Who solved USACO Jan05 Gold problems.
If U solved please post  here your results.

Here is my results:

                               -- case number --
            1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17
cover       *  *  *  *  *  *  t  *  *  *  *  *  t  *  t  t  *
juice       *  *  *  *  *  *  *  t  t  t  t  t  t  t  t
naptime     *  *  *  *  *  *  *  *  *  *  x  t  t  t

I got:
                               -- case number --
            1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17
cover       *  *  *  t  *  *  t  *  *  *  t  t  t  t  t  t  *
juice       *  *  *  *  t  t  t  t  t  t  t  t  t  t  t
naptime     *  *  x  *  *  x  x  x  x  x  x  x  x  x

I used:
Cover: greedy algorithm solution
juice: naive (slow) solution
naptime: I didn't have much time left so I threw something together, hoping to get a few extra test cases.

I think naptime might have a nice DP solution, though.

cover       *  *  *  *  *  *  *  *  *  x  *  *  x  *  x  *  *
juice       *  *  *  *  *  *  *  t  t  t  t  t  t  t
t

cover: greedy, was very fast (0.00 s for each)... if i add in another greedy thing that sorts the coordinates by the most number of coverings, i'll probably pass all the test cases.

juice: naive (slow) solution, but slightly faster than countersixte :D

naptime: couldnt think of anything good, so didnt bother doing

[quote="bugzpodder"]juice: naive (slow) solution, but slightly faster than countersixte :D[/quote]
:lol: Thanks, bugzpodder

[quote="bugzpodder"]naptime: couldnt think of anything good, so didnt bother doing[/quote]
After the contest, I was thinking about this problem.  If you have one sleeping period, then obviously the most utility you can get is 0.  If you have two sleeping periods, then the most utility you can get is the highest single utility.  To go from n to n+1 sleeping periods, you can extend a contiguous block of sleeping periods (either forward or backward) by one period.  To go from n to n+2 sleeping periods, you could extend twice or you could create a new block containing the highest utility not already used.  This might be the basis of a DP solution, but you would have to deal with cases where the contiguous blocks of sleep might overlap and it just becomes too complicated.  I'm eager to find out what others did for this problem, so if anyone has a good solution, let us know!  ;)

I didn't submit the 'cover' solution during the contest. I had an intuision that a bipartite-matching would work.
Just now, I tried to finalize the bipartite-matching stuff and got correct answers pretty quick on USACO inputs. I also tried to submit a solution to SPOJ Online Judge (http://spoj.sphere.pl) for problem MUDDY ... and got AC there.

The bipartite matching model is:
- Left partition contains all nodes representing all possible horizontal boards
- Right partition contains all nodes representing all possible vertical boards
- The horizontal board x which intersects with vertical board y should be connected with capacity = 1.
- Connect the source node to all horizontal board nodes with capacity = 1.
- Connect all the vertical board nodes to sink node with capacity = 1.
Run the mincut/maxflow ... and we'll get the minimum number of boards.

-turuthok-

[quote="turuthok"]
The bipartite matching model is:
- Left partition contains all nodes representing all possible horizontal boards
- Right partition contains all nodes representing all possible vertical boards
- The horizontal board x which intersects with vertical board y should be connected with capacity = 1.
- Connect the source node to all horizontal board nodes with capacity = 1.
- Connect all the vertical board nodes to sink node with capacity = 1.
Run the mincut/maxflow ... and we'll get the minimum number of boards.[/quote]

Nice.  I wouldn't have thought of that.  Too bad I didn't study the network flow algorithms earlier...  I'm working on USACO problem 'ditch', and for some reason, I can't get it to work entirely correctly.  :(  Nice thinking, though.

[quote="KPO"]I found that thread:
https://www.topcoder.com/rtables/viewThread.jsp?forum=7167&thread=440128&message=440127#440127
which describes a dp solution for the naptime problem...[/quote]

Hey, thanks!   :)  As I suspected, a DP solution.  So you do split it into two cases, just take the best one.  Neat!

[quote name="bestzack66" url="/community/p17303948"]
[quote=BananaeGod]reported @above[/quote]

Why are you just reporting a ton of regular posts? Or was it deleted?
[/quote]

My guess at what happened:

1. Post #9 did not follow the ToS
2. BananaeGod reported it
3. It got deleted


Is the AIME floor value system only for those who took AMC10, or to all 10th graders/below?

What was the score you need for AIME as 9th grader?

Suppose I, as a 9th grader, takes AMC-10B and get 120.5 and qualified to AIME. Somehow, I get 7 out of 15 on AIME. Would I be qualified to USAMO?

possibly. it's a pretty good chance, but a 120.5 rarely correlates to a 7 on the AIME

Usually an 8 qualifies you for sure if you are in 9th or 10th grade.

What if i'm a ninth grader taking the AMC 12. Do they still use the floor value system?

[quote="towersfreak2006"]Feiqi:

[url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html]Source[/url]

"The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value."

Hope that helps.  :)[/quote]

Yes, but they also said you have more advantage to take AMC12. If the system works this way, wouldn't you have more advantage taking AMC10 instead?

AMC people are contradicting themselves :?

[quote="tokenadult"][quote="tetrahedr0n"]It's been discussed in previous threads that this quote from the AMC is not true - in general, regardless of having taken AMC 10 or AMC 12, the floor AIME value is sufficient to 10th graders and under. This was true last year and the previous year.[/quote]

I'm sure that AMC headquarters has found empirically that it's pretty darn hard for 10th graders and under to reach the floor value. More power to AoPS in changing that reality over the next few years, as more younger mathletes get better at the AIME at an earlier age.[/quote]

no man, actually there are more sophomore qualifiers than junior qualifiers.

[quote="towersfreak2006"]Feiqi:

[url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html]Source[/url]

"The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value."

Hope that helps.  :)[/quote]

Well, my interpretation of that statement is that it doesn't apply for those who take the AMC-12.  However, [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamoqualstate.html]here[/url] they say "In addition, students in grades 7th through 10th achieving an AIME score of 7 or above
were also selected to write the USAMO" which would imply something different.

The floor value for the AIME applies based only on your grade level, [i]not[/i] based on what exam you've taken.  There is no disadvantage in taking the AMC 12.


function
real analysis
real analysis unsolved

Let $ M_{k}$ be the set of functions $ f$ in $ C[a,b]$ satisfying the Lipschitz condition ( $ |f(t_{1})-f(t_{2})|\leq K|t_{1}-t_{2}|$ for some constat $ K$).
Prove $ M_{k}$ is closed.

(the metric in $ C[a,b]$ is $ d(f,g)=sup{|f(t)-g(t)|;a \leq t \leq b}$)

You have to show that if $ f_{n}$ is a sequence of functions in $ M_{k}$ which converges to a continuos function $ f$ then $ f$ is also Lipschits.

Start with this:

Let $ t_{1}$ and $ t_{2}$ arbitrary taken in $ [a,b]$. Fix $ \epsilon>0$.

Then for sufficiently large $ n$,

$ |f(t_{1})-f(t_{2})|\leq |f(t_{1})-f_{n}(t_{1})|+|f_{n}(t_{1})-f_{n}(t_{2})|+|f_{n}(t_{2})-f(t_{2})|\leq 2\epsilon+K|t_{1}-t_{2}|$

since $ \epsilon$ is arbitrary taken the result follows.


There is a football totalizator. It is allowed to stake on the results of n games. What the smallest number of variants of forecast should we make to be sure that there exist a variant with all n incorrect results? (Every game can be finished by any of three results: winning of the hosts, winning of the guests and draw.)

This problem for n=3 was proposed at the 3rd (regional) stage of All-Russian Mathematical Olympiad 2004. Author: S. Tokarev.

I don't know the solution for arbitrary n. It is not hard to obtain a lower bound. But there are some problems with upper bounds. But I have a conjecture that my lower bound is exact.

Also we may assume that every game have not three but k possible results. And so the question is the same.


symmetry
geometry
geometric transformation
geometry theorems

The spilar similarties is a new chapter for me but i saw this in some problems, can somebody help me? :oops:

The spiral similarity is the product of a dilatation and a translation.

Here's a problem:
If squares, with center $O_{1}$,$O_{2}$,$O_{3}$, are erected externally on the sides $BC$,$CA$,$AB$ of the triangle $ABC$, then the line segments $O_{1}O_{2}$ and $CO_{3}$ are equal and perpendicular.

Yes I know this result(Von Aubel) and i can show this, but how do you apply spiral similarly for example in this problem?

Here is a diagram, when I learn more about spiral symmetry I will post proof :)


geometry
geometric transformation
rotation
analytic geometry
MATHCOUNTS

The triangle with vertices A (6, 1), B (4, 1), and C (4, 4) is rotated 90 degrees counterclockwise about B.  What are the coordinates of the image of C (the point where C is located after the rotation)?

This looks like a mathcounts problem... I would draw it out..  :)

[quote="Ars"][hide](1,1) I drew it[/hide][/quote]

Yeah, that's what i got, pianoforte did point A for us too  :D

[quote="236factorial"]

Yeah, that's what i got, pianoforte did point A for us too  :D[/quote]

oops! :P I guess I should of drawn it out.


Let's calculate $(\sqrt{a}-\sqrt{b}\ )^{2}$ for $a>b>0.$

Using $(\sqrt{a}-\sqrt{b})^{2}=a+b-2\sqrt{ab}=35-10\sqrt{10}$ we make a system of equations.  Since $a,b$ are integers, we know that:
$a+b=35$ and
$2\sqrt{ab}=10\sqrt{10}\Longrightarrow ab=250$.
From here, it is not difficult to see that $(a,b)=(10,25)$.  Then, because the square root must be positive, we have that $a>b$ so
$\sqrt{35-10\sqrt{10}}=\sqrt{25}-\sqrt{10}=5-\sqrt{10}$.

[quote="DiscreetFourierTransform"]Could you be all like
$\sqrt{35-10\sqrt{10}}= x$
Then be all like
$x^{2}-35 =-10\sqrt{10}$
And then
$x^{2}-35+10\sqrt{10}= 0$
And solve that quadratic for X?
Wouldn't that be the simplest solution?[/quote]

How would you solve that quadratic?  :huh: It has irrational coefficients...

Well, if you have $x^{2}-35+10\sqrt{10}= 0$
Then you do

$x = \frac{+-\sqrt{4(35-10\sqrt{10})}}{2}$
Which then can be simplified into $x =+-\sqrt{35-10\sqrt{10}}$ which is what we were looking for in the first place.

Of course, because that just makes the problem so much simpler :P

Oh. Guess I was pwnt. Hehehe. Posted before my brain had time to catch up. That's happening a lot to me ;-). Sorry for misleading info.

can someone go over this step-by-step? I don't quite understand.

samaths explanation was quite thorough, just read over it a couple times if you dont understand and it should click ;)


Euler
modular arithmetic
number theory unsolved
number theory

Prove, that for every $m\geq 2$ and $x\in\mathbb{Z}$ the congruence holds: $x^{m}\equiv{x^{m-\varphi(m)}\pmod{m}}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=123069
just a little below...


algebra
polynomial
number theory
extremal principle

I've recently gotten Engel's [i]PSS [/i]and Barbeau's [i]Polynomials[/i], and I'm wondering what you guys do when you're goinig through the books. Obviously if you've got a long period of time (like summer holidays) you'd want to thoroughly go through the whole thing chapter by chapter, but what if you've got a short amount of time, say, two weeks till a camp or a comp and you want to improve as much as you possibly can in that amount of time? Is it better to skim through each topic in PSS and get a bit of each, or thoroughly understand a few, as much as you can? [i]Polynomials[/i] is more sequential so I'll probably work on that in order and get all the basics first, but what about Engel?

What do you tend to do?

On principle, I would want to go into each chapter in detail, in order to understand things properly. However, I found that when I first got Engel (a week or so before a competition) and I 'skimmed through it' over a couple of days, I found that it actually helped a great deal, more than working through it carefully would have done. Of course, after the competition, I could go back over it more carefully.
If the camp/comp coming up is very important (e.g. you hope to qualify for later rounds/camps by doing well) then take the broader choice. If you can carry on improving after the camp, then start doing things in depth.

On a more general note, I prefer to read through examples and explanations from Engel, but to do questions from past papers (in the resources section or on kalva) rather than questions from the Engel chapters. I don't know why, but that is the way I prefer to work.

Many of the Engel questions are misleadingly easy, in that they usually only seem to require one or two interesting ideas to solve. They certainly led me into a bit of a false sense of security that my Olympiad training was going well (hacking down the Number Theory questions at about 30 per hour), but really it's just because they aren't [i]that[/i] hard. I certainly suggest doing some to get used to the concepts (especially the earlier chapters - for colouring, invariance and extremal principle they provide excellent examples of different places in which this technique crops up), but strictly as part of a balanced diet of real Olympiad class problems.

Yeah, I've definitely got time after, and the camp's important. This one's just training, and it's acceptable for first time seniors (of two categories) to not do that well.

I know the basics of most concepts, but I'll go through them and try examples. Thanks, you guys.


Ellen has five different jobs to be done. She assigns all five jobs to her four kids. Each kid will have at least one job. How many ways can Ellen assign the jobs?

There are $ 5\cdot4\cdot3\cdot2$ ways to assign jobs to the four children such that they are guaranteed to have at least one job. Then, there are $ 4$ children to give the last job to. Finally, we have to divide by $ 2$ to correct for overcounting the number of ways for the child with $ 2$ jobs, since the order obviously does not matter here. This gives us $ \frac{5\cdot4\cdot3\cdot2\cdot4}2\equal{}5\cdot4\cdot3\cdot4\equal{}\boxed{240}$ ways.


geometry
rectangle
function
inequalities
floor function
inequalities unsolved

Hint: How about using some geometry? :wink:
Draw a square with side $k$ and divide some of its sides in segments $a,A,b,B,c,C$ then do some calculations of areas of rectangles, and you will be finished.

Or maybe one more easier solution:
 Write the left side as:
 a*(k-b)+b*(k-c)+c*(k-a)=(a+b+c)*k-(a*b+b*c+c*a)
 which is linear function for variables a,b,c so it has its maximum and minimum when all of a,b,c are 0 or k.After discussing same cases we get that maximum is k^2 q.e.d.

[quote="Fitim"]Let $a,b,c,A,B,C$ are non-negative reals numbers such that:
$a+A=b+B=c+C=k$
Prove that: $aB+bC+cA\leq k^{2}$[/quote]

 We have $k^{3}=(a+A)(b+B)(c+C)$
                     $=(ab+aB+Ab+AB)(c+C)$
                     $= abc+acB+Abc+ABc+abC+aBC+ACb+ABC$
                     $=aB(c+C)+bC(a+A)+cA(b+B)+abc+ABC$
                     $\geq k(aB+bC+cA)$
QED :D

[quote="Fitim"]Let $a,b,c,A,B,C$ are non-negative reals numbers such that:
$a+A=b+B=c+C=k$
Prove that: $aB+bC+cA\leq k^{2}$[/quote]

Such a beautiful inequality! I can't restrain its temptation. Here is its form for 4 variables which I have proved:
Let $a,b,c,d,A,B,C,D$ are non-negative real numbers such that:
$a+A=b+B=c+C=d+D=k$
Prove that $aB+bC+cD+dA \leq 2k^{2}$

[quote="bel.jad5"]we consider triangl ABC (AB=k) equilateral and X in AB such  AX=a ,  Y in BC such  BY=c , Z in AC such  CZ=b
we have S(AXZ)+S(BXY)+S(CYZ)<=S(ABC)
 S(AXZ)+S(BXY)+S(CYZ)=0.5sin(60)(aB+bC+cA)
S(ABC)=0.5sin(60)k²
so aB+bC+cA<=k²[/quote]

 I have to say it is a wonderful solution. Certainly, for 4 variables (my post before), I have found that using square (geometry method) is the best aproaching.
  Thank you for your beautiful solution! :)

[quote="Fitim"]Let $ a,b,c,A,B,C$ are non-negative reals numbers such that:
$ a \plus{} A \equal{} b \plus{} B \equal{} c \plus{} C \equal{} k$
Prove that: $ aB \plus{} bC \plus{} cA\leq k^{2}$[/quote]
Obviously $ A \equal{} k\minus{}a$, $ B \equal{} k\minus{}b$, $ C \equal{} k\minus{}c$ and $ a,b,c \in [0;k]$.

Let $ x \equal{} \frac{a}{k}$, $ y \equal{} \frac{b}{k}$, and $ z \equal{} \frac{c}{k}$, then it is equivalent to:
$ f(x,y,z) \equal{} xy\plus{}yz\plus{}zx \minus{} x \minus{} y \minus{} z \plus{} 1 \ge 0$ when $ x,y,z \in [0;1]$.
But $ f(x,y,z) \equal{} x(y\plus{}z\minus{}1) \plus{} (y\minus{}1)(z\minus{}1)$.
Case $ y\plus{}z > 1$. Then $ f(x,y,z) \ge (y\minus{}1)(z\minus{}1) \ge 0$ with equality iff $ x\equal{}0$ and $ y\equal{}1$ or $ z\equal{}1$. So all permutations of $ (0,k,1)$ where $ 1 \ge k \ge 0$.
Case $ y\plus{}z < 1$. Then $ f(x,y,z) \ge (y\plus{}z\minus{}1) \plus{} (y\minus{}1)(z\minus{}1) \equal{} yz \ge 0$. With equality iff $ x \equal{} 1$ and $ y\equal{}0$ or $ z\equal{}0$. So all permutations of $ (0,k,1)$ where $ 1 \ge k \ge 0$.
Case $ y\plus{}z \equal{} 1$. Then $ f(x,y,z) \equal{} (y\minus{}1)(z\minus{}1) \ge 0$ with equality iff $ y\equal{}1$ or $ z\equal{}1$. Since $ z\plus{}y\equal{}1$ we get equality for the same as before.

For 4 vars generalisation giving by Gibbenergy, it is equivalent to:
$ f(x,y,z,w) \equal{} xy \plus{} yz \plus{} zw \plus{} wx \minus{} x \minus{} y \minus{} z \minus{} w \plus{} 2 \ge 0$
Then $ f(x,y,z,w) \equal{} (x\plus{}z)(y\plus{}w\minus{}1) \minus{} y \minus{} w \plus{} 2 \ge 0$.
If $ y\plus{}w > 1$ then we get $ f(x,y,z,w) \ge 2 \minus{} y \minus{} w \ge 0$ with equality iff $ x\equal{}z\equal{}0$ and $ y\equal{}w\equal{}1$.
If $ y\plus{}w < 1$ we get $ f(x,y,z,w) \ge 2(y\plus{}w\minus{}1) \minus{} y \minus{} w \minus{} 2 \equal{} y \plus{} w \ge 0$ with equality iff $ x\equal{}z\equal{}1$ and $ y\equal{}w\equal{}0$.
If $ y\plus{}w \equal{} 1$ we get $ f(x,y,z,w) \equal{} \minus{} (y\plus{}w) \plus{} 2 \equal{} 1$, so no equality.
Thus equality iff $ x\equal{}z\equal{}0$ and $ y\equal{}w\equal{}1$ or $ x\equal{}z\equal{}1$ and $ y\equal{}w\equal{}0$. QED :P

I conjecture:
Let $ a_1 \plus{} A_1 \equal{} a_2 \plus{} A_2 \equal{} ... \equal{} a_n \plus{} A_n \equal{} k$. where $ a_i,A_i \ge 0$.
Then $ a_1A_2 \plus{} a_2A_3 \plus{} ... \plus{} a_nA_1 \le \left \lfloor \frac{n}{2} \right \rfloor k^2$


a hydrocarbon A($ \ce{C6H10}$) on reduction by $ \ce{H2 /Ni}$,$ \ce{H2}$ Lindlar's Catalyst and $ \ce{Na, liq.NH3}$ gave three different products B,C,D respectively.A on further reacting with $ \ce{NaNH2}$ in inert atmosphere gave a salt C which reacted with an alkyl Iodide to form F which on oxidation gave pentanoic acid as it's only product while one of the products obtained on oxidation of D was Butanoic Acid.Identify A-F with necessary reactions

[hide="Answer"]The index of hydrogen deficiency of A is 2, and conjugating this information with the reactions that were made with A, we may conclude that A is an alkyne. As the only product obtained from F is pentanoic acid, then the structure of F must be

$ F: \,\, CH_3(CH_2)_3 \minus{} C \equiv C \minus{} (CH_2)_3CH_3$.

Since by oxidation D gave butanoic acid (and another product) then

$ D: \,\, E \minus{} CH_3CH_2CH_2CH \equal{} CHCH_3$,

and this is the E isomer because by reduction of an alkyne with an alkali metal and ammonia we get a trans alkene. Also, the hydrogenation of an alkyne with Lyndlar catalyst yields a cis alkene, and so

$ C: \,\, Z \minus{} CH_3CH_2CH_2CH \equal{} CHCH_3$.

Is is now obvious that $ A: CH_3CH_2CH_2 \minus{} C \equiv C \minus{} CH_3$. Also, since from the reduction of an alkyne with hydrogen and a pure transition metal we get an alkane, then

$ B: CH_3CH_2CH_2CH_2CH_2CH_3$.

By reaction with sodium amide, A isomerizes to the terminal alkyne:

$ CH_3CH_2CH_2 \minus{} C \equiv C \minus{} CH_3 \plus{} NaNH_2 \longrightarrow CH_3(CH_2)_3 \minus{} C \equiv C \minus{}Na$

(this is the acetylene zipper reaction that I have mentioned in another post). Now reating this with n-butyl iodide w get compound F.

Question: who is the product E?[/hide]

E is missing  :rotfl:  must report to BanishedTraitor.He knows Holmes  :D (see his signature)


Find the conditions for the system shown in the figure to be in equilibrium.

Please show some of your own work.

1) What are the forces applied to each body?

2) What are the conditions for a system to be in equilibrium?

Well, to be in equilibrium, the tensions in the rope has to be the same everywhere, but then, how can it be possible, since the forces acting on the pulley on the right do not cancel out...


Observemos que \[ \dfrac{\mathrm{d}}{\mathrm{d}x}x^{x^a}\equal{}x^{x^a\plus{}a\minus{}1}(a\log x\plus{}1).\] Luego \[ \int x^{x^a\plus{}a\minus{}1}(a\log x\plus{}1)\mathrm{d}x\equal{}x^{x^a}\plus{}C,\] y en consecuencia \[ 15\equal{}\int_1^2 x^{x^a\plus{}a\minus{}1}(a\log x\plus{}1)\mathrm{d}x\equal{}2^{2^a}\minus{}1.\] Por tanto $ 2^{2^a}\equal{}16$, es decir $ a\equal{}2$ es la única solución.


Miss James has a license plate with six digits.  If commas were placed after the second and fourth digits, it would look like the number is a list of three consecutive even numbers in increasing order.  The first of those numbers has three different prime factors.  What is the least six-digit number that can be on Miss James' license plate?

We have something that looks like __ __, __ __, __ __.  We know that the first set of $ 2$ creates an even 2 digit number that has 3 different prime factors.  Since we want to make the number as small as possible, we'll minimize our prime factors.  The 3 smallest primes are $ 2, 3 \text{ and } 5$.  So that means our first set of $ 2$ is $ 2\cdot3\cdot5\equal{}30$.  Now that we know our first set, we can figure out that our second set is $ 32$ and our final set is $ 34$.  So our final answer is $ \boxed{303234}$


[quote="bodan"]Let $r_j,s_j,t_j,u_j,v_j$ be real numbers not less then 1, for $i=1,2,\ldots,n$ and let $R,S,T,U,V$ be the respective arithmetic means of the $r_j,s_j,t_j,u_j,v_j.$ Prove that \[ \prod_{j=1}^{n}\frac{r_js_jt_ju_jv_j+1}{r_js_jt_ju_jv_j-1}\leq\left(\frac{RSTUV+1}{RSTUV-1}\right)^n.  \][/quote]
Sorry, your inequality is wrong.
It should be \[ \prod_{j=1}^{n}\frac{r_js_jt_ju_jv_j+1}{r_js_jt_ju_jv_j-1}\geq \left(\frac{RSTUV+1}{RSTUV-1}\right)^n.  \]
it's a chinese TST problem that is vey old.


calculus
integration
combinatorics proposed
combinatorics

Peter has three accounts in a bank,each with an integral number of
dollars. He is only allowed to transfer money from one account to
another so that the amount of money in the latter is doubled.
  (a) Prove that Peter can always transfer his money into accounts
  (b) Can Peter always transfer his money into one account?

The solution is avaliable on Kalva website.
But I want to find the different way.

The answer for (b) is easier: the answer is NO. We just take the case 0,1,2 (he has 0 dollars in the first account, 1 dollar in the second and 2 dollars in the third).

Just to clarify, part (a) is "Show that by a series of transfers he can always empty one account"; otherwise if we started with 0, 0, 1 there would be nothing to talk about.

:)  :)  :)
oh ,yes
Obviously , Peter has at least two account > 0


Say we're given a problem like $\sqrt{a+\sqrt{x}}= x$. We square a couple times to get a quartic in $x$. How do we know if we've added extraneous roots?

In general, is there a way to know if you've added extraneous roots?

You can always expect extraneous roots when squaring, and there's no other way to eliminate them than to plug them back in by hand and see which ones work. (Of course, "plugging back in" is often obvious, as when you get complex solutions or a negative quantity whose square root is about to be taken.)

You don't always get extraneous roots though, do you?

So suppose I want to know if I have any extraneous roots given an algebraic expression like above...how would I do it?

squaring the first time, you must restrict $x\ge 0$

the next time, you must restrict $x^{2}-a\ge 0$

this should give all roots that work

You can check for extraneous roots by just plugging back in and checking all of your solutions.


calculus
derivative
LaTeX
search
calculus computations

Hi Guys,
How would I go about solving the problem:

find dy/dx if $y=f(x)=(x^2-1)^4 / (5x+6)^3 ???$

The problem is to find the derivative using the chain rule... It seems to me like you would want to use the quotient rule here but who knows...  How would I go about solving this using the chain rule, the exponents are throwing me...

Thanks in advance!
Chris

$y=f(x)= \dfrac{(x^2-1)^4}{(5x+6)^3}$

Let's let $u = (x^2-1)^4$ and $v = (5x+6)^3$

The formula for the quotient rule would be

$f'(x) = \dfrac{u'v - v'u}{v^2}$

$u' = 4(x^2 - 1)^3 (2x) = 8x(x^2 - 1)^3$ ...this is the application of the chain rule

$v' = 3(5x+6)^2 (5) = 15(5x+6)^2$ ....another application of the chain rule, then simplifying

$v^2 = (5x+6)^5$

Now that we know each of the values, we can just plug it in....

good job with the LaTeX also, you're a fast learner

Well, you do have to use either the quotient rule, or the product rule, or logarithmic differentiation. Do give logarithmic differentiation a look - it's probably the best method here.

Here's the product rule:

$f(x)=(x^2-1)^4(5x+6)^{-3}.$

$f'(x)=\left[4(x^2-1)^3(2x)\right](5x+6)^{-3}+(x^2-1)^4\left[(-3)(5x+6)^{-4}(5)\right]$

$=(x^2-1)^3(5x+6)^{-4}\left(8x(5x+6)-15(x^2-1)\right)$

$=(x^2-1)^3(5x+6)^{-4}(25x^2+48x+15)$

Leaving it as factored as possible is frequently a good idea. That depends on the next question, but it could be a search for critical points, in which case you want it factored.

The chain rule in this is how you differentiate pieces such as $(x^2-1)^4$. If $g(t)=t^2-1$ and $h(t)=t^4$, $(x^2-1)^4=h(g(x))$.

Ankur,
   Thank you very much for your comments.  I think I see it now.

Here we go:

so from what you said,

$f'(x) = {8x(x^2-1)^3 * (5x+6)^3 - 15(5x+6)^2 * (x^2-1)^4}/(5x+6)^6$

$= 8x(5x+6)^3 * (x^2-1)^3 - 15(5x+6)^2 * (x^2-1)^4$

so,

$f' = (x^2-1)^3 * (25x^2 + 48x + 15) / (5x+6)^4$

Edit: Kent & Jmerry sorry I didn't your replies before I posted


calculus
derivative
function
real analysis
real analysis unsolved

let $f(x)=\left(1+x\right)^{\frac 1 x}$
can we prove
$f^{(n)}(x)>0$ for n is even.
and $f^{(n)}(x)<0$ for n is odd.

Yes, we can. Note that it is equivalent to saying that all the coefficients in the Taylor expansion of $f(t-x)$ considered as a function of $x$ for fixed $t$ are non-negative. We shall start with the case $t>0$. Then
\[ \begin{aligned}f(t-x)&=\exp\left\{\frac 1{t-x}\ln(1+t-x)\right\}=\exp\left\{\left(\sum_{k=0}^\infty \frac{x^k}{t^{k+1}}\right)\cdot\left(\ln(1+t)-\sum_{\ell=1}^\infty \frac{x^\ell}{\ell(1+t)^\ell}\right)\right\}\\&=\exp\left\{\sum_{m=0}^\infty\gamma_m x^m\right\}\end{aligned} \]
where
\[ \begin{aligned} \gamma_m&=t^{-m-1}\left(\ln(1+t)-\sum_{\ell=1}^m\frac 1\ell\left(\frac{t}{t+1}\right)^\ell\right)\\&>t^{-m-1}\left(\ln(1+t)-\sum_{\ell=1}^\infty\frac 1\ell\left(\frac{t}{t+1}\right)^\ell\right)=0 \end{aligned} \]
It remains to note that the exponent of a power series with positive coefficients is again a power series with positive coefficients.

Now, once we know it for some fixed $t$, we know it for all smaller $t$ just because redecomposing a power series with positive coefficients at a positive point, we still get a power series with positive coefficients. This observation takes care of all $t>-1$ finishing the proof.

[quote="fedja"]

Now, once we know it for some fixed $t$, we know it for all smaller $t$ just because redecomposing a power series with positive coefficients at a positive point, we still get a power series with positive coefficients. This observation takes care of all $t>-1$ finishing the proof.[/quote]
very nice,fedja. :)
but I don't know this :(
can you explain more,
thanks

Sure. Let's put it another way: if $g(x)=\sum_{k=0}^\infty c_k x^k$ and all $c_k>0$, then, for any $x_0>0$ in the (open) interval of convergence of the series, and any $n\ge 0$, we have $g^{(n)}(x_0)>0$. Indeed, this is true (with $\ge$ instead of $>$) for every term in the series (it is enough to check it for pure powers) and, therefore, for the entire series because we can differentiate it term by term.

Hope this helps :)


Yeah yeah I shoulda made study camp (or worst case of high honors) this year, but I was an idiot and did all of my free response on the scrap paper, not leaving enough time to finish transferring the second half of my responses onto the answer sheet.

I want to make sure I maintain this though...I have Atkins and I'll be reading that the whole year (the text now, not the study guide). I'm also taking organic chemistry at my community college next year. Is there any other book I should get to keep my theory sharp?

You weren't able to turn in the scratch work as your official answer?  Sorry to hear that.

Anyway, your plan of action sounds sufficient for making the study camp, or as sufficient as any method could be.  If you plan on shooting for IChO, don't forget to do the prep problems and look over the syllabus.

by prep problems you mean the ones in Carey and Atkins or just my cache of old IChO problems you know about, or both? (I'm thinking you meant the latter third).

neither--each year, the host country of IChO publishes a set of preparatory problems.  they can be found online around february the year of the competition.  also, im planning on writing a guidebook of sorts for chem olympiad this summer, if it turns out well that could be helpful.

And what books would you recommend to help do the prep problems. Some of them are really hard and I can't do them. Could you please give examples of books for each subtopic of chemistry such for inorganic chem, biochem, physical chem, and analytical chem.

Hmmm... well, I don't have enough expertise to know which books are best.  I can, however, tell you which books are (well have been) used at the US Study Camp.

Organic--Carey
Inorganic--Measler and Tarr
Biochem--Voett, Voett, Pratt
P-Chem--Atkins (although I personally prefer Silbey, Alberty, Bawendi)
Analytical Chem--Skoog, West, Holler, Crouch

Hope that helps!

Keep in mind that none or very few of these problems should be immediately obvious; just keep working at them and ask questions on here or of professors if you need help.

Greg,

Do you know how they select US team at the camp? What topics will the exams test against at the camp? Will they just cover the contents they have gone through in the training sessions?

Thanks!

cl,

I actually have written a document containing several pages worth of description of the camp.  If you pm me your email address, I can send it to you.  To answer these specific questions:

Do you know how they select US team at the camp?
There are a total of 3 exams and about 6 or 7 labs, all of which are scored.  Exams count for 60% of the score and labs 40%.  The team is comprised of the top four scorers.

What topics will the exams test against at the camp?  Will they just cover the contents they have gone through in the training sessions?
Anything covered in lecture (or assigned as reading, especially the orgo) is fair game on the exams.  Also, anything covered in the prep problems, or any level 1 or 2 topic on the syllabus, can show up on an exam.  Usually, the first test is comprised of topics covered in lecture, while the final has a large percentage of topics mentioned only in the prep problems.  (Of course, most of what is covered in lecture is covered in the prep problems.)

Might I ask whether you are a qualifier this year?

I AM...a qualifier...
cl are you a qualifier?

greg: what are the free times like? like, is the time spent during camp devoted solely to studying (-_-) or is there a lot of free time after classes? Also, do they assign problems to do for homework?

Thanks!

Also, I wouldnt mind a copy of that document of yours...hehe

Go ahead and pm me your email address, and I will send it to you.  To answer your questions:

what are the free times like? like, is the time spent during camp devoted solely to studying (-_-) or is there a lot of free time after classes?
Well, the ultimate hope of the camp is that you will spend as much of your free time studying as is humanly possible.  Some people choose to do this, some people do not--at any given time, you can generally find enough people to play a game of cards, for example.  Usually, the people who qualify for IChO spend large amounts of free time studying, but not always.  The only time that is really free on weekdays is from after dinner until bedtime (I can't remember the specific times... 6 to 11 maybe?).  On Saturday, most of the time is taken up by organized activities (often, to which a few campers bring their books), and for the most part Sunday is entirely free.

The general rule of thumb is that breaks from studying are okay, but if the mentors see you spending too much time not studying they might say something.  At no point will they force you to study, however.

Also, do they assign problems to do for homework?
Yes, but none of it is graded or collected.  The point of assigned problems is to give you practice for what will be on the exams.

I took USNCO, but did not qualify for the chem camp this year, but I want to next year. I have Linus Pauling's "General Chemistry" that I have been working through and I think it's pretty useful. The topic poster mentioned Atkins? Would that be helpful, in addition to Pauling? (But I think that Pauling's book is very "hard," somehow.)

Also, Greg, I wouldn't mind a copy of that document, even though I didn't make camp, just for fun :P Should I PM you my email address?

Thanks!!

Nathan

Yes, anyone who wants the document is welcome to PM me with where it should be sent.  About the textbook--I have heard that Pauling's book is very good, but have not actually used it myself.  Since I have little experience with it, I would be curious as to what sort of an experience you have had with it.  Is it up to date?

The book that I used to study for the USNCO National Exam was Atkins.  I think that it is probably preferable over something like say, Zumdahl, or a high school chemistry book.  I can say with certainty that if you are able to do everything in Atkins, you will have no problem on the USNCO National Exam.

[quote="gregx007"]The book that I used to study for the USNCO National Exam was Atkins.  I think that it is probably preferable over something like say, Zumdahl, or a high school chemistry book.  I can say with certainty that if you are able to do everything in Atkins, you will have no problem on the USNCO National Exam.[/quote]
I am currently using Zumdahl to prepare for the olympiad, but I was wondering what is the exact title of the Atkins book you're talking about.

Also, having used Organic Chemistry by Carey, would you recommend it?

Thanks.

The book I was referring to:  Peter Atkins and Loretta Jones's Chemical Principles: The Quest for Insight.  Also, Carey is a pretty good book, although I personally prefer Brown and Foote's Organic Chemistry.  Either one will do the job though--and if you make the Study Camp, you will have to read Carey anyway, so it might be a good bet to use it from the start.


I am searching for a math site(s) or an easy math book that explains the world of polar math at the pre-calculus level.

Any links or book titles?

Thanks

If you mean math using polar coordinate systems, try the Wikipedia page:

[url]http://en.wikipedia.org/wiki/Polar_coordinates[/url]

Also, try any standard around-10th-grade-level trigonometry textbook.


algebra
polynomial
irrational number
number theory unsolved
number theory

Hi.
What are the polynomial that send rational number to rational number ?
Clearly, if all the coefficients of the polynomial are rationnal numbers, then the polynomial send $ \mathbb{Q}$ into $ \mathbb{Q}$.

What about the converse ?
Can we have $ P(X)\equal{}\sum_{k\equal{}0}^{n} a_k X^k$ with some $ a_k$ irrational number such $ P(\frac{p}{q}) \in \mathbb{Q}, \forall \frac{p}{q} \in \mathbb{Q}$
I think the answer is no.
But how to prove it ?

Thanks for your help.

[quote="N.T.TUAN"][quote="zebulon"]

What about the converse ?
[/quote]
Right! It is true by Lagrange's Interpolation Formula.[/quote]
You can use Abel express :
$ f(x)\equal{}a_n(x\minus{}1)..(x\minus{}n)\plus{}...\plus{}a_1(x\minus{}1)\plus{}a_0$

Thank you very much for your help.
It works !


Consider the triangle ABC, with D on the side BC and E on the side AC. Supposed that AD and BE are angle bisectors, and AD=BE. Show that triangle ABC is isosceles.

here you go folks
this doesnt feel right to me

plz help me solve it!

[quote="shobber"]This problem happens to be correct.

You have two simple ways to solve it: One is assume the contrary, the other one is to do a search with highlight。[/quote]

well if you dont mind can u tell me how?
i'm really really dumb and slow so somebody teach me plz!

actually
i did it already
using almost the same method

thx a lot!!!

Theres a very quick solution using areas and contradiction, but i dont quite remember it...

hmm...That's strange...I thought the most popular method is to prove the contrapositive, namely that if the triangle is not isoceles then the angle bisectors will have different lengths.

hey pker

dats how i did it =P

hand shakes


integration
trigonometry
calculus
calculus computations

$ \int_{0}^{n} x\sqrt{1 \plus{} |cosx|}dx$

Hint
$ x\sqrt{1 \plus{} |\cos x|}
\equal{}\left\{\begin{array}{ll}
x\sqrt{1 \plus{} \cos x} & (4i\minus{}1)\frac{\pi}{2} \le x \le (4i\plus{}1)\frac{\pi}{2} \\
x\sqrt{1 \minus{} \cos x} & (4i\plus{}1)\frac{\pi}{2} \le x \le (4i\plus{}3)\frac{\pi}{2}
\end{array}\right.$

[quote="kunny"]I believe the question is $ I \equal{} \int_0^{\pi} x\sqrt {1 \plus{} |\cos x|}\ dx$.

If so, let $ x \equal{} \frac {\pi}{2} \minus{} 2y$, we have $ I \equal{} 2\pi \int_0^ {\frac {\pi}{4}} (|\cos y| \plus{} |\sin y|)\ dy \equal{} 2\pi$.[/quote]

My friend give me this problem. I think it have a mistake at $ n$ as Mr Kunny. If change $ n$ $ \minus{} \minus{} \minus{} >$ $ \pi$ I can done it.

Thank kunny very much.


1. You throw a die and write down the numbers until you have rolled every number at least once. On average, how many rolls until you have marked down every number?

2. What is the probability of it taking [i]n[/i] rolls to mark down each number?

Please provide proof, thanks

To answer your first question, I believe the answer is 14-15 rolls.  I think this is how it is done:
It will take you $ \frac{6}{6}\equal{}1$ rolls to get a new number on the first roll.  It will take you on average $ \frac{6}{5}\equal{}1.2$ rolls to get a new number on your second roll.  It will take on average $ \frac{6}{4}\equal{}1.5$ rolls to get a new number on your third roll, and so, onto $ \frac{6}{1}\equal{}6$ rolls to get a new number on your sixth roll.  Add them and you get 14.7.

You might actually benefit from answering the second question before the first.

Thanks JRav, I'll keep trying at the second question then

Go for it.  I don't know much about probability but the only reason I know the first question is because my friend took a probability course and he actually told me how to do that problem.

I think t0r's suggestion is not good -- the method used by JRav is much nicer here.  For the second question, note that the number of ways to get the six faces in [i]exactly[/i] $ n$ rolls is one-sixth the probability of having [i]exactly[/i] five faces appear in the first $ n \minus{} 1$ rolls.  So the calculation to be made at the base of this problem is to compute the number of ways of partitioning a set into 5 distinguishable, nonempty subsets.


Prove that the supremum of an [b]uncountable[/b] family of measurable real-valued functions can fail to be measurable.

Let $ A$ be a nonmeasurable set.

For $ a\in A,$ define $ f_a(x)\equal{}\begin{cases}1&x\equal{}a\\0&x\ne a\end{cases}.$

Each $ f_a$ is clearly measurable.

However, $ \sup_{a\in A}f_a\equal{}\chi_A,$ which is not measurable.


[b]Task[/b]:

a) Prove the equipollence of the closed interval $\{x\in\mathbb{R}~|~0\le{x}\le{1}\}$ and the open interval $\{x\in\mathbb{R}~|~0<x<1\}$ of the real line $\mathbb{R}$ both using the Schrder-Bernstein theorem and by direct exhibition of a suitable bijection.

b) Analyze the following proof of the Schrder-Bernstein theorem:
$(card~X\le{card~Y})\wedge(card~Y\le{card~X}) \Rightarrow (card~X=card~Y)$.
Proof:
It suffices to prove that if the sets $X,Y,Z$ are such that $X\supset{Y}\supset{Z}$ and $card~X=card~Z$, then $card~X=card~Y$. Let $f:X\rightarrow{Z}$ be a bijection. A bijection $g:X\rightarrow{Y}$can be defined, for example, as follows:
$g(x)=\left\{{f(x),~if~x\in{f^n(X)\setminus{f^n(Y)}~for~some~n\in\mathbb{N},}\atop~{x,~otherwise.}\right}$
Here $f^n=f\circ{...}\circ{f}$ is the [i]n[/i]th iteration of the mapping $f$ and $\mathbb{N}$ is the set of natural numbers. (Remark: N={1,2,3,...} in this terminology)

a) If we want to prove it by Cantor-Bernstein (that's the name I know for it :)), we use the injections $f : (0,1)\to[0,1],\ f(x)=x,\ g : [0,1]\to(0,1),\ g(x)=\frac{x+\frac 12}2$. As for the bijection, just take $h : [0,1]\to(0,1),\ f(0)=\frac 12,f(1)=\frac 14,\ f(\frac 1{2^n})=\frac 1{2^{n+2}},\ \forall n\ge 1,\ f(x)=x$ for any $x$ not of the form $\frac 1{2^k}$.

Regarding b), I don't really know what "analyze the proof" means. Does it mean "check whether it's correct"? I don't know if it's correct, because it's hard for me to visualize it.. :)

[b] Moderator edit: Edited for readability (fixed problems with ":" in math. mode)[/b]


A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of the perpendiculars is
 a. least when point is center of gravity of the traignel
 b. one half the sum of the sides fo the triangle
 c. equal to the altitude
 d. greatest when the point is the center of gravity
 e. none of these

Maddog Poweel wishes to take revenge on both Conquist and Shoemaker. He decides to fill 2 pie crusts with whipped cream to throw at each person. The pie crusts are 8 inches in diamter on the bottom and 10 nches diameter on top. and 3/4 inch deep. What volume in cubic inches of whipped cream is needed to fill all 4 pies?

a. 60 pie
b. 61 pie
c. 244 pie
d. 222 pie

The intersection of 2 perpendicular chords in a circle produce segments 2, 3, 8, and 12. Find the area of the square that can be inscribed in the circle.

A cylinder of radius 5 and height 25 is filled to the top with grape juice. A cone of radius 5 and height 15 and a sphere of radius 5 are dropped int he cylinder. Find the ratio of the grape juice displaced to that still in the cylinder.

[hide="#2 Solution"]There are two ways to do this. The first, more straightforward way, is to extend the pie crust so it makes an upside cone. Then, find the volume of this cone and subtract the volume of the smaller cone formed by the bottom of the crust and part of the big cone's sides. Then, of course, multiply the result by 4.

The second way to do it is to use this formula: $V=\frac{\pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2}\right)}{3},$ where $r_{1}$ and $r_{2}$ are the two radii.

$4V=4\left(\frac{\pi\frac{3}{4}\left(64+100+80\right)}{3}\right)$

$4V=244\pi\Rightarrow\boxed{\text{C}}$[/hide]

[hide="#3 Hint"]The two chords are 14 (2+12) and 11 (3+8).[/hide]

[hide="#4 Solution"]The volume of the cylinder is $625\pi.$ The volume of the cone is $125\pi.$ The volume of the sphere is $\frac{500\pi}{3}.$

The volume of the juice displaced is $\frac{875\pi}{3}.$

The volume of the juice still in the cylinder is $\frac{1000\pi}{3}.$

The ratio is $875: 1000,$ or $\boxed{7: 8}.$[/hide]

thanks.  :lol:

I didn't do as well on this test as I would have liked, but I'm not disappointed either.

1, prove that the sum of the altitudes is equal to the altitudes, by assuming that the sidelength is s and that the altitudes are x, y, and z, and that the area is s^2*sqrt(3)/4

For first -[hide]
The point P in the middle, join it with the  three corners (ABC) to form three small triangles..
(Area = (1/2)* lenght*(altitude))  etc..

what is the (sum of the) area of the three small triangles? If you add all of them- shouldn't the sum be equal to the big triangle. :)  [/hide]

[quote="now a ranger"]thanks

I see how you people did them, and I can say that this thread is now completely answered.  :D[/quote]
well just so you know, it's spelled pi, not pie(pie is those stuff that you throw at ppl's faces)

in that given problem pie meant the thing you throw at ppl's faces.. whip cream  :D

[quote="now a ranger"]Maddog Poweel wishes to take revenge on both Conquist and Shoemaker. He decides to fill 2 pie crusts with whipped cream to throw at each person. The pie crusts are 8 inches in diamter on the bottom and 10 nches diameter on top. and 3/4 inch deep. What volume in cubic inches of whipped cream is needed to fill all 4 pies?

a. 60 [b]pie[/b]
b. 61 [b]pie[/b]
c. 244 [b]pie[/b]
d. 222 [b]pie[/b][/quote]
In the problem, pie is correct. The answers, however, are supposed to be $\pi.$

and it turns out, the correct answer was C. I checked the answer key from what I have, and I don't know what i did wrong on the one about Maddog Poweelll


function
probability
LaTeX
expected value

I missed a couple of my summer school classes (had to go back home)
The class is pretty easy (material I covered in the past), but Ive never seeen the some of the new info that Im being tested on
Please help, thanks

1) One of the following is false
a) 4(side ways U with a straight line in the middle){3,4,5,6}
b) 4 (sideways U) {3,4,5,6}
c){4}(sideways U){3,4,5,6}

2) Given the universal set U={0,1,2,3,4,5,6,7,8,9} and subsets M={1,3,5,7} and N={2,3,4,8}, write out the set A, that represents M'(upsidedown U)N and M'UN'

3) Simplify (C sub n+1,n)/(P sub n,1)  express by a function n

4) Given n(A')=29, n(B')=36, n(A' U B')=59 and n(U)=130, find the following info: n(A), n(B), n(A U B) and n(A (upsidedown U) B)

5) Fill in the blankes with: (sideways U), (sideways U with a slash thru it), (sideways U with a stright line on it), or (sideways U with a straight line and slash)
0 _____ (circle with slash thru it)
(circle with slash) _______ {0}
(circle with slash)________(circle with slash)

6) the probability distribution for the random variable X is:(the following numbers are in a chart)
Xi   -2      1    2     3
pi   .22   .23  .26   .29
What is the expected value of X?

Also got a few word problems that im unsure of how to solve:
1)How many ways can a committe of 5 people with at least 4 democrats be selected from a state legislature that contains 11 democrats and 10 republicans.
2)eight cards are drawn from a deck of 52 cards. what is the probablilty that there are 5 face cards and 3 non-face cards?
3)on three rolls of a single die, you will lose $ 8  if a 3 turns up at least once, and you will win $5 otherwise. What is the expected value of the game?

[quote="joml88"]I suggest you reread the chapter or something (and learn $\LaTeX$).

For example, on  number 1,  $n\subset k$ means $n$ is a subset of $k$.  $n\not\subset k$ just means that $n$ is not a subset of $k$...[/quote]

the chapter does not explain it well at all

a little help would be greatly appreicated here

Ok, a little help :)

[hide="3"]
Did you mean $\frac{{}_{n+1} C_n}{{}_n P_1}$?
If so, then it's simple.
${}_{n+1} C_n=\frac{(n+1)!}{(n!)(1!)}=n+1$
${}_n P_1=n$
So, the answer is $\boxed{\frac{n+1}{n}}$[/hide]

[hide="6"]
The expected value ($EV$) is calculated by
$EV=\sum_{i=1}^{n} X_i P_i=X_1P_1+X_2P_2+...+X_nP_n$

In this case, $n=4$ and thus the expected value of $X$ is $X_1P_1+X_2P_2+X_3P_3+X_4P_4$[/hide]

Ok, then here is basically the rest that you need to know.

$A\cup B$ means the union of $A$ and $B$.  So it's the set that contains all elements that are in either $A$, $B$, or both.

$A\cap B$ is the intersection of $A$ and $B$.  It's the set that contains the elements that are common to both $A$ and $B$.

I believe that $A'$ means the complement of $A$.  So, with respect to the Universal set (whatever that is defined to be), $A'$ contains every element that is not part of $A$ and contains none of the elements of $A$.  Thus $A\cup A'=U$ and $A\cap A'=\emptyset$.

Also, $n(A)$ is the number of elements in $A$.

MathIsTheShiz, you are asking same questions! I solved some in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46008]here here here here here[/url]

[Edit]You say class is pretty easy? I don't understand why you have trouble with these problems if class is pretty easy! Contradiction  :what?:  :huuh:


inequalities
triangle inequality
inequalities solved

Let a1,a2,...,an+1,b1,b2,....,bn+1 be positive real numbers.
Let sqr(x)=x^2.
Prove that
 \sqrt (sqr(a1)+sqr(b1))+ \sqrt(sqr(a2-a1)+sqr(b2-b1))+...++ \sqrt (sqr(an-(an+1))+sqr(bn-(bn+1)))  \geq
\geq \sqrt (sqr(an+1)+sqr(bn+1))

Let x = sqr(x)  :D

 \sqrt (a+b) +  \sqrt (c+d) >=  \sqrt ((a+c)+(b+d)) : triangle inequality

So  \sqrt(a1+b1)+ \sqrt((a2-a1)+(b2-b1)) >=  \sqrt(a2+b2), and so on by induction.


modular arithmetic
number theory proposed
number theory
Fermat s Little Theorem

If $ a^2\minus{}a\minus{}b^2\equal{}m^2$ then $ (2a\minus{}1\plus{}2b)(2a\minus{}1\minus{}2b)\equal{}(2m)^2\plus{}1$. Note that $ (2a\minus{}1\plus{}2b)$ and $ (2a\minus{}1\minus{}2b)$ are of form $ 4k\plus{}3$ therefore $ p|(2m)^2\plus{}1$ for some prime number $ p$ have form $ 4k\plus{}3$. Contradition!

[quote="Atlas_ha78"]Can someone help me? I don't know where the contradition is.....[/quote]

I have edited my post in response to the post of rsa365. The contradiction is there, but I think it needed an explanation to make it clear.

There is a contradiction. The contradiction wouldn't exist if we were using just any sum of squares $a^2+b^2$, but it indeed exists for the sum of squares $(2m)^2+1$. All prime factors of $(2m)^2+1$ are $\equiv 1 \,(mod 4)$:
Just notice that if $p|(2m)^2+1$, then $p$ is odd, and $(2m)^2 \equiv -1 \,(mod\, p) \Rightarrow (\frac{-1}{p})=1 \Rightarrow p\equiv 1 \,(mod \, 4)$.

(a contradiction for $a^2+b^2$ doesn't come so easily because some of $a$ or $b$ can be divisible by $p$, and then all we can get is that if $p\equiv 3 \,(mod\, 4)$, then $p$ appears with an even exponent on $a^2+b^2$).

Oh, you're right, of course.
A more general(and easy to prove) fact: If $p|a^2+b^2 \in \mathbb P, p\equiv 3 \pmod 4$ then $p|a$ and $p|b$.
Proof: Suppose $p \not | a$, then of course $p \not| b$
${-b^2\equiv a^2 \pmod p \implies (\frac{-b^2}{p}})=1 \implies (\frac{-1}{p})=1 \implies p \equiv 1 \pmod 4$, contradiction.

So in our problem, since $p \not |1$ we get a contradiction.

[quote=Valentin Vornicu]Let $a$ and $b$ be natural numbers with $a > b$ and having the same parity. Prove that the solutions of the equation \[ x^2 - (a^2 - a + 1)(x - b^2 - 1) - (b^2 + 1)^2 = 0 \] are natural numbers, none of which is a perfect square.

[i]Albania[/i][/quote]
I will base myself on the solution of [b]@t0rajir0u[/b]
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$$x^2-(a^2-a+1)(x-b^2-1)-(b^2+1)^2=0$$
$$\Rightarrow (x-(b^2+1))(x+(b^2+1))-(a^2-a+1)(x-(b^2+1))=0$$
$$\Rightarrow (x-(b^2+1))(x+(b^2+1)-(a^2-a-1))=0$$
$$\Rightarrow x\in\{b^2+1, a^2-a-b^2 \}$$
$b^2+1$ is not perfect square
Just need to prove that $a^2-a-b^2$ is not perfect square
$\boxed{\textbf{Suppose otherwise:}}$
$$\Rightarrow a^2-a-b^2=c^2$$
Since $a$ and $b$ have the same parity then $m=\frac{a+b}{2}$ and $n=\frac{a-b}{2}$ are integers
$$\Rightarrow (m+n)^2-(m+n)-(m-n)^2=c^2$$
$$\Rightarrow 4mn-m-n=c^2...(\alpha)$$
$\color{red}\boxed{\textbf{Lemma:}}$
$\color{red}\rule{24cm}{0.3pt}$
Let $x$ and $y$ be positive integers
$$\Rightarrow 4xy-x-y \text{ is not a perfect square}$$
$\color{red}\rule{24cm}{0.3pt}$
$\color{red}\boxed{\textbf{Proof:}}$
$\color{red}\rule{24cm}{0.3pt}$
$\boxed{\textbf{Suppose otherwise:}}$
$$\Rightarrow 4xy-x-y=z^2$$
$$\Rightarrow 16xy-4x-4y=(2z)^2$$
$$\Rightarrow 16xy-4x-4y+1=(2z)^2+1$$
$$\Rightarrow (4x-1)(4y-1)=(2z)^2+1$$
$\exists\text{ } p\equiv 3\pmod{4}\text{ } / \text{ } p|4x-1$
$$\Rightarrow p|(2z)^2+1$$
$$\Rightarrow (2z)^2\equiv -1\pmod{p}$$
$$\Rightarrow (2z)^{p-1}\equiv (-1)^{\frac{p-1}{2}}\pmod{p}$$
By Fermat's Theorem:
$$\Rightarrow 1\equiv -1\pmod{p}$$
$$\Rightarrow p|2 \Rightarrow p=2$$
$$\Rightarrow 2\equiv 3\pmod{4} (\Rightarrow \Leftarrow)$$
$$\Rightarrow \text{The assumption is false}_\blacksquare$$
$\color{red}\rule{24cm}{0.3pt}$
In $(\alpha)$ by Lemma:
$$(\Rightarrow \Leftarrow)$$
$$\Rightarrow \text{The assumption is false}_\blacksquare$$
$\color{blue}\rule{24cm}{0.3pt}$


calculus
integration
derivative
complex analysis
function
limit
complex numbers

Hello,

Who could you make a laurent serie of this $ \frac {1}{z^2(z \minus{} 3)^2}$ are ther general rules? Thanks.

Write $ \frac {1}{(z^2)(z \minus{} 3)^2}$ as $ \frac {1}{3(1 \minus{} (1 \minus{} \frac {z}{3}))(z \minus{} 3)^2} \equal{} \frac {1}{3}\sum_{n \equal{} 0}^\infty (\frac {3 \minus{} z}{3})^n \cdot \frac {1}{(z \minus{} 3)^2}$ = $ \frac {1}{3}\sum_{n \equal{} 0}^\infty (\frac {z \minus{} 3}{3})^n( \minus{} 1)^n \cdot \frac {1}{(z \minus{} 3)^2}$=$ \frac {1}{3}\{1 \minus{} (\frac {z \minus{} 3}{3}) \plus{} (\frac {z \minus{} 3}{3})^2 \minus{} \cdots \}\frac {1}{(z \minus{} 3)^2}$ is formallz a Laurant series defined for $ 0<|z\minus{}3|<3$. By the way, the residue of $ \frac {1}{(z^2)(z \minus{} 3)^2}$ is $ b_1 \equal{} 1/9$ and so the line integral
$ \int_C \frac {1}{(z^2)(z \minus{} 3)^2} \equal{} \frac {1}{9}$, where $ C$ is a simple closed contour. How to do this? As Gauß would say "Es kommt vom Hande"--you just have to know (see the book by Churchill and Brown)

Los

I highly doubt calculus goes in this forum...

[quote="ZzZzZzZzZzZz"]That's right.  Calculus should only go in Pre-Olympiad or higher.[/quote] I thought this was complex analysis! :maybe:

$ z^2 \neq 3 \left( 1 \minus{} \left( 1 \minus{} \frac{z}{3} \right) \right) \equal{} z$  :maybe:

Thanks for answering. My excuses for using the wrong forum.

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/1/0/4/10416fd69bd48b2679ee7e566105cfc28fd1f38f.gif[/img]

Yes this must be fault but how can it corrected? The idea is good. Thanks.

i made an error. Correction: write $ \frac{1}{Z} \equal{} \frac {1}{3} \sum_{n \equal{} 0}^\infty {(\frac {z \minus{} 3}{3})^n ( \minus{} 1)^n}$ and term-by-term differentiation gives $ \frac {1}{z^2} \equal{} \frac {1}{3}\sum_{n \equal{} 1}^\infty {n(\frac {z \minus{} 3}{3})^{n \minus{} 1}( \minus{} 1)^n \cdot ( \minus{} 1)} \equal{} \frac {1}{3} \sum_{n \equal{} 0}^\infty {(n \plus{} 1)(\frac {z \minus{} 3}{3})^n( \minus{} 1)^n}$. Then we have  :P $ \frac {1}{z^2(z \minus{} 3)^2} \equal{} \frac {1}{3}\{1 \minus{} 2(\frac {z \minus{} 3}{3}) \plus{} 3(\frac {z \minus{} 3}{3})^2 \minus{} \cdots\}\frac {1}{(z \minus{} 3)^2} \equal{}$ $ \frac {1}{3}\{\frac {1}{(z \minus{} 3)^2} \minus{} 2\frac {1}{3(z \minus{} 3)} \plus{} \frac {1}{3} \minus{} \cdots \}$ defined on 0<|z-3|<3.

Thanks.
But how see you that $ \frac{1}{z}$ is $ \frac{1}{3} \sum (\frac{z\minus{}3}{3}) ^n (\minus{}1)^n$ ?

Thanks a lot for all answers. I go study the reference. Best regards.

[quote="math_music"]Since $ f(z) \equal{} \frac {1}{z}$ is analityc over de disc $ |z \minus{} 3| < 3$ we can expand $ f$ as a Taylor series $ f(z) \equal{} \sum^{\infty}_{n \equal{} 0}\frac {f^{(n)}(3)}{n!}(z \minus{} 3)^n$

You'll find that $ f^{(n)}(z) \equal{} ( \minus{} 1)^nn!\frac {1}{z^{n \plus{} 1}}$ then the Taylor series become $ f(z) \equal{} \sum^{\infty}_{n \equal{} 0}\frac {( \minus{} 1)^n}{3^{n \plus{} 1}}(z \minus{} 3)^n \equal{} \frac {1}{3}\sum^{\infty}_{n \equal{} 0}( \minus{} 1)^n\left(\frac {z \minus{} 3}{3}\right)^n$ for $ |z \minus{} 3| < 3$

More on [url]http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx[/url][/quote] Dumb question number 1: does analytic mean differentiable? If so, is there any reason as to why analytic is used instead of differentiable?

Dumb question number two doesn't need to be asked because I answered it myself! :maybe:

[quote="beans"]Dumb question number 1: does analytic mean differentiable? If so, is there any reason as to why analytic is used instead of differentiable?
[/quote]

The answer is no. Analytic or holomorphic functions must satisfies the Cauchy-Riemann equations on some region $ R$ (open and conex) and the partial derivatives must be continuous on $ R$.

[url]http://en.wikipedia.org/wiki/Analytic_function[/url]
[url]http://en.wikipedia.org/wiki/Holomorphic_function[/url]

[quote="beans"] Dumb question number 1: does analytic mean differentiable? If so, is there any reason as to why analytic is used instead of differentiable?
[/quote]
Why is it a dumb question? A function $ f$ defined in the neigborhood of a point $ a$ is said to be 'analytic' when there exists $ r > 0$ so that $ f$ is infinitely differentiable on $ (a \minus{} r,a \plus{} r)$ and furthermore it converges to its Taylor series.

Analytic is must stronger than differetiability. In complex analysis there is this elegant result  :love: that differentiable functions are analytic and so we simply call them analytic.

EDIT: Can a mod move this to complex analysis computations forum.

The word "differentiable" in this context is ambiguous. math_music seems to be using it to mean differentiable in the sense of functions from $ \mathbb{R}^2$ to $ \mathbb{R}^2,$ in which case you also need the Cauchy-Riemann equations. However if by "differentiable" you mean that the limit

$ \lim_{w\to0}\frac{f(z\plus{}w)\minus{}f(z)}{w}$

exists (with $ w$ taking complex values), then this (or at least this in an open set) is the same as "analytic" or "holomorphic".

- I'm more or less repeating what sylow_theory said (and the topic is now finally in the right place).


abstract algebra
group theory
superior algebra
superior algebra unsolved

The problem I am proposing is the following:

If you take a finite group G and compute the product P of all elements g in G, is the order of that element independent of the way it is written?  Meaning, since the group may not be abelian, can we do the multiplication in any order and always get an element of the same order (in the group theoretic sense, i.e., smallest m such that P^m=e)?

It's not true for just three elements.  ord(abc)=ord(bca)=ord(cab) but may not equal ord(acb).  One can use S_3 to show this.  But could it be true when ALL elements of the group are considered?

This stems from trying to solve a problem that said "Assume you have a finite group G with a unique element f of order two.  Prove the product over all the elements is equal to f."

My feeling is that the writer of the problem left out the word "abelian," b/c then it is truly easy (slide every element next to it's inverse and you are left with f).  But I tried it for two such groups I know of, the quaternions and the non-abelian group of order 12 not isomorphic to A_4 or D_6 (commonly called T).  In either case, I always got the unique element of order two.

So that led me to believe that possibly something the order of this product P is unchanged regardless of how the multiplication is done.  If so, then the above problem in quotes would be solved.

Any thoughts?  Does this make sense?

If the group is non-abelian, the order of the resulting element does depend on the way we arrange the factors. According to Professor Schmidt, $ \mathbf{Q}_{8}$ gives you a minimal counterexample to this conjecture of yours.

If the group is abelian, there is a whole classification to keep in mind...

Let $ G$ be a finite group of order $ n$ and let $ P$ denote the product of all elements of $ G$. Then, we have the following possibilities for $ P$:

- If $ G$ has exactly one element $ i$ of order $ 2$, then $ P \equal{} i$.
- If $ G$ has two or more elements of order $ 2$, then $ P \equal{} e$.
- If $ n \equiv 1 \mod 2$, then $ P \equal{} e.$

[quote="coquitao"]According to Professor Schmidt...[/quote]

Take a look at his last post in [url=http://www.mathlinks.ro/viewtopic.php?t=306792]this[/url] thread.


calculus
integration
logarithms
trigonometry
function
complex numbers
calculus computations

You have a lot more patience than I do to write out that mess.

I can't tell what it is you want me to do, but I'll give you a solution that doesn't use complex numbers.
First pull out $ a$ from the radical in the denominator, so the integral can be re-stated as
$ \frac{1}{a}\int\frac{dx}{\sqrt{1\plus{}\left(\frac{x}{a}\right)^{2}}}$.
The denominator now has a desirable form, which sets you up for a trig substitution, in particular, a tangent substitution.  Let $ \tan\theta\equal{}x/a\Rightarrow a\sec^{2}\theta\,d\theta\equal{}dx$.  Plug everything back in to get
$ \frac{1}{a}\int\frac{a\sec^{2}\theta\,d\theta}{\sec\theta}\equal{}\int\sec\theta\,d\theta\equal{}\ln\left|\sec\theta\plus{}\tan\theta\right|\plus{}C$.
Now we need to substitute in for $ \sec\theta$.  Drawing a triangle here helps, as we see that since $ \tan\theta\equal{}\frac{x}{a}$, the hypotenuse has length $ \sqrt{x^{2}\plus{}a^{2}}$, and we can see that $ \sec\theta\equal{}\frac{\sqrt{x^{2}\plus{}a^{2}}}{a}$.  Therefore, re-inserting everything gives
$ \int\frac{dx}{\sqrt{x^{2}\plus{}a^{2}}}\equal{}\ln\left|\frac{x\plus{}\sqrt{x^{2}\plus{}a^{2}}}{a}\right|\plus{}C$.
I left out some steps, namely the evaluation of the integral of secant and determining what secant was equal to, but those can be checked by you.

Set $ I_{1} \equal{} \int \frac {1}{\sqrt {a^2 \plus{} x^2}} dx$

Setting $ x \equal{} a \sinh \theta$ in $ I_{1}$ gives

$ \boxed {I_{1} \equal{} \sinh^{ \minus{} 1} \frac {x}{a} \equal{} \log |\frac {x \plus{} \sqrt {x^2 \plus{} a^2}}{a}|}$

Another way of obtaining $ I_{1}$ is as follows:

Consider $ I_{2} \equal{} \int \frac {1}{\sqrt {a^2 \minus{} x^2}} dx$

The substitution $ x \equal{} a\sin \theta$ gives $ I_{2} \equal{} \sin^{ \minus{} 1} \frac {x}{a}$.

Now set $ x \equal{} yi$ in both sides of $ I_{2}$, then we get

$ i \cdot I_{1} \equal{} sin^{ \minus{} 1} \frac {yi}{a} \implies \boxed{I_{1} \equal{} \minus{} i\cdot \sin^{ \minus{} 1} \frac {yi}{a}}$

But my question is why:
$ so\,\,why\,\,\int\limits_{1}^{2}{\frac{dx}{\sqrt{{{x}^{2}}\plus{}{{a}^{2}}}}}\equal{}\int\limits_{1}^{2}{\frac{dx}{\sqrt{{{x}^{2}}\minus{}{{\left( \sqrt{\minus{}1}.a \right)}^{2}}}}}\equal{}\int\limits_{1}^{2}{\frac{dx}{\sqrt{\minus{}{{\left( \sqrt{\minus{}1}.x \right)}^{2}}\plus{}{{a}^{2}}}}}\,\,\,\,but$

$ \left. \ln \left| x\plus{}\sqrt{{{x}^{2}}\plus{}{{a}^{2}}} \right|\, \right|_{1}^{2}\ne \left. i.\ln \left| tg\left( \frac{1}{2}.arc\sin \frac{i.a}{x} \right) \right|\, \right|_{1}^{2}\ne \left. \minus{}i\arcsin \frac{i.x}{a} \right|_{1}^{2}\ne \left. \ln \left| tg\left( \frac{1}{2}.arctg\frac{x}{a}\plus{}\frac{\pi }{4} \right) \right| \right|_{1}^{2}$

From thread # 3, we have:
$ \boxed{I_{1}\equal{}\sinh^{\minus{}1}\frac{x}{a} \plus{} C_{1}\equal{}\log |\frac{x\plus{}\sqrt{x^{2}\plus{}a^{2}}}{a}| \plus{} C_{1} \equal{} \minus{}i\cdot\sin^{\minus{}1}\frac{yi}{a} \plus{} C_{2}}$ (note that we have $ y$ not $ x$ in the last part).

Note: In my previous post I made this substitution $ x \equal{} yi$. So $ x \equal{} 1 \implies y \equal{} \minus{} i$ and $ x \equal{} 2 \implies y \equal{} \minus{} 2i$. If you put these limits back in, you should get correct answers. A possible error can arise by setting $ x \equal{} xi$, which leads to $ i \equal{} 1$!

$ when\,\,x\equal{}1\Rightarrow y\equal{}\minus{}i\,\,and\,\,x\equal{}2\Rightarrow y\equal{}\minus{}2i\,\,\,\,\left( x\equal{}i.y \right)$
I can’t see them equal!

$ \left[ \ln \left| x\plus{}\sqrt{{{x}^{2}}\plus{}{{a}^{2}}} \right|\, \right]_{1}^{2}\equal{}\ln \left| 2\plus{}\sqrt{4\plus{}{{a}^{2}}} \right|\minus{}\ln \left| 1\plus{}\sqrt{1\plus{}{{a}^{2}}} \right|\equal{}\ln \left| \frac{2\plus{}\sqrt{4\plus{}{{a}^{2}}}}{1\plus{}\sqrt{1\plus{}{{a}^{2}}}} \right|$

$ \left[ \minus{}i\arcsin \frac{i.x}{a} \right]_{\minus{}i}^{\minus{}2i}\equal{}\minus{}i\left[ \arcsin \frac{2}{a}\minus{}\arcsin \frac{1}{a} \right]$

$ \left[ i.\ln \left| tg\left( \frac{1}{2}.arc\sin \frac{i.a}{x} \right) \right|\, \right]_{\minus{}i}^{\minus{}2i}\equal{}i.\left[ \ln \left| tg\left( \frac{1}{2}.arc\sin \frac{2a}{x} \right) \right|\minus{}\ln \left| tg\left( \frac{1}{2}.arc\sin \frac{a}{x} \right) \right| \right]$

$ \left[ \ln \left| tg\left( \frac{1}{2}.arctg\frac{x}{a}\plus{}\frac{\pi }{4} \right) \right| \right]_{1}^{2}\equal{}\ln \left| tg\left( \frac{1}{2}.arctg\frac{2}{a}\plus{}\frac{\pi }{4} \right) \right|\minus{}\ln \left| tg\left( \frac{1}{2}.arctg\frac{1}{a}\plus{}\frac{\pi }{4} \right) \right|$

Please give me the more specific answer, thanks.

You've already been given 3 solutions, how much more do you want?

I dont understand why that a funtion has 3 different antiderivatives?

They're not three different antiderivatives. They're three different expressions for the same antiderivative, and they're related by trig identities.

Could you give me their relation? Thanks

One of the reasons the complex exponential is so useful is that it gives a relationship between exponential functions and trig functions.

As akech's solution uses the inverses of sine and the hyperbolic sine, here are some relationships:
$ \sin^{\minus{}1}{x}\equal{}\minus{}i\ln\left|ix\plus{}\sqrt{1\minus{}x^{2}}\right|$
$ \sinh^{\minus{}1}{x}\equal{}\ln\left|x\plus{}\sqrt{x^{2}\plus{}1}\right|$

Both of these follow from the complex exponential form of the original functions, followed by some algebraic manipulations.

Yeah, I understood one of four equations.
And I still dont understand the following equations:
$ \left. \ln \left| x \plus{} \sqrt {{{x}^{2}} \plus{} {{a}^{2}}} \right|\, \right|_{1}^{2} \equal{} \left. i.\ln \left| tg\left( \frac {1}{2}.arc\sin \frac {i.a}{x} \right) \right|\, \right|_{1}^{2} \equal{} \left. \ln \left| tg\left( \frac {1}{2}.arctg\frac {x}{a} \plus{} \frac {\pi }{4} \right) \right| \right|_{1}^{2}$

Please help me to explain the equation, thanks.

Have you tried it yourself?  Show us what you've attempted.

But I dont know where I have to begin...

How about using the two identities I gave you?

[quote="chuong01"]
$ \left. \ln \left| x \plus{} \sqrt {{{x}^{2}} \plus{} {{a}^{2}}} \right|\, \right|_{1}^{2} \equal{} \left. i.\ln \left| tg\left( \frac {1}{2}.arc\sin \frac {i.a}{x} \right) \right|\, \right|_{1}^{2} \equal{} \left. \ln \left| tg\left( \frac {1}{2}.arctg\frac {x}{a} \plus{} \frac {\pi }{4} \right) \right| \right|_{1}^{2}$[/quote]
Nobody cant help me?

As I stated a few posts ago, show us what you've attempted.  Tell us which part is tripping you up.  Where are you having trouble?  Which steps have you already attempted?

This is my attempt
$ w \equal{} \arcsin z\Rightarrow z \equal{} \sin w \equal{} \frac {{{e}^{iw}} \minus{} {{e}^{ \minus{} iw}}}{2}\Rightarrow {{e}^{\text{2}i.w}} \minus{} 2z.{{e}^{iw}} \minus{} 1 \equal{} 0\Rightarrow {{e}^{iw}} \equal{} z\pm \sqrt {{{z}^{2}} \plus{} 1}$

$ \Rightarrow i.w \equal{} \ln \left( z\pm \sqrt {{{z}^{2}} \plus{} 1} \right)\Rightarrow w \equal{} \arcsin z \equal{} \minus{} i\ln \left( z\pm \sqrt {{{z}^{2}} \plus{} 1} \right)$

$ *\,\,\int{\frac {dx}{\sqrt {{{x}^{2}} \minus{} {{a}^{2}}}}} \equal{} \ln \left| \tan \left( \frac {1}{2}.\arcsin \left( \frac {a}{x} \right) \right) \right|\,\,\,put\,\,a \equal{} i.b\Rightarrow \int{\frac {dx}{\sqrt {{{x}^{2}} \plus{} {{b}^{2}}}}} \equal{} \ln \left| \tan \left( \frac {1}{2}.\arcsin \left( \frac {i.b}{x} \right) \right) \right|$

$ I\equal{}\int{\frac{dx}{\sqrt{{{x}^{2}}\plus{}{{a}^{2}}}}}\ \ \ \ Put\ x\equal{}a.\tan t\Rightarrow dx\equal{}\frac{a.dt}{{{\cos }^{2}}t},\ \ t\equal{}\arctan \frac{x}{a}$

$ \Rightarrow I\equal{}\int{\frac{a.dt}{{{\cos }^{2}}t\sqrt{{{a}^{2}}.{{\tan }^{2}}t\plus{}{{a}^{2}}}}}\equal{}\int{\frac{a.dt}{a.{{\cos }^{2}}t\sqrt{{{\tan }^{2}}t\plus{}1}}}\equal{}\int{\frac{dt}{{{\cos }^{2}}t\sqrt{\frac{1}{{{\cos }^{2}}t}}}}$

$ \int{\frac{dt}{\cos t}}\equal{}\ln \left( \tan \left( \frac{t}{2}\plus{}\frac{\pi }{4} \right) \right)\Rightarrow I\equal{}\int{\frac{dx}{\sqrt{{{x}^{2}}\plus{}{{a}^{2}}}}}\equal{}\ln \left| tg\left( \frac{1}{2}.\arctan \frac{x}{a}\plus{}\frac{\pi }{4} \right) \right|$

The thing I don't get is why you are making so many convoluted substitutions.  For example, after you solved for $ w$, why did you solve the integral, and then make an immediate substitution using complex numbers?  You already had an answer with purely real numbers?

There is no reason to switch to complex numbers, especially when the problem can be easily solved using purely real methods.

But this is real equality

$ \ln \left| x \plus{} \sqrt {{{x}^{2}} \plus{} {{a}^{2}}} \right| \equal{} \ln \left| tg\left( \frac {1}{2}.\arctan \frac {x}{a} \plus{} \frac {\pi }{4} \right) \right|$
I just want to know why they are equal?
If you are known what you dont know, it's interesting.


How can we prepare 3-methyl-4-hexynoic acid from 2-methyl-4-hexynoic acid?

How yes we can. And in not too many steps. Of course, advanced knowledge is required.

i knew dat we could. i was just kidding. btw i dont suppose neone's gonna post the solution for this, so why dont ya give it yourself

You never know when someone will have a bright idea.

[quote="valeriummaximum"]i think my conversion is taking long steps.......
just see if it is correct..............

1.reduce using LAH
2.reduce using P/HI
3.oxidise using Alk. KMnO4
4.Decarboxylate
5.use HBr(Markanikoff's rule)
6.Use alc.KOH to get the more alkylated alkene
7.Use HBr in the presence of peroxide
8.Use KCN
9.Hydrolyse and reduce using LAH
10.Use P/HI[/quote]

EDIT:
Your synthesis has many flaws. Like you wont be doing wat you wanna do from the 3rd step onwards  :wink:   :D

When writing a synthesis, it would be a good idea to identify what products are expected in each step. Can you do that Valeriummaximum, if you don't mind?


function
real analysis
real analysis unsolved

$f$ is a bounded funtion defined on $\mathbb{R}^1$. Prove that

there exists $g(x)$ which is continuous almost everywhere such that $f(x) = g(x), \hbox{a.e.}, x\in \mathbb{R}^1$ if and only if there exists $E\subset \mathbb{R}^1$ such that $m(\mathbb{R}^1\setminus E)=0$ and $f$ is continuous on $E$.

Hum , it's a collory of Titze-Uryhson Theorem , isn't it?


Ok so in this game someone thinks of something and the other people have to guess what it is by asking questions. whoever correctly identifies the object then thinks of their own object and the game begins again

i'll start... I'm thinking of something....


Few people would know Chebyshev's inequality in Japanese University entrnace exam.


probability
MATHCOUNTS
conditional probability

If a family has 3 children, and we know 2 of them are boys, what is the probability that the other one is a girl?

is it 1/2 or 3/4? or neither?

1/2 since you already know that they are boys.

I put 3/4 for that question on the test-thingy. I think it's 3/4 if you say it like that, but if you phrase it like

A family has 3 children. If the 1st one is a boy, and the 2nd one is also a boy, what is probability that the 3rd one is a girl?

then the ans is 1/2.

but for the other child, its basically just flipping a coin...the probability of the other child is independent of the other two.

Look at it this way: Consider the set of all families with three children.  If you take a sample of eight families, you might expect to find the following results:
ggg
ggb
gbg
bgg
gbb
bgb
bbg
bbb

The first four do not have two boys.  The last four do have two boys, and of those, 3/4 have a girl for the remaining child.  Therefore, of families with three children and at least two boys, 3/4 have two boys and a girl.

As leoxnlin said, if you're given that the [i]first two[/i] children are boys, then only the last two families in the sample count, and so the probability becomes 1/2 that the third child is a girl.

[size=125]I think I did a similiar problem somewhere, and the answer was also 1/2. Both ways, I think it is 1/2...[/size]

[quote="MustItAlwaysBeMe"]Order does matter.  If it is already given that two are boys, then the unknown should be treated as the last child (because that information no longer matters).  So GBB and BGB are irrelevant and there is a 1/2 chance.[/quote]

The only way to test this out is to select 1000 families at random who say they have at least two boys. Then count the number of families out of those who have a girl. If it is close to 500, then the answer is $ \frac 12$. If it is close to 750, then the answer is $ \frac 34$.

hmmm... Monty Hall Problem seems very similar to this one, don't know why.

I think it depends on how it's worded. What your problem says, I think it is 1/2(same reason as the other people that said 1/2)

[quote="ac-king"]Wait... is this a nats prob? (I didn't go nats :()
Because if it is, I'd say 3/4. They wouldn't put a problem on the nats that you could solve in a second (unless it's the #1 question).[/quote]

no, its not, but i have seen questions from mathcounts where it asks similar questions, and the answer is 1/2.

The answer is $ \frac {3}{4}$ because the probability of two boys is greater than the probility of three boys. The probability of two boys is 3/8 and three is 1/8, so $ \frac {\frac {3}{8}}{\frac {3}{8} \plus{} \frac {1}{8}} \equal{} \frac {3}{4}$

[quote="stevenmeow"]i think frost13 can explain it better than im about to...
so the way you say it means the combination of boy/girl has at least 2 boys
the combinations with at least 2 boys are
bbg bgb gbb bbb
3 are repeated but they are not the same, the girl is in a different position
3 out of 4 have a girl, so 3/4[/quote]

Why thank you for the credit. I actually think you did quite a good job explaining it yourself.  You are right by the way.  The wording the way it is means the possible families are

bbg, bgb, gbb, and bbb as stevenmeow said.  From that it is clear the probability is 3/4.  This is classic conditional probability which is rarely used in MC.  I actually did a whole project on this in a college math education class because the college kids had such a big debate about it.  So if you still don't believe me......you aren't the first.  The key is that since you know something about the family already, and the kids are already born, the probability isn't just 1/2 as would be the initial thinking.
Remember probability is defined as $ \frac{number of favorable outcomes}{number of possible outcomes}$

In this case there are only 4 possible outcomes as stevenmeow listed and 3 of them are favorable.

A less confusing example might be something like "I rolled a die and it landed on an even number.  What is the probability it landed on a 4"

Well obviously the probability that a die would land on 4 is $ \frac{1}{6}$, but since we know that the number was even we are only dealing with 3 possibilities and the probability is actually $ \frac{1}{3}$.  You are looking at the probability of something happened under the $ condition$ that something else has already happened.

Hope that helps.

[quote="MustItAlwaysBeMe"][quote="Yongyi781"]The only way to test this out is to select 1000 families at random who say they have at least two boys. Then count the number of families out of those who have a girl. If it is close to 500, then the answer is $ \frac 12$. If it is close to 750, then the answer is $ \frac 34$.[/quote]
Ouch, we're past the generation of Gregor Mendel, you know.  My answer might have been wrong; the "the other one" in the question made me think of it wrong.  The boys are not necessarily the first 2 children, so the answer is 3/4.[/quote]

So it would be close to 750. Interesting...

[quote]If a family has 3 children, and we know 2 of them are boys, what is the probability that the other one is a girl?[/quote]
The wording is very important for this kind of problem, which the discussion above reflects.  This problem is worded poorly.  Since you already told me "2 of them are boys", the other one must be a girl.  Therefore, there is a 100% probability the third one is a girl.

Consider the following independent problems.  [I intend the first four problems to be clear, with a single answer.  The fifth problem is intentionally vague.]:
[quote]My next door neighbors have 8 children.  The first 7 born were boys.  What is the probability the other one is a girl?[/quote]
[quote]My next door neighbors have 8 children.  When registering for summer math camp, the registrar asked for the names of the boys, followed by the names of the girls.  The first seven names were boys.  What is the probability the other one is a girl?[/quote]
[quote]My next door neighbors have 8 children.  When registering for summer math camp, the registrar asked for the names of the girls, followed by the names of the boys.  The first seven names were boys.  What is the probability the other one is a girl?[/quote]
[quote]My next door neighbors have 8 children.  All 8 children lined up in random order to get our city's new library card.  The first seven were boys.  What is the probability the other one is a girl?[/quote]
[quote]My next door neighbors have 8 children.  When registering for summer bible study, the registrar asked for the names in alphabetical order.  If the first seven names were boys, starting with "Jacob" and ending with "Joshua", what is the probability the other child is a girl?[/quote]

[quote="frost13"] Remember probability is defined as $ \frac {number of favorable outcomes}{number of possible outcomes}$[/quote]

Not necessarily. Not to confuse anybody, but as described in Introduction to Counting and Probability, this is only true if what we are dealing with a [b]discrete[/b] situation, in which we are simply counting a finite number of possibilities. However, this definition fails, for example, when we are dealing geometrically, as to say, if a point is randomly chosen within the circle $ x^{2} \plus{} y^{2} \equal{} 36$, what is the probability that it is within the circle $ (x \minus{} 1)^{2} \plus{} (y \minus{} 2)^{2} \equal{} 4$?

Defining probability well is pretty difficult, as far as I know, to the point that some mathematicians don't even believe in the concept of probability.

(edit) also,

[code]$\dfrac{\text{number of favorable outcomes}}{\text{number of possible outcomes}}$[/code]

gives

$ \dfrac{\text{number of favorable outcomes}}{\text{number of possible outcomes}}$.

Oh yeah, as a much simpler counterexample (although of a very different flavor), what if we are flipping a weighted coin? There is still one favorable outcome, and two possible, but the probability is not 1/2.

I agree with stevenmeow.

There are 3 ways to have 1 girl (arrange them), and 1 way to have none.

3/4.

I agree with everyone that said it was 3/4

Ok list the possible ways for the children relative to their ages (youngest to oldest)

bbg
bgb
gbb
bbb

so 3/4


[quote="blackscorpio1120"]1. What is the probability that a multiple of 2 is also a multiple of 3? Express your answer as a common fraction.

2. A bag contains eight white marbles, six red marbles and four blue marbles. How many blue marbles must be added to the bad so that the probability of choosing a blue marble is 1/2?

3. Mary's class took two tests on Friday. Eighty percent of the class passed the math test, 90% of the class passed the English test, and 72% of the class passed both tests. What is the probability that a randomly selected student in Mary's class failed both tests? Express your answer as a percent.

4. Matt will flip a fair coin four times. What is the probability that 1/2 or more of the flips will be Heads? Express your answer as a common fraction.

5. John is about to play Aaron in the best-two-out-of-three chess tournament finals. If the probability that John wins an individual fame agaisnt Aaron is 35%, what is the probability that John will win the tournament? Express your answer as a percent to the nearest tenth.

~*~*~*~

I'll post the next five problems in a couple of days ^.^[/quote]

[hide]1.) $\frac{1}{3}$?
2.) now probability is $\frac{4}{18}$. So if the probability is $\frac{1}{2}$ for drawing a blue marble, it is $\frac{4+x}{18+x}=\frac{1}{2}$, $18+x=8+2x$, $[b: d4950139d6]x=10[/b: d4950139d6]$
3.) I doubt this is right, but $90%+80%-72%=[b: d4950139d6]2%[/b: d4950139d6]
$?
4.) $\frac{1}{2}$??  :?
5.) I have to go, so i'm gonna guess.
35%*35%= 12.25%[/hide]

3 and 4:

[hide]
3. 2%
4.find how many ways you can have 1 or 0 tails. 5. so 16-5/16, ans is 11/16.
[/hide]

[hide]
1. $\boxed{\frac 13}$ The answer is $\frac 13$ to any problem like that (except if the other number is divisible by $3$.
2.  For the probability of choosing blue to be $\frac 12$, there has to be an equal number of blue and any other color marbles. There are $14$ other marbles, so there has to be $14$ blue marbles. There are already $4$ blue marbles so there needs to be $\boxed{10}$
3. Since $72\%$ passed both tests, $90\%-72\%=18\%$ of the people passed just the English test. Since $72\%$ passed both tests, $80\%-72\%=8\%$ just passed the math test. Add $72\%+8\%+18\%=98\%$ passed at least one test. So $\boxed{2\%}$ failed both
4. The ways for there to be less than half heads is exactly $0$ tails of exactly $1$ tails. The number of ways to get $0$ tails is $\binom{4}{0}=1$. The number of ways to get exactly $1$ tails is $\binom{4}{1}=4$. That is a total of $5$ ways. Since there are $2^4=16$ total ways, the probability that there is $\frac 12$ or more will be heads is $\frac{16}{16}-\frac{5}{16}=\boxed{\frac{11}{16}}$
5. The ways for John to win are two consecutive wins, a win then a loss then a win, or a loss than a win then a win. The probability of him winning any game is $\frac{35}{100}=\frac{7}{20}$. So the probability of him winning two consecutive games is $(\frac{7}{20})^2=\frac{49}{400}$. The probability that he wins a game, loses, then wins is $\frac{7}{20} \times \frac{13}{20} \times \frac{7}{20} = \frac{343}{8000}$. The probability for a loss, win, win, is the same as a win. loss, win, because it does not matter what order you multiply in. So the total probability is $\frac{343}{4000}+\frac{49}{400}$. Adding gives $\frac{833}{4000}=\boxed{20.8\%}$[/hide]

[hide]
[hide="1"]
Every 6, there is a multiple of 2 and 3, so $\frac{2}{6}=\boxed{\frac{1}{3}}$[/hide]
[hide="2"]
The probability currently is $\frac{4}{18}$
If you add 10, it becomes $\frac{1}{2}$
$\boxed{10}$[/hide]
[hide="3"]
$80-72=8$
$90-72=18$
$8+18+72=98\%$
So, the percent that failed both tests was $\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{2\%}}}}}}}}}}$[/hide]
[hide="4"]
${4\choose 0}=4$
${4\choose 1}=4$
${4\choose 2}=6$
There is 14 ways, out of $2^4$ ways.
$\frac{14}{16}=\frac{7}{8}$?? I don't think that's right  :? [/hide]
[hide="5"]
The probability that he wins all of the games is $\frac{7}{20}^3=\frac{343}{8000}$
The probability that he wins 2 out of 3 games is $\frac{13}{20}\cdot\frac{7}{20}^2=\frac{637}{8000}$
So, the ansewr is $\frac{980}{8000}=\boxed{\boxed{12.25\%}}$??[/hide][/hide]

pascal,
$\binom{4}{0}=1$. That is your mistake.
Also, for #5, it says best out of 3 meaning that he doesnt have to win three consecutive games, just 2 out of the 3

so here are the answers to the first five

[hide]
1.  1/3

2. 10 marbles

3. 2%

4. 11/16
se
5. 28.2%
[/hide]

next five problems...

~*~*~*~

6. Marcus and Al are playing Rock/Paper/Scissors. The first person to win 10 times does not have to do the dishes. The score is now 9 to 7 in favor of Marcus. If ties are ignored and not counted, what is the probability Marcus will not have to do dishes? Express your answer as a common fraction.

7. If the numerator of a fraction is selected from {1,3,5,7,9} and the denominator of the fraction is selected from {2,3,4,5,6,7}, what is the probabilty that the resulting fraction can be written as a terminating decimal? Express your answer as a common fraction.

8. A six-sided die is weighted so that the probabilities of rolling a 1,2,3 or 4 are equal. The probabilities of rolling a 5 or 6 also are equal to each other. With this dies, Kia is three times more likely to roll a 6 than a 2. What is the probability that she will roll a 4? Express your answer as a common fraction.

9. Lisa has hidden prizes at the ends of some paths in the woods. However, some paths have not prizes at the ends. At each choice in a path, her friends are equally likely to choose either directionsince they can't see what is at the end of any path. According to the map of paths and prizes, what is the probability of a friend getting to a prize? Express your answer as a common fraction.
(i don't know how to post the chart but first it's start then it splits into two then those two split into two on one of the split sides is 'no prize' and prize'. The other side is 'prize' and then splits into two which is then 'prize' and 'no prize'. *hopefully that was clear... sry :oops: *)

10. Marissa, Naomi, Saul, Bob and Carl want to represent their class on the student council. If their teacher randomly chooses two of these students, what are the odds she will choose the two girls? Express your answer in the form a:b

[hide="6"]The probability Marcus doesn't win is the probability Al wins.  The only way Al will win is if he wins 3 times in a row, which has pobability $(\frac12)^3=\frac18$.  That makes the probability Marcus wins $1-\frac18=\frac78$[/hide]

[hide="7"]Look at all of the outcomes that work for each numerator:
$1\Rightarrow2,4,5$
$3\Rightarrow2,3,4,5,6$
$5\Rightarrow2,4,5$
$7\Rightarrow2,4,5,7$
$9\Rightarrow2,3,4,5,6$
There are $5\cdot6=30$ total possibilities, and 20 of those terminate.  That makes the probability $\frac{20}{30}=\frac23$[/hide]

[hide="8"]Assume 1-4 are each weighted once (then they will be all even).  Therefore, 5 and 6 are weighted 3 times each, making a total of $1\cdot4+3\cdot2=10$ weights.  Only 1 of those is a 4, making $\frac1{10}$ the answer.[/hide]

[hide="9"]I will post the solution once I can figure out what the picture is supposed to look like...[/hide]

[hide="10"]There are ${5\choose2}=10$ total outcomes.  Of those, only 1 includes both girls making the odds a girl is chosen $1: 9$.[/hide]

[hide="9"]
i got 3/5.  If its wrong, i probably misinterperated the diagram.[/hide]


$a,b,c\ge 0$
1.Prove that $\sum\frac{a^4}{(2a^2+bc)^2}\ge\frac{1}{3}$
2.Is this ineq true for all $x,y,z,k\ge 0$:$\sum\frac{a^4}{(ka^2+bc)^2}\ge\frac{3}{(k+1)^2}$
If not, find the maximum value of k such that the following ineq is true for all $a,b,c\geq 0$:$\sum\frac{a^4}{(ka^2+bc)^2}\ge\frac{3}{(k+1)^2}$

[quote="scpajmb"]$k_{\max } = \frac{{\sqrt 3 + 1}}{2}$[/quote]
Can you post your solution ? :)

put $x=\frac1{a^3}$ etc. We can assume by homogenity $abc=xyz=1$ and the inequality becomes $\sum \frac1 {(k+x)^2}\ge\frac3{(k+1)^2}$. See http://www.mathlinks.ro/Forum/viewtopic.php?p=397165: this is only true for $k\le\frac{\sqrt{3}+1}2\approx 1.366$


function
functional equation
IMO
IMO 1981
Hong Kong

The function $f(x,y)$, defined on the set of all non-negative integers, satisfies
(i) $f(0,y)=y+1$
(ii) $f(x+1,0)=f(x,1)$
(iii) $f(x+1,y+1)=f(x,f(x+1,y))$
Find f(3,2005), f(4,2005)

I computed $f(1, k), f(2, k)$ and $f(3, k)$ in terms of k respectively by bruteforce.
Any prettier solution?

Well I think this question can only be done by brutal force. Same for $f(4,k)$, a power tower.

[quote="leepakhin"]But mecrazywong said that he used a certain advanced tool to do it. I don't remember the name.[/quote]
No, I just solved it by brute-force...
Maybe you are referring to problem 3 which I solved by isogonal conjugate?
(actually it is totally unnecessary to use isogonal conjugate, but lazy people like me always prefer to write fewer words :P )

Anyway, isogonal conjugate is not really so advanced.

I heard that [url=http://mathworld.wolfram.com/AckermannFunction.html]Ackermann[/url] is featuring the Hongkong retrospect competition. Btw I got to know that he liked this old IMO question as well:

The function $f(x,y)$ satisfies: $f(0,y) = y + 1, f(x+1,0) = f(x,1), f(x+1,y+1) = f(x,f(x+1,y))$ for all non-negative integers $x, y$. Find $f(4, 1981)$.

This is same as Romania 1997
http://www.mathlinks.ro/Forum/viewtopic.php?t=49767

The patterns can basically be found through induction, so no need for brute force. You can basically find $A(k, y+1)=f(A(k, y))$ where $f(x)$ is some sort of function (by using the formulas for $A(1, y), A(2, y)...$), and then you can just do induction on that to find the general formula for $A(k, y)$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=316281#p316281

An elementary solution can be found there.( Also my solution :) )

We will find $f(4,1981)$ for the purpose of solving IMO $1981$ problem $6$.

First we will find $f(1,n)$.

We claim $f(1,n)=n+2$. We will use induction. First we get $f(1,0)=f(0,1)=2$ as a base case.

Now suppose $f(1,k)=k+2$ for all $0\le k\le n$. We will show that $f(1,n+1)=n+3$. We get $f(1,n+1)=f(0,f(1,n))=f(0,n+2)=n+3$, as desired. $\blacksquare$

Now we claim $f(2,n)=2n+3$. We use induction. We have $f(2,0)=f(1,1)=3$ as a base case.

Suppose $f(2,k)=2k+3\forall 0\le k\le n$. We will show that $f(2,n+1)=2n+5$. We get $f(2,n+1)=f(1,f(2,n))=f(1,2n+3)=2n+5$. $\blacksquare$

Now we claim $f(3,n)=2^{n+3}-3$. We use induction. We have $f(3,0)=f(2,1)=5$ as a base case.

Suppose $f(3,k)=2^{k+3}-3\forall 0\le k\le n$. We will show that $f(3,n+1)=2^{n+4}-3$. We get \[f(3,n+1)=f(2,f(3,n))=f(2,2^{n+3}-3)=2^{n+4}-3,\]as desired $\blacksquare$

We claim that $f(4,n)=\underbrace{2^{2^{2^{\ldots}}}}_{n+3\ \text{times}}-3$. We use induction. As a base case, we have $f(4,0)=f(3,1)=13=2^{2^{2}}-3$.

Suppose $f(4,k)=\underbrace{2^{2^{2^{\ldots}}}}_{k+3\ \text{times}}-3$ for all $0\le k\le n$. We will show that $f(4,n+1)=\underbrace{2^{2^{2^{\ldots}}}}_{n+4\ \text{times}}-3$. We get \[f(4,n+1)=f(3,f(4,n))=f(3,\underbrace{2^{2^{2^{\ldots}}}}_{n+3\ \text{times}}-3)=\underbrace{2^{2^{2^{\ldots}}}}_{n+4\ \text{times}}-3,\]as desired $\blacksquare$.

So the answer is $f(4,1981)=\boxed{\underbrace{2^{2^{2^{\ldots}}}}_{1984\ \text{times}}-3}$


geometry
circumcircle
incenter
cyclic quadrilateral
geometry proposed

[quote="kvmedya"]Let $ABCD$ be a quadrilateral which has a circumcircle and an incircle. Let the line $AC$ intersect the line $BD$ at a point $E$, and let $O$ and $I$ be the centers of the circumcircle and of the incircle of the quadrilateral $ABCD$ respectively. Prove that the points $O$, $I$, $E$ are collinear.

[i]Note.[/i] This fact can be rewritten as follows: The circumcenter, the incenter and the point of intersection of the diagonals of a bicentric quadrilateral are collinear. (A [i]bicentric quadrilateral[/i] is a quadrilateral which has a circumcircle and an incircle.)[/quote]

Isn't this the well known property of bicentric quadrilateral?
So far I have seen two proofs, one using inversion, while the other one using area method.

[Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=34217 .]

Then I guess this is a different proof :).

Let $P=AB\cap CD,Q=AD\cap BC$. $PQ$ is the polar of $E$ wrt the circumcircle of $ABCD$ (this is true for all cyclic quadrilaterals), so $OE\perp PQ\ (*)$. On the other hand, $PQ$ is also the polar of $E$ wrt the incircle of $ABCD$ (this is true for any circumscribed quadrilateral), so $IE\perp PQ\ (**)$. From $(*),(**)$ we get what we want.

hey all;

there is another pretty cool proof using Pascal's twice: (it doesn't use any incircle properties)
Let AI intersect the circumcircle at A', similarly define B', C' and D'. Note A'C' and B'D' pass through the circumcentre.
Define Z1 = intersection of CB' and DA'.
By Pascal's, Z1, T and I are collinear. ... (*)
But consider O as intersection fo A'C', B'D'; Z1 as intersectin of CB', DA'; and I as intersectin of DD', CC', we know
by Pascal's that Z1, O and I are collinar. ...(**)
(*) and (**) give us our required result.

I haven't heard of the proof using areas yet, could someone please post it?

In every cyclic quadrilateral, the line connecting the circumcenter with the intersection point of it's diagonals, is perpendicular to the line connecting the intersection points of it's opposite sides.

http://www.mathlinks.ro/Forum/viewtopic.php?t=110887

The same is true about the line connecting the incenter with the diagonals intersection point, in every circumscribed quadrilateral.

http://www.mathlinks.ro/Forum/viewtopic.php?t=139564

Hence, the wanted conclusion, in every bicentric quadrilateral.

Kostas Vittas.

PS. I will poste here next time, the simplest proof I have in mind of this theorem in the configuration of a bicentric quadrilateral, which has already been posted in Hellenic forum.

http://www.mathlinks.ro/Forum/viewtopic.php?t=137281

I use different notations:

[quote="Problem with modified notations"]Let $ ABCD$ be a quadrilateral which has a circumcircle and an incircle. Let the line $ AC$ intersect the line $ BD$ at a point $ M$, and let $ K$ and $ L$ be the centers of the circumcircle and of the incircle of the quadrilateral $ ABCD$ respectively. Prove that the points $ K$, $ L$, $ M$ are collinear.[/quote]

We denote as $ E,$ $ F,$ $ H,$ the orthogonal projections of $ K$ ( =circumcenter ), $ L$ ( =incenter ), and $ M\equiv AC\cap BD$ respectively, on the sideline $ AD$ of $ ABCD$ and as $ E',$ $ F',$ $ H',$ their orthogonal projections on $ BC,$ respectively.

We will prove that $ \frac {EF}{FH} \equal{} \frac {E'F'}{F'H'}$

The line connecting the points $ F,$ $ F',$ as the tangent points of the incircle $ (L)$ to $ AD,$ $ BC$ respectively, passes through the point $ M,$ by [b][size=109]Newton’s theorem[/size][/b], as well.

The triangles $ \bigtriangleup AMD,$ $ \bigtriangleup BMC,$ are similar because of $ \angle ADM \equal{} \angle BCM \equal{} \angle \phi,$ $ \angle DAM \equal{} \angle CBM \equal{} \angle \chi$ and so,

we have that $ \frac {MD}{MC} \equal{} \frac {AD}{BC} \equal{} \frac {DE}{CE'}$ $ ,(1)$ $ ($ because of $ E,$ $ E',$ are the midpoints of $ AD,$ $ BC,$ respectively $ ).$

From $ (1)$ abd because of $ \angle EDM \equal{} \angle E'CM \equal{} \angle \phi,$ we conclude that the triangles $ \bigtriangleup DEM,$ $ \bigtriangleup CME',$ are similar and so, we have that $ \angle DEM \equal{} \angle CE'M \equal{} \angle \rho.$

Hence the right triangles $ \bigtriangleup HEM,$ $ \bigtriangleup H'E'M,$ are similar and from this similarity we conclude that $ \frac {EH}{E'H'} \equal{} \frac {MH}{MH'}$ $ ,(2)$

The right triangles $ \bigtriangleup HFM,$ $ \bigtriangleup H'F'M,$ are also similar, because of $ \angle HFM \equal{} \angle H'F'M \equal{} \angle \omega$ $ ($ as the angles formed by the cord $ FF'$ of the circle $ (L),$ with it’s tangents at points $ F,$ $ F'$ $ ).$

Hence, we have that $ \frac {FH}{F'H'} \equal{} \frac {MH}{MH'}$ $ ,(3)$

From $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {EH}{E'H'} \equal{} \frac {FH}{F'H'}$ $ \Longrightarrow$ $ \frac {EF}{FH} \equal{} \frac {E'F'}{F'H'}$ $ ,(4)$

From $ (4),$ by Thales’s theorem, we can easy to show that the points $ K,$ $ L,$ $ M,$ are collinear and the proof is completed.

$ \bullet$ This proof is dedicated to [b]Giorgos Evangelopoulos.[/b]

Kostas Vittas.

PS. This proof, has already been puplished in mathematical journal [b][size=109]’’Quantum’’[/size][/b] ( Hellenic edition ), volume 2, issue 4, july – August 1995, page 42 and it is the simplest one, I have in mind.

It is known to me, since I was a high school student ( 40 years ago…) and I am interested if there is somewhere, any other reference about it.


This is a somewhat silly way to do it, but it works.  First, count the number of $ 0$'s to the right of the decimal point.  We count $ 5$.  But since we have to move the decimal point between the $ 3$ and the $ 6$, we must add one more, and get $ 6$.  But since we need to get a number smaller than $ 3.6$ when it is multiplied by $ 10^n$, $ n$ must be negative.  So $ n\equal{}\boxed{\minus{}6}$


function
parameterization
algebra proposed
algebra

Find all the functions $f:R\rightarrow R$ stisfying :
$f(x^2)-f(y^2)=(x+y)(f(x)-f(y))$ for all real numbers $x,y$

Write the given relation for any 3 real numbers $x,y,z$  and add them up to obtain that $f(x)=\frac{f(y)-f(z)}{y-z}\cdot{x}+\frac{yf(z)-zf(y)}{y-z}$.So if we fix $y$ and $z$ we get that $f(x)=ax+b$. After this it's easy to check that $f(x)=x$ or $f(x)=b$.

I wrote my solution without any details. Is it understandable? :)

after we check, we find that f(x) = ax + b, (a and b are parameters) so not only f(x) = x or f(x) = b !!

[quote="nhat"]I think it isn't hard
you are easy to prove that $f(x)=b*x+a, (f(0)=a,f(1)=b)$ for all $x \in Q$
after that we get g(x)=f(x)-a for all $x \in Q$
so we have $g(x+y)*(x-y)=g(x-y)*(x+y)$  for all $x,y \in Q$
we get $2*x=(1+k)and 2*y=(1-k)$   for all $k \in R$
so we have g(x)=b*x for all $x \in R$
hence $f(x)=b*x+a$ for all $x \in R$
if i have some mistake ,please correct my solution :lol:  :lol:[/quote]

I couldn't get it, how do you obtain that $g(x+y)*(x-y)=g(x-y)*(x+y)$?


function
limit
integration
real analysis
real analysis unsolved

Let $ a(x)$ and $ r(x)$ be positive continuous functions defined on the interval $ [0,\infty)$, and let \[ \liminf_{x \rightarrow \infty} (x-r(x)) >0.\] Assume that $ y(x)$ is a continuous function on the whole real line, that it is differentiable on $ [0, \infty)$, and that it satisfies \[ y'(x)=a(x)y(x-r(x))\] on $ [0, \infty)$. Prove that the limit \[ \lim_{x \rightarrow \infty}y(x) \exp \left\{ -%Error. "diaplaymath" is a bad command.
\int_0^x a(u)du \right \}\] exists and is finite.

[i]I. Gyori[/i]

This problem was used at ENS Ulm oral examination 2019