geometry
incenter
geometric transformation
reflection
geometry proposed

Let $ ABC$ be a triangle with the incentre $ I$ . Denote $ F\in CI\cap AB$ and $ E\in (AC)$ for

which $ m\left(\widehat { AFE}\right)\equal{}90^{\circ}\minus{}\frac C2$ . Prove that $ \widehat {IBF}\equiv\widehat {IEF}\ \Longleftrightarrow\ B\equal{}3A \ \ \vee\ \ A\equal{}C$ .

Reflect $ B$ in $ CI$ into $ Y \in CA.$ $ \angle IYF \equal{} \angle IBF \equal{} \angle IEF$ $ \Longleftrightarrow$ $ Y \in \odot(IEF)$ $ \Longleftrightarrow$ (1) $ Y \equiv E$ or (2) $ Y \not\equiv E$ and $ IYEF$ is cyclic.

(1)  $ Y \equiv E$ $ \Longrightarrow$ $ BCEF$ is a kite with $ BC \equal{} EC.$ Therefore

$ 180^\circ \equal{} \angle AFE \plus{} \angle EFB \equal{} 90^\circ \minus{} \frac{_1}{^2}\widehat C \plus{} (360^\circ \minus{} 2 \widehat B \minus{} \widehat C)$ $ \Longleftrightarrow$
$ 2 \widehat B \plus{} \frac{_3}{^2} \widehat C \equal{} 270^\circ \equal{} \frac{_3}{^2} \widehat A \plus{} \frac{_3}{^2} \widehat B \plus{} \frac{_3}{^2} \widehat C$ $ \Longleftrightarrow$
$ \frac{_1}{^2} \widehat B \equal{} \frac{_3}{^2} \widehat A$ $ \Longleftrightarrow$ $ \widehat B \equal{} 3 \widehat A$

(2) $ Y \not\equiv E$ and $ IYEF$ is cyclic. $ \Longrightarrow$

$ \angle AEF \equal{} \angle FIY \equal{} \angle BIF \equal{} \angle IBC \plus{} \angle BCI$ $ \Longleftrightarrow$
$ 90^\circ \minus{} \frac{_1}{^2} \widehat C \equal{} \frac{_1}{^2} \widehat B \plus{} \frac{_1}{^2} \widehat C$ $ \Longleftrightarrow$
$ \frac{_1}{^2} \widehat A \plus{} \frac{_1}{^2} \widehat B \plus{} \frac{_1}{^2} \widehat C \equal{} 90^\circ \equal{} \frac{_1}{^2} \widehat B \plus{} \widehat C$ $ \Longleftrightarrow$
$ \frac{_1}{^2} \widehat A \equal{} \frac{_1}{^2} \widehat C$ $ \Longleftrightarrow$ $ \widehat A \equal{} \widehat C$


geometry
parallelogram
cyclic quadrilateral
geometry unsolved

In cyclic quadrilateral $ABCD$, $P,Q,R,S,X,Y$ are midpoints of $AB,BC,CD,DA,AC,BD$ respectively. $N=PR\cap QS$. Show that $NX=NY$.

This result is true, as a basic geometric problem, in every quadrilateral and not necessary cyclic.

So, because of $PQ\parallel= \frac{AC}{2}\parallel= RS,$ $($ from the triangles $\bigtriangleup ABC,$ $ADC$ $),$ we conclude that the quadrilateral $PQRS,$

is a parallelogram and so, the point $N\equiv PR\cap QS,$ bisects the segments $PR,$ $QS.$

Also, because of $PY\parallel= \frac{AD}{2}\parallel= RX,$ $($ from the triangles $\bigtriangleup ABD,$ $ACD$ $),$ we conclude that the quadrilateral $PYRX,$

is a parallelogram and hence, its diagonals $PR,$ $XY,$ bisect each other.

So, we conclude that $NX = NY$ and the proof is completed.

Kostas Vittas.


If $ \mathbb W_1\equal{}<\{(2,\minus{}1,3)\}>$ and $ \mathbb W_2\equal{}\{(x,y,z)\in\mathbb R^3|x\plus{}ky\plus{}z\equal{}0)\}$ are two linear subspaces of $ \mathbb R^3,$ then $ \mathbb W_1\oplus\mathbb W_2\equal{}\mathbb R^3$ iff $ k\equal{}5.$

I know if we have $ \mathbb W_1\oplus\mathbb W_2\equal{}\mathbb R^3$ then we get $ \mathbb W_1\cap\mathbb W_2\equal{}\{(0,0,0)\}$ and $ \mathbb W_1\plus{}\mathbb W_2\equal{}\mathbb R^3,$ but I don't know anyway how to start to solve this problem.

I think you meant "if and only if $ k \neq 5$".

We know $ \mathbb{W}_{1} \equal{} <(2,\minus{}1,3)>$, and since $ (x,y,z) \in \mathbb{W}_{2} \iff z \equal{} \minus{}x\minus{}ky$, we have $ \mathbb{W}_{2} \equal{} <(1,0,\minus{}1);(0,1,\minus{}k)>$.

Now, $ \mathbb{W}_{1}\oplus\mathbb{W}_{2}\equal{}\mathbb{R}^{3}$ if and only if all three generators $ (2,\minus{}1,3);(1,0,\minus{}1);(0,1,\minus{}k)$ are linearly independent.

That means $ \alpha(2,\minus{}1,3) \plus{} \beta(1,0,\minus{}1) \plus{} \gamma(0,1,\minus{}k) \equal{} (0,0,0)$ has the unique solution $ \alpha \equal{} \beta \equal{} \gamma \equal{} 0$, which happens if and only if the determinant of the matrix

$ \left( \begin{array}{ccc}
2 & 1 & 0 \\
\minus{}1 & 0 & 1 \\
3 & \minus{}1 & \minus{}k \end{array} \right)$

is not $ 0$. This happens if and only if $ k\equal{}5$, then any other $ k$ works.

 :D

Ahh!! Well actually the question was meant to say "prove or disprove!"

Thanks for your reply!!


Let $E$ be a normed vectorial space of dimension greater than $N+1$ and $0 <\epsilon <1$.Show that the closed unity ball cannot be covered with $N$ open balls of radius $\epsilon$.Deduce that the closed unity ball in an infinite dimension vectorial space it's not precompact.

Actually, the unit ball of a real normed space of dimension $n$ cannot be covered by $n$ sets of diameter smaller than $2$. Clearly, this is stronger. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Kuratowski&t=72330]this topic[/url].


geometry
analytic geometry
graphing lines
slope

from AHSME.. I'm stuck on this one :(

A point (x,y) in the plane is called a lattice point if both x and y are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to what?

This problem really confuses me really bad. I know what lattice point is but I just can't do it!!! :(

Good help please..

Just a thought, but have you tried pick's theorem?

i put points 0,0  1,0   and   0,1 into the square, and then drawing lines surrounding it with slopes of -1/2 and 2.  the equation of one line was
  -x-2y+3=0   and the opposite side goes thru the point 0,-1.  with point to line distance, i got the distance between the 2 lines as sqrt(5).

so the area is 5

and the other two sides (slope 2) should go thru the points 1,-1  and -1,0  to keep only 3 points in the square

the distance between these also turns out to be sqrt(5)

can you prove that this is the largest area, or did I just not see the proof ( :blush: )?

no, i'm not sure how
is it correct?
does anyone have a proof?


Two persons are dividing a piece of cheese. The first person cuts it into two pieces, then the second person cuts one of these pieces into two, then again the first person cuts one of the pieces into two, and so until they have 5 pieces. After that the first person chooses one of the pieces, then the second person chooses one of remaining pieces and so on until all pieces are taken. For each of the players, what is the maximal amount of cheese he can get for certain, regardless of the other's actions?

Sorry, what I had here before was WRONG- I assumed the cheese was cut into equal pieces

I think something in your reasoning is wrong, since they must not necesarilly cut the pieces in equal parts :?


algorithm
number theory
Euclidean algorithm

[quote="7h3.D3m0n.117"][quote="Ahmed Abdullah"]What is the largest prime that divides 2x+3y and 9x +5y (and for which value of x and y it occurs)?[/quote]

umm I don't understand the question what is $x$ and $y$?[/quote]
Read the part in parenthesis, x and y aren't constants ;)

I am not so sure if this is right:
[hide] We can apply the Euclidean Algorithm on both numbers.  Divide 9x+5y by 2x+3y.  We get a remainder of x-7y.  Now, divide 2x+3y by x-7y.  We get a remainder of 17y.  The largest constant factor of 17y is $17$, which is our answer (if x and y are variables).  However, if x and y could be anything, I could assume y to be a very prime, like 199 or 211, so the answer would have no limit (since there are infinite primes). [/hide]


Exista un concurs cu numele "Cristian S. Calude" in Romania?


Posted before.

...and erased before.

Edited by pbornsztein.
This is one of the current TOT. You have to wait for a couple of weeks.

Can you give me url or post the solution again

[Deleted for the obvious reason by $\zeta X$.]

[quote="Rust"]Edited by pbornsztein.
This is one of the current TOT. You have to wait for a couple of weeks.[/quote]
Which part didn't you understand?

@Beat: no comment on that behaviour, but be sure that repeating it will not be tolerated (in short: last warning after some cheatings).

@Rust: reposting it is not a better behaviour than what Beat did; and don't tell me you didn't see, not reading a thread before posting is not better, either.

i didn`t want to do something wrong. I cannot understand what is the problem to [color=red]RE[/color]post the problem if it is already posted..

1. you (beat) are a very annoying cheater!  :mad: No excuses, this was not the first time  :!:

2. reposting problems is useless (and hasn't happend here), you can always post in the existing thread;
reposting solutions after a moderator has deleted it because it would help scammers and cheaters like beat, and is just an act against the rules.

3. I don't think you didn't want to do something wrong, or why did you post it¿

becouse i will go to this year tournament of towns and i want to prepare well. Don`t call me a cheater, ok?


Does anyone know how to write multiplication problems in $ \text{\LaTeX}$ in column form, like 85*64?

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=xlop]xlop[/url] package.
[code]\usepackage{xlop}
...
\opmul{85}{64}[/code]
gives this:


I guess they both are on graphs and they have similar functions. I don't know much about calculus though.

[quote="Interval"]How is analytic geometry related to calculus?[/quote]

I don't think this type of question should be on this board...


Rectangle $ ABCD$ is 8 cm by 4 cm. $ M$ is the
midpoint of $ \overline{BC}$ , and $ N$ is the midpoint of $ \overline{CD}$.
What is the number of square centimeters in the
area of region $ AMCN$?

[asy]draw((0,0)--(32,0)--(32,16)--(0,16)--cycle);
draw((0,16)--(16,0)--(32,8)--cycle);
label("$A$",(0,16),N);
label("$B$",(32,16),N);
label("$C$",(32,0),S);
label("$D$",(0,0),S);
label("$N$",(16,0),S);
label("$M$",(32,8),E);[/asy]

hello, let $ A(AMCN)$ the area of the region $ AMCN$ then we get
$ A(AMCN)\equal{}32cm^2\minus{}8cm^2\minus{}8cm^2\equal{}16cm^2$
Sonnhard.

if you dont want such a complicated method (like the guy before me used), just use the area of the triangle formula. The base is 8; the height is four. bh/2 = 8*4/2 = 16 :roll:


calculus
integration
logarithms
function
real analysis
real analysis unsolved

The number of integral solutions to:

$ 4\int_0^\infty \frac{\ln t}{x^2 \plus{} t^2}dt \minus{} \pi \ln2 \equal{} 0; x>0$ is?

For $ x>0$: $ \int_0^\infty \frac {\ln t}{x^2 \plus{} t^2}dt \equal{} \frac {\pi}{2}\frac {\log x}{x}$, so we have to solve $ \frac {\log x}{x} \equal{} \frac {\log 2}{2}$.
I guess only $ x \equal{} 2$ and $ x \equal{} 4$ are solutions.

Rewrite that equation as $ 2\ln x\equal{}x\ln 2.$

Let $ f(x)\equal{}2\ln x\minus{}x\ln 2.$ Then $ f''(x)\equal{}\minus{}\frac2x$ and $ f''(x)\ne0$ for all $ x>0.$ Rolle's Theorem then shows that $ f(x)$ has at most 2 zeros. Since $ x\equal{}2$ and $ x\equal{}4$ are zeros of this function, it follows that they are the only zeros.

First make the substitution $ t \equal{} xy.$

$ \int_0^{\infty}\frac {\ln t}{t^2 \plus{} x^2}\,dt \equal{} \int_0^{\infty}\frac {x\ln(xy)}{x^2(1 \plus{} y^2)}\,dy$

$ \equal{} \frac1x\int_0^{\infty}\frac {\ln x}{1 \plus{} y^2}\,dy \plus{} \frac1x\int_0^{\infty}\frac {\ln y}{1 \plus{} y^2}\,dy$

The first integral equals $ \frac {\pi}2\cdot\frac {\ln x}{x}.$

As for the second integral, make the substitution $ y\mapsto \frac1y$ to get

$ \int_0^{\infty}\frac {\ln y}{1 \plus{} y^2}\,dy \equal{} \minus{} \int_0^{\infty}\frac {\ln y}{1 \plus{} y^2}\,dy.$

Which means that integral is zero. This verifies BlueVelvet's claim.

[quote="AndrewTom"]Am bothered by the substitution y to -y: doesn't this give the log of a negative number?[/quote]
No, because you must also change the integration range: $ 0\to 0$ and $ \infty\to \minus{} \infty$.
So $ \minus{} y$ ranges between $ 0$ and $ \minus{} \infty$, which is the same as $ y$ ranging between $ 0$ and $ \infty$.

Or, if you are enthusiastic, you can use the residuum theorem:

$ \int^\infty_0 f(x) \ln x  \,dx \equal{} \frac{1}{2}\text{Re}\left\{\sum_{\text{Re}\{z_0\} < 0}\text{Res}\left(f(z) \ln^2 z, z_0\right)\right\}$

jmerry had it right; I made a typo and it should have been $ y\mapsto\frac1y.$ I'll go back and edit that post now.

I first thought $ y\to \minus{}y$ also works, but I made a mistake (lost one minus sign :)). $ y\to \frac{1}{y}$ does indeed work. For the interested:

$ \eta\to\frac{1}{y}$ gives $ dy\equal{}\minus{}\frac{d\eta}{\eta^2}$. Of course $ y\equal{}0\to\eta\equal{}\infty$ and vice versa.
$ I\equal{}\int_0^\infty \frac{\log y}{1\plus{}y^2}dy\equal{}\int_\infty^0\frac{\log\frac{1}{\eta}}{1\plus{}\left(\frac{1}{\eta}\right)^2}\left(\minus{}\frac{d\eta}{\eta^2}\right)\equal{}\minus{}\int_0^\infty\frac{\log \eta}{\eta^2\plus{}1}d\eta\equal{}\minus{}I$


function
real analysis
real analysis unsolved

Be D the set of the points $|p|\leq 1$. Be $F: D\to D$ a function that is surjective .This function satisfies $|f(p)-f(q)|\leq |p-q|$ . Shows that $f$is a isometry! for any p,q in D


abstract algebra
Ring Theory
vector
superior algebra
superior algebra unsolved

Let $ M$ be a finitely generated module over a PID $ R$.   If $ p$ is a prime in $ R$, define

\[ M[p] \equal{} \{m \in M : pm \equal{} 0 \}\]
and
\[ M(p) \equal{} \{m \in M : p^n m \equal{} 0 , \mbox{for some } n \geq 0 \}\]

[b](i)[/b] Show that the dimension of $ M[p]$ and $ M/pM$ as vector spaces over $ R/pR$ is the same.
[b](ii)[/b] Show that $ M(p)$ is isomorphic to $ M_p$, where $ M_p$ is the localization of $ M$ at the prime ideal $ (p)$

Danken!

Sorry, but are you sure of this all? I can't confirm any of (i) and (ii) for $ R \equal{} \mathbb{Z}$ and $ M\equal{}\mathbb{Z}^n$...

  darij


calculus
integration
LaTeX
calculus computations

hello,why don't you use $ \text{\LaTeX}$? your integral looks like this
$ \int\frac{\pi^x+\pi x^{\pi-1}}{\pi^{x+1}+\pi x^{\pi}}\,dx$
Sonnhard.

hello, i think there is not a closed form for your integral.
Sonnhard.


An urn contains balls of $k$ different colors; there are $n_i$ balls of $i-th$ color. Balls are selected at random from the urn, one by one, without replacement, until among the selected balls $m$ balls of the same color appear. Find the greatest number of selections.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

$ 1 \plus{} \sum_{i \equal{} 1}^k\min\left\{(m \minus{} 1),n_i\right\}$, so that if there are $ m$ or more balls of a given color, until right before the last extraction only $ m \minus{} 1$ balls of that colour have appeared, and if there are less than $ m$ balls of a given color, all the balls of that color have been extracted...


MATHCOUNTS
geometry
rectangle
number theory
relatively prime

[hide]30+35-1=[b]64[/b]? im pretty sure this is wrong though.[/hide]

[hide]
I'm pretty sure this is wrong  :wink: but is it 35? I'll post a solution if I'm right.
[/hide]

[quote="madskillz1"]i have a board that is 30 by 35 units.  If I divide the board into unit squares and I draw a line through it, what is the maximum number of squares this line can pas through?[/quote]

[hide="Is it"]$11 \cdot 6=66$?[/hide]

No correct answers, yet.  [Edit:  There [b]is[/b] a correct answer.]

[hide="hints"]Standard Problem Solving Technique #1:  Try a simpler case.  Can you generalize to the harder case?

Standard Problem Solving Technique #2:  Draw a picture.  Get some graph paper and a straight-edge.  Draw it [i]very[/i] carefully.[/hide]

If I draw a line through from one corner to the opposite one, it passes through $30+35-gcd(30,35)=60$ corners. I don't have a proof that this is optimal solution.

Doesn't the problem say squares though? :|

[hide]

60?

When drawing a line like this, you'd want to go corner to corner.
You can simplify the square into a 6x7, then just multiply your answer by 5.

Drawing a line, you can see it passes through 12. A way to check this answer is that because when it's going from one corner to another, it won't pass through another corner if the dimensions are relatively prime (which they are for the 6x7) So you just have to go R 5 and U 5 (plus the two you started with, since you're drawing on the squares.)

So that's 5*12 = 60

For a formula, I'm thinking that it's dimensions of side1 + side2 - 1 and - the amount of times it goes through a corner. That works for 6+7-1=12, and for 30+35-1 then -4 because it goes through 4 corners before it gets to the other side.

To test the formula I'm drawing a 4*6. 4+6 is ten, -1, then -1 for the time it goes through the center corner. So yes, it appears that forumal works.

Side1+Side2-1-(amount of times it goes through corner when traveling down the middle)= maximum amount of squares a line can pass through.

[/hide]

[quote="rcv"]No correct answers, yet.[/quote]
Whoops!  There [b]was[/b] a correct answer.

[hide]Zenox and MeKnowsNothing have the answer I was looking for, which is correct if you draw a line from corner to corner.  [This has been a MATHCOUNTS question in the past, but maybe with different numbers.]  The key to solving the "corner-to-corner" problem is to notice the line may pass through corners interior to the board.  So, if gcd(30,35)=5, we should expect a line drawn corner-to-corner will pass through five separate 6x7 rectangles.  Figure out how many squares the line passes through in each of the 6x7 rectangles, and multiply by 5.  As you folks have figured out, the line will pass through 6+7-1 square in each of the smaller rectangles.  Times five gives a final answer of 60.

indianamath had the correct answer all along, if you are allowed to draw a line without constraint.  Draw the (infinite) line from corner-to-corner, and then shift the line a tiny bit to the side.  If you haven't shifted the line too far, you'll gain a square every place the 6x7 rectangles previously met at a point.  So, unless I've made another mistake, the correct answer should be 30+35-1=64.
[/hide]
I'm sorry for the confusion.  vamathlete's question of the wording made me do a double-take.  Good job, guys!

I wonder if you shouldn't also count the 2x2, 3x3, 4x4 squares and so on.  The problem doesn't ask for how many "unit squares" but just squares.  Hmmmmm.  If the question just asked for how many squares were on a board 35 by 30 the answer would not be 35x30 = 1050.

Just a thought

Sorry, I mean unit squares

but to follow up on frost, how would you find the number of squares (any size) on a 30 by 35 board?

i got

[hide]
60[/hide]

My Solution:

[hide]35+30-5=60[/hide]

Edit: I'm quite sure mine is right...

i just did the problem in 2 different ways and got 64 then 60.
Both are logical enough.. wait i forget how i got 64
But reflecting back, i don't know why 60 shouldn't be the answer.

Ways i did it and got 60-
1)draw a 7 by 6 rectange and count the number of squares a diagonal could go through. 12. 12 *5 = 60 (64 if you were the one to solve it!)
2)if 30 is the width, it goes 7/6 per 1 width. so for every width there is 2  (7/6 rounded up) squares intersected by the diagonal. 30 *2 = 60

Anyone have a definate answer?

and the number of squares would be so high im not gonna try it (i know how but i don't think my calculator has that many digits to work with)

[quote="bpms"]Eh, note that you can scale this down to a 6 by 7.[/quote]

that only works when the line is the diagonal.


geometry
rectangle
Pascal\u0027s Triangle

In how many ways can racecar be made from the top r to the bottom r in the figure below if only lengths of 6 can be made?
there is a length of one between each letter

ok now its fixed
[img]http://bwsolo.com/racecar.jpg[/img]

(one such solution is to the right)

[hide]1 ways of choosing 1st, 2 of 2nd letter 3 of 3rd 4 of 4th, 3 of 5th, 2 of 6th and 1 of 7th=1*2*3*4*3*2*1=144[/hide]

No, you cant have the first a with the third c doesnt work because that isnt a standard length to make it lengths of 6 i need to add that between each letter is one

there are 2 choices each time except for the last and first 2^5

I'm not exactly following this. What's the final answer?

i get the same as warpedklown, think of it like pascals triangle but with the side parts cut off.

There is a systematic way of solving these kinds of problems.

[hide]You have to start with R. Mark it as 1. Then, since from R you can go to either A, both As should be marked 1. From the A on the left, you can go to either the left C or the centre C; from the A on the right, you can go to either the right C or the centre C. Therefore, left C = right C = 1, and centre C = 2. Keep doing this procedure, marking each element the sum of all previous elements that can connect to it. The final R has value 20, so there are 20 ways in total.[/hide]

It is a piece of Pascals Triangle and yes the answer is [hide]20[/hide]

your argument that it is a piece of pascal's triangle works just fine. however, just for fun, could someone find a way so that it is obvious how to get a combination? i.e. try a combinatorial approach to this instead of a "pascal's triangle" approach.

no, he basically used the Pascal's triangle recursion, not just actually saying "whoa we can do this by choosing k "things" from n "things" to get a binomial coefficient". although, it's basically the same method, just actually post where you get the combination from.

Yes I have a solution for that.  Put the numbers into a rectangle and use 6! / 3!3! combinations over permutations works too

Ah yes. Nicely done SnowStorm. This problem is equivalent to:

You have a 4 by 4 grid. You wish to traverse from the top left corner to the bottom right corner. In each step you may walk one grid to the right or one grid down. In how many ways can you accomplish this?

Solution: [hide]You have to walk 6 steps - 3 steps across, 3 steps down. You choose 3 steps from the 6 steps to walk across, thus there are 6 choose 3 = 20 ways of doing that.[/hide]

blockwalking, ah jes. its just a bijection of how you have to pick 3 ways to one direction, and 3 to another direction which is the same as C(6,3)


[quote="vulgarfraction"]Is anyone using these books?[/quote]
Thousands of students.

[quote="vulgarfraction"]I've trying to study a bit for the national olympiad in early February right now, and at first I just wanted to study enough to go in knowing what to use, and on equal footing, but now I really want to learn all I could. I've got three stabs at this, and I'd love to get as far as the APMO (top 25), or even the IMO selection camps. I'm using [i]Mathematical Toolchest[/i] from the AMT as a reference, but I need more in detail references.

That's just an idea of the level of material I'm looking for. I'm thinking the [i]AoPS Volume 2[/i] would be most suitable. Is that suitable for Olympiad preparation?

Also, what are good sources for online texts? I'm getting those too.

Thanks in advance to anyone who helps! ;)[/quote]I think you should consider looking into the AoPS Olympiad classes (there are two starting just these days): http://www.mathlinks.ro/Classes/AoPS_C_Enroll.php#class3

As for extra practice-problems, you can always use our olympiad resources directory: http://www.mathlinks.ro/Forum/resources.php

[quote="PenguinIntegral"][quote="vulgarfraction"]Is anyone using these books? I'm just trying to see if they are worth ordering.

I've trying to study a bit for the national olympiad in early February right now, and at first I just wanted to study enough to go in knowing what to use, and on equal footing, but now I really want to learn all I could. I've got three stabs at this, and I'd love to get as far as the APMO (top 25), or even the IMO selection camps. I'm using [i]Mathematical Toolchest[/i] from the AMT as a reference, but I need more in detail references.

That's just an idea of the level of material I'm looking for. I'm thinking the [i]AoPS Volume 2[/i] would be most suitable. Is that suitable for Olympiad preparation?

Also, what are good sources for online texts? I'm getting those too.

Thanks in advance to anyone who helps! ;)[/quote]
For your level, AoPS 2 would be good.

From what I have seen you post, the Engel book, Problem solving strategies would be a great help. It contains many great problems, and a review of notation. Stay clear of AacoPS. At your level, most other problem solving instruction material is a waste besides getting more problems. You already seem to know the strategies well.

For formal background reading, [url=http://www.unl.edu/amc/a-activities/a4-for-students/reading-list.html]http://www.unl.edu/amc/a-activities/a4-for-students/reading-list.html[/url]
a list the AMC put out is on that link.

For now until your books come, you might be interested in some [url=http://web.mit.edu/~tmildorf/www/Inequalities.pdf]inequalities[/url] and [url=http://web.mit.edu/~tmildorf/www/MyMockAIMESeries.pdf]some problems.[/url][/quote]

I'll see if I could get those books online, or find the Arthur Engel one locally. But thanks heaps for the help- I'll use those links.

Valentin- I know that, I was more appealing for people who have used the books. :) I had a look at the classes and I just realised that I can't do any of them because they are around 11am my time and school starts in February. Which is a shame. Still, I'll try make do with books.

Actually, I just ordered them last night. (Or rather, my dad did.)

they are really really good. It's quite interesting, and it has good explainations.


calculus
integration
inequalities
real analysis
inequalities unsolved

prove that for every positive integer $ n$ we have
\[ \frac{n\pi}{4}\minus{}\frac{1}{\sqrt{8n}}<  \frac{1}{2}\plus{}\sum_{k\equal{}1}^{n\minus{}1}\sqrt{1\minus{}\frac{k^2}{n^2}}\]

Let integrate $ \sqrt{1\minus{}x^2}$ on [0, 1] by dividing [0,1] by 0, 1/n, 2/n, ..., n/n and form integral sum.

But it will be upper bound ?

I obtained

\[ \sum_{k \equal{} 1}^{n \minus{} 1}\sqrt {1 \minus{} \frac {k^2}{n^2}}<\frac {n\pi}{2}\]

 :|

By numerical evidence it looks like the sequence $ a_n\equal{}\frac {n\pi}{4} \minus{} \frac {1}{\sqrt {8n}} \minus{} \frac {1}{2} \minus{} \sum_{k \equal{} 1}^{n \minus{} 1}\sqrt {1 \minus{} \frac {k^2}{n^2}}$ in monotonous. If this is true then it requires to show that $ a_1<0$.

it may be hard to prove, because $ n$ is in the denominator under the sum  :wink:

Slicing the first quadrant of the unit circle proves the inequality: http://math.stackexchange.com/questions/8042/an-inequality-like-riemann-sum .


Find the residue of the following.

(a) $ \text{Res}(f, z_0\equal{}0)$, where $ f(z)\equal{}\frac{e^z}{\sin(2z^2)}$

(b) $ \text{Res}(f, z_0\equal{}0)$, where $ f(z) \equal{} \frac{\cos(\frac{1}{z})}{e^z}$
We are just beginning residues right now, and I do not know how to proceed. Should I use the Laurent expansion to find the residue for both parts? I am not very familiar with residues yet. Thanks very much.

(a) No, you should write f as a series in powers of z ... you only need the first few terms to see what the residue is.  Note that e^z is regular at zero, e^z = 1 + ...

I was wondering if you can check my work.

For (a) $ \text{Res}(f, z_0\equal{}0)$, where $ f(z)\equal{}\frac{e^z}{\sin(2z^2)}$ we have

$ \frac{1\plus{}z\plus{}z^2/2!\plus{}z^3/3!\plus{}\cdots}{2z^2 \minus{}(2z^2)^3/3!\plus{}(2z^2)^5/5!\minus{} \cdots}$

$ \equal{} \frac{1}{2z^2}\plus{}\frac{1}{2z}\plus{}1/4\plus{}z/12$

So the residue is $ 1/2$.

For (b) $ \text{Res}(f, z_0\equal{}0)$, where $ f(z) \equal{} \frac{\cos(\frac{1}{z})}{e^z}$ we have

$ \cos(\frac{1}{z}) \cdot (1\minus{}z\plus{}z^2/2\minus{}z^3/6\plus{}z^4/24\plus{}\cdots).$

I am stuck. I don't know how to get the residue. I know that $ z_0$ is an essential singularity. I need help on part (b).

b) Try writing $ cos \left(\frac {1}{z} \right )  \equal{} \frac{e^{\frac{i}{z}}\plus{}e^{\frac{\minus{}i}{z}}}{2}$

It will turn out to be $ \left ( 1\minus{}\frac{1}{z^{2}2!} \plus{} \frac{1}{z^{4}4!}..... \right ) \left ( 1\minus{}z\plus{}z^{2}/2\minus{}z^{3}/6\plus{}z^{4}/24\plus{}\cdots \right)$

Residue is $ 1/2$ :maybe:


function
induction
limit
algebra
functional equation
real analysis
real analysis unsolved

Find all $f: {\mathbb R}\to{\mathbb R}$, for which:
1) $\exists f'$
2) $\forall x \in {\mathbb R} \; f(f(x))=f(x)+x$

Also one can prove  that $f$ -monotonic, $f$-biection, and $f(x)f(y)>0$ if $xy>0$, and $f(x)f(y)<0$ when $xy<0$...

[quote="anik_andrew"]i know 2 solutions
$f_{1,2}(x) = \lambda_{1,2}x$ where $\lambda_{1} = \frac{1+\sqrt{5}}{2},\lambda_{2} = \frac{1-\sqrt{5}}{2}$
also $f(0)=0$ and $f(x)=f(y) \Rightarrow x=y$[/quote]
If $f$ is analytic on an interval $(-L,L)$, then $f$ must be one of these functions on this interval.

To see this, differentiate the functional equation: $f'(f(x))f'(x) = f'(x) + 1$. As anik_andrew noted, this implies that $\lambda = f'(0)$ satisfies $\lambda^2 = \lambda + 1$, thus $\lambda$ is one of the two values given above.

Next differentiate again. The result is $f''(f(x))\left(f'(x)\right)^2 + f'(f(x))f''(x) = f''(x)$ and thus $f''(0)\left(\lambda^2 + \lambda - 1\right) = 0$. Since the term in parentheses is not zero, $f''(0) = 0$.

Now one shows that for $k > 2$,
$f^{(k)}(f(x))\left(f'(x)\right)^k + f'(f(x))f^{(k)}(x) + \sum_{i=2}^{k-1} f^{(i)}(f(x))P_{k,i}(x) = f^{(k)}(x)$
where $P_{k,i}(x)$ is analytic. For $k = 3$, this follows by differentiating the equation for $f''$, with $P_{2,2}(x) = 6f'(x)f''(x)$. For $k > 3$ it follows by induction.

Set $x=0$, then an induction argument shows that $f^{(k)}(0) = 0$ for all $k>1$. Indeed, this is true for $k = 2$. To get from $k-1 \ge 2$ to $k$, the formula above shows that $f^{(k)}(0)\left(\lambda^k + \lambda - 1\right) = 0$. Again, one sees that the factor in parentheses can never be zero for any $k \ge 3$, therefore also $f^{(k)}(0) = 0$.

So $f$ is linear and of the form given above ... on the interval where it is analytic.

Intresting solution.
But my friend said that he saw this problem in magazine, and there was condition only that $f\in C({\mathbb R})$, not even diff.
So there must be solution without analytics.

[quote="anik_andrew"]So there must be solution without analytics.[/quote]
OK, here's one (I hope).

Start by setting $g(x) = f(x)-x$, then $g(f(x)) = f(f(x)) - f(x) = x$. This implies $f$ is onto and $f(g(x)) = x$.

Let $f_k(x) = f \circ \dots \circ f(x)$ ($k$ times), and let $F_k$ be the $k$-th Fibonacci number ([i]these have to come into play somewhere, don't they?[/i]), $F_1 = F_2 = 1, \, F_3 = 2, F_4 = 3, \dots$.
We first show that
$f_k(x) = F_kf(x) + F_{k-1}x$
for $k = 2, \, 3, \dots$. This is just an induction proof, using the fact that $f_k(x) = f_{k-1}(x) + f_{k-2}(x)$.

As you observed, we know that $f(0) = 0$ and $f$ is 1-to-1. Let's first consider the case where  $f$ is decreasing. Then $f_2(x) = f(x) + x$ is increasing, thus $|f(x)| \le |x|$ and $|f_k(x)| \le |x|$ for all $k$. It follows that
$|f(x) + \frac{F_{k-1}}{F_k} x| = \frac{1}{F_k}|f_k(x)| \le \frac{1}{F_k}|x|$
for all $k$. Take the limit as $k \to \infty$ to obtain
$f(x) = -\lambda x$, where $\lambda = \lim_{k \to \infty} \frac{F_{k-1}}{F_k} = \frac{\sqrt{5}-1}{2}$, i.e. $f$ is linear. :)

Next consider the case where $f$ is increasing. Then with $g$ as above
$g(g(x)) = f(g(x)) - g(x) = x - g(x)$.
As in the case of $f$, one defines $g_k(x) = g\circ \dots \circ g(x)$ and proves via induction that $g_k(x) = F_{k-1}x - F_k g(x)$, using properties of the Fibonacci numbers.

Now for fixed $x \ne 0$, in this case $|f_k(x)| = |F_{k-1}x + F_kf(x)| \to \infty$. Therefore, for fixed $x \ne 0$, $|g_k(x)| \to 0$. Then
$|\frac{F_{k-1}}{F_k} x - g(x)| \le \frac{1}{F_k}|g_k(x)|$. Take the limit as $k \to \infty$ and conclude that $g$ and therefore also $f$ is again linear, $g(x) = \lambda x$ and $f(x) = \lambda^{-1}x$.  :lol:

Nice proof :lol: . As I see, the existence of $f'$ is not used in the proof. Here is a nice generalization:

http://www.mathlinks.ro/Forum/viewtopic.php?t=72543

Try to prove it too.

I think this problem has to be related with this one :) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=62193

hpe, your solution exactly as in the magazine!  :)  ;)

[quote="anik_andrew"]hpe, your solution exactly as in the magazine! [/quote]Well I didn't read it  ;)

[quote="fedja"]Or to this one (still unsolved!) [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3611[/url] :)[/quote]I think this problem can be solved in exactly the same way. The solution is $f(x) = \lambda x, \, \lambda = \frac{1+\sqrt{1+4k}}{2}$. Continuity is not required if $f$ is assumed to be bijective from the outset. Instead of Fibonacci numbers, one uses another three term recurrence that has the right exponential behavior.

[quote="hpe"]I think this problem can be solved in exactly the same way. The solution is $f(x) = \lambda x, \, \lambda = \frac{1+\sqrt{1+4k}}{2}$. Continuity is not required if $f$ is assumed to be bijective from the outset. Instead of Fibonacci numbers, one uses another three term recurrence that has the right exponential behavior.[/quote]
Well, there is one little difficulty: it is no longer true that $f$ is either increasing or decreasing. On the other hand, now we know that $F(x)>0$ for all $x>0$, which, indeed, seems to be enough to carry out the argument. So, you are right, hpe, it can be solved the same way. :) Actually, harazi's problem is even simpler: all one needs to notice is that, starting with any $x_0=x$, we can construct a sequence of positive numbers $\{x_n\}_{n=-\infty}^\infty$ such that $x_{n+1}=f(x_n)$. This sequence will satisfy the recurrence relation $x_{n+2}=x_{n+1}+kx_{n}$ and, therefore, be of the kind $a\lambda^n+b\mu^n$ where $\mu$ is another (negative) root of the equation $t^2=t+k$. But it is easy to see that it is impossible to have a non-zero coefficient at $\mu^n$ and to have all the terms positive simultaneously. So, $x_{n+1}=\lambda x_n$ for all $n$ and, in particular, $f(x)=\lambda x$.


Here is an inequality:

If $x,y,z  \geq \frac{-1}{4} $ and $ x + y + z = 1$ then: $ \sqrt{4x+1} + \sqrt{4y+1} + \sqrt{4z+1} \leq 5$.

Here are my solutions:
I)we try to create GM $ \leq$ AM and to do this we multiply it by 1 and get:
$ \sqrt{(4x+1)*1} + \sqrt{(4y+1)*1} + \sqrt{(4z+1)*1} \leq \frac{4(x+y+z)+3+3)}{2}=5$.

II) we show that $ f(t) = \sqrt{4t+1}$ is concave on $ [ \frac{-1}{4}, \inf )$
$ f'(t)=2*(4t+1)^{-1} $and $ f''(t)=-16*(4x+1)^{-3} $all done;
we apply Jensen now:
$ \frac{f(x)+f(y)+f(z)}{3} \leq f( \frac{x+y+z}{3}) $
$ \frac{ \sqrt{4x+1} + \sqrt{4y+1} + \sqrt{4z+1}}{3} \leq \sqrt{4* \frac{1}{3}+1} < 5 $ so the inequality is proved!!!

And I have managed to "see" a generalization(can someone check if it is okay) :)
If anyone of you has seen it before then I am not the 1st one to see it, and I hope it is a true generalization! :D :D

here goes nothing! :D :D

If $ a_{1}, \ldots , a_{n} \geq \frac{-1}{4} $ and $ a_{1}+ \ldots + a_{n} =1 (n \geq 3) $ then $ (4  a_{1}+1)^{-n+1} + \ldots +(4 a_{n}+1)^{-n+1} \leq \frac{n^{2}-n+4}{n-1}$ .

The proof is based on the same idea as in the case n=3.

Thank you very much! :D

Everything can become so simple if we true Cauchy sometimes!


Try to subtract 42!  and 68! Without A Calculator
Then find the square root of that.
Then add pi.

68! - 42! Yeah this number has at least twenty digits. to figure it out you need to do at least 101 operations.
the sqrt has at least 15 digits, and I have no clue how to calculate it without memorization of sqrt tables.
However I can easily add pi !

i have totally no idea how to do that.

i know there may be a trick but i dunno how it works :rotfl:

How in the world can you call this easy and nice problem..
I don't think this is even "Classroom Math" level..

Fine use a calculator for the square root part but do the rest by yourself


projective geometry
conics
geometry
similar triangles
geometry unsolved

Let $ ABC\bigtriangleup$ be a triangle and let $ \omega$ be a cicunference which is  tangent to side $ AC$  in  $ Z$ and intersects $ BC$   at $ R$ and $ S$  $ (S\in CR)$ and $ AB$  at  $ P$ and  $ Q$  $ (Q\in AP)$. Let  $ X\equal{}PS\cap QR$ and $ Y\equal{}QC\cap SA$. If $ B$  ,$ Y$  and $ Z$ are collinear prove that  $ B$, $ X$ and  $ Z$ are too collinear.

Daniel

[color=green]Let $ H \equal{} AR\cap CP$.First we may prove that if $ B,Y,Z$ are collinear then so are $ B,H,Z$

since $ AZP$ and $ AQZ$ are similar triangles $ \frac {AZ}{AP} \equal{} \frac {AQ}{AZ}$.

The fact that $ CZS$ and $ CRZ$ are also similar triangles gives us that $ \frac {CR}{ZC} \equal{} \frac {CZ}{CS}$

$ \frac {CR}{RB}*\frac {BP}{PA}*\frac {AZ}{ZC} \equal{} \frac {CR}{RB}*\frac {BP}{AZ}*\frac {AQ}{ZC}$

$ \frac {CR}{RB}*\frac {BP}{AZ}*\frac {AQ}{ZC} \equal{} \frac {CZ}{RB}*\frac {BP}{AZ}*\frac {AQ}{CS}$

now because $ BPR$ and $ BSQ$ are again similar triangles we obtain that: $ \frac {BP}{RB} \equal{} \frac {BS}{BQ}$ so finally

$ \frac {CZ}{RB}*\frac {BP}{AZ}*\frac {AQ}{CS} \equal{} \frac {CZ}{BQ}*\frac {BS}{AZ}*\frac {AQ}{CS}$

i.e.$ \frac {CR}{RB}*\frac {BP}{PA}*\frac {AZ}{ZC} \equal{} \frac {CZ}{AZ}*\frac {AQ}{QB}*\frac {BS}{SC}$

But $ B,Y,Z$ collinear means that $ BZ, CQ, AS$ are concurrent, by Ceva this means $ \frac {CZ}{AZ}*\frac {AQ}{QB}*\frac {BS}{SC} \equal{} 1$

so $ \frac {CR}{RB}*\frac {BP}{PA}*\frac {AZ}{ZC} \equal{} 1$ and $ BZ, CP, AR$ are concurrent, i.e. $ B,H,Z$ are collinear.

Now the rest is trivial, we have that $ B,H,Y,Z$ are collinear, so by Pappus on the lines $ BA$ and $ CB$ we have that $ Y,H,X$ are collinear,

this implies $ B,X,Z$ collinear as we wanted to prove. :blush:

Life is GREEN!  :D

TaUrUs*[/color]

this follow directly by Carnot + Ceva + Pappus Theorems  :wink:

p.s. the problem is true also if we get a conic instead of a circle for the same proof.


Given $ \triangle ABC$, $ w(O)$ is the circumcircle and $ I$ is the incenter. Let $ k(I_c)\cap AB\equal{}N, k(I_c)\cap AC\equal{}K$and $ k(I_c)\cap BC\equal{}M$. If $ F$ is the midpoint of the segment $ KM$ and $ F\in w$ prove that $ N, O$ and $ I$ are colliear.

Hello!

Let $ R$ be the radius of the circumcircle $ \omega(O)$, $ x$ is the radius of the circle $ k(I_{c})$ and $ BC\equal{}a$.
If $ F\in \omega$ then it's easy to prove that $ F\in CI_{c}$ and $ \angle FI_{c}B\equal{}\angle FBI_{c}\equal{}\frac{\angle A}{2}$, so $ FI_{c}\equal{}FB$     (1)
$ sin \angle FKI_{c}\equal{}\frac{FI_{c}}{x}\Rightarrow x\equal{}\frac{FI_{c}}{sin\frac{\angle C}{2}}$     (2)
$ \frac{FB}{sin \frac{\angle C}{2}}\equal{}\frac{BC}{sin \angle BFC}\equal{}\frac{a}{sin \angle A}\equal{}2R\Rightarrow \frac{FB}{sin \frac{\angle C}{2}}\equal{}2R$     (3)
(1), (2), (3) $ \Rightarrow x\equal{}2R$     (4)
$ I_{c}N\bot AB$ and $ OF\bot AB\Rightarrow I_{c}N\parallel{}FO$     (5)
The triangle $ IBI_{c}$ is rectangular and $ FI_{c}\equal{}FB\Rightarrow F$ is the midpoint of $ II_{c}$     (6)
(4), (5), (6) $ \Rightarrow O$ is the midpoint of $ NI\Rightarrow$ the points $ N$, $ O$, $ I$ are collinear.

P.S. It's easy to prove that the quadrilateral $ MIKI_{c}$ is rhomb and the point $ I$ is the orthocenter of the triangle $ MCK$. :)


Five points are situated inside an equilateral triangle whose side has length one unit. Show that two of them may be chosen which are less than one half unit apart. What if the equilateral triangle is replaced by a square whose side has length of one unit?

For the triangle:
Divide it into four equilateral triangles of side $ \frac{1}{2}$(connect the midpoints). By the Box Principle, two points will be in the same little triangle, and so the distance between them is less than $ \frac{1}{2}$.

For the square: is a similar situation:
Divide the square into four squares with side $ \frac{1}{2}$. By the Box principle, two points will be in the same little square, and so the distance between them is less than $ \sqrt{(\frac{1}{2})^2\plus{}(\frac{1}{2})^2}\equal{}\frac{\sqrt{2}}{2}$

 :)


find the number of 6 digit binary strings such that each one is different in at least 3 digits from another.

This is basically a packing problem,
http://en.wikipedia.org/wiki/Bin_packing_problem

You can represent it with n dimensional cubes and travel 3 edges away. If I remember correctly, it's unsolved for something as small as n=15.


For what value of $ k$ is $ 2x\plus{}1$ a factor of $ 2x^2\plus{}7x\minus{}k$?

If $ 2x\plus{}1$ is a factor of $ 2x^2\plus{}7x\minus{}k$, then $ \minus{}1/2$ is a root of the quadratic, thus \[ 2\left( \frac{\minus{}1}{2} \right)^2 \plus{}7\left(\frac{\minus{}1}{2}\right) \minus{}k\equal{} 0 \Rightarrow k\equal{}\boxed{\minus{}3}.\]

how do u find the root of the quadratic with k?

If you know how to divide polynomials into polynomials, then you can easily figure out that $ {\colorbox{green}{\colorbox{yellow}{\colorbox{red}{\colorbox{blue}{\colorbox{green}{\colorbox{yellow}{\colorbox{red}{\colorbox{green}{\colorbox{blue}{\colorbox{yellow}{\colorbox{red}{\colorbox{green}{\colorbox{blue}{\colorbox{yellow}{\text{-3}}}}}}}}}}}}}}}} $ is the answer.

Oh my gosh I silled this problem hard. I forgot the root was negitive, so I got $3$ :wallbash:

[quote=Zanderman]Oh my gosh I silled this problem hard. I forgot the root was negative, so I got $3$ :wallbash:[/quote]

Well, check your answers next time! Wait no cause this is FTW


Let $S=\{(a,b)|a=1,2,\dots,n,b=1,2,3\}$. A [i]rook tour[/i] of $S$ is a polygonal path made up of line segments connecting points $p_1,p_2,\dots,p_{3n}$ is sequence such that

(i) $p_i\in S,$

(ii) $p_i$ and $p_{i+1}$ are a unit distance apart, for $1\le i<3n,$

(iii) for each $p\in S$ there is a unique $i$ such that $p_i=p.$

How many rook tours are there that begin at $(1,1)$ and end at $(n,1)?$

(The official statement includes a picture depicting an example of a rook tour for $n=5.$ This example consists of line segments with vertices at which there is a change of direction at the following points, in order: $(1,1),(2,1),(2,2),(1,2), (1,3),(3,3),(3,1),(4,1), (4,3),(5,3),(5,1).$)

[hide="Short answer, no explanation."]$2^{n-2}$ for $n\ge2.$[/hide]

It's not terribly helpful to post a sentence fragment about how easy the problem is.  If it really is so easy, post a solution so clear that we can all heartily agree with you.  My own solution was drawn out and convoluted so I'd love to read a better one.

Sure.
this is what I did,

any path from 1,1 to n,1
starts by moving right a finite number of times to (i,1) and then it moves up ( 0<=i<=n-1)
as soon as it goes up to (i,2) it has to go left up to (1,2) and go up and then right to (i,3) then we have to find the number of ways
to go from (i,3) to (n,1). Use a funvtion g(n) for this kind of paths and use f(n) for those paths that go from (1,1) to (n,1)
then from what I said:
f(n) = g(n-1) + g(n-2) + ... +g(1)
similarly g(n) = f(n-1) +... +f(1)
thus f(n) = 2^n-2
-Ali

If you want the points for that, go through the motions of proving your formula for $f$ by induction- I hope you did on the real test.

Another idea, based on seeing the answer:

When we first reach the $k$th column at either $(k,1)$ or $(k,3)$, we have a choice: go one step farther horizontally to either $(k+1,1)$ or $(k+1,3)$, or go vertically to $(k,2)$. In the second case, we must then go back along the second row as far as possible (0 steps if $(k-1,2)$ is already touched), then back along the row we haven't followed before until reaching either $(k,3)$ or $(k,1)$ and then either $(k+1,3)$ or $(k,1)$.
Once we reach column $k+1$, we make the choice again. This choice occurs in columns $1,2,\dots,n-1$; there is of course no choice in column $n$.
That gives $2^{n-1}$ tours, but some end at $(n,3)$ instead of $(n,1)$. Each time we choose to move vertically and also in the last column, we switch sides; therefore we end at $(n,1)$ exactly when an odd number of our choices are vertical moves. If $n=1$, this never happens. If $n>1$, we can switch between the two possibilities by reversing the choice in the first column, and exactly half of paths end at $(n,1)$ for a total of $2^{n-2}$.

Here's a sketch of my solution:

The rook can reach all the middle points by upward, downward, or horizontal moved. Every time he moves up or down, the only way he can continue on to his destination and hit all the points is by going back and moving horizontally to cover all the middle points he's passed, and then he'll come out on the "top" (if he had moved up from the bottom) or the "bottom" (if he had moved up from the top).

It is easy to see the rook must move either up or down on the first and last column. He can cover all the other middle points with any combination of up/downs and horizontals. Let $s$ be the number of up/down moves he makes, not counting the two fixed ones. He has $n-2$ places where he can arrange those. So the number of possible tours is $\sum_{s=0}^{n-2}\binom{n-2}{s} = 2^{n-2}$.

It is clear that for $n=1$ there is no solution.

Unfortunately, for reasons I no longer recall, I somehow thought that I needed to subtract one if $n$ was odd. I just hope I get some modicum of partial credit for my solution!

You've got a worse problem- that's the right answer for the wrong reason. There is a choice in the first column, but we have to account for which side we end on.

Oh! Actually, that may be the reason I applied the erroneous -1 for odd $n$.

It's been a while since I last visited this site since I was busy. So I figured I could share the solution I wrote when I took the exam. It is by strong induction, so the professor who coached for the Putnam chose to feature it as an extra credit problem on the Discrete Math final exam.

Proof:

Counting the number of rook tours for $n=2,3,4,5$ carefully gives $1,2,4,8$ for the numbers of tours. This suggests the formula $R(n)=2^{n-2}$ tours for any $n\geq 2$. We will prove just that.

We will call a tour of a width $n$ board [i]primitive[/i] if it has the following form.

Choose an $m$ satisfying $0\leq m\leq n-2$
Go right $m$ spaces.
Go up one space.
Go left $m$ spaces.
Go up one space.
Go right $n-1$ spaces.
Go down one space.
Go left $n-m-2$ spaces.
Go down one space.
Go right $n-m-2$ spaces to finish the tour.

See attached file for an example of a primitive tour with $n=12$ and $m=4$.

Observe that every tour on a board of width $n$ is either primitive or is a chain of connected primitives for narrower boards connected by one rightward move between any two primitives.

We are now ready to set up the recursion.

The base of the recursion is $R(2)=1$, and $R(3)=2$.

There are $n-1$ primitive tours on a board of width $n$.

For any other tour, we start by doing a primitive tour for a board of width $j$ where $2\leq j\leq n-2$, then go right 1 space, and do an arbitrary tour for a board of width $n-j$.

Thus the recursion is
\[R(n)=n-1+\sum_{j=2}^{n-2}(j-1)R(n-j).\]

We are now ready to proceed with strong induction.

[i]Base Case:[/i] $R(2)=1=2^0$ and $R(3)=2=2^1$ agree with the formula we are to prove.

[i]Induction Case:[/i] Assume for some $k\geq 3$, that $R(n)=2^{n-2}$ holds for $2\leq n\leq k$. Then for $n=k+1$,

\begin{align*}
R(n)&=n-1+\sum_{j=2}^{n-2}(j-1)R(n-j)\\
&=n-1+\sum_{j=2}^{n-2}(j-1)2^{n-j-2}\\
&=n-1+\sum_{i=2}^{n-2}\sum_{j=i}^{n-2}2^{n-j-2}\\
&=n-1+\sum_{i=2}^{n-2}\Bigg[2^{n-i-2}\sum_{j=0}^{n-i-2}(1/2)^j\Bigg]\\
&=n-1+\sum_{i=2}^{n-2}\Bigg[2^{n-i-2}\cdot\frac{1-(1/2)^{n-i-1}}{1-1/2}\Bigg]\\
&=n-1+\sum_{i=2}^{n-2}(2^{n-i-1}-1)\\
&=2+\sum_{i=1}^{n-3}2^i\\
&=2+2^{n-2}-2\\
&=2^{n-2}
\end{align*}

Let $A_n$ be the desired answer, and let $B_n$ be the number of paths from $(1,1)$ to $(n,3)$.

Let us first compute $A_n$ in terms of previous values. Note that the only valid starting paths are
[list]
[*] $(1,1)\to(1,2)\to(1,3)\to(2,3)$
[*] $(1,1)\to(2,1)\to\cdots\to(k,1)\to(k,2)\to\cdots\to(1,2)\to(1,3)\to\cdots(k,3)\to(k+1,3)$, for $k=2,\ldots,n-1$.
[/list]
This proves that
\[A_n = B_{n-1} + \cdots + B_1.\]
Subtracting the $n+1$ and $n$ forms of the recursion, we have $A_{n+1}-A_n=B_n$.

Now let's compute $B_n$. Again, the previously mentioned starting paths are the only ones, along with the full path
\[(1,1)\to\cdots\to(n,1)\to(n,2)\to\cdots\to(1,2)\to(1,3)\to\cdots\to(n,3).\]
This proves that
\[B_n = A_{n-1}+\cdots+A_1 + 1.\]
This implies $B_{n+1}-B_n = A_n$. Combining this with what we have above, we see that
\[A_{n+2}-A_{n+1} -A_{n+1} + A_n = A_n,\]
so
\[A_{n+2}=2A_{n+1}.\]
We compute $A_2=1$, so this implies $A_n=2^{n-2}$ for $n\ge 2$, and $A_1=0$.


superior algebra
superior algebra unsolved

Let E be a finite normal extension over Q (the rationals). If F is [i]any[/i] intermediate field, is E normal over F ? i.e. is E a split. field over F?

Two simple proofs (we never need $ Q$ to be the rationals, it can be any field):
a) $ E|Q$ normal $ \iff$ every homomorphism $ f: E\to\bar Q$ fixing $ Q$ is an automorphism. As every homomorphism $ f: E\to\bar F \equal{}\bar Q$ fixes $ q\subset F$, we get that $ E|F$ is normal.
b) If $ E$ is the splitting field of $ f_{1}, ..., f_{k}$ over $ Q$, the same holds for $ E$ over $ F$.


abstract algebra
superior algebra
superior algebra unsolved

Let $ \textrm{M}$ and $ \textrm{N}$ be free $ \textrm{R}$-modules.  Show that $ \textrm{Hom}_{\textrm{R}}\left(\textrm{M},\textrm{N}\right)$ is a free $ \textrm{R}$-module.

Is there a way that we can prove this using the Universal Mapping Property?

we should assume that $ M,N$ are finitely-generated. then $ M$ is a finite direct sum of copies of $ R$ and $ N$ is a finite product of copies of $ R$. now apply the universal properties and you're done.


trigonometry
logarithms
inequalities unsolved
inequalities

prove that :
$ \sqrt [44]{\tan1 \tan2 ...\tan44}<\sqrt{2}-1<\frac{\tan1+ \tan2+ ...+\tan44}{44}$

[i]Edited by Megus: write subjects name next time [/i]

$\tan x$ is convex and $\ln \tan x$ is concave.
 :D

As I remember this is from Austrian-Polish olympiad.


AMC
USA(J)MO
USAMO
AIME
probability
MATHCOUNTS
number theory

I would like to know how you all (talking to anyone who has done well and/or made USAMO) prepared for these tests in the past. Have you taken courses, hired teachers, poured over books, done vast amounts of problems? For me, I can't seem to be able to improve, especially on AIME combanations and counting problems. I have 2 books on counting principles, both by Titu Andresscu(Spelling) and zuming feng. I study those problems, but when I encounter another problem, I am stuck. How can I improve on number theory and counting and probability topics? What do u suggest?

yall, its yall.

ne ways, i haven't made USAMO(but ima 8th grader) but i have done pretty well on mathcounts and i've done pretty well on practice tests, so...just try to learn the concept, then do a ton of problems, then do a ton more problems, then go back and redo all those problems, repeat the doing of problems, wow, i have so many grammer n' spelin misstakes in here, o well, commencing, when u get good and beat me, try not to laugh at me. there we go. can u read dat?

It's all about concentrating when you do problems - look outswide what you would normally think - try and bend and manipulate different formula's on your free time to figure out different patterns when you see them. Do more problems.


calculus
integration
logarithms
trigonometry
function
calculus computations

Evaluate $\int_{1}^{\pi}\left(x^{3}\ln x-\frac{6}{x}\right)\sin x\ dx$.

That is, shall we say, carefully rigged to have a closed-form answer? In particular, that has to be $-\frac6x$ and not any other $\frac{A}x.$

we calculate the integral of x^3ln(x) using integration by parts
concerning 6/x it's obvious

goldbach - would you please write out all of the details? I'd like to see the full, specific version of what you mean by that.

this was an easy and nice integral to solve, but no one seems
to want to do it.  :(

OK, here is the solution (uses a set of 7 equations):

--------------------------------------------------------------------------------
$-(x^{3}\ln x\cos x)'=x^{3}\ln x \sin x-x^{2}\cos x-3x^{2}\ln x\cos x$

$3(x^{2}\ln x \sin x)'=3x^{2}\ln x\cos x+3x\sin x+6x\ln x\sin x$

$6(x\ln x\cos x)'=-6x\ln x\sin x+6\cos x+6\ln x\cos x$

$-6(\ln x\sin x)'=-6\ln x\cos x-\frac{6\sin x}{x}$

$(x^{2}\sin x)'=x^{2}\cos x+2x\sin x$

$5(x\cos x)'=-5x\sin x+5\cos x$

$-11(\sin x)'=-11\cos x$
--------------------------------------------------------------------------------

[b]adding [/b]all 7 of them, and integrating between the limits
$1\;\;\;and\;\;\;\pi$, we eventually get,

$\textbf{I}\;=\;\boxed{5\cos 1-10\sin 1-5\pi-\pi\ln \pi(6-\pi^{2})}$

[ note: [b]adding [/b]top $\rightarrow$ bottom, all of the terms on the
RHS cancel out [u]except [/u] for the 2 terms within $\textbf{I}$ ]

thanks.  :D

You make it sound like misan's answer is really far from correct, but there is just a sign error:

Make it $I=-5\cos(1)+10\sin(1)-5\pi-(6-\pi^{2})\pi\ln\pi$

The key is the fact that $\left( x^{3}\ln x \right)^{(4)}=\frac6{x}$.
Thus, all we have to do is to write
\[\int_{1}^\pi \left( x^{3}\ln x-\frac6{x}\right) \sin x \, dx = \int_{1}^\pi x^{3}\ln x \sin x \, dx-\int_{1}^\pi \frac6{x}\sin x \, dx . \]
As [b]goldbach[/b] pointed out, we need to integrate by parts four times in the first integral (we differentiate $x^{3}\ln x$ and integrate $\sin x$). At the end, we will get some functions to evaluate between $0$ and $\pi$ plus $\int_{1}^\pi \frac6{x}\sin x \, dx-\int_{1}^\pi \frac6{x}\sin x \, dx$ (since, after all, integrating $\sin$ four times - with no constants - gives $\sin$ again).


The outside of this set of four cubic blocks with no space between the blocks is painted. What is the number of painted square faces?
[asy]import olympiad; import geometry_dev; import graph; size(150); defaultpen(linewidth(0.9));
path blockFaces(int side, pair bCorner){
  pair A = bCorner;
  pair B = bCorner + side*dir(20);
  pair C = B + (0,side);
  pair D = C + side*dir(160);
  pair E = bCorner + side*dir(160);
  pair F = E + (0,side);
  pair G = bCorner + (0,side);
  return (A--B--C--G--A--E--F--G--C--D--F);
}
draw(blockFaces(1,origin));
draw(blockFaces(1,(dir(20) + dir(-20)))); draw(blockFaces(1,(dir(20) + (0,1))));[/asy]

There are 5 faces for the 3 cubes that are connected to the one cube, and is invisible. That cube has 3 faces. So there are 18 faces total.

You forgot about the cube that is connecting the other 3 cubes together. There are 3 faces on that cube.


geometry
algebra
polynomial
USAMTS
trapezoid
geometric transformation
rotation

Here are my solutions. I think I got 4 of them right, from the math jam result. Maybe I'll get a 24, or at least more than a 20. :) (Last round I got a 14...) I got both geometry ones right!!! :)

Billy

[quote="Philip_Leszczynski"]I only saw the very end of the Math Jam. Can anyone please post the correct answers for #4?[/quote]
I didn't see the Math Jam either, but I'm certain that the correct answers are $x = \pm (1 \pm \sqrt{3})$ and $x = \pm \sqrt{6}$ (for a total of six solutions).

I was there, and ya, those are the correct answers.

here's mine, I love my solution to the 4th and 5th one, but ignore the third one, I can't believe I made stupid assumptions...

Here goes my 100... :(

[quote="matt276eagles"]Here are my solutions.  My answers to #4 and #5 are incorrect.  I would appreciate feedback on what I did wrong (although I'm pretty sure I know for #4  :P ).  Otherwise, I think my solutions are fine, and I'm hoping for at least 21 for a final score of 90.[/quote]

See, I was tempted to say that they are conjugates too. But if there's not a theorem out there that states so or you can't prove it, I will never make that leap.  (Btw, didn't you figure out that $\pm \sqrt{6}$ work??)

AH, I just read your number five and YOU ARE OFF BY THE VERY LAST STEP!!!! You assume again (as I was tempted) that the midpoint of the major axis lies on the altitude, but this is not so...(As I figured out 1:00h into the problem).  If you can see my parallel lines solution, then you could've got the right answer...(I got that at like 12:30pm of the first day)

Here are my solutions - I got them all right except #5, and I'm hoping for mercy and partial credit for that one since I solved part of the problem correctly and just made an idiotic move figuring out the minor axis length (which isn't hard to do and I was able to do once I re-examined the problem).  I like my solution to #4, but there are probably better ones out there.

HEre are my solutions.  There was some simplification I forgot to do, but I got the right answers.  Anyways, I have to leave for a flight to Japan in a few minutes...so i'll be quick...

[quote="pkerichang"][quote="matt276eagles"]Here are my solutions.  My answers to #4 and #5 are incorrect.  I would appreciate feedback on what I did wrong (although I'm pretty sure I know for #4  :P ).  Otherwise, I think my solutions are fine, and I'm hoping for at least 21 for a final score of 90.[/quote]
See, I was tempted to say that they are conjugates too. But if there's not a theorem out there that states so or you can't prove it, I will never make that leap.[/quote]
The error in matt276eagles' solution for Problem 4/4/17  is that $x$ could be any rational multiple of the conjugate of $-2\pm\sqrt{t}$, not simply the conjugate itself. So $x=u(-2\mp\sqrt{t})$ for some rational number $u$.

Galois Theory states that a product of an algebraic number (algebraic means that it is the root of a polynomial with rational coefficients, so $x$ is algebraic) and all its conjugates is rational. (Only with square roots do you have just a single conjugate.) And you could also prove that any other factor that gives a rational answer would have to be a multiple of the product of the conjugates. However, I did not expect you to know Galois Theory when I wrote Problem 4/4/17.

Fortunately, a proof of that is not necessary. When matt276eagles got that $x(-2\pm\sqrt{t})$ was rational, he could have set up the equation $x(-2\pm\sqrt{t})=v$ and gotten $x = \frac{v}{-2\pm\sqrt{t}}$. Next he would have rationalized the denominator on the right side by multipling numerator and denominator by the conjugate, getting $x=\frac{v(-2\mp\sqrt{t})}{4-t}$. Since $4-t$ is rational, he could have absorbed it into the unknown rational number $v$ as $u=\frac{v}{4-t}$ to say that $x=u(-2\mp\sqrt{t})$. Same result, but via calculation instead of proof.

So matt276eagles had a good start on solving the problem. He simply narrowed down his options too soon.

Erin Schram
USAMTS grader and problem writer

Here are my solutions.

Could someone look at my #5 and tell me what I did wrong? Was there a way to do it similar to my method? (ie finding something in the cross section and rotating it into the semiminor axis)

Also, my solution to #4 (check it out below) was drastically different from the math jam solution (and was correct except for a minor case I overlooked when $x^2-6=0$).
So, did most people have solutions like that of the math jam? Or, was there a huge variety of different approches to that problem?

Thanks for all the replies - I understand my mistakes now. :lol:

Is there a way to delete my own posts?  Can't believe I actually made a double post =p Internet 101:  Do not click submit twice under any circumstances... ever.

Calc_Rulz, I looked over your solutions, and it looks like you assumed that the length of the median of your trapezoid would be the same as the length of the minor axis (I could be wrong about that, as I didn't read/do out the entire proof).  Your statement about the median being a 90 degree rotation of the minor axis is half-true.  Yes, the median is a 90 degree rotation of the minor axis, but it's not a 90 degree rotation over the axis of symmetry of $\mathcal{C}$.  Rather, it's a 90 degree rotation over the midpoint of segment XY (which is offset from the center), so the minor axis is actually shorter than the median.

As for my solutions, I botched #4, even managed to work my way into a circle (my final partial solution is actually an earlier step, oops!)... Should have started that one sooner =p.  I also didn't explicitly state the optimal strategy for #3, but I have the correct expected value/method, so... who knows if I'll get deducted for that... not that it matters, I'm in the 90-95 range for sure, so I'm happy with that.

Open question:  Did anyone manage to generalize #3?  It looked like it would take some hairy calculus to get a function to spit out optimum values for the number of flips to attempt in each "round" of play.  I'm pretty sure it's possible to find a generalization, since each optimal number of flips depends upon only the optimal number of flips assuming you get a tails, but... I didn't have any luck with it...

In any case, here are my solutions:

Yep, I see what I did wrong, I wasn't thinking that the midpoint of XY is not along the altitude of the cone, so it isn't a perfect rotation. My error is easily solved with some basic pythagorean theorem work, that is, finding the distance between the center of the trapezoid median and the midpoint of XY, then PT from the circle point of view and it's fixed.

Regarding #3, it wouldn't be too hard to generalize except for the fact that you can only have integer-value strategies, meaning that one has to take floor functions of number in the process. However, one could express it recursively (I'm fairly sure the optimal strategy is floor function of $2 + \frac{EV}{100}$, where EV is the EV for the case of one less strike).

Here are my solutions. I thought that I did really poorly, but I guess I really didn't do all that bad. I actually typed up complete solutions to four problems, and I'm pretty sure the answers to those are right. On #4, I was doing something really messy and never even had time to finish the problem. Also, I might not get full credit on #3 for not justifying it enough, maybe.

[quote="SamE"]I'd upload my solutions, but I for whatever reason chose not to use the same format, and I have five different TeX's and PDF's. Anyway, I got the first four and half (?) of the last one; I thought I had it completely correct, but apparently the ellipse doesn't have a minor axis of $7\sqrt2$...I'm too lazy to spend an hour figuring out why.

It was pretty cool though, since I found my errors on #'s 2 and 4 while I was writing them up, but still came up with a valid solution for each (and the correct answer(s)!).[/quote]

Here's a quick explanation why: So you thought you could rotate that segment in the cross section that had length $7 \sqrt{2}$ into place as the minor axis of the ellipse? Think again, you must rotate it along the midpoint of the major axis, which is not down the center of the cone, so it gets distorted a bit.

Aaahhh, I see. Yeah, these geometric assumptions I keep making mess me up all the time.


In the diagram, the dots are spaced 1 cm apart both horizontally and vertically. Express in simplest radical form the number of centimeters in the perimeter of the polygon.
[asy]dot((0,0));dot((1,0));dot((2,0));dot((3,0));
dot((0,1));dot((1,1));dot((2,1));dot((3,1));
dot((0,2));dot((1,2));dot((2,2));dot((3,2));
draw((1,0)--(1,1)--(0,1)--(1,2)--(3,2)--cycle);[/asy]

Find the lengths of each side.

The top two are 1 and 1 each.

The two diagonals below them are each $ \sqrt{1^2 \plus{} 1^2} \equal{} \sqrt{2}$.

The third diagonal is also $ \sqrt{2}$.

The drop is one unit.

Thus we add up these numbers to get $ \boxed{3 \plus{} 3\sqrt{2}}$.

You forgot the horizontal in the 2nd row so 4+3sqrt2


function
inequalities
algebra unsolved
algebra

Find all continious functions f(x), for which the inequalities f(2x) – f(x) ≥ 3x² + x and f(3x) – f(x) ≤ 8x² + 2x are true for all x.

Let $g(x) = f(x) - (x+1/2)^2$

Inequalities are now : $g(3x) \leq g(x) \leq g(2x)$

Then $g(x) \leq g(x/3^n), \forall n \in \mathbb{N}$.

Since $g$ is continuous at $0$ (I think continuity at $0$ suffices), $g(x) \leq g(0)$
Doing the same for the lower bound with $g(2x)$ : g is constant.

Solutions are then : $(x+1/2)^2 + Cste$


My teacher said that there isn't just one way to solve these which is my biggest part of confusion. We are currently simplifying them and finding how to make an identity equal a simplified version of it (proving an identity). Can anybody help me as to where I should begin trig identity problems and efficient methods of solving them?

Well, a lot can be proven using triangles (like sums  of angles). After you find out those, you can sub in the negative and find out differences of angles. Then, you can add 2 of the same angles and get the double angle formula, etc.

For example;
You can prove $ \sin(a \plus{} b) \equal{} \sin{a}\cos{b} \plus{} \cos{a}\sin{b}$. Then, you can substitute $ \minus{} b$ for $ b$, and get differences of angles (of course, you might need to know that $ \cos({ \minus{} b}) \equal{} \cos({b})$ and $ \sin({ \minus{} b} )\equal{} \minus{} \sin({b})$). Then, you can let $ a \equal{} b$ and find the double angle identitity. Also, you can find out the half-angle identities using the same method.

Prove that cot^2/csc=csc-sin

Also, I didn't write theta, so just assume it had theta.

here's a nice one: (it's kinda obvious though)

$ \tan (x)\equal{}\frac{\sin (x)}{\cos (x)}$

[quote="Slippy"]thanks for the help but before I do a proof, what is the best way to simplify a trig identity problem.

My math prof told the class that.....

Sin^2 +cos^2=1
tan^2+1=sec^2
1+cot^2=csc^2

and then the basic cos,sin,tan,cot,csc,sec equalities.

Where do you start, and what do you look for to begin these type of problems?[/quote]
I am really bad at explaining, but the idea with these problems is that you want to try and transform them into an equation you know.  Let's do mathgenius's equation as an example:
\[ \frac {\cot^2 \theta}{\csc \theta} \equal{} \csc \theta \minus{} \sin \theta
\]
Multiplying both sides by $ \csc \theta$, we obtain that
\[ \cot^2 \theta \equal{} (\csc \theta \minus{} \sin \theta)(\csc \theta)
\]
Giving us that
\[ \cot^2 \theta \equal{} \csc^2 \theta \minus{} (\sin \theta)(\csc \theta)
\]
$ \csc \theta$ is defined that $ \frac {1}{\sin \theta} \equal{} \csc \theta$, $ \csc \theta\sin \theta \equal{} 1$.  So,
\[ \cot^2 \theta \equal{} \csc^2 \theta \minus{} (\sin \theta)(\csc \theta) \equal{} \csc^2 \theta \minus{} 1
\]
Add $ 1$ to both sides to get that
\[ \cot^2 \theta \plus{} 1 \equal{} \csc^2 \theta
\]
Notice that this is one of the equations that you know (this is one of your professor's equations.)  So, we transformed mathgenius's equation into one that you know and therefore our solution is complete.

Another good strategy is to get common denominators.  For example, it you have $ 1\plus{}\sin \theta$ in the denominator, then see what happens when you multiply by $ \frac{1\minus{}\sin \theta}{1\minus{}\sin \theta}$.

And if you have both $ \sin$ and $ \cos$, sometimes you might want to express it in the form of just sines or just cosines, by using the identities $ \sin\theta \equal{} \cos(90 \minus{} \theta)$, and $ \cos\theta \equal{} \sin(90\minus{}\theta)$.

In trig $ \sin$ and $ \cos$ seem to be the fundamental units off of which all other identities are made.

Whenever I have an identity handed to me and don't immediately see the proof I convert my given equation into just $ \sin$ and $ \cos$ and try to get a common angle. The proof becomes much clearer after this for me.

Just a LateX sidenote: the correct way to typeset $ \sin$ and $ \cos$ is:

[code]$\sin$ and $\cos$[/code]

Not:
[code]$sin$ and $cos$[/code]

LateX interprets the last one as three variables ('s', 'i' and 'n') stuck together so you get the italicised $ sin$ instead of $ \sin$, which looks a bit weird.

thanks for the help! I am getting better at solving these

[quote="The QuattoMaster 6000"]So, we transformed mathgenius's equation into one that you know and therefore our solution is complete.[/quote]

This is a total aside (I think there's been some very good advice on this thread already, and this is a minor point that is probably not going to be too helpful to the original poster), but this last piece of advice is really not so good.  Being able to transform a given equation into an equation you know to be true doesn't imply that the equation you started with is true.  (There's a discussion of one special case of this going on right now in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172149]this[/url] thread, but the details there are not relevant.)  This only works if each of your steps is reversible.  Now, most arithmetic operations (adding something to both sides, multiplying or dividing both sides by something nonzero) that you're likely to carry out in the process of working with trig identities are reversible (so you don't have to worry about this), and Quatto's advice [i]is[/i] very good as problem-solving advice.  However, in terms of writing up a solution, "best practice" would be to turn Quatto's proof upside-down: start with a statement you know to be true and manipulate it until you derive the desired statement from it.


LaTeX
linear algebra
matrix
search
algebra
polynomial
induction

can someone explain this:

Show that if
$ A \equal{}$
\[ \left(\begin{array}{ccccc} 0 & 0 & \ldots & 0 & \minus{} a_{0} \\
1 & 0 & \ldots & 0 & \minus{} a_{1} \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \ldots & 1 & \minus{} a_{n \minus{} 1} \end{array}\right)
\]

then $ p_{A}(\lambda) \equal{} ( \minus{} 1)^{n}(a_{0} \plus{} a_{1} \lambda \plus{} \ldots \plus{} a_{n \minus{} 1} \lambda ^{n \minus{} 1} \plus{} t^{n})$.

p.s. I also don't know why Latex freaked out on me like this. but that matrix is correct

You entered $ \text{\LaTeX}$ mode twice.  You used the dollar signs and you also used \ [.

(I gave your post a slight edit.)

But the title is wrong - this isn't a question about Jordan form at all. It's a question about a companion matrix, and companion matrices are the building blocks of rational canonical form.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=402690566&t=49817]Here[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=402690566&t=131111]are[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127175]some[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28984]links.[/url]

As for finding the characteristic polynomial, what happens if you try to do that directly? Can you find $ \det(A\minus{}\lambda I)$ by minors of some convenient row or column, with an induction?

hi Kent,

I thought we could only use determinants for square matrices, but $ A$ is some    $ m$ x $ n$. So in that way, the question baffles me because how are you suppose to find $ p_{A}(\lambda)$ for $ A$?

$ A$ is an $ n \times n$ matrix.  Count more carefully.

So I read around.

can someone explain how to find the minimal polynomial? It seems like I have to find the eigenvalues and then place in Jordan form. But i m still confused on how to do that exactly. Does anyone have some advice on this?

thanks

I can use induction by using cofactor expansion to show that characteristic polynomial is actually the char polynomial for that matrix right?

One of the virtues of that polynomial is that you can just read off the characteristic polynomial; just expand by the last column, or expand by rows inductively.

As for why it's also the minimal polynomial- that's in the links.


Let $ x_1 , x_2 , ...,x_n >0$ . Prove that :
\[ \sqrt[n]{x_1x_2 \dots x_n} \leq \frac{\sqrt[n]{n^{n\minus{}1}x_1x_2 \dots x_n(x_1\plus{}x_2 \plus{} \dots \plus{} x_n)}}{\sqrt[n]{x_1}\plus{} \dots \plus{} \sqrt[n]{x_n}} \leq \frac{x_1\plus{}x_2\plus{} \dots \plus{} x_n}{n}\] .

[quote="turcas_c"]Let $ x_1 , x_2 , ...,x_n > 0$ . Prove that :
\[ \sqrt [n]{x_1x_2 \dots x_n} \leq \frac {\sqrt [n]{n^{n \minus{} 1}x_1x_2 \dots x_n(x_1 \plus{} x_2 \plus{} \dots \plus{} x_n)}}{\sqrt [n]{x_1} \plus{} \dots \plus{} \sqrt [n]{x_n}} \leq \frac {x_1 \plus{} x_2 \plus{} \dots \plus{} x_n}{n}
\]
.[/quote]
 Left ineq   is  easy.
  equivalent $ ( \sum_{i\equal{}1}^{n}\sqrt[n]{x_i})  \leq n^{n\minus{}1} \sum_{i\equal{}1}^{n} x_i.$ (  this's ineq Hoder's)
 Right ineg.   Let  $ \sum_{1}^{n} x_i \equal{}1$
  then ineq  become  :
 $ \pi_{i\equal{}1}^{n} \leq \frac{1}{n^n}$


The number of distinct pairs (x,y) of real numbers satisfying x = x^3 + y^4 and y = 2xy is:

A)5
B)12
C)3
D)7

hello, from the equation $ y\equal{}2xy$ we have $ y(1\minus{}2x)\equal{}0$. At the first case we assume $ y\equal{}0$. From here we get $ x(1\minus{}x^2)\equal{}0$ with the solutions
$ x_1\equal{}0,x_2\equal{}1,x_3\equal{}\minus{}1$. Now we assume $ x\equal{}\frac{1}{2}$. Inserting this in the first equation we have $ \frac{1}{2}\minus{}\left(\frac{1}{2}\right)^3\equal{}y^4$ which gives us $ y_1\equal{}\frac{\sqrt{2}}{2}, y_2\equal{}\minus{}\frac{\sqrt{2}}{2}$.
So the answer A) is correct.
Sonnhard.


Four points, Q, W, E, R are collinear and are centers of four coplanar circles having radii of lengths b, 1, 2, 6, respectively, where b is a positive real number less than one.  The circle with center E is tangent externally to the circles with centers W and R, and the circle with center W is tangent externally to the circles with centers Q and E.  If each of the four given circles is externally tangent to a fifth coplanar circle, what is the value of b?

Let A, B, C, D, F be the intersections of the 4 circles with their center line k - B is the tangency point of (Q), (W), C is the tangency point of (W), (E), D is the tangency point of (E), (R). AB = 2b, BC = 2, CD = 4, DF = 12 are diameters of (Q), (W), (E), (R). Let O be a point on the center line k (on the ray opposite to AB), such that in an inversion in O, the circles (W), (E) are carried into 2 congruent tangent circles (W'), (E') centered on the line k and intersecting it at points C', B' and D', C', respectively, i.e., C'B' = D'C'. Let r be an arbitrary radius of the inverion circle (O). By the basic inversion properties,

$BC\ \frac{r^2}{OB \cdot OC} = C'B' = D'C' = CD\ \frac{r^2}{OC \cdot OD}$

$\frac{BC}{OB} = \frac{CD}{OD},\ \ \frac{2}{OB} = \frac{4}{OD},\ \ OD = 2 OB$

In addition, OD = OB + BD = OB + BC + CD or OD = OB + 6. Solving these 2 equations gives OB = 6, OD = 12. It follows that OC = OB + BC = 8, OF = OD + DF = 24 and

$C'B' = \frac{2r^2}{6 \cdot 8} = \frac{r^2}{24},\ \ D'C' = \frac{4r^2}{8 \cdot 12} = \frac{r^2}{24}$

In this inversion, the circle (R) is carried into a circle (R') centered on the line k and intersecting it at points F', D'. Again, by the basic inversion properties,

$F'D' = DF\ \frac{r^2}{OD \cdot OF} = \frac{12 r^2}{12 \cdot 24} = \frac{r^2}{24}$

Therefore, the inverted circle (R') is congruent to the inverted circles (W'), (E'). This means that the existing circle (P) tangent to the original circles (Q), (W), (E), (R) is carried into a straight line p' parallel to the line k. It follows that the inverted circle (Q') centered on the line k and touching the line p' parallel to k is also congruent to the inverted circles (W'), (E'), (R'). Hence,

$AB\ \frac{r^2}{OA \cdot OB} = B'A' = C'B' = \frac{r^2}{24}$

$\frac{AB}{OA} = \frac{OB}{24} = \frac 1 4,\ \ OA = 4 AB$

But OA + AB = OB = 6, hence, 5 AB = 6, $AB = 2b = \frac 6 5,\ b = \frac 3 5$.

The fact that the circle (P) tangent to the circles (Q), (W), (E), (R) is inverted into the line p' parallel to k means that this circle is tangent to the line k at the inversion center O. Choosing the arbitrary radius r of the inversion circle (O) to be $r = \frac{C'B'}{2} = \frac{r^2}{48},\ r = 48$ makes the inversion circle (O) tangent to the line p' parallel to k, hence, the circle (P) is tangent to this particular inversion circle. Since the circle (P) also passes through the center O of the inversion circle (O), its radius is $p = \frac r 2 = 24$.

Excellent job, the value of b is correct, I like your approach.  However, there's a simpler way to do it:

Let points Q, W, E, and R lie on line M, and let T be the center of the fifth circle (the big circle).

Let $TH \bot MH$, where the radius of the fifth circle extended to a point H, making it perpendicular to line M.
Let line TH meet the fifth circle at point F.

If $WH - 1 = x$, and $[TF] = R$, then

$(r+1)^2 - (x+1)^2 = [TH]^2$
$(r+2)^2 - (x+4)^2 = (r+6)^2 - (x+12)^2$
implying that:

$0=2r-6x-12$ and
$0=14r-22x-108$

$x=6$ and $r=24$

$[TH]^2 + (6+1)^2 = (24+1)^2$ --> $[TH] = 24 = r$
Then:
$24^2 + (x - b)^2 = (r+ b)^2$ and plugging in r and x values produces $b= \frac 35$

Very nice. What makes your method better is not that it is simpler (that's a subjective opinion of little value), but that it does not depend on the accidental fact of the big circle being tangent to the center line of the 4 circles (Q), (W), (E), (R). Hence, your method is more general.

The inversion method could be adapted for the more general case - if the inverted circles (W'), (E') are congruent, the inverted circles (Q'), (R') are also congruent. Hence, the diameter of the inverted circle (Q') is known, which is sufficient to calculate the diameter of the original circle (Q).

Yetti.


How would you prove $ b^{r \plus{} s} \equal{} b^{r}b^{s}$ for rational $ r$ and $ s$?

You cannot use induction. So I set $ r \equal{} p/q$ and $ s \equal{} m/n$. Now what? I got $ \left(b^{pn\plus{}mq}\right)^{1/qn} \equal{} b^{r}b^{s}$.

Is this correct?

What is your definition of exponentiation?  Based on your discussion, it appears to be $ b^0 \equal{} 1, b_{n\plus{}1} \equal{} b \cdot b^n$ and $ b^{ \frac{p}{q} } \equal{} \sqrt[q]{b^p}$.  Correct?  In that case, you're fine.


Let I be incentre of triangle ABC. Let D, E, F be arbitrary points of segments AI, BI, CI respectively. The perpenndicular bisectors of AD, BE, CF meet at M,N, O. Prove that the circumcentre of MNO coincides with that of ABC iff I is orthocentre of DEF.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=5761 .

  Darij


AMC
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The point of this game is to put the next word that would come after the previous post in the dictionary, but without actually looking in a dictionary. For example, if the word was "gullible" the next poster might put up the word "gully", and the next poster after might use "gum".

Other notes - ignore plurals and past tenses and stuff like that unless the word is irregular.
Also, don't actually look in a dictionary.
And, don't be illiterate and purposely move the game down the alphabet, otherwise your post will just be ignored.

I'll start with just plain "A"

Aa.

It's a word, I think. Some kind of lava... or something like that.

Well, if we're talking about any reasonable dictionary then it'll include acronyms, so I propose that the next entry is simply [b]AAA.[/b]

well, since im sure AAAA doesnt work, i guess the next word (prefixes count, right?) would be ab-

sorry my voc is poor so I go straight to
[b]abate[/b]

acarophobia

i think thats how you spell it

im pretty sure im missing like 10 words, but

acid?

[b]curve[/b]

Note: The word above me was probably meant to be [i]curvature[/i].

Umm No reason to revive this thread. There is already a revival in which you can play the game in.  :)


Wer wann warum wie wohin fährt (genauer: Bremen + Frankfurt/IMO), darf es hier reinschreiben (ist vor allen an Qwerwulff gerichtet).

nach bremen nehme ich wahrscheilich ein flugzeug, damit ich am vormittag noch zur abifeier kann... alles andere hab ich mir noch nicht überlegt


A sequence of 2005 digits has the property that each pair of consecutive digits forms a two digit number which is either divisible by 17 or 23. If the first digit is 6 and the final digit is not 4, then what is the final digit?

Nice one :)

[hide="Solution"]Such numbers are only $17,34,51,68,85$ and $23,46,69,92$. Starting from $6$, we can form two chains: $68517$, which can't be extended any longer, and $6923469234\dots$, which is periodic with period $69234$. Since both chains are five digits long and $5|2005$, it's obvious that the second chain is repeated $400$ times and that the sequence ends with the first chain. Therefore the last digit is $7$.[/hide]

Cool. I was thinking along those lines as well. So this could be expanded to any series whose length is a multiple of 5?

i.e. a 100,000 digit series?


S is a set of positive integers containing 1 and 2002. No elements are larger than 2002. For every n in S, the arithmetic mean of the other elements of S is an integer. What is the largest possible number of elements of S?

That's correct. I really like this problem since it looks quite difficult until you play with it a bit, and then it suddenly becomes really easy. My high school yearbook group asked me for a hard-looking math problem that was really not very hard. I was going to use this one, but then I never got around to giving it to them.

[quote="ComplexZeta"]it looks quite difficult until you play with it a bit[/quote]

I agree  :!: When I saw this problem I thought it will be something difficult with pigeon-holing or some terrific combinatorics (if not linear algebra) :?

In fact, just write the equations, substract first and last, take a square root, and it's (almost) finished.

A very nice problem indeed.  :mrgreen:

I don't quite understand, could someone please type out a full explanation? Thanks

Ok. HEre is a solution:

[hide]We know that the average of any S-1 elements is an integer. So let two arbitrary elements be a_i and a_k. BNow, let P=sum of all elements.

We know that (P-a_i)/(s-1), (P-a_k)/(S-1) are both integers.

NOw, we know that the difference of 2 integers is also an integer.

THerefore, (a_k-a_i)/(S-1) is an integer.

Now, this menas (s-1)l(a_k-a_i)

Since we want to maxmize s, we want to maximize a_k-a_i.

There fore, (s-1)l2002-1

Since all elements are distinct, (s-1)lsqrt[2001]

From this, we can get that s-1l29, 29 being the largest factor of 2001 less than sqrt[2001].

Thus, s-1=29

s=30.[/hide]


i need to learn how to put pictures into the pdf files, but it states that you should use any program that you want, but i dont what program that would be, all i have is paint, and that does not give the pictures in vectors (i.e. when you zoom, the quality gets worse with paint)

so what program can i get to make the pictures?

I use Macromedia Fireworks to make the pictures. The program is usually used by web designers, but is really good for making diagrams.

[quote="Altheman"]thank you...i found another program that makes constructions but it is a pain to do all of the constructions, you get a compass and a straight edge...i wonder if it converts to pdf or not...i will see[/quote].

You don't actually need to convert your images to PDF; you can use the \includegraphics{} command on a JPG file as well.  (And note that when you do use \includegraphics{}, you don't need to put ".PDF" or ".JPG" at the end of the filename; the LaTeX compiler doesn't pay attention to extensions.)

But if you want your images in PDF format, you can use a virtual PDF printer (basically a driver that pretends to be a printer, but instead outputs the stuff that would normally come out of your printer into a PDF file) like Go2PDF or PDF995.

by the way, a very good free geometry construction program is GCLC(before I use paint....)
u just go to google and type it in, then find something that says about geometric constructions.

Wow, it looks cool! :) http://www.matf.bg.ac.yu/~janicic/gclc/

[quote="K81o7"]Hmm...I should have used this...
I drew my diagram as clearly as possible on graph paper, and scanned it.  :D[/quote]
That's what I do, but since I don't have a scanner, I just take pictures of the page with my digital camera. ;)

lol :)

I use GCLC files (.glc) for my diagrams, so should I just send the .pic files that are exported for LaTeX? (Instead of a GIF file you would usually send...)

"Geogebra" was the closest free alternative to G.S.P i found

I HIGHLY recommend GCLC. It makes beautiful diagrams: the exact same quality as in AoPS books, for example. :) And it's very exact, because instead of "point-click" making the diagram, you just tell it to make a point at $(x,y)$, etc. Try it! (Next round... :D)

edit: it's also free! :)

Since I didn't know about this, I made my diagrams in MSPaint... would it put me at any advantage to redo them in a program like this?

they just look better and they are more accurate

[quote="Boy Soprano"]Also, there is the program KSEG, which is free :  [url]http://www.mit.edu/~ibaran/kseg.html[/url]

I downloaded it today.  It is amazing :  you can save your constructions, drag points around each other and watch the diagram contort itself, and probably do all kinds of things I haven't looked at yet.

Before this, I had nothing but Paint.[/quote]

Are there any online directories that let you download pre-made .seg files of common constructions?


[quote="Fanatic"]Help me out here please:

Is the range of a set of integers with lowest value 5 and greatest value 100 equal to 100-5+1 = 96, or is it 100-5 = 95?

Also, what if the set contains decimal numbers also? Is the range defined the same way or is it undefined because there is an infinite number of decimal possibilites?

Thanks![/quote]
[hide]
To answer your first questino, the range is the difference between the numbers, so it's 100-5 = 95.

I'm not sure what you're asking the 2nd time. The set may contain decimal numbers, but the set has to be defined in some way, if there's infinite possibilities then you can't really calculate anything about it...[/hide]

[quote="Fanatic"]Help me out here please:

Is the range of a set of integers with lowest value 5 and greatest value 100 equal to 100-5+1 = 96, or is it 100-5 = 95?

Also, what if the set contains decimal numbers also? Is the range defined the same way or is it undefined because there is an infinite number of decimal possibilites?

Thanks![/quote]
The range has the same definition with decimals and it is 100-5=95.

the range of a set is the largest minus the least wen it's arranged in order from lowest to greatest, so in this case, would be $100-5=95$

For the 2nd part, i dont see how decimal numbers would give infinite possibilities. They would be treated the same way as integers.

[quote="Fanatic"]Help me out here please:

Is the range of a set of integers with lowest value 5 and greatest value 100 equal to 100-5+1 = 96, or is it 100-5 = 95?

Also, what if the set contains decimal numbers also? Is the range defined the same way or is it undefined because there is an infinite number of decimal possibilites?

Thanks![/quote]

[color=red]For the first part, i am very sure that u do 100-5=95.[/color]

For the second part, I understand what u mean because u could say 5.01, 5.001, 5.0001... [size=150]but[/size],

[color=red]I believe that u treat decimals like integers and just subtract the smallest number from the biggest number.[/color]

[quote="kheg_k"]For the second part, I understand what u mean because u could say 5.01, 5.001, 5.0001... [size=150]but[/size][/quote]For an infinite sequence like this, I think the range is $0.01$, because the minimum of the set approaches $5$.

Ok, thanks.

I thought range was just 100-95 but my Java teacher, on an assigment, defined range as Greatest - Least + 1.

Just wasn't sure on the decimals.

Thanks guys!


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$E$ Frechet space, $B\subset E$ is bounded. $T_{k}\in E'$ and $T_{k}\to 0$ weakly. Prove:
$T_{k}(f) \to 0$ uniformly on $B$.

[quote="fedja"]When you wrote "bounded", did you actually mean "completely bounded"?[/quote]
Bounded means, for any defining seminorm $p$ for the Frechet space,
$p(B)$ is bounded.

I know I can use Banach-Steinhaus theorem to get $T_{k}$ is uniformly equicontinuous. But I can't go further.

[quote="fedja"]Yes, that's exactly what "bounded" means and that's why I'm asking whether you misstated the problem because, as stated, it is obviously false even in the separable infinite-dimensional Hilbert space: just take $B$ to be the unit ball and $T_{k}(x)=(x,e_{k})$ where $e_{k}$ is an orthonormal basis.[/quote]Perhaps the intended condition was [i]totally[/i] bounded (i.e. precompact).

Thanks, guys. I think I missed some features in original question.
I tried to understand a Thm (17.8.6) in "Treatise on analysis" by Dieudonne.

$U$ is an open set in $\mathbf{R}^{n}$, $\mathcal{E}(U)$ is the Frechet space of real-valued function of class $C^{\infty}$.

For each compact subset $K$ of $U$, we denote by $\mathcal{D}(U;K)$ the vector subspace of real
-valued function of class $C^{\infty}$ with compact support contained in $K$: This is
clearly a closed subspace of $\mathcal{E}(U)$, thus a Frechet space too.

we denote by $\mathcal{D}(U)$ the union of all subspaces $\mathcal{D}(U;K)$ as $K$ runs though all the
compact subsets of $U$: this is not a Frechet space in general.

We denote by $\mathcal{D}^{'}(U)$ the vector space of linear forms on $\mathcal{D}(U)$, each form is continuous
when restricted to each $\mathcal{D}(U;K)$. (The space of distributions on $U$). Endow this space
with weak topology induced by $\mathcal{D}(U)$.

Then Thm (17.8.6) saids:
If a sequence $(T_{k})$ of distributions belonging to $\mathcal{D}^{'}(U)$ converges weakly to $T$,
then for each compact subset $K$ of $U$ and each bounded subset $B$ of $\mathcal{D}(U;K)$,
the sequence $(<T_{k},\alpha>)$ converges to $<T, \alpha>$ uniformly on $B$.

So now the question is different than what I asked in the begining: $\mathcal{D}(U)$ is not
a Frechet space here (only a subspace). But $B$ is only bounded in his book,
not totally bounded. Any ideas?

[quote="fedja"]Well, here there is little difference between being bounded and being totally bounded: the idea is that a bounded set in $D(U,K)$ is totally bounded in any $C^{m}$ (just control one more derivative) and that, for any weakly convergent sequence $T_{k}$, we have the estimate $|T_{k}(f)|\le C\|f\|_{C^{m}}$ for all $f\in D(U,K)$ with some fixed $m$.[/quote]

I see, the second assersion is actually a previous Thm in his book, and I can understand.
But can you give details why $B$ is totally bounded in any $C^{m}$ ? How to "control one more derivative"? Thanks.

You can apply the [url=http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theorem]Arzelà-Ascoli theorem[/url] to $f^{(k)}$ using the uniform boundedness of $f^{(k+1)}$.


Total dots is $ 21*3\equal{}63$.  visible is $ 6\plus{}10\plus{}6\equal{}22$.  Not visible is then $ 63\minus{}22\equal{}\boxed{41}$


Nan can mow $ \frac15$ of her lawn in an hour. Reagan can mow $ \frac16$ of the same lawn in an hour. If they have two mowers and work together what fractional part of the lawn can they mow in an hour?

Nan can mow 1/5 in an hour and reagan can mow 1/6 in an hour, so in an hour they  can mow 1/5+1/6=11/30


Let $ABC$ be a triangle . Let $A'$ be the symmetric point of $A$ through $BC$ . $B' , C'$ are defined similarly . Suppose $A'B'C'$ is an equilateral triangle . Find $\angle ABC ,\angle BAC , \angle ACB$ ?

[quote="Megus"]$\Delta ABC$ is also equilateral. To see these consider homothety with center in orthocenter of $\Delta ABC$  and coefficient $-1$. You'll get that $(AB,A'B')$ are parallel and similar for other two pairs.[/quote]

Wrong.

  Darij


Daca se poate cineva sa-mi explice, rezolvarea urmatoarei probleme eu am incercat dar nu imi iasa din pacate :(
A da, este chiar prima varianta, stiintele naturii, subiectul IV,

Se considera functiile \[f,g: (0,\infty)\rightarrow\mathbb{R}\] definite prin
\[f(x)=\ln{\frac{x}{x+1}}\] si \[g(x)=f'(x)\] iar la ultimul punct zice sa se arate k \[\frac{n}{2(n+2)}<\ln{\frac{2n+2}{n+2}}\; \forall n\in\mathbb{N^{*}}\]
M-am gandit sa aplic punctele anterioare, dar nu imi iasa,
uitati si linkul pt cei interesati
 http://subiecte2007.edu.ro/bacalaureat/proba_D/filiera_teoretica/profil_real/specializarea_stiintele_naturii/matematicaM1_2/varianta_001.html
Multumiri anticipate!

[i]Metoda I.[/i]
Păi... folosind punctul e) avem:

$g(2) < f(2)-f(1)$
$g(3) < f(3)-f(2)$
$\vdots$
$g(n+1) < f(n+1)-f(n)$.

Adunând aceste inegalităţi obţinem $g(2)+\ldots+g(n+1) < f(n+1)-f(1) = \ln \frac{n+1}{n+2}-\ln \frac12 = \ln \frac{2n+2}{n+2}$.

Apoi folosim punctul f):
\[g(2)+\ldots+g(n+1) = \frac{n+1}{n+2}-\frac12 = \frac{n}{2(n+2)}. \]
[i]Metoda II.[/i]
Scriem inegalitatea sub forma
\[\frac1{2 \left( 1+\frac2{n}\right)}< \ln \frac{2+\frac2{n}}{1+\frac2{n}}. \]
Este de ajuns să demonstrăm că
\[\frac1{2(1+x)}< \ln \left( 1+\frac1{1+x}\right), \, \forall x \in \left( 0, 2 \right] , \]
inegalitate ce e echivalentă cu
\[\frac{y}2 < \ln (1+y), \, \forall y \in \left[ \frac13, 1 \right) . \]
Cu derivate iese destul de uşor  :football:


$m,n\in N$ prove that
$(2^m-1)^2\mid2^n-1 \Leftrightarrow   m(2^m-1)\mid n$


calculus
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Hi, can you tell me how to calculate these integrals:

1. $ \int^{\pi}_0\frac{dx}{(2\plus{}\cos x)^2}$

2. $ \int^{\pi}_0\frac{\sin ^2x}{a^2\sin ^2x\plus{}b^2\cos ^2x}.dx$

for the first integral if you know something about residue theorem then it will helps you.
for the seccond integral:
$ I(sin) \equal{} \int_{0}^{\pi}\frac {\sin^{2}x}{a^{2}\sin^{2}x \plus{} b^{2}\cos^{2}x}dx \equal{} \int_{0}^{\pi}\frac {dx}{a^{2}\sin^{2}x \plus{} b^{2}\cos^{2}x} \minus{} I(cos)$
 where $ I(cos) \equal{} \int_{0}^{\pi}\frac {\cos^{2}x}{a^{2}\sin^{2}x \plus{} b^{2}\cos^{2}x}dx$
on the other hand $ a^{2}I(sin) \plus{} b^{2}I(cos) \equal{} \pi$
thus you just need to calculate $ \int_{0}^{\pi}\frac {dx}{a^{2}\sin^{2}x \plus{} b^{2}\cos^{2}x}$
for example by this way :
$ \int_{0}^{\pi}\frac {dx}{a^{2}\sin^{2}x \plus{} b^{2}\cos^{2}x} \equal{} \int_{0}^{\pi/2}\frac {dx}{a^{2}\sin^{2}x \plus{} b^{2}\cos^{2}x} \plus{} \int_{0}^{\pi/2}\frac {dx}{a^{2}\cos^{2}x \plus{} b^{2}\sin^{2}x} \equal{} \int_{0}^{\pi/2}\frac {dx/\cos^{2}x}{a^{2}\tan^{2}x \plus{} b^{2}} \plus{} \int_{0}^{\pi/2}\frac {dx/\cos^{2}x}{a^{2} \plus{} b^{2}\tan^{2}x} \equal{} \int_{0}^{\infty}\frac {dz}{a^{2}z^{2} \plus{} b^{2}} \plus{} \int_{0}^{\infty}\frac {dz}{a^{2} \plus{} b^{2}z^{2}} \equal{} \frac {\pi}{2}\left( \frac{1}{a^{2}}\plus{} \frac{1}{b^{2}} \right)$

hello
[hide="1"]
i m not applying limits right now .
let $ A \equal{}\frac {sin(x)}{2\plus{}cos(x)}$
$ A' \equal{} \frac{2cos(x)\plus{}1}{(2\plus{}cos(x))^2}$
$ A'\equal{} \frac { 2}{2\plus{}cos(x)} \minus{} \frac{3}{(2\plus{}cos(x))^2}$
Integrating both the sides -
$ A \equal{} \int \frac { 2}{2\plus{}cos(x)} dx \minus{} \int \frac{3}{(2\plus{}cos(x))^2}dx$
$ \int \frac{dx}{(2\plus{}\cos x)^{2}}\equal{} \frac{2}{3}\int  \frac{dx}{2\plus{}cos(x)} \minus{} \frac{A}{3 }$
for $ \int  \frac{dx}{2\plus{}cos(x)}$
write $ cos(x) \equal{} \frac{ 1\minus{}tan^2 (\frac{x}{2})}{1\plus{}tan^2(\frac{x}{2})}$ and
put $ \tan (\frac{x}{2}) \equal{} t$
u will get $ \frac{ 2 arctan(\frac{tan(\frac{x}{2})}{\sqrt(3)})}{\sqrt(3)}$
[/hide]
thank u

I don't think that this is correct answer  ;
since $ \frac{d}{dx} \frac{ 2 arctan(\frac{tan(\frac{x}{2})}{\sqrt(3)})}{\sqrt(3)} \equal{}\frac{1}{2\plus{}\cos x}$

[quote="Extremal"]I don't think that this is correct answer  ;
since $ \frac {d}{dx} \frac { 2 arctan(\frac {tan(\frac {x}{2})}{\sqrt(3)})}{\sqrt(3)} \equal{} \frac {1}{2 \plus{} \cos x}$[/quote]
Hello Extermal
I have given the answer of integration $ \frac{dx}{2 \plus{} cos(x)}$ not of the integration $ \frac {dx}{(2 \plus{} cos(x))^2}$  :| .I thought u people can do the rest as u just need to put this answer in the below relation. For full answer u need to put this in the equation i have given in my first post . i .e.
$ \int\frac {dx}{(2 \plus{} \cos x)^{2}} \equal{} \frac {2}{3}\int\frac {dx}{2 \plus{} cos(x)} \minus{} \frac {A}{3 }$
u know $ A \equal{} \frac {sin(x)}{2 \plus{} cos(x)}$
So the full answer
$ \equal{} \frac { 4arctan(\frac {tan(\frac {x}{2})}{\sqrt(3)})}{3\sqrt(3)} \minus{} \frac {sin(x)}{3(2 \plus{} cos(x)}$  :)
thank u

1) i think we can substitute t = tan x/2,actually i dont know latex so not posting the solution..

@[b]Extremal[/b]

actually what is the use in doing the second integral like that .we want I(sin x) right??please help me..

Actually, I did not  use any great calculation     in doing second integral.
I just get the linear system of equations like this:
$ a^{2}I(\sin) \plus{} b^{2}I(\cos) \equal{} \pi$ and $ I(\sin) \plus{} I(\cos) \equal{} \frac {\pi}{2}\left( \frac {1}{a^{2}} \plus{} \frac {1}{b^{2}}\right)$
Now I think everyone can solve this simplest problem and can find $ I(\sin)$.
$ I(\sin) \equal{} (\pi\minus{}\pi/2 (1\plus{}b^{2}/a^{2}))/(a^{2}\minus{}1)$