geometry
3D geometry
octahedron
ratio
icosahedron
pyramid
trigonometry

Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O$, and that the ratio of the volume of $O$ to that of $C$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers, find $m+n$.

[hide] We have an icosahedron with 8 faces, 2 pyramids, each a triangle. it would be best to investigate a simpler case.  We decide to look at it sideways, yields 2 iscoceles triangles. Let c be a side of the cube.  The portion that extends in one triangle is c/2, while in the other c/2 as well. The diagonal of  the cube, sqrt 2 c, is 2/3 of the height of one of pyramids, because according to similar triangles & centroid.  Let the square base be 3/2 \times \sqrt 2 c. Therefore there are 2 spaces left over whe nyou subtract 2\sqrt c from base.  Then you can find the heght of pyramid using similar triangles and and basically just compare 2 ratios and you find  ratio is 1: 27/6=9/2, so yea the sum of m + n = 11 .  \][/hide]

Yea i need to work on solution writing... I suck at that but o well

Huh...I used that it was a cube, and got 11...
[hide]Let each edge of the octahedron have length 1.  Then the altitude of one of the equilateral faces has length $\frac{\sqrt{3}}{2}$.  By the Pythagorean Theorem, the height of one of the two square pyramids which make up the octahedron is $\sqrt{\frac{3}{4}-\frac{1}{4}}=\frac{1}{\sqrt{2}}$.  Thus, the volume of the octahedron is $(\frac{1}{3})(1)(\frac{2}{\sqrt{2}})=\frac{\sqrt{2}}{3}$.

Now, consider the square formed by the four "middle" vertices of the octahedron.  Let that square be $WXYZ$.  Let the two other vertices of the octahedron be $M$ and $N$.  Let two adjacent vertices of the cube be $A$ and $B$, so that $A$ and $B$ lie on two triangles which share, as a side, one of the edges of that square (WLOG, let the shared edge be $WX$).  Let the midpoint of $WX$ be $C$.  Then we have an obtuse triangle, $ABC$, whose longest side is one edge of the cube, and whose largest angle is twice the angle between $WXYZ$ and $WXM$.  By our previous calculation of the volume of the octahedron, we found a right triangle which contains the angle between the square and the triangle.  We found that angle, $\theta$, is
$\sin^{-1}{\frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}}=\sin^{-1}{\frac{\sqrt{2}}{\sqrt{3}}}}$.
Hence, $\sin{\theta}=\frac{\sqrt{2}}{\sqrt{3}} \implies \cos{\theta}=\frac{1}{\sqrt{3}}$.
Hence, $\cos{2\theta}=2\cos^2{\theta}-1=-\frac{1}{3}$.
Finally, $BC=AC=\frac{1}{3}(Altitude)=\frac{1}{3\sqrt{2}}$
So then, by the Law of Cosines,
$AB^2=BC^2+AC^2-2(AC)(BC)\cos{2\theta}=\frac{1}{12}+\frac{1}{12}-2(\frac{1}{12})(-\frac{1}{3})=\frac{1}{6}+\frac{1}{18}=\frac{2}{9}$
$\implies Area=AB^3=\frac{2\sqrt{2}}{27}$
And now a simple division yields $\frac{9}{2}$
[/hide]
But I didn't even check with the other sides to make sure I didn't mess up.  Oh well.

Yes 11 is what i got, I didnt use trig  :D  before it was 33 because i forgot to simplyfy 27/6=9/2, so i got 33, but yes 11 is correct

Bump...for submitting a solution for the contests section, I have to input a whole post.  Could someone type up a solution for this that contains no extraneous comments (and looks nice, i.e. LaTeX and spacing are good, and is correct) :D

Thanks

[hide]Consider octahedron $ABCDEF$, with $A$ and $F$ as opposite vertices. In other words, the square face is $BCDE$. Extend a perpendicular from $A$ to $G$ on $BCDE$.

Let's consider triangle $ABC$. Extend $A$ to $A'$ on $BC$. We know $AA'$ is divided by the centroid, $C'$ into two congruent portions of ratios $2: 1$.

If we set the side length $BC$ to $2x$, then $AA' = x\sqrt{3}$. Then we realize that by the Pythagorean theorem on right triangle $AA'G$, $AG = \sqrt{(x\sqrt{3})^2 - x^2} = x\sqrt{2}$.

Now draw a perpendicular from centroid $C'$ to $AG$, and call it $H$.

$\bigtriangleup AC'H \approx \bigtriangleup AA'G$, so we can find the side length of the square.

$\frac{C'H}{\frac{2\sqrt{3}}{3}} = \frac{1}{\sqrt{3}} \Rightarrow C'H = \frac{2x}{3}$

$s = C'H\sqrt{2} = \frac{2x\sqrt{2}}{3}$

So $\frac{O}{C} = \displaystyle\frac{\frac{8x^3\sqrt{2}}{3}}{\frac{16x^3\sqrt{2}}{27}} \Rightarrow = \frac{9}{2}$.

$9 + 2 = 11$[/hide]

[hide="Answer"]Set the side length of the octahedron to $1$ for (relative) simplicity. If we cut the octahedron in half into two pyramids, we use a 45-45-90 triangle to see that the height is equal to $\frac{\sqrt{2}}{2}$. Next, the point where the vertices of the cube touch each face of $O$ is the circumcenter of the triangle. In an equilateral triangle, this is $\frac{1}{3}$ of the way up the height of the triangle, and so it is also $\frac{1}{3}$ of the way up the height of the pyramid. The height of the equilateral triangle is $\frac{\sqrt{3}}{2}$, so the portion above the circumcenter has length $\frac{\sqrt{3}}{3}$; this is the hypotenuse of an equilateral triangle with half of a diagonal of the cube as the longer leg, and the shorter leg is $\frac{2}{3}$ of the height of the pyramid, or $\frac{\sqrt{2}}{3}$. Therefore, each side of the cube has length $\sqrt{2}\left(\sqrt{\frac{1}{3}-\frac{2}{9}}\right)=\frac{\sqrt{2}}{3}$. Therefore, the volume of the cube is $\frac{2\sqrt{2}}{27}$, and the volume of the octahedron is $2\left(\frac{1}{3}(1)^2\left(\frac{\sqrt{2}}{2}\right)\right)=\frac{\sqrt{2}}{3}$, and the ratio of the volumes is $\frac{\frac{\sqrt{2}}{3}}{\frac{2\sqrt{2}}{27}}=\frac{9}{2}\Rightarrow 9+2=\boxed{11}$.[/hide]

its amazing how much easier it is the other way around

(octahedron inside of cube)

For a number 13, this is really, really easy.

[quote="K81o7"]And funny how they re-use problems...[/quote]

What prblem are you talking about?

I'm not completely sure, but it reminds me of an AMC12 problem...

Yeah, I'm pretty sure there was an AMC problem with an octahedron inside of a cube, and you had to find the ratio.

Yeah, on the recent AMC12...it was there.

[quote="me@home"]When the problem sais that the vertices of the cube are on the center of each face, how do you know that it is the circumcenter? Why isn't it the point half way up the triangle??[/quote]

It then would not be the center of the face.  The center of an equilateral triangle is either the circumcenter, the centroid, the orthocenter, or incenter.

And on an equilateral triangle, those are all the same point.

That was why i mentioned all of them.

Well, in reality, the "center" would be the centroid, I suppose...physically, it's the center of gravity of a triangle.

Essentially, on an equilateral triangle, it has to be a point symmetric with respect to the three vertices.  Hence, that point.


Let $ x,y,z\in \mathbb{R}$
such that:
$ x \plus{} y \plus{} z \equal{} 5$
$ xy \plus{} yz \plus{} zx \equal{} 3$

Obviously: $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 19$  :lol:

Sorry if I post a mistake solution!
We have:
$ (y\plus{}z)^2\leq 2(y^2\plus{}z^2)\Leftrightarrow (5\minus{}x)^2\leq 2(19\minus{}x^2)\Leftrightarrow (x\plus{}1)(3x\minus{}13)\leq 0\Leftrightarrow \minus{}1\leq x\leq \frac{13}{3}$

x,y,z are roots $ f(t)\equal{}t^3\minus{}5t^2\plus{}3t\minus{}q\equal{}0$ for some q. $ f'(t)\equal{}3t^2\minus{}10t\plus{}3$, therefore $ f_{max}\equal{}f(\frac 13 )\equal{}\frac{13}{27}\minus{}q, \ f_{min}\equal{}f(3)\equal{}\minus{}9\minus{}q$ and $ \minus{}9\le q\le \frac{13}{27}.$ Maximfl root we get when $ q\equal{}\frac{13}{27}$, then $ x\equal{}y\equal{}\frac{13}{27},z\equal{}5\minus{}2*\frac{13}{27}\equal{}4\frac{1}{27}.$

This is moroccan Team Selection Test  :rotfl:


Are there such pairwise distinct natural numbers m , n , p , q satisfying m+ n = p + q and [tex] \sqrt{m} + \sqrt[3]{n} = \sqrt{p} + \sqrt[3]{q} > 2004[/tex]

I solved this one at its time. Check for it.


4.Given is $n \in \mathbb Z$ and positive reals a,b. Find possible maximal value of the sum: $x_1y_1 + x_2y_2 + ... + x_ny_n$ when $x_1,x_2,...,x_n$ and $y_1,y_2,...,y_n$ are in $<0;1>$ and satisfies: $x_1 + x_2 + ... + x_n \leq a$ and $y_1 + y_2 + ... + y_n \leq b$

Hello,hope that the following solution is right.
The maximum value is [a]+{a}*{b} if [a]=[b] else it's equal to min{a,b}
Well we can take n>=a>=b,Sumx_j =a,Sumy_j=b.
Define the sequence “fill in”  (u,v) with (p,q) (0=<u,v,p,q=<1)as changing up+vq with
i)1+(p+u-1)(q+v-1) if p+u>=1,q+v>=1.
ii)u+p if u+p=<1,q+v>=1.
iii)(u+p)(v+q) if u+p=<1,q+v=<1
Then it’s easy to see that the new numbers are not less than the previous.
Now let’s  do this operation to the whole expression E=Sumx_j*y_j  i.e.
First of all “fill in” (x_1,y_1) with (x_n,y_n),then the new number “fill in” with (x_(n-1),y_(n-1)) until the (x_1,y_1) will become (1,1) then “fill in” (x_2,y_2),(x_3,y_3)…
After the operations will end it is easy to see that the new expression E’=< [a]+{a}*{b} if [a]=[b] else min{a,b}.So Sumx_j*y_j=< [a]+{a}*{b} if [a]=[b] else min{a,b} with equality for example when x_j=1 j=1,2,…,a.x_([a]+1)={a}. y_j=1 j=1,2,…,b.y_([b]+1)={b}
Sorry for bad english.

[quote="Megus"]Does nobody want to solve it ? I think it's quite interesting ...[/quote]
Problem is quite simple. For what grade was it proposed?

It is problem from the first stage [there are 3 stages] of Polish Math Olympiad which is basically for high school students

from assumption first stages are simple, in the Polish MO - we don't have something like grades - there is a seperate national contest for high-schools ...and for younger people who can manage to qualify threw consecutive
stages  for high-schools - 3 stages always , 1 is sending solutions , 2 is written
in main cities , 3 is for 60-120 top scorers from the country,organised in one place.


[quote="Inequalities Master"]Let $ a,b,c,d,e > 0$. Prove that:

$  \dfrac{ab}{a + b + c + d} + \dfrac{bc}{b + c + d + e} + \dfrac{cd}{c + d + e + a} + \dfrac{de}{d + e + a + b} + \dfrac{ea}{e + a + b + c}\le \dfrac{1}{4}(a + b + c + d + e) $[/quote]
It is not true. For example $ a = b = c = 1, d, e \rightarrow 0$


Track practice lasts for one hour from 2:30-3:30. At a randomly selected time during track practice, Tania looks at her watch. What is the probability that the minute and hour hand on her watch form an acute angle? Express your answer as a common fraction.

It is easy to see that all times from 3:00 to 3:30 are acute. (3:00 is right, 3:30 is acute, and the hands meet at 3:16-ish).

Also it is easy to see that any other time is obtuse, because 3:00 is right, 2:44-ish is straight, and 2:30 is obtuse.

A picture may help.

Ans: 1/2

2:30 is obtuse, from 2:30 to 3:00 the it reduces gradually to right (at 3:00).  Between 3:00 and 3:30, it is always acute.  So the first half hour is obtuse, and the 2nd half hour is acute and the answer is 1/2.


[quote="NikolayKaz"]$ x_1,x_2,...,x_n \in N$ $ n \in N$

$ x_1 \leq x_2 \leq ... \leq x_n$

Resolve

$ n^{x_1} \plus{} n^{x_2}... \plus{} n^{x_n} \equal{} n^{x_n \plus{} 1}$[/quote]

If $ n\equal{}1$, the equation is verified $ \forall x_1$

If $ n>1$ and dividing by $ n^{x_1}$, we get $ 1 \plus{} n^{x_2\minus{}x_1}... \plus{} n^{x_n\minus{}x_1} \equal{} n^{x_n \minus{}x_1 \plus{} 1}$

And since $ n>1|RHS$, the only solution is $ x_2\minus{}x_1\equal{}x_3\minus{}x_1\equal{}...\equal{}x_n\minus{}x_1\equal{}0$

And so :
$ \forall n\in\mathbb N$, the solution is $ (x_1,x_2,...,x_n)\equal{}(a,a,...,a)$ $ \forall a\in\mathbb N$


Find all function $ f: \mathbb{R}\to\mathbb{R}$ satisfy: $ f(xy)\plus{}f(x\plus{}y)\equal{}f(x)f(y)\plus{}f(x)\plus{}f(y),\forall x,y\in\mathbb{R}$

[hide]$ x \equal{} y \equal{} 0$ gives $ f(0)^2 \equal{} 0$, or $ f(0) \equal{} 0$.

$ x \equal{} 1, y \equal{} \minus{}1$ gives $ f(1)f(\minus{}1) \equal{} \minus{}f(1)$. Either $ f(1) \equal{} 0$ or $ f(\minus{}1) \equal{} \minus{}1$. If $ f(1) \equal{} 0$, then $ y \equal{} 1$ gives $ f(x\plus{}1) \equal{} 0$, meaning $ f$ is identically $ 0$. Assume this is not the case. So $ f(\minus{}1) \equal{} \minus{}1$.

$ x \equal{} y \equal{} \minus{}1$ gives $ f(1) \plus{} f(\minus{}2) \equal{} \minus{}1$. On the other hand, $ x \equal{} 1, y \equal{} \minus{}2$ gives $ f(1)f(\minus{}2) \plus{} f(1) \equal{} \minus{}1$. Combining these gives $ f(\minus{}2) \equal{} 0$ or $ f(1) \equal{} 1$.

Suppose $ f(\minus{}2) \equal{} 0$. Then $ f(1) \equal{} \minus{}1$. Putting in $ y \equal{} 1$ we get $ f(x\plus{}1) \equal{} \minus{}f(x)\minus{}1$. This shows that $ f(x\plus{}2) \equal{} f(x)$. However, $ x \equal{} 2, y \equal{} 1/2$ gives $ f(1) \plus{} f(2\plus{}1/2) \equal{} f(1/2)$, or $ f(1) \equal{} 0$. Contradiction. So $ f(1) \equal{} 1$.

$ y \equal{} 1$ now gives $ f(x\plus{}1) \equal{} f(x)\plus{}1$. We conclude that $ f(x\plus{}n) \equal{} x\plus{}n$ for all integers $ n$ (particularly, $ f(n) \equal{} n$). Then $ y \equal{} n$ gives $ f(nx) \equal{} nf(x)$. From this we easily deduce that $ f(x) \equal{} x$ on the rationals. If $ y$ is a rational number, then $ f(yx) \equal{} yf(x)$.

Put $ x \equal{} y$ in. We get $ f(x^2) \equal{} f(x)^2$. This shows that $ x > 0 \Rightarrow f(x) > 0$.

Let $ a$ be a rational and $ b$ a real such that $ a < b$. Put in $ x \equal{} a, y \equal{} b \minus{} a$. Then $ f(b) \equal{} a \plus{} f(b\minus{}a)$, showing that $ f(b) > a$. On the other hand, if $ a$ is a rational, $ b$ is a real such that $ a > b$, then $ x \equal{} b, y \equal{} a\minus{}b$ gives $ a \equal{} f(b) \plus{} f(a\minus{}b)$, or $ a > f(b)$. Combined, these two show $ f$ is increasing. Since $ f(x) \equal{} x$ on the rationals, $ f(x) \equal{} x$ for all reals.[/hide]

It was not given that $ f$ is continuous, and judging by a [url=http://www.mathlinks.ro/viewtopic.php?t=224241]similar problem[/url] that was published yesterday, it's not necessary to assume that.

Uh, was that directed at my solution? Continuity was never used.

Sorry - my bad. I skipped lightly over your solution and noticed just the final step, which seemed to conclude from $ f(x)\equal{}x$ on the rationals that $ f(x)\equal{}x$ on the reals.

[quote="MellowMelon"]
$ y \equal{} 1$ now gives $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$. We conclude that $ f(x \plus{} n) \equal{} x \plus{} n$ for all integers $ n$ (particularly, $ f(n) \equal{} n$). Then $ y \equal{} n$ gives $ f(nx) \equal{} nf(x)$. From this we easily deduce that $ f(x) \equal{} x$ on the rationals. If $ y$ is a rational number, then $ f(yx) \equal{} yf(x)$.

Put $ x \equal{} y$ in. We get $ f(x^2) \equal{} f(x)^2$. This shows that $ x > 0 \Rightarrow f(x) > 0$.[/quote]

I think at this point, you can just conclude $ f(x) > 0$ for all positive rational number $ x$.

Putting $ x \equal{} y$ into the original equation gives $ f(x^2) \plus{} f(2x) \equal{} f(x)f(x) \plus{} 2f(x)$. The $ f(2x)$ and $ 2f(x)$ cancel by results obtained in the solution. Then if $ x$ is a positive real, $ f(\sqrt{x}^2) \equal{} f(\sqrt{x})^2$, so $ f(\sqrt{x}^2) \equal{} f(x) > 0$. What's wrong here?

I see. There was some thing misunderstood  :D Yours is a perfect solution, MellowMelon!


calculus
integration
logarithms
calculus computations

Find the minimum value of $\int_0^1 \frac{|t-x|}{t+1}dt$

[quote="kunny"][quote="skylarth"]is it at $x = \sqrt{2}-1$ ??

then the minimum value is $3-2\sqrt{2}$

if im right ill post my solution.[/quote]

Yes, you are right, skylarth. :)[/quote]
how can we prove that?

[quote]Find the minimum value of $\int_0^1 \frac{|t-x|}{t+1}\ \mathrm{dt}$ , where $x\in\mathbb R$ .[/quote][b][u]Proof[/u].[/b] Denote $F(t)\in \int\frac{|t-x|}{t+1}\ \mathrm{dt}$ for which $F(0)=0$ , i.e. $F(t)=t-(x+1)\ln|t+1|$ . Thus,

$f(x)\equiv\int_0^1 \frac{|t-x|}{t+1}\ \mathrm{dt}\implies$ $f(x)=\{\begin{array}{ccc}
F(1)-F(0) & \mathrm{if} & x\le 0\\\\
F(0)+F(1)-2F(x) & \mathrm{if} & 0\le x\le 1\\\\
F(0)-F(1) & \mathrm{if} & x\ge 1\end{array}\iff$

$f(x)=\{\begin{array}{ccc}
1-(x+1)\ln 2\ \ \searrow & \mathrm{if} & x\le 0\\\\
1-(x+1)\ln 2-2x+2(x+1)\ln (x+1)\ \ \searrow\ \nearrow  & \mathrm{if} & 0\le x\le 1\\\\
(x+1)\ln 2-1\ \ \nearrow & \mathrm{if} & x\ge 1\end{array}$ ,

where $\{\begin{array}{c}
f(0-0)=f(0+0)=1-\ln 2\\\\
f(1-0)=f(1+0)=2\ln 2-1\end{array}|$ . Therefore, $x_{\mathrm{min}}\in (0,1)$ .

For any $x\in (0,1)$ we have  $f'(x)=\ln\frac {(x+1)^2}{2}$ and $f'(x)=$ $0\iff$ $x=\sqrt 2-1$ .

In conclusion, $x_{\mathrm{min}}=\sqrt 2-1$ and $\min_{x\in\mathbb R}f(x)=f(x_{\mathrm{min}})=f(\sqrt 2-1)=3-2\sqrt 2$ .


Asymptote
analytic geometry
graphing lines
slope
trigonometry
conics
parabola

The problem is early on in the section, so it shouldn't be hard... BUT a few questions about the fundamentals of math come about when looking at the book's six answers....

Furthermore, when graphing one function... it seems at which the points the limits occur can actually construct a parabola... COME ON CALC PEOPLE !!!

?

#15
[hide]
Find '$x$'. Restraints : $[0,2\pi)$

$tan(2x)-cot(x) = 0$, thus...
$\frac{2tan(x)}{1-tan^{2}x}= \frac{1}{tan(x)}$
$2tan(x) = \frac{1-tan^{2}x}{tan(x)}$
$2tan^{2}x = 1-tan^{2}x$
and finally,
$3tan^{2}x-1 = 0$

Solving for $x$, we have
$tan(x) = \pm\frac{1}{\sqrt{3}}$
so,
$x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$

The book says that, in addition to my solutions, $x = \pi, \frac{3\pi}{2}$.
-I don't think so-
another solution, or 'simpler' one (Quotient ID's), is founded on $sin(x)$, giving
$\frac{4sin^{2}x-1}{1-2sin^{2}x}= 0$
Unsurprisingly, the top yields solutions $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$
But, to my understanding, our denominator implies that $x \not= \pi, \frac{3\pi}{2}$
Meanwhile, a graph of $f(x) = tan(2x)-cot(x)$ yields an asymptote at $n\frac{\pi}{6}$

!?!?!?!?

b) Write and equation connecting the points where the slope is $\infty$
    -I think you can, and that it would be a parabola.
[/hide]

Thanks for the help in advance, I have no idea how much time part b might take.... I think theres a formula for it somewhere.... :maybe:

Hesitantly,
                       igiul

a) I think your book may be wrong - $\pi$ and $\frac{3\pi}{2}$ is undefined at $\tan{2x}-\cot{x}$....

b) When you say a slope of infinity, do you mean a changing, increasing slope? If so, then a parabola would work - but so would a selection of a whole myriad of equations! (I know this is advanced for basics but...) an increasing slope is just the derivative of the equation giving a graph that increases. So polynomial (not linear - remember, nor negative),  exponential etc... all work.

Examples -

$\frac{x}{45}^{2}$ , $52x^{3}+89$ , $23^{x}+451$ etc...

--
Unless you mean the equation of a vertical line - which has a slope of infinity...

The formula for slope is $m = \frac{y_{2}-y_{1}}{x_{2}-x_{1})}$ Infinity means denominator is zero. $x_{2}-x_{1}= 0$ $x_{2}= x_{1}$ - only a verticle line slope works - Example $x=\frac{523}{142}$

I'm referring to a parabola. Graph $3tan^{2}x-1$ on a calculator, then go to zoomfit.
It's shockingly clear. You'll see exactly what I mean; a parabola that intercepts 0,0 and those points.

I edited my post - theres another one I forgot that's probably it.

The funciton (and others) cross the x-axis at all of those book - answers actually - so idk- there is no asymptote at pi/6 though... - (never mind previous answer- mistyped into calc)

Hmm.. its not exactly a parabola - but your right there is an asymptote I think. Further more - when I plugged into my equation solver the values of pi/6 and  3p/2 in the original equation - they cam up as exactly the same thing - as with this "parabola you have" - its not an asymptote though - I dont think.

[quote="igiul"]Write and equation connecting the points where the slope is $\infty$[/quote]

I don't follow.  A point on the graph with infinite slope has no defined $y$-coordinate, so how can an equation pass through it?

I think that i termed it incorrectly- i don't think m = $\infty$
so.... that leaves us with a sharp, sharp curve/asymptote.... top one is around 9000...haven't really done many tests, nor does it seem, can I. My calc doesnt want a window over 10k.
I suppose w/out calculus i could count down from about 10k solving for x, and when the equation returns true, thats the max.

I don't want to do that.

can you calculate a limit value that is less than .... -well infinity, for y as x---> value(v) that causes maximum Y value?

Those would be the points i'm referring too i think. graph it on a calculator, you'll see exactly what im talking about; What i'm wondering if a parabola can intersect these 'points'.

clear?

You're using the word "limit" incorrectly, so I don't really know what you're talking about.  Are you trying to

a) find the parabola tangent to a curve at a given point, which for a function $f(x)$ at a point $x_{0}$ is simply $y = f(x_{0})+f'(x_{0}) (x-x_{0})+\frac{f''(x_{0}) (x-x_{0})^{2}}{2}$, or
b) find a parabola that passes through some finite set of points that you have as of yet not defined very well at all and that I still cannot see even after graphing on my calculator?

Sigh. I'm trying. On my TI-83+, sudden juts point up out of the graph of $3tan^{2}x-1$. The tips of these spikes, of which there are three to each side, ascend in height as you move away from the origin. Namely the points to which I am referring, is when these spikes end, creating (including 0,0) what could look like a perfect parabola, if the points of these spikes were to be connected by an equation of quadratic origin.

Some calculators do not graph asymptotes in a complete manner, but rather a fading line, as the distance between dots increases.

[hide]DOES THIS HELP SIR?[/hide]

[quote="igiul"]Namely the points to which I am referring, is when these spikes [b]end[/b][/quote]

This is an artifact of both rounding and pixelation, and doesn't correspond to the actual graph of the function.  Calculating these points, which will differ from calculator model to model, would require knowing technical details about your calculator, and these points have no special meaning.

nonetheless, this misconception remains unnervingly convincing.

lock it.


Prove or disprove for positive numbers $x,y,z$
\[\frac{xy}{x^{2}+xy+yz}+\frac{yz}{y^{2}+yz+xz}+\frac{xz}{z^{2}+xy+xz}\leq1.\]

[quote="ozgurkircak"]Prove or disprove for positive numbers $x,y,z$
\[\frac{xy}{x^{2}+xy+yz}+\frac{yz}{y^{2}+yz+xz}+\frac{xz}{z^{2}+xy+xz}\leq1. \]
[/quote]
It's equivalent to $\sum_{cyc}(x^{3}y^{3}+x^{4}yz-2x^{3}z^{2}y)\geq0,$ which triviality.

[quote="Tomaths"]$\iff \sum \frac{x^{2}+xy}{x^{2}+xy+xz}\geq 2 \iff 2(\sum x^{2}-\sum xy)(\sum \frac{1}{x^{2}+xy+xz}) \geq 3 \times 0$ by chebyshev[/quote]

so is it right?


I need the ancient Greek word for ray as in a graph or related to geometry.  I searched the net but came up short.

Um, why? :? And uh, what's the point of knowing an ancient Greek word?

I just want to know the actual name for a ray in Greek but it is not important.

In new geek we told the ray ray=actina and in Ancient Greek ray=actis


percent
trigonometry
geometry
3D geometry

2 (\frac{\sqrt[3]{8}}{\text{tan } 45^o} \times \frac{\frac{\sqrt[3]{729}}{\sqrt{\pi^2}}}{\frac{\text{circumference of circle}}{\text{diameter of circle}}} \times \text{ the reciprocal of } (\frac{\text{sin } 45^o}{\sqrt{2}}) \times \frac{1}{\sqrt[3]{27^2}})

Does anybody know what this equals?

For the sake of comprehension, I would LaTexify the problem:
[b]Wickedestjr[/b] wrote:
\[ 2 (\frac{\sqrt[3]{8}}{\text{tan } 45^o} \times \frac{\frac{\sqrt[3]{729}}{\sqrt{\pi^2}}}{\frac{\text{circumference of circle}}{\text{diameter of circle}}} \times \text{ the reciprocal of } (\frac{\text{sin } 45^o}{\sqrt{2}}) \times \frac{1}{\sqrt[3]{27^2}}) \]
Does anybody know what this equals?

Simply simplify.

I am $ 99$ percent sure I made a mistake... but... $ \frac {2}{\pi^2}$?

[hide="Hints"]$ \sqrt[3]{8} \equal{} 2$

$ \tan 45^{\circ} \equal{} 1$

$ \sqrt[3]{729} \equal{} 9$

$ \sqrt{\pi^{2}} \equal{} \pi$

circumference / diameter = $ \pi$

reciprocal of $ \frac{\sin 45^{\circ}}{\sqrt{2}} \equal{} 2$

$ \sqrt[3]{27^{2}} \equal{} 3^{2} \equal{} 9$[/hide]

[quote="leoxnlin"]Simply simplify.

I am $ 99%
$ sure I made a mistake... but... $ \frac {2}{\pi^2}$?[/quote]
That seems way off...
but the first step is to simplify the cube root of 8...which is 2  :P

EDIT: Beaten =[

hello, i have
$ 2 \cdot 2 \cdot 9 \cdot \text{the reciprocal}\left(\frac {\frac {\sqrt {2}}{2}}{\sqrt {2}}\right)\cdot\frac {1}{9} \equal{} 2\cdot18\cdot 2\cdot\frac{1}{9} \equal{} 8$.
Sonnhard.

648, hmm, any other answers that are different?

hello, no the right answer is $ \boxed{8}$, i have something misunderstood
what the reciprocal means.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have
$ 2 \cdot 2 \cdot 9 \cdot \text{the reciprocal}\left(\frac {\frac {\sqrt {2}}{2}}{\sqrt {2}}\right)\cdot\frac {1}{9} \equal{} 2\cdot18\cdot 2\cdot\frac {1}{9} \equal{} 8$.
Sonnhard.[/quote]

do you mean
$ 2 \cdot 2 \cdot \boxed{\frac {\frac {9}{\pi}}{\pi}} \cdot \text{the reciprocal}\left(\frac {\frac {\sqrt {2}}{2}}{\sqrt {2}}\right)\cdot\frac {1}{9} \equal{} 2\cdot18\cdot 2\cdot\frac {1}{9} \equal{} \boxed{\frac{8}{\pi^2}}$

i think for the first boxed part you put 9
this boxed part should be $ \frac {9}{\pi^2}$
so i think the answer is indeed $ \frac {8}{\pi^2}$

Hmm, I got that from ernie as his number for AoPS Survivor, but I ddin't know what it equaled, so I told him to send an integer.

Ah... I found my mistake...  :blush:

I meant $ \frac{\text{diameter of circle}}{\text{circumference of circle}}$.  :rotfl:

[quote="ernie"]Ah... I found my mistake...  :blush:

I meant $ \frac {\text{diameter of circle}}{\text{circumference of circle}}$.  :rotfl:[/quote]

Well, that would make it 8/pi.


factorial
ratio
real analysis
real analysis unsolved

show that $ \frac{3^n}{n!}$ converges to 0.

Do you know Stirling's formula? (or even a version without the constant $ \sqrt{2\pi}$?)

$ a_n \equal{} \frac{3^n}{n!} \equal{} \frac{3}{n} a_{n\minus{}1}$

let n > 3, then $ \frac{3}{n} < \frac{3}{4}$

so $ a_n \equal{}\frac{3}{n}\cdot \frac{3}{n\minus{}1} \dots \frac{3}{n\minus{}(n\minus{}4)} \cdot a_3 < \left( \frac{3}{4}\right)^{n\minus{}3} a_3 \to 0$

[quote="Kalle"]That is such ridiculous massive overkill, blahblahblah. Just note that the last $ n \minus{} 3$ factors of $ n!$ are larger than $ 4$, and then it follows by $ (3/4)^n \to 0$[/quote]

it's better (imo) to have a unified method for dealing with expressions involving asymptotic behavior of factorials than to try to find an ad hoc method to approach every problem of this type.

I wouldn't call Kalle's method ad hoc.  It makes perfect sense if the first thing you think of is to try to apply the ratio test, especially if you're thinking within the context of computing the interval of convergence of the power series of $ e^x$.  And I'd definitely call the ratio test a much more general principle than Stirling's formula.

On the other hand, for more difficult problems it's useful to know a more precise asymptotic for $ n!$ and certainly I agree with your point generically about unified methods - I often solve problems using more difficult techniques than necessary to elucidate underlying concepts.  Kalle's point about overkill is more just a general attitude that I'd call "conservation of sledgehammers" - if you don't need to pull out an extremely strong tool to solve a clear problem, don't.  (There is a distinction between "clear" and "straightforward" - computations can be straightforward without being clear.)


algorithm geometry rectangle inequalities

I do not know if this is the right place topost this.

Do you know a method to calculate a square root, like when you times and divide and plus and minus, you have a method calculate it manually.

And I am talking about calculating without a calculator.

See post #10 in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=85500[/url]

The explanation of the square root algorithm on the home-school site is well done. FYI: the geometry behind the method is pretty straightforward.  The given example was to find the sqaure root of 297,504 by paper and pencil.  Geometrically, this means to find a square with area 297,504.

To get you started, first remove a 500x500 square; then in a 'clever' step remove two side rectangles, along with the 'little' square in the corner; then repeat the last step until bored or satisfied.   Give it a try by trying to match the arithmetic to the geometry step-by-step.

There's also an algorithm based on the AM-GM-HM inequality, it's nice!
You have to find the square root of $n$.
First, you find two numbers $a_0$ and $b_0$ such that $a_0b_0=n$, you can take $a_0=1$ and $b_0=n$ if n is prime.

You proceed in this way:
$\displaystyle a_1 = \frac{a+b} 2$
$\displaystyle b_1 = \frac{2} {\frac 1a_0 \frac 1b_0} = \frac{a_0b_0} {a_1}$

$\displaystyle a_2 = \frac{a_1+b_1} 2$
$\displaystyle b_2 = \frac{2} {\frac 1{a_1} \frac 1{b_1}} = \frac{a_1b_1} {a_2}$

and so on.

$a_n$ and $b_n$ converge quickly to $\sqrt{n}$.  :)

We could use Newton's Method, but I think the convergence rate is poor.

http://mathworld.wolfram.com/NewtonsMethod.html


probability trigonometry ratio geometry inequalities

Triangle $ ABC$ is a right triangle with $ \angle ACB$ as its right angle, $ m\angle ABC \equal{} 60^\circ$, and $ AB \equal{} 10$. Let $ P$ be randomly chosen inside $ \triangle ABC$, and extend $ \overline{BP}$ to meet $ \overline{AC}$ at $ D$. What is the probability that $ BD > 5\sqrt2$?

[asy]import math;
unitsize(4mm);
defaultpen(fontsize(8pt)+linewidth(0.7));
dotfactor=4;

pair A=(10,0);
pair C=(0,0);
pair B=(0,10.0/sqrt(3));
pair P=(2,2);
pair D=extension(A,C,B,P);

draw(A--C--B--cycle);
draw(B--D);
dot(P);
label("A",A,S);
label("D",D,S);
label("C",C,S);
label("P",P,NE);
label("B",B,N);[/asy]

$ \textbf{(A)}\ \frac {2 \minus{} \sqrt2}{2} \qquad \textbf{(B)}\ \frac {1}{3} \qquad \textbf{(C)}\ \frac {3 \minus{} \sqrt3}{3} \qquad \textbf{(D)}\ \frac {1}{2} \qquad \textbf{(E)}\ \frac {5 \minus{} \sqrt5}{5}$

it's obvious that BC=5,and find a D on AC so that CD=5,and now BD=5*sqr(2).
the problem then becomes the possibility of P's locating in triangle ABD.
the answer is C.

First, we see that AC=5.square root(2) and BC=5
Let H be a midpoint of AC, so we have: BH^2=2(BC^2+AB^2)-AC^2=50.
Therefore, BH=5.square root(2).
If P is any point which lies on the half space containing AB of BH, we easily prove that BD>5.square root(2)
Now, if P lies on any line on the 2nd half space containing BC of BH, we see that in triangle BDH, angle(BDH) is always the obtue angle, so BH must be the largest side. Therefore, in those cases, BH>BD.
The final result is: 1/2.So D is a correct answer!
It's my mistypo! But I think the problem in my solution is not because of it.You should check it more carefully since I think it must be a very nice solution(no more construction and very natural, I THINK).I'll try to solve it completely again just by this method. :)

[b]ghjk[/b], you are over-complicating things a bit  :wink: Here is a paraphrase of [b]simon_math[/b]'s solution:

Construct point $ Q \in \overline{AC}$ such that $ \overline{CB} \cong \overline{CQ}$

$ \blacktriangleright \; P \in \triangle ABQ \iff BD > 5 \sqrt {2}$
$ \blacktriangleright \; P \in \triangle CBQ \iff BD < 5 \sqrt {2}$

The desired probability is therefore $ \frac {[\triangle ABQ]}{[\triangle ABC]} \equal{} \frac {AQ}{AC} \equal{} \frac {5 \sqrt {3} \minus{} 5}{5 \sqrt {3}} \equal{} \boxed{\textbf{(C)} \; \frac {3 \minus{} \sqrt {3}}{3}}$

[quote="The Zuton Force"][b]ghjk[/b], I think you are complicating things a bit  :wink: Here is a paraphrase of [b]simon_math[/b]'s solution:

$ \blacktriangleright \; P \in \triangle ABD \iff BD > 5 \sqrt {2}$
$ \blacktriangleright \; P \in \triangle CBD \iff BD < 5 \sqrt {2}$

The desired probability is therefore $ \frac {[\triangle ABD]}{[\triangle ABC]} \equal{} \frac {AD}{AC} \equal{} \frac {5 \sqrt {3} \minus{} 5}{5 \sqrt {3}} \equal{} \boxed{\textbf{(C)} \; \frac {3 \minus{} \sqrt {3}}{3}}$[/quote]
Indeed I don't see your solution?How can you infer these properties?
$ \blacktriangleright \; P \in \triangle ABD \iff BD > 5 \sqrt {2}$
$ \blacktriangleright \; P \in \triangle CBD \iff BD < 5 \sqrt {2}$
And I can't find any mistakes in my solution. Can you check it and point it out for me? :wink: I don't think the first out of 2 statements you give is correct since AC=5 square root(2) too.It can lead to contradiction to your statement if you compare BD and AC in the 1st case you showed.

Sorry, I wrote $ D$ in several places where I meant to write $ Q$ -- see the edited version.

Basically, the longer $ CD$ is, the longer $ BD$ is. So, there is some definite point $ Q$ on $ CA$ such that $ BQ \equal{} 5 \sqrt {2}$. If $ D$ is on the "left" side of $ Q$, then $ BD$ is less, and if $ D$ is on the "right" side of $ Q$, then $ BD$ is more.

Finally, notice that $ P$ and $ D$ are on the same side of $ \overline{BQ}$.

The mistake in your solution was the statement $ BH^2\equal{}2(BC^2\plus{}AB^2)\minus{}BC^2\equal{}50$. You add in an extra $ AB^2$ on the right side.

OK! I saw your edited solution!And I think I know where my mistake is :oops: .That is, I considered the probability that BD>[b]=[/b]5*square root(2). Since the equality can't happen, I have to put D on the right of the point H(H is midpoint of AC). And using the laws of cosine to find the estimation of the length of AD when BD>5*square root(2).That will lead to the maximum ratio of area between triangle ABD and ABC, which is equivalent to the probability we need to find.

Never mind! I miscalculated the length of BH :oops:

Here's my official solution :wink: !
Using Law of cosine we have: BD^2=100+AD^2-10*square root(3).AD
We need to find AD in order to satisfy:BD^2>50
And that is equivalent to solve the inequality:50+AD^2-10*square root(3).AD>0
We will have two roots, that is: AD>5.square root(3)+5 or AD<5.square root(3)-5
Since AC=5.square root(3)>AD>0, we choose the 2nd value of AD, that is: AD<5.square root(3)-5
Now, let's use the ratio of area between triangle ABC and ABD we see that the ratio<[square root(3)-1]/square root(3)=[3-square root(3)]/3, which is equivalent to the probability in oder to have BD>5.square root(2)(even the case when equality happens we still reach this answer!More suprising :roll: )
Therefore, C is the correct answer.
PS: The lesson for me:"Don't over complicating simple problems because it can lead me to "disaster"!! :D

What would the probability be if it's $ BD \ge 5\sqrt2$?

It would be the same. A line segment has an infinitely small area, since it has an infinitely small thickness.

You are just calculating the probability that point $ P$ lies in triangle $ \triangle BAD$. The probability is then
\[ \dfrac{ [ABC] \minus{} [BCD] }{ [ABC] }\]
Now, $ \triangle BCD$ is just a 45-45-90 triangle, and its area is $ \frac{25}{2}$. The area of $ \triangle ABC$ is obviously $ \frac{ 25\sqrt{3} }{ 2 }$. So our answer is just
\[ \dfrac{ \frac{ 25\sqrt{3} }{ 2 } \minus{} \frac{25}{2} }{ \frac{ 25\sqrt{3} }{ 2 } }\].
Simplifying this yields
\[ \boxed{ \dfrac {3 \minus{} \sqrt{3} } {3} }\]

The answer is (C).

Use special right triangles and geometric probability. This seems too easy for AMC 12 #22.

Please dont bump old threads. Also note that this was a very old amc 12 so they are easier.

[b]Solution:[/b] Notice that, for $BD > 5\sqrt{2}$, point $P$ would have to be to the right of the line originating from $B$ making a $45$ degree angle with $\overline{BC}$ (let the intersection point of this line with $\overline{AC}$ be $X$). We want to find the probability it is to the right of this said line by comparing the area of $\triangle{ABX}$ to $\triangle{ABC}$. This is a simple computation, and we get C.


if $p\geq 5$ be a prime number and $1\le a,b$ be two integer.Prove that:

$p^3|\binom {pa}{pb}-\binom{a}{b}$

probably use Lucas Correspondence Theorem.

more @ [url=http://mathworld.wolfram.com/LucasCorrespondenceTheorem.html]mathworld.wolfram.com/LucasCorrespondenceTheorem.html[/url]

[sorry , stupid hint]


I am planning on doing it since:

1) My school doesnt offer AP(it offers IB, but no CS)
2) I cant find AP computer scinece course anywhere
3) I still want the credit to go on Transcript.

I have talked to our IB coordinator and she said as well as i have some kind of syllabus to go by, I can do it. Has anyone done this this way???

As someone who has taken it, I can also state that IBCS isn't worth the time you invest in it.  Not at all.

Have you looked at [url=http://epgy.stanford.edu]EPGY[/url] and [url=http://www.eimacs.com]eIMACs[/url] for your self-study of AP computer science?

[quote="tokenadult"]Have you looked at [url=epgy.stanford.edu]EPGY[/url] and [url=http://www.eimacs.com]eIMACs[/url] for your self-study of AP computer science?[/quote]

EPGY don't offer AP. However, I will be looking at eIMACs. Thanks, tokenadult.

eIMACS only offer the review class. aNywhere else??

Apex Learning doesn't offer it either. Maybe check out the Collegeboard's official site for AP COmputer Science and see what textbooks it uses.

http://apcentral.collegeboard.com is a great place to get AP stuff too.

As long as you have experience in another programming language besides Java (maybe C++) and are willing to spend the time to study, then you will have no problem.

I am *trying* to self-study APCS right now with previous knowledge of Visual Basic, and I can tell you that it is extremely frustrating. I have an incompetent first-year teacher who barely knows how to turn the computer on and a worthless textbook that explains things out of order. If I can pull off a 3 on the AP test, I'm happy.

I'd recommend that you borrow a book from the library that teaches the basics before you attempt to try an AP book. Most of these AP books treat you like idiots and expect you to understand the syntax right off the bat.

I used Barron's, but since last year was the first year with Java as the AP CS language, the practice tests were much harder than the actual AP test, which is a good thing I suppose.  :D I got a 5, but I've been programming in C++ and a bit of Java for a couple of years. A suggestion: know the case study, or you're going to spend precious moments during the test learning what each class and method does, while you can just learn all that beforehand, since the College Board releases it.

khan33: Which test did you take? CS A or AB? If it's A, would you mind helping me out? I just went to the library today to look for a book, but since it's April, all the AP books in every subject are checked out. Oh man, the AP test is next month, and I don't understand a thing.

[quote="paladin8"]I self-studied AP Computer Science AB from "Data Structures and Algorithms in Java (2nd Edition)," which can be found [url=http://www.amazon.com/exec/obidos/tg/detail/-/0672324539/qid=1113092617/sr=8-2/ref=pd_csp_2/104-0626265-6944753?v=glance&s=books&n=507846]here[/url].
I got a 5, so it seemed to work pretty well :lol: It's definitely a big book, though, and I don't think you need all the topics discussed in it. My programming background was 2 EPGY courses in C already.[/quote]

which courses???

[quote="Forsaken"]khan33: Which test did you take? CS A or AB?[/quote]

I took AB, which is nearly all data structure stuff in the free response. I'm sure a local bookstore will carry at least one AP CS book. in the meantime, you can download the syllabus from the [url]http://www.collegeboard.com[/url] and there's practice questions included in there. The syllabus is really detailed, so it's helpful to have that at your side when studying. You can also go to [url]http://www.cookienest.com/content/manual-javabasics.php[/url] if you're just getting started with Java.

[quote="xxreddevilzxx"]which courses???[/quote]

The first two C programming ones, [url=http://www-epgy.stanford.edu/courses/cs/]C11A and C11B[/url]; I didn't take the third one because I was taking APCS anyway, where I studied basically the same thing except in Java.


How many $ 15$-letter arrangements of $ 5$ A's, $ 5$ B's, and $ 5$ C's have no A's in the first $ 5$ letters, no B's in the next $ 5$ letters, and no C's in the last $ 5$ letters?

$ \textbf{(A)}\ \sum_{k\equal{}0}^5\binom{5}{k}^3 \qquad
\textbf{(B)}\ 3^5\cdot 2^5 \qquad
\textbf{(C)}\ 2^{15} \qquad
\textbf{(D)}\ \frac{15!}{(5!)^3} \qquad
\textbf{(E)}\ 3^{15}$

[hide="Idea"]
Let each group of 5 characters be denoted $ s_1,s_2,s_3$ respectively. There are no $ a's$ in $ s_1$, no $ b's$ in $ s_2$, no $ c's$ in $ s_3$. Assume are $ k$ $ a's$ in $ s_2$ and $ 5\minus{}k$ $ a's$ in $ s_3$. Since there are no $ b's$ in $ s_2$, the remaining $ 5\minus{}k$ slots are $ c's$, and then there are $ k$ $ c's$ in $ s_1$. The same thing holds for the $ b's$, except there are $ k$ in $ s_3$ and $ 5\minus{}k$ in $ s_1$.
[/hide]

Obviously D and E don't work for our case. C doesn't work because we have limited set of symbols, and at some point of time we cannot choose between 2. B doesn't work either, if we take a simple interpretation to make all possible arrangements out of 1,2,3, so that hundreds digit can be either 1 or 2, tens 2 or 3, units digit 1 or 3 we will be able to obtain 2 different numbers - 123 and 231 not 6. Thereof, the answer is A.

I feel like ranting. I think this problem is way too guessable.

[hide="Useless Guessing Solution"]If we consider the generalizations of the choices, where there are $ n$ of each, with choices:

$ \textbf{(A)}\ \sum_{k \equal{} 0}^n\binom{n}{k}^3 \qquad \textbf{(B)}\ 3^n\cdot 2^n \qquad \textbf{(C)}\ 2^{3n} \qquad \textbf{(D)}\ \frac {(3n)!}{(n!)^3} \qquad \textbf{(E)}\ 3^{3n}$

Now use $ n \equal{} 1$, where the only two possiblities are 2,3,1 and 3,1,2. The choices translate to:

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 27$[/hide]

@veezbo: I think your main error was saying that there are $ \binom{5}{5\minus{}n}$ ways to pick the B's.  The B's are indistinguishable, so there is no need to pick which you use.  I may have misinterpreted you, though.

Here's what I came up with:
[hide="solution"]
As before, divide the spaces into the three zones of 5. Look at Zone 3.  Only A's or B's can go in here, and there can be anywhere from 0 to 5 A's.  So let $ k$ be an arbitrary number from 0 to 5, inclusive, and say there are $ k$ A's in Zone 3.  There are $ \binom{5}{k}$ ways to pick which spaces in zone 3 the A's take up.  After choosing this, the remaining $ 5\minus{}k$ spaces must have B's.

Now, in Zone 2, there are the remaining $ 5\minus{}k$ A's and $ k$ C's.  Again, there are $ \binom{5}{k}$ ways to pick which of the spaces are occupied by C's.

Finally, in Zone 1, there are the remaining $ k$ B's and $ 5\minus{}k$ C's.  So there are $ \binom{5}{k}$ ways to pick the spaces occupied by B's.

Altogether this is $ \binom{5}{k}^{3}$.  Since we picked an arbitrary $ k$, this is the answer in each of the 6 cases, so the final answer is $ \sum_{k\equal{}0}^{5}\binom{5}{k}^{3}$
[/hide]

[quote="cyberspace"]@veezbo: I think your main error was saying that there are $ \binom{5}{5 \minus{} n}$ ways to pick the B's.  The B's are indistinguishable, so there is no need to pick which you use.  I may have misinterpreted you, though.
[/quote]

Sorry, I meant picking which spaces for the B's.

Ok, I understand you're solution, I'm just still trying to figure out why mine doesn't work so I don't make that mistake in the future.

I'll bet it has something to do with my $ \binom{10}{5}$. I don't think that I can just suddenly pick 5 of the spaces from 10, but I have to pick exactly what is defined by my $ n$ (or your $ k$)

[hide="My Real Solution"]

Pick n spaces for n A's to go in Zone 3...$ \binom{5}{n}$

Then, we have 5-n A's left over to place in Zone 2. Pick $ \binom{5}{5 \minus{} n}$ spaces in Zone 2 for these A's.

For the B's, 5-n will go in Zone 3. So, we have to pick n spaces for n B's in Zone 1...$ \binom{5}{n}$

The C's are already determined.

And this just simplifies to:

$ \sum_{n \equal{} 0}^{5}\binom{5}{n}^{3}$

[/hide]

Cool. Oh, and I just realized $ \binom{10}{5}$ does not equal $ \binom{5}{n}$

Place $k$ out of the $5$ $C$'s in the first $5$ spots, and then place the remaining $5-k$ $C$'s in next $5$ spots. This can be done in $\dbinom{5}{k}^2$ ways. Fill in the remaining $5-k$ unfilled spots in the first $5$ letters with $B's$, and then place the remaining $k$ $B$'s in the last $5$ spots in $\dbinom{5}{k}$ ways. Now fill in all the unfilled spots with the $A$'s. There are $\dbinom{5}{k}^3$ such orderings. Now sum over all $k$, so the answer is $A$.

@viperstrike: Your solution is cyberspace's solution verbatim. If you are not contributing an original solution, don't revive old threads.

Oh I though it was a good thing to revive old threads, sorry.

The combinatorial sum method is the best way to go in this case, and the following "solution" actually ends up being combinatorial sums as well, but it is cool new way to look at the problem, and for whomever should come across this problem in the future looking for a combinatorial species solution, I will have hopefully provided some insight!
how many ways are there to arrange AAAAABBBBBCCCCC into 15-lettered words such that there is no A in the first 5, there is no B in the second 5, and there is no C in the third 5. the letters are indistinguishable.
This is isomorphic to species $\widetilde{\text{LinOrd}} (Y+Z) \cdot \widetilde{\text{LinOrd}} (X+Z) \cdot \widetilde{\text{LinOrd}} (X+Y)$, where the tilde over the LinearOrdering species indicates that we are working with the unlabelled generating function. We use unlabelled because the letters are indistinguishable. The species X corresponds to A, Y to B, and Z to C. Recall that the cycle index series of LinearOrdering is $\mathcal{Z}_{\text{LinOrd}} (x_1,x_2,...) = \dfrac{1}{1-x_1}$. Since we want unlabelled composition of species, we perform $\mathcal{\widetilde{F}}_{\text{LinOrd}(X+Y)} = \mathcal{Z}_{\text{LinOrd}}(x+y,x^2+y^2,x^3+y^3,...)=\dfrac{1}{1-(x+y)}$
Since we desire the number of ways to arrange 5 of each species, we actually desire $[x^5y^5z^5] \dfrac{1}{1-(x+y)}\dfrac{1}{1-(y+z)}\dfrac{1}{1-(x+z)}$ From here it is routine algebra to finish the problem.

[quote=veezbo]Sorry for reviving, but I don't understand why I am wrong.
Here's how I did it:

Divide the 15 spots up into zones, $ Z_A$, $ Z_B$, and $ Z_C$, each with 5 spots, and they denote which letter cannot go in the zone.

|_ _ _ _ _ | _ _ _ _ _ | _ _ _ _ _ |

For where the letter A($ Z_B$ or $ Z_C$) can go, we have $ \binom{10}{5}$

Let $ n$ denote how many A's are in $ Z_C$.

Now to satisfy the given conditions, the other $ 5 \minus{} n$ spaces in $ Z_C$ must be B's (otherwise, there will be empty spaces as no C's can be in $ Z_C$)
Thus, there are $ \binom{5}{5 \minus{} n}$ ways to pick the B's.

And for the C's, there are already determined, since they take all the remaining spaces in $ Z_A$ and $ Z_B$

So, my answer gives:

$ \sum_{n \equal{} 0}^{5} \binom{10}{5} \binom{5}{5 \minus{} n} \equal{} 8064$

Can someone please help? What am I doing wrong?[/quote]

I got the same answer, the same exact way. However, I'm not entirely sure what I did wrong... can anyone give me a bit of insight? Thanks in advance! :)

EDIT: Sorry for the revive, by the way! :(

Your cases don't match up. The ${10 \choose 5}$ is the number of ways to arrange all of the $A's$, instead of multiplying by the number of ways to arrange $n$ $A$'s in the third section. As a result, this method overcounts.

Hey guys sorry for revival x2 But why is the answer with the sigma all over k?

In other words why do we sum all over k? Is it just kind of common sense?

Another question- Why isn't it 2^15? It just seems so easy to say its 2^15 since you have 2 choices for the first 5 for each spot (since it can be either B or C, and so on until the last block)

[b]Solution:[/b] Split it up into three parts: first five, middle five, last five. Let there be $x$ Bs in the first five. Therefore, there must be $5-x$ Cs in the first five as well. However, this must mean that the remaining $x$ Cs must go in the middle five, which forces $5-x$ As to go in the middle slot as well. This leaves us with $x$ As and $5-x$ Bs in the last five. Notice that $0 \le x \le 5$ and that for a set value of $x$ the number of possible arrangements is $\binom{5}{x}^3$. Clearly, answer choice $\boxed{A}$ is correct.


今天該開始比賽了，有人有題目嗎？

我现在在成都  我是重庆的队员
题目：
1ST DAY
1. 已知$a^{2005}+b^{2005}$可以被表示为关于a+b和ab的多项式。求该多项式的系数和。

2.PA,PB是圆的切线,A,B是切点。PCD是割线，过B作EF//PA交AC于E,AD于F.求证：BE=BF.

3.集合S={1,2,...,2005},且S中任n个两两互质的不同的数中一定有一个质数，求n的最小值。

4.给定n>2 满足$|\sum^n_{i=1}x_i|>1, |x_i|\le1$
  求证存在k，满足$|\sum^k_{i=1}x_i-\sum^n_{i=k+1}x_i|\le1$

2ND DAY
5.圆O1,O2交于A,B，CD过O1且CD切圆O2于C,D在圆O1上。CA切圆O1于A.过A作AE垂直CD交圆O1于E,AF垂直CE于F.求证：BD平分AF.

6.等腰直角三角形中CA=CB=1,P是三角形边上一点。求PA*PB*PC的最大值。

7.正实数a+b+c=1，求证$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge1$

8.n个人，三人中必有两人认识，4人中必有2人不认识。求n的最大值。

顺便说一句，本人估计第一天全对，第二天5，6全对，7题没做，8题用一个不等式得到最值但没得到相应的图...
考得太丑陋了！

PS:如果某位给出7题的解答（非逐步调整），那么，感谢万分！

nttu gave a muirhead solution. I found a non-muirhead one.
Write the initial inequality as
$15(a+b)(b+c)(c+a)(3(\sum a^2 + \sum ab) -2) \ge 0$
Then it's trivial because $\sum a^2 + \sum ab \ge \frac{2}{3}(a+b+c)^2$ is equivalent to $a^2+b^2+c^2 \ge ab+bc+ca$

To write the inequality in that form, notice the original ineq is equivalent to
$10(\sum a^3 - (a+b+c)^3) - 9(\sum a^5 - (a+b+c)^5) \ge 0$

[quote="Skyward_Sea"]你们都是香港的中学生？
比你们的队员厉害多了，为什么不参加比赛呢？香港队是如何产生的？
考试的时候我前面有一个HongKong 2 Team的，好像是喇沙书院的学生，他最后得了Bronze吧。
这次的冠军是哈萨克斯坦的Kim Andrey，108分，真邪门啊！
本人考的...唉，93分第六名，如果作出这个Inequality...算了，不想了[/quote]

我剛進大學啦。
其實我們拿金獎那個該是失了手，他在2005 的imo拿了金牌的。
之如另外七位，只有一位是去年imo隊的後備，其他都是新人。這是香港一向選人的標準，都是選一些沒有出賽過或者是沒有甚麼經驗的人去，去過的或者是imo隊裏的一般會去cmo，所以一般而言西部奧林匹克香港隊的成績都不好。
選人則是看八月底那個imo 06 tst 1。金獎的yptsoi 在那個test滿分，其他人我就不知道啦

Soaror:
你到底是中学生还是大学生？你似乎还参加了06TST？

[quote="Skyward_Sea"]你们都是香港的中学生？
比你们的队员厉害多了，为什么不参加比赛呢？香港队是如何产生的？
考试的时候我前面有一个HongKong 2 Team的，好像是喇沙书院的学生，他最后得了Bronze吧。
这次的冠军是哈萨克斯坦的Kim Andrey，108分，真邪门啊！
本人考的...唉，93分第六名，如果作出这个Inequality...算了，不想了[/quote]
我是香港的中四學生,應該即是國內的高一年級吧。

[quote="ychjae"]個人認為davidlam和leepakhin的水平比一些香港隊隊員要強，只是他們都在TST1中失手吧[/quote]
你也不弱,我相信你至少能解決兩道題目.........

[quote="shobber"]Soaror:
你到底是中学生还是大学生？你似乎还参加了06TST？[/quote]
只是參加了test 1，玩玩而已。成績也不太好，test 1 的分大概是第五、六。

当时做这个题做了将近两个小时。。。。。。。下来就被我同学洗刷了。。。。

不过最后的题目并不难。。。。是个Ramsey数。。。。。很老的结论

R(3,4)=9, 于是最多8个。。。但是依然要写证明

。。。。。。。。。


quadratics
algebra
number theory solved
number theory

a real hard problem.

I don't know if every step in my proof is right and I don't have the time to write the whole thing down, but I think finding a nontrivial solution(with trivial solutions, I mean those with b=x=0) is equivalent to finding a rational number r such that $2 + 2r + 1/r$ is a square of a rational number which is in turn equivalent to finding a triple (x,y,z) with x,y nonzero such that $2x^4+2x^2y^2+y^4=z^2$ which has already been a problem at mathlinks(it's somehow funny that such a weird equation has already been posted :) ):

http://www.mathlinks.ro/viewtopic.php?t=5944

so, I think there are no nontrivial solutions.

Peter

thanks! that was the problem i was trying to solve (i didnt see it there before). anyway, the problem isnt solved in that forum either (the infinite descent thing doesnt seem to work - we just get that there exists a,b s.t. 2ab and a^2 - b^2 - 2ab are both perfect squares, which then reduces to the problem i posted after some manipulation. i cant get another pythagorean triple from 2ab and a^2 - b^2 being squares) anyway, do you have a proof, peter scholze?

first, I'm wondering about the condition that y is odd, the general case can easily be reduced to that one and my proof doesn't need that condition anyway.
I don't have the whole proof for the equation $2x^4+2x^2y^2+y^4=z^2$ but I'll post it in a minute there. maybe I'll complete the proof while writing.

Peter

now that I finished the proof for $2x^4+2x^2y^2+y^4=z^2$ let me give the proof for that task here. suppose $x\neq 0$. if $x=0$ then $b=0$ and y arbitrarily odd are solutions. the given equation may be written $y^2+4b^2+4x^2=(\frac{x^2}{b})^2$ or, setting $z=\frac{x^2}{b}$, $z^2=4zb+4b^2+y^2$. the discriminant of that quadratic equation in z is $32b^2+4y^2$ and must be a square, thus $8b^2+y^2=t^2$ or $8b^2=(t-y)(t+y)$. Since t-y and t+y have the same parity, we must have that $t-y=2p$ and $t+y=2q$ with $pq=2b^2$. Then $y=q-p$. but one of p and q must be even, so by exchanging p and q if necessary, we may assume that q is even and write(by abuse of notation but I don't have any letters for my variables any more) $y=2q-p$. the original equation now reads $x^4=(4q^2+p^2+4x^2)b^2$. Solving it for x^2, we get $x^2=2b^2+\sqrt{4b^4+(4q^2+p^2)b^2}=b(2b+(2q+p))$. dividing by b^2, we get $(\frac{x}{b})^2=2+2\frac{q}{b} + \frac{p}{b}$. setting $r=\frac{q}{b}$ we must have that $2+2r+1/r$ is the square of a rational number. I will now show that this is never possible.
Let $r=\frac{p}{q}$ with p and q relative prime. then $2+2r+1/r=\frac{2pq+2p^2+q^2}{pq}$. we want to show that the nominator and denominator have no common divisor. actually, $gcd(p,2pq+2p^2+q^2)=gcd(p,q^2)=1$ since p and q are relative prime and $gcd(q,2pq+2p^2+q^2)=gcd(q,2p^2)=gcd(q,2)$. if $q=2q'$ is even, then p is odd and $r'=\frac{q'}{p}$ is another solution(a simple calculation) where nominator and denominator have no common divisor. thus we may assume that nominator and denominator have no common divisor, thus pq is a square. but p and q are relative prime, thus are both squares: $p=s^2$, $q=t^2$. but then the nominator $2p^2+2pq+q^2=2s^4+2s^2t^2+t^4$ is a square, contradiction.

Peter


Let be $ x,y\in \mathbb{Z}$ such that $ x^3\plus{}y^3\plus{}(x\plus{}y)^3\plus{}30xy\equal{}2000$ . Prove that $ x\plus{}y\equal{}10$

[hide="Solution"]$ x^{3} \plus{} y^{3} \plus{} (x \plus{} y)^{3} \plus{} 30xy \equal{} 2000$
implies
$ x^3 \plus{} y^3 \minus{} 1000 \plus{} 30xy \equal{} 1000 \minus{} (x \plus{} y)^3$,
which can be factored to get
$ (x \plus{} y \minus{} 10)(x^2 \plus{} y^2 \plus{} 100 \minus{} xy \plus{} 10x \plus{} 10y) \equal{} (10 \minus{} x \minus{} y)(100 \plus{} 10(x \plus{} y) \plus{} (x \plus{} y)^2)$,
so
$ (x \plus{} y \minus{} 10)[(x^2 \plus{} y^2 \plus{} 100 \minus{} xy \plus{} 10x \plus{} 10y) \plus{} (100 \plus{} 10(x \plus{} y) \plus{} (x \plus{} y)^2)] \equal{} 0$.
$ x\plus{}y\equal{}10$ is a solution. We will show it is the only solution.
Assume for the sake of contradiction that $ x \plus{} y \ne 10$
By the assumption we can divide by $ x \plus{} y \minus{} 10 \ne 0$ to get
$ (x^2 \plus{} y^2 \plus{} 100 \minus{} xy \plus{} 10x \plus{} 10y) \plus{} (100 \plus{} 10(x \plus{} y) \plus{} (x \plus{} y)^2) \equal{} 0$,
which simplifies to
$ 2x^2 \plus{} xy \plus{} 2y^2 \plus{} 20x \plus{} 20y \plus{} 200 \equal{} 0$
and is equivalent to
$ (10 \plus{} x)^2 \plus{} (10 \plus{} y)^2 \plus{} \dfrac{(x \minus{} y)^2 \plus{} 3(x \plus{} y)^2}{4} \equal{} 0$.
Each square must be zero, so $ 10 \plus{} x \equal{} 10 \plus{} y \equal{} x \minus{} y \equal{} x \plus{} y \equal{} 0$.
Solving $ 10 \plus{} x \equal{} 0$ gives $ x \equal{} \minus{} 10$, and solving $ 10 \plus{} y \equal{} 0$ gives $ y \equal{} \minus{} 10$.
We get $ x \plus{} y \equal{} \minus{} 20$, a contradiction, so we are done.
[/hide]

This is JBMO 2000 #1. [url]http://www.mathlinks.ro/resources.php?c=1&cid=21&year=2000[/url]


[hide="Solution"]From $ \tan\theta \equal{} \cot(90 \minus{} \theta)$, $ \tan81 \equal{} \cot9$ and $ \tan63 \equal{} \cot27$. We make use of the following claim:

[b]Claim[/b]: $ \tan\theta \plus{} \cot\theta \equal{} \frac2{\sin2\theta}$
[b]Proof[/b]:
\[ \begin{aligned}\tan\theta \plus{} \cot\theta & \equal{} \frac {\sin\theta}{\cos\theta} \plus{} \frac {\cos\theta}{\sin\theta} \\
& \equal{} \frac {\cos^2\theta \plus{} \sin^2\theta}{\sin\theta\cos\theta} \\
& \equal{} \frac1{\sin\theta\cos\theta} \equal{} \frac2{2\sin\theta\cos\theta} \equal{} \frac2{\sin2\theta}\end{aligned}
\]
So, $ \tan9 \minus{} \tan27 \minus{} \tan63 \plus{} \tan81 \equal{} \frac2{\sin18} \minus{} \frac2{\sin54}$. Let $ A \equal{} \sin18$.

Since $ 54 \equal{} 3\cdot18$, we use $ \sin3\theta \equal{} 3\sin\theta \minus{} 4\sin^3\theta$ (you can prove this using angle sum formulas for sine).

Therefore,
\[ \begin{aligned}\frac2{\sin18} \minus{} \frac2{\sin54} & \equal{} \frac2A \minus{} \frac2{3A \minus{} 4A^3} \\
& \equal{} \frac2A\left(1 \minus{} \frac1{3 \minus{} 4A^2}\right) \equal{} \frac2A\left(\frac {3 \minus{} 4A^2 \minus{} 1}{3 \minus{} 4A^2}\right) \\
& \equal{} \frac2A\cdot\frac {2 \minus{} 4A^2}{3 \minus{} 4A^2} \equal{} \frac {4(1 \minus{} 2\sin^218)}{3\sin18 \minus{} 4\sin^318} \\
& \equal{} \frac {4\cos36}{\sin54} \equal{} 4\end{aligned}
\]
because $ 1 \minus{} 2\sin^218 \equal{} \cos36$ (recall $ \cos2\theta \equal{} 1 \minus{} 2\sin^2\theta$) and $ \cos36 \equal{} \sin54$ (recall $ \cos\theta \equal{} \sin(90 \minus{} \theta)$.)[/hide]

thank you so much.
that's another cool solution from you.
thanks. you are genius.

[quote="harwi"]find the value of  $ E\equal{}\tan 9 \minus{} \tan 27\minus{}\tan 63 \plus{} \tan 81$ (all angles in degrees). [/quote]
Denote $ \tan 9\equal{}x$ , $ \tan 27\equal{}y$ . Thus, $ E\equal{}x\minus{}y\minus{}\frac 1y\plus{}\frac 1x\equal{}\frac {1\plus{}x^2}{x}\minus{}\frac {1\plus{}y^2}{y}$ . Since

$ \sin 2t\equal{}\frac {2\tan t}{1\plus{}\tan^2t}$ obtain : $ E\equal{}2\cdot\left(\frac {1}{\sin 18}\minus{}\frac {1}{\sin 54}\right)\equal{}$ $ 2\cdot \frac {\sin 54\minus{}\sin 18}{\sin 18\sin 54}\equal{}$ $ \frac {4\sin 18\cos 36}{\sin 18\cos 36}\equal{}4$ .

[quote="harwi"]find the value of
tan 9 - tan 27-tan 63 + tan 81
all angle in degrees.
without using a calculator.[/quote]

Hello Virgil Nicula!

Here is not very short, but assured way to solve such problems. Let us try to find the polynomial $ f(x)$ with integer coefficiens, whose root will be $ x\equal{}\tan(9^\circ)$. If $ \alpha\equal{}9^\circ$, then $ \tan(5\alpha)\equal{}\tan(45^\circ)\equal{}1$. Let us use the formula

\[ \tan(5\alpha) \equal{} {\frac {5\,\tan \left( \alpha \right) \minus{}10\,  \tan \left( \alpha
 \right)   ^{3}\plus{} \tan \left( \alpha \right)  ^{5}}{1\minus{}10\,  \tan \left( \alpha \right)  ^{2}\plus{}5\,
\tan \left( \alpha \right) ^{4}}}\]

So, if $ x\equal{}\tan(9^\circ)$, then

\[ 1 \equal{} {\frac {5\,x\minus{}10\,{x}^{3}\plus{}{x}^{5}}{1\minus{}10\,{x}^{2}\plus{}5\,{x}^{4}}}\]
\[ \left( x\minus{}1 \right)  \left( {x}^{4}\minus{}4\,{x}^{3}\minus{}14\,{x}^{2}\minus{}4\,x\plus{}1 \right) \equal{} 0\]

So, we have found, that if $ x\equal{}\tan(9^\circ)$, then $ {x}^{4}\minus{}4\,{x}^{3}\minus{}14\,{x}^{2}\minus{}4\,x\plus{}1\equal{}0$.

What are the other 3 roots of this polynomial? They are $ \tan(81^\circ)$, $ \tan(117^\circ)$ and $ \tan(153^\circ)$ (why?).

Now using the Viète's formulas we can get that sum of all 4 roots is equal to the $ x^3$ coefficient of the polynomial, but with other sign, e.g.

\[ \tan(9^\circ)\plus{}\tan(81^\circ)\plus{}\tan(117^\circ)\plus{}\tan(153^\circ)\equal{}4\]

But $ \tan(117^\circ)\equal{}\minus{}\tan(63^\circ)$ and $ \tan(153^\circ)\equal{}\minus{}\tan(27^\circ)$, so

\[ \tan(9^\circ)\minus{}\tan(27^\circ)\minus{}\tan(63^\circ)\plus{}\tan(81^\circ)  \equal{} \\
\tan(9^\circ)\plus{}\tan(81^\circ)\plus{}\tan(117^\circ)\plus{}\tan(153^\circ)\equal{}4\]

But what to do, if you will need to solve some other problem with $ \tan(9^\circ)$ and similar tangents. For instance, how to find

\[ \tan(9^\circ)\plus{}\tan(27^\circ)\plus{}\tan(63^\circ)\plus{}\tan(81^\circ)  \equal{} \: ?\]

For such tasks it is useful to solve the equation
$ {x}^{4}\minus{}4\,{x}^{3}\minus{}14\,{x}^{2}\minus{}4\,x\plus{}1\equal{}0$ using radicals:

\[ \tan(9^\circ) \equal{} 1\plus{}\sqrt{5}\minus{}\sqrt{5\plus{}2\sqrt{5}}\\
\tan(27^\circ) \equal{} \minus{}1\plus{}\sqrt{5}\minus{}\sqrt{5\minus{}2\sqrt{5}}\\
\tan(63^\circ) \equal{} \minus{}1\plus{}\sqrt{5}\plus{}\sqrt{5\minus{}2\sqrt{5}}\\
\tan(81^\circ) \equal{} 1\plus{}\sqrt{5}\plus{}\sqrt{5\plus{}2\sqrt{5}}\]

Now it's clear that

\[ \tan(9^\circ)\plus{}\tan(27^\circ)\plus{}\tan(63^\circ)\plus{}\tan(81^\circ)  \equal{} 4\sqrt{5}\]

Have a nice day!

Sergey Markelov
markelov@mccme.ru


find all of the fuctionf:R...R such that forx,y member of R we have $f(xf(x)+f(y))=(f(x))^2+y$


complex numbers
superior algebra
superior algebra unsolved

Let $\omega \in \mathbb C \setminus \mathbb Q$ such that $\mathbb Z \left[ \omega \right] = \left\{ a+\omega b : a,b \in \mathbb Z \right\}$ is a ring.
(a) Prove that if $\mathbb Z \left[ \omega \right]$ has exactly four invertible elements, then $\mathbb Z \left[ \omega \right] = \mathbb Z[i]$.
(b) Prove that there are only two rings equal to $\mathbb Z \left[ \omega \right]$ for some $\left| \omega \right| = 1$.

a) $\omega^{2}\in\mathbb{Z}[\omega]$ thus there exist integers $\alpha,\beta$ s.t. $\omega^{2}-\alpha\omega+\beta=0$ and $\alpha^{2}-4\beta$ isn't a perfect square (in particular $\beta\not=0$). Let $\bar{\omega}$ be the $\omega$-algebraic conjugate. $\forall{t\in\mathbb{Z}},\mathbb{Z}[t+\omega]=\mathbb{Z}[\omega]$; thus we may assume that $\alpha\in\{0,1\}$ (because the change $\omega\rightarrow\omega+t$ implies the change $\alpha\rightarrow\alpha-2t$). $1,-1$ are invertible. If $r=u+v\omega\notin\{-1,1\}$ is invertible then $v\not=0$ and $E=\{r,-r,\bar{r},-\bar{r}\}$ is composed with invertible elements ($\bar{r}=u+v\bar{\omega}=u+v\alpha-v\omega$). Moreover $gcd(u,v)=1$ otherwise there exist more than 4 invertible. $\frac{1}{u+v\omega}=\frac{u+v\bar{\omega}}{u^{2}+uv\alpha+v^{2}\beta}\in\mathbb{Z}[\bar{\omega}]$. Thus $u^{2}+uv\alpha+v^{2}\beta$ is a divisor of $u$ and $v$ therefore $u^{2}+uv\alpha+v^{2}\beta=\pm{1}$. Necessarily $cardinal(E)\leq{2}$.
case 1: $\alpha=0$.
Then $u^{2}+v^{2}\beta=\pm{1},v\not=0$; $u\not=0$ otherwise $cardinal(E)>2$. Then $v=\pm{1},\beta=\pm{1},r=\pm\omega$ with $\omega^{2}\pm{1}=0$; we deduce that $\omega=\pm{i}$ and we are done.
case 2: $\alpha=1$.
Then $u^{2}+uv+v^{2}\beta=\pm{1},\beta\not=0,v\not=0$; thus $u^{2}+uv+v^{2}\geq1$ and $\beta<0$. Necessarily $-r=\bar{r}$ i.e. $v=-2u$. Then $u^{2}(-1+4\beta)=\pm{1}$. Therefore $4\beta-1=\pm{1}$ that is impossible.
b) Let $\omega$ s.t.  $|\omega|=1$; $\mathbb{Z}[i]$ admits 4 invertible elements, $\mathbb{Z}[j]$, where $j=e^{\frac{2i\pi}{3}}$, admits 6 invertible elements ($1,-1,j,-j,j^{2},-j^{2}$). It remains to prove that $\mathbb{Z}[\omega]$ is one of these 2 last rings.

b) $\omega\notin{\mathbb{R}}$ thus $\bar{\omega}$ is the $\omega$-complex conjugate and $\beta=1$. The case $\alpha=0$ gives the first ring and the case $\alpha=1$ gives the second ring.


Does anyone know how I can get my hands on a MathCounts handbook, volume 2, for the 2008-2009 school year?

Are you a student or coach? If you're a student your coach should have it, if coach, Mathcounts should send it to you.

I'm not completely sure that we are supposed to openly give it to the public. Oh well

I think it got taken off the website.

It was on a public website.  Argh...I should have bartered with him/her.


logarithms
real analysis
real analysis unsolved

That hint is related to the following theorem: let $a_{n}$ be a sequence such that $a_{n}\neq-1$ for all $n$. Then the infinite product $\prod_{n}(1+a_{n})$ converges absolutely (unconditionally) to a nonzero value if and only if $\sum_{n}a_{n}$ converges absolutely.


this is the last test selection test  this year to prepare to IMO 2008 from Morocco


The bell number $b_n$ is the number of ways to partition the set $\{1,2,\ldots,n\}$. For example $b_3=5$. Find a recurrence for $b_n$ and show that $b_n=e^{-1}\sum_{k\geq 0} \frac{k^n}{k!}$. Using a combinatorial proof show that the number of ways to partition $\{1,2,\ldots,n\}$, such that now two consecutive numbers are in the same block, is $b_{n-1}$.

Post #8 in http://www.mathlinks.ro/Forum/viewtopic.php?t=65305


trigonometry
search
trig identities
cosine double angle
AMC

Source: South Africa Math Talent Search, Round 4

Determine the value of sin:^2: 20 + sin:^2: 40 + sin :^2: 80.

__________________

- mathfanatic

If we let A be that expression and B be the same expression with all the sines replaced by cosines, A+B=3. I have a hunch that A=B, making our expression 1.5, but I can't prove it. B-A can be condensed with the cosine double angle formula but I don't know where to go from there. Help?

Yea, it is 1.5

Try the sum to product formulas..

dangit. I never remeber those. I'm going to make flashcards of them now. *sigh*

Yea, if u use sum to product on the 2 on the ends, then something nice happens, or hopefully happens. Anyway, im just looking at it, and it should work out. Im just too lazy to actually do it....sigh


A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$? (For real $x$, $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.)

[hide="Extremes"]61 1's and 60 1000's or 60 1's and 61 1000's both give a $\lfloor D \rfloor = 495$. There is a unique mode in each, so it satisfies the condition. (Not sure if it's the absolute max though...)[/hide]

[hide="Other extreme"]I got $\lfloor D\rfloor=923$ with $\{1,2,3,\ldots,119,1000,1000\}$[/hide]

Oh, neat. Hadn't realized that, but it does make sense. Also, you could have it on either end only it comes out for 918. Nice solution.

[hide="hmm...no correct answers yet..."]
Basically, we want the mode to be as high as possible and the mean to be as low as possible.  Thus we make 1000 the mode.

If we have two 1000's then the smallest mean comes when the sum of all the numbers is smallest.  So we take as many 1's as possible, then 2's and so forth.  This gives $\{1,2,3,4,\ldots,119,1000,1000\}$ with sum 9140.

Three 1000's we are left with $\{1,1,2,2,\ldots,59,59,1000,1000,1000\}$ with sum 6540.

Four 1000's gives $\{1,1,1,2,2,2,\ldots,39,39,39,1000,1000,1000,1000\}$ with sum 6340

Five 1000's gives $\{1,1,1,1,2,2,2,2,\dots,29,29,29,29,1000,1000,1000,1000,1000\}$ with sum 6740

Six 1000's gives $\{1,1,1,1,1,\ldots,23,23,23,23,23,1000,1000,1000,1000,1000,1000\}$ with sum 7380.

So obviously the smallest mean is with four 1000's.  This gives $D=1000-\frac{6340}{121}\approx 947.7\Rightarrow \lfloor D\rfloor = \boxed{947}.$[/hide]

[quote="cdymdcool"][quote="joml88"][hide="hmm...no correct answers yet..."]
Basically, we want the mode to be as high as possible and the mean to be as low as possible.  Thus we make 1000 the mode.

If we have two 1000's then the smallest mean comes when the sum of all the numbers is smallest.  So we take as many 1's as possible, then 2's and so forth.  This gives $ \{1,2,3,4,\ldots,119,1000,1000\}$ with sum 9140.

Three 1000's we are left with $ \{1,1,2,2,\ldots,59,59,1000,1000,1000\}$ with sum 6540.

Four 1000's gives $ \{1,1,1,2,2,2,\ldots,39,39,39,1000,1000,1000,1000\}$ with sum 6340

Five 1000's gives $ \{1,1,1,1,2,2,2,2,\dots,29,29,29,29,1000,1000,1000,1000,1000\}$ with sum 6740

Six 1000's gives $ \{1,1,1,1,1,\ldots,23,23,23,23,23,1000,1000,1000,1000,1000,1000\}$ with sum 7380.

So obviously the smallest mean is with four 1000's.  This gives $ D \equal{} 1000 \minus{} \frac {6340}{121}\approx 947.7\Rightarrow \lfloor D\rfloor \equal{} \boxed{947}.$[/hide][/quote]

How did u exlucde the possibilities of "mode low, mean high"?[/quote]
This is just replacing $ x$ with $ 1001\minus{}x$, or "flipping" the terms around. The (absolute value) of the difference will remain the same, so we only need to consider one case.


function
search
combinatorics proposed
combinatorics

Compute \[ \sum_{i \equal{} 0}^m \sum_{j \equal{} 0}^{m \minus{} i} \binom{m \minus{} i}{j} \binom{i \plus{} j \plus{} 1}{i}.\]

If possible, give a combinatorial proof.

hello, Mathematica says your sum is equal to $ \sum_{i\equal{}0}^m(1\plus{}i)\text{Hypergeometric2F1}\left[2\plus{}i,i\minus{}m,2,\minus{}1\right]$.
Sonnhard.

:| Which just goes to prove that the symbolic computation capacity of Mathematica leaves something to be desired in this area.   :roll:

[hide="Annoyed rant"]What precisely was the point of that post, anyhow?  I mean, it doesn't even demonstrate an "intelligent plugging-in" -- you've just literally copied the problem into the software and then copied its answer back onto the forum, despite the fact that it has literally done nothing to simplify the problem except give a slightly different (and harder-to-manipulate) expression for the inner sum.  Even printing out the first 5 values of the sequence would have been more useful!  Even if Mathematica [i]had[/i] been able to completely evaluate the sum, the resulting post would have been of no value to anyone whatsoever.  :mad:  Please think more before making such pointless posts![/hide]

JBL, I have been reading most of your posts the last days and I would characterise half of them as agressive and critical. :o I think that Sonnhard just misunderstood your question(he thought that you didn't knew the answer) and he tried to help... Maybe you are a bright mind but such responses do not indicate a bright personality as well. Don't forget that this is a forum not a book!!!!

I don't mean to insult you...

[quote="filosofimenos"]I think that Sonnhard just misunderstood your question(he thought that you didn't knew the answer) [/quote]
But that's the point - Sonnhard's response isn't an "answer" (in the sense of Wilf).  JBL's sum is finite and Sonnhard's is infinite.  There's no comparison in terms of computational complexity.

Anyway, "snake oil" time!

[b]Lemma 1:[/b]  $ \sum_{n \ge 0} {n\plus{}k \choose k} x^n \equal{} \frac{1}{(1 \minus{} x)^{k\plus{}1}}$.

[b]Lemma 2:[/b]  If $ A(x) \equal{} \sum_{n \ge 0} a_n x^n$, then $ \sum a_{n\plus{}1} x^n \equal{} \frac{A(x) \minus{} A(0)}{x}$.

Now, observe that $ i, j$ range over all non-negative integers such that $ {m\minus{}i \choose j}$ isn't equal to zero, so when we write

$ F(x) \equal{} \sum_{m \ge 0} \sum_{i\equal{}0}^{m} \sum_{j\equal{}0}^{m\minus{}i} {m\minus{}i \choose j} {i\plus{}j\plus{}1 \choose i} x^m$

we can freely rearrange this to

$ \sum_{i \ge 0} \sum_{j \ge 0} {i\plus{}j\plus{}1 \choose i} \sum_{m \ge 0} {m\minus{}i \choose j} x^m$.

The inner sum is $ \frac{x^{i\plus{}j}}{(1 \minus{} x)^{j\plus{}1}}$ by Lemma 1, so this gives

$ \sum_{i \ge 0} \frac{x^i}{1 \minus{} x} \sum_{j \ge 0} {i\plus{}j\plus{}1 \choose i} \left( \frac{x}{1 \minus{} x} \right)^j$.

By Lemma 2, the inner sum is $ \frac{ \frac{1}{(1 \minus{} y)^{i\plus{}1}} \minus{} 1}{y}$, where $ y \equal{} \frac{x}{1 \minus{} x}$, which at last gives

$ \frac{1}{y(1 \minus{} x)} \left( \frac{1}{1 \minus{} y} \cdot \frac{1}{1 \minus{} \frac{x}{1 \minus{} y}} \minus{} \frac{1}{1 \minus{} x} \right) \equal{} \frac{1}{x} \left( \frac{1 \minus{} x}{1 \minus{} 3x \plus{} x^2} \minus{} \frac{1}{1 \minus{} x} \right)$

which gives, after some additional computation, $ F(x) \equal{} \frac{1}{(1 \minus{} x)(1 \minus{} 3x \plus{} x^2)}$.  Of course it's routine to convert this into both a closed form and a recurrence.  From the look of the generating functions proof, it looks like you're doing some kind of nested balls-and-urns; care to elaborate on the source?

Okay, nice -- actually the closed form here is very simple, which is what makes me think a combinatorial proof should exist.  (A "hint": [url=http://www.research.att.com/~njas/sequences/?q=1%2C4%2C12%2C33&sort=0&fmt=0&language=english&go=Search]a search at the OEIS[/url] for the first few terms.)  This came out of the work I've been doing on pattern avoidance in permutations, and I don't think it's interesting enough to expound in any sort of detail here :).  (In fact, I'm still trying to decide if it "matters" to me ... a slick combinatorial proof might suggest that it should, for example.)  If you're really curious, drop by my office some time this week ;)


analytic geometry
number theory
relatively prime

[quote="bpms"][quote="4everwise"]Let $R=(8,6)$. The lines whose equations are $8y=15x$ and $10y=3x$ contain points $P$ and $Q$, respectively, such that $R$ is the midpoint of $\overline{PQ}$. The length of $PQ$ equals $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.[/quote]
[hide]So the end points are $(x_{1}, \frac{15x_{1}}{8})$ and $(x_{2}, \frac{3x_{2}}{10})$. Equating the midpoints gives $x_{1}+x_{2}=16$ and $\frac{15x_{1}}{8}+\frac{3x_{2}}{10}=12$. Solving them gives us $x_{1}=\frac{32}{7}$ and $x_{2}=\frac{90}{7}$. The points are $(\frac{32}{7}, \frac{60}{7})$ and $(\frac{90}{7}, \frac{27}{7})$
Applying the distance formula gives us $\frac{\sqrt{4453}}{7}$. It looks like I made a mistake.[/hide][/quote]

Yup. $x_{2}= \frac{80}{7}$, so the answer should be $\frac{60}{7}$.

Very easy question. The answer will be $67$.

What a bad question... Bashy and too easy for AIME P4.

Midpoint formula and distance formula gives us the answer of $\boxed{067}.$


search
number theory proposed
number theory

$a_{1},a_{2}$ are coprime positive integers. Defined a sequence $\{a_{n}\}$ such that if$n\ge{1}$then$a_{n+2}={a}_{n+1}{a}_{n}+1$. Prove that for$%Error. "arbitrary" is a bad command.
{m}\ge{2}$there is$k>m$such that$({a_{m})^{m}}\mid{(a_{k})^{k}}$


function
real analysis
real analysis solved

Does there exist a function $f$ such that $f$ has limit everywhere but it is continuous nowhere?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16089]Here[/url]'s an answer to the question. Read Harazi's second post to see what the problem actually sounds like.


MATHCOUNTS
LaTeX
geometry
rhombus
parallelogram
probability
3D geometry

In honor of the new Amazing Race, I'm making a competition based upon it. First, a standard round will be held to determine the top 12, who will be the competitors. Then, each week, a round will be posted. The lowest score will be eliminated. Just so all know, the perfect competition is officially dead.

[u]List of Contestants (17):[/u]

i_like_pie
goldendomer
SplashD
meenamathgirl
daermon
anirudh
jclarkemathL314159
ragnarok23
bpms
daine
kheg_k
ProtestanT
davidli
moogra
easyas3.14159...
Walk Around The River
kstan013

[quote="perfect628"]Once I have a good # of participants, I'll post Round 0.[/quote]

Is round 0 the one that determines the 12 participants?

I guess I'll do it too.

I don't watch the amazing race, but can I join?

Neither do I, I don't think you need to.

[quote="daermon"]perfect628, what will the format be?[/quote]

[quote="perfect628"]In honor of the new Amazing Race, I'm making a competition based upon it. [b][color=red]First, a standard round will be held to determine the top 12, who will be the competitors. Then, each week, a round will be posted. The lowest score will be eliminated.[/color][/b] Just so all know, the perfect competition is officially dead.[/quote]

oops.  guess I didn't read carefully :lol:

YO. I'm in.

And Ignite, Phil Keoghan doesn't do anything other than telling people that they're eliminated.

[quote="perfect628"]Yes-there will be NO extensions NO MATTER WHAT![/quote]

Alright, don't worry. I'll turn mine in. :)

I wasn't talking to you. I was talking to MOST others.

Dude I'm not sending you anything if you don't answer my question.

Look. #2 is worded badly

I have the same problem as 13375P34K43V312 on #6...

I think there is a typo in #2, it should say what's the probability...the # is..., not the 3 is...

Yeah, please answer the questions about #2 and #6 please.

[quote="anirudh"]I suggest he put on an extension, since not everyone is going to be able to just create 2 solutions in like 15 minutes when they also have other occupations.[/quote]

Yeah, I agree.

Sorry I've been REALLY busy-now I'm not. Because of that, there will be an extension until thursday. On 6, there may not be a nice answer. for 2, i meant the # is divisible by 792. I can't edit it though.

I didn't realize when i said thursday that it was thanksgiving. FINASL EXTENSION IS TO TUESDAY.

oops. yes. Because of that, FINAL FINAL extension is Friday, 6:00, ET.


(1)  Find the maximum value of a positive constant $ a$ such that $ \sqrt{x\plus{}y}\plus{}\sqrt{y}\geq \sqrt{x\plus{}ay}$ for all $ x\geq 0,\ y\geq 0$.

(2)  Let $ a$ be the value found in (1). Find the maximum value of a positive constant $ b$ such that $ \sqrt{x\plus{}y\plus{}z}\plus{}\sqrt{y\plus{}z}\plus{}\sqrt{z}\geq \sqrt{x\plus{}ay\plus{}bz}$ for all $ x\geq 0,\ y\geq 0,\ z\geq 0$.

[quote="kunny"](1)  Find the maximum value of a positive constant $ a$ such that $ \sqrt {x \plus{} y} \plus{} \sqrt {y}\geq \sqrt {x \plus{} ay}$ for all $ x\geq 0,\ y\geq 0$.

(2)  Let $ a$ be the value found in (1). Find the maximum value of a positive constant $ b$ such that $ \sqrt {x \plus{} y \plus{} z} \plus{} \sqrt {y \plus{} z} \plus{} \sqrt {z}\geq \sqrt {x \plus{} ay \plus{} bz}$ for all $ x\geq 0,\ y\geq 0,\ z\geq 0$.[/quote]
(1)
For $ x \equal{} 0$ then $ 2\sqrt {y} \geq \sqrt {ay}$=>$ a \leq 4$
We will prove that $ a \equal{} 4$ or $ \sqrt {x \plus{} y} \plus{} \sqrt {y} \geq \sqrt {x \plus{} 4y}$ for all $ x \geq 0;y \geq 0$
<=>$ x \plus{} 2y \plus{} 2\sqrt {y(x \plus{} y)} \geq x \plus{} 4y$
<=>$ \sqrt {y(x \plus{} y)} \geq y$
<=>$ xy \geq 0$
So $ a \equal{} 4$

[quote="kunny"]
(2)  Let $ a$ be the value found in (1). Find the maximum value of a positive constant $ b$ such that $ \sqrt {x \plus{} y \plus{} z} \plus{} \sqrt {y \plus{} z} \plus{} \sqrt {z}\geq \sqrt {x \plus{} ay \plus{} bz}$ for all $ x\geq 0,\ y\geq 0,\ z\geq 0$.[/quote]
Similar for $ x\equal{}y\equal{}0$ we have
$ 3\sqrt{z} \geq \sqrt{bz}$<=>$ b \leq 9$
We need prove that
$ \sqrt {x \plus{} y \plus{} z} \plus{} \sqrt {y \plus{} z} \plus{} \sqrt {z}\geq \sqrt {x \plus{} 4y \plus{} 9z}$ for all $ x\geq 0,\ y\geq 0,\ z\geq 0$
<=>$ 2\sqrt{z(x\plus{}y\plus{}z)}\plus{}2\sqrt{z(y\plus{}z)}\plus{}2\sqrt{(y\plus{}z)(x\plus{}y\plus{}z)}  \geq 2y\plus{}6z$
Note that $ 2\sqrt{z(x\plus{}y\plus{}z)} \geq 2z$
               $ 2\sqrt{z(y\plus{}z)} \geq 2z$
               $ 2\sqrt{(y\plus{}z)(x\plus{}y\plus{}z)} \geq 2y\plus{}2z$
So $ b\equal{}9$

Your answer is correct, chien than.  :) We can also explain by geometric meaning for (1). Maybe so (2) do.


Seventeen football fans were planning to go to Korea to watch the World Cup football match. They selected 17 matches. The conditions of the admission tickets they booked were such that

- One person should book at most one admission ticket for one match;

- At most one match was same in the tickets booked by every two persons;

- There was one person who booked six tickets.

How many tickets did those football fans book at most?

Discussion should continue at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55858


Let $n$ be a positive integer.

[b](a)[/b] Find the number of all maps $f: \left\{1;\;2;\;...;\;n\right\}\to\left\{0;\;1\right\}$ such that for every $1\leq i\leq n-1$, at least one of the numbers $f(i)$ and $f(i+1)$ equals $0$.

[b](b)[/b] Find the number of all maps $f: \left\{1;\;2;\;...;\;n\right\}\to\left\{0;\;1\right\}$, such that for every $1\leq i\leq n$, at least one of the numbers $f(i)$ and $f(i+1)$ equals $0$. Here, indices are modulo $n$, so that $f\left(n+1\right)=f\left(1\right)$.

  Darij

Everything is in the sources  :P

Pierre.

[quote="darij grinberg"][b](a)[/b] Find the number of all maps $f: \left\{1;\;2;\;...;\;n\right\}\to\left\{0;\;1\right\}$ such that for every $1\leq i\leq n-1$, at least one of the numbers $f(i)$ and $f(i+1)$ equals $0$.[/quote]

Call this number $x_n$.

Let's compute $x_{n+1}$. It includes:

- all the maps of order $n$ and, additionally, $f(n+1)=0$;

- all the maps of order $n-1$ and, additionally, $f(n)=0$, $f(n+1) = 1$.

Thus, $x_{n+1} = x_n + x_{n-1}$. Consequently, $x_n = F_{n+2}$.

So it's indeed well-known. I think I'm too lazy to try (b) :(


vector
linear algebra
matrix
superior algebra
superior algebra unsolved

Hi,

here's a nice problem about group representations:

let $ G$ be a finite Group and $ \rho: G\to GL(V)$ a faithful representation, where $ V$ is a finite-dimensional complex vector space. Prove that every irreducible representation of $ G$ is (isomorphic to) a subrepresentation of $ V^{\otimes k}$ for sufficiently large $ k$.

Have fun,
Hanno

Let $ n$ be the degree of $ \rho$, and consider the character $ \chi$ of $ \rho$.  Let $ \psi$ be the character of some irreducible representation of $ G$.

Let $ f(k) \equal{} \langle \psi, \chi^k \rangle \equal{} \sum_{g\in G} \psi(g) \overline{\chi(g)^k}$, and assume that $ f(k) \equal{} 0$ for all $ k$.

Note that $ f(k)$ is a sum of the form $ \sum_{g\in S} c_g \alpha_g ^k$, where the $ \alpha_g \equal{} \overline{\chi(g)}$ are distinct and nonzero.  Then, by e.g. [url=http://en.wikipedia.org/wiki/Vandermonde_matrix]the Vandermonde matrix[/url] we must have $ c_g \equal{} 0$ for all $ g\in S$.

But $ \chi(g) \neq n \equal{} \chi(1)$ for $ g\neq 1$ (because $ \rho$ is faithful), so $ c_1 \equal{} \psi(1)$, hence $ \psi(1) \equal{} 0$ and $ \psi$ must have degree $ 0$, contradiction.

Thus we must have $ \langle \psi, \chi^k \rangle \neq 0$ for some $ k$, and we're done.


search
calculus
integration
vector
analytic geometry

anyone have a copy that I can sample? pm me

I created a thread where there's some K&K problems.  Use the search enginer.

thanks, that helps, but if anyone has a copy of the whole book or some sample chapters, id like to see how hard it is and what kind of math it incorporates so I can decide whether to buy it (the only place I can find it is amazon who only offers the preface as a preview)

THe highest level of math used is some multiple integrals (easy to pick up on the way), lots of vector calc (like a LOT of it), and some Taylor/Maclaurin approximations, and polar/spherical coordinate (my weakest point).

However, to be able to enjoy the book, you need (previous) mastery of single-variable calculus (especially for the momentum transport problems), and some knowledge of vector calculus, although there's a nice introductory chapter that has everything vector-related you will need to know.

PS: What I like in K&K, as opposed to Irodov, is that the problems' emphasize physical insight/systematic approach, while in irodov, the mathematical computations can get incredibly tedious.


Let's go to http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Surany&t=14008 and see it


probability
function
limit
probability and stats

Show that if $ \mu$ is distribution in $ \mathbb{R}^n$ $ \Longrightarrow \mu_j(B) = \mu (\mathbb{R}\times\mathbb{R}\times...\times\mathbb{R} \times B\times\mathbb{R}\times...\times\mathbb{R})$ where $ (B\in B(\mathbb{R})$ stand on $ j$ place in product) is probability distribution.

It looks kinda obvious, because $ B \subset \mathbb{R}$ and $ B$ is measurable so what could went wrong? But i would like to know what realy should be done here.

[eidt] Okay i get something :) So here is my approach.
Cumulative distribution function of $ \mu_j(B)$ is $ F_X((-\infty,x_k))=\lim_{x_i\to+\infty : i=1,2,...,n ; i\not=k}(x_1,x_2,...,x_k,...,x_n)=\psi((\infty,x_k))$ because Borel sets are measurable and $ \psi$ is right-continious then $ \psi$ is borel function. So $ \psi(X)$ is random variable and $ \mu_j(B)$ is its distribution.

Am i right?

I have't understood your explanation  :oops:
you have probability measure $ \nu$, and you must proove that $ \nu_j$ is also PM.
So you must check the following 3 conditions
1. non-negative - clear
2.sigma-additivity : just write the definiton for $ \nu_j$ and get it from a $ \sigma$-additivity of $ \nu$
3.${ \nu(\mathbb{R})=1}$ -also clear.
And there is no need in talking about DF.

Thank you very much, yeah I have no idea now what i was thinking when I was writting that.


did u use order of operations? coz its hard to remember the answers u get in ur head!

its either 3.5 or 1.5 coz i tried it twice and tatz wat i got! :D

[quote="gr8sk8r"]try to solve this in your head, just for fun
5*6+3/2+3-5*12+30-0.5*2[/quote]

[hide]

5*6+3/2+3-5*12+30-0.5*2

Multiplications first

30+3/2+3-60+30-1

Divisions second

30+1.5+3-60+30-1

Additions third

64.5 - 60 - 1

Subtractions last

4.5 - 1

Answer:

3.5
[/hide]

[quote="gr8sk8r"]how, did you get 407 math92?[/quote]
look two posts up from yours and you'll find out.  i didn't use order of operstions

i got it math92. i didn't read the whole topic, like you sometimes do :lol:

well next time look at the previous posts before you ask a question, because it might have already been answered. :lol:

yeah really, and there all like 1 sentence posts too

yeah its 3.5 like the lsat 5 posts, but... its not "hard" to do in ur head.. just takes like a minute to think


Does the $ E_a$ of a reaction, say A + B => C, change, if we double the moles: 2A + 2B => 2C? Or is it an intrinsic property of the atoms and bonds themselves?

If you double the number of moles of each species, or if make any possible change on them, the equation that describes the reaction is always the same, and so, so as we are at the same temperature, the activation energy remains constant.


floor function
number theory unsolved
number theory

I think it just means when $ a \ne b$ ,there exist 2 different numbers such that they are the divisors.But when $ a \equal{} b$ ,as $ n \ge 4$ ,$ \sqrt n  \le \frac{n}{2}$,that means there are at least 2 different numbers below $ n\minus{}1$ such that $ a \equal{} b\left| k \right.,2k\left( {2k \le n \minus{} 1} \right)$.

Let $ n\equal{}2^k\times m$, where $ m$ is odd.
$ m$ divides $ (n\minus{}1)!$ because $ m$ is either prime less than $ n$, or product of primes less than $ n$.
So, we need to show that $ 2^{k\plus{}1}$ divides $ (n\minus{}1)!$
Among the numbers $ 2,3,...,n\minus{}1$ there are at least $ \frac {n}{2} \minus{}1\equal{}2^{(k\minus{}1)m}\minus{}1$ even numbers so their product is divisible by $ 2^{k\plus{}1}$ because $ 2^{(k\minus{}1)m}\minus{}1>k\plus{}1$ but in the case $ k\equal{}3,m\equal{}1$ it isn't correct, but again $ 2^4$ divides $ 7!$

That's the end of the solution.

[quote="shobber"]If $ n > 4$ is a composite number, then $ 2n$ divides $ (n \minus{} 1)!$.[/quote]
first we can see esaly that $ v_2(n)\le [(n\minus{}1)/2]<v_2((n\minus{}1)!)$ then $ 2n|(n\minus{}1)!\iff n|(n\minus{}1)!$.
1) if we can write $ n\equal{}ab$ such that $ 1<a<b<n$
$ (n\minus{}1)!\equal{}1.2.3...a.(a\plus{}1)...b(b\plus{}1)...n$ then $ n|(n\minus{}1)!$
2) if $ n\equal{}p^2$ for some $ p>2$ prime.
$ (n\minus{}1)!\equal{}1.2.3...p.(p\plus{}1)...(2p)(2p\plus{}1)...n$ then $ n|(n\minus{}1)!$

[quote="shobber"]If $n>4$ is a composite number, then $2n$ divides $(n-1)!$.[/quote]
At first hello
I can't english good

We know that all composite numbers n>4 can write n=ab
when 2=<a,a=<b 2<b
1°We take a=b since n=ab=a²>4 than we have a>2 than  we have 2a>4 and n=a²>2a
well n>2a>4 since n>a>2 and n>2a>4 conspicuous (n-1)*...*2a*...*a*...*2*1 divides
2a*a=2a²=2n
2°Now we take a<b then we have n-1>n-a>n-b>2  the last  inequality is true because
n-b=n-n/a=n(1-1/a)>4*1/2=2     [n>4;1-1/a>=1/2]
Then (n-1)!=(n-1)(n-2)...(n-a)...(n-b)...2*1
This product is conspicuous that divides with
2(n-a)(n-b)=2n(n-a-b+1)
Because of (n-1)! divides with 2n(n-a-b+1) it  divides also with 2n
 :)

PEN [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=462&t=150398]A 30[/url] is similar, but weaker.

The second solution is easily patchable:

If $n=p^2$ with $p$ a prime then $p>2 \implies p^2>2p \implies p^2-1>p^2-p>p$ so $n=2p^2|(p^2-1)!=(p^2-1)\ldots (p^2-p)\ldots p \ldots 2\cdot 1$  since $p^2-p=p(p-2)$, so we have two $p$'s in our product.

Otherwise $n=ab$ with $a,b>1$. So we are required to prove $2ab|(2ab-1)!$
If $a,b \not = 2$ then $(n-1)!=(n-1)\ldots a \ldots b \ldots 2 \cdot 1$ is a multiple of $2ab=2n$.
If one of $a,b$ (say $b$) is equal to $2$ then $n=8$ (checkable) or $(n-1)!=(n-1) \ldots a \ldots 4 \ldots 2 \cdot 1$.

It's also easy to see $4n|(n-1)!$, but difficult to find  $ord_2 \left(\frac{(n-1)!}{n} \right)$

Well, de Polignac's formula gives:

$ord_2 \left(\frac{(n-1)!}{n} \right)=\sum_{i=1}^{\infty} \left\lfloor{\frac{n-1}{2^i}}\right\rfloor-ord_2(n)$

Just for clarification:
I've recently learned that $ord_p(m)$ also denotes the order of $m$ modulo $p$, instead of the exponent $e$ such that $p^e|m$ but $p^e \nmid m $ (or $p^e||m$).
So replace $ord_{2}\left(\frac{(n-1)!}{n}\right)$ with $e_2\left(\frac{(n-1)!}{n}\right)$ if you wish.

Also, while [url=http://en.wikipedia.org/wiki/De_Polignac%27s_formula]Wikipedia[/url] calls it de Polignac's formula, [url=http://www.artofproblemsolving.com/Wiki/index.php/Legendre%27s_Formula]AoPS Wiki[/url] calls it Legendre's formula.

(I was following the free weblication [url=http://www.cpohoata.com/2008/10/17/infinity/]Infinity[/url]- first page).

The question is for when n is composite and clearly, 5 is prime

We shall prove a more stronger result that $4n$ divides $(n-1)!$ for any composite number $n>4$ which will cover the case of problem statement.

Let $n = p.q$ where $q \ge p \ge 2$.

Let us define set $S = \{1, 2, 3 ... n-1\}$

$Case 1: q > p$

First let's note that $p, q \in S$

Now, all multiples of $p$ from $p.1$ to $p.(q-1) \in S$

Since $q > p => q > 2 => q-1 \ge 2$ we have that $p.2 \in S$
Also, since $p \ge 2$ we have that $2 \in S$

So, we have that $2, p.2, q \in S$,
in other words, $4.p.q$ divides $(n-1)!$

$Case 2: q = p$

Now, all multiples of $p$ from $p.1$ to $p.(p-1) \in S$

Since $p > 2 => p-1 \ge 2$ we have that $p.2 \in S$

Also, since $n = p^2 > 4 => p > 2$ so we have that $2 \in S$

So, we have that $2, p.1, p.2 \in S$,
in other words, $4.p^2$ divides $(n-1)!$

Thus $4n$ divides $(n-1)!$.

why is a 13.5 year old thread getting revived....

[quote=fabijan_123]Can't you just make prime factorization of n and we have all prime factors in (n-1)! that has n.[/quote]

yeah but they can appear in n any number of times

$n=ab$ where $a,b \in \mathbb Z_+$ then:

$(n-1)!=(n-1)...(n-a)(n-b)...2*1=2(n-1)...(ab-a)(ab-b)...1=2(n-1)...a(b-1)b(a-1)$, with the a, b, and 2 for a product of 2n in the factors as desired.

I [b]skimmed[/b] the solutions posted above. To me, we should analyze the case when $n=16$ separately. Or do I miss something?!

[quote=S.Das93]No because we simply need to use the fact that $n$ is composite $\Rightarrow n=ab$ for some $a,b>1$ and $\gcd{a,b}>1$. Then splitting into factors of $2$ and looking at appropriate parities gives the result.[/quote]

Could you show a solution a case of which includes $n=16$?
I guess, no need to mention that $16=2\times8=4\times4$


Suppose that the function $ f(a)\equal{}a^a$ and $ g(a)\equal{}a^{2a}$. Which function below is equal to $ f(g(a))$?

$ a^{3a}$  ,
$ a^{a^{2a}}$  ,
$ a^{2a^{2a\plus{}1}}$

Well let's go through it:

$ f(g(a)) \equal{} f(a^{2a}) \equal{} a^{2a\cdot(a^{2a})}\equal{}a^{2(a^{2a\plus{}1})}$

From the rule $ a\cdot a^n\equal{}a^{n\plus{}1}$.

$ 2a(a^{2a})\equal{}2(a\cdot a^{2a})\equal{}2(a^{2a\plus{}1})$.


You can now create your own weblog on the Forum.  To do so, simply click Blog Options (or Blog Control Panel on ML), and you will be able to create your blog.

We are still figuring out all the features of the blog system ourselves, but feel free to post any questions/bug reports here.  We will be adding a few more styles to the blog system, and you can customize the colors of your blog as you see fit.

The blogs do support LaTeX (although currently you won't see a TeX button when you are posting).

when is your blog coming back?  I very much enjoyed the math there.

Hopefully this weekend, using this new blog system.  If not, then later next week.  I still have most of the posts from the old blog, and I'll be sticking them in this blog.

When I try to change the color of my Blog it doesn't seem to work.  I can change the skin but the other options don't seem to be working.

Weird.  Is anyone else having this problem?  I have no problem changing colors.

I will note that if you change the skins, I believe it wipes out all style changes you have made (i.e. if you change the skin and the color, you might have problems if you do it as one step).

I just got it...I was about to delete my post but you were too fast!

Ok, I think I know why I was having trouble.  It doesn't seem to work if you try to change the color while using the ljDigital skin or at least it doesn't show up.

I <3 this blogginess.

Good job, Mr. Ruscyk :) .

When I try changing entry background color it changes the color of the calendar and a few header backgrounds on the right column.  But shouldn't it change the background of where the things I type go?  I'm trying to change that background but can't figure out how.

Thanks.

Which style are you using?  Someone's who's using the same style may be able to let you know which field to mess with.

Since you've made such a wise choice of style (i.e. the one I'm using), I can help you.  It's the non-intuitively named Block Background bit at the bottom that you want to change.  I would have thought what you did first would do it, too.

[quote="Valentin Vornicu"]We'll add in the future the AoPS and MathLinks styles to the Blogs, but for now we're too busy to do that :)[/quote]

That would be the first person singular 'we'.   :lol:

Help...my blog is being very strange. Two and a half posts are down at the way bottom of the page, and the remaining half is at the top. What's up?

Hmm...it's fine in Firefox. Odd. (I was using IE before)


I can solve this problem using regular probablility, but how do you solve it using number theory?

A fair 6-sided die is tossed 8 times.  The sequence 8 results is recorded to form an 8-digit number.  What is the probablility that the number formed is a multiple of 8? (mathcounts).

thank you!!!

[hide]For a number to be divisible by 8, the last 3 digits must be divisible by 8.  Thus, the first 5 digits do not matter, mod 8.

There are $6^{3}=216$ total sets of 3 digits.

Look at the last digit.

If the last digit is 2:
112, 152, 232, 312, 352, 432, 512, 552,632
If the last digit is 4:
144, 224, 264, 344, 424, 464, 544, 624, 664
If the last digit is 6:
136, 216, 256, 336, 416, 456, 536, 616, 656

Thus, the probability is $\frac{27}{216}=\frac{1}{8}$.

Number Theory is used to determine how to count the number that are multiples of 8.[/hide]

[quote="myc"]that's same way I and the intro to probability book solved it, but in  book it says there's a quicker way to solve the problem using number theory.[/quote]
b-flat did use number theory..kind of.

You missed $632$.

[hide="A marginally faster way with more number theory to boot"] $100a+10b+c \equiv 4a+2b+c \bmod 8$

Where $a, b, c \in \{ 1, 2, 3, 4, 5, 6 \}$.  Clearly $c$ must be even.  Let $c = 2d$, so that $d \in \{ 1, 2, 3 \}$.  Then we require that

$2a+b+d \equiv 0 \bmod 4$

[b]Case:[/b]  $a$ is even.  Then $b+d \equiv 0 \bmod 4$.  The equation $b+d = 4$ has three solutions and the equation $b+d = 8$ has two, so there are $5$ ways to choose $b, d$.  Furthermore, there are three ways to choose $a$, for a total of $15$ choices.

[b]Case:[/b]  $a$ is odd.  Then $b+d \equiv 2 \bmod 4$.  The equation $b+d = 2$ has one solution, $b+d = 6$ has three, so there are $4$ ways to choose $b, d$.  There are again three ways to choose $a$ for a total of $12$ choices.

The probability is then $\frac{27}{216}= \boxed{ \frac{1}{8}}$, as expected. [/hide]

[quote="t0rajir0u"]You missed $632$.[/quote]

Whoops; I'll fix that.


function
algebra
polynomial
real analysis
real analysis solved

Let f:[0,1]->R be a continuous function s.t. f(0)=0. Show that there is a g:[0,1]->R s.t. g(0)=0, g is continuous, concave, and g(x)>=f(x) for all x in [0,1].

Consider x(1) \leq x(2) \leq ... \leq x(oo)
the local maximum of f such that
f(x(1)) \leq f(x(2)) \leq .... \leq f(x(max))=max_[0,1]f
Consider a continuous concave curve C_g passing by the points
A_0=( 0 ; 0 ), A_1=( x(1) ; f(x(1) ), ....,Amax=( x(max) ; f(x(max)) )
For an expression of g consider interpolation polynomial passing by
the points A_k.

Much easier than what i had in mind. I looked at F: (0,1]->R, F(x)=f(x)/x and G: (0,1]->R, G(x)=g(x)/x. In order for g to be concave G must be decreasing (it's not hard to prove), and the problem turned into the following:

Given F: (0,1]->R, show that there is a decreasing continuous function G: (0,1]->R s.t. G(x)>=F(x). The interesting case arises when there is a sequence (xn)n>=1 which tends to 0 s.t. F(xn)->00.

Anyway, your soln is MUCH better. Thx!

Thank's for your proof, there is always something to learn in different
solution.

-------------------------------------------------------------------
By symetry there exist h continous convex such that h(0)=0
h(x) \leq f(x) on [0;1]

Well, i didn't really post the complete proof up there (I didn't solve "the interesting case"), but I won't post the solution since it's kind of easy to find. I also used some arguments involving the local maxima, but there are some cases to consider, so I don't want to go through all that trouble now that I know yours. :D