Onomazo to sinolo S (phisikon arithmon) "kalo" ean exei tin idiotita : Gia kathe k kai gia kathe xromatismo tou S me k xromata, iparxoun x,y,z tou idiou xromatos oste x+y=z.

Onomazo to sinolo S (phisikon arithmon) "poli kalo" ean exei tin idiotita : Gia kathe k kai gia kathe xromatismo tou S me k xromata, iparxoun x,y,z,w tou idiou xromatos oste x+y+z=w.

Problima 1 (gnosto, theorima tou Schur) : Na deiksete oti to sinolo ton phisikon arithmon einai "kalo"

Problima 2 (Theorima ton I.Leader kai P.Russell) : Na deiksete oti ean to sinolo S einai "kalo" tote einai kai "poli kalo"

Stergios

Για το πρώτο νομίζω ότι για κάθε $ k$ πρέπει να πάρουμε το σύνολο με τα στοιχεία $ {1,2,..,\left\lfloor en!\right\rfloor}$
to οποίο αν χωριστεί σε $ k$ υποσύναλα (δηλαδή τους χρωματισμούς) και αντιστοιχίζουμε αν το $ rs$ έχει το χρώμα $ c$ τότε
$ |r \minus{} s|\in A_c$ όπου $ A_c$ υποσύνολο του $ S$ . Είναι γνωστό ότι το $ S$ έχει μονοχρωματικό τρίγωνο έστω $ rs,rq,qs$
να έχουν το χρώμα $ c$ .Τότε οι διαφορές ανήκουν στο παραπάνω υποσύνολο και το άθροισμα των δύο δίνουν την τρίτη άρα
η εξίσωση $ x \plus{} y \equal{} z$ είναι επιλύσημη .

Oraia,
Sta combinatorics einai poli simantiko na ksereis pos apodeikniontai ta theorimata, opote gia osous den to kseroun einai oraia askisi na deiksoun oti iparxei monochromatiko trigono.

Ipodeiksei gia to 2) Den einai pio diskolo

Stergios

Elate paidia, perimena megalitero endiaferon gia to 2). Min sas tromazei pou grafo "theorima ton ...", tha itan ena oraio problima gia IMO, p.x. problima 4.

Stergios

Δεν ξέρω αν είναι σωστό αλλά από τα λάθη μαθαίνουμε . Κάνουμε τη διαδικασία όπως και πάνω $ k$ φορές . Αν υπάρχει στοιχείο με ίδιο χρώμα τότε σχηματίζεται μονοχρωματικό τετράπλευρο , οπότε εύκολα παίρνουμε το ζητούμενο (αφού η ύπαρξη μοχοχρωματικού τριγώνου είναι δεδομένη) . Αν όχι , τότε για κάθε χρώμα δημιουργείται μονοχρωματικό τρίγωνο οπότε αν κάνουμε τη διαδικασία άκόμα μία φορά , θα επαναληφθεί ένα χρώμα το οποίο αναγκαστικά προυπάρχει . Έτσι πάλι σχηματίζεται μονοχρωματικό τετράπλευρο , το ζητούμενο

Den mporo na akolouthiso ton silogismo sou. Mporeis na to grapseis ligo pio analitika ? Thanks

Fisika kai apo ta lathi mathainoume, opos sinithizete na lene "ta 'sosta' lathi einai pio simantika apo tis 'lathos' apodikseis"   :)

Stergios

Εννοώ ότι κάνω την ίδια διαδικασία όπως στο πρώτο στα σύνολα τις μορφής

$ \left\{1,...,\left\lfloor en!\right\rfloor\right\}$ , $ \left\{\left\lfloor en!\right\rfloor,...,2\left\lfloor en!\right\rfloor\right\}$

,..., $ \left\{(k \minus{} 1)\left\lfloor en!\right\rfloor,...,k\left\lfloor en!\right\rfloor\right\}$

To S den einai to sinolo to phisikon arithmon, mporei oi arithmoi pou grafeis na min briskonte sto sinolo S.

Στέργιο μπορείς να δώσεις τη λύση σου για το πρόβλημα 2 ? Νομίζω ότι έχει μείνει αρκετό καιρό άλυτο

Esto S ena 'kalo' sinolo kai esto c : S -> {1,2,...,k} enas xromatismos tou me k xromata. Theloume na deiksoume oti iparxoun x,y,z,w tou idiou xromatos, diladi c(x)=c(y)=c(z)=c(w), etsi oste x+y+z=w.
Kataskeuazoume ena neo xromatismo tou S os eksis : Esto x ena stoixeio tou S, ean iparxoun k,l sto S me c(k)=c(l)=c(x) kai k+l=x tote xromatizo to x me to xroma (c(x),1), ean den iparxoun k,l me tis pano idiotites tote xromatizo to x me to xroma (c(x),0).
Epeidi to S einai 'kalo' sinolo iparxoun w,z,f sto S me to idio "neo" xroma oste w=z+f. Alla tote sinepagete oti to "neo" xroma tou w einai to (c(w),1) kai sinepos to "neo" xroma tou f einai (c(f),1) kai ara iparxoun x,y sto S me c(x)=c(y)=c(f) me x+y=f. Ara x+y+z=w me c(x)=c(y)=c(z)=c(w) !

Stergios

Πολύ ωραίο  :lol:  Thanks for sharing Στέργιο


function
inequalities
real analysis
real analysis unsolved

Consider a function f:[a,b]->R,2 times differentiable with f'' continuous.
we denote with M=sup{x\in [a,b]}|f''(x)|, and g:[a,b]->R, g(x)=1/2(x-a)(b-x)

i)using a function f- \lambda g, \lambda \in R, prove that for every x\in [a,b] there is a c_x \in (a,b) s.t. f(x)=-f''(c_x)g(x)

ii)if there exists x_0 \in (a,b) s.t. |f(x)|=Mg(x) then f=Mg or f=-Mg.

Have you left out a condition like $f(a) = f(b) = 0$? I think it's necessary if we want (i) to hold.
(The trick is to find three points at which $f-\lambda g$ is $0$.)

As for (ii), I have come up with a very complicated proof involving an inequality Kent Merryfield posted on the Physics sub-forum (I'm not really sure if my proof is correct and I'm not particularly eager to post it).
Can anyone find a decent solution?

Let's forget my last comment :)

Take some $x_{0}\in (a,b)$ and let $\lambda = \frac{f \left( x_{0}\right)}{g \left( x_{0}\right)}$. Now consider $h : [a,b] \to \mathbb R$, $h(t) = f(t)-\lambda g(t)$. We have that $h(a) = h \left( x_{0}\right) = h(b) = 0$. By Rolle's Theorem there are $c_{1}\in \left( a, x_{0}\right)$ and $c_{2}\in \left( x_{0}, b \right)$ such that $h^\prime \left( c_{1}\right) = h^\prime \left( c_{2}\right) = 0$. Applying Rolle's Theorem again we get that there is $c \in \left( c_{1}, c_{2}\right)$ such that $h^{\prime \prime}(c) = 0$.

Note that we have that $h^\prime (x) = f^\prime (x)+\lambda x-\lambda \cdot \frac{a+b}2$ and that $h^{\prime \prime}(x) = f^{\prime \prime}(x)+\lambda$. Hence, $f^{\prime \prime}(c) =-\lambda$.

Now let's suppose that $\left| \lambda \right| = \sup_{x \in [a,b]}\left| f^{\prime \prime}(x) \right|$. We have that $f^\prime \left( c_{1}\right) = \lambda \left( \frac{a+b}2-c_{1}\right)$ and that $f^\prime \left( c_{2}\right) = \lambda \left( \frac{a+b}2-c_{2}\right)$. Hence,
\[f^\prime \left( c_{2}\right)-f^\prime \left( c_{1}\right) = \lambda \left( c_{1}-c_{2}\right) \, \, \Longleftrightarrow \, \, \frac{f^\prime \left( c_{2}\right)-f^\prime \left( c_{1}\right)}{c_{2}-c_{1}}=-\lambda . \]
From here it's easy to deduce that $f^{\prime \prime}\Big|_{\left[ c_{1}, c_{2}\right]}\equiv-\lambda$. Now let
\[k = \inf_{x \in [a,c_{1}]}\left\{ x : f^{\prime \prime}(t) =-\lambda, \, \forall t \in \left[ x, c_{2}\right] \right\}. \]
$f^{\prime \prime}$ has the Intermediate Value Property, so $f^{\prime \prime}\left( k \right) =-\lambda$.
Hence, $f(x) = \lambda g(x), \, \forall x \in \left[ k, c_{2}\right]$. If $k > a$, then by replacing $x_{0}$ with $k$ we get that there is $\ell < k$ such that
\[f^{\prime \prime}\Big|_{[\ell, c_{2}]}=-\lambda , \]
which contradicts the choice of $k$.
Therefore, $k=a$, so $f(x) = \lambda g(x), \, \forall x \in \left[ a, c_{2}\right]$. By a similar argument we deduce that the formula is good everywhere.

I really hope it's right. The idea for solving (b) came from another problem, but I must admit that I was rather blind not to see it the first time.


modular arithmetic
algebra unsolved
algebra

Let p be a prime number greater than 5. For any integer x, define//
$f_p(x)=\displaystyle\sum_{k=1}^{p-1}\frac{1}{(px+k)^2}$//
Prove that for all positive integers x and y the numerator of $f_p(x)-f_p(y)$, when written in lowest terms, is divisible by $p^3$

It's enough to show that $f_p(t)-f_p(0)$ is divisible by $p^3$ (you can imagine what I mean when I say that a rational is divisible by $p^3$). Write $f_p(t)-f_p(0)$ out explicitely by using $a^2-b^2=(a-b)(a+b)$. In the end, we get something like (all sums are from $i=1$ to $p-1$) $pt\sum\frac 1{i^2(pt+i)^2}+2\sum\frac 1{i(pt+i)^2}\equiv 0\pmod{p^2}$.

The first sum in the $LHS$ is clearly $0\pmod {p^2}$, because we can eliminate a $p$ from both sides, and we're left with $\sum i^{-4}\equiv 0\pmod p$, which is clear because $p>5$, so $4\not |\ p-1$.

We are thus left with $\sum\frac 1{i(pt+i)^2}\equiv 0\pmod{p^2}$. This one is also easy to prove, using first the fact that it's equivalent $\pmod{p^2}$ with $\sum i^{-3}$ (write the difference of the two as the sum of expressions of the type $\frac 1i\cdot(\frac 1{i^2}-\frac 1{(pt+i)^2})$).

[quote]The first sum in the LHS is clearly 0\pmod {p^2}, because we can eliminate a p from both sides, and we're left with \sum i^{-4}\equiv 0\pmod p, which is clear because p>5, so 4\not |\ p-1. [/quote]
I am a bit confused by this. Can you explain why \sum i^[-4} \equiv 0 \pmod p ?


[quote="Ligouras"][quote="Pain rinnegan"][quote="Ligouras"]If $ a, b, c > 0$, then

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\geq \frac {(\sum a^2b)^2}{4(\sum ab^2)} \plus{} \frac {(\sum ab^2)^2}{4(\sum a^2b)} \plus{} \frac {(\sum a)(\sum ab)}{2}$[/quote]

it follows from V.Q.B.C.'s inequality.[/quote]

i don't know  :(   :(   :([/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=234161

[quote="Pain rinnegan"][/quote][quote="Ligouras"][quote="Pain rinnegan"][quote="Ligouras"]If $ a, b, c > 0$, then

$ a^3 + b^3 + c^3 + 3abc\geq \frac {(\sum a^2b)^2}{4(\sum ab^2)} + \frac {(\sum ab^2)^2}{4(\sum a^2b)} + \frac {(\sum a)(\sum ab)}{2}$[/quote]

it follows from V.Q.B.C.'s inequality.[/quote]

i don't know  :(   :(   :([/quote][quote="Pain rinnegan"]

http://www.mathlinks.ro/Forum/viewtopic.php?t=234161[/quote]

thank you very much my friend..... is very nice inequality!!!


What is the maximum number of non-overlapping regions that are formed in a circle by 1999 chords?

[hide] let's look for a sequence:

0 - 1
1 - 2
2 - 4
3 - 7
4 - 11
5 - 16

It seems to be adding 1, then 2, then 3, then 4, then 5, starting at one. That means it is the 1999th triangle number + 1. To calculate the nth triangle number:

$\frac{n(n+1)}2$

so plugging in 1999 gives one less than the answer which is $\boxed {1999001}$
[/hide]

[hide]an explanation for the emergent pattern: think about what each new line does as you add the new chord in.. it divdes each region it traverses into 2, effectively adding the number of regions it traverses. [/hide]


You can download the official .pdf file with the solutions and problems (also from the official IMO site).

Well since the IMO 2006 Questions are already available on the Official site http://imo2006.dmfa.si/ there is no reason for them not to be publised on MathLinks.

[quote="shunka"]Does anyone know which countries proposed the chosen problems?[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?p=573694#573694


number theory
relatively prime
superior algebra
superior algebra unsolved

Suppose $ R$ is a ring and $ I, J$ are relatively prime ideals of $ R$; that is, $ I \plus{} J \equal{} R$. Prove that for any integers $ n, m \leq 1$, $ I^n\plus{}J^m\equal{}R$.

We need to assume that $ R$ has an identity. (otherwise we are not even guaranteed that $ R^2\equal{}R$) Then if $ 1\equal{}x\plus{}y$ with $ x\in I$, $ y\in J$, we have $ 1\equal{}(x\plus{}y)^{m\plus{}n}$, which should expand out into what you want.

Note that commutativity is not necessary.

$ \rm HINT \quad (i^m,j^n) < 1 \implies max \ M \supset (i^m,j^n) \implies M \supset (i,j) \equal{} 1 \ \Rightarrow\Leftarrow$

See also the prior threads
http://www.mathlinks.ro/viewtopic.php?t=217430
http://www.mathlinks.ro/viewtopic.php?t=141604


quadriteal $ABCD$ is inscrbed in circle. Let points $M$ and $N$ be midpoints of $BD$ and $AC$. Prove that if $\angle DNA=\angle ANB$ then $\angle AMB = \angle BMC$

http://www.mathlinks.ro/Forum/viewtopic.php?t=6557
http://www.mathlinks.ro/Forum/viewtopic.php?t=50002

  dg


Okay...I am new at Latex

So i am using the USAMTS latex template but the text is always justified...As in the left and right are BOTH aligned and this creates space issues.  So how do I change it so the text will just align left?  Thanks

Issuing a \raggedright command somewhere should do the trick.

But unless it looks [i]really[/i] bad, don't worry too much about how it looks.

speaking of this, is it possible for their to be an image of how an average template would look like in pdf form where your name and user id are on the top?

-I think I got mine configured but im not sure if it is formatted correctly since the margin looks a little off.

i used to post latex stuff here also but i think it would be better if u posted this stuff in the latex section of the forum which is at the bottom of the forum


group theory
superior algebra
superior algebra unsolved

How do I abelianize a group such as $ D_{12}\times A_{4}$ or $ Q \times S_{4}$ ? (where $ Q$ is the quaternion group, and $ D_{12}$ is the dihedral group on 12 sides)

$ (A \times B)^{ab} \equal{} A^{ab} \times B^{ab}$ (where $ G^{ab}$ is the abelianization of $ G$).

Yes. Moreover $ [D_{12}, D_{12}]$ and $ [A_4, A_4]$ are isomorphic to very nice groups, so you can even figure out what that whole thing is isomorphic to.


Please help me with the following problem:

A ball was floating in a lake when the lake froze. The ball was removed (without breaking the ice), leaving a hole 24 cm across at the top and 8 cm deep. What was the radius of the ball in centimeters? (AHSME 1987)

We can use the pythagorean theorem to say:

\[ x^2 \plus{} 12^2 \equal{} (8\plus{}x)^2\]

Solving shows $ x \equal{} 5$, so the answer is $ 5 \plus{} 8 \equal{} 13$.

why this quetion is not in the resources section sontaining  the ahsme 1987 questionss???


Let $ \{a_n\}$ be a sequence with $ a_n\in (0,1)$ for all $ n$. Prove that the infinite product
\[ \prod_{n \equal{} 1}^{\infty}(1 \minus{} a_n) \equal{} \lim_{n\rightarrow\infty} \prod_{i \equal{} 1}^{n}(1 \minus{} a_i)\]
is positive if and only if the series $ \sum_{i \equal{} 1}^{\infty}a_i$ is convergent.

[color=green]Moderator says: it never makes sense to quote the preceding post in full (especially, if it is the only other post in the thread)[/color]
$ 0 < \prod_{n \equal{} 1}^{\infty}(1 \minus{} a_n) < 1 \ \Longleftrightarrow \ \ \minus{} \infty < \sum_{n \equal{} 1}^{\infty}ln(1 \minus{} a_n) < 0$ . But $ ln(1 \minus{} a_n) < \minus{} a_n < 0$ and it's easy to prove the the requirement :
$ \minus{} \infty < \sum_{n \equal{} 1}^{\infty}ln(1 \minus{} a_n) \Longrightarrow 0 < \sum_{n \equal{} 1}^{\infty}a_n < \infty$ , and, $ \sum_{n \equal{} 1}^{\infty}a_n < \infty \Longrightarrow \minus{} \infty < \sum_{n \equal{} 1}^{\infty}ln(1 \minus{} a_n)$
 :cool:

Moderator, I don't agree  :D .  It makes sense, because some times the "preceding post" can be modified by its autor ..  :cool:


Okay so i have a problem for math and i dont remember exactly how to find the answer.

questions
It takes Jason 6 hours to paint 1 room. Together Jason and Mark can paint 1 room in 4 hours. How long will it take Mark to paint one room?

I thought about putting 6/1=4/1 but that doesnt make any sense, so then i thought i remember something sort of like 1/6+(1/x)=1/4 but im not sure.

Can someone help me?

Don't "remember" anything.  Everything comes down to rates:  Jason paints at the rate of $ \frac{1}{6}$ rooms per hour and Jason and Mark together paint at the rate of $ \frac{1}{4}$ rooms per hour.  So how many rooms per hour does Mark contribute?

Hmm so 1/6 into 1/4 takes means you would have to work at a pace that gets 1/6 into 1/4.....
i set 1/6+1/x=1/4 though that would mean 1/4-1/6, -1/2=1/x so at a -1/2 rate?
EHH

That's not what $ \frac{1}{4} \minus{} \frac{1}{6}$ is.

So -1/12? thats how long it takes Mark?
to check this id add 1/4 and -1/12? which is 2/12 which is 1/6 :) thanks dude apperciate you taking time to help me, sometimes i wonder how im getting a B in calculus haha..

I really like the idea that Mark paints the room in negative time.  To the OP: slow down, work out the computations carefully, and try to keep some sense of physical reality in mind.  You also should write down what everything means: what is $ x$?  What is $ \frac{1}{x}$?  Why are you adding $ \frac{1}{x}$ and $ \frac{1}{6}$ to get $ \frac{1}{4}$?  Etc.  With a bit of care, you will actually be able to solve this.  (In the future, problems of this difficulty are well-suited to High School Basics or one of the Middle School forums.)


Find, with proof, all ordered triples $(x,y,z)$ such that $x^3+y^3+z^3=2006$.

$\in \mathbb{N}$ or $\in \mathbb{Z}$?  Or perhaps $\in \mathbb{Q}$?  :)

[quote="Elemennop"]If $\in \mathbb{Z}$, then no solutions (consider $\pmod 9$). Otherwise, infinite solutions if $\in \mathbb{Q}$.[/quote]

$k^3 \equiv 0, 1, -1 \bmod 9$
$2006 \equiv -1 \bmod 9$

I fail to see the contradiction.   :huh:

[quote="Elemennop"]Ooh, my bad, I thought $2006\equiv\pmod{9}$. In that case, just consider it $\pmod{7}$.  We have $2006\equiv 4\pmod{7}$, but $x^3\equiv0,\pm1\pmod{7}$, and we're done.[/quote]

$2006 \equiv -3 \bmod 7$.  Then we can just have $a^3 \equiv b^3 \equiv c^3 \equiv -1 \bmod 7$.   ;)

Oh, jeez, I'm really off right now.

edit> maybe it has a solution after all? I dunno, if it doesn't there has to be some mod that works here.

[quote="Elemennop"]if it doesn't there has to be some mod that works here.[/quote]
This is not necessarily true.

There are no solutions in $\bf N$.  I'm not sure about $\bf Z$.

yeah the problem wants integer solutions. my bad for forgetting to put that in.

[quote="Elemennop"]maybe it has a solution after all? I dunno, if it doesn't there has to be some mod that works here.[/quote]
There has to be some mod for easy problems, but not in general...

[quote="boxedexe"]How about we use an equation that does have solutions.
\[ x^3+y^3+z^3=3  \]
Masoud Zargar

EDIT: It's $x,y,z\in\mathbb{Z}$.[/quote]
Open problems are always nice, aren't they ;)

[quote="boxedexe"]How about we use an equation that does have solutions.
\[ x^3+y^3+z^3=3  \]
Masoud Zargar

EDIT: It's $x,y,z\in\mathbb{Z}$.[/quote]

are the answers $(1,1,1) and (-5,4,4) \text{permutations}$?

Yes, those are the solutions, btu can you prove that there are only 4 soltuions?

Masoud Zargar


Euler
geometry
geometric transformation
reflection
circumcircle
incenter
complex numbers

This is a very hard result. You can find proofs at

http://www.mathlinks.ro/Forum/viewtopic.php?t=764
http://www.cut-the-knot.org/Curriculum/Geometry/FeuerbachProof.shtml
http://mathforum.org/library/drmath/view/55218.html

as well as a proof in [url=http://reflections.awesomemath.org/2006_2.html]my note [i]From Baltic Way to Feuerbach - A Geometrical Excursion[/i], Mathematical Reflections 2006/2[/url], and a proof using inversion in

  H. S. M. Coxeter, S. L. Greitzer, [i]Geometry Revisited[/i], Toronto - New York 1967,

chapter 5.6. This inversive proof is one of the simplest proofs of the theorem you mentioned. The really simplest proof (I think) is the one presented in paragraph 23.4 of

  Dan Pedoe, [i]Geometry - A comprehensive course[/i], London 1970.

This proof also uses inversion. You can also find an inversive proof (maybe the same) in

  Dan Pedoe, [i]Circles: a mathematical view[/i].

In case you would like to avoid using inversion: Some very short but more algebraic proofs use complex numbers. There are some more elementary proofs, without any inversion and without any complex numbers, but I don't know any good English references for them. In my note "Generalization of the Feuerbach point" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my triangle geometry website[/url], I give two non-inversive proofs of the fact that the nine-point circle touches the incircle (the proof for the excircles is analogous), but these proofs are [b]very[/b] long. Actually, they are interesting not because they are just two more proofs of the theorem that the nine-point circle touches the incircle, but rather because the first one additionally yields some new properties of the point of tangency and the second one proves a strong generalization of this theorem. But I think these are not the proofs you need.

I have written a really elementary proof of the Feuerbach theorem in my paper "Karl Wilhelm Feuerbach, sein Kreis und die Dreiecksgeometrie" (§8) [url=http://www.cip.ifi.lmu.de/~grinberg/Dreigeom/Inhalt.html]on my website again[/url]. The paper is in German, so here is a translation of a part of it that contains the proof. However, you have to download the paper to see the figures used in the proof. These figures are Fig. 4 and Fig. 7-12 in the paper. In fact, the proof depends on the figures.

Now, I translate the text of the proof. Note that this proof is not my invention; I had found it in the booklet

  Emil Donath, [i]Die merkwürdigen Punkte und Linien des ebenen Dreiecks[/i], Berlin 1976.

All my work was to rewrite the proof from this booklet in a more or less comprehensible way (the style of the booklet is really bad). So let's go:

_______________________________________________________________

[b]Theorem 6, the Great Feuerbach Theorem.[/b] The nine-point circle of a triangle ABC always touches the incircle and the three excircles; actually, it touches the incircle internally and the three excircles externally. (See Fig. 7.)

[i]Proof.[/i] We start with two simple lemmas. The first is a well-known converse of the intersecting secants theorem. We state this converse together with the intersecting secants theorem itself:

[b]Theorem 7, the Intersecting Chords Theorem with Converse.[/b] Let AB and CD be two lines intersecting at a point P; assume that this point P lies on the extensions of the segments AB and CD. Then, the points A, B, C, D lie on one circle if and only if $ PA\cdot PB \equal{} PC\cdot PD$. (See Fig. 8.)

[i]Proof.[/i] In the following considerations, we make use of Fig. 10. For other arrangements of points, the proof needs some minor modifications.

[i]Part 1.[/i] If the points A, B, C, D lie on one circle, then $ \measuredangle BCD \equal{} \measuredangle BAD$, hence $ \measuredangle BCP \equal{} \measuredangle DAP$. Furthermore, clearly $ \measuredangle BPC \equal{} \measuredangle DPA$. Hence, the triangles BCP and DAP are similar, and it follows that PB : PC = PD : PA, so that $ PA\cdot PB \equal{} PC\cdot PD$.

[i]Part 2.[/i] If $ PA\cdot PB \equal{} PC\cdot PD$, then PB : PC = PD : PA. Also, we obviously have $ \measuredangle BPC \equal{} \measuredangle DPA$. Hence, the triangles BCP and DAP are similar, and it follows that $ \measuredangle BCP \equal{} \measuredangle DAP$, i. e., $ \measuredangle BCD \equal{} \measuredangle BAD$. Hence, the points A, B, C, D lie on one circle.

Thus, the points A, B, C, D lie on one circle if and only if $ PA\cdot PB \equal{} PC\cdot PD$. Therefore, Theorem 7 is proven.

The next lemma will be:

[b]Theorem 8.[/b] Let P be a common point of two circles k and k'. An arbitrary line g through P meets the circles k and k' at tnhe points Q and Q' (apart from the point P). If the tangents to the circles k and k' at the points Q and Q' are parallel, then the circles k and k' touch each other at P. (See Fig. 10.)

[i]Proof.[/i] Our following proof is adjusted to Fig. 10. Let O and O' be the centers and r and r' the radii of the circles k and k', respectively. Then, OP = OQ = r and O'P = O'Q = r'; hence, the triangles POQ and PO'Q' are isosceles, and $ \measuredangle OPQ \equal{} \measuredangle OQP$ und $ \measuredangle O^{\prime }PQ^{\prime } \equal{} \measuredangle O^{\prime }Q^{\prime }P$. The tangents to the circles k and k' at the points Q and Q' are perpendicular to the radii OQ and O'Q'; since these two tangents are parallel to each other, it follows that the radii OQ and O'Q' are parallel to each other, too, and thus we have $ \measuredangle O^{\prime }Q^{\prime }P \equal{} \measuredangle OQP$. Using $ \measuredangle OPQ \equal{} \measuredangle OQP$ and $ \measuredangle O^{\prime }PQ^{\prime } \equal{} \measuredangle O^{\prime }Q^{\prime }P$, this becomes $ \measuredangle O^{\prime }PQ^{\prime } \equal{} \measuredangle OPQ$, hence $ \measuredangle O^{\prime }PQ \equal{} \measuredangle OPQ$. Thus, the points O', O and P lie on one line, i. e. the point P lies on the central line of the circles k and k'; but as P is a common point of the circles k and k', it follows that the circles k and k' must touch each other at P. This proves Theorem 8.

Now we start with the actual [i]proof of Theorem 6[/i]: Let ABC be a triangle, let A', B', C' be the midpoints of its sides BC, CA, AB, and let $ H_{a}$, $ H_{b}$, $ H_{c}$ be the feet of its altitudes to its sides BC, CA, AB. The points A', B', C', $ H_{a}$, $ H_{b}$, $ H_{c}$ are known to lie on the nine-point circle of triangle ABC.

Let the incircle of triangle ABC touch the side BC at a point X. Let the angle bisector of the angle CAB intersect the line BC at X' and the circumcircle of triangle ABC at a point A" (apart from A). Call O the incenter of triangle ABC (Fig. 11).

Since the point A" lies on the angle bisector of the angle CAB, we have $ \measuredangle CAA^{\prime \prime } \equal{} \measuredangle BAA^{\prime \prime }$. Equal chordal angles in a circle belong to equal chords; hence, we also have CA" = BA". Therefore, the point A" lies on the perpendicular bisector of the segment BC. (Hence, the point A" is the midpoint of the arc BC on the circumcircle.) Now, $ \measuredangle A^{\prime \prime }BC \equal{} \measuredangle A^{\prime \prime }AC$ (chordal angles), hence

$ \measuredangle A^{\prime \prime }BO \equal{} \measuredangle A^{\prime \prime }BC \plus{} \measuredangle CBO \equal{} \measuredangle A^{\prime \prime }AC \plus{} \measuredangle CBO \equal{} \frac {A}{2} \plus{} \frac {B}{2}$,

while the exterior angle theorem in triangle AOB yields

$ \measuredangle A^{\prime \prime }OB \equal{} \measuredangle OAB \plus{} \measuredangle OBA \equal{} \frac {A}{2} \plus{} \frac {B}{2}$.

Thus, $ \measuredangle A^{\prime \prime }BO \equal{} \measuredangle A^{\prime \prime }OB$, and the triangle BA"O is isosceles, so that A"B = A"O.

Since

$ \measuredangle A^{\prime \prime }BX^{\prime } \equal{} \measuredangle A^{\prime \prime }BC \equal{} \measuredangle A^{\prime \prime }AC$        (cyclic)
$ \equal{} \measuredangle A^{\prime \prime }AB$       (angle bisector)

and $ \measuredangle BA^{\prime \prime }X^{\prime } \equal{} \measuredangle AA^{\prime \prime }B$, the triangles A"BX' and A"AB are similar. This implies A"X' : A"B = A"B : A"A, so that $ A^{\prime \prime }X^{\prime }\cdot A^{\prime \prime }A \equal{} A^{\prime \prime }B^{2}$. Since A"B = A"O, this becomes $ A^{\prime \prime }X^{\prime }\cdot A^{\prime \prime }A \equal{} A^{\prime \prime }O^{2}$.

The lines $ AH_{a}$, OX and A"A' are all perpendicular to BC (the line A"A' is perpendicular to BC, since A" lies on the perpendicular bisector of the segment BC); hence, they are parallel to each other; this yields after the Thales theorem that

$ \frac {A^{\prime }X^{\prime }}{A^{\prime \prime }X^{\prime }} \equal{} \frac {A^{\prime }H_{a}}{A^{\prime \prime }A} \equal{} \frac {A^{\prime }X}{A^{\prime \prime }O}$.

Hence, $ A^{\prime }X^{\prime } \equal{} k\cdot A^{\prime \prime }X^{\prime }$, $ A^{\prime }H_{a} \equal{} k\cdot A^{\prime \prime }A$ and $ A^{\prime }X \equal{} k\cdot A^{\prime \prime }O$ for some real k. But since $ A^{\prime \prime }X^{\prime }\cdot A^{\prime \prime }A \equal{} A^{\prime \prime }O^{2}$, we have $ k^{2}\cdot \left( A^{\prime \prime }X^{\prime }\cdot A^{\prime \prime }A\right) \equal{} k^{2}\cdot A^{\prime \prime }O^{2}$, and thus $ \left( k\cdot A^{\prime \prime }X^{\prime }\right) \cdot \left( k\cdot A^{\prime \prime }A\right) \equal{} \left( k\cdot A^{\prime \prime }O\right)^{2}$, so that $ A^{\prime }X^{\prime }\cdot A^{\prime }H_{a} \equal{} A^{\prime }X^{2}$.

(See Fig. 12.) It is well-known that the two tangents from a point to a circle are symmetric to each other with respect to the diameter of the circle through that point. Hence, the tangents from X' to the incircle of triangle ABC are symmetric to each other with respect to the line X'O. One of these two tangents is the line BC; the other is, therefore, the reflection of the line BC in the line X'O. Let this second tangent touch the incircle in Q. Also, let the line A'Q intersect the incircle at a point L (apart from the point Q). Then, the Intersecting Secant and Tangent Theorem shows $ A^{\prime }Q\cdot A^{\prime }L \equal{} A^{\prime }X^{2}$. On the other hand, we have $ A^{\prime }X^{\prime }\cdot A^{\prime }H_{a} \equal{} A^{\prime }X^{2}$. Thus, $ A^{\prime }Q\cdot A^{\prime }L \equal{} A^{\prime }X^{\prime }\cdot A^{\prime }H_{a}$. After Theorem 7, this implies that the points Q, L, X' and $ H_{a}$ lie on one circle; hence, in the cyclic quadrilateral $ QLH_{a}X^{\prime }$, we have $ \measuredangle QLH_{a} \equal{} 180^{\circ} \minus{} \measuredangle QX^{\prime }H_{a}$, i. e. $ \measuredangle A^{\prime }LH_{a} \equal{} 180^{\circ} \minus{} \measuredangle QX^{\prime }H_{a}$.

Since the line X'Q (the second tangent from the point X' to the incircle) is the reflection of the line BC in the line X'O, we have $ \measuredangle OX^{\prime }Q \equal{} \measuredangle OX^{\prime }X$, and thus

$ \measuredangle A^{\prime }LH_{a} \equal{} 180^{\circ } \minus{} \measuredangle QX^{\prime }H_{a} \equal{} 180^{\circ } \minus{} \left( \measuredangle OX^{\prime }Q \plus{} \measuredangle OX^{\prime }X\right)$
$ \equal{} 180^{\circ } \minus{} 2\cdot \measuredangle OX^{\prime }X$
$ \equal{} 180^{\circ } \minus{} 2\cdot \left( \measuredangle X^{\prime }AC \plus{} \measuredangle X^{\prime }CA\right)$       (exterior angle theorem in $ \Delta AX^{\prime }C$)
$ \equal{} 180^{\circ } \minus{} 2\cdot \left( \frac {A}{2} \plus{} C\right) \equal{} 180^{\circ } \minus{} A \minus{} 2C \equal{} \left( A \plus{} B \plus{} C\right) \minus{} A \minus{} 2C \equal{} B \minus{} C$.

Let the tangent to the nine-point circle at A' meet the line CA at R. Then, as a secant-tangent angle, $ \measuredangle B^{\prime }A^{\prime }R$ equals the chordal angle $ \measuredangle B^{\prime }C^{\prime }A^{\prime }$, i. e. it equals $ \measuredangle BCA \equal{} C$, since the triangles ABC and A'B'C' are similar. On the other hand, A'B' || AB, and thus $ \measuredangle B^{\prime }A^{\prime }C \equal{} \measuredangle ABC \equal{} B$. Hence,

$ \measuredangle RA^{\prime }C \equal{} \measuredangle B^{\prime }A^{\prime }C \minus{} \measuredangle B^{\prime }A^{\prime }R \equal{} B \minus{} C \equal{} \measuredangle A^{\prime }LH_{a}$.

But $ \measuredangle RA^{\prime }C$ is the secant-tangent angle of the chord $ A'H_{a}$ in the nine-point circle of triangle ABC; hence, $ \measuredangle RA^{\prime }C \equal{} \measuredangle A^{\prime }C^{\prime }H_{a}$. And thus, $ \measuredangle A^{\prime }C^{\prime }H_{a} \equal{} \measuredangle A^{\prime }LH_{a}$. Consequently, the point L lies on the circle through the points A', C' and $ H_{a}$, i. e. on the nine-point circle of triangle ABC.

We have shown that $ 180^{\circ } \minus{} \measuredangle QX^{\prime }H_{a} \equal{} B \minus{} C$. In other words, $ \measuredangle QX^{\prime }C \equal{} B \minus{} C$. On the other hand, $ \measuredangle RA^{\prime }C \equal{} B \minus{} C$. Thus, $ \measuredangle QX^{\prime }C \equal{} \measuredangle RA^{\prime }C$, and therefore QX' || RA'.

Now, we have the point L lying on the incircle and on the nine-point circle of triangle ABC; the points Q and A' lie on one line through L, where Q lies on the incircle and A' lies on the nine-point circle. Further, the tangents to the incircle and the nine-point circle at the points Q and A', respectively, are parallel (since QX' || RA'). Hence, after Theorem 8, the incircle and the nine-point circle touch each other at the point L, and actually they touch each other internally, as an arrangement consideration shows.

It remains to show that each excircle externally touches the nine-point circle; but this actually means a repetition of the above reasoning with some modifications. I spare this repetition. Theorem 6 is proven.

_______________________________________________________________

It would be a waste of this proof to stop here. In fact, if we look at the proof more carefully, we see that we can conclude something additional:

The point where the incircle of a triangle ABC touches its nine-point circle is called the [b]Feuerbach point[/b] of triangle ABC. (This point is uniquely defined for every non-equilateral triangle ABC, because the incircle and the nine-point circle of a non-equilateral triangle are distinct and thus touch at only one point.) Then we can show:

[b]Additional Theorem.[/b] Let the incircle of triangle ABC touch its sides BC, CA, AB at the points X, Y, Z. Let $ X_{1}$, $ Y_{1}$, $ Z_{1}$ be the reflections of these points X, Y, Z in the angle bisectors of the angles CAB, ABC, BCA, and let A', B', C' be the midpoints of the sides BC, CA, AB of triangle ABC. Then, the lines $ A^{\prime}X_{1}$, $ B^{\prime}Y_{1}$, $ C^{\prime}Z_{1}$ pass through the Feuerbach point of triangle ABC.

[i]Proof.[/i] The two tangents from the point X' to the incircle of triangle ABC touch this incircle at the points X and Q. Thus, these points X and Q are symmetric to each other with respect to the line X'O (since O is the center of the incircle of triangle ABC). In other words, the point Q is the reflection of the point X in the line X'O. Now, since both the point X' and the incenter O of triangle ABC lie on the angle bisector of the angle CAB, the line X'O is the angle bisector of the angle CAB. Hence, the point Q is the reflection of the point X in the angle bisector of the angle CAB. But this was exactly the definition of the point $ X_{1}$. Thus, $ Q \equal{} X_{1}$.

On the other hand, we have shown that the point L is the point where the incircle and the nine-point circle of triangle ABC touch each other. In brief, this means that L is the Feuerbach point of triangle ABC. But by its construction, the point L lies on the line A'Q. Since $ Q \equal{} X_{1}$, the line A'Q is the line $ A^{\prime}X_{1}$, and it thus follows that the Feuerbach point L of triangle ABC lies on the line $ A^{\prime}X_{1}$. Similarly, the Feuerbach point of triangle ABC also lies on the lines $ B^{\prime}Y_{1}$ and $ C^{\prime}Z_{1}$. This proves the Additional Theorem.

  Darij

thanks darij
reading that proof required lots of patience
and i believe that its trranslation must have required another truck load.
nvertheless i am satisfied
thanks once again
aeromsaeroms  :)

Dear Mathlinkers,
reading some old message, I found this one...
After my investigation, Mention is considered as the first geometer who give a "synthetic" proof of this result...
In 2003, I found an enterely synthetic proof that I put on my website in 2007.
http://perso.orange.fr/jl.ayme   (vol. 1  le théorème de Feuerbach) in French
or : revistaoim directed by professor Francisco Bellot Rosado (in spanish).
Did some one known about the life of Mention ?
Sincerely
Jean-Louis

Other quite fast and easy way to prove it is by using Casey's.

Best regards,
sunken rock

[quote="XMG"][b]Feuerbach theorem.[/b] The nine-point circle (also called Euler circle or Feuerbach circle) of any triangle is tangent to the incircle and to the three excircles of the triangle.

Can anyone prove this theorem?[/quote]
This is a very interesting result, [b]here is my approach[/b] using [b]Inversion[/b]:
Consider triangle $ ABC$, with $ (C)$ is its incircle. $ (K)$, respectively, tangents to $ BC, CA, AB$ by the points $ P,Q,R$. The excircle wrt $ \angle C$ is $ (K_c)$, tangents to $ BC,CA,AB$, respectively, at $ Z,Y,X$. The Euler circle $ (\mathcal E)$ passes through midpoints $ L,M,N$ of $ BC,CA,AB$, respectively. Let $ KG$ ($ K\in CA$, $ G\in BC$) be the rest internal tangent of $ (K)$ and $ (K_c)$, which intersects $ (K)$ and $ (K_c)$, respectively, at the point $ H,J$. Denote $ (AB,BC,CA) \equal{} (c,a,b)$. Then, we get
$ AR \equal{} BX \equal{} \frac {b \plus{} c \minus{} a}{2} \Longrightarrow NX \equal{} NR \equal{} \frac {|a \minus{} b|}{2}$.
[b]Case 1[/b]: If $ a \equal{} b$, it is so obvious that $ (\mathcal E)$ will tangent to $ (K)$ and $ (K_c)$ at $ N\equiv R$.
[b]Case 2[/b]: Consider the inversion through pole $ N$, power $ k$, where $ k \equal{} (\frac {a \minus{} b}{2})^2$. Then, it is easy to see that: $ I(N,k): (K)\mapsto (K_c)$. Because $ (C,E,I,I_c) \equal{} \minus{} 1 \Longrightarrow (F,E,R,X) \equal{} \minus{} 1$, (Here, $ I$ is the center of $ (K)$, $ I_c$ is the center of $ (K_c)$, $ F$ is the projection of $ C$ onto $ AB$). Therefore, $ NR^2 \equal{} \overline {NE}.\overline {NF} \equal{} k \Longrightarrow I(N,k): F\mapsto E$. Let $ M_1\equiv MN\cap KG$, it is easy to find out the fact that $ NM \equal{} \frac {a}{2} \longrightarrow \overline {NM_1}. \overline {NM} \equal{} \frac {(a \minus{} b)^2}{4} \equal{} k \Longrightarrow I(N,k): M\mapsto M_1\in KG$, which implies that $ I(N,K): (\mathcal E) \mapsto KG$, such that $ KG$ is the common tangent of $ (K)$ and $ (K_c)$. Therefore, $ (\mathcal E)$ will tangent both $ (K)$ and $ (K_c)$. Argue the same for $ B,A$. Then we will get the result of the problem.
Which ends our proof :D

[quote="mathVNpro"][quote="XMG"][b]Feuerbach theorem.[/b] The nine-point circle (also called Euler circle or Feuerbach circle) of any triangle is tangent to the incircle and to the three excircles of the triangle.

Can anyone prove this theorem?[/quote]
This is a very interesting result, [b]here is my approach[/b] using [b]Inversion[/b]:
Consider triangle $ ABC$, with $ (C)$ is its incircle. $ (K)$, respectively, tangents to $ BC, CA, AB$ by the points $ P,Q,R$. The excircle wrt $ \angle C$ is $ (K_c)$, tangents to $ BC,CA,AB$, respectively, at $ Z,Y,X$. The Euler circle $ (\mathcal E)$ passes through midpoints $ L,M,N$ of $ BC,CA,AB$, respectively. Let $ KG$ ($ K\in CA$, $ G\in BC$) be the rest internal tangent of $ (K)$ and $ (K_c)$, which intersects $ (K)$ and $ (K_c)$, respectively, at the point $ H,J$. Denote $ (AB,BC,CA) \equal{} (c,a,b)$. Then, we get
$ AR \equal{} BX \equal{} \frac {b \plus{} c \minus{} a}{2} \Longrightarrow NX \equal{} NR \equal{} \frac {|a \minus{} b|}{2}$.
[b]Case 1[/b]: If $ a \equal{} b$, it is so obvious that $ (\mathcal E)$ will tangent to $ (K)$ and $ (K_c)$ at $ N\equiv R$.
[b]Case 2[/b]: Consider the inversion through pole $ N$, power $ k$, where $ k \equal{} (\frac {a \minus{} b}{2})^2$. Then, it is easy to see that: $ I(N,k): (K)\mapsto (K)$, $ (K_c)\mapsto (K_c)$. Because $ (C,E,I,I_c) \equal{} \minus{} 1 \Longrightarrow (F,E,R,X) \equal{} \minus{} 1$, (Here, $ I$ is the center of $ (K)$, $ I_c$ is the center of $ (K_c)$, $ F$ is the projection of $ C$ onto $ AB$). Therefore, $ NR^2 \equal{} \overline {NE}.\overline {NF} \equal{} k \Longrightarrow I(N,k): F\mapsto E$. Let $ M_1\equiv MN\cap KG$, it is easy to find out the fact that $ NM \equal{} \frac {a}{2} \longrightarrow \overline {NM_1}. \overline {NM} \equal{} \frac {(a \minus{} b)^2}{4} \equal{} k \Longrightarrow I(N,k): M\mapsto M_1\in KG$, which implies that $ I(N,K): (\mathcal E) \mapsto KG$, such that $ KG$ is the common tangent of $ (K)$ and $ (K_c)$. Therefore, $ (\mathcal E)$ will tangent both $ (K)$ and $ (K_c)$. Argue the same for $ B,A$. Then we will get the result of the problem.
Which ends our proof :D[/quote]

Dear Mathlinkers,
the proof by inversion has been made for the first time by J. P. Taylor in 1900 or by Milne in 1890...
Sincerely
Jean-Louis


geometry
rectangle
incenter
geometry unsolved

$ \triangle{ABC}$ is right-angled at $ B$.Point $ M$ is located on side $ AB$ so that $ AM\equal{}BC$,and point $ N$ on the side $ BC$ so that $ CN\equal{}MB$.Find the angle between $ AN$ and $ CM$.

[hide="Solution"]
Extend $ CB$ to $ X$ so that $ BX \equal{} BM$.  Also, extend $ AB$ to $ Z$ so that $ BZ \equal{} BN$.  Let $ Y$ be the final vertex of rectangle $ BZYX$.  Now, we have that $ YX \equal{} BZ \equal{} BN$ and $ CX \equal{} BC \plus{} BX \equal{} AM \plus{} BM \equal{} AB$.  Furthermore, $ \angle ABN \equal{} 90 \equal{} \angle CXY$, so $ \triangle CXY\cong \triangle ABN$.  This means that $ \angle NAB \equal{} \angle YCX$.  Furthermore, we have that $ MZ \equal{} MB \plus{} BZ \equal{} BN \plus{} NC \equal{} BC$ and $ YZ \equal{} BX \equal{} MB$.  Thus, $ \triangle MZY\cong \triangle CBM$, so $ \angle ZMY \equal{} \angle XCM$.  Yet, $ YX\parallel ZM$, so $ \angle MYX \equal{} \angle XCM$, which means that $ CYXM$ is cyclic.  Now, since $ \angle XBM \equal{} 90$ and $ BX \equal{} BM$, we have that $ \angle BXM \equal{} 45$.  This means that $ \angle YXM \equal{} 90 \plus{} 45 \equal{} 135$, so $ \angle YCM \equal{} 45$.  Hence, $ \angle NAB \plus{} \angle BCM \equal{} 45$. Yet, \[ \angle MCA \plus{} \angle NAC \equal{} \angle BCA \minus{} \angle BCM \plus{} \angle BAC \minus{} \angle NAB \equal{} 90 \minus{} 45 \equal{} 45\] Then, the angle formed by $ CM$ and $ AN$ is $ 45$. [/hide]

Clearly AN and BM intersect on the circle through incenter and A,C.

 M.T.


Assume $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3$.Prove that
$ \frac{(a\plus{}b)^{a\plus{}b}(b\plus{}c)^{b\plus{}c}(c\plus{}a)^{c\plus{}a}}{a^ab^bc^c}\leq 64$.


calculus
integration
function
real analysis
real analysis unsolved

Given a continuous function $ f\colon [a,b]\to\mathbb R$, define $ \mathrm{osc}_f\colon [a,b]\to [0,\infty)$ by $ \mathrm{osc}_f(x)\equal{}\max_{[a,x]}f\minus{}\min_{[a,x]}f$. The function $ \mathrm{osc}_f$ is obviously nondecreasing. Evaluate $ \int_a^b f\,d(\mathrm{osc}_f)$, which exists as a [url=http://en.wikipedia.org/wiki/Riemann-Stieltjes_integral]Riemann-Stieltjes integral[/url].

Here is an idea of how I think it should be solved:

Without loosing the generality let us consider that $ f$ attains the minimum at $ a$ otherwise we can use the translated function instead of f.

Let $ g(x)\equal{}osc_{f}(x)\equal{}\max_{[a,x]}f(t)\minus{}f(a)\equal{}h(x)\minus{}f(a)$.

Also let $ x_{M}$ be the point, located in the interior of $ [a,b]$(to simplify the calculations) where maximum is attained and let $ M$ denote the maximum of f.

We use integration by parts and we get that:

$ I\equal{}f(b)g(b)\minus{}f(a)g(a)\minus{}\int_{a}^{b}g(x)df(x)\equal{}f(b)M\minus{}\int_{a}^{b}h(x)df(x)$

$ \plus{}f(a)(f(b)\minus{}f(a))$.

Thus, we need to calculate $ \int_{a}^{b}h(x)df(x)$, where $ h(x)\equal{}\max_{[a,x]}f(t)$. I think that this integral depends on the number of turning points of $ f$.

Here is what I am trying to say: Assume that $ X_{M}$ is exactly the third turning point, i.e., the function increases from $ f(a)$ to $ f(x_{1})$, i.e., we have a

local maximum at some point $ x_{1}$  then

the function decreases to $ f(x_{2})$ and then increases to its maximum $ M$.

Since $ f$ is continuous, due to Darboux property there is a point $ t$, $ x_{2}<t<x_{M}$,  such that $ f(x_{1})\equal{}f(t)$. Now, in this case we can explicitely find $ h$ as follows:

$ h(x)\equal{}f(x)$, $ a\leq x\leq x_{1}$

$ f(x)\equal{}f(x_{1})$, $ x_{1}\leq x\leq t$

$ h(x)\equal{}f(x)$, $ t\leq x\leq x_{M}$

$ h(x)\equal{}M$, $ x_{M}\leq x\leq b$.

We have that

$ \int_{a}^{b}h(x)df(x)\equal{}\int_{a}^{x_{1}}\plus{}\int_{x_{1}}^{t}\plus{}\int_{t}^{x_{M}}\plus{}\int_{x_{M}}^{b}$

$ \equal{}\frac{f^{2}(x_{1})\minus{}f^{2}(a)}{2}\plus{}f(x_{1})(f(t)\minus{}f(x_{1}))\plus{}\frac{f^{2}(x_{M})\minus{}f^{2}(t)}{2}\plus{}M(f(b)\minus{}f(x_{M}))$.

I used the fact that, when f is an integrable function,  $ \int_{a}^{b}fdf\equal{}1/2(f^{2}(b)\minus{}f^{2}(a))$ and $ \int_{a}^{b}kdf\equal{}k(f(b)\minus{}f(a))$.

I hope I made no computational mistakes.

I was thinking of the following: let $ M_f(x) \equal{} \max_{[a,x]}f$ and $ m_f(x) \equal{} \min_{[a,x]}f$. Then
\[ \int_a^b f\,d(M_f \minus{} m_f) \equal{} \int_a^b f\,dM_f \minus{} \int_a^bf\,dm_f \equal{} \int_a^b M_f\,dM_f \minus{} \int_a^b m_f\,dm_f \equal{} \frac {M_f(b)^2 \minus{} m_f(b)^2}{2}
\]
The second equality holds because the function $ M_f$ is constant on the open intervals where $ f < M_f$ (and the same for $ m_f$).

I see what you are saying, I guess that is a very elegant way of solving the problem.

Here's a harder one: $ \int_a^b \frac{f(x)\minus{}f(a)}{\mathrm{osc}_f(x)}d(\mathrm{osc}_f)$. Or just assume that $ f(a)\equal{}0$.


geometry
geometric transformation
rotation
analytic geometry
quadratics
inequalities
USAMTS

Ok...how difficult does everyone think the round 1 problems were (individually)?
my opinions:

1. quite trivial, except for the fact that it took me about a week to think to massive casework-bash it.
2. took me a while to think that all digits are one larger means x^2-y^2=11111 or whatever.
3. the answer came quite easily. proving it did not. it also was quite hard to set up my solution.
4. aurgh this problem was ridiculous. it took me about 2 weeks to figure out a drawing where the area actually was 40. And i forgot to prove existence when writing up my solution. also, i am terrible at seeing triangle congruencies.
5. wow...i actually thought this was the easiest problem of the bunch. solved part a in 3 minutes, part b in 12. ok fine i used a ti-84 to do experiments, but that was it.

Another question: how would these problems compare to usamo problems?

1. Mieh. It was really easy, but a lot of case-work.
2. Easiest I thought. I figured out how to do the problem in 10 minutes, but then the 6-digit case took me like 1 hour or something to write...
3. This was pretty difficult. If you are used to geometry problems/realized number theory is useful and have enough patience to use it, it was probably all right. The answer was easy to get.
4. I have no idea. It's geo, so I tried it for about 3 minutes before giving up.
5. Darn, this was actually really simple. In fact, this was probably easier than #3, and should have required the least work of the 5 problems. But...I didn't get it because I didn't feel like I could...

I cannot say anything about the comparable difficulty of this round because this is my first year doing usamts.

1. Lot's of casework. But there wer a couple of nice tricks.

2. Easiest problem ever! Got it in like 5 minutes.

3. Hardest one for this year (so far). I brute forced and ended up with a diophantine. However, I didn't send a solution :( .

4. I had no idea. I'm horrible at geometry..

5. part a) was trivial. part b) was not [i]that[/i] hard.

1. Trivial, annoying casework. I got it wrong but I wasn't too surprised (missing 4 stupid cases is pretty much unavoidable)

2. Very easy. Annoying to write up though...

3. Not too bad, I got a mod solution that was relatively nice.

4. Very easy, I saw the folding in the outside triangles immediately.  However, I ended up getting it wrong just because I forgot the stupid 40 case  :P .

5. This was oddly easy for a #5. a) was just manipulation, and b) was just annoying to prove.

So I guess this was a typically easy round 1 (with the exception of #3 which took awhile to prove).

[quote="AIME15"]USAMTS < USAMO by quite a bit.

1. It seemed obvious to use casework (although apparently there was a nice solution).  Pretty easy (except keeping track of the cases was a pain for me)
2. Difference of squares came to mind right away.
3. Same as blackbelt14253, except I never found the proof :)
4. Basically, my trig bash didn't work.  IMO, this was the hardest of the bunch, but that's probably because I suck at geometry.
5. Approximated root with calculator, then used greedy algorithm to find the sequence (pretty easy).[/quote]
how did u prove uniqueness? the proof for 3 was easy when u note that inscribing is the same thing as rotating the smaller square.

I assume you are talking about Problem 5.

Just note that $ r^n > r^{n\plus{}1} \plus{} r^{n\plus{}2} \plus{} \dots$ (which is easy to prove), and the rest follows pretty intuitively.

[quote="AIME15"]USAMTS < USAMO by quite a bit.

1. It seemed obvious to use casework (although apparently there was a nice solution).  Pretty easy (except keeping track of the cases was a pain for me)
[\quote]
my soln to one was pretty easy--
it had like 7 cases.

i did this at 11pm on saturday.
i miscounted in#1. I got #2. I got #3. I got 60 for #4. I got #5a. I fell asleep at 12:30pm.

How do people stand to write up solutions only days before they are due?  I solved all of them the first day, exactly 2 months before they were due, and I wasn't even at my house!  I was traveling home from another state!

Not to sound braggarty, because some of my solutions were incomplete (such as the 40 on the hexagon one) but I fixed them before I sent them in.

Unfortunately, on this set, [EDIT-BLEEPED OUT!!!].  :(

Hey please don't discuss the problems, even if you just say something like that.

Btw, I also solved all of the first round problems on vacation at another state too. :D

Its ok to discuss them now isn't it? I mean the solutions are up....

And this is really weird I solved them all the first day in another state too  :P

#1 and #2 are quite brute-forcy. I attempted to solve a more general version of #1 but it did not seem very simple. For #3, it took me a while to realize what two lattice points coincide mean (rotation of coordinates), but after that it's just simple number theory. #4 was probably the easiest to get, except that I did not know it required explicit construction and just outlined through that part. #5b was a bit trickier than #4 but not very hard to get either.

1. I brute-forced it, and it seemed easy. Somehow, I got 76 though.

2. The easiest problem IMO. All you had to do is factor 11111 and set y+x and y-x equal to the factors. For the second case, it was easier to restrict the factors before starting the casework.

3. Very hard to put in words. I got 8. I didn`t see the connection to NT, and that is because I am very bad at Geo and NT.

4. Only got the flipping thing, 60. I didn`t see any other cases.

5. Pretty easy. Saw it after manipulating a simple equation.

I think I got 2 out of the 5 questions right, with partial credit (not much) for the others.

Hm...this is what i thought...werent too hard
1. (30 minutes)sorta messy
2. (<1 minute)trivial
3. (15 minutes-30 minutes)bashy/messy
4. (30 minutes solving, then realized 40 worked while writing up soln)nice problem
5. (30 minutes)pretty good problem, but a bit easy

[quote="PowerOfPi"]2. The easiest problem IMO. All you had to do is factor 11111 and set y+x and y-x equal to the factors. For the second case, it was easier to restrict the factors before starting the casework.

[/quote]

What do you mean IMO? I've seen a couple of people say this..

IMO=in my opinion.

or I think that's what your question means

I managed to solve all of them except for number 4, which I had no clue on.  Maybe I'm just not used to hard geometry problems.  Out of curiosity, what about the problem inspired you guys to reflect points and such?

@winternights1: The fact that the two outside angles add up to the inside angle (at vertex A).

@modularmarc101: As zapi said, I meant in my opinion.

1 and 2: Very easy to find a solution, then brute-force it (On #2, I listed all solutions to x^2-y^2=111111, not just those without carry.). Takes a bit of time.

3: Interesting  problem, I tried fooling around with a bit of group theory before I performed a coordinate rotation and counted the solutions to the system of diophantine equations. The easiest way to do this was to first realize 1492 HAD to be the answer, then look for a solution.

4: Very cool to see the reflections. So cool, in fact, that I forgot to check for other cases (I only got the 60 case.).

5: Easy if you use the quadratic for (a), then use some inequalities for (b).

In all, this seemed like an easy round. Maybe I'm getting better, but I'm still kinda annoyed I missed the 40 case.

Ugh I only got half of #4....
#5 was the easiest, I thought. #4 was definitely the hardest for me.

agreed......... i totally bombed #4 hoping for 1 point.

I only got an answer for#1.   :(   I only got the problems a couple days before the deadline.   :(   This is my first year doing USAMTS.

I used casework for #1, but somehow got 72.   :(

1. not really hard but ugly with massive caseworks
2. pretty trivial with ugly for me to consider so many pairs of factors
3. pretty nice problem... I solved it using thoughtless coordinates-bash (yes pretty massive in solving system of equations) and then prove there is only one solution for each case except for the first and last by considering the complete residue system
4. pretty cool geo problem (solved it using rotations but missed the second case)
5. a nice problem

so about 3/5/5/3/5 or 21/25??

Just wondering if you mailed your solution last Monday, about when should there be a tick mark next to Round 1 on your individual scoreboard?

imo #4 was the hardest because gaku wrote the question x_x

1: Epic casework bash! (I hope I didn't make a mistake somewhere--my answer is right)
2: I figured out how to solve this but was too lazy to factor stuff until about 2 weeks into the competition.  I think I got it right however.
3: I'm pretty sure I got this one.
4: Almost forgot the 40 case, but remembered it in time.  Unfortunately, when proving existence, I forgot to prove convexity of the resulting hexagons.  (It took me forever to figure out to reflect the points.)
5: Took me the shortest of any problem this round--about 15 minutes for both parts.

mine was very hurriedly written, so:

1: 4 (somehow got 64)
2: 4-5 (good, but I forgot to state that solutions must be integers)
3: 4 (um 1494??)
4: 1 (went nowhere)
5: 4 (some holes in rigor here, I can feel it)

17-18. I would love that score for my meager hours of effort.

1. This one was the first one I got... My solution was basically the official.
2. Difference of squares was quite easy to figure out.
3. The answer and proof were easy to get, but I don't think my proof was the best.
4. Couldn't find 40 case.
5. Didn't do part b.

Overall, I think numbers 1 and 3 were the easiest because I couldn't find a neat way to show that there were no 6-digit cases for number 2.

1. This was pretty easy. I solved it in about 20 minutes.

2. The idea of subtracting and getting 11111 came easily to me. Similarly, I realized quickly to avoid the cases where there were nines in the smaller number. I knew how to solve it within 10 minutes. It took me an extra 30 minutes to actually solve it.

3. It was neither easy nor very hard for me. I figured out how to do it in 10-15 minutes, but I still had to think about the problem while I was doing it. It took me 3 hours total (excluding writing it up.)

4. Very hard. I didn't solve it correctly. I missed the case where the area was 40.

5. The first part was hard. The second was easy once I had done the first.

1. The day I received the problems in the mail, I was sitting at dinner looking at this problem for 5 minutes and thought I had the solution. Around three weeks later, I realized I made a horrible mistake in casework and re-did it. New solution was correct :D. Easiest problem of all.
2. A little more difficult: brute-forced 6-digit #s!
3. I had no idea how to prove it; I only turned the diagram around and thought: "Hey, each corner is a lattice point and every "row" in the square has one lattice point..."
I simply typed in the answer the day before submission (FAIL  :( ). [b]That's one point, right?[/b]
4. Nothing. Absolutely nothing.
5. Used a graphing calc's table function for part a, making my part b proof extremely short.

First year doing USAMTS. Predicted score for 1st round: 12-14. Too much time spent figuring out how to use LATEX.

#3 had a so guessable answer, my proof was like yours...

Heh heh, skipped #4...

1. casework  :|
2. got it pretty quickly; solution in 1 day
3. okay
4. like 2 days of bashing trig (not really) i came up with 40. didnt find the 60, oh well.
5. easy. once i printed the set, i did it and wrote up the proofs.


AoPSwiki Asymptote LaTeX geometry circumcircle incenter

Can Someone Help me???

I've been tried to follow all the steps of aopswiki ...
However, when I saw "import olympiad" and tried to re-create the example, I couldn't.
I followed the instructions of how to install this "package", but in the end my compilation through TecnicCenter was:

You have given so little information that it's imposiible to tell what has happened. Not being mindreaders we can't tell which example you are referring to, what steps you have taken to compile the example nor what you mean by [i]when I saw "import olympiad"[/i]. What do you mean by [i]I followed the instructions of how to install this "package"[/i]? All it says is save the olympiad.asy in the asymptote directory. You don't say if you can compile other asymptote programs which don't use olympiad?

So remember, we can't see what you are doing so you'll need to tell us before we can spot the problem.

You are right ...
Well, I've been using Asymptote in Latex with TexnicCenter. I used the asy environment and it works perfectly. But when I followed the examples from aopswiki (see [url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Useful_commands_and_their_Output[/url])
I had problems just when I reached to the number 7 (the one that needs olympiad package).
So I compile through TexnicCenter and I have no errors, then I use the shortkey for Asymptote's job (export pictures to the .tex folder), and then I compile again (with no errors).
After that, I look over the output and I obtained just what you see in the file above.
I think there must be a problem with olympiad package of asymptote because I did copy-paste the codes from the page of aopswiki ...
Well, I add my .tex right here.

Note: Sorry about my english.

[/hide]

The good news is that I get the same wrong pdf when using the asymptote code. However, it works fine on this site so maybe some package has been updated. I hope fedja, the asymptote expert on AoPS, can help.
Compiling on a PC[code]import olympiad;
pair A,B,C,O,I;
A=origin; B=2*right; C=1.5*dir(70);
O=circumcenter(A,B,C); // olympiad - circumcenter
I=incenter(A,B,C); // olympiad - incenter
draw(A--B--C--cycle);
dot(O);
dot(I);
draw(circumcircle(A,B,C)); // olympiad - circumcircle
draw(incircle(A,B,C)); // olympiad - incircle
label("I",I,W);
label("O",O,S);[/code]gives the wrong diagram as shown in Asymptote_7.pdf in the first post.

On this site however the correct figure is produced. Has olympiad or some other package been updated?
[asy]pair A,B,C,O,I; A=origin; B=2*right; C=1.5*dir(70); O=circumcenter(A,B,C); /* olympiad - circumcenter */ I=incenter(A,B,C); /* olympiad - incenter */ draw(A--B--C--cycle); dot(O); dot(I); draw(circumcircle(A,B,C)); /* olympiad - circumcircle */ draw(incircle(A,B,C)); /* olympiad - incircle */ label("I",I,W); label("O",O,S);[/asy]

Did you use \begin{asy} and \end{asy} and did you have all the asymptote preamble stuff taken care of? I can give you an example of a document with working asymptote images.

jonysatie did exactly that, but I also tried that code in an asy file on its own using the command line and it gave the wrong diagram. Have you tried that specific code? I am beginning to wonder if a package has been updated - the olympiad package itself imports math and graph packages.

You simply forgot to include the unitsize command. I added unitsize(50); to your code and it worked just fine.

Yep, that's it! Thanks very much. I will now correct the code which came from Example 7 in the [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Useful_commands_and_their_Output]AoPSWiki[/url].

So this will improve the aopswiki ... I'm proud of me ...  :P


geometry
incenter
geometric transformation
reflection
circumcircle
analytic geometry
geometry proposed

Let $ ABC $ be a triangle with the incircle $ (w) $ and the incenter $ I $ . $ (w) $ touches $ AB, BC , CA $ at $ C_1 , A_1 , B_1 $ respectively . $ AA_1 $ intersects $ BB_1 $ at $ G $ . $ (w_A) $ is centered at $ A $ with radius $ AB_1 $ . $ (w_B) , (w_C) $ are defined analogously . $ (w_A), (w_B) , (w_C) $ are externally tangent to $ (w_1) $ . $ (w_A) , (w_B), (w_C) $ are internally tangent to circle $ (w_2) $ . Let $ O_1 , O_2 $ be the centers of $ (w_1) , (w_2) $ respectively . $ A_1B_1 $ intersects $ AB $ at $ C_2 $ ,  $ A_1C_1 $ intersects $ AC $ at $ B_2 $ . Prove that : $ I , G , O_1 $ and $ O_2 $ lie on a line $ d $ that is perpendicular to line $ B_2C_2 $

 Remark : one part of this problem 's posted at the topic : About the incenter (author Vornicu)

Mlinkers , do you have a solution for this problem ?? I haven't finished it yet

For $IG\perp B_2C_2$, it is easy by pole and polar.
The inversion with incircle as the reference circle takes $w_1$ to $w_2$, since $w$ is orthogonal to $w_A,w_B,w_C$. This implies $I,O_1,O_2$ are collinear.
That's all I can get till this moment.

[quote="nttu"]Eventually , I had a solution for this problem ![/quote]

please post your solution.  :)

The problem is quite straightforward. I 'm running out of time right now. So, I will give a brief note
 Here is a lemma
         [b]Lemma[/b] Let $ABCD$ be a square and $M$ be any point. The midperpendiculars of $MA, MB, MC, MD$ intersect $DA, CB, DA, CB$ respectively at $O_3, O_4, O_1, O_2$. Then $O_1O_2, O_3O_4, DC$ are concurrent.
 By inversion with center $A_1$ and constant the power of $A_1$ wrt $(w_A)$, we obtain the lemma.

Here is my full solution
  [hide="Solution"]
      Let $X_1, Y_1, Z_1$ be the points of tangency between $(w_1)$ and $(w_a), (w_b), (w_c)$ respectively. Define $X_2, Y_2, Z_2$ analogously.
      We will first show that the line $IG$ is perpendicular to $B_2C_2$. In fact, we will prove a stronger statement : the line $B_2C_2$ is the polar of the Gergonne point $G$ wrt the incircle $(I)$.
      [hide="Proof"] Notice that $G$ is the intersection between $BB_1$ and $CC_1$. Since $C_1A_1$ and $CA$ are respectively the polar of $B$ and $B_1$ wrt the incircle $(I)$, the intersection of $C_1A_1$ and $CA$ (i.e $B_2$) is the pole of $BB_1$. Similar argument yields that $C_2$ is the pole of $CC_1$. Therefore, $B_2C_2$ is the polar of the intersection of $BB_1$ and $CC_1$, i.e the point $G$. [/hide]
      Before showing that the Gergonne point $G$ lies on the line $O_1I$, I would like to introduce you the following lemma
      [b] Lemma[/b] The power of $A_1$ wrt to $(w_a)$ is $2rh_a$.
      This lemma can be proved easily by Stewart 's theorem.
      Now we will show that the Gergonne point $G$ lies on the line $O_1I$.
      [hide="Proof"]
      Consider the inversion center $A_1$ with constant $2rh_a$ (i.e the power of $A_1$ wrt $(w_a)$ from the lemma above). Under this inversion, the circle $(w_a)$ is mapped to itself. The points $B_1$ and $C_1$ are mapped to $B_1^*$ and $C_1^*$. Since the circles $(w_b)$ and $(w_c)$ pass through $A_1$ and extouch the circle $(w_a)$ at $C_1$ and $B_1$, they are mapped to the lines $b$ and $c$ that are perpendicular to $BC$ and tangent to $(w_a)$ at $C_1^*$ and $B_1^*$. This means that
      1) $B_1^*C_1^*$ is a diameter of $(w_a)$ that is parallel to $BC$.
      2) The images of the points $B$ and $C$ are the reflections $B^*$ and $C^*$ of $A_1$ across the lines $b$ and $c$.
      Now the image $(w_1^*)$ of $(w_1)$ is a circle that is tangent to the lines $b$ at $Y_1^*$ and $c$ at $Z_1^*$ and extangent to the circle $(w_a)$ at $X_1^*$. This means the radius of $(w_1^*)$ is equal to that of $(w_a)$. Therefore $B_1^*C_1^*Y_1^*Z_1^*$ is a square.
      Let $O_1^*$ be the image of $O_1$. Since $O_1$ is the intersection of $BY_1$ and $CZ_1$. Hence,  $O_1^*$ is the second intersection of the circumcircles $(A_1B^*Y_1^*)$ and $(A_1C^*Z_1^*)$.
      Call $I^*$ the image of $I$. Then since $IA_1$ is perpendicular to $BC$ and $IA_1 = r$, $I^*$ is the reflection of $A_1$ through $B_1^*C_1^*$.
      Since the point $G$ is the intersection of $BB_1$ and $CC_1$, the image $G^*$ of $G$ is the second intersection of the circumcircles $(A_1B^*B_1^*)$ and $(A_1C^*C_1^*)$.
      Now if we can prove that $A_1, I^*, O_1^*, G^*$ lie on a circle, we are done. This is equivalent to proving that the perpendicular bisectors of $A_1I^*$, $A_1G^*$, $A_1O_1^*$ are concurrent.
       [b]Lemma[/b] Consider the square $B_1^*C_1^*Y_1^*Z_1^*$. Suppose the perpendicular bisectors of $A_1B_1^*$ and $A_1Y_1^*$ intersect the side $Y_1^*C_1^*$ at $C'$ and $C''$. Suppose the perpendicular bisectors of $A_1C_1^*$ and $A_1Z_1^*$ intersect the side $Z_1^*B_1^*$ at $B'$ and $B''$. Then $B'C', B''C'', B_1^*C_1^*$ are concurrent.
       This lemma is very straightforward by coordinates (not computational, I guarantee  :lol: ).
       Notice that the perpendicular bisectors of $A_1I^*$, $A_1G^*$, $A_1O_1^*$ are respectively $B_1^*C_1^*$, $B'C'$, $B''C''$. From the lemma above, we obtain that $A_1, I^*, O_1^*, G^*$ lie on a circle. And we are done. [/hide]
  [/hide]


Let $ (I)$ be the inscribed circle of Triangle $ ABC$ ,  $ D$ be the contact point on $ BC$ and $ P$ be an arbitrary point on $ AD$ .  $ BP$ and $ CP$ meet $ (I)$ at $ S,T$ and $ X,Y$ respectively in order .  Show that both $ BX,CS,AD$ and $ BY,CT,AD$ are concurrent.

Let $ B_0,C_0$ respectively be the tangency points of $ (I)$ with $ AC,AB$. Denote $ P_a\equiv B_0C_0\cap BC$. $ S'\equiv P_aX\cap (I)$, $ S'$ is different from $ X$. $ M_a\equiv P_aX\cap AD$. Since $ AD$ is the polar of $ P_a$ wrt $ (I)\Longrightarrow (P_aM_aXS')\equal{}\minus{}1$. But it is well- known that $ (P_aDCB)\equal{}\minus{}1\Longrightarrow (P_aDCB)\equal{}(P_aM_aXS)\Longrightarrow DM_a,CX,BS'$ are concurrent. But $ AD,PC,BS$ are already concurrent at $ P$. Hence $ BS\equiv BS'\Longrightarrow S'\equiv S$. Thus, $ (P_aM_aXS)\equal{}(P_aM_aSX)\equal{}(P_aDCB)\equal{}\minus{}1$, which implies that $ M_aD,SC,BX$ are concurrent. With the same argument for the second part, we will have $ BY,CT,AD$ are also concurrent. The result is lead as follow.
Our proof is completed then $ \square$


I noticed that in the Olympiad Resources page many competitions have not had any problems posted. I was wondering if there is any way to post them so that these missing problems can be filled.

just post a new topic saying (REMAINDER OLYMPIAD PROBLEMS!!ANSWER PL0X!) or something like that :)

Are you saying that you have the problems and are willing to help put them in the resources section? If so, then click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195579]here[/url].


geometry MATHCOUNTS ARML AMC USA(J)MO USAMO

I think all the Bay Area chapters (Santa Clara Valley, Diablo, San Mateo, etc.) finished yesterday by the latest... so, how'd you guys do?

I went to Santa Clara Valley, which is probably one of the beastliest chapters ever  :(
Tied for 12th (epic fail) but got fourth team, best small school, and most improved school.

[quote="AIME15"]I'm in Diablo chapter, which is arguably the easiest in Cali.  45 score, won by 2 points.  First team also.[/quote]

Wrong. My chapter is the weakest chapter in CA.

45, 2nd, skipped cd to go to arml practice.

That makes you either Debnil Sur or Varun Mohan.

Anyways. I totally screwed up on the tiebreakers. (old news, too). Yet surprisingly I make states.

Wait, was that to me james4l?

Because I am definitely not either Debnil Sur or Varun Mohan, whoever they might be.  Neither is pythag.  I am 100% sure that you know this also.

And, @pythag: 45 is second in ur chapter, and 45 wins by 2 points in my chapter...

That was to Mathematicious.

Anyways. 45 would make you sixth to eleventh in our chapter.

Pretty crazy, huh.

In Peninsula 45 would've gotten you first. 40 would've gotten you 4th :O

Yes but 39 is 3rd in my chapter????

And 1st place is this random guy who probably fails at math.

Our chapter is too hard. 5 perfect scores (and james was among one of them), and I got 45 as well, placing me in [b]7th[/b] place. And I think some people with 44's (maybe a 45 as well) did not make states. This is so unfair...

Yeah, but if the chapter had all usamo probs, that might have worked better.

Hmm... I'm confused... is state 21st or 28th? If its 21st, arml local is the same day.... except arml local is in the afternoon so I might be able to attend...

[quote="Mathmaticious"]Dang, james4l, you're good. I'm the former. And you're Lewis Chen, I assume? Thanks for beating us by 1 point on team. If our third guy got one more target right and our fourth guy didn't get a 38, we would've totally killed you guys :P

Yeah, totally not fair @ dragon. Travis Chen from Harker got a 45 and didn't go. 44's.. Aditya Nandy from Thornton didn't go and the semibeastly Justin Chiang from Redwood didn't make it either :O[/quote]

yeah sure, but how can you assume that I come from Hopkins?

Our team really fails. dragon96 should be in our team.

I am so mad. 45 puts me in 11th place, and I don't get into states by one person

Bowditch fails at everything now, apparently their individual team was 4th in the chapter, while their official team was 6th in the chapter(they were first back when I was in middle school), ensuring that no one made states....

You were first by luck kid, remember that!

Ummm, whatever [i]dad[/i]...just remember you were born by luck....

Might I add that your team failed to make state from the [b]Peninsula[/b] chapter....

[quote="serialk11r"]It doesn't help to have no people who actually do math on the team okay.[/quote]

I'm curious as to who was actually on the team?

[quote="pythag011"]
And 1st place is this random guy who probably fails at math.[/quote]

dang I remember being that random guy when I was in 7th grade

[quote="serialk11r"]It doesn't help to have no people who actually do math on the team okay.[/quote]
Huh did you just forget Calvin's existence?

Also, since when has Peninsula been easy? Only Andrew thinks that. It's libel. Actually, the chapter competition probably was a bit messed up this year because of bad organization, but honestly, we crippled our team with enough bad organization of our own to compensate.

(i.e. We are NEVER EVER going to hold the school sprint after school again. That was pure laziness and stupidity. Judging from previous years' patterns, we missed out on something like half of the school's best people by effectively only testing MathCounts club.)

In any case, in a normal year the ONLY chapter in California more competitive than Peninsula is Santa Clara Valley. I'm pretty sure of that.

No, I meant MY year. Let's just say Bob, Susan, and I all didn't count because we never actually did practice problems.

Peninsula is not super trivial only because there is Ms. Slack and her crazy team.
And yes, we were missing good people but the problem is they refused to offer it to everyone.

DONT ASK
actually ill just tell you

E P I C P H A I L E D >______<
didnt even make states
ruined the history of my school after i switched in this year
BOMBED SPRINT (6 careless mistakes) :wallbash:  :wallbash_red:  :wallbash:
and i got perfect target
but still   :bomb:

[quote="AIME15"]Somebody who got the exact same score (24-16) got like 7th in my chapter :D[/quote]

...and 3rd at mine, except 1 point lower.

[quote="AIME15"]Somebody who got the exact same score (24-16) got like 7th in my chapter :D[/quote]

and 40 would get 30-ish place...

SCV is crazy.

Nooo way.. our fourth team member got a  38 or 39, and he tied for 60th. Insane.

SoCal state is tomorrow. So is ARML Local. I guess I will fail... PLEASE MAKE THE TEST HARD...

Update: 1st Socal.

I wonder if bay area will send 4 people to nats like last year...


I think I'm completely overlooking sometime or just going crazy but is the function $ f(x)\equal{}2^x\minus{}1$ prime for all odd integer values.

This has nothing to do with calculus.

$ 2^9 \minus{} 1 \equal{} (2^3 \minus{} 1)(2^{2 \cdot 3} \plus{} 2^3 \plus{} 1)$   :roll:

Also, in case you're planning on revising your conjecture to "prime exponents," you should go read a bit about Mersenne primes:
http://mathworld.wolfram.com/MersennePrime.html
http://en.wikipedia.org/wiki/Mersenne_prime

x^(2n+1) - 1 = (x-1)(x^2n + x^2n-1 + x^2n-2 ... + x^2 + x + 1)

since x = 2, the first factor will always be 1, not helping us much, but if the exponent is divisible by an odd number, you will be able to factor it in this method

I figured I was just going crazy. Anyway I ment to post it in intermediate- not calculus- sorry about that.


SO PAMO is (almost) officially over.We leave in about 3 hours.  SA came first, no thanks to me.  Carl Veller came 1st, Melissa Munnik tied 2nd, Francois Conradie got the first silver (ie 4th) and i got a bronze (was bout 11th).  Tunisia came 2nd and Benin 3rd.  Well done everyone!!!!!!! We had such an awesome time.  Thanks Senegal for hosting us... hope you enjoy SA in 2008.

Congratulations. Could you post the questions :?:

Mel says she will post them soon in pre-olympiad.theyretoo easy forolympiad forum apparently. carl asks what you think of emma watson (hes obssessed

Here are links to all the problems:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105853]Problem 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105855]Problem 2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105857]Problem 3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105859]Problem 4[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105861]Problem 5[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105863]Problem 6[/url]

Yeah, as the topic says South Africa won, and Tunisia came second. The results are confidential, so I cannot post them here. I do know that Carl Veller (carlv on MathLinks) came first, and me and Tarak (not sure about his surname, he is from Tunisia) were tied for second place.

I'll post something more about my experiences in Senegal later.

[quote="shobber"][quote="Ashegh"]congratulations.but why in pre olympiad??????[/quote]
Because the problems are easy. I posted all the previous year's problems of PAMO in the Pre-Olympiad last summer.[/quote]

yes yes, i rememberd them :wink:


advanced fields
advanced fields unsolved

Let $M$ be the intersection of all montone classes that contains the algebra $A$. Show that $M$ is an algebra.

Let $\mathfrak{M}$ be the intersection of all monotone classes containing $\mathfrak{A}$.Then by definition it is the samllest of such.To show that it is a $\sigma$-algebra generated by $\mathfrak{A}$ we show that it is closed under complement and finite union.
If $A_i\in \mathfrak{M} \implies B_n:=\bigcup_{i=1}^n A_i \in \mathfrak{M}$ and $B_1\subseteq B_2...$ and $\bigcup_{i=1}^{\infty}A_i=\bigcup_{n=1}^{\infty}B_i\in \mathfrak{M}$ Hence it is a $\sigma$-algebra and since it is a monotone class and the smallest it is generated by $\mathfrak{A}$.
Let $A\in \mathfrak{M}$ and let $\mathfrak{\zeta}(A):=\{B \in \mathfrak{M}:A \cup B \in\mathfrak{M}\}$ and let $B_i\in \zeta(A)$ and $B_1\subseteq B_2... \implies (A \cup B_1)\subseteq (A \cup B_2)...$.Hence $A \cup \bigcup_{i=1}^{\infty}B_i=\bigcup_{i=1}^{\infty}(A \cup B_i)\in \mathfrak{M} \implies \bigcup_{i=1}^{\infty}B_i\in \zeta(A)\implies \zeta(A)$ is a monotone class.
If $A\in\mathcal{A}\implies \mathfrak{A}\subseteq\zeta(A)\subseteq \mathfrak{M}$ and since it is an algebra $\zeta(A)=M$ if $A\in\mathfrak{M}\implies \mathfrak{A}\subseteq\zeta(A);B\in \mathfrak{A},\zeta(B)=\mathfrak{M}$.Hence $A\cup B=\mathfrak{M},\zeta(A)=\mathfrak{M}\thinspace \forall A \in \mathfrak{M}$ Let $\zeta:=\{B\in\mathfrak{M}:B^c\in\mathfrak{M}\}$ and let $B^c_1\supseteq B^c_2...\implies (\bigcup_{i=1}^{\infty}B_i)^c=\bigcap_{i=1}^{\infty}B^c_i\in \mathfrak{M}\implies \zeta=\mathfrak{M}$ Hence proved.


At what point does the line containing the points $ (1, 7)$ and $ (3, 11)$ intersect the y-axis? Express your answer as an ordered pair.

First we have to find the equation of the line.  This is $ \frac{11 \minus{} 7}{3 \minus{} 1} \equal{} \frac{4}{2} \equal{} 2$.  So the equation is $ y \equal{} 2x \plus{} b$.  We can plug in $ (1,7)$:

$ 7 \equal{} 2(1) \plus{} b$
$ 7 \equal{} 2 \plus{} b$
$ b \equal{} 5$

So the answer is $ \boxed{(0,5)}$.


${f_{n}(a,x_{n})=K_{n}}$
${f_{n}(b,y_{n})=S_{n}}$

${f_{n+1}(a,x_{n})=y_{n+1}}$
${f_{n+1}(b,y_{n})=x_{n+1}}$

${\{a,b,x,y,S,K,n\}\in \mathbb{Z}}$
${0\leq \{a,b,x,y,S,K\}\leq 1}$

i want a solution for this, find ${f}$

i think its true, but only testing it to see if is true or false

Thales418, think about what you have written...


calculus
integration
real analysis
real analysis unsolved

What is the point of posting questions that do not have elementary closed form solutions?  You've done this several times now.  Post something interesting--and solvable--for a change.


prove that $(36m+n)(36n+m)$ is not power of $2$ for all $m,n \in \mathbb{N}$.

Clearly m,n are even, so dividing by a common factor of 2 we get a contradiction by infinite descent.

Bomb

Please can you explain what do you want to say there dule ...

dule wants us to show that that $(36m+n)(36n+m)$ is never a power of two (of type $2^k$ with $k \in \mathbb{N}_0$) for all positive integers $m,n$.

What's unclear about it  :?

Now i got that thank you very much ... his tryed to write the equaiotn of the form $2^k$


Έστω $ 2n$ πραγματικοί αριθμοί $ a_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n}$ και έστω $ \sum^{n}_{j \equal{} 1}a_{j}\neq 0$ και $ \sum^{n}_{j \equal{} 1}b_{j}\neq 0$. Να αποδείξετε ότι $ \sum^{n}_{j \equal{} 1}a_{j}b_{j} \plus{} \sqrt {\left(\sum^{n}_{j \equal{} 1}a^{2}_{j}\right)\left(\sum^{n}_{j \equal{} 1}b^{2}_{j}\right)}\geq \frac {2}{n}\cdot \left(\sum^{n}_{j \equal{} 1}a_{j}\right)\cdot\left(\sum^{n}_{j \equal{} 1}b_{j}\right)$. Πότε ισχύει η ισότητα?

Μια πρώτη σκέψη: διαχωρισμός ρίζας, με εξέταση περιπτώσεων καταλήγουμεστο ότι για αρνητικό δεύτερο μόλος ισχύει
για θετικό τετράγωνα και στη συνέχει βγαίνει με BCS (ή αναδιάταξη, δεν είμαι σίγουρος γιατί είναι δύσκολο να αναγνωρίσεις τη μορφή όταν έχεις κάνει τα πάντα μέσα στο μυαλό σου).


Can you show this direct consequence ? I can't see it
This is Holder's inequality, isn't it ?
$\sum_{i=1}^{n}a_ib_i \leq (\sum_{i=1}^{n}{a_i}^p)^{\frac{1}{p}} \cdot (\sum_{i=1}^{n}{b_i}^q )^{\frac{1}{q}}$

where $\frac{1}{p} + \frac{1}{q} = 1$

maybe you thought about some generalized Holder's
where $\frac{1}{p_1}+...+\frac{1}{p_n}=1$ and an inequality itself change a bit then it is indeed clear for n=3, but normal Holder ?

It really depends on what you think Holders is.
Look at page 22 of [url=http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/ineqs-080299.ps]A < B[/url], that's the Holder that I'm refering to.

billzhao: I can't open files ended with .ps - can you post it here ?

LevonNurbekian:  Yeah, this is generalized Holder's ineq which I meant and which proves our problem for m=3 and obvious elements

[quote="Megus"]billzhao: I can't open files ended with .ps - can you post it here ?[/quote]

I was just pointing to the definition of Holders in that book, which states it as

If $w_{1}, \dots, w_{n}$ are positive real numbers whose sum is 1, then for any positive real numbers $a_{ij}$,
\[
\prod_{i=1}^{n} \left( \sum_{j=1}^{m} a_{ij} \right)^{w_{i}}
\geq \sum_{j=1}^{m} \prod_{i=1}^{n} a_{ij}^{w_{i}}.
\]


We have an infinitely long ideal conductive cylinder of radius $ R$. We put a plane electromagnetic wave on the cylinder, so that its wavenumber vector, is perpendicular to the axis of the cylinder, and the waves electric field vector is parallel to the axis of the cylinder. Find the electric and magnetic field on the surface of the cylinder, also give the total current on the surface of it.

The problem is quite tough. it takes a while. :D

So no one wants to try?

[hide="Here are a few hints then:"]
First of all choose a usefull coordinate system. I suggest cylindrical coordinates, because then a lots of the components of the fields will be zero.
Then express the magnetic field with the electric field.
Now look for the $ E(r,\varphi)$ electric field of the scattered wave in the :

\[ E(r,\varphi)\equal{}\sum_p R_p(r)\Phi_p(\varphi)\]

form. Here $ R_p$ are solutions of the Bessel equation, and  for $ \Phi_p$ the following equation holds:
\[ \Phi_p'' \plus{}p^2\Phi_p\equal{}0\]

Also use that $ E$ is a cylindrical wave, hence it has to be of a specific form. And the solutions of the Bessel equation of this form are the Hankel functions.
[/hide]

Now this must be more than enough help :)
I said that its not easy :lol:

Still no one interested? Its very nice this problem

Tell me where you got stuck, and I can give more hints then :wink:  :D


On the number line, point $ A$ is located at 0, point $ B$ is located at 4, and point $ C$ is located at 6.
[asy]defaultpen(1);
draw((0,0)--(6,0));
for (int i=0;i<7;++i){
  draw((i,-.1)--(i,.1));
  label(string(i),(i,-.1),(0,-1));
}
label("\(A\)",(0,0),(0,1));
label("\(B\)",(4,0),(0,1));
label("\(C\)",(6,0),(0,1));[/asy]
A dart randomly lands somewhere on the number line between $ A$ and $ C$.  What is the probability that it lands closer to $ B$ than it does to $ A$ and $ C$?

[hide]For it to be closer to be, it would have to be between $ 2$ and $ 5$. This is $ \frac{3}{6}\equal{}\boxed{\frac{1}{2}}$.[/hide]


Given a rectangle $ ABCD,$ let $ AB \equal{} b, AD \equal{} a ( a\geq b),$ three points $ X,Y,Z$ are put inside or on the boundary of the rectangle, arbitrarily. Find the maximum of the minimum of the distances between any two points among the three points. (Denote it by $ a,b$)

I think this one is easy,it's just because putting images is hard?!


Encuentra todos los enteros "n" para los cuales
n divide a 2006^n - 2005^n

Dicen que todos los problemas dejan algo para aprender, pero kreo que este deja bastante!!
Espero sus soluciones

[quote="conejita"]Lo paso al latex, hallar todos los enteros positivos n tales que
$n$ $/$  $2006^{n}-2005^{n}$[/quote]

Tendría que ser $n|(2006^{n}-2005^{n})$, ya que lo que vos escribiste es $\frac{n}{2006^{n}-2005^{n}}$

Jose, eso es lo k kise poner, lo k pasa es k apenas ando aprendiendo a usar el latex.
Mimoide, sinceramente no entiende tu solucion ni el porque del valor que le das a "k".

[quote="Mimoide"]Creo que se me escapa algo, pero bueno. Supongamos que existe un $n$ que cumple la condición. Entonces, si consideramos el mínimo primo  $p$  tal  que $p/n$, que no puede dividir ni a $2005$ ni a $2006$, puesto que entonces no dividiría al numerador, tomando $k=2006\times2005^{-1}\;\;(mod \;p)$,  tenemos  que
$2006^{n}-2005^{n}\equiv 0\leftrightarrow (2006\times2005^{-1})^{n}-(2005\times 2005^{-1})^{n}\equiv 0 \leftrightarrow$ $k^{n}\equiv1 \;\;(mod \; p)$  y que $mcd\;(n,p-1)=1$, por lo que
$k\equiv1 \leftrightarrow (2005+1)\times2005^{-1}\equiv 2005\times2005^{-1}+2005^{-1}\equiv$ $\equiv 1+2005^{-1}\equiv 1 \rightarrow 2005^{-1}\equiv 0 \;\; (mod\;p)$ ,   lo que es imposible.[/quote]

A mi parecer todo el proceso correcto pero lo que si no entendia al inicio era de como afirmabas que : $mcd\;(n,p-1)=1$ Pero con eso de que " p " era el menor primo que divide a  " n " es suficiente .

               ¡ Muy buena solución !


Consider $ \Delta ABC$ with $ AB\equal{}c$,$ BC\equal{}a$, and $ AC\equal{}b$. Draw angle bisector $ AD$ where $ D$ is on $ bc$. Prove \[ AD\equal{} \frac{2bc\cos{A/2}}{b\plus{}c}\]

[hide]consider the areas of triangles

$ 2[ABD]\equal{}ADc\sin (A/2)$
$ 2[ADC]\equal{}ADb\sin (A/2)$
$ 2[ABC]\equal{}bc\sin A\equal{}2bc \sin (A/2)\cos (A/2)$.

thus, $ 2[ABC]\plus{}2[ADC]\equal{}2[ABC] \implies ADc\sin (A/2)\plus{}ADb\sin(A/2)\equal{}2bc\sin (A/2)\cos (A/2)$, and the result follows.[/hide]


linear algebra
matrix
limit
email
induction

Let \[ B_k\equal{}B_{k\minus{}1}\plus{}B_{k\minus{}1}(I\minus{}AB_{k\minus{}1}), k\equal{}1,2,\cdots\]
  Show that if $ \|I\minus{}AB_0\|\equal{}c<1$, then
  \[ \lim \limits_{k\longrightarrow\infty}B_k\equal{}A^{\minus{}1}\]
  and \[ \|A^{\minus{}1}\minus{}B_k\|\leq \frac{c^{2^k}}{1\minus{}c}\|B_0\|.\]
If you have got a solution, please email me
miwalin@163.com

[hide]a) We have :
      $ B_k \equal{} 2B_{k \minus{} 1} \minus{} B_{k \minus{} 1}AB_{k \minus{} 1}$
      $ I_n \minus{} AB_k \equal{} I_n \minus{} 2AB_{k \minus{} 1} \plus{} (AB_{k \minus{} 1})^2 \equal{} (I_n \minus{} AB_{k \minus{} 1})^2$
Then with $ \forall k \geq 1 , k\in N$ (by induction ) We have :
         $ I_n \minus{} AB_k \equal{} (I_n \minus{} AB_{k \minus{} 1})^2 \equal{} ... \equal{} (I_n \minus{} AB_0)^{2^k}$ [b](1)[/b]
I Use : $ \parallel{}XY\parallel{} \leq \parallel{}X\parallel{}|Y\parallel{}|$ and $ \parallel{}X\parallel{} \minus{} \parallel{}Y\parallel{} \leq \parallel{}X \minus{} Y\parallel{}$ , $ \forall X ,Y \in M_n(C)$
 Ww have  : $ \parallel{}I_n \minus{} AB_k\parallel{} \equal{} \parallel{}(I_n \minus{} AB_0)^{2^k}\parallel{} \leq \parallel{}I_n \minus{} AB_0\parallel{}^{2^k} \equal{} c^{2^k}$
 by  $ \lim_{k \to \plus{} \infty}{c^{2^k}} \equal{} 0$ thoerefore $ I_n \minus{} AB_k \to 0$ ,since $ k \to \plus{} \infty$
Implies : $ \lim \limits_{k\longrightarrow\infty}B_k \equal{} A^{ \minus{} 1}$
b)  We  have $ \parallel{}A^{ \minus{} 1}\parallel{}\parallel{}I_n \minus{} AB_0\parallel{} \geq \parallel{} A^{ \minus{} 1}\minus{}B_0\parallel{}$ then
      $ \parallel{}A^{ \minus{} 1}\parallel{} \minus{} \parallel{}A^{ \minus{} 1}\parallel{}\parallel{}I_n \minus{} AB_0\parallel{} \leq \parallel{}A^{ \minus{} 1}\parallel{} \minus{} \parallel{} A^{ \minus{} 1}\minus{}B_0\parallel{} \leq \parallel{}B_0\parallel{}$
From here :  $ \parallel{}A^{ \minus{} 1}\parallel{} \leq \frac {\parallel{}B_0\parallel{}}{1 \minus{} c}$ [b](2)[/b]
and from [b](1),(2)[/b]
         $ \parallel{}A^{ \minus{} 1} \minus{} B^k\parallel{} \equal{} \parallel{}A^{ \minus{} 1}(I_n \minus{} AB_0)^{2^k}\parallel{} \leq \parallel{}A^{ \minus{} 1}\parallel{}\parallel{}I_n \minus{} AB_0\parallel{}^{2^k} \leq \frac {c^{2^k}}{1 \minus{} c}\parallel{}B_0\parallel{}$[/hide]


Prove that for every $n \in \mathbb{N}$ there exist a tripple $(a,b,c)\in \mathbb{N}^3$ satisfing the equation $a^2+b^2+c^2=3abc$, with $a,b,c> n$.

You can have a look at the end of [url=http://www.mathlinks.ro/Forum/post-183940.html&highlight=#183940]this[/url] disscution  :lol:   ;)


Nice  pb ,  we have 2 cases:

case 1: $ a\plus{}b\plus{}c\plus{}d \leq  4$

in that case :  LHS = $ cd(a\plus{}b)\plus{}ab(c\plus{}d) \leq  \frac{ (c\plus{}d)^2(a\plus{}b)\plus{}(a\plus{}b)^2(c\plus{}d)}{4} \equal{}\frac{ (a\plus{}b)(c\plus{}d)(a\plus{}b\plus{}c\plus{}d)}{4} \leq 4$

an the last inequality follows from : $ (1\plus{}a^2)(1\plus{}b^2) \geq (a\plus{}b)^2$  and $ (1\plus{}c^2)(1\plus{}d^2) \geq (c\plus{}d)^2$

case  2 :  $ a\plus{}b\plus{}c\plus{}d  \geq 4$

$ 16\equal{}(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)(d^2\plus{}1) \equal{}( (ab)^2\plus{}a^2\plus{}b^2\plus{}1)( (cd)^2\plus{}c^2\plus{}d^2\plus{}1) \equal{} ( (ab)^2\plus{}a^2\plus{}b^2\plus{}1)(c^2\plus{}(cd)^2\plus{}1\plus{}d^2) \geq  ( abc\plus{}acd\plus{}b\plus{}d)^2$  ( cauchy shwarz)  (*)

now since $ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)(d^2\plus{}1)\equal{}16$ we can't have all the numbers greater than $ 1$ or all of them smaller than $ 1$ thus we may  suppose WLOG that : $ (b\minus{}1)(d\minus{}1) \leq 0$

from (*)  it sufficies to prove that :   $ abc\plus{}acd\plus{}b\plus{}d \geq  abc\plus{}abd\plus{}bcd\plus{}acd$ that its
$ b\plus{}d \geq  bd(a\plus{}c)$
but  $ b\plus{}d \geq 4\minus{}(a\plus{}c)$
hence we left to prove that :   $ 4 \geq  (a\plus{}c)(bd\plus{}1)$  we shall prove the stronger  $ (a\plus{}c)(b\plus{}d) \geq (a\plus{}c)(bd\plus{}1)$ that is:
$ \minus{}(b\minus{}1)(d\minus{}1)\geq 0$ so we are done  :) ,  hope it's true

[quote="2424"]With $ a,b,c,d \geq 0$ such that $ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)(d^2 \plus{} 1) \equal{} 16$
Prove that         $ abc \plus{} bcd \plus{} cda \plus{} abd \leq 4$[/quote]
The following is my proof:

From the Cauchy Schwarz Inequality, we have
$ 16 \equal{} (a^2 \plus{} 1)(1 \plus{} b^2)(c^2 \plus{} 1)(1 \plus{} d^2) \ge (a \plus{} b)^2(c \plus{} d)^2,$
and
$ 16 \equal{} (a^2 \plus{} 1)(b^2 \plus{} 1)(1 \plus{} c^2)(1 \plus{} d^2) \ge [(ab \plus{} 1)(1 \plus{} cd)]^2 \ge \left( \sqrt {ab} \plus{} \sqrt {cd}\right)^4.$
According to these two estimations, we get
$ (a \plus{} b)(c \plus{} d) \le 4,$ and $ \sqrt {ab} \plus{} \sqrt {cd} \le 2.$
Then, it follows that
$ abc \plus{} bcd \plus{} cda \plus{} dab \equal{} ab(c \plus{} d) \plus{} cd(a \plus{} b) \le \sqrt {ab}\cdot \frac {a \plus{} b}{2}\cdot (c \plus{} d) \plus{} \sqrt {cd} \cdot \frac {c \plus{} d}{2}\cdot (a \plus{} b) \equal{} \frac {1}{2}(a \plus{} b)(c \plus{} d)\left( \sqrt {ab} \plus{} \sqrt {cd}\right) \le 4,$
as desired. Note that the equality holds if and only if $ a \equal{} b \equal{} c \equal{} d \equal{} 1.$

[quote="can_hang2007"][quote="2424"]With $ a,b,c,d \geq 0$ such that $ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)(d^2 \plus{} 1) \equal{} 16$
Prove that         $ abc \plus{} bcd \plus{} cda \plus{} abd \leq 4$[/quote]
The following is my proof:

From the Cauchy Schwarz Inequality, we have
$ 16 \equal{} (a^2 \plus{} 1)(1 \plus{} b^2)(c^2 \plus{} 1)(1 \plus{} d^2) \ge (a \plus{} b)^2(c \plus{} d)^2,$
and
$ 16 \equal{} (a^2 \plus{} 1)(b^2 \plus{} 1)(1 \plus{} c^2)(1 \plus{} d^2) \ge [(ab \plus{} 1)(1 \plus{} cd)]^2 \ge \left( \sqrt {ab} \plus{} \sqrt {cd}\right)^2.$
According to these two estimations, we get
$ (a \plus{} b)(c \plus{} d) \le 4,$ and $ \sqrt {ab} \plus{} \sqrt {cd} \le 2.$
Then, it follows that
$ abc \plus{} bcd \plus{} cda \plus{} dab \equal{} ab(c \plus{} d) \plus{} cd(a \plus{} b) \le \sqrt {ab}\cdot \frac {a \plus{} b}{2}\cdot (c \plus{} d) \plus{} \sqrt {cd} \cdot \frac {c \plus{} d}{2}\cdot (a \plus{} b) \equal{} \frac {1}{2}(a \plus{} b)(c \plus{} d)\left( \sqrt {ab} \plus{} \sqrt {cd}\right) \le 4,$
as desired. Note that the equality holds if and only if $ a \equal{} b \equal{} c \equal{} d \equal{} 1.$[/quote]

Very Nice proof  :lol:  , it is simpler than mine  :blush:


geometry
ceiling function
calculus
geometric series
geometric sequence

Let $ S_1$ be a square with side $ a$ units long ($ a>0$), with $ S_1$ circumscribed about a circle $ C_1$. Let $ S_2$ be a square inscribed in $ C_1$ and circumscribed about a circle $ C_2$. Continue in this way with $ S_n$ a square inscribed in circle $ C_{n\minus{}1}$ and circumscribed about a circle $ C_n$ (with $ n$ a positive integer). Find the smallest positive integer value of $ a$ such that the sum of the areas of $ S_1, S_2, \ldots, S_{10}$ and $ C_1, C_2, \ldots C_{10}$ exceeds $ 200$.

[hide="An idea that probably has a few mistakes"]The nth square has area $ \frac {a^2}{2^{n\minus{}1}}$ and then nth circle has area $ \frac {\pi a^2}{2^{n \plus{} 1}}$. We need $ \sum_{n \equal{} 1}^{10}\frac {a^2}{2^{n \minus{} 1}} \plus{} \frac {\pi a^2}{2^{n \plus{} 1}}\ge 200$. Moving some fractions around, $ \sum_{n \equal{} 1}^{10}\frac {a^2(4 \plus{} \pi)}{2^{n \plus{} 1}}\ge 200$. Multiplying both sides by $ 2^{11}$, $ \sum_{n \equal{} 1}^{10}2^{n \plus{} 1}a^2(4 \plus{} \pi) \ge 409600\Leftrightarrow a^2(4 \plus{} \pi)\sum_{n \equal{} 1}^{10}2^{n \plus{} 1} \ge 409600$ $ \Leftrightarrow a^2(4 \plus{} \pi)\ge\frac {409600}{\sum\limits_{n \equal{} 1}^{10} 2^{n \plus{} 1}}\approx 100.097752$ $ \Leftrightarrow a \equal{} \left\lceil \frac {100.097752}{4 \plus{} \pi}\right\rceil \equal{} \boxed{15}$.[/hide]

The answer is [hide="Click me"]8[/hide]
You must've done something wrong. I'm too tired right now to read your solution, but when I did it, I got the right answer, using the geometric series formula.

[quote="Temperal"]$ \sum_{n \equal{} 1}^{10}\frac {a^2(4 \plus{} \pi)}{2^{n \plus{} 1}}\ge 200$. Multiplying both sides by $ 2^{11}$, $ \sum_{n \equal{} 1}^{10}2^{n \plus{} 1}a^2(4 \plus{} \pi) \ge 409600$[/quote]

Not quite.  You're going to need to multiply by $ 2^{13}$ instead...

[quote="Temperal"]$ a^2(4 \plus{} \pi)\ge\frac {409600}{\sum\limits_{n \equal{} 1}^{10} 2^{n \plus{} 1}}\approx 100.097752$ $ \Leftrightarrow a \equal{} \left\lceil \frac {100.097752}{4 \plus{} \pi}\right\rceil \equal{} \boxed{15}$.[/quote]

This should be $ a^2 \ge \frac {100.097752}{4 \plus{} \pi}$ instead.  Of course, that numerator is still unfixed, but that should give you the right answer, I hope.  If we multiply 100.097752 by 4, that gives $ a^2 \ge \frac{400.391008}{4\plus{}\pi}$, and thus, $ a \ge 7.49$, giving 123456789's answer.

Oops, sorry about the silly mistakes.  :oops: It indeed gives the right answer when you do it correctly...

There's a cleaner way that I used (because I don't know calc):

If you start writing out the areas in terms of A for S1, C1, S2, C2 etc.
you see that it is 2 geometric series:

(a^2) + (a^2 pi / 4) + (a ^2 / 2) + (a^2 pi / 8) + (a^2 / 4) + (a^2 pi/16) ... (a^2 / 512) + (a^2 pi / 2048)

NOTE: i'm not sure my last values are correct, but as you will see this won't matter

These 2 geometric sequence factors into:

a^2 [(1 + 1/2 + 1/4 + 1/8....)+(pi/4 + pi/8 + pi/16...)]

Although each sequence is not infinite, you can reguard them each as infinite because after n=10 in the infinite series the values are negligible for this problem.

You solve for the sum of each geometric sequence with (N1) / (1-r)
Giving you the sum of the first sequence as 2, and the second sequence pi/2

so you now have:
a^2 (2+pi/2) > 200

a > ~ 7.48, so the answer is 8.

Uh, what are you talking about? I didn't use calculus, and my solution is actually completely identical to yours.

I guess I don't have any idea what I'm talking about, then.

You both have roughly the same solution. Temperal just used the sigma notation and a bunch of other symbols, lol.


Prove that: for all integer n such that  n>1, then in (n^2-n;n^2) alway exist at least one prime number.

Too strong to be true when compared with Chebyshev's interval $(n; 2n)$ or else I've not understood the question.
Consider for instance $n=28$. Then $n^{2}-n=756, \ \ n^{2}=784$ and the nearest primes are 739 and 829 both outside of the interval $(756; 784)$
P.S. I have simply sticked at random and have got counterexample from the second attempt.

[quote="JBL"]Your counterexample is wrong: the primes 757, 761, 769 and 773 all lie in the given interval.  Also, the problem is not too weak when compared with Chebyshev (aka Bertrand's Postulate) because the intervals in question are far from uniform -- the results aren't comparable.  (For a more obvious example of how this can happen:  the interval $[n, n+2]$ always contains a prime, if we restrict $n$ to the set of numbers one less than a prime.  This result is simply incomparable with Chebyshev.)[/quote]

Nice number of posts (I realize this is complete spam...)

:lol:

(This is my 6667th recorded post, for those wondering what he's talking about.)

[quote="JBL"]Your counterexample is wrong ...  Also, the problem is not too weak when compared with Chebyshev ...  because the intervals in question are far from uniform ...
[/quote]
Oh, shame on my head!  :D

Anyone have a proof?  It doesn't look like an easy result to me... perhaps quadratic residues?

As far as I know, this problem is open.  For related results and conjectures, see
[url]http://en.wikipedia.org/wiki/Prime_gap[/url].

This is an open problem.  In fact, it is stronger than [url=http://mathworld.wolfram.com/LegendresConjecture.html]Legendres Conjecture[/url].


Fun problem that I thought up of recently. It certainly relates to our forum. :)

A spammer decided to aggravate Valentin by freezing up everyone's computer by placing a string of Mr. Green's ( :D ) under a [u]click to reveal hidden  content [/u]thingie. So the spammer is great at math (he must be, right? The spammer is on AoPS.) and realizes he needs to put 1,000,000  :D 's underneath the  [u]click to reveal hidden content [/u] in order to freeze up Valentin's computer. He can only press Mr. Green and/or copy and paste.

What is the quickest way to do this, and how many moves will it take to reach 1,000,000  :D 's (Include pressing Mr. Green as a move, copying as a move, pasting as a move)?

Does there have to be exactely 1 000 000  :D 's or may there be more, just not less?

Please do click at the :D Below!

[color=green][size=75]Important info that will save your time!
Click the smiley below - by chess64 - before clicking evil Mr. Green!![/size][/color]

[hide=":)"]
You'll need to click about 50 times to find out what Gyan eventually reveals.
[/hide]

[hide=":D"][hide=":D"][hide=":D"]Sorry coud not resist it - $2^20>1000000$ But there is one more clever answer                                                                [/hide] [/hide][/hide]

[quote]A spammer decided to aggravate Valentin by freezing up everyone's computer by placing a string of Mr. Green's (  ) under a click to reveal hidden content thingie. So the spammer is great at math (he must be, right? The spammer is on AoPS.) and realizes he needs to put 1,000,000  's underneath the click to reveal hidden content in order to freeze up Valentin's computer.[/quote]

BTW Did you know Valentin is one step ahead,  Much [b]before,[/b] [b]his[/b] computer would freeze,  Forget about 1,000,000.. you  attempt would freeeze at about 10 Mr greens! You can't put more than this number in any one message!!!:D!

[hide]He can click 2 times, copy those, paste once, copy those, paste once (he's now up to 8), copy, paste(16), copy, paste, etc.. For the nth power of 2, it will require 2n clicks. The first power of 2 greater then 1000000 is 2^20, so it will require at least [b]40 clicks[/b][/hide]You can only have 6 smilies in each message , but it doesn't tell you that till you click Submit!:P :D  :D  :D  :D  :P

gyan, one could just quote you and look at the code you typed instead of looking at all the hide things

[quote]For the nth power of 2, it will require 2n clicks. The first power of 2 greater then 1000000 is 2^20, so it will require at least 40 clicks
[/quote]

Math is fun! Can you do it in LESS than 40 clicks? (The answer is yes)

To be clear, Here is how a 'click' is defined:

In the begining there is  $1$  ":D" One can not use a key to type any more ":D"

Only thing you can do is  "select" (This requires NO clicks) ;  "copy" (requires $1$ click) and "paste" (requires $1$click)

What you need is end up with (at least) 1000000 ":D"'s.
[hide="Hint"]
The answer is 38 clicks... question is how? [/hide]

[quote="Melissa"][quote="solafidefarms"]You can only have 6 smilies in each message , but it doesn't tell you that till you click Submit![/quote]
This was not always so. I remember somewhere in the Neverending Story a guy posted a 1000  :rotfl: 's.[/quote]

In the marathon, 10000 smilies.

Nice problem! It was hilarious!

 :rotfl:

By the way, the spammer could click mr green, then c&p, c&p...
or
click mr green 3 times in a row, then c&p, c&p...

Same number of moves.  :)

Or one could click only once and use the keyboard to do the rest.

 :D then hit the combo that selects the previous word. Copy and paste a few times, select all that, copy and paste. Select those words with keyboard etc. etc.

Thats for clicks. For moves.. dunno id have to work it out.

Just to reiterate (to make it clear)

On my problem: What we want is - Do it in less than 40 clicks.
You start with only 1 "smile" - (After that you can not use key board - all you can do is copy & paste)
"Select all" - action is not counted, as far as "clcks" are concerned - that is no "click" is required to "select all"
1 click is required to copy and 1 click is required to paste.