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stupid question....    :blush:

I read something like:
[quote]
consider experiments $ E_1$ and $ E_2$ with probability spaces $ (S_1,\mathfrak A_1,P_1)$ and $ (S_2,\mathfrak A_2,P_2)$...

define the set function $ P$ be the unique measure on "$ \mathfrak A_1\times\mathfrak A_2$" with $ P(A_1\times A_2) \equal{} P_1(A_1)P_2(A_2)$ for all $ A_1\in\mathfrak A_1,A_2\in\mathfrak A_2$ ...

$ B_1 \equal{} A_1\times S_2$ , $ B_2 \equal{} S_1\times A_2$  ...

$ E_1$ and $ E_2$ are said to be independent if $ P(B_1\cap B_2) \equal{} P(B_1)P(B_2)$ for all $ B_1,B_2$
[/quote]

get confused, isn't $ P(B_1\cap B_2) \equal{} P(B_1)P(B_2)$ the same thing as $ P(A_1\times A_2) \equal{} P_1(A_1)P_2(A_2)$ ??

it is section 2.5 in the link below (i didn't quote exactly)

[url]http://books.google.com/books?id=HJmWqrlrdmsC&pg=PA45&vq=%22the+concepts+discussed%22&dq=a+course+in+mathematical+statistics&hl=zh-TW&source=gbs_search_s&cad=0[/url]

StudentA

The thing is that "by construction" the events $ A_1$ and $ A_2$ are independent.

Those sets are used to build the product measure but the product sigma algebra generated by thoses sets contain (usually) events that are not independent.


ABCD is a convex quadrilateral with area 1. The lines AD, BC meet at X. The midpoints of the diagonals AC and BD are Y and Z. Find the area of the triangle XYZ.

[quote="beta"]What happens if ABCD is a parallelogram?[/quote]

Well, ok, then assume AD and BC must intersect and that points Y and Z are distinct.

I got the answer to be [hide]1/4[/hide] by assuming ABCD is a trapezoid, but I got stuck on the proof why it work for all quadrilateral.

JBL wrote:Try it with vectors.

Further hint: [hide]Set X to the origin[/hide]
Further hint: [hide]Use the area of XCD minus the area of XAB (or vice versa)[/hide]
Further hint: [hide]Try using cross-products[/hide]

Hmm...

[hide]Well, say we let X be the origin, and we let XA = a, XB = b, XD = pa, XC = qb (a and b are vectors while p and q are constants). We have 1 = [ABCD] = [XCD] - [XAB]. Since [XCD] = (pa x qb)/2 and [XAB] = (a x b)/2, we find [ABCD] = ((pa x qb) - (a x b))/2 = 1. We also have [XYZ] = (XZ x XY)/2 (where XZ and XY are vectors) so [XYZ] = (pa + b)/2 x (qb + a)/2 = ((pa + b) x (qb + a))/4. [/hide]At this point, because of Fierytycoon's incompetency when it comes to vectors, we yell out to the AoPS forum members for help.

You're right so far.  You just need to know the properties of the cross product.  You're about 1 line away, actually.

[quote]You just need to know the properties of the cross product.[/quote]

Mm...pulling out constants? Expanding products?

Hrm, well...

[hide]Ok, what we got from what I had in my previous post was essentially:

[ABCD] = ((pa x qb) - (a x b))/2 = 1

[XYZ] = ((pa + b) x (qb + a))/4.

If we expand the products in our expression for [XYZ], we have [XYZ] = ((pq)(a x b) + (b x a) + p(a x a) + q(b x b))/4. But a x a = b x b = 0 and a x b = -(b x a), so [XYZ] = (1/4)(pq-1)(a x b).

Uh oh...stuck again... :/[/hide]

I think you had a typo there where you wrote a x b = b x a -- it's antisymmetric, not symmetric (that is, a x b = - b x a), but you fixed it in the next line, so I'm assuming you just left out the " - ."  Anyhow, now just compare the two things you have and you're done.

JBL wrote:I think you had a typo there where you wrote a x b = b x a -- it's antisymmetric, not symmetric (that is, a x b = - b x a), but you fixed it in the next line, so I'm assuming you just left out the " - ."

Yah, sorry, twas a typo.

Quote:Anyhow, now just compare the two things you have and you're done.

Well...

[hide]((pa x qb) - (a x b)) = 2

[XYZ] = (1/4)(pq-1)(a x b)

I don't see the relationship. I somehow have to manipulate the first equation to get something similar to (pq-1)(a x b), but I dunno how to do that.[/hide]


What is the number $\sqrt{1+2\sqrt{1+4\sqrt{1+6\sqrt{1+8...}}}}$ equal to?

Indeed, but the problem is that I do not manage to make any relevant supposition onwhat the limit could be. Any idea?

:blush:  Yes, sorry, the identity $n+1 = \sqrt{1+n(n+2)}$ actually implies the limit $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}} = 3$ instead of this one (the factor of $n(n+2)$ is what deluded me). I presume this was the well-known limit Harazi was referring to?  :lol: (It's attributed to Ramanujan).

Indeed, that well-known limit is not that difficult, but this one seems terribly. Personally, I do not think there is a closed formula for that limit, but it is strange they gave it in an oral examination.

But, since it was given at an oral examination, the limit [i]has[/i] to be expressible in terms of known constants ...  :)

Yes, but who on earth can find these constants? I have already spent hours trying to find the limit with no result.  :(

BTW, if we follow to a logic then the expression must be wtitten in the following way ;)
\[0\sqrt{1+2\sqrt{1+4\sqrt{1+6\sqrt{1+8...}}}}\]

I like the problem, but, to my shame, still cannot solve it. Could someone (say, Myth
or harazi) post a solution or, at least, some more revealing hint? :(

???
No one knows a solution to the original problem. It was harazi's joke in the previous reply. :lol:


who saw the 4x100 freestyle relay? was or was it not (hint, it was) the most baller thing you've ever seen?

OMG that was so amazing I LAUGH AT FRANCE HA HA HA HA HA HA HA HA HA HA HA
HA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HA

ok yeah that was random...

but that was amazing HA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HAHA HA HA HA HA HA HA HA HA HA HA lol.

OMG FRANCE WAS SO COCKY AND THEN THEY GOT PWNED.

THAT WAS AWESOME.

rofl france says- we come to SQUASH U!!

uhhh no u came to get squashed by the US!! :rotfl:

Wow, that was close... and Michael Phelps started going crazy...

Hmm yes the USA won and France got owned by their arrogance but can we get over that and just note how awesome of a race that was? I believe 5 teams broke the previous world record. Watching everyone so far in front of that green line...amazing. And also the last 20 meters...

[quote="splatyango"]did you see the thing about what Michael Phelps eats? (i think it was on NBC, but since we have two olympic channels, idk, we were always doing PIP) i forget, but it was something really random "washed down with three chocolate-chip pancakes". and if that was what he [i]washed it down[/i] with, imagine the main meal...

if anyone else saw that, please refresh me on what that was![/quote]

yea, I saw that. He said he eats 8000-10000 calories a day. :o

[quote="zephyredx"][quote="Nerd_of_the_Ages"]Yeah, I was laughing at France, also (In america! Oh well, nobody on AoPS watches YGOTAS). Although I'd really appreciate it if the reporters stopped worshipping Phelps (next thing we know, they'll be swearing in the name of Phelps). In america!

Dude, the United States are gonna pwn in Beijing, in America. (Ok, I don't think anyone gets these references; i just stop making them).[/quote]

I watched it. Bandit Keith reference.  :)[/quote]
Yeah. Actually, it's rather overused.

Tiny tim delete the video. It doesnt play anymore.

Yep. All the videos of the (AWESOME OH YEAH AMERICA PWNZ) race have been/are being yanked from youtube.


Thank you all the smart people in this great forum.

I have the following problem.  I am using Eukleides to produce chunks of PSTricks code, each one a separate figure, which I then paste into a table (tabular).  This table sits in an item which is part of the enumerate environment.  Though I specify [t] for top alignment of the table, the label of the item is positioned to the left like this

....xxxx
....xxxx
3. xxxx
....xxxx
....xxxx

:huh:  instead of top-left as I would like it:

3. xxxx
....xxxx
....xxxx
....xxxx
....xxxx

where the block of x's is the table that contains PSTricks code and the dots are just blank space I can't reproduce on this form.  Am I supposed to embed the PSTricks code differently, or use a different, more robust tabular environment package?

Thanks in advance for any suggestions.

can you post your code?  It would be easier to try stuff.  Just post a reprensentative part, not 10 pages please  ;)

ledurt

Here's a portion of the unruly code:

[code]
        \begin{enumerate}
            \item
                \begin{tabular}[t]{c@{\hspace{.2in}}c}
% Generated by eukleides 1.0.0
\psset{linecolor=black, linewidth=.5pt, arrowsize=2pt 4}
\psset{unit=0.3000cm}
\pspicture*(-4.0000,-4.0000)(4.0000,4.0000)
\psset{linecolor=red}
\psline[linestyle=dashed](-1.0000,-4.0000)(-1.0000,4.0000)
% ...
\endpspicture
%----------------------------------
                    & % column separator
%----------------------------------
% Generated by eukleides 1.0.0
\psset{linecolor=black, linewidth=.5pt, arrowsize=2pt 4}
\psset{unit=0.3000cm}
\pspicture*(-4.0000,-4.0000)(4.0000,4.0000)
\psset{linecolor=red}
\psline[linestyle=dashed](-4.0000,-1.0000)(4.0000,-1.0000)
% ...
\endpspicture
                \end{tabular}
            \item
[/code]
etc.

Thanks for any suggestions.

I'm not familiar with eukleides (I have no clue what it is and how to use it...) but I tried to put a picture in enumerate.  To have the item number at the top left corner, try this:

\item \raisebox{-\height}{[i]text[/i]}

The command [b]raisebox[/b] has two optinal arguments as felow:

\raisebox{lift}[extend-above-baseline][extend-below-baseline]{text}

to have a bit more info on this command, you can look in "the not so short introduction to $\LaTeXe$" in the section 6.6

Ledurt

Thanks for suggesting to use raisebox.  BTW, Eukleides is just a language that produces PSTricks output, just as METAPOST produces PostScript output.  (METAPOST is naturally much more robust than Eukleides.)

I tried \raisebox{-\height}{...} and it worked like a charm!

Thanks a bunch, Ledurt!


trigonometry induction AMC AIME MATHCOUNTS function

Find the least positive integer $n$ such that \[ \frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}. \]

Try adding appropriate terms after converting appropriate terms into cos. I will give you the step if you still didn't get a breakthrough.

$\{m,n\}\subset N^*,\ m\le n\le 179\Longrightarrow S_{m,n}\equiv \sum\limits_{k=m}^n \frac{1}{\sin k^{\circ}\sin (k+1)^{\circ}}\Longrightarrow$
$\sin 1^{\circ}\cdot S_{m,n}=\sum\limits_{k=m}^n \frac{\sin [(k+1)^{\circ}-k^{\circ}]}{\sin k^{\circ}\sin (k+1)^{\circ}}=\sum\limits_{k=m}^n[\cot k^{\circ}-\cot (k+1)^{\circ}]=$$\cot m^{\circ}-\cot (n+1)^{\circ}.$
$S_{45,133}=\frac{1}{\sin 1^{\circ}}(\cot 45^{\circ}-\cot 134^{\circ})=\frac{1}{\sin 1^{\circ}}(1+\tan 44^{\circ})$ a.s.o.

hint :[hide]

Use $cot(n) - cot(n+1) = sin(1)/(sin(n) \cdot sin(n+1))$[/hide]

[hide]
In this solution, all angle measurements are in degrees.

We have $\sum_{k=23}^{67} \frac{1}{\sin (2k-1)\sin (2k)}$. The partial fraction decomposition is $\frac{1}{\sin 1}\left( \sum_{k=23}^{67} \frac{1}{\sin (2k-1)} - \frac{1}{\sin (2k)} \right)$, which telescopes. We get

$\frac{1}{\sin n} = \frac{1}{\sin 1}\left( \frac{1}{\sin(45)} - \frac{1}{\sin (134)}\right)$.

Now how are you supposed to solve this? :?

BTW, are you sure it's 1/sin n? Because on Kalva it says 2/sin n.
[/hide]

[quote="chess64"]
The partial fraction decomposition is $\frac{1}{\sin 1}\left( \sum_{k=23}^{67} \frac{1}{\sin (2k-1)} - \frac{1}{\sin (2k)} \right)$[/quote]

I don't think so, since $\sin(2k-(2k-1))\neq\sin 2k-\sin(2k-1)$

What? Maybe I made a mistake...

[hide]
$\frac{1}{\sin (2k)\sin (2k-1)} = \frac{A}{\sin (2k)} + \frac{B}{\sin (2k-1)}$
$1 = A\sin (2k-1) + B\sin (2k)$
When $k=0$, we get $1=-A\sin 1\Leftrightarrow A = -\frac{1}{\sin 1}$
When $k=1/2$, we get $1 = B\sin 1 \Leftrightarrow B=\frac{1}{\sin 1}$.
This means that the partial fraction decomposition is $\frac{1}{\sin 1}\left[ \frac{1}{\sin (2k-1)} - \frac{1}{\sin (2k)}\right]$.
[/hide]

I'm confused.....

Your reasoning is wrong, as you can't apply the method of undefined coefficients here

[hide="See why"]\begin{eqnarray*}\frac{1}{\sin(2k-1)\sin 2k} &=& \frac{1}{\sin 1}\frac{\sin(2k-(2k-1))}{\sin(2k-1)\sin 2k}\\ &=& \frac{1}{\sin 1}\frac{\sin 2k\cos (2k-1)-\cos 2k\sin (2k-1)}{\sin(2k-1)\sin 2k}\\ &=& \frac{1}{\sin 1}\left(\frac{\cos (2k-1)}{\sin (2k-1)}-\frac{\cos 2k}{\sin 2k}\right)\\ &=& \frac{1}{\sin 1}(\cot (2k-1)-\cot 2k)\end{eqnarray*}[/hide]

hmmm

[hide]

using the identity:

$\cot (n) - \cot (n+1) = \frac{\sin 1}{(\sin (n))(\sin (n+1))}$, we multiply both sides of our equation by $\sin 1$ to get:

$\frac{\sin 1}{(\sin 45)(\sin 46)} + \frac{\sin 1}{(\sin 47)(\sin 48)} + \ldots + \frac{\sin 1}{(\sin 133)(\sin 134)} = \frac{\sin 1}{\sin n}$

$\cot 45 - \cot 46 + \cot 47 - \ldots + \cot 133 - \cot 134 = \frac{\sin 1}{\sin n}$

Now, using the identity:

$\cot x = -\cot (180 - x)$, we see that all of the terms cross out except $\cot 45$ and $\cot 90$, hence:

$\cot 45 - \cot 90= \frac{\sin 1}{\sin n}$

$1 - 0 = \frac{\sin 1}{\sin n}$

$\sin n = \sin 1$

which implies that the smallest positive integer is actually $n=1$  :huh:

[/hide]

How would you have seen to use that identity? It seems hard to suddenly just realize that you need to multiply by $\sin{1^{\circ}}$

[quote="agolsme"]How would you have seen to use that identity? It seems hard to suddenly just realize that you need to multiply by $ \sin{1^{\circ}}$[/quote]

yea...sorry to revive this thread, but i'm also dying to know how you guys saw to multiply by $ \sin{1^{\circ}}$

Apparently, seeing the product would remind some people of the identify, but the problem is #15, the hardest, and consider that very few people can get a perfect score on the AIME.  But you do have 12 minutes to do it.

[quote="joml88"]Find the least positive integer $ n$ such that
\[ \frac 1{\sin 45^\circ\sin 46^\circ} \plus{} \frac 1{\sin 47^\circ\sin 48^\circ} \plus{} \cdots \plus{} \frac 1{\sin 133^\circ\sin 134^\circ} \equal{} \frac 1{\sin n^\circ}.
\]
[/quote]
An alternative problem is that you can change sin into cos and kill it by the same method! I see it's very nice and slightly tricky although I did it 1 year ago!
I will give you another problem which is harder than this:
Find the sum of: 1/4cos^2(a/2) +1/4^2cos^2(a/2^2)+...+1/4^n.cos^2(a/2^n)
PS: Have fun with this problem! :)

each 1/sin(2n-1)sin(2n) at left is bigger than 1
so left> 44*1> 1/sin2(28.7)
so n<2
so n=1

[quote="ultramanzzy"]so left> 44*1> 1/sin2(28.7)[/quote]

How did you get 28.7?

[quote="ghjk"][quote="joml88"]Find the least positive integer $ n$ such that
\[ \frac 1{\sin 45^\circ\sin 46^\circ} \plus{} \frac 1{\sin 47^\circ\sin 48^\circ} \plus{} \cdots \plus{} \frac 1{\sin 133^\circ\sin 134^\circ} \equal{} \frac 1{\sin n^\circ}.
\]
[/quote]
An alternative problem is that you can change sin into cos and kill it by the same method! I see it's very nice and slightly tricky although I did it 1 year ago![/quote]Could you describe this method?

Well there is no way to automatically do these things but my thought would be that you assume that this is going to be a telescoping problem (or at least you approach the problem by thinking that it is going to telescope).

So we say well $ \frac {1}{\sin x\sin x \plus{} 1} \equal{} f(x) \minus{} f(x \plus{} 1)$ or something like that.

If the problem is going to work out nicely, it would make sense to try $ f(x) \equal{} \frac {g(x)}{\sin x}$ because you then get the correct denominator in a simple way.

From this you get that $ \frac {1}{\sin x\sin x \plus{} 1} \equal{} \frac {\sin (x \plus{} 1)g(x) \minus{} \sin x g(x \plus{} 1)}{ \sin x\sin x \plus{} 1}$ or

$ \sin (x \plus{} 1)g(x) \minus{} \sin (x) g(x \plus{} 1) \equal{} 1$

Then if you are clever, you might notice this to be some sort of sine subtraction formula. You might try $ g(x) \equal{} \cos x$. But this doesn't exactly work so you replace $ g(x) \equal{} A\cos x$ or something along those lines.

[quote="123456789"][quote="ghjk"]
An alternative problem is that you can change sin into cos and kill it by the same method! I see it's very nice and slightly tricky although I did it 1 year ago![/quote]Could you describe this method?[/quote]

From what I understand, ghjk is just mentioning a similar problem that could be solved with the same method. Even if sines are changed into cosines, a similar identity would hold:
[hide]
\[ \tan (n \plus{} 1) \minus{} \tan n \equal{} \frac {\sin 1}{\cos n \cos (n \plus{} 1)}
\]
[/hide]
although the problem wouldn't be valid just by replacing with cosines due to the $ \cos 90^\circ \equal{} 0$ in the denominator.

Here's a "solution" using engineer's induction, which I presented to an AIME class a few months ago and again to my MathCounts team recently.

The main idea is that the increments of $1^{\circ}$ is almost certainly not special in any way. Indeed the degree itself is probably totally irrelevant. So, this motivates us to rewrite the problem as
\[
  \frac{1}{\sin \left( \frac14\pi \right) \sin \left( \frac14\pi+x \right)}
+ \frac{1}{\sin \left( \frac14\pi+2x \right) \sin \left( \frac14\pi+3x \right)}
+ \dots
+ \frac{1}{\sin \left( \frac14\pi-2x \right) \sin \left( \frac14\pi-x \right)}
= \frac{1}{\sin nx} \]
where $x = 1^{\circ}$.

Written in this from, we see that the value of $x$ is not very important.  After all, if $n$ did depend on $x$, we'd somehow have a natural function taking $x = \tfrac{\pi}{180}$ to an integer. This seems very unlikely for an AIME problem.

Exploiting this, we redefine $x=45^{\circ}$, which quickly gives $n = \boxed{001}$.

[hide=Whatever works works]
Notice that
$$\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ} +\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}$$
$$=\frac 1{\cos (90 - 45) ^\circ \cos (90-46)^\circ}+\cdots+\frac 1{\cos (90 - 89)^\circ\cos (90 - 90)^\circ} + \frac 1{\cos (90 - 91)^\circ\cos (90 - 92)^\circ} +\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}$$

$$=\frac 1{\cos 45^\circ \cos 44^\circ}+\frac 1{\cos 43^\circ\cos 42^\circ}+\cdots+\frac 1{\cos 1^\circ\cos 0^\circ} + \frac 1{\cos -1^\circ\cos -2^\circ} +\cdots+\frac 1{\cos -43^\circ\cos -44^\circ}$$
which due to $\cos \theta$ being an even function equals
$$=\frac 1{\cos 0^\circ \cos 1^\circ}+\frac 1{\cos 1^\circ\cos 2^\circ}+\frac 1{\cos 2^\circ\cos 3^\circ} + \cdots + \frac 1{\cos 44^\circ\cos 45^\circ}$$
which rewritten in terms of secant is
$$=\sec 0^\circ \sec 1^\circ + \sec 1^\circ \sec 2^\circ + \sec 2^\circ \sec 3^\circ + \cdots + \sec 44^\circ \sec 45^\circ$$
Now the jankiness begins. Hmm... Notice that $\sec (x) \sec(x + 1^{\circ})$ is god awfully close $\sec^2x$. In fact, since the derivative of secant is $\tan{x}\sec{x}$, which has a maximum value on the interval $[0, \frac{\pi}{4}]$ of $\sqrt{2}$, the error caused at most by approximating $\sec(x + 1^{\circ})$ with $\sec{x}$ is $1^\circ \cdot \sqrt{2}$ or $\frac{\sqrt{2} \pi}{180} \approx 0.025$. We pray to God that this error is negligible. Thus, we approximate that this is about

$$\sec^2{0} + \sec^2{1^\circ} + \sec^2{2^\circ} + \cdots + \sec^2{44^\circ}$$

Then, observe that the summation of $\sec^2{x}$ is simply a left hand Riemann sum approximation of its integral over the region $[0, \frac{\pi}{4}]$ multiplied by $\frac{180}{\pi}$. Once again, we pray that this error is negligible. Conveniently,
$$\frac{180}{\pi} \int_{0}^{ \frac{\pi}{4} } \sec ^2 {x}  \ \text{d}x = \frac{180}{\pi} \tan{x} \ \bigg|_{x=0}^{x=\frac{\pi}{4}} = \frac{180}{\pi} \approx \frac{1}{\sin {n}}$$

Therefore, $\sin{n} \approx \frac{\pi}{180}$. Finally, to end the jank solution with one final janky touch, observe that $\frac{\pi}{180} \approx 0.015$, meaning that the smallest solution to this is near the origin. Near the origin, $\sin{x} \approx x$. Therefore, we (luckily) conclude that $n = \frac{\pi}{180} = 1^{\circ}$, the answer correctly being $\boxed{001}$
[/hide]

First phone solution (thanks mom's free mobile data as the Wi-Fi is being set up)! The fact that my algebraic intuition is strongest still stands :')

By $\sin x=\sin(180^\circ-x)$ (e.g. $\sin 133^\circ=\sin 47^\circ$ and $\sin 134^\circ=\sin 46^\circ$), the given sum rewrites as
\[\frac{1}{\sin 45^\circ\sin 46^\circ}+\frac{1}{\sin 46^\circ\sin 47^\circ}+\dots+\frac{1}{\sin 89^\circ\sin 90^\circ}.\]
After noting that none of the double-angle or product-to-sum gimmicks will work, we shall attempt to force a telescoping sum. Recalling the famous $\textstyle\sum\tfrac{1}{k(k+1)}$ problem, we try writing the summand as $\tfrac{\sin(k+1)^\circ-\sin k^\circ}{\sin k^\circ\sin(k+1)^\circ}$, which readily fails; and yet the vaguely resemblant $\tfrac{\sin(k+1)^\circ\cos k^\circ-\sin k^\circ\cos(k+1)^\circ}{\sin k^\circ\sin(k+1)^\circ}$ comes to mind. In fact,
\begin{align*}
\frac{\sin1^\circ}{\sin k^\circ\sin(k+1)^\circ}&=\frac{\sin(k+1)^\circ\cos k^\circ-\sin k^\circ\cos(k+1)^\circ}{\sin k^\circ\sin(k+1)^\circ}\\
&=\frac{\cos k^\circ}{\sin k^\circ}-\frac{\cos(k+1)^\circ}{\sin(k+1)^\circ},~(\ast)
\end{align*}
and summing from $k=45$ to $k=89$ cancels all the intermediate terms, leaving
\[\frac{\cos 45^\circ}{\sin 45^\circ}-\frac{\cos90^\circ}{\sin90^\circ}=1.\]
Now dividing by the $\sin1^\circ$ multiplied from $(\ast)$ yields $\tfrac{1}{\sin1^\circ}$, ergo $\boxed{001}$.

Pain. Never, ever doing that again

We apply the identity, $\frac{1}{\sin n \sin (n+1)} = \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right)$ in order to do telescoping. This identity is easily derivable using the sine angle subtraction identity on $n+1$ and $n.$ Thus, $$\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).$$ Noting that $\cot (180 - x) = - \cot x,$ the summation simply becomes $\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1}{\sin 1^{\circ}},$ and therefore the answer is $\boxed{001}.$


geometry
circumcircle
trigonometry
geometry proposed

Let a triangle $ ABC$ and $ H$ is the orthocenter of this triangle.Knowing that $ AH^2\plus{}BH^2\plus{}CH^2\equal{}7$ and $ AH.BH.CH\equal{}3$,
Find the radius of the circumcenter of triangle $ ABC$.

we have $ AH^2 \plus{} BH^2 \plus{} CH^2 \equal{} 4R^2 (\cos^2 A \plus{} \cos^2 B \plus{} \cos^2 C) \equal{} x$ and $ AH \cdot BH \cdot CH \equal{} 8R^3 \cos A \cos B \cos C \equal{}y$ and is know that $ \cos^2{A} \plus{}  \cos^2{B} \plus{} \cos^2{C} \plus{} 2 \cos{A} \cos{B} \cos{C} \equal{} 1$ so

$ x\equal{}4R^2(1\minus{}2 \cos A \cos B \cos C) \ \Longleftrightarrow \ x\equal{}4R^2 \left ( 1\minus{} \frac{y}{4R^3} \right ) \ \Longleftrightarrow$

$ \Longleftrightarrow \ 4R^3\minus{}Rx\minus{}y \equal{} 0$

In this case R is the solution of this equation with $ x\equal{}7$ and $ y\equal{}3$ that is $ R \equal{} \frac{3}{2}$


There are 997 points in the plane. Show that they have at least 1991 distinct midpoints. Is it possible to have exactly 1991 midpoints?

Please post separate problems in separate topics.
I split this one.

Pierre.


[quote="Klebian"][hide]
For the second part:
$(a+11)(a-11) = a^2 - 121$

So we have $a^2 - (a^2 - 121)$

121 is our answer.[/hide][/quote]
[hide]
Where a = 690177. (statement of the obvious :P )[/hide]

But knowing that a = 690177 is unneccessary (sp?).
a can be any integer, maybe any rational.


trigonometry
geometry
trig identities
Law of Sines
geometry proposed

An isosceles triangle $ ABC$ has $ AB\equal{}AC$ and $ \widehat{BAC}\equal{}80^o$.$ M\in BC, N\in AC$ such that $ \widehat{MAC}\equal{}\widehat{NBA}\equal{}30^o$. Calculate $ \angle MNC$.

[hide="Solution"][color=darkblue]

$ \hat{MNC}\, = \,\alpha$

If $ \overline{AC}\, = \,\overline{AB}$, then $ \hat{ACB}\, = \,\hat{ABC}\, = \,50$

$ NBC$ and $ AMC$ are similar, so we have the following proportion:

$ \large\frac {\overline{NC}}{\overline{MC}}\large\, = \,\large\frac {\overline{AC}}{\overline{BC}}$

We also have the following proportions, using the law of sines:

$ \large\frac {AC}{\sin50}\, = \,\frac {BC}{\sin80}\large\,\,\Longleftrightarrow\,\,\large\frac {AC}{BC}\, = \,\frac {\sin50}{\sin80}$

And also:

$ \large\frac {MC}{\sin\alpha}\, = \,\frac {NC}{\sin(130 - \alpha)}\large\,\,\Longleftrightarrow\,\,\large\frac {NC}{MC}\, = \,\frac {\sin(130 - \alpha)}{\sin\alpha}$

Substituting:

$ \large\frac {\sin(130 - \alpha)}{\sin\alpha}\, = \,\frac {\sin50}{\sin80}$

Thus $ \alpha\, = \,80^{\0}$

I think that's it.

[/color][/hide]

[color=brown][b]Identity.[/b][/color]
$ \cos{30^\circ} \minus{} 2\sin{20^\circ} \equal{} \frac {1}{2}\tan{20^\circ}$
[hide="Proof"]
$ \boxed{\begin{array}{ccc} \cos{30^\circ} \minus{} 2\sin{20^\circ} \equal{} \frac {1}{2}\tan{20^\circ} \\
 \\
\cos{30^\circ} \minus{} \frac {\sin{40^\circ}}{\cos{20^\circ}} \equal{} \frac {\sin{20^\circ}}{2\cos{20^\circ}} \\
 \\
2\cos{30^\circ}\cos{20^\circ} \minus{} 2\sin{40^\circ} \equal{} \sin{20^\circ} \\
 \\
(\cos{50^\circ} \plus{} \cos{10^\circ}) \minus{} 2\sin{40^\circ} \equal{} \sin{20^\circ}. \\
 \\
\sin{80^\circ} \minus{} \sin{40^\circ} \equal{} \sin{20^\circ} \\
 \\
2\sin{20^\circ}\cos{60^\circ} \equal{} \sin{20^\circ} \\
 \\
2\cos{60^\circ} \equal{} 1 \end{array} }$
[/hide]

[color=darkblue]Let $ T \in AM \cap BN$. Then:
$ \left\| \begin{array}{c} \angle{ABT} \equal{} \angle{MAC} \equal{} 30^\circ \\
\angle{BAT} \equal{} \angle{MCA} \equal{} 50^\circ \\
AB \equal{} AC \end{array} \right\| \Rightarrow \triangle{ABT}\equiv \triangle{CMA} \Rightarrow MC \equal{} AT$

Let $ \angle{NMC} \equal{} \alpha$.

In $ \triangle{ATN}$ and $ \triangle{MTN}$ we have:
$ \left\| \begin{array}{cc} \frac {AT}{\sin{70^\circ}} \equal{} \frac {TN}{\sin{30^\circ}} \\
 \\
\frac {TN}{\sin{(80^\circ \plus{} \alpha)}} \equal{} \frac {MN}{\sin{80^\circ}} \end{array} \right\| \Rightarrow \frac {MC}{MN} \equal{} \frac {AT}{MN} \equal{} \frac {2\sin{70^\circ}\sin{(80^\circ \plus{} \alpha)}}{\sin{80^\circ}}$

In triangle $ MNC$ we have: $ \frac {MC}{MN} \equal{} \frac {\sin{(50^\circ \plus{} \alpha)}}{\sin{50^\circ}}$. So:

$ \boxed{\begin{array}{cc} \dfrac{\sin{(50^\circ \plus{} \alpha)}}{\sin{50^\circ}} \equal{} \dfrac{2\sin{70^\circ}\sin{(80^\circ \plus{} \alpha)}}{\sin{80^\circ}} \\
 \\
\dfrac{\sin{(50^\circ \plus{} \alpha)}}{\cos{40^\circ}} \equal{} \dfrac{\sin{(80^\circ \plus{} \alpha)}\cos{20^\circ}}{2\cos{20^\circ}\sin{20^\circ}\cos{40^\circ}} \\
 \\
2\sin{20^\circ}\sin{(50^\circ \plus{} \alpha)} \equal{} \sin{(80^\circ \plus{} \alpha)} \end{array} }$

Let  $ 180^\circ \minus{} (50^\circ \plus{} \alpha) \equal{} \beta$. Then $ \alpha \equal{} 130^\circ \minus{} \beta$ and:

$ \boxed{\begin{array}{cc} 2\sin{20^\circ}\sin{\beta} \equal{} \sin{(\beta \minus{} 30^\circ)} \\
 \\
2\sin{20^\circ}\sin{\beta} \equal{} \sin{\beta}\cos{30^\circ} \minus{} \cos{\beta}\sin{30^\circ} \\
 \\
2\sin{20^\circ}\tan{\beta} \equal{} \tan{\beta}\cos{30^\circ} \minus{} \sin{30^\circ} \\
 \\
\cot{\beta} \equal{} \frac {\cos{30^\circ} \minus{} 2\sin{20^\circ}}{\sin{30^\circ}} \equal{} 2\cos{30^\circ} \minus{} 4\sin{20^\circ} \end{array} }$

By Identity we obtain that $ \cot{\beta} \equal{} \tan{20^\circ} \equal{} \cot{70^\circ}$. Because $ \beta < 130^\circ$, we obtain that:

$ \beta \equal{} 70^\circ$
$ \alpha \equal{} 60^\circ$
[size=150]$ \boxed{\LARGE{\bf {\angle{MNC} \equal{} 70^\circ}}}$[/size].[/color]

With help from [url]http://www.cut-the-knot.org/triangle/TrigCeva.shtml[/url], which uses an 18-gon I found the answer to be 70. The applet solves quadrilateral BANM. In the applet, let BAN, the 30-80-70 triangle be the red triangle and make the blue arrows the other lines, you will see they intersect at a point, and MNB=NMA=40. Thus, MNC=180-70-40=70. Can someone prove why the lines intersect at a point? This would complete the proof.

On the applet, consider the triangle with two vertices as the blue points directly to each side of another one, and the other vertex at the 80 degree angle of the red triangle. By the converse of Ceva in trig form with this triangle, this is equivalent to proving sin 30$ \cdot$sin 40=sin 20$ \cdot$sin 70, which is easily proven as it is the same as (1/2)(2)(sin 20)(cos 20)=(sin 20)(sin 70). This completes the proof.

Denote $ m(\widehat {MNC}) \equal{} x$ . Thus, $ m(\widehat {AMN}) \equal{} x \minus{} 30$ . Apply an well-known property in the convex quadrilateral $ ABMN$ :

$ \sin\widehat {MAB}\cdot\sin\widehat {NBM}\cdot\sin\widehat {AMN}\cdot\sin\widehat {BNA} \equal{}$ $ \sin\widehat {NAM}\cdot\sin\widehat {ABN}\cdot\sin\widehat {BMA}\cdot\sin\widehat {MNB}$ . Therefore,

$ \boxed {\ \begin{array}{c} \\
\ \sin 50\sin 20\sin (x \minus{} 30)\sin 70 \equal{} \sin^2 30\sin 80\sin (110 \minus{} x) \\
 \\
4\cos 40\sin 20\sin (x \minus{} 30)\cos 20 \equal{} \cos 10\cos (x \minus{} 20) \\
 \\
2\cos 40\sin 40\sin (x \minus{} 30) \equal{} \cos 10\cos (x \minus{} 20) \\
 \\
\sin 80\sin (x \minus{} 30) \equal{} \cos 10\cos (x \minus{} 20) \\
 \\
\cos 10\cos (120 \minus{} x) \equal{} \cos 10\cos (x \minus{} 20) \\
 \\
\cos(120 \minus{} x) \equal{} \cos (x \minus{} 20) \\
 \\
120 \minus{} x \equal{} x \minus{} 20 \\
 \\
x \equal{} 70 \\
\ \end{array}\ }$


I'm apparently doing a module on financial maths which I never even signed up for. So I'm a bit confused to say the least. Here's a question and answer, could someone please explain the terminology and how to get to the answer? Thankyou!

A three-month $960 note with interest at 10% dated July 12 is discounted at 12.5% on August 20. Find the Proceeds and the amount of discount. (ANSWER:
$966.65, $17.55)

Btw I forgot to say it is simple interest, not compounded.


[hide="Solution"]
Well, the sum of the first n squares (the numerator) is rewritten as
$\frac{n(n+1)(2n+1)}{6}$, which evaluates to
$\frac{2n^{3}+3n^{2}+n}{6}$
The limit then becomes
$\lim_{n\to\infty}\frac{2n^{3}+3n^{2}+n}{6n^{3}}=$
$\lim_{n\to\infty}\frac{2n^{2}+3n+1}{6n^{2}}= \frac{1}{3}$

[/hide]

This should be in the calculus forum, or at least intermediate.


3. The positive integers a.b and c are such that a+b+c=20=ab+bc-ca-b^2 Determine all possible value of abc.
4. The sum of 49 positive integers is 624. Prove that three of them are equal to one another.
5. Let A be the larger root of (2004x)^2 -(2003)(2005)x -1=0 and B be the smaller root of x^2 + 2003x - 2004 =0. Determine the value of A-B
6.Determine all integers n such that n^4 - 4n^3 +15n^2-30n + 27 is a prime number
7.Let a,b and c be real numbers such that a+bc = b+ca = c+ab =501. If M is the maximum value of a+b+c and m is the minimum value of a+b+c . Determine the value of M + 2m
8.the positive integers a,b and care such that a+b+c =20=ab + bc -ca-b^2 Determine all positive value of abc.

Just because some people don't answer problems in the first topic doesn't mean that you are allowed to make a second topic. People will answer them eventually.


On the sides $AB, BC, CD, DA$ of a square we consider the points $M, N, P, Q$. Prove that $MP$ and $NQ$ are perpendicular if and only if $AM+CP =  BN + DQ$.

If part:
Let the coordinate of the vertexs are: A: (0,a),  B: (0,0),  C: (a,0),  D: (a,a). And let the equation of the line cut AB and CD be $y=kx+b$. Since the other one is perpendicular to this one, hence we can assume the eqaution of the other line is: $y=-\frac{1}{k}x+c$.

Then we can show that the sum of the two pairs are both $a(k+1)$. And we are done.

As for the only if part...you can draw perpendiculars to form right triangles, and then see that if the sums are equal, then the right triangles are congruent...


Έχουμε συζητήσει τόσα εδώ που δεν θα μπορούσε να λείπει αυτή η κλασσική άσκηση.
http://www.mathlinks.ro/viewtopic.php?search_id=1499362353&t=91822

Λοιπον, εχοντας δει ορισμενα παραδειγματα ασκησεων με αρμονικα και πολικες, θα δοκιμασω να εφαρμοσω αυτα τα θεωρηματα για 1η φορα σε ασκηση, ελπιζω αυτα που θα γραψω να ειναι σωστα.

Θεωρω την εφαπτομενη του κυκλου στο Β, και εστω οτι τεμνει την ΜΚ στο Β' και την ΜΛ στο Β'', τοτε επηδη η ΑΓ ειναι εφαπτομενη στο αντιδιαμετρικο του Β, ειναι ΒΒ'//ΑΓ, οποτε για να τεμνει η ΒΜ την ΑΓ στο μεσο της, θα πρεπει η Β(Β', Α, Μ, Γ) να ειναι αρμονικη, η αρκει η τετραδα (Β', Κ, Μ, Σ) (Σ η τομη της ΒΓ με την ΚΜ) να ειναι αρμονικη, για να ισχυει αυτο αρκει η δεσμη Β''(Β', Κ, Μ, Σ) να ειναι αρμονικη, η πιο απλα αρκει η τετραδα (Β, Τ, Λ, Σ) (Τ η τομη της Β''Κ με τη ΒΓ) να ειναι αρμονικη, αυτο ομως ισχυει διοτι τα σημεια Β,Λ ειναι οι τομες της ΣΤ με τον κυκλο και η ευθεια Β''Κ που παιρναει απο το σημειο Τ ειναι η πολικη του Σ ως προς τον κυκλο αυτο.

Το οτι η Β''Κ ειναι η πολικη του Σ ως προς τον κυκλο προκειπτει ως εξης:
Απο θεωρημα La Hire η ευθεια ΛΒ ειναι πολικη του Β'' λογο του οτι τα σημεια Λ, Β ειναι οι πολοι των εφαπτομενων Β''Μ και Β''Β' του κυκλου, η ευθεια ΣΒ' τωρα ειναι πολικη του Κ, αρα απο La Hire παλι η ευθεια Β''Κ ειναι πολικη του σημειου τομης των ευθειων ΛΒ και ΣΒ' το οποιο ειναι το σημειο Σ.

[quote="Athinaios"]βαζω και εγω την λυση μου αν και στο link η λυση του νικου Ρ. με την αντιπαραλληλο ηταν πραγματικα πολυ καλη!!
Λοιπον ευκολα λογω συμμετροδιαμεσου , $ BKZL$ αρμονικο τετραπλευρο, $ Z$ τομη κυκλου και $ BM$.
αρα $ (DK,DZ;DB,DL) \equal{} \minus{} 1$ . ομως εχουμε τα εξης ζευγη καθετων $ (DK,AB) , (DZ,BM), (DL,BC),(DB,e)$,
 οπου e ευθεια παραλληλη στην $ AC$ που περνα απο το $ B$
Τοτε απο γνωστο λημμα $ (BA,BC;BM,e) \equal{} \minus{} 1$ και απο γνωστη ιδιοτητα η $ BM$ διχοτομει την $ AC\parallel{}e$
qed[/quote]
Μπράβο ρε Ηλία....
Και εγώ έτσι την προσπάθησα με το αρμονικό αλλά δεν μου ήρθε η ιδιότητα με τις κάθετες στις ευθείες αρμονικής δέσμης... :oops:

Well done.

[quote="Sakis"]Εστω $ ABC$ οξυγωνιο τριγωνο. Ο κυκλος με διαμετρο το υψος $ BD$ τεμνει τις $ AB,BC$ στα $ K,L$ αντιστοιχως. Εστω οι εφαπτομενες του κυκλου στα $ K,L$ οι οποιες τεμνονται στο $ M$. ΝΔΟ η $ BM$ διχοτομει την $ AC$.[/quote]

Πώς τη λύσατε στο Μαθ.Σχολείο ; Έτσι ή με πιο στοιχειώδη τρόπο ;

Μπάμπης

Εστω και καθηστερημενα βαζω για τον κυριο Στεργιου μια λυση με πολικες. Η ασκηση μας δοθηκε ,μαζι με αλλες, αλυτη για εξασκηση.

Εστω η $ l$ εφαπτομενη του κυκλου στο $ B$ η οποια τεμνει την $ KL$ στο $ E$.

Τοτε $ BE$ πολικη του $ B$ ως προς τον κυκλο και $ KL$ πολικη του $ M$ ως προς τον κυκλο. Αρα απο θεωρημα La Hire, η $ BM$ ειναι πολικη του $ E$ ως προς τον κυκλο. Αρα αν $ N$ σημειο τομης των $ EK,BM$ τοτε $ (E,L,N,K)\equal{}\minus{}1$.

Τοτε φανερα η δεσμη $ B,BK,BN,BL,BE$ ειναι αρμονικη και αφου $ l // AC$ η $ BM$ διχοτομει την $ AC$.


Let $ n$ be a positive integer and $ A\equal{} \{1,2,...,n \}$. A subset of $ A$ is said to be connected if it consists of one element  or several consecutive elements. Determine the maximum $ k$ for which there exist $ k$ dinstinct subsets of $ A$ such that the intersection of any two of them is connected.

The value of maximum $ k$ is $ n$.

Suppose we start with $ \{1\}$. Let the next set be $ \{1,2\}$. The third set can be either case1:$ \{1,k\}|k \equal{} 3,\dots,n$ or case2:$ \{1,2,k\}|k \equal{} 3,\dots,n$. In case1 we get $ \{1\},\{1,2\},\{1,3\},\dots,\{1,n\}$.

In the second case: If $ k \equal{} 3$, we again have two options. The fourth set is either $ \{1,2,3,4\}$ or $ \{1,2,k\}|k \equal{} 4,\dots,n$ where we can continue as in case1.


integration
function
logarithms
real analysis
real analysis unsolved

I'm having trouble with trying to get a handle on $ L^p$ spaces. When they are nice, I don't really have much trouble. For example,

Define $ \phi(p)\equal{}\int_X |f|^pd\mu$ in other words $ \phi(p)\equal{}\|f\|_p^p$ (and $ 0<p<\infty$). Now define $ E\equal{}\{p: \phi(p)<\infty\}$. Show that if $ 0<r<p<s<\infty$ and $ r,s\in E$, then $ p\in E$. Assume that $ \|f\|_\infty>0$.

Here is where this isn't nice. The two conditions that I'm used to are missing. If I knew that $ \mu(X)<\infty$ and $ p\geq 1$, then I'd have the techniques to do this (use Holder's Ineq). I can't see how to alter it. I'm guessing it has to do with dividing by the essential sup since we know it isn't zero?

Also, I took this from Rudin. I'm trying to familiarize myself more with these interesting spaces, but it all seems so foreign right now like I'm reaching around for stuff without any intuition.

We don't need to restrict to either sets of finite measure or $ r,p,s>1.$

Since $ r<p<s,$ write $ p\equal{}(1\minus{}\theta)r\plus{}\theta s$ for $ 0<\theta<1.$

Now let $ q\equal{}\frac1{1\minus{}\theta}.$ We have $ 1<q<\infty$ and its dual exponent is $ q'\equal{}\frac1{\theta}.$ We use this $ q$ in Hölder's inequality.

$ \int |f|^p\equal{}\int |f|^{(1\minus{}\theta)r}|f|^{\theta s}\le\left(\int |f|^r\right)^{1\minus{}\theta} \left(\int |f|^s\right)^{\theta}$

$ \|f\|_p^p\le\|f\|_r^{(1\minus{}\theta)r}\|f\|_s^{\theta s}$

Taking the $ p$th root and using that $ \theta\equal{}\frac{p\minus{}r}{s\minus{}r}$ we get

$ \|f\|_p\le\|f\|_r^{\frac{s\minus{}p}{s\minus{}r}\cdot\frac rp}\,\|f\|_s^{\frac{p\minus{}r}{s\minus{}r} \cdot\frac sp}.$

I just did this exact computation in a class of mine a few weeks ago. The way I put the motivation in lecture was to "just trust that the exponent arithmetic will work and do what you have to do."

This proves that the set $ E$ is an interval. We can find examples of it being an open interval, a closed interval, or half-open on either end.

I also assigned as homework the case in which $ s\equal{}\infty.$

In fact, here's the homework assignment in question:

[quote]1. Suppose $ p_0 < \infty$ and $ f\in L^{p_0}\cap L^{\infty}.$ (We're working on some measurable set; no assumption is being made about whether the measure of that set is finite or infinite. WLOG you may assume that $ f$ is not the zero function.)

(a) Prove that if $ p_0 < p < \infty,$ then $ f\in L^p$ and $ \displaystyle\|f\|_p\le \|f\|_{\infty}^{1 \minus{} \frac {p_0}{p}}\|f\|_{p_0}^{\frac {p_0}{p}}.$

(b) Prove that $ \displaystyle\lim_{p\to\infty}\|f\|_p \equal{} \|f\|_{\infty}.$

The purpose of this is to justify the "$ \infty$" in $ L^{\infty}.$ Hint: part (a) gets you half of what you need; for the other half, it might help to show for any $ \epsilon > 0,$ there exists a set $ E$ of positive measure such that $ |f| > \|f\|_{\infty} \minus{} \epsilon$ on $ E.$

2. Suppose $ |E| \equal{} 1$ and suppose that $ f\in L^p(E)$ for all $ 0 < p < \infty.$

(a) Show that if $ p < q,$ then $ \|f\|_p\le \|f\|_q.$ (Note: the fact that $ |E| \equal{} 1$ matters.)

(b) Suppose that $ \int_E\ln|f|$ exists as a Lebesgue integral. Show that $ \text{exp}\left(\int_E\ln|f|\right)\le\|f\|_p$ for all $ p > 0.$

Note: the notation $ \text{exp}(x)$ means $ e^x.$ Hint: Jensen.[/quote]
Notation: $ |E|$ denotes the measure of $ E.$

there is a much easier way to do this problem:

note $ X \equal{} E_1 U E_2$ where $ E_1 \equal{} \{X: |f(x)|\le 1\}, E_2 \equal{} \{X: |f(x)| > 1\}$

and $ \int_{E_1} |f(x)|^p \le \int_{E_1} |f(x)|^r \le \int_{X} |f(x)|^r < \infty$ since r < p

and $ \int_{E_2} |f(x)|^p \le \int_{E_2} |f(x)|^s \le  \int_{X} |f(x)|^s < \infty$ since p < s

so $ \int_{X} |f(x)|^p \equal{} \int_{E_1} |f(x)|^p \plus{} \int_{E_2} |f(x)|^p < \infty$

so p is in E

Thanks, that was great. Now to try my hand at it, my attempt at Prof Merryfield's homework 2(a).

Suppose $ p<q$ and $ \mu(X)=1$. Let's look at

\begin{eqnarray*}
\|f\|_p^p & = & \int |f|^p \\
              & = & \int \left(|f|^q\right)^{p/q}\cdot 1 \\
             & \leq & \left(\int |f|^q\right)^{p/q}\left(\int 1\right)^{p/(p-q)}\\
              & = & \|f\|_q^p\mu(X)^{p/(p-q)}=\|f\|_q^p\\
\end{eqnarray*}

Thus, p-th root both sides to get $ \|f\|_p\leq \|f\|_q$.

I have one question that I cannot answer...

[quote="Kent Merryfield"]This proves that the set $ E$ is an interval. We can find examples of it being an open interval, a closed interval, or half-open on either end.[/quote]

What's the example for a closed interval? I can find one for open, and I suppose closed is not that hard to find either...

Try functions of the form $ \frac1{x^a(1\plus{}\log^2x)^b}$ on $ (0,\infty),$ adjusting $ a$ and $ b$ until you get what you want.

Kent, your problem 1(a) was a question on my final last semester.  The question after that was the one HilbertThm90 asked.   :)


The two lines 3x+4y=8 and 8x+ky=5 are perpendicular. find k.


Determine the last three digits of \[2003^{2002^{2001}}.\]

[quote="maxal"]Using CRT, we have $ 2002^{2001}\bmod 400 \equal{} 352.$[/quote]Hmm... how exactly did you compute that?

[quote="ZetaX"]It's eay to check that both are equal $ \mod 25$ and $ \mod 16$. And nothing more is required to be shown.[/quote]Hmm... I need sleep urgently.  :sleeping:


$ a,b>0.$     $ a\plus{}b\plus{}1\equal{}3ab$ prove that   $ \frac{1}{a}\plus{}\frac{1}{b}\geqslant2$  :rotfl:

$ \mbox{Avem: }a \plus{} b \plus{} 1 \equal{} 3ab\Leftrightarrow b \plus{} 1 \equal{} a(3b \minus{} 1)\Leftrightarrow \frac {1}{a} \equal{} \frac {3b \minus{} 1}{b \plus{} 1}\ \ (1)\Rightarrow \\
\Rightarrow\frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {3b \minus{} 1}{b \plus{} 1} \plus{} \frac {1}{b} \equal{} \frac {3b^2 \plus{} 1}{b.(b \plus{} 1)}. \ \ (2) \\
\mbox{Asa ca, tinand seama de relatia (2): }\\
\frac {1}{a} \plus{} \frac {1}{b}\ge 2\Leftrightarrow  \frac {3b^2 \plus{} 1}{b.(b \plus{} 1)}\ge 2\Leftrightarrow 3b^2\plus{}1\ge2b.(b\plus{}1)\Leftrightarrow b^2\minus{}2b\plus{}1\ge 2\Leftrightarrow (b\minus{}1)^2\ge 2.$
[b]Observatie:[/b]
$ \mbox{Pe baza relatiei (1), avem insa: }\\
 \frac {1}{a} \equal{} \frac {3b \minus{} 1}{b \plus{} 1}\equal{}\frac{3.(b\plus{}1)\minus{}4}{b\plus{}1}\equal{}\frac{3.(b\plus{}1)}{b\plus{}1}\minus{}\frac{4}{b\plus{}1}\equal{}3\minus{}\frac{4}{b\plus{}1}<3.\ \ (3)\\
\mbox{In mod analog se arata ca in conditia data avem si: }\frac{1}{b}<3.\ \ (4)\\
\mbox{Asa adunand acum inegalitatile (3) si (4), membru cu membru }\Rightarrow \frac {1}{a} \plus{} \frac {1}{b}<6.\\
\mbox{Avem deci: }a,b>0; a\plus{}b\plus{}1\equal{}3ab\Rightarrow 2\le\frac {1}{a} \plus{} \frac {1}{b}<6.$

hello, we have
$ \frac{1}{a}\plus{}\frac{1}{b}\ge2 \Leftrightarrow a\plus{}b\geq2ab$ from our condition we get
$ a\plus{}b\equal{}3ab\minus{}1$ so we have to show that $ ab\geq1$. This is true after the AM-GM we get
$ \frac{a\plus{}b\plus{}1}{3}\geq\sqrt[3]{ab} \Leftrightarrow ab\geq1$ because $ a\plus{}b\plus{}1\equal{}3ab$
Sonnhard.

dar de ce daca avem $ (ab)^3 \geq \sqrt [3]{ab}$ inseamna ca ab $ \geq$1 ?

Daca $ (ab)^{3}\geq\sqrt [3]{ab}$ inseamna ca $ (ab)^9 \ge ab$ si deoarece $ a,b>0$ rezulta ca $ ab \ge 1$, altfel, daca $ ab<1$ atunci ar rezulta ca $ (ab)^8<1,$ imposibil avand in vedere relatia precedenta ca $ (ab)^9 \ge ab$.

Let $a,b$ be positive real numbers such that $a+b+1=3ab.$ [url=https://artofproblemsolving.com/community/c6h615331p11993318]Prove that[/url]$$ \sqrt{a+1}+\sqrt{b+1}\leq  \sqrt{2}( \sqrt{a}+\sqrt{b}).$$
Let $ a,b>0 $ and $a+b+1=3ab .$ [url=https://artofproblemsolving.com/community/c6h1928647p13385916]Prove that[/url]$$\frac{1}{a(b+1)}+\frac{1}{b(a+1)}\leq 1$$
Let $x,y>0$ such that $ x+y+1=3xy$. [url=https://artofproblemsolving.com/community/c6h1631235p10240844]Prove that[/url]
$$\frac{3x}{y\left ( x+1 \right )}+\frac{3y}{x\left ( y+1 \right )}-\frac{1}{x^{2}}-\frac{1}{y^{2}}\leq 1$$

for $a, b, c, > 0$ $$a+b+c+1=4abc$$ [url=https://artofproblemsolving.com/community/c4h1268491p6618386]is equivalent to[/url] $$\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}=1.$$
[url]https://artofproblemsolving.com/community/c6h2468386p20731943[/url]
[url]https://artofproblemsolving.com/community/c6h1203942p6386705[/url]
[url]https://artofproblemsolving.com/community/c275h306693p1657295[/url]


[b]If the function $ f$ defined by $ f(x)\equal{}\frac{cx}{2x\plus{}3}$, $ x$ is not equal to $ \minus{}\frac{3}{2}$ satisfies $ f(f(x))\equal{}x$ for all real numbers $ x$ except $ \minus{}\frac{3}{2}$ than what does $ c$ equal?[/b]

[b][color=red]Cool problem, man![/color][/b]  :D

anyone want to post a complete solution?

[quote="mihail911"][b]If the function $ f$ defined by $ f(x) \equal{} \frac {cx}{2x \plus{} 3}$, $ x$ is not equal to $ \minus{} \frac {3}{2}$ satisfies $ f(f(x)) \equal{} x$ for all real numbers $ x$ except $ \minus{} \frac {3}{2}$ than what does $ c$ equal?[/b][/quote]

[hide]
\[ f(f(x)) \equal{} \frac{c\left(\frac{cx}{2x\plus{}3}\right)}{2\left(\frac{cx}{2x\plus{}3}\right)\plus{}3} \equal{} \frac{c^2x}{2cx \plus{} 3(2x\plus{}3)} \equal{} x \Longrightarrow c^2 \cdot x \equal{} (2c\plus{}6) x^2 \plus{} 9x\]
from which it follows by comparing coefficients that $ c \equal{} \minus{}3$.
[/hide]

sorry about not posting the complete solution ... yes i did it the same way ... I 'll learn how to use latex soon

[quote="CatalystOfNostalgia"]There is no need to post all of your problems in bold.[/quote]
yeah, i understand that. :-)
but i dont' see a problem with it either.  :wink:


trigonometry
integration
Science Olympiad

Two identical balls, are connected by an inextensible massless string $ AB$, and placed North-South on a horizontal smooth table with the string taut. Suddenly, (and pretty mysteriously too! :D ) the point $ P$ which is the mid-point of  $ AB$ starts moving with constant velocity, $ v$ towards East. Find the minimum value of $ \angle APB$.

[hide]
erm, why wouldn't it be 0? the balls are symmetric, obviously, and there are no forces on them except for the tension force in the string, a part of which is always directed east. so they will move east, since nothing is preventing them from moving that way, until they touch and APB = 0.
[/hide]

This is a very interesting problem, and I have some ideas, but I have not yet been able to solve it.

Let the length of the string be $ 2L$, so that when it's being pulled, it forms two right triangles of side lengths $ L$, $ y$, and $ x$ ($ 2x$ is the distance between the balls).  Differentiating $ L^2 \equal{} x^2 \plus{} y^2$, and noting that $ \dot{L} \equal{} 0$ since the string is inextensible, gives $ x\dot{x} \plus{} y\dot{y} \equal{} 0$.  Say the balls assume their equilibrium position after a time $ \tau$; to find the angle, we can find $ x$ and $ y$ as $ t\rightarrow\tau$.  Call the inward-moving speed of the balls $ v_x$ and the upward-moving speed $ v_y$, so that $ \dot{y} \equal{} v \minus{} v_y$ and $ \dot{x} \equal{} \minus{}v_x$.  Combining these equations,
\[ \tan{\theta/2} \equal{} \frac{x}{y} \equal{} \minus{}\frac{\dot{y}}{\dot{x}} \equal{} \frac{v_y(t) \minus{} v}{v_x(t)}.
\]
Note that since $ v_y \equal{} v$ and $ v_x \equal{} 0$ when $ t \equal{} \tau$, this expression is an indeterminate form as $ t\rightarrow\tau$.  Hence its limit is equal to the limit of the indeterminate form
\[ \frac{a_y(\tau)}{a_x(\tau)},
\]
where these quantities are the accelerations.  Since the final momentum of the balls is $ 2mv$, purely in the upward direction, we have (considering impulse)
\[ \int_0^\tau F_y dt \equal{} 2m\int_0^\tau a_y dt \equal{} 2mv
\]
and
\[ \int_0^\tau F_x dt \equal{} 0 \implies \int_0^\tau a_x dt \equal{} 0
\]
However, their total final kinetic energy is $ mv^2$, so
\[ \int_0^\tau F_y dy \equal{} mv^2.
\]
Furthermore, $ dy \equal{} (v \minus{} v_y)dt$.  Then I tried some things and nothing worked, so maybe what I just said is bunk.  And maybe I'm making this much more complicated than it needs to be.  I'll have to think about this some more.

[quote="bowei"]
erm, why wouldn't it be 0? the balls are symmetric, obviously, and there are no forces on them except for the tension force in the string, a part of which is always directed east. so they will move east, since nothing is preventing them from moving that way, until they touch and APB = 0.
[/quote]
Even I was thinking about that, but IIT-JEE problems are not that easy. So there's surely some hidden catch there. Maybe $ \angle APB \equal{} 0^c$ is the answer, the JEE committee might have decided to put the students in a tizzy. But I don't think it is so.

[quote="aadil"]i think it is 0 because there has to be some force acting on p .if p starts moving the balls overlap each other hence the angle is 0[/quote]
Maybe! But the Indian Science Olympiad published the same question, only that they introduced a higher level of complexity by giving each ball a charge of $ \plus{} q$ and the length of the thread as $ 2l$, they want the minimum distance between the balls.

I must say the options are rather very nasty.
Your options are:

$ A)$ $ \left( \frac {2lq}{q^2 \plus{} 4 \pi \epsilon_0 \cdot mv^2 l} \right)$

$ B)$ $ \left( \frac {2lq^2}{q^2 \plus{} 8 \pi \epsilon_0 \cdot mv^2 l} \right)$

$ C)$ $ \left( \frac {2lq^2}{q^2 \plus{} \pi \epsilon_0 \cdot mv^2 l}\right)$

$ D)$ $ \left( \frac {2lq^2}{q^2 \plus{} 2 \pi \epsilon_0 \cdot mv^2 l}\right)$

$ E)$ None of the above.

I believe this will be easier compared to the JEE '98 problem. The JEE problem requires thinking whereas this one only clerical work-your calculations (and of course correct ones! :wink:  )

That question actually makes sense, because there is a force of repulsion between the two balls.

But in the original problem, there is no such force. So WT* happens in the experiment??

Suppose that they reached equilibrium at an angle $ \neq 0$. Then there will be an x-component of tension force on the balls. But they are in equilibrium, so there must be another force that balances that. What force?

In fact, if the olympiad question is right (and it's not E), just plug in $ 0$ for $ q$ to get a 0 angle :)

[quote="genericme"]OK, I agree that the angle should be 0.[/quote]
Ok. But once the balls attain the velocity of the mid-point, I dont think there should be any reason for the tension to continue existense, as that may accelerate the balls. Don't you think so?


geometry
absolute value
national olympiad

hello¡¡ my name is marko im new here¡¡¡ so for my first topik i will post the 2004 mexikan math olympiad in wich i participated in sadly i only got 3 place so no imo for me :( so anyway here goes:

prob1. find all primes p<q<r such that: 25pq + r=2004, and pqr+1 is a perfect square.+

prob2. what is the maximum number of positive integers that can be found such that for any two of them A  and B   we have:
f(A,B) IS GREATER OR EQUAL THEN AB/100 WHERE F(A,B)= THE ABSOLUTE VALUE OF THE DIFFERENCE BETWEEN a and B.

prob.3 let Z  and Y  be be the points of tangence of the incircle of triangle ABC  with the sides AB and CA respectively.the parallel of YZ through the midpoint M OF BC,  MEETS  ca at N.Let  L  be the point on CA  such that: NL=AB(with L  at the same side of N  than A). the line ML intersects AB  at K. prove that  KA=NC.
 WELL HERE IS THE FIRST DAY I WILL THE SECIND DAY TOMORROW. PLEASE SOLVE THEM IM VERY INTERESTED IN NEW SOLUTIONS PARTICULARLY WITH PROBLEM 2. WICH IS VERY AMBIGOUS ON MY OPINION.

Congrats to the third place and welcome to ML/AoPS.  :)

If you want to see solutions please make up own posts for each problem.  :)

[quote="marko avila"]
prob1. find all primes p<q<r such that: 25pq + r=2004, and pqr+1 is a perfect square.+
[/quote]
It's clear that $p(25p+1)<2004$  and $p$ prime, so $p=2,3,5 \ or \ 7$ but $r$ is prime too, so $p=5$ or $p=7$.

If $p=7$ then $q<12$  and $r=79$ which give that $pqr+1$ is $78^2$.

And if $p=5$ then $q<14$ so $q=7, 11 or 13$, the case $p=5$, $q=7$ gives $r=1129$ prime but $pqr+1$ is NOT a perfect square.
Case $p=5$ and $q=11$ gives $r=629$ NOT prime, and case $p=5$, $q=13$ gives a non prime $r$, so NO solutions.

Then the solution is $p=7$, $q=11$ and $r=79$.
Time: 39 mins aprox.

Best regards.

RX the Crazy Math

[quote="marko avila"]prob2. what is the maximum number of positive integers that can be found such that for any two of them A  and B   we have:
f(A,B) IS GREATER OR EQUAL THEN AB/100 WHERE F(A,B)= THE ABSOLUTE VALUE OF THE DIFFERENCE BETWEEN A and B.
[/quote]
Of course  you can assume that $A\ge B$ so easily you get that $B<100$, so all the elements of the set must be contained into $\{1,2,\dots,99\}$, again if $A\le 99$ then $B\le 9900/199<50$  so the set is contained in $\{1,2,\dots,49\}$ with same technique we get that $A\le 32$ then we discover the technique, we need to find the maximum integer $T\ge 0$ such that
\[\left|\left[\frac{100T}{100+T}\right]-(T-1)\right| \ {\rm minimum}  \quad \iff  T=10.\]
Hence the set of points is contained into $\{1,2,\cdots,10\}$, i.e. the maximum of points must be $10$ points, but we forget that we can choose another point, the last point, for instance as $11$ which satisfies the condition, so the maximum number of points you can consider is $11$.

Time: 52 mins aprox.

RX the Crazy Math

[quote="marko avila"]ARE YOU SAYING THAT 46 IS THE MAXIMUM??? IF SO GIVE AN EXAMPLE PLEASE¡¡¡¡[/quote]
Well a example is every integer into the set  $\{1,2,\dots,11\}.$

Is enough for you?

Good luck
RX TCM

[quote="marko avila"]MMMMM.... I DONT UNDERSTAND ARE YOU SAYING THAT THE MAXIMUM NUMBER OF INTEGERS IS 11???
IF SO YOU ARE WRONG. :([/quote]
I FIXED the answer, I was PARANOIC with the solution  :blush:  :blush:  :blush: I was obtaining the quantity of pairs you can consider, very sorry.

RX TCM

MMM.... YOURE WRONG 11 ISNT THE MAXIMUM THERE CAN BE MORE. TRY CONSTRUCTING THE BIGGEST LIST YOU CAN. THERE IS REALLY NO FORMAL PROOF  THAT  EXPLAINS EXACTLY WHY THE MAXIMUM IS REALLY THE MAXIMUM. REALLY YOU JUST HAVE TO SIT DOWN , SHARPEN YOUR PENCIL AND START USING IT. THATS THE ONLY WAY TO CONJECTURE ON WHAT THE MAXIMUM IS. IF I KNEW THAT MAYBE I WOULD BE GOING TO THE IMO : :(

It is greater than 11,I'll post my solution later.
Well as promised:
Supose that for $A_{m}>A_{m-1}>...>A_{2}>A_{1}$  the maximum is achieved.
We may assume that $A>B$  then $|A-B|=A-B \ge AB/100$ iff $P(A,B)=(A+100)(100-B) \ge 10000$ (it can be seen from factoring it out.)
Now if $A_{m-1}>99$ then $P(A,B) \leq 0$,so $99 \ge A_{m-1}$ implies that  $10000 \leq P(A_{m-1},A_{m-2}) \leq 199(100-A_{m-2})$ from here we get that $[10000/199]+1 \leq 100-A_{m-2} so A_{m-2} \leq 49$.
Proceeding the same way we get $A_{m-3} \leq 32,A_{m-4} \leq 24$....
From this  we find that the set $ S= {100000000,99,49,32,24,19,15,13,11,9,8,7,6,5,4,3,2,1}$ with $|S|=18$ satisfy the conditions of the problem. :)


AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
USAJMO

1.What would happen if I took both AMC 10 and AMC 12, qualified for AIME on both, then took AIME, got an 11, but my index was not high enough on to get me to USAMO? Is this another one of those scenarios where I would be penalized for taking a test?

2.What would happen if I only took the AMC 12, qualified for AIME, got an 11, but my index was not high enough to get to the USAMO? Would I still get in since i'm in 8th grade?

Sorry if these questions have already been asked. I tried to read through most of the stuff in the original thread about the USA(J)MO.

1.  Under the scenario you outline, your index computed from the AMC 12 would be a minimum of 210 (=100 + 11*10).  Based on the 2009 data for AMC 12 and AIME takers that would have been the minimum index necessary to make the USAMO. since I find 255 2009 AMC 12 + AIME takers with index of 210 and higher.  Assuming that 2010 scores are the same as 2009, you would have been selected for USAMO on the basis of the AMC 12 and the AIME.   However, even if the minimum index had been higher, you would have been selected for the USAMO as an AMC 10 taker who scored 11 or higher on the AIME, not already selected for the USAMO.

[quote]What would happen if I only took the AMC 12, qualified for AIME, got an 11, but my index was not high enough to get to the USAMO? Would I still get in since i'm in 8th grade? [/quote]

2.  Using the data from 2009, this appears to be an unlikely scenario, but in the event the minimum index was higher than your index, you would not qualify for the USAMO.  There are no special exceptions for grade in school or for age.  All AMC 12 takers are competing against each other based on score alone.

See also the second item labeled 3 in the first post at :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691

Steve Dunbar
MAA Director, American Mathematics Competitions

and the only way to qualify for USAMO is through AMC12, correct?

You qualify for USAMO if you take AMC 10 only if you get 11+ on AIME.

If you're a student for whom the possibility of making an 11 on the AIME is something you'd take seriously, then you really should be taking the AMC12. If your knowledge and problem solving skills are really at that level, then the AMC10 is an insufficient challenge for you.

Just curious, if that's the case, then why is it even possible to qualify for the USAMO through the AMC 10?

Because surprises could happen. I expect the number of people nationwide qualifying for the USAMO under that provision will be fewer than five. The number of AMC10-only students with 11+ AIME scores in 2009 was greater than that, but then in 2009 there was a perverse incentive for such students to take the AMC10. That incentive no longer exists.

There still is an incentive to take the AMC 10 if you can score 11+ because that guarantees USAMO qualification. A 100/11 might not make USAMO (the index to qualify will probably go up due to former AMC10-only takers taking the AMC 12). "But surely a 100/11 would never happen!" you say. Last year there were indeed a few students with AMC scores near 100 and AIME scores >= 11 (there was even a 102/15).

If you're an underclassman with the potential to score in the double digits on the AIME, you should take the AMC 12 for sure. However, if you can sit both the A and B dates, it might be in your best interest to take both the AMC 12 and the AMC 10 instead of the AMC 12 twice.

That's a good point. Many people think the USAJMO is just for people who are "almost there", and that it isn't a big accomplishment (especially in the AoPS community). It's still a great accomplishment for an underclassman to be one of the top 230 math problem solvers (for 10th and below) in the United States.

[quote] why is it even possible to qualify for the USAMO through the AMC 10? [/quote]

This was already answered in the thread
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691

Specifically:

"Students who want to qualify for the USAMO should take the AMC 12,
since the AMC 12 is the gateway to the USAMO. (The exception is
students who take the AMC 10 only but score 11 or more on the AIME.
Evidence from the last few years indicates that this is typically
about 5 students. These students later excelled at all levels of the
AMC program, and so based on this evidence we have built in this
exception.)"

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="darkdieuguerre"]Taking both the AMC 10 and AMC 12 is preferable if you're unsure about whether you can qualify for USAMO, either with a pure AMC index or getting at least an 11 on the AIME. For one thing, if you don't get into USAMO, you give yourself the change of getting into USAJMO, which is also an accomplishment. It's better if you get into USAJMO then not getting into either olympiad at all.

I feel like many people don't want to get into USAJMO just because it isn't as prestigious as USAMO (or maybe it's just the AoPS community that really wants to get into USAMO), but it's not something to be ignored.[/quote]

This is sort of how I feel. I would really like to qualify for USAMO this year, but if I don't, I would hate to think that [b]maybe[/b] I could have made USAJMO if I had taken the AMC 10. Also, it is plausible that I could get an 11 on the AIME while having an AMC score that just barely gets me to AIME. That's why I'll probably take the 10A and 12B or the 10B and 12A. However, it does not make sense to me why I should get rewarded for taking an easier test. I think that invitations to the USAMO should be extended to [b]all[/b] people in 10th grade or below with an 11 on AIME, regardless of whether they took the AMC 10 or the AMC 12.

Anyway, I should stop wasting my time trying to decide which test to take and hit the books. Thanks for answering my question, Mr. Dunbar.


[quote="lingomaniac88"]When does registration for the 2007-2008 USAMTS start?[/quote]

We'll set up online registration, etc., towards the end of the summer.  We hope to release Round 1 in early August, and will have registration available no later than when we release Round 1.

[quote="rrusczyk"]We hope to release Round 1 in early August.[/quote]One of those days, diophantient?


algebra
polynomial
invariant
algebra unsolved

Find all positive integr n such that there exists a Polynomial $ P$ ,having coefficients,satisfying
$ P(x^{1998}\minus{}x^{\minus{}1998})\equal{}x^n\minus{}x^{\minus{}n}$

It is clearly necessary that $ n$ be a multiple of $ 1998$.  It is both necessary and sufficient that $ \frac {n}{1998}$ be odd (write $ e^{iy} \equal{} x^{1998}$).

Edit:  Let me be more precise.  Write $ u \equal{} x^{1998}$ so that $ P \left( u \minus{} \frac{1}{u} \right) \equal{} u^k \minus{} \frac{1}{u^k}$ where $ k \equal{} \frac{n}{1998}$.  The LHS is invariant under $ u \mapsto \minus{} \frac{1}{u}$, which is only true for the RHS if $ k$ is odd.  Now that we have established necessity, sufficiency follows by, for example, strong induction.

[quote="t0rajir0u"]It is clearly necessary that $ n$ be a multiple of $ 1998$.  It is both necessary and sufficient that $ \frac {n}{1998}$ be odd (write $ e^{iy} \equal{} x^{1998}$).

Edit:  Let me be more precise.  Write $ u \equal{} x^{1998}$ so that $ P \left( u \minus{} \frac {1}{u} \right) \equal{} u^k \minus{} \frac {1}{u^k}$ where $ k \equal{} \frac {n}{1998}$.  The LHS is invariant under $ u \mapsto \minus{} \frac {1}{u}$, which is only true for the RHS if $ k$ is odd.  Now that we have established necessity, sufficiency follows by, for example, strong induction.[/quote]
Sorry t0rajir0u,my EL is quite bad :blush: so i can understand all your post
So Let me ask a quetion,what is your result,thanks

$ n \equal{} 1998(2a\minus{}1), a \in \mathbb{N}$.


[quote="Osud"]Hi!  I'm doing a linguistics project that needs a little help from a native speaker of Vietnamese.  The work would involve reading some Vietnamese sentences and saying if they're grammatical or not.  Can anyone help me out?  If you're interested, please get back to me as soon as possible, preferably not later than the end of the weekend.

Thanks!
Osud[/quote]

yes,I am Vietnamese,so how can I help you??I don't understand your project,could you tell it in detail??


If $A$ is a fifty element subset of the set $1,2,\ldots 100$ such that no two numbers from $A$ add up to $100$, show that $A$ contains a square.

[hide]
If $100 \in A$, we are done.

Suppose $100 \not \in A$.

So $A$ consists of $50$ elements from $\{1,2,\ldots,99\}$.

But if $x \in A$, then $100-x \not \in A$. Pair each $x$ with $100-x$ to get $\{1,99\},\{2,98\},\ldots,\{49,51\}, \{50\}$. There are exactly $50$ sets from which we can only take one element each. However, the set $\{36, 64\}$ contains two squares, so at least one of them must be in $A.$[/hide]

At First we assume that it possible to choose 50 nos. from set A={1,2,3,.......99,100} such that sum of no two number is 100 and the subset do not contain a perfect square.
Lets mark possible nos. in the subset with simple reasoning:-
NO NUMBER in set A add  with !00 or 50 to form 100,but 100 is a perfect square .We can"t take it in the subset.BUT 50 can be there.
Now we have to choose 49 nos. from 98 .Here for every no. k(0<k<50), there is (100-k) in this set A.
For eyery square no. k we can choose ,otherwise (100-k).WEe can not choose both because k+(100-k)=100.
For eg.k=1 is a square.So we choose 100-k=100-1=99 instead.But for k=2,it is not a square,100-k=100-2=98 not a square.HERE we can choose either of them but not BOTH.For k=19 not a square ,100-k=81 a square,we bound to choose i9.
BUT ther will be a problem if k and (100-k) are both squares.THEN we would arrive at a CONTRADICTION OF WE HAVE ASSUMED.THEN the subset MUST ATLEAST CONTAIN A SQUARE.
LETs check:BY using pythagorean triplet method([url]http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html#patterns[/url]) we get
100=6^2+8^2
 k=36,100-k=64
HENCE<PROVED.

On pairing up the numbers which sum up to 100, we get $(1,99);(2,98)$ $\cdots$ $(49,51);50;100$. But 100 is a perfect square, thus we have to choose the 50 elements from these 49 ordered pairs the lone element '50'. If we don't include 50 in our subset then by PHP, atleast 2 of the elements belong to the same ordered pair which is a contradiction. Thus we must include '50' and choose the rest of the 49 elements by taking exactly one element from each ordered pair but we cannot take any element $(36,64)$ as both these elements are perfect squares which again gives a contradiction, thus we are done.


trigonometry
real analysis
real analysis unsolved

the series :
\[\sum_{n=1}^{+\infty}\frac{1}{n}\left(\frac{2+\sin n}{3}\right)^{n}\]
is :
convergent?  divergent ?


[quote="fireforce"]Ever went just web surfing and found something funny?

Well, that's just what i did and I found that if you [b]google [i]failure[/i]; the first thing that shows up is the biography of [i]President George W. Bush[/i][/b] :rotfl: (Try it if u don't believe me)

What funny stuff have u found on the internet?[/quote]

And failure isn't even in the biography.  That's [i][b][u]weird[/u][/b][/i]

[quote="fireforce"]Ever went just web surfing and found something funny?

Well, that's just what i did and I found that if you [b]google [i]failure[/i]; the first thing that shows up is the biography of [i]President George W. Bush[/i][/b] :rotfl: (Try it if u don't believe me)

What funny stuff have u found on the internet?[/quote]
Welcome to 2004.

[url=http://en.wikipedia.org/wiki/Internet_phenomenon]This covers just about everything[/url]

That's explained here : http://en.wikipedia.org/wiki/Miserable_failure :)

Here's another funny one.  Basically it's a copy of wikipedia except eveything is  [i]just a tiny bit off[/i] and it's funny (of course).

[url=http://www.uncyclopedia.org]UNCYCLOPEDIA[/url]

 :rotfl:

[quote="biffanddoc"]yeah, that's funny.
It used to be that when you searched evil microsoft was in the top 10

Funniest thing I ever saw on the internet

http://www.funlol.com/funpages/college-essay-fun.html (its got lots of bad language so beware)

disclaimer: if you kill yourself laughing it's not my fault[/quote]

omg, i almost died laughing :rotfl:  :rotfl:

This is old.  Google "french military victories" and hit "I'm Feeling Lucky."

[url=http://www.idleworm.com/nws/2002/12/aol10.shtml]future internet[/url]
how internet SHOULD be, because hey, why would you need privacy if you're legal?

[url=http://www.thoughton.co.uk/digitallife/weblog/2004/08/be_ready_the_us.php]us tips[/url]
the us government designed some anti terror tips
the pictures are real, this site tries to explain what they really mean ( i find this one very good)

[url=http://www.bush2004.com/]bush site[/url]
surprisingly these people got ahold of this domain name and copied the layout completely of bush's real campaigning site!

[quote="ccy"]Uncyclopedia is pretty funny when it's not racist.  Lol... read the article on Chinese Language.[/quote]

I haven't read it yet, and I probably don't want to.

oh yeah did you know that you can rearrange [b]George Bush[/b] into [b]He bugs Gore[/b]? try it

Ever feel like being insulted? (0.o Don't ask)

Well then,
[url=http://www.getinsulted.com/]www.getinsulted.com/[/url]

On a sign in seafood reasturaunt:
Cod Platter for 4.95

Punk passes sign. Goes away. Sign says: Cold Pet Rat for 9.45
 :rotfl:  :rotfl:  :rotfl:  OOOOOWWWWWWWW!!!!!!!!!!!!!!!!!!!!!!!
I just fell off of my chair.[/code]

:oops_sign:  :oops_sign: forget about the [\code].


[quote="rachid"][b]a,b and c are three reel positive numbers,and:1/ab +1/ac +1/bc =1.
prove that : a/V(a²+1) +b/(V(b²+1) +c/V(c²+1)  <= 3V3/2[/b][/quote]

$ \frac{1}{ab}\plus{}\frac{1}{ac}\plus{}\frac{1}{bc}\equal{}1$ ---> $ a\plus{}b\plus{}c\equal{}abc$  --> $ a\equal{}tanx,b\equal{}tany,c\equal{}tanz$ and $ x\plus{}y\plus{}z\equal{}\pi$ ---> $ \frac{a}{\sqrt{a^{2}\plus{}1}}\plus{}\frac{b}{\sqrt{b^{2}\plus{}1}}\plus{}\frac{c}{\sqrt{c^{2}\plus{}1}}\leq\frac{3\sqrt{3}}{2}$ --->

$ \frac{a}{\sqrt{a^{2}\plus{}1}}\equal{}sinx$ --> $ \frac{a}{\sqrt{a^{2}\plus{}1}}\plus{}\frac{b}{\sqrt{b^{2}\plus{}1}}\plus{}\frac{c}{\sqrt{c^{2}\plus{}1}}\equal{}sinx\plus{}siny\plus{}sinz\leq\frac{3\sqrt{3}}{2}$   ---> $ \frac{1}{3}sinx\plus{}\frac{1}{3}siny\plus{}\frac{1}{3}sinz\leq sin\frac{x\plus{}y\plus{}z}{2}\equal{}sin60^{0}\equal{}\frac{\sqrt{3}}{2}$ ----> $ sinx\plus{}siny\plus{}sinz\leq\frac{3\sqrt{3}}{2}$

$ a\equal{}b\equal{}c\equal{}\frac{\sqrt{3}}{2}$

Let $ x\equal{}\frac{1}{a},y\equal{}\frac{1}{b},z\equal{}\frac{1}{c}$ then $ xy\plus{}yz\plus{}zx\equal{}1$.We need to prove that
$ \sum {\frac{1}{\sqrt{1\plus{}x^2}}}\leq\frac{3\sqrt{3}}{2}$
$ \Leftrightarrow\frac{\sum {\sqrt{x\plus{}y}}}{\sqrt{(x\plus{}y)(y\plus{}z)(z\plus{}x)}}\leq\frac{3\sqrt{3}}{2}$
$ \Leftrightarrow (\sum {\sqrt{x\plus{}y}})^2\leq\frac{27}{4}(x\plus{}y\plus{}z\minus{}xyz)$
$ \Leftrightarrow x\plus{}y\plus{}z\geq 9xyz$.it's true because $ (x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)\geq 9xyz$


What is the average (mean) of all $ 5$-digit numbers that can be formed by using each of the digits $ 1$, $ 3$, $ 5$, $ 7$, and $ 8$ exactly once?

$ \textbf{(A)}\ 48000\qquad
\textbf{(B)}\ 49999.5\qquad
\textbf{(C)}\ 53332.8\qquad
\textbf{(D)}\ 55555\qquad
\textbf{(E)}\ 56432.8$

When you write all of the numbers out and add, each of the numbers 1,3,5,7,8 appears in each of the digits 24 times. Hence, their sum is

$ 240000(1\plus{}3\plus{}5\plus{}7\plus{}8)\plus{}24000(1\plus{}3\plus{}5\plus{}7\plus{}8)\plus{}2400(1\plus{}3\plus{}5\plus{}7\plus{}8)\plus{}240(1\plus{}3\plus{}5\plus{}7\plus{}8)\plus{}24(1\plus{}3\plus{}5\plus{}7\plus{}8)$

$ \equal{}24(240000\plus{}24000\plus{}2400\plus{}240\plus{}24)$. The mean is this over 120, which equals

$ \frac{24(240000\plus{}24000\plus{}2400\plus{}240\plus{}24)}{120}\equal{}\frac{240000\plus{}24000\plus{}2400\plus{}240\plus{}24}{5}\equal{}\frac{24}{5}(11111)$ which is less than $ 55555$, but ends in $ 0.8$, hence it is C.

[hide="Click for solution"]
Each digit appears the same number of times in the $ 1$'s place, the $ 10$'s place, etc.  The average of the digits in each place is $ \frac{1}{5}(1\plus{}3\plus{}5\plus{}7\plus{}8)\equal{}\frac{24}{5}\equal{}4.8$.  Thus, the average of all the numbers is $ 4.8(1\plus{}10\plus{}100\plus{}1000\plus{}10000)\equal{}4.8(11111)\equal{}53332.8$, or $ \boxed{\textbf{(C)}}$.
[/hide]


It is given the equation $ ax^2\plus{}bx\plus{}c\equal{}0\ ,\ a,b,c\in \mathbb{R}\ ,\ a\neq 0$ . If $ a\plus{}b\plus{}c<0$ compare the roots with $ 1$ .

If you replace $ (a,b,c)$ with $ (\minus{}a,\minus{}b,\minus{}c)$, you get an equation with the same roots, but $ a\plus{}b\plus{}c > 0$ now. So basically, the condition says nothing about the roots except that $ a\plus{}b\plus{}c \neq 0 \iff \text{1 is not a root}$

[quote="TZF"]If you replace $ (a,b,c)$ with $ ( \minus{} a, \minus{} b, \minus{} c)$, you get an equation with the same roots, but $ a \plus{} b \plus{} c > 0$ now. So basically, the condition says nothing about the roots except that $ a \plus{} b \plus{} c \neq 0 \iff \text{1 is not a root}$[/quote]

Take two cases for the sign of $ a$ and use Viete's relations .

What happens differently if you do that?

The condition $ a\plus{}b\plus{}c < 0$ means that the quadratic on the LHS is negative at $ x\equal{}1$. If we negate the equation, the roots don't change, but the quadratic on the LHS is now positive at $ x\equal{}1$. Thus, whenever the LHS evaluated at $ x\equal{}1$ is not equal to $ 0$, that is, whenever $ 1$ is not a root, then either that equation or its negation has the property $ a\plus{}b\plus{}c < 0$. This means the only thing the condition says about the roots is that $ 1$ is not a root.

I can illustrate this. I can make an equation with:

Both roots less than 1: $ \minus{} x^2 \equal{} 0$
Both roots greater than 1: $ \minus{} x^2 \plus{} 4x \minus{} 4 \equal{} 0$
One greater one less: $ x^2 \minus{} 2x \equal{} 0$

The only thing I [i]can't[/i] do is make an equation with one or both roots equal to $ 1$.


Find all real numbers $ x$ which satisfy: $ \frac{x^2}{(x\plus{}1\minus{}\sqrt{x\plus{}1})^2}<\frac{x^2\plus{}3x\plus{}18}{(x\plus{}1)^2}.$

How about: $ \minus{}1 < x < 0$ or $ 0 < x < 3$


If $\sin x = 3\cos x$ then what is $\sin x \cos x$?

[hide]I drew a right triangle with lengths $1$,$3$ and $\sqrt{10}$

Thus $\sin{x}*\cos{x}=$
$\frac{3}{\sqrt{10}}*\frac{1}{\sqrt{10}}=$
$\frac{3}{10}$[/hide]

Hurray for Trig!

[hide]Multiply both sides by $\cos x$ to get $\sin x \cos x = 3\cos^2 x$. Square both sides in the original equation to get $\sin^2 x = 9\cos^2 x$, and adding $\cos^2 x$ to both sides we find $1 = 10\cos^2 x, \therefore \cos^2 x = \frac{1}{10} \implies \sin x\cos x = \frac{3}{10}$. $\boxed {}$[/hide]

[hide="Answer"]$\sin ^2x+\cos ^2x=1\Rightarrow 10\cos ^2x=1\Rightarrow \cos x=\frac{1}{\sqrt{10}}$
$\sin x\cos x=\boxed{\frac{3}{10}}$[/hide]

[hide]$\sin x=3\cos x\Rightarrow \tan x=3$.

$\sin x \cos x=\tan x \cos^2 x=\frac{\tan x}{\sec^2 x}=\frac{\tan x}{\tan^2 x+1}=\boxed{\frac{3}{10}}$.[/hide]

[hide]$\sin^2 x=3\sin x\cos x$ and $\cos^2 x=(1/3)\sin x\cos x$

Adding these two equations together yields

$\sin^2 x+\cos^2 x=(1/3+3)\sin x\cos x \Rightarrow 1=10/3 \sin x\cos x \Rightarrow \sin x\cos x=\boxed{3/10}$[/hide]


[img]http://img352.imageshack.us/img352/7360/mathwd7.png[/img]

If anyone can show the step by step process to solving this it would be a great help!!!

To all those who do post a solutions...you guys are legends!![/img]

kiterunner, the mistake you've made is in this step:

the term involving $ c^{100}p^{100}$ appears when 50 - r = [b][i][u]25[/u][/i][/b] and not 100
if 50 - r was = 100, then the term should have a $ c^{400}p^{400}$, which is not possible.
after that step, just proceed normally

hope this helped :)

Would anyone be kind enough to post some fully worked solutions so I come learn from it?

Thanks in advance guys


probability
conditional probability
calculus
calculus computations

I wanted to make sure I had the right answer for a excercise dealing with conditional probability:

1. The probability that an archer hits the target when it is windy is 0.4; when it is not windy, her probability of hitting the target is 0.7. On any shot, the probability of a gust of wind is 0.3. Find the probability that
a) On a given shot there is a gust of wind and she hits the target

P($T$|$W$)=(0.4*0.3)/0.3=0.4

b)hits target on 1st shot

P(target)=P($T$|$W$)*P($W$)+P(T|$W^{c}$)*P($W^{c}$)=0.4*0.3+0.7*0.7=0.61

c)hits target exactly once in two shots

P(hits 1/2)=1-P(hits target on both shots)-P(misses twice))=1-(0.61)^2-(1-0.61)^2=0.48

d)There was no gust of wind on an occasion when she missed.

P($W^{c}$|$T^{c}$)=0.7*0.3/0.3=0.7

I assumed the event P($T$) and P($W$) are indepent so P($T$|$W$)=P($T$).

Also, I need some help with the following two problems:

1a) An event E is indepent of itself. Show that it has probability either 0 or 1.
b)Events A and B are disjoint. Can you say whether they are dependent or independent?

They are dependent because if they are disjoint then $P(A\cap B)=0<P(A)*P(B)$ and so A and B are independent. Is this correct?

c) Prove that if events A and B are independent then so are the events ${A}^{c}$ and B, and the events ${A}^{c}$ and ${B}^{c}$.

Simpson's Paradox:

Two drugs are being tested. Of 200 patients given drug A, 60 are cured; and of 1100 given drug B, 170 are cured. If we assume a homogeneous group of patients, find the probabilities of successful treatement with A or B. Now closer investigation reveals that the 200 patients given drug A were in fact 100 men, if whom 50 are cures, and 100 women of whom 10 were cured. Further, of the 1100 given drug B, 100 were men of whom 60 were cured, and 1000 were women of whom 110 were cured. Calculate the probability of cure for men and women receiving each drug; note that B now seems better than A.

c) $\mathbb{P}(A^{c}\bigcap B^{c})=\mathbb{P}((A\bigcup B)^{c})=1-\mathbb{P}(A\bigcup B)=1-(\mathbb{P}(A)-\mathbb{P}(B)+\mathbb{P}(A\bigcap B))=1-\mathbb{P}(A)-\mathbb{P}(B)+\mathbb{P}(A)\mathbb{P}( B)=\dots=\mathbb{P}(A^{c})-\mathbb{P}(B)\mathbb{P}(A^{c})=\mathbb{P}(A^{c})\mathbb{P}(B^{c})$

The other one,goes more or less the same ,just use $\mathbb{P}(A\bigcap B^{c})=\mathbb{P}(A\bigcap (\Omega \setminus B))$


function
calculus
integration
Support
real analysis
real analysis unsolved

It is 'well known' that $f(x)=\exp(-1/x^{2})$ for $x \neq 0$ and $f(0)=0$ is infinitely many times differentiable but not analytic at $0$.

Can we generalize? for example:

$f(x)=e^{\frac{1}{1-\parallel x\parallel^{2}}}$ for $\parallel x\parallel <1$ and $0$ if $\parallel x\parallel \geq 1$, $x \in \bf R^{n}$

or a better function which is defined on $\bf R^{n}$

Functions which are $C^{\infty}$ and compactly supported are dense in most of the the "nice" function spaces on $\mathbb{R}^{n}.$ Take your function $f,$ only make one small adjustment: multiply it by a constant so that its integral is $1.$ Now, for any $\delta>0,$ define the function $f_{\delta}(x)=\frac1{\delta^{n}}f\left(\frac{x}{\delta}\right).$ The function $f_{\delta}$ is $C^{\infty}$ and the support of $f_{\delta}$ is the ball of radius $\delta.$

Suppose $g$ is continuous on $\mathbb{R}^{n}$ and compactly supported. For a sequence of values of $\delta$ which tend to zero, consider

$f_{\delta}*g(x)=\int_{\mathbb{R}^{n}}f_{\delta}(x-y)g(y)\,dy.$

This gives us a sequence of $C^{\infty}$ compactly supported functions which converge uniformly to $g.$

We can do similar things for functions in $L^{p}$ for $p<\infty.$ We first assume $g$ is compactly supported. (We have to truncate outside some large ball first.) Then $f_{\delta}*g$ converges to $g$ both in $L^{p}$ and almost everywhere.

Essentially what I'm saying is that once you produce one $C^{\infty}$ compactly supported function, which you have, you may then use this to produce $C^{\infty}$ compactly supported approximations to many functions in many function spaces.


A lies on the segment OB and C lies on the segment OD. The line through the midpoints of AD and BC meets the line OB at M and the line OD at N. Show that OM/ON = AB/CD.

Let $ E,F,I$ be the midpoint of $ BC,AD,BD$.
[hide="note"]$ \Delta EIF\sim \Delta NOM$[/hide]


geometry rectangle ratio counting distinguishability LaTeX AoPSwiki

I'm having more question trouble...

1.
Points M and N are the midpoints of sides AB and BC, respectively, of rectangle ABCD. If segments AN and CM intersect at point X, what fraction of the area of rectangle ABCD lies in the quadrilateral BNXM? Express your answer as a common fracton.
[img]http://usera.ImageCave.com/jnma/8.bmp.jpg[/img]

2.
How many three-digit positive integers have the property that the tens digit is greater than or equal to the hundreds digit and the ones digit is greater than or equal to the tens digit?

#2. [hide]We can do casework and find how many there are in the 100s, 200s, and so on.  In the hundreds, the hundreds digit is 1, leaving the tens digit to be 1-9.  The ones digit then would be 1-9 in 11_, 1-8 in 12_, 1-7 in 13_, and so on.  We see that to find how many of those numbers are in the hundreds, it's simply 9+8+7..+1, $ \frac {9\times10}{2} \equal{} 45$.  For the two-hundreds we do the same thing, and find the answer is 8+7+6...+1, and it continues this way all the way to 900.  So we can find all of those and add them together.

$ 45 \plus{} \frac {8\times9}{2} \plus{} \frac {7\times8}{2} \plus{} \frac {6\times7}{2} \plus{} \frac {5\times6}{2} \plus{} \frac {4\times5}{2} \plus{} \frac {3\times4}{2} \plus{} \frac {2\times3}{2} \plus{} \frac {1\times2}{2} \equal{} \boxed{165}$[/hide]

@matt919: You forgot the hundreds digit can [b]equal[/b] the tens, and the tens can [b]equal[/b] the ones.

EDIT: Arthemetic error  :blush:

[quote="007math"]#2. [hide]We can do casework and find how many there are in the 100s, 200s, and so on.  In the hundreds, the hundreds digit is 1, leaving the tens digit to be 1-9.  The ones digit then would be 1-9 in 11_, 1-8 in 12_, 1-7 in 13_, and so on.  We see that to find how many of those numbers are in the hundreds, it's simply 9+8+7..+1, $ \frac {9\times10}{2} \equal{} 45$.  For the two-hundreds we do the same thing, and find the answer is 8+7+6...+1, and it continues this way all the way to 900.  So we can find all of those and add them together.

$ 45 \plus{} \frac {8\times9}{2} \plus{} \frac {7\times8}{2} \plus{} \frac {6\times7}{2} \plus{} \frac {5\times6}{2} \plus{} \frac {4\times5}{2} \plus{} \frac {3\times4}{2} \plus{} \frac {2\times3}{2} \plus{} \frac {1\times2}{2} \equal{} \boxed{225}$[/hide]

@matt919: You forgot the hundreds digit can [b]equal[/b] the tens, and the tens can [b]equal[/b] the ones.[/quote]
My bad 007math thanx

ohhhhhh. thanks 007math! Everything seems a lot simpler when someone else explains it... :)

1) Using similar triangle bashing, the answer is 1/6

Draw $ \overline{BX}$. Do you see any similar triangles (i.e. look for equal heights and ratios of bases)?

Thats a new way tolook at it, NEAT! :10:

Also, for number 2, let the number be $ \overline{abc}$. By the problem, $ c\geq b\geq a$. However, we must also have $ a\geq1$, because the hundreds digit can't be $ 0$. Now, our problem is finding the number of groups of $ 3$ digits, where we can have duplicates, and order doesn't matter. This is because in each of these groups, one arrangement has $ c\geq b\geq a$. We have the digits $ 1 \minus{} 9$ to choose from, and we must choose $ 3$ numbers. Let the number of $ 1$'s we choose be $ a_1$, number of $ 2$'s be $ a_2$, etc. so that $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_9 \equal{} 3$. From balls and urns, we get $ \dbinom{3 \plus{} 9 \minus{} 1}{9 \minus{} 1} \equal{} \boxed{165}$ numbers. In general, there are $ \dbinom{n \plus{} k \minus{} 1}{n \minus{} 1} \equal{} \dbinom{n \plus{} k \minus{} 1}k$ ways to distribute $ n$ distinguishable items to $ k$ indistinguishable items, where duplicates are allowed.

EDIT:
 [geogebra]f6bbc0ffebcc3abfa23ed6ce87f4a8215747d96a[/geogebra]

Does this diagram help?

[quote="math154"]Also, for number 2, let the number be $ \overline{abc}$. By the problem, $ c\geq b\geq a$. However, we must also have $ a\geq1$, because the hundreds digit can't be $ 0$. Now, our problem is finding the number of groups of $ 3$ digits, where we can have duplicates, and order doesn't matter. This is because in each of these groups, one arrangement has $ c\geq b\geq a$. We have the digits $ 1 \minus{} 9$ to choose from, and we must choose $ 3$ numbers. Let the number of $ 1$'s we choose be $ a_1$, number of $ 2$'s be $ a_2$, etc. so that $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_9 \equal{} 3$. From balls and urns, we get $ \dbinom{3 \plus{} 9 \minus{} 1}{9 \minus{} 1} \equal{} \boxed{165}$ [/quote]

So, am I wrong, because I got 225.  I wasn't sure about my answer, but it seemed right.... :maybe:

[quote="math154"]Eh, it's OK. Counting problems are usually pretty tedious.[/quote]

So i'm wrong.... :(  :blush:

[quote="math154"]Oh, checking, you were correct! :D

You just made a silly arithmetic error when you added the first $ 9$ triangular numbers.

In the future, recognize them as triangular numbers first, then the sum is simple:
$ \frac {1\cdot2}2 \plus{} \frac {2\cdot3}2 \plus{} \cdots \plus{} \frac {n(n \plus{} 1)}2 \equal{} \frac {n(n \plus{} 1)(n \plus{} 2)}6$.[/quote]

silly mistake......i feel bad.....this will get me a 0 in math counts......

o_O

Wow. Okay. It's just like how every time I see "prime numbers" I think "perfect squares" in a problem. Thanks everyone! This helped a lot!

Ur welcome
 [geogebra]b41db1e842ce6b45c98159e56d08bba6d5f2c0ca[/geogebra]

latex usually works best. See aopswiki to find out more

[quote="M_M"]Okay for 1 I got 1/6
right?

Oh yeah, and how do you get the fractions upright when you're typing on here...  :blush:[/quote]

[code]$\frac{1}{6}$[/code]
comes out as
$ \frac{1}{6}$


Let $ T_n \equal{} 1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} n$ and
\[ P_n \equal{} \frac {T_2}{T_2 \minus{} 1}\cdot \frac {T_3}{T_3 \minus{} 1}\cdot \frac {T_4}{T_4 \minus{} 1}\cdots \frac {T_n}{T_n \minus{} 1}
\]
for $ n \equal{} 2,3,4,\ldots$.  Find $ P_{1991}$.

[hide="Solution."]
=====
Let:
$ V_n \equal{} \frac {T_n}{T_n \minus{} 1} \equal{} \frac {\frac {n^2 \plus{} n}{2}}{\frac {n^2 \plus{} n}{2} \minus{} 1} \equal{} 1 \plus{} \frac {2}{3}\left(\frac {1}{n \minus{} 1} \minus{} \frac {1}{n \plus{} 2}\right)$.

Now:

$ P_n \equal{} \prod_{i \equal{} 2}^n V_i$

I claim that $ P_n \equal{} \frac {3n}{n \plus{} 2}$. We proceed by induction. For $ i \equal{} 2$, we have $ P_2 \equal{} V_2 \equal{} \frac {3}{2} \equal{} \frac {3\cdot 2}{2 \plus{} 2}$. For the inductive step:

$ P_{n \plus{} 1} \equal{} P_n\cdot V_{n \plus{} 1} \equal{} \frac {3n}{n \plus{} 2}\left(1 \plus{} \frac {2}{3}\left(\frac {1}{n} \minus{} \frac {1}{n \plus{} 3}\right)\right) \equal{} \frac {3n}{n \plus{} 2} \plus{} \frac {6}{(n \plus{} 2)(n \plus{} 3)} \equal{} \frac {3(n \plus{} 1)(n \plus{} 2)}{(n \plus{} 2)(n \plus{} 3)} \equal{} \frac {3(n \plus{} 1)}{n \plus{} 3}$

and our induction is complete. Thus $ P_{1991} \equal{} \frac {3\cdot 1991}{1993} \equal{} \boxed{\frac {5973}{1993}}$.
=====
[/hide]

EDIT:

[hide="Even better."]
=====
Note that $ V_n \equal{} \frac {\frac {n^2 \plus{} n}{2}}{\frac {n^2 \plus{} n}{2} \minus{} 1} \equal{} \frac {n(n \plus{} 1)}{(n \minus{} 1)(n \plus{} 2)}$.

If we write $ P_{1991}$ using this, we find:

$ P_{1991} \equal{} \frac {2\cdot 3}{1\cdot 4}\cdot \frac {3\cdot 4}{2\cdot 5}\cdot \frac {4\cdot 5}{3\cdot 6} \cdot \frac {5\cdot 6}{4\cdot 7} \cdot \dots \cdot \frac {1990\cdot 1991}{1989\cdot 1992} \cdot \frac {1991\cdot 1992}{1990\cdot 1993}$

Now the telescoping is obvious, and we end up with:

$ P_{1991} \equal{} \frac {3\cdot 1991}{1993} \equal{} \boxed{\frac {5973}{1993}}$.
=====
[/hide]


integration
calculus
calculus computations

[quote="shanniepooh2"]1. What is the arc length of h(y) = y^0.5 for y epsilon[c,d] and c=d=12?

a. = 0.158 units
b. 0 units
c. 12.2 units
d. This cannot be determined since c=d

2.What is the arc length of g(y) = y^0.5 for y epsilon [0,5]?
a. =0.16 units
b. =3.33 units
c. =2.20 units
d. =6.33 units[/quote]

Have you made an effort to solve them?

[hide]

$ \text{Arc Length}\equal{}\int_a^{b}\sqrt{1\plus{}f'(x)}dx$

For the first one we need not even do it seeing that the integral will be

$ \int_{12}^{12}\text{irrelevant junk}dx$

Now what is $ \int_a^{a}f(x)dx$?

For the second one seeing that $ y'\equal{}\frac{1}{2\sqrt{x}}$

We have that

$ \text{Arc Length}\equal{}\int_0^{5}\sqrt{1\plus{}\frac{1}{2\sqrt{x}}}d$

[/hide]

[quote="shanniepooh2"]The answers that I came up with were C) 12.2 units for the first question
and C) 2.20 units for the second question[/quote]

$ \int_{12}^{12}f(x)dx\equal{}0$

Mathstud right idea but arclength is $ \int \sqrt{f'(x)^2 \plus{}1} dx$.  Anyway, the first question is just an integral from $ c\equal{}12$ to $ d\equal{}12$, so the answer is clearly $ 0$.

For the second question, we need to do the integral.  $ g'(y) \equal{} \frac{1}{2}y^{\minus{}1/2}$ so we have $ \int_{0}^{5}\sqrt{\frac{1}{4y}\plus{}1}dy$ which I am estimating to be $ 5.67423$... ???

@ the OP.

Sorry for the confusion, it was a  typo.  :blush:


function
integration
real analysis
real analysis unsolved

I was just doing this in class - winging it off the top without having written it out beforehand. I started to go down a blind alley, then remembered how to fix it. It makes a fun little exercise.

Write the measure of a set $ A$ as $ |A|.$ Suppose $ f$ is a measurable function on $ E,$ and $ p>0.$

Let $ S\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}2^{kp}|\{x: |f(x)|>2^k\}|.$

Prove that $ S$ is finite iff $ \int_E|f|^p$ is finite. Furthermore, there exist positive constants $ c$ and $ C$ such that
\[ c\int_E|f|^p\le S\le C\int_E|f|^p\]

I prefer this version
$ \int_X |f|^p \, d\mu \equal{} \int_{0}^\infty pt^{p \minus{} 1}\mu(\{x : |f(x)| > t\}) \, dt$
It's a one-liner almost, but it gives some pretty good information

@Kent below: I was just trying to add a theorem to the topic. I don't want to do the given problem

I know, that's useful too. But how about sticking to the series version?


[quote="paladin8"]Do you mean $x = \prod_{k=0}^{n}\left[1+\left(\frac{1+i}{2}\right)^{2^k}\right]$?[/quote]

yes

[quote="Slizzel"]

I didn't check your work, but the final result can be simplyfied a little.

$(1-i)\left(1-\left(\frac{1+i}{2}\right)^{2^n}\right)=(1-i)\left(1-\left[\left(\frac{1+i}{2}\right)^2\right]^n\right)=(1-i)\left(1-\left(\frac{i}{2}\right)^n\right)$ :D[/quote]
$(1-i)\left(1-\left[\left(\frac{1+i}{2}\right)^2\right]^n\right)=(1-i)\left(1-\left(\frac{1+i}{2}\right)^{2n}\right) \neq (1-i)\left(1-\left(\frac{1+i}{2}\right)^{2^n}\right)$

Slizzel, couldn't you just do the same thing to the original problem, making it $x = \prod_{k=0}^{n}\left[1+\left(\frac{i}{2}\right)^k\right]$

errr im not too good w/ the $\displaystyle\prod $ notation but i multiplied the first few terms out and i noticed that

$\displaystyle\prod_{k=0}^{n} \Big[1+\Big(\frac{1+i}{2}\Big)^{2^k}\Big] = \Big[1+\Big({\frac{1+i}{2}\Big)\Big]+\Big[1+\Big({\frac{1+i}{2}\Big)^2\Big]+\Big[1+\Big({\frac{1+i}{2}\Big)^3\Big]+\Big[1+\Big({\frac{1+i}{2}\Big)^4\Big] ...}}}}$

yeah, normally you end up with a geometric sequence like that when you have the big product and then something like $(a+b^{2^k})$. (I think there's a proof using base 2 or something) the problem is, this geometric sequence is finite so we can't use the formula for the infinite series :(
anyway, I think I made some progress  :)
[hide]
\begin{eqnarray*}
(1-i)\left(1-\left(\frac{1+i}{2}\right)^{2^n}\right)&=&(1-i)\left(1-\frac{(1+i)^{2^n}}{2^{2^n}}\right)\\
&=&(1-i)\left(1-\frac{(2i)^{2^n-1}}{2^{2^n}}\right)\\
&=&(1-i)\left(1-\frac{2^{2^n-1}i^{2^n-1}}{2^{2^n}}\right)\\
&=&(1-i)\left(1-\frac{i^{2^n-1}}{2}\right)\\
&=&\frac{1-i}{2} \text{ or } \frac{3-i}{2}\right)
\end{eqnarray*}
depending on whether n is odd or even
[/hide]

[quote="hmoonster"][hide]
multiply both sides by $\displaystyle\left(1-\frac{1+i}{2}\right)$ and the RHS simplifies to $\displaystyle 1-\left(\frac{1+i}{2}\right)^{2^n}$
so $\displaystyle x= \frac{1-\left(\frac{1+i}{2}\right)^{2^n}}{1-\left(\frac{1+i}{2}\right)}=(1-i)\left(1-\left(\frac{1+i}{2}\right)^{2^n}\right)$
[/hide]
I don't know how to simplify it more because of the $2^n$  :([/quote]

It should be power of 2^(n+1) instead of 2^n .....

Letting 2^n=p then 2^(n+1) = 2p  , hence

{(1+i)/2}^2p = (2i)^p / 2^2p  = i^p / 2^p

when n = odd , i^p / 2^p = - 1/2^p
when n =even , i^p / 2^p = + 1/2^p

Hence x = ( 1 - i ) ( 1 + 1/2^p)     for n = odd
           x= ( 1 - i ) ( 1 - 1/2^p)      for n= even

 :roll:  :)

[quote="Slizzel"][quote="hmoonster"]
multiply both sides by $\displaystyle\left(1-\frac{1+i}{2}\right)$ and the RHS simplifies to $\displaystyle 1-\left(\frac{1+i}{2}\right)^{2^n}$
so $\displaystyle x= \frac{1-\left(\frac{1+i}{2}\right)^{2^n}}{1-\left(\frac{1+i}{2}\right)}=(1-i)\left(1-\left(\frac{1+i}{2}\right)^{2^n}\right)$

I don't know how to simplify it more because of the $2^n$  :([/quote]

I didn't check your work, but the final result can be simplyfied a little.

$(1-i)\left(1-\left(\frac{1+i}{2}\right)^{2^n}\right)=(1-i)\left(1-\left[\left(\frac{1+i}{2}\right)^2\right]^n\right)=(1-i)\left(1-\left(\frac{i}{2}\right)^n\right)$ :D[/quote]

Hi Slizzel!

Is your step equivalent to hmoonster's step?  I doubt.  Please Check.


[quote="Hawk Tiger"]Find all polynomial $P(x)$
such that $P(x^{2}+1)\equiv P(x)^{2}+1$[/quote]
0. If deg(P(x)=0, then $P(x)=\frac{1\pm i\sqrt 3 }{2}$.
1. P(-x)=P(x) or P(x)=P(-x), and leader coefficient $a_{n}=1$. And $x^{n}P(1/x)=P(x)$.
It give $P(x)=x$ if degP(x)=1 and $P(x)=x^{2}+1$ if degP(x)=2.
2. If $P(x)$ is solution, then $P(P(x))$ is solution, because $P(P(x^{2}+1))=P(P(x)^{2}+1)=P(P(x))^{2}+1$.
3. There are not other solutions.

Thank you ,Rust.But Could you post some detail of the proof to "$P(\frac1x)x^{n}=p(x)$" and "3. There are not other solutions."
It is too hard to me. :P
And I think that if $P(x)$is a solution ,$Q(x)$ is  a solution,then $P(Q(x))$ is also solution.

Let y=1/x. We have $P(y^{2}+1)=P(y)^{2}+1$. Therefore $y^{2n}P(y^{2}+1)=y^{2n}P(y^{2}+1$ and if $P(x)$ is solution, then $x^{n}P(1/x)$ is solution.

[quote="Rust"] $P(x)$ is solution, then $x^{n}P(1/x)$ is solution.[/quote]
Sorry,I don't think so.$x^{2}+1$ is a solution,then $(x^{2}+1)^{2}+1=x^{4}+2x^{2}+2$ is a solution,but $2x^{4}+2x^{2}+1$ is not a solution,and $x^{4}+2x^{2}+2\not=2x^{4}+2x^{2}+1$


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that there exists a field $ F$ such that for some $ a$ the polynomial  $ f(X) \equal{} X^p \minus{} X \minus{} a \in F[X]$ is not solvable by radicals .

Nobody can help me?

I also add that $ p$ is a prime, and it has to be $ \mbox{char F} \equal{} p$. :oops:

Yes it is. It's also the charatheristic of F. I edited


Three As, three Bs, and three Cs are placed in the nine paces spaces that each row and column contain one of each letter.  If A is placed in the upper left corner, how many arrangements are possible?

$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 3\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 5\qquad
\textbf{(E)}\ 6$

[hide]There are four spaces left for filling B and C.
And the arrangement of B and C are relatively linked.
Therefore there are 4 type of arrangement.[/hide]