove
$ \frac {a(b \minus{} c)}{a \minus{} b} \plus{} \frac {b(c \minus{} a)}{b \minus{} c} \plus{} \frac {c(a \minus{} b)}{c \minus{} a} \plus{} 2(a \plus{} b \plus{} c) \equal{} 0.$
Am I right, VasC?  :blush:  :maybe:[/quote]
Yes, and no.  :wink:
No, because there is other variants, too. For instant
$ \frac {a(b \minus{} c)}{a \plus{} b \minus{} c} \plus{} \frac {b(c \minus{} a)}{b \plus{} c \minus{} a} \plus{} \frac {c(a \minus{} b)}{c \plus{} a \minus{} b} \equal{} 0.$ :P[/quote]
We still have this from the above  :P since $ (a\minus{}b)(2a\plus{}2b\minus{}3c)\equal{}b(c\minus{}a),$ we obtain
$ a(c\minus{}b) \equal{}2(a\minus{}b)(a\plus{}b\minus{}c),$ or $ \equal{}\frac{a(b\minus{}c)}{a\plus{}b\minus{}c} \equal{}\frac{b\minus{}a}{2},$ which shows that
$ \sum \frac{a(b\minus{}c)}{a\plus{}b\minus{}c} \equal{}\sum \frac{b\minus{}a}{2} \equal{}0.$ :)

But I do mean that your previous question must be prove
$ \frac {a(b \minus{} c)}{a \minus{} b} \plus{} \frac {b(c \minus{} a)}{b \minus{} c} \plus{} \frac {c(a \minus{} b)}{c \minus{} a} \plus{}2(a\plus{}b\plus{}c)\equal{}0$,
not
$ \frac {a(b \minus{} c)}{a \minus{} b} \plus{} \frac {b(c \minus{} a)}{b \minus{} c} \plus{} \frac {c(a \minus{} b)}{c \minus{} a} \equal{}0.$
Am I right?  :oops:  :maybe:

[quote="can_hang2007"] But I do mean that your previous question must be prove
$ \frac {a(b \minus{} c)}{a \minus{} b} \plus{} \frac {b(c \minus{} a)}{b \minus{} c} \plus{} \frac {c(a \minus{} b)}{c \minus{} a} \plus{} 2(a \plus{} b \plus{} c) \equal{} 0$,
not
$ \frac {a(b \minus{} c)}{a \minus{} b} \plus{} \frac {b(c \minus{} a)}{b \minus{} c} \plus{} \frac {c(a \minus{} b)}{c \minus{} a} \equal{} 0.$
Am I right?  :oops:  :maybe:[/quote]
Yes, you are right. :wink:


function
integration
calculus
real analysis
real analysis unsolved

Let $f(x)=\sum_{n=1}^{\infty}\frac{x^n}{\sqrt{n}}$.
a).Prove that $f(x) \sim \frac{\sqrt\pi}{\sqrt{1-x}}$ when $x\to 1^-$.
b).Prove that there exists limit $\displaystyle L=lim_{x\to1^-}(f(x)-\frac{\sqrt\pi}{\sqrt{1-x}})$ and find $L$.

This problem appeared in a quite difficult examination at Ecole Polytechnique. Hopefully, there were some introductory questions in order to solve it. I would really like to see a short solution for this problem.

$f(e^{-t})=\sum \frac{e^{-nt}}{\sqrt{n}}$
first part easily follows from the fact that $g(x)= \frac{e^{-tx}}{\sqrt{x}}$ is a decreasing function .

Thanks for indications, i've already found an official solution(corrige) to this problem.

If $f(x) = \frac{a^x}{\sqrt{x}}, a \in (0,1), x > 0$ then $f'(x) < 0$ so f is decreasing.
Let $a_n = \sum_{k=1}^n f(k) - \int_1^n f(t)dt$. $a_1 = a > 0$.
$a_{n+1} - a_n = f(n+1) - \int_n^{n+1} f(t)dt = f(n+1) - f(c_n), c_n \in (n,n+1)$
But f decrease so $a_n$ decrease.
Then $a_n - f(1) = a_n - a_1 \geq f(n) - f(1), a_n \geq f(n) > 0$. It follows $a_n$ convergent.
From $n^2 (a_{n+1} - a_n) = n^2 f(n+1) - n^2 f(c_n) \to  0 - 0 = 0$ and Stoltz-Cesaro one can find $na_n \to  0$ so $a_n \to  0$.

We have: $\sum_{k=1}^n f(k) \to  A = \int_1^{\infty} f(t)dt$.With the change $a^t = e^{-x}$,
$A = \int_{-ln a}^{\infty} \sqrt{\frac{x}{-ln a}} e^{-x} \frac{1}{-ln a} dx = \frac{1}{\sqrt{-ln a}}\int_{-ln a}^{\infty} x^{-1/2} e^{-x} dx$
When a-> 1-0, last integral is $\int_0^{\infty} x^{-1/2} e^{-x} dx = \Gamma (1/2) = \sqrt{\pi}$
Also, when a-> 1-0, $- ln a \sim 1 - a$ and the first point is demonstrated.

For the second we need to find the limit of: $\frac{\int_0^{- ln x} \frac{e^{-t}}{\sqrt{t}} dt}{\sqrt{1 - x}}$ when x-> 1-0.
But $e^x = 1 + x + x^2/2! + ...$ and $- ln x \sim 1 - x$, so computations give 2. Is this the result ? :blush:


A 1 cm $ \times$ 1 cm square is cut from each of the four corners of a square piece of cardboard with area 64 square centimeters. The
sides are then folded up to make a rectangular box. What is the
volume in cubic centimeters of the box?

The side length of each side is 8. We cut out a 2 cm from each side (1 cm on each corner). This leaves us with a rectangular box with height 1, and length and width 6. Thus our desired volume is $ \boxed{36}$.


ratio
vector
geometry
geometric transformation
reflection

I'm in need of a new computer background so I decided to go to AopS for help.
Does anyone have some good ideas for a background?

Currently i'm thinking of getting one that has to do with the golden ration, like maybe a shell or something. But i havn't found one yet...

i dont know if this is a gud idea or not, but wat i do is i change my desktop image often to go w/ the time of year.

so if its valentine's day i go to google images and find a picture of hearts.

if im having a unit test in world studies, i put a picture of a map.

if its spring break, i put up a "no school!" picture (google image actually has one, its pretty gud 2!)

i hope this helps.

I like simple vector backgrounds personally.

[url=http://img225.imageshack.us/my.php?image=vectorcityor1.png][img]http://img225.imageshack.us/img225/8610/vectorcityor1.th.png[/img][/url]

Also, occasionally the background will reflect my interests.

[url=http://img217.imageshack.us/my.php?image=kirbywallpaperjx5.png][img]http://img217.imageshack.us/img217/5701/kirbywallpaperjx5.th.png[/img][/url]

Or they might just be extremely messed up

[url=http://img209.imageshack.us/my.php?image=rabbitlolwallpapergu8.jpg][img]http://img209.imageshack.us/img209/766/rabbitlolwallpapergu8.th.jpg[/img][/url]

i'm thinking of something more naturalistic like "follow" (the fish one that comes with most windows xp's)

I'm not really into anything right now that i can show my friends without them calling me a super nerd or that my parents would let me put.

Also, no messy colors. It's best to have a picture without too many colors in it.

My current background:
[hide][img]http://img.qj.net/uploads/articles_module/68927/black%20hole_qjpreviewth.jpg[/img][/hide]

Others I've had...
[hide]
[img]http://upload.wikimedia.org/wikipedia/commons/c/c2/Photos-photos_1087592507_Energy_Arc.jpg[/img][/hide]

[hide]
[img]http://upload.wikimedia.org/wikipedia/commons/4/46/Trefoil_knot_arb.png[/img][/hide]

Mm.. only 3 pics allowed in post.
Go to Wikipedia's featured pictures archive.

[quote="Cyrazeno"]My current background:
[hide][img]http://img.qj.net/uploads/articles_module/68927/black%20hole_qjpreviewth.jpg[/img][/hide]

Mm.. only 3 pics allowed in post.
Go to Wikipedia's featured pictures archive.[/quote]

Great Idea for the wikipedia featured pictures.
Right now, i have this
http://upload.wikimedia.org/wikipedia/commons/3/3c/Strix-varia-005.jpg
It's huge, therefore, great quality
[hide][img]http://upload.wikimedia.org/wikipedia/commons/3/3c/Strix-varia-005.jpg[/img][/hide]

Meh, I just took the Vista gradient background and I edited it a bit to look better, since the original one is way too bright.


Let p(Z)=.40, p(Y)=.30, and p(Z or Y)=.58
find the probability of
p(complement of Z or Y)
p(complement of Z and Complement of Y)

If you could please explain to me how you arrived at the answer it would be a big help I am totally lost at this moment.

You can use [url=http://en.wikipedia.org/wiki/De_Morgan's_laws]De Morgan's law[/url]

Well, we can see that $ P(Z \text{and} Y) \equal{} 0.7 \minus{} 0.58 \equal{} 0.12$...I don't exactly understand the problem though

Ok I think that the.58 is supposed to go in the middle but when I do the Y and Z become negative numbers. When I add all of the numbers they gave me it comes out to 128 instead of just 100 is that normal?

By De Morgan's Law, $ P(\overline {Y}\cup \overline{Z})\equal{}P(\overline {Y\cup Z})$
$ \equal{}1\minus{}P(Y\cup Z)\equal{}1\minus{}0.58\equal{}0.42$.


[b]Hints only, please.[/b]

The area of a rectangle is 9 square inches. P and Q are adjacent vertices of the rectangle. If R is the point at which the diagonals of the rectangle meet, then the degree measure of $ \angle PRQ$ is $ 60^\circ$. The length of side PQ is:

(A) $ \sqrt[4]{3}$

(B) $ \sqrt{3}$

(C) $ \sqrt[4]{3^3}$

(D) $ 3$

(E) $ \sqrt[4]{3^5}$

[hide]triangle $ PQR$ has an area of $ \frac{9}{4}$ [/hide]

[hide="Another hint"]
You can find the area of $ \triangle PRQ$ in terms of the length of $ PQ$. You can find the ratio of the area of $ \triangle PQR$ to the area of the rectangle, and you know the area of the rectangle. Therefore, you know the length of $ PQ$.
[/hide]

[hide="hint"]
$ \triangle PQR$ is equilateral.[/hide]


geometry
3D geometry
superior algebra
superior algebra unsolved

Hi,

sorry if this question belongs in the advanced or theorems section

My problem is, suppose you have a classical [url=http://en.wikipedia.org/wiki/Polar_space]polar space[/url]
let us suppose all is finite here

and you have a space of dimension m of that polar space

How many m+1 dimensional in the polar space go through that space?

If you want an easier subcase : consider a nonsingular quadric and an m dimensional space completely on the quadric
How many m+1 dimensonal spaces completely on the quadric go through this?

For m=0 it is pretty doable, but how about the general case?
I am supposed to find that it is closely related to the number of points in a similar polar space , but of lower rank

Surely, all ideas are welcome on this  :|

Well I did some inquiries

I should be able to find this result :

the number of m+1 spaces in the polar space (embedded into s dimensional space) through the m dimensional space is the same as the number of points in a polar space of same 'type' (symplectic, hermitian,...) embeddable in a space of dimension $s-2 m-2$

I don't see why this is so obvious though.  Looks like a shady construction involving tangent space, and singular quadrics,...

Okay, so if I am correct, this should be the same problem as this one :

let $Q$ be a nonsingular quadric, and $W$ an m dimensional subspace on it

consider $Z=T_{W}(Q)$ , the tangent subspace of $W$
it will have dimension $n-m-1$

counting all points in the intersection $Z\cap Q$ , should immediately give me my desired result.
Now that will be a quadric, but not a nonsingular one.  Singular quadrics are always a cone, with as basis a nonsingular quadric, so calculating the points is still easy... if you know the type and dimension of that basis quadric

As may be obvious, my despair  :(   is partially due to this being homework related.  As you can see I am still actively trying to find this on my own, but I doubt any hint can be superfluent.  So please don't hold back!  Any theorems (in books), or urls can help me!

If you think this is the [b]wrong place[/b]then please do tell me!


[quote="Anaxerzia"]Prove:

$ \frac {a \plus{} abc}{1 \plus{} ab \plus{} abcd} \plus{} \frac {1 \plus{} ab}{a \plus{} abc \plus{} abcd} \plus{} \frac {a \plus{} abcd}{1 \plus{} ab \plus{} abc} \plus{} \frac {1 \plus{} abc}{a \plus{} ab \plus{} abcd} \plus{} \frac {ab \plus{} abcd}{1 \plus{} a \plus{} abc}\ge$ $ 10/3$[/quote]

 wrong $ a\equal{}b\equal{}c\equal{}1,d\equal{}2$

[quote="zaya_yc"][quote="Anaxerzia"]Prove:

$ \frac {a \plus{} abc}{1 \plus{} ab \plus{} abcd} \plus{} \frac {1 \plus{} ab}{a \plus{} abc \plus{} abcd} \plus{} \frac {a \plus{} abcd}{1 \plus{} ab \plus{} abc} \plus{} \frac {1 \plus{} abc}{a \plus{} ab \plus{} abcd} \plus{} \frac {ab \plus{} abcd}{1 \plus{} a \plus{} abc}\ge$ $ 10/3$[/quote]

 wrong $ a \equal{} b \equal{} c \equal{} 1,d \equal{} 2$[/quote]

Yes, and you get $ 3.5\ge 10/3$, which is true :), so it still holds...


Yep, you read the title right. I remember there was several years ago a popular gum brand called"Love is", and in each wrapper, there was a sticker with a picture and a phrase, that talked about love. So, we are all romantics ;) , so post about what LOVE IS...

EDIT: And if this forum doesn't get popular, i'll seriously think nearly all of you guys are robots or computer programs that just occasionally post in this forum because the founder of Aops made you do so.

You know, we have a project in health class trying to figure out what love is...very interesting class.  :|  :dry:

Love is quite natural. Do not worry if you love something or somebody - we all do!

[quote="randomdragoon"]$+1$ FTW

(love is t.o.w.)[/quote]
I am kind of getting tired of tug of war.

Anyways, love is  :love:
And, you guys can also write something like "Love is when he/she ..." ;)

Love is like the smell of purple violets....

..... and when love fails...it smells like burnt toast

~tspa

Love is complex :D

The dreams and high feeling u have in love is true

How many of u have been in loooove

Love is a black hole. Time begins and ends in it.

Well that is definitely true, love sucks up all attention, including Hollywood writers, and song writers all over the world.  Hm, I think I'm gonna start a "music should not all be about (being in) love" topic in the Music and Art forum soon :( ..... :mad:

[quote="rem"][quote="lotrgreengrapes7926"][quote="shadysaysurspammed"]unconditional....

something that cannot be defined[/quote]

Love is like $\frac{1}{0}$...undefined :love:  :D[/quote]
And 1/0 is infinity. Love is infinite :lol:[/quote]

Then it means that love is also $\frac{1}{0_{-}}=-\infty$ which means that hate=love  :?

did you notice a orl's signature :

Math is like love. A simple idea but it can get complicated.

just convert the words and you get:

Love is like math. A simple idea but it can get complicated.

love is state of mind, when you indirectly think of passing (spreading) your gens.

no, no, no. that's not true.

i don't know what love is, but i'm sure that :[b]i'm in love!!!!![/b]

Love is a turning point in one's life

and the true essence of romance

and twisted like funnel cake

also like a chain reaction

[quote="langtupanda"][quote="rem"]Love is when you go  :wow:[/quote]
You mean we go on electronic chair, don't you?  :rotfl:[/quote]

murder love??? not sure lol

[quote="langtupanda"][quote="MysticTerminator"]Love tastes like Texas.[/quote]
How could you come up with this equation Love = Texas?[/quote]

Well Texas is....big and so is love...ya

[quote="José"][quote="shadysaysurspammed"]Love is when you sleep dreaming about her[/quote]

Too... ejejej[/quote]

No... I'd call that obsession :roll: .

[quote="Treething"]love tastes like fish

don't eat the fish btw[/quote]

$love$ means the taste of fish.

do u $love$ taste of fish?

$love$ means the taste of fish.

do u $love$ taste of fish?

.....................................

..................................... ;)

Love is when you see an imperfect person perfectly.

Love is when the rest of the world fades away in his presence, and NOTHING matters but him.

Love is the irresistible desire to be irresistibly desired.

Love is when you are willing to wait for this person forever, even when you don't know if what you're waiting for will ever come to you.

Love is...if you had ONE wish, you'd use it on that person.

Love is when you're willing to sacrifice EVERYTHING for this person and don't expect anything back.

Love is unconditional.

That's all I can think of for now... I like this topic so much =)

(From [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=15259]conceive[/url])

Love is everything it's cracked up to be…It really is worth fighting for, being brave for, risking everything for.

[quote="wobster109"]love is like trying to do a math question that begins with [IMO shortlist].

interpret that as you will.[/quote]

Maybe in love with math I suppose no?

[quote="Treething"]actually i hate fish[/quote]

Therefore, you hate love. Therefore, you will not truely marry. Threfore, you cannot have smart, young children that win mathcounts just like you. :evilgrin:

[quote="wobster109"]love is like trying to do a math question that begins with [IMO shortlist].

interpret that as you will.[/quote]

so, it means that i can't be sucessful in love.

LOL

(locks of LOVE) You don't have to marry to love someone.


superior algebra
superior algebra unsolved

Let G be a group.
PautG(automorphism power) is residually finite?
Thanks
Help-me :-((

Perhaps you could try defining some of the terms involved.
What is automorphism power?


calculus integration limit function algebra polynomial trigonometry

Let $g: [0,2\pi]\to\mathbb{R}$ be continuous. Evaluate
\[\lim_{n\to\infty}\int_{0}^{2\pi}g(x)\text{ sin }(nx)\,dx. \]

Is this not the Fourier coefficient of $g$?
Hence the limit is 0.

Or any continuous function can be approximated by polynomials hence consider g to be a polynomial, without loosing the generality let $g=x^{k}$
then by simple calculations

$\int_{0}^{2\pi}x^{k}\sin nx dx\rightarrow 0$
I think a recurrence formula can be found for calculating this integral.

It's called the Riemann-Lebesgue lemma. You don't need to require that $g$ be continuous; all you really need is that $g$ is (Lebesgue) integrable.

All proofs - and there are many - involve writing $g=a+b$ where $a$ is a "good" function, belonging to some set for which we know for some other reason that

$\lim_{n\to\infty}\int_{0}^{2\pi}a(x)\sin(nx)\,dx=0$

and that for any given $\varepsilon>0,$ we can do this so that the "bad" function $b$ has the property that

$\|b\|_{1}=\int_{0}^{2\pi}|b(x)|\,dx<\epsilon.$

The proof works because $\left|\int_{0}^{2\pi}b(x)\sin(nx)\,dx\right|\le\|b\|<\varepsilon.$

didilica has proposed using the set of polynomials as the "good" functions. That works (perhaps not as slickly as some). Other possible sets of "good" functions to consider:

Trigonometric polynomials
$C^\infty$ (infinitely differentiable functions)
$C^{2}$
Piecewise linear continuous functions
Step functions
$L^{2}$ (square integrable functions)

Any one of these sets can be made to work; the differences between the proofs would be the differences in how we prove that we can approximate $g$ arbitrarily well in the $1$-norm.

I found a proof for the Riemann-Lebesgue lemma. But it would be nice to see more details about why the integral being as a Fourier coefficient of $g$ implies that the limit is 0? And what it means that "continuous function can be approximated by polynomials"? Does it mean that if $A$ is a closed subset of reals, then for all $\epsilon>0$ we can find a polynomial $P$ such that
$|g(x)-P(x)|<\epsilon$ for all $x\in A$?

There's a key difference here- you have continuity at the endpoints. With that, you can approximate uniformly by polynomials ([url=http://en.wikipedia.org/wiki/Stone-Weierstrass_theorem]Reference[/url]). The uniform approximation is also good enough in any other standard norm.

Uniform approximation - in other words, that $|g(x)-a(x)|<\varepsilon$ for all $x$ in the interval - suffices for the proof, as jmerry indicated. It's also overkill. You don't need to approximate $g$ uniformly by "good" functions; you only need to approximate in $L^{1}$ norm. In other words, you need

$\int_{0}^{2\pi}|g(x)-a(x)|\,dx<\varepsilon.$

That's a weaker condition, meaning it is more easily satisfied.

You said you found [i]a[/i] proof of the Riemann-Lebesgue lemma. What I was suggesting is that there are as many different proofs of it as there are reasonable candidates for the set of "good" functions - and there are quite a few reasonable candidates for that set.


There $40$ objects on a circle, each of which is white or red. Two players are playing the following game: the first takes all the red objects which have the white neighbour, after that the second takes all the white objects which have the red neighbour, then gain the first takes all the red objects which have the white neibour, etc. The end of the game is when there remains only figures of the color . Is it possible that only one red figure remains on the circle ? Is it possible that two white figures remain on the circle ?  :lol:

[quote="ehsan2004"]which have the white neighbour[/quote]

What does this mean, exactly? Must they have both neighbours white, or is it enough to have one? It's definitely unclear (and gramatically incorrect, if I am not mistaken :)).

[quote][quote]ehsan2004 wrote:
which have the white neighbour
[/quote]What does this mean, exactly? Must they have both neighbours white, or is it enough to have one? It's definitely unclear (and gramatically incorrect, if I am not mistaken ).[/quote]
 :?   :read: [quote]the first one takes[u] all the red objects [/u]which have the white neighbours, after that the second one takes [u]all the white objects [/u]which have the red neighbours[/quote]
I think it's easy to understand and it have no problem in a question :huh: ... I think it have obviously statments and not hard :huh:  :huh:  :?:  :idea:

When you quoted your own post, you modified it :). Your original post said "the red [b]neighbour[/b]", not "the red [b]neighbours[/b]". Anyway, thak you for clarifying it.

I still don't get it, ehsan.. :(

Please tell me which one of the following versions is correct:

The first player takes all the red objects which have [b]at least one[/b] white neighbour

or

The first player takes all the red objects whose [b]both[/b] neighbours are white

I tend to think it's the first one, since I can't see how a player can create, by performing a move, new objects of one color which have both neighbours of the other color, so I don't think the players could make more than one move each in the second situation :?. However, your last message suggests otherwise (i.e. from your last message I understood that the second sentence above is correct version).

On the other hand, in the first case, the problem is very easy:

It's possible to end up with $1$ red object by putting $21$ consecutive red objects on the circle, and then $19$ white ones, and it's also possible to end up with $2$ white objects by placing $20$ consecutive red objects followed by $20$ consecutive white objects on the circle.

I think that means :  ;)
the first one takes all the red objects which have the white neighbours=The first player takes all the red objects which have at least one white neighbour   :lol:  :D

Ok, thanks! :) So this means the problem is solved now?


If chords $AB$ and $CD$ of circle $O$ intersect (inside the circle) at point $E$, then is the following statement true $AE*EB=CE*ED$?
(points $A,B,C,D$ are on the circle).

[quote="Jimmy"]If chords $AB$ and $CD$ of circle $O$ intersect (inside the circle) at point $E$, then is the following statement true $AE*EB=CE*ED$?[/quote]

Yes. See posts #3 and #4 in http://www.mathlinks.ro/Forum/viewtopic.php?t=20168 .

  darij


inequalities
combinatorics unsolved
combinatorics

I think the max is : $n(2n^2+9n+1)/6$ and it is obtained with the following sequence :

$1\ 3\ ...\ 2n-1\ 2n\ 2n-2\ ...\ 2$ in the even case
$1\ 3\ ...\ 2n-1\ 2n+1\ 2n\ 2n-2\ ...\ 2$ in the odd case

A permutation being product of transposition it's enough to prove that if applying a transposition to $1\ 3\ ...\ 2n-1\ 2n\ 2n-2\ ...\ 2$ or $1\ 3\ ...\ 2n-1\ 2n+1\ 2n\ 2n-2\ ...\ 2$decreases the sum.

Then it's a bit cumbersome, we have to consider several cases (swapping two odd or two even numbers, swapping 2n and 2n-1, swapping an odd and an even number, ...). In each case (I haven't checked all cases ;)) the sum decreases.

The formula is then just a matter of elementary computing (hope it's correct nevertheless).

Min value is possibly reached with a similar argument.

Some one can explain more clearly for me ???????? :blush:

I don't see why it's enough to show that applying any single transposition decreases the sum.

It seems like this would be the discrete analogue of only showing that a point is a local maximum instead of a global maximum.  You would show that all "nearby" permutations have a smaller value, but some far away permutation may still be larger.

[quote="Together"]$x_1,x_2,....x_n$ is a  permutation of {1,,2,3...,n}
Find the maximum and min of $P = x_1.x_2 + x_2.x_3 +........x_{n-1}.x_n$[/quote]
may be we can solve it with inequality (if i had no mistake in understand the problem)
we know that $n>n-1>...>1 \rightarrow p=n \times (n-1)+(n-1) \times (n-2)+...+2 \times1$is the max and $p=n\times 1+(n-2) \times 2 +...$ is the min and we can calculate it easily. :D


calculus
integration
real analysis
calculus computations

Suppose $ g(x) \equal{} \begin{cases} f(x) \ \ \ \text{if} \ x \neq c \\ r \ \ \ \ \ \ \ \ \text{if} \ x \equal{} c \end{cases}$
$ f$ is Riemann integrable on $ [a,b]$ and $ c \in [a,b]$. Also $ r \neq f(c)$. Prove that $ g$ is Riemann integrable on $ [a,b]$. Also show that $ \int_{a}^{b} f \equal{} \int_{a}^{b} g$.

So we can consider cases? For $ x \neq c$ we know that $ g(x) \equal{} f(x)$ which is Riemann integrable by definition. So consider case where $ x\equal{}c$? The Riemann sum can take the following values: $ \mathcal{R}(g,P) \equal{} \begin{cases} f(x)(x_{i}\minus{}x_{i\minus{}1}) \ \ \ \ \ \text{if} \ c \neq x_{i}^{*} \ \text{for any} \ i \\ r(x_{i}\minus{}x_{i\minus{}1}) \ \ \ \ \ \ \ \ \ \ \text{if} \ c \equal{} x_{i}^{*} \ \text{for any} \ i \end{cases}$

Is this how we go about it?

And showing $ \int_{a}^{b} f \equal{} \int_{a}^{b} g$ is equivalent to showing that $ \int_{a}^{b} g \equal{} 0$ if $ x\equal{} c$?

easiest thing i can thing of (besides using measure theory) is the darboux integrals

given $ P\equal{}\{a\equal{}x_0<x_1<... < x_{n\minus{}1} <x_n\equal{}b\}$, define $ M_k\equal{}\sup_{x\in [x_{k\minus{}1},x_k]} f(x)$ and $ m_k\equal{}\inf_{x\in [x_{k\minus{}1},x_k]} f(x)$
$ U(f,P)\equal{}\sum_{k\equal{}1}^n M_k(x_k\minus{}x_{k\minus{}1})$ and $ L(f,P)\equal{}\sum_{k\equal{}1}^n m_k(x_k\minus{}x_{k\minus{}1})$

then integrability is if given $ \epsilon >0$ we can find a partition P with $ U(f,P)\minus{}L(f,P)<\epsilon$

so since f is integrable, $ \exists P$ s.t. $ U(f,P)\minus{}L(f,P)<\frac{\epsilon}{2}$

then let $ P'\equal{}P\cup \{c\minus{}\frac{\epsilon}{4\delta},c,c\plus{}\frac{\epsilon}{4\delta}\}$ be a refinement of P

and if we let $ \delta\equal{} \min \{1, \sup_{|x\minus{}c|<\epsilon} |f(x)\minus{}r|\}$
then $ U(g,P')\minus{}L(g,P') \le U(f,P)\minus{}L(f,P)\plus{} \frac{\epsilon}{2} < \epsilon$


Hey, should I take the AMC 12 Problem Series Class or should I take an intermediate/olympiad class?  Would I learn a lot from the AMC 12 class?

If I'm remembering correctly, you qualified for the USAMO last year.  If you want to take a problem series, the AIME is a better fit.  Judging by the work I've seen you do, I'd say 'Take Interm Counting/Probability if you feel weak in counting, otherwise take something at the Interm Trig/Complex level or higher (i.e. Olympiad).'

Is there anyway I can switch out of the AMC problem series class and take intermediate counting/probability at this point?  Like how do I cancel the AMC class?

You can go to your My Classes page and click the 'drop' button.  Then you can simply enroll in whatever other class you like the normal way.

[quote="WarpedKlown1335"]I don't see the drop button...=(.[/quote]

That's because the class hasn't started yet.  If you'd like to drop early, shoot us an email at classes@artofproblemsolving.com and we'll take care of it.


floor function
number theory unsolved
number theory

Let $ n \in N^{*}$ and $ p$ be a prime number.Prove that if $ p^p|n!$ then $ p^{p \plus{} 1}|n!$.

Take the first $ p \minus{} 1$ multiples of $ p$, which are $ 1.p$, $ 2.p$,...,$ (p \minus{} 1).p$. The next multiple is $ p^2$, which has two factors $ p$. So, for any $ p$, if $ p^p|n!$, then $ p^{p \plus{} 1}|n!$, as desired.

One knows that the maximal exponent $ e$ of $ p$ such that $ p^e \mid n!$ (also denoted by $ p^e \parallel{} n!$) is given by Legendre's formula $ \displaystyle e \equal{} \textrm{ord}_p(n) \equal{} \sum_{k \equal{} 1}^{\infty} \left \lfloor \frac {n} {p^k} \right \rfloor$.
If $ n \geq p^2$, then $ \textrm{ord}_p(n) \geq p \plus{} 1$ and we are done.
If, on the contrary, $ n < p^2$, let $ n \equal{} qp \plus{} r$, for some $ 0 \leq r < p$. Then $ \textrm{ord}_p(n) \equal{} q$, and $ p^2 > n \equal{} pq \plus{} r \geq pq$, so $ q < p$, which tells that $ p^p \not | \ n!$.

In other words, we proved that, for $ n < p^2$, one can never even have $ p^p \mid n!$, while for $ n \geq p^2$ one always has $ p^{p \plus{} 1} \mid n!$.

[quote="behdad.math.math"]would u explain your soluton more dear raf92????????[/quote]
Ok. You have to take all the first $ p\minus{}1$ multiples of $ p$ (remember that $ n!\equal{}1.2.3...n$), which are shown in my post above. If $ p^{p\minus{}1}|n!$, so $ n\geq p(p\minus{}1)$. But we have $ p^{p}|n!$, and then $ n$ is greater than or equal to the next multiple of $ p$, and this one is $ p^2$. Therefore, it's easy to notice that in the first case only $ p\minus{}1$ factors $ p$ would divide n!, but now , if $ n \geq p^2$, we have that $ p^{p\plus{}1}|n!$. So, the desired condition will be satisfied.


S13. Let $k$ be an integer and let $n=\sqrt[3]{k+\sqrt{k^2+1}}+\sqrt[3]{k-\sqrt{k^2+1}}+1$. Prove that $n^3-3n^2$ is an integer.

[hide="my approach"]
I used the fact that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$.
But I cant quite finish it.....
[/hide]

This is still an "active" problem in this journal, and so should not be discussed.


Let $ ABC$ be an acute triangle, and let $ D$ and $ E$ be the feet of the altitudes drawn from vertexes $ A$ and $ B$, respectively. Show that if,
\[ Area[BDE]\le Area[DEA]\le Area[EAB]\le Area[ABD]\]
then, the triangle is isosceles.

[hide]Let $ d$ be the distance from $ D$ to $ \overline{AB}$, and let $ e$ be the distance from $ E$ to $ \overline{AB}$. We have $ [BDE] \le [DEA]$, or $ \frac{1}{2} \cdot DE \cdot d \le \frac{1}{2} \cdot DE \cdot e \iff d\le e$. We also have $ [EAB]\le [ABD]$, or $ \frac{1}{2}\cdot AB \cdot e \le \frac{1}{2} \cdot AB \cdot d \iff e \le d$. Thus, $ d\equal{}e$, so $ ABDE$ is a trapezoid. Since it is also cyclic ($ \angle AEB \equal{} \angle ADB \equal{} 90^\circ$), it must be isosceles, so $ \angle EAB \equal{} \angle DBA$, and we are done.[/hide]


Hello everybody,

Here you will find a new variant of sudoku : sudoku++

http://joujoujou.over-blog.com/article-25839168.html

Another more complex : sudokrwazé is on line too.

I like to have your thoughts about the concept so if you can comment do not hesitate.
Welcome to any comment it will help me to improve the game.

Thank you.

It would really help if I could read it.


An electronic oscillator produces 4.5 volts effective AC at a frequency of 1850 kiloherts, feeding it into a 12 picofarad capacitor. How much current flows

$ I=\frac{dq}{dt}$

$ I_{ef}=q_{ef}\omega=cU_{ef}\omega=2\pi CUf$

very nice Namiqfizik. Now all you have to do is plug in these values


trigonometry
geometry
circumcircle
geometry unsolved

⊙I is the inscribed circle of  △ABC; D,E and F are the points of tangency which are in BC、CA and AB;⊙I∩AD=M;⊙P is the circumscribed circle of  △CDM; ⊙P∩DF=N. Now we link the two points C and N and extend it ;the two lines CN and AB meet at the point G.

Prove that CD=3FG.

P.S:I am after a pure geometric solution without  much computation . Thank you!

Very nice problem, now here is my synthetic solution:

Denote $X \in EF \cap CG$, and $T \in EF \cap BC$.

Now because the quadrupoint $(T,B,D,C)$ forms an harmonical division, i.e $(T,B,D,C)=-1$, so does the pencil $F(T,B,D,C)$, and now by intersecting it with $CG$, we obtain that $(X,G,N,C)=-1$. $(1)$.

Now on other hand by applying the Menelaus theorem in $\triangle{BCG}$, for the transversal $\over arrow{D-N-F}$, we obtain that $CD=3GF \iff CN=3NG$.

But from $(1)$, we have that $\frac{CN}{NG}=\frac{CX}{GX}$.

Therefore it is sufficient to prove that $N$ is the midpoint of $(CX)$, from there trivially resulting that $\frac{CX}{NX}=2=\frac{GX}{NX}\Rightarrow CD=3GF$.

Now observe that $\angle{MEX}=\angle{MDF}=\angle{MCX}$, so the quadrilateral $MECX$ is cyclic, so $\angle{MXC}=\angle{MEA}=\angle{ADE}$. $(2)$

And $\angle{MCX}=\angle{ADF}$. $(2')$

Again, $\angle{CMN}=\angle{FDB}$ $(3)$, and $\angle{XMN}=\angle{XMC}-\angle{CMN}=\angle{CEF}-\angle{FDB}=\angle{EDC}$. $(3')$.

Now using $(2)$, $(2')$, $(3)$, $(3')$ in $\frac{NX}{NC}=\frac{\sin{MCX}}{\sin{MXC}}\cdot \frac{\sin{XMN}}{\sin{CMN}}$ we obtain that:

$N$ is the midpoint of $(CX) \iff \frac{\sin{FDA}}{\sin{EDA}}= \frac{\sin{BDF}}{\sin{CDE}}$

But because $DA$ is a symmedian in $\triangle{EDF}$, we have that $\frac{\sin{FDA}}{\sin{EDA}}=\frac{FD}{ED}=\frac{\sin{DEF}}{\sin{DFE}}=\frac{\sin{BDF}}{\sin{CDE}}$.
Therefore $N$ is the midpoint of $(CX)$, so the problem is solved.

Dear pohoatza, I have learned a lot from your solution, thank you for your help.


I)
(a) na deixtei oti to poluonhmo $ z^{2n}\plus{}z^n\plus{}1$ n anoikoi fusikous diaforo tou 0 diairei to poluonhmo$ z^2\plus{}z\plus{}1$ an kai mono an to n dn einai polaplasio tou 3
(b) na bre8ei h anagea sun8ikh pou prepei na ikanopoioun oi fusikoi p,q wste to $ z^p\plus{}z^q\plus{}1$ na diairhte apo to $ z^2\plus{}z\plus{}1$

II) m anoikoi Z kai einai m>0 , na bre8ei to m gia $ 1!3!...(2m\minus{}1)!\equal{} (\frac{m(m\plus{}1)}{2})!$

[quote="KapioPulsar"][quote="Γιώργος"][quote="KapioPulsar"]I)
(a) na deixtei oti to poluonhmo $ z^{2n} \plus{} z^n \plus{} 1$ n anoikoi fusikous diaforo tou 0 diairei to poluonhmo$ z^2 \plus{} z \plus{} 1$ an kai mono an to n dn einai polaplasio tou 3
(b) na bre8ei h anagea sun8ikh pou prepei na ikanopoioun oi fusikoi p,q wste to $ z^p \plus{} z^q \plus{} 1$ na diairhte apo to $ z^2 \plus{} z \plus{} 1$

II) m anoikoi Z kai einai m>0 , na bre8ei to m gia $ 1!3!...(2m \minus{} 1)! \equal{} (\frac {m(m \plus{} 1)}{2})!$[/quote]

gia to II o Bertrand tin skotwnei :lol:[/quote] :D  kai h prwth efkolh einai apla einai bradi twra kai barieme to aode!! :wink:[/quote]

lol :rotfl:  :rotfl:  :rotfl:

[quote="Γιώργος"][/quote][quote="KapioPulsar"][quote="Γιώργος"][quote="KapioPulsar"]I)
(a) na deixtei oti to poluonhmo $ z^{2n} + z^n + 1$ n anoikoi fusikous diaforo tou 0 diairei to poluonhmo$ z^2 + z + 1$ an kai mono an to n dn einai polaplasio tou 3
(b) na bre8ei h anagea sun8ikh pou prepei na ikanopoioun oi fusikoi p,q wste to $ z^p + z^q + 1$ na diairhte apo to $ z^2 + z + 1$

II) m anoikoi Z kai einai m>0 , na bre8ei to m gia $ 1!3!...(2m - 1)! = (\frac {m(m + 1)}{2})!$[/quote]

gia to II o Bertrand tin skotwnei :lol:[/quote] :D  kai h prwth efkolh einai apla einai bradi twra kai barieme to aode!! :wink:[/quote][quote="Γιώργος"]

lol :rotfl:  :rotfl:  :rotfl:[/quote] :D  :lol:

[quote="KapioPulsar"]Βρείτε όλους τους φυσικούς αριθμούς $m$ για τους οποίους ισχύει ότι $1! \cdot 3! \cdot 5 ! \cdot ... \cdot (2m-1)! = (\frac{m(m+1)}{2})!$[/quote] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=462262#p462262]εδώ[/url] (άσκηση 1)