What is your favorite:

1. Bach Prelude and Fugue?

2. Chopin Etude?

3. Beethhoven Piano Sonata?

You know, people wrote piano music [i]after[/i] the 19th century, too :)  What about Ginastera?  Prokofiev?  Keith Emerson?

Well I thought I would ask about MY favourite composers, and unfortunately, I don't much like Ginastera or Prokofiev. Debussy is probably the latest I can go before I find that music takes a turn for the worst. However saying that, I do like a bit of Hindemith, Berg, Tavener and a few lesser known Scottish composers. I doubt many people would have an opinion on which movement of Ludas Tonalis is the best (my favourite is the Preludium). So I decided to restrict myself to composers that people are sure to have an opinion on. I used to attend a specialist music school, so I am not ignorant of any type of music (except from perhaps rap and hip-hop etc.). I think we have pushed he realms of tonality (or lack of tonality) as far it will go. I certainly don't enjoy listening to a solo violinist playing in quarter tone rows. And the fact the the tone rows are inverted and retrograded does not make me appreciate it any more.

Bach - A-flat Major, especially the fugue, one of the darker and slower of the major fugues in WTC.

Chopin - Op. 10, No. 4 (Torrent), really love Richter's recording in particular.

Beethoven - Though I really love the both second movement of Appassionata and the first of Waldstein, there's no complete sonata that I enjoy more than Op. 110.

As far the others mentioned...

Really love Yefim Bronfman's recordings of Prokofiev Sonatas, 3rd one's probably my favorite of his piano pieces.  Horowitz does a pretty good job with them as well.

And for Ginastera, I'd probably pick the Suite de Danzas Criollas, though the second movement of Danzas Argentinas is one of the few pieces I really enjoy playing and listening to.

I'm probably not familiar enough Ludus Tonalis as a whole to pick a favorite section, though I like the twelth fugue...

1. Aufschwung- Schumann

2. Waltz/Valse- Chopin

3. Tarantella- Liszt

4. La Campanella- Liszt

5. Sonata- Chopin

6. Sonata- Bartok

7. Images- Debussy

8. Sonata (especially 'Moonlight' and 'Pathetique' and no. 32)- Beethoven

9. Ballade- Chopin

10. Italian Concerto- Bach

Top 10 favourite!

But what about GOOGOSH? What are your opinions on him? Do you know him? Who is he, anyone knows?

For Sonata I prefer that of Beethoven (especially op.111) and Schubert. In my opinion Beethoven's Sonata is a bit better than Mozart's because it have more pulse and idea(don't know whether this word is correctly used)

For Etude Liszt is the best of all, not only technically but also musically(my favorites are no 3, 8, 11)

Chopin's ballade are brilliant also.

umm.....appasionata i think thats how you spell it

[quote="Identity"][quote="Pul de Algodoncito"]Obviously Chopin Etude!!![/quote]

Meh... I think they're superficial.[/quote]

Amen.

The Beethoven Sonatas are, in my opinion, the deepest and most extraordinary works for the piano every written. Particularly, the last four sonatas (106, 109, 110, 111) illustrates Beethoven's stronghold on music.

In terms of sound, though, sometimes I prefer Mozart, but Beethoven, in his early works (see Op. 10, No. 3!), invokes Mozart's style almost as interestingly as Mozart himself.

On another note, has anyone ever played any Messiaen?  :P

Chopin etudes aren't superficial...unless you can't play them, then they start sounding pretty superficial.

op.10 no.3: Sure, kinda sounds like mozart's K576 sonata (I'm not sure if the number is right, I think it was D major).

Messiaen, Ew.

Me likes counterpoint  :) ... very mathematical.

Any of the Chopin polonaises, Rachmanioff, Mendelssohn, Copland, and Brahms hungarians. I used to like Mozart, now I loathe his music.

The thing with Mozart, IMO, is that since elementary school you learn how great he is, and Beethoven. Then later on, some people discover that those 2 aren't the only composers...yea...and the conclusion that Mozart's music is inferior is quickly reached. Anyways I say that because my general observation is that among people who play piano, people who are noobs like Mozart a lot (I suspect it's because EVERYONE knows just how wonderful he is, only because that's what books say), then people who are more sophisticated in the learning of the art of piano hate Mozart, since they like Liszt, Rachmaninoff, Chopin, Prokofiev, etc. more. Speaking of books, I've never seen a single book (art or history) that praises any romantic era music, or at least not as much as classical era music. My school history textbook has 2 whole pages on Mozart, while Liszt, Rach were neglected, and Chopin had one line saying "He wrote pieces based on Polish music." :(

beethoven sonata and bach preludes and fugues

[quote="kyyuanmathcount"][quote="CatalystOfNostalgia"]Vingt Regards sur l'Enfant Jesus FTW[/quote]
NO[/quote]
lol we've hammered this to death already, don't you think?

my favourite piano music at the moment is the lonely man from hulk, and the entertainer.


erm...whass with the title? [hide]sigh...i didn't learn how to do these sort of problems till i was in calculus...don't ask[/hide]

lol, what does calculus have to do wit it?

Well, people are usually taught how to solve this sort of problem in calculus. You don't need to wait that long though.

People are usually taught partial fractions in calculus.

ah, well i dont know whats in calculus. any takers for the problem though?

this is one of the things you either know how to do or you don't. at someone's request, i have provided a hint.

[hide]this is a branch of something called decomposition by partial fractions. don't be intimidated by that big name, it just means breaking something up into parts. For example, if we are finding 1/2(x-1)-1/2(x+1), we would find a common denominator and add and whatnot, giving us 1/(x-1)(x+1). In decomposition, we do this backwards. We say HEY! we can write 1/(x-1)(x+1) as A/(x-1)+B/(x+1), since those were the factors of the denominator. You can then solve for A and B, etc. I will explain this later after we solve this particular problem. I might also explain why it's useful in calculus, if you reaaaaaaaaaaally want to know. For now, suffice it to know the general idea.[/hide]

Well, with mystic having said that, unless I'm having computational problems, I get

[hide]

100/201

[/hide]

I believe that is right. I'll tell you the rest tomorrow.

Whooo... this problem was in a Chinese 5th grade olympiad book my mom had.

got the same as southie. i always wondered what partial fractions were useful for...


inequalities
geometry
geometric transformation
homothety
geometry proposed

Given two intersecting circles $ c_1$ and $ c_2$.We draw a circle $ c_3$ such that tangent to $ c_1$ and $ c_2$ (such that $ c_1$ and $ c_2$ be in the $ c_3$).$ c_1$ and $ c_2$ intersects in points A,B.$ c_3$ tangent to $ c_1$ in point B and tangemt to $ c_2$ in point C. Are A,B and C collinear?prove your notation. :lol:

First,Jensen Notation should be defined bot proved .
Second you used Letter [b]b[/b] two times but your meaning is clear.
third ,the answer to your problem is [b]No[/b].
[edit]go to post 4.

Yes i knew that,in my sketch they were.so thank you so much Arman!You helped me so  much. :D

no Your problem is not always true.
dear lomos I have read Grobber sulotion but Grobber didn't  mention it or I am erroring.
Would U say how is it ,true ?

Mathx ,you are right .
I was wrong ,I dont know what i was thinking about.
No they are not collinear.
Here is a simple prove :
Problem:let $C_1$ be a circle with center $O_1$. Let $A,B$ be two point on its premeter.
Let $C_2$ be a circle with center $O_2 \in O_1A$,tangent to $C_1$.
Let $AB$ intersects $C_2$ in $F$.
If the above three points be collinear , then all circles like $C_3$ with center $O_3$ on $O_1B$ and tangent to $C_1$ shall passes through $F$ but this is a contradiction ,Since $O_3B$ is not always equal to $O_3F$.

when that line pass trough homothecy center


Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides $ 7$ times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?

$ \textbf{(A)}\ \frac {2}{3}\qquad \textbf{(B)}\ \frac {3}{4}\qquad \textbf{(C)}\ \frac {4}{5}\qquad \textbf{(D)}\ \frac {5}{6}\qquad \textbf{(E)}\ \frac {6}{7}$

[hide] Say he walks x km/hr, his house is h kilometers away from him, and the stadius is s kilometers away from him.  Thus, it will take him $\frac{s}{x}$ hours to get to the stadium from where he is standing.  Now, if he wanted to go back home, it would take him $\frac{h}{x}$ hours.  If he wanted to go to the stadium to his house on his bicylce, it would take him $\frac{h+s}{7x}$ hours since he rides 7 times as fast.  Thus,
\[\frac{h+s}{7x}+\frac{h}{x}=\frac{s}{x}\]
Multiplying both sides by 7x, we get that
\[h+s+7h=7s\Rightarrow 8h=6s\Rightarrow \frac{h}{s}=\frac{3}{4}\]
Sorry about that, I edited from C to B.
Therefore, the answer is B. [/hide]

[hide]Let $a$ be the distance between him and his house, and $b$ be the distance between him and the stadium.

$b=a+\frac{a+b}7$

$7b=7a+a+b$

$6b=8a$

$\frac{a}b=\frac34\Rightarrow\boxed{\text{C}}$[/hide]

There's a typo in your solutions or the problem: [hide]3/4 is B, not C  :P  :wink: [/hide]

[quote="now a ranger"]this problem was problem 13, not 9.  :blush:[/quote]It was 13 on the AMC 10, and 9 on the AMC 12 :wink:


AMC
USA(J)MO
USAMO
algebra theorems
algebra

What are some of the main theorems or formulas needed for USAMO and other NMO's???

As i know, NMO problems are not schoolbook problems,so you don't need formulas to solve them!
I admit there are some results wich can be used but simply knowing them doesn't give you the solution to the problems!


After a gymnastics meet, each gymnast shook hands once with every other gymnast on every team. Afterwards, a coach came down and only shook hands with each gymnast from her own team. There were a total of 281 handshakes. What is the fewest number of handshakes the coach could have participated in?

[quote=Bugi]Obviously 1?
Because if all teams have 1 competitor?[/quote]

No way, because it would be impossible to have 281 handshakes in total!

This is way overrated, it should be level 18 not 23

I think it's way underrated. This took me a while to solve.

However, if the answer [hide=is]5[/hide] and we have 24 gymnasts, how would we divide them into teams? Shouldn't each team have the same amount of gymnasts?

[hide=(For reference, here is the Alcumus solution.)]The number of gymnasts is some integer $n$, so that the number of gymnast-gymnast handshakes is ${n \choose 2}$ for some $n$. Also, the coach must participate in an integer $k<n$ number of handshakes. So, ${n \choose 2} + k = 281$. If we want to minimize $k$, we need the maximal $n$ such that ${n \choose 2} \le 281$, which implies $\frac{n(n-1)}{2} \le 281$ or $n^2 - n - 562 \le 0 $. So, the maximal $n$ is 24. So, $k = 281 - {24 \choose 2} = 281 - 12 \cdot 23 = 281 - 276 = \boxed{5}$.[/hide]

This isn't really a counting question, it can be solved by using mostly algebra:

[hide]

Let there be $n$ gymnasts and $t$ gymnasts in the coach's team, where $n$ and $t$ are positive integers. There are $\frac{n(n-1}{2}$ handshakes between gymnasts, and $t$ handshakes between the coach and the gymnasts, so $\frac{n(n-1)}{2} + t = 281$.

We can multiply by 2 and shift everything to one side to get $n^2 - n + (2t - 562) = 0$, which means $n = \frac{1 \pm \sqrt{2249 - 8t}}{2}$ (by the quadratic formula). We want to minimize $t$ but make $n$ an integer, which means $1 \pm \sqrt{2249 - 8t}$ must be even. We can do this if $2249 - 8t$ is an odd square. Since we are minimizing $t$ we must maximize $2249 - 8t$. The largest odd square smaller than 2249 is $47^2 = 2209$, making $t = 5$.

We can figure out $n = 24$ by plugging $t$ back into the quadratic formula.
[/hide]

[hide=Another Solution (no algebra)]
We see that the sum of the handshakes has to be a triangular number. For reference the first triangular numbers are: 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120,136, 153, 171, 190, 210, 231, 253, 276, 300.

We need the biggest triangular number less than 281, which when we look at the list, in 276.

Hence, the fewest number of handshakes is $281-276=\boxed{5}$.
[/hide]

[hide=sol]
this is equivalent to finding the largest triangular number that is less than $281$
the formula for the handshakes is $\binom{n}{2}=\frac{n(n-1)}{2}$
since $281*2=562$, I estimated $n\approx25$
testing a few numbers around this range, we find that $n=24$ works
finally, $\frac{23(24)}{2}=276$ and $281-276=\boxed{5}$.
[/hide]


Angle PQR is a right angle. The three quadrilaterals
shown are squares. The sum of the areas of the
three squares is 338 square centimeters. What is
the number of square centimeters in the area of
the largest square?

[asy]draw((0,0)--(12,0)--(0,5)--cycle);
dot((0,0));
dot((12,0));
dot((0,5));
draw((0,0)--(0,5)--(-5,5)--(-5,0)--cycle);
draw((0,0)--(0,-12)--(12,-12)--(12,0));
draw((0,5)--(5,17)--(17,12)--(12,0)--cycle);
label("$P$",(0,5),NW);
label("$Q$",(0,0),SE);
label("$R$",(12,0),E);[/asy]

Please see this link: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1166642#1166642

This topic was discussed here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=916557147&t=211607
oops beaten to it


algebra
polynomial
function
limit
search
number theory
relatively prime

Find all polynomials $ p$ of one variable with integer coefficients such that if $ a$ and $ b$ are natural numbers such that $ a \plus{} b$ is a perfect square, then $ p\left(a\right) \plus{} p\left(b\right)$ is also a perfect square.

very very nice problem. I don't have an elementary proof as of now so I would really like to see you solution.
I excluded P from having degree $ \geq 3$ so i'm assuming the only solutions are $ P(x)\equal{}k^2x$

[quote="gagan_goku"]can we say that if a polynomial P(x) takes square values for all integers , then
P(x) = {f(x)}^2
where f(x) = polynomial with integer coefficients[/quote]

Yes, that's a well-known theorem.

See for instance
H. Davenport, D.J. Lewis, A. Schinzel:
Polynomials of certain special types.
Acta Arithmetica 9, 1964, 107-116.

if thats true , then you can say that f(x,y) = P(x) + P(y^2 -x) = g(x,y) ^ 2
maybe this yields something

I've been spending a lot of time mulling this one over in my head. Can anyone solve it?

If $ P$ is not an odd polynomial there exists an $ a$ so that $ P(x)\plus{}P(a)$ is not divisible by $ x\plus{}a$
$ P(x^2\minus{}a)\plus{}P(a)\equal{}Q(x)^2$ so $ P(x)\plus{}P(a)\equal{}Q(\sqrt{x\plus{}a})^2$ and $ P(x)\plus{}P(a)\equal{}R(x)^2$ also $ P(x)\plus{}P(b)\equal{}K(x)^2$
but $ R(x)^2\minus{}K(x)^2\equal{}c$ can not hold as an identity so $ c\equal{}0$ and the polynomial is a constant function $ P(x)\equal{}2m^2$
if P is an odd polynomial $ P(x)\plus{}P(a)\equal{}(x\plus{}a)R(x)^2$ and $ P(x)\plus{}P(b)\equal{}(x\plus{}b)K(x)^2$. how can we show that these last two equations imply some contradiction if the degree of $ P$ is too large?...

[quote="Albanian Eagle"]i'm assuming the only solutions are $ P(x) \equal{} k^2x$[/quote]

Of course there's also $ P(x) \equal{} 2k^2$.

[quote="Albanian Eagle"] and $ P(x) \plus{} P(a) \equal{} R(x)^2$[/quote]why?...can you say more details please?

I will assume the answers to the preceeding questions have already been given, so let me try to answer Albanian Eagle's question. Suppose that $ (X\plus{}a)R(x)^2\minus{}(X\plus{}b)P(X)^2\equal{}c$, a nonzero constant (otherwise the conclusion is obvious...). Then obviously $ P$ and $ R$ are relatively prime and have the same degree. Differentiate the relation to obtain $ (X\plus{}a)2RR'\plus{}R^2\minus{}P^2\minus{}(X\plus{}b)2PP'\equal{}0$, thus $ R$ divides $ P\plus{}2(X\minus{}b)P'$ and thus must be a constant times this thing, by the remark on the degree. This gives you a differential equation on $ P$ and allows finding all solutions, but unfortunately it may very well happen that there are solutions of arbitrarily large degree (and it happens, actually...), but this imposes however very strong conditions on $ P$. Now, by choosing some other value $ d,e,..$ for $ a,b$, I think we can conclude. I'm still puzzled by this apparently extremely easy problem, which however does not seem to have a decent solution.

Oh nice..
so Harazi can substitute "a constant times" with "equal" by considering leading coefficients.
and so we have $ R\equal{}K\plus{}2(x\plus{}b)K'$ but also vice versa so
$ (x\plus{}b)K'\plus{}(x\plus{}a)R'\equal{}0$ and particularly
$ K'\equal{}R'\equal{}...$ by putting different choices So $ P(x)\plus{}P(m)\equal{}(x\plus{}m)(Q(x)\plus{}c_m)^2$ holds for some fixed $ Q$
this is possible only when $ Q$ is constant so $ P$ is linear and we can easily conclude
 :lol:  we don't even need the differential equations.

But I love differential equations! Just kidding, I really hate them, thanks for pointing out this simplification, I had no paper with me to check if indeed it can be made easier. Anyway, great problem.

[quote="fearailhold"][quote="Albanian Eagle"]If $ P$ is not an odd polynomial there exists an $ a$ so that $ P(x) \plus{} P(a)$ is not divisible by $ x \plus{} a$[/quote]

Do you have a proof of this?[/quote]
if for all $ a$, $ P(x)\plus{}P(a)$ is diviseble by $ x\plus{}a$ then $ G(x)\equal{}P(x)\plus{}P(\minus{}x)$ is zero for all $ a$....and P is odd...

:lol:
If the words Iran and Polynomial appear in the same thread it can never disappoint you  :D

1 doubt
how does $ P(x) \plus{} P(a) \equal{} P(\sqrt{x\plus{}a})^2$ imply $ P(x) \plus{} P(a) \equal{} R(x)^2$
secondly , if you have $ P(x) \plus{} P(a) \equal{} R(x,a)^2$ , then i think we can conclude that the degree of $ P$ is 0
but then we dont get the solution $ P(x) \equal{} k^2x$
does this solution come from $ P$ being odd

[quote="gagan_goku"]can we say that if a polynomial P(x) takes square values for all integers , then
P(x) = {f(x)}^2
where f(x) = polynomial with integer coefficients[/quote]
This claim should be slightly modified, cause $ f(x)$ do not need to be with integer coefficients. For example $ P(x)=\frac{x^2(x-1)^2}{4}$.
Then I think I have a elementary proof to this fact.
The most important case is that the degree of $ P$ is even. In this case we let $ \deg P=2n$ and tha cofficent of the $ 2n-th$ power is $ A$.
Consider $ L=\lim_{x \rightarrow \infty}{(\sqrt{P(x+n-1)} -C_{n-1}^1 \sqrt{P(x+n-2)}+\ldots+(-1)^{n-1} \sqrt{P(x)})}$
Here is the most important result:
We claim that $ L$ is a constant.
In fact, we can find a polynomial $ Q(x)$ such that $ \deg Q=n$ and $ \deg(P-AQ^2) \leq n-1$ (This is true due to Newton's formula for symmitric sums.)
Then we have
${ \lim_{x \rightarrow \infty}{\sqrt{P(x)}-\sqrt{A}Q(x)}=\lim_{x \rightarrow \infty}{\frac{P(x)-AQ^2(x)}{\sqrt{P(x)}+\sqrt{A}Q(x)}}}=0$

So
$ L=\sqrt{A} \lim_{x \rightarrow \infty}{(Q(x+n-1) -C_{n-1}^1 Q(x+n-2)+\ldots+(-1)^{n-1} Q(x))}=  \sqrt{A} (n-1)!$
is a constant.
Finally, suppose $ P(x)=a_x^2$ for all $ x \in Z$. Then $ \lim_{x \rightarrow \infty}{(a_{x+n-1}-C_{n-1}^1 a_{x+n-2}+ \ldots +(-1)^{n-1} a_{x})}=L$. Therefore $ L$ is a integer, and for large enough $ x \in Z$, we have ${ a_{x+n-1}-C_{n-1}^1 a_{x+n-2}+ \ldots +(-1)^{n-1} a_{n}}=L$. Note that the sequence $ {a_n}$ has the fomula $ a_n=f(n)$ where $ \deg{f}=n$ for large enough $ n$. So we conclude $ P(x)=f^2(x)$ for large enough $ x \in Z$ and therefore $ P(x)=f^2(x)$ for all $ x$. And it's obvious that $ f(x) \in Q[x]$.

In the case that the degree of $ P$ is odd, let $ P'(x)=P(x^2)$ we do the same thing, and it's easy to claim that such $ P(x)$ don't exist.

[quote="rchlch"][quote="gagan_goku"]can we say that if a polynomial P(x) takes square values for all integers , then
P(x) = {f(x)}^2 where f(x) = polynomial with integer coefficients[/quote]
This claim should be slightly modified, cause $ f(x)$ do not need to be with integer coefficients. For example $ P(x) \equal{} \frac {x^2(x \minus{} 1)^2}{4}$.[/quote]

Aaah, yes indeed. The correct statement is:

If a MONIC polynomial $ P(x)$ takes square values for all integers $ x$,
then $ P(x) \equal{} {f(x)}^2$ where $ f(x)$ is a polynomial with integer coefficients.

[quote="rchlch"][quote="Albanian Eagle"]There are stronger lemmas out there
like http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1576618391&t=148696[/quote]
I think this lemma cannot be used in this problem because no direct evidence shows that $ P(x)$ is a monic polynomial.
And for your lemma, one of my classmates has given a beautiful solution.
Thank you all the same.[/quote]
It was just a reply to test20's post  :lol:

My solution was too big(6 pages). however here is an outline:
[hide]step1. show that a polynomial with integr coefficients taking too many square values at the integer must be the square of a polynomial.
for that check out: http://www.mathlinks.ro/viewtopic.php?t=241684 (a generalisation)

in the following all polynomials P,Q,R,S,T are of integer coeffs.
step2. set ${ a=x^2+x^4,\{   } b=2x^3$ prove $ Q(x)=P(x^2+x^4)+P(2x^3)$ has degree=4n,(if P has deg.=n). and $ Q(x)=(R(x))^2$. also leading coeff of P is a perfect square.

step3. Set $ a=px^2, b=qx^2$ s.t. p+q=a perfect square.. so $ S(x)=P(px^2)+P(qx^2)$=((T(x))^2), equating leading coefficient $ p^n+q^n=r^2,r\in\mathbb{N}$.

step4. prove that if p,q satisfies the property:
            whaenever p+q is a perfect square, $ p^n+q^n$ is also a perfect square for all n$ \in\mathbb{N}$.

        ,then n=1.
step5. ans: $ P(x)=xP(1),$ P(1) being a perfect square.[/hide]

I give this exam and i couldn't solve this problem   :(
in fact only one of students solved this problem, his main idea is : if a+b is square then a*x^2+b*x^2 is square too, so
p(ax^2)+p(bx^2) is perfect square for all integer x so there exist a g(x) such: p(ax^2)+p(bx^2)=g(x)^2
if p(x)=(a_n)*x^n+.... then (a_n)*(a^n+b^n)=u^2 (g(x)=u*x^k+...) for all a,b (a+b=perfect square) let a=b=2 so
(a_n)*2^(n+1)=u^2 and ... it's easy to see that p(x)=(k^2)*(x^2) or 2*(k^2), so we solve this hard problem
 :wink:

Theorem_ let $f$ be a integer polynomial.We have $f(a+tp^s)\equiv {}^{p{}^{s+1}}f(a)+tp^sf{}'(a)$such that $(t,p)=1$.So by above theorem we have for any integer polynomial $Q$ which has not any double root ,there are infinitely many prime $p$ which there exist some $n$ such that $p\mid Q(n)$ but $ Q(n)\not\equiv {}^p{}^{2}0$.now suppose $p(a)+p(x^2-a)=Q(x)$ .(step 1)if $Q$ has not any double root ,we have  $p\mid Q(n)$ but $ Q(n)\not\equiv {}^p{}^{2}0$ a contraction.(step 2)if $Q$ has some double root, we have $Q(x)=R_{a}(x)T_{a}(x)^2$ such that $R_{a}(x)$ has not any double root .$Q(m)$ is a perfect square for any positive integer $m$ so $R_{a}(Y)$ is a perfect square for all $y> M$ and we have a contraction same way to step 1.

[hide=Solution]

As a preliminary note, we exclude the constant solutions, $P(x)=2u^2$ for some integer $u$. Henceforth, degree of $P(x)$ is assumed to be $\ge 1$.

We start off with the following two lemmas. For the sake of completeness, we present their proofs as well making the whole argument quite long.

[i]Lemma 1.[/i] Let $a$ be a given natural number which is not a perfect square. Then there exists infinitely many primes $p$ for which $a$ is a quadratic non-residue.

[i]Proof[/i] By Euler's theorem, $\left(\frac{ab}{p}\right)=\left(\frac{a}{p}\right)\cdot \left(\frac{b}{p}\right)$ and so, we may assume that $a$ is square-free by taking out the largest square divisor of $a$. We can write $a=2^e\cdot p_1\cdot \dots \cdot p_k$ where $p_i \ge 3$ are primes and $e \in \{0,1\}$. In case the product is empty, it shall be assumed to be equal to $1$. If $e=0$ then let $p \equiv 1 (\bmod \, 4p_1\cdot \dots \cdot p_{k-1})$ and $p \equiv s (\bmod \, p_k)$ where $s$ is a quadratic non residue modulo $p_k$. If $e=1$ then let $p \equiv 5 (\bmod \, 8)$ and $p \equiv 1 (\bmod \, p_1\cdot \dots \cdot p_k)$. In any case, we see that $$1=\left(\frac{p}{p_i}\right)=(-1)^{\frac{p-1}{2}\cdot \frac{p_i-1}{2}}\cdot \left(\frac{p_i}{p}\right)=\left(\frac{p_i}{p}\right)$$ for $i=1,\dots, k$. In the former case, however, the equality for $i=k$ is changed to $-1$ by our choice of $s$ and in the latter, as $\left(\frac{2}{p}\right)=(-1)^{\frac{p^2-1}{8}}=-1$; we conclude that $a$ is a quadratic non residue modulo $p$ by using Gauss' Quadratic Reciprocity Theorem. The existence of infinitely many such prime is guaranteed by the Chinese Remainder Theorem and the Dirichlet's Theorem on distributions of primes in an arithmetic progression. The proof is completed.

---

We now present the second crucial lemma.

[i]Lemma 2.[/i] Let $f$ be a non constant polynomial with integer such that $\sqrt{f(n)}$ is an integer for all positive integers $n$. Then there exists a polynomial $g(x)$ with integer coefficients such that $f(x)=g(x)^2$.

[i]Proof[/i] Let us write $f(x)$ as a product of its irreducible divisors. Clearly, we can write $$f(x)=p_1(x)\cdot \dots p_k(x) \cdot \left(R(x)\right)^2$$ where $p_1,\dots, p_k$ are pairwise distinct irreducible divisors of $f$ and all the polynomials $p_1,\dots,p_k,R$ have integer coefficients. If the product $p_1\cdot \dots \cdot p_k$ is a constant then it must be a square and we may conclude. Else, there exists a non-constant polynomial in the set $\{p_1(x),\dots,p_k(x)\}$, which, say, is $p_1(x)$. Then, as $p_1$ being irreducible has no common factor with the product $h=p_2\cdot \dots \cdot p_k$ (which might as well be empty or constant) and $p_1'$ (the derivative of $p_1$), we are assured of the existence of polynomials $A(x),B(x),C(x),D(x)$ with integer coefficients and integers $N_0,N_1 \not= 0$ such that $$A(x)\cdot p_1(x)+B(x)\cdot h(x)=N_0$$ and $$C(x)\cdot p_1(x)+D(x)\cdot p_1'(x)=N_1$$ for all $x$ by the Euclid's Lemma. By Schur's theorem, the set of all primes dividing at least one value taken by $p_1$ over $\mathbb{Z}$ is infinite, hence, we may choose a prime number $p>N=\max(|N_0|,|N_1|)$ and an integer $r$ such that $p \mid p_1(r)$. Since $p \nmid N_0$, we see that $p \nmid h(r)$. Now, choose $x_0=r+ip$ where $0 \le i \le p-1$ is an integer we shall choose later. By the Taylor's Expansion of polynomials, we have $$p_1(x_0) \equiv p_1(r)+ip\cdot p_1'(r) (\bmod p^2)$$ In particular, we let $p_1(r)=p^{j_0}k_0$ and we can select $i$ such that $p^2 \nmid p_1(x_0)$ since the polynomials $p_1$ and $p_1'$ are also coprime $p \mid p_1(x_0),p_1'(x_0)$ will yield $p \mid N_1$, which is false as $p>N$. Thus, $v_p(p_1(x_0))=1$ and $v_p(h(x_0))=0$. This yields that $v_p(f(x_0))$ is an odd number, a contradiction to the fact that it is a square. The result holds.

---

With these two proofs finally completed, we present the main idea of the solution.

Notice that if $(a+b)$ is a perfect square then $(ax^2+bx^2)=(a+b)x^2$ is also a perfect square. Define the polynomial $T(x)=P(x^2)+P((t^2-1)x^2)$ where $t$ is fixed for the moment. Then, $T$ satisfies the conditions of [i]Lemma 2.[/i] and we get that it is the square of a polynomial $Q(x)$ with integer coefficients. Let $d$ be the degree of $P$, $K$ be the leading coefficient of $P$ and $g(t)$ (not a polynomial/function etc) be the leading coefficient of $Q$. The following equation is satisfied by comparing the leading coefficients in the equation $$T(x)=Q(x)^2,$$ $$K(1+(t^2-1)^d)=g(t)^2$$ for all $t$ sufficiently large. Let $K=mc^2$ where $m$ is a square-free number or $m=1$. If $m$ is square-free and $m>1$ then we may write $$(t^2-1)^d=m\cdot \left(\frac{g(t)}{mc}\right)^2-1$$ where, for obvious reasons of divisibility, $$\left(\frac{g(t)}{mc}\right)$$ is an integer. Let $p$ be a prime number and set $t \equiv 1 (\bmod \, p)$,i.e., $p \mid t^2-1$ and so $p \mid m\cdot \left(\frac{g(t)}{mc}\right)-1$ giving that $m$ is a quadratic residue modulo $p$. However, by [i]Lemma 1.[/i] we can choose a sufficiently large prime $p$ with $\left(\frac{m}{p}\right)=-1$ and we get the requested contradiction. If $m=1$ then suppose $d>1$. If $d$ is even then $1$ is a difference of sufficiently big perfect squares, and if $d$ is odd then we have $$\left((t^2-1)^{\frac{d-1}{2}}-1\right)^2<\left(\frac{g(t)}{tc}\right)^2<\left((t^2-1)^{\frac{d-1}{2}}\right)^2$$ which is again a contradiction! Thus, we must have $d=1$. If $d=1$ then we have $P(x)=vx+w$ and if $a=1,b=t^2-1$ then we see that the equation $$z^2-vt^2=w$$ is satisfied for some integer $z$. If $v$ is a perfect square then we get that $w=0$. If not, then it is a Pell type equation, so its solutions must grow exponentially, however, they grow only linearly, as $t$ grows linearly, only to be multiplied by a constant (equal to the largest perfect square divisor of $v$) and we get a contradiction!

Thus, the only solutions to this are the polynomials $P(x)=k^2x$ for some integer $k$ or $P(x)=2u^2$ for some fixed integer $u$. These also satisfy our given equation and we may conclude.

[/hide]

[hide=Credits and remarks] This happens to be a joint solution with [url=https://www.artofproblemsolving.com/community/user/158287]dibyo_99[/url] at the IMO training camp in India, this May. There were other notable contributors to this as well. This is one of my favorite number theory problems! :) [/hide]

[quote=anantmudgal09]and if $d$ is odd then we have $$\left((t^2-1)^{\frac{d-1}{2}}-1\right)^2<\left(\frac{g(t)}{tc}\right)^2<\left((t^2-1)^{\frac{d-1}{2}}\right)^2$$ which is again a contradiction! Thus, we must have $d=1$. [/quote]How can you get the left side?

[quote=anantmudgal09]
[i]Lemma 2.[/i] Let $f$ be a non constant polynomial with integer such that $\sqrt{f(n)}$ is an integer for all positive integers $n$. Then there exists a polynomial $g(x)$ with integer coefficients such that $f(x)=g(x)^2$.[/quote]

I learned another simple way to prove this lemma from lectures on winter camp in Korea around 2018.

Proof) Think of $f(x+m)f(x)$ for some natural number $m$. If $m$ is sufficiently large, we can separate complex roots of $f(x)$ and $f(x+m)$ and therefore $f(x+m)$ and $f(x)$ is relatively prime. By assumption, $f(x+m)f(x)$ is even degree and the maximal coefficient is square. Therefore, $f(x+m)f(x)=h(x)^2$ for some $h(x) \in \mathbb{Z}[x]$. Since $f(x)$ and $f(x+m)$ is prime, you can easily see that $f(x)=g(x)^2$ for some $g(x) \in \mathbb{Z}[x]$.

Same idea is used in Proposition mentioned in https://artofproblemsolving.com/community/u314211h1985879p14247495

Another astonishing proof for Lemma by my student.

Proof) Let $f(x)=a_n x^n +....+a_0$. By assumption, $a_n >0$.
[Case 1] $n$ is odd.
Multiply the term $(a_n x +1)$. There are infinitely many $x$ such that $a_n x+1$ is perfect square. Therefore, infinitely many $x$ such that $(a_n x+1)f(x)$ is perfect square. However, there are also $x$ such that $a_n x +1$ is not a perfect square and contradiction.
[Case 2] $n$ is even, but $a_n$ is not a perfect square.
Multiply the term $(a_n x^2 +1)$. And same as above. Here we used the fact that pell equation $y^2 - a_n x^2 =1$ has infinitely many solution.

Therefore, $n$ is even and $a_n$ is perfect square and $f(x)=g(x)^2$ for some $g(x) \in \mathbb{Z}[x]$.

Here is a smol bren solution.
 Write the polynomial,as $x^d.P(x)$, with $P(0)$ non zero.Assume $P$ non-constant.Then assuming,$P$ is not a square ,write it as a square times a bunch of irreducibles.Consider such a irreducible ,say $h(x)$,and then by Schur ,choose a very large prime factor of the polynomial $h(x^2)$.Then using Bezout,we can conclude that if $p|h(m^2)$,then $p$ does not divide any other irreducible factor of $P$ evaluated at $m^2$.Then,again using bezout but this time with $h'(x)$,we conclude that if $p$,is large enough ,then $p$ also doesn't divide $h'(m^2)$,thus from Hensel,we can uniquely lift the root corresponding to $m^2$ mod $p$ to mod $p^2$ .Due to this unique lift ,we can find such an $n$,such that $n \equiv m^2$ mod $p$ but $p^2$ does not divide $P(n)$.Thus we have that $v_p(P(n))$ is odd.Now choose a $b$ such that $v_p(b)=j$ is even and quite large.As $p$ is large (forgot to mention $p$ doesn't divide $P(0)$,so $p$ doesn't divide $P(b)$).
Now,we use hensel's again.Firstly note that $n$ is a quadratic residue mod $p$.So by hensel's on the polynomial $x^2-n$,we see that the derivative is non zero mod $p$.Thus,we can find a solution to  this poly mod $p^{j+1}$,such that $v_p(x^2-n)$ is exactly $j$.Thus we have found $b+n$ is a perfect square but $v_p(n^d.P(n)+b^d.P(b))$ is odd which is a contradiction.Thus,$P(x)$ is a perfect square.

Using a very similar argument as above,we can prove that $d$ is even(Basically choosing $a,b$,such that $P(a)$ has a large $v_p$,which is obv
 even,except here we choose $b$ to have small odd $v_p$,and obv $p$ doesn't divide $P(b)$.$a+b$ being a perfect square is ensured by the very similar hensel argument as above).
So we have that we can write $P(a)+P(b)$ as $a^{2d}Q(a)^2+b^{2d}Q(b)^2$.So if we fix $a$,we see that $a^dQ(a)$ is a part of infinitely many Pythagorean triples,which is impossible.
Observing the left out cases(which are not that hard),we get our solutions.

As in #26, $f(ax^2)+f(bx^2)$ is always a square for a fixed pair $(a,b)$ with $a+b$ being a square, so $g(x)^2=f(ax^2)+f(bx^2)$ for some integer polynomial $g(x)$. In addition, we consider only the case $\deg(f) \geq 2$.

Next, we consider the leading coefficient of this polynomial; it is of the form $T(a^n+b^n)$ ($T$ is the leading coefficient of $f$), which shall be a square. If $n$ is even, plug $(1,p^2-1),(1,p^2+2p)$ for some prime $p$ to see that $2T, T$ are QR's modulo $p$ for all primes $p$, so $2$ is a QR for all primes $p$, absurd, since $(\frac{2}{p})=(-1)^{\frac{p^2-1}{8}}$. Now we can finish the other case with $(2,2),(3,1)$ : $2^{n+1}T$ is a square, hence $T$ is a square, so $3^n+1=x^2$ is a square, so $3^n=(x-1)(x+1)$ and thus $x-1=1$ and $n=1$, contradiction.

I see pleny of solutions over here depending on this theorem about a polynomial being square too many times. I solved to problem using a totally different approach with pells equations, some estimations of the growth of some terms, and an unexpected use of Norms function at the end. I will write down the solution tomorrow.

Suppose $degP > 0$. If $P$ is constant so $2P(2)$ must be a square, so $P \equiv 2k^2$ for some integer $k$. Indeed all of these work.
Also suppose that the leading coefficient of $P$ is positive. If it's negative then $P(2N^2)$ is negative for big $N$, but is must also be twice a perfect square, contradiction.
Fix some positive integer $t$ such that the equation $b^2 - 2a^2=t$ has a solution.
We know there are infinetely many big $t$'s satisfying this (perfect squares for instance).
By pell's equation we know that, given one solution $b_0^2 - 2a_0^2$, we can find infinetely many solutions of the form
$$b_n + a_n \sqrt2= (b_0 + a_0 \sqrt2)(3+2sqrt2)^n.$$
Just so I'll write quicker, set $ r = 3 + 2 \sqrt2$, we'll use it at lot.
Now we know that, as $2a_n^2 + 2a_n^2$ is the square of an integer, $2P(a_n^2)$ must be the square of an even integer, say, by definition, $4r_n^2$. So $P(a_n^2) = 2r_n^2$.
We also know that $b_n^2 - t + t$ is a perfect square, so $P(b_n^2 - t) = s_n^2 - P(t)$, for some integer $s_n$. But, by definition, $b_n^2 - t = 2a_n^2$, so $2r_n^2$ and $s_n^2 - P(t)$ must be equal. So we have that $s_n^2 - 2r_n^2 = P(t)$.
But this means that $s_n, r_n$ are pell's solutions for $x^2 - 2y^2 = P(t)$. This means that $$s_n + r_n \sqrt2 = (s_0 + r_0 \sqrt2)r^{t_n}$$ for some minimal solution $(s_0,r_0)$ and some integer $t_n$. As there are only finitely many minimal solutions for pell's equation, by the pigeon hole principle one of them must appear infinetely many times, varying $n$. WLOG call it $(s_0,r_0)$.
In other words, we can say that
$$s_n = \frac{s_0 + r_0 \sqrt2}{2}r^{t_n} + \frac{s_0 - r_0 \sqrt2}{2}r^{-t_n}$$
so
$$s_n^2 = \big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2 r^{2t_n} + \big( \frac{s_0 - r_0 \sqrt2}{2} \big)^2 r^{-2t_n} + 2\frac{s_0^2 - 2r_0^2}{4} = (1+o(1))\big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2 r^{2t_n}$$
So this is one estimate for $s_n^2$.
Let $d = degP$ and $c$ the leading coefficient. Now we also know that
$$s_n^2 = P(b_n^2 - t) + P(t) = (1+o(1))P(b_n^2) = (1+o(1))c.b_n^{2d}.$$
As $$b_n = (1+o(1))\frac{b_0+a_0 \sqrt2}{2}r^n,$$
$$s_n^2 = (1+o(1))c.\big( \frac{b_0+a_0 \sqrt2}{2} \big) ^{2d}r^{2nd}$$
Therefore the ratio between the $2$ estimates converges to $1$. So
$$lim_{n \to \infty} \frac{\big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2 r^{2t_n}}{c.\big( \frac{b_0+a_0 \sqrt2}{2} \big) ^{2d}r^{2nd}} = 1.$$
So
$$lim_{n \to \infty} \frac{\big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2}{c.\big( \frac{b_0+a_0 \sqrt2}{2} \big) ^{2d}} r^{2t_n - 2dn} = 1.$$
So, a constant times an integer power of $r$ converges to $1$. Due to the fact the the sequence $C.r^m$ is discrete, if it converges to $1$, it's eventually $1$. So there exists an integer $q$ such that
$$\frac{\big( \frac{s_0 + r_0 \sqrt2}{2} \big)^2}{c.\big( \frac{b_0+a_0 \sqrt2}{2} \big) ^{2d}} r^q = 1.$$
Rearranging things and substituting $r$ back to $3+2 \sqrt2$, we have
$$(s_0 + r_0 \sqrt2)^2 (3+2 \sqrt2)^t = c(b_0 + a_0 \sqrt2)^{2d} 2^{2d-2}$$
Now we are gonna make use of the Norm function defined by $N(m + n \sqrt2) = m^2 - 2n^2$, where $m,n$ are rationals. One can easily prove that this function is well defined and is multiplicative. So we have
$$(s_0^2 - 2r_0^2)^2 . 1^t = c^2(b_0^2 - 2a_0^2)^2 2^{4d-4}$$
So
$$P(t)^2 = c^2 . t^{2d} . 2^{4d-4}$$
As the leading coefficient of $P$ is positive, $P$ is eventually positive, so we can take the squareroot of both sides to conclude that
$$P(t) = c.t^d.2^{2d-2}$$
But notice that this is true for infinetely many big integers $t$'s, so it must be true that
$$P(x) = x^d.c.2^{2d-2}$$
But remember that $c$ is the leading coefficient of $P$, so $c = c.2^{2d-2}$, so $2^{2d-2}=1$, so $d$ must be $1$, and $P(x) = cx$.
Now, as $2P(2)$ is a square, so $4c$ is a square, so $c$ itself is a perfect square. Given that, clearly $a+b$ is a square if and only if ac+bc is a square.
Therefore all solutions are $P(x) = k^2x$ and $ P(x) = 2k^2$, where $k$ is an arbitrary integer.


vector
algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Suppose that F is a field and all F^n are written as column vectors
Let A be an nxn matrice with entry from F, and
W = { B in M_nxn (F) | AB = BA}
Suppose there exists a vector v in F^n such that { v, Av, A^2 v, ........ A^(n-1) n} is a basis for F^n.

Prove that { I, A, A^2, ...., A^(n-1)} is a basis for W.

Above is the question I get stuck. It is easy to prove that W is a subspace and { I, A, A^2, ...., A^(n-1)} is linearly independent.
But how are we going to prove that it is spanning W?
Any idea?

[hide="Hint"]It suffices to show that $ \dim W\le n$. For this purpose consider the evaluation mapping $ W\to F^n$, $ B\mapsto Bv$ which is injective.[/hide]

Alternately (and I'm not sure if this works), write the rows of $ B \in W$ in terms of the basis given.

Why is the linear transformation stated above injective?

If we consider Bv = 0, I can just reach a conclusion that B is non-invertible or B=0 if B is in the span of {I, A,......, A^n-1}. :|
 :|

Assuming that $ Bv\equal{}0$ can you conclude that $ B\equal{}0$ using a basis of $ F^n$ and that $ B\in W$?

Oh! Is that because for every vector in F^n, Bx = 0, therefore B must be equal to zero?

Bv = 0, BAv = 0, BA^2 v = 0 .... BA^(n-1) v =0, hence for every x in F^n, Bx = 0.

Correct.[quote="distantstar"]Oh! Is that because for every vector in F^n, Bx = 0, therefore B must be equal to zero?

Bv = 0, [/quote]implying (together with $ B\in W$)[quote]BAv = 0, BA^2 v = 0 .... BA^(n-1) v =0, hence for every x in F^n, Bx = 0.[/quote]And "hence" stems from the assumption that $ A^mv$, $ m\equal{}0,\ldots,n\minus{}1$, is a Basis of $ F^n$.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113276]here[/url].

The relevant argument:
Suppose the minimal and characteristic polynomials of $ A$ are both equal to $ p$. Consider the algebra* $ V$ of polynomials mod $ p$, and let $ A$ act on this space by $ A(q(x))\equal{}x\cdot q(x)$. In the standard basis of $ V$, the matrix of $ A$ is the companion matrix associated to $ p$, and this is isomorphic to the original matrix form.
Now, we show that any linear transformation $ B$ on $ V$ which commutes with $ A$ commutes with polynomial multiplication: $ B(x^k)\equal{}B(x\cdot x^{k\minus{}1})\equal{}B(A(x^{k\minus{}1}))\equal{}A(B(x^{k\minus{}1}))\equal{}x\cdot B(x^{k\minus{}1})\equal{}\cdots$
$ \equal{}x^k\cdot B(1)$, and $ B(q(x))\equal{}q(x)\cdot B(1)$ by linearity. The map $ B(q(x))\equal{}q(x)\cdot \lambda(x)$ is exactly $ \lambda(A)$, and $ B$ is a polynomial in $ A$.

*An algebra is both a ring and a vector space. In this case, we started with a generic $ n$-dimensional vector space and added ring structure to make things easier.

Hmm, I cannot see why you need all this formalism. All you need here is that $ B\in W$ commutes with $ A^m$, $ m>0$, and that $ F^n$ is $ A$-cyclic (which comes very close what you intend to mean I think). The key fact is that the evaluation mapping $ W\to F^n$ is an isomorphism.

The natural structure on $ W$ is an algebra. The natural structure on $ F^n$ is a vector space. I wouldn't call it an isomorphism without either adding structure on $ F^n$ or removing it from $ W$. I chose the former, while you chose the latter.

Our arguments really are very similar; it's just that the algebra formalism lets me say $ B(q(x))\equal{}q(x)\cdot B(1)$ rather than $ B\cdot (q(A)\cdot v)\equal{}q(A)\cdot (B\cdot v)$, and the finish after that is easier.


1+8+27+............................+8000

Try to solve this. Please show all of your work.

[hide="hint"]Prime factorize each individual term.[/hide]

Source: AoPS(not a post)

if you did it the same way, its still ok to post your solution... :)

[hide]use the formula 20*21/2^2=44100[/hide]

I did it just like the 2nd guy.  :alien:


Jason counts up from 1 to 9 skipping the even numbers, and then immediately counts down without skipping any of the numbers, and then back up to 9 skipping the evens, and so on, alternately counting up and down
\[ (1, 3,5,7,9,8,7,6,5,4,3,2,1,3,5,\ldots ).\] What is
the $ 1000^{\text{th}}$ integer in his list?

[quote=hashtagmath]It's mod 12 because there are 12 elements in each cycle.[/quote]

nice bump lol yeah i know

[hide=bestzack66 (#3)][url=aops.com/community/user/547702][b]bestzack66 (Profile)[/b][/url] $\cdot$ Oct 13, 2020, 12:17 PM $\cdot$ [aops]Y[/aops] 1 $\cdot$ [tip=Avatar][img]https://avatar.artofproblemsolving.com/avatar_547702.jpg?t=1618853880[/img][/tip]
 ----- Who else took mod 13 instead of 12?

whoops 11 year bump $\phantom{Userscript made by WyattB.}$ [/hide]

how is that a whoops bump


trigonometry
probability
function
geometry
algebra
polynomial
conditional probability
\/closed

I'm probably only going to take Intermediate Trig in the fall, and I was wondering if the administrators could post the topic outline for Intermediate Counting.
Also, will AoPS Volume 2 cover most of the material in the counting class? What about AcoPS? Thanks.

[quote="ffdbzathf"]what's AcoPS?[/quote]

I'm guessing that is the book The Art and Craft of Problem Solving by Paul Zeitz, which is a recommended book for some AoPS classes.

[quote="1234567890"]I'm probably only going to take Intermediate Trig in the fall, and I was wondering if the administrators could post the topic outline for Intermediate Counting.
[/quote]LESSON I: Conditional Probability
LESSON II: More Conditional Probability & A Constructive Approach to Counting
LESSON III: Constructive Expectations & 1-1 Counting
LESSON IV: Using Correspondences to Count
LESSON V: Distributions
LESSON VI: Recursion
LESSON VII: The Catalan Numbers
LESSON VIII: Combinatorial Identities
LESSON IX: Principle of Inclusion-Exclusion (PIE)
LESSON X: PIE as a State of Mind
LESSON XI: Generating Functions
LESSON XII: Using Generating Function with Partitions
[quote]Also, will AoPS Volume 2 cover most of the material in the counting class? What about AcoPS? Thanks.[/quote]Chapters 17 and 19 of AoPS Vol 2 cover a lot of material from the class, but in the class we will cover it more thoroughly.  Also, generating functions in particular are covered in significant detail in ACoPS.

Is Mathematics of Choice too easy? Like Intro level?

It goes beyond intro level.  It's a good primer for our Intermediate class.

I went through it, and it looked like it covered intro stuff very thouroghly, but I am not sure about the intermediate class.

I took Intro Counting. Does that mean that Mathematics of Choice won't help much?

It will, I am sure.... there are chapters on generating functions and advanced techniques, but I don't think it will cover it fully.

Will Intermediate Trig and Complex numbers be offered in the spring?

Could you post a syllabus for that as well?

thanks.

intermediate trig and complex numbers will be offered in the fall, and anything that's offered in the fall is not offered in the spring (at least that's how it is this year).  i am not sure if intermediate trig and complex numbers will be offered in the summer though.  do any of the instructors have an idea of that yet?

We will not be considering our Summer 2006 schedule for quite some time.

Intermediate Trigonometry/Complex Numbers:

Lesson I: Basic Trigonometry
Lesson II: Trigonometric Identities I
Lesson III: Trigonometric Identities II
Lesson IV: Laws of Sines and Cosines
Lesson V: Geometry with Trigonometry
Lesson VI: Problem Day!
Lesson VII: Intro to Complex Numbers and the Complex Plane
Lesson VIII: DeMoivre’s Theorem, cis, Re{z}, and Im{z}
Lesson IX: Exponential Form
Lesson X: Roots of Unity and Polynomials
Lesson XI: Geometry with Complex Numbers
Lesson XII: More Problems


I'm interested to learn how it is possible to write huge precision arithmetic programs for example in C++. I know that the numbers are in a linked lists but to do the arithmetic effectively the numbers should be in base 2. But I don't see how to convert a large integer $ n$ to base 2 because I think in some cases one has to approximate $ \operatorname{log}_2n$ more accurate than double precision. I might be wrong of course.

Another question is that how many bits one should put in one cell in the linked list so that one can do the arithmetic effectively? Or is there some other data structures than linked list that one should use in that kind of problems?

A large integer is already stored on a computer in base $ 2$, so I'm not sure what you mean.  A program that receives large integer input (as a string) should read each character as a digit and parse accordingly.  What is the issue?

As for your second question, a different choice of data structure (such as a stack or a queue) will make some operations faster and other operations slower, if I remember correctly.  So what operations do you need?

[quote="t0rajir0u"]A large integer is already stored on a computer in base $ 2$, so I'm not sure what you mean.[/quote]
But if I ask a user an input, he wants it to be in base 10. I can read that input in string or char array, not in longs because there might be overflow. Then if I like to compute something effectively I should convert that string to the corresponding base 2 value and do arithmetic in base 2. Then the user want the result in base 10 so I have to convert it back to base 10.
[quote="t0rajir0u"]As for your second question, a different choice of data structure (such as a stack or a queue) will make some operations faster and other operations slower, if I remember correctly.  So what operations do you need?[/quote]
Nothing in special. I was just thinking how much it takes memory to build a linked list.

Hmm.  I think what you want is a list of [url=http://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations]computational complexities[/url].  The naive base-conversion algorithm I'm thinking of is $ O(n^2)$, which is annoying but doable.

Well, if you want to do arbitrary precision arithmetic, you can use the [url=http://gmplib.org]GNU MP Bignum Library[/url]. Using a library like GMP will allow you to avoid trying to implement arbitrary precision yourself, which could take a while and might be a bit buggy.

On the other hand, if you don't want to use a library, the Wikipedia page t0rajir0u suggested is useful.

In my opinion the best data structure to use would be an array ;). To make it easier to convert from decimal to your internal representation and back, I would use an array of ints but store in each element only values from 0 to 999 999 999. (So it would be a base-one billion representation.) You'll be wasting 2 bits of every 32-bit integer but that's not too bad. The advantage is that you can

1. easily calculate the length of the array you'll need - it's number of digits / 9 (rounded up)
2. easily convert string to your representation and back

i think what he is asking is how big can ya get?

[quote="t0rajir0u"]In the programming languages I'm aware of, I believe the size of an array is an int, which bounds the size of an array of ints.[/quote]
Are you concerned that puuhikki will need an array bigger than 8GB? ;)


AMC
USA(J)MO
USAMO
AMC 10
trigonometry
trig identities

This is not for any of administrators.

The discussion I held resulted out very hard for me.  The reason, I found, was that most people who participated in there were in level of IMO or USAMO, or at least AIME.

So, I need volunteering advanced math chat coach.

If you are interested, please sign-up here

No, Nukular did. And if he's busy, tetrahedr0n will do it. I already talked to with them.

I don't really mind but I would be appreciated if you can join on the easy too  :wink:

Ay.  Well can you give example problems from both?

For the easy one, it's hard for me to give an example problem.. But for the level of the question, it would be like this:

Simplify: (this is very similar to question on AMC10)

[tex]\frac {(2^5)^6}{(4^6)^5}[/tex]

For hard one, I really don't know but since it's going to be in high level, I'll just show you a difficult one

Prove that the sum of the numbers x*sin(x) (x = 2, 4, 6, ..., 180) is 90*cos(1)/sin(1). (degrees)

P.S. I just got from internet, so don't ask me how to do it..

Ay, well the first one is easy...the second...um...

Hint for the second one: [hide]use either telescoping series or complex numbers[/hide]

Er..okay..I will try!

[quote]Prove that the sum of the numbers x*sin(x) (x = 2, 4, 6, ..., 180) is 90*cos(1)/sin(1). (degrees)
[/quote]

Okay, still stuck.  I thought a telescoping series was when the whole sum will have a partial fraction decomposition so that it telescopes into like 2 terms. Like I think AoPS2 gives the example of sum(1, :inf: of 1/n(n+1) so if you do the decomposition it is:
1/n(n+1) = A/n + B/(n+1)
1 = A(n+1) + Bn
1 = An + A + Bn
1 = A
A+B = 0
B=-1
So you want the sum from 1 to  :inf:  of 1/n - (1/(n+1))
You end up w/:
1/1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4...
So the answer =1
However now this confuses me since the limit of 1/n(n+1) as n-> :inf:  seems like it should be 0.  So I have confused myself.
About this problem, I will try writing out some terms:

2sin2+4sin4+6sin6+8sin8+...+180sin(180)
Can factor out 2:
2(sin2+sin4+sin6+...+sin180)
Oh wait, okay I dunno if this proves it but think of the graph of sine:
So:
2(  2(sin2+sin4...+sin88) + sin90 + sin180)
2( 2(sin2+sin4+...+sin88) +1+0)
4(sin2+sin4+...+sin88) +2
Now I am stuck again.

That trig problem is really nice except that it takes about a minute to do it if you see the trick. Someone once said here that college math doesn't help people on the USAMO, but one of my professors would definitely expect us to be able to solve this problem. (We're covering Fourier series right now, and all the problems he does in class are just like this one.) Well, this one is a bit harder than the ones in class, but it is the same idea.

Yes but can you offer a suggestion so I can see the trick?

Perhaps this is an awfully big hint, but [hide]multiply by sin(1) and ask yourself what identities you know about expressions like that[/hide].

[quote="JS1527"]However now this confuses me since the limit of 1/n(n+1) as n-> :inf:  seems like it should be 0.  So I have confused myself.[/quote]

Note that at first, you were calculating the sum of the series.  Then you confused yourself by looking at the limit of the sequence.  In fact, if you want the sum of those things to converge, they better get really small, or else your sum will explode.

JBL - Right.  My teacher acts like the limit of an expression is the same as its sum from 1 to infinity.... So I was correct w/my example w/the partial fraction decomposition?

ComplexZeta - Do you want me to:

[2sin2 + 4sin4 + ... + 180sin180 ]*sin1

or:

S*sin1 = 90*cos(1) ?

[quote="JS1527"]ComplexZeta - Do you want me to:

[2sin2 + 4sin4 + ... + 180sin180 ]*sin1

or:

S*sin1 = 90*cos(1) ?[/quote]

He wants you to take the expression
[2sin2 + 4sin4 + ... + 180sin180 ]*sin1
and manipulate it to show that it is equal to the expression
90*cos(1).

Sorry I didn't abandon this problem, mathfanatic and complexzeta helped me get this one.  Basically, the strategy is to [hide]multiply everything by sin(1) and then use sum-and-difference trig identities so that everything telescopes and you end up w/a simple expression that simplifies to 90*cos(1).  [/hide] Thanks for the help.

Wow you always go the extra step.  I actually remember doing it a little differently, but I think it is essentially the same reasoning:

[hide]

2sin2sin1 + 4sin4sin1 + 6sin6sin1 + ... + 178sin178sin1 + 180sin180sin1 = 90*cos1

since sinA*sinB=1/2 [cos(A-B)  - cos(A+B)]

2(1/2)(cos1 -cos3) + (4)(1/2)(cos3 - cos5) + (6)(1/2)(cos7 - cos5) + ... + (178)(1/2)(cos177 - cos179) +(180)(1/2)(cos179 - cos181) = 90*cos1

cos1 - cos3 +2cos3 - 2cos5 + 3cos7 - 3cos5 + ... + 89cos177 -89cos179 + 90cos179 - 90cos181 = 90*cos1

telescoping to:

cos1 + cos3 + cos5 + cos7 + ... + cos179 + (-90cos181) = 90*cos1

sin1cos1 + cos3sin1 + cos5sin1 + cos7sin1 + ... + cos179sin1 +

-90cos181sin1 = 90*cos1sin1

Since cosAsinB = 1/2 *[sin(A+B) - sin(A-B)]

1/2 * (sin2 - sin0 + sin4 - sin2 + sin6 - sin4 + ... + sin180 - sin178)

-90cos181sin1 =90*cos1*sin1

telescoping to:

1/2 * (-sin0+sin180) -90cos181sin1 = 90*cos1*sin1

-90cos181sin1 = 90*cos1*sin1

-cos181sin1 = cos1sin1

Since cosAsinB = 1/2 *[sin(A+B) - sin(A-B)]

-1/2 * [sin182 - sin180] = 1/2 * [sin2 - sin0]

-sin182 = sin2

Now if you think about the graph of y=sinX, these two are equal.

Ay, this took way more work than Simon's but I just had to do it again so I would remember.  BTW this is not the way they showed me how to do it.[/hide][/hide]


let a,b,c be 3 natural numbers such that a<b<c and gcd(c-a,c-b) = 1

suppose there exixts an integer d such that a+d,b+d,c+d form the sides of a right angled triangle. prove that there exists integers such that
 c+d = l^2 + m^2[/code]

http://www.mathlinks.ro/viewtopic.php?search_id=1967314479&t=126680

As $ (a \plus{} d)^2 \plus{} (b \plus{} d)^2 \equal{} (c \plus{} d)^2$ and $ (c \plus{} d \minus{} (a \plus{} d),c \plus{} d \minus{} (b \plus{} d)) \equal{} 1$, from the link given above we conclude that $ (a \plus{} d,b \plus{} d,c \plus{} d)$ is a primitive solution of the equation $ p^2 \plus{} q^2 \equal{} r^2$. This implies $ a \plus{} d \equal{} l^2 \minus{} m^2, b \plus{} d \equal{} 2lm, c \plus{} d \equal{} l^2 \plus{} m^2$ for relatively prime positive integers $ l,m$ one even and the other odd.

Stronger result: Show that $ c \plus{} d \equal{} l^2 \plus{} m^2$, where $ (l,m) \equal{} 1$.


[color=darkblue] Let $ ABCDE$ be a hexagon with all angles equal. Prove that $ AB \minus{} DE \equal{} EF \minus{} BC \equal{} CD \minus{} FA$.
[/color]

http://www.mathlinks.ro/viewtopic.php?t=16758
1964 < 2000

  darij


also discussed here (without a perfect solution):
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=871033#871033[/url]


for me, I guess have to choose between pool and ping-pong because I can't play the rest due to our rainy weather, and as ping-ping involves much more movement, i choose ping-pong.

Pool owns all sports, except baseball and football. Rack 'em up.

The rain.

unless indoors frisbee with delicate furniture muwahhaha

ping pong FTW...also ping pong for the Xbox 360 isn't that bad  :P

frisbee or basketball. Probably frisbee if I knew someone near here.  :(


geometry
geometric transformation
homothety
ratio
reflection
circumcircle
radical axis

Let $ABC$ be a triangle and $XYZ$ be the Gergonne triangle. Call $M, N, P$ the point on the ray $HA, HB, HC$ such that $HM = HN = HP = R$ where H is the orthocenter and R is the circumradius. Let $R, S, T$ be the intersections of the rays $OM, ON, OP$ and the nine point circle of $ABC$. Prove that the perpediculars from $X, Y, Z$ to $ST, RT, SR$ respectively are concurrent at $K$.

As <MHN = 180-C, we have <HMN = C/2 = 90-<CXY. Hence MN and XY are parallel and the triangles XYZ and MNP are homothetic.
The homothecy (O,1/2) mapping the circle (H,R) to the NP-circle maps the triangle MNP to RST.
Hence XYZ and RST are homothetic and your three lines are the altitudes of XYZ; they concur at the orthocenter of XYZ.
Kind regards

Sorry !!!!! I forgot the main part of the story. I should have not given this problem out as it will be easy for you to do the next problem.
     Let $K$ be the orthocenter of $XYZ$. Let $L$ be the interior similarity point of $(I)$ and the nine point circle. Denote by $T$ the radical point of the three excircles of $ABC$. Then $K, L, T$ are collinear.
     Definitely, you can predict the ratio of $KL$ and $LT$, can't you ?  ;). Thank you Jean Pierrie for the solution to the starter. :)

To avoid two different points T, let's call U the radical center of the excircles = complement of I.
Let E = NP-center.
The reflection wrt E maps M,N,P to the midpoints of the arcs BC,CA,AB of the circumcircle; as I is the orthocenter of the triangle with vertices this midpoints, it follows that the orthocenter V of MNP is the reflection of I wrt E. As U is the midpoint of OV, it follows that U is the orthocenter of RST.
Hence the homothecy (L, -2r/R), mapping RST to XYZ will map U to K.
The result follows with KL/LU = 2r/R.
Kind regards


function
limit
real analysis
real analysis unsolved

f(x,y) is defined on [0,1] x [0,1]. And f is continuous
in each variable seperately. Prove: f is measurable on
[0,1] x [0,1].

(I suspect f is not continuous on [0,1] x [0,1], but I can't think of a counter-example. Maybe f is continuous....? Then of course it is measurable.)

$f$ needs not be continuous. Classical example: $f(0,0)=0$ and $f(x,y)=\frac{xy}{x^{2}+y^{2}}$ at other points.

Construct continuous $f_{n}$ such that $\lim_{n\to\infty}f_{n}(x,y)=f(x,y)$ pointwise(it's very easy, use linear interpolation on variable y.) Use that limit of measurable functions is measurable.


Consider the polynomial equation $q_{1}z^{4}+iq_{3}z^{3}+q_{2}z^{2}+iq_{1}z+q_{0}=0$

If $z=a+bi$ is a solution to this equation, where a and b are real constants and $i^{2}=-1$, which of the following must also be a solution?

A. $-a-bi$
B. $a-bi$
C. $-a+bi$
D. $b+ai$
E. None of these

[quote="jcmtnez"]The correct anwer is (B) since there is a rule that states that complex numbers must come with their conjugates in pairs as a solution to polynomials equations.[/quote]
That's only for the polynomials with [i]real[/i] coefficients.
Notice this polynomial has i's in a couple of the coefficients.

[hide="Hint"]Take the conjugate of the whole thing.[/hide]

this is so stolen from the real amc...

also q_i are real too...


Well, I'm making a project for find the distance between the Earth and the Sun, and I found a cool method for calculate it, but I should also find the distance between Earth and Moon, and all methods that I found for calculate it just can be used when there is a lunar eclypse... and the next is at 2010, and completely useless...

Does anyone know a method for calculate the distance between Earth and Moon without using a lunar eclypse?

yeah.. I know there are a lot of methods... the little trouble is because I can't find any effective method for make the measure without know distance Earth-Sun or using a lunar eclypse...

and, please make more useful posts... I REALLY need help!!!!!!!!

Would the mass of the Moon be known for this?

no  :(

We should use common-use things for make the measure... and find the mass of moon is a little hard if we don't send a satellite to it, and I don't have one  :wink:

I am searching and searching... but I didn't find anything useful  :(

You could use similiar triangles but you would need to know the diameter of the moon.

hi.please see this book:ASTRONOMICAL METHODS &CALCULATIONS; you can find your answer in this book!! :)

[quote="shreyas_patankar"]I think this is the method given in the book.

You need two pictures of the moon, (with the background stars visible) taken a few hours apart. You then calculate the shift in the position of the moon with respect to the stars, and using some math, you can calculate the distance to the moon (of course, you need the radius of the earth). All this needs to be done quite accurately and needs pretty advanced instrumentation.

I don't tink there is any "backyard" method to do this.[/quote]

This is all that I need  :jump:

Thank you

Something that can't be done in a backyard, but the latest efforts towards measuring the earth moon distance is reported at the following location:  http://www.itwire.com/content/view/14993/1066/.

NASA and the NSF both contributed money to this effort.


Can anyone help me prove the following:

Let A be a subset of the ordered set X. If A has an upper bound then A has a least upper bound..

Thank you

I did but all I found was one referring to LUB property implies the GLB property which is different from my question. The other one refers to the LUB axiom which seems to be it but does not give the correct link.

Any other thoughts?

Thank you

For example, if $ X \equal{} \mathbb{Q}$ and $ A \equal{} \{x \in \mathbb{Q} : x^2 < 2\}$, then $ A$ is bounded above by $ 2$,  but $ A$ has no least upper bound.


geometry
trigonometry
trapezoid
perpendicular bisector
geometry proposed

The incircle  $(I, r)$ touches the sides $AB , BC , CA$ of a right  angled triangle  $ABC ( A = 90)$ at points $D, E , F$respectively. Line $BI$ meets $EF$ at $K$ and line $DK$ meetes $AC$ at $L$. Now line $CI$meets $DE$ at $M$ and line $FM$meets $AB$ at $N$. Prove that $BN = CL$ .

[u]Babis[/u]

I will give up the restriction $A=90^{\circ}$ and will add the condition $a>b,\ a>c$. Then

$BN=CL\Longleftrightarrow \{\ b=c\ \}\ \vee \ \{\ (b^2+c^2-a^2)[(a-b)(a-c)-2a(s-a)]=0\ \}.$

[u]Remark.[/u] For example, $a=15,\ b=12,\ c=5:$
$a-b=3,\ a-c=10,\ 2(s-a)=b+c-a=2\Longrightarrow$
$2a^2-2a(b+c)+bc=(a-b)(a-c)-2a(s-a)=0.$

[u]Generally:[/u] $m,n\in N^*,m>n;a=m(m+n),b=2mn,c=m^2-n^2$ and in this
case the $\triangle ABC$ is obtuse ($90^{\circ}<A<120^{\circ}$): $\cos A=-\frac{2m^2+mn+n^2}{4m(m+n)}\ !$
($m: =3,\ n: =2\Longrightarrow a=15,\ b=12,\ c=5$, i.e the previous example).

Do you mean  $BN$ in the begining of the soluiton ?
  This moment I' m in school , but how did you make these calculations ? Please send a [b]hint[/b]. The official solution uses pure synthetic relation - no metrics. I need your solution however.

 [u]Babis[/u]

I am sorry, a letter was wrong ($D: =N$). See above, I corrected. Is interesting matter that exists non right-angle triangle for which $BN=CL$, i.e. the proposed problem isn't true reciprocally.

A synthetic solution:

$BK$ is the perpendicular bisector of $DE$, so $DKE$ is isosceles (with base $DE$). Since $\angle DEF=45^{\circ}$, this means that $DKE$ is a right isosceles triangle, so $DK\perp EF\Rightarrow DK\|CM$. In the same way we prove that $FM\|BK$. This proves that $BKMN,CMKL$ are trapezoids.

$MI\perp EK,KI\perp ME\Rightarrow EI\perp MK$, and since $EI\perp BC$ we get $MK\|BC$, so $\angle MKB=\angle KBE=\angle NBK$, which proves that $BKMN$ is an isosceles trapezoid with $MK=BN$. In the exact same way we show that $MK=CL$, and we're done: both $BN$ and $CL$ are equal to $MK$.

Dear Ph -An , how did you get these relations ?

  Babis

Here is my solution:

    MK  is a part of middle line of triagle ABC.   And as Grobber has shown  this

   segment is projected onto base by two pairs of parallel lines.     A=90  is not part

   of the proof.

   Thank you.

    M.T.


Post here if you plan to compete at Math Prize for Girls this fall!!

Yeah, I am going. I'm from IL and am going to be a sophomore (yeah, skipped 9th grade) for the 2009 MPFG contest. I got a 129 on the AMC 10A this year ( :( darn stupid mistakes) but yet I still qualified, yay.

Also, I'm 14 and going to be 15 in November.

Please also introduce yourselves here. :)

Hiii!

I'm Sara from Illinois.. I go to IMSA, I'm 15, and yah...

I'm actually studying for math team this year because I am sick of barely qualifying for stuff =D

I hope to see you all there! ^_^

btw... i'm 13, going into 8th grade, and I'm from MD. I already know someone who's going. :)

[quote="baozhale"]Alyssa Zisk from MA, I'm going. Going to be a senior, ultra excited about the free Mathematica :)[/quote]

Heh, isn't the cost of Mathematica cheaper than hotel rooms/plane (or gas if you're driving) of going to MPFG (of course, the experience and the possibility of winning 20 grand is worth it)

Oh, and I'm not going due to gender issues.

Sara, you're going? Yay! I'm going too (remember Sarah from MathZoom?) Anyway, I'm Sarah from CA and I'm going.

so you have to be in 11th grade to participate????

No, contestants can be in 12th grade or below.  Back in May, applicants had to be in 11th grade or below, which would presumably put them in 12th grade or below when the contest takes place in November.

well, my family has friends in NYC and we'll probably stay with them. makes it cheaper :)

I'm going. From California, will be a senior. Hopefully I'll have actually done some contest preparation by November.

[quote]I would love to go, but transportation just makes it unfeasible for me.[/quote]

That's too bad. I was hoping to see you there. :(

hm29168, if you don't remember who I am, look me up in the MP 2009 yearbook. :wink:

Hi!  I'm 14, and going to be a freshman this fall at TJHSST. I already know several other people going from my state, everyone is so smart.   :maybe:

hmm i have no idea wut this contest is about??

could anyone tell me more about it? :)

[quote="AIME15"] Oh, and I'm not going due to gender issues.[/quote]

 :rotfl:

Please see the Math Prize web site: [url]http://mathprize.atfoundation.org[/url].  Or view the first thread in the Math Prize forum: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=263427[/url].  If eligible, you can still apply for this year's contest by September 15.

Dang it.

Unfortunately, I just realized I had scheduling conflicts and flight conflicts; therefore, I cannot go. :(

But good luck to everyone who's going there. (especially the people from my state, or actually, the FOUR people from my school!)

I'm a Junior from NYC, and I'm going, too.

It's a two hour train ride away, for me, so not as bad as for most of you. At least one other person is coming from my school, and she lives just few stations away from the place.

Good luck everybody, and don't forget to eat your breakfast :)


Hi I was looking for the ranking of Undergraduate Physics programs. Does anyone have an idea which schools are the best places to study undergraduate physics?

I would assume that Harvard, MIT, Stanford and Princeton are the best.


superior algebra
superior algebra solved

Let $(A,+,\cdot)$ be a commutative ring which have 1, without divisors of 0, and of characteristic $p$ a prime number.
Prove that $(U(A)\cup \{0\},+,\cdot)$ is a field, where $U(A)$ is the set of invertible elements of $A$.

Just not true man!
Look at $\mathbb Z/p\mathbb Z$.

Well, $0$ is missing in his set, but when adding it I see no problem.

how about $Z_p[X,X^{-1}]$ ... Laurent polys, or any local ring


calculus
integration
calculus computations

Assume that a>-1,b>-1,a≠b, prove
                              $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+a)(n+b)}=\frac{1}{b-a}\int_a^b \frac{x^a-x^b}{1-x}dx$

This is absurd. Let $b=a+1$.
This equation asserts that
$\frac1{1+a}=\sum_{n=1}^\infty\frac1{(n+a)(n+a+1)}=\int_a^{a+1}x^a\, dx=\frac{(a+1)^{a+1}-a^{a+1}}{a+1}$.
For large $a$, this last expression is comparable to $a^a$, which gets very large.

Probably, the integral should be from $0$ to $1$ rather than from $a$ to $b$  ;) .

I'm sorry.It is from 0 to 1.But how to do it?

Expand $\frac 1{1-x}=1+x+x^2+x^3+\dots$ and integrate each term $(x^a-x^b)x^n$ separately.


Let a,b,c be positive numbers such that $ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\equal{}1$
Prove that
 $ \frac{a^3}{b^2\plus{}c^2}\plus{}\frac{b^3}{c^2\plus{}a^2}\plus{}\frac{c^3}{a^2\plus{}b^2} \geq \frac{3\sqrt{3}}{2}abc(a^4\plus{}b^4\plus{}c^4)$

[quote="fjwxcsl"]
\[ \frac {a^3}{b^2 + c^2} + \frac {b^3}{c^2 + a^2} + \frac {c^3}{a^2 + b^2} \geq kabc(a^4 + b^4 + c^4),
\]

\[{ k_{max} = \frac {\sqrt {3}}{2}}.
\]
[/quote]
 I don't understand you. Do you want to tell that my ineq is wrong? :(


In the plane is given an infinite set of  points such that any 1*1 square contains finitely many points. Prove that one can find two points A,B from the set such that for every other poit X in the set, min(XA,XB)>=0.999AB.

It works for $k<1$ (here $k=0.999$). Assume there are no two such points. Then choose any two points $X_0,X_1$. We can find $X_2$ s.t. $X_2X_1<kX_1X_0=ka$. We can then find $X_3$ s.t. $X_3X_2<kX_2X_1<k^2a$, and so on. The sequence $(X_n)$ forms a Cauchy sequence, so it converges. Take a unit square centered at the limit point, and we contradict the hypothesis.


Solve the equation $ (x\plus{}a^3)^3\equal{}x\minus{}a$

[quote="aleksandar"]Solve the equation $ (x \plus{} a^3)^3 \equal{} x \minus{} a$[/quote]

Just set $ x\equal{}y\minus{}a^3$ and the equation becomes $ y^3\minus{}y\plus{}a^3\plus{}a\equal{}0$ and use Cardan's formulas


vector calculus geometry inequalities integration derivative probability

May I choose another topic?  I find vector calculus to be pretty interesting, especially differential forms and manifolds.

mine is calculus and co-ordinate geometry

Theory of equations, inequalities, Geometry, number theory,etc...

Number theory is good, too.
[quote="rituraj007"]Differential Calculas [/quote]
What?  Not integral calculus? :(

yes :rotfl:    :blush:

but i don't kno wat is high skul physics..:rotfl:

High school physics is pretty straighforward.  College physics is more interesting.  That's when you really get to use derivatives and integrals.

i've been using it since long
and dat also in life size problems :P  :rotfl:

mine are

Calculus(both)

Probability

Geometry(haven't done after 10th but will resume for RMO)

and functional equation(easy ones,not too hard)


Find all functions $f: \mathbb{R}\to\mathbb{R}$ satisfying:

$\frac{f(xy)+f(xz)}{2} - f(x)f(yz) \geq \frac{1}{4}$  for all $x,y,z \in \mathbb{R}$

Put $ x: \equal{} y: \equal{} z: \equal{} 1$, we have $ f(1) \minus{} (f(1))^2\ge \frac {1}{4}\Leftrightarrow (f(1) \minus{} \frac {1}{2})^2\le 0\Leftrightarrow f(1) \equal{} \frac {1}{2}$
Put $ x: \equal{} y: \equal{} z \equal{} 0$, with the same argument, we also have $ f(0) \equal{} \frac {1}{2}$.
Put $ y: \equal{} z: \equal{} 1\Longrightarrow f(x)(1 \minus{} f(1))\ge \frac {1}{4}\Leftrightarrow f(x)\ge \frac {1}{2}$
But when we put $ y: \equal{} z: \equal{} 0\Longrightarrow \frac {1}{2} \minus{} \frac {1}{2}.f(x)\ge \frac {1}{4}\Longrightarrow f(x)\le \frac {1}{2}$
Hence $ f(x)\equiv \frac {1}{2} \forall x\in \mathbb{R}$

Let $P(x,y,z)$ denote the assertion.
$P(1,1,z)\implies -(f(1)-\frac{1}{2})^2\geq 0 \implies f(1)=\frac{1}{2}.$
$P(x,1,1)\implies f(x)-\frac{1}{2}f(x)\geq \frac{1}{4} \implies f(x)\geq \frac{1}{2}\implies f(\frac{1}{x^2})\geq \frac{1}{2}$
$\implies f(1)-f(x)f(\frac{1}{x^2}) \geq \frac{1}{4} \implies f(x)=\frac{1}{2} ~~\forall x\neq 0.$
$P(x,0,0)\implies f(0)-\frac{1}{2}f(0)\geq \frac{1}{4} \implies f(0)\leq \frac{1}{2} \implies f(0)=\frac{1}{2}.$
Thus, $\boxed{f(x)=\frac{1}{2} ~~\forall x\in \mathbb{R}}.$


show that..........
lim[x^3+1]/[3x^3+x] =1/3 as x goes to infinity by definition of limit

$ \lim_{x\to \infty} \frac{x^{3}\plus{}1}{3x^{3}\plus{}x}\equal{}\lim_{x\to \infty} \frac{x^{\minus{}3}\plus{}1}{x^{\minus{}2}\plus{}3}\equal{}\frac{1}{3}$

$ \lim_{x\to \infty} \frac{1}{x^2}\equal{}0$

$ \lim_{x\to \infty} \frac{1}{x}\equal{}0$

Let's suppose there exists $ M > 0$ for all $ \varepsilon > 0$ such that $ \varepsilon \equal{} \frac {1}{3M^3}$.

When $ x > M$, $ \left|\frac {x^3 \plus{} 1}{3x^3 \plus{} x} \minus{} \frac {1}{3}\right| < \left|\frac {x^3 \plus{} 1}{3x^3} \minus{} \frac {1}{3}\right| \equal{} \frac {1}{3x^3} < \frac {1}{3M^3} \equal{} \varepsilon$. By definition of the limit, $ \lim_{x\rightarrow{\infty}}\frac {x^3 \plus{} 1}{3x^3 \plus{} x} \equal{} \frac {1}{3}$.


Find all polynomial $ P \in\mathbb R[x]$ such that there exists a unique $Q \in \mathbb R[x]$ with $Q(0)=0$ satisfying
\[x+Q(y+P(x)) = y+Q(x+P(y)),\]
for all $x,y \in \mathbb R$.

[quote="hoangclub"]Find all polynomial $ P$  $ \in$ $ R[x]$ such that only have unique $ Q$ $ \in$ $ R[x]$ with $ Q(0) \equal{} 0$ satisfying:
     $ x\plus{}Q(y\plus{}P(x))$=$ y\plus{}Q(x\plus{}P(y))$for all x,y $ \in$$ R$.[/quote]

Let $ p$ an $ q$ be the degree of polynomials $ P$ and $ Q$.

1) $ p\equal{}0$. $ P(x)\equal{}a$ and so $ x\plus{}Q(y\plus{}a)\equal{}y\plus{}Q(x\plus{}a)$. So $ q\equal{}1$, $ Q(x)\equal{}cx$ and $ x\plus{}cy\plus{}ca\equal{}y\plus{}cx\plus{}ca$ and $ c\equal{}1$

2) $ p\equal{}1$. $ P(x)\equal{}ax\plus{}b$ and so $ x\plus{}Q(y\plus{}ax\plus{}b)\equal{}y\plus{}Q(x\plus{}ay\plus{}b)$
Then, if we put $ y\equal{}\minus{}P(x)$ in the original equation, we have $ x\plus{}P(x) \equal{} Q(x\plus{}P(\minus{}P(x)))$ and LHS has degree 0 (if $ a\equal{}\minus{}1$) or 1 and RHS degree 0 (if $ a^{2}\equal{}1$) or q. And so :
2.1) LHS has degree 0 : $ P(x)\equal{}b\minus{}x$ and so $ b\equal{}Q(2b)$ and obviously $ Q(x)$ is not unique.
2.2) LHS has degree 1 and $ q\equal{}1$ : $ P(x)\equal{}ax\plus{}b$ with $ a\neq\minus{}1$ and $ Q(x)\equal{}cx$ (since $ Q(0)\equal{}0$). So :
$ x\plus{}c(y\plus{}ax\plus{}b)\equal{}y\plus{}c(x\plus{}ay\plus{}b)$ and so $ 1\plus{}ac\minus{}c\equal{}0$ and so $ a\neq 1$ and $ c\equal{}\frac{1}{1\minus{}a}$

3) $ p>1$
Then, in $ x\plus{}Q(y\plus{}P(x))$=$ y\plus{}Q(x\plus{}P(y))$, LHS has degree in $ x$ $ \max(1,pq)$ and RHS has degree $ q$ in $ x$ and, since $ p>1$, this is impossible.

And the solutions are :
$ P(x)\equal{}a$ (and unique $ Q$ is $ Q(x)\equal{}x$)
$ P(x)\equal{}ax\plus{}b$ with $ a\notin\{\minus{}1,1\}$ (and unique $ Q$ is $ Q(x)\equal{}\frac{1}{1\minus{}a}x$)


Hi Everyone,

On another thread, people on MathLinks were discussing the benefits and fallbacks of starting out very early on in mathematics.  One common theme among the posts was that one must enjoy what one is doing.

I would like to start a new thread here.  How many people started out mathematics relatively late?  What made you decide to study mathematics, either at school or on your own?  What attracted you to Olympiad problem solving?

I was about 19 when I first discovered that I really liked mathematics.  Before then I always thought that I would be a theoretical physicist.  I ended falling in love with the [i]language[/i] of physics though --- mathematics.   Since I was 19 and wasn't in the normal school system here in Ontario, Canada, my math skills were probably very weak compared to the math skills some of you on MathLinks have or had at the age of 19.   Also not being in the normal school system, I didn't get opportunities to participate in local and provincial mathematics competitions.  I recognized however that Olympiad type problems are a good means to make one learn and absorb "elementary" mathematics.  I also see Olympiad type problems as a way to keep one's mathematical mind sharp, as one most often needs to be very creative when solving Olympiad type problems.

Not having much experience with Olympiad type mathematics problems, I tried the Putnam exam three times.  I didn't do very well as I one would expect, but I enjoyed the experience.  Sitting down and just working on math problems for the hell of it for 6 hours was great, despite my low scores.   I now subscribe to Crux Mathematicurum with Mathematical Mayhem.  Many of you have probably heard about this problem-solving journal published by the Canadian Mathematical Society.  It's a great resource for problems, etc.  See  [url]http://www.cms.math.ca/Publications/CRUXwithMAYHEM.html[/url] for more on the Journal.

I have one chance left to take the Putnam exam.  I hope this time I'll be much more prepared.  In any case, the whole learning process about Olympiad type problems and the fact that I'm not that great at them has made me interested in how to teach Olympiad problem solving to a receptive audience.  Do any of you have any advice, or alternatively stories to tell of your own experiences with Olympiad coaches or others of mathematical influence that may be of use to someone interested in how to teach Olympiad problem solving better?  Can one even teach mathematical Olympiad type problem solving other than by just getting students to solve problem after problem?

Anyway, I look forward to replies to this post.

Have a mathematistic week. :lol:

Yours truly,

mathfreak


Three prisoners Mark, Luke and Thomas find in one cell. Two of they will have to be shoot to you but everyone of they does not know if dovra to die or not. Poiche at least between Luke and Thomas dovra to die Mark asks to one its guard is of the two. The guard answers Luke. To this Mark point rallegra thinking that its probabilita one to die is last from 2/3 to 1/2. She has effectively reason rallegrar?


Jeff has an equal number of nickels, dimes and quarters worth a total of $ \$1.20$. Anne has one more of each type of coin than Jeff has. How many coins does Anne have?

Let $ x$ be the number of each coin Jeff has.

$ 25x\plus{}10x\plus{}5x\equal{}120$
$ 40x\equal{}120$
$ x\equal{}3$

Jeff has 3 nickels, 3 dimes, and 3 quarters.

Therefore Anna has 4 nickels, 4 dimes, and 4 quarters, for a total of $ 3(4)\equal{}\boxed{12}$ coins.


hello
i have a question concerning maple (plots to be precise)
I wanted to ask it here, but didn't find a place (latex help seemed the best spot)

now is there already a place where such questions, in my case about maple commands, can be asked

or, due to maple being a commercial nonfree product, that the majority probably doesn't have here, is it not an option?

So uhm forgive me for pushing it up but is there any maple related areas around here?

You may ask, but it is not certain that anyone will answer :-).  Rather than asking permission to ask questions, you are probably better off just posting some of them -- if there was a really good place for them, some moderator would probably move them there.  This forum is a good first guess.


Please help me perform induction on this problem:

Prove that if n ≥ 12 then n can be written as a sum of 4’s and 5’s. For example, 23 = 5 + 5 + 5 + 4 + 4 = 3 · 5 + 2 · 4. [Hint. In this case it will help to do the cases n = 12, 13, 14, and 15 separately. Then use
induction to handle n ≥ 16.]

12 = 4 + 4 + 4.

Assume it holds for n. If n can be written as a sum of 4's and 5's, then replace a 4 with a 5 and you get n+1. If n only has 5's in its sum, replace 3 5's (n >= 15) with 4 4's and you get n+1.

How about this:

$ n \equal{} 4(\minus{}n\plus{}5t)\plus{}5(n\minus{}4t)$

For any integer $ t$. What we need to do is to find an integer $ t$ which gives $ \minus{}n\plus{}5t\geq 0$ and $ n\minus{}4t\geq 0$, or equivalently $ \frac{1}{5}n\leq t\leq\frac{1}{4}n$. This can be achieved for any $ n\geq 12$

Here is a generalisation:

Prove that every sufficiently large positive integer can be expressed as the sum of $ p$'s and $ (p\plus{}1)$'s where $ p$ is a positive integer.

Thanks, so I'll try two cases for n = 5a + 4b, 1: a 5's and 0 4's, 2: a 5's and b 4's.

The more relevant generalization is the Chicken McNugget theorem.  ;)


geometry
geometric transformation
number theory
modular arithmetic

When n is divided by 5 the remainder is 2, what is the remainder when n+4 is divided by 5?

The local high school band has x musicians. Tube players make up 1/6th of the band. There are 3 more than 3 times as many drummers than tuba players. There are 4 fewer flutists than tuba players. How many musicians in the band are not tuba players, drummers, or flutists?

John works 5 days a week, on Mondays he completes 1/2 of his week's work. On Tuesday he completes 1/4 of the remainder of his work for the week. What fraction of the week remains to be done?

There are 660 feet in one furlong, and 1/2 of a fathom in one yard. how many fathoms are there in one furlong?

[quote="nick42"]1. $ 1$
2.
3. $ 3/8$
4. $ 110$

That's what i think they are. can anybody double-check?  :lol:[/quote]

2. (5/6)x musicans in the band are not Tube players,
(1/2)x -3 musicans in the band are not drummers
(5/6)x+4 musicans in the band are not flutists

[quote="Forte"]Explain your reasoning.  :P[/quote]

T - Tube players
D - drummers
F - flutists

$ T \equal{} \frac {1}{6}x$

$ x \minus{} T \equal{} x \minus{} \frac {1}{6}x \equal{} \frac {5}{6}x$

$ D \equal{} 3T \plus{} 3 \equal{} 3 \cdot \frac {1}{6}x \plus{} 3 \equal{} \frac {1}{2}x \plus{} 3$

$ x \minus{} D \equal{} x \minus{} (\frac {1}{2}x \plus{} 3) \equal{} \frac {1}{2}x \minus{} 3$

$ F \equal{} T \minus{} 4 \equal{} \frac {1}{6}x \minus{} 4$

$ x \minus{} F \equal{} x \minus{} (\frac {1}{6}x \minus{} 4) \equal{} \frac {5}{6}x \plus{} 4$

[quote="nick42"]
2. the number of tuba players is $ 1/6x$, drummers are $ 1/2x \plus{} 3$, and flutists are $ 1/6x \minus{} 4$. just subtract these numbers from $ x$ to get the answers. [/quote]

Statement doesn't say that there are only Tube players, drummers and flutists in the band... there can be some singers and gitarists...

Irrelevant. That's why the value is in terms of x.

Tuba players are $ \frac{1}{6}$ of the band, so the number of Tuba players is $ \frac{1}{6}x$ regardless, and the number that are not Tuba players is $ \frac{5}{6}x$

If we know the number of Tuba players to be $ \frac{1}{6}x$, then 3 more than 3 times that amount is $ 3\left(\frac{1}{6}x\right)\plus{}3\equal{}\frac{1}{2}x\plus{}3$ Drummers, regardless. The number that are not Drummers is then $ \frac{1}{2}x\minus{}3$

If the number of Tuba players is $ \frac{1}{6}x$, then 4 less is $ \frac{1}{6}x\minus{}4$ Flutists, regardless. The number that are not Flutists, then, is $ \frac{5}{6}x\plus{}4$

For each of these proofs, it is irrelevant that there might be other instruments, because each one is based on the first value, which was given to us.

1. We write this in modulo 5. So,$ n\equiv2(\mod5)$. Since we can do addition in modular arithmetic, we add 4 on both sides to get $ n\plus{}4\equiv6\equiv1(\mod5)$ So, the remainder is 1.


How could "change" the view we have about cardinalities if we defined alternatively when 2 sets have the same cardinality? Are there such definitions?

1)  This is not the appropriate forum for such a discussion.

2)  The notion of bijection is so fundamental to anything we could call cardinality that I would hesitate to call any related but distinct concept "cardinality" at all.  On the other hand, if you restrict to the special case that you are comparing sizes of two subsets of the same set - for example, $ \mathbb{N}$ - there are many interesting and useful notions of [url=http://en.wikipedia.org/wiki/Natural_density]density[/url].


Fie triunghiul $ABC$ de afixe $a,b,c$ avand acelasi modul si $a^{3}+b^{3}+c^{3}+abc=0$. Demonstrati ca centrul cercului lui Euler se afla pe cercul circumscris triunghiului $ABC$.

[hide]Un mic ajutor, centrul cercului lui Euler are afixul $\frac{a+b+c}{2}$(Vedeti "Matematica pentru grupele de performanta",clasa a X-a,Vasile Pop)[/hide]


How many integers in base $b$ have $n$ digits? I know the formula - $(b-1)b^{n-1}$ - but I need a reference. Surely it is somewhere!

[hide]$b^n-b^{n-1}$
Say, 3 digits in base 10
1000-100=900
4 digits in base 2
16-8=8
[/hide]

What exactly are you asking for when you state you desire a "reference" ... a formal statement verifying and/or explaining the given formula ...?  If it is the latter, I can proudly say I am qualified for the task! :P

  Take the base b expression of n digits:

$(x_1)b^0 + (x_2)b^1 + (x_3)b^2 ... (x_z)b^{n-1}$

  In the very last term the coefficient $(x_z)$ can be filled with any of the digits from $1$ to $b-1$ a total of $(b-1)-1+1=b-1$ choices.  The rest of the coefficients can be filled with any of the digits $0$ to $b-1$ a total of $(b-1)-0+1=b$ choices.  Since $n-1$ of our terms can choose from $b$ different digits and only one term from $b-1$, there are $(b-1)b^{n-1}$ such numbers.


1) Two runners jog around an oval path at constant rates, but in opposite directions.  One runner completes a lap every 10 minutes, and the other does a lap every 12 minutes.  How many minutes elapse between consecutive meetings of the runners?  Express your answer as a mixed number. [hide]5 5/11[/hide]

[b]Let the perimeter of the path be 600 feet[/b].... Yes I know since i suck at math the only thing i can do is pick numbers for this kind of stuff.....

that means one runner runs 60 feet/min and the other runs 50 feet/min.
you want it to be in the form of "[i]x[/i]feet/ minute b/c the question asks " How many minutes elapse......"

so that means the runners separate at the rate of 110 feet/minute.

since the path is 600 feet, you have 600=110*T (from the D=RT formula)
T is how many minutes elapse......
You find that T=5.555555555555555555555555555555
- drawing a diagram helps

for most people, thats easier, but for people that like to get into algebra, then you can set the distance equal to x. that way helps if you are slow at multipling big numbers and stuff. and you probably want your answer as a fraction, so leaving it as 60/11 is fine i think


We have two points M1(z1), M2(z2). We know that the middle of the segment [M1M2] has the afix M((z1+z2)/2).

The recyproc is true???

You make no sense. What do you mean by afix and recyproc?

This is my interpretation of the problem. There are two points, $ M_{1}$ and $ M_{2}.$ The complex numbers representing them are $ z_{1}$ and $ z_{2}.$ We know that if a point $ M$ is the midpoint of $ M_{1}M_{2},$ then the complex number ("afix") corresponding to $ M$ is $ \frac{z_{1}+z_{2}}{2}.$ Does the converse ("recyproc") hold? That is, if the complex number of $ M$ is $ \frac{z_{1}+z_{2}}{2},$ must $ M$ be the midpoint of $ M_{1}M_{2}$?

The answer is yes. Let $ M'$ be the midpoint of $ M_{1}M_{2}.$ Then the complex number of $ M'$ is $ \frac{z_{1}+z_{2}}{2},$ the same as that of $ M.$ Since $ M$ and $ M'$ have the same complex number, they must be the same point. $ M'$ is the midpoint of $ M_{1}M_{2},$ so $ M$ is the midpoint of $ M_{1}M_{2}$.


geometry
symmetry
ratio
circumcircle
trigonometry
cyclic quadrilateral
geometry proposed

[img]http://i198.photobucket.com/albums/aa269/ConanKudo5/geo.jpg[/img]
$\hat BPC= 90$
Prove that:AP+AE=PD

Let $ EF$ meet $ BC$ at $ L.$ Let $ BP, CP$ meet the incircle $ (I)$ at $ X, Y.$ $ BC$ is polar of $ D$ and $ EF$ is polar of $ A$ with respect to $ (I).$ $ A, P, D$ are collinear $ \implies$ tangent of $ (I)$ at $ P$ goes through $ L.$ Project the circle $ (I)$ to a circle $ (I')$ and the line $ AL$ to infinity, labeling all projecterd points with primes. By symmetry, $ X'Y' \parallel E'F' \parallel B'C'$ $ \implies$ $ XY$ goes through $ L.$ The diagonal intersection $ Q$ of the cyclic quadrilateral $ EFYX$ lies on the polar $ AD$ of $ L \equiv EF \cap XY.$ Let $ XY$ cut $ AD$ at $ Z,$ let $ PF, PE$ cut $ BC$ at $ M, N.$ Projected $ \triangle P'M'N'$ becomes isosceles right and $ B', C'$ midpoints of $ M'D', N'D'$ $ \implies$ cross ratios $ (D, M, B, L) \equal{} (D', M', B', L') \equal{} \minus{} 1$ and $ (D, N, C, L) \equal{} (D', N', C', L') \equal{} \minus{} 1$ are harmonic. Suppose $ P'B'$ intersects $ E'M'$ at $ X^*.$ These are P-, M-medians of isosceles right $ \triangle D'P'M, \triangle P'M'N'$ $ \implies$ $ \measuredangle M'P'B' \equal{} \measuredangle E'M'P'$ $ \implies$ $ \triangle M'P'B \sim \triangle X^*M'B'$ $ \implies$ $ \measuredangle P'X^*E' \equal{} \measuredangle M'X^*B' \equal{} \measuredangle P'M'B' \equal{} \measuredangle P'F'E'\ ( \equal{} 45^\circ)$ $ \implies$ $ X^* \equiv X' \in (I')$ $ \implies$ $ E', X', M'$ are collinear, $ E, X, M$ are collinear and similarly, $ F, Y, N$ are collinear. By central projection of $ BC$ to $ AD$ from $ X,$ the cross ratio $ (D, P, Q, Z) \equal{} (D, M, B, L) \equal{} \minus{} 1$ is also harmonic, $ \frac {\overline{DP}}{\overline{DZ}} \equal{} \minus{} \frac {\overline{QP}}{\overline{QZ}}.$

$ \measuredangle BPC \equal{} 90^\circ$ is right $\iff$ $ XY$ is a diameter of $ (I)$ $\iff$ $ Z$ is the midpoint of $ DP.$ Using the harmonic cross ratio $(D, P, Q, Z) \equal{} -1,$ this is equivalent to $ \frac {\overline{QZ}}{\overline{QP}} \equal{} \minus{} \frac {\overline{DZ}}{\overline{DP}} \equal{} \minus{} \frac {_1}{^2}$ and by the following lemma, to $ \frac {\overline{AD}}{\overline{AP}} \equal{} \left(\frac {\overline{QD}}{\overline{QP}}\right)^2 \equal{} \left(2\ \frac {\overline{QZ}}{\overline{PQ}} \plus{} 1\right)^2 \equal{} 4.$
Since power of $ A$ to $ (I)$ is $\overline{AE}^{2} \equal{} \overline{AP} \cdot \overline{AD} \equal{} 4\ \overline{AP}^{2},$ this is further equivalent to $ \overline{AE} \equal{} 2\ \overline{AP}$ $\iff$ $ \overline{AE} \plus{} \overline{AP} \equal{} 3\ \overline{AP} \equal{} \overline{AD} \minus{} \overline{AP} \equal{} \overline{PD}.$

[color=#0000FF][b]Lemma:[/b][/color] Let $ (I)$ be a circle with diameter $ XY$ and $ Z$ arbitrary point on $ XY.$ Perpendicular to $ XY$ at $ Z$ cuts the circle $ (I)$ at $ P, D.$ Let $ Q \in PZ$ be arbitrary, let $ XQ, YQ$ cut $ (I)$ again at $ F, E,$ respectively, and let tangents to $ (I)$ at $ E, F$ meet at $ A.$
Then $ AQ \perp XY$ and $ \frac {\overline{AD}}{\overline{AP}} \equal{} \left(\frac {\overline{QD}}{\overline{QP}}\right)^2$

Let $ XE, YF$ meet at $ T.$ $ Q$ is orthocenter of $ \triangle XYT,$ $ TQ \perp XY.$ By Pasacal theorem for degenerated cyclic hexagon $ XEEFFY,$ the intersections $ T, Q, A$ are collinear  $ \implies$ $ AQ \perp XY.$ Circumcircle of the right angled kite $ IEAF$ is the 9-point circle $ (K)$ of the $ \triangle TXY$ $ \implies$ $ A$ is the midpoint of $ TQ,$ the circumcenter of right angled quadrilateral $ TEQF,$ let $ (A)$ be its circumcircle. As the angles $ \measuredangle AEI \equal{} \measuredangle AFI$ are right, the circles $ (I) \perp (A)$ are perpendicular. $ XY \perp TQ$ are perpendicular diameters of the perpendicular circles $ (I), (A)$ and $ (I)$ cuts the line $ TQ$ at $ D, P.$ It follows that the points $ D, P$ and the circle $ (A)$ are coaxal and consequently, powers of $ Q \in (A)$ to the points $ D, P$ are in the ratio $ \frac {\overline{QD}^2}{\overline{QP}^2} \equal{} \frac {\overline{AD}}{\overline{AP}}.$

[color=darkblue][b]Very nice problem ![/b]

[u][b]Proof.[/b][/u] Suppose w.l.o.g. that $b>c$. Denote $\{\begin{array}{c}K\in BC\ ,\ AK\perp BC\\\ L\in EF\cap BC\\\ S\in LI\cap AD\\\ \phi=m(\widehat{DLI})\end{array}$. From the well-known property $LI\perp AD$ (see the above [b][u]Yetti's proof[/u][/b])

obtain $L\in PP$ and $\sin\phi=\frac{SD}{LD}=\frac{KD}{AD}$. Prove easily that $KD=\frac{(b-c)(p-a)}{a}$, $LB=\frac{a(p-b)}{b-c}$, $LD=\frac{2(p-b)(p-c)}{b-c}$.

$1\blacktriangleright$ $\{\begin{array}{c}\boxed{PA+AE=PD}\\\\ [AP\cdot AD=(p-a)^{2}]\end{array}\|$ $\Longleftrightarrow$ $\{\begin{array}{c}PD-PA=p-a\\\ PA\cdot (PA+PD)=(p-a)^{2}\end{array}\|$ $\Longleftrightarrow$ $PA\cdot (PA+PD)=(PD-PA)^{2}$ $\Longleftrightarrow$

$PD=3\cdot PA$ $\Longleftrightarrow$ ${\begin{array}{c}AD=2(p-a)\\\ (\sin\phi =\frac{KD}{AD})\end{array}\|}$ $\Longleftrightarrow$ $\boxed{\sin\phi=\frac{b-c}{2a}}$.

$2\blacktriangleright$ Because the division $(B,C,D,L)$ is harmonically obtain : $\boxed{PB\perp PC}$ $\Longleftrightarrow$ $\widehat{BPD}\equiv\widehat{BPL}$ $\Longleftrightarrow$

$\frac{PD}{PL}=\frac{BD}{BL}$ $\Longleftrightarrow$ $2\cdot \frac{DS}{LD}=\frac{BD}{BL}$ $\Longleftrightarrow$ $2\sin\phi=\frac{(p-b)(b-c)}{a(p-b)}$ $\Longleftrightarrow$ $\boxed{\sin\phi =\frac{b-c}{2a}}$.[/color]

[color=darkblue][b][size=134][u]Remark.[/u][/size]

The incircle $C(I,r)$ of an acute triangle $ABC$ is tangent (internally) to the circle with the diameter $[BC]$ if and only if $\boxed{a+h_{a}=b+c}$

and in this case the common tangent point of these circles belongs to the $A$- Gergonne's line if and only if $a=4r$.[/b][/color]


Solve the equation:
$sinx+sin2x+sin3x=1$

[quote="Davidd"]
\[\begin{array}{l}\sin (3x) = 3\sin (x)-4\sin^{3}x \\ \sin (2x) = 2\sin (x)\cos (x) \\ \end{array}\]
Plug in and convert cos to sin and have fun.[/quote]
 :lol: did u try solving it that way?

Here's what you could do:
$\sin x+\sin 2x+\sin 2x \cos x+\cos 2x \sin x=\sin 2x(1+\cos x)+2\sin x \cos^{2}x= \sin 2x (1+\cos 2x)$
So:
$\sin 2x (1+2 \cos x)=1$
form here its trivial. :D


Suppose that $\mid x_{i}\mid <1$ for all $i=1,2,3\dots,n$.

Suppose further that $\mid x_{1}\mid +\mid x_{2}\mid +\mid x_{3}\mid +\dots+\mid x_{n}\mid =19+\mid x_{1}+x_{2}+x_{3}\dots+x_{n}\mid $.

What is the smallest possible value of $n$?

[quote="Andreas"][hide]We have that $19+\mid x_{1}+x_{2}+\cdot\cdot\cdot+x_{n}\mid < n$.
Obviously $-1 < x_{i}< 1$. The smallest value of $n$ occurs when for example all the $x_{i}$ are equal to $0$.
Then $n = 20$.[/hide][/quote]
[hide]Well, looking at the LHS, each of the $\mid x_{i}\mid $ terms is less than 1: $\mid x_{1}\mid <1$, $\mid x_{2}\mid <1$, etc. If we add all these together, we get $\mid x_{1}\mid +\mid x_{2}\mid +\dots+\mid x_{n}\mid <n$, as there are $n$ terms. Since $\mid x_{1}\mid +\mid x_{2}\mid +\dots+\mid x_{n}\mid = 19+\mid x_{1}+x_{2}+\dots+x_{n}\mid $, $19+\mid x_{1}+x_{2}+\dots+x_{n}\mid <n$. The least possible $n$ occurs when $\mid x_{1}+x_{2}+\dots+x_{n}\mid =0$, so we have $19+0<n \longrightarrow \boxed{n=20}$. [/hide]


Consider the positive integers greater than 1 and less than 100. How many of these have a remainder of 2 when divided by 5 and have a remainder of 5 when divided by 6?

[hide="Solution"]$ a \equal{} 5k \plus{} 2\equiv \minus{} k \plus{} 2\equiv 5\bmod 6\implies k\equiv 3\bmod 6$

$ k \equal{} 6j \plus{} 3\implies a \equal{} 30j \plus{} 17$

$ 30j \plus{} 17 < 100\implies j\leq 2$, and clearly $ j\geq 0$, so the only $ j$ that work are $ 0$, $ 1$ and $ 2$.

$ \boxed{a\in\{17,47,77\}}$.[/hide]

Edit: Oops, forgot $ j\equal{}0$. Thanks darkdieguerre.