a 10 by 10 by 10 cube is composed of white and black unit cubes  that are alternating colors.

what is the positive difference between the number of white cubes and black cubes

Hehe, jli, good problem, this beats off lazy people :D I'll let others try it first
[hide="Weird hint"]Zero? Think twice[/hide]

[hide]Uh... wouldn't it be 0? (since there are as many white cubes as black cubes) But it does say "[b]positive[/b] difference," so maybe I'm wrong...[/hide]

[quote="Ignite168"]I still think it's 0. It would be 1 if it was a 9x9x9 cube but it's not,[/quote]

I agree with the 0 part, but not that a 9x9x9 cube would be 1.  On each edge, there would be 5 whites, and 4 blacks, and there are 12 edges (but 8 corners are over counted, leaving only 4 more), which is more than 1.

Actually the top face of the 9 x 9 x 9 cube would look like this (where 1 represents black and 0 represents white):

101010101
010101010
101010101
010101010
101010101
010101010
101010101
010101010
101010101.

The second face looks like this:

010101010
101010101
010101010
101010101
010101010
101010101
010101010
101010101
010101010.

The 1st, 3rd, 5th, 7th, and 9th faces are all the same, and the 2nd, 4th, 6th, 8th faces are all the same. In the first kind, each face has one more black face than white face, while in the second kind, each face has one less black face than white face, so the answer is 5 - 4 = [b]1[/b].

For the 10 x 10 x 10 cube, we calculate it the same way, but now there is an even side length instead of an odd one, so the answer turns out to be [b]0[/b].

sorry if the positive differece thing confused you

[hide]It didn't fool me.  Ok maybe it did.  I get 0.[/hide]

[hide="General Solution"]I think there's a generalization for these kinds of questions:
-If one of the dimensions is even, then the difference between the number of white and black cubes is 0.
-If all of the dimensions are odd, then the difference between the number of white and black cubes is 1.

So then the answer for that specific question would be half of the total number of cubes, or 3 x 4 x 5 / 2 = [b]30[/b]

I'm too lazy to find a proof of the formula though.  :P [/hide]

[quote="b-flat"][quote="jli"]same problem except,

what is the number whites if the cube is instead a 3 by 4 by 5[/quote]

Are you sure that a 3x4x5 figure is a CUBE??[/quote]

He meant rectangular prism.   :roll:

And by the way, why does 10000th user think that it's not 0 for the original problem? What does he think the answer is?

Maybe, he is trying to confuse people and make them think the answer is something else.

Who knows what he thinks the answer is?  Maybe 1, 2, 57, etc.

[quote="jli"][quote="b-flat"][quote="jli"]same problem except,

what is the number whites if the cube is instead a 3 by 4 by 5[/quote]

Are you sure that a 3x4x5 figure is a CUBE??[/quote]

i said INSTEAD of a cube

 ;)[/quote]

Actually, you said if the cube is instead 3x4x5 and the difference would be 0 with 30 white and 30 black.

Hah, I got some of you guys fair and square. The reason I put a "weird hint" is to get your brains working. I didn't want people to put a post like this

[quote]zero[/quote]

or

[quote]0[/quote]

without proper explanations.

[quote="t0rajir0u"]You're all making this problem far too difficult.  ;)

[hide="Generalization"] If the total number of cubes (that is, the product of the sides) is even the positive difference is $0$.  If the total number of cubes is odd the positive difference is $1$.

The reason why should be obvious.  In fact, this can be generalized to $n$ dimensions.[/hide][/quote]

Um... that's exactly what I said...  :roll:  ... and 0 isn't a positive difference  :P


[img]http://photo2.bababian.com/upload1/20090713/F6E53007228FA62FF710C7EF7CD774C0.jpg[/img]

Thanks!! :D

By quoting admire9898, you can see that the URL to the image was:
[url]http://photo2.bababian.com/upload1/20090713/F6E53007228FA62FF710C7EF7CD774C0.jpg[/url]

The problem is:
Find the $ x^{13}$ coefficient of $ \dfrac{(1 \minus{} y^6)(1 \minus{} y^8)(1 \minus{} y^5)(1 \minus{} y^5)}{(1 \minus{} y)^4}$

[hide]If its meant to be a trick question, then the answer is $ 0$ since there is no $ x^{13}$ term when you expand out a function with $ y$ as the only variable.

If the problem was supposed to say "Find the $ y^{13}$ coefficient....", then its more challenging.

One possible way to approach this would be to distribute the $ (1 \minus{} y)^4$ among the four $ (1 \minus{} y^n)$ terms and then use this identity: $ \dfrac{1 \minus{} y^n}{1 \minus{} y} \equal{} 1 \plus{} y \plus{} y^2 \plus{} \cdots \plus{} y^{n \minus{} 1}$

$ \dfrac{(1 \minus{} y^6)(1 \minus{} y^8)(1 \minus{} y^5)(1 \minus{} y^5)}{(1 \minus{} y)^4} \equal{} \left(\dfrac{1 \minus{} y^6}{1 \minus{} y}\right) \left(\dfrac{1 \minus{} y^8}{1 \minus{} y}\right) \left(\dfrac{1 \minus{} y^5}{1 \minus{} y}\right) \left(\dfrac{1 \minus{} y^5}{1 \minus{} y}\right) \equal{}$

$ (1 \plus{} y \plus{} y^2 \plus{} y^3 \plus{} y^4 \plus{} y^5)(1 \plus{} y \plus{} y^2 \plus{} y^3 \plus{} y^4 \plus{} y^5 \plus{} y^6 \plus{} y^7)$ $ (1 \plus{} y \plus{} y^2 \plus{} y^3 \plus{} y^4)(1 \plus{} y \plus{} y^2 \plus{} y^3 \plus{} y^4)$

From here, you'd have to count all the possible ways to multiply one term from each factor to yield $ y^{13}$. This is a bit tedious though. There may or may not be a better way to do this.

[url=http://www.wolframalpha.com]Wolfram Alpha[/url] gives this result:
http://www03.wolframalpha.com/input/?i=(1-y^6)(1-y^8)(1-y^5)(1-y^5)%2F(1-y)^4

$ \dfrac{(1 \minus{} y^6)(1 \minus{} y^8)(1 \minus{} y^5)(1 \minus{} y^5)}{(1 \minus{} y)^4} \equal{}$
$ y^{20} \plus{} 4y^{19} \plus{} 10y^{18} \plus{} 20y^{17} \plus{} 35y^{16} \plus{} 54y^{15} \plus{} 75y^{14} \plus{} 96y^{13} \plus{} 114y^{12} \plus{} 126y^{11} \plus{} 130y^{10} \plus{} 126y^9 \plus{} 114y^8 \plus{} 96y^7 \plus{} 75y^6 \plus{} 54y^5 \plus{} 35y^4 \plus{} 20y^3 \plus{} 10y^2 \plus{} 4y \plus{} 1$

Which has a $ y^{13}$ coefficient of $ 96$.[/hide]

Sorry for some mistakes :blush:
Some questions even unsolved for ,me.
1.how to insert the code like this:       [img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/7/c/4/7c4fa91f7efd16ba086b7d5d487bfb71dbe80247.gif[/img]

2.this original problem was a very difficult problem,but I turned to a superior,he told me that that problem can transform to this one.but the most essential is that how to yield y^(13) in [img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/9/c/f/9cf8d0def070ec82803a9ed2c354c648037f01af.gif[/img][img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/d/2/7/d274c651158cc14a30e7bf4d7a0e26bd438dde28.gif[/img]


integration
advanced fields
advanced fields unsolved

in the applied math class at school they mentioned Laplace's Lemma which helps us get for large t

$ \int_0^\infty e^{\minus{}th(x)}dx \sim  \frac{e^{th(0)}}{\minus{}th'(0)}$ when h is decreasing to negative infinity

similarly we prove that $ \int_R e^{th(x)} g(x)dx \sim  \sqrt{2\pi} \frac{e^{th(0)}}{\sqrt{\minus{}th''(0)}}g(0)$
for large t and the same kind of h with max at zero decreasing to negative  infinity on each end.

this is all well and good, but going over these proofs isnt giving me any inspiration for how these act for small values of t.

i want to know what they are $ \sim $ to as $ t \to 0$

most simply, i want to see how $ \int_0^\infty e^{\minus{}tx}g(x)dx$ acts as $ t\to 0$ or $ \int_0^\infty e^{\minus{}f(t)x}g(x)dx$ as with $ f(0)\equal{}0$ as $ t\to 0$

Well, if $ t\to 0\plus{}$, they converge to $ \int_0^\infty g(x)\,dx$ if the latter exists. What else do you want to know about them? :?

like.. how quickly..?  $ w(f(t))\equal{}\int_0^\infty e^{\minus{}f(t)x}g(x)dx$

is there a nice way to look at $ w(f(t))\minus{}w(f(0))$ and whether the convergence is like $ ct^n$ for some n or $ e^{\minus{}\frac{c}{t}}$ or

$ \ln(1\plus{}t^n)$ or any combination of these things..?


The points $ (x, y)$ represented in this table lie
on a straight line. The point $ (13, q)$ lies on
the same line. What is the value of $ p \plus{} q$?
Express your answer as a decimal to the
nearest tenth.

\[ \begin{array}{c|c}
x & y \\ \hline
2 & \minus{}5 \\
p & \minus{}14 \\
p\plus{}2 & \minus{}17 \\
\end{array}\]

hello, we have $ y\equal{}mx\plus{}n$ and with the point $ (2;\minus{}5)$ we get
$ y\equal{}mx\minus{}5\minus{}2m$ setting the coordinates of the other points in this equation we get
$ \minus{}14\equal{}mp\minus{}5\minus{}2m$
$ \minus{}17\equal{}m(p\plus{}2)\minus{}5\minus{}2m$
solving this system we get
$ m\equal{}\minus{}3/2,p\equal{}8$
and for $ q$ we get $ q\equal{}\minus{}43/2$ so we have $ p\plus{}q\equal{}\minus{}27/2$.

Continuing off DSG's post, we have to express it in a decimal, so it is $\boxed{-13.5}$


Each cell of a $100 * 100$ board is painted with one of four colors, so that each row and each column contains exactly $25$ cells of each color. Show that there are two rows and two columns whose four intersections are all different colors.

OK, I found the solution myself,thank you "pbornsztein". :)

hi i´ve trying it, can you post the solution or the link


[quote="catcurio"]Source: 2005 AMC-10B #12

Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

$\text{(A)} \ \left(\frac{1}{12}\right)^{12} \qquad \text{(B)} \ \left(\frac16\right)^{12} \qquad \text{(C)} \ 2\left(\frac16\right)^{11} \qquad \text{(D)} \ \frac52\left(\frac16\right)^{11} \qquad \text{(E)} \ \left(\frac16\right)^{10}$[/quote]
[hide]I got (E) $\ \left(\frac16\right)^{10}$

There are $6^{12}$ combinations, the only primes possible are $2$, $3$, and $5$. there are $12$ possibilities for $2$, $3$, and $5$. $12\cdot3=6^2$. $\frac{6^2}{6^{12}}=\ \left(\frac16\right)^{10}$
[/hide]

[quote="robinhe"][quote="ch1n353ch3s54a1l"]There are $6^{12}$ combinations, the only primes possible are $2$, $3$, and $5$. there are $12$ possibilities for $2$, $3$, and $5$. $12\cdot3=6^2$. $\frac{6^2}{6^{12}}=\ \left(\frac16\right)^{10}$
[/hide][/quote]

Wait, why are there 12 possibilities for 2, 3, and 5?  :blush:[/quote]

 :P
If one digit is 2, the other 11 are one. There are 12 possibilities because of 12 placements.
2,1,1,1,1, etc., 1,2,1,1,1,1 etc. and...etc. Same with the other three.


Define $\mu(k)$ as the following:

- $\mu(1) = 1$
- $\mu(k) = (-1)^{n}$ for $k$ a product of $n$ distinct primes
- $\mu(k) = 0$ otherwise

[b]One.[/b]

Given an integer $n$, let $\mathcal{D}$ be the set of its positive integral divisors.  Show that

$\sum_{d \in \mathcal{D}}\mu(d) = 0$

[b]Two.[/b]

Show that

$\sum_{d \in \mathcal{D}}\mu(d) \cdot \frac{n}{d}= \varphi(n)$

[hide="Hint"] Refer to the second post in Properties of the Totient, Part Two for a rather large hint about how to go about doing these problems. [/hide]


complex numbers
algebra proposed
algebra


logarithms
number theory unsolved
number theory

[quote="Lousin Garckz"]In the sequence $ 1,2,4,4,8,8,8,8,16,16,16,16,16,16,16,16,32...$, which number is in the position $ 256^{287}$ in the sequence?...
If the binary expresion of the number $ k$ is $ 1010110011010111$, which number is in the position $ k$ in the sequence?....[/quote]

If we consider that $ 1$ is $ x_1$ (first position), then we have $ x_n \equal{} 2^{1 \plus{} [\log_2(n \minus{} 1)]}$ $ \forall n > 1$

Then $ x_{256^{287}} \equal{} 2^{1 \plus{} [\log_2(256^{287} \minus{} 1)]}$ $ \equal{} 2^{1 \plus{} [\log_2(2^{16\times 287} \minus{} 1)]}$ $ \equal{} 2^{16\times 287} \equal{} 2^{4592}$

And since $ 2^{15} < k \minus{} 1 < 2^{16}$, we get $ x_k \equal{} 2^{1 \plus{} [\log_2(k \minus{} 1)]}$ $ \equal{} 2^{1 \plus{} 15}$ $ \equal{} 65536$

[quote="Lousin Garckz"]sorry, i can't figure out the idea from wich you obtained
If we consider that 1 is $ x_1$ (first position), then we have $ x_n \equal{} 2^{1 \plus{} [\log_2(n \minus{} 1)]} \forall n > 1$
.... can you explain it? please  :)[/quote]

It seems that your construction for sequence $ x_n$ is :

$ n\in[2,2]$ $ \implies$ $ x_n\equal{}2$
$ n\in[3,4]$ $ \implies$ $ x_n\equal{}2^2$
$ n\in[5,8]$ $ \implies$ $ x_n\equal{}2^3$

And so : $ n\in[2^k\plus{}1,2^{k\plus{}1}]$ $ \implies$ $ x_n\equal{}2^{k\plus{}1}$

And so : $ n\minus{}1\in[2^k,2^{k\plus{}1}\minus{}1]$ $ \implies$ $ x_n\equal{}2^{k\plus{}1}$

And so : $ \log_2(n\minus{}1)\in[k, k\plus{}1)$ $ \implies$ $ x_n\equal{}2^{k\plus{}1}$

And so : Let $ k\equal{}[\log_2(n\minus{}1)]$, then $ x_n\equal{}2^{k\plus{}1}$

And so $ x_n\equal{}2^{[\log_2(n\minus{}1)]\plus{}1}$


Let k be real number in (-1;2) and a,b,c be three real number .
Prove that
 $ (a^2\plus{}b^2\plus{}c^2\plus{}k(ab\plus{}bc\plus{}ca))(\frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{(b\minus{}c)^2}\plus{}\frac{1}{(c\minus{}a)^2}) \geq \frac{9(2\minus{}k)}{4}$

Let $ k\equal{}\minus{}1\plus{}\frac{3}{1\plus{}t}\in{(\minus{}1,2)},t>0,$then

\[ a^2\plus{}b^2\plus{}c^2\plus{}k(ab\plus{}bc\plus{}ca))(\frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{(b\minus{}c)^2}\plus{}\frac{1}{(c\minus{}a)^2}) \geq \frac{9(2\minus{}k)}{4}\]

is equivalent to

\[ t(a\plus{}b\minus{}2c)^2(b\plus{}c\minus{}2a)^2(c\plus{}a\minus{}2b)^2\plus{}4(a\plus{}b\plus{}c)^2(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^2\ge0.\]

@8825[hide]Ban. la` ai vay.,cung~ o? Ha noi. a`[/hide]


calculus
integration
LaTeX
logarithms
calculus computations

:o  use $ \text{\LaTeX}$, please...

$ \int_0^{\n 6} \frac{dx}{\sqrt{e^x + 3}}= \int_0^{\n 6} \frac{e^x \, dx}{e^x \sqrt{e^x + 3}} \stackrel{t = e^x}{=} \int_1^6 \frac{dt}{t \sqrt{t+3}} = \frac{1}{\sqrt{3}} \ln \left| \frac{\sqrt{3} - \sqrt{t+3}}{\sqrt{3} + \sqrt{t+3}} \right| \Big|_1^6 = \ldots$

Thank you.I am sorry next time i will use latex  :)

[quote="roserose18"][img]http://img176.imageshack.us/img176/8747/adszzq9.png[/img]
thanks for all help :)[/quote]

what a beautiful picture  :rotfl:  :P


Circles $C_1$ and $C_2$ have radii 3 and 7 respectively. The circles intersect at distinct points $A$ and $B$. Point $P$ is outside $C_2$ lies on the line determined by $A$ and $B$ at a distance of 5 from the center of $C_1$. Point $Q$ is chosen on $C_2$ so that $PQ$ is tangent to $C_2$ at $Q$. Find the length of segment $PQ$

[hide]Answer: 4

The circle power of point $P$ with respect to $C_1$ is $(5-3)(5+3)=16$. Since $P$ lies on the radical line $AB$ of $C_1$ and $C_2$, the circle power of $P$ with respect to $C_2$ is also $16$. Because $Q$ is tangent to $C_2$, this implies that $|PQ|=\sqrt{16}=4$.[/hide]


The people that may participate are
Xantos C. Guin
bob123
Hamster1800
chess64
nat mc
Here are the problems:

1.  In how many ways can the four digits 1, 9, 9 and 7 be used as digits to form two numbers which are both prime? (For example, 19 and 97, which are both prime, can be formed, as can 199 and 7.)
-------------------------------------------------------------------------------------------------

2.  Earlier in the week, Green took the Thrust SSC -- the jet-powered car used to set the record -- to 719.137 mph. (Unfortunately, because of damage to the brakes, the car couldn't perform a second run Tuesday; a second run is required to set an official record.) Assuming that his acceleration to 719.137 mph, and his deceleration to stopping, were constant, how long did it take Green to complete the 13-mile run?
-----------------------------------------------------------------------------------------------------
3.  Of the 34 million Americans who traveled this past Labor Day weekend, 90 percent went by automobile, light truck and recreational vehicles. Nearly 9 percent flew, while the remainder traveled by bus, rail or ship. Assuming that those who drove traveled an average of 150 miles and got an average of 20 miles per gallon, what was the total cost of all gasoline used, based on the national average?
--------------------------------------------------------------------------------------------------
4.   McDonald's took much criticism many years ago for using foam containers which were harmful to the environment. Since, they've started using paper containers which are recyclable. Evidence used in this court case consisted of an incredible 40,000 pages of documents, and the testimony of witnesses took an astounding 19,000 pages to record. If 10 pages of used paper can be recycled to make one new sheet, how many sheets will these documents eventually make if they are recycled after each use?
-----------------------------------------------------------------------------------------------------
5.  The average age at which a person begins to smoke is 12. Apparently, smokers need to start that early, though, because a smoker's life is shortened by 15 years due to smoking. Currently, the average smoker spends $700 per year on tobacco. How much would an average smoker spend in his lifetime on tobacco, if the average life is 75 years?
---------------------------------------------------------------------------------------------------------
6.  The Louisiana Superdome is 27 stories tall. One estimate says that ++that's as high as 44 defensive ends stacked head to toe.++ If the average defensive end is 6 ft., 2 in. tall, how tall is one story?
----------------------------------------------------------------------------------------------------------
7.  The exhibit is designed to create a smooth flow of 600 people per hour. If 15 patrons arrive every minute from 8am through 12noon, at what time will the last patron be admitted?
-------------------------------------------------------------------------------------------------------------
8.  The coldest recorded temperature at the South Pole, 90 degrees south latitude, is -71 degrees Celsius. What's the corresponding Fahrenheit temperature?
-----------------------------------------------------------------------------------------------------------
9.  Mainland China supplies 80% of Hong Kong's chickens. They ship 75 000 chickens to Hong Kong every day. In total, then, how many chickens does Hong Kong receive in one day?
---------------------------------------------------------------------------------------------------------
10.  In the American League, the voting for the catcher position occurred as follows:

Ivan Rodriguez, Texas 1,897,774

Sandy Alomar, Cleveland 1,629,674

Jorge Posada, New York 252,739

Charles Johnson, Baltimore 238,587

By what percent did Ivan Rodriguez edged out Sandy Alomar at the starting catcher position?
----------------------------------------------------------------------------------------------------------
YOU HAVE TWO WEEKS TO COMPLETE THIS!!!!
GOOD LUCK!!!

Does anyone want to try it???
If nobody tries it then nobody wins

Okay, okay, you gave us 2 weeks, though, so don't expect any answers until the last day! :roll:

can i particapate? The relay is fun but i can really do the problems.  :?

you can't participate because you didn't get 10/10 on round 2.

[quote="JesusFreak197"]Obviously, you can solve the problems on your own if you want the practice, though.[/quote]
of course. you can't not do the problems if you want

[quote="Xantos C. Guin"]#3. What is the cost of gas? And do we count the people that didn't drive in the total cost?

#4. Do the sheets of paper in the testimony count, or is it just the sheets made onwards? Also, Are the sheets recycled forever, or are they recycled till a full sheet can't be made?

One final question, Are there any problems in which rounding is required? If so, which ones and to what "degree of accuracy"?[/quote]
just forget numbers 3 and 4(don't do them)
and the answer to your final question is to the nearest hudereth.

umm, will you post answers to these after?

i dunno, maybe.
i might lose the answers, because i lose a lot of stuff.

The winner is: [hide][hide][hide]Xantos C. Guin
He was the only one to answer the problems.
Good job!!! :first:
[/hide][/hide][/hide]

Okay, so he really didn't "win",
well....

[quote="math92"]i guess he won by default.  he could have given me only one correct answer and he could still win.[/quote] :rotfl:  :rotfl:  :rotfl:  :winner_first:
Why no one send answers?  :rotfl:  :rotfl:  :rotfl:

there is a ton of contests going on so they might not have had time


logarithms
calculus
calculus computations

Help me to solve the following differential equation:-

 { {y[x ln(y) - y]}/{x[y ln(x) - x]} } = dy/dx

take the equation: multiplying by denominator, you get: \[ (xy\ln x - x^2)y' = xy\ln y - y^2 \Longrightarrow y'\ln x + \frac{y}{x} = \ln y + x\frac{y'}{y} \]
that is $y\ln x = x \ln y + C$, depending on the initial conditions...
but i don't think this is solvable with elementary methods...

[quote="FMako"]$\frac{y(x \ln y - y)}{x (y \ln x - x)} = \frac{dy}{dx}$ ?[/quote]
Yes it is the same

[quote="Tr"]wondering could  change of variables work here ...y=ux  and then solve it..[/quote]

I had tried it but it didn't work.

Bye :ninja:


trigonometry
algebra
polynomial
national olympiad

I was wondering if anyone had the USA Team Selection Test from 2007.
It happened a long time ago, so I don't see any reason why it shouldn't be posted here if someone has it.

$ 1$. Circles $ \omega_1$ and $ \omega_2$ meet at $ P$ and $ Q$. Segments $ AC$ and $ BD$ are chords of $ \omega_1$ and $ \omega_2$ respectively, such that segment $ AB$ and ray $ CD$ meet at $ P$. Ray $ BD$ and segment $ AC$ meet at $ X$. Point $ Y$ lies on $ \omega_1$ such that $ PY \parallel BD$. Point $ Z$ lies on $ \omega_2$ such that $ PZ \parallel AC$. Prove that points $ Q,X,Y,Z$ are collinear.

$ 2$. Let $ n$ be a positive integer and let $ a_1 \leq a_2 \leq \cdots \leq a_n$ and $ b_1 \leq b_2 \leq \cdots \leq b_n$ be two nondecreasing sequences of real numbers such that
\[ a_1 \plus{} \cdots \plus{} a_i \leq b_1 \plus{} \cdots \plus{} b_i \text{ for every } i \equal{} 1,\ldots,n\minus{}1\]
and
\[ a_1 \plus{} \cdots \plus{} a_n \equal{} b_1 \plus{} \cdots \plus{} b_n.\]
Suppose that for any real number $ m$, the number of pairs $ (i,j)$ with $ a_i \minus{} a_j \equal{} m$ equals to the number of pairs $ (k,l)$ with $ b_k \minus{} b_l \equal{} m$. Prove that $ a_i \equal{} b_i$ for $ i \equal{} 1,\ldots,n$.

$ 3$. Let $ \theta$ be an angle in the interval $ (0,\pi/2)$. Given that $ \cos \theta$ is irrational, and that $ \cos k \theta$ and $ \cos[(k\plus{}1)\theta ]$ are both rational for some positive integer $ k$, show that $ \theta \equal{} \pi/6$.

$ 4$. Determine whether or not there exist positive integers $ a$ and $ b$ such that $ a$ does not divide $ b^n \minus{} n$ for all positive integers $ n$.

$ 5$. Triangle $ ABC$ is inscribed in circle $ \omega$. The tangent lines to $ \omega$ at $ B$ and $ C$ meet at $ T$. Point $ S$ lies on ray $ BC$ such that $ AS \perp AT$. Points $ B_1$ and $ C_1$ lie on ray $ ST$ (with $ C_1$ in between $ B_1$ and $ S$) such that $ B_1T \equal{} BT \equal{} C_1T$. Prove that triangles $ ABC$ and $ AB_1C_1$ are similar to each other.

$ 6$. For a polynomial $ P(x)$ with integer coefficients, $ r(2i\minus{}1)$ (for $ i \equal{} 1,2,3,\ldots,512$) is the remainder obtained when $ P(2i\minus{}1)$ is divided by $ 1024$. The sequence
\[ (r(1),r(3),\ldots,r(1023))\]
is called the [i]remainder sequence[/i] of $ P(x)$. A remainder sequence is called [i]complete[/i] if it is a permutation of $ (1,3,5,\ldots,1023)$. Prove that there are no more than $ 2^{35}$ different complete remainder sequences.

Thank you very much! :)

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=982011#982011
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=982016#982016
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=982017#982017
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=982018#982018
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=982020#982020
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=982021#982021

Thanks very much, MellowMelon   :).
Strange that they don't have a single combinatorics problem.
Did you participate in this exam?

I didn't, but I made it to the summer training program, which is where I got the copy of the test.

Were we supposed to give it out?  I recall being told that the problems would be used in various training programs.  Perhaps those were summer programs though...

Yeah I think they tried to make up for the USAMO's large amount of combo and ltitle geo.


Given the side lengths of a triangle, how do you tell if it is obtuse or acute?
Thanks

Let's call the sides a,b, and c such that $ a<b<c$. The triangle is obtuse if $ a^{2} \plus{} b^{2} < c^{2}$

It's acute if $ a^{2} \plus{} b^{2} > c^{2}$ It's right if $ a^{2} \plus{} b^{2} \equal{} c^{2}$

(a,b,c can be equal, I just don't know how to do the sign.)


seemed to me that they were a  little bit too easy :)
i think i have upgraded  the level a little bit in the third round :)


trigonometry
calculus
integration
geometric series
complex analysis
complex analysis unsolved

Let $ f(\theta) \equal{} \sum_{r\equal{}0}^{\infty} a^{2r\plus{}1} \cos (2r \plus{} 1) \theta$ and $ g(\theta) \equal{} \sum_{r\equal{}0}^{\infty} \frac{a^{2r\plus{}1}}{2r\plus{}1} \sin (2r \plus{} 1) \theta$, where $ \minus{}1 < a < 1$.

By considering $ \sum_{r\equal{}0}^{\infty} a^{2r\plus{}1} e^{(2r\plus{}1)i \theta}$, or otherwise, show that \[ f(\theta) \equal{} \frac{a(1 \minus{} a^2) \cos \theta}{(1 \minus{} a^2)^2 \plus{} 4a^2 \sin^2 \theta}\].
Hence or otherwise, find an expression for $ g(\theta)$.

Deduce that:

(i) $ \sum_{r\equal{}0}^{\infty} (\minus{}1)^r \frac{x^{2r\plus{}1}}{2r\plus{}1} \equal{} \arctan x$

(ii) $ \sum_{r\equal{}0}^{\infty} \frac{1}{2r\plus{}1} \sin (2r\plus{}1) \theta \equal{} \frac{\pi}{4}$  if  $ 0 < \theta < \frac{\pi}{2}$.

What is $ \sum_{r\equal{}0}^{\infty} \frac{1}{2r\plus{}1} \sin (2r\plus{}1) \theta$  if  $ \minus{}\frac{\pi}{2} < \theta < 0$?

Where is says "by considering" ?? That thing is a geometric series !!

That means that you can compute the sum $ \sum a^{2r\plus{}1} e^{(2r\plus{}1)x}$ (you are right- this is geometric series), and then by observing $ Re(\sum a^{2r\plus{}1} e^{(2r\plus{}1)x})\equal{}\sum a^{2r\plus{}1} cos (2r\plus{}1)x$ get the desired result :wink:

Yup, that first part is alright.. I'm having a problem for finding the expression for $ g(\theta)$ though  :(

[quote="aidan"] I'm having a problem for finding the expression for $ g(\theta)$ though  :([/quote]
Isn't there some calculus relation that connects $ f(\theta)$ and $ g(\theta)$?

We also have $ \sum a^{2r\plus{}1} \sin (2r\plus{}1)\theta \equal{} Im(\sum a^{2r\plus{}1} e^{(2r\plus{}1)i\theta})\equal{}\frac {a(1\plus{}a^2)\sin \theta} {(1\minus{}a^2)^2 \plus{}4a^2 \sin^2 \theta}$. Now, if we look at $ g(\theta)$ as $ g(\theta, a)$, we also have $ a \frac {\textrm{d}} {\textrm{d}a} g(\theta, a) \equal{} \sum a^{2r\plus{}1} \sin (2r\plus{}1)\theta$. Maybe we can find $ \int \frac {(1\plus{}a^2)\sin \theta} {(1\minus{}a^2)^2 \plus{}4a^2 \sin^2 \theta} \textrm{d}a$ ... as that would provide $ g(\theta)$.


In the row of Pascal's triangle that starts with 1 and then 10, what is the next number?

The first number of the row is $ \binom{10}{0}\equal{}1$, the second is $ \binom{10}{1}\equal{}10$, so the third one must be $ \binom{10}{2}\equal{}\frac{10\cdot 9}{2\cdot 1}\equal{}\boxed{45}$


Solve the inequality. Express your answer in interval notation.

x((1-x²)^4) + (x+3)((1-x²)^4) ≥ 0

[quote="stn0091"]Solve the inequality. Express your answer in interval notation.

$x((1-x^{2})^{4})+(x+3)((1-x^{2})^{4}) \ge 0$[/quote]
[hide] $(2x+3)((1-x^{2})^{4})\ge0$
$(1-x^{2})^{4}$ is always nonnegvative, so we can divide both sides by it.
$2x+3\ge0\Longrightarrow x\ge-\frac{3}{2}$
$\left[-\frac{3}{2},+\infty\right)$ [/hide]


algebra
polynomial
function
linear algebra
matrix
vector
rational function

One of my friends had a very nice problem in oral examination: if a rational fraction (quotient of two complex polynomials) takes rational values for rational values of the variable, then it is a quotient of two polynomials with rational coefficients. This is not hard, but I suspect the following (I do not know if it is true or not):
   If there exists an infinite set of rational numbers A such that if x is in A then f(x) is rational and if f is a rational fraction, then f is the quotient of two polynomials with rational coefficients.

A rational function of degree $(m,n)$ has $m+n+1$ degrees of freedom in the coefficients, and each condition $f(a_i) = b_i$ is a linear equation on these coefficients.  It is natural to expect that this system is (at least generically) nondegenerate.  Perhaps there is also some rational formula similar to Lagrange's interpolation polynomials.

While a solution does always exist to the rational interpolation problem, we can't guarantee that the expression is unique, and I don't see how to get a reasonable formula. The system is a homogeneous system of $m+n+1$ equations in $m+n+2$ unknowns, and the coefficient matrix depends on the values $b_i$ as well as the points $a_i$. The rank of the coefficient matrix is at least $\max(m+1,n+1)$, but it can be that small.

Polynomial interpolation is nice because polynomials of limited degree form a vector space; rational functions of limited degree don't.

Here's an idea: suppose $\frac{p_1(x)}{q_1(x)}$ and $\frac{p_2(x)}{q_2(x)}$ are both interpolating rational functions for the same data (degree at most $(m,n)$, $m+n+1$ points). Then $p_1(x)q_2(x)-p_2(x)q_1(x)$ is a polynomial of degree at most $m+n$ which is zero at all $m+n+1$ points. This forces $p_1(x)q_2(x)-p_2(x)q_1(x)$ to be the zero polynomial, so $\frac{p_1(x)}{q_1(x)}=\frac{p_2(x)}{q_2(x)}$ as rational functions.
This is the uniqueness result for rational interpolation; you may have to reduce to lowest terms after finding an expression, but you do get a unique function. That should finish the problem.

[quote="jmerry"]While a solution does always exist to the rational interpolation problem, we can't guarantee that the expression is unique, [/quote]

I don't think that the interpolation is always possible, but probably it is possible for "generic" data
outside some lower-dimensional set of bad cases.
For example, a degree $(m,n)$ rational function that vanishes at more than $max(m,n)$ points is identically zero, so there is no such function that is zero at (e.g.) all but one of the $m+n+1$ points.

Of course, we don't actually need existence for this problem- just uniqueness.

There is always a nonzero solution to that homogeneous system, which gives us polynomials $p,q$ (not both zero) such that $p(x_i)-a_iq(x_i)=0$. This only gives us trouble if $q(x_i)=0$ for some $x_i$.

If the data includes $a_i=0$ for $1\le i\le m+1$, the false interpolating function we get is $\frac{0}{(x-x_{m+2})(x-x_{m+3})\cdots(x-x_{n+m+1})}.$

I didn't see this problem the first time around; the system believes the equation is true if both the numerator and the denominator are zero at the point.


logarithms
function
integration
inequalities
real analysis
real analysis unsolved

Let $ x_0,x_1,x_2,\cdots$ a sequence of positive numbers with $ \sum_{i\equal{}0}^\infty x_i \equal{} \infty$. Let $ s_n \equal{} \sum_{i\equal{}0}^n x_i$

Prove that:

$ \sum_{i\equal{}0}^\infty \frac{x_i}{s_i} \equal{} \infty$

and that

$ \sum_{i\equal{}0}^\infty \frac{x_i}{s_i^2} <  \infty$

Not a proof, but an example so you can see this thing working:

Let $ x_i$ behave like $ \frac1{i\ln i}$ for large $ i.$ (Fix it near zero so as to not cause any problems.) Then $ s_n$ behaves like $ \ln(\ln n).$

That means $ \sum_i\frac {x_i}{s_i}$ looks like $ \sum_i\frac {1}{i\ln i\ln(\ln i)}$ and diverges,

while $ \sum_i\frac {x_i}{s_i^2}$ looks like $ \sum_i\frac {1}{i\ln i(\ln(\ln i))^2}$ and converges.

(I'm pretty sure this has been posted here before, but it's been quite a while.)

how would you do that in the continuous case ?
you would have a positive nice function $ f$ such that $ \int_0^{\infty} f(t) \, dt \equal{} \plus{}\infty$. You then want to study $ \int_0^{\infty} \frac{f(t)}{F(t)} \, dt \equal{} \int_0^{\infty} \frac{F'(t)}{F(t)} \, dt$, where $ F(t) \equal{} \int_0^t f(u) \, du$. I am sure you know how to conclude from here. The other question is the same. And for sequences, once that you've got the idea in the continuous case, you can do almost the same proof, isn't it ?

Let's be explicit about what alekk is saying.

Let $ y_n \equal{} \ln s_n.$ By the Mean Value Theorem, $ y_n \minus{} y_{n \minus{} 1} \equal{} \frac {x_n}{z_n}$ where $ s_{n \minus{} 1}\le z_n\le s_n.$

Then $ \sum_{i \equal{} 1}^{n}\frac {x_i}{s_i}\ge\sum_{i \equal{} 1}^{n}(y_i \minus{} y_{i \minus{} 1}) \equal{} \ln(s_n)$

And that tends to $ \infty$ as $ n\to\infty$

Repeating the idea: let $ y_n \equal{} \minus{} \frac1{s_n}.$ By the Mean Value Theorem, $ y_n \minus{} y_{n \minus{} 1} \equal{} \frac {x_n}{z_n^2},$ where, again, $ s_{n \minus{} 1}\le z_n\le s_n.$

Now we have $ \sum_{i \equal{} i}^n\frac {x_i}{s_i^2} \equal{} 1 \plus{} \sum_{i \equal{} 2}^{n}\frac {x_i}{s_i^2}\le 1 \plus{} \sum_{i \equal{} 2}^n\frac {x_i}{s_{i \minus{} 1}^2}\le 1 \plus{} \sum_{i \equal{} 2}^{n}(y_i \minus{} y_{i \minus{} 1})$

$ \equal{} 1 \plus{} y_n \minus{} y_1 \equal{} 1 \plus{} \frac1{s_1} \minus{} \frac1{s_n},$ which is bounded as $ n\to\infty.$

The convergence statement can also be proved by writing $ x_i \equal{} s_i \minus{} s_{i\minus{}1}$ and obtaining a telescoping series.

This is known in the literature as Abel-Dini theorem. You may find a lot of materials in the books: 1) Inequalities of Hardy-Littlewood-Polya 2)  An introduction to the theory of infinite series by Bromwich. Also browsing the JStor archive you
may find many papers by T.H.Hildebrand, Rajagopal and others on the subject.


number theory
Diophantine equation
number theory solved

Provided the diophantine equation: x(p-1)! + yp = 1. Let p be an arbitrary prime.
Determine all solutions (x,y) \in Z^2 where Z denotes integer.

This isn't that hard, is it?

We know that (p-1)!=-1 (mod p) (Wilson), so x(p-1)! = -x (mod p), so yp-x=1 (mod p), so x=-1 mod p. The equality (kp-1)(p-1)!+py=1 shows us that y=(1-(kp-1)(p-1)!)/p. The number 1-(kp-1)(p-1)! is divisible by p for any integer k, so the solns are (x, y)= (kp-1, (1-(kp-1)(p-1)!)/p) for any integer k.

Hi grobber,

I didn't say it's hard. I only mentioned in the problem's subject which theorem might be applied. Anyway well done, grobber !  :)


A cylinder's height equals the diameter of its base. What
fraction of the total surface area of the cylinder is the total area
of the two bases?  Express your answer as a common fraction.

The area of the bases of a cylinder is $ 2 \pi r^2$.

The surface area of a cylinder is $ 2 \pi r^2 \plus{} 2 \pi r \times h$.

(where h=height and r=radius)

We can change the second formula in this problem by substituting with $ h \equal{} 2r$. Now we set up the ratio.

$ \frac{2 \pi r^2}{2 \pi r^2 \plus{} 2 \pi r \times 2r} \equal{} \frac{2r^2}{2r^2 \plus{} 4r^2} \equal{} \frac{2r^2}{6r^2} \equal{} \frac{2}{6} \equal{} \frac{1}{3}$


1. There is a number that ends with 2. If You put the number in the beginning, then the number is two times bigger. What is the number? [i] 2xyz is to two time bigger than xyz2[/i]

2. Find the smalest number that You divide with 2 You get a rest 1. If You divide it with 3 You get rest 2. If You divide it with 5 You get the rest 4. But if You divide it with 7 You get a round number. What is the number?

[hide="2"]
we want to solve the system
$n \equiv 1 \mod 2$
$n \equiv 2 \mod 3$
$n \equiv 4 \mod 5$
$n \equiv 0 \mod 7$

From the first and the fourth equation, we obtain clearly $n=7(2k+1)$. Put it back in the second and the third equation,
noting that $7 \equiv 1 \mod 3$ and $7 \equiv 2 \mod 5$. So it must be

$2k+1 \equiv 2 \mod 3 \rightarrow 2k \equiv 1 \mod 3$
$2k+1 \equiv 2 \mod 5 \rightarrow 2k \equiv 1 \mod 5$

Write the conditions as $2k=1+3m$ and $2k=1+5n$. That implies $5n=3m$, and the smallest $n,m$ are obviously $3,5$. So $2k=16$
and the smallest $n=17\cdot 7= 119$
[/hide]

How can we find a number that respect conditions of 1)? Both numbers $2xyz$ and $xyz2$ must have the same number of digits, and so all of those digits must be even. But clearly the first digit is 1, that is odd...

It is a little quicker to solve the second question if you [hide]re-write the first three equations as $n \equiv -1 \mod m$ for $m = 2, 3, 5$.[/hide]

I don't know why you think all the digits have to be even, but there are no solutions.  If we replace "2" in that question by some other digit, there is only a single solution (1111 = 1*1111).  There are no solutions if we "turn it around" (to make it half instead of double), either, even if we replace "2" with some other digit.

Yeah,I agree with your idea.There's no solution for the first one.:pilot:

Math is my life :!:

:wallbash: Oh, how can you get it?

Math is my life :!:

[hide="Hint"]Let $A = \overline {B 2 } =10 \cdot B +2$ where $B =\overline{a_{n}a_{n-1}...a_{1}a_{0}}$. Then $2A= \overline {2B }=...$ [/hide]

I solved the second one a slightly different way:

[hide]

We know that x mod 7 = 0, x mod 2 = 1, x mod 3 = 2, and x mod 5 = 4.

Thus, we can form a vague Idea of what the surroundings of x on the number line is:

multiple of 7 - anynumber - mult. 5 - anynumber - mult. 3 - mult. 2 - x
more concisely:

7 - * - 5 -  * - 3 - 2 - x

Notice that this shows that 2a + 1 = x, 3b + 2 = x, etc.

Now, knowing what we have here on the number line can help us find what the integer after x is. Counting from the multiple of two we also know that (x + 1) mod 2 = 0, (x+1) mod 3 = 0, and (x+1) mod 5 = 0. Thus, x+1 is a multiple of 30. We already know that x is a multiple of 7, so by trial and error we can find a multiple of 30, where one number less is a mutliple of 7.

I quickly found x = 119.

[/hide][/hide]

[hide="1"]looking at 3 digits,

$2(1000x+100y+10z+2)=2000+100x+10y+z$

$1900x+190y+19z=1996$

note that if the RHS were divisible by 19, then the digits would match up to the RHS after we divided out 19
but we need to find a number such that the RHS is divisible by $19$, so looking at n digits

$19|(2*10^{n-1}-4)$

$19|(10^{n-1}-2)$

now we do some mod work:

since 19 is coprime to 10, we apply FLT to see that $10^k-2 \equiv 10^{k+18m}-2 \mod 19$ (${\{k,m}\}\subset\mathbb{Z}$)

beyond this, i think you need to brute force, there does not seem to be a better way, but it does show that if we have one solution, then there are infinitely more, (note that i assumed that there would be enough digits to match up, but that is pretty obvious

[/hide]

[quote="Ramanujan"]there are infinite solutions for 1.

[b][size=150]105263157894736842[/size][/b]

is one of them .  :)[/quote]

Heh, I didn't read the question carefully enough and assumed that it must have exactly four digits.  Nicely done.


modular arithmetic
quadratics
algebra
polynomial
number theory
divisibility tests
number theory unsolved

Find all positive integers $ n\in\{1,2,3,\ldots,2009\}$ such that
\[ 4n^6 \plus{} n^3 \plus{} 5\]
is divisible by $ 7$.

[hide="Solution"]

If n is divisible by 7, obviously the expression isn't.

If n isn't divisible by 7, then by Little Fermat's theorem, $ n^6\equiv 1\pmod 7$, so the given expression is congruent to $ 2\plus{}n^3\pmod 7$.

We have to find all $ n$ such that $ n^3\equiv 5\pmod 7$.

We know that $ n^3\equiv a\pmod 7 \Rightarrow (n\plus{}7k)^3\equiv a\pmod 7$, so we find the solutions in 0,1,2,3,4,5,6

We find that none of the numbers 0,1,2,3,4,5,6 satisfy the divisibility by 7 of the given expression, so there are no solutions.

[/hide]

$ 4n^{6} \plus{} n^{3}\equiv 2\pmod 7$
So all values lies on the series 9,16,23..
a th term:
$ 4n^{6} \plus{} n^{3} \equal{} 7a \plus{} 2$
$ Discriminant \equal{} 33 \plus{} 112a$
You can show that it is never a perfect square.

If $ 7$ divides $ n$, $ 4n^{6}\plus{}n^{3}\plus{}5$ is not divisible by $ 7$.

Otherwise, by Fermat's Little Theorem, $ 4n^{6}\cong 4\pmod 7$.

It is easy to check that  $ n^{3}\cong \pm{1}\pmod 7$.

done  :lol:

My solution.

[list]$ 7\,\mid\,4n^6 \plus{} n^3 \plus{} 5$[/list]
$ \Rightarrow\ 7\,\mid\,4n^6 \plus{} 8n^3 \plus{} 5$

$ \Rightarrow\ 7\,\mid\,\left[2(n^3 \plus{} 1)\right]^2 \plus{} 1$

$ \Rightarrow\ \minus{} 1$ is a quadratic residue $ \bmod\,7$ which is not true. ;)

Hey i got a solution,i think,I am not really sure though if all the steps are correct so if you find any mistake please let me know.

       It suffices to find all n such that $ 7|3n^6 \plus{} 6n^3 \plus{} 2$.
Set $ n^3 \equal{} m$.The quadratic $ 3m^2 \plus{} 6m \plus{} 2 \equal{} 0$ has discriminant $ D \equal{} 12 > 0$ and hence the solutions of these
quadratic are $ \frac {\sqrt {3}}{3} \minus{} 1 and \minus{} \frac {\sqrt {3}}{3} \minus{} 1$.

      Since 7 is prime it has to divide for example $ n^3 \plus{} 1 \plus{} \frac {\sqrt {3}}{3}$ which can never be true since that expression
isn't an integer because it contains the irrational $ \sqrt {3}$

Peace

~Kostas

[quote="new_member"]
      Since 7 is prime it has to divide for example $ n^3 \plus{} 1 \plus{} \frac {\sqrt {3}}{3}$ which can never be true since that expression
isn't an integer because it contains the irrational $ \sqrt {3}$

Peace

~Kostas[/quote]

The statement $ p\mid ab \implies p\mid a$ or $ p\mid b$ is only true when $ a$ and $ b$ are integers. Otherwise the statement would be false: take $ a\equal{}b\equal{}\sqrt p$

[quote="wangsacl"]Find all positive integers $ n\in\mathbb{Z}$ s.t. $ 7 \mid 4n^6 \plus{} n^3 \plus{} 5$.[/quote]
Define the polynomial $ p(x): \equal{}4x^6\plus{}x^3\plus{}5$. Obviusly if $ 7 \mid p(n)$ for some $ n \in \mathbb{Z}$ then $ 7 \nmid n$; then, working in $ (\mathbb{Z}/7\mathbb{Z})^*$ we have $ p(n)\equal{}2\plus{}\left(\frac{n}{7}\right) \in \{1,3\}$. []

[quote="kasym"]When I saw sixth power of 4 I thought about Fermat`s Little Theorem.

Thanks to ValentineA for a perfect solution.[/quote]
Thanks. It was the $ n^6$ and $ n^3$ that made me thinking about completing the square. :)


Real numbers $ a,b,c,d$ satisfies $ (a\minus{}2b\plus{}c)(b\minus{}2c\plus{}d)(a\minus{}b\minus{}c\plus{}d)\ne 0$. Show that if
\[ \frac{ac\minus{}b^2}{a\minus{}2b\plus{}c}\equal{}\frac{bd\minus{}c^2}{b\minus{}2c\plus{}d}\]
then
\[ \frac{ac\minus{}b^2}{a\minus{}2b\plus{}c}\equal{}\frac{bd\minus{}c^2}{b\minus{}2c\plus{}d}\equal{}\frac{ad\minus{}bc}{a\minus{}b\minus{}c\plus{}d}.\]

[quote="MeKnowsNothing"]Real numbers $ a,b,c,d$ satisfies $ (a \minus{} 2b \plus{} c)(b \minus{} 2c \plus{} d)(a \minus{} b \minus{} c \plus{} d)\ne 0$. Show that if
\[ \frac {ac \minus{} b^2}{a \minus{} 2b \plus{} c} \equal{} \frac {bd \minus{} c^2}{b \minus{} 2c \plus{} d}
\]
then
\[ \frac {ac \minus{} b^2}{a \minus{} 2b \plus{} c} \equal{} \frac {bd \minus{} c^2}{b \minus{} 2c \plus{} d} \equal{} \frac {ad \minus{} bc}{a \minus{} b \minus{} c \plus{} d}.
\]
[/quote]

$ \frac {ac \minus{} b^2}{a \minus{} 2b \plus{} c} \equal{} \frac {bd \minus{} c^2}{b \minus{} 2c \plus{} d}$ <-> $ (b\minus{}c)(ac\minus{}b^{2}\plus{}bc\minus{}c^{2}\plus{}bd\minus{}ad)\equal{}0$---->
a) $ ac\minus{}b^{2}\plus{}bc\minus{}c^{2}\plus{}bd\minus{}ad\equal{}0$ ---> $ ad\minus{}bc\equal{}ac\plus{}bd\minus{}b^{2}\minus{}c^{2}$

$ \frac {ac \minus{} b^2}{a \minus{} 2b \plus{} c} \equal{} \frac {bd \minus{} c^2}{b \minus{} 2c \plus{} d}\equal{}k$-->

$ ac\minus{}b^{2}\equal{}(a\minus{}2b\plus{}c)k$
$ bd\minus{}c^{2}\equal{}(b\minus{}2c\plus{}d)k$----> $ ac\minus{}b^{2}\plus{}bd\minus{}c^{2}\equal{}(a\minus{}2b\plus{}c\plus{}b\minus{}2c\plus{}d)k$--->$ (a\minus{}b\minus{}c\plus{}d)k\equal{}ac\plus{}bd\minus{}b^{2}\minus{}c^{2}\equal{}ad\minus{}bc$--> $ k\equal{}\frac{ad\minus{}bc}{a\minus{}b\minus{}c\plus{}d}$

b)$ b\minus{}c\equal{}0$---> $ b\equal{}c$--> $ k\equal{}b$

$ (a \minus{} 2b \plus{} c)(b \minus{} 2c \plus{} d)(a \minus{} b \minus{} c \plus{} d)\ne 0$---> $ a\not \equal{}b$ and $ d\not \equal{}b$--->

$ k\minus{}\frac{ad\minus{}bc}{a\minus{}b\minus{}c\plus{}d}\equal{}b\minus{}\frac{ad\minus{}b^{2}}{a\minus{}2b\plus{}d}\equal{}\frac{ab\minus{}2b^{2}\plus{}bd\minus{}ad\plus{}b^{2}}{a\minus{}2b\plus{}d}\equal{}\frac{(a\minus{}b)(b\minus{}d)}{a\minus{}2b\plus{}d}$-----> $ k\not \equal{}\frac{ad\minus{}bc}{a\minus{}b\minus{}c\plus{}d}$-----> $ \phi$

I had a different solution on my mind. Assume that $ \frac{ac\minus{}b^2}{a\minus{}2b\plus{}2}\equal{}k$. From the first equation we get $ ac\minus{}k(a\plus{}c)\equal{}b^2\minus{}2bk$. Add $ b^2$ to both sides to get $ (a\minus{}k)(b\minus{}k)\equal{}(b\minus{}k)^2$. Therefore $ a\minus{}k,b\minus{}k,$ and $ c\minus{}k$ are in geometrical progression. Similarly, $ b\minus{}k,c\minus{}k,$ and $ d\minus{}k$ are in geometrical progression. Hence $ (a\minus{}k)(d\minus{}k)\equal{}(b\minus{}k)(c\minus{}k)$. But expanding gives $ ad\minus{}(a\plus{}d)k\plus{}k^2\equal{}bc\minus{}(b\plus{}c)k\plus{}k^2$ so $ k\equal{}\frac{ad\minus{}bc}{a\minus{}b\minus{}c\plus{}d}$.


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

Let $ P$ be the the isogonal conjugate of $ Q$ with respect to triangle $ ABC$, and $ P,Q$ are in the interior of triangle $ ABC$. Denote by $ O_{1},O_{2},O_{3}$ the circumcenters of triangle $ PBC,PCA,PAB$, $ O'_{1},O'_{2},O'_{3}$ the circumcenters of triangle $ QBC,QCA,QAB$, $ O$ the circumcenter of triangle $ O_{1}O_{2}O_{3}$, $ O'$ the circumcenter of triangle $ O'_{1}O'_{2}O'_{3}$. Prove that $ OO'$ is parallel to $ PQ$.

[quote="yetti"]Let $ K$ be the circumcenter of the $ \triangle ABC.$ Let $ \triangle P_aP_bP_c, \triangle Q_aQ_bQ_c$ be the pedal triangles of $ P, Q$; the midpoint $ M$ of $ (PQ)$ is their common circumcenter. The midpoint $ N_a$ of $ AQ$ is the circumcenter of the $ \triangle AQ_bQ_c,$ on account of the rigth angles $ \angle QQ_bA, \angle QQ_cA.$ The isogonal $ AP$ of $ AQ \equiv AN_a$ with respect to the angle $ \angle CAB \equiv \angle Q_cAQ_b$ is its A-altitude $ \Longrightarrow$ $ Q_bQ_c \perp AP.$ But $ O_2O_3 \perp AP$ as well $ \Longrightarrow$ $ Q_bQ_c \parallel O_2O_3$ and similarly, $ Q_cQ_a \parallel O_3O_1, Q_aQ_b \parallel O_1O_2.$ The $ \triangle O_aQ_bO_c \sim \triangle O_1O_2O_3$ are centrally similar, having the correspnding sides parallel. $ M, O$ are their corresponding circumcenters and $ Q, K$ are their corresponding points, the concurrency points of their corresponding cevians $ (Q_aQ \parallel O_1K) \perp BC,$ $ (Q_bQ \parallel O_2K) \perp CA,$ $ (Q_cQ \parallel O_3K) \perp AB.$ Their corresponding lines $ QM \parallel KO$ are parallel. In ezactly the same way, $ PM \parallel KO'$ $ \Longrightarrow$ $ OO' \parallel PQ.$[/quote]

이 자의 풀이를 보지 말거라. 짐이 더 좋은 풀이를 올리리라.

Could you please use English instead of Korean here. Thanks.

$ \mbox{Prove that circumcircle of the } \triangle{ABC}, \triangle{O_1O_2O_3} \mbox{ and } \triangle{O'_1O'_2O'_3} \mbox{ coaxial circles.}$

My solution [b]:[/b]

Let $ T $ be the circumcenter of $ \triangle ABC $ .
Let $ D \equiv O_2O_3 \cap O_2'O_3', E \equiv O_3O_1 \cap O_3'O_1', F \equiv O_1O_2 \cap O_1'O_2' $ .

Easy to see $ T=O_1O_1' \cap O_2O_2' \cap O_3O_3' $.
Since $ O_2O_3, O_2'O_3' $ is the perpendicular bisector of $ AP, AQ $, respectively ,
so $ \angle O_2O_3O_3' =180^{\circ} - \angle BAP =180^{\circ} - \angle QAC =\angle O_2O_2'O_3' $ .
i.e. $ O_2, O_3, O_2', O_3' $ are concyclic
Similarly, we can get $ O_3, O_1, O_3', O_1' $ are concyclic and $ O_1, O_2, O_1', O_2' $ are concyclic.

Since $ O_2D \cdot DO_3=O_2'D \cdot  DO_3', O_3E \cdot EO_1=O_3'E \cdot EO_1', O_1F \cdot FO_2=O_1'F \cdot FO_2' $ ,
so $ D, E, F $ are collinear at the radical axis of $ (O_1O_2O_3) $ and $ (O_1'O_2'O_3') $ . i.e. $ \overline{DEF} $ is perpendicular to $ OO' $ [b]...[/b] $ (1) $

Since $ O_2O_3, O_2'O_3' $ is the perpendicular bisector of $ AP, AQ $, respectively ,
so $ D $ is the circumcenter of $ \triangle APQ $ . i.e. $ D $ lie on the perpendicular bisector of $ PQ $
Similarly, we can get $ E, F $ lie on the perpendicular bisector of $ PQ $ . i.e. $ \overline{DEF} $ is the perpendicular bisector of $ PQ $ [b]...[/b] $ (2) $

From $ (1) $ and $ (2) $ we get $ OO' \parallel PQ $ .

Q.E.D

Remark [b]:[/b]

$ (1) $
Since $ TO_1 \cdot TO_1'=TO_2 \cdot TO_2'=TO_3 \cdot TO_3' $ ,
so $ (O_1'O_2'O_3') $ is the image of $ (O_1O_2O_3) $ under the inversion WRT $ (O) $ .
i.e. $ (O), (O_1O_2O_3), (O_1'O_2'O_3') $ are coaxial

$ (2) $
This problem can be generalized to the quadrilateral [b]:[/b]

Let $ P, Q $ be the isogonal conjugate of $ ABCD $ .
Let $ O_1, O_2, O_3, O_4 $ be the circumcenter of $ \triangle ABP, \triangle BCP, \triangle CDP, \triangle DAP $ .
Let $ O_1', O_2', O_3', O_4' $ be the circumcenter of $ \triangle ABQ, \triangle BCQ, \triangle CDQ, \triangle DAQ $ .
Let $ O, O' $ be the circumcenter of $ O_1O_2O_3O_4, O_1'O_2'O_3'O_4' $ .

Then $ OO' \parallel PQ $


Μεταφερω αυτο το προβλημα που μου κινησε την περιεργια απο το High School...
Δεν ξερω εαν εχει δοθει λυση γιατι δεν εχω κοιταξει(το προσπαθω ακομα :blush: )....

Υπαρχουν 3 υπερυπολογιστες οι οποιοι ειναι πανομοιοτυποι(δεν μπορουμε να τους ξεχωρισουμε).Ο πρωτος εγινε στην Αμερικη και λεει παντα την αληθεια,ο δευτερος στην Κινα και λεει παντα ψεματα.Ο τριτος εγινε στην Ρωσια και δινει τυχαιες απαντησεις.Ο καθενας τους, ξερει τα παντα για τους αλλους.
Να βρειτε μια ερωτηση(στην οποια η απαντηση θα ειναι ΝΑΙ ή ΟΧΙ) που θα την υποβαλετε σε εναν απο τους υπολογιστες(χωρις να ξερουμε σε ποιον αφου ειναι ιδιοι)ετσι ωστε να βεβαιωθειτε οτι ΔΕΝ θα διαλεξετε τον υπολογιστη που εγινε στην Κινα....

yparxei kapoio hint gia ayto to provlima na dothei mias kai den mporo na kano proodo.
Pantos thimamai ekeino to rosiko vivlio to kotopoulo apo to minsk to opoio eixe paromoia provlimata
me ayta edo...

Σκεφτηκα μια ερώτηση που σου δίνει με βεβαιότητα 1 υπολογιστή που ΔΕΝ είναι από τη ρωσσία.. Δηλαδή δεν δίνει τυχαίες απαντήσεις.. Τώρα για τον κινέζικο θέλω άλλη μια :Ρ

Μου φαίνεται πως δεν γίνεται. Ίσως κάπου να κάνω λάθος. Αν το εντοπίσετε ίσως βοηθήσει να φτιάξουμε την ερώτηση.

Ας αριθμήσουμε τους υπολογιστές 1,2 και 3. Μπορούμε να υποθέσουμε πως ρωτάμε κάτι τον 1. Η στρατηγική μας θα πρέπει να είναι

-Αν ο 1 απαντήσει ΝΑΙ τότε διαλέγω τον υπολογιστή νούμερο Α
-Αν ο 1 απαντήσει ΟΧΙ τότε διαλέγω τον υπολογιστή νούμερο Β

Υπάρχουν άλλες στρατηγικές; Πως θα μοιάζουν αυτές;

Αν Α=Β=1 τότε χάσαμε. (Υπάρχει η περίπτωση ο 1 να είναι ο Κινέζος.) Ας υποθέσω ότι Α=/=1. Έστω Α=2. Αλλά αν 1 = Ρώσος, 2=Κινέζος και 3=Αμερικάνος και ο 1 απαντήσει ΝΑΙ πάλι χάσαμε. Ομοίως αν Α=3 ή Β=/=1.

Που κάνω λάθος;

Αφού τους αριθμήσεις (στην τύχη) 1,2 και 3 μπορείς να ρωτήσεις τον 1: "Αν σε ρωτούσα "Είναι ο 2 αυτός που απαντάει τυχαία" θα έλεγες "ναι"?
Με αυτή την ερώτηση παίρνεις σίγουρη πληροφορία για το ποιος δεν απαντάει στην τύχη. Γιατί αν απαντήσει "ναι" τότε:
-Ο 1 είναι ο τυχαίος
-Αν ο 1 είναι αυτός που λέει αλήθεια οπότε ο 2 είναι αυτός που απαντάει τυχαία (θα απάνταγε ναι και μας είπε ότι θα απάνταγε ναι)
-Αν ο 1 είναι ο ψεύτης τότε πάλι ο 2 είναι αυτός που απαντάει τυχαία, γιατί με την απάντηση "ναι" σημαίνει ότι θα μας έλεγε όχι στο αν ο 2 απαντάει τυχαία δηλαδή ως ψεύτης ο 2 είναι ο τυχαίος.
Άρα σε κάθε περίπτωση με το "ναι" ο 3 ΔΕΝ είναι ρώσικος.

Αντίστοιχα αν απαντήσει "οχι" θα καταλήξουμε ότι ο 2 δεν είναι ρώσικος..

Τώρα για τον κινέζικο θα το προσπαθήσω κι άλλο, εφόσον ο Dimitris X επιβεβαιώσει πως το πρόβλημα λέει όντως για κινέζο και δεν είναι τυπογραφικό  :P

[quote="lemonidas"]
Τώρα για τον κινέζικο θα το προσπαθήσω κι άλλο, εφόσον ο Dimitris X επιβεβαιώσει πως το πρόβλημα λέει όντως για κινέζο και δεν είναι τυπογραφικό  :P[/quote]
Δες το και απο εδω http://www.artofproblemsolving.com/Forum/viewtopic.php?t=264300  :wink:

[quote="Demetres"]Αν Α=Β=1 τότε χάσαμε. (Υπάρχει η περίπτωση ο 1 να είναι ο Κινέζος.) Ας υποθέσω ότι Α=/=1. Έστω Α=2. Αλλά αν 1 = Ρώσος, 2=Κινέζος και 3=Αμερικάνος και ο 1 απαντήσει ΝΑΙ πάλι χάσαμε. Ομοίως αν Α=3 ή Β=/=1.

Που κάνω λάθος;[/quote]

Δεν σε πολυ-πιανω... Τεσπα και εγω νομιζω οτι δεν γινεται
,να γιατι:

Εστω οτι υπαρχει μια τετοια ερωτηση κ, στην οποια ξερουμε τι θα απαντουσε ο αμερικανος, τι ο κινεζος και τι ο ρωσσος, και η απαντηση του αμερικανου ειναι α. Τοτε στην καλυτερη περιπτωση(λεγοντας αυτο ενοω οτι θα μπορουσε ισως να ειναι και του κινεζου α)του κινεζου θα ειναι ά, ωπου ά αντιθετο του α. Αν πεσουμε στον ρωσσο(εμεις οταν τον ρωταμε δεν ξερουμε οτι πεσαμε σε αυτον) τοτε αυτος μπορει να μας πει ειτε α ειτε ά(τυχαια παντα). Αν μας πει α τοτε ολα καλα και τον διαλεγουμε αφου σιγουρα δεν ειναι ο κινεζος. Αν ομως μας πει ά τοτε εχουμε προβλημα αφου και ο κινεζος και ο ρωσσος θα μπορουσε να δωσει αυτη τη απαντηση και αρα δεν μπορουμε να πουμε σιγουρα για καποιον οτι δεν ειναι ο κινεζος.
Αρα τετοια ερωτηση δεν υπαρχει.

Νομιζω οτι το επιχειρημα ειναι πληρως θεμελιωμενο... :maybe:

Η ερώτηση θα πρέπει να είναι ως εξής "Η επόμενη λέξη σου θα είναι ναι??? " αν χρειάζεται περαιτέρω εξήγηση πείτε . (Νομίζω είναι εμφανές.)

Ναι, ετσι ο κινεζος ειναι ο μονος που αδυνατει να απαντησει επιτρεπεται ομως να υποβαλουμε ερωτηση στην οποια καποιος δεν μπορει να δωσει απαντηση(το ενδεχομενο αυτο δεν το ελαβα υπ' οψην στην αποδειξη μου)?   :?:

Θα μπορουσες να εξηγησεις πως το σκεφτηκες?

Ειναι εμφανες πως για να εντοπιστει καποιος που δεν ειναι κινεζος πρεπει η απαντηση του κινεζου να ειναι διαφορετικη απο των αλλων 2. Και εφοσον ο ρωσος απανταει τυχαια πιθανον να απαντησει το αντιθετο απο αυτο που θα απαντησει ο αμερικανος. Αρα πρεπει να υπαρχει και 3τη δυνατη ανταποκριση στην ερωτηση ωστε να αποφυγουμε τον κινεζο, και εφοσον οι μονες δυνατες απαντησεις ειναι το "ναι" και το "οχι" προφανως η τριτη ανταποκριση θα ειναι η μη απαντηση. Αρα πρεπει να κατασκευαστει μια ερωτηση οπου ο κινεζος και μονον ο κινεζος να μην μπορει να απαντησει. Και γιατι να μην μπορει να απαντησει? Γιατι, προφανως αν πει ναι λεει αληθεια και αν πει οχι λεει παλι αληθεια.

Μεχρι εκει καταλαβαινω τον τρπο σκεψης η ερωτηση αυτη ομως πως σου ηρθε?   :?:

Filosofimene ( αλήθεια το μικρό σου ) το σκέφτηκα ως εξης :
Για να βρουμε ποιός είναι ο κινέζικος θα πρέπει όπως αναφέρεις και εσυ να ο κινέζικος να δίνει διαφορετική απάντηση από τους άλλους ή είναι ο μοναδικός που δεν απαντάει . Τώρα επειδή η ερώτηση θέλουμε να δέχεται ως απάντηση ΝΑΙ ή ΟΧΙ συμπεραίνουμε ότι θα πρέπει να παίξουμε με τις απαντήσεις ( ο λόγος είναι η παραπάνω αναφορά για τον κινέζικο ) . Επομένως πρέπει να προσδιορίσουμε μια ερώτηση που να μην μπορεί να απαντήσει ο κινέζικος ( η περίπτωση του να δίνει διαφορετική απάντηση είναι δυσκολη από το γεγονός ότι και ο ρώσικος θα μπορουσε να πει ψέμματα ) . Έτσι , η ερώτηση που θα σχετίζεται με το ναι ή με το όχι θα είναι ως εξής :" Η επόμενη λέξη σου θα είναι όχι ?( αντίστοιχα και με το ναι ) " όμως παρατηρουμε ότι με το όχθ υπάρχει τρόπος να απαντήσει και ο κινέζικος άρα αναγκαστικά καταφευγουμε στην ερώτηση με το ναι όπου παρατηρουμε ότι ο κινέζικος είναι ο μόνος που δεν μπορεί να δωσει απάντηση . Από εδώ σημαίνει ότι τελείωσε .

( Μήπως θα μπορουσαμε να βρουμε και μια άλλη ερώτηση χρησιμοποιόντας το γεγονός ότι ο κάθε υπολογιστής γνωρίζει τα πάντα για τον άλλο???)

[quote="kritikos"] ( αλήθεια το μικρό σου ) [/quote]

Με λενε Χρήστο.

[quote="kritikos"]( Μήπως θα μπορουσαμε να βρουμε και μια άλλη ερώτηση χρησιμοποιόντας το γεγονός ότι ο κάθε υπολογιστής γνωρίζει τα πάντα για τον άλλο???)[/quote]

Mα για αυτο ρωταω, γιατι μεχρι το σημειο που μας οδηγει η καθαρη λογικη δεν ειναι σιγουρο οτι η ερωτηση ειναι μοναδικη. Εσυ κατασκευαζεις μια ερωτηση που ικανοποιει τις συνθηκες αλλα οχι ακολουθοντας καθ' ολη την διαρκεια συνεπαγωγικους συλλογισμους.

[quote="kritikos"] Επομένως πρέπει να προσδιορίσουμε μια ερώτηση που να μην μπορεί να απαντήσει ο κινέζικος ( η περίπτωση του να δίνει διαφορετική απάντηση είναι δυσκολη από το γεγονός ότι και ο ρώσικος θα μπορουσε να πει ψέμματα ).  [/quote]

Εδω τελειωνουν οι συνεπαγωγικοι συλλογισμοι (ή κανω λαθος/?).

Μετα "φανταζεσαι" την ερωτηση και ομοιως ισως μπορουμε να "φανταστουμε" και αλλες ερωτησεις.


If a, b, c are sides of a triangle ABC and R_1, R_2, R_3 are the
distances from a point P in plane of ABC to the respective
vertices A, B, C then prove that
 cyclic  \sum a*R_1[a^4 +(R_1)^4]  \geq 10abcR_1R_2R_3

c'mon guys,try this one
i have a proof which uses complex numbers.and btw i found out this inequality but i thought this whould be known.that's why i have put the source as unknown.
awaiting other beautiful solutions :)


calculus integration inequalities trigonometry parameterization function geometry

Let's denote by f(n) our integral $f(n)=I_n=\int^\frac{\pi}{2}_0 {x^n}sin^nx dx $  then as follows from known theorem about diffrentiation integral with parameter $D\int f(n)=\int Df(n)$)---we can apply it cause our function is continious in rectangle $[0,\frac{\pi}{2}]*[0,\frac{\pi}{2}]$ as a function with two variables.
And by diffrentiating it is easy to see that f`(n) >0 , so $I_{2004}>I_{2003}$

Hello!eugene!
In this case,can we use the fundamental theorem of calculus?

kunny

hello kunny

i meant that we can apply Leibniz integral rule  $ D\int f(x,n)=\int Df(x,n)$
where $f(x,n)={x^n}sin^nx $cause $f(x,n)$ is continious on both variables and its derivative by n is continious too

Here's a completely different approach. The thing to note is that we can write $I_n=\int g^n$, where $g(x)=x\sin(x)$ and we understand that all integrals in this problem are over the interval $[0,\pi/2]$. Note that this is a nonnegative function; that will be important.
By direct calculation, $I_n=1$. This will be very important as we proceed.
By a somewhat more painful, but still elementary, calculation, $I_2=\frac{\pi^3}{48}+\frac{\pi}8.$ This is approximately 1.0387; what matters to us is that it is greater than 1.

Claim: For all $n\ge 2, 1<I_n<I_{n+1}.$

Suppose that we know that $I_n>1.$ Now we employ Hlder's Inequality with exponents $p=n$ and $p'=\frac n{n-1}$:

$\int g^n=\int g^{\frac1n}g^{\frac{n^2-1}n} \le \left(\int g\right)^{\frac1n}\left(\int g^{n+1}\right)^{\frac{n-1}n}$.
Since $\int g=I_1=1$, we have $I_n<(I_{n+1})^{\frac{n-1}n}$. If we first raise this inequality to the $\frac{n}{n-1}$ power and use the fact that $I_n>1$, we see that $I_{n+1}>1$. Then from the original inequality and the fact that $y^{\frac{n-1}n}<y$ for $y>1$, we have that $I_n<I_{n+1}.$

I would also add that it is clear that $\lim_{n\to\infty}I_n=\infty$. Let $a$ be the point at which $a\sin a=1$. Then $I_n>\int_a^{\pi/2}g^n$. By the Monotone Convergence Theorem, this integral tends to infinity.

Thank you,eugene.

I didn't  know Leibniz integral rule. I' m sorry!

To Kent Merryfield,

your answer is great! I didn't notice in using Holder's inequality.

I will show you my solutuion as follows.

The calculation of  $I_1 ,   I_2 $ is exactly that of yours.

In the latter part, I will show  $I_{n+1}>I_n$ as follows.

$I_{n+2}-2I_{n+1}+I_n=\int^\frac{\pi}{2}_0 x^n{sin^n}x(xsinx-1)^2 dx>0$, since  $I_2>I_1 $,  we have $I_{n+2}-I_{n+1}>I_{n+1}-I_n.$

Therefore by induction we deduce $I_{n+1}-I_{n}>0.$

Finally the desired answer is $I_{2004}>I_{2003}.$

Thank you.
kunny

Eugene:

Differentiating with respect to $n$ leads to the integral
$\int_0^{\pi/2}x^n(\sin x)^n\ln(x\sin x) dx$.
It may well be true that this is positive, but that is by no means obvious, since the function being integrated takes on both positive and negative values.

For you to say, "It is easy to see that $f'(n)>0$" - that's not enough. You've got to provide more argument than that. It would be possible to turn your approach into a proof, but that proof would need some quantitative details that you didn't show.

You're trying to directly show that $I_n$ is an increasing sequence. If you examine kunny's proof and my proof, you'll see that we did not show that directly. What kunny showed is that the sequence is concave upward; what I showed is also related to convexity. You'll notice that we both had to have the independent direct (and non-obvious) calculation that $I_2>1$ before we could proceed to higher $n$.

Ok, sorry for my mistake,i'm fully agree with Kent Merryfield:
i was incorrect with calculating $\int Df(x,n)$ - that is of course not obvious, and i think that this approach will not go in this case


geometry
3D geometry
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number theory solved
number theory

Show that the cube roots of three distinct prime numbers cannot be terms in an arithmetic progression

We just need to show that $\frac{\sqrt[3]{p_3}-\sqrt[3]{p_1}}{\sqrt[3]{p_2}-\sqrt[3]{p_1}}$ is not rational. That shouldn't be too hard..:)

Assume for a contradiction that the three pairwise distinct primes $p,q,r$ are such that their respective cube roots belong to a same arithmetic progression. Let $d$ be the ratio of this sequence. Then there exist non-zero integers $m,n$ such that, say :
$\sqrt[3] p - \sqrt[3]  q = md$ and $\sqrt[3] p - \sqrt[3] r = nd$,
which gives :
$\sqrt[3] p - \sqrt[3] r = \frac n m (\sqrt[3] p - \sqrt[3]  q)$
Thus, there exist non-zero rationals $\alpha , \beta$ such that :
$\alpha \sqrt[3] p + \beta \sqrt[3]  q = \sqrt[3] r$
Cubing, it leads to :
$r = \alpha ^3 p + \beta ^3 q + 3 \alpha \beta \sqrt[3] {pqr}$.
It follows that $\sqrt[3]  {pqr}$ is rational, which is clearly impossible since $p,q,r$ are distinct.

Pierre.

Let $\alpha=\sqrt[3]{p_{1}^{2}p_{3}}$, $\beta=\sqrt[3]{p_{1}^{2}p_{2}}$. Consider two cubic fields $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$. The polynomial $x^{3}-p_{1}^{2}p_{3}$ is Eisenstein polynomial according to $p_{3}$, hence the first field is totally ramified at $p_{3}$, unramified at $p_{2}$ if $p_{2}\neq 3$ and possibly ramified at $p_{2}$ if $p_{2}=3$. In the same manner, the second field is totally ramified at $p_{2}$ and unramified at $p_{3}$ with the same exception. Since both $p_{2}$ and $p_{3}$ cannot be equal to $3$, therefore, these fields are not equal, and since both have degrees $3$, they are disjoint. Let $\rho$ be primitive cubic root of unity. So, the field $\mathcal{K}=\mathbb{Q}(\alpha, \beta)$ posseses an isomorphic embedding $\sigma:\mathcal{K}\rightarrow\mathbb{C}$ with properties $\sigma\alpha=\alpha$ and $\sigma\beta=\rho\cdot\beta$. Therefore, $\frac{\alpha-p_{1}}{\beta-p_{1}}$ is mapped by $\sigma$ to $\frac{\alpha-p_{1}}{\rho\beta-p_{1}}$. Since they are not equal (the first expression is real, the second is imaginary), it cannot be rational number.
P.S. The proof without changes is valid for all radicals $\sqrt[q]{...}$, q=2,3,5,..., a prime number, where only $\rho$ should be replaced by the primitive $q$-th root of unity.
P.S. Trivially, the statement is wrong without assumtion of primality of $p_{2}$ and $p_{3}$. In fact, $\sqrt[3]{p_{1}}$, $\sqrt[3]{8p_{1}}$, $\sqrt[3]{27p_{1}}$ form an arithmetic progression.

[quote="Myth"]How will be executioner :ninja:? Peter? Pierre? Grobber? Me (:o :huh: I am peaceful kind man!!!)?[/quote]

Is it true ? I remember Myth fighting for valiowk...  :D

we already know the executioner... giedrius :lol:

No, he is not an executioner! He is a bloody problem's butcher  :lol:  :lol:  :lol:


function
calculus
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trigonometry
complex numbers
complex analysis

Hi,I have a problem that I can't seem to solve.Using the Cauchy-Riemann equations,I have to construct an analytic function f(z).I am given that v(x,y)=y/(x^2+y^2),and from that and the C-R eqations,I have to find u(x,y) to make the analytic function f(z)=u(x,y)+iv(x,y).I am finding the integration of dv/dx and dv/dy extremely hard.

The approach I took may be wrong.I diffrentiated v(x,y) w.r.t y and got (x^2-y^2)/(x^2+y^2)^2, and then integrated using 'A' as a substitute for '(x^2+y^2)'.For that I got -y*ln(x^2+y^2).
Doing the same after diffrentiating v(x,y) w.r.t x,I integrated to get 1/2y*ln(x^2+y^2)+y/a
Any help would be much appreciated.Thanks. :P

[quote="jeff-844"]The approach I took may be wrong.I diffrentiated v(x,y) w.r.t y and got (x^2-y^2)/(x^2+y^2)^2[/quote]
Yes. It is right and $\int \frac{x^2-y^2}{(x^2+y^2)^2}dx = -\frac{x}{x^2+y^2}+c$.
(Hint: $\int \frac{x^2-1}{(x^2+1)^2}dx = \int \frac{\tan^2 t - 1}{\sec^4 t}(\sec^2t) dt = \int (\sin^2t-\cos^2t)dt = -\int cos 2t\ dt = -\frac{1}{2}\sin 2t + c = -\sin t\cos t + c = -\frac{\tan t}{sec^2 t}+c = -\frac{x}{x^2+1} +c$)

In fact, you can guess the answer out quickly from the symmetry.

Thanks for that Liyi.
I just worked out the other half of the problem there,(integrating dv/dx=-2xy/(x^2+y^2)^2) using the substition A=(x^2+y^2), and i got -x/(x^2+y^2) as the solution so I know I am on the right track.Is it possible to  use this substition to integrate the half of the problem that you solved in your last post?I find trig identity's hard to understand  :blush:

I forget to say that $x=\tan t$ in my post...  :blush:  :blush:


MATHCOUNTS
AMC
AIME
quadratics
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probability
AMC 10

I am going to purchase from AOPS either Volume 1 or Volume 2, I am in 7th grade so I will still do MATHCOUNTS next year and if everything goes well, and by well i mean extremely well I am planning on qualifying for NATs. On the other hand I am going to do Honors Algebra II next year as well as take the AMC 10 & 12, and possibly qualify for AIME (although highly unlikely). Currently I know the Mathcounts curriculum, (simple Algebra,quadratics,combinatorics,geometry etc.) The one thing that I am concerned about if I purchase Volume I is that I will already know most of it, but on the other hand im not very familiar with modular arithmetic, or logarithms.
Considering all of this, which book should I purchase Volume I or II?
Thanks in Advance  :D

I don't know you might want a different 1 than those b/c vol. 1 is very easy and only 1 chapter has logs.  On the other hand Vol. 2 is much harder than MC material but might help w/ the AIME.  I would suggest one of the following:
Intro. 2 counting and probability
Intro. 2 number theory
Intermediate counting and probability
 :)

Volume 1 has material suited for MC, while volume 2 is meant for AMC and above.

The geometry and probability sections are quite nice for MC practice in volume 1  :) .

Volume 1 and 2 are pretty helpful.
I would recommend you buy Vol 1 before you buy any other books.


The radius of the inscribed circle is 6 cm. What is the number of centimeters in the length of $ AB$? Express your answer in simplest radical form.

[img]http://img515.imageshack.us/img515/5494/2001nationalsprint22nj1.gif[/img]

[hide="Solution"]
[img]http://img408.imageshack.us/img408/8223/2001nationalsprint22solha1.gif[/img]
We can label the point where there is a right angle as $ C$ for reference. We can then see that by 45-45-90 triangles, the distance from the center of the inscribed circle to $ C$ is $ 6\sqrt{2}$. This means that the altitude to $ AB$ is $ 6\plus{}6\sqrt{2}$. We can now label the point where the altitude to $ AB$ touches segment $ AB$ as $ M$.

We can see that $ BCM$ is a $ 30\minus{}60\minus{}90$ triangle. this means that $ BC\equal{}2\left(\dfrac{6\plus{}6\sqrt{2}}{\sqrt{3}}\right)$. Rationalizing the denominator results in $ 2\left(2\sqrt{3}\plus{}2\sqrt{6}\right)\implies 4\sqrt{3}\plus{}4\sqrt{6}$. Factoring out the $ 4$ helps: $ 4(\sqrt{3}\plus{}\sqrt{6})$. We need to multiply this by $ \sqrt{3}$ in order to find the value of $ AB$. We then get $ 4\sqrt{3}(\sqrt{3}\plus{}\sqrt{6})\implies 4(3\plus{}\sqrt{18})\implies 4(3\plus{}3\sqrt{2})\implies\boxed{12\plus{}12\sqrt{2}}$.

[/hide]


A set of five distinct positive integers has a mean of $ 1000$ and a median of $ 100$.  What is the largest possible integer that could be included in the set?

Sum of those numbers is 1000*5=5000, and if one is the largest possible, others are smallest possible, so those are 1,2,100,101 and 4796 as the middle number is 100.


function
limit
geometric series
algebra unsolved
algebra

Let $f: \mathbb R \to \mathbb R$ be a function such that \[|f(x + y) - f(x)- f(y)|\leq 1\] for all $ x,y \in \mathbb R$. Prove that there exists a function $g: \mathbb R \to \mathbb R$ with $ |f(x) - g(x)| \leq 1$ and $ g(x +y)= g(x)+g(y)$ for all reals $x$ and $y$.

[quote="kihe_freety5"]g(x)=$ \lim_{n\to\infty}{\frac {f(2^nx)}{2^n}}$[/quote]
can you post full solution?

Ps Khai:Anh nho bai nay thay Quoc giao roi ma tim mai khong thay cai to loi giai dau.

setting x=y=$ 2^nx$ we attain |$ \frac {f(2^{m \plus{} 1}x)}{2^{m \plus{} 1}} \minus{} \frac {f(2^{m}x)}{2^m}$|$ \le \frac{1}{2^{m \plus{} 1}}$
so sum of $ \frac {f(2^{m \plus{} 1}x)}{2^{m \plus{} 1}} \minus{} \frac {f(2^{m}x)}{2^m}$ with m starts 0 to infinite
because the values of the terms are bounded by the geometric series $ \frac {1}{2}, \frac{1}{4}...$ which sums to 1, this sum converges absolutely and is bounded by 1 as well. on the other hand sum equals
$ \lim_{n\to\infty} {\frac {f(2^{n}x)}{2^n}} \minus{} f(x)$
so we attains |$ f(x) \minus{} g(x)$|$ \le 1$
 prove  g(x+y)=g(x)+g(y)
$ g(x \plus{} y) \minus{} g(x) \minus{} g(y) \equal{} \lim_{n\to\infty} {\frac {f(2^{n}{x \plus{} y})}{2^n}} \minus{} \lim_{n\to\infty} {\frac {f(2^{n}x)}{2^n}} \minus{} \lim_{n\to\infty} {\frac {f(2^{n}y)}{2^n}} \equal{} \lim_{n\to\infty} {\frac {f(2^{n}{x \plus{} y})}{2^n} \minus{} \frac {f(2^{n}x)}{2^n} \minus{} \frac {f(2^{n}y)}{2^n}}$
from the given | $ f(2^{n}{x \plus{} y}) \minus{} f(2^{n}x) \minus{} f(2^{n}y)$| $ \le 1$ for any n
so we have g(x+y)=g(x)+g(y)

[quote="kihe_freety5"]setting x=y=$ 2^nx$ we attain |$ \frac {f(2^{m \plus{} 1}x)}{2^{m \plus{} 1}} \minus{} \frac {f(2^{m}x)}{2^m}$|$ \le \frac {1}{2^{m \plus{} 1}}$
so sum of $ \frac {f(2^{m \plus{} 1}x)}{2^{m \plus{} 1}} \minus{} \frac {f(2^{m}x)}{2^m}$ with m starts 0 to infinite
because the values of the terms are bounded by the geometric series $ \frac {1}{2}, \frac {1}{4}...$ which sums to 1, this sum converges absolutely and is bounded by 1 as well. on the other hand sum equals
$ \lim_{n\to\infty} {\frac {f(2^{n}x)}{2^n}} \minus{} f(x)$
so we attains |$ f(x) \minus{} g(x)$|$ \le 1$
 prove  g(x+y)=g(x)+g(y)
$ g(x \plus{} y) \minus{} g(x) \minus{} g(y) \equal{} \lim_{n\to\infty} {\frac {f(2^{n}{x \plus{} y})}{2^n}} \minus{} \lim_{n\to\infty} {\frac {f(2^{n}x)}{2^n}} \minus{} \lim_{n\to\infty} {\frac {f(2^{n}y)}{2^n}} \equal{} \lim_{n\to\infty} {\frac {f(2^{n}{x \plus{} y})}{2^n} \minus{} \frac {f(2^{n}x)}{2^n} \minus{} \frac {f(2^{n}y)}{2^n}}$
from the given | $ f(2^{n}{x \plus{} y}) \minus{} f(2^{n}x) \minus{} f(2^{n}y)$| $ \le 1$ for any n
so we have g(x+y)=g(x)+g(y)[/quote]

That's very nice!

I had a doubt about the existence of $ \lim_{n\to\infty} {\frac {f(2^{n}x)}{2^n}}$ but |$ \frac {f(2^{m \plus{} 1}x)}{2^{m \plus{} 1}} \minus{} \frac {f(2^{m}x)}{2^m}$|$ \le \frac {1}{2^{m \plus{} 1}}$ indeed is enough to show this existence.

Congrats !

[quote="kihe_freety5"]setting x=y=$ 2^nx$ we attain |$ \frac {f(2^{m \plus{} 1}x)}{2^{m \plus{} 1}} \minus{} \frac {f(2^{m}x)}{2^m}$|$ \le \frac {1}{2^{m \plus{} 1}}$
so sum of $ \frac {f(2^{m \plus{} 1}x)}{2^{m \plus{} 1}} \minus{} \frac {f(2^{m}x)}{2^m}$ with m starts 0 to infinite
because the values of the terms are bounded by the geometric series $ \frac {1}{2}, \frac {1}{4}...$ which sums to 1, this sum converges absolutely and is bounded by 1 as well. on the other hand sum equals
$ \lim_{n\to\infty} {\frac {f(2^{n}x)}{2^n}} \minus{} f(x)$
so we attains |$ f(x) \minus{} g(x)$|$ \le 1$
 prove  g(x+y)=g(x)+g(y)
$ g(x \plus{} y) \minus{} g(x) \minus{} g(y) \equal{} \lim_{n\to\infty} {\frac {f(2^{n}{x \plus{} y})}{2^n}} \minus{} \lim_{n\to\infty} {\frac {f(2^{n}x)}{2^n}} \minus{} \lim_{n\to\infty} {\frac {f(2^{n}y)}{2^n}} \equal{} \lim_{n\to\infty} {\frac {f(2^{n}{x \plus{} y})}{2^n} \minus{} \frac {f(2^{n}x)}{2^n} \minus{} \frac {f(2^{n}y)}{2^n}}$
from the given | $ f(2^{n}{x \plus{} y}) \minus{} f(2^{n}x) \minus{} f(2^{n}y)$| $ \le 1$ for any n
so we have g(x+y)=g(x)+g(y)[/quote]
nice  :D
Thank you very much.


geometry
AMC
AIME
3D geometry
tetrahedron
ratio
trigonometry

I have only had v1 + v2 geom and i want to do better on aime geom problems. I was thinking this could be a thread for medium level geometry problems. Also any recommendations? I have Zeitz perhaps that will be enough.

First Problem:
A tetrahedron is constructed by connecting the centers of the 4 sides of a regular tetrahedron. What is the ratio of the smaller tetrahedron's volume to the smaller's volume?

I'm pretty sure the answer is $ \frac{1}{3}$, just by drawing a diagram, but I have no way to prove it.

Zeitz seriously lacks in geometry. [url=http://www.amazon.com/exec/obidos/ASIN/0486691543/artofproblems-20]This[/url] might just be what you are looking for.

Well, since both of the tetrahedron's are regular, they are similar. So, if we can find the length of the small tetrahedron relative to the large tetrahedron, we can find the ratio.

[hide]Let the large tetrahedron be $ ABCD$ and the small tetrahedron be $ WXYZ$, with $ W$ on $ ABC$, $ X$ on $ BCD$, $ Y$ on $ ABD$, and $ Z$ on $ ACD$. We'll attempt to find WX. Let the side length of $ ABCD$ be $ 1$, just for convenience. Now, let the foot of the altitude from W to BC be M. $ WM\equal{}XM\equal{}\sqrt{3}/6$, by the Pythagorean theorem. Since $ \cos WMX\equal{}\cos AMX\equal{}XM/AM\equal{}\frac{\frac{\sqrt3}{6}}{\frac{\sqrt3}{2}}\equal{}1/3$, by Law of Cosines, we have: $ WX\equal{}\sqrt{1/12\plus{}1/12\minus{}2(1/12)(1/3)}\equal{}1/3$. Hence, the ratio of the volume of the small tetrahedron to the large tetrahedron is $ (1/3)^3\equal{}1/27$[/hide]

I think that's right, but I'm not completely sure.

[quote="Poincare"]new problem:

In triangle ABC, the circumradius is 9 times the AB. The Length of the angle bisector is 20 times BC. The cosine of angle C is 1/2. Solve the triangle's sides (bonus points if you can prove you can't get the angles or you get the angles)[/quote]

This may or may not matter but can you specify which angle bisector you're talking about?

Since the angle bisector problem isn't specific...I'll provide another solution to the tetrahedron one:

[hide="Without Trig"]Let the sides of the bigger tetrahedron have length "s". So, you have tetrahedron ABCD.  Let the centers of triangles ABC and ACD be E and F, respectively. So, draw AE and AF to BC and CD to G and H, respectively.  We know by similar triangles that GH=1/2(BD)=(1/2)s.In triangle AGH, we see that EF=2/3(GH) since AE/EG=2/3 (since F, as the "center," is the centroid).  Thus, EF=1/2(2/3)s=1/3s.  Thus, the ratio of their areas is: [(1/3s)/s]^3=(1/3)^3=[u]1/27[/u][/hide]

sorry i meant bisector of angle ABC. sorry.

[hide]
I don't think this triangle exists?
Clearly, $ \angle C \equal{} 60$. So the area of $ \triangle ABC$ is
$ \frac {1}{2}ab \sin C \equal{} \frac {\sqrt {3}}{4} ab$
Also note that the area is $ \frac {abc}{4R}$ where R is the circumradius. So the area is $ \frac {abc}{36c} \equal{} \frac {ab}{36}$. From this we conclude that $ \frac {\sqrt {3}}{4} \equal{} 1/36 ???$

Did I do something wrong?

[/hide]

Answer Poincare? If not, Darklegacy52 should post a new problem. I'll post one to keep things moving  :P
Given $ \triangle ABC$ with $ AB\perp BC$, $ BD$ is the altitude to $ AC$, $ AF$ bisects $ \angle BAC$, $ AP \equal{} 12$, and $ PF \equal{} 8$. Find $ tan\angle BAF$ and find $ PD$.
Source: AOPS Volume I

Is P just a random point? What's significant about it?

Whoops, that's the problem with copying from a problem with a diagram. $ P$ is the intersection of $ AF$ and $ BD$. $ F$ lies on $ BC$.


2)  Let x,y,z be positive real numbers such that x+y+z=1. Prove that
\[ \frac {x^4}{y^3} \plus{} \frac {y^4}{z^3} \plus{} \frac {z^4}{x^3}\ge 9(x^3 \plus{} y^3 \plus{} z^3)
\]

[quote="Marius Mainea"]1) Let x,y,z be positive real numbers such that x+y+z=1. Prove that
\[ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2}\ge 9(x^3 \plus{} y^3 \plus{} z^3)
\]
[/quote]
AM-GM kills it. :wink:
$ 2)$ follows from $ 1).$

[quote="arqady"][quote="Marius Mainea"]1) Let x,y,z be positive real numbers such that x+y+z=1. Prove that
\[ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2}\ge 9(x^3 \plus{} y^3 \plus{} z^3)
\]
[/quote]
AM-GM kills it. :wink:
[/quote]

 arqady, can you explain?

 Thanks :)

[quote="Marius Mainea"]1) Let x,y,z be positive real numbers such that x+y+z=1. Prove that
\[ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2}\ge 9(x^3 \plus{} y^3 \plus{} z^3)
\]
[/quote]
$ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2}\ge 9(x^3 \plus{} y^3 \plus{} z^3)\Leftrightarrow(x \plus{} y \plus{} z)^2\sum_{cyc}\frac {x^3}{y^2}\geq9\sum_{cyc}x^3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {x^5}{y^2} \plus{} \frac {x^3z^2}{y^2} \plus{} \frac {2x^4}{y} \plus{} \frac {2x^4z}{y^2} \plus{} \frac {2x^3z}{y} \minus{} 8x^3\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {x^5}{y^2} \plus{} \frac {y^3x^2}{z^2} \plus{} \frac {2x^4}{y} \plus{} \frac {2x^4z}{y^2} \plus{} \frac {y^3x}{z} \plus{} \frac {x^3z}{y} \minus{} 8x^3\right)\geq0.$
But by AM-GM we obtain:
$ \sum_{cyc}\left(\frac {x^5}{y^2} \plus{} \frac {y^3x^2}{z^2} \plus{} \frac {2x^4}{y} \plus{} \frac {2x^4z}{y^2} \plus{} \frac {y^3x}{z} \plus{} \frac {x^3z}{y}\right)\geq8\sum_{cyc}\sqrt [8]{\frac {x^{27}}{y^3}}\geq8\sum_{cyc}x^3.$

We prove that $ \frac {x^4}{y^3} \plus{} \frac {y^4}{z^3} \plus{} \frac {z^4}{x^3}\ge \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2}$

[b]Solution:[/b] Since  $ x \plus{} y \plus{} z\le \frac {x^2}{y} \plus{} \frac {y^2}{z} \plus{} \frac {z^2}{x}$ , suppose that $ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2} < \frac {x^2}{y} \plus{} \frac {y^2}{z} \plus{} \frac {z^2}{x}$ then , adding up $ 2\sum {\frac {x^2}{y}}\le \sum {\frac {x^2}{y}(\frac {y}{x} \plus{} \frac {x}{y})} < 2\sum{\frac {x^2}{y}}$ wich is false.

 So
\[ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2}\ge\frac {x^2}{y} \plus{} \frac {y^2}{z} \plus{} \frac {z^2}{x}
\]
Similarly
\[ \frac {x^4}{y^3} \plus{} \frac {y^4}{z^3} \plus{} \frac {z^4}{x^3}\ge \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2}
\]


[quote="jensen"]find the value of $\frac{sinxcos3x}{sin2x}-cos2x$ for $x=\frac{\pi}{8}$[/quote]
we have
$sin2x= 2sinxcosx$ and  $cos3x=4cos^3{x}-cosx$
then
 $\frac{sinxcos3x}{sin2x}-cos2x$= $\frac{sinx(4cos^3{x}-cosx)}{2sinxcosx}-cos2x$
$=2cos^2{x}-\frac{1}{2}-cos2x$
$=cos2x-cos2x+\frac{1}{2}$
$=\frac{1}{2}$ for all x

[quote="Jan"]Nice solution but I don't agree. \[ \cos{3x}=4\cos^3{x}-3\cos{x}  \]
So you should get $\frac{-1}{2}$ for all $x$. But of course this is just a detail, your solution was very nice.  :)[/quote]
thanx
[hide="correction of my solution"] we have
$sin2x= 2sinxcosx$ and  $cos3x=4cos^3{x}-3cosx$
then
 $\frac{sinxcos3x}{sin2x}-cos2x$= $\frac{sinx(4cos^3{x}-3cosx)}{2sinxcosx}-cos2x$
$=2cos^2{x}-\frac{3}{2}-cos2x$
$=cos2x-cos2x-\frac{1}{2}$
$=-\frac{1}{2}$ for all x [/hide]

[quote="jensen"]nice solution Femat-Euler. :D[/quote]
thanx jensen :jump:


Why is it there is some power, n, that a permutation matrix is raised to and equals the identity?

The permutation matrices form a finite group- every power of a permutation is another permutation, so the sequence of powers must repeat. When it does, we get some power that's the identity.


Solve equation

\[ \sqrt{\frac{1\minus{}\sin x\sin 3x}{2}}\equal{}\sin \frac{\pi\minus{}4x}{2}.\]

\[ \sqrt {{{1 \minus{} \sin x\sin 3x} \over 2}}  \equal{} \sin ({{\pi  \minus{} 4x} \over 2}) \equal{} \cos (2x)\,\]
\[ \Leftrightarrow \,\,2{\cos ^2}(2x) \equal{} 1 \minus{} \sin x\sin 3x\,\,\,\,,\,\cos (2x) \ge 0\]
\[ \Leftrightarrow \,\,\cos (4x) \equal{} \cos (\pi  \minus{} 2x)\,\,\,\,\,,\,\cos (2x) \ge 0\]
\[ S \equal{} \left\{ {{{(6k \plus{} 1)\pi } \over 6},{{(6k \plus{} 5)\pi } \over 6}/k \in Z} \right\}\]
I hope It's right!  :D

[quote="Evariste-Galois"]\[ S \equal{} \left\{ {{{(6k \plus{} 1)\pi } \over 6},{{(6k \plus{} 5)\pi } \over 6}/k \in Z} \right\}\]
I hope It's right!  :D[/quote]

I think it is :)
I first thought you missed some cases in the last line but I forgot the constraint $ \cos 2x \ge 0$ and so your last line is OK :)

hello, i have found
$ x=2\arctan(\sqrt{7-4\sqrt{3}})+2k\pi$
$ x=-2\arctan(\sqrt{7+4\sqrt{3}})+2k\pi$
$ x=2\arctan({\sqrt{4+4\sqrt{3}})+2k\pi}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have found
$ x = 2\arctan(\sqrt {7 - 4\sqrt {3}}) + 2k\pi$.[/quote]
So $ x=\frac{\pi}6+2k\pi$
[quote="Dr Sonnhard Graubner"]$ x = - 2\arctan(\sqrt {7 + 4\sqrt {3}}) + 2k\pi$.[/quote]
So $ x=-\frac{5\pi}6+2k\pi$
[quote="Dr Sonnhard Graubner"]$ x = 2\arctan({\sqrt {4 + 4\sqrt {3}}) + 2k\pi}$.[/quote]
I think this one is wrong

And I think you missed $ x = 2\arctan(\sqrt {7 - 4\sqrt {3}}) + (2k+1)\pi$ and $ x = - 2\arctan(\sqrt {7 + 4\sqrt {3}}) + (2k+1)\pi$


function
algebra
polynomial
binomial theorem

Find all value(s) of N such that the equation

$2X+2Y+Z = N$

has exactly 28 different solutions for which $X, Y and Z$ are all positive integers

OR

Show that no such N exists.

[hide="Most of the way"]
We can use generating functions backward.

Let c represent the number of possibilities for a sum.  In this case, it is 28.

$C(x) = \frac{1}{(1-x^{2})(1-x^{2})(1-x)}= (1-x)^{-3}(1+x)^{2}$

Now for our solution, we use the generalized binomial theorem:

$c = \sum_{k=0}^{n}\binom{-3}{k}\binom{-2}{n-k}$

Which can be rewritten as:

$2c = \sum_{k=0}^{n}(-1)^{k}(k+1)(k+2)(n-k+1)$

We're looking for the value of n ($N$) that makes the sum equal to 56.

I'm sure from here this can be split into two sums and telescoped into a polynomial, but I'll leave that to someone else!  Anyway, we can check that 12 and 13 satisfy the sum, hence they are the only two solutions.
[/hide]

[hide="Remark"]
There is probably an easier way to do this, but this leads to some good generalizations that can be applied in other scenarios.
[/hide]

[quote="Potato Theory"]
There is probably an easier way to do this ...[/quote]
[hide="Probably this one?"]
Consider it as Diophantin equation. It seems to be more convinient to renote $x: =x-1, y: =y-1, z: =z-1$
Now we consider equation $2x+2y+z=n=N-5$ with nonnegative integers $x,y,z$
Let $x+y=u$ then  $z=n-2u$. Evidently we have the following possibilities:
$u=0,1, ... , m=[\frac{n}{2}]$ and for each $u$ we have $u+1$ solutions:
$(x,y)=(0,u), \ (1,u-1), \ (2,u-2), \ ... , \ (u-1,1), \ (u,0)$
Hence the total number of solutions is
$1+2+...+m+(m+1)=\frac{(m+1)(m+2)}{2}$
For the case $\frac{(m+1)(m+2)}{2}=28$ we have $m=6$ so $n=12$ or $n=13$ and $N=17$ or $N=18$. [/hide]

It is an old AHSME problem..

x + y = (n-z) over 2  < n over 2 ...

I am sorry for no LATEX..

AFTER korean pre-olympiads (07/5/12) , I will learn LATEX. :)

if $N$ is even.
the numbre of solution of,
$2x+2y+z=N$ is $card\{(a,h-a,N-2h)|\frac{N}{2}\ge h\ge a\ge 0\}=\frac{(N+2)(N+4)}{8}=28$
$N=12$

if $N$ is odd.
the numbre of solution of,
$2x+2y+z=N$ is $card\{(a,h-a,N-2h)|\frac{N-1}{2}\ge h\ge a\ge 0\}=\frac{(N+1)(N+3)}{8}=28$
$N=13$


Prove that don't the exist roots $ A \in M_2(\mathbb{C})$ such that  equation :
              $ e^{2A} \plus{} \left[\begin{array}{cc}2 & 2009 \\
0 & 2\end{array}\right]e^A \minus{} I_2 \equal{} 0$
Here $ I_2$ is $ 2x2$ identity matrix .
----------------------------------------------------
[color=blue]I edited :[/color]
Prove that don't the exist roots $ A \in M_2(\mathbb{C})$ such that  equation :
              $ e^{2A} \plus{} \left[\begin{array}{cc}2i & 2009 \\
0 & 2i\end{array}\right]e^A \minus{} I_2 \equal{} 0$
Here $ I_2$ is $ 2x2$ identity matrix .
Here $ i^2\equal{}\minus{}1$
----------------------------------------------------
I am sorry everyone! :blush:

Let $ U\equal{}\left(\begin{array}{cc}2 & 2009 \\0 & 2\end{array}\right)$ and $ TAT^{\minus{}1}\equal{}J$ be the Jordan normal form of $ A$, then the equation is equivalent to $ e^{2J}\minus{}Ue^J\minus{}I_2\equal{}0$. Clearly, $ J$ cannot be of diagonal form, since $ Ue^J$ would have a non-zero entry in the upper right corner and $ e^{2J}\minus{}I_2$ is diagonal, i.e. we can assume $ J\equal{}\left(\begin{array}{cc}\lambda & 1 \\0 & \lambda\end{array}\right)$ and $ e^J\equal{}I_2\plus{}\sum_{n\equal{}1}^{\infty}\frac{J^n}{n!}\equal{}I_2\plus{}\sum_{n\equal{}1}^{\infty}\frac{\left(\begin{array}{cc}\lambda^n & n\lambda^{n\minus{}1} \\0 & \lambda^n\end{array}\right)}{n!}\equal{}e^{\lambda}\left(\begin{array}{cc}1 & 1 \\0 & 1\end{array}\right)$. Substitution into the equation yields:

$ e^{2\lambda}\left(\begin{array}{cc}1 & 2 \\0 & 1\end{array}\right)\minus{}e^{\lambda}\left(\begin{array}{cc}2 & 2011 \\0 & 2\end{array}\right)\minus{}I_2\equal{}0$ which is equivalent to $ (2e^{\lambda}\minus{}2)^2\equal{}8$ and $ 2e^{\lambda}\equal{}2011$. From this we conclude that there is no solution.

Carefull with change Expression ! you have mistake in since you change!   :(  :flex:
$ Te^{2A}T^{\minus{}1}\plus{}T\left[\begin{array}{cc}2&2009\\0&2\end{array}\right]e^AT^{\minus{}1}\minus{}I_2\equal{}0$ ???

We study the equation $ e^{2A}\plus{}(\alpha{I}\plus{}\beta{N})e^A\minus{}I\equal{}0$ where $ N$ is a non zero nilpotent matrix (I don't know if the Quybac's choice is correct!).
We transform the equation by a left multiplication by $ N$ ; we obtain $ P(e^A)\equal{}N$ where  $ P$ is a polynomial. Thus $ A$ is not diagonalizable and $ A\equal{}\lambda{I}\plus{}U$ where $ U$ is nilpotent and $ UN\equal{}NU$ because $ e^A\equal{}e^\lambda(I\plus{}U)$. Therefore $ A\equal{}\lambda{I}\plus{}\mu{N}$ where $ \mu\not\equal{}0$. We introduce this value of $ A$ in the first equation and by identification we obtain these 2 equations in the unknowns $ \lambda,\mu$: $ e^{2\lambda}\plus{}\alpha{e^\lambda}\minus{}1\equal{}0,e^{2\lambda}2\mu\plus{}e^\lambda(\beta\plus{}\mu\alpha)\equal{}0$.
Thus $ e^\lambda\equal{}\dfrac{\minus{}\beta\minus{}\mu\alpha}{2\mu}$ and if $ \alpha\not\equal{}\pm{2i}$ then $ \mu^2\equal{}\dfrac{\beta^2}{\alpha^2\plus{}4}$.
If $ \alpha\equal{}\pm{2i}$ and $ \beta\not\equal{}0$ then there is no solution.

Follwoing [b]loup blanc's [/b]steps

Let $ e^{2A}\plus{}(\alpha I\plus{}N)e^A\minus{}I\equal{}0$ where $ N$ is a non-zero, nilpotent matrix (remark: we do not need the $ \beta$-factor, as long as $ \beta\not\equal{}0$ holds). Since the dimension is $ 2$, there is only one eigenvector space of $ N$ and it is determined by $ Ker\,N$, i.e. $ N\bar{w}\in \,<\bar{w}>$ is equivalent to $ w\in Ker\,N$. Now, let $ \lambda$ be an eigenvector of $ A$ with corresponding eigenvector $ \bar{v}\not\equal{}\bar{0}$. This implies $ (e^{2\lambda}\plus{}\alpha e^{\lambda}\minus{}1)\bar{v}\equal{}\minus{}e^{\lambda}N\bar{v}$. This implies $ N\bar{v}\in\, <\bar{v}>\,\Longrightarrow \bar{v}\in Ker\,N$. As a consequence we have $ e^{2\lambda}\plus{}\alpha e^{\lambda}\equal{}1\Longleftrightarrow 2e^{\lambda}\equal{}\sqrt{4\plus{}\alpha^2}\minus{}\alpha$ and $ A$ cannot be diagonalized. Its unique eigenvector space is given by $ Ker\, N$ and we can determine a vector $ \bar{u}\not\in Ker\,N$ with $ A\bar{u}\equal{}\lambda\bar{u}\plus{}\bar{v}$. More generally, we have $ A^m\bar{u}\equal{}\lambda^m\bar{u}\plus{}m\lambda^{m\minus{}1}\bar{v}$ and $ e^{tA}\bar{u}\equal{}e^{t\lambda}(\bar{u}\plus{}t\bar{v})$. Substitution into the original matrix equation yields $ N\bar{u}\equal{}\minus{}\sqrt{4\plus{}\alpha^2}\,\bar{v}$. Since $ N\bar{u}\not\equal{}\bar{0}$ must hold, we get $ 4\plus{}\alpha^2\not\equal{}0$ as a necessary criterion for the existence of a solution. Conversely, if $ 4\plus{}\alpha^2\not\equal{}0$, we can determine a vector $ \bar{w}\not\in Ker\,N$ with $ N\bar{w}\equal{}\minus{}\sqrt{4\plus{}\alpha^2}\,\bar{v}$, since $ N$ is a non-zero, nilpotent matrix. If we define $ A\bar{v}\equal{}\frac{\sqrt{4\plus{}\alpha^2}\minus{}\alpha}{2}\,\bar{v}$ and $ A\bar{w}\equal{}\frac{\sqrt{4\plus{}\alpha^2}\minus{}\alpha}{2}\,\bar{w}\plus{}\bar{v}$, then $ A$ is a solution.


WA divides rectangle WXYZ into two parts whose areas are in the ratio of 7:1. What is the ratio of $ \frac{XA}{AY}$? Express your answer as a common fraction.

Make your own diagram. (Reminds me of our school's 8th grade Geometry teacher).

All basic rules apply (Points in clockwise/counter-clockwise order).

2001 State Sprint Round

O.K. So make the distance of xy as n. 1+7=8. Make xa at a 45 degree angle to xy, and a the point that intersects yd on the 45 degree  line that has a distance of n(square root of 2) this means that yz will be 4n, because 8 triangles of the area of 1/2(n) equal a rectangle n wide and 4 n long for a total area of 4(n squared). The line xa is n times the square root of two, and line ya equals n, so xa/ya equals the square root of two.

I think the answer varies depending upon the degree of the angle yxa.

[hide="Rectangle Problem"]We draw the altitude from A to WZ to B, splitting the area of WXYZ into two congruent right triangles (by SSS), and a rectangle.  Since ∆WXA has area 1/8[WXYZ], so too does our new congruent triangle, WBA.  Hence, the rectangle has area 3/4[WXYZ].  Now, we see that WBA and BAYZ share a side, so we try to relate their areas.  If a rectangle has base and altitude b and a, respectively, and a triangle has base and altitude c and a, then the ratio of their areas is:
ba/ca/2=2b/c
Hence, we have:
[WBA]/[BAZY]=
WB/2BZ=
1/8[WXYZ]/3/4[WXYZ]=
1/6==>Hence, WB/BZ=XA/AY=[u]1/6[/u][/hide]


We see that $ f' \equal{} e^x \plus{} x e^x \equal{} (x \plus{} 1) e^x > 0$ for $ x \geq 0$, so $ f$ is strictly increasing. Then, as $ f(0) \equal{} 0$ we have that the range of $ f$ is equal to $ [0,\infty)$.

See above, it's strictly increasing and continuous. It's obviously surjective onto its image, and it is injective because of the monotonicity.

$ f(1) \equal{} 1\cdot e^1 \equal{} e$ by inspection.

We have $ \frac {d}{dx}\left(f^{ \minus{} 1}(x)\right) \equal{} \frac {1}{f'(f^{ \minus{} 1}(x))}$ by the [url=http://en.wikipedia.org/wiki/Inverse_function_theorem]inverse function theorem[/url] (its conditions are satisfied; check this).

So $ (f^{ \minus{} 1})'(e) \equal{} \frac {1}{f'(1)} \equal{} \frac1{2 e}$ as we had just seen that $ f^{ \minus{} 1}(e) \equal{} 1$.

thanks man, i really apreaciate it. I'm new to these stuff of calculus.


inequalities
rotation
function
geometry
geometric transformation
dilation

What does it mean for an inequality to be homogenous? What does this allow you to do?

What does it mean for an inequality to be symmetric?

For more info, read Kiran Kedlaya's Notes on Inequalities. I think a homogenous inequality is one that has equal degrees on both sides. (All terms, maybe?) This allows for easy simplification. We can deal with the god of inequalities: Karamata and Muirhead (which is kinda cheap) in a much easier fashion this way. An inequality is symmetric if the variables rotate:

$\frac{a^{3}}{b+c}+\frac{b^{3}}{a+c}+\frac{c^{3}}{a+b}$

is a symmetric equation, for example.

If an inequality is homogeneous, we can let the sum of the variables equal 1, or the product of the variables equal 1 (but not both!)

Some clarification.

We say that $f(a_{1}, a_{2}, ... a_{n})$ is a [b]homogeneous function with degree $k$[/b] if

$f(ta_{1}, ta_{2}, ... ta_{n}) = t^{k}f(a_{1}, a_{2}, ... a_{n})$ for some (integer? real?) $k$.

When the function involved in some inequality is homogeneous with degree $0$, this means that

$f(ta_{1}, ta_{2}, ... ta_{n}) = f(a_{1}, a_{2}, ... a_{n})$.

Therefore, we can "dilate" the variables by an arbitrary real to do useful things such as assume WLOG that $a_{1}+a_{2}+...+a_{n}= 1$ (since we can dilate by the inverse of the sum).

We say that $f(a_{1}, a_{2}, ... a_{n})$ is a [b]symmetric[/b] function if

$f(a_{1}, a_{2}, ... a_{n}) = f(a_{ \delta(1)}, a_{ \delta(2)}, ... a_{ \delta(n) })$

Where $\delta(i)$ is any arbitrary permutation of $\{ 1, 2, ... n \}$.

This allows us to arbitrarily change the order of the variables, so we can do useful things such as assume WLOG that $a_{1}\le a_{2}\le ... \le a_{n}$.

[quote="t0rajir0u"]$f(ta_{1}, ta_{2}, ... ta_{n}) = t^{k}f(a_{1}, a_{2}, ... a_{n})$ for some (integer? real?) $k$. [/quote]
Real. Usually not 0 (in that case there's probably a more precise way to describe the function).


1)Find all x,y: $ {5^x}\equal{}3y\minus{}1$.
2)consider a set of 50 distinct numbers at random from the set of positive integers greater than zero whose total sum is 3000,what is the least amount of even numbers possible?.
[hide]For number one i expressed $ {5^x}$ as $ {5k}$ and proceeded to solve the question with the aim of some final touches but i am not true with it.[/hide]

$ 1)$if $ 2\nmid x$  $ \implies$  $ 5^{x}\equiv \minus{} 1\pmod{3}$  maybe this question $ 5^{x} \equal{} 3^{y} \minus{} 1$ ?

for all odd values of $ x$, 1) has integer solutions.
If I have interpreted 2) correctly the answer is [b]0[/b]. An example of such a set is {11,13,15,....109}

[quote="21 7 15"]
2)consider a set of 50 distinct numbers at random from the set of positive integers greater than zero...
[/quote]
This is redundant; as they are from the set of positive integers,
THEN this means that they are greater than zero.


I drew an image with Geometer's Sketchpad, but it is a .gsp file. I can't get it into my $\LaTeX$ document because it is not .png

What should I do?

In order to import images into $\LaTeX$, you must make them PDF files. If you don't already have one, you should download a PDF printer (Go2PDF). To do this, go onto the $\LaTeX$ page of AoPS, then go to the Downloads page and scroll down to the bottom. There should be a link to Go2PDF. Click here and install with the instructions provided. After a short time, the installation will be finished. When you want to convert a file into PDF format, just go to Print and scroll down on your printer menu to the Virtual PDF Printer. Does this help?


what is a quartile deviation, because, on http://mathworld.wolfram.com/QuartileDeviation.html , I don't understand it.  :(  :?  :huh:  :|

Isn't it in your high school curriculum?

Anyway, from the given formula, of course you have to find the value of $Q_3$ and $Q_1$, which is the third and first quartile respectively.

i don't really think it is really getting started level.  :(

Isn't it? Well, I think it's what high school students should be able to do

This is purely statistical. We don't really do much of that around here. And it's not in my high school cirriculum by the way.

true, this kind of stuff while potentially useful, is not covered in amc/aime/etc. therefore, nobody really knows this stuff, maybe if i take statistics, then i will do this