trigonometry
complex numbers
algebra unsolved
algebra

Hi,  please solve this:

w=cos40 + isen40

 /w +2w^2 + 3w^3 + ..... 9w^9/^(-1) = ?

And  a nice solution for this:

 z+ /z/= 2 + 8i

/z/^2 = ?

without just compare real real imaginary imaginary! :)

thanks!

$ w \equal{} \cos 40^0 \plus{} i\sin 40^0 \equal{} \cos\frac {2\pi}{9} \plus{} \sin\frac {2\pi}{9}$
Let $ S \equal{} w \plus{} 2w^2 \plus{} 3w^3 \plus{} .... \plus{} 9w^9$
    $ wS \equal{} w^2 \plus{} 2w^3 \plus{} 3w^4 \plus{} ....8w^9 \plus{} 9w^{10}$

$ S \minus{} wS \equal{} w \plus{} w^2 \plus{} w^3 \plus{} .... \plus{} w^9 \minus{} 9w^{10}$

$ S(1 \minus{} w) \equal{} \frac {w(1 \minus{} w^9)}{1 \minus{} w} \minus{} 9w^{10} \equal{} \minus{} 9w^{10} \equal{} \minus{} 9w$          (since  $ w^9 \equal{} 1$)

$ S \equal{} \frac { \minus{} 9w}{1 \minus{} w} \equal{} \frac {9}{4\sin20^0}( \minus{} \sqrt {3} \plus{} i)$

Let $ z \equal{} x \plus{} iy$

$ x \plus{} iy \plus{} \sqrt {x^2 \plus{} y^2} \equal{} 2 \plus{} 8i$...........................(1)

Equating real and imaginary part
$ x \plus{} \sqrt {x^2 \plus{} y^2} \equal{} 2;y \equal{} 8$

Now you may solve for $ x$.

But if you want to do without comparing then take conjugate on both sides of (1)
$ x \minus{} iy \plus{} \sqrt {x^2 \plus{} y^2} \equal{} 2 \minus{} 8i$...........................(2)

Subtract (2) from (1)

you easily get $ y\equal{}8$.Now solve for $ x$


[quote="anirudh"]When is it going to work? :maybe:[/quote]It's working right now, but I have to fix some display (CSS) issues first.


algorithm
combinatorics unsolved
combinatorics

Given an n rows and n columns table, we write numbers from 1 to n² in this order :
start everywhere you want with 1, then write 2 at a distance (verticaly, horizontaly or diagonaly) of 2 squares, then write 3in the same way, and this until n², if you can !

Could you give an algorithm to get a complete table ?
How many different complete tables can you get ?

[quote="Darkil73"]a distance (verticaly, horizontaly or diagonaly) of 2 squares[/quote]

What does this mean?  On a $ 5 \times 5$ table, which squares are at most this distance from the central square?  All of them?

Along the same line:

Which pairs of squares could possibly be $ n^2$ apart in order to serve as your $ n^2\minus{}1$ and $ n^2$?

Salam aleykum aghaye.
Why do you need to take n rows and n columns specially?

I understood.
Now let's have again the nearest neighbour algorithm once more. Assume not to take the nearest two squares but just the nearest. Then obviously...

Then obviously the number of the ways we can edit such a table will decrease, and for this you can have a look at Applicative Mathematics For MATLAB by E.K.M..
So now we are looking for if it is possible to configure such a table or not.
For this we must look at the least element you need, i.e. as satisfying the rules you have said.

And go to research and then inform me, okay. I'm at doing something like it will be good for all the world people.
Okay. If my solution or at least, my manner is not good enough for you, then leave me and don't inform me anything. Don't send just a reply.
I'm Iranian.

And I want to remember that the least element I have just spoken about is a table with 5 columns and 5 rows.

[quote="kevinatcausa"]Which pairs of squares could possibly be $ n^2$ apart in order to serve as your $ n^2 \minus{} 1$ and $ n^2$?[/quote]

Oh, I assumed that "in the same way" meant that each distance was to be 2 or less.


function
vector
induction
analytic geometry
algebra unsolved
algebra

On the Cartesian plane consider the set $V$ of all vectors with integer coordinates. Determine all functions $f : V \rightarrow \mathbb{R}$ satisfying the conditions:
(i) $f(v) = 1$ for each of the four vectors $v \in V$ of unit length.
(ii) $f(v+w) = f(v)+f(w)$ for every two perpendicular vectors $v, w \in V$
(Zero vector is considered to be perpendicular to every vector).

We'll prove by induction that:
(*) $f([n,0])=f([0,n])=f([-n,0])=f([0,-n])=n^2$.
Prove. For $n=1$ we have it from (i).
Now let (*) be true for $k\leqslant n$. We have: $f([n,1])=f([1,-n])=f([-n,-1])=f([-1,n])=n^2+1$, because $f([n,1])=f([n,0])+f([0,1])=n^2+1$ and similarly for remaining vectors. Further we have: $2n^2+2=f([n,1])+f([1,-n])=f([n+1,1-n])=$
$=f([n+1,0])+f([0,-(n-1)])=f([n+1,0])+(n-1)^2$, whence:
$f([n+1,0])=n^2+2n+1=(n+1)^2$.
Similarly it goes that: $f([0,n+1])=f([-(n+1),0])=f([0,-(n+1)])=(n+1)^2$.
So we have (*) for all $n\in \mathbb{N}$.
Now let $u\in V$. Then $u=[m,n]$, where $m,n$ are integer. By (*) we have:
$f(u)=f([m,n])=f([m,0])+f([0,n])=m^2+n^2$.
So there's only one function satisfying the desired conditions.

Now, solve the same problem but without the first condition.

Pierre.

Well, I didn't think about it deeper, but I suppose it all depends on the values for unit vectors. If you don't determine them, you'll have infinitely many functions, but if you determine them, you'll have only one function for each set of the four values for the four unit vectors. That's what I suppose, but, as I said, I did not investigate it further.

Sure, but we can say more. The set of solution is a vector space generated by some particular ones...

Pierre.

intersting problem!!!
to tram:I think your solution isn't complete. becasue you didn't prove that your answer is suffient to the condition(though it is easy to prove).
To pbornsztein:
I have solved it. all the function si that $f(m,n)=(t_{m+1}+t_{n})(f(1,0)+f(0,1))+(t_{m}+t_{n-1})(f(0,-1)+f(-1,0))$ when  $m,n>0$
$f(m,n)=(t_{m+1}+t_{n})(f(1,0)+f(0,-1))+(t_{m}+t_{n-1})(f(0,1)+f(-1,0))$for $m<0,n>0$
$t_{n}=[\frac{n^{2}}{4}]$
the other two case is similar.


If the polynomial $ x^4\plus{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ does not admit real roots and $ d>0$, prove that $ c^2\plus{}a^2d<4bd$

This isn't true of $ (x\plus{}1)^4\plus{}x^2\equal{}x^4\plus{}4x^3\plus{}7x^2\plus{}4x\plus{}1$.

a=b=c=0; d=1 is a simple counter example

i think they meant to say that the polynomial can't have ALL real roots.

Let $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d \equal{} (x^2 \plus{} px \plus{} q)(x^2 \plus{} rx \plus{} s)$, where $ p^2 < 4q$ and $ r^2 < 4s$.

We have

$ p^2s^2 < 4qs^2$ and $ q^2r^2 < 4q^2s\ \Rightarrow\ p^2s^2 \plus{} q^2r^2 < 4qs^2 \plus{} 4q^2s$.

Also, since $ qs \equal{} d > 0$,

$ p^2qs < 4q^2s$ and $ qr^2s < 4qs^2\ \Rightarrow\ p^2qs \plus{} qr^2s < 4qs^2 \plus{} 4q^2s$

Finally, we have

$ p^2q^2 < 4q^3$ and $ r^2s^2 < 4s^3\ \Rightarrow\ p^2s^2q^2r^2<16(qs)^3\ \Rightarrow\ \minus{} 4d\sqrt {d} < pqrs < 4d\sqrt {d}$.

Now if you expand $ (x^2 \plus{} px \plus{} q)(x^2 \plus{} rx \plus{} s)$ and compare coefficents with $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$, you get

$ c^2 \plus{} a^2d \equal{} p^2s^2 \plus{} q^2r^2 \plus{} p^2qs \plus{} r^2qs \plus{} 4pqrs$
and
$ 4bd \equal{} 4q^2s \plus{} 4qs^2 \plus{} 4pqrs$

Hence

$ c^2 \plus{} a^2d \minus{} 4pqrs < 8bd \minus{} 8pqrs$
$ \Rightarrow$
$ c^2 \plus{} a^2d < 8bd \minus{} 4pqrs < 8bd \plus{} 16d\sqrt {d}$

Maybe that’s the inequality Hell_ever is looking for.

The truth is i found this question in a book and attempted to prove it in many ways but never even tried to verify it using an example...no wonder i never got it....


linear algebra
matrix
geometry
geometric transformation
rotation

how can i show that
$ \varepsilon^{ijk} R_{im} R_{jn} R_{kr} \equal{} \varepsilon^{mnr}\varepsilon^{ijk} R_{i1} R_{j2} R_{k3}$
where i'm using einstein summation and levi-civita permutation symbol and $ R_{ij}$ is the $ i,j$ entry of a matrix $ R$ (which is the matrix of a rotation, but i don't think this is useful here)?

I'm pretty sure that this is an identity for the matrix $ R$, so I think we actually need $ R$ to prove the identity.

$ R$ is a rotational 3x3 matrix, nothing else;

Being a rotation means that $ R$ is orthogonal: that $ RR^T\equal{}I.$

This notation is not the preferred vocabulary of most mathematicians. Let me try to make some sense of it. I guess $ (R^T)_{jk}\equal{}R_{kj}.$ So we should have $ R_{ij}R_{kj}\equal{}\delta_{ik}$ where by $ \delta_{ik}$ I mean the Kronecker delta, which is the entries of the identity matrix.


Let $ f(x)$ be a polynomial with the following properties:
(i)  $ f(0)\equal{}0$; (ii) $ \frac{f(a)\minus{}f(b)}{a\minus{}b}$ is an integer for any two different integers $ a$ and $ b$. Is there a polynomial which has these properties, although not all of its coefficients are integers?

[hide]Let $ f(x) \equal{} (x^4 \plus{} x^2)/2$.  Then $ f(0) \equal{} 0$, and
\[ \frac{f(a) \minus{} f(b)}{a \minus{} b} \equal{} \frac{1}{2} (a \plus{} b)(a^2 \plus{} b^2 \plus{} 1).\]
If $ a$ and $ b$ have the same parity, then $ (a \plus{} b)/2$ is an integer.  If $ a$ and $ b$ have opposite parity, then $ (a^2 \plus{} b^2 \plus{} 1)/2$ is an integer.  Hence, $ \frac{f(a) \minus{} f(b)}{a \minus{} b}$ is always an integer.[/hide]

[quote="coolyogi"]yes there can exists such a polynomial.
polynomial will be without constant term and degree will be greater than or equal to 2.[/quote]
Do you mean that any polynomial with at least one non integer coefficient, without constant term, and with degree greater than or equal to $2$ fits all the problem requirements ?

Or just that some (you dont precise what) such polynomials are solutions ?


An acute triangle $ABC$ has altitudes $BB': CC'$.Let $L;M;N$ be the midpoints of $B'C';C'B;B'C$,respectively.Let $K$ be the intersection of the line through $M$ perpendicular to $LC$.Prove that $KB=KC$

Please clarify your problem.

Kostas Vittas.

[quote="Davidd"]I drew it in a simple CAD program, exactly as described, and they are clearly not equal in general.  There is a simple check, drawing a circle center K, which meets only one of B or C.[/quote]
No,I think my problem is true!

Dear toan hoc.

Something it doesn't fit to the notations as your problem states.

Please, see at the figure I presented and check if it's OK with your wording.

If yes, I am sorry but we can't help you, because the problem is not correct.

Kostas Vittas.

Toan,

A Google web search, using keyword [free cad] will allow you to do a drawing for yourself.  You might also consider presenting that drawing here instead of trying to describe it in words.  I don't know how to do that, but vittasko might help.

Sorry,this problem is:
An acute triangle $ABC$ has altitudes $BB';CC'$.Let $L;M;N$ be the midpoints of $B'C';C'B;B'C$,respectively.Let $K$ be the intersection of the line through $M$ perpendicular to $LB$ and the line through $N$ perpendicular to $LC$.Prove that $KB=KC$
Help me,please!

I'm scared $KB$ can't be equal to $KC$ in all cases. I made a drawing which shows $KB$ more than 2 times longer than $KC$  :!:

I think it's true now ...
$KM \perp BL \Leftrightarrow MB^{2}-ML^{2}= KB^{2}-KL^{2}\Leftrightarrow C'B^{2}-B'B^{2}+4KL^{2}= 4KB^{2}\\ \text{Similarly, }B'C^{2}-C'C^{2}+4KL^{2}\\ \text{So }KB = KC \Leftrightarrow C'B^{2}-B'B^{2}+4KL^{2}= B'C^{2}-C'C^{2}+4KL^{2}\\ \Leftrightarrow C'B^{2}-B'B^{2}= B'C^{2}-C'C^{2}\Leftrightarrow C'B^{2}+C'C^{2}= B'C^{2}+B'B^{2}\\ \Leftrightarrow BC^{2}= BC^{2}\text{ (which is true)}$

[quote="PDatK40SP"]I think it's true now ...
$KB^{2}-KL^{2}\Leftrightarrow C'B^{2}-B'B^{2}+4KL^{2}= 4KB^{2}\\$[/quote]

Can you prove this equivalence? I have no idea  :oops:


Prove that for each positive real numbers x,y,z satisfying condition x+y+z = xyz holds the following inequality:

xy + yz + xz >= sqrt(x^2 + 1) + sqrt(y^2 + 1) + sqrt(z^2 + 1) + 3

What a pity that I do not have enough time to type the full solution. Having commented that, I will give a small hint
Set $ a\equal{}\frac{1}{xy} ....$

[quote="Anthony2"]Prove that for each positive real numbers x,y,z satisfying condition x+y+z = xyz holds the following inequality:

$ xy \plus{} yz \plus{} xz$≥$ \sqrt{x^2 \plus{} 1}\plus{}\sqrt{y^2 \plus{} 1}\plus{}\sqrt{z^2 \plus{} 1}\plus{}3$[/quote]
I have another idea.Let x=tanA,y=tanB,z=tanC(tanA,tanB,tanC>0) where ABC is a acute-angled triangle.The ineq is equivalent to:
$ tanAtanB\plus{}tanBtanC\plus{}tanCtanA$≥$ \frac{1}{cosA}\plus{}\frac{1}{cosB}\plus{}\frac{1}{cosC}\plus{}3$.This is easy to prove.


Tommorrow at 12:00 PM I will post the mock ARML contest I have composed. It will consist of 8 questions, each of which contains an integral answer. For those who want to take it, PM me the answers to them BY 7:00 PM. We will discuss the answers and solutions at around 7:30 PM. Although your testing obviously cannot be monitored, try taking it like the real ARML. The first two problems, then the second two, etc. You will have 12 minutes to complete each set of two as opposed to the normal 10 minutes because these problems will be tougher than the usual ARML problems. And NO CALCULATOR is allowed. Thank you.

We will put the top scorer and one honorable mention (for the second highest scorer) up after i have tallied the results. Hopefully we have atleast 10 people taking it, excluding me ofcourse.

Is there a prize?  You could give out prizes, since the winner *cough Joe cough* will most likely be at ARML.

Also, 7.5 hours isn't very much to work on the thing.  I don't get home until 6:00 some days, and then I don't go straight to my computer.  We're not all on Spring Break, Shobhit.  You should extend the deadline to give more people a chance to compete.

yea, i won't get home till 5:30 (or later if i go do something) maybe more people could do it if you waited till the weekend or extended the time period. and is this just for those who check out the SC section?

[quote="solafidefarms"]I'll do it so I can get just below HM. :D[/quote]

Huh?  Who's HM?

I won't be home until god knows when tomorrow since it's a track meet day.  And then I'll be working on this collaborative competition thingie...sorry Shobby.

HM=honorable mention

[quote="mysmartmouth"]
Huh?  Who's HM?[/quote]
Honorable Mention. and you were making fun of shobhit earlier  :roll:

edit: looks like joe beat me to it  :mad:

Okay then, answers can be submitted by Thursday!

but if people wanna submit them tommorrow, that's cool

[quote="furious"][quote="mysmartmouth"]David, don't make fun of people, even Shobhit.  Let's try to be nice and encourage the new SC members.  OK?[/quote]
i wasn't making fun i shobhit, i was making a pointed statement at you. but none the less, i will be nicer  :P[/quote]

Am I missing something? Is there something I did wrong? :huh:  :|

Anyway, my mock arml is posted and i've decided that people can turn them in any time this week, although you only have an hour to work on it, but Sean and Billy have turned their answers in so far and both of them got perfect scores on the relay :) :lol:

Whoa, no smiley limit anymore?

Sorry for all the smilies!  (I deleted them already).  I was testing Shobhit's comment on the smilie limit, and then AoPS got so slow it died.  I kept hitting submit, and then got distracted and left.  When I came back, I saw about 300 smilies, just waiting to be deleted.  :blush: Oooops!


In the quadrilateral $ ABCD$, $ N\in BC$ and $ Q \in AD$ so that $ \frac{NB}{NC}\equal{}\frac{QA}{QD}\equal{}\frac{AB}{CD}.$ Prove that $ NQ$ is parallel with the bisector of the angle formed by $ AB$ and $ CD.$

[hide]A good case for a solution without words.  :wink: [/hide]
Kostas Vittas.


The following questions are extended to people who have knowledge on psychology, specifically in reference to rational decisions and game theory

1. Do you think that the conscious decisions we make are rational or irrational (And thus the unconscious desicisions would be accordingly irrational or rational)?

2. (To those who have knowledge about the updates in the field) What is the more supported answer to the above question from the professional field?

I hope I'm not helping  you do homework by answering this. The short answer is that there is now extensive research on "cognitive illusions," (Google that term), and there is abundant evidence that most people are systematically IRrational in their decisions about many issues.

But you would know that if you read history, so what's new about this?

No, this isnt homework help. I recently read Mero's book Moral Calculuations, and the last few chapters really highlight this question.

Anyone else have any input?

And TokenAdult, by history, do u mean the history of psychology, or general history?

Unfortunately, I don't think most people know about this psychology stuff.  Could you explain the terms, or will it require a lot of background?

I finished reading Six Degrees by Duncan J Watts a few days/weeks ago. It said that people are inclined to go with the majority when making decisions.

Some guy did this experiment. There are 8 people in a movie theater-like room with a screen. On the screen there is a question, like 'which line is closer?', and the questions are so designed so that the answer is obvious. 1 of the 8 is a participant, and the other 7 are the guy's assistants. The 7 people say the 'wrong' answer, and most people (I forgot the fraction) hesitated and decided to go with the majority (pick the wrong answer). But, when there were fewer people in the room (2 or 3, maybe), the participants rarely hesitated.

For the attached picture, the answer the 7 people would say is A, and the participant would say A also, despite him/herself.

I think the term "rational" has to be carefully defined, because it has a lot of implications tied into it.  For example, I don't believe that an "economically rational" agent is really a rational agent.

Even without a good definition, however, I would tend to agree with Tokenadult that people are supremely irrational.

[quote="Duck-Billed Platypus"]I finished reading Six Degrees by Duncan J Watts a few days/weeks ago. It said that people are inclined to go with the majority when making decisions.

Some guy did this experiment. There are 8 people in a movie theater-like room with a screen. On the screen there is a question, like 'which line is closer?', and the questions are so designed so that the answer is obvious. 1 of the 8 is a participant, and the other 7 are the guy's assistants. The 7 people say the 'wrong' answer, and most people (I forgot the fraction) hesitated and decided to go with the majority (pick the wrong answer). But, when there were fewer people in the room (2 or 3, maybe), the participants rarely hesitated.

For the attached picture, the answer the 7 people would say is C, and the participant would say C also, despite him/herself.[/quote]
Isn't C closer to B in the picture? So wouldn't the wrong answer be A?

For the original question, it depends on what rational means. I think being rational means doing what you think is best to do. By my definition, the people in the experiment could have been rational. It may have made more sense not to alienate themselves from the group rather than to answer the question correctly. If a large sum of money was offered for a correct answer, I think more people would have chosen the correct answer.

I agree that the term rational needs some clarification. But from a philosophical point of view, Kant argued that each individual is first and foremost an emotion being. That is, our primary motivations for things are for emotional comfort, approval and satisfaction. So we make our decisions accordingly.

Intellectual, or rational, decisions can sometimes appeal to our senses of emotion; but the underlying motivations are always emotional, and they aren't always necessarily rational.

And then groups of people, collectively, can be extremely irrational, because there's some level of limited coordination of information, values, and motivation among the people. Plus a strong tendency for people to seek social acceptance.

[quote="confuted"][quote="JBL"]...I don't believe that an "economically rational" agent is really a rational agent.[/quote]
Why not? Isn't that saying a rational actor will always select the good with the highest marginal value per dollar to him at the time? I don't see how that's not rational, but I may have confused definitions.[/quote]

Well, you've got the right definition of economically rational: an economically rational person as one who always acts in his own best interest.  I don't think that this is a good definition at all, however.  Most notably, it often leads to a situation in which what is rational is strongly and directly to be contrasted with what is good.  In order to accept that people should always act rationally, I would require a definition of rational that also does not destroy basic moral principles.  I wouldn't want to spend any time around an economically rational person.

[quote="Yasha"]Isn't C closer to B in the picture? So wouldn't the wrong answer be A?[/quote]
Yes, sorry. It was a typo. I fixed it...


function
integration
advanced fields
advanced fields unsolved

1. Prove that if $ u \in W^{1,p}(0,1)$ for some $ 1 \leq p < \infty$, then $ u$ is equal a.e. to an absolutely continuous function, and $ u'$ (which exists a.e.) belongs to $ L^p(0,1)$.

2. Prove that if $ u \in W^{1,p}(0,1)$ for some $ 1 < p < \infty$, then

\[ |u(x) \minus{} u(y)| \leq {|x\minus{}y|}^{1 \minus{} \frac{1}{p}} {\left( \int_{0}^1 {|u'|}^p dt \right)}^{\frac{1}{p}}\]

1. $ v(x) : \equal{} \int^x_0 u'(t)\,dt$ is absolutely continuous on $ (0,1)$...

2. $ u(x) \minus{} u(y) \equal{} \int^x_y u'\,dt$ a.e. by 1., now use Hölder's inequality.


Could anyone help to show the expression for Log(z1/z2) is not always valid to Log z1 - Log z2 ?

I think this is an interesting question :) I'll think about it.

Also, you might want to have a look at here : http://mathforum.org/library/drmath/view/65417.html

if $z_1$ and $z_2$ get very close to $-1$ in the upper half plane, then $arg(z_1)+arg(z_2)\approx 2\pi$, hence $\neq arg(z_1z_2)$ which is supposed to be in $(-\pi,\pi]$ i guess. so $Log(z_1z_2)\neq Log(z_1)Log(z_2)$.

We can also have $z_1/z_2\in\mathbb{R}_{<0}$ with neither of $z_1,z_2\in\mathbb{R}_{<0}$, so that one is defined while the other two are not.  The reverse can also happen.


Find sequence real numbers $ x_{1},x_{2},...,x_{n}$ satisfying the equation:
$ (1 \minus{} x_{1})^2 \plus{} (x_{1} \minus{} x_{2})^2 \plus{} ... \plus{} (x_{n \minus{} 1} \minus{} x_{n})^2 \plus{} x_{n}^2 \equal{} \frac {1}{n}$

Let $ x_0\equal{}1, x_{n\plus{}1}\equal{}0,y_i\equal{}x_{i\minus{}1}\minus{}x_i,$, then $ \sum_i y_i\equal{}1$, because \[ \sqrt{\frac{1}{n\plus{}1} \sum_{i\equal{}1}^{n\plus{}1} y_i^2}\ge \frac{1}{n\plus{}1}\sum_{i\equal{}1}^{n\plus{}1}y_i\equal{}\frac{1}{n\plus{}1}\]
We get, that $ \sum_i y_i^2\ge \frac{1}{n\plus{}1}$. If we solve $ (1\minus{}x_1)^2\plus{}...\plus{}x_n^2\equal{}\frac{1}{n\plus{}1}$, then we get unique solution $ x_i\equal{}\frac{n\plus{}1\minus{}i}{n\plus{}1}$.
In your case we get infinetely many solutions.


function algebra domain calculus integration logarithms trigonometry

Assume we have a function defined interval-wise such that it is differentiable within the intervals. But one does not know of the differntiability at the end points.

Similarly f'(x) is defined interval wise. It is found that f'(x) is continuous throughout the domain of definition.

Then can we always say that f(x) is differntiable at the endpoints. I am having this doubt mainly because differntiability is not based on what the tangent tends but on the secant which tends to the tangent.

[quote="valeriummaximum"]i don think we can speak abt differentiability when considering a closed interval...

this is because even when the function is continous in closed interval.. continuity does not always imply differentiabilty..

i think this is why we cant speak abt diff at end points[/quote]

What are you blabbering da? I don't think you read the question properly. Obviously you can see LHD and RHD.

Come on guys.. One of the math genies answer this no???

[quote="Poincare"]How old are you CyberShreyas?[/quote]

102.7534732897432823 to be precise :D .He walks around with a walking stick on his own even at that age :o :10:

[quote="Poincare"]How old are you CyberShreyas?[/quote]

That's the weirdest and most irrelavant question, I've ever seen in this serious forum. But just to set the record straight. I'm in Grade 9 and just finished learning calculus especially integration, Gamma Functions, etc.. But I have such fundamental doubts. So I'm sorry.

Is anybody actually interested in answering???? Pardesi??

[quote="Cyber-Shreyas"][quote="Poincare"]How old are you CyberShreyas?[/quote]

That's the weirdest and most irrelavant question, I've ever seen in this serious forum. But just to set the record straight. I'm in Grade 9 and just finished learning calculus especially integration, Gamma Functions, etc.. But I have such fundamental doubts. So I'm sorry.
[/quote]

And you are Misan'd most beloved student arent you? :D And didnt you teach him a bit more abt Gamma fns? :D :rotfl:

[hide="counter-example"]
i suppose this is ur claim $ f$ is a function such that
1)$ f(x)$  is continious in intervals $ [x_{i},y_{i}]$
2)$ f'(x)$ exist in $ (x_{i},y_{i})$
3)$ f'(x)$ is continious in $ (x_{i},y_{i})$
then $ f'(x)$ is differentiable throught it's domain

consider the function $ f(x) \equal{} \sqrt {(x \minus{} 1)(x \minus{} 2)}$ ,$ x \in ( \minus{} \infty,1)\cup (2,\infty)$
clealry $ f(x)$ is continious wherever it is defined even at the end points
also $ f'(x)$ exists at all ponts except the endpoints and is continious
but $ f'(x)$ doesn't exist at the end points[/hide]

Here's one more good question:

Find $ \frac {d(\ln{(\ln{\sin{x}})})}{dx}$

chemrock is the ans the derivative doesnot exist? :maybe:

[because log of a negative number is not defined.......]

|Sinx| lies between (0,1) and lnsinx is obviously less than zero. So the given function has its domain as a null set.

Yes right. Summa was hoping for someone to fall into the trap.

This was the key point I was looking for.

$ x^2 \sin \frac {1}{x}$ (and defined to be 0 @ x=0)

 is differntiable at x=0 but its derivative is discontinuous at x=0. Clearly just because the derivative is discontinuous it does not mean that it does not exist. Simply because derivate is limit of the secant and not of the tangent.


When $ (a\plus{}b\plus{}c\plus{}d)^{10}$ is expanded and like terms combined, how many terms are in the result?

Thanks in advance!

We need to find the number of integer solutions to $ a\plus{}b\plus{}c\plus{}d\equal{}10$, because each unique term will have a number of terms whose exponents add to $ 10$.  We can use balls and urns to figure this out: $ \binom{10\plus{}4\minus{}1}{4\minus{}1}\equal{}\binom{13}{3}\equal{}\boxed{286}$ terms.

I don't understand, be more clear please.

What exactly don't you understand, enndb0x?

Each term of the expansion will be of the form $ C(a^xb^yc^zd^w)$ for some constant $ C$ and nonnegative integers $ x,y,z,w$ such that $ x\plus{}y\plus{}z\plus{}w\equal{}10$ (Why? This is the key insight.). The number of solutions to the described equation is $ \binom{10\plus{}4\minus{}1}{4\minus{}1}$, as AIME15 stated.

Could someone explain balls and urns to me? It's not in the wiki.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1323456#1323456]Balls and Urns. (Case 4)[/url]

Can anyone of you help me find more of these problems because I haven't gotten the handle of solving them yet.
I used the search button but I wasn't lucky to find something really good.


[quote="tennis123"]I am not sure where this post would go, so I am posting it here.
A while ago, I read this article on Yahoo! that was about some guy who found a way to quickly multiply big numbers in your head. He would mail you the explanation if you mailed $ \$$6. I was wondering if anyone knew anything about this. Sorry for the bad explanation, I don't remember that much.[/quote]

Don't trust him. I don't think he will give you the explanation.

Hey I can do that for free!! Ha ha ha!!


M is a 8-digit base ten positive integer having the form ABCDEFGH that uses each of the nonzero digits from 2 to 9 exactly once, and satisfies all of these conditions:

(i) AB is divisible by 3.
(ii) BC is divisible by 4.
(iii) CD is divisible by 5.
(iv) DE is divisible by 6.
(v)  EF is divisible by 7.
(vi) FG is divisible by 8.
(vii) GH  is divisible by 9.

Determine all possible value(s) that M can assume.

Some of the possible values that 'M' can assume are given below:

M : [b] 7 2 8 5 4 9 6 3 [/b]

M : [b] 8 7 2 5 4 9 6 3 [/b]

These are two values of 'M' I had obtained that conform to all the 'rules' as stated in the problem.

For obvious reasons, D=5. Then, E=4 (54), F=9 (49) or F=2 (42), G=6 if F=9, G=4 if F=2. But H=5 if G=4, contradiction.

So, the numbers we have are:

54963

with 2,7,8 left.

C is not 7, for obvious reasons.
If C=2, then B=7, A=8
If C=8, then B=2, A=7

Giving us the answers Gen8 has posted.


Hi everyone,

This is not an official announcement, rather something we are just throwing out there that you should look for in the coming months. The Lexington High School Math Team is thinking of sponsoring a Middle School Math Competition, in a similar style to HMNT/HMMT (more like HMNT). Perhaps with a PUMaC-style (taken abroad across the week or so before) power round. A lot of things have to be worked out, but a very optimistic date that we are looking to is Saturday, February 27.

I reiterate: this is not an official announcement, and don't bother marking your calendars. Just be aware that this is happening, and spread the word. There are many things we need to work out, but we're hoping as much as possible that this will ultimately work (we won't start any serious work on this until after our senior captains finish their January 1 College Applications).

Also, Phillips Exeter Academy is running a similar competition on (I believe) January 31. This is all I know about it - contact someone there, I guess, if you're interested.

This is great to hear.  Please keep us posted here.

Tentative date: April 3, 2010 (Saturday)

Hopefully, our website will be up and running soon. We will then probably open something resembling 'pre-registration', so we can get an idea of how many teams are interested and thus set the cost (which will hopefully be at most something like $ \$50$ per team of 6 and $ \$10$ per individual). We wlll look for some sponsors to cut this cost.

Can someone kindly post here when the website or registration is open?  I've been checking http://www.lhsmath.co.cc/ regularly.

lhsmath.co.cc/lmt

Registration has just been opened.

Thank you, Carl (and the rest of the LHS team).

Would you care to compare the intended difficulty with the recent Exeter Math Club Competition ( http://peamath.dyndns.org/emc2/problems.php ) and/or any levels of AMC or MATHCOUNTS?  Many were somewhat taken aback by the difficulty of the EMC2 problems.  I don't mean that's a bad thing, just good know in advance, especially when recruiting teams.

We're aware that EMC2 was too hard and some EMC2 staff agreed. We've made a conscious effort to ensure that the difficulty will be gentler, while keeping a challenge for the most advanced students.

Congratulations to the LHS team for planning a fabulous tournament and executing it beautifully.  I never imagined that it would go that well, from the excellent mix of beautiful problems, to the very hospitable staff, to the smoothness with which you carried it all off, making it [i]look [/i]almost effortless.

I don't know if Saif has recovered from his all-nighter preparing for the tournament or his other work which surely got backed up.  Do you know when results, problems, or solutions might be posted to your website?  Do you know when individual team results will be or were email to coaches?

Thank you, again.  I expect that next year there will be two to four times as many teams entering, as there is a strong local buzz about how much fun your tournament was.

I'm glad to hear it, and we appreciate the comments. All results (not places, unfortunately, as I can't get them from the database without counting by hand) have been e-mailed to coaches. Solutions don't exist yet, and the problem czar (me) is working on getting them during the week of April 19. Problems and top scorers, i.e. the ones that were announced, should go on the website soon, but Saif may want to wait for everything to go up at the same time. Also during the April 19 we're planning some kind of brute force party in which we'll do an exhaustive item analysis on everything and produce pretty graphs. Stay tuned.

But here are the results that we read and a little more. Note that the overall team places were wrong when we read them; fields got screwed up and we had the correct scores, but the wrong order of teams because we somehow got the order of total individual scores for the teams. Awards will be adjusted accordingly (basically, we'll buy Diamond a 4th place plaque). Anyway...

Individual Round
1. Jonathan Tidor, Old Clarke (54)
2. Zachary Polansky, Sage (48)
T3. Zach Lowry, Old Clarke (45)
T3. Ying Gao, Bigelow (45)

Theme Round
1. Jonathan Tidor, Old Clarke (56)
T2. Shohini Stout, Diamond's Best Friends (48)
T2. Noah Golowich, Diamond 1 (48)
T2. Anna Ellison, Bigelow (48)

Overall Individual
1. Jonathan Tidor, Old Clarke (110)
2. Zachary Polansky, Sage (92)
3. Anna Ellison, Bigelow (90)
T4. Ying Gao, Bigelow (89)
T4. Zach Lowry, Old Clarke (89)
6. Rohil Prasad, Old Clarke (86)
7. Arthemy Lugin, AMSA (83)
8. Katie Fraser, Old Clarke (82)
T9. Noah Golowich, Diamond 1 (81)
T9. Shohini Stout, Diamond's Best Friends (81)

Team Round
1. Sage (190)
2. Old Clarke (187)
3. Bigelow (175)
I also note a very impressive showing by Individual 1, who scored 174 and also had the highest score on Part 2 (the long answer part).

Guts Round
1. Old Clarke (252, which is a perfect score minus 34-36)
2. Clockers (204)
3. Bigelow (196)

Overall Team
1. Old Clarke (957.5)
2. Bigelow (779.333)
3. Sage (756)
4. Diamond 1 (703.333)
5. New Clarke (702.5)
6. AMSA (646.167)

I notice something rather fishy about Anna Ellison's scores.

yeah ... Anna should have 93 points and 2nd place(according to the individual and theme round scores)

hmm nice you did a middle school math contest

and you seem to have pulled it off better than my school did last year :P good job

Good catch, she actually had a 42 on the individual round (manual copying error on my part).


Let a,b be positive integers and $\{x_n\}$ sequence s.t.:
$x_0=0,x_1=1$ and $x_{2n}=ax_{2n-1}-x_{2n-2},x_{2n+1}=bx_{2n}-x_{2n-1}$
for $n\geq 1$.Show that:for any $m,n$ then $x_{n+m}x_{n+m-1}...x_{n+1}$ is divisible by $x_mx_{m-1}$

in fact,this is a question for 2005 china national team members.
to prove it,we need to prove such a proposition:
  if a series of numbers{Xn} have the the charcter that for any prime number "p" and positive interger"k",
  the set{i:m is divisor of Xi}(m=p*p*……*p, there are "k"  'p's in total) equals to either nonnull or
  the set{d,2d,3d……},where d is a positive interger in relation to "p" and"a".
  then X1*X2*……Xm is divisor of  X(m+n)*X(m+n-1)*……*X(n+1),for any positive interger"m"and"n".
after proving this proposition,we just need to prove that the series of numbers suit all the reqirement
in the proposition.
(since my English is poor and I don't know how to type mathematical marks on the internet,
 I have to write briefly,sorry)  :blush:

[quote="newcomer"]in fact,this is a question for 2005 china national team members.[/quote]
But SPB MO 2001 was earlier, than China TST 2005 ;)

You really make me interest!I will be very glad to see its solution...


inequalities
function
search
induction
inequalities open

Is this inequality true ? \[\sum (\frac{a}{b+c})^{n}\geq \frac{3}{2^{n}}\] where $a,b,c>0$ ,$n\in \mathbb N$
If it's true ,could anyone give me a link (i did't find anything) or post a solution ?

yes i suppose

let's normalise to $a+b+c$ $=$ $1$ $\imp a,b,c<1$
define a function

$f(x)$ $=$ $\frac{x^{n}}{(1-x)^{n}}$ for $x$ $(0,1)$

now
observe $f$ is convex in the above interval

so

$\frac{f(a)+f(b)+f(c)}{3}$ $\geq$ $f(\frac{a+b+c}{3})$ $=$ $\frac{1}{2^{n}}$

[quote="Svejk"]But how can you assume that $a+b+c=1$ ?[/quote]

since the equation is homogeneous u can normalize it

$a+b+c$ $=$ $n$

so,
 $\sum_{cyc}\frac{a}{n}=1$

put $\frac{a}{n}=u$
      $\frac{b}{n}=v$
      $\frac{c}{n}=w$
so the inequality becomes

$\sum_{cyc}\frac{u^{n}}{(1-u)^{n}}\geq \frac{3}{2^{n}}$

where ,$u+v+w = 1$

instead we can write without any loss of generality as

$\sum_{cyc}\frac{a^{n}}{(1-a)^{n}}\geq \frac{3}{2^{n}}$

I'm not very familiar with this kind of proof.Very nice .Thanks a lot

welcome :lol:.well if u really won't to read abou inequalities read it from HOOJOO LEE's book it's available in pdf format on the net .do a google search for it

$\left( \frac{a}{b+c} \right)^{n} +\frac{n-1}{2^n}\ge n\sqrt[n]{\left( \frac{a}{b+c} \right)^{n}\left( \frac{1}{2^n}\right)^{n-1}}=\frac{n}{2^{n-1}}\cdot \frac{a}{b+c}$

$\implies\sum_{cyc} \left( \frac{a}{b+c} \right)^{n} \ge \sum_{cyc} \left( \frac{n}{2^{n-1}}\cdot \frac{a}{b+c}- \frac{n-1}{2^n}\right)$

$=\frac{n}{2^{n-1}}\cdot\sum_{cyc} \frac{a}{b+c} - \frac{3(n-1)}{2^n}\ge \frac{n}{2^{n-1}}\cdot \frac{3}{2} - \frac{3(n-1)}{2^n}= \frac{3}{2^n}$.

$\implies(\frac{a}{b+c})^{n}+(\frac{b}{c+a})^{n}+(\frac{c}{a+b})^{n}\geq \frac{3}{2^{n}}$

The following inequalities are also true.
Let $a, b, c, p, q$ be positive real numbers, $n$ be positive integer. Prove that:
$\biggl(\frac{a}{pb+qc}\biggl)^n+\biggl(\frac{b}{pc+qa}\biggl)^n+\biggl(\frac{c}{pa+qb}\biggl)^n\geq \frac{3}{(p+q)^n}$

[quote="sqing"]Let $a, b, c, p, q$ be positive real numbers, $n$ be positive integer. Prove that:
$\biggl(\frac{a}{pb+qc}\biggl)^n+\biggl(\frac{b}{pc+qa}\biggl)^n+\biggl(\frac{c}{pa+qb}\biggl)^n\geq \frac{3}{(p+q)^n}$[/quote]
$\left(\frac{a}{pb+qc}\right)^n+\left(\frac{b}{pc+qa}\right)^n+\left(\frac{c}{pa+qb}\right)^n\ge\frac{1}{3} (a^n+b^n+c^n) \left[\left(\frac{1}{pb+qc}\right)^n+\left(\frac{1}{pc+qa}\right)^n+\left(\frac{1}{pa+qb}\right)^n\right]$

$\ge \frac{(a+b+c)^n}{3^n}\frac{\left(\frac{1}{pb+qc}+\frac{1}{pc+qa}+\frac{1}{pa+qb}\right)^n}{3^{n-1}}$

$\ge \frac{3(a+b+c)^n}{9^n} \frac{9^n}{(p+q)^n(a+b+c)^n}=\frac{3}{(p+q)^n}$

[quote="sqing"]The following inequalities are also true.
Let $a, b, c, p, q$ be positive real numbers, $n$ be positive integer. Prove that:
$\biggl(\frac{a}{pb+qc}\biggl)^n+\biggl(\frac{b}{pc+qa}\biggl)^n+\biggl(\frac{c}{pa+qb}\biggl)^n\geq \frac{3}{(p+q)^n}$[/quote]

my dear friend seems that your inequality holds for all real n such that

$n \in (-\infty,-1] u [1,+\infty)u \{0\}$

[quote="joybangla"][quote="sqing"]The following inequalities are also true.
Let $a, b, c, p, q$ be positive real numbers, $n$ be positive integer. Prove that:
$\biggl(\frac{a}{pb+qc}\biggl)^n+\biggl(\frac{b}{pc+qa}\biggl)^n+\biggl(\frac{c}{pa+qb}\biggl)^n\geq \frac{3}{(p+q)^n}$[/quote]
Lemma:
$\frac{a}{pb+qc}+\frac{b}{pc+qa}+\frac{c}{pa+qb}$ $=\frac{a^2}{pab+qca}+\frac{b^2}{pbc+qab}+\frac{c^2}{pca+qbc}$ $\ge \frac{(a+b+c)^2}{(p+q)(ab+bc+ca)}$ $\ge \frac{3}{p+q}$ since    $(a+b+c)^2\ge 3(ab+bc+ca)$.
Now coming to the problem use Holder's inequality for the sequences $\left(\frac{a}{pb+qc},\frac{b}{pc+qa},\frac{c}{pa+qb}\right)$ and $(1,1,1)$ with exponents $\left(n,\frac{n}{n-1}\right)$ So the inequality becomes:
$\left(\frac{a^n}{(pb+qc)^n}+\frac{b^n}{(pc+qa)^n}+\frac{c^n}{(pa+qb)^n}\right)^{\frac{1}{n}}$ $\cdot 3^{\frac{n-1}{n}}$ $\ge \left(\frac{a}{pb+qc}+\frac{b}{pc+qa}+\frac{c}{pa+qb}\right)$ $\implies \left(\frac{a^n}{(pb+qc)^n}+\frac{b^n}{(pc+qa)^n}+\frac{c^n}{(pa+qb)^n}\right)\cdot 3^{n-1}$ $\ge \left(\frac{3}{p+q}\right)^n$ $\implies \left(\frac{a}{pb+qc}\right)^n+\left(\frac{b}{pc+qa}\right)^n+\left(\frac{c}{pa+qb}\right)^n \ge \frac{3}{(p+q)^n}$.Hence proved.[/quote]

@joybangla ur proof is nice. but u could do it easily with power mean.

$\left[\left(\frac{a}{pb+qc}\right)^n+\left(\frac{b}{pc+qa}\right)^n+\left(\frac{c}{pa+qb}\right)^n\right] \ge \frac{\left(\frac{a}{pb+qc}+\frac{b}{pc+qa}+\frac{c}{pa+qb}\right)^n}{3^{n-1}} \ge \frac{3}{(p+q)^n}$


The trisectors of angles $ B$ and $ C$ of scalene triangle $ ABC$ meet at points $ P$ and $ Q$, as shown. Angle $ A$ measures 39 degrees and angle $ QBP$ measures 14 degrees. What is the measure of angle $ BPC$?

[asy]draw((0,0)--(1.2,1.5)--(1,0)--cycle,linewidth(1));
draw((0,0)--(.8,.6)--(1,0),linewidth(1));
draw((0,0)--(.7,.2)--(1,0),linewidth(1));
label("B",(0,0),W);
label("A",(1.2,1.5),N);
label("C",(1,0),E);
label("Q",(.8,.6),N);
label("P",(.7,.2),N);[/asy]

If $ QBP$ is $ 14$, $ ABC$ is $ 42$.  Thus, $ ACB$ is $ 99$.  Divide this by $ 3$, for the answer of $ \boxed{33}$.

EDIT: Wrong angle.

Ok, so we have a triangle with $ 14^o$ and $ 33^o$ degree angles.  Thus, the third angle is ${{ \boxed\boxed\boxed{133}}}$.


geometry
rectangle
analytic geometry
trigonometry
inequalities
triangle inequality
similar triangles

In rectangle ABCD, we have AB=6 and BC=4. Draw a semicircle with DC as its diameter. P is a point on the semicircle.

Find the length of DP when AP is maximized.

is the semicircle drawn inside the rectangle or outside?
if inisde P will coincide with C and DP =6

coordinate bash, neh?
a(-3,4), b(3,4), c(3,0), d(-3,0)
so P is at (3cos x, 3sin x) with pi<=x<=2pi. Then AP=$ \sqrt{(\minus{}3\minus{}3\cos{x})^2\plus{}(4\minus{}3\sin{x})^2}\equal{}\sqrt{34\plus{}18\cos{x}\minus{}24\sin{x}}$. The maximum value of this is $ \sqrt{34 \plus{}6k}$ , where k is the maximum value of 3cosx -4sin x (pi<=x<=2pi). This is achieved when cosx/sinx=-3/4, and cos x=3/5, and sinx=-4/5.
So DP=$ \sqrt{(\minus{}3\minus{}3\cos{x})^2\plus{}(0\minus{}3\sin{x})^2}\equal{}\dfrac{12\sqrt{5}}{5}$
I hope that's right...

Lol.

[hide="Alternatively"]Draw the whole circle with diameter $ CD$, and let the second point of intersection with the circle (closer to $ A$) be $ X$. By Power of a Point, $ AX\cdot AP\equal{}AD^2\equal{}16\implies AP\equal{}\frac{16}{AX}$, so we want to minimize $ AX$ to maximize $ AP$. The shortest distance from a point outside a circle to a circle is the distance to the center minus the radius (so the point is the intersection closer to the point of the center and the point; we can use the Triangle Inequality to show that the distance is greater if any other point on the circle is chosen). Thus, $ AXP$ passes through the center of the circle (call it $ O$), $ AX\equal{}2$, $ XO\equal{}3$, $ OP\equal{}3$. Drop the perpendicular from $ P$ to $ AD$ can call the foot $ E$. Similar triangles give $ DE\equal{}\frac{12}5$ and $ PE\equal{}\frac{24}5$, so $ DP\equal{}\boxed{\frac{12\sqrt5}5}$.[/hide]


How many times does the digit 1 appear in base 10 in the number $ N=9+99+999\cdots999\cdots99$, if the last addend has 1998 digits?

N = 9 + 99 + 999 + ... + 99...99999 where the last addend has 1998 digits
N = (10 - 1) + (100 - 1) + (1000 - 1) + ... + (100...00000 - 1)
N = 10 + 100 + 1000 + ... + 100...00000 - 1998
N = 111...11110 (1999 digits all being 1 except the one's digit) - 1998
N = 111...09112 (A number with 1999 digits as stated above, all but three being 1)
N will have 1996 occurrences of 1, with the exceptions being in the one's, thousand's, and ten-thousand's places.


A number multiplied by three is seven more than half the number. What is the number:?:  :(

Two numbers sum to 36. The difference between the numbers is 71. What are both numbers :?:

The lines 12(x - y) = 7 and 30x -12y + 23 = 0 intersect in which quadrant. Show me your work.

Give an equation of the line that passes through the point (1, 4) and is parallel to y- x = 0

Variables...? :huh:

[hide="Solutions"]
1. Let $m$ denote the number we seek. Given: $3m-7=\frac{1}{2}m$. Solving this yields $m=\frac{14}{5}$, which is true by plugging in, if you want to.

2. Call the two numbers $a$ and $b$. Given: $a+b=36$, $a-b=71$. Adding those two yields $a=\frac{107}{2}$. Plugging that into one of the two equations, $b=-\frac{35}{2}$.

3. $12x - 12y - 7 = 0$ and $30x -12y + 23 = 0$. Adding the additive inverse of the first to the second yields $18x=-30 \equiv x=-\frac{5}{3}$. Plugging that in: $-20 - 12y - 7 = 0 \equiv y=-\frac{27}{12}$. The point $\left(-\frac{5}{3},-\frac{27}{12}\right)$ lies in the third quadrant.

4. Parallel to $x=y$, so it's of the form $(y-y_1)=(x-x_1)$. Filling in the point $(1,4)$ yields $y=x+3$.
[/hide]


Each person marked the table with an X on the day(s) he/she would not be able to attend a meeting.  On which day can the most people attend a meeting?

\begin{tabular}{|l|c|c|c|c|c|}
\multicolumn{6}{c}{}\\\hline
&Mon&Tues&Wed&Thurs&Fri\\\hline
Anna&X&&X&&\\\hline
Bill&&X&&X&X\\\hline
Carl&X&X&&X&X\\\hline
\end{tabular}

Since X means that a person can't go to the meeting, you want the least # of X's for a day.
There are 2 X's for Mon, Tues, Thurs, and Fri and 1 X for Wednesday.

[b]WEDNESDAY[/b]

$ \begin{tabular}{|l|c|c|c|c|c|}
\multicolumn{6}{c}{}\\\hline
&Mon&Tues&Wed&Thurs&Fri\\\hline
Anna&X&&X&&\\\hline
Bill&&X&&X&X\\\hline
Carl&X&X&&X&X\\\hline
\end{tabular}$

For your convenience.


Alright, I have a lot of questions from past contests that I'm unsure of how to do. Some of them are very trivial and I'm not getting them...

1) A particular compound contains only C, H, and O. A 1.5g sample of this compound was burned completely in oxygen to give 1.738g CO2 and 0.711g H2O. What is the empirical formula of this compound?

Answer is C2H4O3, but I don't understand why. I found the moles of carbon dioxide and water, both come out to around 0.035. Unsure of what to do after that though...

2) In which salt do the ions have the  same total number of electrons?
A. MgS
B. BeO
C. NaCl
*D. MgO
E. KBR

I have no idea what this question is asking.

-------
Thanks in advance. I'll probably have more questions later on lol.

2. The way it was worded was a bit tricky, but basically in which of the compounds do the ions that make up the compound have the same number of electrons? MgO is made up of Mg2+ and O2-, both which have the same number of electrons as Ne.


Show that any polyhedron has 2 faces with same number
of edges.

I think I solved this before, though I'm not sure. From what I remember Dirichlet's principle works.

Consider the face (if more than one , then one of them) which have the maximal number of vertices , and let n be this number. Then there are n faces which have a common edge with this one , and together with this one that makes at least n+1 faces of the polyhedron , so there are at least two with the same no of vertices.


Euler
function
number theory proposed
number theory

Let k is a natural number
Conside a sequence as follow
$ a_{1}\in N$
$ a_{n\plus{}1}\equal{}\varphi{(ka_{n})}$
Find all k such that for each $ a_{1}$ the sequence $ a_{n}$ bound.

Let $ P\equal{}\{p_{i}\}$ set of primes, suth that $ p_{i}|k$ or $ p_{i}|\phi (k)$ or ... $ p_{i}|\phi^{(l)}(k)$ ( l - times).
Then sequence bounded if and only if $ g(k)\equal{}k\prod_{i}(1\minus{}\frac{1}{p_{i}})\le 1$. Obviosly g(k) is multiplicative and if $ k|m$, then $ g(k)\le g(m)$, therefore sufficiently to chek k=p (p prime). If $ p>3$ g(p)>1, therefore work only k=1,2,3.

A same problem
$ a_{n\plus{}1}\equal{}\varphi({a_{n}})$
Find all $ a_{1}$ satisfy the sequence doesn't contain any perfect square.

Because $ \phi(n)<n$ exist k, suth that $ a_{k}\equal{}1$ - perfect square. :lol:

I am sorry
$ a_{n\plus{}1}\equal{}\tau (a_{n})$
Determine all $ a_{1}$ such that the sequence doesn't contain any perfect square.


Let $ \mu$ be a measure on an algebra $ \mathfrak{A}$ and $ E$ a set with $ \mu^{*}E<\infty$. If $ \beta$ is any real number with $ \mu_{*}E \leq \beta \leq \mu^{*}E$, extend $ \mu$ to a measure $ \bar{\mu}$ on a $ \sigma$-algebra containing $ \mathfrak{A}$ and $ E$ such that $ \bar{\mu}E\equal{}\beta$.

$ \mu^{*}$ refers to the associated outer measure and $ \mu_{*}$ refers to the associated inner measure.


Guys, one again help me do this.

I find it really complicated.

Kindly show the representations and also the complete solution.

THANKS ;]

[hide="hint"]
Define the velocity of emptying the tank for each pipe. Now if they work together what is the velocity the tank is being emptied?[/hide]
[hide="solution"]Let $ t_1$ and $ t_2$ be the required times.

$ (1) \; t_1\equal{}t_2\plus{}10$

$ v_1\equal{}\frac{V}{t_1} \\
\\
v_2\equal{}\frac{V}{t_2}$

Where $ V$ is the volume of the water tank.

Let $ T$ be the time the pipes need to empty the tank together. Then:

$ v_1\cdot T \plus{} v_2\cdot T\equal{}V$

$ T(v_1\plus{}v_2)\equal{}V$

$ T(\frac{V}{t_1} \plus{} \frac{V}{t_2}) \equal{} V$

$ T(\frac{1}{t_1} \plus{} \frac{1}{t_2})\equal{}1$

$ (2) \; \frac{1}{t_1} \plus{} \frac{1}{t_2}\equal{} \frac{1}{12}$

Now solve $ (1)$ and $ (2)$ to get:

$ \boxed{t_1\equal{}20, \;
t_2\equal{}30}$[/hide]

Now try this one, its a little bit harder...

At first two different pipes fill a tank in $ \frac{20}{3}$ min, then both pipes are adjusted.

When running alone and at their new rate, the first pipe fills the tank $ 1$ min faster and the second pipe fills the tank $ 2$ min solwer.

Together at their new rates, the pipes fill the tank in $ 7$ min.

How long does it take each pipe to fill the tank alone at their new rates?


Let $ ABC$ be an triangle with $ AB\equal{}AC$, $ \angle BAC\equal{}\alpha$ and $ M$ the midpoint of $ BC$. Let be $ E \in AM$ such that $ M\in [AE]$ and $ \angle CBE\equal{} \frac{\alpha}{4}$. Fiind $ \alpha$.

M is the middle of AE too?If yes angle A=90


Could you please introduce some good books or websites which talk about the history of geometry in a brief way?

[url=http://en.wikipedia.org/wiki/History_of_geometry]Wikipedia[/url] has a good article that covers everything from 3000 BC to now. Another online history can be found at the [url=http://plato.stanford.edu/entries/geometry-19th/]Stanford Encyclopedia of Philosophy[/url]. Although the latter site only covers the 19th century, it contains a pretty big list of resources and other suggested readings.


geometry
3D geometry
dodecahedron
trigonometry
perimeter
analytic geometry

How would one find the area of a dodecahedron or such figures??

Depending on what kind of dodecahedron it is, you could divide it into triangles, and find the area of those, or if it is a regular dodecahedron, there's a nice proof shown in one of the wordless proofs under the left bar of the page.

However, this all depends on how much information you have. As the figures get more and more complex, it takes more and more info. to find the area.

If you would like to find the area of a regular polygon with a high number of sides, a good approximation is the area of the circumscribed circle. This approximation has less and less error as the number of sides increases.

For a $ n$-gon, the area is the distance from the center to one of the sides times half the perimeter.

that is assuming it is regular
(all angles and sides are the same)
here is the ultimate formula for the area of a regular n-gon with side x
$ \frac{nx^2}{4tan(\frac{180}{n})}$
this is because the distance from the center of a n-gon to the midpoint of a side (called the apothem) is
$ \frac{x}{2tan(\frac{180}{2})}$ (found by applying trig rules and the fact that an angle measures $ 180 \cdot \frac{n\minus{}2}{n}$)
and half the perimeter is $ \frac{nx}{2}$ and multiplying them gives the first formula mentioned

Would you elaborate on what you wrote under the formula?

2 points:

Wouldn't the apothem change, depending on which side you measure on an irregular n-gon?

And what does your thing in the parentheses mean?

Well if it's on a coordinate, you can use pick's.
shoelace is also good


Show that if $ a|b$, $ b|c$, and $ c|a$, then $ a\equal{}b\equal{}c$.

Yeah, that seems really easy, im just stupid.

Well, if $ a$ is a divisor of $ b,$ and $ b$ is a divisor of $ c,$ then $ a$ is a divisor of $ c$.  It is stated that $ c$ divides $ a$, so if $ a$ divides $ c$ and $ c$ divides $ a$, $ a \equal{} c$.  If $ a$ divides $ b$, and $ b$ divides $ c$, because $ a\equal{}c$, we can substitute $ a$ for $ c$ and get: "if $ a$ divides $ b$, and
$ b$ divides $ a.$"  For this to be true, they must be equal.

[quote="t0rajir0u"]This is only true if you require that $ a, b, c$ all have the same sign.[/quote]

That's a good point.

[quote="AdithyaGanesh"]Well, if $ a$ is a divisor of $ b,$ and $ b$ is a divisor of $ c,$ then $ a$ is a divisor of $ c$.  It is stated that $ c$ divides $ a$, so if $ a$ divides $ c$ and $ c$ divides $ a$, $ a \equal{} c$.  If $ a$ divides $ b$, and $ b$ divides $ c$, because $ a \equal{} c$, we can substitute $ a$ for $ c$ and get: "if $ a$ divides $ b$, and
$ b$ divides $ a.$"  For this to be true, they must be equal.[/quote]

Thanks :)

[quote="Thunder365"]Show that if $ a|b$, $ b|c$, and $ c|a$, then $ a \equal{} b \equal{} c$.
[/quote]

Yes, mandate that all of these quantities must have the same sign.
Suppose we look at the case where $ a, b,$ and $ c$ are all
positive integers.

$ a|b$ means that $ b \equal{} ak_1$, for a positive integer $ k_1$.  . . . (*)

$ b|c$ means that $ c \equal{} bk_2$, for a positive integer $ k_2$.

Then, using (*), $ c \equal{} bk_2$ becomes

$ c \equal{} ak_1k_2$. . . . (**)

$ c|a$ means that $ a \equal{} k_3c$, for a positive integer $ k_3$.

Then, using (**), $ a \equal{} k_3(a)k_1k_2$, or

$ a \equal{} ak_1k_2k_3$.

Divide by $ a$ on each side:

$ 1 \equal{} k_1k_2k_3$.

Because the product of three positive integers on one side equals one,
then it must be the situation that $ k_1 \equal{} k_2 \equal{} k_3 \equal{} 1$.

Then $ b \equal{} ak_1$ becomes $ b \equal{} a$, and $ c \equal{} bk_2$ becomes $ c \equal{} b$.

Because of transitivity, $ b \equal{} a$ together with $ c \equal{} b$ means that $ c \equal{} a$.

Therefore, $ a \equal{} b \equal{} c$.


calculus
integration
function
trigonometry
limit
logarithms
complex analysis

Without the use of complex methods or any non-elementary functions compute:
\[ \int_{0}^{\infty} \cos{x^2} dx\]

I'd sure like to see a more elementary solution... The most simple way I could remember is:

partial integration of $ \frac {1}{2}\int_{0}^{\infty} \frac {\cos x}{\sqrt {x}}dx$ with taking into account $ \frac {1}{\sqrt {x}} \equal{} \frac {1}{\sqrt {2\pi}}\int_{0}^{\infty}e^{ \minus{} xy^2}dy$. If this last integral is a problem (as you said, you don't want any non-elementary functions), after all, it's the gaussian integral, provable in many ways...

[quote="beezees222"]Without the use of complex methods or any non-elementary functions compute:
\[ \int_{0}^{\infty} \cos{x^2} dx
\]
[/quote]
is this acceptable to you ?   :D i did not use the residue theorem in complex analysis .

we know,

$ \int_{\minus{}\infty}^{\infty} e^{\minus{}a x^2} dx \equal{} \sqrt{\frac{\pi}{a}},\qquad \forall a\in\mathbb C$

put $ a\equal{}\imath$ to get

$ \int_{\minus{}\infty}^{\infty} e^{\minus{}\imath x^2} dx \equal{}   \int_{\minus{}\infty}^{\infty}\; \left[ \cos (x^{2}) \minus{} \imath\; \sin (x^{2}) \right]\;dx \equal{} \sqrt{\frac{\pi}{\imath}} \equal{} \sqrt \pi \cdot \sqrt {\minus{}\imath }$

$ \equal{} \sqrt \pi \cdot e^{\minus{}\imath \;\frac{\pi}{4} } \equal{}\sqrt \pi \; \frac{1}{ \sqrt 2} ( 1\minus{} \imath )$

equate [b]real[/b] parts to get

$ \int_{\minus{}\infty}^{\infty}\;  \cos (x^{2}) \;dx \equal{} \sqrt \pi \; \frac{1}{ \sqrt 2}$

$ \Rightarrow \int_0 ^{\infty}\;  \cos (x^{2}) \;dx \equal{} \sqrt \pi \; \frac{1}{ 2\sqrt 2}\equal{}\boxed{\sqrt { \frac{\pi}{8}} }$

All right, that's not a convincing solution. In particular, that identity isn't valid for all complex numbers, since convergence matters. It's only valid on the right half-plane, and even that requires complex analysis to justify. Convergence on the boundary is another complication- that's an Abel-type theorem. That means you also need a separate argument that the integral converges at all.

This calculation requires at least multivariable techniques; either real multivariable tricks as in hsiljak's post or complex analysis, most simply a contour integral of $ e^{ix^2}$ around a $ 90^{\circ}$ wedge. You're not going to get it any simpler than that, since the function doesn't have an elementary antiderivative.

lol, i understand this can be done using complex methods (using residue theorem or not). i have a solution that does not use this in anyway but is a little bit more long winded. let me know if you want to see a peek.
and yes. multivariable is a part of my solution and in my opinion does fall under the realm of elementary methods.

[hide]
Let ${ I = \int_{0}^{\infty} \cos{x^2}} dx$
First a lemma.
Lemma:
We make note of the famous Gaussian Integral
\[ \int_{0}^{\infty} e^{-y^2} dy = \frac{\sqrt{\pi}}{2}\]
A proof may be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213041
or here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212555.

Now we will consider the double integral which represents the volume lying between the surface $ z = e^{-y^2} \cos{x^2}$ and $ x,y=0$.
\[ V = \iint_{{\Re^+}^2} e^{-y^2} \cos{x^2} dx dy =\int_{0}^{\infty} \left(\int_{0}^{\infty} e^{-y^2} dy\right) \cos{x^2} dx = I\frac{\sqrt{\pi}}{2} \]
which is given by the lemma.
Next we proceed to calculate the same volume by changing to polar coordinates.
\[ V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} e^{-\sin^2{\theta}} \cos{(\cos^2{\theta})} r dr d\theta = \int_{0}^{\frac{\pi}{2}} \sec^2{\theta} \int_{0}^{\infty} e^{-u\tan^2{\theta}} \cos{u} du d\theta\]
Now by integrating by parts with respect to u we end up with
\[ V = \int_{0}^{\frac{\pi}{2}}  \sec^2{\theta} \frac{\tan^2{\theta}}{1 + \tan^4{\theta}} d\theta.\]
Let $ v = \tan{\theta}$ to get
\[ \int_{0}^{\infty} \frac{v^2}{1 + v^4} dv\]
This integral can be broken up using high school (maybe harder) partial fraction decomposition methods to yield
\[ V = \lim_{b \rightarrow \infty} \left(\frac{1}{2\sqrt{2}} \arctan{\frac{x^2-1}{x\sqrt{2}}} + \frac{1}{4\sqrt{2}} \ln{\frac{x^2-x\sqrt{2} + 1}{x^2-x\sqrt{2} + 1}}\right) ^{b}_{0}= \frac{\pi \sqrt{2}}{8}\]
We equate and end up with $ I = \frac{\sqrt{2\pi}}{4}$.
QED
[/hide]

An important issue with that: you don't have absolute convergence. Without absolute convergence, you don't have Fubini's theorem, and the change of variables might not work.

absolute convergence with respect to which integral?

Absolute convergence of the whole double integral- that's what matters.

One way to get around this is to integrate by parts first, transforming the conditionally convergent integral you started with into something with $ x^2$ in the denominator.


function
logarithms
limit
complex analysis
calculus
calculus computations

People were posting a problem, which included as part of it,
the lim (as x -> 0) of x^x, in another site.  However, that
includes the lim (as x -> 0-) of x^x.  I took that not to
exist because there is, supposedly, at least one value
different from 1 (that we get on the right-hand side limit)
that is being approached on the left-hand side of x = 0.

I am okay with the lim (as x-> 0+) of x^x side of it, but not
with trying to take the limit on the other side.

$ x^x$ isn't even well-defined as a (edit: [i]real[/i]) function to the left of $ 0$.  It clearly cannot take real values if $ x$ is a negative rational with even denominator, and trying to assign it a value if $ x$ is irrational is hopeless.  ($ e^{x \ln x}$ isn't defined either.)

If you define $ x^x\equal{}e^{x\ln x}$ and $ \ln x \equal{} \ln|x| \plus{} \pi i$ for $ x < 0$ (a rather 'natural' definition from the point of view of complex analysis), then the limit is ok, and is equal to $ 1$ from the left just as it is from the right.  More generally, it's true that $ \lim_{z\to 0}z^z \equal{} 1$ in the complex sense.

The basic, though non-trivial, point is that the limit is what you define it to be. :)


Let $a,b,c > 0$, real numbers
such that: $a^2 + b^2 + c^2 = abc$.

Prove that:

\[{{{ \sqrt{\frac{ac}{b} + b^2} } + \sqrt{\frac{ab}{c} + c^2} } + \sqrt{\frac{bc}{a} + a^2} } \geq \sqrt{abc} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} + \sqrt{\frac{bc}{a}}  \]

Carlos Bravo :)
Lima-PERU

all sums cyc.

==========

sub  bc/a = x  ; note xy = c^2.

consraint becomes $\sum xy = xyz$

prob is $\sum \sqrt{x + yz} \ge \sqrt{xyz} + \sum \sqrt{x}$.

see http://www.kalva.demon.co.uk/apmo/asoln/asol024.html


Una mente prodigiosa tras los numeros !!!

En este link les dejo algo inaudito.

En este blog no se vende nada,ni se hace ninguna publicidad ni tampoco se pide donativo con el fin de no obstaculizar  la difusión de la Verdad.
ADVERTENCIA Todo aquel que obstaculice o haga impedimento en la difusión del contenido de este  blog, será con seguridad reo de la Justicia Divina

http://matematicas-misticas.blogspot.com/

GRACIAS

autorización…
¿usted apenas está anunciando este Web site?
¿o hay algo que usted desea pedir?

y utilicé un traductor, así que algo de esto no pudo parecerse tan fluido


define positive integers   $ x, y$, such that      $ x^2  \minus{} 3y^4 \equal{} 1$

it's directly pell's equation

first non-trivial solution

$ x_0\equal{}2, b_0\equal{}1$

where $ b_0\equal{}y^2$

$ x_n\plus{}b_n\sqrt{3}\equal{}(x_0\plus{}b_0\sqrt{3})^{n}$

then we should only find all n for which the coefficient of $ \sqrt{3}$ in the RHS is a perfect square and then solve for y


What is the minimum number of comparisons necessary between pairs of objects so that we can arrange the five objects in order of their size?

Take the first object, compare it to the other four, break those into 2 categories: larger and object 1 and smaller than object 1.  4 comparisons done.

Now we have 3 cases: 0/4, 1/3, or 2/2 on either side of the 1st object. (the / denotes the 1st object)

In the 0/4 case, we need 3 comparisons on the right with one of those objects, resulting in either 0/3 or 1/2 arrangements around a 2nd object.
-The 0/3 case needs 2 comparisons with a 3rd object on the right, which will give either a 0/2 or 1/1 arrangement (finished).  In the 0/2 case an additional comparison with a 4th object is needed, for a total of [b]10[/b] comparisons.
-the 1/2 case needs  an additional comparison the the right, for 8 total comparisons.

in the 1/3 case, we need 2 comparisons on the right, resulting in either a 0/2 case or a 1/1 case (in which case we are finished with 6 comparisons).  The 0/2 case needs an additional comparison to finish, for a total of 7 comparisons.

in the 2/2 case, we need 1 comparison on each side and then the order is determined in 6 comparisons.

So we need at least 6 comparisons, and are guaranteed to be able to determine the order in 10.

To determine the order of $ n$ objects we need $ n(n\minus{}1)/2$ comparisons to guarantee that we know the order (sum of the first $ n\minus{}1$ numbers) because each step to order $ n$ elements we need an additional $ n\minus{}1$ comparisons, and the worst case scenario is that we are comparing the numbers to either the largest or the smallest, and thus end up with the 0/(n-1) case.

$ \binom{n}{2}$ isn't right, because I can order four things with only five comparisons:

[hide]Choose three of them and use three comparisons to order them, say $ a < b < c$.  Now compare $ d$ with $ b$ -- if $ d$ is larger, comparing $ d$ with $ c$ is all we need, while if $ d$ is smaller then comparing $ d$ with $ a$ is all we need.[/hide]

facis made the tacit assumption that the optimal solution involves comparing a single element to all others, then doing the same with a second element, and so on.  So, together we've shown that this is [i]not[/i] generally a best strategy ;)


Find the minimum value of

$ \frac{\sin A\sin B}{\sin^2{\frac{C}{2}}}\plus{}\frac{\sin B\sin C}{\sin^2{\frac{A}{2}}}\plus{}\frac{\sin C\sin A}{\sin^2{\frac{B}{2}}}$,

where $ A,B,C$ are angles of a triangle.

[quote="easternlatincup"]Find the minimum value of

$ \frac {\sin A\sin B}{\sin^2{\frac {C}{2}}} \plus{} \frac {\sin B\sin C}{\sin^2{\frac {A}{2}}} \plus{} \frac {\sin C\sin A}{\sin^2{\frac {B}{2}}}$,

where $ A,B,C$ are angles of a triangle.[/quote]

$ \frac {\sin A\sin B}{\sin^2{\frac {C}{2}}}\equal{}\frac{(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)}{c^{2}}\equal{}(\frac{a\plus{}b}{c})^{2}\minus{}1$

----->

 $ \frac {\sin A\sin B}{\sin^2{\frac {C}{2}}} \plus{} \frac {\sin B\sin C}{\sin^2{\frac {A}{2}}} \plus{} \frac {\sin C\sin A}{\sin^2{\frac {B}{2}}}\equal{}(\frac{a\plus{}b}{c})^{2}\minus{}1\plus{}(\frac{c\plus{}b}{a})^{2}\minus{}1\plus{}(\frac{a\plus{}c}{b})^{2}\minus{}1\geq 3\sqrt[3]{(\frac{a\plus{}b}{c})^{2}(\frac{c\plus{}b}{a})^{2}(\frac{a\plus{}c}{b})^{2}}\minus{}3\geq3\sqrt[3]{(\frac{2\sqrt{ab}}{c})^{2}(\frac{2\sqrt{cb}}{a})^{2}(\frac{2\sqrt{ac}}{b})^{2}}\minus{}3\equal{}3*4\minus{}3\equal{}9$

$ a\equal{}b\equal{}c$ <->  $ A\equal{}B\equal{}C\equal{}60^{0}$

[quote="easternlatincup"][quote="zaya_yc"]

$ \frac {\sin A\sin B}{\sin^2{\frac {C}{2}}} \equal{} \frac {(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)}{c^{2}}$

[/quote]

Well, why is this?[/quote]

$ \frac {\sin A\sin B}{\sin^2{\frac {C}{2}}}\equal{}\frac {\frac {ab}{4R^{2}}}{\frac {1 \minus{} cosC}{2}}\equal{}\frac {ab}{2R^{2}(1 \minus{} cosC)}\equal{}\frac {ab}{2R^{2}(1\minus{}\frac {a^{2} \plus{} b^{2}\minus{}c^{2}}{2ab})}\equal{}\frac {(ab)^{2}}{R^{2}(2ab \minus{} a^{2} \minus{} b^{2} \plus{} c^{2})}\equal{}\frac {(ab)^{2}}{R^{2}(c^{2} \minus{} (a \minus{} b)^{2})}\equal{}$
$ \frac {(ab)^{2}}{R^{2}(c \plus{} b \minus{} a)(c \minus{} b \plus{} a)}\equal{}\frac {(ab)^{2}}{(\frac {abc}{4S})^{2}(c \plus{} b \minus{} a)(c \plus{} a \minus{} b)}\equal{}\frac {16S^{2}}{(c \plus{} b \minus{} a)(c \plus{} a \minus{} b)c^{2}}\equal{}\frac {16p(p \minus{} a)(p \minus{} b)(p \minus{} c)}{(c \plus{} b \minus{} a)(c \plus{} a \minus{} b)c^{2}}\equal{}\frac {(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(c \plus{} b \minus{} a)}{(c \plus{} b \minus{} a)(c \plus{} a \minus{} b)c^{2}}\equal{}\frac {(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)}{c^{2}}$


:oops:  :DI think  heron's formula would be useful, though i have confusion regarding a final step


In this game you give two or more clues whose intersection gives a single word.

Example:

(1) Target
(2) Grade

The answer would be [b]mark[/b].

The next person posts the answer to the previous one and gives a new set.

I'll start off with an easy-ish one

(1) A body part
(2) Give

(Final note: The clues given are crossword-style clues, which means they can be a little crazier than standard dictionary definitions. Be creative!)

hand??????

1) correct
2) one of ur hands

ring???

If so,

1) This clue
2) This clue
3) This clue
4) It's -------!!!

And neither was mine. :|

BOGTRO's is one/won, if I'm not mistaken.

I think yongyi781's was bad luck, but continue with 1=2's clues.

Is BOGTRO's 21? (Card game)  Or Phase 10?

[quote="BOGTRO"]Mine is NOT one/won.

1) Number
2) Game.
EXTRA CLUE TO COME IF NO-ONE CAN GET IT...[/quote]

24

Ok, all good guesses, but...

1) Game
2) Number
3) Group

Huge hint to come tomorrow.

Uno is a good answer, you don't need to be dumb and state it like that- you could merely say that that is not the answer you are looking for

Theory is the only one that fits.

I will restate mine.

1) This clue
2) This clue
3) This clue
4) It's -------!!!

Think about it. The dashes stand for one word, one letter per dash. It's a phrase(a not very nice one at that)

[quote="Klebian"]trivial?[/quote]

That has the same number of letters as obvious... ok that's also an answer.

ok now do mine

easy

1) Sandman
2) Do Not

wake up...

ok
1)FTW
2)president's

thats probably uber obvious...i couldnt think of anything better

What does "president's" have to do with speed?

mine is rating duh
ftw rating
presidential rating as in approval rating

guys, you can't make yours super-general or else there will be the problem of multiple solutions.

As a side note, Presidents is a popular card game (at least, in some parts of the country) where speed is an important factor (at least in some parts of the country that play with jump in completions)

mz94's was never answered. The correct answer is "Enter". Enter Sandman is the title of a song, and "Do Not Enter" is obvious.

@ 1=2: Money?

[quote="1=2"]speed.

1) You can't take it with you.
2) Grampa Munster did.
3) Greed.[/quote]

death???

1) golfers
2) gnat
3) downer
4) priest
5) naive

erabel got mine right. If you ever watched The Munsters, that would be easy.


Let x_1,...,x_n>0 such that  \sum x_i= \sum 1/x_i. Prove that  \sum 1/(n-1+x_i)  \leq 1.

i have a solution for your inequality
suppose that $\displaystyle \sum_{i=1}^n \frac 1{n-1+x_i} > 1$
we will prove that
$\displaystyle \sum x_i \leq \sum \frac 1{x_i}$
in fact, we have to prove it with the condition $\displaystyle \sum_{i=1}^n \frac 1{n-1+x_i} = 1$
since $\displaystyle \sum_{i=1}^n \frac 1{n-1+x_i} = 1$, we can suppose that
$\frac 1{n-1+x_i} = \frac {a_i}{S}$ where $S=\displaystyle \sum{i=1}^n a_i$ and $a_i>0$
so $x_i=\frac{S-(n-1)a_i}{a_i}$ and $S-(n-1)a_i$ must be possitive
let $S-(n-1)a_i = b_i > 0$
so the inequality will be equivalent to
$\displaystyle \sum_{i=1}^n \frac{(n-1)^2 b_i}{P -b_i} \leq \sum_{i=1}^n \frac{P-b_i}{b_i}$ (*) where $\displaystyle P=\sum_{i=1}^n b_i$
but (*) can be proved easily by this inequality
$\displaystyle \sum_{i=1}^n \frac 1{z_i} \geq \frac {n^2}{\sum z_i}$

 :D

Yeah, this is also my solution and I just love it and the problem.

Sorry, but I don't understand these:

1) in fact, we have to prove it with the condition sum 1/(n-1+x_i)=1

2) we can suppose that a_i >0

Thank you very much.

dear manlio
sorry because i didn't reply soon
for the first
suppose that we have the group of numbers ($x_1,x_2,...,x_n$) so that $\sum \frac 1{n-1+x_i} > 1$
if we replace it by ($x_1,x_2,...,x_{n-1},y$) so that $\sum \frac 1{n-1+x_i} + \frac 1{n-1+y} = 1$ (*) then $y > x_n$
from (*) if we have $\sum x_i + y \leq \sum \frac 1{x_i} + \frac 1{y}$
then $\sum x_i + x_n \leq \sum \frac 1{x_i} + \frac 1{x_n}$
and the second
if we have $z_1, z_2, ..., z_n$ are possitive having sum $k > 0$ (**)
chosse a possitive number $S$
$z_i = \frac {k.y_i}{S}$    (of course $y_i > 0$)
so from (**) we get $S = \sum y_i$

now, replace $z_i = \frac 1{n-1+x_i}$ and $k = 1 $

i hope this is what you need  :D

[b]Solution.[/b] Suppose that $ \sum_{cyc}\frac{1}{n-1+x_{i}}=\frac{1}{p}$, with $ p<1.$ Let's note $ a_{i}=\frac{p}{n-1+x_{i}}, i\in\overline{1;n}.$

So $ \displaystyle\sum^n_{i=1}a_{i}=1.$ The condition from hypothesis is equivalent with: $ \displaystyle\sum^n_{i=1}(\frac{p}{a_{i}}-n+1)=\displaystyle\sum^n_{i=1}\frac{1}{\frac{p}{a_{i}}-n+1}.$ While $ p<1$ we can write that:

\[ \displaystyle\sum^n_{i=1}(\frac{1}{a_{i}}-n+1)>\displaystyle\sum^n_{i=1}(\frac{p}{a_{i}}-n+1)=\displaystyle\sum^n_{i=1}\frac{1}{\frac{p}{a_{i}}-n+1}>\displaystyle\sum^n_{i=1}(\frac{1}{\frac{1}{a_{i}}-n+1}).\]

Remains to prove that for every $ a_{1},\dots,a_{n}$, with $ \displaystyle\sum^n_{i=1}a_{i}=1$ the inequality is satisfied: $ \displaystyle\sum^n_{i=1}(\frac{1}{a_{i}}-n+1)\leq\displaystyle\sum^n_{i=1}(\frac{1}{\frac{1}{a_{i}}-n+1})$, which is equivalent with: $ \displaystyle\sum^n_{i=1}\frac{a_{i}}{1-(n-1)a_{i}}\geq\displaystyle\sum^n_{i=1}\frac{1-(n-1)a_{i}}{a_{i}}.$ Now we'll make another substitution: $ b_{i}=1-(n-1)a_{i}, i\in\overline{1;n}.$

$ \Rightarrow \displaystyle\sum^n_{i=1}b_{i}=1).$

$ \displaystyle\sum^n_{i=1}\frac{a_{i}}{1-(n-1)a_{i}}\geq\displaystyle\sum^n_{i=1}\frac{1-(n-1)a_{i}}{a_{i}}\Leftrightarrow\displaystyle\sum^n_{i=1}\frac{1-b_{i}}{b{i}}\geq (n-1)^2(\displaystyle\sum^n_{i=1}\frac{b_{i}}{1-b_{i}}).$

The last inequality followings immediately from: $ \frac{b_{i}}{1-b_{i}}=\frac{b_{i}}{\sum_{i\neq j}b_{j}}=\frac{1}{\sum_{j\neq i}\frac{1}{\frac{b_{i}}{b_{j}}}}\leq \frac{1}{(n-1)^2}\cdot(\sum_{j\neq i}\frac{b_{i}}{b_{j}}).$ ([color=blue]AM-HM inequality[/color]).

When we adding up these inequalities we get that: $ (n-1)^2\cdot (\displaystyle\sum^n_{i=1}\frac{b_{i}}{1-b_{i}})\leq \sum_{i}\sum_{j\neq i}\frac{b_{i}}{b_{j}}=\sum_{j\neq i}\sum_{i}\frac{b_{i}}{b_{j}}=\displaystyle\sum^n_{i=1}\frac{1-b_{j}}{b_{j}}.$


Find all pairs $ (n,m)$ of positive integers such that $ n | m^2\plus{}1$ and $ m | n^2\plus{}1$

posted before, use that condition equavalent to $ mn|m^2\plus{}n^2\plus{}1$.


The blog has x-hundred topics.  However, it only has 140 views...?

Your point?

#H34N1 used to have over 200 entries and over 100k views. He deleted all his entries and his views started over.

It's not a big deal  :wink: .

Oh ok, I thought this was a bug.  :blush:  :D


We have 2 circles $S_1$ and $S_2$. They intersect at $A$ and $B$. A line through $A$ intersects $S_1$ at $X$ and $S_2$ at $Y$. We choose any point $M$ on $XY$. The line parallel to $BY$ through $M$ intersects $BX$ at $S$. The line parallel to $BX$ through $M$ intersects $BY$ at $T$. The line through $S$ perpendicular to $XB$ intersects $S_1$ at a point $E$ on the opposite side of $A$ to $XB$. Similarly, the line through $T$ perpendicular to $BY$ intersects $S_2$ at a point $F$ on the opposite side of $A$ to $BY$.
Prove:
[b](i)[/b] $AMEF$ is inscribed quadrilateral.
[b](ii)[/b] $\angle EAF=90^\circ$.

[color=red][Moderator edit: This problem was again discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=22201 .][/color]

Note that $ES\perp MT$ and $SM\perp TF$. It implies that $\angle ESM=\angle MTF$ [b](*)[/b].

We will prove that $\triangle ESM \sim \triangle MTF$ [b](**)[/b].
Indeed, $BX:BY=R_{S_1}:R_{S_2}$, since $BX=2R_{S_1}\sin BAX$ and $BY=2R_{S_2}\sin BAY$. Moreover, $BT:TY=XM:MY=XS:SB$. It follows [b](!)[/b] that chord $BY$ and point $T$ for circle $S_2$ is the same as chord $BX$ and point $S$ for circle $S_1$. Thus we immediately conclude $\angle TBF=\angle SXF'=\angle SEB$. But $BS:SE=\tan SEB$ and $BS=MT$, $FT:TB=\tan TBF$ and $TB=MS$. Thus $ES:MT=SM:TF$.

Using [b](*)[/b] and [b](**)[/b], we conclude that $EM\perp MF$.
Moreover, $\angle EAF=\angle EAB+\angle FAB=\frac{1}{2}(\breve{EB}+\breve{XF'})=90^\circ$.

Finally, $\angle EAF=\angle EMF=90^\circ$, so $EMAF$ is an inscribed quadrilateral.


abstract algebra
function
integration
real analysis
real analysis unsolved

Suppose $ \phi$ is an integrable function on $ \mathbb{R}^d$ with $ \int_{\mathbb{R}^d} \phi(x)dx\equal{}1.$
Set $ K_{\delta} (x)\equal{}\delta^{\minus{}d}\phi(x/\delta), d>0.$
Prove that:
(1)$ \int_{\mathbb{R}^d} |K_{\delta}(x)|dx \le A$, where $ A$ is a constant independent of $ \delta.$
(2)For every $ \rho$ > 0,
 $ \int_{|x|\ge \rho} |K_{\delta}(x)|dx \to 0$ as $ \delta \to 0.$
-------------------------------------------------------------------

(1) My Solution
Since $ \phi \in L^1$, set $ \int_{\mathbb{R}^d} |\phi(x)|dx\equal{}A.$
Then
$ \int_{\mathbb{R}^d} |K_{\delta}(x)|dx$
$ \equal{}\int_{\mathbb{R}^d} |\delta^{\minus{}d}\phi(x/\delta)|dx$
$ \equal{}\int_{\mathbb{R}^d} \delta^{\minus{}d}|\phi(x/\delta)|dx$
$ \equal{}\int_{\mathbb{R}^d} \delta^{\minus{}d}|\phi(y)|\delta^{d}dy$
$ \equal{}\int_{\mathbb{R}^d} |\phi(y)|dy$
$ \equal{}A$

Is my solution right?

(2)Please, give me a few useful hints.

make the same change of variable (dillatation) and use the dominated convergence theorem, for example.

another hint for $ 2$ is as follows, although i like alekk's way better:

fix $ \varepsilon\in (0,A)$ and pick $ M$ big enough that
\[ \int_{B(0,M)} |\phi(x)|dx > A \minus{} \varepsilon
\]
the same calculation as you have done above shows that
\[ \int_{B(0,M\delta)} |K_\delta(x)|dx \equal{} \int_{B(0,M)} |\phi(x)|dx > A \minus{} \varepsilon
\]
Now what happens when $ M\delta<\rho$?


function
inequalities unsolved
inequalities

[quote="AndrewTom"]Reading a proof of AM-GM using elementary functions, I notice that the authors apply x >= ln (1+x) for x>=0 to x = a(i)/A - 1, where A is the arithmetic mean of the a(i), i =  to n. However this latter quantity is > -1. Can you prove that x >= ln(1+x) for x > -1?[/quote]
$ f(x)\equal{}x\minus{}ln(1\plus{}x)$, $ f'(x)\equal{}\frac{x}{1\plus{}x}$
1) $ \minus{}1<x\le0$ then $ f'(x)\le 0$ so $ f(x)\ge f(0)\equal{}0$
2) $ x\ge 0$ then $ f(x)\ge 0$ so $ f(x)\ge f(0)\equal{}0$ its done!


I vote for a proof by contradiction.  (The proof of (b) follows immediately from the proof of (a).)

I have to also find the time $t$ when $x(t)$ and $\dot{x}(t)$ are $0$. Here is how I showed that they are $0$ at most once:

$c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}= 0$
$c_{1}e^{(r_{1}-r_{2})t}+c_{2}= 0$

$e^{(r_{1}-r_{2})}=-\frac{c_{2}}{c_{1}}$

The exponential function only has one root where $x(t) = 0$ (from definition of exponential function). From this does it suffice that $x(t)$ is $0$ at most once?

$c_{1}r_{1}e^{r_{1}t}+c_{2}r_{2}e^{r_{2}t}= 0$
$c_{1}r_{1}e^{(r_{1}-r_{2})t}+c_{2}r_{2}= 0$

$\frac{r_{1}}{r_{2}}e^{(r_{1}-r_{2})t}=-\frac{c_{2}}{c_{1}}$

From this does it suffice that $\dot{x}(t)$ is $0$ at most once?

How would I find the time $t$ that this occurs?

Thanks


The polynomial $ x^3\minus{}2004x^2\plus{}mx\plus{}n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $ n$ are possible?

$ \textbf{(A)}\ 250,\!000 \qquad
\textbf{(B)}\ 250,\!250 \qquad
\textbf{(C)}\ 250,\!500 \qquad
\textbf{(D)}\ 250,\!750 \qquad
\textbf{(E)}\ 251,\!000$

[quote="ocha"][hide="solution"]

Ok, let $ x^3 \minus{} 2004x^2 \plus{} mx \plus{} n \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c)$
We are told that $ c\in \mathbb(Z)$, $ a,b \not\in \mathbb(Z)$ and $ a \plus{} b \equal{} c$

From Vieta's Formula; $ a \plus{} b \plus{} c \equal{} 2004 \longrightarrow 2c \equal{} 2004 \Rightarrow c \equal{} 1002$

So $ a \plus{} b \equal{} 1002$ and $ 1002ab \equal{} n$. Since $ n$ is and intiger we need the number of positive intiger solutions to the equation

$ y \equal{} a(1002 \minus{} a)$
This has a maximum at $ a \equal{} 501 \Longrightarrow y\in[0,(501)^2]$, but since $ a$ is not an intiger we need
$ 501^2 \minus{} 501 \equal{} 250500 \longrightarrow \boxed{C}$[/hide][/quote]

Yes, you are right, it should be $ y \equal{} 1002a(1002 \minus{} a)$, this part is wrong if you go straight to it, since he never used the fact that $ m$ must be an integer. Note that $ m \equal{} ab \plus{} bc \plus{} ca \equal{} bc \plus{} a(b \plus{} c) \equal{} bc \plus{} (b \plus{} c)^2$ must be an integer. But we know $ (b \plus{} c)^2 \equal{} 1002^2$ is an integer, so $ bc$ must be an integer. This is where we can get the equation $ y \equal{} a(1002 \minus{} a)$ must be an integer as he said, and this will clearly also make $ n$ an integer as $ n\equal{}1002ab$.

[quote="ocha"]

So $ a\plus{}b\equal{}1002$ and $ 1002ab\equal{}n$. Since $ n$ is and intiger we need the number of positive intiger solutions to the equation

$ y \equal{} a(1002\minus{}a)$

This has a maximum at $ a\equal{}501 \Longrightarrow y\in[0,(501)^2]$, but since $ a$ is not an intiger we need
$ 501^2\minus{}501\equal{}250500 \longrightarrow \boxed{C}$[/quote]

Shouldn't $-n= 1002ab$? Where does the variable y come from? Why is it that we need need the number of positive integer solutions to the equation

$ y \equal{} a(1002\minus{}a)$ ?

I'm really confused. Can someone help clarify? Thanks in advance.

i am confuse about that too, could somebody answer?

[quote="Dr Sonnhard Graubner"]hello, why are you confused?
Sonnhard.[/quote]

About ocha's mistake.
By the way , which problem it is?

[quote="ocha"][hide="solution"]

Ok, let $ x^3\minus{}2004x^2\plus{}mx\plus{}n \equal{} (x\minus{}a)(x\minus{}b)(x\minus{}c)$
We are told that $ c\in \mathbb(Z)$, $ a,b \not\in \mathbb(Z)$ and $ a\plus{}b\equal{}c$

From Vieta's Formula; $ a\plus{}b\plus{}c\equal{}2004 \longrightarrow 2c\equal{}2004 \Rightarrow c\equal{}1002$

So $ a\plus{}b\equal{}1002$ and $ 1002ab\equal{}n$. Since $ n$ is and intiger we need the number of positive intiger solutions to the equation

$ y \equal{} a(1002\minus{}a)$

This has a maximum at $ a\equal{}501 \Longrightarrow y\in[0,(501)^2]$, but since $ a$ is not an intiger we need
$ 501^2\minus{}501\equal{}250500 \longrightarrow \boxed{C}$[/hide][/quote]

So yeah, I get most of the solution but why do we subtract 501 from 501^2 ? (i know a is not an integer...)

Because if $a=501$ is max, then we have already counted $a=1,2, \cdots 501$ as possible $a's$. All of these values of $a$ give us a distinct integer $y$, which is what we want to count. However, $a$ cant be an integer, so we have to take of $501$ from the total possible number of values.

Hope this helps.

We never proved that all of these solutions allow $m$ to be a integer.

[quote=Target_cmi]since a = (1,2,...501) , for what value of a will y=0 ? it is not possible... shouldn't it be y = [b]([/b]0, (501)^2 ] instead of y =[b] [[/b] 0, (501)^2 ]....[/quote]

Exactly correct !!

I don't understand why y=a(1002-a) leads to 501^2. Isn't that 501^2 just the product of the biggest value of a (a=501)?

Let $a,b,c$ be the roots of the polynomial $x^3 - 2004x^2 + mx + n$. Then $a+b+c = 2004$, and by the fact that $c=a+b$, we see $c=1002$. Note that $m = ab + bc + ca = c(a+b) + ab = 1002^2 + ab \iff ab \in \mathbb{Z}$. Thus $ab = a(c-a) = a(1002-a) = k$ where $k \in \mathbb{Z}$, $a \leq 1002$. The maximum of this equation is at $a = 501 \iff k = 501^2$. Note that the quadratic is symmetric around $a = 501$. The minimum of $k$ is $1001$ when $a = 1$. Thus we can make $k$ any integer between $1$ and $501^2$, meaning $k$ takes up $501^2$ values. However when $a$ is an integer, which is not allowed, $k$ is an integer. There are $501$ values of $a$ which make this happen. Thus, the total amount of values of $n = -abc = -1002k$ is $501^2 - 501 = 250,500$.


Ok, clearly I've created that by fixing the other stuff. I'll look into it Monday. The code is good, it's just displayed badly in the post.


function
real analysis
real analysis unsolved

Let $ f$ be a differentiable real function and let $ M$ be a positive real number. Prove that if \[ |f(x\plus{}t)\minus{}2f(x)\plus{}f(x\minus{}t)| \leq Mt^2 \; \textrm{for all}\ \;x\ \; \textrm{and}\ \;t\  , \] then \[ |f'(x\plus{}t)\minus{}f'(x)| \leq M|t|.\]

[i]J. Szabados[/i]

For any $n\in \mathbb{N}$ We have
\[|(f(x+nt)-f(x+(n-1)t))-(f(x)-f(x-t))|\]
\[=|\sum_{i=0}^{n-1} f(x+(i+1)t)-2f(x+it)+f(x+(i-1)t)|\]
\[\leq\sum_{i=0}^{n-1} | f(x+(i+1)t)-2f(x+it)+f(x+(i-1)t)|\]
\[\leq nMt^2\]
Let $nt=y$. By dividing by $t$ we get
\[\big|\frac{f(x+y)-f(x+y-t)}{t}-\frac{f(x)-f(x-t)}{t}\big|\leq My\]
Letting $y$ constant and $n \rightarrow \infty$ it follows that $t \rightarrow 0$ and therefore
\[|f'(x+y)-f'(x)|\leq  My\]


Solve the equation  [sqrt(x)]=sqrt[x] for real  numbers.

you already posted this ... (anyway since [sqrt(x)] is integer, sqrt [x] must be integer so [x] is perfect square, so x is the of the form a^2 + b, where 0<=b<1 and a is integer)


Let $ a,b,c>0$. Prove that: $ \sum\frac{b\plus{}c}a\ge\sum\frac{2bc}{a(b\plus{}c)}\plus{}\frac{3(\sum a^2)}{\sum bc}$

[quote="dduclam"]Let $ a,b,c > 0$. Prove that: $ \sum\frac {b \plus{} c}a\ge\sum\frac {2bc}{a(b \plus{} c)} \plus{} \frac {3(\sum a^2)}{\sum bc}$[/quote]
We see that the Ineq is
$ (\sum a(b \plus{} c))(\sum \frac {b^{2} \plus{} c^{2}}{a(b \plus{} c)}) \geq 3(\sum (b^{2} \plus{} c^{2}))$
Is true

[quote="dduclam"]Let $ a,b,c > 0$. Prove that: $ \sum\frac {b \plus{} c}a\ge\sum\frac {2bc}{a(b \plus{} c)} \plus{} \frac {3(\sum a^2)}{\sum bc}$[/quote]

One solution for it. :)


ceiling function
real analysis
real analysis unsolved

Is ceiling(a/b) equivalent to ceiling(ceiling(a)/b) if a and b are real number?
May I know how to prove it or disprove it, thanks!

if $ a$ is very small, then you would expect the second one to be much larger.

Pick $ a \equal{} 0.1$ and $ b \equal{} 0.1$;

\[ \left\lceil \frac{a}{b} \right\rceil \equal{}  \lceil 1 \rceil \equal{} 1 \not\equal{} 10 \equal{} \lceil 10 \rceil \equal{} \left\lceil \frac{1}{0.1}\right\rceil \equal{} \left\lceil \frac{\lceil a\rceil}{b} \right\rceil
\]

Sorry my mistake, the condition is wrong, a is real number but b is integer, both a and b is larger than 0.

As $ \lceil a \rceil < a\plus{}b$, we see that

\[ \left\lceil \frac{\lceil a\rceil}{b} \right\rceil \leq \left\lceil \frac{a\plus{}b}{b}\right\rceil \equal{} \left\lceil \frac{a}{b}\plus{}1 \right\rceil \equal{} \left\lceil \frac{a}{b}\right\rceil \plus{} 1
\]

can we ever have equality?

Possible to give an example to show there are not equal?

but they are ...

maybe this is simpler.  Write $ a \equal{} m \plus{} b$, where $ 0\leq b < 1$.  If $ a$ is an integer, then $ b \equal{} 0$ and there is nothing to show.  If $ a$ is not an integer, then $ \lceil a \rceil \equal{} m \plus{} 1$, and, if we say that $ \lceil a/b \rceil \equal{} q$ for some integer $ q$, then $ qb \minus{} b < a \leq qb$, and, since $ a$ is not an integer and $ qb$ is, and since $ m \plus{} 1$ is the smallest integer greater than $ a$, we get $ m \plus{} 1 \leq qb$ as well.  Clearly $ qb \minus{} b < a < m \plus{} 1$, so that $ qb \minus{} b < m \plus{} 1 \leq qb$.  But this shows that $ \lceil \lceil a\rceil/b\rceil \equal{} q$ ...


This is an earnest request to those appearing for RMO frm Karnataka
I wud request them to post the link to some website frm where I may be able to download previous years KRMO papers. :)
[hide]plz post the link bfore tis too late...post the link by atleast a week bfore the big event :( [/hide]

http://www.isid.ac.in/~rbb/rmoqpapers.html

the rmo papers in evry state are the same.

ummmm....sry....fforgot to mention :P
I got them,the papers of 5-6 years,here's the link
http://www.isibang.ac.in/~sury/krmosol95-00.pdf
Maharashtra and Karnataka have different papers.


I am sure this would have been posted before, but could not find it using search. so apologies in advance.

1. Find $a, b, c$ such that $a! \times b! = c!$ where $1 < a$ $\le$ $b < c-1$

2. Find $a, b, c, d$ such that $a! \times b! = c! \times d!$ where $c+1 < a$ $\le$ $b < d-1$

Note the following:
a) if $b=d-1$, the divide by $b! \times d!$ so $a!/c! = d$ which will work for any $a, c$ with $a < c$
b) if $c+1=a$, the divide by $b! \times c!$ so $a = d!/b!$ which is impossible since $d>b$ and $d>a$
c) If $c=1$, then problem $2$ is same as problem $1$

[hide]
$3!\cdot 5! =6!$, but this doesn't satisfy the condition.
[/hide]

[hide="Oops... new answer"]
You know I meant (6, 7, 10)
[/hide]

[quote="xxxyyyy"][quote="maniacman384"][hide]if you try out 6 for a and 7 for b, you notice that 6!=720 and 10! = 7!*720 so, we have a triplet of numbers, (6, 7, 10)[/hide][/quote]

Can we have a general solution for this equation?[/quote]

The original problem is modified to include two problems now... didnt want to start a new thread. :blush:


[quote="236factorial"]I also have one more thing to ask. When i see my post count, it's some number, but then when i search for all my posts, it returns a number less than the post count. I thought when moderators delete messages, the post count is supposed to go down too.[/quote]Well, there is a small bug with the post count. Since I personally don't care that much about post counts in general, I didn't put any time into solving the bug yet. But I can resync the post counts whenever I want (takes 5 mins off the site), but as I said, I don't care that much in general ;)

[quote="236factorial"]I understand, and I don't care that much either, and when you say 5 mins. off the site, does that mean no one can post for 5 min.?[/quote]Yeah, it means all the site must be shutdown for 5 minutes or so. Obviously not a nice thing for most people :P

I believe it was one day in April or May on the weekends, but the site was closed for something like 3 hours. Was that time to resync the post counts?

Topic split. The answer is ... I really don't remember, but I think not :)


[quote="DaReaper"]How many distinguishable combinations of the word $RACECAR$ exist?[/quote]
[hide]There are 7 letters in total, 2 a's and a e's. So its just $\frac{7*6*5*4*3*2*1}{2*1*2*1}=1260$.[/hide]

[quote="DaReaper"]ye...there are 2 a's 2 e's and you forgot the 2 r's ..and the answer is 630.. so you were right but forgot the r's  :lol:[/quote]
Oops. :blush: I overlooked the problem.

[quote="DaReaper"]How many distinguishable combinations of the word $RACECAR$ exist?[/quote]
[hide]$\frac{7!}{2!2!2!}=\boxed{630}$[/hide]

Please explain your solutions. Most of the people in Classroom Math post problems for the [b]method[/b], not the answer. (I hope I am not being arrogant...)

[hide="Anyway, here is the solution (not answer)  :wink: "]There are $2$ $C's$, $R's$, and $A's$. There is only $1$ $E$. This is all we really care about. What if both $C's$, $R's$, and $A's$ are distinguishable? Then we have $7!$ ways. But we have [b]overcounted[/b]. This is because the repeated letters are indistinguishable. We have counted each repeated letter $2!=2$ times (why?), so we must divide everything by $2!$ three times to get: $\frac{7!}{2!\cdot 2!\cdot 2!}= \frac{7\times6\times5\times4\times3\times2\times1}{2\times1\times 2\times1\times 2\times 1}= 7\times 6\times 5\times 3 = 30\times 21=630$. Thus, the answer is $\boxed{630}$.[/hide]

just wondering, but n! is the number of letters in this problem, which is 7. bu how do you get r? maybe i'm being stupid to some of you, but i really want to know...

nPr= n!/(n-r)!
when n=r
0!=1 for denominator
the (2!)^3 on bottom is for repeating letters not r(though you probably knew that but in case...)


Have all the MOP invitations been sent out?
Maybe I might not make it this year...

You can tell if you made MOP by looking at your score [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/09_OfficialUSAMOScores.pdf]here[/url] and the cutoffs [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276193]here[/url].


Discuss here.

IL came in 14th place team with an 8 on the team round. The team score was 45.

Rankings of Individuals:

Megan Chen 20th place
Patrick Liscio 62nd place
James Tao 100th place
David Wang 102nd place

:)

@aryth and whitesoxfan: fortunately our cheer was not lame at all :)

Hey the banquet is still going on, so we won't know the spirit winner until I rush back to my phone and report what happened...

Oh ok well I'll be waiting for the report!

man if you please hurry I wan't to know who acually won.
I don't know.

To our surprise, Virgin Islands won the spirit stick, how weird :o

What the??  The Virgin Islands won the spirit stick??  They're not even a state.

What did they do to get the spirit stick?

[quote="Aryth"]Chuck Norris visited the Virgin Islands. After he left, they were just called "The Islands".[/quote]

rofl.  :rotfl:

EDIT: after the unfortunate scoring debacle, IL is now down to 15th place team after the GA scoring messup :(

Aww, that sucks.  :(  It's always awesome when graders mess up on your test and then you bounce up places but it'll make the people you pass feel really bad lol :(

dang that's horrible i hate it when ppls mess up grading!! but good job!!!!


By standard notations prove the following:

    a MA+b MB +c MC=(a+b+c) MI where MA,MB,MC,MI are vectors.

** MA+b MB +c MC=(a+b+c) MI
$a(A-M)+b(B-M)+c(C-M)=aA+bB+cC-(a+b+c)M$
so when the equality(!) hold we should have $Aa+Bb+Cc=0$ !!
i think you have lost one of the statments.