geometry
3D geometry
prism
pyramid
geometric transformation
reflection
rotation

The solid shown has a square base of side length $s$.  The upper edge is parallel to the base and has length $2s$.  All other edges have length $s$.  Given that $s = 6 \sqrt{2}$, what is the volume of the solid?
[asy]
import three;
size(170);
pathpen = black+linewidth(0.65);
pointpen = black;
currentprojection = perspective(30,-20,10);
real s = 6 * 2^.5;
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);
draw(F--B--C--F--E--A--B);
draw(A--D--E, dashed);
draw(D--C, dashed);
label("$2s$", (s/2, s/2, 6), N);
label("$s$", (s/2, 0, 0), SW);
[/asy]

[hide]I think you can complete the figure into a triangular prism and then subtract off the volumes of the two extra pyramids on the ends of the prism.  I got 288 as well.[/hide]

Yes, that's exactly what I did.

[tex]6\sqrt{2}=|AB|=|BC|=|CD|=|CA|=b,\ \ \ \ 12\sqrt{2}=|EF|=a[/tex]

Then the volume: [tex]V=\frac{1}{12}\cdot (a+2b)\cdot b\cdot \sqrt{2b(a+b)-a^2}[/tex]

If you reflect the figure over the plane of the square and rotate it 90 degrees about the axis perpendicular to the square through the center of the square, it and the original figure form a regular tetrahedron with side 12sqrt2. Find its volume and take half of it to get 288.

[hide="Solution"]First, we find the height of the figure by drawing a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle is the median of equilateral triangle $ADE$ one of the legs is $3\sqrt{2}$. We apply the pythagorean theorem to find that the height is equal to $6$.

Next, we complete the figure into a triangular prism, and find the area, which is $\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432$.

Now, we subtract off the two extra pyramids that we included, whose combined area is $2\cdot \left( \frac{12\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144$.

Thus, our answer is $432-144=288$.[/hide]

Edit by mod: I think that this solution is fine here. Just don't click if you don't want to see it.  :)
[hide="Click to see a Calculus Solution"]From the $30^\circ-60^\circ-90^\circ$ relationship of $\bigtriangleup AEM$, where $M$ is the midpoint of $AD$, we find that $EM=3\sqrt{6}$, since $AD=6\sqrt{2}$, as a side of square $ABCD$. Since $BCF$ and $AED$ are both equilateral triangles and $EF$ is parallel to $ABCD$, then $E$ extends $\frac{1}{2}(12\sqrt{2}-6\sqrt{2})$ out from $AD$. Thus, the height of the figure is $\sqrt{(3\sqrt{6})^2-(3\sqrt{2})^2}=6$. (This is just the same process as before).

Since $BCF$ and $ADE$ are equilateral, at any given height in the figure, the rate of change in the width and length of the figure must be the same at any given height. Let $z=0$ be ground level and $z=6$ be the height of $\overline{EF}$. Thus, at $z=0$, the length is $6\sqrt{2}$ and the width is $6\sqrt{2}$. At $z=6$, the length is $12\sqrt{2}$ and the height is $0$. Therefore, at height $z$, $0\le z\le6$, the cross sectional area is $(6\sqrt{2}-\sqrt{2}z)(6\sqrt{2}+\sqrt{2}z)=72-2z^2$. Summing the cross sectional area over all possible heights, \[ \int_0^672-2z^2=(72z-\frac{2}{3}z^3)|_{z=0}^{z=6}=432-144=\boxed{288}.  \][/hide]

[hide]Let $M$ be the midpoint of $EF$
Split the figure up into tetrahedra $MADE$ and $MCBF$, and square pyramid $MABCD$.
The volume of each of the tetrahedra is $\frac{(6\sqrt{2})^3\sqrt{2}}{12}$
The volume of the square pyramid is $\frac{(6\sqrt{2})^2h}{3}$
The height can be found by taking a perpendicular plane cross section through $BC$, resulting in a $3\sqrt{6}-3\sqrt{6}-6\sqrt{2}$ isosceles triangle with height $6$.
Thus ${(6\sqrt{2})^2(\frac 63 + 2(\frac{6\sqrt{2}\sqrt{2}}{12}}))=72(2+2) = \boxed{288}$[/hide]

I feel so bad to rely on calculus for some of the problems... it seems like cheating on the AIME because it requires no ingenuity... that's how I did the problem..

I did it using the pythagorean theorem 2 or 3 times, which requires just as little ingenuity.

There's no reason to feel guilty for using calculus - 90% of the ingenuity of a problem is to find an easy way to do it.

Does anyone have the image for this problem? Thanks!

[img]http://img211.yfrog.com/img211/1109/imgh.gif[/img]
Not to scale obviously. It's a square base, two equilateral triangles, and two isosceles trapezoids.

The image is on aopswiki: http://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems

[b]Solution:[/b] You can cut the solid perpendicular to the edges of the square base of side length $s$ and put the two cut-off parts together to obtain a triangular prism and a regular tetrahedron. Notice that the area of the triangle base in the triangular prism is $\frac{s}{2} \cdot \frac{s}{\sqrt{2}} = \frac{s^2}{2\sqrt{2}}$, which makes the total volume of the prism $\frac{s^3}{2\sqrt{2}}$. The volume of the regular tetrahedron is just $\frac{s^3}{6\sqrt{2}}$ (which I should probably memorize). Substituting $s=6\sqrt{2}$ and adding gives us our final answer of $216 + 72 = \boxed{288}$.

[u]Motivation:[/u] We see symmetry, we use symmetry.

This is just a trianglular prisim, and a tetrahedron, so this has area $\frac{s^3}{2\sqrt{2}}+\frac{s^3}{6\sqrt{2}}$, now plug in the given value of $s$, if we do this, we get $216+72=\boxed{288}.$

[hide=11]We want to split this weird shape into smaller shapes we can find the volume of easily. We note that if we split the shape as follows: [url]https://gyazo.com/b7f27151c0f7669a2ed84d2c966e69b1[/url], it becomes a triangular prism, and two triangular ones. To find the volumes of these shapes, we first want to find the height of the solid. If we draw lines like so: [url]https://gyazo.com/93ab730434215f3b5ed48af7521431b6[/url] we can see that the height is a leg of a right triangle with the other leg being $\frac{s}{2}$, and the hypotenuse being the height of an equilateral triangle with sidelength $s$, which is $\frac{s\sqrt{3}}{2}$. Therefore, by pythagorean theorem, the height is $\frac{s\sqrt{2}}{2}=6$. Now, we can find the area of the pink face here [url]https://gyazo.com/66effe2c01cdc89a7b8d1ad562b6b99b[/url] and multiply it by the height of the prism, which is $s$. The area of the pink face is $s$ times the height of the solid we found earlier over two, which is $18\sqrt{2}$. Therefore, the area of the prism is $216$.

To find the volume of the triangle pyramids, we want to take the area of the pnk face and multiply it by the height, which is $\frac{s}{2}$: . Using the height we found earlier, we know the area of this triangle face is $\frac{s\cdot \frac{s\sqrt{2}}{2}}{2}=18\sqrt{2}$. So the area of each of the pyramids is $\frac{18\sqrt{2}\cdot \frac{s}{2}}{3}=36$. Finally, the total area is $216+72=288$.

Key idea: Recognizing how to split the shape into easier shapes to calculate volume [/hide]


Let $ ABC$ be a triangle in which $ AB\equal{}AC$. Let $ D$ be the midpoint of $ BC$ and $ P$ be a point on $ AD$. Suppose $ E$ is the foot of perpendicular from $ P$ on $ AC$. Define
\[ \frac{AP}{PD}\equal{}\frac{BP}{PE}\equal{}\lambda , \ \ \ \frac{BD}{AD}\equal{}m , \ \ \ z\equal{}m^2(1\plus{}\lambda)\]
Prove that
\[ z^2 \minus{} (\lambda^3 \minus{} \lambda^2 \minus{} 2)z \plus{} 1 \equal{} 0\]

Hence show that $ \lambda \ge 2$ and $ \lambda \equal{} 2$ if and only if $ ABC$ is equilateral.

This easily comes from $ kavat sergey seramov's theorem$ he is a russian mathematician.

[quote="fallinlovewithmaths"]This easily comes from [b]kavat sergey seramov's[/b] theorem he is a russian mathematician.[/quote]
Please post the theorem (as its not well known) along with the whole solution fully please.

Bump, can anyone solve it? I'm bumping the topic since this seems to be an immensely bashy problem.

A solution can be found in this [url=http://olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/09/inmo-2007.pdf]pdf[/url].


integration
logarithms
calculus
calculus computations

[url=http://img85.imageshack.us/i/questionl.jpg/][img]http://img85.imageshack.us/img85/8667/questionl.th.jpg[/img][/url]

I am stuck with $ \int (3x\minus{}5)e^{4ln(x)} dx$

But $ e^{4 \ln x} \equal{} e^{\ln x^4} \equal{} x^4$

$ \int 3x^5\minus{}5x^4 \equal{} \frac{x^6}{2} \minus{} x^5 \plus{} C$ then divide all of this by $ x^4$ (integrating factor)  I will get $ \frac{x^2}{2} \minus{} x \plus{} \frac{C}{x^4}$

put in the initial condition which is y(1) = 5,. I can solve for C which is 11/2 then plug it back in I will have $ y \equal{} \frac{x^2}{2} \minus{} x \plus{} \frac{11}{2x^4}$ as a particular solution, true or not?

If it is the solution then it has to satisfy y(1) = 5 and the differential equation as well. Can you check those two conditions please?

yes it does satisfy, however web assign won't let me enter this answer

That sounds like a technical problem- what's the format of the answer entry? There are many ways to write that answer.

That answer agrees with my answer. You just need to double check how you are entering the problem into the computer. Some programs have special ways of entering operations.


geometry
inequalities
vector
geometry solved

Let ABC is a triangle and D,E,F are respectively midpoints of BC,CA,AB.
$3(AB^2 +BC^2 +CA^2 )\geq 4(AD^2 +BE^2 +CF^2)$
When this inequlity is true?

Using Vector, We can get the proof.

Let $A=\vec{a}, B=\vec{b}, C=\vec{c}$

Then $3\sum AB^2 = 3\sum (\vec{a}-\vec{b})^2 = 6 \sum\vec{a}^2 - 6 \sum \vec{a}\vec{b}$.

and $ 4\sum AD^2 = 4 \sum (\vec{a}- \frac{\vec{b}+\vec{c}}{2})^2 = 6 \sum \vec{a}^2 - 6 \sum \vec{a}\vec{b}$.

So. We get Identity.

$Q.E.D.$

I wrote problem wrongly.  :blush:
When next inequality is true?
$2(AD+BE+CF)\geq\sqrt{3}(AB+BC+CA)$

It's not inequality,it's identity.It follows easily from the median formula.

I'm talking about the first one you have posted.

[quote="Parkdoosung"]When next inequality is true?
$2(AD+BE+CF)\geq\sqrt{3}(AB+BC+CA)$[/quote]

Hmm, which kind of condition do you want for the triangle ABC in order to make this inequality true? All I can say is that there are lots of triangles (including acute-angled and obtuse-angled ones) for which it is true and lots of triangles (including acute-angled and obtuse-angled ones) for which it is wrong. Actually, if the inequalty $2\left(AD+BE+CF\right)\geq\sqrt3\left(AB+BC+CA\right)$ holds for a triangle ABC, then it doesn't hold for the medians triangle of triangle ABC (this is the triangle whose sidelengths are the medians of triangle ABC), and conversely, if it doesn't hold for a triangle ABC, then it holds for the medians triangle of triangle ABC. This is quite obvious, since the median triangle of the median triangle of triangle ABC is similar to triangle ABC with factor of similitude $\frac34$ (this is a rather well-known fact).

[quote="Parkdoosung"]Let ABC is a triangle and D,E,F are respectively midpoints of BC,CA,AB.
$3(AB^2 +BC^2 +CA^2 )\geq 4(AD^2 +BE^2 +CF^2)$[/quote]

Actually, as Chang Woo-JIn pointed out in post #3, this inequality is actually an identity. Using the notations a = BC, b = CA, c = AB for the sidelengths and $m_a=AD$, $m_b=BE$, $m_c=CF$ for the medians of triangle ABC, this identity rewrites as follows:

$4\left(m_a^2+m_b^2+m_c^2\right)=3\left(a^2+b^2+c^2\right)$.

Indeed, this identity can be proven by writing up the well-known formulas for the lengths of the medians of a triangle,

$4m_a^2=2b^2+2c^2-a^2$;
$4m_b^2=2c^2+2a^2-b^2$;
$4m_c^2=2a^2+2b^2-c^2$,

and adding them up:

$4m_a^2+4m_b^2+4m_c^2=\left(2b^2+2c^2-a^2\right)+\left(2c^2+2a^2-b^2\right)+\left(2a^2+2b^2-c^2\right)$;

this simplifies to the required identity $4\left(m_a^2+m_b^2+m_c^2\right)=3\left(a^2+b^2+c^2\right)$.

  Darij

[quote="darij grinberg"]$2\left(AD+BE+CF\right)\geq\sqrt3\left(AB+BC+CA\right)$ holds for a triangle ABC, then it doesn't hold for the medians triangle of triangle ABC (this is the triangle whose sidelengths are the medians of triangle ABC), and conversely, if it doesn't hold for a triangle ABC, then it holds for the medians triangle of triangle ABC. This is quite obvious, since the median triangle of the median triangle of triangle ABC is similar to triangle ABC with factor of similitude $\frac34$ (this is a rather well-known fact).
[/quote]

median triangle == triangle with vertices the midpoints of the sides (in romanian it is so)?? if yes, that looks like $\frac14$

[quote="perfect_radio"]median triangle == triangle with vertices the midpoints of the sides (in romanian it is so)?? if yes, that looks like $\frac14$[/quote]

No. My notations are as follows:

Medial triangle of triangle ABC = the triangle whose vertices are the midpoints of the sides of triangle ABC;
Medians triangle of triangle ABC = the triangle whose sidelengths are equal to the medians of triangle ABC.

  Darij


Mientka Publishing Company prices its bestseller [i]Where's Walter?[/i] as follows: \[C(n) \equal{}
\begin{cases}
   12n, &\text{if } 1 \le n \le 24\\
   11n, &\text{if } 25 \le n \le 48\\
   10n, &\text{if } 49 \le n
\end{cases}\] where $ n$ is the number of books ordered, and $ C(n)$ is the cost in dollars of $ n$ books. Notice that $ 25$ books cost less than $ 24$ books. For how many values of $ n$ is it cheaper to buy more than $ n$ books than to buy exactly $ n$ books?

$ \textbf{(A)}\ 3\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 8$

[hide="Click for solution"]
The cost of $ 25$ books is $ 25 \times 11\equal{}275$.  The cost of $ 24$ books is $ 24 \times 12\equal{}288$, and $ 23$ books cost $ 276$ while $ 22$ books costs $ 264$.  Also, $ 49$ books costs less than $ 45$, $ 46$, $ 47$, or $ 48$ books.  In these $ 6$ cases the total cost is reduced by ordering more books.  There are no other cases, so the answer is $ \boxed{\textbf{(D)}}$.
[/hide]

But the cost of 26 books is less than the cost of 24 books. Likewise 50 costs less than 48. The answer is E

Pay attention to the question, crazypianist.

For how many values of n is it cheaper to buy more than n books than to buy exactly n books?

Both 24 and 48 were already counted as possible values of n.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\frac{\pi}{2}} \frac{\cos \theta \minus{}\sin \theta}{(1\plus{}\cos \theta)(1\plus{}\sin \theta)}\ d\theta$

$ I\equal{}\int_{0}^{\pi /2}\frac{\cos \theta \minus{}\sin \theta }{\left( 1\plus{}\cos \theta \right) \left( 1\plus{}\sin \theta \right) }d\theta$

Set $ u\equal{}\frac{\pi }{2}\minus{}\theta$

$ I\equal{}\allowbreak \int_{0}^{\pi /2}\frac{\sin u\minus{}\cos u}{\left( 1\plus{}\sin u\right) \left( 1\plus{}\cos u\right) }\,du\equal{}\int_{0}^{\pi /2}\frac{\sin \theta \minus{}\cos \theta }{\left( 1\plus{}\cos \theta \right) \left( 1\plus{}\sin \theta \right) }d\theta$

Hence $ 2I\equal{}0\Rightarrow I\equal{}0$

Alternatively,

$ I\equal{}\int_{0}^{\pi /2}\frac{\cos \theta \minus{}\sin \theta }{\left( 1\plus{}\cos \theta \right) \left( 1\plus{}\sin \theta \right) }d\theta$

Set $ u\equal{}\theta \minus{}\frac{\pi }{4}$

$ I\equal{}\allowbreak \minus{}\sqrt{2}\int_{\minus{}\frac{1}{4}\pi }^{\frac{1}{4}\pi }\frac{\sin u}{\allowbreak 1\plus{}\sqrt{2}\cos u\plus{}\frac{1}{2}\cos 2u}\,du\equal{}0$

Since the integrand is an odd function.

That's right. :)

The background of the problem:

$ \int_0^{\frac {\pi}{2}} \frac {\cos \theta}{1 \plus{} \cos \theta}\ d\theta \equal{} \int_0^{\frac {\pi}{2}} \frac {\sin \theta}{1 \plus{} \sin \theta}\ d\theta$. Now I think that you could see what we should do next.

Yes... but we should just leave it alone :roll:

[quote="kunny"]
The background of the problem:

$ \int_0^{\frac {\pi}{2}} \frac {\cos \theta}{1 \plus{} \cos \theta}\ d\theta \equal{} \int_0^{\frac {\pi}{2}} \frac {\sin \theta}{1 \plus{} \sin \theta}\ d\theta$. Now I think that you could see what we should do next.[/quote]

Shouldn't it be $ \int_0^{\frac {\pi}{2}} \frac {1}{1 \plus{} \sin \theta}\ d\theta \equal{} \int_0^{\frac {\pi}{2}} \frac {1}{1 \plus{} \cos \theta}\ d\theta$ ?


We have to construct a cube of side $ 3$ cm using a wrapping paper of 3 cm width. What's the minimum possible lenght of the paper if we are not allowed to cut it during the construction?

I really can't do these kind of problems, because I don't think it's possible

@ alanchou - You can still fold it. :wink:

[hide]I'm pretty sure this isn't the minimum, but it's definitely possible with length $ 24\text{ cm}$.[/hide]

Yeah, and each face doesn't need to be the same thickness.
And yeah, 24 is possible, but I think there's another way, not using the conventional way of folding.
Wait, 21 is also possible.

Can you describe the method of construction, please?(for $ 21$ of course)

PLease ZhangPeijin, post your solution for $ 21$

You know how a normal cube is made with a cross?
Make one with a T.
It's 21.
Oh wait dang.
I messed up.
It's 24.

24 is indeed the minimum. There are 6 faces to cover, 1 for folding/turning the 'direction' of wrapping, and 1 overlapping face.


$a_{n}$ is a sequence of positive integers such that, for every $n\geq 1$, $0<a_{n+1}-a_{n}<\sqrt{a_{n}}$. Prove that for every $x,y\in{R}$ such that $0<x<y<1$ $x< \frac{a_{k}}{a_{m}}<y$ we can find such $k,m\in{Z^{+}}$.

It is true, because $a_{n}$ strongly increase integer sequence. For any N exist k, suth that $a_{k}>N$. Then $a_{i+1}-a_{i}<\sqrt{\frac{a_{k}}{x}}$ while $i\le m, a_{m}<\frac{a_{k}}{x}$, therefore exist m, suth that $x<\frac{a_{k}}{a_{m}}<y$ if $a_{k}>\frac{xy^{2}}{(x-y)^{2}}$.

Note that, $a_{n+1}>a_n$, thus, the sequence is strictly increasing; in particular, $a_n$ diverges. Moreover, using the second part of inequality, we have,
$$
a_{n+1}-a_n <\sqrt{a_n}\implies 1-\frac{a_n}{a_{n+1}}<\frac{\sqrt{a_n}}{a_{n+1}}<\frac{1}{\sqrt{a_n}}\implies \frac{a_n}{a_{n+1}}>1-\frac{1}{\sqrt{a_n}}.
$$
In particular, since $a_n$ diverges, for any given $\epsilon>0$, there exists an $N$ such that, for every $n\geq N$,
$$
\frac{a_n}{a_{n+1}}>1-\epsilon.
$$
We will now argue through contradiction. Suppose that, there exists a pair $0<x<y<1$, for which, for every $k,m$;
$$
\text{either}\quad \frac{a_k}{a_m}<x \quad\text{or}\quad \frac{a_k}{a_m}>y.
$$
Now, we select an $\epsilon>0$, for which,
$$
\epsilon<1-\frac{x}{y}<1-x.
$$
The reasoning will become clear soon. First, observe that this indeed makes sense, since, $x<x/y$ as $y<1$. Now, for this epsilon; select $N_\epsilon$ such that,
$$
n\geq N_\epsilon \implies \frac{a_n}{a_{n+1}}>1-\epsilon>x.
$$
Since, $1-\epsilon>x$; we have $\frac{a_n}{a_{n+1}}>x$; hence,
$$
\frac{a_n}{a_{n+1}}>y.
$$
We now keep multiplying,
$$
\frac{a_n}{a_{n+1}}\cdot \frac{a_{n+1}}{a_{n+2}}\cdots \frac{a_{n+t-1}}{a_{n+t}}=\frac{a_n}{a_{n+t}}<\frac{a_n}{a_{n+1}}.
$$
In fact, as $t\to \infty$, $\frac{a_n}{a_{n+t}}\searrow 0$. Hence, one should eventually go below $y$; and therefore, by hypothesis, also below $x$. Let, $k$ be the first time when this holds, namely,
$$
\frac{a_n}{a_{n+k-1}}>y \quad \text{but}\quad \frac{a_n}{a_{n+k}}<x.
$$
Observe that,
$$
x>\frac{a_n}{a_{n+k}}=\underbrace{\frac{a_n}{a_{n+k-1}}}_{>y}\underbrace{\frac{a_{n+k-1}}{a_{n+k}}}_{>1-\epsilon}>y(1-\epsilon) \implies \epsilon>1-\frac{x}{y},
$$
contradicting with the selection of $\epsilon$.


calculus
integration
function
limit
real analysis
real analysis unsolved

I have two problems, both of which I have an attempted solution so I'd just like verification.

1) Suppose that $ f$ and $ (f_{n})$ are non-negative measurable functions, that $ (f_{n})$ decreases pointwise to $ f$, and that $ \int f_{k}<\infty$ for some $ k$.  Prove that $ \int f\equal{}\lim_{n\rightarrow\infty}\int f_{n}$.
[hide="My Solution"]Since $ \int f_{k}<\infty$ for some $ k$ and $ (f_{n})$ is decreasing, $ \int f_{n}<\infty$ for all $ n\geq k$.  Fix this $ k$.  Now $ f_{k}\minus{}f_{n}$ increases to $ f_{k}\minus{}f$ as $ n\rightarrow\infty$ so $ \int f_{k}\minus{}f_{n}$ increases to $ \int f_{k}\minus{}f$.  Then by the monotone convergence theorem and linearity,
$ \int f_{k}\minus{}f\equal{}\lim_{n\rightarrow\infty}\int f_{k}\minus{}f_{n}\equal{}\lim_{n\rightarrow\infty}\int f_{k}\minus{}\lim_{n\rightarrow\infty}\int f_{n}\equal{}\int f_{k}\minus{}\lim_{n\rightarrow\infty}\int f_{n}\equal{}\int f_{k}\minus{}\int f$.
By rearranging the last two equalities, we get the desired result.[/hide]
2) If $ f$ is non-negative and integrable, show that $ \int_{\minus{}\infty}^{\infty}f\equal{}\lim_{n\rightarrow\infty}\int_{\minus{}n}^{n}f\equal{}\lim_{n\rightarrow\infty}\int_{\{f\geq (1/n)\}}f$.
[hide="My Solution"]Let $ E_{n}\equal{}[\minus{}n,n]$.  Then $ \int_{E_{n}}f\equal{}\int_{\minus{}n}^{n}f$ and so $ \lim_{n\rightarrow\infty}\int_{E_{n}}f\equal{}\lim_{n\rightarrow\infty}\int_{\minus{}n}^{n}f\equal{}\int_{\bigcup_{n\equal{}1}^{\infty}E_{n}}f\equal{}\int_{\mathbb{R}}f\equal{}\int_{\minus{}\infty}^{\infty}f$, which yields the first result.  Now since $ f$ is non-negative and measurable, $ \int_{\minus{}\infty}^{\infty}f\equal{}\int_{\mathbb{R}}f\equal{}\int_{\{f\geq 0\}}f\equal{}\int_{\{f>0\}}f\plus{}\int_{\{f\equal{}0\}}f\equal{}\int_{\{f>0\}}f$.  Since $ \{f>0\}\equal{}\bigcup_{n\equal{}1}^{\infty}\{f\geq\frac{1}{n}\}$, we can define $ E_{1}\equal{}\{f\geq 1\}$ and $ E_{n}\equal{}\{\frac{1}{n\minus{}1}>f\geq\frac{1}{n}\}$ for $ n\geq 2$.  Hence $ \bigcup_{n\equal{}1}^{\infty}E_{n}\equal{}\{f>0\}$ so $ \int_{\{f>0\}}f\equal{}\int_{E_{1}}f\plus{}\int_{E_{2}}f\plus{}\cdots\equal{}\lim_{n\rightarrow\infty}\int_{\{f\geq (1/n)\}}f$, as desired.[/hide]

hi
i saw that is very good !!! but where is your problem??
_________________________

[quote="JRav"]No one can check these?[/quote]Part 1 looks OK to me. Try to use monotone convergence also for part 2.

[quote="JRav"]Which part in particular?[/quote]

note that for $ \Omega\subset\mathbb{R}$, $ \int_\Omega fdx \equal{} \int_\mathbb{R} f\chi_\Omega dx$.  For example, if we apply this with $ \Omega_n \equal{} [\minus{}n,n]$, then as long as $ f$ is nonnegative, you have that $ f\chi_{\Omega_n} \uparrow f$, so you can use the monotone convergence theorem.  It works for the other part of 2) as well.

By "other part" are you referring to when I define $ E_{n}\equal{}\{\frac{1}{n\minus{}1}>f\geq\frac{1}{n}\}$?  So then $ f\chi_{E_{n}}\uparrow f$ and therefore by MCT, $ \int_{\{f\geq (1/n)\}}f\equal{}\int_{E}f\equal{}\int_{\bigcup_{i\equal{}1}^{\infty}E_{n}}f\equal{}\int_{\mathbb{R}}f\chi_{E}\equal{}\int_{\minus{}\infty}^{\infty}f$.

I think you don't need MCT here, since $ f$ is nonnegative and integrable, $ E\rightarrow \int_E f$ is a measure,  both $ [ \minus{} n,n]$ and $ \{f \geq \frac {1}{n}\}$ are a sequence of increasing set, so by continuty of measure with respect to increasing sets, we have both claims.

Edit: Ups  :blush: , I think I just went in circle with that claim, to prove that $ E\rightarrow \int_E f$ is measure, we need MCT. Sorry about that.

[quote="JRav"]By "other part" are you referring to when I define $ E_{n} \equal{} \{\frac {1}{n \minus{} 1} > f\geq\frac {1}{n}\}$?  So then $ f\chi_{E_{n}}\uparrow f$ and therefore by MCT, $ \int_{\{f\geq (1/n)\}}f \equal{} \int_{E}f \equal{} \int_{\bigcup_{i \equal{} 1}^{\infty}E_{n}}f \equal{} \int_{\mathbb{R}}f\chi_{E} \equal{} \int_{ \minus{} \infty}^{\infty}f$.[/quote]

Not quite.

For example, in the example I gave you, you have

\[ \lim_{n\to\infty} \int^n_{\minus{}n} fdx \equal{} \lim_{n\to\infty} \int_{\Omega_n} fdx \equal{} \lim_{n\to\infty} \int_\mathbb{R} f\chi_{\Omega_n} \equal{} \int_\mathbb{R} fdx
\]

The MCT is the last equality.

The other one is similar; instead you have something like

\[ \lim_{n\to\infty} \int_{\{f\geq n^{\minus{}1}\}} fdx \equal{} \lim_{n\to\infty} \int_\mathbb{R} f\chi_{\{f\geq n^{\minus{}1}\}} \equal{} \int_\mathbb{R} fdx
\]

since, of course, $ f\chi_{\{f\geq n^{\minus{}1}\}} \uparrow f$ as well.

[quote="JRav"]That makes sense.  It seems like I used a different theorem, which is basically a corollary to MCT.[/quote]

It's not exactly clear what theorem you are using to justify interchanging limits and integrals.  I've marked for a graduate course on measure and integration before and I wouldn't give full marks for your solution to 2.

I was using a theorem that says that if you have disjoint measurable sets, then the integral over the union is equal to the sum of the individual integrals.  But yes, I should have included a mention of MCT to justify moving the limit inside.


Let $ a$, $ b$, $ c$, $ d$ be positive reals. Prove that
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}d}\plus{}\frac{c}{d\plus{}a}\plus{}\frac{d}{a\plus{}b}\geq 2$.

[quote="drapt@"]Let $ a$, $ b$, $ c$, $ d$ be positive reals. Prove that
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} a} \plus{} \frac {d}{a \plus{} b}\geq 2$.[/quote]
It's Shapiro's inequality.
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} a} \plus{} \frac {d}{a \plus{} b} \equal{}$
$ \equal{} \sum_{cyc}\frac {a^{2}}{ab \plus{} ac}\geq\frac {(a \plus{} b \plus{} c \plus{} d)^{2}}{ab \plus{} ac \plus{} bc \plus{} bd \plus{} cd \plus{} ac \plus{} ad \plus{} bd}\geq2.$

[quote="me@home"]What Shapiro is this named after? It seems like a very trivial inequality :lol:
(Well I guess it is harder than Nesbitt's and Nesbitt's got a name...)[/quote]
Shapiro proposed a harder inequality:
http://planetmath.org/encyclopedia/ShapirosInequality.html
This is just a case of it -- one of the most trivial ones.

okay I figured it was something like that...
for instance there's the "common" popoviciu's inequality then the actual functional inequality he gave...

[quote="drapt@"]Let $ a$, $ b$, $ c$, $ d$ be positive reals. Prove that
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} a} \plus{} \frac {d}{a \plus{} b}\geq 2$.[/quote]
Put $ S\equal{}\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} a} \plus{} \frac {d}{a \plus{} b}$
$ M\equal{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}d}\plus{}\frac{d}{d\plus{}a}\plus{}\frac{a}{a\plus{}b}$
$ N\equal{}\frac{c}{b\plus{}c}\plus{}\frac{d}{c\plus{}d}\plus{}\frac{a}{d\plus{}a}\plus{}\frac{b}{a\plus{}b}$
We have $ M\plus{}N\equal{}4$
By AM-GM:
$ M\plus{}S\equal{}\frac{a\plus{}b}{b\plus{}c}\plus{}\frac{b\plus{}c}{c\plus{}d}\plus{}\frac{c\plus{}d}{d\plus{}a}\plus{}\frac{d\plus{}a}{a\plus{}b} \geq 4$
$ N\plus{}S\equal{}\frac{a\plus{}c}{b\plus{}c}\plus{}\frac{b\plus{}d}{c\plus{}d}\plus{}\frac{a\plus{}c}{d\plus{}a}\plus{}\frac{b\plus{}d}{a\plus{}b} \geq \frac{4(a\plus{}c)}{a\plus{}b\plus{}c\plus{}d}\plus{}\frac{4(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d} \equal{}4$
=>$ M\plus{}N\plus{}2S \geq 8$
=>$ S \geq 2$
Equality occurs when $ a\equal{}b\equal{}c\equal{}d$

we can assum $ a\plus{}b\plus{}c\plus{}d\equal{}4$
By AM-GM we have:$ \frac{a}{b\plus{}c}\plus{}\frac{a(b\plus{}c)}{4}\geq a$
Hence $ LHS\geq 4\minus{}\frac{ab\plus{}ac\plus{}bc\plus{}bd\plus{}cd\plus{}ca\plus{}da\plus{}db}{4}\geq 4\minus{}2\equal{}2$


For what $n$, a graph with order $n$ (the number of vertices) isomorph to its compliment?

It's necessary $n$ be of the form $4k$ or $4k+1$ for some integer $k$ ,but what is the sufficient condition?

This question is discussed in almost every olympiad book. It can be shown that this condition is also sufficient, by introducing a graph for each case!

Sorry to intrude,

Hi every1,

Friends, i am sorry but can u kindly explain it to me how the above condition is both necessary and sufficient...

Thanx and have a nice day......

That is you mean every graph of order $4k$ or $4k+1$ ?  :?:
It seems that it isn't true !!!
Are you sure?

Let $G$ and $H$ be two graphs. $G==H$ is used to denote the graph obtained from adjoining of their vertices. For $n=4k$ consider $K_t==\bar K_t==\bar K_t==K_t$. And for $n=4k+1$ construct a graph by adding a single vertex to the structure above in a pentagonal form.
Am I clear enough?

Oh, I found that you want to construct a graph for  each case.($4k$ or $4k+1$)
but my question is "Do every graph with $4k$ or $4k+1$ is isomorph to its complement?"
Is it clear?


So what did you get on the Tokutomi Math Contest? I got $20$.

20 in junior or senior?  ... yea, i got a 14 in the senior division.

[quote="mathgeniuse^ln(x)"]Got al kink for this?[/quote]

 :huh:   Oh yeah and I dont think they have any links to past problems from Tokutomi.

This is [i]seriously[/i] late, because I just saw this post, but I got 19 (I totally suck...) :(

19 because of number 1. I read it wrong. :|

Would've gotten first because my explanation for the bonus pwned.

i would love to say 20, but 19 because i put my answer to one of them as a mixed number, because one of the proctors said i could, but the graders said i couldnt.

It's a competition held at SJSU every...January? I think it is. It's 20 problems, and all participants are split up into 2 groups: junior division or senior division. Lots of prizes.

Hi. Its the annoying kid that keeps annoying vishal and alvin :) I got 19 :(

Stupid number 17 :(

Oh yeah, what did alvin get? he said he got 18 but I dont believe him


Ring Theory
superior algebra
superior algebra unsolved

A problem was posed to me to find a ring $ R$ and ideal $ I$ of $ R$ such that the element-wise product of two cosets is not again a coset.

I first considered $ a,b\in R$ and $ A \equal{} a \plus{} I,B \equal{} b \plus{} I$ as the cosets.  Then, elementwise,
\[ AB \equal{} \left\{\left(a \plus{} x\right)\left(b \plus{} y\right)\ : \ x,y\in I\right\}.
\]
The elements in $ AB$ are of the form $ ab \plus{} z$ where $ z\in I$.  How do we find $ R$ and $ I$ such that this isn't a coset?  Thank you.

I don't see how you get $ ab\plus{}z$ with $ z\in I$. Is $ I$ a two-sided ideal?

[quote="JSteinhardt"]$ \{0,3,6\} \subset Z_9$[/quote]

Why?

$ (2 \plus{} \{0,3,6\})(1 \plus{} \{0,3,6\}) \equal{} 2 \plus{} 2\{0,3,6\} \plus{} \{0,3,6\}1 \plus{} \{0,3,6\}^2 \equal{} 2 \plus{} \{0,3,6\}$

Anyway $ ab \plus{} z$ with $ z \in I$ is always a coset because the left cosets form a partition of $ R$.

We know that if $ I$ is a two-sided ideal, the quotient ring $ R/I$ is isomorphic to the image of a ring homomorphism. In this case the surjective homomorphism $ f: \mathbb{Z}_9 \minus{} > \mathbb{Z}_3$.
“Find a ring $ R$ and ideal $ I \triangleleft R$ such that the element-wise product of two cosets is not again a coset” is like to ask to find a homomorphism in which $ f(R)$ is not a subring of the codomain ring because the multiplication is not closed in $ f(R)$ and it's non-sense.

In the case $ \{0,3,6\} \triangleleft Z_9$, $ \mathbb{Z}_9/\{0,3,6\} \cong \mathbb{Z}_3$ (a field)

The element-wise product of $ a+I$ and $ b+I$ need not be $ ab+I$. In my example, I was considering ${ (0+\{0,3,6})(0+\{0,3,6\}) = 0 + \{0,3,6\}^2 = \{0\}$.

[quote="JSteinhardt"]The element-wise product of $ a \plus{} I$ and $ b \plus{} I$ need not be $ ab \plus{} I$. In my example, I was considering $ (0 \plus{} \{0,3,6\})(0 \plus{} \{0,3,6\}) \equal{} 0 \plus{} \{0,3,6\}^2 \equal{} \{0\}$.[/quote]

 ok, you're right


geometry
3D geometry
pyramid
geometry unsolved

Let SABC be a regular triangular pyramid. Find the set of all points $ D (D! \equal{} S)$ in the space satisfing the equation $ |cos ASD \minus{} 2cosBSD \minus{} 2 cos CSD| \equal{} 3$.

Did you mean (D!= S) for D ≠ S?

Please clarify!

Anyone? Very interesting problem in my opinion.

a better restatement of the problem

Let ${SABC}$. be a regular triangular pyramid (${SA=SB=SC}$. and ${AB=BC=CA)}$ ). Find the locus of all points ${D \, (D\ne S)}$. in the space that satisfy the equation ${ |cos \delta_A -2cos \delta_B - 2cos \delta_C | = 3 }$. where the angle ${\delta_X=\angle XSD}$ for each ${X \in \{ A,B,C\} }$.


geometry
geometric transformation
homothety

I have a question regarding methods of showing collinearity. I saw in several olympiad-level problems homothety being used to show collinearity in certain triangle problems. When is this a better, more efficient method, and how often is it better to use this over let's say Menelaus's theorem. ( I know the obvious answer: when one involves given lengths :wink:)

I dunno, I'd think (with my limited knowledge) that homothety is harder to apply but easier to use.  Maybe this should be moved to Intermediate or up?

does anyone have any other input to add?

[quote="Altheman"]It is a specialized technique. There are only about three problems that I think it is useful for...
i) tangent circles
ii) incircle - excircle homothety
iii) 9 - point circle[/quote]

thanks for the input altheman!
in each of these situations, what is a tip-off to use homothety? this is kinda a general problem too. i know homothety is a very powerful tool, and yet im having a little trouble knowing when to use it.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that every group of order 255 is cyclic.

These proofs are from http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=0514.0001.0001

Let $ H$ be the unique Sylow 17-subgroup of $ G$. Then $ N_G(H)/C_G(H)$ is isomorphic to a subgroup of $ Aut(H)$, so $ |N_G(H)/C_G(H)|$ divides both 255 and 16. Hence $ N_G(H) \equal{} C_(H) \equal{} G$, so $ H$ is central, in which case $ |Z(G)|$ has order a multiple of 17 and is a divisor of 225. It follows $ |G/Z(G)| \in \{ 1, 3, 5, 15\}$, hence $ G$ must abelian. Finally, since the order of $ G$ is a product of distinct primes, it follows from the fundamental theorem of finite abelian groups that $ G$ is cyclic.

Or

Since there's a unique Sylow 17-sbgp P, it is normal. There can't be both 51 Sylow 5-sbgps. and 85 Sylow 3-sbgps. (why?), so either there's unique and thus normal Sylow 5-sbgp $ F$ or 3-sbgp T ==> anyhow, we have a sbgp. FT of G of order 15, and since clearly P /\ FT = 1, we get G = P*FT ==> G is the semidirect product of P and FT ==> there exists a homom. FT --> Aut(P), but this homom. HAS to be the trivial one (why?) ==> the above product is actually direct, and since both P and FT are abelian you're done.


[quote="mr. math"]i need help solving this problem. it doesn't seem very straightforward:

an entomologist places a centipede with 100 feet on a scale, but notes that only 80 of its feet touch the scale; the rest are being raised in the air. if the scale reads 0.16 grams what is the true weight of the centipede in grams.

 :ninja:  :arrow:  :huh:[/quote]

[hide]$(0.16)(\frac{100}{80})=\boxed{0.2}$ grams[/hide]

[quote="mr. math"]thats what i got, but the answer says 0.16 grams. i can't find the trick in the problem. :mad:[/quote]

Well, that is true actually. Do you weigh the same on a scale whether you're standing on 1 leg or 2?

Never mind this post I got confused.  :wallbash_red:  :blush:

if his other 20 legs are on something that's not on the scale, and he supports himself equally on all legs, then his real mass is .2 grams, but if he's just holding his legs up, that doesn't change his mass on the scale and he is still 0.16 grams.
wow this is sort of like a physics problem. wouldn't it be scary if I stood on one leg on the scale and the weight it registered was only half the wieght I really am because oly half my legs were on the scale? same thing.

[quote]Well, that is true actually. Do you weigh the same on a scale whether you're standing on 1 leg or 2?[/quote]
Or for that matter stand on two feet and put other two (some may call it hands but if you call it feet - the weight will not change) in air :)
(Or in other words, just call those 20 'feet' which are in air 'hands' :) )

[quote="t0rajir0u"]"Weight," as it is commonly used, is really a measure of the amount of force that your body is exerting on the ground.  This amount of force depends solely on the strength of gravity and on your mass, and not on contact area.[/quote]
Exactly.

$w = F_{g}= mg$

NOTE:
That is what a scale measure by a set of springs that push you upward. Since we don't move the net force is zero. Then:
$|F_{s}| = F_{g}= mg$

EXAMPLE:
Imagine you have a box whose dimensions are $3 \times 9 \times 27$. If you drop the box to a position where a $9 \times 27$ face is in contact with the scale will it weight less then if the face that is in contact with the scale is an $3 \times 9$?

[quote="Teki-Teki"]Centipedes don't have 100 legs.  Thus these centipedes must be from outer space where there is far, far more potential for somatic damage and thus their true mass is most likely not 0.16 grams.[/quote]

I thought centipedes did have 100 legs and millipedes had a thousand

[quote="http://en.wikipedia.org/wiki/Centipede"]Despite their name, which stems from the Latin words 'centum' (meaning 'hundred') and 'ped' (meaning 'foot'), they normally have around half that number of legs, though it is possible to find centipedes with over 200 legs.[/quote]

[quote="http://en.wikipedia.org/wiki/Millipede"]Despite their name, these creatures do not have a thousand legs, although the rare species Illacme plenipes have up to 750[1]. However, common species have between 80 and 400 legs.[/quote]

Sorry--this is getting really off topic.

If the centipede has the other 20 legs in the [b]air[/b], then it doesn't make any difference in its weight, just like if you stand on one leg while on a scale with the other leg in the air. However, if the other legs are on the ground or leaning against something else, then it gets more complicated.

yea, if there were ropes attached to the other 20 legs, and was slightly holding some of the centipedes weight, then yes the centipede wouldnt be .16 grams, since  the question didnt say there were 20 ropes holding the centipede, then i assume the centipedes just wants to hold its legs in the air for no apparent reason.

let's say i stand on a scale with my feet flat on it. and then what happens if i stand tippy toed on it?....i get the same result. as a result the weight of the 20 legs in the air falls on the other legs..thereby keeping the weight the same. Am i right?


limit
integration
LaTeX
logarithms
real analysis
calculus
calculus computations

If $n$ is a positive integer then $\lim_{n\to\infty}\frac{1}{n}\binom{\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdot\cdot\cdot+\frac{1}{1+\frac{n}{n}}}$.

$\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdot\cdot\cdot+\frac{1}{1+\frac{n}{n}}\right)$ is a Riemann sum for $f(x)=\frac{1}{1+x}$ on $[0,1]$. Therefore, it converges to $\int_{0}^{1}\frac{dx}{1+x}$.

The answer is $\ln{(x+1)}$ evaluated, which yields
the answer $\ln 2$

[quote="jmerry"]A $\LaTeX$ note: \left( and \right) work much better for what you were doing than \binom.[/quote]

Ah okay thank you. I am not very experienced with Latex symbols.


limit algebra polynomial counting distinguishability function calculus

So on the other hand we need to find the sum:
$\sum_{k = 0}^{\infty}\frac{k^{100}}{k!}$

I do not know of any trick for finding the above sum;-)

The next few terms are 203, 877, 4140.

No, it's not at all clear. I don't even see a reasonable recursion (that doesn't involve the whole polynomial) yet.

As it can be seen the sum can be written in the form $e^{x}P_{n}(x)=\sum_{k=0}^{\infty}\frac{(k+1)^{2}x^{k}}{k!}$ where $P_{n}\in Z[X],\deg P_{n}=n$.
The recursion that I got is $P_{n+1}(x)=(x+1)\cdot P_{n}(x)+x\cdot P'_{n}(x)$. Are you sure that this can be explicitated?

It's the [url=http://www.research.att.com/~njas/sequences/A000110]Bell numbers[/url], the number of ways of placing $n$ distinguishable balls into $n$ indistinguishable boxes (possibly with some left empty).  The coefficients of the polynomial are the Stirling numbers $S_{n, k}$, the number of ways of placing $n$ distinguishable balls into $k$ indistinguishable boxes (with none left empty).

It's easy to see that the coefficients of the polynomials obey the same recurrence relation, $S_{n+1, k}= S_{n, k-1}+kS_{n, k}$, that the Stirling numbers do.  It's also easy to see that the Bell number $B_{n}= \sum_{k = 1}^{n}S_{n, k}$, so our sequence is the Bell numbers.  Then from the combinatorial definition one can argue that $B_{n+1}= \sum_{k = 0}^{n}B_{k}\binom{n}{k}$.

yes sir !!  :thumbup:

http://en.wikipedia.org/wiki/Bell_numbers

till today, i don't honestly understand how to derive the [b]generating[/b] function for the Bell numbers:  :(

$e^{e^{x}-1}\;=\;\sum_{k=0}^{\infty}\;\left( \frac{\mathbb{B}_{k}}{k!}\;\;x^{k}\right)$

I don't know what you mean -- the solution to your problem together with sos's observation is a derivation.  (Maybe not a really natural one, but take what you can get.)