logarithms
advanced fields
advanced fields unsolved

I don't know where to post this one, if it is not a good place please move it:

$ \log 10$ (log means log with Euler's $ e$ as a base) is easily irrational.

Is it transcendental?

Where (please recommend me a good book, website or something else)
can I find a complete, clearly written, proof of transcendence of some numbers ?

I found clear proofs only for $ e$ and $ \pi$ ..............

If $ a \neq 0$ is algebraic, then $ e^a$ is transcendental (special case of Lindemann-Weierstrass-Theorem).
So if $ \log 10$ would be algebraic, then $ e^{\log 10}\equal{}10$ would be transcendental, which is not the case. Thus $ \log 10$ is transcendental itself.


inequalities
floor function
combinatorics proposed
combinatorics

hmm... can we use some inequality?

like

$(m n)! > m! n!$
$(m n)! > m!^{n}$

?!!?

we just need to verify that the numerator is every greater than the denominator.

since the numerator have all numbers than 1 to mn. the denominator have just from 1 to m.  (just the primes...)

so the denominator does not have a prime number that does not appear in numerator

we can make a more ellegant inequality from this $(m n)! > m!^{n}$

no!?

hmm.... i will try think more about this tomorrow

and thanks for the good problem

and sorry if i saied some stupid things... im poor on math

[quote="rem"]I remember there was a problem that states that the product of $m!$ consecutive positive integers is always divisible by $m$, and the result follows immidiately.[/quote]
No, must is product of $m$ consecutive positive integers is always divisible by $m!$

[quote="sinajackson"]For $m,n\in N$ prove that $(mn)!$ is divisible by $(m!)^{n}$.[/quote]
It is easy indeed.
The question is to prove that for every prime factor $p$ of $(m!)^{n}$ ($p^{a}|(m!)^{n}, p^{a+1}\nmid(m!)^{n}$), we have $p^{a}|(mn)!$
This is equivalent to prove
$n\sum^{\infty}_{l=1}\lfloor\frac{m}{p^{l}}\rfloor\leq \sum^{\infty}_{l=1}\lfloor\frac{mn}{p^{l}}\rfloor$
which is true because$\lfloor nx\rfloor\geq n\lfloor x\rfloor$ :)


I am not sure this is the most appropriate place in the forum for asking some kind of questions
but please somebody write down the LATEX CODE for the image I attached . I don't know how to write in latex the "down-array"(if I can name it like this) , and I couldn't find it in my Latex Tutorial .

                        Thank you

\underbrace{11\dots 11}_{n \text{ digits}} $ \underbrace{11\dots 11}_{n \text{ digits}}$


inequalities
function
inequalities unsolved

I know this will be so ugly that most of you won't even try, but I really don't know what to do with this inequality, hence I post it here...

If a, b, c are three positive numbers, then prove or disprove the inequality

$3\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq \left( a+b+c\right) \left( a^{3}\left( a^{2}+ab+b^{2}\right) +b^{3}\left( b^{2}+bc+c^{2}\right) +c^{3}\left( c^{2}+ca+a^{2}\right) \right)$.

Equivalent form:

$3\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq \left( a+b+c\right) \left( a^{3}b^{2}+b^{3}c^{2}+c^{3}a^{2}+b^{4}c+c^{4}a+a^{4}b+a^{5}+b^{5}+c^{5}\right)$.

Just to tell you how I came up with this nightmare:

I was searching for a proof of Fuzzylogic's inequality

$\displaystyle \frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}}\geq \frac{1}{3}$,

where a, b, c are positive reals with a + b + c = 1.

I got rid of the a + b + c = 1 restraint by "de-homogenizing" the inequality into the form

$\displaystyle \frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}}\geq \frac{a+b+c}{3}$.

Then I defined $S=a+b+c$ and $T=a^{2}+b^{2}+c^{2}+bc+ca+ab$, transforming this inequality into

$\displaystyle \frac{a^{3}}{T-Sc}+\frac{b^{3}}{T-Sa}+\frac{c^{3}}{T-Sb}\geq \frac{a+b+c}{3}$.

Defining the function $\displaystyle f\left( x\right) =\frac{1}{T-Sx}$, this becomes

$\displaystyle a^{3}f\left( c\right) +b^{3}f\left( a\right) +c^{3}f\left( b\right) \geq \frac{a+b+c}{3}$.

The function $\displaystyle f\left( x\right) =\frac{1}{T-Sx}$ is positive and convex in the interval $\displaystyle \left[ 0;\;\frac{T}{S}\right]$; the numbers a, b, c lie in this interval (since $\displaystyle f\left( a\right) =\frac{1}{b^{2}+bc+c^{2}}$, $\displaystyle f\left( b\right) =\frac{1}{c^{2}+ca+a^{2}}$, $\displaystyle f\left( c\right) =\frac{1}{a^{2}+ab+b^{2}}$ are positive), and hence we can apply the weighted Jensen inequality to the numbers a, b, c with the respective weights $b^3$, $c^3$, $a^3$, and get

$\displaystyle a^{3}f\left( c\right) +b^{3}f\left( a\right) +c^{3}f\left( b\right) \geq \left( a^{3}+b^{3}+c^{3}\right) f\left( \frac{a^{3}c+b^{3}a+c^{3}b}{a^{3}+b^{3}+c^{3}}\right)$.

Now, in order to be able to deduce the desired inequality

$\displaystyle a^{3}f\left( c\right) +b^{3}f\left( a\right) +c^{3}f\left( b\right) \geq \frac{a+b+c}{3}$

from this one, we must have

$\displaystyle \left( a^{3}+b^{3}+c^{3}\right) f\left( \frac{a^{3}c+b^{3}a+c^{3}b}{a^{3}+b^{3}+c^{3}}\right) \geq \frac{a+b+c}{3}$.

But I couldn't finish the proof since I don't know whether this one is true or not. Actually, I don't really need it since I have a much simpler proof now, but I am still curious, as this inequality is somewhat tighter than Fuzzylogic's one (in case it is true!).

  Darij

Using the techniques I described in this thread
   http://www.mathlinks.ro/Forum/viewtopic.php?t=14418
I find that the inequality is true since the difference between its right- and left-hand sides can be expressed as a sum of squares (however, the actual sum I got is too messy to be included here, and can probably be simplified).

Thanks, now at least I know it's true...

  Darij

:Sigma: (a^3/(a^2+ab+b^2))- :Sigma: (b^3/(a^2+ab+b^2))= :Sigma: (a-b)=0
So we only need to prove  :Sigma: ((a^3+b^3)/(a^2+ab+b^2))>=2/3, which is much simpler.


let f is a function from R to R
such that:
f(x) f(1-x)= f(ax+b)

and   f is injective .

1) prove that a=0.

2) prove that f(1-b)=1.

3) prove that f isn't surjectif.

a) $ x\equal{}0$ give $ f(0)f(1)\equal{}f(b)$
    $ x\equal{}1$ give $ f(1)f(0)\equal{}f(a\plus{}b)$

From injective we have $ a\plus{}b\equal{}a\rightarrow a\equal{}0$

So $ f$ is such that $ f(x)f(1\minus{}x)\equal{}f(b)$

b) $ x\equal{}b$

We have $ f(b)f(1\minus{}b)\equal{}f(b)$

$ f(b)$ can't be 0 cause $ f$ wouldn't be injective.Then $ f(1\minus{}b)\equal{}1$

c) $ 0\not\in Im(f)$ cause $ f(x)\equal{}0\rightarrow f(b)\equal{}0$,absurd.

There is nothing wrong with the solution to (3), and this is a double post.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=921833

Apparently a new Robertux interferes! This is the last warning: any other double post will be deleted together with the initial post (this concerns for now only this new user).


This problem is probably really easy for most of you, but i have never taken a Physics C or a Calculus class before, so this one took some effort for me. Well, let's see if i did this correctly.

A model rocket is launched upward and it accelerates at 50m/s^2 for 2 seconds and then it's engine burns out. It then coasts upwards. What is the maximum height it reaches?

[hide] First, i will use the equation h = v0t + 0.5a(t)^2 where v0 = initial velocity. In this case, the rocket started from rest (v0 = 0), so the equation becomes:

h = 0.5a(t)^2. Plugging in the values given in the problem will give the height the rocket reaches right when it's engine burns out, so doing this yields:

h = 0.5(50 m/s^2)(2.0 sec)^2 = 100 m. So during the time the rocket is accelerating, it reaches a height of 100 m.

Now, we need to find the velocity at t = 2 seconds, and we can find that by taking a derivative.

(dh/dt) = (d/dt)0.5at^2 = at = v.

so the velocity at 2 seconds is (50 m/s^2)(2 s) = 100 m/s. This value will be used for our NEW initial velocity in the new h equations that will be used. Now we can use the full h = v0t + 0.5a(t)^2 equation where v0 = 100 m/s and a = -g = -9.8m/s^2 because it is now accelerating in the opposite direction of it's motion (gravity pulls it downward while it coasts upward). In order to find the time, we have to find how long it takes for the rocket to slow down to v = 0. We can take another derivative to find this:

v = (dh/dt) = (d/dt)v0t + 0.5g(t)^2 = v0 + gt. So setting (dh/dt) = 0 will enable us to find the time.

v0 + gt = 0
gt = -v0
t = (-v0/g) = (-100 m/s)/(-9.8 m/s^2) = approximately 10. 2 sec. Thus. the TOTAL height the rocket reaches is

height (while accelerating) + height(while coasting)

= 100m + (100m/s)(10.2 sec) - 0.5(9.8m/s^2)(10.2 sec)^2

= 100m + approximately 510m = 610m
 [/hide]

Yes, your answer is correct.  Have you learned about potential and kinetic energy yet?  If so, here's a slightly easier solution:

From your solution, we know that the rocket reaches a height of 100 meters, at a velocity of 100 meters per second when the engine shuts off, so we have

 $\text{KE}= \frac{1}{2}mv^{2}= \text{PE}= mgh \Rightarrow h \approx 510 \ \text{m}$

and thus the total height is $510+100 = 610 \ \text{m}$.

yeah i took AP Physics B so i know about energy. Too bad i didn't think of using that  :blush:


My brother's team registered around Sept 28 for the contest, but they haven't received anything in the mail.  He called the people at MOEMS and they said that they are a little late sending things out.  It is now the end of October, has anyone received anything for MOEMS?

I registered recently and still haven't gotten anything eiter.  I know they extended the deadline, so maybe that's impacting.

I have a question for you guys can you still register?

Teams can still enroll.

Our apologies for the slowness of the process this year. Technological problems, all one-time events, hurt us badly.
[list](1) The phone company insisted on replacing their copper cables with fiber-optics and accidentally cut our phone/fax lines. It took 5 days to correct the problem and several more to handle the resulting calls and emails.

(2) The web server we employ accidentally locked us out of our main email accounts for 3-4 weeks. Enough messages came through our other account to fool us. When we found and fixed this problem, we got about 2000 messages to handle! We're down to about 75 returned passwords, due mostly to school districts' email filters.

(3) Illness limited the number of man-hours available to process enrollments and books. [/list]
I ask for your patience and suggest you postpone contest 1 for 1-3 weeks if necessary, so as to prepare your students properly. Mostly new teams were affected. All teams will be entered and, wherever we can, helped to start properly.

Again, I apologize for the concern this has caused.

Richard Kalman
Executive Director, MOEMS

[quote="funcia"]Is there a website on which we would enroll? I had done this contest but am not sure how to now...[/quote]

If you are asking whether your school can enroll a team, go to http://www.moems.org/enroll.htm. If you are asking about the "MOEMS contest" or "MOEMS marathon" on the AoPS site, those are by individual students using the MOEMS Forum.

[quote="funcia"]Yeah I was asking for my school thanks for the link BTW if people don't want to participate other than me can I do it myself?[/quote]

Only schools can enroll, not individuals. The upper limit on team size is 35 students. There is no lower limit.

My brother's coach just got the materials today.  Are they really allowed to delay contest 1 for 1-3 weeks?  I think they will take advantage of that  :D


algorithm
analytic geometry
abstract algebra
articles
logarithms

One is given many coordinates in a Cartesian plane. What is the most efficient algorithm of determining which pair of coordinates are closest together (shortest distance between them) among the given coordinates?

[url=http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=lineSweep]This article[/url] may be of interest to you. It gives an $O(n \log{n})$ algorithm to your problem.


geometry
parallelogram
angle bisector
geometry unsolved

Let ABC be a triangle. Let D and E be the midpoints of its sides AB and AC, and let M be the point where the internal angle bisector of the angle CAB meets the line BC. Assume that DM*EM=DA*EA. Prove that AB = AC.

[quote]Let ABC be a triangle. Let D and E be the midpoints of its sides AB and AC, and let M be the point where the internal angle bisector of the angle CAB meets the line BC. Assume that $DM\cdot EM=DA\cdot EA$. Prove that AB = AC.[/quote]

I think I have solved it:

we know  S(DME)=S(DAE)

so Sin(A)=Sin(M)

1- A+M=180   => ...
2- A=M
 AD/DM = ME/AE
 Sin(A/2)^2=Sin(M1)*Sin(A-M1)
...    A/2=M1 , A/2=M2 => parallelogram
ADM :    A = M   =>  AD=DM
AEM :    AE = EM   => ...


let $ f$ be a diff function such that $ f'$ is continuous and :
$ \forall x\in \mathbb{R} f(x\plus{}T)\equal{}f(x)$ with $ T>0$

prove that there existe $ x$ and $ y$ in $ [0,T)$ such that :

$ f(x)f'(y)\equal{}f(y)f'(x)$

[quote="mehdi cherif"]let $ f$ be a diff function such that $ f'$ is continuous and :
$ \forall x\in \mathbb{R} f(x \plus{} T) \equal{} f(x)$ with $ T > 0$
[/quote]

$ f$ is a function with period $ T$, and within $ [0,T)$, $ x \equal{} y$ is the only possibility.

It will be true if there are at least two distinct points in $ [0, T)$ where $ f'$ vanishes.

If $ f$ is constant, there is nothing to prove, so we assume otherwise. Since $ f$ is continuous on $ [0, T]$, there is $ x_0, x_1 \in [0, T)$ such that $ f(x_0) \equal{} \max_{0\leq x \leq T} f(x)$ and $ f(x_1) \equal{} \min_{0\leq x \leq T} f(x)$. Since these two value disagree, $ x_0 \neq x_1$. Then this completes proof, since $ f'$ vanishes at $ x_0$ and $ x_1$.


Results from today's WOCOMAL varsity meet are now online.
Team results: http://www.wocomal.org/statistics/varsity/200607/200607v_team.html
Individual results: http://www.wocomal.org/statistics/varsity/200607/200607v_individual.html

Also, the results from last Thursday's MML competition were posted this afternoon: http://www.geocities.com/mmlleague/February.rtf

Rick Yanco
Worcester Academy

Over the weekend I updated the "unofficial MML website" to show team and individual statistics through the fifth meet.

Individual:
http://www.massmathleague.org/statistics/mml/200607/200607v_individual.html
(sorted by top four scores, since the final ranking will only consider the top five)

Team:
http://www.massmathleague.org/statistics/mml/200607/200607v_team.html

If you see any misspelled names, let me know.

Rick Yanco
Worcester Academy


Vieta
algebra
polynomial
algebra theorems

Does anyone know the Vieta's formulas for polionyms with degree>=3 :?:
Thanks!

[url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta's_formulas]http://www.artofproblemsolving.com/Wiki/index.php/Vieta's_formulas[/url]
Did you even try to look for them ? I believe there are plenty of sources on the internet

[quote="arkhammedos"]hy.may be that is what you're looking for.
[url]http://planetmath.org/?op=getobj&from=objects&id=7190[/url][/quote]
Yes that's it
Thans :)

general formula for the kth symmetric sum of a polynomial of degree n: $ \minus{}1^k \frac{a_{n\minus{}k}}{a_n}$
so for the sum of the roots taken 1 at a time its $ \minus{}a_{n\minus{}1}$


[quote="dheardie"]Find all integers [tex]x[/tex] such that [tex]x^4 + 6x^3 + 11x^2 + 3x + 31[/tex] will be a perfect square of an integer.[/quote]
the squares this is nearest to are
we could not possibly set it equal to squares
(x^2)^2
or squares
((x+3)^2)^2 so we have to set it equal to all the squares in between and solve

which gives us ugly cubics, which hopefully are factorable

[quote="bugzpodder"]this is a Chicago ARML individual? question with 10 as answer.
and set that thing to (x^2+ax+b)^2 will NOT guarentee all the answers, since the range of x^2+ax+b is not R[/quote]

Actually 10 is not the only answer.  I know the answer from my calculator, but not how to do it though.

If someone completes the square for me for the expression, I'd be willing to tackle the rest of the problem  :D

compeleting the square yields fractional coefficients, so i dont know how it helps.  what other answer is there?  somebody ran a program and found 10 as the only answer.  i think the answer to the question is 10 only also.  (obviously i dont know the solution)

How do you complete the square for polynomial functions of degree greater than 2?


MATHCOUNTS
LaTeX
number theory
prime factorization

Ok, this problem is from Mexico Olympiad and because it's Olympiad problem, most people will think this is not a problem for middle schooler.

But I think middle schoolers can handle this because there is formula (I used it when I was in mathcounts) to solve this kind of problem.

[size=142]How many positive integers divide 20!?[/size]

hm....

[hide="hint"]prime factorizing can help tremendously[/hide]

[hide]Prime factorize

$2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19$

The number of factors is then

$19 \cdot 9 \cdot 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 41040$[/hide]

Serious Latex problem and your answer is incorrect.

shinwoo has right idea.

[hide]$\displaystyle2^{20} \cdot 3^7 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19$

$\displaystyle 21 \cdot 8 \cdot 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32156$[/hide]

20/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1

therefore there are 18 twos

20/3 = 6
6/3 = 2

thus there are 8 threes

20/5 = 4

thus there are 4 fives

20/7 = 2

thus there are 2 sevens

20/11 = 1

thus there is only 1 11, 1 13, 1 17, and 1 19.

so that gives us:

$20! = 2^{18}\cdot3^8\cdot5^4\cdot7^2\cdot11\cdot13\cdot17\cdot 19$

The number of factors is: $19\cdot9\cdot5\cdot3\cdot2\cdot2\cdot2\cdot2 = \boxed{41040 \text{ factors}}$

So that means I agree with tarquin

Yes, tarquin is right.

I'm proud of you guys for solving Olympiad problem. :first:

lol do it my way. Every other factor in $20!$ is divisible by 2.

20 /2 = 10

every other one of those is divisible by antoher two

10/2=5

every other one of those is divisible by yet another two.

5/2 = 2 (no frac

2/2 = 1

so there are 18 2s. do that with the other primes below 20

your way is a lot better, yes I mean a lot. At least it didn't take $200\pi$ hours to type  :lol:

:blush:  :rotfl:  :blush:

i should've done it your way.

You don't have to do that dividy thing
$20!=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18\cdot19\cdot20$
Then you can easily see how many 2's 3's 5's and so on

yes but it's not as easy to see that 16 has 4 2s that way or that 20 has 2 2s that way. The "dividy" thing gets the exact number very quickly


Prove that the number  permutations $ \pi$ of $ {1,2,...,n}$ such that there isn't exists
$ i<j<k$ such that $ \pi(i) >\pi(k)>\pi(j)$ are the $ n$-th Catalan number.

Restricted permutations are fun.  I posted this question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162478]here[/url], along with another.

Given any such permutation $ \pi \in S_n$, suppose $ \pi(j) \equal{} 1$.  If $ i < j < k$ then we cannot have $ \pi(i) > \pi(k) > \pi(j) \equal{} 1$, so we must have $ \pi(i) < \pi(k)$, i.e. we have the set $ \{\pi(1), \ldots, \pi(j \minus{} 1)\} \equal{} \{2, \ldots, j\}$.  In addition, the permutations $ \pi(1) \cdots \pi(j \minus{} 1)$ and $ \pi(j \plus{} 1)\cdots \pi(n)$ must also contain no triples of the noted type.  Moreover, given any two permutations $ \pi_1 \in S_{j \minus{} 1}$ and $ \pi_2 \in S_{n \minus{} j}$ of the desired type, we can form a permutation $ \pi \in S_n$ by having the first $ j \minus{} 1$ values be $ \{2, \ldots, j\}$ and be arranged order-isomorphic to $ \pi_1$, then have a 1, then have the last $ n \minus{} j$ values be order-isomorphic to $ \pi_2$.

In other words, there's a bijection between permutations of the given sort of length $ n$ and pairs of permutations of the given sort of length totaling $ n \minus{} 1$.  Thus if $ C(n)$ is the number of such permutations, we have $ C(n) \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} C_i C_{n \minus{} i \minus{} 1}$.  But this is exactly the recursion for the Catalan numbers, so we're done.


trigonometry
geometry
analytic geometry
AMC
AIME
AMC 10
USA(J)MO

I thought DPatrick said that we can't discuss any part of it.

Please do not discuss [b]any[/b] aspect of the AMC 10A/12A until Wednesday morning.

How about projected scores?  That's not really revealing anything important.

115.5 for me (18 correct, 2 wrong, 5 unanswered) on the 12.

I've heard a few higher numbers around school.  Nothing above 132 though.  How did everyone else do?

I totally bombed the AMC 10 because of stupid arithemtic errors. I was quite happy I was able to solve 23 of the problems though, the cross problems was very nice, but the paintbrush one was just an ugly smash-through.

What were some of the harder trig problems you had on the 12?

:(  :(  :(

Mistakes cost me 120.

I got a 114.0 on the 10A.

my 1st AMC, and I totally bombed it.

Did you have any stupid brute force questions? 25 on the 10 was idiotic. I don't get how, even if you use all the available shortcuts, can cut the solve time down to under 3 minutes. Same for the paintbrush one. I didn't answer it because I didn't want to start forcing an answer out I thought would be wrong, but it turns out that was what I was supposed to do...

24 was a good problem though. I always miss the easiest things on geometry problems.

BTW, on the cross problem, like 18 or 17 I think, did anyone try coordinates and shoelace instead of similar triangles? It was the first idea that came to mind but I thought it would be too slow...

@now a ranger: you're lucky you got >100. At least 4 wrong because of idiot mistakes, 23 answered, you do the math...

[quote="white_horse_king88"]The only hard trigo. one was number twenty four, which I conveniently skipped during the test.[/quote]
Can you post it and it's solution in this thread, please?

Yeah...except I'm a retard and got it wrong even though my work got it right. One word: f---.

[quote="Xaero"]Yes, there is a school winner pin.

Also, I think number 25 on the 10 was easy with some weak bounds.[/quote]

That was #22 on ours.

It seemed much easier compared to last years the AMC 12.

I got 144 on it this year.

I totally underestimated the time factor. Had the first 15 problems done with no errors when I realized I would have to work faster to meet 140 and the 75 minute mark and things went to f*cking pieces...

I seriously almost cried after the test  :oops: . It was THAT BAD.

[quote="dkushnir"]Do you guys think lower then 120 will qualify for AIME this year by ANY chance? Everybody's scores went down because of scoring change. I guess 115nish will be in the top 2-3 percentile though.
So as far as 115.5 - probably no way.
Any brownies for 8th graders making 115.5?
Is there a school winner pin for AMC10?[/quote]

ehh...

me= 8th grader

114 here.

My friend says the percentile won't let the cutoff be under 120.

[quote="dkushnir"] Everybody's scores went down because of scoring change.[/quote]
Not true. A lot of people's scores are going up. Perfect scores are never going down.

According to the preliminary results, which are when the test takers circle answers on the test sheet and then the teachers check them, I got a perfect 150.  On the last two problems, though, I had erased bubbles that the teachers and I could not completely erase, so the machine might misread.  On that last problem, I wrote a calculator program as usual.  That is one shortcut that would take you less than 3 minutes (if you program quickly!)  Then, while the program was running on my 84+ SE, I used my 83+ to do other problems.  I was stupid, though, and made it check all numbers from 1 to 2007, instead of about 1960 to 2007, and it took a long time.  The answer was 4, by the way.  Also, what is the cutoff for the USAMO, and how is it calculated from the AMC 10 and AIME?  Central doesn't administer the B round of the tests, I think, so I can't take the AMC 12.

[quote="PathofNeo"]This year's going to end up pretty much like the last.  Less than a 120/100 (10/12) probably won't qualify you.
I would like to post my solution to #12 on the 12.
[i]12. Integers a, b, c, and d, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive.  What is the probability that ad-bc is even?[/i]
Best solution ever: use the 89.
[code]prob12()
Prgm
0→t
0→e
   Loop
   t+1→t
   rand(2007)→a
   rand(2007)→b
   rand(2007)→c
   rand(2007)→d
      If (a*d-b*c)/2=floor((a*d-b*c)/2) Then
      e+1→e
      EndIf
   Disp e/t
   EndLoop
EndPrgm[/code]
This prints all possible solutions, each variants of (4,6,9)...so obviously the answer is 9.  I'm curious how everyone else did this one.[/quote]

Why write a code when there is a pretty simple solution?

ad-bc is even if ad and bc are both odd or both even. The chance of a product being even is 3/4 while the chance of it being odd is 1/4 so it's 3/4*3/4 + 1/4*1/4 = 5/8.

Penguin:
Yeah, I used that method on the cross prob (which was 18, fyi)

Problem #25 has some sort of "mathematically correct" method which allows you to check only 8 possibilities (see attached)

For 19 I tried cutting up the center of the cross and attaching to the ends to form 4 congruent rectangles w/ length= square diagonal-1/2 brush width. Wasn't able to finish within the time limit, though.

In general, it wasn't that bad, but I made some stupid mistakes.

I think the bound can be further improved. If n, s(n), and s(s(n)) are all divisible by 9, then their sum is divisible by 27, which 2007 is not.

What... $n+s(n)+s(s(n)) \equiv 3n \equiv 0\pmod{9}\iff n\equiv 0\pmod{3}$.

[quote="ragnarok23"]eh it was pretty easy
24 was the best problem on the test
25 was retarded

on the ad-bc one you didnt even need to solve it you could tell by the answer choices

there's obviously a lot more even than odd so 5/8
too bad i realized that after i solved the problem with ugly casework[/quote]

What about 9/16

[quote="BFG"]Did anybody know what score Lucy Chen got?[/quote]
Ask her.  [url]http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=25977[/url].

I'm in 8th grade, and I took the 10 last year, didn't do too well, but at least this year I qualified for the AIME!! :lol:  i got, unofficially, 122. I have been doing practice ones from past years, and I think this one was harder than last year and previous years. :P

For 19 I just set a leg of each of the isosceles right triangles to 1 and solved for the brush stroke length. I ended up missing it though; my TI-89T said $2(\sqrt{2}+1)$, and I put (A) $2\sqrt{2}+1$...
For 18, instead of all the similarity, I think the easiest way is just to use coordinate geometry to brute force.

[quote="BFG"]Did anybody know what score Lucy Chen got?[/quote] I believe she got 138; at least that's what she told me.

For # 19, you can just assign a number to the side length and proceed from there, it wasn't too bad.

For # 25, here is how you can do it in under 3 min.
Instantly S(n) resembles mod 9, so let's look at it mod 9.
3n=0 mod 9
n=0 mod 3
So n would need to be divisible by three.
Above 2000, 2001 is clearly the only working one.
Now we know that the thousands digit is 1 and the hundreds digit is 9.
Let's call the last two digits a and b.
Case 1: a+b<10
1900+10a+b+10+a+b+1+a+b=2007
96=12a+3b
32=4a+b
This gives (8,0), (7,4) doesn't work (the sum is over 10), and (6,8) doesn't work.
Now a+b>10
1900+10a+b+10+a+b+2+a+b-10=2007
105=12a+3b
35=4a+b
(8,3), (7,7)
work.
So there are four numbers, 1977, 1983, 1980, 2001.

UGG!
The B test was so easy, yet i only got 126, i couldve gotten 144 :(


A function f:R->R sastisfying : $|f(x+y)-f(x-y)-2f(y)|\le 1 \forall x,y\in R$
Prove that there is only one function g:R->R such that :
$g(x+y)=g(x)+g(y) \forall x,y\in R$
and $|f(x)-g(x)|\le \frac{3}{2} \forall x\in R$

It's quite resemble to Tuymadar problem.Here,is $f$ continous,badboy??


Has anybody even purchased merchandise from : http://www.cheap-nike-sneaker.com or http://www.i-nikejordan.com ?

If it was advertised from this site, then no.
We hate advertisements with all our hearts.


Aflati numarul de patrate perfecte $p$ a.i. $2^{m}\leq p\leq 2^{m+1}$.

[hide="hmm..."]Se vede ca oricare ar fi $m \in \mathbb{N}$ ( banuiesc ca in enunt e natural  :blush:  ) una dintre puterile alea a lui $2$ va fi patrat perfect .

 Odata , daca $m$ - par , atunci $m = 2k$ , cu $k \in \mathbb{N}$ atunci :

 $(2^{k})^{2}\in$ [$2^{2k}, 2^{2k+1}$] si vrem sa gasim numarul $n$ , a. i

  $(2^{k}+n)^{2}\le 2^{2k+1}$ =>

$n^{2}+2^{k+1}n-2^{2k}\le 0$ si din semnul functiei de gradul 2 , obtinem $n \le 2^{k}(\sqrt{2}-1)$ deci $n_{max}=$[$2^{k}(\sqrt{2}-1)$] .  Unde [$x$] reprezinta partea intreaga a lui $x$ .

 Deci in total inte $2^{2k}$ si $2^{2k+1}$ vor fi  $n_{max}+1$ patrate perfecte . Analog se trateaza si pentru $m$ - impar.

[/hide]


geometric transformation
geometry
circumcircle
incenter
homothety
Euler
power of a point

Prove that: $P, I, S$ are collinear and $\frac{PI}{PS}=\frac{R}{R_0}$
Where:
$P$ = exsimilicenter of $(O,R)$ and $(O_0,R_0)$
$(O,R)$ = circumcircle
$(O_0,R_0)$ = Apollonian hull circle of excircles.
$I$ = incenter
$S$ = Spieker's point

Could you, please, explain what

a) exsimilicenter

b) Apollonian hull circle of excircles

c) Spieker's point

are?

Of course:
exsimilicenter = external center of similitude
Apollonian hull circle of excircles = outer Apollonius circle of excircles (externally tangent to each one of them)
Spieker's point= incenter of medial triangle.

Let $ M_a,M_b,M_c$ be the midpoints of $ BC,CA,AB$ and $ U,V$ the tangency points of the excircles $ (I_b),(I_c)$ with $ BC.$ From the fact that $ \triangle ABC$ and $ \triangle M_aM_bM_c$ are homothetic through the  the centroid $ G$ of $ \triangle ABC$ with factor $ \minus{} \frac {_1}{^2},$ it follows that $ M_aS \parallel AI,$ where $ I$ is the incenter of $ \triangle ABC,$ i.e. $ M_aS \perp I_bI_c$ and since $ M_aU \equal{} M_aV \equal{} \frac {_1}{^2}(b \plus{} c)$ we deduce that $ M_aS$ is the radical axis of $ (I_b),(I_c).$ Similarly, $ M_bS$ and $ M_cS$ are radical axes of $ (I_a),(I_c)$ and $ (I_a),(I_b)$ $ \Longrightarrow$ $ S$ is the radical center of $ (I_a),(I_b),(I_c).$

Inversion with center $ S$ and power equal to the power of $ S$ WRT $ (I_a),(I_b),(I_c)$ transforms them into themselves and maps the 9-point circle $ (N)$ of $ \triangle ABC$ into the Apollonius circle $ (O_0).$ Since $ (N)$ is inside $ (O_0),$ then $ S$ is the insimilicenter of $ (N) \sim (O_0).$ By Monge & d'Alembert Theorem, we conclude that the insimilicenter $ S$ of $ (N) \sim (O_0),$ the insimilicenter $ G$ of $ (O) \sim (N)$ and the exsimilicenter $ P$ of $ (O) \sim (O_0)$ are collinear and obviously $ I \in GS,$ since $ S$ is the image of $ I$ under the referred homothecy $(G, \minus{} \frac {_1}{^2}).$ By Menelaus' theorem for $ \triangle PSO_0$ cut by the Euler line $ NGO$ we have

$ \frac {\overline{NG}}{\overline{NO}} \cdot \frac {\overline{OO_0}}{\overline{PO_0}} \cdot \frac {\overline{PS}}{\overline{GS}} \equal{} 1.$  Because of $ \frac {\overline{NO}}{\overline{NG}} \equal{} 3 \ , \ \frac {\overline{GS}}{\overline{IS}} \equal{} \frac {1}{3}$

$ \Longrightarrow \frac {\overline{OO_0}}{\overline{PO_0}} \equal{} \frac {\overline{IS}}{\overline{PS}} \equal{} \frac {\overline{OO_0} \minus{} \overline{PO_0}}{\overline{PO_0}} \equal{} \frac {\overline{IS} \minus{} \overline{PS}}{\overline{PS}} \Longrightarrow \ \frac {\overline{PI}}{\overline{PS}} \equal{} \frac {\overline{PO}}{\overline{PO_0}}$

$ \frac {\overline{PO}}{\overline{PO_0}}$ is the factor of the direct homothecy taking $ (O)$ into the Apollonius circle $ (O_0)$

$ \Longrightarrow \frac {\overline{PI}}{\overline{PS}} \equal{} \frac {\overline{PO}}{\overline{PO_0}} \equal{} \frac {R}{R_0}.$


$x,y\in\mathbb{N}$ and $3x^2+x=4y^2+y$, prove $x-y$ is a perfect square.

[hide="hint"]
$3x^2+x=4y^2+y \Rightarrow 3x^2-3y^2+x-y=y^2$
$\Rightarrow ....$[/hide]

In your question you have $3x^2-x$ but in your hint you have $3x^2+x$...

:oops: yes it was a typo in the original problem, $3x^2+x$ is correct... :)

Your second case analysis is flawed because when you say, in a similar fashion, it is not actually a similar fashion. In the second case, you get $x^2<(4/3)y^2$, but that doesn't say anything about the relationship between x and y, ie, you cannot conclude from that alone that x-y>0.

Also if x-y=0, then y=x, which is true only for x=0, y=0, [u]not[/u] natural numbers.

[quote="Singular"]$3x^2 + x = 4y^2 + y$

$3x^2 - 4y^2 = y-x$

case 1:  $3x^2 \geq 4y^2$

Then LHS is positive, and implies $x^2 \geq (4/3)y^2 \geq y^2$.  So $3x^2 - 4y^2 \geq 0$ and $y-x \leq 0$, equality everywhere when $x = y = 0$.

case 2:  $3x^2 \leq 4y^2$

In a similar fashion, the LHS is negative, and the RHS positive, with equality when x=y=0.

So x-y = 0 = 0^2 a perfect square.[/quote]
that's not right,
i also said: $x.y\in\mathbb{N}$ $0\not\in\mathbb{N}$
the scond case, is totally false... ;)

[hide="another hint"]
$3x^2+x-3y^2+x-y=y^2 \Rightarrow 3(x-y)(x+y)+(x-y)=y^2 \Rightarrow  y^2=(x-y)(3x+3y+1)$
now if you can prove that $x-y$ and $3x+3y+1$ are pirimitive, and since the mulipleication of two primitive numbers has resulted in a perfect square both them have to be perfect squares so now prove:
$(x-y,3x+3y+1)=1$  :w00tb: [/hide]

I have another way I will post it later I have to go to school soon.

Acutally I didnt read amirthlusa's last hint I did the problem the same way as him.

amirhtlusa: your second hint is wrong. take x=9, y=4
x-y=5
3x+3y+1=40
(5,40)=5
(or maybe i am just sleepy :) )

the hint was that if x-y and 3x+3y+1 are relatively prime and their product is a perfect square, they are each perfect squares...I think you thought it meant x and y are relatively prime...

the hint was ambiguous, because the fact that the product of x-y and 3x+3y+1 is a perfect square is already being used after you prove that (x-y, 3x+3y+1)=1. so i guessed( and i still am not sure wether i was wrong or not) he meant that it should be proven that (x-y, 3x+3y+1)=1 and that's all with no other condition. Otherwise something is fishy in using this information(that the product is a perfect square) both to prove those 2 numbers are primitive and then to finish the proof also.

if you prove that those numbers are pirimitive, then the only way that $(x-y)(3x+3y+10$ is a perfect square is that both of them be perfect squres.
for example let $9*4=36$ since $9$ and $4$ are primitive, then because thier product is a perfect square, then each one of them must be perfect squares,(as they are...)


[hide=" BE!! VERY CAREFUL!!"] I'll just point out what you need to watch out for,,,
convert feet to yards
[/hide]

[quote="MCrawford"]The rectangular stage at Radio City Music Hall in New York City measures 144 feet wide and 60 feet deep.  What is the number of square yards in its area?[/quote]

[hide]
144 ft. x 60 ft.
=48 yards x 20 yards
=960 square yards
[/hide]

[hide]144 ft = 48 yards
60 ft = 20 yards
48 yds * 20 yds = 960 yrds 2[/hide]


If a person with genotype $AaBbCc$ is crossed with a person of genotype $AaBbCc$, then what is the probabily that the final genotype of the hybrid is $AaBbCc$? Note that each parent can only give one of each letter (ie, either A or a) to the offspring.

The correct probability is 20/64. But for some very wierd reason, my mind is not working right now. Can anyone show me how to get this answer?Please?

Have you learned about Punnett Squares yet?
Set up a three by three table and set up each genotype on the top and left.

For each cell, divide it into four subcells.
Each subcell will have one of the four (nondistinct) combinations of the letter.
For example, we set up the table and see that for the first gene,  the first parent has $Aa$, and the second also has $Aa$.
Thus the four possible combinations are $(A,A);(A,a);(a,A);(a,a)$
Note that the middle two are effectively the same.
We see that the probability of getting the same values for letter  A is $\frac{2}{4}$, and this is the case for each letter.
Now apply your math skills to the greater punnett square. I'll leave that as an excersize for you to do.

Make each subcell like this for each letter.
Then count the subcells that are of the desirable outcome.
You should reach the correct solution.


geometry
circumcircle
trigonometry
geometry unsolved

There are $n$ different points  $A_1, \ldots , A_n$ in the plain and each point $A_i$ it is assigned a real number $\lambda_i$ distinct from zero in such way that $(\overline{A_i A_j})^2 = \lambda_i + \lambda_j$ for all the $i$,$j$ with $i\neq{}j$}
Show that:

(1) $n \leq 4$
(2) If $n=4$, then $\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}+ \frac{1}{\lambda_4} = 0$

Here is my solution for this problem:

At first we will show that $n\le 4$, so for sake of contradiction let $n>4$. We will be using the criterion for perpendicular segments, namely the one that $AB\bot CD\Leftrightarrow AC^2-BC^2=AD^2-BD^2$. It is trivial using Pythagoras' theorem. Let us suppose that some three of the points ($A_1, A_2, A_3$) lie on a single line. Then if all points lie on this line without loss of generality we may assume the line is horizontal and they are ordered from left to right $A_1, A_2, \ldots, A_n$. We see that \[ A_1A_{n-1}^2-A_nA_{n-1}^2=\lambda_1+\lambda_{n-1}-\lambda_n-\lambda_{n-1}=\lambda_1+\lambda_{2}-\lambda_n-\lambda_{2}= A_1A_{2}^2-A_nA_{2}^2<A_1A_{n-1}^2-A_nA_{n-1}^2  \] and we get a contradiction since $A_2 \neq A_{n-1}$. So let $A_{4}$ be a point that does not lie on that line. We get $A_1A_4^2-A_1A_2^2=\lambda_1+\lambda_4-\lambda_1-\lambda_2=\lambda_3+\lambda_4-\lambda_3-\lambda_2=A_3A_4^2-A_3A_{2}^2$ and hence $A_1A_3\bot A_2A_4$. Analogously we obtain $A_1A_2\bot A_3A_4$ which gives a contradiction. Hence no three of our points are collinear. But noting that for fixed $A_1, A_2, A_3$ and distinct from them $A_i$ in the same way as above we get $A_1A_2\bot A_3A_i$ and since no three of the points are collinear we get that $n\le 4$.
Next suppose that $n=4$. Say our four points $A_1, A_2, A_3, A_4$ form a convex quadrilateral (with perpendicular diagonals as seen from the proof above). If the feet of a perpendicular from $A_1$ to $A_3A_4$ does not belong to $\overline{A_3A_4}$ then we easily get that all such feets of perpendiculars do not lie on respective sides of a quadrilateral and thus all the angles of a quadrilateral are $>90^o$ which is impossible. But in the other case we get that we have a square $A_1A_2A_3A_4$ for which $\lambda_1=\lambda_3=-\lambda_2=-\lambda_4$ and the result follows. In the case when $A_1A_2A_3A_4$ is not convex using the criterion we get that $A_i$ is an orthocenter of the triangle formed by the other three points. Alas I do not know how to finish the proof in this case.

[hide="Solution"][b]Lemma[/b]: If $ A,B,C,D$ are distinct points, then $ (AB^2\plus{}CD^2\equal{}BC^2\plus{}DA^2) \Leftrightarrow (AC\perp BD)$
[hide="Proof"] We have that:
$ AB^2\plus{}CD^2\equal{}BC^2\plus{}DA^2\Leftrightarrow (A\minus{}B)^2\plus{}(C\minus{}D)^2\equal{}(B\minus{}C)^2\plus{}(D\minus{}A)^2 \Leftrightarrow$
$ \Leftrightarrow AB\plus{}CD\equal{}BC\plus{}DA \Leftrightarrow$
$ \Leftrightarrow (A\minus{}C)(B\minus{}D)\equal{}0 \Leftrightarrow AC \perp BD$ (since $ A\neq C$, $ B\neq D$)[/hide]

Let $ 1\leq i,j,k,l\leq n$ be distinct integers, and assume that $ n\geq 4$.

$ A_iA_j^2\plus{}A_kA_l^2\equal{}(\lambda_i\plus{}\lambda_j)\plus{}(\lambda_k\plus{}\lambda_l)\equal{}(\lambda_i\plus{}\lambda_k)\plus{}(\lambda_j\plus{}\lambda_l)\equal{}A_iA_k^2\plus{}A_jA_l^2$

Using the lemma, we have that $ A_iA_l \perp A_kA_j$. Analogously we get $ A_jA_l\perp A_iA_k$, $ A_kA_l\perp A_iA_j$, so $ A_l$ is the orthocenter of $ \triangle A_iA_jA_k$. In particular, since the orthocenter of a triangle is unique, we get that $ n\leq 4$.

Now let the points be $ A,B,C,H$ with $ H$ in the interior of $ \triangle ABC$, $ BC\equal{}a,CA\equal{}b,AB\equal{}c$, and $ R$ is the circumradius of $ \triangle ABC$.

$ \lambda_A\plus{}\lambda_B\equal{}AB^2\equal{}c^2$
$ \lambda_A\plus{}\lambda_C\equal{}AC^2\equal{}b^2$

Subtracting, $ \lambda_B\minus{}\lambda_C\equal{}c^2\minus{}b^2$. But $ \lambda_B\plus{}\lambda_C\equal{}BC^2\equal{}a^2$, so $ \lambda_B\equal{}\frac{a^2\plus{}c^2\minus{}b^2}{2}\equal{}ac\cos(\beta)$, by the cosine law, and analogous equalities for $ A,C,H$.

By the sine law, $ \frac{AH}{AB}\equal{}\frac{\sin(\frac{\pi}{2}\minus{}\alpha)}{\sin(\gamma)} \Leftrightarrow AH\equal{}2R\cos(\alpha)$, and $ BH\equal{}2R\cos(\beta)$.

We have to prove that:

$ \frac{1}{\lambda_A}\plus{}\frac{1}{\lambda_B}\plus{}\frac{1}{\lambda_C}\plus{}\frac{1}{\lambda_H}\equal{} 0\Leftrightarrow$
$ \Leftrightarrow \sum_{cyc}\frac{1}{ab\cos(\gamma)}\equal{}\minus{}\frac{1}{AH\times BH \cos(\pi\minus{}\gamma)}\equal{}\frac{1}{AH\times BH \cos (\gamma)}\Leftrightarrow$
$ \Leftrightarrow \sum_{cyc}\frac{1}{4R^2\sin(\alpha)\sin(\beta)\cos(\gamma)}\equal{}\frac{1}{4R^2 \cos (\alpha)\cos(\beta)\cos (\gamma)}\Leftrightarrow$

$ \Leftrightarrow \tan(\alpha)\plus{}\tan(\beta)\plus{}\tan(\gamma)\equal{}\tan(\alpha)\tan(\beta)\tan(\gamma)$

This last equality is true because $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$, as desired.[/hide]

There are $n$ different points $A_1,... , A_n$ in the plane and  at each point $A_i$ is assigned a real number $\lambda_i$, distinct from zero, in such way that $(\overline{A_i A_j})^2 = \lambda_i + \lambda_j$ for all $i,j$ with $i\ne j$ . Show that:
(1) $n \le 4$
(2) If $n=4$, then $\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}+ \frac{1}{\lambda_4} = 0$


conics
parabola
geometry
analytic geometry

[quote="linzk3"]A point moves so that its distance from the point (-3,2) is always equal to its distance from the line y=-6.  Derive the equation of its locus.[/quote]

[hide]It's a parabola, with focus at (-3,2) and directrix at y=-6, so the vertex is at (-3,-2) and the equation is $ (x \plus{} 3)^2 \equal{} 16(y \plus{} 2)$

You could use analytic geometry to justify this, but I used the formula $ (x \minus{} h)^2 \equal{} 4a(y \minus{} k)$ for parabolas opening upward.[/hide]


Let $ x_{1},x_{2},...,x_{n}$ be positive real numbers such that $ \prod^{n}_{i\equal{}1}x_{i}\equal{}1$. Prove that $ \sum^{n}_{i\equal{}1}x^{n}_{i}(1\plus{}x_{i})\geq \frac{n}{2^{n\minus{}1}}\prod^{n}_{i\equal{}1}(1\plus{}x_{i})$.

[quote="Γιώργος"]Let $ x_{1},x_{2},...,x_{n}$ be positive real numbers such that $ \prod^{n}_{i \equal{} 1}x_{i} \equal{} 1$. Prove that $ \sum^{n}_{i \equal{} 1}x^{n}_{i}(1 \plus{} x_{i})\geq \frac {n}{2^{n \minus{} 1}}\prod^{n}_{i \equal{} 1}(1 \plus{} x_{i})$.[/quote]
by Chebyshev : $ LHS \ge \frac{1}{n}(x_1^n\plus{}...\plus{}x_n^n)(n\plus{}x_1\plus{}...x_n) \ge (\frac{x_1\plus{}...x_n}{n})^n(n\plus{}x_1\plus{}...x_n)$;

$ x_1\plus{}...x_n\equal{}t \ge n$;

$ \frac{t^n(n\plus{}t)}{n^n} \ge \frac{n}{2^{n\minus{}1}}(\frac{n\plus{}t}{n})^n \ge \frac{n}{2^{n\minus{}1}}\prod^{n}_{i \equal{} 1}(1 \plus{} x_{i})$  :)


Find $PQ$ such that, red area $=$ blue area

$AB=12$

$PQ\parallel AB$

[quote="SM4RT"]I think this is right..correct me if i am wrong..
[hide="Solution"]
since [color=red]RED[/color]=[color=blue]BLUE[/color] therefore
$\text{Let }\triangle{PCQ}=x\Rightarrow\triangle{ACB}=2x$
Clearly $\triangle{PCQ}\sim\triangle{ACB}$ since $PQ\parallel{AB}$
The ratio of the sides is the squareroot value of the ratio of the area...so
$\frac{x}{2x}=\frac{1}{2}\implies$ratio for the sides is $\frac{1}{\sqrt{2}}$
$\overline{PQ}=\frac{12}{\sqrt{2}}=\frac{12\sqrt{2}}{2}=\boxed{6\sqrt{2}}$
[/hide][/quote]

You are right!! :lol:


If a person gets 150/6 on AIME II, then will that affect AIME I takers too or just AIME II takers. Also, will the floor be different? and will there be different indices for A and B tests? Thanks

They will have split indexes and floors if there is a large disparity in difficulty between the two tests. It happened in 2005. Otherwise, the index and floor will be unified.


when f(x)+f(y)=f(x+y) for all realnumber x,y and f(f(x))=x
prove that f(x)=x or -x.

Assume the values of f are also reals.

Can you prove f(f(x)) = x for rational x?  That will be the hard part... the rest easy.

that's original problem is CMO2008-2
I proved that f(f(x))=x for all realnumber x and f(x)+f(y)=f(x+y).
but I can't prove that f(x)=x or -x  :blush:

I think you've dropped too much of the original problem statement; I believe this problem is false. Doing the standard construction of a noncontinuous solution to the Cauchy equations, pairing up the basis elements to satisfy $ f(f(x)) \equal{} x$ as needed, gives a working solution.

If you are talking about Canada 2008/2... that function is on the rationals. Why'd you say reals?

oh sorry!
one translator wrote rational to R -_-


HI

I need help obtaining an approximation for f'''(Xo) using:

f'(Xo) = (f(Xo + h/2) - f(Xo - h/2))/h

Thanks

f''(x) = (f(x + h/4) - f(xx - h/4))/(h/4)

But I cant seem to come up with an approximation for f'''(x).

Anyone?

[quote]f''(x) = (f(x + h/4) - f(xx - h/4))/(h/4) [/quote]
I don't even know what that means, with the "xx" in there, but it doesn't look like an approximation of the second derivative.

It's also beside the point - in this problem, we won't have any intermediate result involving the second derivative.

The key to many things in numerical analysis is Taylor polynomials. Assume that $f$ is sufficiently smooth to justify the following calculations.

$f\left(x_0+\frac h2\right)=f(x_0)+\frac h2f'(x_0)+\frac{h^2}8f''(x_0)+ \frac{h^3}{48}f'''(x_0) +ah^4f''''(x_0)+O(h^5)$

$f\left(x_0-\frac h2\right)=f(x_0)-\frac h2f'(x_0)+\frac{h^2}8f''(x_0)- \frac{h^3}{48}f'''(x_0) +ah^4f''''(x_0)+O(h^5)$

Subtract and divide by $h:$

$\frac{f\left(x_0+\frac h2\right)-f\left(x_0-\frac h2\right)}h=f'(x_0)+\frac{h^2}{24}f'''(x_0) +O(h^4).$

Now double $h$ in that formula:

$\frac{f\left(x_0+h\right)-f\left(x_0-h\right)}{2h}=f'(x_0)+\frac{h^2}{6}f'''(x_0) +O(h^4).$

Subtract these two formulas:

$\left[\frac{f\left(x_0+h\right)-f\left(x_0-h\right)}{2h}\right]- \left[\frac{f\left(x_0+\frac h2\right)-f\left(x_0-\frac h2\right)}h\right]= \frac{h^2}8f'''(x_0)+O(h^4)$

Now multiply that last equation by $\frac8{h^2},$ and use a common denominator:

$\frac{4\left[f\left(x_0+h\right)-2f\left(x_0+\frac h2\right)+2f\left(x_0-\frac h2\right) +f\left(x_0-h\right)\right]}{h^3}=f'''(x_0)+O(h^2).$


Ring Theory
superior algebra
superior algebra solved

$f:R\to R'$ is an epimophism. If $L$ is a maximal ideal of $R$, is $L'=f(L)$ a maximal ideal of $R'$?
If $L'=f(L)$ is a maxmial ideal of $R$, is $L$ a maximal ideal of R?

I think the answer to the first question is yes. The image of $L$ is clearly an ideal, and it's maximal: let $a'\in R'\setminus L'$. We have $a'=f(a),\ a\in R\setminus L$, so $[a,L]=R\Rightarrow [a',L']=R'$, where $[S]$ is the ideal generated by the set $S$.

The answer to the second question is not always yes, I think. If $L$ does not contain $\ker f$, for instance. Let $a\in\ker f\setminus L$. We get $f([L,a])=L'\ne R'\Rightarrow [L,a]\ne R$.

$f^{-1}(L')$ is, however, a maximal ideal.


geometry
trigonometry
Pythagorean Theorem

[quote="ch1n353ch3s54a1l"][quote="jli"]there is a hexagonal doghouse is 10 feet on all sides.  a dog is leashed to one corner of it and the leash is 40 feet long.  what is the area that the dog can be on[/quote]

[hide]...well can't the dog reach all parts of the hexagon? So isn't it just the area of the hexagon..$150\sqrt{3}$...[/hide][/quote]

sorry...the do is outside

if the dog is tethered with a 40 ft leash then dog would be able to reach all the vertices if it wants, isnt it ??? :(

the leash would fold around the sides of 10 ft each. ??

So the dog is on the outside of the dog house and cannot cut through the dog house (assuming its enclosed). Can the rope "go over" (cut through) the dog house (as the dog goes around)? Should we assume a two-dimensional plane?

Come on guys, you are all making this so complicated. It's not hard if you draw appropriate diagrams. Just find the sweeping area and that's it.

[quote="LordoftheMorons"]What's really annoying me is the overlap when the rope wraps around the shed.[/quote]

I can see you are almost to the solution, yes take away the overlapping area.

yes...you guys are making it way to complicated.  just draw a diagram and youll be able to see how to  calculate the overlapping area

I don't think so, tz.

The thing that's throwing me is the overlapping area that you get depending on which way the dog decides to run. His leash is too long -_-"

[quote="tz"][hide]Is it the area of the circle subtracted by the area of the hexagon?
In that case, $1600\pi-150\sqrt{3}$ is about $4766.74$[/hide][/quote]

no

[quote="VenDiddy"]I don't know if this is right because there were so many steps and I could have made a mistake somewhere or could have misdrawn my diagram.

[hide]I drew a diagram and noticed that all of the areas that the dog could cover were circle sectors. When the rope wraps around a vertex of the hexagon its starts a new circle sector with a radius which is 10 feet less. I think that the two sectors of radius 20 feet overlap and that the sector of radius 10 is completely covered so it can be ignored.

The area of the 40 foot sector is:
$\left(\frac{240}{360}\right)\pi(40)^2=\frac{3200\pi}{3}$

The total area of the two 30 foot sectors is:
$2\left(\frac{60}{360}\right)\pi(30)^2=300\pi$

The two 20 foot sectors overlap, and this region can be broken into two triangles and two sectors. The base of the triangle is $5\sqrt{13}-5$. The height of the triangle is $5\sqrt{3}$. The area of the two triangles would be $2\left(\frac{5\sqrt{3}}{2}\right)(5\sqrt{13}-5)=25\sqrt{26}-25\sqrt{3}$.

To find the area of the sector in the overlapping region, you need its angle, which turns out to be $\sin^{-1}{\left(\frac{5\sqrt{3}}{20}\right)}=\sin^{-1}{\left(\frac{\sqrt{3}}{4}\right)}$. The area of both sectors is $2 \left( \frac{\sin^{-1}{\left(\frac{\sqrt{3}}{4}\right)}}{360} \right) \pi(20)^2 = \left( \frac{20\pi}{9} \right) \sin^{-1}{\left(\frac{\sqrt{3}}{4}\right)}$.

Now just add up all of these areas to get:
$\frac{4100\pi}{3} + 300\pi + 25\sqrt{26}-25\sqrt{3} + \left( \frac{20\pi}{9} \right) \sin^{-1}{\left(\frac{\sqrt{3}}{4}\right)} = 4579.3416484884 \ldots$[/hide][/quote]

im not sure wha you did but you were only about 10 off

here is a diagram that i made.

...

well, to bad because i dont know how to put it on.  (it says "Sorry, you have reached your maximum Upload Quota Limit of 0 MB")

what does that mean?

[quote="10000th User"]The answer is
[hide]$T=\frac{4700}3\pi-400\cos^{-1}\left(\frac{\sqrt3}4\right)+25\sqrt3(\sqrt{13}-1)$[/hide]

jli: don't even try, it's not fixed yet. Peter VDD tried uploading attachments before but it wouldn't let him. The admins already know about this problem and hopefully they will fix it.[/quote]

which equals 4585.  My answer was 4589.

What is your [u]exact[/u] answer, jli? I'm pretty sure I got it right unless my diagram is off? :whistling:

okay...heres how i did it.

First calculate the area that the rope doesnt have to bend on a corner.  it is:
$\frac{2}{3} \cdot \pi \cdot 40^2=3351.0322$

Then calculate when the rope bends once.  It is:
$\frac{1}{3}\cdot \pi \cdot 30^3 = 942.4778$

Then comes the tricky part and i sorry if i cant explain it well without a diagram.  The rope now bends 2 times.  It forms two 1/6 circle of radius 20.  The volume of one of them is
$\frac{1}{6}\pi \cdot 20^2=209.4395$

now shade one of the 1/6 circles.  That leaves us with an equilateral triangle with side 10 and another shape.  the other shape should look like an equilateral triangle with one side bulging out and another bulging it.  you can mover the bulging side and fit it into the caved it side and you get another equilateral triangle with side 10.  The area of the two equilateral trianglwes combined is:
$2\cdot \frac{10^2\cdot \sqrt{3}}{4}=86.6025$

Add them together to get $\boxed{4589.552}$

[quote="10000th User"]The last part is not correct jli, because the intersection point where the two arcs of the 1/6 circles of radius 20 intersect is not the midpoint of the intersecting arcs. No wonder the area differs by a small amount from mine because the angle is actually off by about 4 degrees! I'll try to make a picture tonight and show you where you missed.[/quote]

oh yeah...i see now.

how did you do it?

[quote="10000th User"]Ok I know this might sound stupid but I had to create a MySpace.com to upload the figure. Don't bother viewing my MySpace.com, it only has a figure, the jli area figure.

[img]http://myspace-952.vo.llnwd.net/00852/25/92/852322952_l.jpg[/img]

Now to the math!

OK, points P,R,Q are vertices of the hexagon. M is the midpoint of PQ. X is the intersection point of the two arcs made by the 1/6 circles of radius 20. PB=QA=QX=20.

The 40 and 30 radius part are easy and their areas are $\frac{4100\pi}3$.

Add the 20 radius part and we find that it is $\frac{400\pi}3$.

So, the overlapping area is the region ARBX.

To simplify things, all we need to do is find the area of the region ARX. From the picture, [ARX]=[AQX]-[RQX]

[b]Area AQX[/b]
To find the area of this region, first recall that the area of a section is $\frac12r^2\theta$ where $\theta$ is the central angle with center at vertex of angle. So, $[AQX]=\frac12(20)^2\cdot\angle AQX$.

Consider right triangle MQX. We know $MQ=5\sqrt3$ and QX=20. Use definition of cosine and we find that

$\cos\angle MQX=\frac{MQ}{QX}=\frac{\sqrt3}4\Rightarrow \angle MQX=\cos^{-1}\left(\frac{\sqrt3}4\right)$

Right triangle MQR is a 30-60-90 triangle, so

$\angle AQX=\angle MQX-\angle MQR=\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6$

Put everything together to get

$[AQX]=\frac12(20)^2\left(\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6\right)$

[b]Area RQX[/b]
Area of RQX is simple, we only need to find the length of the base RX since the height is MQ. By Pythagorean theorem, $MX=5\sqrt{13}$. We know MR=5 (MQR is 30-60-90), so $RX=5(\sqrt{13}-1)$. Thus,

$[RQX]=\frac12(5\sqrt3)(5(\sqrt{13}-1))=\frac{25}2\sqrt3(\sqrt{13}-1)$

[b]Area ARX and ARBX[/b]
Combining the two previous result, we have

$[ARX]=\frac12\left(20^2\left(\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6\right)-25\sqrt3(\sqrt{13}-1)\right)$

$[ARBX]=2[ARX]=20^2\left(\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6\right)-25\sqrt3(\sqrt{13}-1)$

The total area in the 20 radius part is

$\frac{400\pi}3-20^2\left(\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6\right)+25\sqrt3(\sqrt{13}-1)$ or

$\frac{600\pi}3-400\cos^{-1}\left(\frac{\sqrt3}4\right)+25\sqrt3(\sqrt{13}-1)$

The area that the dog can be is

$T=\frac{4700\pi}3-400\cos^{-1}\left(\frac{\sqrt3}4\right)+25\sqrt3(\sqrt{13}-1)$

PS: Wow, sorry for the long long long solution :D[/quote]

Now that is a sweet solution, who cares if it is long, you were the one who got an answer.

P.S.  I also got that answer too, so cheers to the people who got the answer correctly.

[quote="10000th User"]Ok I know this might sound stupid but I had to create a MySpace.com to upload the figure. Don't bother viewing my MySpace.com, it only has a figure, the jli area figure.

[img]http://myspace-952.vo.llnwd.net/00852/25/92/852322952_l.jpg[/img]

Now to the math!

OK, points P,R,Q are vertices of the hexagon. M is the midpoint of PQ. X is the intersection point of the two arcs made by the 1/6 circles of radius 20. PB=QA=QX=20.

The 40 and 30 radius part are easy and their areas are $\frac{4100\pi}3$.

Add the 20 radius part and we find that it is $\frac{400\pi}3$.

So, the overlapping area is the region ARBX.

To simplify things, all we need to do is find the area of the region ARX. From the picture, [ARX]=[AQX]-[RQX]

[b]Area AQX[/b]
To find the area of this region, first recall that the area of a section is $\frac12r^2\theta$ where $\theta$ is the central angle with center at vertex of angle. So, $[AQX]=\frac12(20)^2\cdot\angle AQX$.

Consider right triangle MQX. We know $MQ=5\sqrt3$ and QX=20. Use definition of cosine and we find that

$\cos\angle MQX=\frac{MQ}{QX}=\frac{\sqrt3}4\Rightarrow \angle MQX=\cos^{-1}\left(\frac{\sqrt3}4\right)$

Right triangle MQR is a 30-60-90 triangle, so

$\angle AQX=\angle MQX-\angle MQR=\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6$

Put everything together to get

$[AQX]=\frac12(20)^2\left(\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6\right)$

[b]Area RQX[/b]
Area of RQX is simple, we only need to find the length of the base RX since the height is MQ. By Pythagorean theorem, $MX=5\sqrt{13}$. We know MR=5 (MQR is 30-60-90), so $RX=5(\sqrt{13}-1)$. Thus,

$[RQX]=\frac12(5\sqrt3)(5(\sqrt{13}-1))=\frac{25}2\sqrt3(\sqrt{13}-1)$

[b]Area ARX and ARBX[/b]
Combining the two previous result, we have

$[ARX]=\frac12\left(20^2\left(\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6\right)-25\sqrt3(\sqrt{13}-1)\right)$

$[ARBX]=2[ARX]=20^2\left(\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6\right)-25\sqrt3(\sqrt{13}-1)$

The total area in the 20 radius part is

$\frac{400\pi}3-20^2\left(\cos^{-1}\left(\frac{\sqrt3}4\right)-\frac{\pi}6\right)+25\sqrt3(\sqrt{13}-1)$ or

$\frac{600\pi}3-400\cos^{-1}\left(\frac{\sqrt3}4\right)+25\sqrt3(\sqrt{13}-1)$

The area that the dog can be is

$T=\frac{4700\pi}3-400\cos^{-1}\left(\frac{\sqrt3}4\right)+25\sqrt3(\sqrt{13}-1)$

PS: Wow, sorry for the long long long solution :D[/quote]

 :huh:

wow...nice solution, i would never have thought of that


Let  $ \frac{1}{1\plus{}(n\minus{}1)a_i}\equal{}x_i \to  a_i\equal{} \frac{1\minus{}x_i}{(n\minus{}1)x_i}$
 Purpose that: $ \sum_{i\equal{}1}^{n} x_i \equal{}1$  then we  need prove that: $ \sum_{i\equal{}1}^{n}\frac{1\minus{}x_i}{(n\minus{}1)x_i}  \geq 1$
 We  use  AM-GM  ineq :
$ (1\minus{}x_i) \geq (n\minus{}1)\sqrt[n\minus{}1]{\prod_{j \ne i} x_i}$
 hence  the problem is proved.

[quote="Inequalities Master"]Actually, i don't understand your solution.It seems for me that you are proving another problem :( .
But if I am wrong can someone help me to understand all this? :?:[/quote]
 i write  poorly  hence i can't  prove clearly, but this's method  in book  Hungkhtn's

Now I understood the solution. :)
If someone wants, I can write Hong Quy's solution again, well translated.
Again, an easy inequality.... :D

i want...i dun know what he is saying..so sorry.


logarithms
real analysis
real analysis unsolved

[quote="micael"]$ \ln{\left(\frac{1+\frac{i}{(n+1)^2}}{1+\frac{i}{n^2}} \right)}=\ln{\left(1-i\cdot\frac{2n+1}{n^2(n+1)^2} \right)} $ [/quote]
Unfortunately, the correct formula is
$ \ln{\left(\frac{1+\frac{i}{(n+1)^2}}{1+\frac{i}{n^2}} \right)}=\ln{\left(1-i\cdot\frac{2n+1}{(n^2+i)(n+1)^2} \right)} $, which invalidates micael's nice argument... :(  ;)

Sorry for the mistake, I have deleted the post


Let  $a,b,c>0$. Prove that  $\dfrac{ab}{1+ab}+\dfrac{bc}{1+bc}+\dfrac{ca}{1+ca}\geq\dfrac{9abc}{a+b+c+3abc}$


I wrote the Junior CCC. Unfortunately my score was only 49. Bleh
I'm still not sure how you would solve question 4 either.

I wrote the senior test and I found it to be much, much, much harder compared to the questions I looked at from last years senior exam. I finished 2/5 problems, when I could have done 4/5 easily on last years exam.

The junior problems didn't look too hard, so I was kicking myself after the 3 hours were up that I didn't do that exam instead. (I did this for admission reasons, and I'm not sure how unimpressed Waterloo is going to be about my terrible performance  :oops:)

So now  it comes down to the Euclid test, which I will have to force myself to actually study for.  :(

How did everyone else find the senior problems?


I'm not quite sure how to solve this problem.
I've attached the picture.
here's the question: create a net force equation for each object, suppose there's friction, tension in rope and the system accelerates to the right.
Thank you so much in advance.

i get it but why did you have m1 in the net force formula in m2a2. isn't that:
m2a2= T12- m2gu- T23?
Thank you so much for your help!

Typo.  It should indeed be $ m_{2}$. :oops:


A real function $ f$ satisfies $ 4f(f(x))\minus{}2f(x)\minus{}3x\equal{}0$ for all real numbers $ x$. Prove that $ f(0)\equal{}0$.

Cnyd's solution is hard to read, so let's write his solution properly.

Assume that $ f(0)\equal{}c$. Substituting $ x\equal{}0$ to the equation we get $ 4f(f(0))\minus{}2f(0)\equal{}0$, so $ 4f(c)\equal{}2c$. Therefore $ f(c/2)\equal{}c/4$. Substituting $ x\equal{}c$ to the original equation we get $ 4f(f(c))\minus{}2f(c)\minus{}3c\equal{}0$. But $ f(c)\equal{}c/2$ so we get $ 4f(c/2)\minus{}c\minus{}3c\equal{}0$, or $ f(c/2)\equal{}c$. Thus $ c\equal{}c/2$ so $ c\equal{}0$.


function
real analysis
real analysis unsolved

Abolutely continuous (for a real function) requires finite set of disjoint intervals.

If we remove "disjoint", will it be STRICTLLY stronger?

I guess so , but have not found an example.

Can anybody show me one example?

Thx!

I think the stronger version is equivalent to Lipschitz continuity.
Example: $ f(x)\equal{}\sqrt{x}$. Take the intervals $ [0,1/n^2]$.


function
quadratics
analytic geometry
graphing lines
slope

Find all the real values of $\lambda$ such that $\lambda x^2-2(\lambda-1)x+\lambda+2>0$ for any $x\in[0,3]$.

We can rewrite the given equation as $\lambda (x-1)^2+2(x+1)>0\ \cdots[\#]$,so we are to find the range of $\lambda$ for which $[\#]$ holds for any $x\in{[0,3]}$.

Case 1: For $\lambda=0$; $[\#]\Longleftrightarrow 2(x+1)>0$,since the linear function $y=2(x+1)$ is increasing in the range of $0\leq x\leq 3$.

Case 2: For $\lambda >0$; $[\#]\Longleftrightarrow -(x-1)^2<\frac{2}{\lambda}(x+1)$,so finding the condotion that the quadratic function $y=-(x-1)^2$ is lower than the linear function $y=\frac{2}{\lambda}(x+1)$ passing through the fixed point $(-1,0)$ in the range of $0\leq x\leq 3$, we have $\frac{2}{\lambda }>0\Longleftrightarrow \lambda >0\ (\because\lambda >0)$.

Case 3:For $\lambda <0$; $[\#]\Longleftrightarrow -(x-1)^2>\frac{2}{\lambda}(x+1)$, so finding the condition that the quadratic function $y=-(x-1)^2$ is upper than the linear function $y=\frac{2}{\lambda}(x+1)$ in the range of $0\leq x\leq 3$, since the slope of the line is $\frac{2}{\lambda}=-1$ when the line passes through the points $(0,-1)$ and $(3,-4)$ simultaneously,

 we have $\frac{2}{\lambda}<-1\Longleftrightarrow -2<\lambda <0\ (\because\lambda <0)$. Finally the desired answer is $\boxed{\lambda >-2}$


inequalities
triangle inequality
inequalities proposed

hi,

let $a,b,c$ be nonnegative reals .

prove that :

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

It's easy with the geometric interpretation.

[quote="toetoe"]could you be more clair ,please ? :|[/quote]
First Fermat point and triangle inequality.

My not sure solution is write a non-regular pentagon $PQRST$ which has sides $PQ=a,QR=b,RS=c,ST=a,\angle PQR=60^{\circ},\angle QRS=60^{\circ},RST=120^{\circ}$. Then use law of cos and Triangle Inequality. We have $\sqrt{a^{2}+b^{2}-ab}+\sqrt{b^{2}+c^{2}-bc}=PR+QS$. Let $PR \cap QS = U$ then $PU+UQ+QU+US>a+c=RS+ST>RT=\sqrt{a^{2}+c^{2}+ac}$

This is picture :D
[img]http://img61.imageshack.us/img61/2500/ineqpu5.jpg[/img]

I think this is simpler.

Construct quadrangle $ABCD$, with $\angle ADB=\angle BDC=60^{o}$ and $AD=a, BD=b,CD=c$

Then $AB=\sqrt{a^{2}+b^{2}-ab}$, $BC=\sqrt{b^{2}+c^{2}-bc}$, and $CA=\sqrt{c^{2}+a^{2}+ac}$

By triangle inequality, we can get that $AB+BC \geq CA \Rightarrow \sqrt{a^{2}+b^{2}-ab}+\sqrt{b^{2}+c^{2}-bc}\geq \sqrt{c^{2}+a^{2}+ac}$

Actually, I have seen the solution long time ago  :P


algebra
polynomial
algorithm
number theory
greatest common divisor
quadratics
function

Let $k\in N$ and the polynomial $P_{k}(X)=X^{k-1}+X^{k-2}+...+X+1$. Prove that if $m,n\in N, (m,n)=1$ then $P_{n}(X)\cdot P_{m}(X)$ divides $P_{nm}(X)$. (The almost fun part would be proving it without using the complex roots of $1$)

[hide="Quick"] Equivalently,

$(X^{n}-1)(X^{m}-1) | (x-1)(X^{nm}-1)$

It's clear that each factor divides $X^{nm}-1$.  By Euclidean Algorithm on polynomials, $gcd(X^{n}-1, X^{m}-1) = X-1$, so the conclusion follows.    [/hide]

t0rajir0u used the lemma that: $( x^{m}-1,x^{n}-1)=x^{ (m,n)}-1$.

I think it is easier to prove by considering the set of the roots of both polynomials. They are both roots of unity. So we can compare roots by looking at their complex argument. That seems more straightforward than using the euclidean algorithm. I don't see how that conclusion directly follows from that method of argument.

[b]Lemma:[/b]  $a | c, b | c, gcd(a, b) = 1 \implies ab | c$

This can be proved by Chinese Remainder Theorem.  And yes, all of this reasoning is as valid with polynomials as it is with naturals; now, since I showed that

$gcd(P_{n}, P_{m}) = 1, P_{n}| P_{nm}, P_{m}| P_{nm}$

It follows from the lemma that $P_{n}P_{m}| P_{nm}$.  QED.

I prefer this argument over the argument by roots of unity if only because it uses nothing more than the same ideas as integer number theory, just applied to polynomials.

Thats interesting. I have no concrete source for these things...so I like to learn a bit more.

Can we look at this from a field extension type of thing? In a sense, we are looking at $R[x]$, and we need to have our common factor be also from $R[x]$.

In general, can we say

if $F$ is a field, and $a,b,c\in F$, if we have $\frac{c}{a}\in F$, $\frac{c}{b}\in F$, and $a$, $b$ have no common factor in $F$ that is not a unit, then $\frac{c}{ab}\in F$

[does that "unit" part make sense? in $R[x]$, can we have a constant as a common factor?],

Is our gcd uniquely defined? it seems that 2x and x both divide x^2
Is it convention to say: whatever the euclidean algorithm produces is the gcd? What if it is difficult to define the euclidean algorithm in our field? for polynomials, it seems that we multiply one factor by $kx^{t}$, where $k$, $t$ are constant, etc. What about $R[x][y]$?
Does our field need to have some ordering? WOP for integers, degrees of terms for polynomials, etc

So:

What if we take $Z[\sqrt{2}]$? can we still use the same lemma? Are there certain quadratic field extensions of the integers where that lemma breaks down?

Certain properties of the field that make the lemma break down?

I seem to remember glancing over something about: unique prime factorization down to units.

[quote="Altheman"] if we have $\frac{c}{a}\in F$, $\frac{c}{b}\in F$[/quote]
What exactly do you understand by those fraction? Is it $c\cdot a^{-1}$? I think that the answers for your question can be found in the Superior Algebra forum because from what I can understand your asking about prime elements in fields which makes a vaste domain.

You don't want to use fields (in a field, every non-zero element is a unit and division is always possible, except by $0$, so your property is pretty much redundant), what you're after are [url=http://en.wikipedia.org/wiki/Integral_domain]Integral Domains[/url]. You can scroll down to the "Divisibility, prime and irreducible elements". And, of course, you can search for related stuff.

[quote="Altheman"]What if it is difficult to define the euclidean algorithm in our field? [/quote]

In fact, this is a valid concern.  The Euclidean Algorithm is equivalent to the division algorithm, and, for example, for many natural values of $D$ it is not possible to define a convenient division algorithm in $\mathbb{Q}[ \sqrt{-D}]$ (take, for example, $D = 5$).  In fact, what you're really after is a [url=http://mathworld.wolfram.com/PrincipalIdealDomain.html]principal ideal domain[/url].

Thanks for all the help. I will look more into this subject.


trigonometry
inequalities
inequalities unsolved

Let $ x,y\in [0,\frac{\pi}{2}]$. Prove that:
$ \cos x \plus{}\cos y \le 1\plus{}\cos xy$

see here http://www.mathlinks.ro/viewtopic.php?t=305377 for reference


function
inequalities
algebra proposed
algebra

Suppose that $S$ is set of $f:N \rightarrow  R$ such that:

$f(1)=2$
$f(n+1)\ge f(n) \ge \frac{n}{n+1}f(2n)$

Find the minimum $M \in N$ such that $f(n)\le M$ for all $n \in N$ and $f \in S$.

From the condition we have $f(2n)\leqq \frac{n+1}{n}f(n) $,since $2^n\geqq n+1$ we get

$f(n)\leqq \frac{\frac{n}{2}+1}{\frac{n}{2}}=f\left(\frac{n}{2}\right)\leqq \frac{n+2}{n}\cdot \frac{n+2^2}{n}f\left(\frac{n}{2^2}\right)$

$\leqq \cdots\cdots\ \leqq \frac{n+2}{n}\cdot \frac{n+2^2}{n}\cdot \frac{n+2^3}{n}\cdots\cdots \frac{n+2^n}{n}f\left(\frac{n}{2^n}\right)$

$\leqq \frac{n+2}{n}\cdot\frac{n+2^2}{n}\cdot \frac{n+2^3}{n}\cdots\cdots \frac{n+2^n}{n}f\left(\frac{n}{n+1}\right)$

I also don't have it, but simply because I'm too lazy to make some computations. The idea is the following:
   From $ f(2n)\leq\frac{n+1}{n} f(n)$ you will find that $f(n)\leq 4(1+1/2)(1+1/4)(1+1/8)...(1+1/2^{n-1})$. This can be done either by substituting $ n=2^k$ and multiplying the relations or directly, by induction. Now, the fact that $f(n)$ is bounded by some M is equivalent to the fact that $f(2^k)$ is bounded by some M, due to monotony of the function. Since one can take very well the function $ f(n)=4(1+1/2)(1+1/4)...(1+1/2^{n-1})$ which is in S, we can easily see that we are asked to compute the largest natural N such that $ 4\cdot\prod_{n\geq 1}{1+1/2^k)}<N$. And now comes the principal "quality" of a chinese TST problem: computations!  One could compute the first 7-8-9-10 terms  :D  to see what is the possible N and then the idea could be to write $ (1+1/2^k)(1+1/2^{k+1})...<e^{\frac{1}{2^{k-1}}}$ and to choose some $k=7-8-9-10$  :D  so that you can prove it.


Euler
geometry
geometric transformation
reflection
circumcircle
ratio
perpendicular bisector

Let ABC be a triangle and P a point on its Euler line. Denote by X, Y, Z the reflections of the point P in the perpendicular bisectors of the segments BC, CA, AB, respectively. Prove that the lines AX, BY, CZ concur at one point, and the isogonal conjugate of this point with respect to triangle ABC lies on the Euler line of triangle ABC.

This is a result from Hyacinthos message #2211. I haven't posted it here earlier because just now I found a satisfying synthetic proof. Can you find one too?

  darij

Let $H$ and $O$ be respectively the orthocenter and the circumcenter of $ABC$. Then $P$ lies on the line $OH$. Call $K$ the deLongchamp point of $ABC$ (i.e the reflection of $H$ at the point $O$). Let $A'B'C'$ be the tangential triangle of $ABC$ and $MNP$ be the median triangle of $ABC$.

Here is the key lemma:

[i]Lemma[/i] : Consider points $X', Y', Z'$ dividing $OA', OB', OC'$ respectively in the ratio $\frac{OX'}{OA'} = \frac{OY'}{OB'} =\frac{OZ'}{OC'} = t$. Then the lines $AX', BY', CZ'$ are concurrent at the isogonal conjugate of the point $S$ on the line Euler line OH in the ratio $\frac{OS}{SH} = \frac{1}{2t}$.

For the proof of this lemma, please see theorem 3 in http://forumgeom.fau.edu/FG2005volume5/FG200502.pdf

Thanks to this lemma, we have the light to approach the problem. Let $A_1, B_1, C_1$ be the reflections of $A, B, C$ at the perpendicular bisector of $BC, CA, AB$ respectively. Denote by $X'$ the intersection of $A_1P$ and $AX$. Then $X'$ lies on $OA'$. Analogously, we define the points $Y', Z'$. If we can show that $\frac{OX'}{OA'} = \frac{OY'}{OB'} =\frac{OZ'}{OC'} = t$, then from the lemma, the problem is solved.

It is sufficient to prove that $OX'.OM = OY'.ON =OZ'.OP$ since $OM.OA' = ON.OB'=OP.OC' = R^2$. It is easy to see that $KA_1$ is perpendicular to $BC$. Hence $KA_1$ is parallel to $OX'$. Thus $\frac{OX'}{KA_1} = \frac{PO}{PK}=k$. As a result, we obtain

$OM.OX' = \frac{PO}{PK}OM.KA_1=k.RcosA (-2Rcos(B-C) + 2RcosA)= -2kR^2cosAcosBcosC$.

By similar argument, we obtain

$OM.OX'=ON.OY'=OP.OZ'=-2kR^2cosAcosBcosC$.

Therefore, the problem is solved.

[b]Remark[/b] Let $\frac{PO}{PH}=m$. Then $k = \frac{m}{2m-1}$. Thus $t = \frac{2m}{1-2m}.cosAcosBcosC$. Hence the concurrent point of $AX, BY, CZ$ is the isogonal conjugate of the point $S$ which lies on $OH$ in the ratio $\frac{OS}{SH} = \frac{1}{2t} = \frac{1-2m}{4m.cosAcosBcosC}$.

Luis's proof here (post #5):

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=587985

Here is a "nice" proof for this problem. :) The main idea begin from [url=https://artofproblemsolving.com/community/c374081h1362657]this problem[/url].

Let $O$ and $H$ be the circumcentre and orthocentre of triangle $ABC$, respectively. $AX$ meets $OH$ at $T$, the isogonal line of $AX$ wrt $\angle $$ABC$ meets $OH$ at $S$. Note that $AO$, $AH$ are isogonal wrt $\angle$$ABC$ so $AX$, $AS$ are isogonal wrt $\angle$$ABC$. By Steiner theorem we get that : $\tfrac{SH}{SO}$ $=$ $\tfrac{AH^2}{AO^2}$ $\cdot$ $\tfrac{TO}{TH}$ imply $\tfrac{SH}{SO}$ $=$ $\tfrac{AH^2}{AO^2}$ $\cdot$ $ \tfrac{TO}{TH}$. Let $G$, $R$ be the intersection of $XO$ and $AH$, circle $(ARO)$ and $AO$. Apply Menelaus theorem fro $\triangle OGH$ with $\overline{ATX}$ we get that : $\tfrac{TO}{TH}$ $=$ $\tfrac{XO}{XG}$ $\cdot$ $\tfrac{AG}{AH}$ imply $\tfrac{SH}{SO}$ $=$ $\tfrac{AH\cdot XO\cdot AG}{AO^2\cdot AG}$. Note that $\tfrac{XO}{XG}$ $=$ $\tfrac{OP}{OP+OH}$ so $\tfrac{SO}{SH}$ $=$ $\tfrac{AH\cdot OP\cdot AG}{AO^2\cdot (OP+OH)}$. Since $HGRO$ is a cyclic quadrilateral so $AG\cdot AH=AR\cdot AO$, note that $\angle HRO$ $=$ $\angle AHO$ so $\triangle HRO$ $\sim $ $\triangle AHO$. Thus $OR$ $=$ $\tfrac{OH^2}{AO}$, hence we conclude that : $\tfrac{SH}{SO}$ $=$ $(1-\tfrac{OH^2}{OA^2})$ $\cdot$ $\tfrac{OP}{OP+OH}$. Similarly with $B$, $C$ and we are done. $\square$

$\textbf{Simple proof :}$

Consider variety of points $P_t$ on Euler line of $ABC$. Also define reflections $X_t, Y_t, Z_t$. Let $AX_t\cap BY_t = Q_t$. From projective geometry (cross-ratios properties) it's well known that while point $P_t$ moving on Euler line, point $Q_t$ move on some conic $\mathcal{H}$. By assuming some particular cases we get that $A, B, C, H, O\in\mathcal{H}$. So now it's easy to check that $Q\in CZ_t$ for every $t$. So lines $AX, BY, CZ$ are concurrent at point on conic $\mathcal{H}$ which goes through $A, B, C, O, H$. Also from well known property we get that isogonal conjugation of $Q$ wrt $ABC$ lies on Euler line. $\Box$


LaTeX
inequalities
real analysis
real analysis unsolved

Suppose that E, A, B ⊂ ℜ and E = A ∪ B. Prove that if E has a supremum and both A and B are nonempty, then sup A and sup B both exist, and sup E is one of the numbers sup A or sup B.

You can use $ \text{\LaTeX}$ for math expressions.

Let $ s =\sup S$. Then $ s$ is an upper bound of both $ A$ and $ B$. So both $ A$ and $ B$ has supremum.

Let's denote them by $ a =\sup A$, $ b =\sup B$ and $ m =\max(a, b)$. Then $ m\leq s$.

Also, note that for all $ x\in S = A\cup B$, $ x\leq m$. This means that $ m$ is an upper bound of $ S$. Hence $ m\geq s$.

Combining two inequality gives $ s = m$.


inequalities
function
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that

$ \sqrt{(5a^2\plus{}3)(5b^2\plus{}3)} \plus{}\sqrt{(5b^2\plus{}3)(5c^2\plus{}3)}  \plus{}\sqrt{(5c^2\plus{}3)(5a^2\plus{}3)} \ge 24$.

[quote="secrets"]I think we can prove:
$ (5a^2 \plus{} 3)(5b^2 \plus{} 3)(5c^2 \plus{} 3) \ge 512$  :)[/quote]
After homogenization your inequality is wrong when $ a\rightarrow\plus{}\infty.$  :wink:

It is quite a cool inequality but I don't think that it is particularly hard... WLOG $ a\ge b\ge c$. Then

$ f(a,b,c)\ge f(a,c,c)\equal{}f(3\minus{}2c,c,c)$ which is easy enough...

$ 2\sqrt{(3\plus{}5(3\minus{}2c)^2)(3\plus{}5c^2)}\plus{}3\plus{}5c^2\ge 24\iff$

(note $ 21\minus{}5c^2\ge 21\minus{}5>0$)

$ 4(3\plus{}5(3\minus{}2c)^2)(3\plus{}5c^2)\minus{}(21\minus{}5c^2)^2\equal{}15(c\minus{}1)^2(5c\minus{}3)^2\ge 0$

The equality cases follow. Due to the nature of the function, we can effectively equalize two of the variables...

[quote="Altheman"]
$ f(a,b,c)\ge f(a,c,c) \equal{} f(3 \minus{} 2c,c,c)$ which is easy enough...
[/quote]

Alex, i think you should check $ f(a,b,c)\ge f\left(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2}\right)$. Your smoothing is not taking into account the constrain. Either way, I think it's still true, but I don't feel like checking that.


modular arithmetic
induction
pigeonhole principle
combinatorics unsolved
combinatorics

The smallest possible $n$ is $7$. To show that a smaller $n$ is not enough, take $7$ senators $0, 1, 2, 3, 4, 5, 6$ such that senator $i$ hates senators $i+1, i+2, i+3 \pmod{7}$. Then in any pair of these senators, one hates the other, so no two can be together in a commitee, and we need at least $7$ commitees.

To show that $7$ is enough, let $k=51$ and do induction on $k$. The base case is obvious. If the assertion is proved for $k$, take $k+1$ senators; there are exactly $3k+3$ instances of someone hating someone else, so by pigeonhole there's a senator hated by at most $3$ other senators. Take him out and put the other $k$ senators into $7$ good commitees (obviously we can do this by IH); now, since the left out senator is hated by at most $3$ people and hates exactly $3$, at most $6$ commitees are forbidden and there's at least one commitee for him, so we're done.

we have a generalization of it:
for every directed graph that for every vertice of G outest degree is $\leq k$ we can partition the graph to $2k+1$ independent set.
it can be easily prove by induction on number of vertices and with use of cancellation of the vertice with minimum inner degree.

[quote=MithsApprentice]There are 51 senators in a senate. The senate needs to be divided into $n$ committees so that each senator is on one committee. Each senator hates exactly three other senators. (If senator A hates senator B, then senator B does [i]not[/i] necessarily hate senator A.) Find the smallest $n$ such that it is always possible to arrange the committees so that no senator hates another senator on his or her committee.[/quote]

We prove a generalized version of the problem :\\
[quote=Generalized version]
Let $G$ be a DG on $n$ vertices such that no vertex has an out-degree greater than $k$. Show that the vertices of $G$ can be colored in $2k+1$ colors such that no two vertices of the same color are connected.
[/quote]

[u]Proof:[/u] We proceed by induction on $n$. The base case for $n=3$ is trivial. So suppose that the assertion is true for a graph on $n$ vertices. Now consider a graph on $n+1$ vertices such that no vertex has an out-degree greater than $k$. Notice that this graph, $G'$, must contain a vertex having in-degree $\le k$ i.e. $\exists \nu \in \text{V} (G')$ such that $\delta \nu \le 2k$. If not, then notice that the total number of edges would be greater than $(n+1)k$, a contradiction to the assertion. Thus, we can partition $G'$ into $\nu \cup G'\setminus \nu$. By the induction hypothesis,$G'\setminus \nu$ can be colored in $2k+1$ colors. Now $\nu$ has at most $2k$ neighbors in the other set and hence at least one color will always be available for $\nu$ different from that of its neighbors.


Suppose that a and b are two distinct positive integers such that ab(a+
b) is a multiple of a2 + ab + b2. Prove that |a − b| > p3 ab.

Let d = gcd(a;b). So a = dx and b = dy with x, y $ \in N$  ; (x,y) = 1.
 Hence, xy(x+y)$ d^3$ divisible by $ d^2(x^2 \plus{} xy \plus{} y^2)$. Therefore, xy(x+y)d  divisible by $ x^2 \plus{} xy \plus{} y^2$.
 Because (x;y) = 1, $ (xy; x^2 \plus{} xy \plus{} y^2) \equal{} 1$ and  $ (x \plus{} y; x^2 \plus{} xy \plus{} y^2) \equal{} 1$.
So d divisible by $ x^2 \plus{} xy \plus{} y^2$. Hence d > xy $ \rightarrow$ $ d^3{|x \minus{} y|}^{3} > d^2xy$ $ \rightarrow$ $ |a \minus{} b|^3 > ab$.
So we have the proof.

P/S: WALID RUSHDY, please use Latex.


A thin ring is of mass $M$ and radius $R$.
Find the moment of inertia of this ring about its diameter.

My calculation gives me the answer that $\large I=\frac{MR^2}{2}$

$dM=R\lambda d\theta$

$\large dI=dMr^2=\lambda R^3\cos^2\theta d\theta$

$\large I=2\int_{0}^{\pi}\lambda R^3\cos^2\theta d\theta=\pi R^3\lambda$

But $\large M=2\pi\lambda R$

$\Longrightarrow I=\frac{MR^2}{2}$


I just wonder to know whether the following one is ture or false?
For $ a,b,c>0$
\[ \frac{1}{4a^2\minus{}ab\plus{}4b^2}\plus{}\frac{1}{4b^2\minus{}bc\plus{}4c^2}\plus{}\frac{1}{4c^2\minus{}ca\plus{}4a^2}\]
\[ \geq\frac{9}{16(ab\plus{}bc\plus{}ca)\minus{}18abc(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a})}\]

[quote="Tourish"]I just wonder to know whether the following one is ture or false?
For $ a,b,c > 0$
\[ \frac {1}{4a^2 \minus{} ab \plus{} 4b^2} \plus{} \frac {1}{4b^2 \minus{} bc \plus{} 4c^2} \plus{} \frac {1}{4c^2 \minus{} ca \plus{} 4a^2}\]

\[ \geq\frac {9}{16(ab \plus{} bc \plus{} ca) \minus{} 18abc(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a})}\]
[/quote]

[size=200]The inequality holds[/size]

I got this inequlity some days ago when I try to show a problem of Vasc
\[ \frac{1}{4a^2\minus{}ab\plus{}4b^2}\plus{}\frac{1}{4b^2\minus{}bc\plus{}4c^2}\plus{}\frac{1}{4c^2\minus{}ca\plus{}4a^2}\geq\frac{9}{7(a^2\plus{}b^2\plus{}c^2)}\]
But I am still fail.In fact,Both
\[ 16(ab\plus{}bc\plus{}ca)\minus{}18abc(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a})\leq 7(a^2\plus{}b^2\plus{}c^2)\]
and \[ 16(ab\plus{}bc\plus{}ca)\minus{}18abc(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a})\geq 7(a^2\plus{}b^2\plus{}c^2)\]
are false!!

[quote="Tourish"]I got this inequlity some days ago when I try to show a problem of Vasc
\[ \frac {1}{4a^2 \minus{} ab \plus{} 4b^2} \plus{} \frac {1}{4b^2 \minus{} bc \plus{} 4c^2} \plus{} \frac {1}{4c^2 \minus{} ca \plus{} 4a^2}\geq\frac {9}{7(a^2 \plus{} b^2 \plus{} c^2)}\]
But I am still fail.In fact,Both
\[ 16(ab \plus{} bc \plus{} ca) \minus{} 18abc(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a})\leq 7(a^2 \plus{} b^2 \plus{} c^2)\]
and
\[ 16(ab \plus{} bc \plus{} ca) \minus{} 18abc(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a})\geq 7(a^2 \plus{} b^2 \plus{} c^2)\]
are false!![/quote]

you are right

$ 16(ab \plus{} bc \plus{} ca) \minus{} 18abc(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a})\leq 7(a^2 \plus{} b^2 \plus{} c^2)$

and

$ 16(ab \plus{} bc \plus{} ca) \minus{} 18abc(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a})\geq 7(a^2 \plus{} b^2 \plus{} c^2)$

$ \frac {9}{4a^{2} \minus{} ab \plus{} 4b^{2} \plus{} 4b^{2} \minus{} bc \plus{} 4c^{2} \plus{} 4c^{2} \minus{} ca \plus{} 4a^{2}}$
$ \geq\frac {9}{16(ab \plus{} bc \plus{} ca) \minus{} 18abc(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a})}$

does not hold.


I know this probably is something discussed extensively. I was searching in the forum and, in any case, I couldn't find the answer to the questions
Why is it that the problems of the Chinese National Olympiad are so hard to find?
Are they in the web in chinese sites? If so, could somebody translate them? I remember they were on the kalva site but it seems that they have been no longer there for some time now.

Daniel

well,we chinese have a big forum in mathematical olympiad

http://www.aoshoo.com/bbs1

we discuss problems in chinese,and most of our teachers are not excellent in english.

just wait,my friend.some of them will translate these problems and solutions.


ratio
geometry
perimeter
modular arithmetic
algebra
polynomial
number theory unsolved

If you have a partition of the integers (pairwise disjoint subsets which together are the integers) in arithmetic sets, i.e. {a, a+d, a-d, a+2d, a-2d, ...} (with some d>1) then there
are two sets which have the same d.

Does anybody know a solution to this problem? Unfortunately, I have forgotten it. :(

I think this can done as follows :

Let $d_1, \cdots , d_p$ be the respective ratio of the arithmetical sequences. Let $n= d_1 \cdot d_2 \cdots d_p$, and for each $i$, let $n=k_id_i$.
Now consider a circle with radius $R = \frac n {2 \pi}$. Starting at some arbitrary point, say $O$, as origin on the circle, let's consider the usual correspondance between the integers and points on the circle. Since the perimeter of the circle is a multiple of each $d_i$, it follows that for each $i$, the integers which belong to the arithmetical sequence with ratio $d_i$ form a regular $k_i$-gon on the circle. That is : the vertices of the regular $n$-gon one of which being $O$ are divided into $p$ disjoint subsets, each of which formed by the vertices of a regular polygon.

Now, the problem is to prove that there are two of these polygons which have the same number of vertices that is $k_i = k_j$ for some $i \neq j$).

This last problem is an old and beautiful problem from All-soviet union. I'm quite sure that a solution already appeared here, but I do not find it. If nobody find this thread, I will write it again.

Pierre.

Alex-S, you forgot to mention that the partition is into [i]finitely[/i] many arithmetic progressions (an example of a partition into infinitely many progressions in which the assertion fails is $2^{i} \pmod{2^{i+1}}, i=1,2,\ldots$.)

Basically, the maximal of the differences must appear at least twice; this is stronger than the assertion that some two $d$'s are equal. In other words, if $1 < d_1 \leq d_2 \leq \cdots \leq d_k$ are the differences of the progressions, then $d_{k-1}=d_k$. To show this, let $a_i \pmod{d_i}$ be the given arithmetic progressions, $1 \leq i \leq k; 0 \leq a_i < d_i$. In terms of formal power series, the condition that these progressions form a disjoint partition of the integers is equivalent to
\[
\sum_{i=1}^k \frac{x^{a_i}}{1-x^{d_i}} = \frac{1}{1-x}.
\]
Assume to the contrary that $d_k>d_{k-1}$. The above identity holds for $|x|<1$. By clearing denominators, we obtain a polynomial identity which holds for all $|x|<1$, and therefore for all $x \in \mathbb{C}$.In particular, it holds for $x=\omega$, where $\omega$ is a primitive $d_k$th root of unity. This is a contradiction, because $\omega^{d_k} - 1=0$ and $\omega^{d_i} - 1 \neq 0$ for $1 \leq i < k$. Hence, $d_{k-1}=d_k$.

--Vesselin

I like to not have to write long solutions.

Pierre.

Thanks to vess for his nice solution. With the hint that it holds even dk-1=dk I found another one. ( I take the same letters like you) Assume the statement is wrong. We take the smallest counter example (where k is minimal). Let p be a prime which divides dk.

Case 1)
p devides all di,
then we can consider those subsets which have ai ≡ ak mod p (there are at least two of them as dk ≠ p because p | dk-1 <= dk=p would yield to dk-1 =dk). Those are a partition of the integers ≡ ak mod p, isomorph to another counter example of the statement and k was not minimal.

Case 2)
Not all di are divisible by p.
Let dl be the greatest of the numbers which are not divisible by p and dm the greatest of the numbers which are divisible by p and where am ≢ ak mod p (exists because all mod classes of the integers "have the same number of numbers"). By assumption, we have dk ≠ dm. So dm ≠ p * dl or dk ≠ p * dl, respectively. If we look now on the integers ≡ am mod p respectively ≡ ak mod p then we get another counter example (if you think a bit) so that k was not minimal.

I want to ask the relationship of this problem with my problem in [url]http://www.mathlinks.ro/viewtopic.php?t=152442[/url].
Mine is also partition with each partition has a finite number of elements, but $n$ isn't infinite. The problem here says integer which is infinite.
So, can we "really" say that it's equivalent to my problem ?  :maybe:

And then, about Vess' solution. In my thought, that solution works if the number of elements in each partition is infinite. Since $\frac{x^{a_{i}}}{1-x^{d_{i}}}= \sum_{k=0}^{\infty}x^{a_{i}+k \cdot d_{i}}$.


geometry
trigonometry
geometric transformation
reflection
circumcircle
radical axis
geometry unsolved

Let $ABC$ be an acute triangle and let $k_{1},k_{2},k_{3}$ be the circles with diameters $BC,CA,AB$, respectively. Let $K$ be the radical center of these circles. Segments $AK,CK,BK$ meet $k_{1},k_{2},k_{3}$ again at $D,E,F$, respectively. If the areas of triangles $ABC,DBC,ECA,FAB$ are $u,x,y,z$, respectively, prove that \[u^{2}=x^{2}+y^{2}+z^{2}.\]

Since the triangle sides are diameters of the circles $k_{1}, k_{2}, k_{3},$ their pairwise radical axes are the triangle altitudes and their radical center K is the triangle orthocenter. Let $h_{a}= AA',\ h_{b}= BB',\ h_{c}= CC'$ be the A-, B-, C-altitudes of $\triangle ABC$ and let $h_{d}= DA',\ h_{e}= EB',\ h_{f}= FC'$ be the D-, E-, F-altitudes of $\triangle BCD,\ \triangle CAE,\ \triangle ABF.$

$h_{a}= b \sin C = c \sin B,\ h_{b}= c \sin A = a \sin C,\ h_{c}= a \sin B = b \sin A$

$h_{a}^{2}= bc \sin B \sin C,\ h_{b}^{2}= ca \sin C \sin A,\ h_{c}^{2}= ab \sin A \sin B$

Powers of A', B', C' to the circles $k_{1}, k_{2}, k_{3}$ are

$h_{d}^{2}= BA' \cdot CA' = bc \cos B \cos C$

$h_{e}^{2}= CB' \cdot AB' = ca \cos C \cos A$

$h_{f}^{2}= AC' \cdot BC' = ab \cos A \cos B$

$\frac{x^{2}+y^{2}+z^{2}}{u^{2}}= \frac{h_{d}^{2}}{h_{a}^{2}}+\frac{h_{e}^{2}}{h_{b}^{2}}+\frac{h_{f}^{2}}{h_{c}^{2}}=$

$= \cot B \cot C+\cot C \cot A+\cot A \cot B = 1$

for $\angle A+\angle B+\angle C = 180^\circ.$

[color=darkred][b]Lemma 1[/b] (well-known). Let $ABC$ be a triangle $ABC$ with the orthocenter $H\ .$

Denote the points $D\in AH\cap BC$ and the reflection $D_{1}$ of the point $H$ w.r.t. the line $BC\ .$

Then the point $D_{1}$ belongs to the circumcircle of the triangle $ABC$ and exists the relation $\boxed{\ DB\cdot DC=AD\cdot HD\ }\ .$

[b]Lemma 2.[/b] Denote the area $[XYZ]$ of the triangle $XYZ\ .$ Let $ABC$ be an [b][u]acute[/u][/b] triangle with the orthocenter $H$.

Denote the point $D\in AH\cap BC$ and $D'\in (AH)$ so that $D'B\perp D'C\ .$ Then $\boxed{\ [BD'C]^{2}=[BAC]\cdot [BHC]\ }\ .$[/color]

[b]Proof.[/b] $\frac{[BD'C]^{2}}{[BAC]\cdot [BHC]}=$ $\frac{[BD'C]}{[BAC]}\cdot\frac{[BD'C]}{[BHC]}=$ $\frac{D'D}{AD}\cdot \frac{D'D}{HD}=\frac{DB\cdot DC}{AD\cdot HD}=1$ $\Longrightarrow$ $[BD'C]^{2}=[BAC]\cdot [BHC]\ .$

[b]Remark.[/b] Analogously define the pairs of points $\{\ E\in BH\cap CA\ ;\ E'\in (BH)\ ,\ E'C\perp E'A\ \}\ ,$

$\{\ F\in CH\cap AB\ ;\ F'\in (CH)\ ,\ F'A\perp F'B\ \}$ $\Longrightarrow$ $\boxed{\ \sum_{\mathrm{cyc}}[BD'C]^{2}=[ABC]\cdot \sum_{\mathrm{cyc}}[BHC]=[ABC]^{2}\ }\ .$


modular arithmetic
number theory proposed
number theory

Find all the pairs of positive integers $ (a,b)$ such that $ a^2 \plus{} b \minus{} 1$ is a power of prime number $ ; a^2 \plus{} b \plus{} 1$ can divide $ b^2 \minus{} a^3 \minus{} 1,$ but it can't divide $ (a \plus{} b \minus{} 1)^2.$

I found a pair that satisfies: $ (a,b)\equal{}(3,8)$.
Someone has a solution.

[i]I have worked with this pro three days and don't have anything new

     Until someone post a solution , I will try Gaussian numbers :) [/i]

Let $ a^2\plus{}b\minus{}1\equal{}p^k$, $ b^2\minus{}a^3\minus{}1\equal{}mp^k$
=> $ b \equiv 1\minus{}a^2 (mod p^k)$
=> $ b^2 \equiv (a^2\minus{}1)^2 \equiv a^4\minus{}2a^2\plus{}1 (mod p^k)$
=> $ a^4\minus{}2a^2 \equiv a^3 (mod p^k)$
=> $ a^2(a\minus{}2)(a\plus{}1) \equiv 0 (mod p^k)$ (1)

$ (a\plus{}b\minus{}1)^2 \equiv (a\plus{}1\minus{}a^2\minus{}1)^2 \equiv a^2(a\minus{}1)^2 (mod p^k)$ (2)
Consider the case $ p>3$
(1),(2),and $ a^2<p^k\plus{}1$ imply $ a\equal{}2$=>$ b\equal{}p^k\minus{}3$,where $ p>3,k>0$

We need to consider $ p\equal{}2,p\equal{}3$,but this case is not easy for me :maybe:

[quote="Fang-jh"]Find all pairs of positive integers $ (a$,$ b)$ such that $ a^2 \plus{} b \minus{} 1$ is a power of a prime number, $ a^2 \plus{} b \plus{} 1$ divides $ b^2 \minus{} a^3 \minus{} 1$, but it doens't divide $ (a \plus{} b \minus{} 1)^2$.[/quote]
Before I start solving the problem, is there anyone who knows if there is a typo in the statement of the problem? Actually I'm not sure about the following: "[i]$ a^2 \plus{} b \plus{} 1$ divides $ b^2 \minus{} a^3 \minus{} 1$, but it doens't divide $ (a \plus{} b \minus{} 1)^2$[/i]". Shouldn't it be "[i]$ a^2\plus{}b\minus{}1$ divides...[/i]"?

Actually this is the right one:
Find all the pairs of positive integers $ (a,b)$ such that $ a^2 \plus{} b \plus{}1$ is a power of prime number $ ; a^2 \plus{} b \plus{} 1$ can divide $ b^2 \minus{} a^3 \minus{} 1,$ but it can't divide $ (a \plus{} b \minus{} 1)^2.$
BTW,it's different from this one:
http://www.mathlinks.ro/viewtopic.php?p=1806011#1806011  This is a problem in 2003.

[quote="litongyang"]Find all the pairs of positive integers $ (a,b)$ such that $ a^2 \plus{} b \plus{} 1$ is a power of prime number $ ; a^2 \plus{} b \plus{} 1$ can divide $ b^2 \minus{} a^3 \minus{} 1,$ but it can't divide $ (a \plus{} b \minus{} 1)^2.$[/quote]Let $ a^2 \plus{} b \plus{} 1 \equal{} p^k$ where $ p$ is a prime and $ k$ is a positive integer. We have $ p^k|b^2 \minus{} a^3 \minus{} 1\equiv(a^2 \plus{} 1)^2 \minus{} a^3 \minus{} 1 \equal{} a^4 \minus{} a^3 \plus{} 2a^2$ and $ p^k\not|(a \plus{} b \minus{} 1)^2 \equal{} (a \plus{} p^k \minus{} a^2 \minus{} 2)^2\equiv(a^2 \minus{} a \plus{} 2)^2$. Let $ v_p(a) \equal{} d$. Since $ a^2 < p^k$, then $ d < \frac {k}2$. Since $ p^k|a^2(a^2 \minus{} a \plus{} 2)$, then $ v_p(a^2 \minus{} a \plus{} 2)\ge k \minus{} 2d$. But $ p^k\not|(a^2 \minus{} a \plus{} 2)$ so $ v_p(a^2 \minus{} a \plus{} 2) < \frac {k}2$. Thus $ \frac {k}2 > k \minus{} 2d$, which gives $ 4d > k > 0$, hence $ p|a$. If $ p$ is odd, then $ p$ cannot divide $ a^2 \minus{} a \plus{} 2$ otherwise $ p$ divides 2. Thus, in case $ p$ is odd, $ \nu_p(a^2 \minus{} a \plus{} 2) \equal{} 0\ge k \minus{} 2d$, which is a contradiction to the fact that $ d < \frac {k}2$. Therefore $ p \equal{} 2$. Since $ \gcd(a,a^2 \minus{} a \plus{} 2) \equal{} \gcd(a,2)\le2$, then $ \min\{\nu_p(a),\nu_p(a^2 \minus{} a \plus{} 2)\} \equal{} 1$.

(i) $ \nu_p(a) \equal{} 1$
So $ a\equiv2\pmod4$, which implies $ 4|a^2 \minus{} a \plus{} 2$, hence $ k > 2$. But $ k < 4d \equal{} 4$, so $ k \equal{} 3$. We get $ a^2 \plus{} b \plus{} 1 \equal{} 2^3 \equal{} 8$, so $ a \equal{} 2$ and $ b \equal{} 3$. However, in this case, $ a^2 \plus{} b \plus{} 1 \equal{} 8$ divides $ 16 \equal{} (2 \plus{} 3 \minus{} 1)^2$.

(ii) $ \nu_p(a^2 \minus{} a \plus{} 2) \equal{} 1$
Since $ p^k|a^2(a^2 \minus{} a \plus{} 2)$, then $ 2d \plus{} 1\ge k$, hence $ d\ge\frac {k \minus{} 1}2$. But we also have $ \frac {k}2 > d$, hence we must have $ d \equal{} \frac {k \minus{} 1}2$ where $ k$ is odd. Let $ k \equal{} 2n \plus{} 1$. So $ a \equal{} 2^nm$ where $ m$ is odd. Since $ a^2 < 2^k$, we get $ 2^{2n}m^2 < 2^{2n \plus{} 1}$, so $ m \equal{} 1$. So we get $ a \equal{} 2^n$, $ b \equal{} 2^k \minus{} a^2 \minus{} 1 \equal{} 2^{2n \plus{} 1} \minus{} 2^{2n} \minus{} 1 \equal{} 2^{2n} \minus{} 1$, it is easy to check that $ (2^n,2^{2n} \minus{} 1)$ satisfies the given conditions for all positive integers $ n > 1$ (for $ n \equal{} 1$ it's the same as case i).

Answer: $ (2^n,2^{2n} \minus{} 1)$ for positive integer $ n > 1$.


Is there exists $a,b$ in positive rational numbers, such that $3\mid a+b+\frac 1a+\frac 1b$. :D

It's been discussed several times. Baltic easy problem.


function
limit
trigonometry
calculus
calculus computations

How do you determine if a function defined as a limit is continuous?
for example, how do you know that
$ f(x) \equal{} \lim_{n\to \infty} \sum_{k \equal{} 0}^{n}{\frac {\sin(x(2k \plus{} 1))}{2k \plus{} 1}}$ is not continuous, but
$ f(x) \equal{} \lim_{n\to \infty} \sum_{k \equal{} 0}^{n}{\frac {\sin(x(2k \plus{} 1))}{(2k \plus{} 1)^2}}$ is?
what do you do if you get something like
$ f(x) \equal{} \lim_{n\to \infty} \sum_{k \equal{} 0}^{n}{\frac {\sin(xe^{k})}{e^k}}$
 or $ f(x) \equal{} \lim_{n\to \infty} \sum_{k \equal{} 0}^{n}{\frac {\sin(xk^2)}{k^2}}$?

The most important theorem is this: if $ f_n$ is a sequence of continuous functions on some set $ S$ and if $ f_n$ converges [b]uniformly[/b] to $ f$ on $ S,$ then $ S$ is continuous.

And here's a useful theorem to get you to that point in some cases, with a catchy name: The [b]Weierstrass M-test:[/b]

Suppose $ g_n$ is a sequence of functions on some set $ S,$ and suppose that there exists a sequence of constants $ M_n$ such that $ |g_n(x)|\le M_n$ for all $ x\in S,n\in\mathbb{N},$ and $ \sum_{n\equal{}1}^{\infty}M_n<\infty.$ Then the series $ \sum_{n\equal{}1}^{\infty}g_n$ converges absolutely and uniformly on $ S.$

If the $ g_n$ are themselves continuous functions, we then conclude that the sum of the series is continuous on $ S.$

Since sines and cosines have obvious uniform bounds, the Weierstrass M-test is especially useful for trigonometric series.

$ \sum_{k\equal{}0}^{\infty}\frac{\sin((2k\plus{}1)x)}{(2k\plus{}1)^2}:$ Since $ \sum_{k\equal{}0}^{\infty}\frac{1}{(2k\plus{}1)^2}$ converges, we can apply the M-test and conclude that the original trigonometric series converges uniformly to a continuous limit.

You can also apply the M-test to your third and fourth examples; each of them has a continuous limit.

For $ \sum_{k\equal{}0}^{\infty}\frac{\sin((2k\plus{}1)x)}{2k\plus{}1},$ we cannot use the M-test since $ \sum_{k\equal{}0}^{\infty}\frac1{2k\plus{}1}$ diverges.

That does not necessarily mean the failure of uniform convergence, and the failure of uniform convergence does not necessarily mean that the limit function is discontinuous. However in this case, we know what the limit function is by direct computation. It's the "square wave" and it's discontinuous at both $ 0$ and $ \pi.$ Hence, the contrapositive of the first theorem that I stated proves that the convergence could not have been uniform.


[quote="zaizai-hoang"]Let $a,b,c$ are positive real number such that $ab+bc+ca+abc=4$. Prove that:
$a+b+c+abc\ge 4$[/quote]
$ab+bc+ca+abc=4\Leftrightarrow\sum_{cyc}\frac{1}{a+2}=1.$
Let $\frac{1}{b+2}+\frac{1}{c+2}-\frac{1}{a+2}=x,$ $\frac{1}{a+2}+\frac{1}{c+2}-\frac{1}{b+2}=y$
and $\frac{1}{a+2}+\frac{1}{b+2}-\frac{1}{c+2}=z.$ Hence,
$x=\frac{1}{b+2}+\frac{1}{c+2}-\frac{1}{a+2}=1-\frac{2}{a+2}=\frac{a}{a+2}>0.$
Similarly, $y>0$ and $z>0.$
Hence, $x+y+z=1$ and $\frac{2}{a+2}=y+z.$
Id est, $a=\frac{2}{y+z}-2=\frac{2-2y-2z}{y+z}=\frac{2x}{y+z}.$
Similarly, $b=\frac{2y}{x+z}$ and $c=\frac{2z}{x+y}.$
$a+b+c+abc\ge 4\Leftrightarrow\sum_{cyc}\frac{2x}{y+z}+\frac{8xyz}{(x+y)(x+z)(y+z)}\geq4\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}x(x+y)(x+z)+4xyz\geq2(x+y)(x+z)(y+z)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)\geq0,$ which Schur. :)

Generalisation:
If $a,$ $b,$ $c$ and $k$ are positive numbers, such that
$ab+ac+bc+kabc=\frac{4}{k^{2}}$ then the following inequality holds
$a+b+c+k^{2}abc\geq\frac{4}{k}.$
But this same.

nice solution. My solution use mixing variable  :D
$f(a,b,c)-f(a,\frac{2}{a+1},\frac{2}{a+1})$
And we can use U.M.V theorem:
$f(a,b,c)\ge min[f(a,\frac{2}{a+1},\frac{2}{a+1}), f(a,\frac{4}{a},0)]$


I felt like sharing to all of you that I am a UC Berkeley incentive award finalist.  All that means is that I have to compete with three other kids to get a scholarship to Berkeley that is basically a full ride.  I am just excited and felt like sharing.

yea, any major you can get into Tare...I am most likely going to be in the school of letters and sciences to study computer science and minor in math.  That is if I get it and go there.

Wow. Congratulations! Beat the other three. You can do it.

So only one gets to go to UC Berkeley...Woot, we'll be cheering...er, at the computer :lol:
Anyways, good luck!