geometry
geometric transformation
reflection
3D geometry
prism

Ok, I know from physic's their is 2 types of waves Mechanical waves and ElectroMagnetic waves. Mechanical waves must propogate thru a medium such as water, air , ..etc.
ElectroMagnetic waves can propogate thru a vacuum.

The velocity of a wave is given by v = f w where ( f = frequency ,w= wavelength, v = velocity)
Since ElectroMagnetic waves all move at the speed of light c (negleting the medium factor )
we have c = f w  which means that the frequency determines the wave length (and vise versa)
for an ElectroMagnetic wave.

All waves carry energy and momentum.
The energy of a wave is given by E = f h (where h = plank's constant)

Since smaller wavelength implies greater frequency implies greater energy.
This is why ultravoilt ray's are worse on the body then visible light.
(the greater the energy the more harmful on the body)

My question is knowing all this stuff. Is their away to determine what electromagnetic waves
pass thur an object as opposed to bounce off the object.

For example
Take visible light. The color of an object is determined by what wavelength of visible light is reflected off the object back to your eyes.

So I quess I am asking the general question of what determines an ElectroMagnetic wave
from passing thru (refracting in the case of visible light) or bouncing back (reflecting in the case of visible light )

Think about the case when you drive in a tunnel you sometimes lose reception.
But if you where in a closed room you can still listen to your radio with no noticeable reception lose.

So I am assuming it has to do with the Energy carried by the ElectroMagnetic waves ??
And probably the energy of a given atom or substance it is trying to pass thru
But I am not sure. :huh:
Probably you need more energy to pass thru cement then wood.

What I would like to know is it possible to determine if ElectroMagnetic wave
will pass (thru an object / not pass thru ) just be knowing some wave property such as the
frequency of the wave?

And in the case of visble light if you have total refraction (i.e  no reflection ) then is the object
just black?

For visible light  say you have 2 different wavelength's of light reflected from an object what color would you see???

Thanks for any help on this one. :maybe:

Yes, its called [url=http://en.wikipedia.org/wiki/Absorption_spectroscopy]absorption spectroscopy[/url]. It's the study of what wavelengths (frequencies) of light are absorbed/reflected by a material. It has a theoretical basis rooted in quantum chemistry. The absorption spectrum of many basic elements can be determined theoretically. But for more complex molecules the absorption spectrum is known through experiment or computer modeling.

An object that absorbes all wavelengths of light is called a [url=http://en.wikipedia.org/wiki/Black_Body]black body[/url].

A so it's a whole subject in itself.

Does the subject absorption spectroscopy only apply to the study of visible light or does the
study also apply to all electromagnetic waves.

Would this subject also answer my questions on when an electromagnetic wave can pass thru
and when it won't pass thru an object. (not only visible light stuff)

Also I should have mentioned that I am a mathematician interested in getting into the theoretical physics for a while.
Is absorption spectroscopy not fully developed like do they know just by doing the math
at the atomic level what wavelength will be reflected. I would assume their is alot of approximation and factors at this level.

Also I had a question on what if 2 different wavelengths of light where reflected from the same object. Would the color be a mix changing at the speed of light. Or can you only have on wavelength of light reflect per object or  per section of an object. Or could a point on an object
ever reflect 2 different wavelengths of light at the same time.

I am going to look more into absorption spectroscopy.

Ok another question but now say we are dealing with mechanical waves.
For Example
Sound waves, if you close the door of the bed room you can still here the person from with in it.
Assuming the walls are within reason. So sound must pass thru some objects and not pass thru others does this also fall into a question that absorption spectroscopy  would answer.
Or does absorption spectroscopy  only deal with electromagnetic waves.

Another example would be water waves.

Anyway I would like to beable to understand why a wave gets reflected back or not on a given object.

Also with rainbows I have read that the colors of the rainbow are formed by light hitting the raindrop and reflecting/refraction out of the raindrop at a certain angle. Like they say blue light is something like 42 degree angle. But I don't get where the angle (degrees) corospondes to the wavelength of blue. Prisms use the same kind of angle thing except I thought it was the wavelength that determined the color how does the angle come into play????

Note I know snells law ,and I can do the math of optical geometry,..etc etc.
Any help with this one would be great as well. :blush:  :maybe:  :maybe:

[quote]The color is a light characteristic in a point of a sensor ( eye, film, etc.) so it's a time and not a space property: frequency and not wavelength.
[/quote]

I don't get why we cann't just classify color by the wavelength of light being reflected.
As opposed to using it's frequency. (c = f w so changing frequency changes wavelength accordingly)
Either way my question still stands just change wavelength to frequency.

I still don't understand how the angle effects the frequency/wavelength and gives the different colors. In the case of the rainbow problem.

You did say color is a light characteristic. Is it a characteristic of the frequency of light.
Or  is color just a characteristc made by the eye's interpertation of the frequency of light.
Also you did say color was time not space depended but if you have the frequency then
you know the wavelength which is space depended.

Maybe the distinction comes from the fact that c = f w only truely holds in a vacuum and that
we would have to find out the speed of light in the different medium first before classifying it by wavelength. c_m = f w  where c_m = speed of light in the medium under consideration.
That is why we use frequency to classify color because the same wavelength can have 2 or more colors associated  to it? :maybe:

Thanks for any clarity

When an electromagnetic wave passes from a medium to another, in general, there is a change in its velocity, therefore in the wavelength (the "color" or frequency doesn't change). The change in the propagation's direction, "bending",  is due to boundary conditions for Electric and Magnetic Fields.

Ok so color is determined by the frequency reflected back to your eye's.
But I still don't get your statement fully yet.
[quote]The change in the propagation's direction, "bending", is due to boundary conditions for Electric and Magnetic Fields.
[/quote]

In the case of a rainbow something like 42 degrees is blue. But why cann't a frequency that gives red be bent at a 42 degree angle as well.

Maybe it lies in snell's law. I will look back into it tonight. If I remember correctly it is
sin a/sin b = c_b/c_a but I think where I will find my answer is in the other form.
The one with the index of refractions.

But obviously I can reflect light any way I want and it won't change the color.
I think it is the combo of refraction and reflection thru different medium that gives the color
seperation. Which is known as dispersion and is what a prism uses to seperate color's.
However you need a refracting material to do this. What house hold materials allow this if I wanted to test it out.

Anyway please correct me if I am wrong or if you have more to add to it.
Thanks
Still unsure though  :maybe:

You can understand the bending using Snell's law, indeed: the speed is inversely proportional to the refraction index that varies slightly with the frequency (normally, reflexion doesn't contributes to dispersion or decomposition of light).
This is geometric optics and it's OK. But actually, Snell's law can be deducted from the ondulatory optics. That was what I had in mind when I mentioned boundary conditions.


function
algebra
polynomial
advanced fields
advanced fields unsolved

I found the following results in Knuth's Concrete Mathematics, and the proofs are supposed to be envolving with generating function. Does anyone know how to prove them?

Let

$ B_t(z) = \sum_{k\geq 0} \frac{z^k}{k!} (tk)^{\underline{k-1}} $

where

$x^{\underline{k}} = x(x-1)...(x-k+1)$

Prove that:

$B_t(z)^r = \sum_{k\geq 0} { tk + r \choose k } \frac{r}{tk+r} z^k$

It means that the sequence of polynomials $P_k (r) = r (r + tk - 1)^{\underline{k-1}}$ is of binomial type:  $\sum {n \choose k} {P_k}(x) {P_{n-k}}(y) = {P_n}(x+y)$.
Equivalently, $P$ should enumerate functions $[k] \to [r]$ with a given type of decoration on the fibers.  What type of decorations? The number of decorations on a set of size $k$ is $P_k(1) = (tk)^{\underline{k-1}}$.  So we are reduced to finding a counting problem whose answer for a set of size $k$ is $(tk)^{\underline{k-1}}$.
I don't know a nice answer, but it should be analogous to the case $(tk)^{k-1}$ which enumerates $t$-colored rooted trees.

Thank you Fleeting. Actually I don't understand what you are saying :)

By the way, For $t=2$,  $C(t) = B_2(t)$ is the generating function of $C_n$ where $C_n$ is the $n$-th Catalan number. Using the identity

$C^2(t) = 1/t(C(t)-1)$

we can show that

$B_t(z)^r = \sum_{k\geq 0} { tk + r \choose k } \frac{r}{tk+r} z^k$ holds for all $r\in Z$ where $t = 2$.

Even though this is not a strong result, but I am quite happy with it. Do I have a different approach to solve the problem for $t$ and $r$ real? Graph approach is good but I don't think we can use it when $r$ is real.

Catalan numbers enumerate rooted trees with $t=2$ (possibly empty) subtrees at each vertex.
The $P_k(r)$, for $t=1$, enumerate maps $[k] \to [r]$ with a linear ordering on each fiber.  A linear ordering is the same as a rooted tree with $t=1$ subtree under each vertex.  So probably $P_k(r)$ enumerates rooted trees with an ordered $t$-tuple of (possibly empty) subtrees at each vertex.  If $C = {C_t}(z)$ is the generating function for such structures, then

$C(z)^t = 1/{z(C(z)-1)}$.

Your proof for $t=2$ may then generalize.

If the identity is true for positive integer $r$ and $t$, it is true for all values of $r$ and $t$,
because for each $n$ it says that some polynomial $Q_n(r,t)$ in two variables is equal to zero, and this is certainly true if it $Q=0$ at integer points.


If $a,b,c>0$  and $abc=1$  then prove or disprove that

$(a+3)(b+3)(c+3)\geq (a+1)^2(b+1)^2(c+1)^2$

If I'm not mistaken, it doesn't hold for $a=b={1\over 2}, c=4.$ And I think it generally doesn't hold for triplets $({1\over n},{1\over n},n^2)$ where $n\geqslant 2$


The larger root of the equation $x^2+x-12=0$ is:

answer : 3. ( the solutions are -4 and 3)

[hide]when you factor it, it becomes (x-3)(x+4), so x = {3, -4} so the answer to your question is 3[/hide]

[size=150][i][b]( X - 3 )( X + 4 ) = 0
since the answer is X = 3 [/b][/i][/size]


number theory
relatively prime
number theory unsolved

[quote="enndb0x"]Let $ p$ be a prime number and $ n$ is nonegative integer. The numbers $ x_1 , x_2 , x_3 , ...,x_s$, are all numbers of the set :$ \{1,2,...,p^n\}$,who are relatively prime with $ p^n$.Prove the equality :

 $ x_1^3 \plus{} x_2^3 \plus{} ... \plus{} x_s^3 \equal{} \frac {1}{4}\cdot p^{2n}\cdot (p \minus{} 1)(p^{2n \minus{} 1} \minus{} 1)$

Sorry for english ![/quote]

We have:
$ \sum_{i\equal{}1}^{p^n}{i^3}\equal{}\frac{p^{2n}(p^n\plus{}1)^2}{4}$
$ \sum_{i\equal{}1}^{p^{n\minus{}1}}{p^3i^3}\equal{}p^3\cdot \frac{p^{2n\minus{}2}(p^{n\minus{}1}\plus{}1)^2}{4}$
So
$ x_1^3 \plus{} x_2^3 \plus{} ... \plus{} x_s^3 \equal{}\frac{p^{2n}(p^n\plus{}1)^2}{4}\minus{}p^3\cdot \frac{p^{2n\minus{}2}(p^{n\minus{}1}\plus{}1)^2}{4}\equal{}\frac{p^{2n}(p^{2n}\plus{}2p^n\plus{}1)}{4}\minus{}\frac{p^{2n}(p^{2n\minus{}1}\plus{}2p^n\plus{}p)}{4}\equal{}\frac {1}{4}\cdot p^{2n}(p^{2n}\minus{}p^{2n\minus{}1}\minus{}(p\minus{}1))\equal{}        \frac {1}{4}\cdot p^{2n}\cdot (p \minus{} 1)(p^{2n \minus{} 1} \minus{} 1)$


Given is a sequence $a_n$ which goes to 0, there is a decreasing sequence $b_n$ of nonnegative numbers such that $\sum {a_n b_n}$ converges absolutely.

Are you sure it must be increasing? What if $a_n=\frac 1n$?

You're right again. Grobber, for some of the simple problems I post, please let the others enjoy solving them and not a solution in 5 minutes.  ;)


I don't know where to put this one, and let it be here ;)
hm  this  is a inernational math forum, so it's not easy  to find here people from such little countries as for example Georgia.  I'm from Georgia and I want to find kids from my countrey who like math. There is no georgian math forum at all because too many kids are not interested in math  here. I just want to have new friends who like math.

If anybody is here from Georgia, tell me please.

send me a PM or contact me on the sky pe :)
sky pe: anika_555

[quote=".Ana."]There is no georgian math forum[/quote]

Edit: My bad. But you should try posting here:

http://www.artofproblemsolving.com/Forum/index.php?f=242

thank you, but I did not mean that Georgia which is in the USA.   It's a countrey !!!  we, georgians, call it "saqartvelo" have you heared about it??

...yeah, I edited. New link is where you should probably post your question...

thank you :) I'll post this topic there then


Every open connected set, $ U \subset \mathbb{R}^2$ is path connected.

I am trying to show that $ U \equal{} A \cup A^{c}$ where $ A \equal{} \{y | y$ is path connected to some x in U} and trying to show that both disjoint sets are open. I was able to show that A is open, but how do you show this for A complement when we're not sure that U has elements in which not connected in which I can say $ A^{c}$ is open?

On $ U$, define the relation $ \sim$ such that $ x \sim y$ if there exists a path in $ U$ joining $ x$ and $ y$. It is easy to see that $ \sim$ is an equivalence relation. Now consider equivalence classes $ P_i$ in $ U$. It is easy to show that each $ P_i$ is open connected. Thus $ U \equal{} \bigsqcup_i P_i$. This shows that if there are two different (hence disjoint) equivalence classes in $ U$, $ U$ cannot be connected.

[quote="mstrip"]So I don't have to show that $ A^{c}$ is open?[/quote]

Well, actually what I've showed is that $ A^c$ is open, but in slightly indirect way. Combining your notation and my notation, we must have $ A \equal{} P_i$ for some index $ i$, and then $ \textstyle A^c \equal{} \bigsqcup_{j \neq i} P_j$. Since $ A^c$ is a union of open sets, $ A^c$ is open.

How do we know that $ A^{c}$ is a union of open sets? Isn't that what we're trying to prove? I can see the equivalence relation holding for $ A$, but I'm not seeing how the relation applies to the union of disjoint sets I was going for.

[quote="mstrip"]How do we know that $ A^{c}$ is a union of open sets?[/quote]

As long as you admit the fact that $ \textstyle A^c \equal{} \bigcup_{j \neq i} P_j$, then it's over. Notice that each $ P_j$ is open.

Ok, thanks. This is definitely something to think about.

Yes.

It's also a shame you didn't spend more than 4 minutes thinking about either of sos440's posts before responding.  Perhaps you should try this "thinking about it" step a little earlier in the process next time  :wink:


Find the requested function value of secq, if sin q = 14/15

Wow, I'll be your homework help for today.

Let leg $a$ be opposite $\theta$ and $c$ be the hypotenuse. Leg $b=\sqrt{15^{2}-14^{2}}=\sqrt{29}$
So $\sec{\theta}=\frac{1}{\cos{\theta}}=\frac{1}{\frac{\sqrt{29}}{15}}=\frac{15}{\sqrt{29}}$

That's exactly what I said, except that I rationalized it.


[quote="kstan013"]What Math Leagues/multiple meet math competitions are you in? How are you doing in them?

I'll start by saying that I am in the CWML (Central Wisconsin Math League) and I am currently in 6th place out of 500-ish.[/quote]

Welll, I don't know if this is the same as MML, but I am currently tied for 1st with 5 other people with a 12/12.

I'm tied for 4th with I don't know how many people with a 10/12 :mad: in that. Nobody in Wisconsin has a perfect score.

Yeah...The math league I am in is way way different. It's just kind of thrown together by a university but like I said, 500-ish people. There's just a test with 10 questions. The first question has 5 true/falses and you get 2 points for each one of those that are right. There are 9 other questions worth 10 ppoints and out of those there are 3-5 multiple choice. You get 55 minutes to do it. They split the thing way up though. They divide it into 3 categories (Geometry, Algebra, and (this is my favorite one) Advanced Topics  :rotfl: ) Past that they divide it further into school sizes 1,2,3 and 4. So I'm first in Advanced topics school size 2, but 6th in Advanced topics overall :wink: . Anyway, on the last test I got a 88. It's got some Calculus on there and I got one of the Calculus problems wrong. Yeah...(note that I'm only in 8th grade so I have 4 more years to do it! :D )

The "official" national math league contest refers to http://www.mathleague.com

yes, but here I am referring to any leagues of math, not necessarily the official one :wink:


search
LaTeX
geometry
inequalities
IMO Shortlist

[quote="MithsApprentice"]The scariest phrase you've ever seen is "Source: IMO Shortlist".[/quote]  :D  :D

[quote="MithsApprentice"]The scariest phrase you've ever seen is "Source: IMO Shortlist".[/quote]

There's only one reason that might not make the shirt -- at some point that phrase will cease to frighten you!

[quote="furious"]
....if you don't like english because you can't "prove it".[/quote]

Once I read about Kurt Gödel that initially he discarded the American constituion as he found logical flaws in it which would allows dictators to take over the country.

Regarding the IMO Shortlist you also can use IMO LongList. Waiting for the ML sayings...  :)

[quote="orl"]Regarding the IMO Shortlist you also can use IMO LongList. [/quote]

True, I just thought the Shortlist is a scarier thought. :-) I'd actually attempt a Longlist problem since there are some very do-able ones on there in my opinion. But a Shortlisted problem? Psh. *Skips*

I say A - ops also...but most ppl don't know what that is so I have to say Art of Problem Solving afterwards.

[quote="Treething"]What's the IMO Shortlist?

You know you're an AoPSer if you start capitizing the first "e" and the "o" in "everyone". (see upper left)[/quote]

[url=http://www.artofproblemsolving.com/Forum/index.php?f=177]Look here.[/url]

[quote="joml88"]

[quote="Treething"]What's the IMO Shortlist?

You know you're an AoPSer if you start capitizing the first "e" and the "o" in "everyone". (see upper left)[/quote]

[url=http://www.kalva.demon.co.uk/short.html]Look here.[/url][/quote]

 :mad: ... if I am on the forum for almost one year and I not yet figured out that we have a much more comprehensive [url=http://www.mathlinks.ro/Forum/index.php?f=177]IMO ShortList/LongList section here.[/url]

A lot of us are still not used to having so many boards, sub-boards, you name it. It's just all very confusing. I used to look at problems from boards I knew were at my level. Now, it's like there's 2 complete SECTIONS of the forum I don't go to because they're just way over my head and it takes about 15 minutes to just click on the threads and read all the new stuff.

Okay, this isn't a bad thing necessarily. But with such a big change, it's understandable that we forget "oh yea, we have one of those right here too". :-)

Yes I admit that there are many forums, but I think that they are ordered well :) I was merely thinking about shortlists. Orlando and some other members have been putting a lot of effort in the Shortlisted project and it's not good too see other people reffering to another site when people are asking about  the shortlists. Even Google gives AoPS- MathLinks credit (being the first place when searching for shorlists), so it's kinna sad that some old members do not :(

[quote="Valentin Vornicu"][quote="joml88"][url=http://www.kalva.demon.co.uk/short.html]Look here.[/url][/quote]I always wonder why old members of the site always link to Kalva instead of showing the [b]much more[/b] stuff about the Shortlists which is on this site right here: http://www.mathlinks.ro/Forum/index.php?f=177  :?  ...  maybe there isn't much publicity about this? Google seems to think otherwise: http://www.google.com/search?hl=en&q=imo+shortlists&btnG=Google+Search  :?[/quote]

And I want to stress that the ML/AoPS IMO ShortList/LongList (ISL/ILL) section has some real advantages vs. the Kalva site:

1.) you can download the ISLs/ILLs as distribute to team leaders during the IMO, either as .pdf and sometimes even as $\LaTeX$

2.) you can get an overview of the problems in $\LaTeX$ and this looks MUCH better than plain HTML/text, look at the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119224#119224]IMO ShortList 2001.[/url] Just browse down the site at this link.

3.) On the sites as mentioned in 2.) you find links to the threads where the problems were discussed on the forum.

4.) You often obtain several solutions: those in the official solution booklet plus those given at the forum discussions.

5.) The forum solutions are often correct as there are given by mature IMO medal winners/trainers and some of them have special knowledge as Darij Grinberg and grobber for geometry, Myth/Pierre Bornsztein/Iura for combinatorics, manlio/vasc for inequalities, billzhao/harazi for algebra and Peter Scholze/Valentin Vornicu for number theory. Obviously many other examples can be given.

6.) The ML/AoPS forum does not only have IMO ShortList but [b]also[/b] has IMO LongLists for many years, e.g. you can find the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15574]IMO ShortList and LongList 1987 as .pdf-files here.[/url]

7.) The ML/AoPS forum simply has much more content. Past content among the years 1971-1980 is missing completely on Kalva and the years before 1971 are very imcomplete.

8.) And the ML/AoPS forum also has more new content as the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15621]IMO ShortList 2003.[/url]

[b]Do you need any more reasons ? [/b] :lol:

[quote="Treething"]What's the IMO Shortlist?

[/quote]

This just again proves as Valentin usually states that nobody reads any annoucements or stickies in the forums, e.g. this one  on [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2111]sources.[/url]

:( ....I'm sorry.  I never visit the Shortlist section.

I will change the link.

[quote="orl"]Once I read about Kurt Gödel that initially he discarded the American constituion as he found logical flaws in it which would allows dictators to take over the country.[/quote]

The story I heard is that after he found a contradiction in the constitution, his friends tried to get him not to bring it up when he went to his citizenship interview. Of course, he ignored this advice. However, the examiner was so delighted at meeting Einstein (who was one of Godel's witnesses) that Godel was naturalized anyway.

[quote="ComplexZeta"]

The story I heard is that after he found a contradiction in the constitution, his friends tried to get him not to bring it up when he went to his citizenship interview. Of course, he ignored this advice. However, the examiner was so delighted at meeting Einstein (who was one of Godel's witnesses) that Godel was naturalized anyway.[/quote]

This is in short form what I read in his biography.  :)

@joml88: Ok. I forgive you.  :)  Do you have any suggestions to make the ISL/ILL section better known, especially on the AoPS parts ?

apparently not? come on joe, lend a hand!

[quote="orl"]Once I read about Kurt Gödel that initially he discarded the American constituion as he found logical flaws in it which would allows dictators to take over the country.[/quote]

What is the harm in allowing dictators to take over the country?

Dictators tend to be selfish after some time bothering only about their very own interest. But of course having a smart dictator always acting in favour of the common goals would be the best form of governing but unfortunately people are not like that.

True. State is a concession to human weakness.

Could someone explaing the difference between a shortlist and a longlist?

The most obvious would be that a shortlist is short, and a longlist is long. The list part is same in both.


inequalities
calculus
calculus computations

[url=http://bildr.no/view/490095][img]http://bildr.no/thumb/490095.jpeg[/img][/url]
[url=http://bildr.no/view/490096][img]http://bildr.no/thumb/490096.jpeg[/img][/url]
[url=http://bildr.no/view/490097][img]http://bildr.no/thumb/490097.jpeg[/img][/url]

I can not understand what is Stress Testing.

thanks

[quote="Duelist"]I don't quite understand what it is either, but for someone else trying to figure it out here's the link to the whole text.

http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_HilbertandCompensatingDifficulties.pdf

Edit: I think this belongs in an algebra forum, not a calculus computations and tutorial forum.[/quote]

Yes , I am reading that book.


function
algebra
functional equation
algebra unsolved

[quote="vietnamesegauss89"]Find all the function $f: R_+^*\to R_+^*$satisfly:
                     $g(x).g(\frac{y}{g(x)})=g(x+y) \forall x,y>0$[/quote]

If there is such $a$ that $1>g(a)$ then let $y=\frac{g(a)a}{1-g(a)} \Rightarrow$
$a+y=\frac{y}{g(a)} \Rightarrow g(a)g(\frac{y}{g(a)})=g(\frac{y}{g(a)}) \Rightarrow$
as $g(x)>0 \Rightarrow g(a)=1$  that is contradiction to the assumptiom that $1>g(a)$
So there is no such a.

Thus for any x,y>0
$g(x+y)=g(...)g(x) \geq g(x)$

If there is such $a$ that $g(a)=1 \Rightarrow g(a)g(\frac{y}{g(a)})=g(a+y) \Rightarrow$
$g(y)=g(a+y)=g(a+2y)=...=g(a+ky)$ for any positive integer k.
If there is such x,y that $g(x)<g(y) \Rightarrow x<y$
But there exist such positive integer $t : \ \; y<x+t*a \Rightarrow$
$g(x)<g(y) \leq g(x+t*a)=g(x)$ - contradiction
So the funtion is constant in this case $\Rightarrow g(x) \equiv 1$

In another case we have ${g(x)>1 \mbox{ for any real } x>0} \Rightarrow$
$g(x+y)>g(x) \mbox{ for ane real } x,y>0 \Rightarrow$
$x<y \Leftrightarrow g(x)<g(y) \Rightarrow$

$y \rightarrow yg(x) \Rightarrow g(x)g(y)=g(x+yg(x))=g(y+xg(y)) \Rightarrow$
$x+yg(x)=y+xg(y) \Rightarrow g(x)=\frac{a+x(g(a)-1)}{a}$ for some real a
${\Rightarrow g(x)=c+kx \mbox{ for some real } c,k>0 \Rightarrow}$

Finally we have two solutions of this functional equation:

$g_1(x)\equiv1$
$g_2(x)=1+x$

V:)

$g_1(x)\equiv1$
$g_2(x)=1+x$

V:)[/quote]
g(x)=1+kx is solution for any k>=0.

[quote="Rust"]
g(x)=1+kx is solution for any k>=0.[/quote]

Yes, i'm sorry :blush:
(I had no time so i did such stupid mistake)

So let me finish my solution:

We'v got $g(x)=c+kx$ for some real nonnegative x,y.
$\Rightarrow (c+kx)(c+k\frac{y}{c+kx})=c+k(x+y) \Leftrightarrow c^2+ckx+ky=c+kx+ky \Leftrightarrow$
for every pair of real positive x,y.
$\Leftrightarrow c^2+ckx=c+kx \Rightarrow c=1 (\mbox{ as } c \neq 0)$
and for any $k\geq 0$ the function $g(x)=1+kx$ satisfies our condition.

So the answer is $g(x)=1+kx$ for any $k\geq 0$
V:)


function
integration
real analysis
real analysis unsolved

Determine the coninous functions $f: R \to R$ which satisfy:
  $\int_{a}^{b}f(t)dt+\int_{c}^{d}f(t)dt=\int_{a+c}^{b+d}f(t)dt$, $\forall a,b,c,d \in R$

I think that $f$ must be a constant function.
 To see this we have that, ($a=c$, $b=d=x$):

$2\int_{a}^{x}f(t)dt=\int_{2a}^{2x}f(t)dt$.

Differentiating we get that:

$2f(x)=2f(2x)$, or $f(x)=f(2x)$. Now, if $2x=y$ we get that

$f(y)=f(\frac{y}{2})$, and hence $f(y)=f(\frac{y}{2^{n}})$.

Letting $n$ converge to $\infty$ we get , since $f$ is continuous, that

$f(y)=f(0)$.


There are 4 different pairs of gloves. 4 right handed gloves are given randomly to four persons then 4 left handed gloves are given.how many ways are possible such that nobody gets the right (correct)pair of gloves.pl explain the fact .the answer is 4x 9! but how

[hide="HINT"]All possible arrangements - Arrangements in which atleast 1 gets the right pair.[/hide]

[quote="Conjurer"]Ye, thats because its wrong.Srry about that.

Use derarrangements formula i.e

Possible arrangements - Arrangements in which only 1 person gets the right pair - Arrangements in which only 2 persons gets the right pair[/quote]


analytic geometry
rotation
geometry
geometric transformation
graphing lines
slope
MATHCOUNTS

Point O, with coordinate (0,0), and point A, with coordinate (6,8), are on a Cartesian plane.  If segment OA is rotated about point O in a counterclockwise direction to from a new segment OA', where the coordinate of point A' is (-8,6), how many square units are "swept over" by this rotation?

[hide]
The first thing that jumps to my mind:
Pythagorean triples!
So I draw some right triangles, and I find that the sweep is a 90 degree sector of a circle with radius 10.  So the answer is $25\pi$[/hide]

I feel like it might be a good idea to explain why [hide]the sector is a 90 degree one[/hide]

Just use distance formula and prove its a right triangle...

I just used the fact that the slope of the first line segment was $\frac{3}{4}$, and the slope of the other was $\frac{-4}{3}$ (or something like that), so the segments were perpendicular.

It's also pretty easy to cheat using the dot product...or if the angle isn't nice you could easily own it with trig.

But with this kind of problem it's pretty easy to draw a rough sketch and guess that it's right. More than half of Mathcounts problems have things that look like right triangles and turn out to be right triangles.

I don't get what you mean by right triangles. I just used the distance formula to get the length of the line segment, then found the area of 1/4 of the circle.

Sorry if it was confusing, I just have a habit of drawing lines parallel to the axes.
One way or another, that was the way I found the length of the segment.


algorithm
geometry
geometric transformation
rotation
linear algebra
linear algebra unsolved

Hi
I work in algorithm to compute a common eigenvector of two matrices for my project to obtain a DEA ( master) and I need applications to test what I’ve done

Does any one know an application for the common eigenvector of two matrices?
Does any one know an application for the simultaneous triangularization of two matrices?
thnx

[quote="king_ae"]Does any one know an application for the common eigenvector of two matrices?[/quote]
In 3 dimensions it is the rotation axis of your transformation (for square matrices), I think that is extedable to more general conditions. (if you call that an application)


This is the thread where we shall hand out accurate news of the world cup matches.
For now I am able to do that myself, but I want some enthusiastic reporters, as I may miss some matches myself (especially those starting at 7 pm GMT). Please inform me if you are in a postion to watch all the matches.

Anyway, here are the first news items:

Match 1:

GERMANY - $4$
Philipp LAHM 6'
Miroslav KLOSE 17' 61'
Torsten FRINGS 87'

COSTA RICA - $2$
Paolo WANCHOPE 12' 73'

Bookings -
Danny FONSECA (Yellow) 30' [COSTA RICA]

Match 2:

POLAND - $0$

ECUADOR - $2$
Carlos TENORIO 24'
Agustin DELGADO 80'

Bookings -
Ivan HURTADO (Yellow) 31' [ECUADOR]
Ebi SMOLAREK (Yellow) 37' [POLAND]
Edison MENDEZ (Yellow) 70' [ECUADOR]

These are from my personal archive. But I can't possibly watch [i]every[/i] match. So I need your help.

And by the way, we shall also have some game analyses here if others are interested. And I hope others [i]are[/i] interested.

Well, I'd like to post the results of Group C

And here we go:

Argentina 2 C. Ivoire 1

Goals: Crespo (Arg)
Saviola (Arg)
Drogba (Civ)

What did you think of Argentina vs. Ivory Coast game?
I was sort of hoping Ivory Coast would win....they had some really good chances but they were always wide.
I loved that goal made by Saviola, how he owned the defense by not being offsides.

Right now, Netherlands and Serbia and Montenegro are playing against eachother.

Lets see who wins.

Netherlands has, and right now the Mexicans are getting ready to take on Iran (in fact it is starting now ... in fact it [i]has[/i] started!) I'll get out the official stats for matches 3,4,5 tomorrow.

http://sportsillustrated.cnn.com/soccer/worldcup/2006/matchcast/2006/06/11/2006061110186/live/index.html
The site automatically updates itself for play by play action  :D
Iran seems to have the upper hand right now

Yay, Netherlands beat Serbia and Montenegro.  1-0.  Breakaway goal scored by Robben in the 18th minute.

Right now after 22 minutes, the game between Iran and Mexico is tied 0-0.

Mexico 1- 0 Iran  :P
Omar Bravo scored 37 min into the game

Mexico won 3-1.
Nice job Mexico, I am impressed. :D
Portugal wins 1-0 against Angola
Nice try Ronaldo, but , but you couldn't pull off that goal.  ;)

Match 3:

ENGLAND - $1$
Carlos GAMARRA 4' (Own Goal)

PARAGUAY - $0$

Bookings -
Steven GERRARD (Yellow) 18' [ENGLAND]
Nelson VALDEZ (Yellow) 21' [PARAGUAY]
Peter CROUCH (Yellow) 63' [ENGLAND]

Match 4:

TRINIDAD AND TOBAGO - $0$

SWEDEN - $0$

Bookings -
Avery JOHN (Yellow) 15' [TRINIDAD AND TOBAGO]
Avery JOHN (Yellow) 46' [TRINIDAD AND TOBAGO]
Avery JOHN (Red) 46' [TRINIDAD AND TOBAGO]
Henrik LARSON (Yellow) 90' [SWEDEN]

I'll give the rest later. For now, must rush to watch Australia vs Japan.

GOOOOOOOOOOOOOOOAL!

Japan scores! one nil to japan after 21 minutes!

Wow, who would've thought that Australia would win the match against Japan.

Australia wins 3-1 over Japan.

Australia
Bookings
(Yellow)Grella 33 min
(Yellow)Moore 57 min
(Yellow)Cahill 68 min
(Yellow)Aloisi 77 min
Goals
Cahill 83 min
Cahill 88 min (Aloisi)
Aloisi 91 min (Aloisi)

Japan
Bookings
(Yellow)Miyamoto 30 min
(Yellow)Takahara 39 min
(Yellow)Nakamura 67 min
Goals
Nakamura 25 min

Lets see what happens between US and Czech Republic!

[quote="ragnarok23"]Bleh....3-0 US lost....horrible game for them, but a very nice one for Czech Republic, they dominated everything.[/quote]Man, i wanted the US to win... :(

Brazil is out. Thay played horrible, and were eliminated

Almost everyone thought that Brazil and/or England would make it to the semi-finals.

But look what happened :roll:

[quote="nutz_for2.718281828"][quote="nutz_for2.718281828"]I found some of the best talents in the world cup. Here is the link. Arrow to the "UNbelievable skills" movie and you'll see some awesome stuff!
[url=http://www.google.com/ig]http://www.google.com/ig[/url]

Really! try it! It's fun! One guy kicked a curveball![/quote]

Won't ANYBODY watch it???!!![/quote]
It leads to my google homepage, check your link.
And most people here have probably seen it.

italy 2   Germany 0

i really enjoyed...

Germany should eliminated sooner than this time.

At half time, the score is 1-1. After Zidane converted a controversial penalty, Materazzi headed in the equalizer from Camoranesi's brilliant corner.

Zidane's PK is the skill of oxidezed silver.

Now the second half is about to begin. I'm exciting, but I'm sleepy. :huh:

Goes into extra time.
I am excited as well, and sleepy as well.

[quote="mathcountsfreak"]I don't understand why Zidane headbutted that guy. It was his last game, why go out on a red card????[/quote]

He won't have to miss additional games :D

Yes, bet he let the hole world a bad image

the headbutt was shocking. But zizzou was frustrated after Buffon's brilliant save. That is why he is not Maradona. That is why Maradona is one and only one in footballing history.

Italy deserved a win though.

[quote="Ashegh"]by the way zidan has been the best player of the world cup2006.[/quote]

Maybe 2002 but I don't think so for 2006. He did do well though.


Does there exist two infinite sets $A,B$ such that every number can be written uniquely as a sum of an element of $A$ and an element of $B$?

Do they have to be disjoint?
I presume that they must only contain non-negative integers? May they contain $0$?

Pierre.

$A$ and $B$ are not necessarily disjoint, and "natural number" := "non-negative integer"

Ok, so the answer is no.
I'm a bit in a hurry so I will post a solution later.

Pierre.

:maybe:  And what if you define  $A=\{n | n=\sum_{i \in I}2^{i},\;\; I\subset \{m\in \mathbb{N}\;|m\equiv0\; (mod\; 2)\}\}$;  $B=\{n | n=\sum_{i \in I}2^{i},\;\; I\subset \{m\in \mathbb{N}\;|m\equiv1 \;(mod \;2)\}\}$ ?

:blush:  Hmmm...wrong computations... I have to sleep a bit...

Pierre.


probability
combinatorics proposed
combinatorics

let Sn be a sequence  with n objects
where every object is a natural number <=n

what is the probability  that Sn is monotonically non-decreasing?

Total cases: $n^{n}$ cases
Monotonically non-decreasing cases:
Let $S_{n}=(a_{1},a_{2},....,a_{n})$, and $b_{1}=a_{1}-1, b_{2}=a_{2}-a_{1}, ....., b_{n}=a_{n}-a_{n-1}, b_{n+1}=n-a_{n}$.
Then, when $S_{n}$ monotonically non-decreases,
$b_{1},b_{2},...,b_{n+1}\ge0$.
And, $b_{1}+b_{2}+....+b_{n+1}=n-1$, so the case of this condition is
$\binom{n+n-1}{n}$.
So, the probability is $\frac{\binom{2n-1}{n}}{n^{n}}$.


[size=59]Dei, what is this? You should do your own homework[/size]

 :D

Strange, I doubt if i am right...But i think i am pretty much sure there's a cyclic ether

Ene reaction is also a possible pathway.

[quote="hell_ever"][size=59]Dei, what is this? You should do your own homework[/size]
[/quote]
 :lol:

most of these problems are from a worksheet VARUNARASU gave me....it has close to 100 probs....but no answers  :(

Hence wherever i am not too sure(or dont know)abt the products i post them here  :P

can you give/state a rough mechanism for your product?  :(

Step 1: $ \ce{H2C\equal{}O} \plus{} \ce{H\plus{}} \rightleftharpoons \ce{H2C\equal{}OH\plus{}}$

Step 2: now carbon 2 of the alkene forms a bond with the carbon of protonated formaldehyde, forming a secondary carbocation.

Step 3: hydrogen migration to yield a tertiary carbocation.

Step 4: the oxygen used one of its lone pairs to make a bond with the positive carbon, yielding an oxonium ion.

Step 5: an acid-base equilibrium with the solvent removes the hydrogen from oxygen.

In the link given [url=http://en.wikipedia.org/wiki/Ene_reaction]here[/url], under the section Variations, specifically, carbonyl ene reactions, what does the wavy bond in the product side indicate?

[quote="hell_ever"]In the link given [url=http://en.wikipedia.org/wiki/Ene_reaction]here[/url], under the section Variations, specifically, carbonyl ene reactions, [b]what does the wavy bond in the product side indicate[/b]?[/quote]

i think it means indefinite stereochemistry......  :maybe:

Yes, the wavy bond means that the stereochemistry at that stereocenter is not known, or it doesn't matter.


inequalities
trigonometry
inequalities unsolved

Let $ ABC$ is a triangle. Prove that:

 $ \left ( 1 + \cos^2{\frac {A}{2}} \right) \left ( 1 + \cos^2{\frac {B}{2}} \right) \left ( 1 + \cos^2{\frac {C}{2}} \right) < \left ( 1 + \frac {\sqrt {3}}{4} \right)^{3\sqrt {3}}$

To prove this inequality we use the following:

1.Bernoulli's Inequality: $ (1+x)^p$ < 1+px , when 0 <p <1

2.In any triangle   ABC ,       $ \sin^2\frac{A}{2}$ + $ \sin^2\frac{B}{2}$ +$ \sin^2\frac{C}{2}$ =1 -   2$ \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ , and

3.    $ \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ $ \leq$ $ \frac{1}{8}$ .

Now $ (1+\cos^2\frac{A}{2})^{\frac{1}{\sqrt3}}$ $ <$  $ 1+\frac{1}{\sqrt3}\cos^2\frac{A}{2}$ by .....(1)

Therefore $ \prod(1+\cos^2\frac{A}{2})^{\frac{1}{\sqrt3}}$     $ <$      $ \prod(1+\frac{1}{\sqrt3}\cos^2\frac{A}{2})$

<$ (\frac{3+\frac{1}{\sqrt3}\sum\cos^2\frac{A}{2}}{3})^3$ by AM-GM inequality.

=$ (\frac{3 +\frac{1}{\sqrt3}(3-1 +2\prod\sin\frac{A}{2}}{3}))^3$ using ............(2)

$ \leq$  $ (\frac{3+\frac{1}{\sqrt3}(2+\frac{1}{4})}{3})^3$ using .............(3)

=$ (1+\frac{\sqrt{3}}{4})^3$ and therefore we have

$ \prod(1+\cos^2\frac{A}{2})^{\frac{1}{\sqrt3}}$ <$ (1+\frac{\sqrt3}{4})^3$ which becomes the inequality to be proved after raising both sides to an exponent of $ \sqrt3$.


function
calculus
derivative
trigonometry
real analysis
real analysis solved

Hello,

Any example of a non-primitivable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f$ satisfies the intermediate value property on any interval ? It may be trivial, I haven't thought about it yet.

Try searching the forum for an example of non-measurable function which attains all real values on every interval (or even better, try constructing this yourself; that's always more rewarding than reading someone else's proof :)). That should do it, since a derivative must be measurable.

how about $\begin{cases}\sin\frac{1}{x}, &x\neq 0\\ 1/2, &x=0\end{cases}$.

[quote="cimabue"][quote="grobber"] That should do it, since a derivative must be measurable.[/quote]

are you sure?[/quote]

The dreivative $f'$ of $f$ is the pointwise limit of the sequence of continuous and thus measurable functions $f_n(x)=\frac 1n\left(f\left(x+\frac 1n\right)-f(x)\right)$. Now let's prove that if $(f_n)_n,\ f_n: \mathbb R\to\mathbb R$ is a sequence of measurable functions converging pointwisely to a function $u: \mathbb R\to\mathbb R$ ($f'$ in our case, but we're not concerned with this particular setting anymore), then $u$ is measurable:

Let $D\subset\mathbb R$ be an arbitrary open subset of the real line. We have $u^{-1}(D)=\bigcup_{m\ge 1}\bigcap_{n\ge m}f_n^{-1}(D)$, and since each $f_n^{-1}(D)$ is measurable, we find $u^{-1}(D)$ to be measurable. Since $D$ was arbitrary, we have the desired result: $u$ is measurable.

So yeah, I'm sure :).

P.S.

You can edit your posts, you know :).

really?
this is silly: you are right of course, i just asked a question :)

but i still think my example works unless i have the wrong definition of primitivable.

Thank you for your examples. cimabue, your example works fine and is even measurable.


I was wondering, who actually writes the IMO tests? :P  :D

There's a whole thread for that: http://www.mathlinks.ro/Forum/viewtopic.php?p=13981#p13981
Also all the IMO problems are here: http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16

As far as I know, every team (the leaders) gets to propose some X problems, making up the longlist. Then the IMO people make a shortlist from that and then make the IMO.

[quote="Valentin Vornicu"]It's not the leaders proposing the problems. They are rather the currier.[/quote]

In the thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]Introduction to the IMO ShortList/LongList project[/url] you can find a bit more background and history of the IMO proposals.


integration
trigonometry
LaTeX
calculus
calculus computations

Compute $ \int \frac{\sin x \cos x}{\sin^4x \plus{} \cos^4x}\,dx$.

[hide="Alternatively !"]$ \left[\frac 12\cdot\arctan\left(\tan^2x\right)\right]' \equal{}$ $ \frac 12\cdot\frac {2\tan x\left(1 \plus{} \tan^2x\right)}{1 \plus{} \tan ^4x} \equal{} \frac {\sin x\cos x}{\sin^4x \plus{} \cos^4x}$ $ \implies$

$ \boxed {\int\frac {\sin x\cos x}{\sin^4x \plus{} \cos^4x}\ \mathrm {dx} \equal{} \frac 12\cot\left(\tan^2x\right) \plus{} \mathcal C}\ .$[/hide]

Very nice, but from where did come your idea of 1/2 [cot(tan^2x)]' ?

[quote="Carcul"]Very nice, but from where did come your idea of $ \left[\frac 12\cdot\arctan\left(\tan^2x\right)\right]'$ ?[/quote]
The answer is at the first your reply (the your hidden proof - "Alternatively") from http://www.mathlinks.ro/viewtopic.php?t=173983 !

On a second thought, $ \frac {1}{2}\left[\cot(\tan^2x)\right]' \equal{} \minus{} \frac {\tan x \sec^2x}{\sin^2(\tan^2x)} \neq \frac {\sin x \cos x}{\sin^4x \plus{} \cos^4x}$. What do you have to say? Or what am I missing?

[color=darkblue]All these are due to LaTeX and to the English language !
Sorry, no [b]COT[/b], but [b]ARCTAN[/b] ! Here is the "indirect" proof :[/color]

$ \int\frac {\sin x\cos x}{\sin^4x \plus{} \cos^4x}\ \mathrm {dx} \equal{}$ $ \int\frac {\sin x\cos x\left(\sin^2x \plus{} \cos^2x\right)}{\sin^4x \plus{} \cos^4x}\ \mathrm {dx} \equal{}$

$ \int\frac {\tan x\left(1 \plus{} \tan ^2x\right)}{1 \plus{} \tan^4x}\ \mathrm {dx} \equal{}$ $ \int\frac {\tan x\left(\tan x\right)'}{1 \plus{} \tan^4x}\ \mathrm {dx} \equal{}$ $ \frac 12\cdot\int\frac {\left(\tan^2x\right)'}{1 \plus{} \left(\tan^2x\right)^2}\ \mathrm {dx} \implies$

$ \boxed {\int\frac {\sin x\cos x}{\sin^4x \plus{} \cos^4x}\ \mathrm {dx} \equal{} \frac 12\cdot\arctan\left(\tan^2x\right) \plus{} \mathcal C}\ .$

Ok then, good job. Here's my answer:

$ \sin^4x \plus{} \cos^4x \equal{} \sin^4x \plus{} 2 \sin^2x \cos^2x \plus{} \cos^4x \minus{} 2 \sin^2x \cos^2x \equal{}$
$ \equal{} (\sin^2x \plus{} \cos^2x)^2 \minus{} 2 \sin^2x \cos^2x \equal{} 1 \minus{} 2 \left(\frac{\sin 2x}{2}\right)^2 \equal{}$
$ \equal{} 1 \minus{} \frac{1}{2} \sin^2(2x) \equal{} 1 \minus{} \frac{1}{2}[1 \minus{} \cos^2(2x)] \equal{} \frac{1}{2} [1 \plus{} \cos^2(2x)]$.

So: $ P \equal{} 2 \int \frac{\sin x \cos x}{1 \plus{} \cos^2(2x)}\,dx$. But [cos(2x)]' = -2sin(2x) = -4sin(x)cos(x), and so

$ P \equal{} \minus{} \frac{1}{2} \int \frac{\minus{}4 \sin x \cos x}{1 \plus{} \cos^2(2x)}\,dx \equal{} \minus{} \frac{1}{2} \arctan[\cos(2x)] \plus{} C$.

[quote="Carcul"]$ P \equal{} \minus{} \frac {1}{2} \arctan[\cos(2x)] \plus{} C$.[/quote]

[b]Remark.[/b] Exists $ k\in\mathcal R$ so that for any $ x\not\in \left(\frac {\pi}{2}\plus{}\pi Z\right)$ , $ \arctan(\cos 2x) \plus{} \arctan\left(\tan^2x\right) \equal{} \frac {\pi}{4}\ .$


Solve this ecuation:
$ 4^x\plus{}2\equal{}3 \cdot 2^{x^2}$

Set $ f(x)\equal{}3.2^{x^2}\minus{}4^x\minus{}2$
We have $ f'(x)\equal{}2ln2(3x2^{x^2}\minus{}2^{2x})$
$ f''(x)\equal{}2ln2(3.2^{x^2}\plus{}6ln2.x^2.2^{x^2}\minus{}2^{2x}.2.ln2)$
We have $ 3.2^{x^2}>2^{2x}.2.ln2$ so $ f''(x).0$ so $ f(x)$ has max $ 2$ root
Note that $ 0$ and $ 1$ is root of this equation
I hope it true

[quote="Differ"]Does x have to be an integer?[/quote]
$ x$ can be real number.


Dear friends, I look for a simple way to prove that

$ \Delta \equal{}(b\plus{}c)^2 a^3 \minus{}(b\minus{}c)^2(b\plus{}c)a^2 \minus{}(b \minus{} c)^2 (b^2 \plus{} 4 b c \plus{} c^2) a \plus{} (b \minus{} c)^2 (b \plus{} c) (b^2 \plus{} c^2).$

is positive when $ a, b, c$ are the sides of a triangle.


function
induction
modular arithmetic
algebra proposed
algebra
binary representation

Consider the function $ f: \mathbb{N}_0\to\mathbb{N}_0$, where $ \mathbb{N}_0$ is the set of all non-negative
integers, defined by the following conditions :

    $ (i)$ $ f(0) \equal{} 0$; $ (ii)$ $ f(2n) \equal{} 2f(n)$ and $ (iii)$ $ f(2n \plus{} 1) \equal{} n \plus{} 2f(n)$ for all $ n\geq 0$.

$ (a)$ Determine the three sets $ L \equal{} \{ n | f(n) < f(n \plus{} 1) \}$, $ E \equal{} \{n | f(n) \equal{} f(n \plus{} 1) \}$, and $ G \equal{} \{n | f(n) > f(n \plus{} 1) \}$.
$ (b)$ For each $ k \geq 0$, find a formula for $ a_k \equal{} \max\{f(n) : 0 \leq n \leq 2^k\}$ in terms of $ k$.

[quote="BaBaK Ghalebi"][hide="(a)"]
note that for $ n > 1$ we have:

$ f(2n \plus{} 1) \equal{} 2f(n) \plus{} n > 2f(n) > f(2n)$

and for $ n \equal{} 0$ we have $ f(0) \equal{} f(1) \equal{} 0$

hence:

$ L \equal{} \{2n\mid n\in\mathbb{N}\}$

$ E \equal{} \{0\}$

$ G \equal{} \{2n \plus{} 1\mid n\in\mathbb{N}\}$[/hide][/quote]

You're wrong  :maybe:

[quote=mathpk]
You're wrong  :maybe:[/quote]

I think you're also wrong.
for example $f (1 ) = f ( 2 ) = 0$

Very nice APMO problem from my birth year :D

to me, it felt like the main idea here is about understanding the structure of $f$ and using a better interpretation of it (binary appending) to prove the results.

although the solution is quite long, i felt like a lot of the steps are quite motivated and it was relatively painless to break down the problem into small claims

Also, no posts for 3 years?

The way to think of the function $f$ is, think of the input in binary. The value of $0$ and $1$ are both $0$. When a new digit is appended, if the appended bit is $0$, the value doubles, and if it is $1$, the value doubles and then is increased by the original input.\\

We actually solve part (b) first, given this clearly the optimal strategy to maximize the value given a fixed number of bits is to append $1$ at every step, meaning that $f(2^k-1)$ is maximal. The answer for $k=0$ and $k=1$ is clearly $0$, focus on $k\geq 2$ from now on. We claim that $$f(2^k-1)=(k-2)\cdot 2^{k-1}+1$$ for $k\geq 2$ This is true for $k=2$, and we will use induction. We have $$f(2^{k+1}-1)=2f(2^k-1)+2^k-1=(k-1)\cdot 2^k+1,$$ completing the proof. Thus, the answer is $$(k-2)\cdot 2^{k-1}+1.$$

Now for part (a). It stays the same from 0 to 1. We claim that beyond this,

-Going from $1\pmod{4}$ to $2\pmod{4}$ does not change the value,\\
-going from $2\pmod{4}$ to $3\pmod{4}$ increases the value,\\
-going from $3\pmod{4}$ to $0\pmod{4}$ decreases the value,\\
-going from $0\pmod{4}$ to $1\pmod{4}$ increases the value.\\

The second and fourth claims are clear since we are appending a $1$ instead of a $0$ at the end with everything else equal.\\

To prove the first claim, we show that appending $01$ and appending $10$ to something has the same effect. Suppose we start with $f(n)=v$. If we append $0$ then $1$, then we have $f(2n)=2v$ and $f(4n+1)=4v+2n$. If we append $1$ then $0$, then we have $f(2n+1)=2v+n$ and $f(4n+2)=4v+2n$, as desired.

Finally, we will show the third claim where going from 3 mod 4 to 0 mod 4 decreases the value. Suppose that the 3 mod 4 number in binary ends with $01\dots 1$ for some number (at least two) of 1s, and call this the "ending". When we go to $n+1$, it rolls over to $10\dots 0$. Note that this ending of $n$ starts with $01$, and the ending of $n+1$ starts with $10$, everything prior to the ending is equal for both $n$ and $n+1$. We have shown that the $01$ and $10$ has the same effect. However, $n$ will continue appending 1s, while $n+1$ can only append zeroes, hence shown as 3 mod 4 means there is at least one more $1$.

Thus, the answer is:

$L$ consists of $n\equiv 2\pmod{4}$ as well as $n\equiv 0\pmod{4},n\neq 0$.\\
$E$ consists of $n=0$ and $n\equiv 1\pmod{4}$.\\
$G$ consists of $n\equiv 3\pmod{4}$.


[color=blue][i] Μια άσκηση στα πλαίσια του Αρχιμήδη:[/i][/color]

 Σε ένα τρίγωνο $ ABC$ είναι $ \angle A \equal{} 60 ^o$. Αν $ O,H$ είναι το περίκεντρο και το ορθόκεντρο και η ευθεία $ OH$ τέμνει τις πλευρές  $ AB,AC$ στα σημεία $ K,L$ , να αποδείξετε ότι  $ HK \equal{} OL$

  [b] ΣΧΟΛΙΟ[/b]

[color=blue][i]Κατά λάθος την έβαλα εδώ , ενώ την είχα για τους ''μικρούς''.[/i][/color]

Ωραία άσκηση κ. Στεργίου, αλλά κάτι μου θυμίζει :huh:

Τέλος πάντων να η λύση μου:

[hide]Κατ' αρχάς $ \angle{AHK} \equal{} AOL \equal{} 90 \minus{} B$

Τώρα έχουμε, an $ M$ to meso tis $ BC$

$ AH \equal{} 2\cdot{OM} \equal{} 2\cdot{OB}\cdot{\cos{BOM}} \equal{} 2R\cdot{\cos{60}} \equal{} R \equal{} AO$

Απο αυτό το σημείο είναι τετριμμένο να αποδείξουμε ότι τα τρίγωνα $ AHK \equal{} AOL\iff HK \equal{} OL$

Η άσκηση αυτή έχει πολλά επιπλέον ενδιαφέροντα πορίσματα όπως π.χ. αν N το κέντρο του κύκλου Εuler $ AN\perp{OH}$[/hide]


calculus
integration
calculus computations

\documentclass[11pt]{article}
\begin{document}
Evaluate the sum $\displaystyle\sum_{i=3}^n i^6$ * $ i (i-2) $
\end{document}


let $ f : S^{1}\to S^{1}$ by $ f(z)\equal{}z^{2}$, let $ X\equal{}(S^{1}\times [0,1])/(z,1)\thicksim (f(z),0)$. Give an explicit  CW decomposition, with attaching maps, for $ X$, and use it to compute $ \pi_{1}(X)$ and $ H_{1}(X)$.

Any idea is highly appreciated. Thanks

Well, my guess is that $ \pi_1(X)\equal{}\langle a, b | ab\equal{}b^2a\rangle$ just by rewriting the space as a square with sides identified (note the space is just a torus, by making a cylinder and attaching the two ends of the cylinder by doubling). Then $ H_1(X)$ is just the abelianization of that group.


linear algebra
matrix
linear algebra unsolved

we have a $n*n$ matrix (it's elements are $0,1$ )that is symmetric due the main diogonal and  it's main diogonal is fulled completely with $1$.
is it true  that it's determinant is $1$?.(note that calculations are in mod $2$)
(note that no two rows are equal.)

is there a solution using cute linear algebra?

It's difficult to help to solve a problem whose statement is false: if you have a $2$ by $2$ matrix consisting of all ones, for example, then the determinant is $0$, not $1$.


geometry
circumcircle
radical axis
power of a point
perpendicular bisector
angle bisector
geometry proposed

S is a incircle of  the triangle ABC.S touch side BC at M.AM cuts S at M and K.N is a midpoint of MK.Prove that /_BNM=/_CNM.

If $I$ is center of circle $(S)$ and $J,L$ are the touch points of $(S)$ with $AB,AC$.The line $IN$ is orthogonal to $AM(KM)$ and cut $BC$ at $P$ then $P$ is the radical center of $(S)$, circumcircle $(S_1)$ of $INM$ and circumcircle $(S_2)$ of pentagon $AJNIL$ due to the line $IN$ is radical axis of circumcircle $(S_1)$ and $(S_2)$ and also $(S_1)$ and $(S)$ are tangent at $M$, now the line $JL$ (radical axis of $(S_1)$ and $(S_2)$)passes trough $P$.Further $P,M$ are harmonic conjugates of $B,C$, i.e. $\angle BJM=\angle MJC$.

sorry, the last part:
i.e. $\angle BNM= \angle MNC$

I have another solution. First I change some notations:

[color=blue][b]Problem.[/b] Let S be the incircle of a triangle ABC, and let X be the point where this incircle S touches the side BC. The line AX intersects the circle S at a point K (different from X). Let N be the midpoint of the segment XK. Prove that < BNX = < CNX.[/color]

[i]Solution.[/i] Let I be the center of the circle S. It is well-known that the perpendicular bisector of a chord of a circle passes through the center of the circle. Thus, the perpendicular bisector of the chord XK of the circle S passes through the center I of this circle S. Hence, as the midpoint of the chord XK is the point N, we have $NI\perp XK$.

We will use directed segments on the lines BC, CA, AB; hereby, these lines should be directed in such a way that the segments BC, CA, AB are positive (and thus, the segments CB, AC, BA are negative). Let the incircle S of triangle ABC touch the sides CA and AB at the points Y and Z. Let the line YZ meet the line BC at a point R. Then, by the Menelaos theorem, applied to the triangle ABC and the three collinear points R, Y, Z on its sidelines BC, CA, AB, we have

$\frac{BR}{RC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$.

But since the two tangents from a point to a circle are equal in length, we have YA = AZ, ZB = BX and CY = XC, so that this becomes

$\frac{BR}{RC}\cdot\frac{XC}{AZ}\cdot\frac{AZ}{BX}=-1$.

In other words, $\frac{BR}{RC}\cdot\frac{XC}{BX}=-1$, so that $\frac{BR}{RC}=-\frac{BX}{XC}$. Thus, the points R and X divide the segment BC harmonically.

The points Y and Z are the points where the two tangents from the point A to the circle S touch S. Hence, the line YZ is the polar of the point A with respect to the circle S. The line BC is the tangent to the circle S at the point X; thus, the line BC is the polar of the point X with respect to the circle S. Since the lines YZ and BC are the polars of the points A and X with respect to the circle S, their point of intersection R is the pole of the line AX with respect to the circle S. Thus, as I is the center of this circle S, we have $IR\perp AX$. In other words, $IR\perp XK$. But on the other hand, we know that $NI\perp XK$. Thus, the points I, R and N lie on one line perpendicular to the line XK. Hence, the two lines NR and NX are perpendicular to each other. On the other hand, these two lines divide the angle BNC harmonically (since the points R and X divide the segment BC harmonically). But if two perpendicular lines divide an angle harmonically, they must be the two angle bisectors of this angle (the internal and the external angle bisector). Hence, the lines NR and NX are the two angle bisectors of the angle BNC. In particular, the line NX must be the internal angle bisector of the angle BNC, since the point X lies inside the segment BC. Hence, < BNX = < CNX, and the problem is solved.

  Darij

Let $B' ,C'$ be the tangency point of $S$ on $CA, AB$. Then $MB'KC'$ is a harmonic quadrilateral. Hence the intersection $U$ of $B'C'$ and $BC$, which is harmonic to $M$ with respect to $BC$, lies on the tangent at $K$ of $S$. Hence the midpoint $N$ of $MK$ lies on $OU$, or $\measuredangle{UNM} = 90^0$. From the property of Apollonius, we obtain that $NM$ is the bisector of $\measuredangle{BNC}$.


LaTeX
trigonometry
geometry
angle bisector
geometry unsolved

Let $ ABC$ be an isosceles triangle with $ AB \equal{} BC \equal{} a$ and $ \angle ABC \equal{} 140^o$.Let $ AN$ and $ AH$ respectively be the angle-bisector and the altitude from $ A$.Prove that $ 2BH.CN \equal{} a^2$

[quote="friendlist"] [color=darkred]Let $ ABC$ be an isosceles triangle with $ AB \equal{} BC \equal{} a$ and $ \angle ABC \equal{} 140^{\circ}$.Let $ AN$ and $ AH$

respectively be the angle-bisector and the altitude from $ A$.Prove that $ 2\cdot BH\cdot CN \equal{} a^2$ .[/color] [/quote]

[b]Proof I ( entertainment in LaTeX ! ).[/b]

$ \boxed {\ \begin{array}{c} \\
\ 2\cdot BH\cdot CN \equal{} a^2 \\
 \\
2\cdot\frac {BH}{BA} \equal{} 1 \plus{} \frac {NB}{NC} \\
 \\
2\cdot \sin 50 \equal{} 1 \plus{} \frac {AB}{AC} \\
 \\
2\cdot\sin 50 \equal{} 1 \plus{} \frac {\sin 20}{\sin 40} \\
 \\
2\cdot\sin 50 \equal{} 1 \plus{} \frac {1}{2\cdot\cos 20} \\
 \\
4\cdot\cos 40\cdot\cos 20 \equal{} 2\cdot\cos 20 \plus{} 1 \\
 \\
2\cdot (\cos 60 \plus{} \cos 20) \equal{} 2\cdot\cos 20 \plus{} 1 \\
 \\
2\cdot\cos 60 \equal{} 1 \\
 \\
2\cdot \frac 12 \equal{} 1 \\
 \\
1 \equal{} 1 \\
 \\
\mathrm {Q.E.D.} \\
\ \end{array}\ }$

[quote="gauravpatil"]extend $ BH$ to $ K$ such that  $ BH \equal{} HK$
as $ BH \equal{} HK$ we get $ AB \equal{} AK$
now observe that the angle $ BAN \equal{}$ angle $ NKA$
this implies by tangent secant rule $ BA$ is tangent to circle $ NAK$ and thus
power of $ B \equal{} BN \cdot BK \equal{} 2 BN \cdot BH \equal{} AB^{2} \equal{} a^{2}$[/quote]

Haha, you're fall or there are some mistakes! Please check your solution!


MATHCOUNTS trigonometry AMC AIME ARML USA(J)MO USAMO

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107528&start=43

since PenguinIntegral suggested it

A few statements:

1) MathCounts is a computation contest.

2) MathCounts sucks the souls out of 8th graders who have better  things to do.

When I was in 8th grade, I never understood what Arnav meant.  But now I really do.  MathCounts forces middle school students to practice middle school math.  In middle school, I wanted to learn trig, but it wasn't necessary for MathCounts, so I didn't do it until the summer after 8th grade.

MathCounts is a good foundation, perhaps the problem is that we shouldn't be practicing for [b]it[/b].  Perhaps more, diverse, middle school contests would solve the problem, and encourage students to practice for different types of problems.

Maybe they should step up this AMC 8 thing.

Most students preparing for MathCounts have not fully mastered the material.  For the top 20 or students (random estimate) it probably is a computation competition.

However, I don't see why a computation competition is that bad...particularly in today's world where many students rely on calculators to do everything.  Furthermore, it's helpful to have a solid grounding and proficiency in the basics before moving on to higher stuff.

Also remember that MathCounts is not the only competition that one can participate in in middle school.

Hi - I'm the coach of a bright student who during practice tests on MC scored 144 consistently but only scored a 137 on actual, thereby NOT making States -- but then she took the AMC 10 -- not 8 (though she did that too) but on AMC 10, she got over 136 so how do you reconcile that??? To me, it makes no sense when AMC's seem SO much more difficult -- it seems the harder the problems (short of calc and trig), the better this person does.  Can anyone out there relate .. or explain this to me?

@life1:

Mathcounts questions are open response.  AMC's are multiple choice.  Bright students will recover from many more errors with the multiple choice format.  If the AMC's were open response, scores would be significantly lower overall than they are now.

However, I'm not convinced that the AMC 10 questions are more difficult than the National MathCounts questions.  If anything, I'd say the opposite was true.  Top students in large states can typically answer just half of the questions correctly at National MathCounts.  When those same students take the AMC 10, they can generally get over 80% correct.

I'd say the State-level Mathcounts questions are closer to the level of the AMC 10.

Remember also that AMC problems allow twice as much time as MC sprint problems- that makes a big difference, and it's hard to compare directly. A comparison of MC target problems to AMC problems would be fairer.

[quote="life1"]Hi - I'm the coach of a bright student who during practice tests on MC scored 144 consistently but only scored a 137 on actual, thereby NOT making States [/quote]

If this is a score from this year's MC, it should not be here (although it does not look like a MC score at all...)

[quote="JSteinhardt"]Well I mean, MathCounts can be pretty useful to people who aren't practicing for it, because then it's actually a challenge. So if no one practiced for MathCounts, then everything would be fine and dandy. I think the main thing is to get the top students who really have better things to do, to do those better things.

This isn't really a problem unique to MathCounts. Yes, it is sort of pointless for people who could probably already make AIME to be practicing MathCounts problems. But it is also sort of pointless to practice AMC-12 problems / ARML problems if you can already qualify for the USAMO. If it sort of pointless (usually) to practice (most) AIME problems if you have already qualified for MOP. It is sort of pointless to practice USAMO #1/#4 if you have already qualified for Black MOP. etc. etc.

I think if anything, we should take this "problem" as indicative of a larger syndrome, and not fall into the "<insert contest that is too easy for <insert person here> here> trap".

@life1: This isn't that abnormal. First of all, your student scored lower on the AMC than MathCounts (well okay, by 1 point, but still, they were scoring 144's on the practices, as you said), they only scored higher relatively. Perhaps your student is finding MathCounts to be too easy for her, and is focusing more energy on the AMC? Also, as mentioned before, computational accuracy is much more important on MathCounts than on the AMC. On the AMC, problem solving ability is much more important. When your student takes the AIME, problem solving ability/preparation will be even more important. So it just may be that your student is good at problem solving but not at making calculations.[/quote]

bah. people who have mastered AIME and similar contests KNOW they can move on to the USAMO

at mathcounts level, there is simply not an emphasis on moving on

it's all just about well hey why don't you go do mathcounts

and I'm like no dude wtf

Seriously. How do you get 137? That's not even a possible score!

A lot of middle school students could be doing more productive things than Mathcounts, but do only Mathcounts. A lot of middle school students could be doing more productive things than Mathcounts, and do. Tens of thousands more middle school students could [b]not[/b] be doing more productive things than Mathcounts.

A lot of high school students could be doing more productive things than AMC's, but do only AMC's. A lot of high school students could be doing more productive things than AMC's, and do. Tens of thousands more high school students could [b]not[/b] be doing more productive things than AMC's.

Aargh! Ban the AMC's!!! Everybody must take the AIME!

Point made.

Mathcounts isn't a waste of time; it's just something everyone tries do do good at and spends a lot of time on, but really depends somewhat on external factors, i.e. if you got enough sleep the day before, or if you ate a good breakfast. AMC and AIME depend more on if you know the material.

At least, that's my opinion, according to what I've been through.

Dude it is pretty fun to bash Mathcounts@@ but actually it was helpful to get me excited about math... like the first time I made states I was like WOOHOO I AM PRETTY EXCITED and then next year I made nationals and I was like WOOHOO I AM STILL PRETTY EXCITED and I definitely learned a lot at least in that first year (like in 6th grade... I didn't really know anything? so it was actually vaguely tricky figuring out those mathcounts problems)

then I practiced a lot for nationals (remember I was EXCITED) and usually got 40+/46 except for some reason I screwed up on test day? so then I was no longer excited

but then luckily I started doing AIME the next year so really the trick is to get out quickly... so yeah as long as you are at the national level it is probably a good idea to escape mathcounts asap...

also they pay for your expenses for mathcounts... so that is pretty fun... although not as fun as mop... but still I read Count Down by Steve Olson during my last year of mathcounts nationals

(dude I think I actually stayed up past midnight reading that, and it was the night before the competition... that was a pretty good idea...)

and then I was EXCITED again which was pretty awesome... see it's good to be EXCITED and now this year I am pretty EXCITED again dude it is so fun to be EXCITED

Last year, I didn't know the significance of AMC, AIME, USAMO, HMMT, or any other of these acronyms.
The only competition I knew of was mathcounts.
After I made nationals last year, my coach told me about AMC/AIME.
A couple months later, I looked at my first AIME and was hooked. I solved problems like crazy, my scores shot up.

No where in my city was the AMC offered.
I got my school to do it.
This year, in 8th grade, I qualified for USAMO (probably).

I believe that mathcounts is good.

mathcounts, for top middle schoolers, is simply too much emphasis on computation. however, for most people it is harder and is a good way to get them into math.
personally, i dont practice for mathcounts

wwooww...
i don't know what to say..
you guys are making a big deal out of all this mathcounts stuff....
and mathcounts isn't a waste of time, just memorize a lot of stuff...
and i don't know anything about the mathcounts scores..
i didn't even get mine from chapter... (i guess it was because i wasn't there when they gave them out, but oh well..)

:)

I think that despite what people say, a large percent of the top people at mathcounts do well on the olympiads as well. So clearly, these top 12-20 aren't wasting all their time practicing for computational contests... for example, 2 years ago, of the top 12 people at nationals, 8 people qualified for red mop this yea rand 1 person was a point a way from qualifying. (Then, from 39th to 57th place, 7 more people qualified for red mop). So, there definitely is a correlation.

Though you can argue that the people who spend the most time on this 'computational' contest are the ones who do well but fail to succeed in higher levels of mathematics, I think that the numbers show that this really isn't as true as you're trying to say it is.


integration
calculus
calculus computations

Hi everyone,

Could you explain to me how to find the antiderivative of e^-2x, please? I know that the answer is e^-2x/2, but I don't understand how you get this.

I think you have to do a u substitution, but when you do that you get e^-2x/4.

Thank you very much

Try differentiating your alleged antiderivative. What happens to the constants?

The substitution rule is just the chain rule reinterpreted.

[quote="chocolatelover"]
Could you explain to me how to find the antiderivative of e^-2x, please?[/quote]
$ (e^{\minus{}2x})'\equal{}\minus{}2e^{\minus{}2x}.$

Now integrate both sides,

$ e^{\minus{}2x}\plus{}k_1\equal{}\minus{}2\int e^{\minus{}2x}\,dx.$

Divide by $ \minus{}2$ and we happily get $ \minus{}\frac12e^{\minus{}2x}\plus{}k\equal{}\int e^{\minus{}2x}\,dx.$

Thank you very much

Could you also do it with a u subtitution?

u=e^-2x
du=e^2xdx
du/2=1/2e^2dx

int. e^-2x=

int. (u^2x)(1/2du)=
1/4e^2x

I'm not sure why I end up getting 1/4e^2x instead of 1/2e^2x. Does anyone see my mistake?

Thank you very much

[quote="Chocolatelover"]Does anyone see my mistake?[/quote]

Here it is:

[quote="Chocolatelover"]u=e^-2x ... du=e^2xdx[/quote]

If $ u \equal{} e^{\minus{}2x}$ then $ \frac{du}{dx} \equal{} \minus{}2e^{\minus{}2x}$, and so $ dx \equal{} \minus{}\frac{du}{2e^{\minus{}2x}} \equal{} \minus{} \frac{du}{2u}$, and

$ \int e^{\minus{}2x}\,dx \equal{} \int u \cdot \left(\minus{} \frac{du}{2u}\right) \equal{} \minus{} \frac{1}{2} \int du \equal{} \minus{} \frac{u}{2} \equal{} \minus{} \frac{e^{\minus{}2x}}{2} \plus{} C$.


Find all postive integer $\ x$ such that $\ x^{n}+2^{n}+1\mid x^{n+2}+2^{n+2}+1$.

Is this supposed to be for all n? Or is it more likely find all pairs of integers x and n?

[quote="stancioiu sorin"]$\forall n\in\mathbb{R}$[/quote]

I am sorry: is it $\forall n\in\mathbb{R}$ or $\forall n\in\mathbb{N}?$

Well we can't talk about divisiblity with irrationals can we? (Unless one is an exact multiple of the other I guess...)

yes, I think you're right.Then take $\ n\in\mathbb{N}$.


trigonometry
analytic geometry
AMC
USA(J)MO
USAMO
USAMTS

It's now OK to discuss Round 4 of the USAMTS.

We'll have the Round 4 Math Jam on Friday, March 14 (that's today!), at 7:30 PM Eastern / 4:30 PM Pacific.  We'll discuss solutions to all 5 problems on Round 4.  [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here for the complete Math Jam schedule[/url].

The Round 4 Math Jam transcript is now available on the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]Math Jam transcripts page[/url].

I particularly direct your attention to the solution to #5 if you haven't seen it.  No trig, no coordinates, nothing icky.  Just basic geometry.

[quote="tenniskidperson3"]Hmm... is there any reason why after 7:40 nobody but DPatrick is saying anything? Just wondering.[/quote]Yeah...because the transcripter broke. :(

We'll look into it.

[quote="Ubemaya"]Do you know when Round 4 scores will be available?[/quote]Probably the 3rd week of April (April 14-18).

arhGA. Now I have to be suspenseful of my USAMO qualification and my USAMTS score. lol


let $ ABCD$ a convex quadrilateral. $ M$ is the  intersection point of  [$ AC$] and [$ BD$]. the bisector of the angle $ \widehat{ACD}$ intersect $ (AB)$ in $ K$. prove that  $ MA$ x $ MC \plus{} MA$ x $ CD \equal{} MB$ x $ MD$ $ \Leftrightarrow$ $ \widehat{BKC} \equal{} \widehat{BDC}$

since B is outside the angle <ACD, the subject angle bisector cannot meet the segment (AB).

Best regards,
sunken rock

greatestmath, you yourself posted this problem before.
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=198133[/url].


X is a subset of eight elements of {1, 2, 3, ... , 17}. Show that there are three pairs of (distinct) elements in X with the same difference.

whatever it was, this is its solution:
suppose there was a group of 8 for which it's not like that

if the problem has no solutions for {1,...,17} then it will clearly not have solutions for {2,...,17} (d'uh, it's part of it) so that means we can assume the first number (1) would be in the group if there was one.

write them in ascending order and count the difference each time
the optimal will be 4x each prime number, as low as possible in total
here that means: 4 times +2 and 3 times +3
for a total of +17

example: 1, 3, 5, 7, 9, 12, 15, 18 for the lowest option. So 18 elements would be required for that. (which there aren't)


Hi, okay I need help!

Question is:
Find all integers, x and y that:
$2xy+\frac{2x^{3}+2y^{3}}{3}= 4014$

this is problem 9 in the [b]currently running [/b]awesome math admissions test

please lock this topic

Pathetic, quote Relevation"this is from a chinese school textbook"... Anyways, #9 is actually quite easy when you figure out the solution, the next step is the most important. In my opinion, #9 is much more easier than #4.

Again, [quote="Altheman"] this is problem 9 in the [size=200]currently running[/size] awesome math admissions test  [/quote]
Mods, lock this topic and delete the previous posts, please.


[quote="mps"]$a+b+c = 1, a,b,c \geq 0$

$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}} +\frac{c}{\sqrt{a+b}}\geq \sqrt \frac{3}{2}$[/quote]

It is just Weight Jensen.
Let $f(x)= \sqrt{\frac{1}{x}}$  $f''(x)=\frac{3}{4x^{2.5}} \geq 0$
$\sum \frac{a}{a+b+c} \sqrt{\frac{1}{b+c}} \geq \sqrt{\frac{1}{\sum a(b+c)}}=\frac{(a+b+c)}{\sqrt{2(ab+bc+ca)}} \geq \sqrt{\frac{3}{2}}$ :)

Phew...I'm starting to learn Jensen's...I used it correctly...and for the first time!
My solution was basically the same...

$f(x)=\frac{1}{\sqrt{x}}$ is concave up, so...
\[ af(b+c)+bf(a+c)+cf(a+b) \ge f(a(b+c)+b(a+c)+c(a+b)) \] \[ =\frac{1}{\sqrt{2(ab+bc+ac)}} \]
Multiply both sides by $\sqrt{2}$ and take the reciprocals, so the original inequality follows if we prove
\[ \sqrt{ab+bc+ac} \le \frac{1}{\sqrt{3}} \]
\[ ab+bc+ac \le \frac{1}{3} \]
By Cauchy,
\[ ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2} \le \frac{(a+b+c)^2-\frac{1}{3}(a+b+c)^2}{2} \] \[ =\frac{1}{3}(a+b+c)^2=\frac{1}{3} \]
and the proof is complete.

Thanks very much But Does anybody have solution without solution?

[quote="mps"]Thanks very much But Does anybody [b]have solution without solution[/b]?[/quote]

What do you mean ???  :D  :D  :)

[quote="mps"]Thanks very much But Does anybody have solution without function
?[/quote]

Thanks This is true

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=71060[/url]

 :)

[quote="mps"]Let $x,y,z>0,\ x+y+z=3$. Prove:
\[ \frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \ge \frac{3}{\sqrt{2}}  \] [/quote]
By AM-GM:
\begin{eqnarray*} LHS=\sum_{cyc} \frac{x}{\sqrt{y+z}} \\ &=& 2\sqrt{2}\cdot \sum_{cyc} \frac{x}{2\sqrt{2}\sqrt{y+z}} \\ &\ge& 2\sqrt{2} \sum_{cyc} \frac{x}{2+y+z} \\ &\ge& \sqrt{2} \frac{(x+y+z)^2}{x+y+z+xy+yz+zx} \\ &=& \cdots \end{eqnarray*}

[quote="mps"]$a+b+c = 1, a,b,c \geq 0$

$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}} +\frac{c}{\sqrt{a+b}}\geq \sqrt \frac{3}{2}$[/quote]
$\sum \frac{a}{\sqrt{b+c}}\geq \frac13 \sum a \sum\frac{1}{\sqrt{b+c}}\Rightarrow \frac13 \frac{3\sqrt{3}}{\sqrt{2(a+b+c)}}=\sqrt\frac32$

[quote="mps"]$a+b+c = 1, a,b,c \geq 0$

$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}} +\frac{c}{\sqrt{a+b}}\geq \sqrt \frac{3}{2}$[/quote]

    Another way:
  $\sum\frac{a}{\sqrt{b+c}}=\sum\frac{a^2}{\(a\sqrt{b+c}}\geq\frac{(a+b+c)^2}{\(a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{b+a}}=\frac{1}{\(a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{b+a}}$
    So now we try to solve that:
  $\frac{1}{\(a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{b+a}}\geq\sqrt\frac{3}{2}$
    Equivalent to
  $\(a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{b+a}\leq\sqrt\frac{2}{3}$
    By Cauchy Schwarz we have:
  $\(a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{b+a}=\sqrt{a}\sqrt{ab+ac}+\sqrt{b}\sqrt{bc+ab}+\sqrt{c}\sqrt{ac+bc}\leq\sqrt\(2\((a+b+c)\((ab+bc+ca)\=\sqrt{2\(ab+bc+ca)}\=\sqrt{2\(ab+bc+ca)}$
    Since $\(ab+bc+ca\leq\frac{(a+b+c)^2}{3}=\frac{1}{3}$
    So we have:
  $\(a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{b+a}\leq\sqrt\frac{2}{3}$
    and the proof is complete.

Yes, and you can prove the inequalitiy
$\frac{a}{\sqrt{b+c}}+\frac{a}{\sqrt{b+c}}+\frac{a}{\sqrt{b+c}} \geq \sqrt{\frac{3}{2}(a+b+c)}$ where a,b,c be positive reals

[code]I'm sorry, i didn't view Ramanujan post[/code]


At the beginning of the school year, Lisa's goal was to earn an A on at least $80\%$ of her $50$ quizzes for the year. She earned an A on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?

$\text{(A)}\ 1 \qquad\text{(B)}\ 2 \qquad\text{(C)}\ 3 \qquad\text{(D)}\ 4 \qquad\text{(E)}\ 5$

[hide]Step 1: We first need to find 80% of 50. 50*.8=40.
Step 2: Subtract 22 from 40(40-22)which is 18.
Step 3: Since she has 20 more tests, she has to get an "A" on 18 of those test. So we subtract 18 from 20(20-18) which is 2 and [size=150]B[/size]. I pretty sure I'm right. :) [/hide]

[hide]If she is to achieve 80% of 50 quizzes level A, 40 have to be A
So far her score is 22/30 with 20 quizzes left.
If she was to achieve 20 more A's,
then she has a total of 42 which is 2 more than her goal (not that it's bad to surpass your goal though)
[b]That means Lisa can get 2 questions wrong in order to maintain her goal![/b]
 :arrow: [b](B)[/b][/hide]

[hide]
She needs to get an A on at least $80\%\cdot50=.8\cdot50=40$ quizzes.
That means she can miss a maximum of 10 quizzes.
She has already missed 8, so the most she can miss after now is 2.
The answer is thus 2
$\boxed{B}$[/hide]

[quote="easyas3.14159..."]This problem's already been posted in the MOEMS forum. :D
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Lisa%5C%27s+quizzes&t=61661]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Lisa%5C%27s+quizzes&t=61661[/url][/quote]

Sorry, I'm just posting the AMC-10 from the same year.


Post your work and we'll see what you did wrong  :D

[quote="Adam4444"]Hmpfh. I get the wrong answer on the following task:

Calculate the value of the expression $3x-4x(3-5x)-2(3x+2)$ for $x =-2$.

I don't know what I'm doing wrong, but I should manage to solve this one by my own, but I get the wrong answer all the time. :S[/quote]

[hide="Answer"]-6+8(3+10)-2(-6+2)
-6+24+80+12-4
106[/hide]

You could possibly have forgotten to switch the negative signs to positive when multiplying -2(-6....)

[quote="funcia"]Calculate the value of the expression $3x+(-4x)(3-5x)+(-2)(3x+2)$
when x=-2

does that help to put it like this?[/quote]Yes, actually.  :o
Now I get it to 106. I counted wrong at the first parenthesis.

Thanks!

Your mistake could be in the order of your operations. Remember P (lease) E(xcuse) M(y) D(ear) A(unt) S(ally). First perinthisies(sorry about spelling) then exponents (including factorials when you get to that) then multiply/ divide then add and subtract. :)  hope this helps


A sealed envelope contains a card with a single digit on it. Three of the following statements are true, and the other is false.

I. The digit is 1.
II. The digit is not 2.
III. The digit is 3.
IV. The digit is not 4.

Which one of the following must necessarily be correct?

$ \textbf{(A)}\  \text{I is true.} \qquad \textbf{(B)}\  \text{I is false.}\qquad \textbf{(C)}\ \text{II is true.} \qquad \textbf{(D)}\ \text{III is true.} \qquad \textbf{(E)}\ \text{IV is false.}$

Apprentice, please hide your answers.  :!:

[hide]We know that I and III cannot both be true, since the number can't be both 1 and 3. Since 3 of the statements are true, II and IV must be true. Thus, the answer is [b]C[/b].[/hide]

hold on catcurio... the digit cannot be 1 or 3; the other two answers can be true...
So, either $I$ is wrong or $III$ is wrong...
The question asked for what statement is correct,
The only choice that says an answer is incorrect from 1 and 3 is B, do you follow or am I wrong?  If so please elaborate... :?

[hide]One of I and III must be true and the other false. Thus, the possible true statements are I, II, and IV, or III, II, and IV. However, we don't know which of I and III is true. You said that I is false. This would make III true, which is also an answer choice. The only thing we know for certain is that II and IV are true, and the choice that we are given is that II must be true.[/hide]

[quote="Apprentice"]
The question asked for what statement is correct,
[/quote]

Careful reading! The question DIDN'T ask for what statement is correct, but "what of the FOLLOWING is correct".

Oops! I follow what you are saying...   :idea:

[quote="ritchjp"][hide]
I believe that the answer has to be $B$

for option A: if $I$ was true:

 If the 3 correct answers were $I, \ II, \ III$ there would be a contradiction, so that eliminates answer A.

for option C: if $II$ was true:

 Then also there would be the same contradiction.

for option D: if $III$ was true:

 Again the same contradiction.

for option E: if $IV$ was false:

 Again the same contradiction

But if for option B: if $I$ was false:

 There is no way that the other three statements could contradict each other.

[/hide][/quote]

And how did you get I, II, and III to all be true? (hint: you can't!!!)

[hide]In order for three of I, II, III, and IV to be true, either I or III must be true (otherwise at most two statements would be true).

So the digit is 1 or 3. Either way, the digit is not 2, so II is true $\Rightarrow \boxed{C}$.[/hide]

I'll try to explain in a little more detail on how I got my answer.

[hide]

First, let's take answer $A$.

We are saying that $I$ is true. Let's assume that it is.
There are three possible scenarios that we can get.
They are: $I, \ II, \ III$ are true, $I, \ II, \ IV$ are true or $I, \ III, \ IV$ are true.
$I$ and $III$ can't both be true but $I, \ II, \ IV$ COULD be true.
This raises doubt about whether $A$ is the answer or not.

Second, let's take answer $C$.

We are saying that $II$ is true. Again, let's assume that it is.
The three scenarios we get are: $I, \ II, \ III$ are true, $I, \ II, \ IV$ are true or $II, \ III, \ IV$ are true.

As in answer $A$, $I$ and $III$ can't both be true so that eliminates $I, \ II, \ III$ and as in answer $A$, $I, \ II, \ IV$ COULD be true. The last scenario, $II, \ III, \ IV$ could also be true. But there is still the possibility that $C$ may not be the answer because of $I, \ II, \ III$.

Next, let's take answer $D$.

This time we are saying that $III$ is true. As before, let's assume that it's true.
This time the three scenarios: $I, \ II, \ III$ are true (which we know can't be possible), $II, \ III, \ IV$  are true or $I, \ III, \ IV$ are true (which we also know can't be true). So even if $II, \ III, \ IV$ could be true, there is still that chance that $D$ may not be the right answer.

Now let's take answer $E$, $IV$ is false.

This is a no-brainer because if $IV$ was false, then $I, \ II, \ III$ are all true, but three times we have seen that this combination is impossible. So answer $E$ is definitely out.

That leaves only one answer and that is $B$. To verify this assume $I$ is false. That leaves $II, \ III, \ IV$ to be true and that is the ONLY scenario for answer $B$.

For $A$ there is a $\frac{1}{3}$ chance of $A$ being right.

For $C$ there is a $\frac{2}{3}$ chance of $C$ being right.

For $D$ there is a $\frac{1}{3}$ chance of $D$ being right.

For $E$ there is no chance.

But for $B$ there is a $100$ percent chance that it is right.

That's why I chose $B$

[/hide]

[quote="4everwise"]A sealed envelope contains a card with a single digit on it. Three of the following statements are true, and the other is false.

I. The digit is 1.
II. The digit is not 2.
III. The digit is 3.
IV. The digit is not 4.

Which one of the following must necessarily be correct?

$\text{(A)} \text{I is true.} \qquad \text{(B)} \text{I is false.}\qquad \text{(C)}\text{II is true.} \qquad \text{(D)}\text{III is true.} \qquad \text{(E)}\text{IV is false.}$[/quote]
[quote="Hokkage"]I'm sorry, but I'm pretty sure that's completely wrong. Three of the statements
I. The digit is 1.
II. The digit is not 2.
III. The digit is 3.
IV. The digit is not 4.
must be true. The only such groupings such that you have 3 correct statements are I, II, IV and II, III, IV since any grouping with I and III is self-contradictory (the card cannot have two numbers). Therefore II and IV are true, thus C is the correct answer. Simple logic.[/quote]

I also got C.. obviously II or IV can't be wrong and the only one with the right answer as either II or IV is C. but couldn't IV also be true? even if IV is true but it's not an answer choice, you can safely assume that the answer is C?

just read the question carefully, it says that "which of these statements are true?"

we have already established that either I or III must be false, so II and IV are true, and therefore C is the answer, you guys are overthinking the problem, there are plenty of other statements that are true, but we are choosing from the answer choices, not all possible statements

If C is the right answer and we know that II and IV are true do you know which one of the other statements is true and which one is false?

Isn't the answer to tell what THREE statements are true and which one is false?

With B as the answer there is no question as to which statement is false and what three statements are true.

[quote="4everwise"]A sealed envelope contains a card with a single digit on it. Three of the following statements are true, and the other is false.

I. The digit is 1.
II. The digit is not 2.
III. The digit is 3.
IV. The digit is not 4.

Which one of the following must necessarily be correct?

$\text{(A)} \text{I is true.} \qquad \text{(B)} \text{I is false.}\qquad \text{(C)}\text{II is true.} \qquad \text{(D)}\text{III is true.} \qquad \text{(E)}\text{IV is false.}$[/quote]
[quote="ccy"]But does I [i]have[/i] to be true?  I could be false and III true and that wouldn't cause any contridictions.[/quote]

[b]I [/b] can be false , meaning that the digit is 3  :| but at the same time[b] III [/b] could also be false, making the digit 1
but "I is false" is not "necessarily" correct (question asked) because [b]I [/b] can also be true, so it doesnt really answer the question.. unless i misunderstood something :dry:

Now, the answer shall be provided by the person who has the answer. :P

[hide="Answer"]Are you ready for the real answer??  :rotfl: [hide]$\text{(C)}$!![/hide][/hide]


Two players write alternatively some integers on the blackboard. The rules are the following :
- The first player write $1$.
- At each of the other turns, the player has to write $a+1$ or $2a$ where $a$ is any number already wrote in the blackboard and $2a \leq 1000.$
- One cannot write a number which has already been written, and no number is erased.
- The player who writes $1000$ is the winner.

Determine which player has a winning strategy.

Pierre.

bumping this up

haven't had too much time to think about it, but at first thought, it seems like the first player can always write an odd integer on the board.

@tequila: I think you are incorrectly assuming that the $a$ must be the previous number written on the board.

If $a$ can indeed be any number already written, then the first player loses.

To lose, one must be forced to write either 500 or 999.  No one is going to have to do that until all the numbers 1-499 are written, because $a+1$ is always an option.  Since 502 is also an option as soon as 251 is written, 502-998 must also be written.
Thus, after all integers from 1 to 998, excluding 500 and 501, are written on the board, the person whose turn it is loses.

996 integers => even => first player's turn.

That's it Renshaw. :)
But, for Tequila, I have made the same mistake as you did at first attempt. Some contestants have made it too and, unfortunately for them, they did not correct their answer (but I did  :D  ).

Pierre.


Hi Mr. beaty. I have a question. What is the golden mean??

Isn't golden mean like golden ratio in this case?

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

I think this site gives a clearer explanation of the golden ratio:

[url]http://www.mathsisfun.com/numbers/golden-ratio.html[/url]


If $ a$ is real, find the sum of all real roots of the equation $ x^5\plus{}x^4\plus{}(1\minus{}a^2)x^3\plus{}(1\minus{}a^2)x^2\plus{}(1\minus{}a^2)x\plus{}1\equal{}0$

$ x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1\ \equal{} \ a^2x(x^2 \plus{} x \plus{} 1)$

Multiply by $ (x \minus{} 1)$:

$ x^6 \minus{} 1\ \equal{} \ a^2x(x^3 \minus{} 1)$

$ x^3 \minus{} 1 \equal{} 0$ or $ x^3 \plus{} 1 \equal{} a^2x$

The former has no real roots except $ x \equal{} 1$. And $ x \minus{} 1$ is a root of the original equation $ \Leftrightarrow\ a \equal{} 0$. So the other real root(s) will be the real solution(s) to the cubic equation for $ a\neq0$.

work with the factorisation $ (1\plus{}x\plus{}x^2)[x^3\plus{}(1\minus{}a^2)x\plus{}1]$

:oops_sign:

I misread the question; I thought we were supposed to find all the real roots. (Apparently I didn’t notice the thread title either.)


The question states:
Show that if the Kelvin Planck statement of the 2nd law were not true,a violation of the Clausius statement would be possible.

My attempts:(Please confirm if I am correct)

Kelvin Planck statement says that it is impossible for any system to operate in thermodynamic cycle and deliver a net amount of work to its surroudnings while receiving energy by heat transfer from a single thermal reservoir.
Qcool>0 and efficiency <1
Kelvin Planck statement is not true because you can't the net amount of work to its surroundings while receiving energy by heat transfer from a single thermal reservoir.

we know the second law of thermodyanics cycle is that you have hot reservior and cool reservoir. and you can have work output from the hot reservoir by extract heat from and reject heat to cool reservoir.

W_net=Q_h - Q_c

Clausius machine can be put together with a kelvin machine and an indirect carnot machine.
hence if kelvin planck statement is not true then clausius statement isnt true either :D

So,.........
What about my logic?
Is it correct?


If $\log_7 \Big(\log_3 (\log_2 x) \Big) = 0$, then $x^{-1/2}$ equals

$\displaystyle \text{(A)} \ \frac{1}{3} \qquad \text{(B)} \ \frac{1}{2 \sqrt 3} \qquad \text{(C)} \ \frac{1}{3 \sqrt 3} \qquad \text{(D)} \ \frac{1}{\sqrt{42}} \qquad \text{(E)} \ \text{none of these}$

[hide="Solution"]
Log of 1 equals zero, therefore log_3( log_2(x) ) = 1
3^1 = log_2(x)
2^3 = x
x = 8
x^-0.5 = 1 / sqrt( x ) = 1 / sqrt ( 8 ) = 1 / ( 2 * sqrt ( 2 ) ) therefore the answer is E none of these.[/hide]

[hide="Answer"]$\log_2z=0\Rightarrow z=1$, so then $\log_3y=1\Rightarrow y=3$, so then $\log_2x=3\Rightarrow x=8$, and therefore $\frac{1}{\sqrt{8}}=\frac{1}{2\sqrt{2}}\Rightarrow \boxed{E}$.[/hide]

[hide="Solution"]
Yeah, two have already been posted, and likely mine is the same, but here goes:

$\log_2 \Big(\log_3 (\log_2 x) \Big) = 0$

So, $\Big(\log_3(\log_2 x) \Big) = 1$.

And, $\Big(\log_2 x \Big) = 3$.

Which means, $x = 8$.

$x^{-1/2}$ is $\frac{1}{\sqrt{8}}$, which is not given. So the answer is E.[/hide]


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope

reflect y=1/x on y=2x, the image equation is 12(x^2)-7xy-12(y^2)+25=0.

Can someone explain to me how to get to the image equation???
Thanks

First find the symmetrical S  of the point A(a , 1/a) with respect to line y = 2x ( find the slope of the perpendicular from A to y = 2x , the equation of this perp. and the common point with y = 2x , solving the system ect) and then prove that the coordinates of S sutisfy the given equation. You will need some good calculatins to do all this.This is the normal way.

 Babis


Let a,b,p,q,r,s be naturals s.t. qr-ps=1 and p/q<a/b<r/s. Prove that b>q+s-1.

Bomb

You can check here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=40806


The ball has to go $ 4$ times to the left and $ 4$ times to the right, of which there are $ \binom84\equal{}\boxed{70}$ ways to do.

There are $ 2$ choices for each turn of $ 8$, so it is out of $ 2^8\equal{}256$, so the answer is $ \frac{70}{256}\equal{}\boxed{\frac{35}{128}}$.

[hide="Where Am I wrong"]The first four deflections does not matter. In first four deflections it can go anyway. But in the next four it have to go in a selected way. So probability is $\frac{1}{2^4}$ or ${\boxed{\frac{1}{16}}}$.[/hide]

Actually, it does not have to go in a selected way for the last four. For example, left-right-left-right yields the same result as left-left-right-right, and they are considered different.

Instead, treat the board like a sideways grid and solve for the possible "paths."

For the chip to end up back in the center, the chip must deflect left 4 of the 8 times, and right the other 4 times. There are $\binom{8}{4}$ ways to pick which 4 of the rows to deflect the chip left and then the other 4 to deflect it right. Each deflection is to the left with probability $\frac12$ and to the right with probability $\frac12$, so each possible path to the bottom occurs with probability $\left(\dfrac{1}{2}\right)^{\!8}$. Thus, the probability that 4 out of 8 deflections go left and ball goes in the middle is \[\binom{8}{4}\frac{1}{2^8}=\boxed{\frac{35}{128}}.\]

Beautiful problem. Very satisfying for me to solve, as 18 of my 32 problems wrong are C&P.

[quote=pog]isnt this just the generic path counting questions[/quote]

Me: Beautiful math problem
pog: uh generic
Me: *roasted*  :blush: yeah, that's what I meant

i mean uh it's worded interestingly???

edit: idk what to say i am not the pure judger of problem quality on aops considering i can barely solve any difficult problems

This is a great problem. Sidenote but I was at a fair once (before COVID), and they had the exact same layout as in the problem (it's a real game). And the if you got it into one of these holes (red in diagram), they would give you a free mint (actually as many as you won). I think I got two mint packets in total. The fair person was actually off center of the holes, so it would be harder for you to get it in the hole, and consequently, harder to get a free mint, but I started from the middle/center, which is where you actually maximize your chances (like in the problem). The fair was great. And the mints.  :juggle:

[quote name="RTB" url="/community/p28152924"]
Maybe I'm stupid or something but this was the first way I thought to solve it. I hope you can read the numbers on the diagram...
[hide=Solution]There are $70$ different pathways to the centre slot and a total of $256$ to the bottom. $\frac{70}{256} = \frac{35}{128}$[/hide]
[/quote]

orz pascals triangle sol :omighty:


Hi, I have a pde:

xu_x+(x+y)u_y = 1

So using the method of characteristics..

dx/ds=x
dy/ds=x+y
dz/ds=1

Hence x=Ae^s, the boundary condition (not shown for simplicity), gives A=1
Hence x=e^s

But this is where I'm stuck...

we now have...

dy/ds = e^s + y

I don't see how we can solve this to get y, since the variables won't separate (will they?)

can anyone help?

Thanks very much

Ok, I think I figured it out, it can't be solved by separation of variables, but rearranging...

dy/ds - y = e^s

so using an integrating factor, we get:

y(s) = (s+c)e^s

And thats it.


trigonometry
integration
calculus
logarithms
modular arithmetic
complex numbers
real analysis

Study the convergency of the following  series.

\[f(t)=\sum_{k=1}^{\infty}\frac{\sin kt}{k}\]

A stronger statement was made by Dr. Merryfield in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134868]this thread[/url].

For the particular series here, it's worth asking for the exact value of the sum. That is a feasible question.

Also worth asking: what is $\frac1{2\pi}\int_{0}^{2\pi}[f(t)]^{2}\,dt$ - computed both directly and as a sum on $k.$

$\frac1{2\pi}\int_{0}^{2\pi}[f(t)]^{2}\,dt = \frac1{2\pi}\sum_{m}\pi \frac{1}{m^{2}}dt = \frac{\pi^{2}}{12}$, because $\{sin(nt)\}$ is orthonormal system in $[0,2\pi]$.
$f(t) = \frac{\pi-t}{2}$ (actually quessing  :blush: , for strict answer we need to look at Fourier coeficients), when $t \in (0,2\pi)$ Integral of "this" $f^{2}$ , $\frac1{2\pi}\int_{0}^{2\pi}[f(t)]^{2}\,dt$, also equals to $\frac{\pi^{2}}{12}$.

Here's a non-Fourier approach to pou's second claim. In order to do this, I must first show that the series converges for all $t.$ The link in paladin8's post has me claiming that, although the actual proof is elsewhere. Summation by parts plays a prominent role in the argument.

$\sum_{n=1}^{\infty}\frac{\sin n\theta}{n}=\text{Im}\left(\sum_{n=1}^{\infty}\frac{e^{in\theta}}{n}\right).$

Now examine $\sum_{n=1}^{\infty}\frac{z^{n}}{n}.$ This power series converges for $|z|<1$ to $-\log(1-z).$

The imaginary part of the series is then $-\arg(1-z).$ In order to make sense of this, we need a branch of the logarithm that is continuous on the set $\{(1-z): |z|<1\},$ which lies in the right hand half plane, and for which $\log 1=0.$ It's enough to say that $-\frac{\pi}2<\arg(1-z)<\frac{\pi}2.$

We must treat the point $z=1,$ meaning $\theta\equiv 0\pmod{2\pi},$ as special - the rest of the argument below won't work for that point. But there, the original series has zero terms and sums to zero.

For every other point, we quote Abel's Theorem: since we know the series converges at each point on the boundary circle, it's value equals the limit from inside the circle along a radial path. Hence, by continuity,

$\text{Im}\left(\sum_{n=1}^{\infty}\frac{e^{in\theta}}{n}\right)=-\arg(1-e^{i\theta}).$

The finish is a purely geometric argument, an angle chase. Draw the isosceles triangle with vertices $0,1,$ and $e^{i\theta}.$ The angle of this triangle at $0$ is $\theta,$ whereas the angle $\alpha$ of the triangle $1$ (which you can see most easily by extending the edges and looking on the outside) is $\alpha=-\arg(1-e^{i\theta}).$ Since this is an isosceles triangle, we have $2\alpha+\theta=\pi$ or $\alpha=\frac{\pi-\theta}2.$

Conclusion: for $0<\theta<2\pi,\ \sum_{n=1}^{\infty}\frac{\sin n\theta}{n}=\frac{\pi-\theta}2.$