calculus
integration
logarithms
calculus computations

Please solve:
$ I \equal{} \int\frac{e^{\minus{}ax}\minus{}1}{e^{\minus{}ax}\plus{}1}\,dx$

$ \frac {e^{ \minus{} ax} \minus{} 1}{e^{ \minus{} ax} \plus{} 1} \equal{} \minus{} \frac {e^{ax/2} \minus{} e^{ \minus{} ax/2}}{e^{ax/2} \plus{} e^{ \minus{} ax/2}} \equal{} \minus{} \tanh\frac {ax}{2}$.  So $ I \equal{} \minus{} \int \tanh\frac {ax}{2}dx \equal{} \frac {2}{a}\log\cosh\frac {ax}{2} \plus{} C$ for $ a\neq 0$, and $ C$ otherwise.  I think the above computations are correct, but I haven't checked them.

hello
$ I \equal{} \int\frac {e^{ \minus{} ax} \minus{} 1}{e^{ \minus{} ax} \plus{} 1}\,dx$
$ I \equal{} \int\frac {1 \minus{} e^{ax}}{1 \plus{} e^{ax}} dx$
$ I \equal{} \int \frac {1 \plus{} e^{ax} \minus{} 2e^{ax}}{1 \plus{} e^{ax}} dx$
$ I \equal{} \int 1 \minus{} \frac {2e^{ax}}{1 \plus{} e^{ax}} dx$
$ I \equal{} x \minus{} \frac {2log(1 \plus{} e^{ax})}{a(1 \plus{} e^{ax})} \plus{} C$
Thank u


induction
limit
algebra unsolved
algebra

A sequence of numbers $a_n, n = 1,2, \ldots,$ is defined as follows: $a_1 = \frac{1}{2}$ and for each $n \geq 2$
\[ a_n =  \frac{2 n - 3}{2 n}  a_{n-1}. \]
Prove that $\sum^n_{k=1} a_k < 1$ for all $n \geq 1.$

My ideal
Let $b_{n} = \sum^n_{k=1}a_k, we have: a_{n} = b_{n} - b_{n-1}$
From conditional we have: $b_{n} = \frac{4n-3}{2n}.b_{n-1} - \frac{2n-3}{2n}.b_{n-2}$ .

By simple iteration we get $a_n = \dfrac {1} {2n-1} \prod_{k=1}^n \dfrac {2k-1} {2k}$. It is well-known (and easily proved by induction) that $\prod_{k=1}^n \dfrac {2k-1} {2k} \leq \frac {1} {\sqrt{3n+1}}$. However, $\sum^{\infty}_{n=1} \frac {1} {(2n-1)\sqrt{3n+1}} > 1$, not by much, but enough so we cannot infer that $\sum^{\infty}_{n=1} a_n \leq 1$. Stronger methods are needed.

Very nice. So now, knowing that $\frac{1}{\sqrt{4n}} \leq \prod_{k=1}^{n}\frac{2k-1}{2k}< \frac{1}{\sqrt{3n}} $, and in fact $\lim_{n\to \infty} \sqrt{n} \prod_{k=1}^{n}\frac{2k-1}{2k} =  \frac{1}{\sqrt{\pi}}$, we actually get that $\sum_{k=1}^{\infty} a_k = 1$, with the estimation $\lim_{n\to \infty} \sqrt{n} \left ( 1 - \sum_{k=1}^{n} a_k\right) =  \frac{1}{\sqrt{\pi}}$.


geometry
3D geometry
combinatorics proposed
combinatorics

The vertices of a cube are arbitrarily labeled from $ 1$ to $ 8$ . Prove that there exists a three-edge path joining two diagonally opposite vertices, along which the assigned numbers(end-points included) add up to at least $ 21$ .

Let us name set $ X \equal{} (a,b,c,d)$ where $ X\in {1,2,...8}$

In fact there are $ 11 * 4! \equal{} 264$ number of sets X (including all permutations) that satisfy $ a \plus{} b \plus{} c \plus{} d \ge 21$. I do not understand what you want us to prove? :huh:

Let us consider the cube $ ABCDA\prime B\prime C\prime D\prime$ and consider the follow labels: $ A-1,B-2,C-3,D=4,A\prime -5, B\prime -6, C\prime-7, D\prime-8$. The vertices $ A\prime$ and $ C$ are joined by a three-edge path formed by $ A\prime -D\prime-C\prime-C$ and the sum of assigned numbers is $ 5+8+7+3=23>21$ .

I hope you can understand me.

I think it is too obvious that there exists a three-edge path joining two diagonally opposite vertices, along which the assigned numbers add up to at least 21.  I dont understand why you want to prove this? :?:

Write your proof, please!  :)  I`m really intrested in solving this problem.


A rhombus has half the area of the square with the same side-length. What is the ratio of the long diagonal to the short one?

[hide="Solution"]Let $ d_1$ and $ d_2$ be the diagonals of the rhombus and $ s$ be the side length of the square. Then

$ \frac{1}{2} d_1 d_2 \equal{} \frac{1}{2} s^2$

$ d_1d_2 \equal{} s^2$

Also, since they have the same sidelength, we know that $ s \equal{} \sqrt{ (\frac{d_1}{2})^2 \plus{} (\frac{d_2}{2})^2}$. So,

$ d_1d_2 \equal{} (\frac{d_1}{2})^2 \plus{} (\frac{d_2}{2})^2$

$ 0 \equal{} d_1^2 \minus{} 4d_1d_2 \plus{} d_2^2$

$ 0 \equal{} \frac{d_1}{d_2} \minus{} 4 \plus{} \frac{d_2}{d_1}$

$ 0 \equal{} t \minus{} 4 \plus{} \frac{1}{t}$

$ 0 \equal{} t^2 \minus{} 4t \plus{} 1$

$ t \equal{} \frac{4 \pm 2\sqrt{3}}{2}$
Since the ratio is greater than one, we have that $ \boxed{t \equal{} \frac{4 \plus{} 2\sqrt{3}}{2}}$.[/hide]

To make the answer shorter, seing that the altitude is half of the side length, the rhomus has an angle of $30^\circ$, hence the required ratio will be $tan\ 75^\circ$.


inequalities
trigonometry
inequalities proposed

In any  triangle  ABC  prove  that
 (p*pi)/2 < 4R+r<2p

where  p  is  semiperimeter , pi = 3,14.... , R  is  circumaradius , r inradius.
THANKS  AND HAPPY  NEW YEAR!!

[i]unknown source[/i]

I have doubts about the second part holding for EVERY triangle. Consider this situation: We take A, B, C on a fixed circle of radius R. We can move A, B, C anywhere we want. We move them really close to each-other s.t. 2p=AB+BC+CA becomes really small (it can be as small as we want). We will eventually reach a point when AB+BC+CA=2p<4R<4R+r, so the second part doesn't hold in this case.

However, I think it does hold for the case ABC-acute triangle. I'll just post the outline. Let D, E, F be the midpts of BC, CA, AB, respectively. Then R+r=OD+OE+OF and 3R=OA+OB+OC, so we must prove BC+CA+AB>OA+OB+OC+OD+OE+OF. it should be enough to prove the 3 ineqs:

OA+OF<EA+EF;
OB+OD<FB+FD;
OC+OE<DC+DE.

By adding them up we would get what we want, since EA+FD=CA; FB+DE=AB; DC+EF=BC.

The inequality OA+OF<EA+EF (the others are proven analogously) shouldn't be too hard to prove. We just have to remember that OEAF is cyclic and then consider the following problem:
Let M, N be fixed pts on a fixed circle. X is a variable pt on the same circle. What happens to the sum XM+XN as X moves closer/farther from the midpt of the larger arc MN? I think it becomes larger when X is closer and smaller when it goes farther. This is easy to prove with some sines and stuff.


calculus integration trigonometry function algebra polynomial algorithm

Definite Integral ((x^2)-1)/((x^2)+1) dx?

May be you wanted to say Indefinite integral, unless you forgot to put the integration limits.

Anyway your problem is easy: make $ x^2\minus{}1\equal{}x^2\plus{}1\minus{}2$ and the rest follows.

If it's definite, where are it's boundaries? Still, here's the solution to the INdefinite integral:

[hide="Click"]$ \frac {x^2 \minus{} 1}{x^2 \plus{} 1} \equal{} \frac {x^2 \plus{} 1}{x^2 \plus{} 1} \minus{} \frac {2}{1 \plus{} x^2} \equal{} 1 \minus{} \frac {2}{1 \plus{} x^2}$

so:

$ \int \frac {x^2 \minus{} 1}{x^2 \plus{} 1}\,dx \equal{} \int \left(1 \minus{} \frac {2}{1 \plus{} x^2}\right)\,dx \equal{} \int dx \plus{} \int \frac {2}{1 \plus{} x^2}\,dx \equal{} x \minus{} 2 \tan^{ \minus{} 1}x \plus{} C$[/hide]

$ \int \frac{x^2\minus{}1}{x^2\plus{}1}\,dx$

This is the integral of a rational function (a polynomial divided by a polynomial). In the decision tree for deciding what to do about it, the very first question is this: "Is it a proper fraction?" The definition of proper fraction is a case in which the degree of the numerator is strictly less than the degree of the denominator. Therefore, this is not a proper fraction.

If it's not a proper fraction, perform the division algorithm for polynomials to obtain a polynomial plus a proper fraction. We know how to integrate the polynomial; the proper fraction gets referred to the next question in the decision tree.

In this case, $ \frac{x^2\minus{}1}{x^2\plus{}1}\equal{}1\minus{}\frac{2}{x^2\plus{}1}.$

Can you take it from there?

Another, less clever way to do it is using long division in hopes of getting a term $ \frac{a}{x^{2}\plus{}1}$ since the integral of that appeals to the well known $ a\tan^{\minus{}1}(x)$

That was exactly what I was saying: "Division algorithm for polynomials" and "long division" mean the same thing.

There's no particular point in saying "I hope I get a certain form." You will get whatever comes. For the denominator $ x^2\plus{}a^2,$ we will in general get $ \frac{bx\plus{}c}{x^2\plus{}a^2}.$ Some specific cases may, accidentally, have either $ b\equal{}0$ or $ c\equal{}0.$ But we do have to know how to handle the integrals of both $ \frac{x}{x^2\plus{}a^2}$ and $ \frac{1}{x^2\plus{}a^2}.$


function
calculus
derivative
calculus computations

Let x and y be functions of time t that are related by the equation y^2 = 2 +xy. At time t = 5, the value of y is 3 and dy/dx = 6. Find the value of dx/dt at time t = 5.

If $ y^2\equal{}2\plus{}xy$ and if $ x$ and $ y$ are differentiable functions of $ t,$ then

$ 2y\frac{dy}{dt}\equal{}x\frac{dy}{dt}\plus{}y\frac{dx}{dt}.$ (Chain rule and product rule.)

Question: What derivative is supposed to be $ 6?$ Is it $ \frac{dy}{dx}$ or is it $ \frac{dy}{dt}\,?$


AMC AIME probability counting distinguishability Harvard college

A licence plate is 3 letters followed by 3 digits. If all possible licence plates are equally likely, what is the probability that a plate has either a letter palindrome or a digit palindrome (or both)?

I thought AIME had integer solutions? Anyways, I thought this was pretty easy. I got

[hide]

37/260

[/hide]

I might be wrong, though.

Um, i got a different answer.

Show your work, please

[hide]

Well, there are 26:^2: * 10:^2: liscence plates that are both palindromes. There are 26:^2: * 10:^3: + 26:^3:*10:^2: liscence plates that have at least one palindrome.....

CRAP! IT OVERLAPS!!!!! ARGHHHHHHHHH!!!

[/hide]

I hate stupid mistakes.

Yea, u gotta subtract the: [hide]overlaps[/hide]

[color=cyan]I don't think licence plates have the digit 0.[/color]

[color=cyan]They have a distinguishable 0 and O?  In NY, we only have one of them.  I don't know which one.  I think a 0, actually, and no O.[/color]

Yes, in California we have distinguishable 0 and O. I know someone who has a license plate with an O and a 0.

[hide]

The total possible number of numberplates is 26^3 x 100^3. The probability we *don't* have either palindrome is 26*26*25 (after we choose the first two letters we have 25 choices for the last) * 100*100*99 (same reasoning). So the probability there isn't one is 25*99/(26*100)=99/104.. So I get 5/104.

[/hide]

i dont understand any of that :(, arent the the 3 digits and 3 letters separate entities?

but arent the answers supposed to be integers between 0-999 or somat?

the question is probably: the probability is of the form m/n; find m+n. That's how the AIME almost always does it. Anyways, it doesn't really matter.

ah, so i guess that darkq reworded it. aaaaaaah

Daah I meant 10^3 not 100^3. .

So that gives [hide]25/26 * 9/10 = 45/52 so the final answer is 7/52. [/hide]

yeah, i was confused by that, but i didnt want to reply, cause i thought i would be wrong.

Yes TripleM.

Your revised answer is correct

Does three of the same numbers/letters count as a palindrome? (like 555 or LLL)

I thought those count as a palindrome, so I came up with [hide]2/26*2/10=1/65[/hide]...

and crap is a good word, right? So I could say "Oh crap!" riiiiiiight?

show your solution so we can see where you messed up.

As I said before, I counted three same numbers/alphabets such as 666 and KKK as palindromes. The outlines are the same as TripleM's, except for the number of possible palindromes are 2*26*26*2*10*10, not 1*26*26*1*10*10 (or as TripleM writes, the number of number plates that don't have a palindrome is 24*26*26*8*10*10 in my calculation instead of 25*26*26*9*10*10) After cancelation of terms I got 4/260 so I got 1/65.

I think it depends on how you define palindrome...(Most define it as the same read from the front and the back, for example "a man, a plan, a canal, Panama!") I thought "IIIIIIIIIIIIII" is also a palindrome. (Same letters throughout)

Yeah they are counted as palindromes - two things:

One: Why have you got 2 instead of 1? Just consider the 3 letters for example. There are 26 ways of choosing the first letter, 26 ways of choosing the 2nd letter, and then *1* way of choosing the last letter - it has to be the same as the first, there aren't two choices.

Two - you have worked out the probability that there is *both* a letter palindrome and a number palindrome. You want the probability of either, thats why you work out the probability of neither and then subtract that from 1.

Oh, I see. whoops. screwed up, as usual :)
I'm going into a routine: see a problem :arrow: works on it (for a long time) :arrow: post solution :arrow: finds out I screwed up
... :lol:  :cry:

[color=cyan]Hey, I just saw a car from Virginia that has an ampresand (you know, a &) on its license plate.  Is that a number or a letter?

If I were in Virginia, and if I knew how to drive, and if I could afford a car, I would get the licence plate &&& &&&.[/color]

but i thought ur driving to harvard???

but u just said u don't know how to drive???

[color=cyan]Being driven, not driving.[/color]


Solve for x,y
$ x^3\minus{}2x^2\minus{}1\equal{}y^3\minus{}3y$
$ x^3\plus{}y^3\minus{}2\equal{}3y(xy\minus{}1)$

hello, it has the solutions
$ \{x\equal{}1,y\equal{}1\}$ and $ \{x \approx \minus{}0.42273,y \approx 0.52622\}$.
Sonnhard.


topology
real analysis
real analysis unsolved

Is an open subset on the complex plane a union of open balls--by balls I mean circles in 2D fully shaded except the line that drew the circle, you know ;)  ?

sorry for this really dumb question.

Yes, but not necessarily a finite union. Recall that for every point $ x$ in an open set $ U$, there exists $ r_x>0$ such that $ D(x,r_x)\subset U$. Then $ U\equal{}\bigcup_{x\in U}D(x,r_x)$.


function
real analysis
real analysis unsolved

[quote="mlok"]Have you tried equating $dS/da$ to zero?[/quote]

right. :blush:

Also my book has a typo: it should be $a$ satisfy $M_{0}=M_{1}a^{1/p_{1}-1/p_{0}}$, then
everythings clicks.


Find the sum of the next three numbers of 1, 2, 5, 12, 27, 58, ___, ___, ___.

this is tricky :/

[hide]
take differences:

1 3 7 15 21 53 117....

take differences of THOSE:

2 4 8 16....  :)

see a apattern?  next two terms are 111, 228 i think.[/hide]


When will we find out if we qualify for the USAMO from the AIME I? Thanks!

I think AMCDirector said beginning of April, because they need to tally the results from the AIME II.

Exactly.  I think around the 6th.  If I understand correctly, as soon as the AIME-II gets sent in, there's a frenzy of scoring, followed by a frenzy of making qualifier lists.

Please read all the announcement posts at the top of this forum before asking questions like this.  Thanks.


what is the measure of the smaller angle formed by the vectors (-5,-2) and (5,7) to the nearest tenth of a degree.
149.2
146.5
147.3
139.6
none of these

[quote="davidyko"][hide]Well, they can be obtuse as long as they aren't reflex angles.
If you sketch the vectors on the Cartesian system, it is apparent that said angle is equal to $ (180^{\circ} - \tan^{ - 1}{(\frac {7}{5})}) + \tan^{ - 1}{(\frac {2}{5})} = \boxed{147.3^{\circ}}$
(I do hope we can use calculators on this.)[/hide][/quote]

Could explain what you just did? I did understand anything that you just wrote. I dont' know much about vector stuff yet.

Thanks a lot.

Sure. Basically, the angle formed between any line through the origin and the x-axis is given by $ \tan^{\minus{}1}{y/x}$ because if one drops a perpendicular from the point to the x-axis, a right triangle is created.
So if you look at the pic below, you can see that $ \angle ACD \equal{} \tan^{\minus{}1}{7/5}, \angle BCE \equal{} \tan^{\minus{}1}{2/5}$, so the angle we're looking for, $ \angle ACB \equal{} 180^{\circ} \minus{} \tan^{\minus{}1}{7/5} \plus{} \tan^{\minus{}1}{2/5}$.

[hide="The angle between two tail to tail vectors is"]
$ \cos \theta \equal{} \frac {\vec{\mathbf{u}} \cdot \vec{\mathbf{v}}}{\left\| \vec{\mathbf{u}} \right\| \left\| \vec{\mathbf{v}} \right\|}$.
$ \cdot$ is the dot or scalar product.
$ \|\|$ is the magnitude.
[/hide]

[quote="Bictor717"]Seeing as how all of the choices are obtuse, I don't see how any of them can be the [i]smaller[/i] angle.[/quote]

Example: 120 and 240.


integration
calculus
limit
calculus computations

[quote="countdownmath"]If $f$ is non-negative and $\int_{1}^{\infty}f(x)dx$ exists as an improper integral, must $\lim_{x\to \infty}f(x) = 0$?
Must $f$ be bounded?
Additionally, what can you say if $\lim_{x\to \infty}f(x)$ exists?[/quote]
You didn't say if $f(x)$ is integrable on $[1,+\infty]$ can we assume that?

Well let's say $\lim_{x\to \infty}f(x)$ is not zero, either it doesn't exist (ignore that case for now) or its above zero (because $f\geq 0$).

Say,
$\lim_{x\to \infty}f(x) = L$ where $L$ is positive and finite (ignore $L=+\infty$ for now).

Then pick $0<M<L$ then $f(x) > M$ for sufficiently large $x$.

But then,

$M<f(x)$
And,
$+\infty=\int_{1}^{\infty}M dx \leq \int_{1}^{\infty}f(x) dx$
Thus,
$\int_{1}^{\infty}f(x) dx =+\infty$ a contradiction.

If $L=+\infty$ then we have a similar situation.

But, can it be the case that $\lim_{x\to \infty}f(x)$ just not to exist?
Im note sure, there is a chance.

So again, IF the limit exists IT must be zero.
----

$f(x)$ DOES NOT need to be bounded on this infinite interval.
Define $f(x)=0$ for all non-integer points and $f(n)=n$ for positive integers. So we see $f(x)$ is unbounded. But yet it is continous on countable many points thus it is integrable and is zero.

That example can be made continuous; use thin peaks (width comparable to $\frac1n\cdot a_{n}$ for some convergent series $\sum a_{n}$) instead of single points.


Find the constant term in the expansion of $ \left(1\plus{}x\plus{}\frac{1}{x^{2}}\plus{}\frac{1}{x^{3}} \right)^{10}$

[quote="Dr Sonnhard Graubner"]hello, using the multinomial theorem we  get $ 11851$ as the constant term.
Sonnhard.[/quote]

Please give a detailed solution!!!

This method uses strictly the binomial theorem.  There is another way to do it via multinomial theorem, but the other guy can post that if he wants to.

[hide]Factor $ \frac{1}{x^3}$ out of the right terms of the equation to leave us with $ ( 1 \plus{} x \plus{} \frac{1}{x^3}(1 \plus{} x))^{10}\equal{} (1\plus{}x)^{10} (1 \plus{} \frac{1}{x^3})^{10}$

Now use binomial theorem.  Notice that to get the constant term we need $ \displaystyle\sum_{i\equal{}0}^{3}\dbinom{10}{3i}\dbinom{10}{i}$
Since for every $ \frac{1}{x^3}$ term we need three $ x$ terms.[/hide]


Find the last six digits of $ 1999^{1999}$ AND explain how you found them.

Do you know the binomial theorem?

$ (a \plus{} b)^{n} \equal{} \sum_{k \equal{} 0}^{n}\binom{n}{k}a^{n \minus{} k}b^k$.

What happens when $ a \equal{} 2000$ and $ b \equal{} \minus{}1$?

Edit: fixed the issues, thanks. Guess my mind was just elsewhere that day :)

[quote="grn_trtle"]Do you know the binomial theorem?

$ (a \plus{} b)^{n} \equal{} \sum_{k \equal{} 0}^{n}\binom{n}{k}a^{n \minus{} k}y^k$.

What happens when $ a \equal{} 2000$ and $ b \equal{} 1$?[/quote]

Small typos, should this be:

$ (a \plus{} b)^{n} \equal{} \sum_{k \equal{} 0}^{n}\binom{n}{k}a^{n \minus{} k}b^k$.

What happens when $ a \equal{} 2000$ and $ b \equal{} \minus{}1$?


function
integration
real analysis
real analysis unsolved

Is primitive of a periodic function periodic?

Please help! :help:

If by "primitive" you mean "antiderivative," note that a constant function is periodic.

In order for the antiderivative of a periodic function $ f$ to be periodic, there's a necessary condition. We must have
\[ \int_0^Pf(x)\,dx\equal{}0\]
where $ P$ is the period of $ f.$

Note that if $ g(x)\equal{}\int_a^xf(t)\,dt$ so that $ g'(x)\equal{}f(x)$ and $ g$ is periodic, then the Fourier coefficients are related by
\[ \widehat{f}(k)\equal{}\frac{2\pi ik}{P}\,\widehat{g}(k)\].
This has no chance of working unless $ \widehat{f}(0)\equal{}0,$ which is another way of saying that the average value of $ f$ over a period is zero.

We then get that $ \widehat{g}(k)\equal{}\frac{P\widehat{f}(k)}{2\pi ik}$ for $ k\ne 0,$ but we need to separately compute $ \widehat{g}(0).$


let p(x) be any polynomial with integer coefficients such that p(21)=17,p(32)=-247,p(37)=33.prove that if p(n)=n+51,then n=26

Well, first we take this n and we use Bezout's theorem to find it. If p(n) = n+51 then n-37 divides p(n)-33. Here checking yields that only solutions are n=48, 92,-18, 32,36 or 26. Now checking for this values (by bezout) gives that only solution is n=26.
(for example if n=48 then 48-21 divides p(48)-p(21) but this gives us no solution. You can continue as i did.)


The classic binomial expansion of $ (x \plus{} y)2008$ is written. How many of the $ 2009$ terms will have odd coefficients. :D

yes gaurav, sry did not use latex properly

@gaurav patil
Are u the INMO topper {for the whole country} in this year's INMO?
The name is same. :)

unfortunately and fortunately so, I am :)

Could you pls tell me what was the highest score(=your score) and the lowest score(=cut-off score) of this year's INMO?

u can work out the method and post it
but i am just posting the final answer

[hide]Answer:
128
[/hide]

highest INMO score = 73
cuttoff =52(i think)

Aravind i am pretty sure the answer is 128.
Could you please post  your method  as  to how you got the  answer as 126.

I am not able to post it in latex.
But I will check my workng again.

This problem can be tackled using Lucas theorem.
Lucas theorem states that
 $ (xC_y) = (x{}_1C{}_{y1})(x{}_2C{}_{y2}).........(x{}_nC{}_{yn})$ $ mod p$ where $ p$ is any prime.
 Here $ {x{}_1x{}_2........x_n}  = x$ and  $ {y{}_1y{}_2........y_n}  = y$ in base $ p$ representation.
All the coefficients are of the form $ (2008C_n)$ where $ n$ ranges from $ 0$ to $ 2008$.
We first write $ 2008$ in base two since all odd coefficients are of the form $ 2k +1$
$ 2008 = 11111011000_2$
Now according to lucas theorem to find the remainder when the coefficients are divided by $ 2$ we write $ n$ ranging from $ 0$ to $ 2008$. in base $ 2$.
Let $ n = a{}_1a{}_2a{}_3a{}_4.........a_11$ in base 2.
Then the remainder when the coefficient  $ (2008C_n)$ is divided by $ 2$ is the remainder when $ (1C{}_a{}_1)(1C{}_a{}_2)......(0C{}_a{}_{11})$ is divided by 2. Here each of the corresponding digit of $ 2008$ in base $ 2$ is written in $ (aC_b)$ form where corresponding $ b$ is the corresponding digit in $ n$ base $ 2$.
So since we have 4 digits in 2008 base 2 as zero for odd coefficients we should have corresponding digits in $ n$ base 2 as $ 0$.(Since or else $ (0C_1)$ is conventionally zero and $ (0C_0)$ is conventionally 1.)
Now in the corresponding $ 7$ other digits the digits can either be $ 0$ or $ 1$
This is because $ (1C_0)$ and $ (1C_1)$ are both 1.
This implies all the seven corresponding places have two choices $ 0$ or $ 1$.
So the answer is $ 2^{7} = 128$
$ QED$
If this solution is unclear or if you want to know more about Lucas theorem refer to the link below.
[url]http://en.wikipedia.org/wiki/Lucas%27_theorem[/url]
P.S I really like this problem because it is an application of a number theoretic result in combinatorics. :D


The parallel sum is defined as A:B = (A^-1 + B^-1)^-1. I need to show that A:B = A - A((A + B)^-1)A. So far I got it down to A:B= A((B + A)^-1)B = A((B + A)^-1)(A + B -A). Not sure where to go from here. Any help would be greatly appreciated. Thanks!!


The three symmetric means of the positive quantities $ a,b,c$ are $ \sigma_1\equal{}(a\plus{}b\plus{}c)/3,$ $ \sigma_2\equal{}(ab\plus{}ac\plus{}bc)/3,$ and $ \sigma_3\equal{}abc.$

Prove that $ 6\sigma_1^3\plus{}\sigma_3 \geq 7 \sigma_1 \sigma_2$.  It is easy to do this if you substitute, because you just get $ \sum a^3 \geq \sum a^2b,$ where both sums are symmetric, and this is an immediate consequence of Muirhead’s Theorem.  Can you prove it using Newton’s and Maclaurin’s Inequalities of symmetric means?  And is there a general method of approaching inequalities relating symmetric means?

Yes, there is a general method of approaching inequalities relating symmetric means. I called it is "p,q,r method" which $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$
For the multi-variables, for example having at least 4 variables in symmetric form, we should use Roll's theorem to reducing the inequality into less variables :wink:

Thank you for your reply, zaizai-hoang.  Can you illustrate the method on this particular problem?  And do you have another post where you have illustrated the method in detail?  Thank you for your help.

dabug, if you know Russian then see here my post ( the problem number 565 ):
http://kvant.mirror1.mccme.ru/1980/04/resheniya_zadachnika_kvanta_ma.htm

maybe this would help:
[url]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=11002[/url]

I've figured out how to solve the problem, although I had to use Schur's Inequality in addition to Newton's Inequality of Symmetric Means.

Schur's Inequality $ x(x\minus{}y)(x\minus{}z)\plus{}y(y\minus{}x)(y\minus{}z)\plus{}z(z\minus{}x)(z\minus{}y) \geq 0$ can be written in the form $ 3 \sigma_1 ^3 \plus{}\sigma_3 \geq 4 \sigma_1 \sigma_2.$  Newton's Inequality says that $ \sigma_1 ^2 \geq \sigma_2,$ which is equivalent to $ 3\sigma_1 ^3 \geq 3\sigma_1 \sigma_2.$  When Schur is added to this last inequality, we get the desired $ 6\sigma_1 ^3 \plus{} \sigma_3 \geq 7 \sigma_1 \sigma_2.$

Michael, thanks for the article on proving Newton's Inequalities.  I did take Russian for a few years, so I could stumble through it.  There's another proof in Steele's book on The Cauchy-Schwartz Master Class.


inequalities
geometry
trigonometry
geometry proposed

Given are $\triangle ABC$ with any $M$ prove that
$(MB\cdot MC+MC\cdot MA+MA\cdot MB)\left(\frac{a}{MA}+\frac{b}{MB}+\frac{c}{MC}\right)\ge 12S$ with $S$ is area of $ABC$
equal when $ABC$ is equaliteral triangle and $M$ is center of it!

As promised, here is how I do it...

[color=blue][b]Problem.[/b] For any point M in the plane of a triangle ABC, prove that

$\left(MB\cdot MC+MC\cdot MA+MA\cdot MB\right)\left(\frac{a}{MA}+\frac{b}{MB}+\frac{c}{MC}\right)\geq 12S$,

where S is the area of triangle ABC.[/color]

[i]Solution.[/i] With the notation $\sum$ for cyclic sums, the inequality we have to prove becomes $\sum MB\cdot MC\cdot\sum\frac{a}{MA}\geq 12S$.

By the Cauchy-Schwarz inequality,

$\sum MB\cdot MC\cdot\sum\frac{a}{MA}\geq\left(\sum\sqrt{\left(MB\cdot MC\right)\cdot\frac{a}{MA}}\right)^{2}=\left(\sum\sqrt{\frac{MB\cdot MC\cdot a}{MA}}\right)^{2}$.

By the inequality $\left(x+y+z\right)^{2}\geq 3\left(yz+zx+xy\right)$ which holds for any three reals x, y, z, we have

$\left(\sum\sqrt{\frac{MB\cdot MC\cdot a}{MA}\right)^{2}\geq 3\sum\sqrt{\frac{MC\cdot MA\cdot b}{MB}}\cdot\sqrt{\frac{MA\cdot MB\cdot c}{MC}}}$
$=3\sum MA\cdot\sqrt{bc}$.

By the GM-HM inequality, $\sqrt{bc}\geq\frac{2bc}{b+c}$. But by the Mollweide formulas, $\left(b+c\right) / a=\cos\frac{B-C}{2}/ \sin\frac{A}{2}$, and since $\cos\frac{B-C}{2}\leq 1$ (as every cosine is $\leq 1$), this yields $\left(b+c\right) / a\leq 1/ \sin\frac{A}{2}$, so that $b+c\leq a/ \sin\frac{A}{2}$, and thus $\frac{2bc}{b+c}\geq\frac{2bc}{a/ \sin\frac{A}{2}}=\frac{2bc\sin\frac{A}{2}}{a}$. Together with $\sqrt{bc}\geq\frac{2bc}{b+c}$, this becomes $\sqrt{bc}\geq\frac{2bc\sin\frac{A}{2}}{a}$. Hence,

$3\sum MA\cdot\sqrt{bc}\geq 3\sum MA\cdot\frac{2bc\sin\frac{A}{2}}{a}=12\sum MA\cdot\frac{\frac12 bc\sin\frac{A}{2}}{a}$.

But the area S of triangle ABC equals $S=\frac12 bc\sin A$; hence, $\frac12 bc=S/ \sin A$. Hence,

$12\sum MA\cdot\frac{\frac12 bc\sin\frac{A}{2}}{a}=12\sum MA\cdot\frac{S/ \sin A\cdot\sin\frac{A}{2}}{a}=12\sum MA\cdot\frac{S\sin\frac{A}{2}}{a\cdot\sin A}$
$=12\sum MA\cdot\frac{S\sin\frac{A}{2}}{a\cdot 2\sin\frac{A}{2}\cos\frac{A}{2}}=12\sum MA\cdot\frac{S}{a\cdot 2\cos\frac{A}{2}}=6S\sum\frac{MA}{a\cos\frac{A}{2}}$.

But [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=593372#593372]http://www.mathlinks.ro/Forum/viewtopic.php?t=63462 post #4[/url] Theorem 4 yields $\sum\frac{MA}{a\cos\frac{A}{2}}\geq 2$. Hence,

$6S\sum\frac{MA}{a\cos\frac{A}{2}}\geq 6S\cdot 2=12S$.

Combining all the above inequalities, we get

$\sum MB\cdot MC\cdot\sum\frac{a}{MA}\geq 12S$,

qed..

  Darij

ok thanks for very good solution! Darij
In your solution there is a nice result:
$\sqrt{bc}MA+\sqrt{ca}MB+\sqrt{ab}MC\ge 4S(*)$
and a nice corollary
$(b+c)MA+(c+a)MB+(a+b)MC\ge 8S$
$\Leftrightarrow (a+b+c)(MA+MB+MC)\ge 8S+(aMA+bMB+cMC)\ge 12S\Rightarrow MA+MB+MC\ge 6r$
through it we see $MA+MB+MC\ge 6r$ is weaker than $(*)$ quite more!  :)


$ a,b,c\,\geq\,0$   with   $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
prove that :

$ \frac {1}{3 \plus{} a^2 \minus{} 2bc} \plus{} \frac {1}{3 \plus{} b^2 \minus{} 2ac} \plus{} \frac{1}{3 \plus{} c^2 \minus{} 2ab}\,\preceq\,\frac {9}{8}$

$ \sum(\frac{1}{3\plus{}a^2\minus{}2bc})\,\geq\,\sum(\frac{1}{3\plus{}a^2\minus{}b^2\minus{}c^2}\equal{}\sum\frac{1}{2(a^2\plus{}1}$

now we have to prove that :

$ \sum\frac{4}{a^2\plus{}1}\,\preceq\,9$

someone can continue ??  :maybe:

mehdi cherif that inequality you got is wrong take $ a\equal{}b\equal{}(1/2)^{1/2}$, $ c\equal{}0$

Mmmm...I wonder if we could parameterise the inequality using spherical polars...

i dont understand what you want to say. explain plz

Mehdi Cherif // your solution is wrong.
First of all, on the first line, there should be a $ \leq$ instead of a $ \geq$ sign.
Moreover, $ \sum \frac{4}{a^2 \plus{} 1} \geq 9$ (Easy by AM-HM or Cauchy Inequality).


Heres a question that im stuck on:

In a certain area, an average of 1.5 people die each week.

What is the probability that 5 people will die in a week?

If there is enough info given to answer the question, then it seems it has to be a Poisson distribution.

Depends if the marine corps was involved! :lol:


wow, i never knew this forum was here!  um, so, anyways, can anyone sponsor a hillsborough team to compete at the chiles invitational?  it's a little far for my sponsor to take us, and it's probably an overnight thing for us too if we go.  if anyone is willing to be our official sponsor, we'd really appreciate it!  we'll behave ourselves, and we promise to make curfew!


function
LaTeX
advanced fields
advanced fields unsolved

Hi!

Let f and g be entire functions of finite order $\lambda$ and suppose that
$f(a_n) = g(a_n)$ for a sequence ${a_n}$ s.that $\sum{|a_n|^{(-(\lambda+1))} =\infty}$.
Show that $f = g$.

Ok, so if $h(x)=f(x)-g(x)$, then to show is that $h = 0$.
Since f and g are of finite order, h is also of finite order.
So by Hadamard's Factorization Theorem, $h$ has a finite genus.

Mmh... But has somebody an idea how to show that $h = 0$?

Thanks!

Since your post seems to indicate you know some $\LaTeX$, I'll just say this: put dollar signs around your code when posting here.


Let $ a,b,c$ be the positive number and $ a\plus{}b\plus{}c\equal{}1$
Prove that:
$ \sqrt{a^{2}\plus{}b}\plus{}\sqrt{b^{2}\plus{}c}\plus{}\sqrt{c^{2}\plus{}a} \ge 2$

It's not so hard but I hope there'll get some nicer solutions.

[quote="zjlywjh001"]Let $ a,b,c$ be the positive number and $ a \plus{} b \plus{} c \equal{} 1$
Prove that:
$ \sqrt {a^{2} \plus{} b} \plus{} \sqrt {b^{2} \plus{} c} \plus{} \sqrt {c^{2} \plus{} a} \ge 2$
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106782


Prove that for any positive reals $x, y, z$ the following inequality is true: \[\frac{1}{(x+y+z)^{2}}\cdot\left(5-\frac{4xyz}{(x+y)(y+z)(z+x)}\right)^{2}-\frac{4(x+y+z)}{(x+y)(y+z)(z+x)}\geq\frac{9}{4(xy+yz+zx)}.\]


group theory
abstract algebra
function
linear algebra
matrix
Euler
induction

For a free abelian group $ G$, let $ a_n(G)$ be the number of subgroups of $ G$ with index equal $ n$.
1. Prove that $ a_n(\mathbb{Z}\oplus \mathbb{Z}) \equal{} \sigma (n)$.
Futhermore let $ \zeta_{\mathbb{Z}^d}(s) \equal{} \sum_{n \equal{} 1}^\infty a_n(G)n^{ \minus{} s}$ for $ s\in\mathbb{C}$.
2. Prove that $ \zeta_{\mathbb{Z}^d}(s) \equal{} \zeta (s)\zeta (s \minus{} 1)\cdots \zeta (s \minus{} d \plus{} 1)$ and this function has an Eulerian product as in numbertheoretic and grouptheoretic sense.

1. is wrong.

let $ U$ be a subgroup of $ \mathbb{Z}^2$ with finite index $ k > 1$. then $ U$ is free abelian of rank $ 1$, say $ U \equal{} \langle (m,n) \rangle$ with integers $ m,n$. if $ m \equal{} 0$, the quotient $ \mathbb{Z}^2 / U \equal{} \mathbb{Z} \oplus \mathbb{Z}/n$ is not finite, contradiction. thus $ m \neq 0$ and $ n \neq 0$. we may assume $ m > 0$. then $ \langle (m,n) \rangle$ can be identified with the pair $ (m,n)$. now, if $ m,n$ are coprime, there are integers $ u,v$ such that $ um \plus{} vn \equal{} 1$. then $ \begin{pmatrix} m & n \\
\minus{} v & u \end{pmatrix} \in SL_2(\mathbb{Z})$. thus $ \mathbb{Z}^2 / U \cong \langle ( \minus{} v,u) \rangle \cong \mathbb{Z}$. hence, for general $ m,n$, we have $ \mathbb{Z}^2 / U \equal{} \mathbb{Z}^2 / gcd(m,n) \langle (m/gcd(m,n),n/gcd(m,n)) \rangle \cong \mathbb{Z} / gcd(m,n)$. thus $ U$ is of finite index $ > 1$ if and only if $ m,n$ are not coprime, and then the index is $ gcd(n,m)$.

thus for $ k > 1$ we have $ a_k(\mathbb{Z}^2) \equal{} \#\{(m,n) \in \mathbb{Z}_{ > 0} \times \mathbb{Z} : gcd(m,n) \equal{} k\}$, which is infinite.

I cannot follow you here:
[quote="-oo-"]let $ U$ be a subgroup of $ \mathbb{Z}^2$ with finite index $ k > 1$. then $ U$ is free abelian of rank $ 1$, ...[/quote]If $ U$ has finite index in $ \mathbb Z^2$, then it is automatically free of rank 2.

On the other hand problem 1 could be tackled by the fact that every $ G$ of index $ n$ contains $ n\mathbb Z^2$ such that one has to count the subgroups of $ \mathbb Z^2/n\mathbb Z^2$ by - just guessing - using the Chinese remainder theorem or the structure theorem of finite abelian groups.

Problem 2 should drop out of the Euler product formulae - still guessing. ;)

[quote="Third Edition"]then it is automatically free of rank 2.[/quote]
no, look at the subgroups mentioned above. but in fact I missed the fact that there are proper subgroups of rang $ 2$ and they may also be of finite index.

[quote="Third Edition"]On the other hand problem 1 could be tackled by the fact that every $ G$ of index $ n$ contains $ n\mathbb Z^2$ such that one has to count the subgroups of $ \mathbb Z^2/n\mathbb Z^2$ by - just guessing - using the Chinese remainder theorem or the structure theorem of finite abelian groups.[/quote]

I tried to follow this approach. I want to summarize my progress:
- both sides are multiplicative (with respect to a product of coprime numbers), so that we may assume that $ n \equal{} p^k$ is a prime power.
- there are $ p^k \plus{} p^{k\minus{}1}$ cyclic subgroups of $ (\mathbb{Z}/p^k)^2$ of order $ p^k$. in particular, we know the claim in the case of $ k\equal{}1$.
- the non-cyclic subgroups correspond to subgroups of $ (\mathbb{Z}/p^{k\minus{}1})^2$, but since the order stays $ p^k$ we cannot use an induction following this idea.

[quote="-oo-"][quote="Third Edition"]then it is automatically free of rank 2.[/quote]
no, look at the subgroups mentioned above. but in fact I missed the fact that there are proper subgroups of rang $ 2$ and they may also be of finite index.[/quote]Well, I still cannot follow your reasoning. So here the argument that every $ U$ of finite index $ k$ in $ \mathbb Z^2$ has rank 2: Since $ \mathbb Z^2/U$ is annihilated by $ k$, we have the inclusions, say, of $ \mathbb Z$-modules:
\[ k\mathbb Z^2 \subset U\subset \mathbb Z^2
\]
Then it immediately follows that $ U$ has rank 2 (e.g., by tensoring with $ \mathbb Q$ - which is an exact functor as localization) - or, alternatively, $ ke_1, ke_2$ are linearly independent elements in $ U$ which shows that the rank of $ U$ is 2 at least.

ok my first post is wrong. correct calculations of Z^2/<(m,n)> can be found here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=112335

here is the full proof. the missing part was found by a friend of me.

[b]lemma 1.[/b] if $ G$ is an abelian group of order $ nm$ where $ n,m$ are coprime natural numbers, then $ G \equal{} Tor_n(G) \oplus Tor_m(G)$, where $ Tor_n(G) \equal{} \{x \in G : nx \equal{} 0\}$. furthermore, we have $ |Tor_n(G)| \equal{} n$.

proof: the order of every element in $ Tor_n(G) \cap Tor_m(G)$ divides $ n$ and $ m$, thus the sum is direct. now choose integers $ u,v$ such that $ un \plus{} vm \equal{} 1$. since $ nm G \equal{} 0$, we have $ nG \subseteq Tor_m(G)$ and $ mG \subseteq Tor_n(G)$, thus $ G \subseteq Tor_n(G) \plus{} Tor_m(G)$. if $ p$ is some prime divisor of $ |Tor_n(G)|$, cauchy's theorem shows that $ p$ is a divisor of $ n$. thus $ |Tor_n(G)|$ is a divisor of $ n$ and $ |Tor_m(G)|$ is a divisor of $ m$, but equality must hold since the product must be $ |G| \equal{} nm$.

[b]lemma 2.[/b] if $ G$ is an abelian group and $ a_n(G)$ denotes the number of subgroups of index $ n$, we have for coprime natural numbers $ n,m$:

$ a_{nm}(G) \equal{} a_n(G) * a_m(G)$

proof: we construct inverse bijections

$ \{\text{subgroups of index } nm\} \leftrightarrow \{\text{subgroups of index } n\}\times \{\text{subgroups of index } m\}.$

if $ U,V$ are subgroups such that $ |G: U| \equal{} n$ and $ |G : V| \equal{} m$, we have $ G \equal{} U \plus{} V$ since $ |G : U \plus{} V|$ divides $ n$ and $ m$. the group-theoretic generalization for the chinese remainder theorem shows $ G/(U \cap V) \cong G/U \times G/V$. thus $ U \cap V$ has index $ nm$. remark that $ G/U \equal{} Tor_n(G/(U \cap V))$ and $ G/V \equal{} Tor_m(G/(U \cap V))$.

now assume that $ W$ is a subgroup of index $ nm$. by lemma 1, we have $ G/W \equal{} Tor_{n}(G/W) \oplus Tor_{m}(G/W)$. writing $ Tor_{n}(G/W) \equal{} U/W$, $ U$ is a subgroup of index $ [G: U] \equal{} [G : W]/[U : W] \equal{} n$ by lemma 1. in the same way we get a subgroup $ V$ of index $ m$. finally, remark that $ W \equal{} U \cap V$ since $ W \subseteq U \cap V$ and both groups have the same finite index.

[b]now[/b] we are ready to prove $ a_n(\mathbb{Z}^2) \equal{} \sigma(n)$, where $ \sigma(n)$ is the sum of divisors of $ n$. it is clear that $ \sigma$ is multiplicative on coprime numbers, and the same is true for $ a_n$ (lemma 2). thus we may assume that $ n \equal{} p^k$ is a prime power. the case $ k \equal{} 0$ being trivial, we assume $ k \geq 1$. lagrange shows that subgroups of index $ n$ in $ \mathbb{Z}^2$ correspond to subgroups of index $ n$ in $ \mathbb{Z}^2 / n \mathbb{Z}^2 \equal{} (\mathbb{Z}/n)^2$.

now let $ U$ be a [i]cyclic[/i] subgroup of index $ n$, i.e. order $ n$ in $ (\mathbb{Z}/n)^2$. then $ U$ is determined by its generator, namely an element $ (a,b)$ of $ (\mathbb{Z}/n)^2$ such that $ a$ or $ b$ generates $ \mathbb{Z}/n$. there are $ 2 * (n \minus{} \phi(n)) * \phi(n) \plus{} \phi(n) * \phi(n) \equal{} (p^k \plus{} p^{k \minus{} 1}) \phi(n)$ such elements.  they generate the same subgroup iff they are multiplied with a unit in $ \mathbb{Z}/n$, of which there are $ \phi(n)$. thus there are $ p^k \plus{} p^{k \minus{} 1}$ cyclic subgroups of order $ p^k$ in $ (\mathbb{Z}/n)^2$. in particular, we have proven the claim in the case of $ k \equal{} 1$, so let's assume $ k \geq 2$.

in the following, to count the non-cyclic subgroups $ U$, we use an induction on $ k$. since $ \mathbb{Z}$ is noetherian, $ U$ is generated by two elements $ (a,b),(c,d)$. since the subgroups of $ \mathbb{Z}/n$ are totally ordered, we may assume that $ a$ is a multiple of $ c$ and then, by subtracting a multiple of $ (c,d)$ from $ (a,b)$, that $ a \equal{} 0$. since $ U$ is not cyclic, we have $ b \neq 0$, so that $ (0,p^{k \minus{} 1})$ is contained in $ U$. in the same manner, we see that $ (p^{k \minus{} 1},0)$ is contained in $ U$. hence, $ U$ contains the (non-cyclic) subgroup $ \langle (0,p^{k \minus{} 1}) , (p^{k \minus{} 1},0) \rangle \cong (\mathbb{Z}/p)^2$. hence, the non-cyclic subgroups of order $ p^k$ in $ (\mathbb{Z}/n)^2$ correspond to subgroups of order $ p^{k \minus{} 2}$ in $ (\mathbb{Z}/n)^2 / (\mathbb{Z}/p)^2 \cong (\mathbb{Z}/p^{k \minus{} 1})^2$. but these lie in the $ p^{k \minus{} 2}$-torsion subgroup $ (\mathbb{Z}/p^{k \minus{} 2})^2$, so that we may apply the induction hypothesis and get the desired result.


Given a sequence x1 = 1 + 1/x2, x2 = 1 + 1/X3 ... Xn = 1+1/X(n+1) and xn = 1+1/x1. Solve the sequence. Find x145.

Hi dreamer.ande.

I've not done this carefully but I hope that what I have done will help you.

Beginning with $ x_{n} \equal{} 1 \plus{} \frac{1}{x_{1}}$, and substituting, I get a continjued fraction with (I think) $ n$ "levels".

Then working backwards from the bottom, I get $ x_{n} \equal{} \frac{F_{n\plus{}2}\plus{}F_{n}x_{n}}{F_{n\plus{}1}\plus{}F_{n\minus{}1}x_{n}}$

where $ F_{n}$ is the $ n$-th Fibonacci number, and sustiuting for $ x_{1}$, I have

$ x_{n} \equal{} \frac{F_{n\plus{}2} \plus{} F_{n} \plus{} \frac{F_{n}}{x_{1}}}{F_{n\plus{}1} \plus{} F_{n\minus{}1} \plus{} \frac{F_{n\minus{}1}}{x_{1}}}$

I hope this is of some use to you.

Can show me how u get xn = Fn+1... Things. the workings. I dun get it why it is related to fibonacci number.

[quote="dreamer.ande"]Xn = 1+1/X(n+1) and xn = 1+1/x1. [/quote]
How come this has two different values?


I know this has been a problem of the forum, but I can't find it!! Could anyone link me to the solution?

Leta $ABC$ be a triangle inscribed in a circle, and consider the tangent to the circle passing through the point $A$. This tangent meets $BC$ in $P$. Now, take $M$ as the middlepoint of the segment $PA$, and $X$ the intersection between the circle and the segment $BM$. And now call $Y$ the intersection between $PX$ and the circunference. Show that $YC$ is parallel to the line $PA$.

Thanks

This problem was posted before as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59213]Prove parallel[/url].


limit
inequalities
logarithms
function
algebra
binomial theorem

[quote="duy_nguyen"]Help Me:  Find: $\large\lim_{x\to n}\large$[/quote]

In your current formulation, you are asking for the limit of the constant sequence
$<\sqrt[n]{n},~\sqrt[n]{n},~\sqrt[n]{n},~...>$

[quote="duy_nguyen"]Help Me:  Find: $\large\lim_{x\to n}\large\sqrt[n]{n}$[/quote]
Maybe are you mean to find $\large\lim_{n\to{+\infty}}\large\sqrt[n]{n}$ $?$
If so, use the following inequality $1\leq\sqrt[n]n<1+\sqrt{\frac{2}{n}}.$

this can just be "l'hopital"-ized

e^ (ln n)/n=e^( (1/n)/(1) )=e^( 0)=1

Or simply done using the squeeze theorem.

[quote="arqady"]$1\leq\sqrt[n]n<1+\sqrt{\frac{2}{n}}.$[/quote]

how do you prove that?

The (extended) binomial theorem works wonders: $\left(1+\sqrt{\frac{2}{n}}\right)^{n}> 1+n \cdot \sqrt{\frac{2}{n}}+\frac{n(n-1)}{2}\cdot \frac{2}{n}> 1+\sqrt{2n}+(n-1) > n$ for $n \ge 2$.

I solved this like altheman.

$\lim_{n \rightarrow \infty}\sqrt[n]{n}= \lim_{n \rightarrow \infty}e^{\frac{\ln n}{n}}$

By L'Hôpital's rule,
$\lim_{n \rightarrow \infty}\frac{\ln n}{n}= \lim_{n \rightarrow \infty}\frac{\frac{1}{n}}{1}= 0$

Therefore:
$\lim_{n \rightarrow \infty}\sqrt[n]{n}= \lim_{n \rightarrow \infty}e^{\frac{\ln n}{n}}= e^{0}= 1$

uhm.Sorry to every one.n to infinity.sr again!!!!!!!!!!!!!!

daofeishi, are $\ln n$ and $n$  differentiable functions, or  sequences ?

If he treats them as differentiable functions and finds a limit, it follows that as sequences they share the corresponding limit.  (The converse, of course, is not true.)

It's a "very very hard" solution of a very simple problem.
Using the equivalence between Cauchy's and Heine's definition of limit, one uses, in fact, the Axiom of Choice. It's not necessary.

Randomizer: what?  Are you claiming that the step \[\lim_{\substack{x \to \infty\\x\in \bf R}}f(x) = L \Longrightarrow \lim_{\substack{n \to \infty \\ n\in \bf N}}f(n) = L\] requires the axiom of choice, or something else?

[quote] the step ... requires the axiom of choice, or something else? [/quote]
Certainly not.  The set of all "real sequences" (that is, sequences with real numbers) contains the set of all "positive integer sequences", therefore JBL´s argument ("limit exist, sequences share limit") applies (through a little technical proof, for sure).      :)

[Randomizer said "It's not necessary." , by the way]

I interpreted randomizer's comment to mean, "daofeishi's solution requires the axiom of choice.  However, there exists a solution to this question that does not rely on it."  I know that the axiom of choice is not required for the step I isolated, but I'm trying to figure out which step of daofeishi's proof randomizer thinks it [i]is[/i] required for.

[quote="JBL"]Are you claiming that the step
\[\lim_{\substack{x \to \infty\\ x\in \bf R}}f(x) = L \Longrightarrow \lim_{\substack{n \to \infty \\ n\in \bf N}}f(n) = L \]
requires the axiom of choice, or something else?[/quote]
Indeed, I mistaked implications, what a shame. AC is needful to obtain ($f(x)\to L$ if $x\to\infty$) from ($\forall_{(a_{n})\to \infty}f(a_{n})\to L$).

How about this?

[hide]Let $y=n^{\frac{1}{n}}$. So we have that
\[\ln y = \left(\frac{1}{n}\right)(\ln n)\]
Then
\begin{eqnarray*}\lim_{n \rightarrow \infty}\ln y &=& \lim_{n \rightarrow \infty}\left(\frac{1}{n}\right)(\ln n) \\ &=& \lim_{n \rightarrow \infty}\frac{\ln n}{n}\\ &\stackrel{L}{=}& \lim_{n \rightarrow \infty}\frac{1}{n}\\ &=& 0 \end{eqnarray*}
Therefore,
\[\lim_{n \rightarrow \infty}\ln y = 0 \ \Rightarrow \ \lim_{n \rightarrow \infty}y=\boxed{\lim_{n \rightarrow \infty}n^{\frac{1}{n}}=1}\][/hide]


conics
email
geometry
geometric transformation
reflection
parameterization
power of a point

I can't remember the number of the message, but I remember seeing this nice problem in one of the emails received from Hyacinthos:

Given four points $A,B,C,D$ in the plane and another point $P$, the polars of $P$ wrt the conics passing through $A,B,C,D$ pass through a fixed point (well, unless $P$ is one of $AB\cap CD,AD\cap BC,AC\cap BD$, in which case the polar is fixed).

[quote="grobber"]Given four points $A,B,C,D$ in the plane and another point $P$, the polars of $P$ wrt the conics passing through $A,B,C,D$ pass through a fixed point (well, unless $P$ is one of $AB\cap CD,AD\cap BC,AC\cap BD$, in which case the polar is fixed).[/quote]

As for now, I have only an atomic bomb solution:

We work in the complex projective plane. We will consider only the situation when the points C and D are the circular points at infinity (since all other situations can be mapped to this one by an appropriate projective transformation - we just have to find a projective transformation mapping two given points C and D to the circular points at infinity). In this situation, all conics passing through these points C and D are circles; in particular, the conics passing through the points A, B, C, D are simply the circles passing through the points A and B. And thus, we have to show that the polars of the point P with respect to all circles through the points A and B pass through a fixed point.

This directly follows from the following theorem:

[color=blue][b]Theorem 1.[/b] Let P, A, B be three points in the plane, and let u be the circle through these three points P, A, B. The tangent to the circle u at the point P meets the line AB at a point S; let R be the reflection of the point P in this point S. Then, if k is an arbitrary circle through the points A and B, then the polar of the point P with respect to the circle k passes through the point R.[/color]

Since k can be an arbitrary circle through the points A and B, and the point R does not depend on the circle k, this point R is the required fixed point where the polars of the point P with respect to all circles through the points A and B pass through; and thus, the problem follows trivially from Theorem 1. So, it remains to prove Theorem 1.

[i]Proof of Theorem 1.[/i] Since the point R is the reflection of the point P in this point S, the point S is the midpoint of the segment PR. Thus, the circle with diameter PR has center S. This center S lies on the line AB, which is the radical axis of the circles u and k (in fact, the line AB is the line joining the two common points A and B of the circles u and k, and thus it is the radical axis of these two circles). Since the point S lies on the tangent to the circle u at the point P, the line SP touches the circle u at the point P; thus, the circle with diameter PR is orthogonal to the circle u.

Now, it is well-known that if a circle is orthogonal to one of two given circles, and its center lies on the radical axis of these two circles, then it must also be orthogonal to the other one of the two circles. Hence, considering the circle with diameter PR, which is orthogonal to the circle u, and whose center S lies on the radical axis of the circles u and k, it follows that this circle with diameter PR must also be orthogonal to the circle k. Hence, the points P and R are reciprocal conjugates with respect to the circle k. In other words, the polar of the point P with respect to the circle k passes through the point R. This proves Theorem 1.

Thus, the problem is solved.

  Darij

Note that, as the conics through A,B,C,D depend linearly from a parameter, it is the same for the polar of P. Thus this polar goes through a fixed point.
Now, if Da,Db,Dc is the cevian triangle of D and Pa,Pb,Pc the precevian triangle of P, the polar of P wrt the pair of lines (AB,CD) is the line DcPc,...
We conclude that the fixed point is the perspector of the triangles DaDbDc and PaPbPc; this point is called the D-ceva conjugate of P (or cevian quotient D/P) wrt ABC in
http://faculty.evansville.edu/ck6/encyclopedia/glossary.html
Kind regards. Jean-Pierre


linear algebra
matrix
linear algebra unsolved

Assume that $ A \equal{} (a_{ij})_{n*n}$ and $ A^2 \equal{} I$ . Prove that $ r(I \plus{} A) \plus{} r(I \minus{} A) \equal{} n$

[quote="Kent Merryfield"]May we assume that $ I$ and $ E$ were both supposed to have been the same letter?[/quote]

oh, I had a mistake. I have corrected it.

Assuming that $ 2\neq 0$ in the field $ K$, show that \[ \ker(A\minus{}E)\oplus \ker(A\plus{}E)\equal{}K^n\,.\]

Following up on this along the lines suggested by Third Edition.

Suppose $ x\in K^n$ is arbitrary.

Define $ u=\frac12(x+Ax)$ and $ v=\frac12(x-Ax).$ Note that $ u+v=x$ and $ Au=u,Av=-v.$

Now define $ U=\{u\in K^n\,|\,Au=u\}$ and $ V=\{v\in K^n\,|\,Av=-v\}.$

We have just shown directly that $ U+V=K^n.$ It is also clear from the definitions of $ U$ and $ V$ that $ U\cap V=\{0\}.$ That is, $ U\oplus V=K^n.$

That makes it clear that if we adjoin a basis of $ U$ to a basis of $ V,$, the result spans $ K^n$ and is still independent; hence is a basis for $ K^n.$ That means that $ \dim U+\dim V=n.$

But $ U=\text{Nul}\,(I-A)$ and $ V=\text{Nul}\,(I+A).$

By the rank-nullity theorem, $ \text{rank}\,(I-A)=n-\dim U$ and $ \text{%Error. "rank" is a bad command.
}\,(I+A)=n-\dim V.$ Add those together to get that

$ \text{%Error. "rank" is a bad command.
}\,(I-A)+\text{%Error. "rank" is a bad command.
}\,(I+A)=2n-\dim U-\dim V=2n-n=n.$

Now, about that restriction to $ \text{char}\,K\ne 2:$ where did I use that? In two places. Right at the beginning, I defined $ u$ and $ v$ using division by $ 2$ as part of the formula; you can't do that in characteristic $ 2.$ Later on I asserted that $ U\cap V=\{0\}$ because $ Au=u$ and $ Au=-u$ can't both be true except for $ u=0.$ But that fails utterly in characteristic $ 2.$ In fact, in characteristic $ 2,$ $ U=V.$

For the stupid example to prove the point: Let $ A$ be the $ n\times n$ identity matrix in a field of characteristic $ 2.$ Then $ A^2=I$ is satisfied. But $ I+A$ and $ I-A$ are both the zero matrix, and hence both have rank zero, so the sum of the ranks is not $ n.$

[hide]For any square matrices P, Q of order n we have

$ rank (P\plus{}Q) \leq rank(P) \plus{} rank(Q) \leq rank(PQ) \plus{} n$.

Consider P = I+A, Q = I-A, and we get

$ n \leq rank(I\plus{}A) \plus{} rank(I\minus{}A) \leq n. \Diamond$[/hide]


The roots of $f(x)=0$ are $2,3,5,7,9$
The roots of $g(x)=0$ are $3,5,7,8,-1$
Find all possible solutions of the equation $\frac{f(x)}{g(x)}=0$.

[hide]I think it is just 2 and 9 since 3, 5, and 7 will make g(x)=0 and make it undefined.[/hide]

[quote="Iversonfan2005"]well, obviously $f(x)=0$ so it's simply the roots of $f(x)$ which will satisfy the equation.[/quote]

Hmmm... if x= 3, then g(x)=0. Then we have an undefined operation.

[quote="Mint"]Then that would just make the answers be all of the roots of f(x) such that g(x) does not equal 0, right?[/quote]

correct


let ABCD be a quad with an inscribed circle. let AC and BD meet at P, prove the incenters of APB, BPC, CPD, and DPA are concyclic

An old problem http://www.mathlinks.ro/Forum/viewtopic.php?t=21758


given 0<a,b<1.Prove that:$ a^b$ +$ b^a$>1

OH! my friend know my password and he post this topic  :mad: .Sorry everyone .Admin,Can you delete this topic ? :P

If this was a proper question why we didn't solve it?

[hide] Since Bernoulli’s inequalities,we have:
$ a^b$=$ (1\plus{}a\minus{}1)^b$ $ \ge$ 1+ab-b and $ b^a$=$ (1\plus{}b\minus{}1)^b$ $ \ge$ 1+ab-a. Add two inequality,we have:$ a^b  \plus{} b^a  > 2 \plus{} 2ab \minus{} a \minus{} b \equal{} 1 \plus{} (1 \minus{} a)(1 \minus{} b) \plus{} ab > 1$(because a,b<1)

 [/hide]


I'm a high school student in Jacksonville in Precalculus Honors.  It probably wouldn't be impossible for me to travel to Tallahassee to take some sort of qualifying test, if needed.

yeah saiem, how DO you get on the team? =P

uh yeah, if you guys want to get on the team for princeton. just lemme know. i need a couple people for the second team.

What is the Princeton competition?  It sounds interesting.

http://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition

Good luck Florida


Does there exist an infinate positive sequence ${a_{n}}$ such that    $\displaystyle \sum a_{n}$   and    $\sum {\frac{1}{n^{2}{a_{n}}}}$    are convergant?

Lovely!  [hide]Apply Cauchy-Schwarz to the product of the partial sums of the two sequences.  We see that their product is larger than the partial sum of the harmonic sequence, which diverges.  So at least one of the two sequences must diverge.[/hide]

Nice one! I used the sum intsead of the product and by using the AM GM got:
[hide]$\displaystyle \sum (a_{n} + {\frac{1}{n^{2}{a_{n}}}})\geqq\sum{\frac{2}{n}} = \infty\ $[/hide]


[quote="PHONGVAN"]Let $ a,b,c > 0$Prove  or disprove that
a,$ \frac{a^{3}}{a^{3}\plus{}abc\plus{}b^{3}}\plus{}\frac{b^{3}}{b^{3}\plus{}abc\plus{}c^{3}}\plus{}\frac{c^{3}}{c^{3}\plus{}abc\plus{}a^{3}}\leq\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca}$
[/quote]

$ \frac{a^{3}}{a^{3}\plus{}abc\plus{}b^{3}}\plus{}\frac{b^{3}}{b^{3}\plus{}abc\plus{}c^{3}}\plus{}\frac{c^{3}}{c^{3}\plus{}abc\plus{}a^{3}}\geq\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca}$

not :

$ \frac{a^{3}}{a^{3}\plus{}abc\plus{}b^{3}}\plus{}\frac{b^{3}}{b^{3}\plus{}abc\plus{}c^{3}}\plus{}\frac{c^{3}}{c^{3}\plus{}abc\plus{}a^{3}}\leq\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca}$

Prove by Cauchy-Schwarz

 :wink:

[quote="quangpbc"]

$ \frac{a^{3}}{a^{3}\plus{}abc\plus{}b^{3}}\plus{}\frac{b^{3}}{b^{3}\plus{}abc\plus{}c^{3}}\plus{}\frac{c^{3}}{c^{3}\plus{}abc\plus{}a^{3}}\geq\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca}$

Prove by Cauchy-Schwarz

 :wink:[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0^{\plus{}}.$ :wink:

The equality in c/ does not occur , doesn't it ?


China
combinatorics
combinatorics solved
Probabilistic Method
double counting
China TST
maximum

[quote]16 students took part in a competition. All problems were multiple choice style. Each problem had four choices. It was said that any two students had at most one answer in common, find the maximum number of problems.[/quote]

A student is allowed to NOT answer a question (leave it blank as an answer), right?

[quote]A student is allowed to NOT answer a question (leave it blank as an answer), right?[/quote]
No, the answer to every problem has only four choices, blank is not permitted.

Suppose there are $m$ problems $P_1,P_2...,P_m$ for problem $P_i$, suppose $a_i$ students answer first choice, $b_i$ secound choice, $c_i$ third choice and $d_i$ fourth choice. Since $a_i+b_i+c_i+d_i=16$ we have at least $4\binom{4}{2}=24$ pairs with an answer in common for each problem. Since there are $m$ problems there are at least $24m$ such pairs, but on the other hand there are at most $\binom{16}{2}=120$ such pairs. So we have that $m$ is at most $5$. Now can equality hold? since for $m=5$ we need stronger conditions (namely for every problem every choice is selected by 4 students, and every pair has a common choice for some problem) I think such construction is possible since i dont fin any other constraint. ;)

I think I found an example
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{student} & Q1 & Q2 & Q3 & Q4 & Q5 \\ \hline 1 & A & A & A & A & A \\ \hline 2 & A & B & B & B & B \\ \hline 3 & A & C & C & C & C \\ \hline 4 & A & D & D & D & D \\ \hline 5 & B & A & C & D & B \\ \hline 6 & B & C & A & B & D \\ \hline 7 & B & D & B & A & C \\ \hline 8 & B & B & D & C & A \\ \hline 9 & C & A & D & B & C \\ \hline 10 & C & D & A & C & B \\ \hline 11 & C & B & C & A & D \\ \hline 12 & C & C & B & D & A \\ \hline 13 & D & A & B & C & D \\ \hline 14 & D & B & A & D & C \\ \hline 15 & D & C & D & A & B \\ \hline 16 & D & D & C & B & A \\ \hline \end{array}  \]

Here's a way to motivate my construction (which I did by trial-and-error back then).

Consider the grid of 16 points $ \mathbb{Z}_{4}\times \mathbb{Z}_{4}$ (a finite affine plane). Consider all families of 4 parallel lines (each passing through 4 points). There are 5 such families, each corresponding to a unique slope.

Each family of lines corresponds to one problem. For a given family, color the four lines with four different colors (corresponding to the 4 different multiple-choice answers). The 16 points correspond to the 16 questions.

This gives the desired construction.

Oops, I meant $ \mathbb{F}_{4}\times \mathbb{F}_{4}$, where $ \mathbb{F}_{4}$ is the field with 4 elements.

Sorry, but I don't quite understand what you mean. Would it be possible to draw a diagram?

@ Pascual: I have another way to prove that there cant be more than 5 questions. Suppose the Q1(question 1) was answered as option 'a' by p1 (participant 1),p2,p3 and p4 .(only: There cant be a p5 as all of them (p1 to p5) have to opt for different options of Q2 and it would not be possible by pigeon hole principle).
So 1 can have common options with {2,3,4}{5,6,7}{8,9,10}{11,12,13}{14,15,16}
Hence not more than 5 questions. But we need a construction to say this is surely possible and billzhao does it(I have not checked it but).

@billzhao: I only get the feel of your 4*4 affine plane construct, but not able to get the complete proof. Would you explain in detail? I feel this problem somehow connects with Kirkman schoolgirls problem, any idea? You created a example, can we enumerate the number of such possible constructs?

A more clear solution (indicating double-counting) is as follows. Suppose, there are $m$ tasks in the contest. Count the total number of triples of form $(C_i,C_j,Q_k)$, where contestants $1\leq i <j\leq 16$ agreed on problem $k\in\{1,2,\dots,m\}$. From the perspective of each pair, there is at most one such triple, hence, the number is upper bounded by $\binom{16}{2}$. Now, from the perspective of any problem, if $x,y,z,t,$ are the number of people selected first choice, ..., fourth choice, then there is precisely $\binom{x}{2}+\binom{y}{2}+\binom{z}{2}+\binom{t}{2}$ pairs, agreed on problem $k$. Using convexity of Binomial coefficients with Jensen's inequality, and the fact that $x+y+z+t=16$, the total number of contestants, we get, $\binom{x}{2}+\binom{y}{2}+\binom{z}{2}+\binom{t}{2}\geq 24$. This yields, $\binom{16}{2}\geq 24m$, namely, $m\leq 5$. An example of $m=5$ is easy to construct, and is provided above.

A very much similar solution to the one above.
[hide=Solution]
Let, T denote number of triples $(S_i,S_j,Q_k)$ such that students $S_i,S_j$ answered the same for Question $Q_k$.

[color=#f00][b]Lemma:[/b][/color] If $a+b=p$, then $\binom{a}{2}+\binom{b}{2}\geq \frac{(p-1)^2}{4}$
[color=#00f][b]Proof:[/b][/color]

$\binom{a}{2}+\binom{p-a}{2} \geq \binom{(\frac{p-1}{2})}{2}+\binom{\frac{p+1}{2}}{2}\geq \frac{(p-1)^2}{4}$

There are $\binom{16}{2}$ ways to choose 2 students and atmost one way to choose question.

$\implies\boxed{T \leq \binom{16}{2}}$

Let, $x,y,z,w$ students choose first option , second option and so on... and there are m questions in total.

Applying Lemma twice and using Cauchy Schwartz inequality yields :

$\binom{x}{2}+\binom{y}{2}+\binom{z}{2}+\binom{t}{2}\geq 24$

$\implies\boxed{T \geq 24.m}$

Solving both inequalities yields $\boxed{m\leq 5}$
[/hide]

Let the students be $S_1, S_2, \ldots, S_{16}$ and the questions be $Q_1, Q_2, \ldots, Q_n$. Define $X$ as the number of triples $(S_i, S_j, Q_k)$ where $1 \le i < j \le 16$ such that $S_i$ and $S_j$ got the same answer for $Q_k$.

Because any two students have at most one answer in common, we have $$X \le 1 \cdot \binom{16}{2} = 120.$$ Now, we find a lower bound for $X$ by analyzing each question individually. WLOG, pick $Q_1$.

For $Q_1$, suppose $a$ people answered "A", $b$ people answered "B", $c$ people answered "C", and $d$ people answered "D". Since $a+b+c+d = 16$, Cauchy-Schwarz implies $$\binom{a}{2} + \binom{b}{2} + \binom{c}{2} + \binom{d}{2} = \frac{a^2 + b^2 + c^2 + d^2 - (a+b+c+d)}{2}$$ $$\ge \frac{(a+b+c+d)^2}{8} - 8 = 24.$$ Thus, there are at least $24$ triples of the form $(S_i, S_j, Q_1)$. Now, symmetry implies $$24n \le X \le 120 \implies n \le 5$$ as required. $\blacksquare$

[b]Remarks:[/b] Also see [url=https://artofproblemsolving.com/community/c6h1479862p8630483]here[/url] and [url=https://artofproblemsolving.com/community/c4h1130999p5253870]here[/url].


Show that $ \sin 1\cdot \sin 2\cdot \sin3\cdots \sin89\cdot \sin90 \& = \frac {6 \sqrt {10}}{4^{45}}$

THE question has been corrected. ($ 2^{45}$ change -> $ 4^{45}$)

Mmm, Zed, I think is not possibile to repeat the process after this:

$ \frac{\sqrt2}{2^{67}} \cdot \sin(4) \sin(8) \dots \sin(88)$

Because the 1st term and the last don't add to 90

[quote="Zed_darkx"]Be cause we have 44 pairs of SinX into sin 90-X pairs we divide by 2 ^45 and Multiply by the same
to get and expression  of sin2*sin4*sin6....sin88* we have in here 22 pairs so we repeat the process. And maybe then we'd reach the answer. I havent tried it on paper yet but im sure this is how we proceed.[/quote]

thats what i thought first too,
however at the third round, things got ugly...
sin(4) sin(8)...sin(44) [b]sin(48)[/b]...sin(88)

sin(48) = cos (42)
etc.
:(

Isnt there any formula for continous products then?

[quote="Zed_darkx"][quote="Shaon0"]Since; sin(90-x)=cos(x)
LHS={sin(1).sin(2)...sin(44).cos(44)...cos(2).cos(1)}sin(45)
={sin(1)cos(1).sin(2)cos(2)...sin(44)cos(44)}(sqrt(2)/2)
={0.5.sin(2)*0.5sin(4)...0.5sin(88)}(sqrt(2)/2)
={sin(2)sin(4)...sin(88)}(sqrt(2))/(2^(45))
LHS={sin(2)sin(2+2)...sin(2+43(2))}(sqrt(2))/(2^(45))
Since, there are 44 terms.
={{sin(45/2)sin(46)}/sin(1)}(sqrt(2))/(2^(45))[/quote]
Not working.[/quote]
I finally put pen to paper. And the series should yield;
{sin(46)/sin(1)}*{(2)^(45)} but it clearly doesn't work. So my approach is wrong.
Does anyone have the solution?

UP!

I'm curious to see the solution  :|

See part 1 of this article for a solution to this problem - http://www.artofproblemsolving.com/Resources/Papers/TrigIdent.pdf

[Note of warning: The proof of the main result in the article uses some advanced mathematical notation, complex numbers, and some calculus. I would not consider it at the "intermediate" level. But if you just want to see how you would solve this problem, you will find it here.]

[quote=sarih]Hi, sorry for reviving.
Could someone share a link or upload the file "TrigIdent.pdf" mentioned in the above reply ?
Thanks in advance.[/quote]

https://web.archive.org/web/20060824173143/http://artofproblemsolving.com/Resources/Papers/TrigIdent.pdf


limit
integration
calculus
real analysis
real analysis unsolved

Prove that \[ \lim_{k \to \infty}\int_{\mathbb{R}^n}f(x) \delta_k(x)dx\equal{}f(0)\] where $ (\delta_k)$ is a Dirac sequence  and $ f \in \mathcal{C}_c(\mathbb{R}^n)$

Umm... what is the definition of "Dirac sequence"? If you ask me to offer what I think a Dirac sequence is, I would define it as satisfying the limit above - and that would make the whole problem circular.

Sorry.A Sequence $ \delta_k$ in $ \mathcal{L}_1(\mathbb{R}^n)$ is Dirac sequence wenn $ \delta_k \geq 0$ and $ \int_{\mathbb{R}^n}\delta_k(x)dx\equal{}1$ and \[ lim_{k \to \infty}\int_{|x|\geq \xi}\delta_k(x)dx\equal{}0\]  $ \forall \xi>0$

The definition suggests to split the integral of $ f\delta_k$ into $ |x|\le \xi$ and $ |x|\ge \xi$, then argue that the second part has limit zero. In the first integral you use the continuity of $ f$ in the form $ f(0)\minus{}\epsilon\le f(x)\le f(0)\plus{}\epsilon$.

Yet another example of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=963840#963840]mechanization in math[/url].


This table shows the number of classes
at Wilson H.S. within various categories
of class-size range. Of these
109 classes, what is the greatest
possible number of students that could
be the median class size?

\begin{tabular}{ |c | c|}
   \hline \textbf{\# of Students in Class} & \textbf{\# of Classes} \\ \hline
  1-10 & 10 \\ \hline
  11-20 & 25 \\ \hline
  21-30 & 58 \\ \hline
    31-40 & 16 \\ \hline
\end{tabular}

$ begin{tabular}{ |c | c|}
   \hline \textbf{\# of Students in Class} & \textbf{\# of Classes} \\ \hline
  1-10 & 10 \\ \hline
  11-20 & 25 \\ \hline
  21-30 & 58 \\ \hline
    31-40 & 16 \\ \hline
\end{tabular}$

It must lie in the $ 21-30$ range.  Thus, the answer is $ \boxed{30}$.


(1) Find all solutions to the following system of equations:
x1/(x1+1) = x2/(x2+3) = x3/(x3+5) .. = x(1001) / x(1001)+2001
x1+x2 ... + x(1001) = 2002

(2) Determine all x,y that satisfy x!+24=y^2

(3) For each pair (k,l) of integers determine all infinite sequences of integers a(1),a(2), a(3) ... sucht that the sum of every 28 consecutive terms equals k and the sum of every 15 consecutive terms equals l.

(4) Determine all functions f with the following properties:
(i) f is defined on all reals.
(ii) |f(x)| <= 2002 <= | x*f(y) - y*f(x) / x-y |
for all unequal x,y.

for 2)
We verify each value of $x\in \{0,1,2,3,4,5\}$ and we find that $(x,y)\in\{(0,5),(1,5),(5,12)\}$ are solutions.
For $x\ge 6$ we have $\displaystyle 24\left(\frac{x!}{4!}+1\right)=y^2$ but the right factor of left member is odd and 24 is divisible by $2^3$ which is not a perfect square. Results that we have no solutions for $x\ge 6$.
So the solutions for the equation $x!+24=y^2$ are $(x,y)\in\{(0,5),(1,5),(5,12)\}$.

for (IV)

Put $y=-x\leq 0$, we have \[|f(x)|\leq 2002\leq \left|\frac{xf(-x)+xf(x)}{2x}\right|=\left|\frac{f(x)+f(-x)}{2}\right|\leq \frac{|f(x)|+|f(-x)|}{2}\qquad (*)\]

$|f(x)|\leq \frac{|f(x)|+|f(-x)|}{2} \Longrightarrow |f(x)|\leq |f(-x)|$, $\forall x\neq 0$.

Replace $x$ by $-x$, we obtain $|f(-x)|\leq |f(x)|$
thus $|f(x)| = |f(-x)$. Substitute this into (*), we get $|f(x)|=2002, \forall x\neq 0$.

Put $x=0$, we get $|f(0)| \leq 2002 \leq |\frac{-yf(0)}{-y}| = |f(0)|$, thus $|f(0)|=2002$.

Now we know that it holds that $|f(x)|=2002$ for all $x$.

If there exists $x$, $y$ s.t. $f(x)=2002$ and $f(y)=-2002$, we have
\[2002\leq \left|\frac{-2002x-2002y}{x-y}\right| = 2002\left|\frac{x+y}{x-y}\right|\]
or \[|x+y|\geq |x-y|,\] hence $xy\geq 0$.

Now suppose there is a $x_0>0$ such that $f(x_0)=2002$. We know that $f(x)=2002$ for all negative $x$. Thus there is a $x_1<0$ such that $f(x_1)=2002$, and this makes $f(x)=2002$ for all $x>0$.

Similarly, we can prove that if there is a $x_0$ such that $f(x_0)=-2002$, we have $f(x)=-2002$ for all $x\neq 0$.

Thus the functions we are looking for is
(i) $f(0)=\pm 2002$, $f(x)=2002,\, x\neq 0$
or
(ii) $f(0)=\pm 2002$, $f(x)=-2002,\, x\neq 0$

for 3)
Since the sum of every 15 terms of the sequence have the same sum l, we conclude that $(a_{15n+k'})_{n\ge 0}$ is constant for every $1\le k'\le 14$.
Same the sequence $(a_{28n+k'})_{n\ge 0}$ is constant for every $1\le k'\le 27$.
We have $a_{k_1}=a_{15n_1+k_1}$ and $a_{k_2}=a_{28n_2+k_2}$, with $n_1,n_2\in\mathbb{N}$ and $0<k_1,k_2<15, k_1\ne k_2$.
Since the equation $15n_1+k_1=28n_2+k_2$ has infinitely many solutions in natural numbers (the variables are $n_1,n_2$) we conclude that $a_{k_1}=a_{k_2}$ for every $0<k_1,k_2<15$, $k_1\ne k_2$.
So the sequence $(a_n)_{n\ge 1}$ is constant.
We must have $l=15p,k=28p,p\in\mathbb{Z}$ and thus $a_n=p,\forall n\in\mathbb{N}^*$.


There's an excellent description in this MathJam:

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=35


inequalities
geometry
AMC
AIME
rearrangement inequality

This is the thread where COP 5th Round 2 will be posted. I was very busy, and I'll try to write up the Round 2 this week.

The people who are eligible to participate are:

eryaman
Jan
dk2
dakyru
Andreas
JesusFreak197
236factorial
Drunken_Math

I have to say this COP is so far the hardest COP and hence, there aren't much qualifiers. But this doesn't mean the Round 2 will appear easy. It will be approximately level of problems you'll see on later AMC-12's.

I'll post the problems this weekend so be excited for it!  :D

We have to get all of the problems right, right?

No, not necessarily.

Top two scores (not just two people!) will go to Round 3. This may be a person who got 6 and 5 other people who get 4.

 :)

Here are problems. I'm posting this in 10/11/05 in my time so you'll have TWO weeks to send me ANSWERS. Hence, the deadline is 10/25/05.

If you have any questions concerning the rule, read the thread "COP (Contest of Opportunity)." For problems, contact me or post here (as long as you're not giving the answer away). Also, the people on the first post are only ones who are eligible to send the answers.

1. In "cute" arrangement, one can randomly pick any 6 consecutive numbers from the set $\{x_1,x_2,x_3,...x_n\}$ where $x_p= x_{p-1} + 1$ so that the chosen 6 consecutive numbers' product is ALWAYS divisible by the greatest integer $m$, regardless of what $x_1$ and $x_n$ is. Find what $m$ is.

2. In $x^5+2x^4+x^3-x^2-2x = 1$, there exists an imaginary root called $w$ which satisfy the equation true. Let $f(x) = w^x - 1$. What is the second smallest $x$ such that $f(x) = 0$?

3. A hexagon is inscribed in a circle. Two of its sides are 5 while rest of four sides are 24. If this circle has a radius of $\frac{r + \sqrt s}{t}$, find a positive integer $u$ such that $u^2 = s - r - t$.

4. Compute the sum: $\sum^{50}_{n=0} \frac{(m+n)!}{n!}$ when $m = 2$.

5. Find the number of solutions for positive integer $p$ and $q$ in equation:

\[ p^3+18q^3 = 2005 \]

6. The three unique sequences are defined in interesting way:

$\{a_k\} = 1+2+3+\cdots k$
$\{b_k\} = \frac{a_2}{a_2 - 1} \cdot \frac{a_3}{a_3 - 1} \cdot \frac{a_4}{a_4 - 1} \cdots \frac{a_k}{a_k - 1}$
$\{c_k\} = \frac{\{a_k\}}{\{b_k\}}$

Find $c_{2005}$.

7. For positive real numbers $a,b,c$, we have an inequality:

$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} + \frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} \geq M$

where $M$ is the minimum. If part $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ of the original inequality is minimized, what should be the value for $\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c}$ so that the whole sum is exactly $M$?

So, if we get 6 or higher, we automatically qualify? I should have three of them already...

[b]EDIT:[/b] [quote]there exists an imaginary root called w which satisfy the equation true.[/quote]

What does that mean? Just that $w$ is a root of the original quintic?

IS there more than one imaginary root?  :?

$w$ is the complex root of the original equation. You should be able to figure out how many are imaginary and how many are real from the original equation. I'm not sure exactly what to tell you except giving the answer or big hint away.

Also, dk2 PMed me and I have to add this important thing about #2. $x$ is a positive integer greater than 0.

PM me.  :)

It would kind of help to know, especially for #3... because otherwise, it reminds me quite a bit of #6 on the previous Round 2. ;)

[quote="JesusFreak197"][b]EDIT:[/b] So, is $x=0$ the smallest, and we're supposed to find the next one, and that's what you meant by $x>0$, or are we supposed to find the second one with $x>0$?

Also, on #3, is $\frac{r+\sqrt{s}}{t}$ necessarily in simplest form?[/quote]

The second option. The second one such that both first and second are $> 0$.

I don't think you should get a fraction that requires a simplification.. Hmm.. Let me check..

Okay, you're probably right, so I must have made a mistake somewhere. I just wanted to know, because I didn't want to spend forever testing possible multiples of the number when I just didn't have the right value for $R$ in the first place...

[b]EDIT:[/b] Yeah, you were right; I made a small mistake near the end that threw my value of $s-r-t$ off enough to not get a perfect square, but I fixed it. :) I have all the answers now, but I want to check over and make sure I didn't make any dumb mistakes, so I'll probably send them in tomorrow...

I'm just going to add the deadline to this Thursday not Tuesday because I received a message from one of the contestants that he has huge exams in Wednesday.

I hope adding two more days is okay with everyone. As you know, school always come first.  :)

I normally make separate threads but since it's not in my moderating area, I'll just post the results in this thread. I can always use this specific post for results anyway.

RESULTS:
JesusFreak197: 6
dakyru: 6
Andreas: 5
dk2: 5
236factorial: 4

Since this is a top two scores, the qualified people are JesusFreak197, dakyru, Andreas, and dk2. I think if there was a Honorable Mention in this contest, this certainly goes to 236factorial since he's only in middle school and did a nice job on this test. Let's give appaluse to 236factorial and people who qualified!

Here are solutions (not hidden)

1. There are many ways to do this problem. First, understand that six chosen integers are consecutive integers namely $x,x-1,x-2,x-3,x-4,x-5$. You can find how many are odd, even, divisible by 3, etc but there is a clever solution. Note that number of ways to choose 6 elements of total $n$ elements can be written as $\binom{n}{6}$ which equals $\frac{n \cdot n-1 \cdot n-2 \cdot n-3 \cdot n-4 \cdot n-5}{6!}$ and note that for any $n$, the result is integer since number of ways to do this has to be integer. So, $m = 6! = 720$. I really thought this problem was neat to solve this way.  :)

2. With easy inspection, we find that the  $(x^3-1)(x^2+2x+1)$. The right one can be factored as $(x+1)(x+1)$ so the roots are real. $(x^3-1) = (x-1)(x^2+x+1)$ and the right part has complex root. Now, since $x = w$, we get $w^2+w+1 = 0$ and hence, $w^3-1 = 0$. But we don't want exponent to be 3 so we want the one comes next it. The smallest possible answer is when we multiply $w^3-1$ by $w^3+1$ which gives us result of $w^6-1 = 0$. Hence, the answer is 6. Note that we never actually found what $w$ is. We could but there was no need for it.

For 3-6, I'll quote the explanation from JesusFreak197. He kindly posted solutions along his answers (although you don't need to) and they were all right.

[quote="JesusFreak197"]
[b]3.[/b] We begin by drawing the hexagon with sides of length $5$ opposite each other, then drawing the perpendiculars to those sides from the center. We label this length $x$, then we split the hexagon into two identical pentagons made of a rectangle with width $2x$ and height $2.5$ and an isosceles triangle with base $2x$ and other sides $24$. From here, we can express the area in two ways: 1) The area of the rectangle is $2.5\cdot 2x=5x$, and the area of the triangle can be found by Heron's formula to be $x\sqrt{576-x^2}$, and 2) Cut the pentagon in half, flip one half around onto the other to make a rectangle of height $R+2.5$ and width $x$, and calculate the area to be $x(R+2.5)$. We also make use of the fact that $x^2+(2.5)^2=R^2\Rightarrow x^2=R^2-(2.5)^2$ (see the original figure). Therefore, we have $5x+x\sqrt{576-x^2}=x(R+\frac{5}{2})\Rightarrow 5+\sqrt{576-x^2}=R+\frac{5}{2}\Rightarrow$
$\sqrt{576-(R^2-(\frac{5}{2})^2)}=R-\frac{5}{2}\Rightarrow \sqrt{2329-4R^2}=2R-5$
$(2R-5)^2=4R^2-20R+25=2329-4R^2\Rightarrow 8R^2-20R-2304=0$
$R=\frac{5+\sqrt{4633}}{4}\Rightarrow u^2=4633-5-4=4624\Rightarrow \boxed{u=68}$

[b]4.[/b] $\sum_{n=0}^{50}\frac{(n+2)!}{n!}=\sum_{n=0}^{50}(n+1)(n+2)=\sum_{n=1}^{51}n(n+1)=\sum_{n=1}^{51}n^2+\sum_{n=1}^{51}n$
$\frac{51(52)(103)}{6}+\frac{51(52)}{2}=\boxed{46852}$

[b]5.[/b] $p$ must be odd so that $2005-p^3$ can be even and therefore divisible by $18$, so our only possibilities are $p={1,3,5,7,9,11}$. None of these work, so there are $\boxed{0}$ possible pairs.

[b]6.[/b] ${a_k}=\frac{k(k+1)}{2}=\frac{(k+1)((k+1)-1)}{2}$
$a_k-1=\frac{(k+1)((k+1)-1)}{2}-1=\frac{(k+1)^2-(k+1)-2}{2}=\frac{((k+1)+1)((k+1)-2)}{2}$
${b_2}=\frac{a_2}{a_2-1}=\frac{\frac{(k+1)((k+1)-1)}{2}}{\frac{((k+1)+1)((k+1)-2)}{2}}=\frac{(k+1)((k+1)-1)}{((k+1)+1)((k+1)-2)}=\frac{3\cdot 2}{4\cdot 1}$
Bottom left cancels out with top left from the next term, and top right cancels out with bottom right from the next term. In this way, everything will cancel out except for the first $3$ and the $2005$ and $2007$ from the last term, and we have $b_{2005}=\frac{3\cdot 2005}{2007}$.
Therefore, $c_{2005}=\frac{a_{2005}}{b_{2005}}=\frac{\frac{2005(2006)}{2}}{\frac{3\cdot 2005}{2007}}=\boxed{671007}$. [/quote]

Number 7 was my favorite one. Only one person got this problem right and that was dakyru.

First, to find the minimum of whole thing, we use AM-GM on six variables.

$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} + \frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} = 6 \cdot \sqrt[6] {1}$. So, $M = 6$. If you do not understand AM-GM, you might want to learn this because this is very useful inequality used in many contests including AIME as well. For practice, look for 1983 AIME #9.

Second, the minimum of $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$.

This is definitely where most people stuck I think. Using Rearrangement Inequality, we get two sets of inequalities:

$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{b}{b+c} + \frac{a}{a+c} + \frac{a}{a+b}$
$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{c}{b+c} + \frac{c}{a+c} + \frac{b}{a+b}$

Adding these two gives:

$2 \Big(\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}\Big) \geq 3$

Dividing by 2 tells that the minimum is $\frac{3}{2}$. Hence, the exact value for the rest is $6 - \frac{3}{2} = \frac{9}{2}$.

Now, just for sidenote, the minimum of $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ is called Nesbitt's Inequality. I found at least 6 or 7 proofs to this in internet and maybe in some angle, it was not useful to use so well-known inequality as problem. But instead of other problems where you are asked to prove, my problem #7 asked you to derive it and I found a nice way to do this using Rearrangement.

Any questions? Please post here or you can simply leave comment here. Thank you.

P.S. This will be the last COP for at least 6 or 7 months. I'm going to do some independent work so I'll have to let go of COP for while. But don't worry, it will come back soon!

Another message: The Round 3 will be posted soon. I'll post whenever I can.

My dumb mistake on #7 was that I didn't realize that the minimum of $M$ was $6$. :roll:

Well, that qualified two more people for the next round. :P

[quote="eryaman"]Oops, I didn't participate.  I was sick all week and swamped with hw/tests, so I didn't have a chance to.[/quote]

 :(

Hope you feel better now.  :)

[quote="Silverfalcon"][quote="dk2"]Uh.. silverfalcon, are there actually values of a, b, and c such that M = 6? Because AM-GM min only works when all terms are equal, so that means that a/b+c = b+c/a, so a=b+c, and similarly we can find that b=a+c and c=a+b.. but the only values of a,b, and c which satisfy that are (0,0,0).. which obviously don't work. I might be wrong, but if I am can you give values of a, b, and c such that M=6?[/quote]

To be honest, I'm not sure when the equality holds but I used AM-GM since $a,b,c$ are all positive reals and that's only thing I really considered.

I know what you're asking but to be honest, I'm not sure..  :? Although it doesn't matter I think.[/quote]

Hmmm...
Well, I think I have a pretty decent proof M can't equal 6.

If you know how to prove AM-GM it is pretty easy to show that

AM(a_1,a_2,a_3,a_4...a_n)>=GM(a_1,a_2,a_3,a_4,a_5,...a_n), with equality holding iff a_1=a_2=a_3=a_4=...=a_n.
(There have been other topics on this before.. see http://www.artofproblemsolving.com/Forum/topic-30544.html post 2).

In order to find the minimum, you assumed the equality case.
Thus,

(a+b)/c=(b+c)/a=(c+a)/b=c/(a+b)=a/(b+c)=b/(c+a).
We are mainly interested in the equalities:
(a+b)/c=c/(a+b)
(b+c)/a=a/(b+c)
(c+a)/b=b/(c+a)
Since the numbers are equivalent to their reciprocals, and they're positive reals, they must equal one. Therefore:
(a+b)/c = 1, a+b=c
And similarly,
b+c=a.
c+a=b.
Adding them all together we get:
(a+b+c)=0.
So adding c to both sides of our first equation, we find that,

a+b+c=2c, therefore 2c=0, so c=0.
Already the value is undefined, but we can further show that a=0, and b=0.

So I'm quite sure M is not 6.
What I am confused about, though, is what M actually is.
I let a=b=c to get M=15/2, but I have no idea if it actually is 15/2 or if there is another value... symmetrically I'd think that a=b=c, but I'm not sure..

If anyone sees anything useful for this problem could you please post?

If the first part is minimized, the second part should equal $\frac{9}{2}$ in order for the whole sum to equal 6.  But I don't think that's possible.  However, it said should, so I just did $6 - \frac{3}{2} = \frac{9}{2}$.  Problem #3 totally evaded me, lol.

[quote="JesusFreak197"]If $a=b=c$, then $M=\frac{15}{2}$...[/quote]
Oops, thanks for the correction.

If I knew the inequalities, I probably would've had more clue on number 7  :blush:

So what is the rearrangement inequality?

[quote="Silverfalcon"]Moderator, can you unsticky this now?

This round is officially over so I don't see why it should be stickied anymore.

Thank you!  :)[/quote]I'll unsticky it for you.

All done.  :)


function
topology
algebra proposed
algebra
Fixed point

Suppose $A\subseteq \mathbb R^m$ is closed and non-empty. Let $f:A\to A$ is a lipchitz function with constant less than 1. (ie there exist $c<1$ that $|f(x)-f(y)|<c|x-y|,\ \forall x,y \in A)$. Prove that there exists a unique point $x\in A$ such that $f(x)=x$.

Have a look at the fixed-point theorem ( the result holds for any f: M-> M where M is a complete metric space and f is such that |f(x)-f(y)| <= K*|x-y| where K<1 ).

It is quite classical am I wrong?

Unicity : If $x \neq y$ are such that $f(x)=x$ and $f(y)=y$ then $|x-y| = |f(x)-f(y)| < c|x-y| < |x-y|$ which is absurd.

Existence :
Let $a \in A$. Let $x_0 = a$ and $x_{n+1} = f(x_n).$ (1)
Thus $|x_{n+1}-x_n | < c |x_n - x_{n-1}|.$
Using that $c<1$, it is now easy to deduce that the sequence has the Cauchy property. Since $\mathbb {R}^m$ is a complete space, we deduce that the sequence has a limit, say $L$. Since $A$ is closed and each of the $x_i$ is in $A$, it follows that $L$ is in $A$.
That $L = f(L)$ is obvious from (1).

Pierre.


geometry
parallelogram
geometry proposed

Hi,

I have just started learning about inversion techniques.. :D
But I'm not entirely certain to which questions I can apply inversion to, due to the shortage of inversion examples..
If anyone has problems at olympiad level that can be solved by inversion, can you post them?

Thanks.

Let $ABCD$ be a tangential quadrilateral and $E, F, G, H$ the tangency points of its incircle with the sides $AB, BC, CD, DA$. Prove that $EG$ is perpendicular to $FH$ iff the quadrilateral $ABCD$ is also cyclic.

[quote="jpark"]what's tangential quadrilateral?

A quadrilateral with inscribing circle?  :?[/quote]

Yes, exactly, a quadrilateral with an incircle.

  darij

[quote="jpark"]Thanks.

Any other problems?  :D[/quote]

Construct a parallelogram $PQRS$ with the diagonal intersection $X$, inscribed in a given parallelogram $ABCD$, given its area $S$ (i.e., a square with the same area) and the angle $\phi = \angle PXQ$ formed by its diagonals $PR, QS$.

Edit: Previously, I forgot to add "inscribed in a given parallelogram $ABCD$".

Thanks,

It feels like I'm finally getting inversion!!  :D

[quote="yetti"]Edit: Previously, I forgot to add "inscribed in a given parallelogram $ABCD$".[/quote]

Haha, this is really funny..
I automatically assumed that statement and solved it.. I never even realized that it was missing  :D  :D  :D


A number $ k$ is randomly chosen from the
set $ \{0, 1, 2, 3, \ldots 10\}$. What is the probability that $ x^2 \plus{} kx \plus{} 12 \equal{} 0$ has two integer solutions? Express your answer as a common fraction.

The quadratic has solution $ x\equal{}\frac{\minus{}k\pm\sqrt{k^{2}\minus{}48}}{2}$.  Hence we know immediately that $ k\geq 7$ or else the solutions are non-real.  To generate two integer solutions, we need $ k^{2}\minus{}48$ to be a perfect square, and this is accomplished with $ k\equal{}7,8$.  So the probability is $ \frac{2}{11}$.

We can also think of $k$ by the opposite of the sum of the two solutions (by Vieta's).  The product of the solutions is $12$ by Vieta's and we know they must be integers.  Therefore, $k$ can be $1+12=13$, $2+6=8$, and $3+4=7$.  However, $13$ is not in the given set, and only $2$ are included in that set.  There is a total of $11$ elements in the given set therefore the probability is $\boxed{\frac{2}{11}}$.


induction
combinatorics solved
combinatorics
graph theory
Enumeration

We are given $n$ vertices (unjoined). How many trees can we form by joining them? A tree is a graph without cycles.

Does the order of the vertices matter or not ? I mean , are they labelled ?

No
It doesn't matter.
If u label them, some trees may be identical, so they will count only once

For example, for n=2 we have 1 tree.
for n=3, we have 3 trees 1-2-3   1-3-2   2-3-1
notice that 2-3-1=1-3-2 :)

Well for starters , you are looking for THE NUMBER OF LABELLED TREES with n vertices . If they weren't labelled , then for n=3 you would only have one tree (all three trees you wrote down are isomorphic to one another).

The recurrence relation for the numbers is : if f(n) is the searched-for number , then :

f(n+1) = 1/2n * sum(k=1;n) C(n+1,k) * f(k) * f(n+1-k) * k * (n+1-k)

It seems that the number we are looking for is f(n)=n^(n-2) by studying the first small values . The formula above can be justified in this manner :

To make up a tree with n+1 vertices , for fixed k , take a tree with k vertices and another with n+1-k vertices . You can unite these two with an edge in exactly k * (n+1-k) ways . The k vertices can be chosen in C(n+1,k) ways and the tree with k vertices can be chosen in f(k) ways . The n+1-k vertices are determined by the choice of the k vertices and the tree with n+1-k vertices can be chosen in f(n+1-k) ways . Finally , we count each tree 2n times , because a tree with n+1 vertices has exactly n edges and so , by "taking out" each edge we are left with a tree with k vertices and another with n+1-k vertices , for some k . Also , we can either choose the k vertices or choose the n+1-k vertiecs , so we count the same thing 2n times . Thus the formula .

I am trying to prove it by induction now , I'll post the solution here if I manage to make up one . Hope this recurrence relation helps and that it's correct .

Is it TR(n) when $n$ is the number of veritices? Just came to this problem

where did you find this post after 14 years. wow.


LaTeX
inequalities unsolved
inequalities

Please help me:
'Find the maximum and the minumum value of the expression: \[P=x^{y^z}\] where x,y,z are integer numbers such that : x,y,z>=3 and x+y+z=20"

What is x+y+z=20" ??? :?  And why did you use $LATEX$ for the first part and not for the second :?:

Lets see how $P$ changes when we switch $x \rightarrow x-1$ and $z \rightarrow z+1$.
$x^{y^z} \bullet (x-1)^{y^{z+1}} \Leftrightarrow y^ zln(x) \bullet  y^{z+1}ln(x-1)$ or
$ yln(x-1) \bullet ln(x)$ And it is easy to see that the $LHS$ is greater but the case when $x=y=3$. So it means that when we are decreasing $x$ and increasing $z$ no matter what $y$ is $P$ is getting larger and viceversa. So the maximum value will be reached when $x=3$. So I guess we are left to see when $y^z$ takes minimum or maximum values. :?

Lemma:
If $x\ge y\ge e$.Then $x^y\leq y^x$,
Proof:
The problem is equivalent to prove that
$\frac{lny}{y}\ge \frac{lnx}{x}$.Put $f(x)=\frac{lnx}{x}$.
It is easy to see that $f'(x)\leq 0$ iff  $ \frac{(1/x)\cdot x-lnx}{x^2}\leq 0$
So $f(x)$ is decreasing for $x\ge e$.The proof is complete.
RaduSorici proved that while changing $x\rightarrow x-1$ and $z\rightarrow z+1$ the expression increases.
So $z$ must take it's minimum value to expression be smaller i.e $z=3$.
Also $x$ must take it's minimum value to expression be larger i.e $x=3$.
Now from the above Lemma  $14^{3^3}\leq x^{y^z}\leq 3^{3^{14}}$.


geometry
trapezoid
geometric transformation
reflection
trigonometry
trig identities
Law of Sines

ABCD is a trapezoid with AB parallel to CD. In addition, AC = AD, BD = CD, and angle CAD is right. Find the measure of angle ADB.

[hide="Solution"]
Since $ AC = AD$ and $ m\angle{CAD} = 90^\circ$, we know $ \triangle{CAD}$ is isosceles right. Let $ AC = AD = x$. By Pythagorean, $ CD = x\sqrt2$. We are given $ CD = DB$ so $ DB = x\sqrt2$.

Let $ H$ be the foot of the altitude from $ A$ to $ \overline{CD}$. Reflect $ B$ over $ \line{AH}$ to $ E$. Since the reflection of $ D$ over $ \line{AH}$ is $ C$, we can conclude that $ \triangle{ECA}\cong\triangle{BDA}$. We know that the sum of the measures of $ \angle{EAC}$, $ \angle{CAD}$, and $ \angle{DAB}$ is $ 180^\circ$, and that $ \angle{ECA}\cong\angle{BAD}$, so we can solve for $ m\angle{BAD}$ and get $ 45^\circ$.

Now we use Law of Sines to solve for $ \sin\angle{ABD} = k$.

$ \frac {BD}{\sin45^\circ} = \frac {AD}{k}$

$ \frac {x\sqrt2}{\frac {\sqrt2}{2}} = \frac {x}{k}$

$ k = \frac {1}{2}$

$ \sin{\angle{ABD}} = \frac {1}{2}$

$ m\angle{ABD} = 30^\circ$, since it obviously can't be $ 150^\circ$

Now we know that $ m\angle{BAD} + m\angle{ADB} + m\angle{DBA} = 180^\circ$.
Solving for $ m\angle{ADB}$, we have ${ 45^\circ + m\angle{ADB} + 30^\circ = 180^\circ}$

$ m\angle{ADB} = 105^\circ$
[/hide]

[quote="FightTheTide"][hide="Solution"]
Since $ AC = AD$ and $ m\angle{CAD} = 90^\circ$, we know $ \triangle{CAD}$ is isosceles right. Let $ AC = AD = x$. By Pythagorean, $ CD = x\sqrt2$. We are given $ CD = DB$ so $ DB = x\sqrt2$.
[/hide][/quote] Based on my diagram and your solution, I think at this point you switched points C and D.
[img]http://i193.photobucket.com/albums/z278/Anaxerzia/Trapezoid2-1.png[/img][quote="Fight the Tide"][hide]
Let $ H$ be the foot of the altitude from $ A$ to $ \overline{CD}$. Reflect $ B$ over $ \line{AH}$ to $ E$. Since the reflection of D over $ \line{AH}$ is $ C$, we can conclude that $ \triangle{ECA}\cong\triangle{BDA}$. We know that the sum of the measures of $ \angle{EAC}$, $ \angle{CAD}$, and $ \angle{DAB}$ is $ 180^\circ$, and that $ \angle{ECA}\cong\angle{BAD}$, so we can solve for $ m\angle{BAD}$ and get $ 45^\circ$. [/hide][/quote]

$ m\angle{BAD}$ is 135, and not 45, in the original diagram that you used in the previous quote. $ m\angle{BAC}$ is 45, however...

Since you used information from before switching it with switching it, the solution begins to contain inaccuracies. Here is how it should be (as if there was no switching):

[quote="Fight the Tide (revision)"][hide]
Now we use Law of Sines to solve for $ \sin\angle{ABD} = k$.
$ \frac {BD}{\sin135^\circ} = \frac {AD}{k}$

$ \frac {x\sqrt2}{\frac {\sqrt2}{2}} = \frac {x}{k}$

$ k = \frac {1}{2}$

$ \sin{\angle{ABD}} = \frac {1}{2}$

$ m\angle{ABD} = 30^\circ$, since it obviously can't be 150^\circ

Now we know that $ m\angle{BAD} + m\angle{ADB} + m\angle{DBA} = 180^\circ$.
Solving for $ m\angle{ADB}$, we have ${ 135^\circ + m\angle{ADB} + 30^\circ = 180^\circ}$

$ m\angle{ADB} = 15^\circ$ [/hide][/quote]

Minor error though; other wise, good job  :) .


First note that function $f(x)=\frac{x}{1+x}$ is increasing on $[0,\infty )$ because $f'(x)=\frac{1}{(1+x)^{2}}\geq 0$ and that $g(x)=\frac{f(x)}{x}$ is convex on this set : $g''(x)=2(1+x)^{-3}\geq 0.$ But we know that $|a+b+c|\leq |a|+|b|+|c|$ by triangular inequality. Hence
\[\frac{|a+b+c|}{1+|a+b+c|}\leq \frac{|a|+|b|+|c|}{1+|a|+|b|+|c|}. \]
Now apply weighted Jensen's inequality to $g$ and get that
\[\frac{|a|}{|a|+|b|+|c|}g(|a|)+\frac{|b|}{|a|+|b|+|c|}g(|b|)+\frac{|c|}{|a|+|b|+|c|}g(|c|)\geq g\left( \frac{a^{2}+b^{2}+c^{2}}{|a|+|b|+|c|}\right) \]
So
\[\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}+\frac{|c|}{1+|c|}\geq \frac{|a|+|b|+|c|}{1+\frac{a^{2}+b^{2}+c^{2}}{|a|+|b|+|c|}}\geq \frac{|a|+|b|+|c|}{1+|a|+|b|+|c|}\geq \frac{|a+b+c|}{1+|a+b+c|}\]
because $(|a|+|b|+|c|)^{2}\geq a^{2}+b^{2}+c^{2}$ with equality iff two among $a,b,c$ are 0.

Note that this solution can be used to prove following (stronger):

Let $n$ be a natural number and $x_{1},x_{2},...,x_{n}$ real numbers. Prove that \[\sum_{i=1}^{n}\frac{|x_{i}|}{1+|x_{i}|}\geq \frac{\sum_{i=1}^{n}|x_{i}|}{1+\sum_{i=1}^{n}|x_{i}|}\] with equality iff there is at most one $x_{j}\neq 0$.

I might be wrong but:

[hide]
Note that when $x > 0$, $\frac{x}{1+x}$ is increasing.

Then if all $a,b,c > 0$ we have to prove:

$\frac{|a+b+c|}{1+|a+b+c|}=\frac{a+b+c}{1+a+b+c}\le \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}= \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}+\frac{|c|}{1+|c|}$

but now just note that $\frac{a}{1+a+b+c}\le \frac{a}{1+a}$, substitute $b$ and $c$ for $a$ and sum all to get the result.

If all $a,b,c < 0$ we can multiply each of them by $(-1)$ without changing the value on either side and have the same result as previously.

Now suppose that one of them (let's assume it's $c$) is negative and other two are positive:

If the sum $a+b+c$ is positive we have (using the first result):

$\frac{|a+b+c|}{1+|a+b+c|}=\frac{a+b+c}{1+a+b+c}\le \frac{a+b-c}{1+a+b-c}\le \frac{a}{1+a}+\frac{b}{1+b}+\frac{-c}{1-c}= \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}+\frac{|c|}{1+|c|}$.

If the sum is negative we similarly have:

$\frac{|a+b+c|}{1+|a+b+c|}=\frac{-a-b-c}{1-a-b-c}\le \frac{a+b-c}{1+a+b-c}\le \frac{a}{1+a}+\frac{b}{1+b}+\frac{-c}{1-c}= \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}+\frac{|c|}{1+|c|}$.

If two of them are negative then we can (again) multiply by $(-1)$ to take use of the previous result.

[/hide]


Prove that $\forall n \in \mathbb{N}$ $\dfrac{\prod \limits^{4n}_{k=1} k!}{(2n)!}$ is a square.

So that is 1!.2!.3!. ... .(2n-2)!.(2n-1)!.(2n+1)!.(2n+2)!. ... (4n-1)!.(4n)!

And it's easy to "check" that every factor appears an even amount of times, so we get a thing of the form $p_1^{2n_1}\cdots p_k^{2n_k}$ with k,n naturals

square root of this is $p_1^{n_1}\cdots p_k^{n_k}$ which is thus always a square :)


[hide]Use Vendermonte's Identity.

${\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}}$.

Let $m=k=n$ so that we have ${\sum_{i=0}^{k}\binom{n}{i}\binom{n}{n-i}={\sum_{i=0}^{k}\binom{n}{i}\binom{n}{i}={\sum_{i=0}^{k}(\binom{n}{i})^{2}=\binom{2n}{n}}}}$[/hide]

[hide]I believe this is equivilant to $\binom{2n}{n}$.[/hide]

[quote="drunner2007"][hide]Use Vendermonte's Identity.

${\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}}$.[/hide][/quote]

This is a general statement of which this problem is a specific case, and is moreover very easy to prove.  So prove it.  ;)

[quote="t0rajir0u"]Evaluate

$\sum_{k=0}^{n}{n \choose k}^{2}$[/quote]

$\binom{n}{k}^{2}=\binom{n}{k}\binom{n}{n-k}$

then consider the coefficient of x^n in $(x+1)^{n}*(x+1)^{n}$

[quote="Jumbler"]Anyone wants to give a combinatorial proof?  :)[/quote]

Sure. It's the same idea as if you had two piles, one of $a$ things and one of $b$ things and you wanted to choose $c$ of them total. The number of ways to do this would be

$\binom{a+b}{c}= \sum_{k=0}^{c}\binom{b}{k}\binom{a}{c-k}$,

where each term represents the number of ways we can choose $k$ of the $c$ from the pile of $b$ things and $c-k$ from the pile of $a$ things. Letting $a = b = c = n$ and we have the desired sum and its equivalent form. This is in fact the identity used by drunner2007 in the first post.

[quote="Jumbler"]Anyone wants to give a combinatorial proof?  :)[/quote]
The combinatorial proof uses in fact the same idea as the polynomial proof.
[hide]
We choose $n$ persons form a group of $2n$ people consisting of $n$ mathematicians and $n$ physicists. There are of course $\binom{2n}{n}$ ways to make such choice, but we can take $k$ mathematicians and $n-k$ physicists in $\binom{n}{k}\binom{n}{n-k}= \binom{n}{k}^{2}$ ways for any $0\leqslant k\leqslant n$. Thus $\sum_{0\leqslant k\leqslant n}\binom{n}{k}^{2}=\binom{2n}{n}$. [/hide]

Edit: I see I was late.

[hide]Proof: Consider the number of paths from $\binom{0}{0}$ to $\binom{2n}{n}$ on Pascal's Triangle. Every path passes through the $n+1$st row, with entries $\binom{n}{0}$, $\binom{n}{1}, \ldots, \binom{n}{n}$.
The number of paths from $\binom{0}{0}$ to $\binom{n}{k}$ is $\binom{n}{k}$ for $0\leq k \leq n$ since each of these paths has length $n$ and exactly $k$ of the segments of the path move down and to the right. Also, the number of paths from $\binom{n}{k}$ to $\binom{2n}{n}$ is $\binom{n}{n-k}$ since each of these paths has length $n$ and exactly $n-k$ of the segments of the path move down and to the right. Thus the number of paths from $\binom{0}{0}$ to $\binom{2n}{n}$ passing through $\binom{n}{k}$ is $\binom{n}{k}\binom{2n}{n-k}.$
Summing as $k$ ranges from $0$ to $n$ gives the number of paths from $\binom{0}{0}$ to $\binom{2n}{n}$ as
$\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+\ldots+\binom{n}{n}\binom{n}{0}=\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}=\sum_{k=0}^{n}(\binom{n}{k})^{2}=\binom{2n}{n}$.[/hide]

[quote="t0rajir0u"]Those are the proofs I've come up with so far, too. :)

Now, my friend tells me there's a [b]geometric[/b] proof.  Who's up for finding that one?[/quote]

i'll be looking for one


[quote=now or never]Find the best constant for this inequality, with $ a,b,c$ are positive reals :
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}k \geq (k\plus{}3)\sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}}$[/quote]
The $uvw$'s technique kills it.
$k_{max}=1.812...$